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Wholesaling Properties This is the continuation of the Flip2Freedom podcast transcription. In this episode, Sean interviews the nation’s #1 wholesaler, Kent Clothier, and finds out how he created a wholesaling empire – averaging over 50 deals a month. Sean: I know when I first got started I had this fear of if I get a property under contract and someone is a buyer, or portrays to be a buyer and is kind of messing me around and backs out on the closing day, what happens? I was always scared of that, but if you know someone is proven, and you know that they have actually closed transactions prior, within the same specific area, that is a no-brainer if you can find those people. Now, you call them up, right? I’m sure you call probably five to ten different buyers, or do you just call a group of two or three and it’s a first come first served type of thing? Kent: It is a first-come, first-served. What we don’t do is we don’t just turn around and throw properties out across a list of 10, 20, 30 buyers. We are picking up the phone calling a very specific buyer, because of this very specific scenario that fits into what they are looking for. In other words, we are not trying to just say, “Hey, by the way, I’ve got this house, I think you might be interested.” It’s, “We’ve got this house, you told me that you are looking for this kind of cash flow, you told me you were looking for this kind of return; this one fits into that scenario. We need to move fast; you have got about an hour before this is going to be gone.” And, because our buyers know that our buyers react. I think that’s very important. Because, any time that you are in this business, one of the things that I think is critical is that you can establish with your buyers, very early on and make them take you very seriously, if they know you’re not going to waste their time. But, if you’re just going to throw stuff at them and just kind of hope and cross your fingers that it is going to sell, then you are probably not going to get the kind of reaction that you’re looking for. We have really dialed in on this, our buyers jump when we need them to jump. Sean: So, you train your buyers on a specific expectation on, “Here is a property. It fits your criteria you have purchased before. We are calling other investors. You’ve got about an hour to make a decision on this thing, yes or no?” Kent: You got it. See “F2F Kent Clothier Interview” Part 8… Click the Join Now Button Below to Get Your 30 Day Risk Free Trial of Flip2Freedom Academy	141,052
Mark R. Goldschmidt Partner Overview Mark R. Goldschmidt is a partner in Holland & Knight's Financial Services Practice Group. He practices largely in the area of financial services with a strong focus on public and private financings, securities issuances, mergers and acquisitions, bank regulatory issues, general business transactions and strategic business initiatives. Mr. Goldschmidt primarily represents public and private companies in a range of industries on a variety of matters, including securities law compliance, corporate governance matters, executive compensation and general corporate matters. He regularly advises management teams, boards of directors and public company committees with regard to fiduciary duties, governance and securities law disclosure as well as compliance issues. Additionally, he represents investment banks, as placement agents, underwriters and financial advisers, in connection with the public offering and private placement of securities and mergers and acquisitions. Mr. Goldschmidt is also well-versed in overseeing multi-disciplinary legal teams of licensing and consumer regulatory, tax, data compliance and other subject matter experts in the development and execution of innovative and complex strategic business transactions, including (but not limited to) innovative automotive industry initiatives. He has recently structured and advised automotive industry clients on the development of state-of-the-art consumer vehicle subscription, leasing and fleet leasing programs and related aspects of these programs, including novel automotive telematics and driver and vehicle data collection and monetization programs. Furthermore, Mr. Goldschmidt has extensive experience working with early-stage technology companies and entrepreneurs, advising in such matters as incorporation and start-up, equity and debt financings, and technology licensing and transfers. He has restructured global custody and securities lending agreements with a number of high-profile financial institutions, and he has represented one of the world's largest private equity investors in numerous international private equity transactions. While in law school, he was the managing editor of the Environmental Affairs Law Review at the Boston College Law School. Before law school, Mr. Goldschmidt served as an infantry officer in the United States Army, primarily in Airborne and Ranger units. He left active duty at the rank of Captain. Prior to his service in the United States Army, Mr. Goldschmidt attended the United States Military Academy at West Point, where he was elected chairman of the Cadet Honor Committee, which is responsible for overseeing the functioning of West Point's Honor Code and System. Credentials - Boston College Law School, J.D., cum laude - United States Military Academy, West Point, B.S. - Colorado - Massachusetts - The Best Lawyers in America guide, Banking and Finance Law, 2018-2021 - Holland & Knight Pro Bono All-Star, 2020	187,162
Regions bank locations in tennessee - REGIONS BANK IN TENNESSEE LIST OF CITIES AND BRANCHES IN TN Farrport 2734 Airport Highway, Alcoa 37701 Midland 216 South Calderwood Street, Alcoa 37701 Antioch Athens Bartlett 6231 Stage Road, Bartlett 38134 Berry Hill Melrose 2608 8th Avenue South Suite 101, Berry Hill 37204 Bolivar Bolivar 201 W. Market Street, Bolivar 38008 Brentwood Bristol Brownsville Camden 120 S Forrest Ave, Camden 38320 Chattanooga North River 345 Frazier Avenue, Suite 109, Chattanooga 37405 Highland 3894 Hixson Pike, Chattanooga 37415 Brainerd 5515 Brainerd Road, Chattanooga 37411 Clarksville Riverside 1001 South Riverside Drive, Clarksville 37040 Cleveland Clinton Clinton 245 North Main Street, Clinton 37716 Collierville Columbia Shadybrook 884 S. James Campbell Boulevard, Columbia 38401 Cookeville Cordova Covington Crossville Dickson Dickson 116 Mathis Drive, Dickson 37055 Dover Dover 705 Donelson Parkway, Dover 37058 Dyersburg Bypass 580 Us Highway 51 Bypass E, Dyersburg 38024 Erin Erin 4657 W. Main Street, Erin 37061 Franklin Franklin 1206 Murfreesboro Road, Franklin 37064 Germantown Goodlettsville Henderson Henderson 611 North Church Street, Henderson 38340 Hendersonville Glenbrook 1042 Glenbrook Way, Hendersonville 37075 Hermitage Hixson Hohenwald Humboldt Vinings 1861 N. Highland, Jackson 38305 Jefferson City Johnson City Jonesborough Knoxville Magnolia 1503 Magnolia Avenue, Knoxville 37917 Farragut 11513 Kingston Pike, Knoxville 37934 West Town 7821 Kingston Pike, Knoxville 37919 Knollwood 151 Major Reynolds Place, Knoxville 37919 Gay Street 465 S. Gay Street, Suite 101, Knoxville 37902 Sequoyah 4600 Kingston Pike, Knoxville 37919 Powell 6601 Clinton Highway, Knoxville 37912 Northshore 1935 Pinnacle Pointe Way, Knoxville 37922 La Vergne 5021 Murfreesboro Road, La Vergne 37086 Lakeland Lebanon Lebanon 715 West Main Street, Lebanon 37087 Lenoir City Lexington Lexington 26 E. Church Street, Lexington 38351 Madison 500 S. Gallatin Road, Madison 37115 Madisonville Martin Martin 844 University Street, Martin 38237 Maryville Walland 2214 East Lamar Alexander Parkway, Maryville 37804 Mc Kenzie Mc Kenzie 15820 Highland Drive, Mc Kenzie 38201 McMinnville Memphis Lamar 2800 Lamar Avenue, Memphis 38114 Berclair 4383 Summer Avenue, Memphis 38122 Poplar 3307 Poplar Avenue, Memphis 38111 Southwind 3577 Hacks Cross Road, Memphis 38125 Milan Morristown Mount Juliet Munford Munford 86 S. Tipton Street, Munford 38058 Murfreesboro Innsbrooke 115 Innsbrooke Blvd, Murfreesboro 37128 Plus Park 831 Murfreesboro Road, Nashville 37217 Woodbine 2511 Nolensville Road, Nashville 37211 Village 21 1624 21st Avenue South, Nashville 37212 Donelson 2409 Lebanon Road, Nashville 37214 100 Oaks 729 Thompson Lane, Nashville 37204 Bordeaux 3701 Clarksville Highway, Nashville 37218 Music Row 1600 Division Street, Suite 100, Nashville 37203 Norris Norris 30 West Norris Street, Norris 37828 Old Hickory Oliver Springs Ooltewah Ooltewah 5596 Little Debbie Parkway, Ooltewah 37363 Powell Red Bank 3401 Dayton Boulevard, Red Bank 37415 Ripley Rockwood Rockwood 240 West Rockwood Street, Rockwood 37854 Selmer 116 South Third Street, Selmer 38375 Sewanee Sewanee 69 University Avenue, Sewanee 37375 Shelbyville Signal Mountain Smithville Smithville 120 South Public Square, Smithville 37166 Smyrna Smyrna 301 South Lowry Street, Smyrna 37167 Soddy Daisy Somerville Spring Hill Springfield Sweetwater Sweetwater 401 North Main Street, Sweetwater 37874 Trenton Tullahoma Tullahoma 1200 North Jackson Street, Tullahoma 37388 Union City 1702 West Reelfoot Avenue, Union City 38261 Waverly 201 Waverly Plaza, Waverly 37185 White House Winchester Winchester 121 South College Street, Winchester 37398 Woodbury Woodbury 200 West Water Street, Woodbury 37190 Routing number regions il routing number regions il 062000019 – Alabama. 064000017. Collinsville, Illinois 62234 There are several routing numbers for Regions Bank reported in our bank database. Find the nearest Regions Bank near you. Jun 30, 2021 · The Southern Illinois routing number for U. Bank is 081202759. Please make sure this is the correct routing number for your branch! Sep 03, 2021 · Regions Alabama Routing Number: 062000019. There is a total of 34 unique routing numbers in the USA as per the states. Regions' routing number for domestic wire transfers is 062005690, and its international wire transfer routing number is UPNBUS44. This routing number is used for ACH and wire money transfer from Regions Bank Indianapolis to other banks in United States of America. Godfrey, Illinois 62035 There are several routing numbers for Regions Bank reported in our bank database. AL. Regions Bank Florida Routing Number - 063104668. Today we are going to provide you with the regions bank routing numbers of Florida, Alabama, Arkansas, Illinois, Texas, North Carolina, South Carolina, Georgia, and many more. Sponsored Links Routing Numbers Alabamas — 062000019 Arkansas — 082000109 Florida — 063104668 Georgia — 061101375 Illinois — 071122661 Indiana Sep 30, 2021 · Regions’ routing number for wire transfers is 062005690, and international wire transfer routing number is UPNBUS44. BOX 681 BIRMINGHAM, AL 35201-0000 800 734-4667. Dec 19, 2002 · 062005690. Regions Bank Illinois Routing Number - 071122661. REGIONS BANK. The Regions Bank wire transfer routing number is 062005690 for any transfers inside of the United States. After you provide a few specific details to identify yourself, a representative will be able to confirm your account’s routing number. Make sure to take note of these numbers, especially the international Wire Transfer routing number. Regions Bank Georgia Routing Number - 061101375. Please call the number next to a routing number below to confirm your Regions Bank Collinsville Facility routing number for wire transfer, reorder checks or setting up direct deposit. Oct 09, 2017 · Here is the image of a check of the Regions Bank where the routing number is given in the left side of the check. You can then choose, 'See full account number' next to your account name and a box will open to display your account and routing number. The routing number is of 9 digits and therefore, it starts from 11 as shown in the image that is for example. Account Number. The ACH routing number will have to be included Nov 04, 2005 · Regions Bank A routing number is a nine digit code, used in the United States to identify the financial institution. GET DIRECTIONS. 061101375 – Georgia. BOX 681,BIRMINGHAM, AL - 35201, Telephone : 800-734-4667 Apr 29, 2021 · The ACH routing number for Regions Bank is 062000019. For International Wire Transfer: 062005690. Sometimes, banks have multiple routing numbers for different branches or uses. Zip. For Wire Transfer Routing Number: 062000019. Busey Bank Checking Routing Number. The checking and saving account routing number and the ACH routing number for PNC varies state by state, you can find these in the table above. 111900785. It's also referred to as RTN, routing transit number or bank routing number. Short for Automatic Clearing House, ACH numbers are unique to each bank in the US. Tennessee. Sep 03, 2021 · Regions Alabama Routing Number: 062000019. Regions Bank’s online transfer FAQ is located here. To see whether a check is a local or a nonlocal check, look at the routing number on the check: Personal Check Business Check If the first four digits of the routing number (1234 in the examples above) are (list of local numbers), then the check is a local check. Otherwise, the check is a nonlocal check. Virginia. Please call the number next to a routing number below to confirm your Regions Bank Godfrey Branch routing number for wire transfer, reorder checks or setting up direct deposit. The business offers services such as retail banking, commercial banking, trust, stock brokerage, and mortgage . For wire transfers inside of the United States, the routing number is 062005690. Routing number 071109338 is assigned to ILLINOIS NATIONAL BANK located in SPRINGFIELD, IL. In the example, you would use 34 to determine your routing number using the chart below. O. The real-time regions bank routing number in the USA is a world-class example of banking safety and cybersecurity. Texas. This number system is used by the US banks for carrying out Automatic Clearing House and wire transfers. Please call the number next to a routing number below to confirm your Regions Bank Wood River Branch routing number for wire transfer, reorder checks or setting up direct deposit. Regions Bank Routing Numbers Regions Financial Corporation is a bank holding company with headquarters in Birmingham, Alabama, in the Regions Centre. Sep 29, 2021 · If you have a U. You will be using this address for any wire transfers even if you bank at a different location. For ACH transfers to any Regions Bank account, you would need an ACH routing number. You see, the routing number or the ABA routing number is available on your checkbook. Regions Bank Iowa Routing Number - 073900438. Bank any time at 800-872-2657 to find the routing number for your account. Regions Bank's routing number (the leftmost number on the bottom of a check) is 53012029. Routing Number of Regions Bank. Bank Name. Aug 11, 2020 · When making a Regions Bank wire transfer, you’ll need to use routing number 062005690 and also supply the bank’s name, your name and account number, and address: Regions Bank. The ACH number and your bank account number are used by banks and transfer apps like Zelle or Cash App to identify the exact account payments should be taken from and sent to. Regions Bank’s customer service number is 800-734-4667. Regions Bank Indiana Routing Number - 074014213. Wire transfers made outside of the United States to a USA account have a routing number of 062005690. Belleville, Illinois 62226 There are several routing numbers for Regions Bank reported in our bank database. Regions Bank allows users to transfer funds in many ways, including by wire transfer, Western Union, or through Personal Pay, a streamlined transfer service operated by Popmoney. Routing numbers to Regions Bank in Belleville are collected manually from the banks official website or provided by the Federal Reserve Financial Services Database. Please call the number next to a routing number below to confirm your Regions Bank Belleville Main Branch routing number for wire transfer, reorder checks or setting up direct deposit. Thus, these routing numbers are on the left most side of the below corner of the check. Birmingham, AL 35203. Find your routing and account number by signing in to chase. While sending a domestic ACH transfer in a Regions Bank account, you will need to use the ACH routing number, which differs from state to state. Routing Number. Bank checking account, you can also find your routing number on a check — the check routing number is the first nine numbers in the lower left corner. . Copy to Clipboard. Routing number : 081001387, Institution Name : REGIONS BANK, Delivery Address : P. 053201814 – South Carolina. The routing number of the Regions Bank is the customary 9-digit number, which is needed for setting up several kinds of banking transactions like the direct deposits, wire transfers, electronic payments, etc. valid routing number. Routing numbers are used by Federal Reserve Banks to process Fedwire funds transfers, and ACH(Automated Clearing House) direct deposits, bill payments, and other automated transfers. Also ask questions and discuss related issues here. to the correct bank branch. 74014213. 1900 Fifth Avenue North. com and choosing the last four digits of the account number that appears above your account information. What is the Regions Bank routing number of Illinois? The Regions Bank routing number of Illinois is 071122661. Click here to find U. Wood River, Illinois 62095 There are several routing numbers for Regions Bank reported in our bank database. You can find your account number in the top of the right column of a bank statement. 062202150. Jun 25, 2020 · You can find your Regions Bank routing number on its website or by doing one of the following: Call customer service. Sep 30, 2021 · Regions Routing Numbers for Wire Transfers. Routing/Transit/RTN number is a nine-digit code that is used to identify financial institution involved in a transaction. The SWIFT code for wire transfers made outside of the United States is UPNBUS44. It is also important that you ensure the Regions Bank routing number of your account in Illinois is correct and safe to be used in all the possible ways and for all kids of transfers. Is 043000096 a PNC routing number? 043000096 is the domestic and international wire transfer routing number for PNC. Please make sure this is the correct routing number for your branch! Today we are going to provide you with the regions bank routing numbers of Florida, Alabama, Arkansas, Illinois, Texas, North Carolina, South Carolina, Georgia, and many more. Routing Numbers, also known as ABA Numbers or Routing Transfer Numbers are 9 digit numbers used by the banking system in the United States for identifying banks and financial institutions. Routing Number for Regions Bank in Illinois A routing number is a 9 digit code for identifying a financial institute for the purpose of routing of checks (cheques), fund transfers, direct deposits, e-payments, online payments, etc. 515405. Bank: REGIONS BANK: Routing Number : 081001387: Telegraphic name : UPB BELLEVILLE IL: City : SAINT LOUIS: State : MO: Funds transfer status : eligible: Funds Mar 03, 2018 · Regions Routing Numbers: The transfers that are made through the wire, paper or electronic became even more accessible by the Region Banks. Address. 071122661 – Illinois. Sep 03, 2021 · Illinois: 071000013: Indiana: 074000010: Kentucky: 083000137: You may also check our website for a separate discussion on Regions routing number, Wells Fargo Sep 28, 2021 · Head to the Account Information and Settings menu to find your account’s routing number. 34 corresponds to the routing number 074000078. The states that are not appointed with any unique routing number uses 091000022 as the routing number for recording transactions location-wise. S. One can do fund transfers or direct deposits through this Routing Number. Call Customer Service: Call PNC Bank at (888) 762-2265. Get Regions Bank routing numbers, swift codes, locations, financial information and etc on 58 branches in IL. You might not have a check handy, however, so you can also call U. 051009296. There is a lot of reasons as to why the Routing Numbers of the bank is necessary for the transfer of funds. Feb 10, 2021 · Regions Bank Arkansas Routing Number - 082000109. ABA routing number 071109338 is used to facilitate ACH funds transfers and Fedwire funds transfers. The Routing Number is also termed as the routing transit number or RTN, ABA number, and ACH number. Regions Bank [REGIONS BK] P. What Is a Routing Number? A routing number is a nine-digit code on checks and is sometimes referred to as the ABA number or check routing number. [info] 4/19/2011. The SWIFT code for Regions Bank is UPNBUS44. 053012029 – North Carolina. Below is a list of Banks with ACH Departments in the State of IL. Regions’ routing number for wire transfers Regions Bank routing number for ACH transfers. Bank: REGIONS BANK: Routing Number : 081001387: Telegraphic name : UPB BELLEVILLE IL: City : SAINT LOUIS: State : MO: Funds transfer status : eligible: Funds Sep 30, 2021 · Regions’ routing number for wire transfers is 062005690, and international wire transfer routing number is UPNBUS44. Please refer to the Miscellaneous Deposit Fees section in the Regions Pricing Schedule for a complete list of wire transfer fees. In Illinois, the ACH routing number is the same, also being 071122661. Bank routing numbers in other states. Please note: Beginning on April 23, 2021, Busey resumed sending Notices of Change (NOCs) to the companies with whom customers have established automatic payments or deposits using legacy routing numbers, advising them to update the routing number to 071102568. 073900438 – Iowa. You’ll supply the bank’s name, your name and account number, and the following bank address: Regions Bank. 074014213 – Indiana. A routing number is a nine-digit numeric code printed on the bottom of checks that is used to facilitate the electronic routing of funds (ACH transfer) from one bank account to another. Your Regions transit/routing number is the first 9 digits located at the bottom corner of your deposit slips and/or checks. Jul 19, 2020 · Look up Bank Routing Numbers by Bank Name or State. Regions Bank Routing Number for ACH transfers. Complete list of the 1,325 Regions Bank locations with address, financial information, reviews, routing numbers etc. 082000109 – Arkansas. Aug 27, 2021 · Here is a list of Regions Bank routing numbers for each state where the institution operates. routing number regions il Here’s Your State’s Regions Bank Routing Number Banking / Banks How to Find Your Routing Number There are a few ways to find a Regions Bank routing number. If you remember which state you opened your account in, just use the chart. If you have a Regions checking account, you can also find your routing number on a check — the check routing number is the first nine numbers in the lower left corner. You might not have a check handy, however, so you can also call Regions.. Regions Routing Numbers for Wire Transfers Regions’ routing number for wire transfers is 062005690, and international wire transfer routing number is UPNBUS44. You’ll. Information accurate as of Sept. 30, 2021.. Your Checking. Your Story. Android is a trademark of Google Inc. Insurance Products, Investments & Annuities: Not A Deposit (513) 777-5900 Community Development: 9577 Beckett Road, Suite 100 4 Replies to “Regions bank locations in tennessee” Tujh pe bharosa kar ke tera channel subscribe kiya tha but ab unsubscribe kar rha hu No need of character certificate Schwab has average to lower fees than most robos Sir Clark ka Full preparation bata do Plz.....Kya kya Padhe ?	350,416
The online series follows food trends in five cities and features a popular young talent who hosts a show on the Scripps-owned Food Network called Good Deal With Dave Lieberman. Because Mr. Lieberman is in his mid-20s, Scripps hopes that his contemporaries, who are about 15 years younger than the middle-aged audience the cable network draws, will watch the Web series and then discover the cable programming. "We definitely push back to the network and get the audience as young as it can be," said Beth Higbee, vice president for online for Foodnetwork.com, Fineliving.com and GACTV.com. Scripps will place two or three parts of the cooking series on Foodnetwork.com a few days before Thanksgiving, when traffic typically spikes. Last November, the site received 7.4 million visitors (about 2 million more than non-holiday months), according to Nielsen/NetRatings. Target: 8 million viewers This year Ms. Higbee expects to surpass 8 million visitors. Each segment is between two and five minutes long. "We asked people point blank in focus groups if they would watch 30-minute programming," she said. "They said their screens and connections are not on a par they would need." The Foodnetwork.com informational content lends itself to what people are accustomed to seeking out online, Ms. Higbee added. "If there's something educational about it, they will stick around to watch it." Experts say consumers watch mostly sports, news and movie trailers on broadband. Lifestyle offerings like Food Network's nestle nicely between being educational and entertaining. It may take a how-to nugget to convert Web users to watch video online, Ms. Higbee said.	221,647
\begin{document} \maketitle \begin{abstract} We study rational generating functions of sequences $\{a_n\}_{n\geq 0}$ that agree with a polynomial and investigate symmetric decompositions of the numerator polynomial for subsequences $\{a_{rn}\}_{n\geq 0}$. We prove that if the numerator polynomial for $\{a_n\}_{n\geq 0}$ is of degree $s$ and its coefficients satisfy a set of natural linear inequalities then the symmetric decomposition of the numerator for $\{a_{rn}\}_{n\geq 0}$ is real-rooted whenever $r\geq \max \{s,d+1-s\}$. Moreover, if the numerator polynomial for $\{a_n\}_{n\geq 0}$ is symmetric then we show that the symmetric decomposition for $\{a_{rn}\}_{n\geq 0}$ is interlacing. We apply our results to Ehrhart series of lattice polytopes. In particular, we obtain that the $h^\ast$-polynomial of every dilation of a $d$-dimensional lattice polytope of degree $s$ has a real-rooted symmetric decomposition whenever the dilation factor $r$ satisfies $r\geq \max \{s,d+1-s\}$. Moreover, if the polytope is Gorenstein then this decomposition is interlacing. \end{abstract} \section{Introduction} For integers $r\geq 1$ we consider the operator $\Urd{r}{d}\colon \mathbb{R}[t]\rightarrow \mathbb{R}[t]$ defined in such a way that \[ \sum _{n\geq 0} a_{rn}t^n =\frac{\Urd{r}{d} h(t)}{(1-t)^{d}}\quad \text{ whenever } \quad \sum _{n\geq 0} a_{n}t^n =\frac{h (t)}{(1-t)^{d}} \] for polynomials $h(t)\in \mathbb{R}[t]$. We investigate properties of $\Urd{r}{d} h (t)$ as a function of $r$. We study symmetric decompositions of $\Urd{r}{d+1} h(t)$ and their roots for polynomials $h(t)=h_0+h_1t+\cdots h_st^s$, $s\leq d$, with nonnegative coefficients that satisfy both the following sets of inequalities \begin{itemize} \item[\hypertarget{H}{(H)}] $h_0+h_1+\cdots + h_{i}\geq h_d+h_{d-1}+\cdots +h_{d-i+1}$ for all $i$, and \item[\hypertarget{S}{(S)}] $h_0+h_1+\cdots + h_{i}\leq h_s+h_{s-1}+\cdots +h_{s-i}$ for all $i$, \end{itemize} where $h_i:=0$ for all $i\not \in \{0,1,\ldots, s\}$. Our motivation comes from Ehrhart theory, and more generally, commutative algebra. If $\sum _{n\geq 0}a_nt^n$ is the Hilbert series of a graded ring $R$ then $\sum _{n\geq 0}a_{nr}t^n$ is the Hilbert series of its $r$-th Veronese subalgebra $R^{\langle r\rangle}$. In Ehrhart theory, which is concerned with the enumeration of lattice points in dilates of lattice polytopes, $\Urd{r}{d+1} h^\ast _P(t)$ is equal the $h^\ast$-polynomial of the $r$-th dilate of a $d$-dimensional lattice polytope $P$. The Properties \hyperlink{H}{(H)} and \hyperlink{S}{(S)} naturally appear in this context: Property \hyperlink{S}{(S)} and the nonnegativity of the coefficients of $h(t)$ were proved by Stanley~\cite{StanleyS} for Hilbert series of semistandard graded Cohen-Macaulay domains. Property \hyperlink{H}{(H)} was proved to hold by Hibi~\cite{HibiH} for $h^\ast$-polynomials of lattice polytopes, using methods from commutative algebra. The limiting behavior of $\Urd{r}{d} h(t)$ was studied by Beck and Stapledon~\cite{Beck2010}, Brenti and Welker~\cite{BrentiWelker}, and Diaconis and Fulman~\cite{Diaconis2} who proved under mild conditions on $h(t)$ that $\Urd{r}{d} h (t)$ eventually has only real roots. In~\cite{jochemko2018} it was shown that for every polynomial $h(t)$ with nonnegative coefficients $\Urd{r}{d} h(t)$ has only real roots whenever $r\geq \deg h$ thereby proving a uniform bound conjectured by Beck and Stapledon~\cite{Beck2010}. See also~\cite{Zhang} for related work. In the present article we provide a uniform bound on $r$ for which the symmetric decomposition of $\Urd{r}{d} h(t)$ has only real zeros. With basic linear algebra it can be observed that for every polynomial of degree at most $d$ there are uniquely determined symmetric polynomials $p(t)=t^dp(1/t)$ and $q(t)=t^{d-1}q(1/t)$ such that $h(t)=p(t)+tq(t)$. The pair $(p,q)$ is called the \textbf{symmetric decomposition} (with respect to degree $d$) of $h(t)$. In case $h(t)$ is the $h^\ast$-polynomial of a lattice polytope with an interior lattice point Betke and McMullen~\cite{BetkeMcMullen} gave a combinatorial interpretation for $p(t)$ and $q(t)$ from which the nonnegativity of their coefficients follows. Stapledon~\cite{stapledon2009inequalities} considered more general decompositions (see Lemma~\ref{lem:extendedStapledon}) in order to study inequalities amongst the coefficients of the $h^\ast$-polynomial of arbitrary lattice polytopes. In particular, he obtained purely combinatorial proofs for the Properties \hyperlink{H}{(H)} and \hyperlink{S}{(S)} in case $h(t)$ is the $h^\ast$-polynomial of a lattice polytope. Br\"and\'en and Solus~\cite{branden2018symmetric} recently initiated a systematic study of symmetric decompositions for which $p$ and $q$ have only nonnegative coefficients and only real roots. Such decompositions are called \textbf{nonnegative} and \textbf{real-rooted}. Real-rooted symmetric decompositions and real-rooted polynomials in general are of current particular interest in combinatorics (see, e.g., \cite{athanasiadis2020,haglund2019real,hlavacek2020subdivisions,savage2015s}), and especially also in Ehrhart theory (see, e.g., \cite{jochemkozonotopes,ferroni2020ehrhart,higashitani19,SolusSimplices}), due to their applicability to unimodality questions. A polynomial $h(t)=h_0+h_1t+\cdots + h_dt^d$ is called \textbf{unimodal} if $h_0\leq h_1 \leq \cdots \leq h_k \geq \cdots \geq h_d$ for some $k$. A central open conjecture in Ehrhart theory going back to Stanley~\cite{Stanleyunimodal} states that the $h^\ast$-polynomial of every lattice polytope having the integer decomposition property (IDP) is unimodal. Schepers and Van Langenhoven~\cite{Schepers} strenghtened this conjecture for IDP polytopes with interior lattice points by asking if every such polytope has an alternatingly increasing $h^\ast$-polynomial. The alternating increasing property is equivalent to both parts $p(t)$ and $q(t)$ in the symmetric decomposition being nonnegative and unimodal~\cite{jochemkozonotopes}. One way of proving that a polynomial with nonnegative coefficients is unimodal is by showing that it has only real roots. In fact, even stronger, if $h(t)=h_0+h_1t+\cdots + h_dt^d$ is real-rooted and has only nonnegative coefficients then it is \textbf{log-concave}, that is, $h_{i-1}h_{i+1}\leq h_i^2$ is satisfied for all $i$ (see, e.g., \cite[Lemma 1.1]{Brandenunimodal}). For further reading on real-rooted, log-concave and unimodal polynomials and their applications we recommend~\cite{Brandenunimodal,Braununimodal,Brentiunimodal,Stanleyunimodal}. By a result of Bruns, Gubeladze and Trung~\cite{Bruns} every lattice polytope eventually becomes IDP under dilations. Hering~\cite{Hering} moreover showed that if $P$ is a lattice polytope whose $h^\ast$-polynomial has degree $s$ then $rP$ is IDP whenever $r\geq s$. On the other hand, $r=d+1-s$ is the smallest integer such $rP$ has an interior lattice point. In light of Schepers and Van Langenhoven's question~\cite{Schepers} it is natural to study the symmetric decomposition of the $h^\ast$-polynomial under dilation and, in particular, for dilation factors $r\geq \max\{s,d+1-s\}$. This corresponds to studying $\Urd{r}{d+1} h^\ast _P(t)$ for $r\geq \max\{s,d+1-s\}$. In Section~\ref{sec:symmetricdecomp} we derive an explicit formula for the symmetric decomposition of $\Urd{r}{d+1}h(t)$ (Proposition~\ref{prop:formula}) by generalizing the decompositions studied by Stapledon in \cite{stapledon2009inequalities} (Proposition~\ref{prop:decomp2}). In Section~\ref{sec:realrootedness}, we study the roots of these symmetric decompositions. The following is our main result. \begin{thm}\label{thm:main} Let $h(t)=h_0+h_1t+\cdots +h_st^s$ be a polynomial of degree $s\leq d$ with nonnegative coefficients such that \begin{itemize} \item[(H)] $h_0+h_1+\cdots + h_{i}\geq h_d+h_{d-1}+\cdots +h_{d-i+1}$ for all $i$, and \item[(S)] $h_0+h_1+\cdots + h_{i}\leq h_s+h_{s-1}+\cdots +h_{s-i}$ for all $i$. \end{itemize} Then the polynomial $\Urd{r}{d+1} h(t)$ has a nonnegative real-rooted symmetric decomposition whenever $r\geq \max \{s, d+1-s\}$. \end{thm} This theorem strengthens results by Higashitani~\cite{higashitani2009} who proved that under the same conditions $\Urd{r}{d+1} h(t)$ has a log-concave symmetric decomposition whenever $r\geq \max \{s, d+1-s\}$. To prove Theorem~\ref{thm:main} we use the method of interlacing polynomials, a powerful method that gained a lot of recent attention due to its role in the proof of the longstanding Kadison Singer Conjecture by Marcus, Spielman and Srivastava~\cite{KadisonSinger}. A real-rooted polynomial $f\in \mathbb{R}[t]$ with roots $s_k\leq s_{k-1}\leq \cdots \leq s_1$ is said to \textbf{interlace} a real-rooted polynomial $g\in\mathbb{R}[t]$ with roots $t_m\leq t_{m-1}\leq \cdots \leq t_1$ and we write $f\preceq g$ if \[ \ldots \leq s_2\leq t_2\leq s_1\leq t_1 \, . \] In Section~\ref{sec:preliminaries} we collect necessary preliminaries on interlacing polynomials, the operator $U_r^d$ as well as symmetric decompositions. We call a symmetric decomposition $(p,q)$ \textbf{interlacing} if $q\preceq p$. In Section~\ref{sec:interlacing} we investigate interlacing symmetric decompositions of $\Urd{r}{d+1}h(t)$. We provide a simple characterization (Proposition~\ref{lem:essential}) and use it to prove the following. \begin{thm}\label{thm:main2} Let $h(t)$ be a polynomial of degree $s\leq d$ with nonnegative coefficients and such that $h(t)=t^{s}h(1/t)$. Then $\Urd{r}{d+1} h(t)$ has a nonnegative interlacing symmetric decomposition for all $r\geq \max\{s,d+1-s\}$. \end{thm} We furthermore show that $\Urd{r}{d+1} h(t)$ has a nonnegative interlacing symmetric decomposition for all polynomials $h(t)$ with nonnegative coefficients and $\deg h(t)\leq \frac{d+1}{2}$ whenever $r\geq d+1$ (Proposition~\ref{prop:small}). We conclude in Section~\ref{sec:Ehrharttheory} by applying our results to $h^\ast$-polynomials of dilated lattice polytopes. In particular, we obtain that the $h^\ast$-polynomial of every dilation of a $d$-dimensional lattice polytope of degree $s$ has a real-rooted symmetric decomposition whenever the dilation factor $r$ satisfies $r\geq \max \{s,d+1-s\}$ (Corollary~\ref{cor:Ehrrealrooted}). If the polytope is Gorenstein then this decomposition is also interlacing (Corollary~\ref{cor:Ehrinterlacing}). \section{Preliminaries}\label{sec:preliminaries} \subsection{Symmetric Decompositions} For a polynomial $h(t)\in \R[t]$ of degree at most $d$, we define \[ \mathcal{I}_d (h(t)) = t^dh(1/t) \, . \] If $\mathcal I _d (h(t))=h(t)$ then $h(t)$ is called \textbf{symmetric} or \textbf{palindromic} (with center of symmetry at $d/2$.) From basic linear algebra it follows that for every polynomial $h(t)$ of degree at most $d$ there exist uniquely determined symmetric polynomials $p(t) = t^d p(1/t)$ and $q(t) = t^{d-1}q(1/t)$ such that $h(t) = p(t) + tq(t)$. We call $(p,q)$ the \textbf{symmetric decomposition} of $h(t)$. The following was observed by Br\"and\'en and Solus~\cite{branden2018symmetric}. \begin{lem}[{\cite{branden2018symmetric}}]\label{lem:solus} Let $h(t)$ be a polynomial of degree at most $d$. Then its symmetric decomposition $(p,q)$ is given by \[ p(t) = \frac{h(t)-t\mathcal{I}_d (h)}{1-t} \quad \text{ and } \quad q(t) = \frac{\mathcal{I}_d(h)-h(t)}{1-t} \, . \] \end{lem} In~\cite{stapledon2009inequalities} Stapledon considered also more general symmetric decompositions. For a polynomial $h(t)$ of degree $s\leq d$ the \textbf{co-degree} of $h(t)$ is defined by $\ell:=d+1-s$. \begin{lem}[{\cite{stapledon2009inequalities}}]\label{lem:extendedStapledon} Let $h(t)=\sum _{i=0}^s h_it^i$ be a polynomial of degree $s\leq d$ and let $(p,q)$ be its symmetric decomposition. Then there are uniquely determined symmetric polynomials $p_\ell(t)=\sum_{i=0}^{d}p_{\ell,i}t^i$ and $q_\ell(t)=\sum_{i=0}^{s-1}q_{\ell,i}t^i$ satisfying $p_\ell(t)=t^{d}p(1/t)$ and $q_\ell(t)=t^{s-1}q_\ell(1/t)$ such that \begin{equation}\label{eq:extendedStapledon} (1+t+\cdots +t^{\ell-1})h(t)=p_\ell(t)+t^\ell q_\ell(t) \, , \end{equation} namely \[ p_{\ell,i} \ = \ h_0+\cdots + h_i - h_{d}-\cdots - h_{d+1-i} \, \] and \[ q_{\ell,i} \ = \ -h_0-\cdots -h_i + h_s+\cdots + h_{s-i} \, , \] where $h_i := 0$ whenever $i\not \in \{0,1,\ldots,s\}$. In particular, $p_\ell (t)$ equals $p(t)$ in the symmetric decomposition of $h(t)$. \end{lem} Underlying the arguments in~\cite{stapledon2009inequalities} is the following observation. \begin{obs}\label{obs:obs3} The polynomial $h(t)$ as given in Lemma~\ref{lem:extendedStapledon} satisfies the Properties \hyperlink{H}{(H)} and \hyperlink{S}{(S)} if and only if $p_\ell (t)=p(t)$ and $q_\ell (t)$ have nonnegative coefficients. \end{obs} \subsection{Interlacing polynomials} A real-rooted polynomial $f\in \mathbb{R}[t]$ with roots $s_k\leq s_{k-1} \leq \cdots \leq s_1$ is said to \textbf{interlace} a real-rooted polynomial $g\in\mathbb{R}[t]$ with roots $t_m\leq t_{m-1} \leq \ldots \leq t_1$, and we write $f\preceq g$, if \[ \ldots \leq s_2 \leq t_2 \leq s_1 \leq t_1 \, . \] In particular, $\deg g=m$ is equal to either $\deg f=k$ or $\deg f +1 =k+1$. For technical reasons we also set $0\preceq f$ and $f\preceq 0$ for all real-rooted polynomials $f$. The following lemma collects some basic facts about interlacing polynomials that can, for example, be found in~\cite{wagner1992total}. \begin{lem}[{\cite[Section 3]{wagner1992total}}]\label{lem:basicsinterlace} Let $f,g,h\in \R [t]$ be real-rooted polynomials with positive leading coefficients. Then \begin{itemize} \item[(i)] $g\preceq f$ if and only if $cg\preceq df$ for all $c,d\neq 0$. \item[(ii)] $h\preceq f$ and $h\preceq g$ implies $h\preceq f+g$. \item[(iii)] $f\preceq h$ and $g\preceq h$ implies $f+g\preceq h$. \item[(iv)] $g\preceq f$ if and only if $f\preceq tg$, if $f$ and $g$ have only nonpositive roots. \end{itemize} \end{lem} A sequence of polynomials $(f_1,\ldots, f_m)$ is called an \textbf{interlacing sequence} if $f_i\preceq f_j$ whenever $i\leq j$. The following results are due to Br\"and\'en~\cite{Petter2,Brandenunimodal}. \begin{lem}[{\cite[Lemma 8.3]{Brandenunimodal}}]\label{lem:commoninterlacer} If $(f_1,f_2,\ldots, f_m)$ and $(g_1,g_2,\ldots, g_m)$ are two sequences of interlacing polynomials with positive leading coefficients then \[ f_1g_m+f_2g_{m-1}+\cdots + f_mg_1 \] is real-rooted. \end{lem} In particular, if $(f_1,f_2,\ldots, f_m)$ is an interlacing sequence of polynomials with positive leading coefficients then \[ c_1f_1+c_2f_2+\cdots + c_mf_m \] is real-rooted for every choice of $c_1,\ldots,c_m\geq 0$. \begin{lem}[{\cite[Lemma 2.3]{Petter2}}]\label{lem:interlacingends} Let $f_1,\ldots, f_m$ be polynomials such that $f_i\preceq f_{i+1}$ for all $1\leq i\leq m-1$ and such that $f_1\preceq f_m$. Then $f_i\preceq f_j$ for all $1\leq i\leq j\leq m$, that is, $(f_1,\ldots,f_m)$ is an interlacing sequence. \end{lem} In~\cite{branden2018symmetric} Br\"and\'en and Solus studied real-rooted and interlacing symmetric decompositions. A symmetric decomposition $(p,q)$ is called \textbf{real-rooted} if $p$ and $q$ are both real-rooted. If moreover $q\preceq p$ then the symmetric decomposition is called \textbf{interlacing}. The following theorem by Br\"and\'en and Solus~\cite{branden2018symmetric} characterizes interlacing symmetric decompositions. \begin{thm}[{\cite[Theorem 2.7]{branden2018symmetric}}]\label{thm:solus2} Let $h(t)$ be a polynomial of degree at most $d$ and let $(p,q)$ be its symmetric decomposition. If $p(t)$ and $q(t)$ have only nonnegative coefficients, then the following are equivalent. \begin{itemize} \item[(i)] $q\preceq p$. \item[(ii)] $p\preceq h$. \item[(iii)] $q\preceq h$. \item[(iv)] $\mathcal I _d (h)\preceq h$. \end{itemize} \end{thm} \subsection{Two operators} A fundamental fact about generating function is that for every sequence $\{a_n\}_{n\geq 0}$ that eventually agrees with a polynomial of degree at most $d$ there is a unique polynomial $h(t)$ such that \[ \sum _{n\geq 0}a_{n}t^n = \frac{h(t)}{(1-t)^{d+1}} \, . \] Vice versa, via this equation every polynomial $h(t)$ defines such a sequence (see, e.g.,~\cite[Corollary 4.3.1]{EC1}). If $\{a_n\}_{n\geq 0}$ that eventually agrees with a polynomial of degree at most $d$ then the same holds for $\{a_{rn}\}_{n\geq 0}$. Thus there is a unique polynomial $\Urd{r}{d+1} h (t)$ such that \[ \sum _{n\geq 0}a_{rn}t^n = \frac{\Urd{r}{d+1} h (t)}{(1-t)^{d+1}}. \] We observe that the map $h(t)\mapsto \Urd{r}{d+1} h (t)$ defines a linear operator $\R[t]\rightarrow \R[t]$. Moreover, $h(t)$ has degree at most $d$ if and only if $\{a_n\}_{n\geq 0}$ agrees with a polynomial of degree at most $d$ for all $n\geq 0$. In particular, in this case also $\{a_{rn}\}_{n\geq 0}$ is given by a polynomial of degree at most $d$, and therefore also $\Urd{r}{d+1} h (t)$ has degree at most $d$. In particular, the symmetric decomposition of $\Urd{r}{d+1} h (t)$ exists. To every Laurent series $f(t)=\sum _{n=-\infty}^\infty a_nt^n$ and any natural number $r\geq 1$ there are uniquely defined Laurent series $f_0,f_1,\ldots,f_{r-1}$ such that \[ f(t) \ = \ f_0(t^r)+rf_1(t^r)+\cdots + t^{r-1}f_{r-1}(t^r) \, . \] For all $0\leq i\leq r-1$ we define $f\hri{r}{i}=f_i$. Then $\hri{r}{i}$ defines a linear operator on Laurent series. This operator extends the definition given in~\cite{jochemko2018} from formal power series to Laurent series. (Also compare~\cite{GilRobins} where these operators have been studied in a more general setup.) The following result relates the two operators $\Urd{r}{d+1} $ and $\hri{r}{0}$. \begin{lem}[{\cite{Beck2010,BrentiWelker}}]\label{lem:dilatedh} For integers $r\geq 1$ and $d\geq 0$ and any polynomial $h(t)$ \[ \Urd{r}{d+1} h (t) = (h(t)(1+t+\cdots +t^{r-1})^{d+1})\hri{r}{0} \, . \] \end{lem} A key element in~\cite{jochemko2018} was the following set of polynomials: For $r\geq 1$, $d\geq 0$ and $0\leq i\leq r-1$ let \[ a_d \hri{r}{i} (t)=((1+t+\cdots + t^{r-1})^d)\hri{r}{i} \, . \] The following result was proved in~\cite{jochemko2018} (also compare Fisk~\cite[Example 3.76]{fisk2006polynomials}). \begin{prop}[{\cite[Proposition 3.4]{jochemko2018}}]\label{prop:ans} For all integers $r,d\geq 1$ the polynomials \[ \left(a_d\hri{r}{r-1}, a_d\hri{r}{r-2},\ldots, a_d \hri{r}{0}\right) \] form an interlacing sequence. \end{prop} Equivalently, by Lemma~\ref{lem:basicsinterlace}, \[ \left(a_d \hri{r}{0},ta_d\hri{r}{r-1}, ta_d\hri{r}{r-2},\ldots, ta_d \hri{r}{0}\right) \] is an interlacing sequence of polynomials. With this definition, Lemma~\ref{lem:dilatedh}, can be reformulated as follows. \begin{lem}[{\cite{jochemko2018}}]\label{lem:reform} For integers $r,d\geq 1$ and any polynomial $h(t)$ \[ \Urd{r}{d+1} h (t)=h\hri{r}{0}a_{d+1}\hri{r}{0}+h\hri{r}{1}ta_{d+1}\hri{r}{r-1}+\cdots +h\hri{r}{r-1}ta_{d+1}\hri{r}{1} \, . \] \end{lem} \section{Symmetric decomposition of $U_r^{d+1} h (t)$}\label{sec:symmetricdecomp} In this section we provide explicit formulas for the symmetric decomposition of $U_r ^{d+1}h(t)$. We need the following slight generalization of Lemma~\ref{lem:extendedStapledon}. \begin{prop}\label{prop:decomp2} Let $h(t)=\sum _{i=0}^s h_it^i$ be a polynomial of degree $s\leq d$, let $(p,q)$ be its symmetric decomposition and let $r \geq \ell=d+1-s$ be an integer. Then there are uniquely determined symmetric polynomials $v_r(t)=\sum_{i=0}^{d}v_{r,i}t^i$ and $w_r(t)=\sum_{i=0}^{r+s-\ell -1}w_{r,i}t^i$ satisfying $v_r(t)=t^{d}v_r(1/t)$ and $w_r(t)=t^{r+s-\ell -1}w_r(1/t)$ such that \begin{equation}\label{eq:extendedStapledon} (1+t+\cdots +t^{r-1})h(t)=v_r(t)+t^\ell w_r(t) \, , \end{equation} namely \[ v_{r,i} \ = \ h_0+\cdots + h_i - h_{d}-\cdots - h_{d-i+1} \, \] and \[ w_{r,i} \ = \ -h_{\ell -r}-h_{\ell +1 -r}-\cdots -h_{\ell + i-r} + h_s+h_{s-1}+\cdots + h_{s-i} \, , \] where $h_i := 0$ whenever $i\not \in \{0,1,\ldots,s\}$. In particular, $v_r(t)=p(t)$ for all $r\geq \ell$. \end{prop} \begin{proof} The argument goes along the lines of the proof of Lemma~\ref{lem:extendedStapledon} given in~\cite{stapledon2009inequalities}. Let $f(t)=\sum _i f_i t^i=(1+t+\cdots + t^{r-1})h(t)$. Then $f(t)$ is a polynomial of degree $r+s-1$ with \[ f_i = h_i+h_{i-1}+\cdots +h_{i-r+1} \, , \] where $h_i := 0$ whenever $i\not \in \{0,1,\ldots,s\}$. Since $d+1=\ell +s$, for all $i\geq \ell$ we obtain \begin{eqnarray} v_{r,i}+w_{r,i-\ell}&=&h_0+\cdots + h_i - h_{d}-\cdots - h_{d-i+1}\\&&-h_{\ell -r}-h_{\ell +1 -r}-\cdots -h_{i-r} + h_s+h_{s-1}+\cdots + h_{s-i+\ell}\\ &=&h_0+\cdots + h_i - h_{d}-\cdots - h_{s+\ell -i}\\&&-h_{\ell -r}-h_{\ell +1 -r}-\cdots -h_{i-r} + h_s+h_{s-1}+\cdots + h_{s-i+\ell}\\ &=&h_{i-r+1}+\cdots +h_i\\ &=&f_i \, , \end{eqnarray} and \[ v_{r,i}=h_0+\cdots +h_i = f_i \] for $i<\ell$. Thus, Equation~\eqref{eq:extendedStapledon} is satisfied. Moreover, by taking $v_{r,i}$ and $w_{r,i}$ as defined above, \begin{eqnarray} v_{r,i}-v_{r,d-i}&=&h_0+\cdots +h_i -h_d-\cdots -h_{d-i+1}-h_0-\cdots -h_{d-i}+h_{d}+\cdots +h_{i+1}\\ &=&0\, \end{eqnarray} and \begin{eqnarray} w_{r,i}-w_{r,r+s-\ell -1 -i}&=&-h_{\ell-r}-h_{\ell -r+1}-\cdots - h_{\ell -r+i}+h_s+h_{s-1}+\cdots +h_{s-i}\\ &&+h_{\ell -r}+h_{\ell -r+1}+\cdots + h_{s-1-i}-h_s-h_{s-1}-\cdots -h_{\ell -r +1+i}\\ &=&0 \, \end{eqnarray} which shows $v_r(t)=t^{d}v_r(1/t)$ and $w_r(t)=t^{r+s-\ell -1}w_r(1/t)$, respectively. Uniqueness of $v_r(t)$ and $w_r(t)$ with the assumed properties is easily verified. \qedhere \end{proof} \begin{prop}\label{prop:representation}\label{prop:formula} Let $h(t)$ be a polynomial of degree $s\leq d$ and let $(p,q)$ be its symmetric decomposition. Let $r\geq 1$ and let $(\tilde{p},\tilde{q})$ be the symmetric decomposition of $U_r^{d+1} h (t)$. Then \[ \tilde{p}(t)=(p(t)(1+t+\cdots + t^{r-1})^d)\hri{r}{0} \] and \[ \tilde{q}(t)=\frac{1}{t}((h(t)(1+t+\cdots + t^{r-1})-p(t))(1+t+\cdots + t^{r-1})^d))\hri{r}{0} \, . \] If furthermore $r\geq \ell=d+1-s$ then \[ \tilde{q}(t)=\frac{1}{t}(t^\ell w_r(t)(1+t+\cdots + t^{r-1})^d))\hri{r}{0} \, , \] where $w_r(t)$ is defined as in Proposition~\ref{prop:decomp2}. \end{prop} \begin{proof} Let $f(t)$ denote $\Urd{r}{d+1} h(t)$. Then, by Lemma~\ref{lem:solus}, \[ \tilde{p}(t)=\frac{f(t)-t\mathcal{I}_d (f(t))}{1-t} \, . \] We have \begin{eqnarray} f(t)&=&((p(t)+tq(t))(1+t+\cdots + t^{r-1})^{d+1}))\hri{r}{0} \end{eqnarray} and \begin{eqnarray} t\mathcal{I}_d (f(t))&=&t^{d+1}(h(1/t)(1+1/t+\cdots + 1/t^{r-1})^{d+1}))\hri{r}{0}\\ &=&(t^{r(d+1)}(p(1/t)+1/tq(1/t))(1+1/t+\cdots + 1/t^{r-1})^{d+1}))\hri{r}{0}\\ &=&(t(p(t)+q(t))(1+t+\cdots + t^{r-1})^{d+1}))\hri{r}{0} \, . \end{eqnarray} We thus obtain \begin{eqnarray} f(t)-t\mathcal{I}_d (f(t))&=&((p(t)-tp(t))(1+t+\cdots + t^{r-1})^{d+1}))\hri{r}{0} \\ &=&((1-t^r)p(t)(1+t+\cdots + t^{r-1})^{d}))\hri{r}{0} \\ &=&(1-t)(p(t)(1+t+\cdots + t^{r-1})^{d}))\hri{r}{0} \end{eqnarray} which proves the claimed formula for $\tilde{p}(t)$. For $\tilde{q}(t)$ we observe that \begin{eqnarray} t\tilde{q}(t)=f(t)-\tilde{p}(t)=((h(t)(1+t+\cdots +t^{r-1})-p(t))(1+t+\cdots + t^{r-1})^d)\hri{r}{0} \, . \end{eqnarray} If $r\geq \ell$ then, by Proposition~\ref{prop:decomp2}, \[ h(t)(1+t+\cdots +t^{r-1})-p(t)=t^\ell w_r(t) \] and the claim follows. \end{proof} \begin{cor}\label{cor:nonnegativity} Let $h(t)$ be a polynomial of degree at most $d$ and let $(\tilde p, \tilde{q})$ be the symmetric decomposition of $\Urd{r}{d+1} h(t)$. \begin{itemize} \item[(i)] If $h(t)$ satisfies Property \hyperlink{H}{(H)} then $\tilde p$ has nonnegative coefficients. \item[(ii)] If $h(t)$ has nonnegative coefficients, satisfies Property \hyperlink{S}{(S)} and $r\geq \ell$ then $\tilde q$ has nonnegative coefficients. \end{itemize} \end{cor} \begin{proof} By Observation~\ref{obs:obs3}, $p(t)$ has nonnegative coefficients whenever $h(t)$ satisfies Property \hyperlink{H}{(H)}. From Proposition~\ref{prop:formula} we see that in this case also $\tilde p(t)$ has nonnegative coefficients. If $r\geq \ell$, then by Proposition~\ref{prop:decomp2} \[ w_{r,i}=-h_{\ell -r}-h_{\ell -r +1}-\cdots -h_{\ell+i-r}+h_s+h_{s-1}+\cdots +h_{s-i}\geq -h_0-h_1-\cdots -h_i+h_s+h_{s-1}+h_{s-i}\geq 0 \] where we used that $h_i\geq 0$ for all $i$ and that $h(t)$ satisfies Property \hyperlink{S}{(S)}. Thus, again by Proposition~\ref{prop:formula}, we see that $\tilde q(t)$ has nonnegative coefficients. \end{proof} The bound given in Corollary~\ref{cor:nonnegativity} (ii) is optimal. To see this consider a polynomial $h(t)$ of degree $s$ with $h(0)>0$ and symmetric decomposition $(p,q)$. Let $(\tilde{p},\tilde{q})$ be the symmetric decomposition of $\Urd{r}{d+1}h$. By Proposition~\ref{prop:decomp2}, \[ t\tilde{q}=(((1+t+\cdots + t^{r-1})h(t)-p(t))(1+t+\cdots +t^{r-1})^d)\hri{r}{0} \, . \] If $r<\ell$, then the leading coefficient of $(1+t+\cdots + t^{r-1})h(t)-p(t)$ equals $-p_d=-p_0=-h_0<0$ which is equal to the leading coefficient of $t\tilde{q}$. \section{Real-rootedness}\label{sec:realrootedness} The goal of this section is to prove Theorem~\ref{thm:main}. We use the following lemma. \begin{lem}\label{lem:arealrooted} Let $g(t)=\sum _{i=0}^d g_it^i$ be a polynomial of degree at most $d$ with nonnegative coefficients and let $\ell\geq 1$ be such that \[ g_{0}\leq g_{1}\leq \cdots \leq g_{\ell -1} \, \quad \text{ and } \quad g_{d+1-\ell}\geq \cdots \geq g_{d-1}\geq g_{d} \, . \] Then \[ \Urd{r}{d} g (t)=\left(g(t)(1+t+\cdots + t^{r-1})^d\right)\hri{r}{0} \] is real-rooted for all $r\geq \max\{d+1-\ell,\frac{d+1}{2}\}$. \end{lem} \begin{proof} By Theorem~\ref{lem:reform} \begin{eqnarray} \Urd{r}{d} g (t)&=&g\hri{r}{0}a_{d}\hri{r}{0}+g\hri{r}{1}ta_{d}\hri{r}{r-1}+\cdots +g\hri{r}{r-1}ta_{d}\hri{r}{1}\, . \end{eqnarray} Since $r\geq \frac{d+1}{2}$, $g\hri{r}{i}$ has degree at most $1$ for all $i$. More precisely, \[ g\hri{r}{i}=\begin{cases}g_i+g_{r+i}t \quad & \text{ if } 0\leq i\leq d-r\\ g_i \quad & \text{ if } d-r+1\leq i\leq r-1 \end{cases} \] Since $d-r\leq \ell -1$, by assumption we have $g_{0}\leq g_{1}\leq \cdots \leq g_{d-r}$ and $g_{r}\geq \cdots \geq g_{d-1}\geq g_{d}$. From that we see that \[ (g\hri{r}{r-1}, g\hri{r}{r-2},\ldots ,g\hri{r}{0}) \] is an interlacing sequence. Since, by Proposition~\ref{prop:ans}, $(a_{d}\hri{r}{0},ta_{d}\hri{r}{r-1},\ldots, ta_{d}\hri{r}{1})$ is an interlacing sequence, the claim follows by Lemma~\ref{lem:commoninterlacer}. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:main}] Let $(p,q)$ be the symmetric decomposition of $h(t)$ and $(\tilde{p},\tilde{q})$ be the symmetric decomposition of $\Urd{r}{d+1}h(t)$. By Proposition~\ref{prop:representation}, \[ \tilde{p}(t)=\left(p(t)(1+t+\cdots + t^{r-1})^d\right)\hri{r}{0} \, . \] By Lemma~\ref{lem:extendedStapledon}, \begin{equation}\label{eq:proofmain1} p_i=h_0+h_1+\cdots + h_i-h_d-h_{d-1}-\cdots -h_{d-i+1}\, . \end{equation} Therefore, since $h(t)$ satisfies Property \hyperlink{H}{(H)}, $p(t)$ and thus also $\tilde{p}(t)$ have only nonnegative coefficients. Since $h(t)$ has nonnegative coefficients and $h_i=0$ for all $i\geq s+1$ we furthermore obtain from equation \eqref{eq:proofmain1} \[ p_{0}\leq p_{1}\leq \cdots \leq p_{\ell -1}\quad \text{ and } \quad p_{d+1-\ell}\geq \cdots \geq p_{d-1}\geq p_{d}\, , \] where $\ell=d+1-s$ denotes the co-degree as usual. Since $\max \{s,\frac{d+1}{2}\}=\max \{d+1-\ell,\frac{d+1}{2}\}$ we obtain that $\tilde{p}(t)=\Urd{r}{d} p(t)$ is real-rooted for all $r\geq \max \{s,\frac{d+1}{2}\}$ by Lemma~\ref{lem:arealrooted}. In particular, $\tilde{p}(t)$ is real-rooted for all $r\geq \max \{s,d+1-s\}\geq \max \{s,\frac{d+1}{2}\}$. To see that $\tilde{q}(t)$ is real-rooted we recall that since $r\geq \ell$, by Proposition~\ref{prop:representation}, \[ t\tilde{q}(t)=\left(t^\ell w_r(t)(1+t+\cdots + t^{r-1})^d\right)\hri{r}{0} \, . \] Let \[ f(t)=t^\ell w_r(t) \, . \] Then $f(t)=\sum _{i=0}^{r+s-1}f_i t^i$ is a polynomial of degree $r+s-1$ and has nonnegative coefficients as was argued in the proof of Corollary~\ref{cor:nonnegativity}. In particular, also $\tilde{q}(t)$ has only nonnegative coefficients. We distinguish two cases: If $s\leq \ell$ then \[ f\hri{r}{i}=\begin{cases} f_{i+r}t & \text{ if } 0\leq i \leq s-1\\ f_i & \text{ if } s\leq i \leq r-1 \, . \end{cases} \] In particular, $(f\hri{r}{r-1},f\hri{r}{r-2},\ldots,f\hri{r}{0})$ is an interlacing sequence. Therefore, by Lemma~\ref{lem:commoninterlacer}, \[ t\tilde{q}(t)=f\hri{r}{0}a_d\hri{r}{0}+f\hri{r}{1}ta_d \hri{r}{r-1}+\cdots +f\hri{r}{r-1}ta_d \hri{r}{1} \] is real-rooted. If $\ell \leq s$ then \begin{equation}\label{eq:case2} f\hri{r}{i}=\begin{cases} f_{i+r}t & \text{ if } 0\leq i \leq \ell -1\\ f_i+f_{i+r}t & \text{ if } \ell \leq i \leq s-1\\ f_i & \text{ if } s\leq i \leq r-1 \, . \end{cases} \end{equation} We observe that by Proposition~\ref{prop:decomp2} \begin{eqnarray} f_i&=&\begin{cases}0 & \text{ if }i\leq \ell -1\\ w_{r,i-\ell} & \text{ if } i\geq \ell \end{cases}\\ &=&\begin{cases}0 & \text{ if }i\leq \ell -1\\ -h_{\ell -r}-h_{\ell +1-r}-\cdots -h_{i-r}+h_s+h_{s-1}+\cdots+h_{s-i+\ell} & \text{ if } i\geq \ell \end{cases} \end{eqnarray} In particular, since $r\geq s$ we have \[ f_\ell \leq f_{\ell +1} \leq \cdots \leq f_{s-1} \] and therefore by symmetry of $w_r (t)$ also \[ f_{r+\ell}\geq f_{r+\ell +1}\geq \cdots \geq f_{r+s-1} \, . \] Together with equation~\eqref{eq:case2} this implies that $(f\hri{r}{r-1},f\hri{r}{r-2},\ldots,f\hri{r}{0})$ is an interlacing sequence. Therefore, again as in the first case, it follows by Lemma~\ref{lem:commoninterlacer} that $t \tilde{q}$ and thus $\tilde{q}$ is real-rooted. This finishes the argument. \end{proof} \begin{rem} The proof of Theorem~\ref{thm:main} actually shows something slightly stronger if $h(t)$ has small degree: If $\deg h\leq \frac{d+1}{2}$ then $\tilde{p}(t)$ in the symmetric decomposition of $\Urd{r}{d+1}h(t)$ is already real-rooted whenever $r\geq \frac{d+1}{2}$. \end{rem} \section{Interlacing Symmetric Decompositions}\label{sec:interlacing} In this section we investigate when $\Urd{r}{d+1} h(t)$ has an interlacing symmetric decomposition. We obtain the following characterization. \begin{prop}\label{lem:essential} Let $h(t)$ be a polynomial of degree $s\leq d$ with nonnegative coefficients that satisfies Property \hyperlink{H}{(H)} and \hyperlink{S}{(S)}. For all $r\geq \ell=d+1-s$, $\Urd{r}{d+1} h(t)$ has an interlacing symmetric decomposition if and only if \[ \Urd{r}{d+1} h(t)\preceq \Urd{r}{d+1} (\mathcal I _{d+1} h(t)) \, . \] \end{prop} \begin{proof} Since $r\geq \ell$, by Theorem~\ref{thm:solus2} and Corollary~\ref{cor:nonnegativity}, $\Urd{r}{d+1} h(t)$ has an interlacing symmetric decomposition if and only if $\mathcal{I}_d (\Urd{r}{d+1} h(t))$ interlaces $\Urd{r}{d+1} h(t)$. Since all coefficients of $\Urd{r}{d+1} h(t)$ are nonnegative this is the case if and only if $\Urd{r}{d+1} h(t)$ interlaces $t\mathcal{I}_d (\Urd{r}{d+1} h(t))=\mathcal{I}_{d+1} (\Urd{r}{d+1} h(t))$ which equals \begin{eqnarray} \mathcal I _{d+1} (\Urd{r}{d+1} h(t))&=&t^{d+1}(h(1/t)(1+1/t+\cdots +1/t^{r-1})^{d+1})\hri{r}{0}\\ &=&(t^{r(d+1)}h(1/t)(1+1/t+\cdots +1/t^{r-1})^{d+1})\hri{r}{0}\\ &=&(t^{(d+1)}h(1/t)(1+t+\cdots +t^{r-1})^{d+1})\hri{r}{0}\\ &=&\Urd{r}{d+1} (\mathcal I_{d+1} h(t))\, . \end{eqnarray}\qedhere \end{proof} This characterization is used to prove the following result and Theorem~\ref{thm:main2}. \begin{prop}\label{prop:small} Let $h(t)$ be a polynomial of degree $s\leq \frac{d+1}{2}$ with nonnegative coefficients that satisfies Property \hyperlink{S}{(S)}. Then for all $r\geq d+1$, $\Urd{r}{d+1} h(t)$ has an interlacing symmetric decomposition. \end{prop} \begin{proof} By Lemma~\ref{lem:reform}, we have \[ \Urd{r}{d+1} h(t)=h_0a_{d+1}\hri{r}{0}+h_1ta_{d+1}\hri{r}{r-1}+\ldots + h_sta_{d+1}\hri{r}{r-s} \] and \[ \Urd{r}{d+1} (\mathcal I _{d+1} h(t))=h_0ta_{d+1}\hri{r}{r-(d+1)}+h_1ta_{d+1}\hri{r}{r-d}+\ldots + h_sta_{d+1}\hri{r}{r-(d+1-s)} \, . \] Since $s\leq d+1-s$, the polynomials $(a_{d+1}\hri{r}{0},ta_{d+1}\hri{r}{r-1},\ldots, ta_{d+1}\hri{r}{r-s},ta_{d+1}\hri{r}{r-(d+1-s)},\ldots,ta_{d+1}\hri{r}{r-(d+1)})$ form an interlacing sequence and thus we obtain $\Urd{r}{d+1} h(t)\preceq \Urd{r}{d+1} (\mathcal I _{d+1} h(t))$ by repeated application of Lemma~\ref{lem:basicsinterlace}. Furthermore, since Property \hyperlink{H}{(H)} is automatically satisfied whenever $\deg h \leq \frac{d+1}{2}$, $\Urd{r}{d+1} h(t)$ has an interlacing symmetric decomposition by Lemma~\ref{lem:essential}. \end{proof} To prove Theorem~\ref{thm:main2} we use the following generalization of Lemma~\ref{prop:ans}. This should be compared with Fisk~\cite[Proposition 3.72 \& Example 3.76]{fisk2006polynomials}. Compare also with Br\"and\'en~\cite[Section 8]{Brandenunimodal}. \begin{lem}\label{lem:ans2} Let $h(t)$ be a polynomial and for all $0\leq i<r$ and $d\geq 0$ let \[ a_{h,d}\hri{r}{i}:=(h(t)(1+t+\cdots +t^{r-1})^d)\hri{r}{i} \, . \] Then \[ a_{h,d+1}\hri{r}{i}=a_{h,d}\hri{r}{0}+a_{h,d}\hri{r}{1}+\cdots +a_{h,d}\hri{r}{i}+ta_{h,d}\hri{r}{i+1}+\cdots +ta_{h,d}\hri{r}{r-1} \, . \] In particular, if $h(t)$ is of degree $s$ and has only nonnegative coefficients then \[ (a_{h,d}\hri{r}{r-1}, a_{h,d}\hri{r}{r-2},\ldots, a_{h,d} \hri{r}{0}) \] is an interlacing sequence of polynomials for all $r\geq s$. \end{lem} \begin{proof} The proof combines the ideas of the proofs of \cite[Lemma 3.2]{jochemko2018} and \cite[Proposition 3.4]{jochemko2018}. Let $f(t)=(1+t+\cdots +t^{r-1})$ and $g(t)=h(t)(1+t+\cdots + t^{r-1})^{d}$. We compute \begin{eqnarray} h(t)(1+t+\cdots +t^{r-1})^{d+1}&=&f(t)g(t)\\ &=&(f\hri{r}{0}(t^r)+tf\hri{r}{1}(t^r)+\cdots +t^{r-1}f\hri{r}{r-1}(t^r))\\ &&\cdot(g\hri{r}{0}(t^r)+tg\hri{r}{1}(t^r)+\cdots +t^{r-1}g\hri{r}{r-1}(t^r))\\ &=&\sum _{i=0}^{r-1}\left(t^i\sum _{k+\ell=i}f\hri{r}{k}(t^r)g\hri{r}{\ell}(t^r)+t^{i+r}\sum _{k+\ell=i+r}f\hri{r}{k}(t^r)g\hri{r}{\ell}(t^r)\right)\\ &=&\sum _{i=0}^{r-1}t^i\left(\sum _{k+\ell=i}a_{h,d}\hri{r}{\ell}(t^r)+\sum _{k+\ell=i+r}(ta_{h,d}\hri{r}{\ell})(t^r)\right) \, . \end{eqnarray} In particular, \begin{equation}\label{eq:recursion} a_{h,d+1}\hri{r}{i}=a_{h,d}\hri{r}{0}+a_{h,d}\hri{r}{1}+\cdots +a_{h,d}\hri{r}{i}+ta_{h,d}\hri{r}{i+1}+\cdots +ta_{h,d}\hri{r}{r-1} \, . \end{equation} Now we assume that $h(t)$ has only nonnegative coefficients and degree $s$. To prove the second part of the claim we use induction on $d$. If $d=0$ then \[ a_{h,0}\hri{r}{i}=\begin{cases} h_i & \text{ if } 1\leq i \leq r-1\\ h_0 & \text{ if } i=0 \text{ and } r>s\\ h_0+th_s & \text{ if } i=0 \text{ and }r=s \, . \end{cases} \] where $h_i:=0$ if $i\not \in \{0,1,\ldots,s\}$. Thus, in all cases, $(a_{h,d}\hri{r}{r-1}, a_{h,d}\hri{r}{r-2},\ldots, a_{h,d} \hri{r}{0})$ is trivially an interlacing sequence. If we now assume by induction that $(a_{h,d}\hri{r}{r-1}, a_{h,d}\hri{r}{r-2},\ldots, a_{h,d} \hri{r}{0})$ is an interlacing sequence then, by applying a standard interlacing preserving map (see, e.g., \cite[Proposition 2.2]{jochemko2018} or \cite[Theorem 2.3]{savage2015s}) together with Equation~\eqref{eq:recursion}, also $(a_{h,d+1}\hri{r}{r-1}, a_{h,d+1}\hri{r}{r-2},\ldots, a_{h,d+1} \hri{r}{0})$ is an interlacing sequence. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:main2}] Since $r\geq s$, by Lemma~\ref{lem:ans2}, $(a_{h,d}\hri{r}{0},ta_{h,d}\hri{r}{r-1},\ldots,ta_{h,d}\hri{r}{1},ta_{h,d}\hri{r}{0})$ is an interlacing sequence. We furthermore observe \[ \Urd{r}{d+1}(t^kh(t))=(t^kh(t)(1+t+\cdots +t^{r-1})^{d+1})\hri{r}{0}=ta_{h,d+1}\hri{r}{r-k} \] for $1\leq k\leq r$. In particular, since $d+1-s\leq r$, \[ \Urd{r}{d+1}h(t)=a_{h,d}\hri{r}{0} \preceq ta_{h,d}\hri{r}{r-d+s-1}=\Urd{r}{d+1}(t^{d-s+1}h(t))=\Urd{r}{d+1}(\mathcal I _{d+1} h(t)) \, . \] We moreover observe that Properties \hyperlink{H}{(H)} and \hyperlink{S}{(S)} are automatically satisfied for symmetric polynomials with nonnegative coefficients. Thus, by Proposition~\ref{lem:essential}, $\Urd{r}{d+1}h(t)$ has an interlacing decomposition for $r\geq \max\{s,d+1-s\}$. \end{proof} The bound in Theorem~\ref{thm:main2} is optimal. Indeed, if we consider polynomials of the form $h(t)=1+t^s$, and let $(\tilde{p},\tilde{q})$ be the symmetric decomposition of $\Urd{r}{d+1}h(t)$ then, by Corollary~\ref{cor:nonnegativity}, $\tilde{p}$ has only nonnegative coefficients for all $r\geq 0$. If $\tilde{q}$ interlaces $\tilde{p}$ then also the coefficients of $\tilde{q}$ are nonnegative as all roots have to be nonpositive. At the end of Section~\ref{sec:symmetricdecomp} we argued that $r$ thus needs to be greater or equal to $d+1-s$. Furthermore, if $\tilde{q}$ interlaces $\tilde{p}$, then $\Urd{r}{d+1}h(t)$ is real-rooted. In~\cite[Section 5]{jochemko2018} it was shown that for $h(t)=1+t^s$ the bound $r\geq s$ is sharp for $\Urd{r}{d+1}h(t)$ to be real-rooted. We have seen that $\Urd{r}{d+1} h(t)$ has an interlacing decomposition for all $r\geq d+1$ if $h(t)$ has small degree (Proposition~\ref{prop:small}) and for $r\geq \max\{s,d+1-s\}$ if $h(t)$ is symmetric (Theorem~\ref{thm:main2}). Based on computational experiments we believe that these bounds might constitute general uniform bounds. \begin{quest} Let $h(t)$ be a polynomial of degree $s\leq d$ with nonnegative coefficients that satisfies Properties \hyperlink{H}{(H)} and \hyperlink{S}{(S)}. Does $\Urd{r}{d+1}h(t)$ have an interlacing symmetric decomposition for all $r\geq d+1$ (possibly even for $r\geq \max\{s,d+1-s\}$)? \end{quest} \section{Ehrhart theory}\label{sec:Ehrharttheory} A lattice polytope in $\mathbb{R}^d$ is defined as the convex hull of finitely many points in the integer lattice $\mathbb{Z}^d$. Ehrhart theory is concerned with counting lattice points in lattice polytopes and their integer dilates. For a comprehensive introduction to Ehrhart theory we recommend~\cite{BR}. In~\cite{ehrhartRational} Ehrhart proved that the number of lattice points $\vert nP\cap \mathbb{Z}^d\vert$ in the $n$-th dilate of a lattice polytope $P\subset \R^d$ agrees with a polynomial $\Ehr_P (n)$ of degree $\dim P$ for all integers $n\geq 0$. The polynomial $\Ehr_P(n)$ is called the \textbf{Ehrhart polynomial} of $P$. The $h^\ast$-polynomial $h^\ast _P(t)$ of a polytope $P$ encodes $\Ehr _P(n)$ in a particular basis. If $P$ is $d$-dimensional, then the relation between $h^\ast _P(t)$ and $\Ehr _P(n)$ is given by \[ \sum _{n\geq 0} E_P (n)t^n \ = \ \frac{h^\ast _P (t)}{(1-t)^{d+1}} \, . \] In particular, the degree of $h^\ast _P$ is at most $d$. The coefficients of the Ehrhart polynomial can be negative and rational in general. In contrast, Stanley~\cite{Stanley78} showed that the $h^\ast$-polynomial has only nonnegative integer coefficients. Subsequently, Hibi~\cite{HibiH} and Stanley~\cite{StanleyS} proved two fundamental families of inequalities for the coefficients of the $h^\ast$-polynomial, namely Property \hyperlink{H}{(H)} and Property \hyperlink{S}{(S)}. In summary, the $h^\ast$-polynomial satisfies all conditions of Theorem~\ref{thm:main}. Furthermore, for any $d$-dimensional lattice polytope $P$ and any dilation factor $r\in \mathbb{Z}_{>0}$ \[ \frac{h^\ast _{rP} (t)}{(1-t)^{d+1}} \ = \ \sum _{n\geq 0} E_{rP} (n)t^n \ = \ \sum _{n\geq 0} E_{P} (nr)t^n \ = \ \frac{\Urd{r}{d+1}h^\ast _{P} (t)}{(1-t)^{d+1}} \, , \] that is, $\Urd{r}{d+1}h^\ast _{P} (t)$ is equal to the $h^\ast$-polynomial of the $r$-dilate of the lattice polytope $P$. The following corollary is therefore an immediate consequence of Theorem~\ref{thm:main}. \begin{cor}\label{cor:Ehrrealrooted} Let $P$ be a $d$-dimensional lattice polytope and let $h^\ast _P (t)=h_0+h_1t+\cdots +h_st^s$, $\deg h^\ast _P=s\leq d$, be the $h^\ast$-polynomial of $P$. Then $h^\ast _{rP}(t)$ has a real-rooted symmetric decomposition for all integers $r\geq \max\{s,d+1-s\}$. \end{cor} A lattice polytope $P$ in $\R^d$ is called \textbf{reflexive} if \[ P \ = \ \{\mathbf{x}\in \R^d\colon \mathbf{Ax}\leq \mathbf{1}\} \, , \] up to a translation by a vector in $\mathbb{Z}^d$, where $\mathbf{A}$ is an integer matrix and $\mathbf{1}$ denotes the all $1$s vector. A polytope $P$ is called \textbf{Gorenstein} if $P$ has an integer dilate $\ell P$ that is reflexive. Hibi~\cite{Hibireflexive} showed that a polytope $P$ is reflexive if and only if $h_P^\ast(t)=t^dh^\ast _P (1/t)$. Stanley~\cite{StanleyGorenstein} extended this characterization to Gorenstein polytopes. As an immediate consequence of Theorem~\ref{thm:main2} we obtain the following. \begin{cor}\label{cor:Ehrinterlacing} Let $h^\ast _P(t)$ be the $h^\ast$-polynomial of a $d$-dimensional Gorenstein lattice polytope. Then $h^\ast _{rP} (t)$ has an interlacing decomposition whenever $r\geq \max\{s,d+1-s\}$. \end{cor} \textbf{Acknowledgements:} The author wants to thank Petter Br\"and\'en and Liam Solus for interesting discussions, and Florian Kohl as well as the anonymous referee for careful reading and many helpful comments on the manuscript. This article was finished while the author was a participant at the semester program ``Algebraic and Enumerative Combinatorics'' at Mittag-Leffler-Institute in Djursholm in spring 2020. This work was partially supported by the Wallenberg AI, Autonomous Systems and Software Program (WASP) funded by the Knut and Alice Wallenberg Foundation, as well as by grant 2018-03968 of the Swedish Research Council. \bibliographystyle{siam} \bibliography{SymmetricDecomp} \end{document}	27,353
. a.m. CT on Saturday, April 27. In addition to Thursday night’s appearances by players and personnel, Curly’s Pub will provide Draft Party food and drink specials, as well as contests and door prizes throughout the weekend..	186,831
TITLE: Is Bohmian mechanics an interpretation of quantum mechanics? QUESTION [4 upvotes]: I don't understand why some people argue that Bohmian quantum mechanics is just an interpretation of quantum mechanics. In addition to the usual Schrödinger equation, we have the following deterministic equation (in 1d): $$ m\dot q=\hbar\nabla_q \Im (\ln\psi(q,t)). $$ And we have more information than standard quantum mechanics. If we solve the Schrödinger equation, the paths of particles are known. The above equation has not an intrinsic importance in quantum mechanics and we may crudely write the current as $J\sim m\dot q$ which is equal to $\hbar\nabla_q \Im (\ln\psi(q,t))$. REPLY [3 votes]: In my vocabulary an interpretation, an extended deterministic theory , should reproduce the mathematics of quantum mechanics as accepted by mainstream physics, so that the successes of quantum mechanics to fit and predict data are reproduced. In that sense the non relativistic Bohm theory is an interpretation of mainstream QM. It fails in relativistic quantum mechanics to be an interpretation. From the link: Like nonrelativistic quantum theory, of which it is a version, Bohmian mechanics is not compatible with special relativity, a central principle of physics: Bohmian mechanics is not Lorentz invariant. Nor can it easily be modified to accommodate Lorentz invariance.	148,443
Bellator 31 Weigh-Ins (Updated) MMAPlayground.com » MMA General » General MMA Talk » Bellator 31 Weigh-Ins (Updated) DCRage 9/29/10 4:04:40PM Results: Zolia Frausto has missed weight. She has an undetermined amoung of time to lose .2 pound. ( Update : She has now made weight on her 3rd try) Herman (241.6) vs. Michael Kita (226.2) Mayhem13 9/30/10 4:36:55PM WAR LOZANO!! Hope he messes Yoshida up! Related Topics Angela Magana is latest addition to Bellator's season-three women's tourney Bellator 21 results Upcoming knee surgery sidelines Bellator middleweight Jared Hess up to six months Bellator 115-LBs Women's Tourney Taking Shape Sengoku 4 Discussion: Post Weigh Ins, FOTN, KOTN, and SOTB and Predict the LW Tourny Winner Does anybody have a link for the ufc 82 weigh-ins?	342,503
CRL's Global Resources Partnership for News takes a different approach than that of the other major domains. Access to news in its many forms requires coordination and participation of a variety of stakeholders. As one of the foremost news collections in North America and the world, CRL works with its member institutions to systematically identify critical collections; establish consensus on preservation, archiving, and digitization needs; and negotiate licenses and partnerships to obtain access to materials available electronically. World Newspaper Archive CRL pursues systematic digitization of historic global newspapers in partnership with Readex. The World Newspaper Archive is a community-driven effort to provide persistent electronic access to newspapers from the holdings of CRL and other major newspaper repositories. Licensing and acquisition of electronic news resources CRL aligns the interests and collective influence of its membership to obtain favorable terms of access to historical (digitized) and current (born-digital) news sources. CRL pursues relationships with media organizations (such as the New York Times), commercial providers, and academic partners to ensure the persistence and accessibility of these resources. International Coalition on Newspapers (ICON) CRL's partnership with the major news repositories around the world focuses on preservation and improved access to historic news collections. Today, ICON places increasing emphasis on assessment of sustainable digital news repositories, and the ICON Database of newspapers stands as one of the most comprehensive sources for information on digitized newspapers in trustworthy repositories to assist libraries in making informed collection decisions. Print retention and preservation CRL continues to support the identification and acquisition of rare print news resources. CRL negotiates deposits from its members, on a case-by-case basis, to support retention needs. CRL augments its collection of print news material with holdings for which microfilm or digital surrogates are not available. CRL is considering strategic solutions for the mass reformatting of these endangered collections to support widespread access. CRL also continues to preserve key global newspapers on microfilm for libraries' ongoing collection needs.	153,341
Thank you for a fantastic response to new branch MAY I just say a big thank you to the editor for printing my letter about the launch of the Royal Navy Electrical Branch Association last year, and also to your readers as well. The response has been nothing less than fantastic; we have had ex-RN elect MAY I just say a big 'thank you' to the editor for printing my letter about the launch of the Royal Navy Electrical Branch Association last year, and also to your readers as well. The response has been nothing less than fantastic; we have had ex-RN electricians coming out of the woodwork seeking their old shipmates and with some wonderful stories ranging from the involvement in the Korean War, Suez and the Gulf to name some of the conflicts, to the time that HMS Bermuda went to Bermuda and they all enjoyed a few days fun. An unexpected line of help came from the readers who had relatives living abroad who were ex-Royal Navy Electrians; consequently we have members who live in the USA, Australia and in Spain as a result of the local newspaper being sent on to them with my letter highlighted. The highlight of any association is the annual reunion and we have our first one booked. If there are any ex-Royal Navy electricians out there who are wonder what happened to their old shipmates, then please drop me line and tell me when you served and, if you can remember it, your old service number. You may also want to watch: INFO: Mike Crowe, Royal Naval Electrical Branch Association, 7 Heath Road, Lake, Sandown, Isle of Wight. PO36 8PG. If you include a stamp it will help with branchIKE CROWE	346,638
Skip table of contents <Developers/> search PRODUCTS Payments Converge Fusebox viaConex Terminals Commerce SDK Simplify Reports Payments Core 365 Accounts eBoarding Partner API Security EMV 3D Secure2 EMV 3D Secure Overview Getting Started Authentication Integration Options Direct integration to 3DS Server Lookup - Sample Request and Response Authentication - Sample Request and Response Challenge Result - Sample Request and Response Validate - Sample Request and Response Integrate using the Web SDK Web SDK Integration Prerequisites Installing the Web SDK Using the All-in-one Workflow of the Web SDK Using the Toolkit Workflow of the Web SDK Web SDK Demo Application Other Features 3RI Overview 3RI: Lookup - Sample Request and Response 3RI Authentication - Sample Request and Response 3DS 1 Fallback Travel industry message extension 3D Secure 2 for merchants not processing with Elavon Test your integration Error Codes Common Error Scenarios HTTP response status codes Release notes API Reference Appendices Product Security and Best Practices UI Mockups of cardholder screens Glossary Glossary Acronyms used in the documentation in the context of EMV 3D Secure Acronym Expansion 3DS 3D Secure 3RI 3DS Requestor Initiated ACS Access Control Server AFS Advanced Fraud Services AReq Authentication Request ARes Authentication Response BIN Bank Identification Number CNP Card Not Present CReq Challenge Request CRes Challenge Response DS Directory Server FSG Fraud Services Gateway NPA Non-Payment Authentication OTP One-time Passcode OOB Out-of-Band PA Payment Authentication PAN Primary Account Number RReq Result Request RRes Result Response Glossary of terms used in the documentation in the context of EMV 3D Secure Term Explanation 3D Secure (3DS) An e-commerce authentication protocol that enables the secure processing of payment, non-payment, and account confirmation card transactions. 3DS Client The consumer-facing component that allows the cardholder to interact with other components and vice versa. 3DS Method A scripting call provided by the 3DS Integrator that is placed on the 3DS Requestor website. It allows for additional browser information to be gathered by an ACS before receipt of the AReq message to help facilitate the transaction risk assessment. The use of the 3DS Method by an ACS is optional. 3DS Requestor Initiator of the 3DS authentication request. 3DS Requestor App An App on a consumer device that can process a 3D Secure transaction through the use of Elavon’s Mobile SDKs (iOS or Android). 3DS Server Elavon’s server that handles online transactions and facilitates communication between the 3DS Requestor and the DS. 3DS Integrator Also referred to as ‘Integrator’ in Elavon’s 3DS documentation. An EMV 3D Secure participant that facilitates and integrates the 3DS Requestor Environment, and optionally facilitates integration between the Merchant and the Acquirer. 3DS Requestor Initiated (3RI) 3D Secure transaction initiated by the 3DS Requestor to confirm that an account is still valid or for Cardholder authentication. The main use case of a 3RI transaction is recurrent transactions (TV subscriptions, utility bill payments, etc.) where the merchant wants to perform a payment transaction to receive authentication data for each bill or a non-payment transaction to verify that a subscription user still has a valid form of payment. The second main use case is when the 3DS Requestor requests Decoupled Authentication as a method to authenticate the Cardholder. Access Control Server (ACS) A component of the Issuer Domain that verifies whether authentication is available for a card number and device type, and authenticates specific Cardholders. Authentication request (AReq) Message requesting authentication of the cardholder. Might contain cardholder, payment, and device details used in the transaction. Authentication response (ARes) ACS’s response if the transaction has been authenticated or needs further interaction to complete the authentication. Authorisation A process by which an Issuer, or a processor on the Issuer’s behalf, approves a transaction for payment. Challenge flow A 3D Secure flow that requires further cardholder authentication to process the transaction. Challenge request (CReq) Initiates cardholder interaction in a challenge flow. Sent by the SDK in an app-based scenario and sent by 3DS Server in a browser-based scenario. Challenge response (CRes) ACS response to indicate the result of cardholder authentication. In an app-based scenario, the CRes has the necessary elements to generate and display the UI for the challenge. Decoupled Authentication Decoupled Authentication is an authentication method whereby authentication can occur independently from the cardholder’s experience with the 3DS Requestor (browser/SDK). For example, a push notification to a banking app that completes authentication and then sends the results to the ACS (issuer). Device Information Data provided by the Consumer Device that is used in the authentication process. Directory Server (DS) A server that performs several functions that include: authenticating the 3DS Server, routing messages between the 3DS Server and the ACS, and validating the 3DS Server, the Mobile SDK, and the 3DS Requestor. Frictionless flow ACS authenticates the transaction without a challenge. Mobile SDK (Releasing soon) Software Development Kits (SDK) for iOS and Android. You must integrate the SDK into the merchant app so that it can be used for the 3D Secure (3DS) transaction authentication process. This SDK supports both EMV 3D Secure 2.1 and 2.2. Integrator An integrator can be an MSP (merchant services provider), a PSP (payment service provider), or other partners who use the 3DS 2 solution by Elavon and handle the payment relationship directly with the merchant. Merchant Entity that contracts with Elavon as an acquirer to accept payment cards. Manages the online shopping experience with the Cardholder, obtains the card number, and then transfers control to the 3DS Server, which conducts payment authentication. Merchant alias An alias for the merchantId of the service provider merchant. The integrator can use the merchant alias and their API key (the integrator’s API key) to retrieve the authentication token and send 3DS 2 related API requests on behalf of the merchant. Message Category Indicates the category of the EMV 3D Secure message. Either: Payment (01-PA) or Non-Payment (02-NPA) Out-of-Band (OOB) A Challenge activity that is completed outside of, but in parallel to, the 3D Secure flow. The final Challenge Request is not used to carry the data to be checked by the ACS but signals only that the authentication has been completed. Result request (RReq) Communicates the authentication or verification result sent by the ACS to the 3DS Server. The ACS sends this message only in a challenge flow. Result response (RRes) Receipt acknowledgment of the RReq message from the 3DS Server to the ACS. Present only in a challenge flow. Service provider merchant Merchant who does not process with Elavon and wants to use 3D Secure solution by Elavon. Such merchants get registered in Elavon’s database through an integrator, who could be their acquirer. Whitelisting In this specification, the process of an ACS enabling the cardholder to place the 3DS Requestor on their trusted beneficiaries list. Related topics 3D Secure 2 integration options Direct integration to 3DS Server 3D Secure 2 for merchants who do not process their payment with Elavon 3DS Web SDK Integration 3DS requestor initiated authentication UI Mockups of cardholder screens	43,557
A: I took a “Women In Islam” course in my final year of university completing my BFA in Photography from Ryerson. A: The Sisters Project is a one woman team. Alia Youssef created The Sisters Project in December 2016 and has since been working independently taking the photographs, conducting interviews, and on occasion making videos for the project. A: Although the series is largely self-funded, some side "projects" have been made possible by generous funders. In Summer 2018, The Sisters Project went across Canada, photographing 80+ women in 12 cities from coast to coast, and that was made possible by a ChangeUp grant given by the Inspirit Foundation. The Sisters Project was exhibited at the Ryerson Image Centre Fall 2018 and that exhibition was funded by Ryerson University's Faculty of Communication and Design and the Office of the Provost. The series was also loaned equipment by Ryerson's Office of the President. A:.	169,324
“We have some of the finest talent in fashion’s history from this very country.” London Fashion Week (LFW) is perhaps one of the most prominent events on any designer-loving-fashionista’s calendar. It is the equivalent to popular fashion weeks held by the world’s remaining fashion capitals: New York, Milan and Paris. Open to the general public, the fashion event regularly craves media attention and associates itself with some of the biggest names in fashion and the celebrity sphere. Taking place between September 13 and 17, 2013, much of the support came from British celebrities: Sienna Miller, Ellie Goulding and Pixie Lott were all on board to cheer on British fashion. From Tom Ford to Vivienne Westwood, the catwalk boasted hot new fashion trends from some of the world’s most reputable designers. Sparkling sequins to floral prints – almost anything and everything was on the agenda for this year’s LFW. Day 1 With the public queuing in the rain, the show had to be worth the wait. With months, days, and hours of planning, the first day of London Fashion Week had finally arrived. The beginning of the week started with a blast as designers such as Fyodor Golan exposed their Spring/Summer 2014 collection. The garments on show oozed summer charm as pastel colours were the designer’s ultimate colour palette. The designer fused elegance with feathers, ruffles and sweater styles to produce something unimaginable. Certain aspects of Golan’s collection could be perceived as an explosion of creativity whilst others were more subtle everyday wear. The designer was also a big fan of high necklines and transparent skirts. Designers alike showcased similar styles as see-through materials and ruffles were common amongst other collections. Felder and Felder were big fans of net and billowing silhouettes. Blues, barely blacks and whites were the designers’ favourite shades. Day 2 With a great start to the week, day 2 had to be even bigger and better. Celebrities and familiar faces that attended the event complemented the designers and their collections. Television presenter, Jameela Jamil said: “We have some of the finest talent in fashion’s history from this very country.” To prove her right, designers were all set to showcase their creativity and talent. Julien Macdonald’s collection astounded the audience as his warrior-like embellishments and red carpet wear had graced the catwalk again and again. Pixie Lott had already dubbed Julien’s collection as ‘performance wear’ – and it was exactly that! The show-stopping collection teased the audience with nude colours, risqué cuts and plunging necklines, and that left them wanting more. In contrast to Macdonald, Zoë Jordan’s collection was a little less glamorous and exposed the ‘female with an edge’. For the streetwise, confident city chick, Zoë created ensembles with graffiti art at the forefront of her designs. Her collection was bold and spoke for itself. Her colour palette was minimal with reds, blacks and whites dominating her collection. Ashish Gupta, the only British-Asian designer showcasing his designs at Somerset House played around with sequins and all things sparkly for his latest collection. His collection included menswear, which is very rare for designers at LFW—those too were covered with glitz and glamour, but luckily Ashish pulled it off! Day 3 Top British model, Cara Delevingne also joined the catwalk to help showcase designer brands. She was spotted strutting in well-renowned fashion label, Mulberry. Mulberry also incorporated a fun and light-hearted feel to their show, as models were walking dogs up and down the catwalk as their very own mascots! The British label played around with soft colours, all adding a ladylike touch to the collection. Greek designer, Mary Katrantzou’s unimaginable designs looked like they were manufactured from out of space. Her hyperrealist collection rocked the metal stage and backdrop. Although the prints were a little eccentric they were nonetheless wearable and definitely workable. Mary’s collection worked with an array of colours, from deep purple to pink. Actress Sienna Miller quoted London as the ‘epicentre of fashion’, and this week it certainly was! Day 4 LFW was almost coming to a close, but that didn’t stop the huge hype and excitement before each show. And for once, the weather was on board too! Celebrities like Harry Styles and the Beckhams were out and about on day 4 to watch the catwalk as Burberry, Christopher Kane and other designers put on a show. Christopher Kane’s Spring/Summer 2014 collection endorsed metallic dresses and unusual peardrop shaped cuts. The collection made the audience feel the immediate presence of Spring as greens, whites and lilacs dominated the show. Pixie Geldoff was in awe of the collection as she referred to Kane as a ‘complete genius’. It was fair to say that he was exactly that as he implemented cut out shapes and intricate designs wherever he could! Burberry also worked with Spring shades, green, white and lilac whilst also revamping nude garments with embellishments and sparkly waistlines. Burberry created outfits for the day and night; light and pastel shades were set for daytime glamour, whilst sequins and embellishments welcomed evening/night style. Day 5 The final day of LFW wrapped the week up in appropriate style. Anya Hindmarch’s collection was definitely one of the must see shows of the entire week. Innovative, fun and a high level of creativity was one way of summing up the show. The show involved floating handbags and models walking in the air. Stella McCartney’s adidas collection was equally unusual. The typical block colours were exchanged for daisy and marble prints. The new look added a feminine feel to the sportswear collection in addition to the choice of colours: burgundy, baby blues and oranges. It was clear to see that this year’s LFW had transformed the fashion-obsessed woman; from sequins to sporting jumpers, the collections on show were a mix of elegance and attitude. With interest from around the globe, the atmosphere at London Fashion Week was definitely buzzing. The week has served its purpose, as international orders are set to strengthen the economy once again. All that is left is to look ahead to LFW 2014!	15,417
TITLE: Linear diophantic equation QUESTION [1 upvotes]: My task requires some euclidian work beforehand gcd(4386, 89744) = 4386 x + 89744 y I've confirmed that gcd(4386, 89744) = 2 Does that mean that 4386 x + 89744 y = 2? I ask because this format is the basis of a youtube tutorial. REPLY [2 votes]: It does clearly not mean that for any $x,y$ : $4386 x + 89744 y = 2$ if it's what you meant. What is true is that the equation $4386 x + 89744 y = 2$ has an integer solution, that is: there are $x,y$ so that $4386 x + 89744 y = 2$ (namely $x = 22569, y = -1103$). In general: If you have gcd(a,b)=c then the equation $ax +by = c$ has an integer solution.	12,600
883/162 Vilhelm Hammershøi: Portrait of the artist's sister Anna Hammershøi seen from the back. Unsigned. Oil on canvas laid on panel. 38×28 cm. - Description Portrait of the artist's sister Anna Hammershøi seen from the back. Unsigned. On the reverse certified by the painter's wife Ida Hammershøi: “Malt af Vilh. Hammershøi, attesteres: Ida Hammershøi” (Painted by Vilh. Hammershøi, certified: Ida Hammershøi). Oil on canvas laid on panel. 38×28 cm. In the profound and newly published book on Hammershøi, the authors describe this motif of the woman with her back turned, which is so characteristic of Hammershøi’s paintings, and they describe the close connection of the subject with contemporary photography, which greatly fascinated Hammershøi, and which he to a large degree used in his works. “Like the photographers Félex Nadar (1820–1910), Onésipe Aguado de las Marismas (1830–1893) and many other contemporary artists, Hammershøi was interested in motifs with necks and figures with their back turned. The motifs often appear enigmatic because we can get so close without seeing the figure's face. And apparently without the figure seeing us. The neck motifs are particularly oriented towards a sense of desire in a photograph because it includes the awareness of the photographer, who can take the picture while the model cannot see what is going on. Hammershøi transfers this intense photographic situation to for instance his painting…” motif on this painting – the woman seen from the back – occurs in a number of Hammershøi’s interior paintings. Provenance: Alfred Bramsen, his daughter the violinist Karen Bramsen and her husband museum director Gustav Falck, thence by descent, until the sale of the collections of Alfred Bramsen and Gustav Falck, Bruun Rasmussen auction 576, 1992 no. 141, reproduced p. 110. Here described as his wife Ida seen from behind, but it is rather the characteristic features of the sister – her smooth hair, ear and neck as seen in other portraits of her by Hammershøi – than those of his wife. Alfred Bramsen (1857–1932) was actually a dentist, but he is today mainly known as an art collector and more specifically as the patron for and collector of works by Vilhelm Hammershøi. In 1918, he published, together with Sophus Michaëlis, the work “Vilhelm Hammershøi, Kunstneren og hans Værk” (Vilhelm Hammershøi, the Artist and His Work). Bramsen's works by Hammershøi were inherited by his daughter Karen Bramsen (1877–1970), and she married Gustav Falck (1874–1955). Gustav Falck was an art historian, and in 1925 he became Karl Madsen's successor as director of the National Gallery of Denmark. The museum can thank him for the purchasing of portraits of Tizian and Frans Hals. Falck had a large private collection of an exquisite quality, which in addition to many works by Vilhelm Hammershøi included works by many other artists. - Condition Condition report on request. Contact fine-art@bruun-rasmussen.dk - Auction Paintings, 27 November 2018 - Category - - Artist Vilhelm Hammershøi (b. Copenhagen 1864, d. s.p. 1916) - Estimate 1,500,000–2,500,000 kr. - Sold - Price realised 1,500,000 kr.	291,637
TITLE: Rotation of a matrix QUESTION [3 upvotes]: Sorry for boring you my friends before the holiday. I am haunted by a question of rotation of a matrix. Suppose that we have a special matrix $\Omega$ takes the form of: $\Omega = \left[ \begin {array}{ccc} 0&-\omega_{{3}}&\omega_{{2}} \\ \omega_{{3}}&0&-\omega_{{1}}\\ -\omega_{{2}}&\omega_{{1}}&0\end {array} \right] $ and there is a abstract rotation matrix $R$ which verifies the property of $R^T=R^{-1}$. I would like to find out the geometry meaning behind the following equation: $R^T\cdot \Omega \cdot R$. I tried to give an answer. This is the trail: Firstly, I extracted a vector $\omega$ from $\Omega$ and it is given: $\omega = \left[ \begin {array}{c} \omega_{{1}}\\\omega_{{2} }\\\omega_{{3}}\end {array} \right] $ , Then, I multiplied $\omega$ by $R^{T} $ on the left side and got $\bar{\omega}=R^{T}\cdot\omega$, the $\bar{\omega}$ takes the form of: $\bar{\omega} = \left[ \begin {array}{c} \bar{\omega}_{{1}}\\\bar{\omega}_{{2} }\\\bar{\omega}_{{3}}\end {array} \right] $ Finally, I returned the $\bar{\omega}$ back into the matrix form as $\bar{\Omega}$: $\bar {\Omega} = \left[ \begin {array}{ccc} 0&-\bar{\omega}_{{3}}&\bar{\omega}_{{2}} \\ \bar{\omega}_{{3}}&0&-\bar{\omega}_{{1}}\\ -\bar{\omega}_{{2}}&\bar{\omega}_{{1}}&0\end {array} \right] $ Several simple numerical applications have been conducted and verified the above mathematical conjecture which is $\bar {\Omega} = R^T\cdot \Omega \cdot R$. But I failed to demonstrate this conjecture mathematically, or probably the conjecture is wrong. Thank you in advance for taking a look. Nice holiday! REPLY [2 votes]: The matrix $\Omega$ has matrix elements of the following form: $$\Omega_{i,j} = \sum_{k=1}^3 \epsilon_{i,j,k} \omega_k $$ where $\epsilon$ is the absolutely anti-symmetric Levi-Civita tensor. For column vectors $x$, $y$ and $z$, the determinant of the matrix with these columns is $$\det[x \mid y \mid z] = \sum_{i,j,k} \epsilon_{i,j,k} x_i y_j z_k$$ As such, the Levi-Civita tensor remain invariant under rotations: $$ \left(R \Omega R^t\right)_{i,j} = \det \left[ R_{i,\cdot}, R_{j,\cdot}, \omega \right] = \det \left[ R_{i,\cdot}, R_{j,\cdot}, R^t \left(R \omega\right) \right] = \det \left[ e_{i}, e_{j}, R \omega \right] $$ That is the vector $\omega$ under conjugation of $\Omega$ by rotation matrices simply gets the same rotation. Confirming in Mathematica: In[45]:= With[{axisVec = {1, 0, 0}}, RotationMatrix[\[Theta], axisVec].(LeviCivitaTensor[3].{Subscript[w, 1], Subscript[w, 2], Subscript[w, 3]}).RotationMatrix[-\[Theta], axisVec] - (LeviCivitaTensor[3].RotationMatrix[\[Theta], axisVec].{Subscript[ w, 1], Subscript[w, 2], Subscript[w, 3]}) // Simplify] Out[45]= {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}	6,329
TITLE: How to prove indirectly that if $42^n - 1$ is prime then n is odd? QUESTION [10 upvotes]: I'm struggling to prove the following statement: If $42^n - 1$ is prime, then $n$ must be odd. I'm trying to prove this indirectly, via the equivalent contrapositive statement, i.e. that if $n$ is even, then $42^n - 1$ is not prime. By definition, for every even number $n$ there exists an integer $k$ with $n = 2k$. We substitute and get $$42^n - 1 = 42^{2k} - 1 = (42^2)^k - 1.$$ Now, how do I prove that $(42^2)^k - 1$ isn't a prime number? Is this even the right way to approach this proof? REPLY [11 votes]: Note that $$42^{2k}-1=(42^k)^2-1=(42^k-1)(42^k+1)$$ where $1\lt 42^k-1\lt 42^k+1$. REPLY [7 votes]: Note that $$\begin{align}42^n - 1 &\equiv 1 - 1 \\&\equiv 0 \pmod{41}\end{align}$$ so the only way for $42^n - 1$ to be a prime is for $n$ to be $1$. In general, for $a^n - 1$ to be a prime, where $a, n \in\mathbb{Z}^+$, either $a = 2$ or $n = 1$. (Not sure if this counts as indirect, but you could turn it into some form of contradiction) REPLY [6 votes]: By the binomial theorem, $42^n = (43-1)^n=43a+(-1)^n$. If $n$ is even, then $42^n-1$ is a multiple of $43$. On the other hand, $42^n = (41+1)^n=41b+1$, and so $42^n-1$ is always a multiple of $41$. Thus, $42^n-1$ is not prime if $n>1$, regardless of the parity of $n$.	213,499
Information This is a bezel for Buick and Cadillac cars. This bezel fits 22 General Motors car models which have been manufactured for US market. This part is genuine General Motors part with GM OEM part number 05966256. Browse to find other vehicles 05966256 bezel fits Parts related to 05966256 bezel: -:	75,888
Plan to Attend World Gay Rodeo Finals®! Two full days of rodeo action, October 27 & 28, 2018 at Mesquite Rodeo Arena in Mesquite Texas. The World Gay Rodeo Finals® is the culminating event of the IGRA rodeo year.® presented by IGRA. In addition, IGRA holds its annual Royalty Competition where Mr. Miss, Ms and MsTer are selected from association title holders to represent IGRA nationally and to raise funds for our designated charities.	384,205
\begin{document} \maketitle \begin{abstract} In Section 1 the author defines the notion of harmonic map between generalized Lagrange spaces. Section 2 analyses the particular case when the generalized Lagrange spaces are Lagrange spaces of electrodynamics. In Section 3 it is proved that for certain systems of differential or partial differential equations, the solutions are harmonic maps between certain generalized Lagrange spaces, in the sense of Section 1. Section 4 describes the main properties of the generalized Lagrange spaces constructed in Section 3. \end{abstract} \par \noindent {\bf Mathematics Subject Classification:} 53C60, 49N45, 35R30\\ {\bf Key words:} generalized Lagrange spaces, harmonic maps, geodesics, differential equations, partial differential equations. \section{Introduction} \hspace{5mm}Let $(M^m,\;g_{\alpha\beta})$ and $(N^n,\;h_{ij})$ be two generalized Lagrange spaces, where $m$, respectively $n$, is the dimension of $M$, respectively $N$. The manifold $M$, respectively $N$, is coordinated by $(a^\alpha)$, respectively $(x^i)$. On $M\times N$, the first $m$ coordinates are indexed by $\alpha,\beta,\gamma,\ldots$ and the last $n$ coordinates are indexed by $i,j,k,\ldots$. The fundamental metric tensors are expressed locally by i) $g_{\alpha\beta}=g_{\alpha\beta}(a,b),\;\forall\;\alpha ,\beta= \overline {1,m}$, where $(a,b)=(a^\mu ,b^\mu )$ are adapted coordinates on $TM$. ii) $h_{ij}=h_{ij}(x,y),\;\forall\;i,j=\overline{1,n}$, where $(x,y)=(x^k,y^k)$ are adapted coordinates on $TN$. On $M\times N$, we consider an arbitrary tensor of type $(1,2)$, denoted by $P$, with all components null except $P^\beta_{\alpha i}(a, x)$ and $P^j_{\alpha i}(a, x)$, where $\alpha ,\beta = \overline{1,m},\;i,j=\overline{1,n}$, which will be called {\it tensor of connection}. The connection tensor $P$ allows one to build the directions $b$ and $y$ of the metric tensors $g_{\alpha\beta}(a,b)$ and $h_{ij}(x,y)$ used in the construction of the energy functional $E$ whose extremals will be the harmonic maps between the generalized Lagrange spaces $M$ and $N$. In conclusion, this tensor makes the connection between the metric structures of the spaces $M$, respectively $N$, and the harmonic maps that we will define. We assume that the manifold $M$ is connected, compact, orientable and endowed with a Riemannian metric $\varphi_{\alpha\beta}$. These conditions assure the existence of a volume element and, implicitly, of a theory of integration on $M$. Using the generalized Lagrange metrics $g_{\alpha\beta}$ and $h_{ij}$ and the tensor of connection $P$, we can define the following $\left( \begin{array}{ccc} &P&\\ g&\varphi&h \end{array}\right)$ {\it-energy functional}, $$ E=E^{\scriptstyle P}_{g\varphi h}:C^\infty (M,N)\to R, $$ $$ E^{\scriptstyle P}_{g\varphi h}(f)=\displaystyle{{1\over 2}\int_M g^{\alpha\beta}(a,b(a,f^k,f^k_\gamma)) h_{ij}(f(a),y(a,f^k,f^k_\gamma))f^i_\alpha f^j_\beta\sqrt\varphi\;da,} $$ where $\left\{\begin{array}{l}\medskip \displaystyle{ f^i=x^i(f),\;f^i_\alpha ={\partial f^i\over\partial a^\alpha},\; \varphi=det(\varphi_{\alpha\beta})},\\\medskip \displaystyle{b(a,f^k,f^k_\gamma)=b^\gamma (a) \left.{\partial\over\partial a^\gamma}\right\vert_a \stackrel{\hbox{def}}=\varphi^{\alpha\beta}(a)f^i_\alpha (a)P^\gamma_{\beta i}(a,f(a)) \left.{\partial\over\partial a^\gamma}\right\vert_a },\\ \displaystyle{ y(a,f^k,f^k_\gamma)=y^k(a)\left.{\partial\over\partial x^k}\right\vert_{f(a)} \stackrel{\hbox{def}}=\varphi^{\alpha\beta}(a)f^i_\alpha (a)P^k_{\beta i}(a,f(a)) \left.{\partial\over\partial x^k}\right\vert_{f(a)}} .\end{array}\right.$ {\bf Definition.} A map $f\in C^\infty(M,N)$ is $\left( \begin{array}{ccc} &P&\\ g&\varphi&h \end{array}\right)$ {\it-harmonic} iff $f$ is a critical point for the functional $E^{\scriptstyle P}_{g\varphi h}$. {\bf Particular cases.} i) If $g_{\alpha\beta}(a,b)=\varphi_{\alpha\beta}(a)$ and $h_{ij}(x,y)=h_{ij}(x)$ are Riemannian me\-trics and the connection tensor is an arbitrary one, it recovers the classical definition of a harmonic map between two Riemannian manifolds \cite{1}. We remark that, in this case, the definition of harmonic maps is independent of the connection tensor field $P$. ii) If we take $N=R,\;h_{11}=1$ and the tensor of connection is of the form $P=(\delta^\alpha_\beta,P^1_{\beta 1})$, we obtain $C^\infty(M,N)={\cal F}(M)$ and the energy functional becomes $$ E^{\scriptstyle P}_{g\varphi 1}(f)=\displaystyle{ {1\over 2}\int_Mg^{\alpha\beta}(a,grad_\varphi f)f_\alpha f_\beta\sqrt\varphi \;da\;,\;\forall f\in{\cal F}(M).} $$ iii) If we consider $M=[a,b]\subset R,\;\varphi_{11}=g_{11}=1$ and the tensor of connection is\linebreak $P=(P^1_{1i},\delta^k_i)$, we obtain $C^\infty (M,N)=\{x:[a,b]\to N\;\vert\;x-C^\infty\;\hbox{differentiable} \}$. Denoting $C^\infty(M,N) =\Omega_{a,b}(N)$, the energy functional should be $$ E^{\scriptstyle P}_{11h}(x)=\displaystyle{{1\over 2}\int_a^bh_{ij}(x(t),\dot x(t)){dx^i\over dt}{dx^j\over dt}dt,\;\forall x\in\Omega_{a,b}(N).} $$ In conclusion, the $\left( \begin{array}{ccc} &P&\\ 1&1&h \end{array}\right)$-harmonic curves are exactly the geodesics of the generalized Lagrange space $(N,h_{ij}(x,y))$ \cite {2}. \section{Harmonic maps between Lagrange spaces of electrodynamics} \hspace{5mm}Let $(M^m,L_M)$ and $(N^n,L_N)$ be Lagrange spaces with the Lagrangians $$ L_M(a,b)=g_{\alpha\beta}(a)b^\alpha b^\beta+g_{\alpha\beta}(a)U^\alpha (a)b^ \beta +F(a), $$ $$L_N(x,y)=h_{ij}(x)y^iy^j+h_{ij}(x)V^i(x)y^j+G(x), $$ where \begin{itemize} \item $g_{\alpha\beta}$ (resp. $h_{ij}$) is Riemannian metric on $M$ (resp. $N$) representing the gravitational potentials on $M$ (resp. $N$). \item $U^\alpha$ (resp. $V^i$) is a vector field on $M$ (resp. $N$) representing the electromagnetic potentials. \item $F$ (resp. $G$) is a smooth function on $M$ (resp. $N$) representing the potential function. \end{itemize} The fundamental metric tensors of these Lagrangians are $$ \displaystyle{g_{\alpha\beta}(a)= {\partial^2L_M\over\partial b^\alpha\partial b^\beta},\; h_{ij}(x)={\partial^2L_N\over\partial y^i\partial y^j}.} $$ Taking an arbitrary tensor of connection $P$, the energy functional becomes $$ E_{g\varphi h}(f)=\displaystyle{ {1\over 2}\int_Mg^{\alpha\beta}(a)h_{ij}(f(a))f^i_\alpha f^j_\beta\sqrt\varphi\;da}. $$ We remark that, in this case, the energy functional is independent of the connection tensor $P$. The Euler-Lagrange equations will be, obviously, the equations of harmonic maps, namely $$\displaystyle{ g^{\alpha\beta}\left\{f^k_{\alpha\beta}-\left [G^\gamma_{\alpha\beta}+{1\over 2} {\partial\over\partial a^\alpha}\left(\ln{g\over\varphi}\right)\delta^\gamma_ \beta\right]f^k_\gamma+H^k_{ij}f^i_\alpha f^j_\beta\right\}=0\;,\;\forall\;k= \overline{1,n},} $$ where \begin{itemize} \item $f^k_{\alpha\beta}=\displaystyle{{\partial^2f^k\over\partial a^\alpha\partial a^\beta}\;,\;g=det(g_{\alpha\beta})\;,\;\varphi =det(\varphi_{\alpha\beta})}$. \item $G^\gamma_{\alpha\beta}$ are the Christoffel symbols of the metric $g_{\alpha\beta}$. \item $H^k_{ij}$ are the Christoffel symbols of the metric $h_{ij}$. \end{itemize} {\bf Remarks.} i) If $g_{\alpha\beta}=\varphi_{\alpha\beta}$, we recover the classical equations of harmonic maps between two Riemannian manifolds. ii) Denoting $\displaystyle{\Delta^\gamma_{\alpha\beta}= G^\gamma_{\alpha\beta}+ {1\over 2}{\partial\over\partial a^\alpha}\left(\ln{g\over\varphi}\right)\delta^\gamma_ \beta}$, we remark that $\Delta^\gamma_{\alpha\beta}$ represent the components of a linear connection induced by the metrics $g_{\alpha\beta}$ and $\varphi_ {\alpha\beta}$. By the last remark, we can give the following {\bf Definition.} Let $g,\varphi$ be Riemannian metrics on $M$. The curve $c:I\to M$, expressed locally by $c(t)=(a^\alpha (t))$, is a $(g,\varphi)$ -{\it geodesic} iff $c$ is an autoparallel curve of the connection $\Delta^\gamma_ {\alpha\beta}$ induced by the metrics $g$ and $\varphi$, namely $$\displaystyle{{d^2a^\gamma\over dt^2}=-\Delta^\gamma_{\alpha\beta} {da^\alpha\over dt}{da^\beta\over dt}}. $$ {\bf Remarks.} i) If $g=\varphi$, then we recover the classical definition of a geodesic on the Riemannian manifold $(M,g=\varphi)$. ii) It is obviously that a $(g,\varphi)$-geodesic is a reparametrized geodesic of the metric\nolinebreak $\;g$. {\bf Theorem.} {\it Let $f:(M,L_M)\to (N,L_N)$ be a smooth map which carries $(g,\varphi)$-geodesics into $h$-geodesics. Then $f$ is harmonic map.} {\bf Proof.} Let $c:I\subset R\to M\;,\;c(t)=(a^\alpha(t))$ be a $(g,\varphi)$-geodesic. Then we have $\displaystyle{{d^2a^\gamma\over dt^2}=-\Delta^\gamma_{\alpha\beta}{da^\alpha \over dt}{da^\beta\over dt}}$. Because $\overline{c}(t)=f(c(t))$ is $h$-geodesic, it follows that $\displaystyle{{d^2\overline{c}^k\over dt^2}+H^k_{ij} {d\overline{c}^i\over dt}{d\overline{c}^j\over dt}=0}$. But $\displaystyle{ {d\overline{c}^k\over dt}=f^k_\alpha(c(t)){da^\alpha\over dt}\Rightarrow {d^2\overline{c}^k\over dt^2}=f^k_{\alpha\beta}{da^\alpha\over dt}{da^\beta \over dt}+f^k_\alpha{d^2a^\alpha\over dt^2}}$. In conclusion we obtain $$\displaystyle{{d^2a^\alpha\over dt^2}f^k_\alpha+f^k_{\alpha\beta}{da^\alpha\over dt} {da^\beta\over dt}+H^k_{ij}f^i_\alpha f^j_\beta{da^\alpha\over dt}{da^\beta\over dt}=0\Rightarrow} $$ $$\displaystyle{\left(f^k_{\alpha\beta}-\Delta^\gamma_{\alpha\beta}f^k_\gamma+ H^k_{ij}f^i_\alpha f^j_\beta\right){da^\alpha\over dt}{da^\beta\over dt}=0\;, \;\forall k=\overline{1,n}}\;\Box. $$ \section{Geometrical interpretation of solutions of certain PDEs of order one} \hspace{5mm}The problem of finding a geometrical structure of Riemannian type on a manifold $M$ such that the orbits of an arbitrary vector field $X$ should be geodesics, was intensively studied by Sasaki. The results were not satisfactory, but, in his study, Sasaki discovered the well known almost contact structures on a manifold of odd\linebreak dimension \cite{6}. After the introduction of generalized Lagrange spaces by Miron \cite{2}, the same problem is resumed by Udri\c ste \cite{8,9}. This succeded to discover a Lagrange structure on $M$, depending of the vector field $X$ and an associated (1,1)-tensor field, such that the orbits of $C^2$ class should be geodesics. Moreover, he formulated a more general problem \cite{9}, namely 1) Are there structures of Lagrange type such that the solutions of certain PDEs of order one should be {\it harmonic maps}? 2) What is a {\it harmonic map} between two generalized Lagrange spaces? A partial answer of these questions is offered by author in his paper \cite{4}, using the notion of harmonic map on a direction between a Riemannian manifold and a generalized Lagrange manifold. The notion of {\it harmonic map} between two generalized Lagrange spaces introduced in \cite{7} allows one to extend the results of previous papers \cite{4}, \cite{8}, \cite{9} and to obtain a beautiful geometrical interpretation of the solutions of the certain PDEs of order one. For every smooth map $f\in C^\infty(M,N)$, we use the following notation $$\displaystyle{ \left.\delta f=f^i_\alpha da^\alpha\vert_a \otimes{\partial\over\partial y^i}\right\vert_{f(a)}\in \Gamma(T^M\otimes f^{-1}(TN))}.$$ On $M\times N$, let $T$ be one tensor of type $(1,1)$ with all components equal to zero except $(T^i_{\alpha} )_{i=\overline{1,n}\\ \atop \alpha=\overline{1,m}}$. \medskip Let the system of partial differential equations $$\displaystyle{ \delta f=T\;\hbox{expressed locally by}\;{\partial f^i\over\partial a^\alpha} =T^i_\alpha(a,f(a)). }\leqno(E)$$ If $(M,\varphi_{\alpha\beta})$ and $(N,\psi_{ij})$ are Riemannian manifolds, we can build a scalar product on $\Gamma(T^M\otimes f^{-1}(TN))$ by $<T,S>=\varphi^{\alpha\beta}(a)\psi_{ij}(f(a))T^i_\alpha S^j_\beta$, where $\displaystyle{T=T^i_\alpha da^\alpha\otimes{\partial\over\partial y^i}}$ and $\displaystyle{S=S^j_\beta da^\beta\otimes{\partial\over\partial y^j}}$. Under these conditions, we can prove the following {\bf Theorem.} {\it If $(M,\varphi),(N,\psi)$ are Riemannian manifolds and $f\in C^\infty(M,N)$ is a solution of the system $(E)$, then $f$ is a solution of the variational problem asociated to the functional ${\cal L}_T:C^\infty(M,N)\backslash\{f\;\vert\;\exists\;a\in M\;\hbox{such that}\; <\delta f,T>(a)=0\}\to R_+$, $$ \displaystyle{{\cal L}_T(f)={1\over 2}\int_M{\Vert\delta f\Vert^2\Vert T\Vert^2 \over <\delta f,T>^2}\sqrt{\varphi}\;da={1\over 2}\int_M{\Vert T\Vert^2\over < \delta f, T>^2}\varphi^{\alpha\beta}\psi_{ij}f^i_\alpha f^j_\beta\sqrt{\varphi}\;da}. $$} {\bf Proof.} In the space $\Gamma(T^M\times f^{-1}(TN))$, the Cauchy inequality for the scalar product $<,>$ holds. It follows that the following inequality is true, \linebreak$<T,S>^2\leq\nolinebreak\Vert T\Vert^2\Vert S\Vert^2\;,\;\forall\; T,S\in\Gamma(T^M\times\nolinebreak f^{-1}(TN))$, with equality if and only if there exists ${\cal K}\in{\cal F}(M)$ such that $T={\cal K}S$. Consequently, for every $f\in C^\infty (M,N)$ we have $$\displaystyle{{\cal L}(f)={1\over 2}\int_M{\Vert\delta f\Vert^2 \Vert T\Vert^2\over<\delta f,T>^2}\sqrt{\varphi}\;da\geq{1\over 2}\int_M\sqrt {\varphi}\;da={1\over 2}Vol_\varphi(M)}. $$ Obviously, if $f$ is a solution of the system $(E)$, we obtain $\displaystyle{{\cal L}_T(f)={1\over 2}Vol_\varphi(M)}$, that is, $f$ is a global minimum point for ${\cal L}_T$ $\Box$. {\bf Remarks.} i) In certain particular cases of the system $(E)$, the functional ${\cal L}_T$ becomes exactly a functional of type $\left( \begin{array}{ccc} &P&\\ g&\varphi&h \end{array}\right)$-energy. ii) The global minimum points of the functional ${\cal L}_T$ are solutions of the system $\delta f={\cal K}T$, where ${\cal K}\in{\cal F}(M)$, not necessarily with ${\cal K}=1$. iii) Replacing the Riemannian metric $\psi_{ij}$ by a pseudo-Riemannian metric, the preceding theorem survives because the form of the Euler-Lagrange equations remains unchanged. The difference is that the solutions of the system $(E)$, in the pseudo-Riemannian case, are not the global minimum points for the functional ${\cal L}_T$. Moreover, the statement (ii), of above, does not hold. {\bf Fundamental examples.} {\bf 1. Orbits} For $M=([a,b],1)$ and $T=\xi\in\Gamma(x^{-1}(TN))$, the system $(E)$ becomes $$\displaystyle{{dx^i\over dt}=\xi^i(x(t)),\;x:[a,b]\to N},\leqno{(E_1)}$$ that is the system of orbits for $\xi$, and the functional ${\cal L}_\xi$ is $$\displaystyle{{\cal L}_\xi(x)={1\over 2}\int^b_a{\Vert\xi\Vert^2_\psi\over [\xi^b(\dot x)]^2}\psi_{ij}{dx^i\over dt}{dx^j\over dt}dt}, $$ where $\xi^b=\xi_idx^i=\psi_{ij}\xi^jdx^i$. Hence the functional ${\cal L}_\xi$ is a $\left( \begin{array}{ccc} &P&\\ 1&1&h \end{array}\right)$-energy (see (iii) of first particular cases of this paper), where $$h_{ij}:TN\backslash\{y\;\vert\;\xi^b(y)=0\;\hbox{for some}\;y\}\to R$$ is defined by $$\displaystyle{ h_{ij}(x,y)={\Vert\xi\Vert^2_\psi\over[\xi^b(y)]^2}\psi_{ij}(x)=\psi_{ij}(x) \exp{\displaystyle{\left[ 2\ln{\Vert\xi\Vert_\psi\over\vert\xi^b(y)\vert}\right]} } }.$$ This case is studied in other way by Udri\c ste in \cite{8}-\cite{9}. {\bf 2. Pfaffian systems} For $N=(R,1)$ and $T=A\in\Lambda^1(T^M)$, the system $(E)$ becomes $$df=A,\;f\in{\cal F}(M),\leqno{(E_2)}$$ that is a Pfaffian system, and the functional ${\cal L}_T$ reduces to $$ \displaystyle{{\cal L}_A(f)={1\over 2}\int_M{\Vert A\Vert^2_\varphi\over[A( grad_\varphi f)]^2}\varphi^{\alpha\beta}f_\alpha f_\beta\sqrt{\varphi}da}. $$ Hence, the functional ${\cal L}_A$ is a $\left( \begin{array}{ccc} &P&\\ g&\varphi&1 \end{array}\right)$-energy (see (ii) of first particular cases of this paper), where $g_{\alpha\beta}:TM\backslash\{b\;\vert\; A(b)=0\;\hbox{for some}\;b\}\to R$ is defined by $$\displaystyle{ g_{\alpha\beta}(a,b)={[A(b)]^2\over\Vert A\Vert^2_\varphi} \varphi_{\alpha\beta}(a) =\varphi_{\alpha\beta}(a)\exp{ \displaystyle{ \left[2\ln{\vert A(b)\vert\over\Vert A\Vert_\varphi} \right]}} }.$$ {\bf 3. Pseudolinear functions} We suppose that $T^k_\beta (a,x)=\xi^{k}(x)A_\beta (a)$, where $\xi^k$ is vector field on $N$ and $A_\beta$ is 1-form on $M$. In this case the system $(E)$ is $$\displaystyle{{\partial f^k\over\partial a^\beta}=\xi^k(f)A_\beta (a)} \leqno{(E_3)} $$ and the functional ${\cal L}_T$ is expressed by $$\displaystyle{{\cal L}_T(f)={1\over 2}\int_M{\Vert\xi\Vert^2_\psi\Vert A \Vert^2_\varphi\over[A(b)]^2}\varphi^{\alpha\beta}\psi_{ij}f^i_\alpha f^j_\beta \sqrt\varphi da= }$$ $$\displaystyle{={1\over 2}\int_Mg^{\alpha\beta}(a,b)h_{ij}(f(a))f^i_\alpha f^j _\beta\sqrt\varphi da}, $$ where $h_{ij}(x)=\Vert\xi\Vert^2_\psi\psi_{ij}(x)$, the tensor of connection is $P^\gamma_{i\beta}(x)=\delta^\gamma_\beta\xi_i(x),\linebreak b^\gamma= \varphi^{\alpha\beta}f^i_\alpha P^\gamma_{i\beta}$ and the Lagrange metric tensor $$g_{\alpha\beta}:TM\backslash\{b\;\vert\;A(b)=0\;\hbox{for some}\;b\}\to R$$ is defined by $$\displaystyle{g_{\alpha\beta}(a,b)={[A(b)]^2\over\Vert A\Vert^2_\varphi} \varphi_{\alpha\beta}(a)=\varphi_{\alpha\beta}(a) \exp{\displaystyle{\left[2\ln{\vert A(b)\vert\over\Vert A\Vert_\varphi}\right]}} }.$$ It follows that the functional ${\cal L}_T$ becomes a $\left( \begin{array}{ccc} &P&\\ g&\varphi&h \end{array}\right)$-energy.\medskip {\bf Remark.} Take $M$ to be an open subset in $(R^n,\varphi=\delta)$ and $N=(R,\psi=1)$, the system from the third example is $$\displaystyle{{\partial f\over\partial a^\alpha}=\xi(a)A_\alpha(f(a)),\; \forall\;\alpha=\overline{1,m}.}\leqno{(PL)} $$ Supposing that $(grad\;f)(a)\ne 0,\;\forall\;a\in M$, the solutions of this system are the well known {\it pseudolinear functions} \cite{5}. These functions have the following property, $-$for every fixed point $x_0\in M$, the hypersurface of constant level $$M_{f(x_0)}=\{x\in M\;\vert\;f(x)=f(x_0)\}$$ is totally geodesic \cite {5} (i. e. the second fundamental form vanishes identically).\medskip In conclusion, the pseudolinear functions are examples of harmonic maps between the generalized Lagrange spaces $\displaystyle{\left(M,g_{\alpha\beta}(a,b)={[A(b)]^2\over\Vert A\Vert^2} \delta_{\alpha\beta}\right)}$ and $\displaystyle{(R,h(x)=\xi^2(x))}$. For example, we have the following pseudolinear functions \cite{5}:\medskip {\bf 3. 1.} $f(a)=e^{<v,a>+w}$, where $v\in M,\;w\in R$, is solution for the system $(PL)$ with $\xi(a)=1$ and $A(f(a))=f(a)v$.\medskip {\bf 3. 2.} $\displaystyle{f(a)={<v,a>+w\over<v',a'>+w'}}$, where $v,v'\in M \;,\;w,w'\in R$, is solution for $(PL)$ with $\displaystyle{\xi(a)={1\over<v',a>+w'}}$ and $A(f(a))=v-f(a)w$. {\bf Remark.} The preceding cases appear also in \cite{7} and, from another point of view, in \cite{4}. The following case is the main novelty of this paper. {\bf 4. The general case} If we have $T^i_\alpha(a,x)=\sum_{r=1}^t\xi_r^i(x)A^r_\alpha(a)$, where $\{\xi_r\}_{r=\overline{1,t}}\subset{\cal X}(N)$ is a family of vector fields on $N$ and $\{A^r\}_{r=\overline{1,t}}\subset\Lambda^1(T^M)$ is a family of 1-forms on $M$, the system of equations $(E)$ reduces to $$\displaystyle{{\partial f^i\over\partial a^\alpha}=\sum_{r=1}^t\xi_r^i(f)A^r_\alpha(a) }.\leqno{(E_4)}$$ Without loss of generality, we can suppose that $\{\xi_r\}_{r=\overline{1,t}}\subset{\cal X}(N)$ (resp. $\{A^r\}_{r=\overline{1,t}}\subset\Lambda^1(T^M)$) are linearly independent. In these conditions, we shall have $t\leq\min\{m,n\}$, where $m=dim\;M$ and $n=dim\;N$. {\bf 4. 1.} Assume that $\{\xi_r\}_{r=\overline{1,t}}\subset{\cal X}(N)$ is an orthonormal system of vector fields with respect to the Riemannian metric $\psi_{ij}$ on $N$. Let $B\in\Lambda^1(T^M)$ be an arbitrary unit 1-form on $M$. With our assumptions, by a simple calculation, we obtain $$\Vert T\Vert^2=\varphi^{\alpha\beta}\psi_{ij}\xi^i_rA^r_\alpha\xi^j_sA^s_\beta= \sum^t_{r,s=1}<\xi_r,\xi_s>_\psi<A^r,A^s>_\varphi=\sum_{r=1}^t\Vert A^r\Vert^2_\varphi,$$ $$<\delta f, T>=\varphi^{\alpha\beta}\psi_{ij}f^i_\alpha\xi^j_rA^r_\beta B^\mu B_\mu.$$ Defining the tensor of connection by $P^\gamma_{i\beta}(a,x)=\psi_{ij}(x)\xi^j_r(x)A^r_\beta(a)B^\gamma(a)$ and \linebreak $b^\gamma=\varphi^{\alpha\beta}f^i_\alpha P^\gamma_{i\beta}$, the functional ${\cal L}_T$ takes the form $$\displaystyle{{\cal L}_T(f)={1\over 2}\int_M{\sum_{r=1}^t\Vert A^r\Vert^2_\varphi\over[B(b)]^2} \varphi^{\alpha\beta}\psi_{ij}f^i_\alpha f^j_\beta\sqrt{\varphi}\;da= {1\over 2}\int_Mg^{\alpha\beta}(a,b)\psi_{ij}(f(a))f^i_\alpha f^j_\beta\sqrt {\varphi}\;da },$$ where the Lagrange metric tensor $g_{\alpha\beta}:TM\backslash\{b\;\vert\;B(b)=0\;\hbox{for some}\;b\} \to R$ is expressed by $$\displaystyle{ g_{\alpha\beta}(a,b)={[B(b)]^2\over\sum_{r=1}^t\Vert A^r\Vert^2_\varphi} \varphi_{\alpha\beta}(a)=\varphi_{\alpha\beta}(a) \exp{\displaystyle{\left[2\ln{\vert B(b)\vert\over \sqrt{\sum_{r=1}^t\Vert A^r\Vert^2_\varphi}}\right]}} }.$$ Consequently, the functional ${\cal L}_T$ is a $\left( \begin{array}{ccc} &P&\\ g&\varphi&\psi \end{array}\right)$-energy. {\bf 4. 2.} As above, we assume that the system $\{A^r\}_{r=\overline{1,t}}$ of 1-forms is orthonormal with respect to the metric $\varphi^{\alpha\beta}$ and we choose an arbitrary unit vector field $X\in{\cal X}(N)$. By analogy to {\bf 4. 1} we shall have $$\Vert T\Vert^2=\sum_{r=1}^t\Vert\xi_r\Vert^2_\psi\;\hbox{and}\; <\delta f,T>=\varphi^{\alpha\beta}\psi_{ij}f^i_\alpha\xi^j_rA^r_\beta X^kX_k. $$ Using the notations $P^k_{i\beta}(a,x)=\psi_{ij}(x)\xi^j_r(x)A^r_\beta(a)X^k(x)$ and $y^k=\varphi^{\alpha\beta}f^i_\alpha P^k_{i\beta}$ we \linebreak obtain the following expression of the functional ${\cal L}_T$, $$\displaystyle{{\cal L}_T(f)={1\over 2}\int_M{\sum_{r=1}^t\Vert\xi_r\Vert^2 _\psi\over[X^b(y)]^2}\varphi^{\alpha\beta}\psi_{ij}f^i_\alpha f^j_\beta\sqrt {\varphi}\;da= {1\over 2}\int_M\varphi^{\alpha\beta}(a)h_{ij}(f(a),y)f^i_\alpha f^j_\beta \sqrt{\varphi}\;da },$$ where the Lagrange metric tensor $h_{ij}:TN\backslash\{y\;\vert\;X^b(y)=0\;\hbox{for some}\;y\} \to R$ is $$\displaystyle{ h_{ij}(x,y)={\sum_{r=1}^t\Vert\xi_r\Vert^2_\psi\over[X^b(y)]^2} \psi_{ij}(x)=\psi_{ij}(x)\exp{\displaystyle{ \left[2\ln{\sqrt{\sum_{r=1}^t\Vert\xi_r\Vert^2_\psi}\over\vert X^b(y)\vert}\right]}} }.$$ Obviously, the functional ${\cal L}_T$ is a $\left( \begin{array}{ccc} &P&\\ \varphi&\varphi&h \end{array}\right)$-energy. {\bf Remarks.} i) For the use of the above, we assume {\it a priori} a Riemannian metric $\varphi$ or $\psi$ on $M$ or $N$ such that the system of covectors $\{A^r\}_{r=\overline{1,t}}$ or of vectors $\{\xi_r\}_{r=\overline{1,t}}$ is orthonormal. This fact is always possible. In conclusion, our assumptions on the orthonormality of these systems do not restrict the generality of problem. ii) In the particular case, $n=r=1\;,\;N=R\;,\;\psi_{11}=1\;,\;\xi= \displaystyle{d\over dx}$, and $A\in\Lambda^1(T^M)$ is an arbitrary 1-form on $M$, we recover the Pfaffian system $df=A$. In this situation, taking $B\in\Lambda^1(T^M)$ to be an arbitrary unit 1-form, we obtain, for the functional ${\cal L}_T$, the expression $$\displaystyle{{\cal L}_T(f)={1\over 2}\int_M{\Vert A\Vert^2\over[B(b)]^2} \varphi^{\alpha\beta}f_\alpha f_\beta\sqrt{\varphi}\;da={1\over 2}\int_M {\Vert A\Vert^2\over[A(grad_\varphi f)]^2}\varphi^{\alpha\beta}f_\alpha f_\beta \sqrt{\varphi}\;da },$$ where the tensor of connection is defined by $P_{1\beta}^\gamma=A_\beta B^\gamma$ and the direction $b$ is\linebreak $b^\gamma=\varphi^{\alpha\beta}f^i_\alpha A_\beta B^\gamma=B^\gamma A(grad_\varphi f)$. Since the last integral is the functional ${\cal L}_A$ from the example {\bf 1}, we remark that the solutions of the Pfaffian system $df=A$ can be regarded in an infinity manner as $\left( \begin{array}{ccc} &P&\\ g&\varphi&1 \end{array}\right)$-harmonic maps. This fact appears because the tensor of connection $P$ and the generalized Lagrange metric $g$ are dependent of the arbitrary unit covector field $B$. iii) Analog to ii), if we take $m=r=1\;,\;M=[a,b]\;,\; g_{11}=\varphi_{11}=1\;,\; A=dt$ and $\xi\in{\cal X}(N)$ is an arbitrary vector field on $N$, we find the system of orbits for $\xi$, that is, $$\displaystyle{{dx^i\over dt}=\xi^i(x(t)),\;x:[a,b]\to N .}$$ Starting with $X\in{\cal X}(N)$ an arbitrary unit vector field, the functional ${\cal L}_T$ becomes $$\displaystyle{{\cal L}_T(x)={1\over 2}\int_a^b{\Vert\xi\Vert^2_\psi\over[X^b(y)]^2} \psi_{ij}{dx^i\over dt}{dx^j\over dt}dt={1\over 2}\int_a^b{\Vert\xi\Vert^2_\psi \over[\xi^b(\dot x)]^2}\psi_{ij}{dx^i\over dt}{dx^j\over dt}dt },$$ where the tensor of connection is $P_{i1}^k=\xi_i^bX^k$ and $y^k=\xi^b(\dot x)X^k$. Because we can vary the tensor of connection $P$ and the generalized Lagrange metric $h$ by the arbitrary unit vector field $X$, we remark that the trajectories of the vector field $\xi$ can be also regarded in an infinity manner as $\left( \begin{array}{ccc} &P&\\ 1&1&h \end{array}\right)$-harmonic maps. \section{Lagrange geometry asociated to PDEs of order one} \hspace{5mm}We first remark that, in all above cases, the solutions of $C^2$ class of the system $\delta f=T$ becomes harmonic maps between generalized Lagrange spaces, in the sense defined in this paper. Moreover, the above generalized Lagrange structures are of type $(M^n,e^{2{\sigma(x,y)}}\gamma_{ij}(x))$, where $\sigma :TM\backslash\{ \hbox{Hyperplane}\}\to R$ is a smooth function. Using the ideas exposed in \cite{2}, in these spaces, we can construct a Lagrange geometry and field theory. This geometrical Lagrange theory will be regarded as a natural one associated to the PDE system $\delta f=T$, in the sense of the first Udri\c ste's question. Now, we assume that a generalized Lagrange space $(M^n,g_{ij}(x,y))$ satisfies the following axioms: a. 1. The fundamental tensor field $g_{ij}(x,y)$ is of the form $$ g_{ij}(x,y)=e^{2\sigma(x,y)}\gamma_{ij}(x). $$ a. 2. The space is endowed with the non-linear connection $$ N^i_j(x,y)=\Gamma^i_{jk}(x)y^k, $$ where $\Gamma^i_{jk}(x)$ are the Christoffel symbols for the Riemannian metric $\gamma_{ij}(x)$. Under these assumptions, our space verifies a constructive-axiomatic formulation of General Relativity due to Ehlers, Pirani and Schild \cite{2}. This space represents a convenient relativistic model, since it has the same conformal and projective properties as the Riemannian space $(M,\gamma_{ij})$. In the Lagrangian theory of electromagnetism, the electromagnetic tensors $F_{ij}$ and $f_{ij}$ are $$\displaystyle{ F_{ij}=\left(g_{ip}{\delta\sigma\over\delta x^j}-g_{jp}{\delta\sigma\over \delta x^i}\right)y^p\;,\;f_{ij}=\left(g_{ip}{\partial\sigma\over\partial y^j}- g_{jp}{\partial\sigma\over\partial y^i}\right)y^p }.$$ Developping the formalism presented in \cite{2}, \cite{3} and denoting by $r^i_{jkl}$ the curvature tensor field of the metric $\gamma_{ij}(x)$ , the following Maxwell equations of the electromagnetic tensors hold $$\left\{\begin{array}{lll} \displaystyle{F_{ij\vert k}+F_{jk\vert i}+F_{ki\vert j}=-\sum_{(ijk)}g_{ip} r^h_{qjk}{\partial\sigma\over\partial y^h}y^py^q},\\ F_{ij}\vert_k+F_{jk}\vert_i+F_{ki}\vert_j=-(f_{ij\vert k}+ f_{jk\vert i}+f_{ki\vert j}),\\ f_{ij}\vert_k+f_{jk}\vert_i+f_{ki}\vert_j=0, \end{array}\right. $$ where ${_\vert}_i$ (resp. $\vert_a$) represents the $h-$ (resp. $v-$) covariant derivative induced by the non-linear connection $N^i_j$. Suppose $\sigma=\sigma(x)$. Then the $v-$ electromagnetic tensor is $f_{ij}=0$, the $h-$ covariant operator "${_\vert}_i$" becomes the covariant derivative with respect to Levi-Civita connection of the metric $g_{ij}(x,y)=e^{2\sigma(x)}\varphi_{ij}(x)$, the $h-$ electromagnetic tensor $F_{ij}$ is the same with the classical electromagnetic tensor and the Maxwell's equations reduce to the classical ones. In the construction of the gravitational field equations, we shall use the notations $$\left\{\begin{array}{l}\medskip r_{ij}=r^k_{ijk},\;r=\gamma^{ij}r_{ij},\;\displaystyle{ {\delta\over\delta x^i}={\partial\over \partial x^i}- N^j_i{\partial\over\partial y^j}},\\ \sigma^{\scriptscriptstyle {H}}= \displaystyle{\gamma^{kl}{\delta\sigma\over\delta x^k}{\delta\sigma\over\delta x^l},\;\sigma^{\scriptscriptstyle {V}}=\gamma^{ab}{\partial\sigma\over\partial y^a}{\partial\sigma\over\partial y^b},\;\overline\sigma=\gamma^{ij}\sigma_{ij}, \;\dot\sigma=\gamma^{ab}\dot\sigma_{ab}}, \end{array}\right. $$ where $\displaystyle{ \sigma_{ij}={\delta\sigma\over\delta x^i}\vert_j+{\delta\sigma\over\delta x^i}{\delta\sigma\over \delta x^j}-{1\over 2}\gamma_{ij}\sigma^{\scriptscriptstyle H}},\; \displaystyle{ \dot\sigma_{ab}=\left.{\partial\sigma\over\partial y^a}\right\vert_b+ {\partial\sigma\over\partial y^a}{\partial\sigma\over\partial y^b}-{1\over 2} \gamma_{ab}\sigma^{\scriptscriptstyle V}}.$\\ In these conditions, the Einstein's equations of the space $(M,g_{ij}(x,y))$ take the form $$\left\{ \begin{array}{ll}\medskip \displaystyle{r_{ij}-{1\over 2}r\gamma_{ij}+t_{ij}={\cal K} T^{\scriptscriptstyle H}_{ij}}\\ (2-n)(\dot\sigma_{ab}-\dot\sigma\gamma_{ab})={\cal K}T^{\scriptscriptstyle V} _{ab},\\ \end{array}\right. $$ where $T^{\scriptscriptstyle H}_{ij}$ and $T^{\scriptscriptstyle V}_{ab}$ are the $h-$ and the $v-$ components of the energy momentum tensor field, ${\cal K}$ is the gravific constant and $$ t_{ij}=(n-2)(\gamma_{ij}\overline\sigma-\sigma_{ij})+\gamma_{ij}r_{st}y^s \gamma^{tp}{\partial\sigma\over\partial y^p}+{\partial\sigma\over\partial y^i} r^a_{tja}y^t-\gamma_{is}\gamma^{ap}{\partial\sigma\over\partial y^p} r^s_{tja}y^t. $$ It is clear that the general metric $g_{ij}(x,y)=e^{2\sigma(x,y)}\varphi _{ij}(x)$ implies Einstein equations which differ from the classical ones by the additional tensor $t_{ij}$. Finally, we remark that, in certain particular cases, it is posible to build a generalized Lagrange geometry and field theory naturally attached to a system of partial differential equations. In these geometrical structures, the solutions of $C^2$ class of the PDE system become harmonic maps. This idea was suggested by Udri\c ste in private discussions and in \cite{9}, \cite{10}. {\bf Open problem.} Because the generalized Lagrange structure constructed in this paper is not unique, it arises a natural question: $-$Is it possible to build a unique generalized Lagrange geometry naturally asociated to a given PDEs system? An answer to this question will be offered by author in a subsequent paper, using a more general Lagrange geometry, naturally attached to a multidimensional Lagrangian defined on the jet fibration of order one. {\bf Acknowledgements.} I would like to express my gratitude to the reviewer of Southeast Asian Bulletin of Mathematics and Prof. Dr. C. Udri\c ste for their valuable comments and very useful suggestions.	9,546
\begin{document} \date{October 30, 2020} \subjclass[2000]{32V05, 32H35, 53A30} \thanks{Duong Ngoc Son was supported by the Austrian Science Fund, FWF-Projekt M 2472-N35. Bernhard Lamel was supported by the Austrian Science Fund, FWF-Projekt I3472.} \begin{abstract} We study a CR analogue of the Ahlfors derivative for conformal immersions of Stowe \cite{stowe2015ahlfors} that generalizes the CR Schwarzian derivative studied earlier by the second-named author \cite{son2018schwarzian}. This notion possesses several important properties similar to those of the conformal counterpart and provides a new invariant for \edit{spherically} equivalent CR maps from strictly pseudoconvex CR manifolds into a sphere. The invariant is computable and distinguishes many well-known sphere maps. In particular, it vanishes precisely when the map is spherically equivalent to the linear embedding of spheres. \end{abstract} \maketitle \section{Introduction} The main purpose of this paper is to extend the notion of \edit{the} CR Schwarzian derivative for CR diffeomorphisms \cite{son2018schwarzian} to the case of CR immersions. For conformal immersions of Riemannian manifolds, the Ahlfors derivative of Stowe \cite{stowe2015ahlfors} generalizes the Schwarzian derivative of Osgood--Stowe \cite{osgood1992schwarzian} in a similar way and goes back to Ahlfors \cite{ahlfors1988cross}. Precisely, we shall define, for each CR immersion $f\colon (M,\theta) \to (N,\eta)$ between pseudohermitian manifolds, a tensor denoted by $\ahlfors_{\theta}(f)$. This tensor reduces to the Schwarzian tensor introduced in \cite{son2018schwarzian} for CR diffeomorphisms in the equidimensional case. We shall call this tensor the CR Ahlfors \edit{derivative (or tensor)}\editb{. We} refer the reader to Stowe's paper for further discussions regarding the history and motivations in the conformal case; however, it turns out that the CR setting has some special properties not present in the conformal setting, which we shall point out as we go. The tensor which we are going to construct satisfies a ``chain rule'' described as follows: For a chain of CR immersions $(M,\theta) \xrightarrow{~F~} (N,\eta) \xrightarrow{~G~} (P,\zeta)$, it holds that \begin{equation}\label{e:cr} \ahlfors(G\circ F) = \ahlfors(F) + F^{\ast} \ahlfors(G). \end{equation} It was shown in \cite{son2018schwarzian} that if $(M,\theta)=(N,\eta)$ is the sphere with its standard pseudohermitian structure, then $\ahlfors(F)$ (which reduces to the CR Schwarzian derivative as already noted) vanishes identically if $F$ is a CR automorphism of the sphere. Therefore, the chain rule \cref{e:cr} implies that $\ahlfors$ is an invariant for spherically equivalent CR maps into spheres of higher dimensions. This invariant property is a main motivation for us to extend the notion of \edit{the} CR Schwarzian derivative to the case of higher dimensional targets. We shall in fact apply the Ahlfors derivative to study equivalence of sphere maps, a problem which has been studied extensively; we can mention only several papers \cite{d1988proper,d1991polynomial,d2016homotopy} and refer the readers to numerous references therein. To construct the CR Ahlfors derivative, we shall follow Stowe's construction for the conformal case. First, we define a notion of CR second fundamental form for the ``isopseudohermitian'' immersions and the $(1,0)$-mean curvature vector (this step was not needed in the equidimensional case). Precisely, let $(N,\eta)$ be \edit{a} pseudohermitian manifold and let $\iota \colon (M,\theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold of $N$. This means the standard inclusion $\iota$ is CR and $\theta = \iota^{\ast} \eta$. We denote by $\nabla$ and $\widetilde{\nabla}$ the Tanaka-Webster connections on $(M,\theta)$ and $(N, \eta)$, respectively, introduced by Tanaka \edit{and Webster} \cite{tanaka1975differential}. For any two vector fields $X, Y \in \Gamma(\C TM )$ extended to smooth sections $\widetilde{X}, \widetilde{Y}$ of $\C TN$, we define the pseudohermitian \textit{second fundamental form} by the Gauß formula, namely, \begin{equation}\label{e:sffdef} \sff(X, Y) = \sff_M^N(X, Y) := \widetilde{\nabla}_{\widetilde{X}} \widetilde{Y} - \nabla_XY, \end{equation} This notion was \edit{previously} studied by many authors, see, e.g., \cite{webster1979rigidity} for the codimension one case and \cite{dragomir1995pseudohermitian,ebenfelt2004rigidity} for the case of pseudohermitian immersions (i.e., when the Reeb field of $\eta$ is tangent to $\iota(M)$.) For our applications, we shall make no assumption on the Reeb field of the target. Due to the presence of the torsion, $\sff$ is not \edit{necessarily} symmetric and thus we also consider the symmetrized second fundamental form, i.e., \begin{equation} \Sym \sff (X,Y) = \tfrac{1}{2}\left(\sff(X,Y) + \sff(Y,X)\right). \end{equation} In most situations, we shall consider the second fundamental form $\sff$ as a tensor on the ``horizontal'' space $T^{0,1}M \oplus T^{1,0}M$ (the ``good directions'') where it behaves quite well. In particular, we define the $(1,0)$-mean curvature vector to be the trace of $\sff$ on the horizontal subspace: \begin{equation} H := \sum_{\alpha = 1}^{n} \sff (Z_{\abar}, Z_{\alpha}) \end{equation} where $\{Z_{\alpha} \colon \alpha = 1,2,\dots ,n\}$ is an orthonormal frame of $T^{1,0}M$. The trace of $\Sym\sff$ is denoted by $\mu$, so that $\mu = \Re H$. Let us point out that the consideration here is similar to \cite{son2019semi} in which we consider the case of CR immersions into a Kähler manifold. Moreover, when the target is the standard sphere, the second fundamental form \eqref{e:sffdef} is closely related to the one for CR immersions into complex euclidean space. Similar to \cite{stowe2015ahlfors}, we define the tensor $\nu = \nu_M^N$ as a symmetric real tensor on $T^{1,0}M \oplus T^{0,1}M$ via the formula \begin{equation} \nu(X,Y) = 2\left\langle \Sym \sff (X,Y) , \mu \right\rangle - \langle X,Y\rangle \|\mu\|^2. \end{equation} Moreover, we define, for each smooth function $u$ on $M$, \begin{align} \mathcal{H}_{\theta}(u) = \Sym \nabla \nabla u -\partial_b u \otimes \partial_b u - \bar{\partial}_b u \otimes \bar{\partial}_b u + \frac{1}{2}\|\bar{\partial}_b u\|^2 L_\theta. \end{align} \edit{Here, $L_{\theta}(Z,\Wba): = -i d\theta(Z,\Wba)$ ($Z,W \in T^{1,0}M$) is the Levi form.} We refer the reader to \cref{ss:changecontact}, in particular \eqref{e:dudecomp}, for the (standard) notation used here. We can now introduce the CR analogue of the Ahlfors derivative as follows. \begin{definition}[cf. \cite{stowe2015ahlfors}] Let $(M^{2n+1},\theta)$ and $(N^{2d+1},\eta)$ be strictly pseudoconvex pseudohermitian manifolds and let $F\colon M\to N$ be a CR immersion. Let $u$ be the smooth function on $M$ such that $F^{\ast} \eta = e^{u} \theta$. We define the CR Ahlfors derivative \edit{(or CR Ahlfors tensor)} of $F$ to be \begin{equation} \ahlfors(F) := \mathcal{H}_{\theta}(u) + F^{\ast}\left(\nu^{N}_{F(M)}\right) +\frac12 F^{\ast}(J_{\Theta} L_{\Theta}) - \frac12 J_{\theta} L_{\theta}. \end{equation} where $J_{\theta} = R_{\theta}/(n(n+1))$ and $J_{\Theta} = R_{\Theta}/(d(d+1))$ are the normalized Webster scalar curvatures on $M$ and $N$, respectively, and \edit{$L_{\Theta}$ and $L_{\theta}$ are the corresponding Levi forms.} \end{definition} As mentioned above, the Ahlfors tensor $\ahlfors$ generalizes the CR Schwarzian tensor for CR diffeomorphisms in \cite{son2018schwarzian} in the same spirit the conformal Ahlfors generalizes the Schwarzian of Osgood--Stowe. We shall explain this in the next section. As briefly discussed, our first result of the paper is the following chain rule. \begin{theorem}\label{thm:crintro} For CR immersions $F\colon (M,\theta) \to (N,\eta)$ and $G\colon (N,\eta) \to (P,\zeta)$, we have \begin{equation}\label{e:chainrule} \ahlfors(G\circ F) = \ahlfors(F) + F^{\ast}\ahlfors (G). \end{equation} \end{theorem} We point out that although this theorem is analogous to Theorem 1 in \cite{stowe2015ahlfors}, the chain rule in CR case is in fact simpler than its conformal counterpart: the excess term $\epsilon$ does not appear in \edit{the} CR case. Specializing this chain rule to the case of CR maps into \edit{the sphere}, we obtain a new tensorial invariant for spherically equivalent classes of such maps. This invariant property is a consequence of the fact that the Schwarzian tensor of a CR automorphism of the sphere with \edit{its} standard pseudohermitian structure vanishes identically \cite{son2018schwarzian}. Recall that two CR maps $F$ and $G$ from $M$ into $\mathbb{S}^{2N'+1}$ are said to be (left) spherically equivalent if there exists a CR automorphism $\phi$ of $\mathbb{S}^{2N' +1}$ such that $G = \phi \circ F$. When $M=\mathbb{S}^{2N+1}$ is also a sphere, we can use the CR automorphisms of $M$ to define a weaker version of spherical equivalence. Namely, we say that $F$ and $G$ are spherically equivalent if there exist CR automorphisms $\gamma$ (of $\mathbb{S}^{2N+1}$) and $\phi$ (of $\mathbb{S}^{2N'+1})$ such that $G\circ \gamma = \phi \circ F$. In the following, the unit spheres are always equipped with their standard pseudohermitian structures. \begin{corollary}\label{cor:se} Let $(M,\theta)$ be a strictly pseudoconvex pseudohermitian manifold. \begin{enumerate}[(i)] \item Suppose that $F\colon M \to \mathbb{S}^{2N+1}$ is a CR immersion and $\phi \colon \mathbb{S}^{2N+1} \to \mathbb{S}^{2N'+1}$ ($N' \geq N$) is a totally geodesic embedding, then \begin{equation}\label{e:invariant} \mathcal{A}(F) = \mathcal{A}(\phi \circ F). \end{equation} In particular, if $F$ and $G$ are left spherical equivalent CR maps from $M$ into $\mathbb{S}^{2N+1}$, then \begin{equation} \ahlfors(F) = \ahlfors(G). \end{equation} \item Suppose that $G \colon (\mathbb{S}^{2n+1}, \Theta) \to (M,\theta)$ is a CR immersion and $\gamma \colon N \to \mathbb{S}^{2n+1}$ is a totally geodesic embedding, then \begin{equation}\label{e:invariant2} \gamma^{\ast}\ahlfors(G) = \ahlfors(G \circ \gamma). \end{equation} \end{enumerate} \end{corollary} \begin{remark} In Part (ii), if $N$ admits a totally geodesic embedding into a sphere, then it is \edit{necessarily} CR spherical. This follows from \cite{ebenfelt2004rigidity} for the case $\dim_{\mathbb{R}} N \geq 5$ and \cite{son2019semi} for the case $\dim_{\mathbb{R}} N = 3$. \end{remark} In view of \cref{cor:se}, an interesting question that arises is whether the Ahlfors derivative distinguishes the spherical equivalent classes of sphere maps. Although we can check that this is the case for maps between \edit{spheres} of ``low'' codimension, we do not know the answer to this question in the full generality. However, we prove that the CR Ahlfors distinguishes the totally geodesic CR map: an arbitrary CR map into the sphere with vanishing \edit{CR Ahlfors derivative} must be a totally geodesic embedding. \begin{theorem} Let $F\colon (M,\theta) \to (\mathbb{S}^{2d+1}, \Theta)$ be a CR immersion. If $\ahlfors(F) = 0$, then $M$ is CR spherical and $F$ is spherically equivalent to the linear mapping. \end{theorem} It is \edit{also} natural to ask under which conditions the Ahlfors derivative is a nonzero functional multiple of the Levi metric. We shall discuss this question after analyzing several examples \edit{in} the last section; see \cref{q:1}. The paper is organized as follows. In section 2, we study the geometry of the CR second fundamental form for CR immersions. In section 3, we prove \cref{thm:crintro}. In section 4, we study the maps with vanishing CR Ahlfors derivatives. We study the case when the source is of \edit{dimension three} in section 5. In section 6, we provide an explicit \edit{formula} for the Ahlfors derivative which is used to analyze various examples in section 7. \noindent{\bf Acknowledgment.} The authors would like to thank an anonymous referee for very careful reading of the manuscript and pointing out many, many typographical errors that we were not aware of. \section{Immersions of CR manifolds and the second fundamental form} \subsection{The second fundamental form} Let $\iota \colon (M,\theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold, i.e., \edit{$\iota$ is CR} and $\theta = \iota^{\ast} \eta$, where $\iota$ is the inclusion. In this case, $\iota$ is ``isopseudohermitian'' in the sense of \cite{dragomir1995pseudohermitian}. This notion is more general than that of ``pseudohermitian immersions,'' as the latter requires that the Reeb field of $\eta$ is tangent to $M$ \edit{along $M$}. In the latter case, \edit{the pair} $(\theta, \eta)$ is admissible in the sense of \cite{ebenfelt2004rigidity}. For any two vector fields $X, Y \in \Gamma(\C TM )$ extended to smooth sections $\widetilde{X}, \widetilde{Y}$ of $\C TN$, we recall that the \textit{second fundamental form} is defined by $\sff(X, Y) = \widetilde{\nabla}_{\widetilde{X}} \widetilde{Y} - \nabla_XY$ (see \eqref{e:sffdef}) where $\widetilde{\nabla}$ and $\nabla$ is the Tanaka-Webster connection on $(N ,\eta)$ and $(M , \theta)$, respectively. We summarize the basic properties of $\sff$ as follows (cf. \cite{son2019semi} which treats a similar situation), where $T$ and $\widetilde{T} $ denotes the Reeb field of $(M,\theta)$ and $(N,\eta)$, respectively. \begin{proposition} The second fundamental form $\sff$ is well-defined, tensorial, and satisfies the following properties for all $(1,0)$-vectors $Z$ and $W$: \begin{align} \sff(\Zbar, \Wbar) & = \overline{\sff(Z,W)}, \label{e:ssf1}\\ \sff(Z,\Wbar) & = \overline{\sff(\Zbar, W)},\label{e:ssf2}\\ \sff(Z,W) & = \sff(W,Z), \label{e:ssf3} \\ \sff(Z,\Wbar ) & = \sff(\Wbar , Z) - i \langle Z, \Wbar \rangle_{\theta} (T - \widetilde{T}), \label{e:ssf4} \\ \sff(Z, T) & = \widetilde{\nabla}_Z (T - \widetilde{T}), \label{e:ssf5} \\ \sff(T,Z) & = \widetilde{\nabla}_{T - \widetilde{T}} Z + [Z , T - \widetilde{T}] + \widetilde{\tau} Z - \tau Z. \label{e:ssf6} \end{align} Here $ \widetilde{\tau} Z : = \mathbb{T}_{\widetilde{\nabla}}(\widetilde{T}, Z)$ is the pseudohermitian torsion of $\widetilde{\nabla}$ and similarly for $\tau$. Moreover, $\sff$ is symmetric if and only if $\iota$ is pseudohermitian (i.e. $\widetilde{T} = \iota_{\ast} T$). \end{proposition} \begin{proof} That $\sff$ is well-defined and tensorial follows from standard arguments. Equations \cref{e:ssf1,e:ssf2} follow from the reality of the Tanaka-Webster connection. Equation \cref{e:ssf3} follows from the equation $\mathbb{T}_\nabla(Z,W) = 0$ for $(1,0)$-vectors $Z$ and $W$ \edit{on $M$ and similarly for $N$}. Proof of \cref{e:ssf4} uses the fact that $\mathbb{T}_{\nabla}(Z, \Wbar ) = i\langle Z,\Wbar \rangle T$. Precisely, by \cite{tanaka1975differential} \begin{equation} \nabla_Z \Wbar - \nabla_{\Wbar} Z - [Z, \Wbar] = \mathbb{T}(Z, \Wbar) = i\langle Z , \Wbar \rangle T, \end{equation} and similarly for $\widetilde{\nabla}$ and thus \cref{e:ssf4} follows. To prove \cref{e:ssf5}, observe that $\nabla T = 0$ and $\widetilde{\nabla}\widetilde{T} = 0$ \cite{tanaka1975differential}, and hence $\sff (Z,T) = \widetilde{\nabla}_ZT - \nabla_ZT = \widetilde{\nabla}_Z(T - \widetilde{T})$, as desired. The proof of \cref{e:ssf6} also follows from direct calculations. We omit the details. Assume that $\widetilde{T} = \iota_{\ast} T$, then $\iota$ is called a pseudohermitian immersion \cite{dragomir1995pseudohermitian} and the pair $(\theta, \eta)$ is said to be an admissible pair \cite{ebenfelt2004rigidity}. In this case, it follows from \cref{e:ssf5} that $\sff(Z,T) = 0$. On the other hand, from \cref{e:ssf6}, $\sff(T,Z) = \widetilde{\tau} Z - \tau Z$ and hence both sides vanish by type consideration. Similarly, from \edit{\cref{e:ssf4}}, $\sff(Z,\Wbar) = \sff(\Wbar,Z)$ and thus both sides vanish. Thus, $\sff$ is symmetric. Conversely, if $\sff$ is symmetric, then it follows from \cref{e:ssf4} that $\widetilde{T} = \iota_{\ast} T$. The proof is complete. \end{proof} \begin{definition} Let $\iota \colon (M, \theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold, $\iota^{\ast} \eta = \theta$. The $(1,0)$-\emph{mean curvature} vector of $M$ in $N$ is the $(1,0)$-vector defined by \begin{equation} H:= \frac{1}{n}\sum_{\alpha = 1}^n \sff(Z_{\abar} , Z_{\alpha}), \end{equation} where $\{Z_{\alpha} \colon \alpha = 1,2,\dots , n\}$ is an orthonormal frame of $T^{1,0}M$ and $Z_{\abar} = \overline{Z}_{\alpha}$. We also define $\mu = \mu_M^N$ to be the trace of $\Sym\sff$, i.e. $\mu = \Re H$. \end{definition} Let $\iota \colon (M, \theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold, $\iota^{\ast} \eta = \theta$. Then $\iota_{\ast}$ sends $T^{1,0}M$ into $T^{1,0} N$. We can define the normal bundle $N^{1,0}M$ as a subundle of $T^{1,0} N$ as usual. Here the orthogonality only depends on the CR structure, but not on the pseudohermitian structure. We also define $N^{0,1}M$ similarly. \begin{proposition}\label{prop:ttprime} Let $\iota \colon (M, \theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold, $\iota^{\ast} \eta = \theta$. If $T$ and $\widetilde{T}$ are the Reeb fields corresponding to $\theta$ and $\eta$, respectively, then for all tangent vectors $Z$, $W$ in $T^{1,0}M$, \begin{align} H - \overline{H} & = i(T - \widetilde{T}), \label{e:s1}\\ \sff(Z,\Wbar) & = \langle Z , \Wbar \rangle \overline{H}, \label{e:s2}\\ \widetilde{\tau} Z - \tau Z &= \sff(T,Z) - \sff(Z,T) = -i \widetilde{\nabla}_Z\overline{H}, \label{e:s3} \\ \sff(T,Z) & = -i \widetilde{\nabla}_Z H. \end{align} Moreover, $H \in N^{1,0}M$ and $\sff(Z,W) \in N^{1,0}M$. \end{proposition} \begin{proof} By direct calculation using \eqref{e:ssf4}, one has \begin{align} H & = \frac{1}{n} \sum_{\alpha = 1}^n \sff(Z_{\abar}, Z_{\alpha}) \notag \\ & = \frac{1}{n} \sum_{\alpha = 1}^n\left(\sff(Z_{\alpha} , Z_{\abar} ) + i \langle Z_{\alpha} , Z_{\abar}\rangle_{\theta}(T - \widetilde{T})\right) \notag \\ & = \overline{H} + i( T - \widetilde{T}). \end{align} \edit{Combining with \cref{e:ssf4}, we have} \edit{$\sff(Z,\Wbar) - \sff(\Wbar , Z) = -i \langle Z , \Wbar \rangle (T - \widetilde{T}) = \langle Z , \Wbar \rangle \left(\overline{H} - H\right)$}. Taking the (1,0) and (0,1) parts, we obtain \cref{e:s2}. To show that $H\in N^{1,0}M$, observe that $d\theta = d(\iota^{\ast}\eta) = \iota^{\ast}(d\eta)$. Thus, for every $X \in T^{1,0}M$, \begin{equation} 0 = d\theta(T,X) = d\eta(T,X). \end{equation} Therefore, if $X$ is tangent to $M$, \begin{equation} \langle \overline{H}, X \rangle = \langle \overline{H} - H , X \rangle = -i \langle T - \widetilde{T} , X \rangle = 0. \end{equation} This implies that $\overline{H} \in N^{0,1}M$ and $H \in N^{1,0}M$. Finally, for $\Zbar, \Wbar \in T^{0,1}M$ and $X \in T^{1,0}M$, it holds that \begin{equation} \langle \sff(\Zbar , \Wbar) , X \rangle = -\langle \Wbar , \sff(\Zbar, X) \rangle = -\langle \Zbar , X \rangle \langle \Wbar , H \rangle =0. \end{equation} This implies that $\sff(\Zbar , \Wbar) \in N^{0,1}M$ and $\sff(Z,W) \in N^{1,0}M$. The proof is complete. \end{proof} \subsection{Change of contact forms}\label{ss:changecontact} Let $\iota \colon (M, \theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold, $\iota^{\ast} \eta = \theta$. The total differential $du$ \editb{of a smooth function $u$} can be decomposed into $(1,0)$, $(0,1)$, and the transverse parts as follows: \begin{equation}\label{e:dudecomp} du = \partial_b u + \bar{\partial}_b u + (T^{\eta} u) \eta. \end{equation} This decomposition depends on the choice of pseudohermitian structure~$\eta$. We then define \begin{equation} \grad^{1,0} u = u^{\gamma} Z_{\gamma}, \quad \grad^{0,1} u = \overline{\grad^{1,0} \bar{u}}. \end{equation} If $\iota \colon M\hookrightarrow (N,\eta)$ and $\theta = \iota^{\ast} \eta$, then we have \begin{equation} \grad^{1,0}_N u = \grad^{1,0}_{N,M} u + (\grad^{1,0}_{N,M} u)^\perp, \end{equation} and similarly for $\grad^{0,1} u$. Here the orthogonal complements in $T^{1,0} N$ and $T^{0,1} N$ are defined using the Levi metric corresponding to any pseudohermitian structure on $N$. \begin{proposition}\label{prop:conch} Let $\iota \colon (M, \theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold, $\iota^{\ast} \eta = \theta$. Suppose that $P$ is the manifold $N$ with $\widetilde{\eta} = e^{u} \eta$. Put $\widetilde{\theta}: = e^{u\circ \iota} \theta$. Let $\widetilde{\sff}$ be the second fundamental form of the inclusion $\iota\colon (M,\widetilde{\theta}) \hookrightarrow (P,\widetilde{\eta})$. Then \begin{equation}\label{e:crsff} \widetilde{\sff}(Z, W) = \sff(Z, W), \quad Z, W \in T^{1,0}M, \end{equation} \begin{equation}\label{e:221} \widetilde{\sff}(\Zbar, W) = \sff(\Zbar , W) - \langle \Zbar , W\rangle_{\eta} (\grad^{1,0}_{N,M} u)^\perp, \quad \Zbar \in T^{0,1}M, \ W \in T^{1,0}M, \end{equation} and \begin{equation}\label{e:224} e^{u}\mu_M^P = \mu_M^N - \Re (\grad^{1,0}_{N,M} u)^\perp . \end{equation} \end{proposition} \begin{proof} The first two identities follow from Lee's formulas for the pseudoconformal change of the metrics \cite{lee1986fefferman}. Precisely, on $M$, we have \begin{equation} \widetilde{\nabla}_{Z}W = \nabla_ZW + Z(u) W + W(u) Z, \end{equation} and \begin{equation}\label{e:224a} \widetilde{\nabla}_{\Zbar}W = \nabla_{\Zbar}W - \langle W,\Zbar\rangle_{\theta}\, \grad^{1,0}_N u. \end{equation} Similar formulas hold on $M$ \edit{and hence \cref{e:crsff,e:221} follow immediately. The last identity \cref{e:224} also follows by taking the trace of \cref{e:221} and its conjugate.} \end{proof} \begin{remark} In view of \cref{e:crsff}, $\sff(Z,W)$, \edit{where $Z,W$ are $(1,0)$-vectors}, is called the CR second fundamental form of the CR immersion. It can be computed by any pair of pseudohermitian structures $\theta = \iota ^{\ast} \eta$, not necessary admissible. This notion has been extensively used in the study of the CR immersions \cite{dragomir1995pseudohermitian,ebenfelt2004rigidity,ebenfelt2014cr}. \end{remark} \subsection{The Gauß and Weingarten equations} \begin{proposition}[Pseudohermitian Weingarten Equation] If $N$ is a section of $N^{1,0}M \oplus N^{0,1}M$, then \begin{equation} \langle\widetilde{\nabla}_ X N , Y \rangle = -\langle N , \sff(X,Y) \rangle \end{equation} for all sections $X,Y$ of $T^{1,0}M \oplus T^{0,1}M$. \end{proposition} In the following, we shall use the following convention for the curvature operator: \begin{equation} R(X,Y)Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X,Y]}Z. \end{equation} Then for $X, Y, Z$ tangent to $M$, \begin{align} \widetilde{R}(X,Y)Z & = R(X,Y)Z + \sff(X, \nabla_{Y} Z) - \sff(Y, \nabla_{X} Z) \notag \\ & \qquad + \widetilde{\nabla}_X (\sff (Y, Z)) - \widetilde{\nabla}_Y (\sff (X, Z)) - \sff([X,Y], Z). \end{align} Here, $R$ and $\widetilde{R}$ are the curvature on $(M,\theta)$ and $(N,\eta)$ respectively. \begin{proposition}[Pseudohermitian Gauß equation]\label{prop:gauss} Let $\iota \colon (M, \theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold, $\iota^{\ast} \eta = \theta$. Then the Gauß equation holds, i.e., \begin{align}\label{e:gauss} \langle \widetilde{R} (X ,\overline{Y}) Z , \Wbar \rangle & = \langle R (X ,\overline{Y}) Z , \Wbar \rangle + \left\langle \sff(X, Z) , \sff(\overline{Y}, \Wbar) \right\rangle \notag \\ & \quad - \|H\|^2 \left( \langle \overline{Y} , Z \rangle \langle X , \Wbar \rangle + \langle X , \overline{Y} \rangle \langle Z , \Wbar \rangle \right). \end{align} Moreover, \begin{equation}\label{e:tor} \langle \widetilde{\tau} Z , W \rangle = \langle \tau Z , W \rangle + i \langle \sff(Z,W) , \overline{H} \rangle. \end{equation} \end{proposition} \begin{remark} For the special case of pseudohermitian \edit{immersions}, this is Eq. (5.3) or Proposition 5.1 in \cite{ebenfelt2004rigidity} as in this case, $H=0$; see also \cite{son2019semi} for similar equations. It is sometimes helpful to write the Gauß equations using index notations, that is \begin{equation} \widetilde{R}_{\alpha\bbar\gamma\bar{\sigma}} = R_{\alpha\bbar\gamma\bar{\sigma}} + \omega_{\alpha\gamma}^{a}\, \omega_{\bbar\bar{\sigma}}^{\bar{b}}\, g_{a\bar{b}} - \|H\|^2 \left(g_{\gamma\bbar} g_{\alpha\bar{\sigma}} + g_{\alpha\bbar} g_{\gamma\bar{\sigma}} \right), \end{equation} and \begin{equation} \widetilde{A}_{\alpha\beta} = A_{\alpha\beta} + i \omega_{\alpha\beta}^{a} H^{\bar{b}} g_{a\bar{b}}. \end{equation} \edit{Here $\omega_{\alpha\beta}^a$ is the components of the second fundamental form in a local frame, i.e. $\sff(Z_{\alpha},Z_{\beta}) = \omega_{\alpha\beta}^a Z_a$, $H = H^{\bar{b}}Z_{\bar{b}}$, $A_{\alpha\beta} = \langle \tau Z_{\alpha}, Z_{\beta}\rangle$, and so on. These hold} for all isopseudohermitian immersions. \end{remark} \begin{proof}[Proof of \cref{prop:gauss}] The proof is analogous to the case of CR immersions into Kähler manifolds considered in \cite{son2019semi}. Namely, for $X, Z \in T^{1,0}M$ and $\overline{Y}, \Wbar \in T^{0,1}M$, \begin{align}\label{e:tem} \langle \widetilde{R}(X,\overline{Y}) Z, \Wba\rangle & = \langle R(X,\overline{Y}) Z, \Wba\rangle - \langle \sff ([X,\overline{Y}], Z), \Wba \rangle \notag \\ & \quad + \langle \widetilde{\nabla}_X (\sff (\overline{Y}, Z)) , \Wba \rangle - \langle \widetilde{\nabla}_{\overline{Y}}(\sff (X,Z)) , \Wba \rangle \notag \\ & = \langle R(X,\overline{Y}) Z, \Wba\rangle - \langle \sff ([X,\overline{Y}], Z), \Wba \rangle \notag \\ & \quad - \langle \sff (\overline{Y}, Z) , \sff (X,\Wba) \rangle + \langle \sff (X,Z) , \sff (\overline{Y},\Wba) \rangle \notag \\ & = \langle R(X,\overline{Y}) Z, \Wba\rangle + \langle \sff (X,Z) , \sff (\overline{Y},\Wba) \rangle \notag \\ & \quad - \langle \overline{Y}, Z \rangle \langle X,\Wba \rangle \|H\|^2 - \langle \sff ([X,\overline{Y}], Z), \Wba \rangle . \end{align} On the other hand, by the defining properties of the Tanaka-Webster connection \edit{\cite[Proposition~3.1]{tanaka1975differential}}, we have \begin{equation} [X , \overline{Y}] = \nabla_X \overline{Y} - \nabla_{\overline{Y}} X - i\langle X , \overline{Y} \rangle T. \end{equation} Therefore, using \cref{e:s3}, \begin{equation} \langle \sff([X , \overline{Y}] , Z) , \Wbar \rangle = -i\langle X , \overline{Y} \rangle \langle \sff(T, Z) , \Wbar\rangle = \langle X , \overline{Y} \rangle \langle Z , \Wbar \rangle \|H\|^2. \end{equation} Putting this into \cref{e:tem} we obtain \cref{e:gauss}. Equation \cref{e:tor} follows easily from \cref{e:s3} and we omit the detail. \end{proof} \section{The chain rule: Proof of \cref{thm:crintro}} Let $\iota \colon (M, \theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold, $\iota^{\ast} \eta = \theta$. Recall that the tensor $\nu = \nu_M^N$ introduced in the introduction is the symmetric real tensor on $T^{1,0}M \oplus T^{0,1}M$ defined by \begin{equation} \nu(X,Y) = 2\left\langle \Sym \sff(X,Y) , \mu \right\rangle - \langle X,Y\rangle \|\mu\|^2, \end{equation} where $\mu$ is the trace of $\Sym \sff$, i.e., $\mu = \Re H$. The components of $\nu$ in the ``horizontal'' directions are given in local frame by \begin{equation} \nu_{\alpha\beta} = \overline{\nu_{\abar\bbar}} = \omega_{\alpha\beta}^{a} H^{\bar{b}} g_{a\bar{b}}, \quad \nu_{\alpha\bbar} = \nu_{\bbar\alpha} = \tfrac{1}{2}\|H\|^2 g_{\alpha\bbar}. \end{equation} \edit{Here the Greek indices $\alpha$ and $\beta$ run from $1$ to $n:=\dim_{C\!R}M$, the lowercase indices $a$ and $b$ run from $n+1$ to $\dim_{C\!R} N$, and $g$ is the Levi metric.} \begin{proposition} Let $\iota \colon (M, \theta) \hookrightarrow (N, \eta)$ be a pseudohermitian submanifold, $\iota^{\ast} \eta = \theta$. Suppose $P$ is the manifold $N$ with the pseudohermitian structure $\widetilde{\eta} = e^u \eta$ and \editb{put} $\widetilde{\theta} = e^{u\circ \iota} \theta$. Then for any sections $X, Y$ of $T^{1,0} M \oplus T^{0,1}M$, \begin{equation} \nu_M^P(X,Y) = \nu_M^N(X,Y) - 2 \langle \Sym \sff (X,Y) , \xi \rangle_{\theta} + \langle X , Y \rangle_{\theta} \|\xi\|^2_{\theta} \end{equation} where $\xi = \Re (\grad^{1,0}_N u)^{\perp}$. \end{proposition} \begin{proof} Observe that \begin{equation} \widetilde{\sff}_M^P(X,Y) = \sff_M^P(X,Y) - \langle X,Y\rangle_{\theta} (\grad^{1,0}_N u)^{\perp}, \end{equation} and thus, with $\xi = \Re (\grad^{1,0}_N u)^{\perp}$, \begin{equation} \Sym \widetilde{\sff}(X,Y) = \Sym \sff(X,Y) - \langle X,Y\rangle_{\theta} \xi. \end{equation} Taking the trace, we obtain \begin{equation} e^u \mu_M^P = \mu_M^N - \xi. \end{equation} Therefore, \begin{align} \nu_M^P(X,Y) & = 2 \langle \Sym \widetilde{\sff}(X,Y), \mu_M^P \rangle_{\widetilde{\theta}} - \langle X,Y \rangle_{\widetilde{\theta}} \|\mu_M^P\|^2_{\widetilde{\theta}} \notag \\ & = 2\langle \Sym \sff(X,Y) - \langle X,Y \rangle_{\theta}\, \xi , \mu_M^N - \xi\rangle_{\theta} - \langle X,Y \rangle_{\theta} \|\mu_M^N - \xi \|^2_{\theta} \notag \\ & = \nu_M^N(X,Y) - 2 \langle \Sym \sff (X,Y) , \xi \rangle_{\theta} + \langle X , Y \rangle_{\theta} \|\xi\|^2_{\theta}. \end{align} The proof is complete. \end{proof} Recall that the operator $\mathcal{H}(u)$ on a pseudohermitian manifold $(M,\theta)$ is defined, for horizontal vectors, by \begin{align} \mathcal{H}_{\theta}(u) = \Sym \nabla\nabla u - \partial_b u \otimes \partial_b u - \bar{\partial}_b u \otimes \bar{\partial}_b u + \tfrac{1}{2}\|\bar{\partial}_b u\|^2 L_{\theta}. \end{align} The tensor $\mathcal{H}_{\theta}(u)$ is closely related to the CR Schwarzian tensor \cite{son2018schwarzian}. In the notations of \cite{son2018schwarzian}, \begin{equation} B_{\theta}\left(\tfrac{1}{2}u\right) = \mathcal{H}_{\theta}(u) + \frac{1}{2n} \left( \Delta_b u - n \|\bar{\partial}_b u \|^2 \right)L_{\theta}. \end{equation} is the traceless part of $\mathcal{H}$. Here \edit{$L_{\theta}(X,Y) = \langle X, Y\rangle$ is the Levi metric and} we use the convention that $\Delta_b$ is a nonnegative operator. We can sometimes write $u$ for $u\circ \iota$ as a function on $M$. \begin{proposition}[cf. \cite{son2018schwarzian}]\label{prop32} Let $u, v \colon M \to \mathbb{R}$ be smooth functions on $M$. Then \begin{align}\label{e:1cocycle} \mathcal{H}_{\theta}(u + v) = \mathcal{H}_{\theta} (u) + \mathcal{H}_{\hat{\theta}} (v) \end{align} where $\hat{\theta} = e^u \theta$. \end{proposition} We point out that the equality of the traceless parts of both sides was proved in \cite{son2018schwarzian}. \begin{proof} We need to verify \cref{e:1cocycle} for each pair of vectors of $(1,0)$ and each pair of one $(1,0)$ and one $(0,1)$-vector. First, for any vector $Z,W$ of type $(1,0)$, we have from \cref{e:224} \begin{align} \mathcal{H}_{\hat{\theta}}(v)(Z, \Wbar) & = \Sym \widehat{\nabla} v (Z, \Wbar) + \tfrac12\|\bar{\partial}_b v\|^2_{\hat{\theta}} \, \langle Z, \Wbar \rangle _{\hat{\theta}} \notag \\ & = \Sym \nabla v (Z, \Wbar) + \left(\Re \langle \bar{\partial}_b{u} , \partial_b v\rangle + \tfrac12 \|\bar{\partial}_b v\|^2_{\theta}\right) \langle Z, \Wbar \rangle _{\theta} \end{align} Thus, \begin{align} \mathcal{H}_{\theta}(v)(Z, \Wbar) + \mathcal{H}_{\hat{\theta}}(v)(Z, \Wbar) & = \Sym \nabla u(Z, \Wbar) + \tfrac{1}{2}\|\bar{\partial}_b u\|^2 \langle Z, \Wbar \rangle _{\theta} \notag \\ & \quad + \Sym \nabla v (Z, \Wbar) + \left(\Re \langle \bar{\partial}_b{u} , \partial_b v\rangle + \tfrac12 \|\bar{\partial}_b v\|^2_{\theta} \right) \langle Z, \Wbar \rangle _{\theta} \notag \\ & = \Sym \nabla (u + v)(Z, \Wbar) + \tfrac12 \|\partial_b u + \partial_ b v \|^2_{\theta} \langle Z, \Wbar \rangle _{\theta} \notag \\ & = \mathcal{H}_{\theta}( u + v). \end{align} This verifies \cref{e:1cocycle} for any pair of vectors of mixed type. The identity for vectors of pure type is exactly the same as \cite{son2018schwarzian}. We omit the details. \end{proof} \begin{proposition} \editb{If} $(M,\theta) \subset (N,\eta)$ is a pseudohermitian submanifold, $X, Y \in \Gamma (T^{1,0} \oplus T^{0,1} M)$, and $u \in \mathcal{C}^{2}(N)$, then \begin{align} \mathcal{H}_{\theta} (u\|_M)(X,Y) - \mathcal{H}_{\eta}(u)(X,Y) = (du)(\Sym \sff (X,Y)) - \tfrac{1}{2} \left\|(\grad_{N,M}^{1,0} u )^{\perp}\right\|^2 \langle X, Y \rangle _{\theta}. \end{align} \end{proposition} \begin{proof} On $M$, we have \begin{equation} \nabla^2 (u\|_M)(X,Y) = X(Y(u)) - d(u\|_M)(\nabla_XY), \end{equation} and similarly for $\widetilde{\nabla}^{2} u(X,Y)$ on $N$. Thus, \begin{equation} \nabla^{2} u(X,Y) - \widetilde{\nabla}^{2} u(X,Y) = (du)(\sff(X,Y)). \end{equation} Consequently, \begin{align} \mathcal{H}_{\theta}(u\|_M)(X,Y) - \mathcal{H}_{\eta}(u)(X,Y) & = \tfrac{1}{2} (du)(\sff(X,Y) + \sff (Y,X)) \notag \\ & \quad +\tfrac{1}{2}\left(\|\partial_b (u\|_M)\|^2 - \|\partial_ b u\|^2 \right) \langle X, Y \rangle_{\theta} \notag \\ & = (du)(\Sym \sff(X,Y)) - \tfrac{1}{2} \|(\grad_{N,M}^{1,0} u )^{\perp}\|^2 \langle X, Y \rangle _{\theta}. \qedhere \end{align} \end{proof} \begin{proposition}\label{prop:e} \editb{For any} tower of pseudohermitian submanifolds $M\subset N \subset (P,\eta)$ \editb{ it holds that } \begin{align}\label{e:mean} \mu_M^{P} & = \mu_M^N + \mu_N^P, \\ \label{e:nu} \nu_M^P - \nu_M^N & = (\iota_M^N)^{\ast} \nu_N^P. \end{align} \end{proposition} \begin{proof} For $Z\in T^{1,0}M$ and $\Wbar \in T^{0,1}M$, we have that \begin{equation} \langle \Zbar , W \rangle H_M^P = \sff_M^P(\Zbar, W) = \sff_N^P (\Zbar,W) + \sff_M^N(\Zbar,W) = \langle Z , \Wbar \rangle (H_N^P + H_M^N) \end{equation} Taking $Z = W \ne 0$, we immediately obtain $H_M^P = H_M^N + H_N^P$ and hence \cref{e:mean} follows. Plugging this into the definition of $\nu$, we obtain \cref{e:nu}. The proof is complete. \end{proof} \begin{proof}[Proof of \cref{thm:crintro}] The idea of the proof is essentially the same as in \cite{stowe2015ahlfors} and based on the calculations above. The formula in the CR case turns out to be simpler than its conformal counterpart because of \cref{prop:e} above. Indeed, assume that $F \colon (M,\theta) \to (N,\eta)$ and $G\colon (N,\eta) \to (P,\zeta)$. The chain rule for $\ahlfors$ is equivalent to the analogous assertions for $\ahlfors'$ where \begin{equation} \mathcal{A}' \colon F \mapsto \mathcal{H}_{\theta} (u) + F^{\ast} \left(\nu_{F(M)}^N\right). \end{equation} In the equidimensional case, this and \cref{e:chainrule} reduce to an analogous statement for \edit{the} CR Schwarzian that was proved in \cite{son2018schwarzian}. Next, we \editb{assume that} $P$ and $N$ have the same dimension. In this case, we can suppose that $P$ is the manifold $N$ with a pseudohermitian structure $\zeta = e^{v} \eta$ and $G$ is the identity map. Since $e^u\theta = F^{\ast} \eta$, by \cref{prop32}, \begin{align} \mathcal{H}_{\theta} (u + v \circ F) & = \mathcal{H}_{\theta}(u) + \mathcal{H}_{e^{u} \theta} (v \circ F) \\ & = \mathcal{H}_{\theta}(u) + F^{\ast}\mathcal{H}_{(F(M),\iota^{\ast} \eta)} (v). \end{align} Therefore, \begin{align} \ahlfors'(G \circ F) - \ahlfors'(F) - F^{\ast} \ahlfors'(G) & = \mathcal{H}_{\theta}(u + v\circ F) + F^{\ast} \left(\nu_{F(M)}^{(N, e^v \eta)}\right) \\ & \quad - \left(\mathcal{H}_{\theta}(u) + F^{\ast} \left(\nu_{F(M)}^{(N,\eta)}\right)\right) - F^{\ast} \left(\mathcal{H}_{\eta}(v)\right)\notag \\ & = F^{\ast} \left(\mathcal{H}_{(F(M),\iota^{\ast} \eta)} (v) - \mathcal{H}_{\eta} (v) + \nu_{F(M)}^{(N, e^v \eta)} - \nu_{F(M)}^{(N,\eta)}\right) \notag \\ & = 0. \end{align} To conclude the proof, we consider the general case when $G$ is a CR immersion, $G^{\ast} \zeta = e^{v}\eta$. We have \begin{align} \mathcal{A}'(G\circ F) - \mathcal{A}'(F) - F^{\ast} (\mathcal{A}'(G)) & = \mathcal{H}_{\theta}(u + v \circ F) + (G\circ F)^{\ast}(\nu_{F(G(M))}^P) \notag \\ & \quad - \left(\mathcal{H}_{\theta}(u) + F^{\ast}(\nu_{F(M)}^N)\right) - F^{\ast}\left(\mathcal{H}_{\eta}(v) + G^{\ast}(\nu_{G(N)}^P\right) \notag \\ & = (G\circ F)^{\ast}\left(\nu_{G(F(M))}^P - \nu_{G(N)}^P\right) - F^{\ast} (\nu_{F(M)}^N) \notag \\ & = 0. \end{align} At the last step, we have used \cref{prop:e}. The proof is complete. \end{proof} \section{CR maps with vanishing Ahlfors derivative} It is well-known that the Schwarzian derivative of a conformal diffeomorphism measures the change of the traceless component of the Ricci tensor, see \cite{osgood1992schwarzian}. The same is true for the CR analogue of the Schwarzian \cite{son2018schwarzian}. The case of immersions is a bit different due to the presence of the second fundamental form. We point out that \edit{in the CR case a certain} part of the Ahlfors \edit{derivative} actually measures how the pseudohermitian torsion changes when going from the original structure to the pull-back. We denote $\ahlfors_{\alpha\beta}(F) = \ahlfors(F)(Z_{\alpha}, Z_{\beta})$ the ``holomorphic'' components and $\ahlfors_{\alpha\bbar}(F) = \ahlfors(F)(Z_{\alpha}, Z_{\bbar})$ the ``mixed type'' components of the Ahlfors \edit{derivative}. \editc{As usual, $h_{\alpha\bbar}$ denotes the Levi metric on $(M,\theta)$, i.e. $h_{\alpha\bbar} = \langle Z_{\alpha} , Z_{\bbar}\rangle = -i d\theta(Z_{\alpha}, \overline{Z_{\beta}})$, and similarly, $g_{A\bar{B}}$ denotes the one for $(N,\eta)$.} \begin{proposition} Let $F \colon (M,\theta) \to (N,\eta)$ be a CR immersion. Let $A_{\alpha\beta}$ be the \edit{pseudohermitian} torsion of $(M,\theta)$ and $\widetilde{A}_{AB}$ the pseudohermitian torsion of $(N,\eta)$. Also let $R_{\alpha\bbar}$ be the Ricci tensor of the Tanaka-Webster connection associated to $\theta$. In an adapted coframe \edit{the following hold\editb{:}} \begin{enumerate}[\rm (i)] \item The ``holomorphic'' components of the Ahlfors \edit{are} \begin{equation}\label{e:holomorphicpart} \ahlfors_{\alpha\beta}(F) = -i(\widetilde{A}_{\alpha\beta} - A_{\alpha\beta}). \end{equation} \item The tracefree components of \editb{mixed type} \edit{are} \begin{equation} \tf \ahlfors_{\alpha\bbar} (F) = \tf R_{\alpha\bbar} - \tf \hat{R}_{\alpha\bbar}, \end{equation} where $\hat{R}_{\alpha\bbar}$ is the Ricci tensor associated to $\hat{\theta} = F^{\ast}\eta$. \item If $(N,\eta)$ is the sphere with its standard pseudohermitian structure, then \begin{equation}\label{e:traceahlfors} \trace_{\theta}\ahlfors(F):= h^{\alpha\bbar} \ahlfors_{\alpha\bbar}(F) = \tfrac{1}{2(n+1)} e^u \|\sff^{2,0}\|^2, \end{equation} where $\|\sff^{2,0}\|^2$ is the \edit{squared norm} of the CR second fundamental form. \end{enumerate} \end{proposition} \begin{remark} This proposition exhibits some new features of the CR analogue of the Ahlfors derivative compared to its conformal counterpart. Observe that (1) and (2) generalize similar formulas for the CR Schwarzian in \cite{son2018schwarzian} while (3) says that the Ahlfors is traceless if and only if $F$ has vanishing CR second fundamental form. \edit{Moreover, part (iii) also holds when $(N,\eta)$ has the Tanaka-Webster curvature tensor of the form \[ R_{A\bar{B}C\bar{D}} = c\left(g_{A\bar{B}} g_{C\bar{D}} + g_{A\bar{D}} g_{C\bar{B}}\right), \ c\in \mathbb{R}. \] In particular, it holds when $(N,\theta)$ is the Heisenberg hypersurface with \editb{a} pseudohermitian structure of vanishing curvature and torsion.} \end{remark} \begin{proof} We identify $M$ with $F(M)$ and consider $\iota \colon (M,\hat{\theta}) \to (N,\eta)$ as \editb{the inclusion of} a submanifold and thus $\hat{\theta}: = \iota^{\ast}\eta = e^u\theta$ for some smooth function~$u$. Let $\hat{A}_{\alpha\beta}$ be the pseudohermitian torsion of $\hat{\theta}$ on $M$. By the Gauß equation, we have \begin{equation} \widetilde{A}_{\alpha\beta} = \hat{A}_{\alpha\beta} + i\, \omega_{\alpha\beta}^a H^{\bar{b}} g_{a\bar{b}}. \end{equation} On the other hand, $A_{\alpha\beta} = \hat{A}_{\alpha\beta} - i(u_{\alpha,\beta} - u_{\alpha} u_{\beta})$, by \cite{lee1986fefferman}. \edit{Here the indices preceded by a comma indicate covariant derivatives.} Therefore, \begin{align} \ahlfors_{\alpha\beta}(F) &= u_{\alpha,\beta} - u_{\alpha} u_{\beta} + \omega_{\alpha\beta}^a \xi^{\bar{b}} g_{a\bar{b}} \notag \\ &= u_{\alpha,\beta} - u_{\alpha} u_{\beta} -i(\widetilde{A}_{\alpha\beta} - \hat{A}_{\alpha\beta}) \notag \\ &= -i(\widetilde{A}_{\alpha\beta} - A_{\alpha\beta}). \end{align} Thus, (1) is proved. From the definition, the \editb{mixed type} components are given by \begin{equation}\label{e:tracepart} \ahlfors_{\alpha\bbar} (F) = \tfrac{1}{2}\left(u_{\alpha,\bbar} + u_{\bbar,\alpha}\right) + \tfrac{1}{2} \left(\|\bar{\partial}_b u\|^2 + e^u \|H_{F(M)}\|^2 \circ F + e^u J_{\eta} \circ F - J_{\theta}\right) h_{\alpha\bbar}. \end{equation} Therefore, using \cite{lee1986fefferman}, we obtain \begin{equation} \tf \ahlfors_{\alpha\bbar} (F) = \tfrac{1}{2}\left(u_{\alpha,\bbar} + u_{\bbar,\alpha}\right) + \frac{1}{2n} \Delta_b u\, h_{\alpha\bbar} = \tf R_{\alpha\bbar} - \tf \hat{R}_{\alpha\bbar}, \end{equation} where $\hat{R}_{\alpha\bbar}$ is the Ricci curvature of $\hat{\theta} = e^u \theta = F^{\ast}\eta$. Suppose that $(N,\eta)$ is a sphere with the standard pseudohermitian structure, then \edit{the Gauß equation} implies that \begin{equation} J_{\eta} \circ F = J_{\hat{\theta}} - \|H_{F(M)}\|^2 \circ F + \frac{1}{n(n+1)}\|\sff_{F(M)}^{0,2}\|^2 \circ F. \end{equation} On the other hand, \begin{equation} e^u J_{\hat{\theta}} = J_{\theta} +\tfrac{1}{n} \Delta_b u - \|\bar{\partial}_b u\|^2. \end{equation} Putting them together, we obtain that \begin{equation} e^uJ_{\eta} \circ F - J_{\theta} + \|\bar{\partial}_b u\|^2 + e^u \|H_{F(M)}\|^2 \circ F -\frac{1}{n} \Delta_b u = \frac{1}{n(n+1)} e^u \|\sff^{2,0}_{F(M)}\|^2 \circ F. \end{equation} \editb{It thus holds that} \begin{equation}\label{e:traceahlfors} \trace_{\theta}{\ahlfors(F)} = h^{\alpha\bbar} \ahlfors_{\alpha\bbar}(F) = \frac{1}{2(n+1)} e^u \|\sff^{2,0}\|^2, \end{equation} \editb{and the} proof is complete. \end{proof} \begin{definition} Let $\iota \colon (M,\theta) \hookrightarrow (N,\eta)$ be a pseudohermitian submanifold, $\iota^{\ast} \eta = \theta$. We say that \editb{$\iota$ (or $M$ if the embedding is understood)} is umbilic at $p$ if $\sff_M^N(p)$ is a multiple of the Levi metric, i.e., $\sff(Z,W) = 0$ for all $(1,0)$ vectors $Z$ and $W$ at~$p$. \end{definition} We remark that \edit{the pseudohermitian total umbilicity does not depend on the choice of pseudohermitian structures}. In fact, if $\iota$ is pseudohermitian umbilic, then it is CR totally geodesic for any admissible pair of contact forms. This \editb{holds} because the CR second fundamental form (as defined in \cite{ebenfelt2004rigidity}) associated to any admissible pair of pseudohermitian structures coincides with the ``holomorphic'' components of the pseudohermitian second fundamental form. \begin{corollary} Let $(M,\theta)$ be a pseudohermitian manifold and \editb{assume that} $F \colon (M,\theta) \to (\mathbb{S}^{2N+1},\Theta)$ a CR immersion into a sphere. Then $\trace\ahlfors(F)\geq 0$, \editb{and} equality holds at \editb{a point} $p\in M$ iff $(M,\theta)$ is umbilic at~$p$. If $\trace\ahlfors(F) = 0$ identically, then $M$ is CR spherical and $F$ is spherically equivalent to the linear embedding. \end{corollary} \begin{proof} That $\trace \ahlfors(F) \geq 0$ follows directly from \cref{e:traceahlfors}. The equality occurs precisely when the CR second fundamental form vanishes. This implies that $M$ must be CR spherical by \cite{ebenfelt2004rigidity} for $n\geq 2$ and \cite{son2019semi} for the case $n=1$. Moreover, $F$ is equivalent to the linear embedding by \cite{ji2010flatness} for the case $\dim M \geq 5$ and \cite{son2019semi} for the three-dimensional case. \end{proof} \section{Maps from three-dimensional CR manifolds into spheres} Suppose that $(N, \eta) = (\mathbb{S}, \Theta)$ where $\Theta$ is the standard pseudohermitian structure \edit{on the sphere} \editb{$\mathbb{S} := \mathbb{S}^{2N+1}$} . If $F = \phi \circ G$ for some CR automorphism $\phi$ of the target sphere, then $\ahlfors(F) = \ahlfors(G)$ by \cref{thm:crintro} and, in particular, $\ahlfors^{\theta}_{\alpha\bbar}(F) = \ahlfors^{\theta}_{\alpha\bbar}(G)$. This last equality implies that $u_{\alpha,\bbar} = v h_{\alpha\bbar}$ for some function $v$. When $n\geq 2$, \edit{using Lee's characterization of CR-pluriharmonicity \cite{lee1988pseudo}, we \editb{can infer} from the last condition that $u$ must be CR-pluriharmonic}. This argument does not work in the case $n=1$. We shall deduce the CR-pluriharmonicity from a simpler argument which also works in the case $n=1$ as follows. \begin{proposition}\label{prop:crph} Let $M$ be a strictly pseudoconvex CR manifold and assume that \editb{$F$ and $G$ are CR immersions from $M$ into the sphere $\mathbb{S}$}. Let $\theta = F^{\ast} \Theta$ and $\theta' = G^{\ast} \Theta$, where $\Theta$ is the standard pseudohermitian structure on $\mathbb{S}$. Put $\theta' = e^{u}\theta$. If $F = \phi \circ G$, for some CR automorphism $\phi$ of $\mathbb{S}$, then $u$ is CR-pluriharmonic. \end{proposition} \begin{proof} If $G = \phi\circ F$ for some CR automorphism $\phi$ of $\mathbb{S}$, then this automorphism extends to \edit{a} biholomorphic automorphism of the ball $\mathbb{B}$ which we will also denote by $\phi$. Let $\phi(0) = a$ and let $\rho = \\|z\\|^2 - 1$ be the defining function for $\mathbb{S}$. Then by \cite{rudin1980function}, Theorem 2.2.2, \begin{equation} (\rho \circ \phi)(z) = \|\phi(z)\|^2 - 1 = \left(\frac{1 - \|a\|^2}{\|1 - z\cdot \bar{a}\|^2}\right) \cdot \rho(z). \end{equation} Observe that $\Theta = -\iota^{\ast}( i\partial \rho)$ and hence $\phi^{\ast}\Theta = -\iota^{\ast}(i \partial (\rho \circ \phi)) = e^{\varphi}\Theta$, where \begin{equation} \varphi (z) = \log (1 - \|a\|^2) - \log \left\|1 - z\cdot \bar{a}\,\right\|^2, \quad z \in \mathbb{S}. \end{equation} In particular, $\varphi$ is CR-pluriharmonic and so is $u = \varphi \circ F$. \end{proof} In view of this proposition, we can construct another invariant which plays the role of the traceless part of $\ahlfors(F)$ as follows. By \cite[Proposition~3.4]{lee1988pseudo}, $u$ is CR-pluriharmonic if and only if \begin{equation} P u : = \left(u_{\bar{1},}{}^{\bar{1}}{}_{1} + iA_{11} u^1\right)\, \theta^1 =0. \end{equation} Here $A_{11}$ is the component of the pseudohermitian torsion. This suggests the following notion: For each CR immersion $F$ from the $3$-sphere into a sphere, define the $(1,0)$-form $\ahlfors_1(F)$ by \begin{equation} \mathcal{A}_{1}(F) = P(u), \end{equation} where as before $u$ is determined by the relation $ F^{\ast} \Theta = e^u\theta$. We formulate the discussion above as follows. \begin{proposition}\label{prop:3dinvariant} Let $(M,\theta)$ be a three-dimensional strictly pseudoconvex pseudohermitian manifold \editb{and} assume that $F$ and $G$ are CR immersions of $M$ into \editb{the} sphere \editb{$\mathbb{S}$}. Then $\mathcal{A}_{1}$ \edit{is invariant with respect to } \editb{ spherical equivalence} in the following sense: \begin{enumerate}[\rm(1)] \item If $\phi$ is a CR automorphism of the target, then \begin{equation} \mathcal{A}_{1}(\phi \circ F) = \mathcal{A}_{1}(F). \end{equation} \item If $M$ is \editb{the 3-sphere} and $ \gamma$ is a CR automorphism of \editb{$M$}, then \begin{equation}\label{e:ahlforsprime} \ahlfors_1(F\circ \gamma) = \gamma\, ^{\ast}\left(e^{\psi}\mathcal{A}_{1}(F) \right) \end{equation} where $\psi$ is determined by the relation $\gamma^{\ast} \theta = e^{\psi}\theta$. \end{enumerate} \end{proposition} \begin{proof} We write $F^{\ast}\Theta = e^u\theta$ and $G^{\ast}\Theta = e^v\Theta$. Suppose that $G = \phi \circ F$, then \cref{prop:crph} implies that $u - v$ is CR pluriharmonic and hence \begin{equation} \ahlfors_1(\phi \circ F) = P(v) = P(u) = \ahlfors_1(F), \end{equation} as desired. Suppose that $M$ is a 3-sphere, $\gamma$ a CR automorphism of $M$, and $\theta$ is the standard pseudohermitian structure on $M$. Suppose that $G = F\circ \gamma$, then \begin{equation} v = u\circ \gamma + \psi. \end{equation} Under the change of contact form $\widetilde{\theta} = e^{\psi}\theta$, the operator $P$ changes as follows: We write $P^{\theta}$ for the operator $P$ associated to $\theta$ and similarly for $\widetilde{\theta}$. Then \begin{equation} e^{\psi}P^{\widetilde{\theta}}(u) = P^{\theta}(u) + 2i \left\langle P^{\theta}(u) , \bar{\partial}_b \psi \right\rangle \, \theta. \end{equation} Since $P(\psi) = 0$, we have that \begin{equation} P^{\theta}(v) = P^{\theta}(u \circ \gamma) = \gamma^{\ast}\left(P^{(\gamma^{-1})^{\ast}\theta} (u)\right) = e^{\psi} \gamma^{\ast} \left(P^{\theta}(u)\right) \mod \theta. \end{equation} This completes the proof. \end{proof} \section{Explicit calculations in local coordinates} \subsection{The Ahlfors tensor $\ahlfors$ in general dimensions} The purpose of this section is to give an explicit formula for the hermitian part of the Ahlfors tensor. The formula will be explicit in terms of the defining functions of the source and target and will be simplified when the target is the sphere. Let $M \subset \mathbb{C}^{n+1}$ be a strictly pseudoconvex real hypersurface defined by $\rho = 0$. It is well-known that there exists \editb{a unique} $(1,0)$ vector field $\xi$ satisfying the following two conditions (see, e. g. \cite{graham1988smooth,li--son}): \begin{equation} \xi \, \rfloor \, i\partial\bar{\partial} \rho = ir \bar{\partial} \rho, \quad \partial\rho(\xi) = 1. \end{equation} The function $r \editb{=r[\rho]}: = \rho_{j\kbar}\xi^j \xi^{\kbar} $ is called the transverse curvature of $M$ and $\rho$ \cite{graham1988smooth}. When $\rho$ is strictly plurisubharmonic, then $r^{-1} = \|\partial \rho\|^2$ in the Kähler metric with potential $\rho$. \begin{proposition}\label{prop:mcnt} Let $M$ be a strictly pseudoconvex real hypersurface in $\mathbb{C}^{n+1}$ and $F\colon M \to N \subset\mathbb{C}^{d+1}$ a CR immersion. Suppose that $F$ extends holomorphically to a neighborhood of $M$. Let $\widetilde{\rho}$ be a strictly plurisubharmonic defining function for $N$ \editb{and let} $\eta = i\bar{\partial}\rho$, $\rho = \widetilde{\rho} \circ F$, $\theta = i\bar{\partial}\rho$, \editb{so that } $F^{\ast} \eta = \theta$. Then \begin{equation} \|H_{F(M)}\|^2 \circ F = r[\rho] - r[\widetilde{\rho}] \circ F. \end{equation} In particular, $r[\widetilde{\rho}] \circ F \leq r[\rho]$ on $M$. The equality holds if and only if $(\theta,\eta)$ is an admissible pair for the CR immersion $F$. \end{proposition} \begin{proof} \editb{As already noted,} $F^{\ast} \eta = \theta$ and \editb{also note} that the Reeb vector field is $T = i(\xi - \bar{\xi})$. We shall compute $F_{\ast}{\xi}$ as follows. \edit{In local coordinates} \editb{$(z_1, \dots, z_{n+1})$} we write $\xi = \xi^j \partial_j$ (\editb{using} summation convention). By direct calculations (see, e.g., \cite{lee--melrose,li--son}) \editb{we obtain} \begin{equation} \xi^j = r \rho^j = r \rho^{j\bar{k}} \rho_{\bar{k}}, \end{equation} \edit{where} $\rho_j = \partial\rho /\partial z_j$, $\rho_{j\bar{k}} = \partial^2 \rho/\partial \bar{z}_k \partial z_j$, and $\rho^{j\bar{k}}$ is the inverse \edit{transpose} matrix of $\rho_{j\bar{k}}$. Thus, for $p\in M$ and $q = F(p) \in M\subset N$, and in local coordinates $z'^A$, \begin{equation} F_{\ast} (\xi_p) = F^A_j (p)\,\xi^j_p\, \partial_A. \end{equation} On the other hand, the Reeb vector field on $N$ is given by $\widetilde{T} = i(\xi' - \overline{\xi'})$, with $\xi' = \xi'^A\partial_A$. Thus, by \cref{prop:ttprime}, \begin{equation} H_{F(M)}\|_{F(p)} = \left(F^A_j (p) \xi^j_p - \xi'^A\right) \partial_A. \end{equation} By direct calculations, \begin{equation} \rho_j = \widetilde{\rho}_A F^A_j, \quad \rho_{j\bar{k}} = F^A_j F^{\bar{B}} _{\bar{k}} \widetilde{\rho}_{A\bar{B}}. \end{equation} Therefore, \begin{equation} F^A_j \xi'^{\bar{B}}\widetilde{\rho}_{A\bar{B}} = \widetilde{r}F^A_j\widetilde{\rho}_{A} = \widetilde{r}\rho_j. \end{equation} Consequently, \begin{align} \|H_{F(M)}\|^2 \circ F & = \left(\xi'^A - F^A_j \xi^j \right) \left(\xi'^{\bar{B}} - F^{\bar{B}}_{\bar{k}} \xi^{\bar{k}}\right) \widetilde{\rho}_{A\bar{B}}\notag \\ & = \xi'^A\xi'^{\bar{B}} \widetilde{\rho}_{A\bar{B}} -2 \Re \left(F^A_j \xi^j \xi'^{\bar{B}}\widetilde{\rho}_{A\bar{B}} \right) + F^A_jF^{\bar{B}}_{\bar{k}}\xi^j\xi^{\bar{k}} \widetilde{\rho}_{A\bar{B}}\notag \\ & = \widetilde{r} -2\Re \left(\widetilde{r} \xi^j \rho_j \right) + \xi^j\xi^{\bar{k}} \rho_{j\bar{k}} \notag\\ & = \widetilde{r} -2 \widetilde{r} + r \notag \\ & = r - \widetilde{r}. \end{align} The proof is complete. \end{proof} \editb{We also need the following computational result.} \begin{proposition}[\cite{son2019semi}]\label{prop:fun} Let $M$ be a strictly pseudoconvex real hypersurface defined by $\rho = 0$, with $d\rho\ne 0$ along $M$. Let $\sigma$ be a smooth function in a neighborhood of $M$ and $\hat{\rho} = e^{\sigma} \rho$. Then \begin{equation} e^{\sigma} \hat{r} = r + 2\Re(\xi) \sigma - \|\bar{\partial}_b \sigma\|^2. \end{equation} Here, $\bar{\partial}_b$ is the Cauchy-Riemann operator associated to $\theta: = -i\partial \rho$ and $\xi$ is the transverse vector field of $\rho$. \end{proposition} Suppose that $\rho$ and $\widetilde{\rho}$ are defining functions for $(M,\theta)$ and $(N,\eta)$ such that $\theta = i\bar{\partial}\rho$ and $\eta = i\bar{\partial} \widetilde{\rho}$. Let $F\colon M \to N$ be a CR immersion, extended as a holomorphic immersion in a neighborhood of a point $p\in M$, and let $Q$ be such that \begin{equation} \widetilde{\rho} \circ F = Q \rho. \end{equation} By a transversality argument, $Q\ne 0$ on $M$. Thus, we may assume that $Q> 0$ on $M$. Then $F^{\ast} \eta = e^u\theta$ with $u = \log Q\|_M$. If $r[\rho]$ and $J[\rho]$ are the transverse curvature of $\rho$ and the Levi-Fefferman determinant (see \cite{li--son}), i.e. \begin{equation} J(\rho) = -\det \begin{bmatrix} \rho & \rho_{\bar{k}} \\ \rho_{j} & \rho_{j\kbar} \end{bmatrix}, \quad r(\rho) = \frac{\det[\rho_{j\kbar}]}{J(\rho)}, \end{equation} then by the Li-Luk formula for the Webster scalar curvature \cite{li--luk} (see also \cite[Proposition~4.1]{li--son}), \begin{equation} J_{\theta} = r(\rho) + P_{\rho} \log J(\rho), \end{equation} where \begin{equation} P_{\rho} : = \frac{1}{2n(n+1)}(\xi^j\xi^{\kbar} - \psi^{j\kbar})\partial_j \partial_{\kbar}. \end{equation} Then by using \cref{prop:mcnt,prop:fun} and the formulas above, we obtain that, in terms of the local frame $Z_\alpha:= \partial_{\alpha} - (\rho_{\alpha}/\rho_w)\partial_w$, the \editb{mixed type} \edit{components} of $\ahlfors$ are given by \begin{align}\label{e614} \ahlfors_{\alpha\bbar}(F) = \frac{1}{2}\left(u_{\alpha,\bbar} + u_{\bbar,\alpha}\right) + \frac12 \bigl(N_{\rho} u\bigr) h_{\alpha\bbar} + \frac{1}{2} \bigl(P_{\widetilde{\rho}} \log J(\widetilde{\rho}) - P_{\rho} \log J(\rho) \bigr) h_{\alpha\bbar}, \end{align} where $u = \log Q$. Following \cite{li--luk}, we define the second order operator \begin{equation} D^{\rho}_{\alpha\bbar} = \partial_{\bbar}\partial_{\alpha} -(\rho_{\alpha}/\rho_{w}) \partial_w \partial_{\bbar} - (\rho_{\bbar}/\rho_{\wbar}) \partial_{\wbar}\partial_{\alpha} + (\rho_{\alpha}\rho_{\bbar}/\|\rho_w\|^2) \partial_{w}\partial_{\wbar}, \end{equation} which satisfies \begin{equation} D^{\rho}_{\alpha\bbar} \varphi = \varphi_{Z\Zbar}(Z_{\alpha}, Z_{\bbar}). \end{equation} \editb{We can now give a completely explicit formula for the \editb{mixed type} components of the Ahlfors tensor.} \begin{proposition}\label{prop:explicitahlfors} With the notations \editb{introduced} above, it holds that \begin{equation} \ahlfors_{\alpha\bbar}(F) = D^{\rho}_{\alpha\bbar} \log Q - \frac{1}{2}\bigl(P_{\widetilde{\rho}} \log J(\widetilde{\rho}) - P_{\rho} \log J(\rho) \bigr) h_{\alpha\bbar}. \end{equation} \end{proposition} \begin{proof} This follows from \cref{e614} and a well-known formula for the Christoffel symbols of the Tanaka-Webster connection \cite{li--luk}. \end{proof} We point out that this formula involves both tangential and normal derivatives of the quotient $Q$ and the Fefferman \edit{determinants} on the source and the target. The interesting case is when $\rho$ and $\widetilde{\rho}$ are approximate Fefferman defining functions for the source and target of order $3$, i.e., when $J(\rho) = 1 + o(\rho^{2})$ and $J(\widetilde{\rho}) = 1 + o(\widetilde{\rho}^{2})$, because then the formula simplifies to \begin{equation} \ahlfors_{\alpha\bbar}(F) = D^{\rho}_{\alpha\bbar} \log Q.\end{equation} In particular, we have a particularly simple formula when both source and target are spheres. \begin{corollary} Let $(M,\theta)$ be a strictly pseudoconvex pseudohermitian manifold, $F$ and $G$ nonconstant holomorphic maps sending $M$ into $\mathbb{S}^{2N+1}$, and \begin{equation} \\|F\\|^2 - 1 = Q_F\cdot \rho, \quad \\|G\\|^2 - 1 = Q_G \cdot \rho. \end{equation} where $\rho$ is a defining function for $M$, $\theta: = i\bar{\partial}\rho$. If $F = \gamma \circ G$ for some $\gamma \in \aut(\mathbb{S}^{2N+1})$, then \begin{equation}\label{e:a} \iota^{\ast} \partial\bar{\partial} \log (Q_F/Q_G) = 0. \end{equation} Here $\iota \colon M \to \mathbb{C}^{n+1}$ is the inclusion. \end{corollary} \begin{proof} As explained above, in the frame $Z_{\alpha} = \partial_{\alpha} - (\rho_{\alpha}/\rho_w)\partial_w$, we have that \begin{equation} \ahlfors_{\alpha\bbar}(F) = D^{\rho}_{\alpha\bbar} \log Q_F + \frac{1}{2}P_{\rho} \log J(\rho) h_{\alpha\bbar}, \end{equation} and similar for $\ahlfors_{\alpha\bbar}(G)$. Thus, if there exists such a $\gamma$, then by \cref{thm:crintro}, we have that \begin{equation} D^{\rho}_{\alpha\bbar} \log (Q_F/Q_G) = 0. \end{equation} The last equality is clearly equivalent to \cref{e:a}, \editb{ and the proof} is complete. \end{proof} Note that if $F$ and $G$ are nonconstant then $Q_F$ and $Q_G$ are nonvanishing on $M$ by a Hopf Lemma (a transversality result). The quotient $Q_F$ defined as above has been used extensively in the study of proper holomorphic maps between balls or sphere maps; see, e.g., \cite{d1993several,d2017symmetries} and the references therein. The explicit formula can be used to deduce a necessary condition, which is simple to check, for a map to be equivalent to a monomial map as follows. \begin{corollary}\label{cor:monomial} Let $F$ be a monomial map between spheres in the standard coordinates, then $\trace \ahlfors(F)$ only depends on the moduli \editb{ of the } $\|z_j\|$, $j=1,2,\dots, n + 1$. Moreover, for each~$j$, $\ahlfors_{\alpha\bbar}(F)$ is tracefree on the set $\|z_j\| = 1$. Consequently, if $F$ is equivalent to a monomial map, then the umbilical locus of $F(\mathbb{S}^{2n+1})$, if not empty, is invariant under a torus action. \end{corollary} We mention a similar result characterizing the monomial maps $F$ in terms of the invariant group $\Gamma_F$ in a recent paper by D'Angelo and Xiao \cite{d2017symmetries}. Precisely, they proved that $F$ is \edit{spherically equivalent to a monomial map} iff the group $\Gamma_F$ contains an $n$-torus. \begin{proof} Suppose that $F$ is a monomial map, then $\\|F\\|^2 -1$ is a multivariate polynomial in $\|z_j\|^2$, $j=1,2,\dots , n+1$: There is a polynomial $f(t_1,\dots , t_{n+1})$ such that \begin{equation} \\|F\\|^2 - 1 = f(\|z_1\|^2,\dots , \|z_{n+1}\|^2). \end{equation} Since $F$ sends sphere into sphere, $f(t)$ is divisible by $t_1 + t_2 +\dots + t_{n+1} - 1$ in the polynomial ring $\mathbb{R}[t]$. Thus, the quotient $g(t_1,\dots, t_{n+1})$ is a polynomial in $t_j$ with real coefficients. As $\log Q_F(z) = \log g(\|z_1\|^2, \dots , \|z_{n+1}\|^2)$, we can write at point \editb{for wich} $w = z_{n+1}\ne 0$, that \begin{equation} \ahlfors_{\alpha\bbar}(F) = D^{\\|z\\|^2 - 1}_{\alpha\bbar} \log Q_F = \phi_{\alpha}\delta_{\alpha\beta} + \zbar_{\alpha} z_{\beta} \,\phi_{\alpha\beta}, \end{equation} where \begin{align} \phi_{\alpha} & = \frac{\partial \log g}{\partial t_{\alpha}}\bigl\|_{t_j = \|z_j\|^2}, \quad \text{and}\\ \phi_{\alpha\beta} & = \left(\frac{\partial^2 }{\partial t_{\alpha} t_{\beta}} - \frac{\partial^2 }{\partial t_{\beta} \partial t_{n+1}} - \frac{\partial^2 }{\partial t_{\alpha} \partial t_{n+1}} + \frac{\partial^2 }{\partial t_{n+1}^2}\right) \log g(t)\bigl\|_{t_j = \|z_j\|^2}, \end{align} are real-valued. Since the inverse of \edit{the Levi matrix} is $h^{\alpha\bbar} = \delta_{\alpha\beta} - z_{\alpha}\, \zbar_{\beta}$, we obtain that $\ahlfors_{\alpha\bbar}(F)$ is tracefree on $\|w\|=1$. Moreover, \begin{equation} \trace \ahlfors(F) = \sum_{\alpha} \phi_{\alpha} + \sum_{\alpha} (\phi_{\alpha\alpha} - \phi_{\alpha}) \|z_{\alpha}\|^2 - \sum_{\alpha,\beta} \phi_{\alpha\beta} \|z_{\alpha}\|^2 \, \|z_{\beta}\|^2. \end{equation} These prove the corollary. \end{proof} \subsection{The $(1,0)$-form $\ahlfors_1$ in dimension three} As discussed in the last section, in dimension three, the $(1,0)$-form $\ahlfors_1$ \edit{can be thought of as} a replacement for the tracefree part of the Ahfors \edit{derivative}, which is trivial in \editb{in this case}. It is possible to give an explicit formula for $\ahlfors_1$ in general, but we shall focus on the case where $M = \mathbb{S}^3$ is the \edit{3-sphere} with the standard pseudohermitian structure $\Theta$. Since the pseudohermitian torsion vanishes, we have $P(u) = u_{\bar{1},}{}^{\bar{1}}{}_{1}$ \edit{(covariant derivatives)}. Put \begin{equation}\label{e:standardframe} L = \wbar \partial_{z} - \zbar \partial_{w}, \quad \theta^1 = w dz - z dw. \end{equation} Then clearly, \begin{equation} P(u) = (LL\overline{L}\, u)\, \theta^1. \end{equation} In the case of the \edit{3-sphere}, the conjugate of $P$ reduces to the operator characterizing the CR pluriharmonic \editb{functions introduced} by Bedford \cite{bedford1974dirichlet}. Let $F, G \colon \mathbb{S}^3 \to \mathbb{S}^N$ be CR maps with \begin{equation} \\|F\\|^2 - 1 = Q_F\cdot \rho, \quad \\|G\\|^2 - 1 = Q_G \cdot \rho. \end{equation} Then \begin{equation} \ahlfors_1(F) = (LL\overline{L} \log Q_F)) \theta^1, \end{equation} and similarly for $G$. Therefore, if $F = \phi \circ G$, then \begin{equation} (LL\overline{L}\, \log Q_F) = (LL\overline{L}\, \log Q_G) \end{equation} Moreover, if $F \circ \gamma = \phi \circ G$, then \begin{equation} (LL\overline{L}\, \log Q_F) \circ \gamma = e^{-2\psi}(LL\overline{L}\, \log Q_G) \end{equation} where $\psi$ is determined by $\gamma^{\ast} \theta = e^{\psi}\theta$. {Similar to \cref{cor:monomial}, we have the following characteristic of the $\ahlfors_1$ of monomial maps. For a monomial map $F$ from \edit{a 3-sphere} into another sphere, $\ahlfors_1(F)$ expressed in the standard coframe \cref{e:standardframe}, must have the following form: \begin{equation} \ahlfors_1(F) = \bar{z} \bar{w} \editb{p}(\|z\|^2, \|w\|^2) / Q_F^3, \end{equation} where $\editb{p}$ is a polynomial of $\|z\|^2$ and $\|w\|^2$ with real coefficients. In particular, the vanishing locus of $\ahlfors_1(F)$, which is an invariant for equivalent sphere maps, must \edit{contain} at least two circles. The proof of this fact is similar to that of the aforementioned corollary. We omit the \edit{details}. \section{Examples and a question} We calculate the Ahlfors tensor $\ahlfors$ and \edit{the $(1,0)$-form} $\ahlfors_1$ for various sphere maps that have \editb{previously appeared} in the literature. For the sphere case, the calculations are simple by \cref{prop:explicitahlfors}. All calculations can be done by hand, but \edit{some tedious calculations} can also be done by a computer algebra system. \begin{example}\rm In \cite{d1988proper}, D'Angelo provided a list of 13 discrete examples and two 1-parameter analytic families of monomial maps from $\mathbb{S}^3$ to $\mathbb{S}^7$ which includes 4 trivial extensions of maps from $\mathbb{S}^3 \to \mathbb{S}^5$ of Faran's list. Later, Watanabe \cite{watanabe1992proper} found another map (numbered 16 in the \cref{t1}). We compute the trace of the Ahlfors derivative of each map. We then locate the umbilical points of the images of $\mathbb{S}^3$ in $\mathbb{S}^7$ via the maps. There are four types of umbilical loci that occur: the empty set, the whole sphere, one circle, \edit{and} the union of two circles; this is predicted in \cref{cor:monomial}. To simplify the notations, we put $S_1 = \{(e^{it},0) \colon t\in \mathbb{R}\}$ and $S_2 = \{(0,e^{it}) \colon t\in \mathbb{R}\}$. In \cref{t1}, the expressions in the Ahlfors column are the traces of the Ahlfors derivatives which only depend on $\|z\|^2$, and thus we put $s=\|z\|^2$. This trace determines the ``hermitian part'' \edit{(or mixed-part)} of the Ahlfors derivative as $n=1$. The ``holomorphic'' parts vanish in all \edit{cases} since the standard spheres have vanishing pseudohermitian torsions. Moreover, the norms of the CR second fundamental forms $\|\sff^{CR}\|$ can be computed easily from these results. Observe that the Ahlfors derivatives of these 16 maps are all different. Thus, two different maps are not left equivalent. Moreover, the umbilical loci can also be used to distinguish equivalent classes. For examples, the maps in the Faran's list are pairwise nonequivalent, since their umbilical loci are not congruent under the CR automorphisms of $\mathbb{S}^3$. We point out that in \cref{t1}, the maps numbered 1, 3, and 5, are special cases of the homogeneous maps. The trace of \edit{the Ahlfors derivatives} are constant while the $\ahlfors_1$'s vanish identically. More generally, for homogeneous maps of degree $d$ from $\mathbb{S}^{2n+1}$ with $n\geq 1$, the Ahlfors derivatives are nonzero multiples of the Levi metric. \begin{landscape} \small \renewcommand{\arraystretch}{1.5} \begin{table} \begin{tabular}{\|l\|l\|l\|l\|l\|} \hline ~ & $F$ & $\trace\ahlfors(F)\quad s = \|z\|^2$ &$\ahlfors_1(F)\quad s= \|z\|^2$& Umbilic \\ \hline 1 & $(z,w,0,0)$ & $0$ &0& $\mathbb{S}^3$ \\ 2 & $(z,zw,w^2,0) $ & $\dfrac{s}{\left(s-2\right)^2}$ &$-\dfrac{4 \bar z \bar w}{\left(s-2\right)^3}$& $S_2$ \\ 3 & $(z^2,\sqrt{2}zw,w^2,0)$ & $ \dfrac12$ &$0$& $\emptyset$ \\ 4 & $(z^3,\sqrt{3}zw,w^3,0)$ & $-\dfrac{3 s \left(s-1\right)}{\left(s^2 -s+1\right)^2}$ &$\dfrac{2 \bar z \bar w \left(s-2\right) \left(s+1\right) \left(2 s-1\right)}{\left(s^2-s+1\right)^3}$& $S_1 \cup S_2$ \\ 5 & $(z^3 , \sqrt{3} z^2w , \sqrt{3} zw^2 , w^3)$ & 1 &$0$& $\emptyset$ \\ 6 & $(z^3,z^2w,zw,w)$ & $\dfrac{\left(1-s\right) \left(s^2 +4 s+1\right)}{\left(s^2+s+1\right)^2}$ &$-\dfrac{18 \bar z \bar w s \left(s+1\right)}{\left(s^2+s+1\right)^3}$& $S_2$ \\ 7 & $(z^2,z^2w,zw^2,w)$ & $\dfrac{s^3+4 s^2-5 s+2}{\left(-s^2+2 s+1\right)^2}$ &$-\dfrac{8 \bar z \bar w \left(s^3+3 s-2\right)}{\left(s^2-2 s-1\right)^3}$& $\emptyset$ \\ 8 & $(z^2, \sqrt{2}z^2w , \sqrt{2} zw^2, w^2)$ & $\dfrac{7 s^2-7 s+3}{2 \left(-s^2+s+1\right)^2}$ &$-\dfrac{2 \bar z\bar w \left(2 s-1\right) \left(s^2-s+4\right)}{\left(s^2-s-1\right)^3}$& $\emptyset$ \\ 9 & $(z^3, \sqrt{3}z^2w , \sqrt{2}zw^2, w^2)$ & $\dfrac{9 s^2 -6 s+6}{\left(-s^2+2 s+2\right)^2}$ &$-\dfrac{12 \bar z \bar w \left(s^3+6 s-4\right)}{\left(s^2-2 s-2\right)^3}$& $\emptyset$ \\ 10 & $(z , z^2w , \sqrt{2}zw^2 , w^3)$ & $\dfrac{2 s^2-3 s+3}{\left(3-2 s\right)^2}$ &$-\dfrac{12\bar z \bar w}{\left(2 s-3\right)^3}$& $\emptyset$ \\ 11 & $(z^4, z^3w , \sqrt{3}zw , w^3)$ & $-\dfrac{3 s \left(s^4-4 s^3+12 s^2-9\right)}{\left(s^3+3 s^2-3 s+3\right)^2}$ &$-\dfrac{36 \bar z \bar w \left(s^2-2 s-1\right) \left(s^4-2 s^3-6 s+3\right)}{\left(s^3+3 s^2-3 s+3\right)^3}$& $S_1 \cup S_2$ \\ 12 & $(z^4 , \sqrt{3}z^2w , zw^3 , w)$ & $-\dfrac{3 \left(9 s^4-12 s^2 +4 s-1\right)}{\left(3 s^3-3 s^2+3 s+1\right)^2}$ &$\dfrac{36 \bar z \bar w \left(s^2+2 s-1\right) \left(3 s^4-6 s^3-2 s+1\right)}{\left(3 s^3-3 s^2+3 s+1\right)^3}$& $S_1$ \\ 13 & $(z^5 , \sqrt{5}z^3w , \sqrt{5}zw^2 , w^5)$ & $ \dfrac{s \left(s^6-7 s^5+15 s^4-25 s^3+35 s^2-27 s+9\right)}{\left(s^4-3 s^3+4 s^2-2 s+1\right)^2}$ &$\mathcal{A}_1 (F)$& $S_2$ \\ 14 & $(z , tw , \sqrt{1-t^2} zw, \sqrt{1-t^2} w^2)$ & $-\dfrac{\left(t^2-1\right) s}{\left(t^2 \left(1-s\right)+s-2\right)^2}$ &$\dfrac{2 \bar z \bar w \left(t^4-3 t^2+2\right)}{\left(\left(t^2-1\right) s-t^2+2\right)^3}$& $S_2$ \\ 15 & $(z^2, \sqrt{1+t^2} zw , tw^2 , \sqrt{1-t^2} w)$ & $\dfrac{t^2 \left(s+1\right)-s+1}{\left(t^2 \left(1-s\right)+s+1\right)^2}$ &$\dfrac{4\bar z \bar w(1- t^4)}{\left(\left(t^2-1\right) s-t^2-1\right)^3}$& $\emptyset$ \\ 16 & $(z^2,\sqrt{2}zw,zw^2,w^3)$ & $\dfrac{-7 s^2 +6 s+3}{\left(s^2-2 s+3\right)^2}$ &$\dfrac{8\bar z \bar w \left(s^3-9 s+6\right)}{\left(s^2-2 s+3\right)^3}$& $\emptyset$ \\ \hline \end{tabular} \medskip \caption{The \edit{CR Ahlfors derivative} $\ahlfors$ and \edit{the $(1,0)$-form} $\ahlfors_1$ of monomial sphere maps from $\mathbb{S}^3 \to \mathbb{S}^7$.} \label{t1} \end{table} \end{landscape} The formula for the $\ahlfors_1$ tensor of the $13^{\mathrm{rd}}$ map is too big to fit in the table. Precisely, \begin{align} \ahlfors_1(F_{13}) = 2 \bar z \bar w \frac{2 s^9-6 s^8+18 s^7-64 s^6+117 s^5-114 s^4+42 s^3+24 s^2-27 s+6}{\left(s^4-3 s^3+4 s^2-2 s+1\right)^3}. \end{align} \end{example} \begin{example}\rm Consider the following cubic map (appeared earlier in \cite{d1991polynomial}) \begin{equation} \left(\frac{z^2 - z^2w}{\sqrt{2}} , \frac{zw - zw^2}{\sqrt{2}} , \frac{z+zw}{\sqrt{2}} , w^2 \right). \end{equation} Then the trace of the \edit{CR Ahlfors derivative} is \begin{equation} \trace \ahlfors(F) = \frac{\|w\|^6 + 4\|w\|^4 + 2 \Re(w)(1-3\|w\|^2) - 5\|w\|^2 + 4}{Q}\biggr\|_{\mathbb{S}^3} \end{equation} for some polynomial $Q$ positive on $\mathbb{S}^3$. Observe that this trace vanishes if and only if $w=1$, i.e., the umbilical locus is a \textit{singleton}. This immediately implies that $F$ is not equivalent to any map in \cref{t1} whose umbilical locus is either empty or of positive dimension. In fact, by \cref{cor:monomial}, it is not equivalent to any monomial map regardless of the target dimension, a fact that was first observed by D'Angelo in \cite{d1991polynomial} for the target dimension 4. \end{example} \begin{example}\rm Consider the following map which was discussed in \cite{faran2010rational}, Proposition~3.3, \begin{equation} F(z,w) = \left(\frac{\sqrt{3}}{9} (z^2+4z-2), \frac{\sqrt{6}}{9}(z^2+z+1) , \frac{\sqrt{3}}{12} w(3z+5) , \frac{\sqrt{6}}{6} w^2 , \frac{\sqrt{13}}{12} w(z-1) \right) \end{equation} Observe that this is not monomial and does not send $0$ to $0$. Then the trace of \editb{its CR} Ahlfors \editb{derviative} is \begin{equation} \trace\ahlfors(F) = \frac{30\|z\|^2 + 24\Re z + 18}{\|z\|^4 - 16\|z\|^2\Re z + 32\Re(z^2) + 272\Re z + 289}. \end{equation} Since this trace does not vanish on $\mathbb{S}^3$, $F(\mathbb{S}^3)$ is a submanifold of $\mathbb{S}^9$ without umbilical point. Moreover, \begin{equation} \ahlfors_1(F) = \frac{264\, \wbar (1 + 4 \zbar + \zbar^2)}{(17 + 8\, \Re z - \|z\|^2)^3}. \end{equation} \edit{Thus}, $F$ is not left equivalent to any monomial map from $\mathbb{S}^3$ into $\mathbb{S}^N$, $N\geq 3$. \end{example} \begin{example}[D'Angelo's maps \cite{d1988proper}]\rm For each $t$, put $c = \cos(t)$ and $s = \sin (t)$ and consider the maps \begin{equation} F_t(z,w) = (z_1, \dots , z_n, cw, sz_1w, \dots , s z_n w, sw^2). \end{equation} Then $F_t$ maps $\mathbb{S}^{2n+1}$ into $\mathbb{S}^{4n+3}$. It is clear that $F_0 = L$ is the linear embedding and $F_{\pi/2} \cong (\Wm, 0)$ where $\Wm$ is the complex Whitney map from $\mathbb{S}^{2n+1}$ \editb{into} \edit{$\mathbb{S}^{4n+1}$}. \editb{We compute that} \begin{equation} Q_t = 1+ s^2\|w\|^2, \quad F_t^{\ast} \Theta = i\bar{\partial} \rho = e^{u_t} \theta, \quad u_t = \log (1 + s^2 \|w\|^2). \end{equation} In \editb{the} local frame $\widetilde{Z}_{\alpha} = \wbar \partial_{\alpha} - \zbar_{\alpha} \partial_w$, we have that \begin{equation} \ahlfors_{\alpha\bbar}(F_{\editb{t}}) = \frac{s^2 \zbar_{\alpha} z_{\beta}}{(1+s^2\|w\|^2)^2}, \end{equation} is non-negative and of rank one except at the umbilical points. Taking the trace, we have \begin{equation} \trace \ahlfors(F_{\editb{t}}) = \frac{s^2(1-\|w\|^2)}{(1+s^2\|w\|^2)^2}\biggr\|_{\mathbb{S}^3}. \end{equation} Thus, $F_t$ \editb{and $F_{t'}$} are not left equivalent for $t\neq t'$. Moreover, if $\Psi \in \aut(\mathbb{S}^{2n+1},F_t)$ for $t\ne 0$, then $\Psi(0,w) = (0, \psi_{n+1}(0,w))$. The inclusion $F(\mathbb{S}^{2n+1}) \subset \mathbb{S}^{4n+3}$ is umbilic along the locus $\{F(0, e^{iy})\}$. \end{example} The following example appears originally in Webster \cite{webster1979rigidity}. \begin{example} \label{ex:7}\rm Let $M\subset \mathbb{C}^{n}$ be the strictly pseudoconvex real hypersurface defined by $\rho = 0$, with \begin{equation} \rho = \|z\|^2 + b(z) + \overline{b(z)} -1. \end{equation} Put $\theta: = i\bar{\partial} \rho$ and let \begin{equation} F(z) = \frac{1}{1-b(z)}\left(z_1 , \dots , z_n , b(z)\right). \end{equation} Then $F$ maps $M$ into the unit sphere in $\mathbb{C}^{n+1}$, \editb{actually, }one can compute $\\|F\\|^2 - 1 = \|1-b(z)\|^{-2} \rho.$ Since $D_{\alpha\bbar}^{\rho} \log \|1 - b(z)\|^{-2} = 0$, we see that \begin{equation} \tf \ahlfors_{\alpha\bbar}(F) = 0. \end{equation} \editb{Let} $L$ and $\mathcal{W}$ be the linear embedding and Whitney map from $\mathbb{S}^{n+1}$ into $\mathbb{S}^{2d+1}$ with $d\geq 2n-1$. Then $L \circ F$ and $\mathcal{W} \circ F$ are inequivalent CR immersions from $M$ into $\mathbb{S}^{2d+1}$. (In general, post composing with inequivalent maps may still yield equivalent maps. For example, take $g$ to be the linear embedding of $\mathbb{S}^{2n-1}$ into $\mathbb{S}^{2n+1}$, then $L\circ g = \mathcal{W}\circ g$ is the linear embedding.) To compute the trace part, we need to compute the Fefferman determinant $J[\rho]$. In a special case $b(z) = \frac{1}{2} \sum_{k=1}^{n} z_k^2$, we can easily compute $J[\rho] = \rho + 2$ and thus, \begin{equation} \ahlfors_{\alpha\bbar}(F) = (n+1)^{-1} h_{\alpha\bbar}. \end{equation} In particular, the imbedding image has no umbilical point. \end{example} \begin{question}\label{q:1}\rm Among the examples discussed above, we have encountered three families of CR maps with the property that their Ahlfors derivatives are constant multiples of the Levi metric: linear embeddings between spheres, the homogeneous maps, and the ones considered in \cref{ex:7}. Motivated by these examples, we pose the following question: Suppose that $F$ is a nonconstant CR map between spheres and that $\ahlfors(F)$ is a nonzero constant (resp. functional) multiple of the Levi metric (i.e., $\ahlfors_{\alpha\beta}(F) = 0$ or $\ahlfors_{\alpha\bbar}(F) = g h_{\alpha\bbar}$, $g \ne 0$, respectively). Does it \edit{follow} that $F$ is spherical\editb{ly} equivalent to a homogeneous monomial map? \end{question}	115,748
Honda FCX Clarity - Hydrogen Powered Car If you live in California and can spare $600 per month, you can drive the FCX zero emission car Look - no exhaust pipe! Just a vertical gas flow cell structure The styling of the car is fairly restrained, optimised for aerodynamic efficiency Now A tuner car, is what we call an import, which is a sporty, economical and a very practical car that is built for the purpose of daily usage, decent gas mileage and comfort, but has the potential of being a high performance car, which requires engine modifications. That’s what sets it apart from other High performance cars or sports cars, is that anyone can buy a tuner car and unleash its true potential. Tuners differ from sports cars in many different areas such as price, engine displacement, horsepower and many other factors. They also differ from typical sedans or SUV import cars; they are lighter, faster and have better handling..	267,820
TITLE: Rewrite the following 2nd order linear differential equation as a linear system in the form y'=Ay. QUESTION [1 upvotes]: Question: Rewrite the following 2nd order linear differential equation as a linear system in the form y'=Ay. Solve both the original DE and the linear system you derived. How do the two solutions compare? ${2y''+3y'-6y=0}$ Here's what I have done so far: ${2y''+3y'-6y=0}$ $y''+ \frac{3y'}{2} - 3y = 0$ Changed the formula to this $r^2+ \frac{3r}{2} - 3 = 0$ Found the zeros to be $Ae^{\sqrt(57)/2 -3/4} + Be^{-\sqrt(57)/2 -3/4} = y$ I used the wolfram calculator to check, and apparantly my answer is correct but I'm not sure what I'm supposed to be checking my solution with? Thanks so much for any and all help. REPLY [0 votes]: To do this as a system of equations $y'' = 3y - \frac 32 y'\\ y'=x\\ x' = 3y - \frac 32 x$ $\begin{bmatrix} {x\\y}\end{bmatrix}' = \begin{bmatrix}-\frac 32&3\\1&0\end{bmatrix}\begin{bmatrix} x\\y\end{bmatrix}$ $\mathbf x' = A \mathbf x\\ \mathbf x = e^{At}\mathbf x_0$ Diagonalize A, the chartaristic polynomial is the same one you found the more tradtional way.	58,205
I have done every search I can and can't seem to pull up any info on the P Bass Jr here on the boards. They only started making them in Jan 2004 so maybe they have not been around long enough for many people to try out. One would think that this bass would have more bottom than the mustang. The fact that it does not have a tone knob kind of turns me off. Heck the Mustang has a tone knob, why not the P Jr? I'm trying to find another short scale that has phat tone like an Allen Woody Epi but with a fender style bridge. As soon as I win the lottery I'm going to buy an Alembic. I got to noodle around on a Stanly Clark model once. Its too bad short scale basses seem to come in two classes. The $5000 n over uber-bass and cheap-O's	287,922
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EAI before B2B? While EAI and B2B are distinct activities, it makes sense that the larger an enterprise becomes, the more it will benefit from an EAI program. The requirement for EAI is driven by the volume and periodicity of transactions. It also makes sense that B2B becomes easier if there is fully automated transaction processing on the back end. Fortunately, there is likely to be a convergence of B2B and EAI technology - driven by real-time requirements on both sides of the firewall and by the emergence of XML-based standards in both inter-company and intra-company interfaces. Web Services (XML+UDDI+WSDL+SOAP) creates the possibility that the same technologies that are applied to solve B2B problems can be used for EAI applications as well. In this case, it is possible to initiate both activities with a reasonable assurance that the result will be able to exchange data between systems, even if they use different technology paths (i.e., Sun vs Microsoft) for implementation. In conclusion, it makes sense that EAI should come before B2B in the sense that a company should have its house in order before it conducts business with the outside world. EAI and B2B face the same fundamental challenge, which is non-technical. The corporate head office must elevate its master data dictionary to the same level as the global chart of accounts and sponsor a department of end user specialists to maintain it. The primary concern of this department should be "what" and not "how" even if preferred technology solutions are supported. The success of both EAI and B2B hinges on the acceptance of lightweight protocols that allow system independent transaction processing. These technologies are emerging today in the form	74,134
Golden – Fall Out Boy Sheet Music Album Cover Artist: Fall Out Boy Song Details: "Golden" is a song recorded by Fall Out Boy and included in the band's third studio album, "Infinity on High", released on 5 Feb 2007. It is a pop rock piano ballad, written by band members Patrick Stump and Pete Wentz and Wesley Eisold. It was produced by Neal Avron. Sheet Music Print and download options may vary. something else?	64,977
NEWS FLASH In Wake Of Voter Purge, Rick Scott’s Approval Rating Drops To 31% \| Florida Gov. Rick Scott’s continued push to remove hundreds of eligible voters is unpopular with both Democrat and Republican voters, according to Public Policy Polling. Just 31 percent approve of Scott’s performance and 56 percent disapprove. Support has dropped off among Republicans as well, which reflects the bipartisan opposition to his voter suppression efforts, including both Republican and Democrat local election officials. The voter purge especially targets minority voters, and more Hispanics (14 percent) and African-Americans (8 percent) said they knew someone who was removed from the voter.	88,393
TITLE: Let $X$ be an infinite dimensional normed space. If $Y$ is a nonzero normed space,then there is some $F:X\to Y$ which is linear but not closed QUESTION [0 upvotes]: Let $X$ be an infinite dimensional normed space. If $Y$ is a nonzero normed space,then there is some $F:X\to Y$ which is linear but not closed. I know that every infinite dimensional normed space has a discontinous linear functional. But how to find a linear functional which is linear but not closed. REPLY [0 votes]: If $f$ is a dis-continuous linear functional then there is a sequence $x_n \to 0$ such that $f(x_n) \to \infty$. [See details below]. Let $y_n=\frac {x_n} {f(x_n)}$. Then $y_n \to 0$ and $f(y_n)=1$ for all $n$. So $(y_n,f(y_n))$ is a sequence in the graph of $f$ which converges to $(0,1)$ and $(0,1)$ is not in the graph of $f$. Hence, $f$ is not a closed linear map (in the FA sense). [ There exist $u_n$ with $ u_n \to 0$ such that $f(u_n)$ does not tend to $0$. There is a subsequence $(w_n)$ of $(u_n)$ and $r>0$ such that $f(w_n) \geq r$ for all $n$. [You may have to replace $u_n$ by $c_nu_n$ with $\|c_n\|=1$ if necessary). Take $x_n=\frac {w_n} {\sqrt {\\|w_n\\|}}$]. $f$ is not closed in the topological sense either! We can modify above argument to get $y_n\to 0$ but $f(y_n)=1-\frac 1 n$. In that case $\{0,y_1,y_2,..\}$ is a clsoed set whose image is not closed (since $1$ is a limit point which is not in the image.	82,915
TITLE: A property of countable dense subsets QUESTION [0 upvotes]: Let $X$ be a non discrete topological space and let $D$ be a countably infinite dense subset. Is it true that $D$ less a finite subset is still dense? thank you REPLY [1 votes]: Firstly, non-discrete is not enough. A space $X$ can have isolated points, i.e. $x \in X$ such that $\{x\}$ is open. These always have to be in any dense set $D$ (as $D$ has to intersect the non-empty open set $\{x\}$). So we want $X$ to be "dense in itself", or $X$ has no isolated points. Second, we need $T_1$-ness of $X$ (which is pretty mild), which is equivalent to all finite sets being closed. If $D$ is dense and $X$ is $T_1$ without isolated points, then no finite non-empty subset $F$ can be open (check this) and then, if $O$ is open and non-empty, so is $O \setminus F$, and so $D$ must also intersect $O \setminus F$, which is equivalent to $O$ intersecting $D \setminus F$, so the latter set is dense whenever $D$ is. If $X$ is not $T_1$ this might fail, and finite sets could be dense (and not omitted). E.g. take the natural numbers and let $O$ be open iff it is empty or it contains both $0$ and $1$. Then $\{0,1\}$ is dense and there are no isolated points. So the answer is yes: for spaces without isolated points, and that are also $T_1$, and both conditions are necessary.	74,030
AL-NOR Commercial complex opens soon here in the city. So far this is the biggest commercial complex ever built here in the city. I would say a one stop shop. The front portion of the building would composes of pharmacies, 24 hours convenient store, water refilling stations, money transfer company, boutiques, internet cafes, beauty parlor, bioessence branch 2 and pizza hut (wow, I wish this is the one that am imagining of). At the back portion would be the hotel, convention center, bowling center, fitness gym, badminton court, and the convention hall for wedding, seminars and among others. 1 comment: its really good to know....	25,595
Greetings from the Seattle airport (free Wi-Fi is a Christmas present from The Google. I've never been to Seattle. I don't really think the airport counts as a visit, though. I have an incredible view of the snowy mountains from where I'm sitting. I'm here way early but didn't have much choice re: flights from Portland. From here, I'll go to Paris, and then to Belgrade, Serbia. A North Carolina fan and I just tried to take over a TV in a restaurant in this terminal, but one freakin' football fan wouldn't budge. In addition, when I initially asked if we could change the channel to CBS, the waitress chirped, "Oh, there's no game on CBS." I said, yes, UK was playing UNC, and she said, "Oh, basketball..." It's soooooooo hard living in a region that's got American football so far up its bum. Oh, well, I'm sure my alliance with the Tarheel would have splintered as soon as the game came on. The woman at the Delta desk for my flight to Paris keeps coming on the intercom, and I could listen to her beautiful Caribbean French accent all day... si beau... By contrast, the high pitched perky newscaster's voice on the nearby TV is making me insane. Attention women TV newscaster's: talk in a deeper voice and stop the uptalk. Or speak with a Caribbean French accent. I would love to take a photo of the Tai airline stewardess having their pre-flight orientation because they look so freakin' adorable, but I just know I'll get in trouble for it for some reason. In 2007, I lead pilgrims returning from Hadj through the oh-so-confusing Frankfurt airport to make their connecting flights. Today: I lead a woman from Japan through the oh-so-confusing Seattle airport terminals to make her connecting flight to Baltimore. So, I've now got Islam and Buddhism covered re: airport karma. Here's why I'm going to Serbia. I'm thrilled to be going -- I've been to most of the countries around it, but not to Serbia. And I have another consulting job in Australia in March. But so far, I haven't gotten even one gig because I've moved back to the USA; every little gig I've had I either got while I was in Germany or got because of a connection elsewhere. So, moving back to the USA has yet to pay off for me at all. Except for learning to ride a motorcycle. Still no title for the Africa Twin; our guy checked the wrong box on the paperwork, and had to resubmit it all. That means another six weeks of waiting. Sigh. All I want for Christmas is my husband to get something back that he's loves so much... As soon as I figure out how, I'm making this my cell phone ring tone. It is AWESOME! (Here's how I found out about it). I cannot stand Snake Oil Salesman Deepak Chopra, and this blog by Phil Plait explains why. Religious/spiritual people out there: please don't tell me what I do or don't believe as a skeptic/atheist. Don't tell me how I think. You truly have no idea. Speaking of me, and of atheism, how weird was it that I defended Islam the other day? Someone had made one of those classic "All Muslims think such-and-such" statements and I went off. Not that I don't believe in criticizing religion; I can do it, as most of you have know! I just loathe the "My religion is enlightened and peaceful while yours is violent and dogmatic" argument I've heard Christians say against Muslims. Islam is just as nuanced as Christianity or any other religion. Yes, there are large factions of Muslims who believe some rather dreadful things and engage in some horrific practices regarding women -- and there are factions of Christians who do the same (look at the forced child marriages in various Christian cults throughout the USA, the shootings and bombings by Christian fundamentalists at least every year in the name of God, the hate spewed by people like Mullah Pat Robertson, the insults non-Christian students often endure from classmates and their parents, etc.). Heck, I defend Scientology when someone attacks it and they admit to being religious themselves -- how is it really any different than any other religion out there? Having been exposed at length to a variety of Christian and Islamic cultures, I can sincerely say that they are much more alike than different. Having read the Bible cover to cover (twice, actually) and the Koran, I can also sincerely say that the former is a heck of a lot more violent and horrific in how it defines appropriate treatment of women than the latter. Imagine no religion... I'm trying, John, I'm trying... Just saw that Kentucky won! Hurrah! It's God's will! Go Big Blue!	160,183
TITLE: $(n-1)$-alternative tensor on E are decomposable QUESTION [1 upvotes]: $E$ is a real vector space with dimension $n$ and $E^$ is dual space of $E$. Assume $\alpha \in Λ^{n-1}(E)$ Show that there exists $\alpha_1,\alpha_2,...,\alpha_{n-1} \in E^$ such that $$\alpha=\alpha_1\wedge\alpha_2\wedge ...\wedge \alpha_{n-1}$$. REPLY [1 votes]: Let me make my life easier by just proving this for $E = \mathbb R^n$; I'm going to use the standard basis $e_1, \ldots, e_n$ for that, and the dual basis $\phi_1, \ldots, \phi_n$, where $\phi_i(v) = e_i \cdot v$, that's defined by the usual inner product. I'm also going to show that there are functionals $\alpha_i$ such that $$ \alpha = r \alpha_1 \wedge \ldots \wedge \alpha_{n-1} $$ i.e., I'm going to end up with a real factor $r$ which needs to be combined with one of the $\alpha_i$ to get the result in the form you asked for. The dimension of $\Lambda^{n-1} (E)$ is $n$. There's a nice map $F$ from $\Lambda^1(E)$ to $\Lambda^n(E)$ defined by $$ \phi \mapsto \alpha \wedge \phi. $$ The codomain -- alternating $n$-forms on $R^n$ -- is one dimensional, generatoed by the determinant function. For any $\phi$, there's a constant $c(\phi)$ with $F(\phi) = c(\phi) \cdot \det$. The function $c: \Lambda^{n-1}(E) \to \mathbb R$ is evidently linear. If it's zero, then $\alpha$ is also zero, and I'm gonna leave that case to you. :) So assuming $c$ is nonzero, it's got a kernel, spanned by an orthonormal basis $v_1, \ldots v_{n-1}$. Let $\alpha_i$ be the dual of $v_i$, i.e., the linear functional such that $$ \alpha_i(u) = v_i \cdot u. $$ Then $\alpha_1 \wedge \ldots \wedge \alpha_{n-1} (v_1, \ldots, v_{n-1}) = 1$. And in fact, $\alpha = r \alpha_1 \wedge \ldots \wedge \alpha_{n-1}$ where $$ r = \alpha(v_1, \ldots, v_{n-1}). $$ Let me do an example: on $R^3$, look at $\alpha = dxdy + dy dz$. For this, we have $$ F(dx) = 1 \\ F(dy) = 0 \\ F(dz) = 1. $$ So my $v$-basis will be $s (-1, 0, 1), (0, 1, 0)$ (where $s = \sqrt{2}/2$). That means (up to constants) that $$ \alpha_1 = (-dx + dz) \\ \alpha_2 = dy $$ so I'm claiming that $$ dx dy + dy dz = r (-dx + dz) \wedge dy $$ which is correct (for $r = -1$). Now that I've done the construction, I leave the proof of correctness to you. It really amounts to checking that on a basis for $R^n$ provided by the $v_i$s, together with a final unit vector $w$ orthogonal to all of them, i.e., their "cross product", the product of the $\alpha_i$s and the $n-1$-form $\alpha$ give the same values.	201,301
\begin{document} \maketitle \begin{abstract} We present a complete list of groups $G$ and fields $F$ for which: (i) the group of normalized units $V(FG)$ of the group algebra $FG$ is locally nilpotent; (ii) the group algebra $FG$ has a finite number of nilpotent elements and $V(FG)$ is an Engel group. \end{abstract} \section{Introduction} Let $V(FG)$ be the normalized subgroup of the group of units $U(FG)$ of the group algebra $FG$ of a group $G$ over a field $F$ of characteristic ${\rm char}(F)=p\geq 0$. It is well known that $U(FG)=V(FG)\times U(F)$, where $U(F)=F\setminus \{0\}$. The group of normalized units $V(FG)$ of a modular group algebra $FG$ has a complicated structure and was studied in several papers. For an overview we recommend the survey paper \cite{Bovdi_survey}. An explicit list of groups $G$ and rings $K$ for which $V(KG)$ are nilpotent was obtained by I.~Khripta (see \cite{Khripta_1} for the modular case and \cite{Khripta_2} for the non-modular case). In \cite{Bovdi_solvable} it was completely determined when $V(FG)$ is solvable. It is still a challenging problem whether $V(FG)$ is an Engel group. This question has a long history (see \cite{Bovdi_engel, Bovdi_Khripta_1, Bovdi_Khripta_2,Bovdi_Khripta_4, Bovdi_Khripta_3, Shalev_I}). The non-modular case was solved by A.~Bovdi (see \cite{Bovdi_engel}, Theorem 1.1, p.174). For the modular case there is no complete solution (see \cite{Bovdi_engel}, Theorem 3.2, p.175), but there is a full description of $FG$ when $V(FG)$ is a bounded Engel group (see \cite{Bovdi_engel}, Theorem 3.3, p.176). It is well known (for example, see \cite{Traustason}) that the Engel property of a group is close to its local nilpotency. A locally nilpotent group is always Engel (see \cite{Traustason}). However these classes of groups do not coincide (see the Golod's counterexample in \cite{Golod}). The following results are classical (see \cite{Traustason}): each Engel profinite group (see \cite{Wilson_Zelmanov}), each compact Engel group (see \cite{Medvedev}), each Engel linear group (see \cite{Suprunenko_Garashchuk}), each 3-Engel and 4-Engel group (see \cite{Heineken} and \cite{Havas_Vaughan-Lee}) and all Engel groups satisfying max (see \cite{Baer}) are locally nilpotent. A group $G$ is said to be {\it Engel } if for any $x,y\in G$ the equation $(x,y,y, \ldots,y)=1$ holds, where $y$ is repeated in the commutator sufficiently many times depending on $x$ and $y$. We shall use the left-normed simple commutator notation $(x_1,x_2)=x_1^{-1}x_2^{-1}x_1x_2$\quad and \[ (x_1,\ldots,x_{n})=\big((x_1,\ldots,x_{n-1}),x_n\big), \qquad\quad (x_{1},\ldots, x_n\in G). \] A group is called {\it locally nilpotent} if all its f.~g. (finitely generated) subgroups are nilpotent. Such a group is always Engel (see \cite{Traustason}). The set of elements of finite orders of a group $G$ (which is not necessarily a subgroup) is called {\it the torsion part} of $G$ and is denoted by $\mathfrak{t}(G)$. We use the notion and results from the book \cite{Bovdi_book} and the survey papers \cite{Bovdi_survey, Traustason}. In several articles, M.~Ramezan-Nassab attempted to describe the structure of groups $G$ for which $V(FG)$ are Engel (locally nilpotent) groups in the case when $FG$ have only a finite number of nilpotent elements (see Theorem 1.5 in \cite{Ramezan-Nassab_1}, Theorems 1.2 and 1.3 in \cite{Ramezan-Nassab_2} and Theorem 1.3 in \cite{Ramezan-Nassab_3}). The following theorem gives a complete answer. \begin{theorem}\label{T:1} Let $FG$ be the group algebra of a group $G$. If $FG$ has only a finite number of non-zero nilpotent elements, then $F$ is a finite field of ${\rm char}(F)=p$. Additionally, if $V(FG)$ is an Engel group, then $V(FG)$ is nilpotent, $G$ is a finite group such that $G=Syl_p(G)\times A$, where $Syl_p(G)\not=\gp{1}$, $G'\leq Syl_p(G)$ and $A$ is a central subgroup of $G$. \end{theorem} The next result plays a technical role. \begin{theorem}\label{T:2} Let $G$ be a group such that $G'$ is a locally finite p-group and the $p$-Sylow subgroup $P=Syl_q(G)$ of $G$ is normal in $G$. If $V(FG)$ is an Engel group and ${\rm char}(F)=p>0$ then the following conditions hold: \begin{itemize} \item[({i})] $V'\leq 1+\mathfrak I (G')$ and $1+\mathfrak I (P)=Syl_p( V(FG))$; \item[(ii)] for each $q\not=p$, the group $Syl_q(G)$ is central; \item[(iii)] the group $\mathfrak{t}(G)=P\times D$, where $D= \times_{\substack{q\not=p}} Syl_q(G)$; \item[(iv)] let $M$ be the subgroup of $G/P$ generated by the central subgroup $DP/P$ and by all the elements of infinite order in $G/P$. Then $V(FG)/(1+\mathfrak I (P))\cong V(FM)$ and $V(FM)/V(FD)$ is a torsion-free group; \item[(v)] $G/\mathfrak{t}(G)$ is an abelian torsion-free group and, if $D$ is finite, then \begin{equation}\label{E:1} \begin{split} V(FM)\cong V(FD) \times (\underbrace{G/\mathfrak{t}(G)\times\cdots\times G/\mathfrak{t}(G)}_n),\\ \end{split} \end{equation} where $n$ is the number of summands in the decomposition of $FD$ into a direct sum of fields. \end{itemize} \end{theorem} The next two theorems completely describe groups $G$ with $V(FG)$ locally nilpotent. Some special cases of Theorem 3 were proved by I.~Khripta (see \cite{Khripta_2}) and M.~Ramezan-Nassab (see Theorem 1.2 in \cite{Ramezan-Nassab_2} and Corollary 1.3 and Theorem 1.4 in \cite{Ramezan-Nassab_1}). \begin{theorem}\label{T:3} Let $FG$ be a modular group algebra of a group $G$ over the field $F$ of positive characteristic $p$. The group $V(FG)$ is locally nilpotent if and only if $G$ is locally nilpotent and $G'$ is a $p$-group. \end{theorem} \begin{theorem}\label{T:4} Let $FG$ be a non-modular group algebra of characteristic $p\geq 0$. The group $V(FG)$ is locally nilpotent if and only if $G$ is a locally nilpotent group, $\mathfrak{t}(G)$ is an abelian group and one of the following conditions holds: \begin{itemize} \item[(i)] $\mathfrak{t}(G)$ is a central subgroup; \item[(ii)] $F$ is a prime field of characteristic $p=2^t-1$,\quad the exponent of $\mathfrak{t}(G)$ divides $p^2-1$ and $g^{-1}ag=a^p$ for all $a\in \mathfrak{t}(G)$ and $g\in G\setminus C_G(\mathfrak{t}(G))$. \end{itemize} \end{theorem} As a consequence of Theorems \ref{T:2} and \ref{T:3} we obtain the classical result of I.~Khripta. \begin{corollary}\label{C:1} Let $FG$ be a modular group algebra of positive characteristic $p$. The group $V(FG)$ is nilpotent if and only if $G$ is nilpotent and $G'$ is a finite $p$-group. The structure of $V(FG)$ is the following: the group $\mathfrak{t}(G)=P\times D$, where $P$ is the $p$-Sylow subgroup of $G$, $D$ is a central subgroup, $1+\mathfrak I (P)$ is the $p$-Sylow subgroup of $V(FG)$ and $V'\leq 1+\mathfrak I (G')$. Moreover, if $D$ is a finite abelian group, then we have the following isomorphism between abelian groups \[ \begin{split} V(FG)/\big(1+\mathfrak{I}(P)\big) \cong V(FD) \times (\underbrace{G/\mathfrak{t}(G)\times\cdots\times G/\mathfrak{t}(G)}_n),\\ \end{split} \] where $n$ is the number of summands in the decomposition of $FD$ into a direct sum of fields. \end{corollary} \section{Proofs} We start our proof with the following important \begin{lemma}\label{L:1} Let $FG$ be the group algebra of a group $G$ such that $V(FG)$ is an Engel group. If $H$ is a subgroup of $\mathfrak{t}(G)$ and $H$ does not contain an element of order ${\rm char}(F)$, then $H$ is abelian, each subgroup of $H$ is normal in $G$ and each idempotent of $FH$ is central in $FG$. \end{lemma} \begin{proof} If $a\in \mathfrak{t}(G)$ and $g \in G\setminus N_{G}(\gp{a})$, then $x=\widehat{a}g(a-1)\not=0$ and $1 + x\in V(FG)$. A straightforward calculation shows that \[ (1+x, a, m) =1+x(a -1)^m, \qquad (m\geq 1). \] Since $V(FG)$ is an Engel group, $x(a-1)^s\ne 0$ and $x(a -1)^{s+1}=0$ for a suitable $s\in \mathbb{N}$. It follows that $x(a -1 )^sa^i =x(a-1)^s$ for each $i\geq 0$ and \[ x(a-1)^s\cdot \|a\|=x(a-1)^s(1+a+a^2+\cdots +a^{\|a\|-1})=0. \] Since ${\rm char}(F)$ does not divide $\|a\|$, we have that $x(a -1 )^s=0$, a contradiction. Hence every finite cyclic subgroup of $H$ is normal in $G$, so $H$ is either abelian or hamiltonian. If $H$ is a hamiltonian group, then using the same proof as in the second part of Lemma 1.1 of \cite{Bovdi_Khripta_3} (see p.122), we obtain a contradiction. Hence $H$ is abelian and each of its subgroups is normal in $G$. We claim that all idempotents of $LT$ are central in $L\gp{T,g}$, where $T$ is a finite abelian subgroup of $H$, $g\in G$ and $L$ is a prime subfield of the field $F$. Assume on the contrary that there exist a primitive idempotent $e\in LT$ and $g\in G$ such that $geg^{-1}\ne e$. The element $b=g^{-1}a^{-1}ga \neq 1$ for some $a \in T$ and \[ W=\gp{c^{-1}(g^{-1}cg)\mid c\in T}=\gp{T,g}'\vartriangleleft \gp{T,g} \] is a non-trivial finite abelian subgroup of $T$. Obviously each subgroup of $T\leq H$ is normal in $\gp{T,g}$, so the idempotent $f= \frac{1}{\|W\|}\widehat{\gp{W}}\in L[T]$ is central in $L\gp{T, g}$ and can be expressed as $f=f_1+\cdots+f_s$ in which $f_1,\ldots, f_s$ are primitive and mutually orthogonal idempotents of the finite dimensional semisimple algebra $L[T]$. The idempotent $e$ does not appear in the decomposition of $f$. Indeed, otherwise we have $ef=e$. If $e=\sum_i\alpha_it_i$, where $\alpha_i\in L$ and $t_i\in T$, then \[ g^{-1}(ef)g=g^{-1}egf=\sum_i\alpha_ig^{-1}t_igf=\sum_i\alpha_it_i (t_i, g)f=ef \] so $e=fe$ is central, a contradiction. If $ef\not=0$, then $ef=e$, again a contradiction. Thus $ef=0$. Consider $f_= \frac{1}{\|b\|}\widehat{\gp{b}}\in L[T]$. Since $\gp{b}\subseteq W$, the idempotent $f_$ appears in the decomposition of $f$, so $ef_=0$. Furthermore $geg^{-1}\in L[T]$ is primitive as an automorphism's image of the primitive idempotent $e$, so $ege=e(geg^{-1})g=0$. Evidently $(1 + eg)^{-1}=1- eg$ and \[ \begin{split} (1+eg,a)&=(1-eg)(1+ea^{-1}ga)\\&=(1-eg)(1+egb)=1+eg(b-1). \end{split} \] Now an easy induction on $n\geq 1$ shows that $(1 +eg,a,n) =1+eg(b -1)^n$. However $V(FG)$ is Engel, so there exists $m$ such that \[ eg(b - 1)^m\ne 0\qquad \text{and} \qquad eg(b -1)^{m+1}= 0. \] Thus $eg(b-1)^mb^i = eg(b -1)^m$ for any $i\geq 0$, which leads to the contradiction \[ \begin{split} eg(b -1)^m = eg(b -1)^m f_= eg(b -1)^{m-1}\big((b-1) f_\big)= 0. \end{split} \] Therefore, all idempotents of $FH$ are central in $FG$. \end{proof} \smallskip \begin{proof}[\underline{Proof of Theorem \ref{T:1}}] Let $N_$ be the (finite) set of all nilpotent elements of $FG\setminus 0$. If $x\in N_$ then the subset of nilpotent elements $\{\lambda x\mid \lambda\in F\}$ is finite, so $F$ is a finite field of ${\rm char}(F)=p$. Assume that there exist $u\in N_$ and $g\in G\setminus \mathfrak{t}(G)$ such that $gu=ug$. Since the right annihilator $L$ of the left ideal $\frak{I}_l(\gp{g})=\gp{g^i-1\mid i\in \mathbb{Z}}_{FG}$ is different from zero if and only if $\gp{g}$ is a finite group (see Proposition 2.7, \cite{Bovdi_book}, p.9), the set $\{ u(g^i-1)\mid i\in \mathbb{Z}\}$ is an infinite subset of $N_$, a contradiction. Let $0\not=u\in N_$. Clearly $g^{-i}ug^i\not=0$ and $g^{-i}ug^i\in N_$ for any $g\in G\setminus \mathfrak{t}(G)$ and $i\in \mathbb{Z}$. Since $N_$ is a finite set, at least for one $i\in \mathbb{Z}$ there exists $j\in \mathbb{Z}$ ($j\not=i$) such that $g^{-i}ug^i=g^{-j}ug^j$, so $g^{i-j}u=ug^{i-j}$, a contradiction. Consequently $G=\mathfrak{t}(G)$, i.e. $G$ is a torsion group. If $Syl_p(G)=\gp{1}$, then each subgroup of $\mathfrak{t}(G)(=G)$ is abelian and normal in $G$ by Lemma \ref{L:1}. Thus $FG$ is a direct sum of fields, so $N_*=\varnothing$, which is impossible. Let $P=Syl_p(G)\not=\gp{1}$. Evidently $\{h-1\mid h\in P\}$ is a finite subset of nilpotent elements, so $P$ is a finite subgroup in $G$. Assume that $P$ is not normal in $G$. Since $P<V(FG)$ is Engel and nilpotent, some element of $P$ must have a conjugate $w\in N_G(P)\setminus P$ (see \cite{Plotkin_book}, Lemma V.4.1.1, p.379 [p.307 in the English translation]). Consequently $P\lneqq\gp{P, w}$ and $\gp{P, w}$ is a finite $p$-group, a contradiction. Hence $P$ is a normal subgroup in $G$ and the subset of nilpotent elements \[ \{(h-1)g\lambda \mid h \in P,\; g\in G,\; \lambda \in F \} \] is finite, so $G$ is finite as well. Since every finite Engel group is nilpotent by Zorn's theorem (see \cite{Robinson}, 12.3.4, p.372), $G$ is a finite nilpotent group which is a direct product of its Sylow subgroups (see \cite{Hall_book}, 10.3.4, p.176) and $G'$ is a finite $p$-group (see \cite{Bovdi_engel}, Theorem 3.2, p.175). Moreover any $q$-Sylow subgroup ($q\ne p$) is an abelian normal subgroup in $G$ and each of its subgroups is also normal in $G$ by Lemma \ref{L:1}. Consequently the finite group $G=P\times A$, where $A=\cup_{q\not=p} Syl_q(G)$ is an abelian normal subgroup of $G$. Moreover, if $g\in G$ and $z\in A$, then \[ (z,g)=z^{-1}(g^{-1}zg)\in P\cap A=\gp{1} \] so $A$ is central. Since $FG$ is a finite algebra, $V(FG)$ is nilpotent (see \cite{Robinson}, 12.3.4, p.372). \end{proof} \smallskip \begin{proof}[\underline{Proof of Theorem \ref{T:2}}] Clearly $G'$ is locally nilpotent and the ideal $\mathfrak I (P)$ is locally nilpotent, hence $1+\mathfrak I (P)$ is a locally nilpotent $p$-group which coincides with the $p$-Sylow normal subgroup of $V(FG)$. In view of $G'\subseteq P$ we obtain that \begin{equation}\label{E:2} V(FG)/\big(1+\mathfrak I (P)\big)\cong V(F [G/P]),\quad V(FG)/\big(1+\mathfrak I (G')\big)\cong V(F [G/G']) \end{equation} are abelian groups. From the second isomorphism in (\ref{E:2}) it follows that $V'\leq 1+\mathfrak I (G')$. Let $Syl_q(G)$ be a Sylow $q$-subgroup ($q\not=p$) of $G$. According to Lemma \ref{L:1}, $Syl_q(G)$ is a normal abelian subgroup in $G$. Moreover, if $g\in G$ and $z\in Syl_q(G)$, then \[ (z,g)=z^{-1}(g^{-1}zg)\in P\cap Syl_q(G)=\gp{1} \] so $Syl_q(G)$ is central. Consequently, the group $\mathfrak{t}(G)$ is equal to $P\times D$ where $D=\times_{\substack{q\not=p}} Syl_q(G)\subseteq \zeta(G)$ and $\zeta(G)$ is the center of $G$. Define the subgroup $M$ of $G/P$ to be the group generated by the central subgroup $DP/P$ and by all elements of infinite order in $G/P$. Evidently, each element of $M$ can be expressed as a product of an element from $DP/P(\cong D)$ and an element of infinite order. Consequently, the subgroup $M$ of $G/P$ has no element of order $p$ and $M/D$ is a free abelian group. The algebra $F[M]$ can be expressed as a crossed product $S(K, F[D])$ of the group $K\cong M/D$ over the group algebra $F[D]$ (see \cite{Bovdi_book}, Lemma 11.1, p.63). Since $F[D]$ isomorphic to a central subalgebra of $S(K, F[D])$, the algebra $S(K, F[D])$ is a twisted group ring with $G$-basis $\{t_g\mid g\in G\}$ such that $t_gt_h=t_{gh}\mu_{g,h}$ for all $g,h\in G$, where $\mu_{g,h}\in \mu=\{\mu_{a,b}\in U(FD)\mid a,b\in G\}$ is the factor system of $S(K, FD)$ (see \cite{Bovdi_book}, Chapter 11). Any unit $u\in FM$ can be written as $u=\sum_{i=1}^s\alpha_it_{g_i}\in S(K, FD)$,\quad where $\alpha_i\in FD$. Define \[ B=\gp{supp(\alpha_1), \ldots, supp(\alpha_s)}\leq D. \] If $e_1,\ldots, e_n$ is a complete set of primitive, mutually orthogonal idempotents of the f.d. commutative algebra $FB$ such that $e_1+ \cdots +e_n=1$, then \[ FB=FBe_1\oplus \cdots \oplus FBe_n \quad \text{and} \quad FM=FMe_1\oplus \cdots \oplus FMe_n. \] The field $FBe_i$ is denoted by $F_i$. It is easy to see that \[u\in S(K, FB)=S(K, F_1\oplus \cdots\oplus F_n)=S(K, F_1)\oplus \cdots \oplus S(K, F_n). \] Since $K$ is an ordered group, all units in $S(K, F_i)$ are trivial (see \cite{Bovdi_Crossed}, Lemma 3, p.495) and $ue_i=\beta_ie_it_{h_i}$ for some $h_i\in K$ and $\beta_i\in FB$. Consequently \[ u=(\beta_1 e_1)t_{h_1}+(\beta_2 e_2) t_{h_2}+\cdots+(\alpha_n e_n) t_{h_n},\qquad (\beta_i\in FB,\, h_i\in K) \] so $U\big(S(K, F_i)\big)\cong U(F_i)\times K$. Note that the description of the group $V(FM)$ depends only on the subgroup $D$. Let us give all invariants of $V(FM)$ in the case when the abelian group $D$ is finite. Set $B=D$. Since $U(S(K, F_i))\cong U(F_i)\times K$ for $i=1,\ldots, n$, \[ \begin{split} U(F[M])& \cong U(S(K, F_1))\times \cdots \times U(S(K, F_n))\\ & \cong (U(F_1) \times K)\times \cdots \times (U(F_n)\times K)\\ & \cong (U(F_1) \times \cdots \times U(F_n))\times L,\\ \end{split} \] where $L$ is a direct product of $n$ copies of $K$. It follows that $U(FM)$ is an extension of the torsion-free abelian group $L$ by a central subgroup $U(FD)$. \end{proof} \smallskip \begin{proof}[\underline{Proof of Theorem \ref{T:3}}] Since $V(FG)$ is a locally nilpotent group, $G$ is also locally nilpotent. Let $S=\gp{f_1,\ldots, f_s\mid f_i\in V(FG)}$ be an f.~g. subgroup of $V(FG)$. Clearly \[ H=\gp{supp(f_1),\ldots, supp(f_s)} \] is an f.~g. nilpotent subgroup of $G$ and $S\leq V(FH)< V(FG)$. Hence we may restrict our attention to the subgroup $V(FH)$, where $H$ is an f.~g. nilpotent subgroup of $G$. Let $H$ be an f.~g. nilpotent group containing a $p$-element and let $g, h\in H$ such that $(g,h)\not=1$. Obviously $\mathfrak{t}(H)$ is finite (see \cite{Hall_2}, 7.7, p. 29) and the finite subgroup $Syl_p(\mathfrak{t}(H))$ is a direct factor of $\mathfrak{t}(H)$ (see \cite{Hall_book}, 10.3.4, p.176). Consequently, there exists $c\in \zeta(H)$ of order $p$, and $\widehat{c}=\sum_{i=0}^{p-1}c^i$ is a square-zero central element of $FG$. Since $L=\gp{g,h, 1+g\widehat{c}}$ is an f.~g. nilpotent subgroup of $V(FG)$, so $L$ is Engel and there exists $m\in \mathbb{N}$ such that the nilpotency class $cl(L)$ of $L$ is at most $p^m$. Let us show that $(g,h)$ is a $p$-element. Indeed, if $q=p^m$ then we have \[ 1=\big(1+g\widehat{c},h, q\big)= 1+\widehat{c}\sum_{i=0}^q (-1)^i\textstyle\binom {q}{i} g^{h^{q-i}}=1+\widehat{c}(g^{h^{q}}-g), \] because $\binom {q}{i} \equiv 0 \pmod{p}$ for $0 <i< q$. It follows that $\widehat{c}(g^{h^{q}}-g)=0$ and $(g,h^{q})\in supp(\widehat{c})$, which yields that $g^{h^{q}}=c^ig$ for some $0\leq i<p$. Hence \[ \big(\underbrace{h^{-q}\cdots h^{-q}}_p) g(\underbrace{h^{q}\cdots h^{q}}_p\big)=(c^i)^pg=g \] and $h^{p^{m+1}}\in C_G(g)$, so $h^{p^{m+1}}\in \zeta(L)$ for all $h\in L$ and $L'$ is a finite $p$-group by a theorem of Schur (see \cite{Robinson}, 10.1.4, p.287). Consequently $G'$ is a $p$-group. Conversely, it is sufficient to prove that if $H$ is an f.~g. nilpotent subgroup of $G$ such that $H'$ is a $p$-group then $V(FH)$ is nilpotent. We known that all subgroups and factor groups of $H$ are also f.~g. groups (\cite{Robinson}, 5.2.17, p.132). Moreover $\mathfrak{t}(H)=P\times D$ is finite (see \cite{Robinson}, 12.1.1, p.356) and $H/\mathfrak{t}(H)$ is a direct product of a finite number of infinite cyclic groups. Let $Syl_q(H)$ be a Sylow $q$-subgroup ($q\not=p$) of $H$. It is well known (see \cite{Hall_2}, 7.7, p. 29 and \cite{Hall_book}, 10.3.4, p.176) that $Syl_q(H)$ is normal in $H$. Moreover, if $g\in H$ and $z\in Syl_q(H)$, then \[ (z,g)=z^{-1}(g^{-1}zg)\in P\cap Syl_q(G)=\gp{1} \] so $Syl_q(H)$ is central. Consequently, $\mathfrak{t}(H)=P\times D$, where $D=\times_{\substack{q\not=p}} Syl_q(G)\subseteq \zeta(H)$ and (\ref{E:1}) holds by Theorem \ref{T:4}({\rm v}). Hence $V(F[G/P])$ is an extension of an f.~g. abelian group $L$ by a central subgroup $V(FD)$. By Lemma 2.4 from \cite{Bovdi_Khripta_2} (or by Lemma 2.1 in \cite{Bovdi_engel}, on p.176) every extension of a nilpotent group $1+\mathfrak I (P)$ by an f.~g. abelian group $L$ is nilpotent. Since $U(FD)$ is central so the group $U(FH)$ is nilpotent.\end{proof} \smallskip \begin{proof}[\underline{Proof of Theorem \ref{T:4}}] If $V(FG)$ is locally nilpotent, then it is Engel, so by Theorem 1.1 from \cite{Bovdi_engel}, the locally nilpotent group $G$ satisfies conditions (i)--(ii) of our theorem. Since a locally nilpotent group is an u.~p. group, the converse of our theorem follows from Theorem 2 in \cite{Khripta_2} and Theorem 1.1 in \cite{Bovdi_engel}. \end{proof}	181,471
TITLE: Proving that if $a,b$ are even, then $\gcd(a,b) = 2 \gcd(a/2, b/2)$ QUESTION [3 upvotes]: Prove that if $a, b$ are both even then $\gcd(a,b) = 2\cdot\gcd(a/2,b/2)$. Little confused here. I have tried the following but it's basically just repeating the proof unfortunately: $a = 2 \cdot a_1 $ and $ b = 2 \cdot b_1$ (Take factor of 2 out) $d = \gcd(a,b)$ $d = 2\gcd(a_1, b_1)$ $a/2 = a_1$ and $b/2 = b_1$ I've even confused myself! Could someone help! REPLY [1 votes]: With the notation $\rm\ (x,y) := gcd(x,y),\ $ we have $\rm\:2(m,n)\mid 2m,2n\:\Rightarrow\:\color{#c00}{2(m,n)\mid (2m,2n)},\:$ $\rm\ (2m,2n)\mid 2m,2n\:\Rightarrow\: (2m,2n)/2\mid m,n\:\Rightarrow\:(2n,2m)/2\mid (m,n)\:\Rightarrow\:\color{#c00}{(2n,2m)\mid 2\,(m,n)}.$ Remark $\ $ Note that $\rm\:(2m,2n)/2\in\Bbb Z\:$ since $\rm\: 2\mid 2n,2m\:\Rightarrow\:2\mid (2n,2m)$ The proof exploits to the hilt the fundamental universal property of the gcd i.e. $$\rm a\mid b,c\iff a\mid(b,c)$$ Note that the same proof works if we replace $2$ by any integer $\ne 0,\,$ yielding the GCD Distributive Law $\rm\ \ \ (ac,bc)\, =\, (a,b)c$ See here for a few proofs (by linearity (Bezout); universality; prime factorization), e.g. the following more efficient version of the original proof above. Theorem $\rm\ \ (a,b)\,c\, =\, (ac,bc)$ Proof $\rm\quad\: d\mid(a,b)\iff d\ \|\ a,b \iff dc\ \|\ ac,bc \iff dc\ \|\ (ac,bc) \iff d\ \|\ (ac,bc)/c$	54,829
\begin{document} \maketitle \footnote[0]{\hspace{-4ex}{\it Keywords}: Linear algebra, Invariant of matrices, Congruence, Zeropotent algebras} \footnote[0]{\hspace{-4ex}{\it 2010 Mathematics Subject Classification}: Primary 15A15; Secondary 15A72} \footnote[0]{\hspace{-4ex}Research of the first author is supported in part by JSPS KAKENHI Grant Number JP18K11172.} \footnote[0]{\hspace{-4ex}Research of the second author was supported in part by JSPS KAKENHI Grant Number JP25400120.} \begin{abstract} For a nonsingular matrix $A$, we propose the form $Tr(^t\!A A^{-1})$, the trace of the product of its transpose and inverse, as a new invariant under congruence of nonsingular matrices. \end{abstract} \section{Introduction} Let $K$ be a field. We denote the set of all $n\times n$ matrices over $K$ by $M(n,K)$, and the set of all $n\times n$ nonsingular matrices over $K$ by $GL(n,K)$. For $A, B\in M(n,K)$, $A$ is {\it similar} to $B$ if there exists $P\in GL(n,K)$ such that $P^{-1}AP=B$, and $A$ is {\it congruent} to $B$ if there exists $P\in GL(n,K)$ such that $^t\!PAP=B$, where $^t\!P$ is the transpose of $P$. A map $f: M(n,K) \rightarrow K$ is an {\it invariant under similarity of matrices} if for any $A\in M(n,K)$, $f(P^{-1}AP)=f(A)$ holds for all $P\in GL(n,K)$. Also, a map $g: M(n,K) \rightarrow K$ is an {\it invariant under congruence of matrices} if for any $A\in M(n,K)$, $g(^t\!PAP)=g(A)$ holds for all $P\in GL(n,K)$, and $h: GL(n,K) \rightarrow K$ is an {\it invariant under congruence of nonsingular matrices} if for any $A\in GL(n,K)$, $h(^t\!PAP)=h(A)$ holds for all $P\in GL(n,K)$. As is well known, there are many invariants under similarity of matrices, such as trace, determinant and other coefficients of the minimal polynomial. In case the characteristic is zero, rank is also an example. Moreover, much research on similarity (or {\it simultaneous conjugation}) has been done from the viewpoint of invariant theory for a group action on tuples of matrices, for example, see \cite{procesi}. On the other hand, few invariants under congruence of matrices or nonsingular matrices are known except for rank. Restricted to symmetric matrices over the reals, another exception would be from Sylvester's law of inertia, which states that the numbers of positive, negative, and zero eigenvalues are all invariants under congruence of real symmetric matrices. In this note, we propose a new invariant under congruence of general nonsingular matrices. The form is $\sigma(A)=Tr(^t\!A A^{-1})$, where $Tr$ denotes trace. \section{Invariance and other properties of $\sigma$} First of all, let us show that $\sigma$ is an invariant under congruence of nonsingular matrices. In fact, for any nonsingular matrix $X$, \begin{eqnarray} \sigma(^t\!XAX)&=&Tr(^t(^t\!XAX) (^t\!XAX)^{-1})\\ &=&Tr((^t\!X\,^t\!AX)(X^{-1}A^{-1}(^t\!X)^{-1}))\\ &=&Tr(^t\!X(^t\!AA^{-1})(^t\!X)^{-1})\\ &=&\sigma(A). \end{eqnarray} As an immediate corollary, any polynomial in $\sigma(A)$ over $K$ is also an invariant under congruence of nonsingular matrices. Moreover, any rational function in $\sigma(A)$ over $K$ is an invariant under congruence of nonsingular matrices for which its denominator does not vanish. Next, let us describe some other properties of $\sigma$. \begin{prop}\label{prop:properties} \noindent For any $n\times n$ nonsingular matrix $A$ over $K$, the following equalities hold. \begin{enumerate} \renewcommand{\labelenumi}{\rm{(\arabic{enumi})}} \item\label{enu:symmetry} $\sigma(^t\!A)=\sigma(A)$. \item\label{enu:inverse} $\sigma(A^{-1})=\sigma(A)$. \item\label{enu:adj} $\sigma(adj(A))=\sigma(A)$, where $adj(A)$ is the adjugate of $A$. \item\label{enu:scalar} $\sigma(cA)=\sigma(A)$ for any $c\in K\setminus\{0\}$. \item\label{enu:symmetric} $\sigma(A)=n$ if $A$ is symmetric. \end{enumerate} \end{prop} \noindent{\it Proof.} (\ref{enu:symmetry}) \begin{eqnarray} \sigma(^t\!A)&=&Tr(^t(^t\!A)\, (^t\!A)^{-1})\\ &=&Tr(A\,^t(A^{-1}))\\ &=&Tr(A^{-1}(^t\!A))\\ &=&\sigma(A) \end{eqnarray} (\ref{enu:inverse}) \begin{eqnarray} \sigma(A^{-1})&=&Tr(^t(A^{-1})\,(A^{-1})^{-1})\\ &=&Tr(^t(A^{-1})\,A)\\ &=&\sigma(A) \end{eqnarray} (\ref{enu:adj}) \begin{eqnarray} \sigma(adj(A))&=&Tr(^t(adj(A))\,(adj(A))^{-1})\\ &=&Tr(adj(^t\!A)\,(adj(A))^{-1})\\ &=&Tr((\|^t\!A\|(^t\!A)^{-1})\,(\|A\|^{-1}A))\\ &=&Tr((^t\!A)^{-1}\,A)\\ &=&Tr(^t(A^{-1})\,A)\\ &=&\sigma(A) \end{eqnarray} (\ref{enu:scalar}) For any $c\in K\setminus\{0\}$, \begin{eqnarray} \sigma(cA)&=&Tr(^t(cA)\,(cA)^{-1})\\ &=&Tr(c\,^t\!A\,c^{-1}A^{-1})\\ &=&\sigma(A) \end{eqnarray} (\ref{enu:symmetric}) If $^t\!A=A$, then \begin{eqnarray} \sigma(A)&=&Tr(AA^{-1})\\ &=&Tr(E)\\ &=&n \end{eqnarray} \section{Background on $\sigma$} Let us briefly describe the background that led to the discovery of $\sigma$. The authors of the present note with Sin-Ei Takahasi and Makoto Tsukada tried to classify three-dimensional {\it zeropotent} algebras up to isomorphism, where a zeropotent algebra is a nonassociative algebra in which the square of any element is zero, see \cite{KSTT,STTK}. We there expressed an algebra by its structure constants. For a three-dimensional zeropotent algebra $A$, let $\{e_1, e_2, e_3\}$ be a linear base of $A$. By zeropotency of $A$ it suffices to consider the structure constants for $e_2e_3, e_3e_1, e_1e_2$, namely, the algebra can be identified with the $3\times 3$ matrix $A$ such that $^t(e_2e_3\,\, e_3e_1\,\, e_1e_2)=A\,^t(e_1\,\, e_2\,\, e_3)$. We hereafter use the same symbol $A$ both for the matrix and for the algebra. The following proposition gives a criterion for two zeropotent algebras to be isomorphic. \begin{prop}[\cite{KSTT}]\label{prop:equiv} Let $A$ and $A'$ be three-dimensional zeropotent algebras over $K$. Then, $A$ and $A'$ are isomorphic if and only if there is a nonsingular matrix $X$ satisfying \begin{equation}\label{eq:equiv} A'=\frac{1}{\|X\|}\,\, \\^t\!XAX. \end{equation} In particular, if $K$ is algebraically closed, then $A$ and $A'$ are isomorphic if and only if there is a nonsingular matrix $X$ satisfying $A'=\,^t\!XAX$, that is, $A$ and $A'$ are congruent. \end{prop} Our strategy of classification was to divide zeropotent algebras into {\it curly} algebras and {\it straight} algebras. A zeropotent algebra is curly if the product of any two elements lies in the space spanned by these elements, otherwise straight. Let $K$ be an algebraically closed field. A straight algebra of rank 3 can be expressed by the {\it canonical form} \[ A(a, b, c)=\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix} \] with $a, b, c\in K$. During the course of classifying the straight algebras, we encountered the quantity $D=a^2+b^2+c^2-abc$, which seems strange because it is not homogeneous with respect to $a$, $b$, and $c$. That is, in case where $D\ne 0$ and $a^2+b^2-abc\ne 0$, we showed that $A(a,b,c)$ is congruent to $A(\sqrt{D},0,0)$. From this, it turned out that $D$ is an invariant under congruence of some $A(a,b,c)$-type matrices. Inspired by this fact, we got interested in elucidating the origin of $D$. From $D$, later, by backward reasoning from the canonical form to a general matrix using computer, the second author derived a more general homogeneous quantity $\dfrac{1}{\|A\|}\{ a_{13}^2a_{22}+3a_{12}a_{23}a_{31}-a_{21}a_{23}a_{31}+a_{22}a_{31}^2+a_{11}(a_{23}-a_{32})^2-a_{12}a_{31}a_{32}-a_{21}a_{31}a_{32}-a_{13}(a_{21}a_{23}+2a_{22}a_{31}-3a_{21}a_{32}+a_{12}(a_{23}+a_{32}))+a_{12}^2a_{33}-2a_{12}a_{21}a_{33}+a_{21}^2a_{33}\}$, for a $3\times 3$ matrix $A=(a_{ij})$ of rank 3. Let us denote this quantity by $\kappa(A)$. By computer we confirmed that $\kappa$ is certainly an invariant under congruence of general nonsingular matrices, but could not prove it mathematically. Since the original expression of $\kappa(A)$ was awkward and complicated, by an enormous amount of transformations of trial and error, the first author obtained the more beautiful form $3-\dfrac{1}{\|A\|}\displaystyle\sum_{i,j}(a_{ij}\widetilde{a_{ji}})$, where $\widetilde{a_{ij}}$ denotes the $(i,j)$ cofactor, that is, the product of $(-1)^{i+j}$ and the $(i,j)$ minor. Denoting $\dfrac{1}{\|A\|}\displaystyle\sum_{i,j}(a_{ij}\widetilde{a_{ji}})$ by $\sigma(A)$, by further reasoning he had $\sigma(A)=\dfrac{1}{\|A\|}Tr(A\, adj(^t\!A))$, and finally by using $^t\!A(adj(^t\!A))=\|^t\!A\|E$, reached the simple form $Tr(^t\!A A^{-1})$ as initially defined and then $\kappa(A)=3-Tr(^t\!A A^{-1})$. Through this expression we can now mathematically prove that $\kappa$ is an invariant under congruence of nonsingular matrices. Beyond the topic of three-dimensional zeropotent algebras, we found that in general, for any integer $n\geq 1$, $\sigma$ is an invariant under congruence of $n\times n$ nonsingular matrices over a general field. \bibliographystyle{amsplain}	24,327
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\begin{document} \title{The Landau-Pekar equations: Adiabatic theorem and accuracy } \author{Nikolai Leopold, Simone Rademacher, Benjamin Schlein and Robert Seiringer } \maketitle \begin{abstract} \noindent We prove an adiabatic theorem for the Landau-Pekar equations. This allows us to derive new results on the accuracy of their use as effective equations for the time evolution generated by the Fr\"ohlich Hamiltonian with large coupling constant $\alpha$. In particular, we show that the time evolution of Pekar product states with coherent phonon field and the electron being trapped by the phonons is well approximated by the Landau-Pekar equations until times short compared to $\alpha^2$. \end{abstract} \noindent \section{Introduction} We are interested in the dynamics of an electron in a ionic crystal. For situations in which the extension of the electron is much larger than the lattice spacing, Fr\"ohlich \cite{froehlich} derived a model which treats the crystal as a continuous medium and describes the polarization of the lattice as the excitations of a quantum field, called phonons. If the coupling between the electron and the phonons is large it is expected that the dynamics of the system can be approximated by the Landau-Pekar equations, a set of nonlinear differential equations which model the phonons by means of a classical field. The coupling parameter of the Fr\"ohlich model enters into the Landau-Pekar equations leading to a separation of times scales of the electron and the phonon field. This phenomenon, often referred to as adiabatic decoupling \cite{teufel}, is believed to be responsible for the classical behaviour of the radiation field.The physical picture one has in mind is that the electron is trapped in a cloud of slower phonons which increase the effective mass of the electron \cite{liebseiringer}. The goal of this paper is to compare the time evolution generated by the Fr\"ohlich Hamiltonian with the Landau-Pekar equations and to give a quantitative justification of the applied approximation. In particular, we will consider the evolution of factorized initial data, with a coherent phonon field and an electron trapped by the phonons and minimizing the corresponding energy. For such initial data, we show that the Landau-Pekar equations provide a good approximation of the dynamics, up to times short compared to $\alpha^2$, with $\alpha$ denoting the coupling between electron and phonons (space and time units are chosen so that the electron is initially localised in a volume of order one and has speed of order one). This result improves previous bounds in \cite{frankschlein, frankgang}, which only holds up to times of order $\alpha$ (but for more general initial data). Also, it extends the findings of \cite{griesemer}, which show a result similar to ours but only for initial data minimizing the Pekar energy functional (in this case, the solution of the Landau-Pekar equations remains constant). To prove our bound, we establish an adiabatic theorem for the solution of the Landau-Pekar equations. The idea of considering states with the electron trapped by the phonon field and to show an adiabatic theorem was first proposed in \cite{frankgang2,frankgang3}, where an adiabatic theorem is proved for a one-dimensional version of the Landau-Pekar equations. Apart from the restriction to the one-dimensional setting, the adiabatic theorem in \cite{frankgang2,frankgang3} differs from ours, because it compares the full solution of the Landau-Pekar equations with the solution of a limiting system of equations, independent of $\alpha$ (in Theorem \ref{thm:adiabatic} below, on the other hand, we only compare the electron wave function with the ground state of the Schr\"odinger operator associated with the phonon field determined by the Landau-Pekar equations; this is sufficient for our purposes). \section{Model and Results} We consider the Fr\"ohlich model which describes the interaction between an electron and a quantized phonon field. The state of the phonon field is represented by an element of the bosonic Fock space $\mathcal{F} \coloneqq \bigoplus_{n \geq 0} L^2(\mathbb{R}^3)^{\otimes_s^n}$, where the subscript $s$ indicates symmetry under the interchange of variables. The system is described by elements $\Psi_t \in \mathcal{H}$ of the Hilbert space \begin{align} \mathcal{H} \coloneqq L^2(\mathbb{R}^3) \otimes \mathcal{F}. \end{align} Its time evolution is governed by the Schr\"odinger equation \begin{align} \label{eq: Schroedinger equation} i \partial_t \Psi_t = H_{\alpha} \Psi_t \end{align} with the Fr\"ohlich Hamiltonian \begin{align} \label{eq: Froehlich Hamiltonian} H_{\alpha} \coloneqq - \Delta + \int d^3k \, \abs{k}^{-1} \big( e^{ik \cdot x} a_k + e^{-ik \cdot x} a_k^* \big) + \int d^3k \, a_k^* a_k . \end{align} Here, $a_k^$ and $a_k$ are the creation and annihilation operators in the Fock space $\mathcal{F}$, satisfying the commutation relations \begin{align} [a_k, a_{k'}^] &= \alpha^{-2} \delta(k - k') , \quad [a_k, a_{k'}] = [a_k^, a_{k'}^] = 0 \quad \text{for all} \, k, k' \in \mathbb{R}^3, \end{align} for a coupling constant $\alpha >0$. One should note that the Hamiltonian is written in the strong coupling units, which gives rise to the $\alpha$ dependence in the commutation relations. These units are related to the usual ones by rescaling all lengths by $\alpha$ and time by $\alpha^2$, see \cite[Appendix A]{frankschlein}. We will be interested in the limit $\alpha \rightarrow \infty$. Motivated by Pekar's Ansatz, we consider the evolution of initial states of product form \begin{align} \label{eq:product} \psi_0 \otimes W(\alpha^2 \varphi_0) \Omega. \end{align} Here $\Omega$ denotes the vacuum of the Fock space $\mathcal{F}$ and $W(f)$ for $f \in L^2 ( \mathbb{R}^3)$ denotes the Weyl operator given by \begin{align} \label{eq: definition Weyl operator} W(f) = \exp \left[ \int d^3k \, \left( f(k) a_k^* - \overline{f(k)} a_k \right) \right] . \end{align} Note that the Weyl operator is unitary and satisfies the shifting property with respect to the creation and annihilation operator, i.e. \begin{align} \label{eq:Weyl_comm} W^(f)a_k W(f) = a_k + \alpha^{-2} f(k), \quad W^(f)a^_k W(f) = a^_k + \alpha^{-2} \overline{f}(k) \end{align} for all $f \in L^2( \mathbb{R}^3)$. Due to the interaction the system will develop correlations between the electron and the radiation field and the solution of \eqref{eq: Schroedinger equation} will no longer be of product form. However, for an appropriate class of initial states we will show that it can be approximated up to times short compared to $\alpha^2$ (in the limit of large $\alpha$) by a product state $\psi_t \otimes W(\alpha^2 \varphi_t) \Omega$, with $(\psi_t,\varphi_t)$ being a solution of the Landau-Pekar equations \begin{align} \label{eq: Landau Pekar equations} \begin{cases} i \partial_t \psi_t &= \left[ - \Delta + \int d^3k \, \abs{k}^{-1} \left( e^{ ik \cdot x} \varphi_t(k) + e^{- ik \cdot x} \overline{\varphi_t(k)} \right) \right] \psi_t(x) , \\ i \alpha^2 \partial_t \varphi_t(k) &= \varphi_t(k) + \abs{k}^{-1} \int d^3x \, e^{- ik \cdot x} \abs{\psi_t(x)}^2 \end{cases} \end{align} with initial data $(\psi_0,\varphi_0)$. We define for $\varphi \in L^2(\mathbb{R}^3)$ the potential \begin{align} \label{eq: definition potential} \; V_{\varphi} (x) = \int d^3k \, \|k\|^{-1} \left( \varphi (k) e^{ik \cdot x} + \overline{\varphi (k)} e^{-ik \cdot x} \right) . \end{align} We are interested, in particular, in initial data of the form \eqref{eq:product} where the phonon field $\varphi$ is such that the Schr\"odinger operator \begin{align} \label{eq: hphit} h_{\varphi} &\coloneqq - \Delta + V_{\varphi} \end{align} has a non-degenerate eigenvalue at the bottom of its spectrum separated from the rest of the spectrum by a gap, and the electron wave function $\psi$ is a ground state vector of \eqref{eq: hphit}. \begin{assumption} \label{assumptions} Let $\varphi_0 \in L^2(\mathbb{R}^3)$ such that \begin{align} e(\varphi_0) \coloneqq \inf \lbrace\langle \psi, h_{\varphi_0} \psi \rangle : \psi \in H^1( \mathbb{R}^3), \\| \psi \\|_2 =1 \rbrace <0 . \end{align} \end{assumption} This assumption ensures the existence of a unique positive ground state vector $\psi_{\varphi_0}$ for $h_{\varphi_0}$ with corresponding eigenvalue separated from the rest of the spectrum by a gap of size $\Lambda (0) >0$. If we then consider solutions of \eqref{eq: Landau Pekar equations} with initial data $(\psi_{\varphi_0},\varphi_0)$ the spectral gap can be shown to persist at least for times of order $\alpha^2$. \begin{lemma} \label{lemma: spectral gap} Let $\varphi_0$ satisfy Assumption \ref{assumptions} and let $( \psi_t, \varphi_t) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$ denote the solution of the Landau-Pekar equations with initial value $( \psi_{\varphi_0}, \varphi_0 )$. Moreover, let \begin{align} \Lambda(t) := \inf_{\substack{\lambda \in \mathrm{spec}( h_{\varphi_t})\\ \lambda \not= e(\varphi_t)}} \vert e(\varphi_t) - \lambda \vert. \end{align} Then, for all $\Lambda$ with $0 < \Lambda < \Lambda(0) $ there is a constant $C_{\Lambda} >0$ such that, for all $\|t\| \leq C_\Lambda \alpha^2$, the Hamiltonian $h_{\varphi_t}$ has a unique positive and normalized ground state $\psi_{\varphi_t}$ with eigenvalue $e( \varphi_t)<0$, which is separated from the rest of the spectrum by a gap of size $\Lambda (t) \geq \Lambda$. \end{lemma} The Lemma is proven in Subsection \ref{sec: minimizer}. Using the persistence of the spectral gap, we can prove the following adiabatic theorem for the solution of the Landau-Pekar equations \eqref{eq: Landau Pekar equations}. As mentioned in the introduction, the idea of such result is based on \cite{frankgang2,frankgang3}, where an adiabatic theorem for the Landau-Pekar equations in one dimension is proved. \begin{theorem} \label{thm:adiabatic} Let $T >0$, $\Lambda >0$ and $( \psi_t, \varphi_t) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$ denote the solution of the Landau-Pekar equations with initial value $( \psi_{\varphi_0}, \varphi_0 ) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$. Assume that the Hamiltonian $h_{\varphi_t}$ has a unique positive and normalized ground state $\psi_{\varphi_t}$ and a spectral gap of size $\Lambda(t) > \Lambda$ for all $\|t\| \leq T$. Then, \begin{align} \label{eq: adiabatic theorem} \\| \psi_t - e^{- i \int_0^t du \; e( \varphi_u)} \psi_{\varphi_t} \\|_2^2 \leq C \Lambda^{-4} \alpha^{-4} \left( 1 + \left( 1 + \Lambda^{-1} \right)^2 \alpha^{-4} \|t\|^2 \right), \hspace{0.3cm} \forall \|t\| \leq T. \end{align} \end{theorem} \begin{remark} \label{remark: adiabatic theorem for short times} One also has $ \\| \psi_t - e^{- i \int_0^t du \; e( \varphi_u)} \psi_{\varphi_t} \\|_{2}^2 \leq C \alpha^{-2} \Lambda^{-1} \abs{t} $ for all $\abs{t} \leq T$. \end{remark} \begin{remark} Note that the proof of the theorem only requires the existence of the spectral gap $\Lambda>0$. Assuming $\Lambda$ to be of order one for times of order $\alpha^4$, the theorem shows that $\psi_t$ is well approximated by the ground state $\psi_{\varphi_t}$ for any $\|t\| \ll \alpha^{4}$. \end{remark} \begin{remark} Lemma \ref{lemma: spectral gap} shows that the existence of the ground state and the spectral gap for all times $\abs{t} \leq C_{\Lambda} \alpha^2$ can be inferred from Assumption \ref{assumptions}. In this case, \eqref{eq: bound main theorem} is valid for all $\abs{t} \leq C_{\Lambda} \alpha^2$ without any assumptions on $h_{\varphi_t}$ and $\Lambda(t)$ at times $t >0$. Theorem \ref{thm:adiabatic} implies that there exists $C_\Lambda, \widetilde{C}_\Lambda >0$ such that \begin{align} \label{eq: adiabatic theorem improved bound} \\| \psi_t - e^{- i \int_0^t du \; e( \varphi_u)} \psi_{\varphi_t} \\|_{2}^2 &\leq \widetilde{C_\Lambda} \alpha^{-4} \end{align} for all $\|t\| \leq C_\Lambda \alpha^2$. \end{remark} Using Theorem \ref{thm:adiabatic} we can show that the Landau-Pekar equations (\ref{eq: Landau Pekar equations}) provide a good approximation to the solution of the Schr\"odinger equation (\ref{eq: Schroedinger equation}), for initial data of the form (\ref{eq:product}), with $\varphi_0$ satisfying Assumption \ref{assumptions} and with $\psi_0 = \psi_{\varphi_0}$ being the ground state of the operator $h_{\varphi_0}$ defined as in (\ref{eq: hphit}). \begin{theorem} \label{theorem: main theorem} Let $\varphi_0$ satisfy Assumption \ref{assumptions} and $\alpha_0 >0$. Let $( \psi_t, \varphi_t) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$ denote the solution of the Landau-Pekar equations with initial data $( \psi_{\varphi_0}, \varphi_0 ) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$ and \begin{align} \omega(t) \coloneqq \alpha^2 \Im \scp{\varphi_t}{\partial_t \varphi_t} + \norm{\varphi_t}_2^2 . \end{align} Then, there exists a constant $C>0$ such that \begin{align} \label{eq: bound main theorem} \norm{e^{-i H_{\alpha} t} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega - e^{ - i \int_0^t du \, \omega(u)} \psi_t \otimes W(\alpha^2 \varphi_t) \Omega} &\leq C \alpha^{-1} \abs{t}^{1/2} \end{align} for all $\alpha \geq \alpha_0$. The constant $C>0$ depends only on $\alpha_0>0$ and the initial condition, i.e. $( \psi_{\varphi_0}, \varphi_0 ) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$ and the spectral gap $\Lambda (0)$ of $h_{\varphi_0}$. \end{theorem} \begin{remark} Theorem \ref{theorem: main theorem} shows that the Pekar ansatz is a good approximation for times small compared to $\alpha^2$. Note that even though $(\psi_t, \varphi_t)$ stay close to their initial values for these times (as shown in Theorem \ref{thm:adiabatic}), it is essential to use the time-evolved version in \eqref{eq: bound main theorem}. This is due to the large factor $\alpha^2$ in the Weyl operator $W(\alpha^2 \varphi_t)$, which leads to a very sensitive behavior of the state on $\varphi_t$. It remains an open problem to decided whether the Pekar product ansatz remains valid also for times of order $\alpha^2$ and larger. \end{remark} A first rigorous result concerning the effective evolution of the Fr\"ohlich polaron was obtained in \cite{frankschlein}, where the product $\psi_t \otimes W(\alpha^2 \varphi_0) \Omega$, with $\psi_t$ solving the linear equation \begin{align} i\partial_t \psi_t = -\Delta \psi_t + \int dk \, \|k\|^{-1} \left( e^{ik \cdot x} \varphi_0 (k) + e^{-ik \cdot x} \overline{\varphi_0 (k)} \right) \psi_t (x) \end{align} was proven to give a good approximation for the solution $e^{-i H_\alpha t} \psi_0 \otimes W (\alpha^2 \varphi_0) \Omega$ of the Schr\"odinger equation (\ref{eq: Schroedinger equation}), up to times of order one\footnote{In fact, a simple modification of the Gronwall argument in \cite{frankschlein} leads to convergence for times $\|t\| \ll \alpha$.}. This result was improved in \cite{frankgang}, where convergence towards (\ref{eq: Landau Pekar equations}) was established for all times $\|t\| \ll \alpha$ (the analysis in \cite{frankgang} also gives more detailed information on the solution of Schr\"odinger equation (\ref{eq: Schroedinger equation}); in particular, it implies convergence of reduced density matrices). Notice that the results of \cite{frankschlein,frankgang} hold for general initial data of the form (\ref{eq:product}), with no assumption on the relation between the initial electron wave function $\psi_0$ and the initial phonon field $\varphi_0$. Theorem \ref{theorem: main theorem} shows therefore that, under the additional assumption that $\psi_0 = \psi_{\varphi_0}$ the Landau-Pekar equations (\ref{eq: Landau Pekar equations}) provide a good approximation to (\ref{eq: Schroedinger equation}) for longer times (times short compared to $\alpha^2$). In this sense, Theorem \ref{theorem: main theorem} also extends the result of \cite{griesemer}, where the validity of the Landau-Pekar equations was established for times short compared to $\alpha^2$, but only for initial data $(\psi_0, \varphi_0)$ minimizing the Pekar energy functional (in this case, the solution of (\ref{eq: Landau Pekar equations}) is stationary, i.e. $(\psi_t, \varphi_t) = (\psi_0, \varphi_0)$ for all $t$). In fact, similarly to the analysis in \cite{griesemer}, we use the observation that the spectral gap above the ground state energy of $h_{\varphi_t}$ allows us to obtain bounds that are valid on longer time scales (it allows us to integrate by parts, after (\ref{eq:adia_1}) and after (\ref{eq:parts2}); this step is crucial to save a factor of $t$). The classical behaviour of a quantum field does not only appear in the strong coupling limit of the Fr\"ohlich polaron but has also been studied in other situations. In \cite{ginibrenironivelo} it was shown in case of the Nelson model that a quantum scalar field behaves classically in a certain limit where the number of field bosons becomes infinite while the coupling constant tends to zero. The emergence of classical radiation was also proven for the Nelson model with ultraviolet cutoff \cite{falconi, leopoldpetrat}, the renormalized Nelson model \cite{ammarifalconi} and the Pauli-Fierz Hamiltonian \cite{leopoldpickl} in situations in which a large number of particles weakly couple to the radiation field. The articles \cite{davies_1979, hiroshima_1998, teufel2} revealed in addition that quantum fields can sometimes be replaced by two-particle interactions if the particles are much slower than the bosons of the quantum field. \section{Preliminaries} In this section, we collect properties of the Landau-Pekar equations that are used in the proofs of Theorem \ref{thm:adiabatic} and Theorem \ref{theorem: main theorem}. For $\psi \in L^2( \mathbb{R}^3)$ we define the function \begin{align} \label{eq: sigma} \sigma_{\psi}(k) = \|k\|^{-1} \int d^3x \, e^{- i k \cdot x} \|\psi (x)\|^2 . \end{align} Next, we notice that the first and second derivative of the potential $V_{\varphi_t}$ are given by\footnote{We use the notation $\dot{f}$ to denote the derivative of a function $f$ with respect to time.} \begin{align} \label{eq: time derivative of the potential} \partial_t V_{\varphi_t}(x) = V_{\dot{\varphi}_t}(x) &= - \alpha^{-2} \int d^3k \, \abs{k}^{-1} \big( e^{i k \cdot x} i \varphi_t(k) + e^{- ik \cdot x} \overline{i \varphi_t(k)} \big) \nonumber \\ &\quad - i \alpha^{-2} \int d^3k \, \abs{k}^{-1} \big( e^{i k \cdot x} \sigma_{\psi_t}(k) - e^{- i k \cdot x} \sigma_{\psi_t}(-k) \big) = - \alpha^{-2} V_{i \varphi_t}(x) \end{align} and \begin{align} \label{eq: derivative potential with i-varphi} \partial_t V_{i \varphi_t}(x) &= V_{i \dot{\varphi}_t}(x) = V_{\alpha^{-2} ( \varphi_t + \sigma_{\psi_t} )}(x) = \alpha^{-2} V_{\varphi_t}(x) + \alpha^{-2} V_{\sigma_{\psi_t}}(x) . \end{align} Then we define the energy functional $\mathcal{E}: H^1(\mathbb{R}^3) \times L^2(\mathbb{R}^3) \rightarrow \mathbb{R}$ \begin{align} \mathcal{E} ( \psi, \varphi ) = \langle \psi, h_{\varphi} \psi \rangle + \\| \varphi \\|_2^2. \end{align} Using standard methods one can show that the Landau-Pekar equations are well posed and that the energy $\mathcal{E} ( \psi_t , \varphi_t )$ is conserved if $(\psi_t, \varphi_t)$ is a solution of \eqref{eq: Landau Pekar equations}. For a proof of the following Lemma see \cite[Appendix C]{frankgang}. \begin{lemma}[\cite{frankgang}, Lemma 2.1] \label{lemma: well posedness LP} For any $ ( \psi_0, \varphi_0) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$, there is a unique global solution $( \psi_t, \varphi_t)$ of the Landau-Pekar equations \eqref{eq: Landau Pekar equations}. The following conservation laws hold true \begin{align} \\| \psi_t \\|_{2} = \\| \psi_0 \\|_{2} \hspace{0.3cm} and \hspace{0.3cm} \mathcal{E} ( \psi_t, \varphi_t ) = \mathcal{E} ( \psi_0, \varphi_0 ) \hspace{0.3cm} \forall t \in \mathbb{R}^3. \end{align} Moreover, there exists a constant $C$ such that \begin{align} \\| \psi_t \\|_{H^1( \mathbb{R}^3)} \leq C, \hspace{0.3cm} \\| \varphi_t \\|_{2} \leq C \end{align} for all $\alpha >0$ and all $t \in \mathbb{R}$. \end{lemma} The next Lemma (also proven in \cite[Appendix B,C]{frankgang}) collects some properties of quantities occurring in the Landau-Pekar equations. \begin{lemma} \label{lemma: Potental} For $V_\varphi$ being defined as in \eqref{eq: definition potential} there exists a constant $C>0$ such that for every $ \psi \in H^1( \mathbb{R}^3)$ and $ \varphi \in L^2( \mathbb{R}^3)$ \begin{align} \\| V_\varphi \\|_6 \leq C \\| \varphi \\|_2 \hspace{0.3cm} and \hspace{0.3cm} \\| V_\varphi \psi \\|_2 \leq C \\| \varphi \\|_2 \; \\|\psi \\|_{H^1( \mathbb{R}^3)} . \end{align} Furthermore, for every $\delta >0$ there exists $C_\delta >0$ such that \begin{align} \pm V_\varphi \leq - \delta \Delta + C_\delta, \end{align} thus there exists $C>0$ such that \begin{align} \label{eq: bound for h-varphi} - \frac{1}{2} \Delta - C \leq h_{\varphi} \leq - 2 \Delta + C. \end{align} Let $\sigma_\psi$ be defined as in \eqref{eq: sigma}. Then, there exists $C>0$ such that \begin{align} \label{eq: bound sigma-psi} \\| \sigma_\psi \\|_{2} \leq C \\| \psi \\|_{H^1( \mathbb{R}^3)}^2 . \end{align} \end{lemma} \begin{remark} Let $T >0$, $\Lambda >0$ and $( \psi_t, \varphi_t) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$ denote the solution of the Landau-Pekar equations with initial value $( \psi_{\varphi_0}, \varphi_0 ) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$. Assume that the Hamiltonian $h_{\varphi_t}$ has a unique positive and normalized ground state $\psi_{\varphi_t}$ and a spectral gap of size $\Lambda(t) > \Lambda$ for all $t \leq T$. Lemma \ref{lemma: Potental} then implies the existence of constant such that \begin{align} \label{eq: h1 minimizer} \\| \psi_{\varphi_t} \\|_{H^1( \mathbb{R}^3)} \leq C \quad \forall \abs{t} \leq T. \end{align} \end{remark} \begin{proof}[Proof of Lemma \ref{lemma: Potental}] Recall the definition \eqref{eq: definition potential} of the potential $V_\varphi$. We write \begin{align} V_{\varphi} (x) = 2^{3/2} \pi^{-1/2} \; \Re \int \frac{d^3y}{\|x-y\|^2} \check{\varphi} (y) , \end{align} where $\check{\varphi}$ denotes the inverse Fourier transform defined for $\varphi \in L^1( \mathbb{R}^3)$ through \begin{align} \check{\varphi}(x) = (2 \pi)^{-3/2} \int d^3k \, e^{ ik \cdot x} \varphi(k). \end{align} The first inequality follows directly from the Hardy-Littlewood-Sobolev inequality \begin{align} \\| V_\varphi \\|_6 \leq C \\| \varphi \\|_2. \end{align} In order to prove the second inequality we use the first one and the H\"older inequality \begin{align} \\| V_\varphi \psi \\|_2 \leq \\| V_\varphi\\|_6 \; \\| \psi \\|_3 \leq C \\| \varphi \\|_2 \; \\| \psi \\|_3 . \end{align} Since the interpolation inequality together with the Sobolev inequality implies \begin{align} \\| \psi \\|_3 \leq \\| \psi \\|_2^{1/2} \; \\| \psi \\|_6^{1/2} \leq \\|\psi \\|_2^{1/2} \; \\|\nabla \psi \\|_2^{1/2}, \end{align} we obtain \begin{align} \\| V_\varphi \psi \\|_2 \leq C \\| \varphi \\|_2 \; \\| \nabla \psi \\|_2^{1/2} \; \\| \psi \\|_2^{1/2} \leq C \\| \varphi \\|_2 \\| \psi \\|_{H^1( \mathbb{R}^3)} . \end{align} The second operator inequality follows again from the Sobolev inequality. For this let $\psi \in H^1( \mathbb{R}^3)$, then for $\varepsilon >0$ \begin{align} \langle \psi, V_\varphi \psi \rangle \leq C \\| V_\varphi \\|_6 \\| \psi \\|_{12/5}^2 \leq C \\| V_\varphi \\|_6 \left( \varepsilon \\| \nabla \psi \\|_2^2 + \varepsilon^{-1} \\| \psi \\|_2^2 \right), \end{align} where we used the interpolation inequality. The first inequality of the Lemma implies \begin{align} \pm \langle \psi, V_\varphi \psi \rangle \leq \langle \psi, \left( - \varepsilon \Delta + C_\varepsilon \right) \psi \rangle, \end{align} and \eqref{eq: bound for h-varphi} follows. The last inequality of the Lemma follows from the observation that \begin{align} \\| \sigma_\psi \\|^2_{2} &= \int \frac{dk}{\|k\|^2} \int dx dy \;\| \psi (x)\|^2 \| \psi (y)\|^2 e^{ik \cdot (x-y)} \nonumber \\ &= 2 \pi^2 \int dxdy \frac{\| \psi (x)\|^2 \| \psi(y)\|^2}{\|x-y\|} \leq C \\| \|\psi \|^2 \\|_{6/5}^2 , \end{align} where we used again the Hardy-Littlewood-Sobolev inequality. As before, the interpolation and the Sobolev inequality imply \eqref{eq: bound sigma-psi}. \section{Proof of the adiabatic theorem} \subsection{The ground state $\psi_{\varphi_t}$} \label{sec: minimizer} Before proving the adiabatic theorem, we show that the spectral gap of $h_{\varphi_t}$ does not close for times of order $\alpha^2$. In particular, we show the existence of the ground state $\psi_{\varphi_t}$. \begin{lemma} \label{lemma: existence minimizer} Let $\varphi_0$ satisfy Assumption \ref{assumptions}. Then, there exists a unique positive and normalized ground state $\psi_{\varphi_0}$ of $h_{\varphi_0} = - \Delta + V_{\varphi_0}$. Let $( \psi_t, \varphi_t) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$ denote the solution of the Landau-Pekar equations \eqref{eq: Landau Pekar equations} with initial value $( \psi_{\varphi_0}, \varphi_0 )$. There exists $C>0$ such that for all $\|t\| \leq C \alpha^2$, there exists a unique, positive and normalized ground state $\psi_{\varphi_t}$ of $h_{\varphi_t} = - \Delta + V_{\varphi_t}$ with corresponding eigenvalue $e( \varphi_t) < 0$. It satisfies \begin{align} \partial_t \psi_{\varphi_t} = \alpha^{-2} R_t V_{i \varphi_t} \psi_{\varphi_t} \hspace{0.3cm} \mathrm{with} \hspace{0.3cm} R_t = q_t (h_{\varphi_t} - e(\varphi_t))^{-1} q_t , \end{align} where $q_{t} = 1- \| \psi_{\varphi_t} \rangle \langle \psi_{\varphi_t} \|$ denotes the projection onto the subspace of $L^2 ( \mathbb{R}^3)$ orthogonal to the span of $\psi_{\varphi_t}$. \end{lemma} \begin{proof} Lemma \ref{lemma: well posedness LP} and Lemma \ref{lemma: Potental} imply that $V_{\varphi_t} \in L^6( \mathbb{R}^3)$ for all $t \in \mathbb{R}$. The existence of the ground state $\psi_{\varphi_t}$ at time $t=0$ then follows from the negativity of the infimum of the spectrum (see \cite[Theorem 11.5]{liebloss}). In order to prove the existence of the ground state $\psi_{\varphi_t}$ of $h_{\varphi_t}$ at later times, it suffices to show that $e( \varphi_t)$ is negative. For this we pick the ground state $\psi_{\varphi_0}$ at time $t=0$ and estimate \begin{align} \label{eq: bound lowest eigenvalue} e( \varphi_t) &\leq \langle \psi_{\varphi_0}, ( - \Delta + V_{\varphi_t} ) \psi_{\varphi_0} \rangle = e( \varphi_0) - \alpha^{-2} \int_0^t ds \; \langle \psi_{\varphi_0}, V_{i \varphi_s} \psi_{\varphi_0} \rangle \nonumber \\ &\leq e( \varphi_0) + C \int_0^t ds\; \alpha^{-2} \\| \varphi_s \\|_2 \; \\| \psi_{\varphi_0} \\|_{H^1( \mathbb{R}^3)}^2 \leq e( \varphi_0) + C \|t\| \alpha^{-2}, \end{align} by means of \eqref{eq: time derivative of the potential}, Lemma \ref{lemma: well posedness LP} and Lemma \ref{lemma: Potental}. Thus if we restrict our consideration to times $\|t\| < C^{-1} \|e( \varphi_0)\| \alpha^{2} $, we conclude that $e( \varphi_t) <0$. The ground state $\psi_{\varphi_t}$ satisfies \begin{align} \label{eq:deriv minimizer 1} 0= \left(h_{\varphi_t} - e( \varphi_t ) \right) \psi_{\varphi_t} \end{align} Differentiating both sides of the equality with respect to the time variable leads to \begin{align} \label{eq:deriv minimizer 2} 0 = \left( \dot{h}_{\varphi_t} - \dot{e}( \varphi_t) \right) \psi_{\varphi_t} + \left(h_{\varphi_t} - e( \varphi_t ) \right) \dot{\psi}_{\varphi_t}. \end{align} On the one hand $\dot{h}_{\varphi_t} = V_{\dot{\varphi}_t} =- \alpha^{-2} V_{i \varphi_t}$ by means of \eqref{eq: time derivative of the potential} and on the other hand, the Hellmann-Feynman theorem implies \begin{align} \label{eq: time-derivative lowest eigenvalue} \dot{e}( \varphi_t) = \langle \psi_{\varphi_t} , \dot{h}_{\varphi_t} \psi_{\varphi_t} \rangle =- \alpha^{-2} \langle \psi_{\varphi_t} ,V_{i \varphi_t} \psi_{\varphi_t} \rangle \end{align} so that \eqref{eq:deriv minimizer 2} becomes \begin{align} 0= -\alpha^{-2} q_t V_{i \varphi_t} \psi_{\varphi_t} + \left(h_{\varphi_t} - e( \varphi_t ) \right) \dot{\psi}_{\varphi_t}. \end{align} Since $\psi_{\varphi_t}$ is chosen to be real and normalized for all $t \in \mathbb{R}$, it follows that $\scp{\psi_{\varphi_t}}{\dot{\psi}_{\varphi_t}} = 0$ for all $t \in \mathbb{R}$. Hence, \begin{align} \dot{\psi}_{\varphi_t} = \alpha^{-2} q_t ( h_{\varphi_t} - e( \varphi_t))^{-1} q_t V_{i \varphi_t} \psi_{\varphi_t} = \alpha^{-2} R_t V_{i \varphi_t} \psi_{\varphi_t} . \end{align} \end{proof} Using the Lemma above, we prove Lemma \ref{lemma: spectral gap}. \begin{proof}[Proof of Lemma \ref{lemma: spectral gap}.] By the min-max principle \cite[Theorem 12.1]{liebloss}, the first excited eigenvalue of $h_{\varphi_t}$ (or the bottom of the essential spectrum) is given by \begin{align} \label{eq:e1} e_1(t) = \inf_{\substack{A \subset L^2( \mathbb{R}^3) \\ \mathrm{dim} A=2}} \sup_{\substack{\psi \in A \\ \\| \psi \\|_2 =1} } \langle\psi, h_{\varphi_t} \psi \rangle . \end{align} For any $\psi \in L^2( \mathbb{R}^3)$ with $\\| \psi \\|_2 =1$ we have by Lemma \ref{lemma: Potental} \begin{align} \langle \psi, h_{\varphi_t} \psi \rangle =& \langle \psi, h_{\varphi_0} \psi \rangle - \alpha^{-2} \int_0^t ds \; \langle \psi, V_{i \varphi_s} \psi \rangle \nonumber \\ \geq& \langle \psi, h_{\varphi_0} \psi \rangle - C \|t\| \alpha^{-2} \sup_{\|s\| \leq \|t\|} \\| \varphi_s \\|_2 \\| \nabla \psi \\|_2^2 - C \|t\| \alpha^{-2} \sup_{\|s\| \leq \|t\|} \\| \varphi_s \\|_2 \\| \psi\\|_2^2 \nonumber \\ \geq& (1-2 C \|t\| \alpha^{-2} ) \langle \psi, h_{\varphi_0} \psi \rangle - C \|t\| \alpha^{-2} \end{align} Inserting in \eqref{eq:e1}, we conclude that \begin{align} e_1(t) \geq (1-2C\|t\|\alpha^{-2}) e_1(0) - C \|t\| \alpha^{-2} . \end{align} With $e(\varphi_t) \leq e(\varphi_0) + C \|t\| \alpha^{-2}$ (see \eqref{eq: bound lowest eigenvalue}), we obtain \begin{align} \Lambda (t) \geq \Lambda (0) - C \|t\| \alpha^{-2} . \end{align} \end{proof} Using the persistence of the spectral gap, the resolvent $R_t = q_t \left( h_{\varphi_t}- e( \varphi_t) \right)^{-1} q_t$ can be estimated as follows. \begin{lemma} \label{lemma: resolvent} Let $T >0$, $\Lambda >0$ and $( \psi_t, \varphi_t) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$ denote the solution of the Landau-Pekar equations with initial value $( \psi_{\varphi_0}, \varphi_0 ) \in H^1( \mathbb{R}^3) \times L^2( \mathbb{R}^3)$. Assume that the Hamiltonian $h_{\varphi_t}$ has a unique positive and normalized ground state $\psi_{\varphi_t}$ with $e(\varphi_t) < 0$ and a spectral gap of size $\Lambda(t) > \Lambda$ for all $t \leq T$. Then, for all $\|t\| \leq T$ \begin{align} \\| R_t \\| \leq \Lambda^{-1}, \hspace{0.5cm} \\| ( - \Delta +1)^{1/2} R_t^{1/2} \\| \leq C ( 1 + \Lambda^{-1})^{1/2}, \end{align} and \begin{align} \\| \dot{R}_t \\| \leq C \Lambda^{-3/2}\alpha^{-2} ( 1 + \Lambda^{-1})^{1/2} \end{align} with $C >0$ depending only on $\varphi_0$. \end{lemma} \begin{proof} Since the spectral gap is at least of size $\Lambda>0$ for times $\|t\| \leq T$, it follows that \begin{align} \\| R_t \\| \leq \Lambda^{-1}. \end{align} To prove the second inequality, we estimate for arbitrary $\psi \in L^2( \mathbb{R}^3)$ \begin{align} \\|( - \Delta +1)^{1/2} R_t^{1/2} \psi \\|^2_2 &= \langle \psi, R_t^{1/2} ( - \Delta +1) R_t^{1/2} \psi \rangle \leq C \langle \psi, R_t^{1/2} \left( h_{\varphi_t} + 1\right) R_t^{1/2} \psi \rangle,\label{eq:form bound p^2} \end{align} where the last inequality follows from Lemma \ref{lemma: Potental}. Thus, \begin{align} \\|(- \Delta +1)^{1/2} R_t^{1/2} \psi \\|^2_2 &\leq C \langle \psi, R_t^{1/2} \left( h_{\varphi_t} - e(\varphi_t) + e(\varphi_t) + 1\right) R_t^{1/2} \psi \rangle \nonumber \\ &= C \langle \psi, \big( q_t + ( e(\varphi_t) + 1 ) R_t \big) \psi \rangle \nonumber \\ &\leq C \langle \psi, \big( 1 + R_t \big) \psi \rangle \end{align} since $e( \varphi_t) <0$ for all $\|t\| \leq T$ by assumption. The gap condition then implies \begin{align} \label{eq: resolvent sqrt p } \\|(- \Delta +1)^{1/2} R_t^{1/2}\psi \\|^2_2 \leq C ( 1 + \Lambda^{-1} ) . \end{align} In order to prove the third bound of the Lemma we calculate (with $p_t = 1 - q_t$) \begin{align} \dot{R}_t &=- \alpha^{-2} p_t V_{i\varphi_t} R_t^2 - \alpha^{-2} R_t^2 V_{i \varphi_t} p_t + q_t \left( \partial_t ( h_{\varphi_t} - e( \varphi_t))^{-1} \right) q_t \end{align} by means of the Leibniz rule and Lemma \ref{lemma: existence minimizer}. With the resolvent identities, \eqref{eq: time derivative of the potential} and \eqref{eq: time-derivative lowest eigenvalue} this becomes \begin{align} \label{eq: derivative rho} \dot{R}_t &=- \alpha^{-2} p_t V_{i\varphi_t} R_t^2 - \alpha^{-2} R_t^2 V_{i \varphi_t} p_t - q_t ( h_{\varphi_t} - e( \varphi_t))^{-1} \left( \dot{h}_{\varphi_t} - \dot{e}( \varphi_t) \right) ( h_{\varphi_t} - e( \varphi_t))^{-1} q_t \nonumber \\ &=- \alpha^{-2} p_t V_{i\varphi_t} R_t^2 - \alpha^{-2} R_t^2 V_{i \varphi_t} p_t + \alpha^{-2} R_t \left( V_{i \varphi_t} - \langle \psi_{\varphi_t}, V_{i \varphi_t} \psi_{\varphi_t} \rangle \right) R_t . \end{align} Hence, Lemma \ref{lemma: Potental} leads to \begin{align} \\| \dot{R}_t \\| \leq& C \alpha^{-2} \\| V_{i \varphi_t} R_t \\| \; \\| R_t\\| + C \alpha^{-2} \\| R_t \\|^2 \; \\| \psi_{\varphi_t} \\|_{H^1( \mathbb{R}^3)}^2 \nonumber \\ \leq& C \alpha^{-2} \\| (- \Delta +1)^{1/2} R_t \\| \; \\| R_t \\| + \alpha^{-2} \\| R_t \\|^2 \; \\| \psi_{\varphi_t} \\|_{\Hone}^2 \nonumber \\ \leq& C \Lambda^{-3/2}\alpha^{-2} ( 1 + \Lambda^{-1} )^{1/2} + C \alpha^{-2} \Lambda^{-2} , \end{align} for all $\|t\| \leq T$, where we used \eqref{eq: h1 minimizer} and the second bound of the Lemma. \end{proof} \subsection{Proof of Theorem \ref{thm:adiabatic}} In the following we denote $\widetilde{\psi}_{\varphi_t} = e^{-i \int_0^t du \; e( \varphi_u)} \psi_{\varphi_t}$. The fundamental theorem of calculus implies that \begin{align} \\| \psi_t - \widetilde{\psi}_{\varphi_t} \\|_{2}^2 =& -\int_0^t ds \; \frac{d}{ds} \; 2 \; \Re \langle \psi_s, \widetilde{\psi}_{\varphi_s} \rangle \nonumber \\ =&-2 \int_0^t ds \; \Re \langle -i h_{\varphi_s} \psi_s, \widetilde{\psi}_{\varphi_s} \rangle \nonumber \\ & - 2 \int_0^t ds \; \Re \langle \psi_s, \left( -i e ( \varphi_s) + \alpha^{-2} R_s V_{i \varphi_s}\right) \widetilde{\psi}_{\varphi_s} \rangle \nonumber \\ =& -2 \alpha^{-2} \int_0^t ds \; \Re \langle \psi_s, \;R_s V_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \rangle, \label{eq:adia_1a} \end{align} where we used that $h_{\varphi_s} \widetilde{\psi}_{\varphi_s} = e( \varphi_s) \widetilde{\psi}_{\varphi_s}$ and Lemma \ref{lemma: existence minimizer} to compute the derivative of the ground state $\psi_{\varphi_s}$. Using the Cauchy-Schwarz inequality, Lemma \ref{lemma: spectral gap} and Lemma \ref{lemma: Potental} together with Lemma \ref{lemma: resolvent} and \eqref{eq: h1 minimizer}, we obtain the inequality from Remark \ref{remark: adiabatic theorem for short times}, i.e. a bound of order $ \alpha^{-2} \|t\|$. In the following we shall improve this bound. We define $\widetilde{\psi}_s := e^{i \int_0^s d\tau \; e ( \varphi_\tau)} \psi_s$ satisfying \begin{align} i\partial_s \widetilde{\psi}_s = \left( h_{\varphi_s} - e( \varphi_s) \right) \widetilde{\psi}_s \end{align} and write \eqref{eq:adia_1a} as \begin{align} \\| \psi_t - \widetilde{\psi}_{\varphi_t} \\|_{2}^2 =&-2 \alpha^{-2} \int_0^t ds \; \Re \langle q_s \widetilde{\psi}_s, \; R_s V_{i \varphi_s} \psi_{\varphi_s} \rangle . \label{eq:adia_1} \end{align} Then, we exploit that the time derivative of $\widetilde{\psi}_s$ is of order one while the time derivatives of $R_s$, $V_{i \varphi_s}$ and $\psi_{\varphi_s}$ are of order $\alpha^{-2}$ (compare also with \cite[p.9]{teufel}). We observe that \begin{align} i \partial_s \left( \widetilde{\psi}_s - \psi_{\varphi_s} \right) = \left( h_{\varphi_s} - e(\varphi_s) \right) \widetilde{\psi}_s + i \alpha^{-2} R_s V_{i \varphi_s} \psi_{\varphi_s} \end{align} Hence, \begin{align} q_s \widetilde{\psi}_s = R_s i \partial_s \left( \widetilde{\psi}_s - \psi_{\varphi_s} \right) - i \alpha^{-2} R_s^2 V_{i \varphi_s} \psi_{\varphi_s}. \end{align} Plugging this identity into \eqref{eq:adia_1} and using that $\Im \scp{\psi_{\varphi_s}}{V_{i \varphi_s} R_s^3 V_{i \varphi_s} \psi_{\varphi_s}} = 0$, we obtain \begin{align} \\| \psi_t - \widetilde{\psi}_{\varphi_t} \\|_{2}^2 =& - 2 \alpha^{-2} \int_0^t ds \; \Im \langle \partial_s ( \widetilde{\psi}_s - \psi_{\varphi_s} ) , \; R_s^2 V_{i \varphi_s} \psi_{\varphi_s} \rangle . \end{align} Integrating by parts and using the initial condition $\psi_0 = \psi_{\varphi_0}$ lead to \begin{align} \\| \psi_t - \widetilde{\psi}_{\varphi_t} \\|_{2}^2 =&2\alpha^{-2} \int_0^t ds\; \Im \langle R_s \left( \widetilde{\psi}_s - \psi_{\varphi_s} \right) , \; \partial_s \left( R_s^2 V_{i\varphi_s} \psi_{\varphi_s}\right) \rangle \nonumber \\ & -2 \alpha^{-2} \Im \langle \left( \widetilde{\psi}_t - \psi_{\varphi_t} \right) , \;R_t^2 V_{i\varphi_t} \psi_{\varphi_t} \rangle. \end{align} The Leibniz rule with \eqref{eq: derivative potential with i-varphi} and Lemma \ref{lemma: existence minimizer} leads to \begin{subequations} \begin{align} \\| \psi_t - \widetilde{\psi}_{\varphi_t} \\|_{2}^2 =& - 2 \alpha^{-2} \Im \langle R_t \left(\psi_t - \widetilde{\psi}_{\varphi_t} \right) , \;R_t V_{i \varphi_t} \widetilde{\psi}_{\varphi_t} \rangle \label{eq:adiab_2,1}\\ &+ 2 \alpha^{-2} \int_0^t ds \; \Im \langle R_s \left( \psi_s - \widetilde{\psi}_{\varphi_s} \right) , \; \left( \partial_s R_s^2 \right) V_{i \varphi_s} \widetilde{\psi}_{\varphi_s}\rangle \label{eq:adiab_2,2} \\ &+ 2 \alpha^{-4} \int_0^t ds \; \Im \langle R_s \left( \psi_s - \widetilde{\psi}_{\varphi_s} \right) , \; R_s V_{\varphi_s + \sigma_{\psi_s}} \widetilde{\psi}_{\varphi_s}\rangle \label{eq:adiab_2,4} \\ &+ 2 \alpha^{-4} \int_0^t ds \; \Im \langle R_s \left( \psi_s - \widetilde{\psi}_{\varphi_s} \right) , \; \left(R_s V_{i \varphi_s} \right)^2 \widetilde{\psi}_{\varphi_s}\rangle \label{eq:adiab_2,5} . \end{align} \end{subequations} Using Lemma \ref{lemma: Potental} the first term can be estimated by \begin{align} \vert \eqref{eq:adiab_2,1} \vert &\leq C \alpha^{-2} \\| R_t \\|^2 \\| V_{i \varphi_t} \widetilde{\psi}_{\varphi_t} \\|_2 \norm{\psi_t - \widetilde{\psi}_{\varphi_t} }_2 \nonumber \\ &\leq C \alpha^{-2} \\| R_t \\|^2 \\| \widetilde{\psi}_{\varphi_t} \\|_{H^1( \mathbb{R}^3)} \\| \varphi_t \\|_2 \norm{ \psi_t - \widetilde{\psi}_{\varphi_t} }_2. \end{align} On the one hand Lemma \ref{lemma: well posedness LP} and \eqref{eq: h1 minimizer} show that $\\| \varphi_t \\|_2$ and $ \\| \psi_{\varphi_t} \\|_{\Hone}$ are uniformly bounded in time. On the other hand Lemma \ref{lemma: resolvent} implies that the resolvent $R_t$ is bounded for all times $\|t\| \leq T$, so that we obtain \begin{align} \vert \eqref{eq:adiab_2,1} \vert &\leq C \Lambda^{-2} \alpha^{-2} \norm{ \psi_t - \widetilde{\psi}_{\varphi_t} }_2 \leq \frac{1}{2} \norm{ \psi_t - \widetilde{\psi}_{\varphi_t}}_2^2 + C \Lambda^{-4} \alpha^{-4} , \hspace{0.5cm} \forall \; \|t\| \leq T . \end{align} Similarly, we bound the second term by \begin{align} \vert \eqref{eq:adiab_2,2} \vert &\leq C \alpha^{-2} \int_0^t ds \; \\| R_s \\| \; \\| V_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \\|_2 \; \\| \dot{R}_s \\| \norm{ \psi_s - \widetilde{\psi}_{\varphi_s} }_2 \nonumber \\ &\leq C \alpha^{-4} \Lambda^{-5/2} ( 1 + \Lambda^{-1} )^{1/2} \int_0^t ds \, \norm{ \psi_s - \widetilde{\psi}_{\varphi_s} }_2 \end{align} for all $\|t\| \leq T$, using Lemma \ref{lemma: well posedness LP}, Lemma \ref{lemma: Potental} and Lemma \ref{lemma: resolvent}. The third term \eqref{eq:adiab_2,4} can be bounded using $\\| \sigma_{\psi_t} \\|_2 \leq C \\| \psi_t \\|_{\Hone}^2 \leq C$ by Lemma \ref{lemma: well posedness LP}. We find \begin{align} \vert \eqref{eq:adiab_2,4} \vert \leq C \Lambda^{-2} \alpha^{-4} \int_0^t ds \, \norm{ \psi_s - \widetilde{\psi}_{\varphi_s} }_2, \hspace{0.5cm} \forall \; \|t\| \leq T . \end{align} Using the same ideas we estimate the last term by \begin{align} \vert \eqref{eq:adiab_2,5} \vert &\leq C \alpha^{-4} \int_0^t ds \; \\| R_s \\| \; \\| V_{i \varphi_s} R_s \\|^2 \norm{ \psi_s - \widetilde{\psi}_{\varphi_s} }_2 \nonumber \\ &\leq C \Lambda^{- 1} \alpha^{-4} \int_0^t ds \; \\| V_{i \varphi_s} R_s \\|^2 \norm{\psi_s - \widetilde{\psi}_{\varphi_s} }_2 . \end{align} Lemma \ref{lemma: Potental} implies that for all $\psi \in H^1( \mathbb{R}^3)$ \begin{align} \\| V_{i \varphi_s} \psi \\|_2 \leq C \\| \varphi_s \\|_2 \; \\| \psi \\|_{H^1( \mathbb{R}^3)} \end{align} and therefore that $ \\| V_{i \varphi_s} (- \Delta +1)^{-1/2} \\| \leq C$ for all $\|t\| \leq T$ by Lemma \ref{lemma: Potental}. Hence \begin{align} \\| V_{i \varphi_s} R_s \\| \leq C \\| (- \Delta +1)^{1/2} R_s \\| \leq C \Lambda^{-1/2} ( 1 + \Lambda^{-1})^{1/2}, \end{align} for all $\|t\| \leq T$ where we used Lemma \ref{lemma: resolvent} for the last inequality. Thus, \begin{align} \vert \eqref{eq:adiab_2,5} \vert \leq C \alpha^{-4} \Lambda^{-2} ( 1 + \Lambda^{-1}) \int_0^t ds \, \norm{\psi_s - \widetilde{\psi}_{\varphi_s} }_2, \hspace{0.5cm} \forall \; \|t\| \leq T, \end{align} In total, this leads to the estimate \begin{align} \\| \psi_t - \widetilde{\psi}_{\varphi_t} \\|_{2}^2 &\leq C \Lambda^{-4} \alpha^{-4} + C \Lambda^{-2} \left( 1 + \Lambda^{-1} \right) \alpha^{-4}\int_0^t ds \, \\| \psi_s - \widetilde{\psi}_{\varphi_s} \\|_{2} . \end{align} A Gronwall type argument then leads to \begin{align} \\| \psi_t - \widetilde{\psi}_{\varphi_t} \\|_2^2 \leq C \Lambda^{-4} \alpha^{-4} + C \Lambda^{-4} \left( 1 + \Lambda^{-1} \right)^2 \alpha^{-8} \|t\|^2 . \end{align} \end{proof} \section{Accuracy of the Landau-Pekar equations} \subsection{Preliminaries} For notational convenience we define \begin{align} \Phi_x = \int d^3 k \; \|k\|^{-1} \left( e^{i k \cdot x}a_k + e^{- ik \cdot x} a_k^* \right) = \Phi_x^+ + \Phi_x^- \end{align} with \begin{align} \Phi_x^+ = \int d^3 k \; \|k\|^{-1} e^{i k \cdot x} a_k, \hspace{0.5cm} \text{and} \hspace{0.5cm} \Phi_x^- = \int d^3 k \; \|k\|^{-1} e^{-i k \cdot x} a_k^* . \end{align} In addition we introduce for $ f \in L^2( \mathbb{R}^3)$ the creation operator $a^(f)$ and the annihilation operator $a(f)$ which are given through \begin{align} a^(f)= \int d^3k \; f(k) a_k^, \hspace{0.5cm} a(f) = \int d^3k \; \overline{f(k)} a_k \end{align} and bounded with respect to the number of particles operator $\mathcal{N}= \int d^3k\; a_k^ a_k$, i.e. \begin{align} \label{eq:bounds_a,a} \\| a(f) \xi \\| \leq \\| f \\|_2 \; \\| \mathcal{N}^{1/2} \xi \\|, \hspace{0.5cm} \\| a^(f) \xi \\| \leq \\| f \\|_2 \\| ( \mathcal{N}+\alpha^{-2})^{1/2} \xi \\| \end{align} for all $\xi \in \mathcal{F}$. Moreover, recall the definition \eqref{eq: definition Weyl operator} of the Weyl operator $W(f) = e^{a^(f) - a(f)}$. For a time dependent function $f_t \in L^2( \mathbb{R}^3)$ the time derivative of the Weyl operator is given by \begin{align} \label{eq:Weyl_deriv} \partial_t W(f_t) = \frac{\alpha^{-2}}{2} \left( \langle f_t, \partial_t f_t\rangle - \langle \partial_t f_t,f_t\rangle \right) W(f_t) + \left( a^(\partial_t f_t) - a(\partial_t f_t) \right) W(f_t). \end{align} The proof of this formula can be found in \cite[Lemma A.3]{frankgang}. \subsection{Proof of Theorem \ref{theorem: main theorem}} It should be noted that \eqref{eq: bound main theorem} is valid for all times which are at least of order $\alpha^2$ because both states in the inequality have norm one. To show its validity for shorter times we split the norm difference by the triangle inequality into two parts and use Remark \ref{remark: adiabatic theorem for short times} to estimate \begin{align} &\norm{e^{-i H_{\alpha} t} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega - e^{ - i \int_0^t du \, \omega(u) } \psi_t \otimes W(\alpha^2 \varphi_t) \Omega}^2 \nonumber \\ &\quad\leq 2 \norm{e^{-i H_{\alpha} t} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega - e^{ - i \int_0^t du \, \omega(u) } \widetilde{\psi}_{\varphi_t} \otimes W(\alpha^2 \varphi_t) \Omega}^2 \nonumber \\ &\quad\quad + 2\norm{ \widetilde{\psi}_{\varphi_t} \otimes W(\alpha^2 \varphi_t) \Omega - \psi_t \otimes W(\alpha^2 \varphi_t) \Omega}^2 \nonumber \\ &\quad\leq C \alpha^{-2} \abs{t} + 2 \norm{e^{-i H_{\alpha} t} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega - e^{ - i \int_0^t du \, \omega(u) } \widetilde{\psi}_{\varphi_t} \otimes W(\alpha^2 \varphi_t) \Omega}^2 \end{align} for all $\|t\| \leq C_\Lambda \alpha^2$ where we used the notation $\widetilde{\psi}_{\varphi_t} = e^{-i \int_0^t du \; e( \varphi_u)} \psi_{\varphi_t}$. Therefore it remains to estimate the second term. For this, we introduce \begin{align} \xi_s &= e^{ i \int_0^s du \, \omega(u)} W^(\alpha^2 \varphi_s) e^{-i H_{\alpha} s} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega \end{align} to shorten the notation and compute \begin{align} W^(\alpha^2 \varphi_t) H_{\alpha} W(\alpha^2 \varphi_t) &= h_{\varphi_t} + \norm{\varphi_t}_2^2 + \mathcal{N} + \Phi_x + a(\varphi_t) + a^(\varphi_t) . \end{align} Using \eqref{eq:Weyl_comm}, this leads to \begin{align} \label{eq: time derivatice xi-s} i \partial_s \xi_s &= \big( i \partial_s W^(\alpha^2 \varphi_s) \big) W(\alpha^2 \varphi_s) \xi_s + \Big( W^(\alpha^2 \varphi_s) H_{\alpha} W(\alpha^2 \varphi_s) - \omega(s) \Big) \xi_s \nonumber \\ &= \Big( h_{\varphi_s} + \Phi_x - a(\sigma_{\psi_s}) - a^(\sigma_{\psi_s}) + \mathcal{N} \Big) \xi_s . \end{align} Note that $\xi_0 = \psi_{\varphi_0} \otimes \Omega $. We apply Duhamel's formula and use the unitarity of the Weyl operator $W(\alpha^2 \varphi_t)$ to compute \begin{align} &\norm{e^{-i H_{\alpha} t} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega - e^{ - i \int_0^t du \, \omega(u) } \widetilde{\psi}_{\varphi_t} \otimes W(\alpha^2 \varphi_t) \Omega}^2\nonumber \\ &= \norm{ \xi_t - \widetilde{\psi}_{\varphi_t} \otimes \Omega}^2 = \int_0^t ds \, \partial_s \norm{\xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega}^2 = - 2 \Re \int_0^t ds \, \partial_s \scp{ \xi_s}{\widetilde{\psi}_{\varphi_s} \otimes \Omega} . \end{align} Recall that $\Phi_x = \int d^3k \|k\|^{-1} \left( e^{i k \cdot x} a_k + e^{- i k \cdot x} a_k^\right)$. We obtain \begin{align} &\norm{e^{-i H_{\alpha} t} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega - e^{ - i \int_0^t du \, \omega(u) } \widetilde{\psi}_{\varphi_t} \otimes W(\alpha^2 \varphi_t) \Omega}^2 \nonumber \\ &= 2 \Im \int_0^t ds \, \Big( \scp{i \partial_s \xi_s}{\widetilde{\psi}_{\varphi_s} \otimes \Omega} - \scp{\xi_s}{i \partial_s \widetilde{\psi}_{\varphi_s} \otimes \Omega} \Big) \nonumber \\ &= 2 \Im \int_0^t ds \, \scp{\xi_s}{\Big( h_{\varphi_s} - e(\varphi_s) + \Phi_x - a(\sigma_{\psi_s}) - a^(\sigma_{\psi_s}) + \mathcal{N} \Big) \widetilde{\psi}_{\varphi_s} \otimes \Omega} \nonumber \\ &\quad - 2 \alpha^{-2} \Re \int_0^t ds \, \scp{\xi_s}{ R_s V_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \otimes \Omega} \nonumber \\ &= 2 \Im \int_0^t ds \, \scp{\xi_s}{ \big( \Phi_x^- - a^(\sigma_{\psi_s} ) \big) \widetilde{\psi}_{\varphi_s} \otimes \Omega} - 2 \alpha^{-2} \Re \int_0^t ds \, \scp{\xi_s}{ R_s V_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \otimes \Omega} . \end{align} Here we used the definition $R_s = q_s ( h_{\varphi_s} - e( \varphi_s) )^{-1} q_s$ and Lemma \ref{lemma: existence minimizer}. Thus if we insert the identity $1= p_{s} + q_{s}$ and note that $q_s a^( \sigma_{\psi_s}) \widetilde{\psi}_{\varphi_s} \otimes \Omega = 0$ and $q_s \widetilde{\psi}_{\varphi_s} = 0$, we get \begin{subequations} \begin{align} &\norm{e^{-i H_{\alpha} t} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega - e^{ - i \int_0^t du \, \omega(u) } \widetilde{\psi}_{\varphi_t} \otimes W(\alpha^2 \varphi_t) \Omega}^2 \nonumber \\ \label{eq: gronwall c} &\qquad \quad = - 2 \alpha^{-2} \Re \int_0^t ds \, \scp{\xi_s}{ R_sV_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \otimes \Omega} \\ \label{eq: gronwall a} &\qquad \qquad +2 \Im \int_0^t ds \, \scp{\xi_s }{ p_s \big( \Phi_x^- - a^(\sigma_{\psi_s} ) \big) \widetilde{\psi}_{\varphi_s} \otimes \Omega} \\ \label{eq: gronwall b} &\qquad \qquad + 2 \Im \int_0^t ds \, \scp{\xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega}{ q_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}. \end{align} \end{subequations} \noindent We observe that the first term \eqref{eq: gronwall c} is already of the right order, namely $\alpha^{-2} t$. To be more precise, \begin{align} \abs{\eqref{eq: gronwall c}} &\leq 2 \alpha^{-2} \int_0^t ds \, \norm{\xi_s} \norm{R_s} \norm{V_{i \varphi_s}\widetilde{\psi}_{\varphi_s}}_2 \nonumber \\ &\leq 2 \alpha^{-2} \int_0^t ds \\| \varphi_s \\|_2 \norm{R_s} \norm{\widetilde{\psi}_{\varphi_s}}_{\Hone} \leq C \alpha^{-2} \abs{t} \end{align} for all $\|t\| \leq C_\Lambda \alpha^2 $ where we used Lemma \ref{lemma: Potental}, Lemma \ref{lemma: resolvent}, Lemma \ref{lemma: well posedness LP} and \eqref{eq: h1 minimizer}. We estimate the remaining two terms \eqref{eq: gronwall a} and \eqref{eq: gronwall b} separately. \subsection{The term \eqref{eq: gronwall a} } We have \begin{align} \abs{\eqref{eq: gronwall a}} &\leq 2 \int_0^t ds \, \abs{ \scp{\xi_s }{\int d^3k \, a_k^* \abs{k}^{-1} \Big( \scp{\widetilde{\psi}_{\varphi_s}}{ e^{- i k \cdot} \widetilde{\psi}_{\varphi_s}} - \scp{\psi_s}{e^{- i k \cdot}\psi_s} \Big) \widetilde{\psi}_{\varphi_s} \otimes \Omega} } \nonumber \\ &\leq 2 \int_0^t ds \, \norm{ \int d^3k \, a_k^* \abs{k}^{-1} \Big( \scp{\widetilde{\psi}_{\varphi_s}}{ e^{- i k \cdot} \widetilde{\psi}_{\varphi_s}} - \scp{\psi_s}{e^{- i k \cdot}\psi_s} \Big) \widetilde{\psi}_{\varphi_s} \otimes \Omega} \nonumber \\ &= 2 \int_0^t ds \, \norm{ \int d^3k \, a_k^* \abs{k}^{-1} \Big( \scp{\widetilde{\psi}_{\varphi_s}}{ e^{- i k \cdot} \big( \widetilde{\psi}_{\varphi_s} - \psi_s \big)} + \scp{\big( \widetilde{\psi}_{\varphi_s} - \psi_s \big)}{e^{- i k \cdot}\psi_s} \Big) \widetilde{\psi}_{\varphi_s} \otimes \Omega} . \end{align} Since $\\| a^( f) \psi \otimes \Omega \\| = \alpha^{-1} \\| f \\|_2 \; \\| \psi \\|$ for all $f \in L^2( \mathbb{R}^3)$, we find \begin{align} \abs{\eqref{eq: gronwall a}} &\leq C \alpha^{-1} \int_0^t ds \Bigg[ \int d^3k \, \abs{k}^{-2} \Big( \abs{\scp{\widetilde{\psi}_{\varphi_s}}{ e^{- i k \cdot} \big( \widetilde{\psi}_{\varphi_s} - \psi_s \big)}}^2 + \abs{\scp{\big( \widetilde{\psi}_{\varphi_s} - \psi_s \big)}{e^{- i k \cdot}\psi_s}}^2 \Big) \Bigg]^{1/2} . \end{align} With the help of $\widehat{\|\cdot\|^{-2}} (x) = \pi^{-1} \|x\|^{-1}$ and the inequalities of Hardy-Littlewood-Sobolev and H\"older we obtain \begin{align} \int d^3k \, \abs{k}^{-2} & \abs{\scp{\widetilde{\psi}_{\varphi_s}}{ e^{- i k \cdot} \big( \widetilde{\psi}_{\varphi_s} - \psi_s \big)}}^2 \nonumber \\ &= C \int d^3x \int d^3y \, \abs{x-y}^{-1} (\widetilde{\psi}_{\varphi_s} - \psi_s)(x) \overline{\widetilde{\psi}_{\varphi_s}(x)} \widetilde{\psi}_{\varphi_s}(y) \overline{ (\widetilde{\psi}_{\varphi_s} - \psi_s)(y)} \nonumber \\ &\leq C \norm{\widetilde{\psi}_{\varphi_s} (\widetilde{\psi}_{\varphi_s} - \psi_s)}_{6/5}^2 \leq C \norm{\widetilde{\psi}_{\varphi_s} - \psi_s}_2^2 \norm{\widetilde{\psi}_{\varphi_s}}_3^2 \nonumber \\ & \leq C \norm{\widetilde{\psi}_{\varphi_s} - \psi_s}_2^2 \norm{\widetilde{\psi}_{\varphi_s}}_{H^1( \mathbb{R}^3)}^2 \leq C \norm{\widetilde{\psi}_{\varphi_s} - \psi_s}_2^2 \end{align} for all $\|t \| \leq C_\Lambda \alpha^2$ by \eqref{eq: h1 minimizer}. Similarly, \begin{align} \int d^3k \, \abs{k}^{-2} & \abs{\scp{\big( \widetilde{\psi}_{\varphi_s} - \psi_s \big)}{e^{- i k \cdot}\psi_s}}^2 \leq C \norm{ \psi_s}_{H^1( \mathbb{R}^3)}^2 \norm{\psi_s - \widetilde{\psi}_{\varphi_s}}_2^2 \leq C \norm{\psi_s - \widetilde{\psi}_{\varphi_s}}_2^2 \end{align} by Lemma \ref{lemma: well posedness LP}. Hence, \begin{align} \abs{\eqref{eq: gronwall a}} &\leq C \alpha^{-1} \int_0^t ds \norm{\psi_s - \widetilde{\psi}_{\varphi_s}}_2 \end{align} for all $\|t\| \leq C_\Lambda \alpha^2$. Applying Theorem \ref{thm:adiabatic} leads to \begin{align} \abs{\eqref{eq: gronwall a}} \leq C \alpha^{-2} \abs{t} \quad \text{for all} \, \abs{t} \leq C_\Lambda \alpha^2 . \end{align} \subsection{The term \eqref{eq: gronwall b}} In order to continue we note that \cite[Theorem X.71]{reedsimon}, whose assumptions can easily shown to be satisfied by Lemma \ref{lemma: Potental}, guarantees the existence of a two parameter group $U_{h}(s;\tau)$ on $L^2(\mathbb{R}^3)$ such that \begin{align} \frac{d}{ds} U_h(s; \tau) \psi = - i h_{\varphi_s} U_h(s; \tau)\psi , \quad U_h(\tau; \tau) \psi = \psi \quad \text{for all} \, \psi \in H^1(\mathbb{R}^3). \end{align} Moreover, we define \begin{align} \widetilde{U}_{h}(s; \tau) &= e^{ i \int_{\tau}^s du \, e(\varphi_u)} U_h(s; \tau) . \end{align} We then have for all $s \in \mathbb{R}$ \begin{align} \frac{d}{ds} \widetilde{U}_{h}^(s; \tau) = \widetilde{U}_{h}^(s; \tau) i \big(h_{\varphi_s} - e(\varphi_s) \big) \end{align} and \begin{align} \label{eq:parts2} \widetilde{U}^_{h}(s; \tau) q_s f_s &= - i \frac{d}{ds}\left[ \widetilde{U}_{h}^(s; \tau) R_s f_s\right] + i \widetilde{U}_{h}^(s; \tau)\dot{R_s} f_s + i \widetilde{U}_{h}^(s; \tau) R_s \partial_s f_s, \end{align} for $f_s \in L^2( \mathbb{R}^3)$. This allows us to express \begin{align} \eqref{eq: gronwall b} &= 2 \Im \int_0^t ds \, \scp{\xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega}{ q_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega} \nonumber \\ &= 2 \Im \int_0^t ds \, \scp{ U_h^(s;0) \big( \xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega \big)}{ \widetilde{U}_{h}^(s; 0) q_s \Phi_x^- \psi_{\varphi_s} \otimes \Omega} \end{align} by three integrals which contain a derivative with respect to the the time variable. Note that we absorbed the phase factor of $\widetilde{\psi}_{\varphi_s}$ in the dynamics $\widetilde{U}_{h}^(s;0)$. Thus, \begin{align} \abs{\eqref{eq: gronwall b}} &\leq 2 \abs{\int_0^t ds \, \scp{U_h^(s;0) \big( \xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega \big)}{ \frac{d}{ds} \left[ {U}_{h}^(s;0)R_s \Phi_x^- \psi_{\varphi_s} \otimes \Omega\right] }} \nonumber \\ &\quad + 2 \abs{\int_0^t ds \, \scp{U_h^(s;0) \big( \xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega \big)}{ \widetilde{U}_{h}^(s;0)\dot{R}_s \Phi_x^- \psi_{\varphi_s} \otimes \Omega}} \nonumber \\ &\quad + 2 \abs{\int_0^t ds \, \scp{U_h^(s;0) \big( \xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega \big)}{ \widetilde{U}_{h}^(s;0) R_s \Phi_x^- \partial_s \psi_{\varphi_s} \otimes \Omega}} . \end{align} In the first term we integrate by parts and we use $\xi_0 = \widetilde{\psi}_{\varphi_0} \otimes \Omega$. We find \begin{align} \abs{\eqref{eq: gronwall b}} &\leq 2 \abs{ \scp{U_h^(t;0) \big( \xi_t - \widetilde{\psi}_{\varphi_t} \otimes \Omega \big) }{U_h^(t;0) R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega}} \nonumber \\ &\quad + 2 \abs{ \int_0^t ds \, \scp{\frac{d}{ds} \left[ U_h^(s;0) \big( \xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega \big) \right] }{U_h^(s;0) R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \nonumber \\ &\quad + 2 \abs{\int_0^t ds \, \scp{\big( \xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega \big)}{\dot{R}_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \nonumber \\ &\quad + 2 \abs{\int_0^t ds \, \scp{ \xi_s }{ R_s \Phi_x^- \partial_s \psi_{\varphi_s} \otimes \Omega}} . \label{eq:92_1} \end{align} In order to compute the time derivative occurring in the second summand, we use \eqref{eq: time derivatice xi-s} and the notation \begin{align} \label{eq: definition of delta H} \delta H_s = \Phi_x - a(\sigma_{\psi_s}) - a^(\sigma_{\psi_s}) + \mathcal{N} \end{align} and get \begin{align} \label{eq:hf1} \frac{d}{ds} \left[ U_h^(s;0) \big( \xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega \big)\right] &= - i U_h^(s;0) \delta H_s \, \xi_s - \alpha^{-2} U_h^(s;0) R_s V_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \otimes \Omega \end{align} as well as \begin{align} \label{eq:hf2} U_h^(t;0) \big( \xi_t - \widetilde{\psi}_{\varphi_t} \otimes \Omega \big) &= - i \int_0^t ds \, U_h^(s;0) \delta H_s \, \xi_s - \alpha^{-2} \int_0^t ds \, U_h^(s;0) R_s V_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \otimes \Omega. \end{align} Applying \eqref{eq:hf2} to the first term of the r.h.s. of \eqref{eq:92_1} and \eqref{eq:hf1} to the second, we obtain \begin{subequations} \begin{align} \label{eq: gronwall b1} \abs{\eqref{eq: gronwall b}} &\leq 2 \abs{\int_0^t ds \, \scp{\delta H_s \, \xi_s}{R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \\ \label{eq: gronwall b2} &\quad + 2 \abs{\int_0^t ds \, \scp{\delta H_s \, \xi_s}{ U_h^(t;s) R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega}} \\ \label{eq: gronwall b3} &\quad + 2 \alpha^{-2} \abs{\int_0^t ds \, \scp{R_s V_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \otimes \Omega}{ R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \\ \label{eq: gronwall b4} &\quad + 2 \alpha^{-2} \abs{\int_0^t ds \, \scp{R_s V_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \otimes \Omega}{U_h^(t;s) R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega}} \\ \label{eq: gronwall b5} &\quad + 2 \abs{\int_0^t ds \, \scp{\big( \xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega \big)}{\dot{R}_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \\ \label{eq: gronwall b6} & \quad + 2 \abs{\int_0^t ds \, \scp{ \xi_s }{ R_s \Phi_x^- \partial_s \psi_{\varphi_s} \otimes \Omega}} . \end{align} \end{subequations} \paragraph{The term \eqref{eq: gronwall b1}:} According to the definition of $\delta H$, we decompose \eqref{eq: gronwall b1} as \begin{subequations} \begin{align} \label{eq: gronwall b1a} \eqref{eq: gronwall b1} &\leq 2 \int_0^t ds \, \abs{\scp{\xi_s}{ \mathcal{N} R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega }} \\ \label{eq: gronwall b1b} &\quad + 2 \int_0^t ds \, \abs{\scp{\xi_s}{\big( a(\sigma_{\psi_s}) + a^(\sigma_{\psi_s}) \big) R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega }} \\ \label{eq: gronwall b1c} &\quad + 2 \int_0^t ds \, \abs{\scp{\xi_s}{ \Phi_x R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega }}. \end{align} \end{subequations} We notice that $\big[ \mathcal{N} , R_s \big] =0$ and that $\mathcal{N} \Psi = \alpha^{-2} \Psi$ if $\Psi \in \mathcal{H}$ is a one-phonon state and write the first line as \begin{align} \eqref{eq: gronwall b1a} &= 2 \alpha^{-2} \int_0^t ds \, \abs{ \scp{\xi_s}{R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega }} \nonumber \\ &= 2 \alpha^{-2} \int_0^t ds \, \abs{ \scp{\xi_s}{R_s \int d^3k \, \abs{k}^{-1} e^{- ikx} a_k^ \widetilde{\psi}_{\varphi_s} \otimes \Omega }}. \end{align} By means of Lemma \ref{lemma: resolvent} and Lemma \ref{lemma: frankgang 3.1} this becomes \begin{align} \eqref{eq: gronwall b1a} &\leq C \alpha^{-3} \int_0^t ds \, \norm{\xi_s} \norm{R_s(- \Delta + 1)^{1/2}} \norm{\psi_{\varphi_s}}_{2} \leq C \alpha^{-3} \abs{t} \end{align} for all $\|t\| \leq C_\Lambda \alpha^2$. In a similar way, we calculate $\big[ a(\sigma_{\psi_s}), a_k^* \big] = \alpha^{-2} \overline{\sigma_{\psi_s}(k)}$ for all $k \in \mathbb{R}^3$ and estimate \begin{align} \eqref{eq: gronwall b1b} &= 2 \int_0^t ds \, \abs{\scp{\xi_s}{\big( a(\sigma_{\psi_s}) + a^(\sigma_{\psi_s}) \big) R_s \int d^3k \, \abs{k}^{-1} e^{- i k \cdot x} a_k^ \, \widetilde{\psi}_{\varphi_s} \otimes \Omega }} \nonumber \\ &\leq 2 \int_0^t ds \, \abs{\scp{\xi_s}{R_s \int d^3k \, \abs{k}^{-1} e^{- i k \cdot x} a^(\sigma_{\psi_s}) a_k^ \widetilde{\psi}_{\varphi_s} \otimes \Omega }} \nonumber\\ &\quad + 2 \alpha^{-2} \int_0^t ds \, \abs{\scp{\xi_s}{ R_s \int d^3k \, \abs{k}^{-1} e^{- i k \cdot x} \overline{\sigma_{\psi_s}(k)} \, \widetilde{\psi}_{\varphi_s} \otimes \Omega }} . \end{align} Applying Lemma \ref{lemma: frankgang 3.1}, Lemma \ref{lemma: Potental} and Lemma \ref{lemma: resolvent} to the first line and using the same arguments as in Lemma \ref{lemma: frankgang 3.1} for the second line this becomes \begin{align} \eqref{eq: gronwall b1b} &\leq C \alpha^{-2} \int_0^t ds \, \norm{\xi_s} \norm{R_s (- \Delta +1)^{1/2}} \norm{\sigma_{\psi_s}}_{2} \norm{\psi_{\varphi_s}}_{2} \nonumber \\ &\leq C \alpha^{-2} \abs{t} \quad \text{for all} \, \abs{t} \leq C_\Lambda \alpha^2 . \end{align} Since $\Phi_x = \Phi_x^{+} + \Phi_x^{-}$ we have \begin{subequations} \begin{align} \eqref{eq: gronwall b1c} &= 2 \int_0^t ds \, \abs{\scp{\xi_s}{\Phi_x R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \nonumber \\ \label{eq: gronwall b1c1} &\leq 2 \int_0^t ds \, \abs{\scp{\xi_s}{\Phi_x^+ R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \\ \label{eq: gronwall b1c2} &\quad + 2 \int_0^t ds \, \abs{ \scp{\Phi_x^+ \xi_s}{ R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}}. \end{align} \end{subequations} Making use of Lemma \ref{lemma: frank schlein lemma 10} the first line can be estimated by \begin{align} \eqref{eq: gronwall b1c1} &\leq C \int_0^t ds \, \norm{(- \Delta + 1)^{1/2} \mathcal{N}^{1/2} R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega} \nonumber \\ &= C \int_0^t ds \, \norm{(- \Delta + 1)^{1/2} R_s \mathcal{N}^{1/2} \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega} . \end{align} Since $\\| (- \Delta +1)^{1/2} R_s (- \Delta +1)^{1/2} \\| \leq C$ for all $\|t\| \leq C_\Lambda \alpha^2$ by Lemma \ref{lemma: resolvent} and $\mathcal{N}^{1/2} \Psi = \alpha^{-1} \Psi$ if $\Psi \in \mathcal{H}$ is a one-phonon state, we find \begin{align} \eqref{eq: gronwall b1c1} &\leq C \alpha^{-1} \int_0^t ds \, \norm{ (- \Delta + 1)^{-1/2} \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega} \end{align} for all $\abs{t} \leq C_\Lambda \alpha^2.$ With Lemma \ref{lemma: frankgang 3.1} we arrive at \begin{align} \eqref{eq: gronwall b1c1} &\leq C \alpha^{-2} \int_0^t ds \, \norm{\psi_{\varphi_s}}_{2} \leq C \alpha^{-2} \abs{t} \end{align} for all $\abs{t} <C_\Lambda \alpha^2.$ In similar fashion we use Lemma \ref{lemma: frank schlein lemma 10}, Lemma \ref{lemma: frankgang 3.1} and $\mathcal{N} R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega = \alpha^{-2} R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega$ to estimate \begin{align} \label{eq: estimates on the difficult term 1} \eqref{eq: gronwall b1c2} &= 2 \int_0^t ds \, \abs{ \scp{\Phi_x^+ \xi_s}{ R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \nonumber \\ &= 2 \int_0^t ds \, \abs{ \scp{ (\mathcal{N} + \alpha^{-2})^{-1/2} \Phi_x^+ \xi_s}{ (\mathcal{N} + \alpha^{-2})^{1/2} R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \nonumber \\ &\leq 2 \int_0^t ds \, \norm{ (\mathcal{N} + \alpha^{-2})^{-1/2} \Phi_x^+ \xi_s} \norm{(\mathcal{N} + \alpha^{-2})^{1/2} R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega} \nonumber \\ &\leq C \alpha^{-1} \int_0^t ds \, \norm{(- \Delta +1)^{1/2} \xi_s} \norm{R_s \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega} \nonumber \\ &\leq C \alpha^{-1} \int_0^t ds \, \norm{(- \Delta +1)^{1/2} \xi_s} \norm{R_s (- \Delta +1)^{1/2}} \norm{(- \Delta + 1)^{-1/2} \Phi_x^- \widetilde{\psi}_{\varphi_s} \otimes \Omega} \nonumber \\ &\leq C \alpha^{-2} \int_0^t ds \, \norm{(- \Delta +1)^{1/2} \xi_s} \norm{\psi_{\varphi_s}}_{2} \nonumber \\ &= C \alpha^{-2} \int_0^t ds \, \norm{(- \Delta +1)^{1/2} e^{- i H_{\alpha} s} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega} \end{align} for all $\abs{t} < C_{\Lambda} \alpha^2$. Thus, if we now use $- \Delta + 1 \leq C (H_{\alpha} +1)$ (see Lemma \ref{lemma: bound for the Froehlich Hamiltonian}) this becomes using the properties \eqref{eq:Weyl_comm} of the Weyl operators \begin{align} \label{eq: estimates on the difficult term 2} \eqref{eq: gronwall b1c2} &\leq C \alpha^{-2} \int_0^t ds \, \norm{(H_\alpha + 1)^{1/2} e^{- i H_\alpha s} \psi_{\varphi_0} \otimes W( \alpha^2 \varphi_0) \Omega}\notag \\ &= C \alpha^{-2} \int_0^t ds \, \norm{(H_\alpha + 1)^{1/2} \psi_{\varphi_0} \otimes W( \alpha^2 \varphi_0 ) \Omega} \nonumber \\ &= C \alpha^{-2} \int_0^t ds \, \big( e(\varphi_0) + \norm{\varphi_0}^2 + 1 \big)^{1/2} = C \alpha^{-2} \abs{t} \end{align} for all $\abs{t} < C_\Lambda\alpha^2$. In total, we obtain $\eqref{eq: gronwall b1c} \leq C \alpha^{-2} \abs{t} $ and hence $\eqref{eq: gronwall b1} \leq C \alpha^{-2} \abs{t}$ for all $\abs{t} < C_\Lambda\alpha^2$. \paragraph{The term \eqref{eq: gronwall b2}:} For the next estimate, we recall the notation \eqref{eq: definition of delta H} to write \eqref{eq: gronwall b2} as \begin{align} \eqref{eq: gronwall b2} &\leq 2 \int_0^t ds \, \abs{ \scp{ \xi_s}{ \mathcal{N} U_h^(t;s) R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega}} \nonumber \\ &\quad + 2 \int_0^t ds \, \abs{ \scp{ \xi_s}{ \big( a(\sigma_{\psi_s}) + a^(\sigma_{\psi_s} \big) U_h^(t;s) R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega}} \nonumber \\ &\quad + 2 \int_0^t ds \, \abs{ \scp{\xi_s}{ \Phi_x U_h^(t;s) R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega}} . \end{align} Using $[\mathcal{N}, U_h^(t;s)] = [ a(\sigma_{\psi_s}), U_h^(t;s)] = [a^(\sigma_{\psi_s}), U_h^(t;s)] = 0$ allows us to estimate the first two lines in exactly the same way as \eqref{eq: gronwall b1a} and \eqref{eq: gronwall b1b} and leaves us with \begin{align} \eqref{eq: gronwall b2} &\leq C \alpha^{-2} \abs{t} + 2 \int_0^t ds \, \abs{ \scp{\xi_s}{ \Phi_x U_h^(t;s)) R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega}} \end{align} for all $\abs{t} < C_\Lambda \alpha^2$. The difficulty of this term is the fact that the operators $\Phi_x$ and $U_h^(t;s)$ do not commute. Nevertheless, we can use $\Phi_x = \Phi_x^+ + \Phi_x^-$ to get \begin{subequations} \begin{align} \eqref{eq: gronwall b2} &\leq C \alpha^{-2} \abs{t} + 2 \int_0^t ds \, \abs{\scp{\Phi_x^+ \xi_s}{ U_h^(t;s) R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega}} \\ \label{eq: gronwall b2b} &\quad + 2 \int_0^t ds \, \abs{ \scp{\xi_s}{ \Phi_x^+ U_h^(t;s) R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega}}. \end{align} \end{subequations} Using the same estimates as in \eqref{eq: estimates on the difficult term 1} and \eqref{eq: estimates on the difficult term 2} we bound the first integral by \begin{align} 2 \int_0^t ds \, &\abs{\scp{\Phi_x^+ \xi_s}{ U_h^(t;s) R_t \Phi_x^- \tilde{\psi}_{\varphi_t} \otimes \Omega}} \nonumber \\ &= 2 \int_0^t ds \, \abs{\scp{(\mathcal{N} + \alpha^{-2})^{-1/2} \Phi_x^+ \xi_s}{ U_h^(t;s) (\mathcal{N} + \alpha^{-2})^{1/2} R_t \Phi_x^- \tilde{\psi}_{\varphi_t} \otimes \Omega}} \nonumber \\ &\leq 2 \int_0^t ds \, \norm{(\mathcal{N} + \alpha^{-2})^{-1/2} \Phi_x^+ \xi_s} \norm{(\mathcal{N} + \alpha^{-2})^{1/2} R_t \Phi_x^- \tilde{\psi}_{\varphi_t} \otimes \Omega} \nonumber \\ &\leq C \alpha^{-2} \abs{t} \quad \text{for all} \, \abs{t} \leq C_{\Lambda} \alpha^2. \end{align} For the second term Lemma \ref{lemma: frank schlein lemma 10} and $U_h^(t;s) = U_h(s;t)$ imply \begin{align} \eqref{eq: gronwall b2b} &\leq C \int_0^t ds \, \norm{ (- \Delta + 1)^{1/2} \mathcal{N}^{1/2} U_h(s;t) R_t \Phi_x^- \tilde{\psi}_{\varphi_t} \otimes \Omega} . \end{align} It follows from Lemma \ref{lemma: Potental} and \eqref{eq: time derivative of the potential} that for $\xi \in L^2( \mathbb{R}^3) \otimes \mathcal{F}$ \begin{align} \langle \xi, U^_h(s;\tau) (- \Delta +1) U_h(s;\tau) \xi \rangle \leq& C \langle \xi, U^_h(s;\tau) ( h_{\varphi_s} + 1) U_h(s;\tau) \xi \rangle \notag \\ =& C \langle \xi, ( h_{\varphi_{\tau}} + 1) \xi \rangle - \alpha^{-2} \int_\tau^s d\tau ' \; \langle \xi, U^_h(\tau ';\tau) V_{i \varphi_{\tau '}} U_h(\tau ';\tau) \xi \rangle\notag \\ \leq& C \langle \xi, (- \Delta +1)\xi \rangle + \alpha^{-2} \int_\tau^s d\tau ' \; \langle \xi, U^_h(\tau ';\tau) (- \Delta +1) U_h(\tau ';\tau) \xi \rangle. \end{align} The Gronwall inequality yields \begin{align} \\| (- \Delta +1)^{1/2} U_h(s;\tau ) \xi \\| \leq C e^{ \alpha^{-2} \|s - \tau\|} \\| (- \Delta +1)^{1/2}\xi \\| \leq C\\| (- \Delta +1)^{1/2}\xi \\| \end{align} for all $\|s - \tau\| \leq C_\Lambda \alpha^2$. Thus \begin{align} \eqref{eq: gronwall b2b} &\leq C \int_0^t ds \, \norm{ (- \Delta + 1)^{1/2} \mathcal{N}^{1/2} R_t \Phi_x^- \widetilde{\psi}_{\varphi_t} \otimes \Omega} \leq C \alpha^{-2} \|t\| \end{align} for all $\|t\| \leq C_\Lambda \alpha^{-2}$, where we concluded by Lemma \ref{lemma: frankgang 3.1} and Lemma \ref{lemma: resolvent} as for the term \eqref{eq: gronwall b1c1}. \paragraph{The terms \eqref{eq: gronwall b3} and \eqref{eq: gronwall b4}:} With the help of Lemma \ref{lemma: Potental}, Lemma \ref{lemma: frankgang 3.1}, Lemma \ref{lemma: resolvent} and \eqref{eq: h1 minimizer} one obtains \begin{align} \eqref{eq: gronwall b3} &= 2 \alpha^{-2} \abs{ \int_0^t ds \, \scp{R_s V_{i \varphi_s} \widetilde{\psi}_{\varphi_s} \otimes \Omega}{ R_s \int d^3k \, \abs{k}^{-1} e^{- ik \cdot x} a_k^ \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \nonumber \\ &\leq 2 \alpha^{-2} \int_0^t ds \, \norm{R_s} \norm{\psi_{\varphi_s}}_{H^1( \mathbb{R}^3)} \norm{R_s (- \Delta +1)^{1/2}} \norm{(- \Delta + 1)^{-1/2} \int d^3k \, \abs{k}^{-1} e^{- ikx} a_k^* \widetilde{\psi}_{\varphi_s} \otimes \Omega} \nonumber \\ &\leq C \alpha^{-3} \abs{t} \end{align} and $ \eqref{eq: gronwall b4} \leq C \alpha^{-3} \abs{t} $ for all $\abs{t} < C_\Lambda\alpha^2$. \paragraph{The term \eqref{eq: gronwall b5}:} Applying Lemma \ref{lemma: frankgang 3.1} once more we estimate \begin{align} \eqref{eq: gronwall b5} &= 2 \abs{ \int_0^t ds \, \scp{\big( \xi_s - \widetilde{\psi}_{\varphi_s} \otimes \Omega \big)}{ \dot{R}_s \int d^3k \, \abs{k}^{-1} e^{- ik \cdot x} a_k^* \, \widetilde{\psi}_{\varphi_s} \otimes \Omega}} \nonumber \\ &\leq 4 \int_0^t ds \, \norm{\dot{R}_s\big( - \Delta + 1 \big)^{1/2}} \norm{( - \Delta + 1 )^{-1/2} \int d^3k \, \abs{k}^{-1} e^{- ik \cdot x} a_k^* \, \widetilde{\psi}_{\varphi_s} \otimes \Omega} . \end{align} From \eqref{eq: h1 minimizer}, \eqref{eq: derivative rho} and Lemma \ref{lemma: resolvent} we get \begin{align} \eqref{eq: gronwall b5} &\leq C \alpha^{-3} \abs{t} \quad \text{for all} \, \abs{t} < C_\Lambda\alpha^2. \end{align} \paragraph{The term \eqref{eq: gronwall b6}:} With the help of Lemma \ref{lemma: existence minimizer} and Lemma \ref{lemma: Potental} we get \begin{align} \eqref{eq: gronwall b6} &= 2 \alpha^{-2}\abs{\int_0^t ds \, \scp{ \xi_s }{R_s \Phi_x^- R_sV_{i \varphi_s} \psi_{\varphi_s} \otimes \Omega}} \nonumber \\ &\leq 2 \alpha^{-2} \int_0^t ds \, \norm{R_s(- \Delta +1)^{1/2}} \norm{(- \Delta +1 )^{-1/2} \Phi_x^- R_s V_{i \varphi_s} \psi_{\varphi_s} \otimes \Omega} \nonumber \\ &\leq C \alpha^{-3} \int_0^t ds \; \norm{R_s(- \Delta +1)^{1/2}} \\| R_sV_{i \varphi_s} \\| \; \\| \psi_{\varphi_s} \\|_2 \leq C \alpha^{-3 } \|t\| . \end{align} Here we used again Lemma \ref{lemma: frankgang 3.1} and Lemma \ref{lemma: resolvent}. In total, we obtain \begin{align} \abs{\eqref{eq: gronwall b}} &\leq C \alpha^{-2} \abs{t} \quad \text{for all} \, \abs{t} < C_\Lambda \alpha^2. \end{align} Summing up, we have shown that \begin{align} &\norm{e^{-i H_{\alpha} t} \psi_{\varphi_0} \otimes W(\alpha^2 \varphi_0) \Omega - e^{ - i \int_0^t du \, \omega(u) } \widetilde{\psi}_{\varphi_t} \otimes W(\alpha^2 \varphi_t) \Omega}^2 \leq C \alpha^{-2 } \|t\| , \end{align} for all $\|t\| \leq C_\Lambda \alpha^2$. \appendix \section{Auxiliary estimates} \begin{lemma} \label{lemma: frankgang 3.1} There exists a constant $C>0$ such that for all $u \in L^2(\mathbb{R}^3)$ and $f \in L^2(\mathbb{R}^3)$ \begin{align} \norm{( - \Delta + 1)^{-1/2} \int d^3k \, \abs{k}^{-1} e^{- ik \cdot x} a_k^* u \otimes \Omega } \leq C \alpha^{-1} \norm{u}_{2} , \\ \norm{(- \Delta + 1)^{-1/2} \int d^3k \, \abs{k}^{-1} e^{- ik \cdot x} a^(f) a_k^ u \otimes \Omega } \leq C \alpha^{-2} \norm{u}_{2} \norm{f}_2 . \end{align} \end{lemma} \begin{proof} The commutation relations imply \begin{align} \\| ( - \Delta + 1)^{-1/2} &\int d^3k \; \|k\|^{-1} e^{- ik \cdot x} a_k^* u \otimes \Omega \\|^2 \notag\\ =& \int d^3k \int d^3k' \; \|k\|^{-1} \|k'\|^{-1} \langle e^{- ik \cdot x} a_k^* u \otimes \Omega , ( - \Delta + 1)^{-1} e^{- ik' \cdot x} a_{k'}^* u \otimes \Omega \rangle \notag\\ =& \alpha^{-2} \int d^3k \; \|k\|^{-2} \langle e^{- ik \cdot x} u , ( - \Delta +1)^{-1} e^{- ik \cdot x} u \rangle \notag\\ =& \alpha^{-2} \int d^3k \; \|k\|^{-2} \langle u , ( (- i \nabla -k)^2 +1)^{-1} u \rangle\notag \\ =& \alpha^{-2} \int d^3p\; \| \hat{u}(p)\|^2 \int d^3k \; \frac{1}{((p-k)^2+1) \|k\|^2} . \end{align} Since $\| \cdot \|^{-2}$ and $(\|\cdot\|^2 +1)^{-1}$ are radial symmetric and decreasing functions we have \begin{align} \sup_{p \in \mathbb{R}^3} \int d^3k \; \frac{1}{((p-k)^2+1) \|k\|^2} &= \int d^3k \; \frac{1}{(k^2+1) \|k\|^2} < \infty \end{align} by the rearrangement inequality. Hence, \begin{align} \\| (- \Delta + 1)^{-1/2} &\int d^3 \|k\|^{-1} e^{- ik \cdot x} a_k^* u \otimes \Omega \\|^2 \leq C \alpha^{-2} \int d^3p \; \| \hat{u}(p) \|^2 = C \alpha^{-2} \\| u \\|_2 . \end{align} The second bound of the Lemma follows from the first one and the bounds \eqref{eq:bounds_a,a} for the creation and annihilation operators. \end{proof} \begin{lemma}[Lemma 4, Lemma 10 in \cite{frankschlein}] \label{lemma: frank schlein lemma 10} Let $\Phi_x^+ = \int d^3k \, \abs{k}^{-1} e^{ik \cdot x} a_k$ and $\mathcal{N} = \int d^3k \, a^_k a_k$. Then \begin{align} \norm{\Phi_x^+ \Psi} &\leq C \norm{(- \Delta +1)^{1/2} \mathcal{N}^{1/2} \Psi} \quad \text{and} \quad \norm{(\mathcal{N} +\alpha^{-2})^{-1/2} \Phi_x^+ \Psi} \leq C \norm{(- \Delta + 1)^{1/2} \Psi}. \end{align} \end{lemma} \begin{proof} We split the operator $\Phi_x^+ = \Phi_x^{+, >} +\Phi_x^{+, <} $, where \begin{align} \Phi_x^{+, >} = \int_{\|k\|> \kappa} \frac{d^3k}{\|k\|} e^{ik \cdot x} a_k, \hspace{0.5cm}\Phi_x^{+, <} = \int_{\|k\|< \kappa} \frac{d^3k}{\|k\|} e^{ik \cdot x} a_k \end{align} for a constant $\kappa >0$ of order one. Then, we deduce from \cite[Lemma 10]{frankschlein} that \begin{align} \norm{\Phi_x^{+,>} \Psi} &\leq C \norm{(- \Delta +1)^{1/2} \mathcal{N}^{1/2} \Psi} \quad \text{and} \quad \norm{(\mathcal{N} +\alpha^{-2})^{-1/2} \Phi_x^{+,>} \Psi} \leq C \norm{(- \Delta + 1)^{1/2} \Psi} , \end{align} where the constant $C>0$ depends only on $\kappa$. Since the function $f_< : \mathbb{R}^3 \rightarrow \mathbb{R}$ given through $f_<(k) =\|k\| \chi_{\|k\| \leq \kappa }$ is in $L^2( \mathbb{R}^3)$, \cite[Lemma 4]{frankschlein} implies \begin{align} \norm{\Phi_x^{+,<} \Psi} &\leq C \norm{\mathcal{N}^{1/2} \Psi} \quad \text{and} \quad \norm{(\mathcal{N} +\alpha^{-2})^{-1/2} \Phi_x^{+,<} \Psi} \leq C \norm{ \Psi} . \end{align} for a constant $C>0$ depending only on $\kappa$. \end{proof} \begin{lemma}[\cite{frankschlein}, p.7] \label{lemma: bound for the Froehlich Hamiltonian} Let $\alpha_0 >0$. Let $H_\alpha$ denote the Fr\"ohlich Hamiltonian defined in \eqref{eq: Froehlich Hamiltonian} and $\varepsilon >0$. There exists a constant $C_\varepsilon$ (depending on $\alpha_0$) , such that \begin{align} ( 1-\varepsilon) ( - \Delta + \mathcal{N} ) - C_\varepsilon \leq H_\alpha \leq ( 1+\varepsilon) ( - \Delta + \mathcal{N} ) +C_\varepsilon . \end{align} for all $\alpha \geq \alpha_0$. \end{lemma} The proof is given in \cite{frankschlein} (see Lemma 7 and p.7) and relies on arguments of Lieb and Yamazaki \cite{liebyamazaki} (see \cite[p.12]{liebthomas} for a concise explanation). \section*{Acknowledgments} N.\,L.\ and R.\,S.\ gratefully acknowledge financial support by the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme (grant agreement No 694227). B.\,S.\ acknowledges support from the Swiss National Science Foundation (grant 200020\_172623) and from the NCCR SwissMAP. N.\,L.\ would like to thank Andreas Deuchert and David Mitrouskas for interesting discussions. B.\,S. \ and R.\,S.\ would like to thank Rupert Frank for stimulating discussions about the time-evolution of a polaron.	151,002
TITLE: Conjugacy classes generating an infinite group QUESTION [1 upvotes]: It is well-known that a finite group can be generated by arbitrary representatives of each conjugacy class, see e.g.: https://mathoverflow.net/questions/26979/generating-a-finite-group-from-elements-in-each-conjugacy-class Now, for infinite groups this fails in general, for example the upper triangular matrices are a conjugate-dense subgroup of $GL_n(K)$ if $K$ is an algebraically closed field. So in this example there is some choice of representatives of conjugacy classes such that these elements do not generate the whole group. I am looking now for other examples: Is there an infinite group $G$ such that for any choice $S$ of representatives of conjugacy classes, the subgroup generated by $S$ is proper. As in 1. and additionally requiring that $G$ is finitely generated. As in 2. and additionally requiring that $G$ has finitely many conjugacy classes. REPLY [4 votes]: One easy way to get examples of type (1) is to take an uncountable group with countably many conjugacy classes. For instance, let $S$ be an uncountable set and take the group $G$ of finite-support permutations of $S$. Then $G$ has only countably many conjugacy classes, since permutations with the same cycle structure are conjugate. So if you pick one element from each conjugacy class, they can only generate a countable subgroup of $G$. There also exist infinite groups in which all non-identity elements are conjugate, so a subgroup generated by representatives of the conjugacy classes would be cyclic and not the whole group. Here's the idea: start with an infinite torsion-free group $G_0$, then for each pair of non-identity elements of $G_0$, adjoin a new generator to your group which conjugates one into the other. This gives a new group $G_1$, which can be shown to still be torsion free. Repeat this process infinitely many times and take the direct limit, and you get a group in which any two non-identity elements are conjugate. See Infinite group with only two conjugacy classes for a bit more detail and some references.	44,223
HUBBARD The 5th annual Charity Clinic Day to benefit The Bummer Fund will take place from noon to 3 p.m. Saturday at the Animal Care Hospital, 47 Hall Ave. The Bummer Fund is a nonprofit organization working in Mahoning, Trumbull, and Columbiana counties to provide financial assistance to qualified pet owners in a time of crisis. At the event, regular priced veterinary exams and routine procedures, such as nail trimming, ear cleaning and vaccinating will be done by appointment, and the proceeds will go to The Bummer Fund. Appointments are available from noon to 4 p.m. The attending vet is Dr. Charles Sung. A gift basket raffle is scheduled. Call the clinic at 330-534-2000 for an appointment or basket and ticket information. For information about The Bummer Fund call Susan Sexton at 330-519-3152. Don't Miss a Story Sign up for our newsletter to receive daily news directly in your inbox.	2,419
Scharner set for 6m Fulham move as Wigan turn to Swansea star Gomez Fulham are set to sign Paul Scharner from Wigan for £6m. The Austria international's agent Valentin Hobel revealed Scharner has rejected Aston Villa because they wanted to play him in defence rather than midfield, and hopes to conclude a deal with Fulham shortly. Roy Hodgson wants the 29-year-old for Fulham's Europa League campaign next season, and Scharner said: 'My chapter at Wigan is closed. I want to leave for a club that is competing in Europe.' New Wigan boss Roberto Martinez has made Jordi Gomez his first signing, as revealed by Sportsmail last week. Gomez, who spent last season working with Martinez on loan at Swansea, is signing on a four-year contract after a £1.7m fee was agreed with Espanyol. Sponsored links advertisement Related Aston Villa News advertisement advertisment	344,385
The story of Ape's heroism and dedication to his human team is here at the New York Times. Long time readers of Pure Florida know I am a dog guy. So you know any dog story, happy or sad, tugs at my heart, but this ... There is no other animal on the planet, none, that can do what we ask these tactical dogs to do. Whether it's a military dog or a rural deputy's police dog, we ask them to go first, to take the point like Ape did. When a human officer puts on a gun and a badge, they are making a conscious decision and are aware of the possible consequences. They know that they are duty bound to respond to any emergency and that when things go bad, the public will turn to them for protection. I know this to be true from my 8 years in law enforcement. Dogs don't have that luxury of being able to ponder the possible outcomes and consequences of their service. They simply want to perform as they were trained and instinctively act to protect "their" humans. I think it is that innocence of the dog, even a highly trained, powerful police or military dog that makes these losses so painful to me. These tactical dogs have saved thousands of lives over the years through their keen senses, intelligence, bravery, and awesome physical abilities. We owe them our respect and gratitude for the work they do and, like Ape, the sacrifices they make. But words are cheap, and canine protective vests ... "bulletproof vests" ... are not. A single K-9 protective vest costs a little over $1000.00. Please take a moment and make a donation to VESTED INTEREST IN K-9S, Inc. If a hundred of us donated ten dollars in Ape's memory, we could outfit another loyal police dog with a protective vest. We can't bring Ape back, but we can honor his sacrifice with a little sacrifice of our own. 15 comments: So sorry to read about Ape, he is a hero! You are absolutely right. Folks who have not worked "up close and personal" with police work dogs have no real concept of how hard they (and their partners) train, how incredibly versatile and useful K-9's are, or how short their careers and how much they give for love of their partners. I wasn't aware of the vests for K-9's program. A donation is on the way. Thanks for the reminder. Bill Happened only 40 miles from here...the whole incident went a long way toward destroying the small town innocence of the area. People we know were in lockdown. I like what you are doing here. There is a groupon for this right now. You pay 10 dollars for a 20 dollar donation towards bullet proof doggy vests! My friend, Joe, raised a beautiful Rottweiler who went in to police work. Chief is still remembered fondly by the Las Vegas PD, killed in the line of duty protecting his humans. Thank you friends. Yes, Anon (Emma?) that groupon is active and you can access it at the Vested site pretty easily. That's what I did this morning. That's a very sad story. Ape was such a beautiful dog. FC, is one of yours at UCF? Robin, Very painful to think about not only Ape, but the officer who was his partner. I know he or she is feeling a terrible loss. Kathy A, No, two are graduated from USF and Junior is at UF. He wants to be a lawyer. Hi FC, This is so sad. My union bought a vest for this group. We need to protect all members of the force. I can't find the Groupon. Do you have the link to email to me? P Just read about this and made a donation. Such a tragedy for all those involved. Hey FC, Just Swagbucked the offer and found it. Donated and posted on my Facebook page. p Thank you for your donations to this good cause! I made a donation. Wish it could have been more. Hey! The goal was 100 vests and the groupon deal bought 160 vests!!!! AWESOME!	129,829
Ducati India has launched two new bikes in India. These two cars are Monster 797 and Multistrada 950. Both bikes have been priced at Rs.7.77 lakh and Rs.16.6 lakh (ex-showroom, Delhi) respectively. Multistrada 950's name is in Off-Road-Bike, which was first introduced at the 2016 EICMA show held in Milan. On the other hand, Monster 797 is the cheapest bike in this series. Multistrada 950: it has 937cc, testastretta, twin cylinder engine which generates 113 bhp power and 96Nm maximum torque. For transmission, it has 6-speed gearboxes. Talking about the features, there are anti-lock braking systems, 4 riding modes, ride by wire technology and multiple Ducati traction control systems. Customers will be able to buy this bike in the classic Ducati Red and stylish Star White Color option in India. Monster 797: it has 803CC, testastretta, twin cylinder engine. Which generates 75bhp power and 69Nm maximum torque. Indian customers will be able to buy this bike in three color options including Star White Silk, Smart Dark Stealth and Classic Ducati Red. Also Read: This is Audi's most powerful convertible car so far Pre GST Sale: These cars are available at discount of Rs. 10 lakhs Datsun's new Redi Go is likely to launch next month To get all latest updates on Cricket like us on Facebook and follow us onTwitter or download Android App	244,031
! Need that backup like everyone use clips functionality and switch phone. Please! Would be perfect. I have the same problem too! I use "pinned clipboard" elements for shortcuts of text snippets, special characters and emojis, all thanks to the "shortcut" option :3 It's REALLY useful and I think it should be included 🌟 ( otherwise I take screenshots of the pinned clipboard and re-enter them one by one later...) Hi, for me Microsoft SwiftKey (using beta) is by far the best Keyboard available... However, the lack of this SYNC feature, regarding pinned clipboard is really a black stain on the overall superb quality of Microsoft SwiftKey App... How much more complains would be necessary, so that giant Microsoft can implement such an obvious feature, that is really missing on this first class App.? Candido Rodrigues, 👍 you're absolutely right. They update languages, that's what we see as their "focus" - they (Microsoft) do NOT listen us, users, and do NOT focus on making that beautiful feature (Pinned Clipboard) better. 😔 Sync, export, must be added. 👍 We save essential data there! Just to lose it and start all over again, in case of accidents that DO happen Please, all, share this link on social media. Microsoft might respond if it gets more 👍s, I dunno PS. On one of my smartphones, Pinned Clipboard is BUGGY TOO. I cant "Manage" it, it force closes! Microsoft needs to fix bugs + add new features. (sync) Pinned 📋 is one of main reasons why I love and use SwiftKey. It's essential FEATURE that deserves much more attention from the Microsoft app developers team. :(((((( Please, Microsoft team: Listen. We gave you good suggestions here. 💞 Improve Pinned Clipboard 📋. Add cloud sync (Google Drive or whatever), add import export && fix bugs! I'm telling you, i have so many saved items but i cant ACCESS AND MANAGE THEM, Force close happens. I need sync or export, reinstall, then import. And that bug that makes large amount of saved items inaccessible must be solved. Pinned Clipboard is an awesome feature - - what's the reason for you to NOT improve it, why? Everyone who is a member of community here : PLEASE, vote up this. By clicking arrow on the FIRST POST, this: This issue is on page 2: Just because it has 5 votes (I voted 👍 already) on the original post by Mrs Helen Walberg : Some LESS IMPORTANT things have more votes. We need 30 votes for this to be able to successfully reach page/spot 1. So pleaseee, people, do this: 1. 👍 the Helen Walberg' (first post) : 2. Share this on social media and ask friends to contribute, by just clicking 👍 (up arrow) - - SYNC MUST be added, both for iOS and Android. It should of been added and improved YEARS AGO. The original post is posted TWO years ago and still nothing :(((((( Support the good cause! Hi Ana Malden thank you for your feedback. Please submit feature requests via Also, regarding the issue you are experiencing managing your clipboard, please also include details on the issue and we'll be able to investigate further. Thanks. Denis K, I just shifted my phone and lost 5 years of establishing & maintaining my Clipboard 📋 items history; pinned ones (with custom SHORTCUTS) especially being heartbreaking to lose 😢 How hard is it, Microsoft? See the image - How hard is it to listen to your user's feedback? What should we do? Submit the damn feature request to your programmers - this is an essential yet non-existent FEATURE THAT MAKES US LOSE OUR WORK! 😭 😩 😔 Hi Mark Maček 🇸🇮¦🇷🇸 thank you for your feedback. I will pass your feedback on however we do ask to submit feature requests - this way we can track them & inform you of any updates.	86,262
TITLE: Is every 3-dim self-dual Galois representation a symmetic square of 2-dim representation? QUESTION [7 upvotes]: Basically, my question as in the title. Here the Galois representation I consider is an $\ell$-adic Galois representation (comes from geometry). And by the word "self-dual" I mean that representation is isomorphic to its dual representation (transpose inverse in language of matrix) up to a Tate twist. If the answer is positive, please let me know the idea or reference. If the answer is negative, I am wondering if there is any extra condition(s) to make my statement to be true? Thanks Edit: Thanks for the comments from @David Loeffler and @David E Speyer, I think the "correct" statement should be: Is any essentially self-dual 3 dimensional $\ell$-adic Galois representation isomorphic (up to a quadratic character) to a symmetric square of a 2-dimensional Galois representation? REPLY [17 votes]: The correct statement is: any 3-dimensional selfdual Galois representation is isomorphic to a quadratic twist of the adjoint of some 2-dimensional representation. (The quadratic twist is really necessary, as one can see from a variant of David Speyer's example.) If $\rho: G_F \to GL_3(\overline{\mathbf{Q}}_\ell)$ is a Galois representation, where $F$ is a number field, and $\rho \cong \rho^$, then the image of $\rho$ has to be contained in the subgroup $O_3(\overline{\mathbf{Q}}_\ell)$ of matrices preserving an orthogonal form. The quotient $O_3 / SO_3$ has order 2, so there is some quadratic character $\nu$ such that $\rho \otimes \nu$ takes values in $SO_3$. There is an exceptional isomorphism $Ad: PGL_2 \cong SO_3$, so $\rho$ must be $Ad(\tau)$ for some 2-dimensional projective representation $\bar\tau$. You want to know if $\bar\tau$ lifts to an actual representation; the obstruction to this lies in some $H^2$, and by a theorem of Tate this vanishes, so $\bar\tau$ is the image in $PGL_2$ of some $\tau: G_F \to GL_2(\overline{\mathbf{Q}}_\ell)$. (Note that $\tau$ is non-unique, and it is far from obvious a priori that $\tau$ can be chosen to be geometric if $\rho$ is. This kind of lifting-with-local-conditions question has been studied in detail in Patrikis' thesis.) This statement has an analogue on the automorphic side: Theorem A of Ramakrishnan's paper "An exercise concerning the selfdual cusp forms on GL(3)" states: Let F be a number field, and $\Pi$ a cuspidal, selfdual automorphic representation of $GL_3(\mathbb{A}_F)$. Then there exists a non-dihedral cuspidal automorphic representation $\pi$ of $GL_2/F$, and an idele class character $\nu$ of F with $\nu^2 = 1$, such that $\Pi \cong Ad(\pi) \otimes \nu$. EDIT. I just realised I had missed something: you are taking a (slightly unusual) definition of "selfdual". You aren't requiring that $\rho^ = \rho$ but just that $\rho^* = \rho \otimes \kappa^n$ for some $n$ (where $\kappa$ is the cyclotomic character), so $\rho$ is "selfdual up to Tate twists". However if you make the even more general assumption that $\rho^* = \rho \otimes \chi$ for any character $\chi$ (this is what people call "essentially selfdual"), then comparing determinants we have $\chi^3 \det(\rho)^2 = 1$, and hence $\chi$ is a square in the group of characters of $G_F$. So we can twist $\rho$ to make it self-dual on the nose, and apply the previous argument to that; and the conclusion is still that $\rho$ is a twist of the adjoint of something (or, if you prefer, a twist of the symmetric square of something). EDIT 2. (Let's not keep doing this, if you want to change the question then open a new question.) You now ask if any essentially selfdual 3-d representation is a quadratic twist of a symmetric square. This is not clear to me. What is clear from the above arguments is the following: if $A_\ell$ is the abelian group of characters $G_F \to \overline{\mathbf{Q}}_\ell^\times$, and $\Sigma$ is any set of representatives for the quotient $A_\ell / 2 \cdot A_\ell$, then any ess. selfdual $\rho$ can be written as $\operatorname{Sym}^2(\tau) \otimes \chi$ for some 2-dimensional $\tau$ and some (unique) $\chi \in \Sigma$. So your new question reduces to a purely one-dimensional one: is every class in $A_\ell / 2A_\ell$ represented by a quadratic character? But I don't see any reason why this should be true in general: I don't see why $A_\ell[2]$ should surject onto $A_\ell / 2A_\ell$.	126,430
\begin{document} \maketitle \renewcommand{\thesubsection}{\arabic{subsection}} \subsection{Introduction} Let $G^{\vee}$ be a reductive algebraic group defined over $\RR$. The local Langlands correspondence \cite{L} describes the set of equivalence classes of irreducible admissible representations of $G^{\vee}(\RR)$ in terms of the Weil-Deligne group and the complex dual group $G$. Roughly, it partitions the set of irreducible representations into finite sets (L-packets) and then describes each L-packet. Since Langlands' original work, this has been refined in several directions. Most relevant to this document is the work of Adams-Barbasch-Vogan \cite{ABV}. The key construction in \cite{ABV} is that of a variety (the parameter space) on which $G$ acts with finitely many orbits; each L-packet is re-interpreted as an orbit, and representations in the L-packet as the equivariant local systems supported on it. Adams-Barbasch-Vogan further demonstrate that the parameter space encodes significant character level information. In \cite{So} W. Soergel has outlined a conjectural relationship between the geometric and representation theoretic categories appearing in \cite{ABV}. This relationship, roughly a type of Koszul duality, yields a conceptual explanation for the phenomena observed in \cite{ABV}.\footnote{ The localization theorem of Beilinson-Bernstein also establishes a relationship between representation theory and geometry. However, Soergel's approach is very different: localization leads to geometry on the group itself; Soergel's approach results in geometry on the dual group. } The current note was born in an attempt to settle Conjectures 4.2.2, 4.2.3, 4.2.6 in \cite{So}, and Soergel's `Equivariant Formality' conjecture (implicit in \cite{So}; see \S\ref{s:formality} and \cite[\S0.2]{Lun}). These conjectures describe the structure of the geometric categories appearing in \cite{ABV}; the current document has little to say about Soergel's Basic Conjecture (relating the geometric categories to representation theory). We `\emph{almost}' succeed (see \S\ref{s:formality}). Soergel's conjectures are formulated in graded versions of our categories (in the sense of \cite[\S4]{So}). These `graded representation theories' are not constructed here because of a rather frustrating reason: we use the language of Hodge modules, and the category of Hodge modules is too large for the purposes of graded representation theory (see \S\ref{s:formality} and footnote \ref{note}). The main result is Theorem \ref{main} describing the Hodge structure on equivariant $\Ext^{\bullet}$ between simple perverse sheaves on the parameter space.\footnote{ The parameter space is never explicitly mentioned. The translation between the symmetric varieties $G/K$ appearing below and the parameter space is provided by \cite[Proposition 6.24]{ABV}. The parameter space is essentially a disjoint union of varieties of the form $G/K$. } It yields a host of ancillary results which are of independent interest: Corollary \ref{extparity} and \ref{icparity} (`parity vanishing'); Theorem \ref{mqstable} and Corollary \ref{positivity} (`positivity' of a Hecke algebra module). An informal discussion regarding Soergel's graded categories is contained in \S\ref{s:formality}. \textbf{Acknowledgments:} This work was conceived while I was visiting the Albert-Ludwigs-Universit\"at, Freiburg, in the summer of 2012. I am indebted to W. Soergel for his hospitality, his continuing explanations and patience.\footnote{\label{note} In January, 2010, W. Soergel explained to me: \emph{``In a way, there should be a better category than what we work with, sort of much more motivic, where these problems disappear. Think about Grothendieck's conjecture: the action of Frobenius on the \`etale cohomology of a smooth projective variety should be semisimple! So this non-semisimplicity is sort of due to the fact we are not working with motives, but with some rather bad approximation, I suggest."} At my glacial pace it has taken me four years to appreciate this (see \S\ref{s:formality}). } I am further grateful to W. Soergel and M. Wendt for sharing their beautiful ideas on the use of motivic sheaves in representation theory. \subsection{Conventions}\label{s:conventions} Throughout, `variety' = `separated reduced scheme of finite type over $\mathrm{Spec}(\CC)$'. A \emph{fibration} will mean a morphism of varieties which is locally trivial (on the base) in the \`etale sense. Constructible sheaves, cohomology, etc. will always be with $\RR$ or $\CC$ coefficients, and with respect to the complex analytic site associated to a variety. Given an algebraic group $G$, we write $G^0$ for its identity component. Suppose $G$ acts on $X$. Then we write $G_x$ for the isotropy group of a point $x\in X$. Given a principle $G$-fibration $E\to B$, we write $E\qtimes{G}X \to B$ for the associated fibration.\footnote{ Generally, $E\qtimes{G} X$ is only an algebraic space. It is a variety if, for instance, $X$ is quasi-projective with linearized $G$-action; or $G$ is connected and $X$ can be equivariantly embedded in a normal variety (Sumihiro's Theorem). One of these assumptions will always be satisfied below. } We write $D_G(X)$ for the $G$-equivariant derived category (in the sense of \cite{BL}), and $\Perv_G(X)\subseteq D_G(X)$ for the abelian subcategory of equivariant perverse sheaves on $X$. Perverse cohomology is denoted by $\pH^$. Change of group functors (restriction, induction equivalence, quotient equivalence, etc.) will often be omitted from the notation. All functors between derived categories will be tacitly derived. Both the functor of $G$-equivariant cohomology as well as the $G$-equivariant cohomology ring of a point will be denoted by $H^_G$. \subsection{$\mathbf{B\backslash G/K}$}\label{s:bgk} Let $G$ be a connected reductive group, $\theta\colon G\to G$ a non-trivial algebraic involution, $T$ a $\theta$-stable maximal torus, and $B\supseteq T$ a $\theta$-stable Borel containing it (such a pair $(B,T)$ always exists, see \cite[\S7]{St}). Write $W$ for the Weyl group. Let $K = G^{\theta}$ denote the fixed point subgroup. Then \begin{enumerate} \item $K$ is reductive (but not necessarily connected, see \cite[\S1]{V}); \item $\|B\backslash G/K\| < \infty$ (a convenient reference is \cite[\S6]{MS}); \item $K$-orbits in $G/B$ are affinely embedded (see \cite[Ch. H, Proposition 1]{M}); \item for each $x\in G/B$, the component group $K_x/K_x^0$ has exponent $2$ \cite[Proposition 7]{V}. \end{enumerate} Our primary concern is the category $D_{B\times K}(G)$, for the $B\times K$-action given by $(b,k)\cdot g = bgk^{-1}$. The evident identification of $B\times K$-orbits in $G$, with $B$-orbits in $G/K$, and with $K$-orbits in $G/B$, respects closure relations. There are corresponding identifications: $D_B(G/K) = D_{B\times K}(G) = D_K(G/B)$. These identifications will be used without further comment. Let $s\in W$ be a simple reflection, $P\supseteq B$ the corresponding minimal parabolic, and $v$ a $B$-orbit in $G/K$. Then the subvariety $P\cdot v \subseteq G/K$ contains a unique open dense $B$-orbit $s\star v$. Let $\leq$ denote the closure order on orbits, i.e., $v\leq w$ if and only if $v$ is contained in the closure $\overline{w}$. \begin{thm}[{\cite[Theorem 4.6]{RS}}]\label{order} If $w\in B\backslash G/K$ is not closed, then there exists a simple reflection $s\in W$, and $v\in B\backslash G/K$ such that $v\lneq w$ and $s\star v= w$. \end{thm} Let $\pi\colon G/B \to G/P$ be the evident projection. Let $x\in G/B$. Set $y=\pi(x)$, and $L_x^s = \pi^{-1}(y)$. Note: $L_x^s\simeq \PP^1$. \begin{equation}\label{defpi}\tag{$\ast$}\begin{gathered} \xymatrix{ \xy (0,0)\xycircle(8,8){-}; (0,0)\ellipse(8,2){.}; (0,0)\ellipse(8,2)__,=:a(-180){-}; \endxy\ar[d]\ar[r]^-{\sim} & L_x^s\ar[d]\ar[r] & G/B\ar[d]^-{\pi} \\ \bullet\ar[r] & \{y\}\ar[r] & G/P }\end{gathered} \end{equation} The $K$-action induces an isomorphism $K \qtimes{K_y} L_x^s \mapright{\sim} K\cdot L_x^s$. Thus, \[ D_K(K\cdot L_x^s) = D_K(K\qtimes{K_y} L_x^s) = D_{K_y}(L_x^s) = D_{K_y}(\PP^1).\footnote{ This is the analogue of the Lie theoretic principle that `local phenomena is controlled by $SL_2$'. } \] As $\|B\backslash G/K\|<\infty$, the image of $K_y$ in $\mathrm{Aut}(L^s_x)$ has dimension $\geq 1$. Identify $\PP^1$ with $\CC\sqcup \{\infty\}$. Modulo conjugation by an element of $\mathrm{Aut}(L^s_x)\simeq PGL_2$, there are four possibilities for the decomposition of $\PP^1$ into $K_y$-orbits: \begin{description} \item[Case G] $\PP^1$ (the action is transitive); \[ \xy (0,0)\xycircle(8,8){-}; (0,0)\ellipse(8,2){.}; (0,0)\ellipse(8,2)__,=:a(-180){-}; \endxy \] \item[Case U] $\PP^1 = \CC \sqcup \{\infty\}$; \[ \xy (0,8){\bullet}; (0,0)\xycircle(8,8){-}; (0,0)\ellipse(8,2){.}; (0,0)\ellipse(8,2)__,=:a(-180){-}; \endxy \] \item[Case T] $\PP^1 = \{0\} \sqcup \CC^* \sqcup \{\infty\}$; \[ \xy (0,8){\bullet}; (0,-8){\bullet}; (0,0)\xycircle(8,8){-}; (0,0)\ellipse(8,2){.}; (0,0)\ellipse(8,2)__,=:a(-180){-}; \endxy \] \item[Case N] $\PP^1 = \{0,\infty\}\sqcup\CC^$; both $\{0\}$ and $\{\infty\}$ are fixed points of $K_y^0$. \[ \xy (0,8){\bullet}; (0,-8){\bullet}; *\crv{~=<2mm>{.} (-10,-12) & (-20,0) & (-10, 12)} ?(.5)\dir{>}; (0,8); (0,-8); \crv{~=<2mm>{.} (10,-12) & (20,0) & (10, 12)} ?(.5)\dir{<}; (0,0)\xycircle(8,8){-}; (0,0)\ellipse(8,2){.}; (0,0)\ellipse(8,2)__,=:a(-180){-}; \endxy \] \end{description} We will say that $w$ is of type \textbf{G}, \textbf{U}, \textbf{T} or \textbf{N} relative to $s$ depending on which of these decompositions actually occurs. Given an irreducible equivariant local system $V_{\tau}$ on a $K$-orbit $j\colon w\hookrightarrow G/B$, set \[ \cL_{\tau} = j_{!}V_{\tau}[d_{\tau}], \quad\mbox{where}\quad d_{\tau} = \dim(w). \] Call $\cL_{\tau}$ \emph{clean} if $\cL_{\tau} \simeq j_!V_{\tau}[d_{\tau}]$. Call $\cL_{\tau}$ \emph{cuspidal} if for each simple reflection $s$, each $v\neq w$ with $s\star v = w$, and each $K$-equivariant local system $V_{\gamma}$ on $v$, the object $\cL_{\tau}$ does \emph{not} occur as a direct summand of $\pH^(\pi^\pi_\cL_{\gamma})$, where $\pi$ is as in \eqref{defpi}.\footnote{ The term `cuspidal' has a very specific meaning in representation theory. It is not clear to me whether the terminology is completely justified in the current geometric setting.} \begin{lemma}[{\cite[Lemma 7.4.1]{MS}}]\label{clean} Cuspidals are clean. \end{lemma} \begin{proof} As indicated, this is \cite[Lemma 7.4.1]{MS}. Regardless, the language employed in \cite{MS} is a bit different from ours, so we sketch the argument in order to orient the reader. Let $j\colon w\hookrightarrow G/B$ be a $K$-orbit, and $V_{\tau}$ a local system on $w$ such that $\cL_{\tau}$ is cuspidal. Write $\overline{w}$ for the closure of $w$. To demonstrate the assertion we need to show that $(j_V_{\tau})\|_v = 0$ for each orbit $v$ in $\overline{w} - w$. If $s$ is a simple reflection such that $s\star w = w$ and $P_s\cdot w$ contains an orbit other than $w$, then as $\cL_{\tau}$ is cuspidal, $w$ must be of type \textbf{T} or \textbf{N} relative to $s$. In the language of \cite{MS}, this means that each such $s$ is `of type IIIb or IVb for $w$'. Let $I$ be the set consisting of simple reflections $s$ as above and let $P_I$ be the parabolic subgroup of $G$ containing $B$ and corresponding to $I$. Then in \cite[\S7.2.1]{MS} it is shown that $P_I\cdot w = \overline{w}$. Now if $v$ is an orbit of codimension $1$ in $\overline{w}$, then there exists $s\in I$ such that $s\star v = w$. Inspecting the cases \textbf{T} and \textbf{N} yields the required vanishing in this case. For arbitrary $v$, proceed by induction on codimension. Let $s\in I$ be such that $s\star v > v$. Let $\pi$ be as in \eqref{defpi}. As $\cL_{\tau}$ is cuspidal, $\pi_(j_V_{\tau}) =0$. Furthermore, if $(j_V_{\tau})\|_v\neq 0$, then $\pi_((j_V_{\tau})\|_v)\neq 0$. All of this follows by inspection of the cases \textbf{G}, \textbf{U}, \textbf{T}, \textbf{N}. Combined these vanishing and non-vanishing statements imply $(j_V_{\tau})\|_v\neq 0$ only if $(j_V_{\tau})\|_{s\star v}\neq 0$. Thus, applying the induction hypothesis yields the result. \end{proof} \subsection{Mixed structures}\label{s:mixed} Given a variety $X$, write $M(X)$ for the category of $\RR$-mixed Hodge modules on $X$, and $DM(X)$ for its bounded derived category \cite{Sa}. If a linear algebraic group acts on $X$, write $DM_G(X)$ for the corresponding mixed equivariant derived category. When dealing with mixed as well as ordinary categories, objects in mixed categories will be adorned with an $\vphantom{V}^H$. Omission of the $\vphantom{V}^H$ will denote the classical object underlying the mixed structure. A mixed Hodge structure is called \emph{Tate} if it is a successive extension of Hodge structures of type $(n,n)$. A mixed Hodge module $\cA^H\in M(X)$ will be called \emph{$\ast$-pointwise Tate} if, for each point $i\colon \{x\}\hookrightarrow X$, the stalk $H^(i^\cA^H)$ is Tate. Call $\cA^H\in DM(X)$ $\ast$-pointwise Tate if each $\pH^i(\cA^H)$ is so. An object of $DM_G(X)$ is $\ast$-pointwise Tate if it is so under the forgetful functor $DM_G(X)\to DM(X)$. \begin{lemma}\label{s:tatepistable} Let $\pi$ be as in \eqref{defpi}. Then $\pi^\pi_$ preserves the class of $\ast$-pointwise Tate objects. \end{lemma} \begin{proof}Use the notation surrounding \eqref{defpi}. Then the assertion reduces to the claim that if $\cA^H\in M_{K_y}(L_x^s)$ is $\ast$-pointwise Tate, then $H^(L_x^s; \cA^H)$ is Tate.\footnote{ The core of this argument is due to R. MacPherson (in a `parity vanishing' context for Schubert varieties), see \cite[Lemma 3.2.3]{So00}. } This is immediate from the possible $K_y$-orbit decompositions \textbf{G}, \textbf{U}, \textbf{T} and \textbf{N}. \end{proof} Each irreducible $B\times K$-equivariant local system $V_{\tau}$, on an orbit $w$, underlies a unique (up to isomorphism) polarizable variation of Hodge structure of weight zero. Denote this variation by $V^H_{\tau}$. Taking intermediate extension, we obtain a pure (equivariant) Hodge module $\cL^H_{\tau}$ of weight $d_{\tau} = \dim(w)$, i.e., \[ \cL^H_{\tau} = j_{!}V^H_{\tau}[d_{\tau}], \] where $j\colon w \hookrightarrow G$ is the inclusion. \begin{prop}\label{ptwise}$\cL_{\tau}^H$ is $\ast$-pointwise Tate. \end{prop} \begin{proof}Work in $G/B$. The statement is true for cuspidals, since they are clean (Lemma \ref{clean}). The general case follows by induction (employing Theorem \ref{order}) and Lemma \ref{s:tatepistable}. \end{proof} \begin{prop}\label{contracting}Let $i\colon v\hookrightarrow G$ be the inclusion of a $B\times K$-orbit. Then $i^\cL^H_{\tau}$ is pure. \end{prop} \begin{proof} Work in $G/K$. According to \cite[\S6.4]{MS} (also see the comments at the end of \S1 in \cite{LV}), each $B$-orbit admits a contracting slice in the sense of \cite[\S2.3.2]{MS}. This implies purity (see \cite[\S2.3.2]{MS} or \cite[Lemma 4.5]{KL} or \cite[Proposition 1]{So89}). \end{proof} Given an algebraic group $L$ acting on a variety $X$, set \[ \Ext^{i}_{L}(-,-) = \Hom_{D_L(X)}(-,-[i]).\] The Hodge modules $\cL^H_{\nu}$ endow each $\Ext^{\bullet}_{B\times K}(\cL_{\tau}, \cL_{\gamma})$ with a Hodge structure. \begin{thm}\label{main}$\Ext_{B\times K}^{\bullet}(\cL_{\tau}, \cL_{\gamma})$ is Tate and pure of weight $d_{\gamma}-d_{\tau}$.\footnote{ \textbf{Warning:} the non-equivariant analogue of this result is false! } \end{thm} \begin{proof} Work in $G/K$. Filtering $\Ext^{\bullet}_B(\cL_{\tau}, \cL_{\gamma})$ by the orbit stratification, one sees that it suffices to argue that $\Ext_B^{\bullet}(i^\cL_{\tau},i^!\cL_{\gamma}))$ is pure and Tate for each $B$-orbit inclusion $i\colon u\hookrightarrow G/K$. Proposition \ref{ptwise} and Proposition \ref{contracting} imply that both $i^\cL_{\tau}^H$ and $i^!\cL_{\gamma}^H$ are direct sums of (shifted) one dimensional variations of Hodge structure. As $H^_L$ is pure and Tate, for any linear algebraic group $L$ (cf. \cite[\S9.1]{D}), the assertion is immediate. \end{proof} \begin{cor}\label{extparity} $\Ext_{B\times K}^{i}(\cL_{\tau}, \cL_{\gamma}) = 0$ unless $i = d_{\tau} + d_{\gamma}\mod 2$. \end{cor} \begin{cor}\label{icparity} $H^_{B\times K}(G; \cL_{\tau})$ vanishes in either all even or all odd degrees. \end{cor} Some remarks are in order: \begin{enumerate} \item Corollary \ref{extparity} should be compared with \cite[Conjecture 4.2.6]{So}; also see \S\ref{s:formality}. \item Corollary \ref{icparity} is essentially contained in \cite{LV}. Lusztig-Vogan work in the non-equivariant $\ell$-adic setting, but the Hecke algebra computations in \cite{LV} can be used to obtain Corollary \ref{icparity}; also see \S\ref{s:hecke}. Note that Lusztig-Vogan rely on explicit calculations with the Hecke algebra and arguments from representation theory. An argument analogous to \cite{LV}, involving Hecke algebra computations (but no representation theory), can be found in \cite{MS}. \item Let $P\supseteq B$ be a parabolic subgroup. One should be able to obtain similar results for the analogous $P\times K$-action on $G$ using the technique of \cite{So89}. \end{enumerate} \subsection{Hecke algebra}\label{s:hecke} Let $L\subseteq G$ be a closed subgroup (we are mainly interested in $L=B$ or $K$). Let $B\times L$ act on $G$ via $(b,l)\cdot g= bgl^{-1}$. Define a bifunctor \[ - \conv - \colon DM_{B\times B}(G) \times DM_{B\times L}(G) \to DM_{B\times L}(G), \] called \emph{convolution}, as follows. For $M\in DM_{B\times B}(G)$, $N\in DM_{B\times L}(G)$, the object $M\boxtimes N$ descends to an object $M\ttimes N \in DM_{B\times L}(G\qtimes{B}G)$. Set $M \conv N = m_!(M\ttimes N)$, where $m\colon G\qtimes{B} G \to G$ is the map induced by multiplication. This operation is associative in the evident sense. As $m$ is projective, convolution adds weights and commutes with Verdier duality (up to shift and Tate twist). Taking $L=B$ yields a monoidal structure on $DM_{B\times B}(G)$. For each $w\in W$, set \[ \T_w = j_{w!}\const{Bw B} \quad \mbox{and}\quad \C_w = (j_{w!}\const{Bw B}[\dim(BwB)])[-\dim(BwB)], \] where $j_w\colon Bw B \hookrightarrow G$ is the inclusion, and $\const{B w B}$ denotes the trivial (weight $0$) variation of Hodge structure on $B w B$. The unit for convolution is $\11 = \T_e$. \begin{prop}\label{braid}The $\T_w$ satisfy the braid relations. That is, if $\ell(vw) = \ell(v) + \ell(w)$, then $\T_v\conv \T_w = \T_{vw}$, where $\ell\colon W\to \ZZ_{\geq 0}$ is the length function. \end{prop} \begin{proof} Multiplication yields an isomorphism $B v B \qtimes{B} B w B \mapright{\sim} B v wB$. \end{proof} \begin{prop}\label{s:convpi} Let $s\in W$ be a simple reflection, and let $\pi$ be as in \eqref{defpi}. Then, under the equivalence $DM_{B\times K}(G) \mapright{\sim} DM_K(G/B)$, convolution with $\C_s$ is identified with $\pi^\pi_$. \end{prop} \begin{proof} Left to the reader (see \cite[Lemma 3.2.1]{So00}). \end{proof} Let $\HH_q\subseteq DM_{B\times B}(G)$ be the triangulated subcategory generated by the $\C_w$, $w\in W$, and Tate twists thereof. \begin{prop}$\HH_q$ is stable under convolution. \end{prop} \begin{proof} This follows from \cite[Proposition 3.4.6]{So}. Alternatively, note \[ DM_{B\times B}(G) \mapright{\sim} DM_B(G/B) \mapright{\sim} DM_{G}(G\qtimes{B} G/B) \mapright{\sim}DM_G(G/B\times G/B).\] This puts us in the setting of the previous sections (with group $G\times G$ and involution $\theta(g_1,g_2)=(g_2,g_1)$). Now use Lemma \ref{s:tatepistable} and Proposition \ref{s:convpi}. \end{proof} \begin{cor}Each $\T_w$ is in $\HH_q$. \end{cor} \begin{proof}In view of Proposition \ref{braid}, it suffices to prove this for each simple reflection $s$. In this case we have a distinguished triangle $\11[-1] \to \T_s \to \C_s \leadsto$ \end{proof} Let $\MM_q\subseteq DM_{B\times K}(G)$ be the triangulated subcategory generated by the $\cL^H_{\tau}$ and Tate twists thereof. \begin{thm}\label{mqstable}$\MM_q$ is stable under convolution with objects of $\HH_q$. \end{thm} \begin{proof} Combine Lemma \ref{s:tatepistable} with Proposition \ref{s:convpi}. \end{proof} Let $H_q=K_0(\HH_q)$ and $M_q = K_0(\MM_q)$ be the respective Grothendieck groups. These are free $\ZZ[q^{\pm 1}]$-modules via $q[\cA] = [\cA(-1)]$, where $(-1)$ is the inverse of Tate twist. Convolution makes $H_q$ a $\ZZ[q^{\pm 1}]$-algebra, and $M_q$ an $H_q$-module. \begin{cor}\label{positivity} The coefficients $c^w_{\gamma, \tau}(q)$ in the expansion \[ (-1)^{d_{\tau}}[\C_w\conv \cL^H_{\tau}] = \sum_{\gamma} (-1)^{d_{\gamma}}c^w_{\gamma,\tau}(q)[\cL^H_{\gamma}], \] are polynomials in $q^{\pm 1}$ with non-negative coefficients. \end{cor} The algebra $H_q$ is isomorphic to the Iwahori-Hecke algebra associated to $W$. That is, $H_q$ is isomorphic to the $\ZZ[q^{\pm 1}]$ algebra on generators $T_w$, $w\in W$, with relations: $T_vT_w=T_{vw}$ if $\ell(vw) = \ell(v)+\ell(w)$; and $(T_s+1)(T_s-q)=0$ if $\ell(s)=1$. The isomorphism is given by $T_w\mapsto [\T_w]$. The convolution product $\C_w\conv \cL^H_{\tau}$, the groups $H^_{B\times K}(G; \cL_{\tau})$, $\Ext^{\bullet}(\cL_{\tau}, \cL_{\gamma})$, $\Ext^{\bullet}(j_!V_{\tau}, \cL_{\gamma})$, can all be explicitly computed via the module $M_q$: $\C_w\star \cL^H_{\tau}$ because of Theorem \ref{positivity}; the rest because they are pure and Tate (Theorem \ref{main}) and can consequently be recovered from their weight polynomials. For an explicit description of $M_q$, see \cite{LV} and \cite{MS}. Corollary \ref{positivity} appears to be new (although it might be possible to deduce it from the results of \cite{LV}). It is a generalization of the well known positivity result for the Kazhdan-Lusztig basis (the classes $[\C_w]$) in the Hecke algebra. \subsection{Informal remarks}\label{s:formality} Let $\cA$ be an abelian category, $\mathrm{Ho}(\cA)$ the homotopy category of chain complexes in $\cA$, and $D(\cA)$ the derived category of $\cA$. Given a collection of (bounded below) complexes $\{T_i\}$ each of whose components are injectives, set $T=\bigoplus_i T_i$. The complex $\cE = \End^{\bullet}_{\cA}(T)$ has an evident dg-algebra structure. Let $e_i\in\cE$ denote the idempotent corresponding to projection on $T_i$. The functor $\Hom^{\bullet}(\cE, -)$ yields an equivalence between the full triangulated subcategory of $D(\cA)$ generated by the $T_i$ and the full triangulated subcategory of the dg-derived category $\dgDer{\cE}$ (of right dg $\cE$-modules) generated by the $e_i\cE$. The dg-algebra $\cE$, and hence $D(\cA)$, becomes significantly more tractable if $\cE$ is \emph{formal}, i.e., quasi-isomorphic to its cohomology $H^(\cE)$ (viewed as a dg-algebra with trivial differential). In general, it can be difficult to establish formality. However, there is a criterion due to P. Deligne: if $\cE$ is endowed with an additional $\ZZ$-grading $\cE^i = \bigoplus_{j\in \ZZ} \cE^{i,j}$ which is respected by the differential, and each $H^i(\cE)$ is concentrated in degree $i$ (for the additional grading), then $\cE$ is formal. In the setting of the previous sections, let $\cL = \bigoplus_{\tau}\cL_{\tau}$ be the direct sum of the simple objects in $\Perv_{B\times K}(G)$. Let $\cE = \Ext^{\bullet}_{B\times K}(\cL,\cL)$, viewed as a dg-algebra with trivial differential. Assume that the category $\MM_q$ of the previous section is the derived category of an abelian category containing enough injectives. Further, assume that the forgetful functor $\MM_q\to D_{B\times K}(G)$ yields a grading (via the weight filtration) in the sense of \cite[\S4]{BGS}. Then, modulo some finiteness adjectives, Theorem \ref{main} and Deligne's criterion yield $D_{B\times K}(G) \simeq \dgDer{\cE}$ (this is Soergel's Formality Conjecture). Conjectures 4.2.2, 4.2.3 and 4.2.6 of \cite{So} also follow. Now $D_{B\times K}(G)$ is not the derived category of an abelian category, but this is not a serious problem for implementing the above argument. However, $\MM_q\to D_{B\times K}(G)$ simply does not yield a grading. The category of Hodge modules is too large. The issue is already visible over a point, since the category of Tate mixed Hodge structures is larger than the category of graded vector spaces. Further, isolating a suitable subcategory of $\MM_q$ seems to be quite difficult (cf. \cite[\S4.5]{BGS}). W. Soergel has explained to me how combining the arguments of this note with a joint project of his and M. Wendt's on `motivic representation theory' (see \cite{SW}) should allow this basic idea to be carried through (also see footnote \ref{note}). This perspective is also explicit in \cite[\S4]{BGS} and \cite[\S G]{B}.	119,357
I’m curious as to how the graphical aspects of the Mark II are going to work. By now we’ve all seen the images, and watched the videos. There’s some very cool examples of Mycroft helping with cooking and setting timers. How are those kinds of features going to work with Picroft units (assuming there will still be the possibility of Picrofts after the Mark II is in production)? How are those features going to work with desktop installations? In a perfect world, we could add a display to our Picrofts with a custom printed enclosure and use the existing graphics there, or have desktop widgets for our PC based installations. Is there an API available (planned to be available) for those kinds of interactions, or will each graphical display need to be part of the skill development? I guess I’m just curious as to how that’s going to work and what will be available to modders.	359,357
TITLE: Prove Abelian groups with two elements of order 2 have a subgroup of order 4. QUESTION [2 upvotes]: Prove Abelian groups with two elements of order 2 have a subgroup of order 4. (From Gallian Algebra). My proof is below, as an answer. Can you verify, critique, or improve my proof, or the proof writing; or provide an alternate or simpler proof? I proved this "manually," via construction. Is there a simpler, more direct, or more abstract approach? My proof is limited to the exact question asked. Is there a more universal approach that would work in similar questions as well? REPLY [4 votes]: Your proof is fine. More generally, if $G$ is an Abelian group with $H$ and $K$ subgroups thereof, the set $$ HK=\{hk:h\in H,k\in K\} $$ is a subgroup of $G$ (this holds also in general groups, provided one of the subgroups is normal). In the case of Abelian groups, though, there is a further interesting property: the map $$ \mu\colon H\times K\to HK,\qquad \mu(h,k)=hk $$ is a group homomorphism (easy verification); the domain is the product group. What's the kernel? We have $\mu(h,k)=e$ if and only if $k=h^{-1}$, hence the kernel is $$ \ker\mu=\{(x,x^{-1}):x\in H\cap K\} $$ By the homomorphism theorem, we know that, in case of finite groups, $$ \|HK\|=\frac{\|H\times K\|}{\lvert\ker\mu\rvert}=\dfrac{\|H\|\,\|K\|}{\|H\cap K\|} $$ Note that $H\cap K$ is not $\ker\mu$, but they have the same order. Now specialize this to $H=\langle a\rangle$ and $K=\langle b\rangle$, where $a,b\in G$ have order $2$ and $a\ne b$. Then $\langle a\rangle\cap \langle b\rangle=\{e\}$, because it cannot contain $a$ and is a subgroup of $\langle a\rangle$. Hence $$ \|\langle a\rangle\langle b\rangle\|=\frac{2\cdot2}{1}=4 $$ Since $e,a,b,ab\in\langle a\rangle\langle b\rangle$, this is the full list of elements. Can we generalize it? Yes, try the case of two elements of order the prime $p$; however, $a\ne b$ is no longer sufficient and something more has to be required, precisely that neither is a power of the other.	18,281
It’s not just Brixton paying tribute. One of the most acclaimed periods of David Bowie’s career came when he was based in Berlin in the 1970s, releasing the Low, Lodger and Heroes albums. Bowie fans are currently leaving flowers at Hauptstrasse 155, the location of the flat that Bowie shared with Iggy Pop during his time in Berlin. The German foreign office have already thanked Bowie for his help in bringing down the Berlin wall (his song ‘Heroes’ is about two lovers separated by it). Tributes are also springing up in Bowie’s birthplace of Brixton. The singer died yesterday following an 18-month battle with cancer.	411,301
TITLE: Relationship between polynomials and the symmetric algebra. QUESTION [8 upvotes]: My question is motivated by the following observation: If one considers a representation of $S_n$ on $\mathbb R^n$ given by permutation of co-ordinates, we can see that there are two invariant subspaces, generated by $$x_1+...x_n=0$$ and its orthogonal complement, spanned by the single vector $(1,...,1)$. The definition for the symmetric algebra on $S^2A:=\{a_1 \otimes a_2+a_2 \otimes a_1=0\}$ which is also invariant under permutation of indices etc. (action of $S_2$) I'm not used to the symmetric algebra, but one can maybe expect in general to find a similar relationship between $S_n$ and $S A$ on one hand, but what is less clear is what the orthogonal complement of $S^nA$ looks like (is it still just the collection of tensor elements that commute? This doesn't seem likely.) My question: is there anyway to build up an analogy between polynomials and the symmetric algebra? What would the indeterminates "look like"? I expect degree $\dim V^n$ with indeterminates in the basis elements for the tensor product, but I'm not so sure. REPLY [9 votes]: Indeed are the same algebra, the dimension of the vectorial space is equal to the number of variables of the polynomial algebra, i.e. $\mathbb{K}[x]$ is obtained by the symmetric algebra of the 1-dimensional vector space over the field $\mathbb{K}$, $\mathbb{K}[x,y]$ is obtained by the symmetric algebra of the 2-dimensional vectorial space, etc.. I will use the notation of generators and relations to outline the thing. If it's not something you're used to, just tell me and I will change notation to a more familiar one. Let's do an example with 2 variables and then you can generalize easily. In this case $V$ has to be a 2 dimension space. Let's call the vector of the base {$x,y$}. Then the simmetric algebra is obtained by $$S\left(V\right)=T\left(V\right)/\left\langle v{\scriptstyle \otimes}w-w{\scriptstyle \otimes}v\right\rangle,$$ where $T(V)$ is the usual tensorial algebra of tensors of any degree $T\left(V\right)\cong\mathbb{K}\oplus V\oplus\left(V\otimes V\right)\oplus\left(V\otimes V\otimes V\right)\oplus\ldots\,\,\,$. Using the base ${x,y}$ then any element in $S(V)$ is of the form $$p=a_1+a_2x+a_3y+a_4x\otimes x+a_5x\otimes y+a_6y\otimes y+....$$ or using another notation $p$ il the polynomial $$p=a_1+a_2x+a_3y+a_4x^2+a_5xy+a_6y^2+....$$ So in general if you have an $n$-dimensional vector space $V$ over $\mathbb{K}$ with basis {$x_1,x_2,....x_n$}, then the symmetric algebra is isomorfic to the algebra of polynomials with n variables $$S\left(V\right)\cong\mathbb{K}[x_1,x_2,....x_n]$$	10,912
TITLE: Cyclic amalgamation of solvable groups QUESTION [6 upvotes]: Given a nonempty set of groups $\{G_\lambda\|\lambda \in \Lambda\}$ together with a group $H$ which is isomorphic with a subgroup $H_\lambda$ of $G_\lambda$ by means of monomorphism $\phi_\lambda: H \to G_\lambda$. There is a frequently used term known as the free product of the $G_\lambda$'s with the amalgamated subgroup $H$. Roughly speaking, this is the largest group generated by the $G_\lambda$'s in which the subgroups $H_\lambda$ are identified with $H$ by $\phi_\lambda$. Such groups are known as generalized free products. My question concerns a result proved by Kahorbaei and Majewicz claiming the following. Question. I'm uncertain about this proof. Why must the cosets $a\delta_{m+1}A$ and $b\delta_{n+1}B$ have the same order in its respective factor group? I'm worried about the infinite cyclic subgroup. Here is the definition for residually solvable group. Edit: I suspect there are counterexamples to the statement (or maybe I misunderstood their definition of the generalized free product). It seems that it requires extra hypothesis such as cosets of same order or something called compatible filtration in Proposition 1 of Baumslag (as suggested by Carl-Fredrik Nyberg Brodda). REPLY [3 votes]: The elements do not have to have the same order it would seem. According to the paper ON TORSION-FREE METABELIAN GROUPS WITH COMMUTATOR QUOTIENTS OF PRIME EXPONENT by Narain Gupta and Said Sidki there are torsion-free metabelian groups whose commutator quotients are nontrivial with prime exponent for any odd prime $p$. So choosing examples $A$ and $B$ with different primes, say $3$ and $5$, gives infinite order elements $a\in A$ and $b\in B$ (not in the commutator subgroup) that have orders $3$ and $5$, respectively, modulo the commutator subgroup.	125,138
TITLE: About a theorem in Beauville's book. QUESTION [7 upvotes]: Look at the following theorem (due to Castelnuovo) taken from Beauville's book on complex surfaces: I have a question about the behavior of $f$ on the exceptional curves generated by $\eta$. Suppose that $E$ is an exceptional curve coming from a blow up of $\eta$ at a point $p\in S$, then what is $f(E)$? I think that $f(E)$ can't be a point, otherwise I could define $\phi(p):=f(E)$. This is a contradiction with the definition of rational map. Hence I conclude that $f(E)$ is a curve on $X$. Is my reasoning correct? Edit: Maybe the above reasoning is wrong, but at least I'd like to know which are the exceptional curves (coming from blow-ups) that are not contracted by $f$ (if there is any). I'm sure that I've read somewhere the following thing: The exceptional curve of the last blow-up can't be contracted by $f$. Can you explain this sentence? REPLY [0 votes]: Let think of a rational map $\phi: S\dashrightarrow X$, and $\phi$ is not defined at an isolated point $p\in X$. What happened such that $\phi$ can not be extend to p? there must be 2 points near p such that their images are far away to each other. Otherwise by continuity, the map can be extended. So take a neighborhood of p, the image of this ball(removing its centre p) will be quite stretched in X. Let's seem a concrete example: $\phi : \mathbb{C}^2 \longrightarrow S^3$, where $S^3$ is the unite 3-sphere $\phi(r_1e^{i\theta_1},r_2e^{i\theta_2})=(e^{i\theta_1}, e^{i\theta_2})$ is not defined at (0,0). However any small neigborhood of origin (removing the origin) has image equals the whole $S^3$. If you blow up the origin, then the image of exceptional curve can not be any point. because any point infinitely near the origin is also infinitely near each point on the exceptional divisor. If you want the rational map has an analytical continuation to the exceptional divisor, then you must let the image of the exceptional curve as stretched as the image of the neigborhood (removing the origin).	172,412
11 Copycat Recipes for Baking Ingredients: Recipes for Homemade Bisquick & More Bonus: Get our newsletter & special offers for free. We will not share or sell your email address. View our Privacy Policy Trail Mix Cookie Mix We are adding the recipe to your Recipe Box. This was added to your Recipe Box. You must be logged in to add a recipe. Login \| Register Ingredients - 1/2 cup packed brown sugar - 1/2 cup white sugar - 3/4 cup wheat germ - 1/3 cup quick cooking oats - 1 cup raisins - 1/3 cup packed flaked coconut - 1/2 cup all-purpose flour - 1 teaspoon baking powder Instructions	275,769
If I were a more ambitious man, I would save the first part of this review for a White Paper that I would sell to the music industry. It is for the label executives who are going hungry in the street having the bread torn from their children’s mouths by dastardly file sharers, CD burners, and MP3 traders that I write this. I’ve entitled it “The Successful Marketing and Selling of a Band”. It was nearly two years ago when, while taking a break from the drudgery of the workplace, I discovered the Clean while visiting the Merge Records website. I first became interested in Merge because it was founded and run by members of Superchunk, one of my all time favorite bands. At first I was strictly a Chunk man, however, driven by my insatiable musical curiosity I began to look into other albums that bore the Merge label. A college chum recommended I check out the Magnetic Fields, which led to a brief period in my life where I refused to listen to any music not sung by Stephin Merritt. My fascination with all things Merge led me to two bands that would forever alter my perception of music, the Neutral Milk Hotel and Lambchop. Both bands presented music in a format that was vastly different than anything I had ever heard before and both led me down paths that would open my eyes to worlds that I never before would have found. Having never steered me wrong, I developed a trust with Merge that would bring me to purchase records by bands I had never heard. Because of my success with Merge I often visited their website. On their MP3 page, which offers free downloadable music, I listened to a track by the Clean. Smitten with their unique brand of psychedelic pop, I went out and purchased their album Getaway. The Clean struck me as a developed version of the Soft Boys, and when I heard that they had been existence for over 20 years, I yearned for more. I returned to my trusty Merge Records website and found out that they had put out a solo album, A Feather in the Engine, by Clean founder David Kilgour, which I ordered online. I also discovered the Bats, featuring Clean member Robert Scott (their records can be found on Mammoth). By building up consumer trust through reliable products that consistently meet or exceed expectations Merge continues to thrive. So instead of investing billions in flashy videos or commercial campaigns, labels would be wise to invest in bands that actually have talent. Pleasing consumers doesn’t hurt either. The Clean were formed by Hamish and David Kilgour in New Zealand in 1978. In 1979, they finalized their lineup when they added bassist Robert Scott. The trio went on to release two 12-inch EPs, two 7-inch singles, a cassette compilation, Odditties, and a live EP Live Dead Clean, on Flying Nun Records. Had they come from New York City or England, they would most likely be household names joining the ranks of seminal bands like XTC, Sonic Youth and even Velvet Underground. Instead, they’ve been relegated to cult status embraced by bands that follow them and the few enlightened fans that were lucky enough to get clued into them. Thankfully, Merge Records has seen fit to compile everything the band has ever recorded up to Getaway and release it as a double disc anthology. The first disc is made up of all of the group’s early recordings for Flying Nun records. Disc two is made up of the band’s three full lengths Vehicle, Modern Rock, and Unknown Country, all released between 1990 and 1996. As a bonus, the second disc also features four songs that appeared on two rare 7” singles.. “Outside the Cage” and “Safe in the Rain” are the two most optimistic VU tracks that Lou Reed never wrote. Since none of this is still in print it is all essential. The second disc showcases a more mature and focused band. As the band progresses from Vehicle to Modern Rock, they begin to develop a sound akin to the one Yo La Tengo has been toying with for the last five years. The tracks from Unknown Country are mostly instrumental and lack the spirit of earlier efforts. Music lovers should log off the web (or go to the Merge website) and purchase this immediately, it’s that important. The Clean are one of those rare bands whose sound can be heard in the best of the bands who came after them. The amazing guitar sound shared by Yo La Tengo, Pavement, Superchunk and Archers of Loaf is realized here, only years before those bands came along. For 20 years, too many people have been overlooking one of the most forward thinking, melodic bands ever, missing out on the treasure that is the Clean. Anthology is an audiophile’s treasure, delivered to the masses by the fine people at	44,625
R2-D2 and BB-8 Towel, Star Wars: The Force Awakens now £9.00 reg £14.99 , you save: 39% Item no. 427265191982 FREE SHIPPING on orders over £50 R2-D2 and BB-8 are ready to join in the poolside fun with this characterful Star Wars towel! Made from super soft fabric, it features bold artwork of the loveable droids with Star Wars logo detail.See more See less It's added to your favourites!View Your Favourites Added to {0} Gift List!View Your Gift List	243,372
TITLE: For every binary number x, is there a power of 3 beginning in x? QUESTION [2 upvotes]: I noticed that for the binary number $k=(1,10,11)$, $3^k$ begins with the digits of k. That relationship immediately breaks down, but it doesn't seem hard to find a power of 3 beginning similar to any binary number. E.g., $3^2$ starts with 100, $3^4$ starts with 101, etc. Is this actually the case? Can every binary number be found at the start of some power of 3? REPLY [4 votes]: Yes. For each finite binary string starting with $1$ there is an interval $[a,b) \subset [0,1]$ such that $3^n$ begins with that string iff the fractional part of $\log_2(3^n)$ is in that interval. But because $\log_2(3)$ is irrational, the fractional parts of $\log_2(3^n) = n \log_2(3)$ are dense in $[0,1]$.	173,492
TITLE: Geometric interpretation for the row space QUESTION [3 upvotes]: I have a clear understanding of how the column space relates to a transformation of the basis vectors, but I really do not see the connection between the column space and the row space of the matrix. If we have $3 \times 3$ transformation matrix $A$ defined by $$ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} $$ We have the basis for the column space given by $$ \left( \vec{c_1}\begin{pmatrix} a_{11} \\ a_{21} \\ a_{31} \end{pmatrix}, \vec{c_2}\begin{pmatrix} a_{12} \\ a_{22} \\ a_{32} \end{pmatrix}, \vec{c_3}\begin{pmatrix} a_{13} \\ a_{23} \\ a_{33} \end{pmatrix} \right) $$ Then the operation on a $3 \times 1$ column vector $\vec{v}$ is defined as $$ A \cdot \vec{v} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} xa_{11} + ya_{12} + za_{13} \\ xa_{21} + ya_{22} + za_{23} \\ xa_{31} + ya_{32} + za_{33} \end{pmatrix} $$ Which is equivalent to the linear combination of the basis vectors for the column space $\left( \vec{c_1}, \vec{c_2}, \vec{c_3} \right)$ such as $$A \cdot \vec{v} = x\vec{c_1} + y\vec{c_2} + z\vec{c_3}$$ Or the dot product with the basis vectors for the row space as shown in the matrix-vector product. We have the basis for the row space given by $$ \left( \vec{r_1}\begin{pmatrix} a_{11} \\ a_{12} \\ a_{13} \end{pmatrix}, \vec{r_2}\begin{pmatrix} a_{21} \\ a_{22} \\ a_{23} \end{pmatrix}, \vec{r_3}\begin{pmatrix} a_{31} \\ a_{32} \\ a_{33} \end{pmatrix} \right) $$ And the resulting matrix-vector product given by $$ A \cdot \vec{v} = \begin{pmatrix} \vec{r_1} \cdot \vec{v} \\ \vec{r_2} \cdot \vec{v} \\ \vec{r_3} \cdot \vec{v} \end{pmatrix} $$ My question concerns the relationship between the linear span of the column space and the linear span of the row space. Is there any geometric relationship between the column space and the row space apart from this algebraic manipulation that emphasizes the fact the linear combination of the basis vectors for the column space $\left( \vec{c_1}, \vec{c_2}, \vec{c_3} \right)$ is equal to the projection onto the basis vectors for the row space $\left( \vec{r_1}, \vec{r_2}, \vec{r_3} \right)$? I am looking for a more geometric interpretation rather than the standard algebraic relationship. REPLY [1 votes]: One nice interpretation of this phenomenon comes out of studying the bilinear form $f:\Bbb R^3 \times \Bbb R^3 \to \Bbb R$ defined by $f(x,y) = x^TAy$. Note that for any fixed $x \in \Bbb R^3$, the function $f(x,\cdot): \Bbb R^3 \to \Bbb R$ defined by $f(x,\cdot)(y) = f(x,y)$ gives us a linear map (an element of the dual space to $\Bbb R^3$, if you prefer). We can identify the column space of $A$ with all maps of the form $f(x,\cdot)$; each column of $A$ corresponds to plugging in a standard basis vector for $x$. Similarly, we can identify the row space of $A$ with all maps of the form $f(\cdot,y)$; each row of $A$ corresponds to plugging in a standard basis vector for $y$. In a sense, your observation amounts to the statement that $$ f(x, \cdot)(y) = f(\cdot,y)(x). $$	139,946
Interested in this property? Ground floor Service quarters Gym laundry room 1st floor Wide Sitting & dining area fully fitted ,open plan kitchen Guest Clock room with basin 2nd floor exquisite Master bedroom with en-suite bathroom & dressing area 3 bedrooms with en-suite bathrooms Exterior Marvelous Garden Parking area *Security..	6,380
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Nathan Garrett joined the law firm of Graves Bartle Marcus & Garrett, LLC after a highly distinguished career in both state and federal law enforcement. Prior to entering private practice, Nathan was a Supervisory Assistant United States Attorney, serving as the Chief Prosecutor of the National Security and Terrorism Section in the United States Attorney’s Office for the Western District of Missouri. In this capacity, Nathan was responsible for coordinating and managing complex intelligence and criminal investigations and prosecuting the District’s national security cases, which included terrorism financing and support, money laundering, fraud, immigration violations, and violations of the International Emergency Economic Powers Act (security controls related to import/export activity, specially designated states and nationals, and Presidential Executive Orders). From 2002 through 2005 (and by special detail through 2007), Nathan served as an Assistant United States Attorney for the Northern District of Texas, assigned to the Terrorism and Cybercrime Section. While working out of the United States Attorney’s Office in Dallas, Nathan coordinated, investigated and prosecuted complex national security cases in addition to computer and intellectual property related offenses and fraud schemes. Nathan also served as a Special Assistant United States Attorney from 2000 through 2002 on special detail to the United States Attorney’s Office in Dallas, Texas while serving as a Special Agent for the FBI. In this unique capacity, Nathan was responsible for transitioning traditional national security investigations into criminal prosecutions. Nathan was assigned at the request of the Chief of the Counterterrorism Section at the Department of Justice to the post-9/11 created Terrorism Financing Task Force in Washington D.C. This assignment involved assisting the Department of Justice in identifying and coordinating national security investigations around the United States. Nathan served as a Special Agent in the FBI from 1998 through 2002, and was assigned to the Bureau’s International Terrorism Squad and the Violent Crime and Major Offender Squad in Dallas, Texas. Nathan’s work in the national security arena led to extensive foreign travel and cooperative interaction with all levels and divisions of government within the United States and abroad. Nathan has been the keynote speaker at numerous government and private seminars focusing on national security, international investigations, immigration, trade restrictions, terrorism and money laundering. Prior to Nathan’s work in the FBI and Department of Justice, he served as a Missouri State Trooper and State Prosecutor. Nathan was born and raised in West Plains, Missouri. He received his Bachelor of Arts degree with a double major in History and Philosophy from Cornell College located in Mount Vernon, Iowa. Nathan received his Juris Doctorate from the University of Tulsa, School of Law. Nathan’s private practice is focused on federal and high-level state white-collar crime and regulatory enforcement, international and domestic corporate compliance and policy, internal investigations and due diligence. Nathan F. Garrett Voice Work Pref Office (816) 256 3181 Graves Bartle Marcus & Garrett, LLC 1100 Main St Suite 2700 Kansas City, MO 64105 © 2009 Graves Bartle Marcus & Garrett, LLC;	157,422
The Knox City Council reviewed a new Park Board Ordinance at its meeting this week. “This was brought about when we were doing our five year plan,” said Mayor Chambers. “Our five year plan was sent to OCRA for approval and they found that our original park plan was outdated by state statute. It had to be newer than what ours was so Dave [Matsey, City Attorney] has put together this ordinance and it is almost identical to the old one. There’s very little difference. It’s right out of the state statute books.” The only changes in the new ordinance had to do with establishing a Capital Improvement Fund and language for accepting gift funds. The size of the Board would remain the same with four members appointed by the Mayor and one each by the Library Board and School Board. Two questions came out of the discussion that delayed approval of the ordinance. One had to do with how long the Library Board and School Board member’s terms were, and the other was whether the Council needed to approve the acceptance of gift funds. The ordinance will be revisited after research done on these two issues.	261,884
Exceptions – it’s not you, it’s me. A lot of grammar nonsense comes from labels that we use and that we assume are sufficient explanation in themselves to generate their own correct examples. Then, when students attempt to produce examples in accordance with these labels only to find out that they sound ‘strange’ to a teacher’, they then often start asking questions about examples that fail to fit the labels. Teachers then respond in one of three ways: there’s the easy (woolly liberal) ‘Oh, that’s an exception’, the more dogmatic ‘It’s wrong/bad English’, or there’s the extended ‘subtle’ explanation that tries to encompass these more complex uses. Of these three options, our preference would be ‘the exception’, because at least this is less likely to bring about feelings of failure in students. Exceptions are down to the idiosyncracies / curiosities / stupidities of the English language (delete as appropriate). They’re essentially linguistic versions of the ‘it’s-not-you-it’s-me’ break-up line, which may be annoying and disappointing, but at least isn’t laying the blame at your uselessness in the way that the ‘it’s bad English’ response is. Nor will it bore you to death and and ultimately confuse students like the extended explanation almost always does. Usage above meaning However, it may be the case that sometimes we’d be better off just avoiding the label in the first place. Students need to accept ambiguity to be successful in language learning (and perhaps in life!). I think one of the key elements of a lexical view of language is that the meanings we give to any pieces of vocabulary or grammar can only ever be partial – and rather than giving more explanation, more ‘meaning’, more labels, we would be better off simply giving more examples of usage (and getting students to read and listen more to language in use). The case of stative verbs Take stative verbs. For those of you not familiar with this particular description, ‘stative’ is a label that tends to be given to a group of verbs that (supposedly) don’t use the present continuous tense. So here are a few explanations from coursebooks which shall remain nameless: - Some verbs express a state – not an activity – and are usually used in the present simple only. For example: like, know, think, agree, understand, love. - We cannot normally use some verbs (stative verbs) in the continuous form. For example: agree, belong, cost, know, like, love, matter, mean, need, seem, understand, want. - We don’t use stative verbs (be, have, like, love, hate, want) in the present continuous. Of course, as you may well be aware, many of these verbs can actually be used (and are used!) in the continuous form. I’m loving it has become incredibly widespread, perhaps thanks to the McDonald’s slogan, but then the slogan no doubt came from advertisers picking up on usage. Here are some other common examples: - So if I’m understanding this right … - It’s costing me an arm and a leg! - I’ve been meaning / wanting… to do it for ages - I’m thinking of … leaving. - He’s having … a crisis of confidence. - Ignore me, I’m just being silly. So what is a stative verb? The problem of circularity Part of the problem with using the term ‘stative verbs’ or ‘verbs that express a state’ is that it suggests the verbs are somehow infused with this sense all the time. In fact, at best we can say that some verbs when they express the meaning of a state are not usually expressed with a continuous form. But even then, does the example of cost above contradict that? Or is it not a state here? Which brings us to the rather bigger problem of what the hell a ‘state’ is anyway? Think about your Facebook status: job, relationship, friends, likes etc. Certainly, we would normally only say the following in the present simple: - I’m unemployed. - I have a girlfriend. - I hate my brother. - I love swimming. But the following could also be expressions of these same ‘states’: - I’m not working. - I’m seeing someone. - I’m not speaking to my brother ever again. - I’m really loving my swimming. So why are these verbs not seen as stative? Because they are used in the continuous form which is a mark of being non-stative! And so we enter a rather circular and pointless version of grammar rules – rather than a generative one. We don’t use stative verbs with the present continuous because stative verbs are verbs which aren’t used in the present continuous! Do we actually need a new rule? Interestingly, one of the grammar explanations quoted above, also gives the following example as an example which is different to the explanation about stative verbs: - Frazer comes from Scotland - NOT Frazer is coming from Scotland. The first thing to say about this is that the example is clearly not wrong if seen in terms of the verb come (there is no further explanation of why it’s wrong). The first sentence explains the permanent fact of his birthplace/nationality, the second could be telling us where he is travelling from. Come therefore can be used to express a ‘state’, but I’ve never seen it listed as a stative verb! Why not? Essentially, it’s because the way come is used here is in keeping with the normal meanings which we attach to the present simple and the present continuous. However, isn’t the same true of other ‘stative’ verb?. Rather like we discussed with reported speech, we seem to have actually created a new category of rule where none is needed. If we take the idea that the present simple expresses ideas about now that we consider permanent or complete, or facts about ourselves, compared to continuous forms which are essentially temporary and unfinished or in progress, then both the ‘stative’ and the continuous use of all the verbs so far mentioned fit these meanings more or less without creating any new categories for students to worry about. Present simple for present and complete actions/states I think part of the issue here is also that we forget the ‘present and complete’ use of the present simple as in ‘He takes on Stones . . . He shoots and scores’. Neither the past simple nor the present continuous fit here when commentating on events you are watching and the same is true when we say things like I know / I understand / I agree. They happen at this moment, but we see them as complete in the moment. It won’t solve everything, but maybe it’s one less problem! This is not to say, of course, that our meanings for the present simple and present continuous are unambiguous and will never lead to student ‘errors’. However, we would suggest that more explanation and extended lists of meanings will not actually help. In the end, usage that corresponds with what the vast majority of people say can only come from learners experiencing the language of the vast majority of people! In terms of study and learning, students are also probably better off trying to expand the meanings they are able to make in English through learning more vocabulary. Want to learn more about grammar nonsense and what to do about it? Buy our e-book on the subject here. As ye guys well know, Micheal Lewis tackles this thorny issue brilliantly in his book, “The English Verb”. He sees it as a semantic issue: does the person intend to desribe sth that’s changing for them(continous= i’ve been having second thoughts about going.) or desribing sth that’s seen as complete, a fact at that moment (present simple= i think i’ll go) Interestingly, I found out that “loving/liking” is often used “I’m loving/liking this album more and more, it’s growing on me bit by bit”. Indeed, though as Andrew says above, in this sense, these uses are not in any way special or unusual. The example you give of I’M LOVING THIS LP is a case in point: it’s just a normal continuous usem whereas I LOVE THE NEW YORK DOLLS is something that’s always true, it’s a permanent fact. very appropriate example. i’d say in your case Hugh, it definitely IS a permanent fact!!!!! The New York Dolls are great I got caught out in class with ‘love’. In the MacDonald’s slogan it really means ‘enjoy’ so can be used in the continuous form I found out (after a bit of Googling). I’ve also noticed that it’s quite trendy among hip young people to overuse present continuous, I guess this is why those clever people in MD’s marketing department chose it and messed about with the spelling and punctuation, to make it sound like ‘teen speak’? Hi Neil. Thanks for taking the time to write. Appreciated. I think ‘love’ is a classic case of a fairly useless rule creating extra problems for all concerned. As you note, it’s often more to do with different meanings of words tending towards continuous uses – or not. In a sense, though, whether you think I’M LOVIN’ IT means they’re enjoying it – or that they are actually really are loving it, as they say, is irrelevant because I’m loving it simply means at the moment, started not finished – just as It’s raining means at the moment and If I’m understanding you correctly does! As for young folk and their ‘overuse’ of the present continuous . . . I’m assuming that by ‘overuse’ you’re simply meaning they use it more than you do, right? Policemen, eh? They’re getting younger every year. 🙂 Yes indeed, by overuse I didn’t mean any negative connotation, just something I heard when I taught in an FE college full of apprentices, and then heard my nieces and nephews doing it too, think it’s called being ‘street’ LOL (as they kids would say) PS Loving your posts/site Look at you and your hip young speak. 🙂 I don’t know if my solution is bulletproof (I doubt that), but I use general semantic labels like REPORTING OPINIONS (‘I guess you’re right’ vs. ‘I’m just guessing.’ / ‘I think that’s true.’ vs. ‘Hold on, I’m thinking.’ etc.) and they seem to work fine. Other labels would include PERCEIVING PROPERTIES (i.e. passive, involuntary input vs. active examination, e.g.: ‘I see her now.’ vs. ‘I’m looking at her now.’ / ‘I smell the fish now.’ vs. ‘I’m smelling the fish now.’), DESCRIBING PROPERTIES (‘The soup tastes delicious.’, ‘The fish smells awful.’, but also ‘The beach stretches on and on.” vs. “She’s stretching her arms.’ / ‘The tower rises to an enormous height.’ vs. ‘The balloon is rising quickly.’, the latter two examples showing, I believe, this class is potentially open), AFFILIATION (e.g. by ‘ownership’: ‘I’m having a shower now.’ vs. ‘I have a shower now.’, but also ‘part-whole’ sort of relationships: ‘The soup contains meat.’ vs. ‘They’re just containing the enemy.’), and a couple of others. I always stress it is actually a particular meaning/use rather than a verb (represented by, say, a lemma in learners’ minds) itself that requires/implies stative interpretation, thus usually blocking the continuousness/progressivity that would otherwise normally apply in that particular situation. My students seem to graps rather quickly that once a verb can be assigned one of these lables in a particular situation, it can be interpreted as being ‘stative’. On the other hand, I also remember to warn them that the verb’s ‘stative behaviour’ may be ‘overriden’, e.g. for special emphasis. Thanks for the comment. I have to say I’m not sure I fully follow all of your descriptions! What level do you teach this at? I think there would be a problem with teaching this language to low levels. If you are using L1, I guess that would help, but personally I would prefer to spend time teaching something they might use outside the class. Some of these examples may be helpful from that point of view – it smells awful, it contains meat – but others seem to make use of infrequent meanings or unusual examples (they’re containing the enemy, she’s stretching her arms . Who would you say this second example to and why?). I also thinking that saying the meaning of a verb ‘blocks’ the use of the continuous aspect is an odd way of looking at things. To me that suggests somehow it should be continuous, but just because it’s this verb (meaning) it isn’t! When we say ‘the soup contains meat’, it’s reasonable to think of this as neither temporary nor unfinished, but complete and part of the soup’s ‘identity’: hence we use the present simple. There is no need to provide a new category of verb or meaning that is separate from the general rules for the simple tense and progressive aspect. Having said that, and as I said in the post, this does not mean my simplified explanation will result in greater accuracy. I basically think that only comes from exposure and interaction over time. to sum it up – don’t tell the students about stative verbs at all, just teach them simple/continuous difference as ‘something that is always true, routine vs something happening now or temporary’? And this rule applies to ALL verbs? Did I get your idea right? It sounds quite revolutionary))) (is sounding?))) Hi Olga – Pretty much, yes. It keeps life much simpler and clearer for students! There will be the odd verb here and there (know, own, belong) which you may want to say don’t GENERALLY get used in the continuous form, but other than that . . . Thanks for answering, for me, as a not-native-speaker, it’s kinda difficult to explain sometimes))) I just FEEL that this is OK to say, and that is not, I’ve always dreamt for a universal rule, which doesn’t obviously exist))) which verbs to mark as ‘odd’ that’s the question))) as the old rule with a lot of exceptions doesn’t help students))	280,062
Fix #1 - May 2014 Fix #2 - June 2014 Fix #3 - July 2014 Fix #4 - August 2014 Fix #5 - September 2014 Fix #6 - October 1, 2014 Fix #7 - October 29, 2014 Fix #8 - November 2014 Fix #9 - December 2014 Fix #1 gives you all the information about the Stitch Fix service, so I won't reiterate here. This month, I requested: - More long-sleeved tops and sweaters - The elusive 'colorful cardigan' - A less fitted skirt option and - A 'statement necklace.' This time around, I was ready to invest in a cute piece of jewelry, and I looked forward to what my stylist would send. My box arrived yesterday. It was a larger box size, so I knew it would have some bulky items in it. I was hoping for a heavy sweater or cardigan. I mean, it's January, and I can practically see Canada from my living room, so I need some more warm stuff! I opened it up, and had 4 separate wrapped packages. (Usually you get everything wrapped up in one bundle, so this was intriguing). I took out all 4 bundles and snapped a photo of them. Can you guess what my stylist may have done for me? Yes, I was pretty certain my stylist had sent me TWO pieces of jewelry to choose from. I wasn't sure if I was sad about 2 of my 5 items being jewelry or happy. I went for 'happy' because after the holidays, I was feeling pretty broke and knew I wouldn't be spending much this month. I unwrapped each bundle and took in the beauty that is Stitch Fix: I hung up everything for picture-taking and then started the try-on phase. I never look at prices until I am all done giving my evaluations of what's in my box. I want to choose items based on likeability, not cost. Plus, I sometimes have kept something despite the price and have been very glad I did so! 1) The first item I tried on was the jeans. They were the Denna Skinny Jean by Kut From the Kloth for $58.00. During my November fix I discovered that Kut From the Kloth jeans fit me very well. ADVICE: If you are pear-shaped or have a 'curvy' lower half, this is the jean brand for you. Very forgiving in the thigh and butt areas. If other jeans from Stitch Fix have not worked, I highly recommend this brand. I was not in the market for a new pair of jeans, but if these babies fit well, I was up for adding them to my wardrobe. A woman does not turn down a nice-fitting pair of jeans! At the time I hung them up and tried them on, I did not know the price was so awesome. Most pants I have purchased from Stitch Fix have been between $75 and $100. I slipped them on with a plain black t-shirt. The fit was fantastic! It was a mid-rise jean with a great waistband that didn't seem to overstretch. The legs on these 'skinny' jeans were a little bit loose, giving me a different fitting jean. I liked the comfort of these jeans. The length was perfect (I'm almost 5'10") I knew these were a KEEP item. I had to have these. By the way, I wore these all day yesterday, and the waistband never stretched out. Really a great pair of pants. 2) The next item I tried on was the Moni Stud Detailed 3/4 Sleeve Blouse by 41Hawthorn for $58.00. I will be honest. I've seen this blouse before in other people's fixes online and really wasn't over the moon about it. I've never seen it in white. It was a bit sheer, so it came with its own white tank. I didn't have high hopes for this one. I was also disappointed in the thinness of the material. I needed warm stuff, not sheer blouses. A little bit of sad crept into my day. But I try everything on, people! I do not set something aside in the "NO" pile until it has been tried on. Here we go! a) This was actually really cute and flattering. b) It was in white, which is a good neutral for me (off-white is bad, bad, bad). c) The sleeves that I thought I would find irritating weren't half-bad. e) This would be wearable with many pants & skirts I own. Those are the good things. If it were spring, I might've kept this one. Here's the bad: a) It was very sheer fabric. I have 18 inches of snow outside, so I wouldn't be wearing this any time soon. b) The tank that came with this was too large and I had to yank it backwards so it actually covered me properly. So the good did outweigh the bad. I put this in the "MAYBE" pile in case the price was so incredible that it was too good to pass up. 3) The next item was the Reeda Quilted Puffer Jacket by Andrew Marc for $118.00 in navy blue. Let me start by saying I received a lovely red puffer coat with a hood and adjustable waist for Christmas from my husband. I love that coat. It is my dream coat. Very flattering with a fantastic hood. So, needless to say, I was not in the market for a new jacket. But let's hang it up, shall we? I really liked the navy blue color. That is one of my 'neutrals,' so that struck me right off the bat. It also looked like it had some shape to it. Time to try that puppy on. But, hey, we already knew I wasn't looking for or needing a new jacket, so it didn't pain me at all to put this in the NO pile. 4) Sadly, we are now done with the clothing portion of this review. We move to the jewelry! Next I tried the Genie Faceted Teardrop Stone Statement Necklace by Romolo for $38.00. It was okay. I was not blown away by this necklace at all. It still had the look of cheapness to it. However, the worst part was the clasp. Holy heck, who designed this thing? The clasp was very chunky. The links you had to clip it into were pretty small. The main part of the clasp was too thick to fit through the links. Only the tiny sliding clasp piece fit through the hole of the link (if that makes sense). So, unless I perfectly aligned the retracted clasp end with the middle of the link, there was no way to successfully close the clasp. In fact, the first time I tried it, I thought I had it closed properly. But then the necklace fell off, and I took a closer look. NEARLY IMPOSSIBLE TO CLOSE. And the necklace is too short to slip over your head. You HAVE to use the clasp to wear it. That sealed the deal for me. NO. 5) The last item was another piece of jewelry - the Carly Multi-Stone Layering Necklace by Bancroft for $36.00. It's hard to see in the box how long this necklace is. The green of the 'stones' appealed to me. I thought this necklace might be better. At least with this one, I didn't have to worry about the clasp. It looked okay, and I probably could've found things to wear it with. However, the quality was so cheap, I really couldn't envision myself spending money on it, unless it was incredibly discounted. This was a NO. So, all in all not a great box for me. I kept 1 item. However, I will remind you that I didn't have much money to spend...so it was actually GOOD that I didn't like more in this box. For me, the best part was looking at the prices and finding out the jeans I was keeping were only $58! I'd already spent $20 on the styling fee, so an additional $38 for a great pair of jeans wasn't that bad at all. In fact, I was pretty pleased. And, yes, I decided to send back the blouse...the one 'maybe' in the mix. When I saw the price ($58), I just couldn't pull the trigger. Maybe if it had been $38, I would've done it. But it was too hard to justify buying a summery-type item in the middle of January! My stylist told me in the note that came with my fix that she couldn't find a cardigan in my size. And since I didn't receive much of anything I asked for, my guess is this was an inventory dump box. I possibly got such a good deal on the jeans b/c they were trying to unload them. Which is all right with me. The biggest disappointment was the jewelry. I thought both pieces were incredibly overpriced for the quality. I even told Stitch Fix so in my remarks about this Fix. I have bought better quality jewelry in the $5 sale at my local department store. I think if Stitch Fix is going to include jewelry, they need to up the quality or lower the price. I don't know if I'd ask for jewelry again, honestly. Definitely still enjoying my Stitch Fix experience. My wardrobe is being transformed by my Stitch Fix items, and I find myself weeding out old stuff as I accumulate more up-to-date and stylish pieces. Can't wait for February to see what the Stitch Fix Fairy brings my way. If you are interested in trying out Stitch Fix, can I interest you in using my referral link? Full disclosure: I will receive a $25 credit after you make your first order. Thanks for stopping by!	323,975
Davies: It’s a challenge but we’re quietly confident Despite having won twice in Dublin on their way to lifting the Guinness PRO12 title last season the Welsh regions hopes of appearing in their first European final has been dismissed by the bookmakers. But the 32-times capped Wales international scrum-half insists being written off doesn’t bother Wayne Pivac’s side. “Leinster are rightly favourites going into the game but we as a group of players quite like going into games as underdogs,” said the 27-year-old. “Games don’t get bigger than Leinster in Dublin in a Champions Cup semi-final. It’s going to be a massive challenge but it’s one that we are looking forward to. “Leinster are one of the best teams in Europe. They’ve finished top of our conference in the Guinness PRO14 by a good number of points, while they had a great victory over Saracens a couple of weeks ago, but we are quietly confident.” Prior to their championship winning exploits last season the Scarlets’ end of season run-in would typically consist of scrapping for Champions Cup qualification. But nowadays the benchmark has been set much higher with the West Walians fighting for two trophies. Davies admits it is a tough ask for the squad especially with injuries piling up but insists the West Walians have the quality to bring more silverware back to Llanelli. “Every game from now until the end of the season is knockout rugby,” he said. “It’s been mentioned once or twice but we know how big of an achievement it is for us to have got this far having not won anything for a long time before last season. But we as a group of players want to take that extra step and make it to the final. We set a target at the start of the season to be involved in the knockout stages in both the league and the cup. I think that’s an achievement for ourselves to have got that far but we are hungry for more. The Scarlets have some unfinished business with this stage of the competition having been denied at the death on two occasions. Their most infamous semi-final defeat came in 2002 at Nottingham’s City Ground against Leicester Tigers when a Tim Stimpson penalty from 60 metres hit the post and the cross bar to deny Llanelli. But Davies is adamant the players won’t be giving their previous European semi-final failures much thought. Neither will the squad be drawing on last season’s stunning PRO12 semi-final triumph over Leinster at the RDS. “I’m aware of what happened with Tim Stimpson’s kick knocking us out,” he said. “It has been mentioned a few times but we aren’t going to look too much into the past because it’s about the here and now. It’s all about this group of players and hopefully we can make a bit of history on Saturday. “Lat season’s PRO12 semi-final was a good day for us and I thought we played really well but that’s nearly a year ago now. A lot has changed since then and both sides have a few different players that have come in. “We won’t be looking back too much at that game because it’s a completely different challenge for us this weekend.” The visitors have won many admirers for their free-flowing brand of rugby which blew away both Leinster and Munster in last season’s PRO12 play-off’s. Often when sides reach the latter stages of competitions they abandon the style of rugby which got them there in the first place opting to play a low-risk game instead. But the Scarlets are different. “We will stick to our usual game and not change anything,” insists Davies. “We are approaching this game as we would approach any other game. We’ve showed over the last 18 months that we want to play an expansive brand of rugby. “We’ve been training well so hopefully we can bring that brand of rugby on the weekend.”	110,877
TITLE: How can I prove that no limit exists for this function over this particular interval? QUESTION [1 upvotes]: I was given the following function: $$ f(x) = \begin{cases} \frac{1}{q} &\text{if }x = \frac{p}{q} \text{ is rational, reduced to lowest terms} \\ 0 &\text{if }x \text{ is irrational}\end{cases}$$ And was asked to attempt to prove or disprove whether or not, for some $j$ on the interval $(0,1)$, $\lim \limits_{x \to j}{}f(x)$ exists. Intuitively, I know that I can find an irrational number between any two rationals which would obviously result in some discontinuity. I considered the idea that maybe it could be possible assuming that among the infinitely many rationals on the interval which also share a common denominator and could be used to find the limit, but that obviously would not work. However, I cannot formulate any proper line of reason to prove the limit does not exist. How can I prove this, as formally as possible? REPLY [1 votes]: Recall the definition of $\lim_{x \to j}f(x)=a$: $$\mbox{Given}\ \varepsilon > 0\ \mbox{there exists}\ \delta > 0\\ \mbox{such that}\ \left\| f(x) - a\right\|<\varepsilon\\ \mbox{whenever}\ 0<\left\| x-j \right\| <\delta$$ Now for any $j$, given $\varepsilon>0$, if we choose $\delta$ small enough we can ensure that the denominator $q$ of any fraction $\frac{p}{q}$ (in its lowest terms) in the interval $\left(j-\delta,j+\delta\right)$ satisfies $q>\frac{1}{\varepsilon}$, except possibly for $j$ itself, if it is rational. So that $0\le\left\| f(x)\right\|<\varepsilon$ whenever $0<\left\|x-j\right\|<\delta$. Plugging this into the definition of a limit we see: $$\lim_{x\to j}f(x) = 0\ \mbox{for all}\ j\in\left(0,1\right)$$ Now a function $g$ is continuous at $j$ iff $\lim_{x\to j}g(x)=g(j)$. It follows that $f$ is discontinuous at rational points and continuous at irrational points.	57,544
\begin{document} \author{Alan Frieze\thanks{Department of Mathematical Sciences,Carnegie Mellon University,Pittsburgh PA 15213, Research supported in part by NSF grant DMS1661063}, Wesley Pegden\thanks{Department of Mathematical Sciences,Carnegie Mellon University,Pittsburgh PA 15213. Research supported in part by NSF grant DMS}, Gregory Sorkin\thanks{Department of Mathematics, London School of Economics, London,}, Tomasz Tkocz\thanks{Department of Mathematical Sciences,Carnegie Mellon University,Pittsburgh PA 15213. Research supported in part by the Collaboration Grants from the Simons Foundation.}} \title{Minimum-weight combinatorial structures under random cost-constraints} \maketitle \begin{abstract} Recall that Janson showed that if the edges of the complete graph $K_n$ are assigned exponentially distributed independent random weights, then the expected length of a shortest path between a fixed pair of vertices is asymptotically equal to $(\log n)/n$. We consider analogous problems where edges have not only a random length but also a random cost, and we are interested in the length of the minimum-length structure whose total cost is less than some cost budget. For several classes of structures, we determine the correct minimum length structure as a function of the cost-budget, up to constant factors. Moreover, we achieve this even in the more general setting where the distribution of weights and costs are arbitrary, so long as the density $f(x)$ as $x\to 0$ behaves like $cx^\gamma$ for some $\gamma\geq 0$; previously, this case was not understood even in the absence of cost constraints. We also handle the case where each edge has several independent costs associated to it, and we must simultaneously satisfy budgets on each cost. In this case, we show that the minimum-length structure obtainable is essentially controlled by the product of the cost thresholds. \end{abstract} \begin{footnotesize} \noindent {\em 2010 Mathematics Subject Classification.} Primary 05C80; Secondary 05C85, 90C27. \noindent {\em Key words.} Minimum weight, cost constraint, shortest path, matching, TSP. \end{footnotesize} \section{Introduction} Let the edges of the complete graph be given independent random edge weights $w(e)$ and a random cost $c(e)$ for $e\in E(K_n)$. We are interested in the problem of estimating the minimum weight of a combinatorial structure $S$ where the total cost of $S$ is bounded by some value $C$. More generally, we allow $r$ costs $\bc(e)=(c_i(e),i=1,2,\ldots,r)$ for each edge. The distribution of weights $w(e)$ will be independent copies of $Z_E^\a$ where $Z_E$ denotes the exponential rate one random variable and $\a\leq 1$. The distribution of costs $c_i(e)$ will be $Z_E^\b$ where $\b\leq 1$. In Section \ref{mgd} we will see that, since we are allowing powers of exponentials, a simple coupling argument will allow us to model a very general class of independent weights and costs, where we require just that the densities satisfy $f(x)\approx cx^\g,\g\geq 0$ as $x\to 0$; here we mean that $cx^\g/f(x)\to 1$ as $x\to 0$. Suppose we are given cost budgets of $\bC=(C_i,i=1,2,\ldots,r)$ and we consider the following problem: let $\cS$ denote some collection of combinatorial strutures such as paths, matchings, Hamilton cycles, we would like to solve \[ Opt(\cS,\bC):\text{Minimise }w(S)\text{ subject to }S\in \cS\text{ and }c_i(S)\leq C_i,i=1,2,\ldots,r, \] and let \[ \text{$w^(\bC)$ denote the minimum value in $Opt(\cS,\bC)$.} \] We remark that Frieze and Tkocz \cite{FT1}, \cite{FT2} have considered finding minimum weight spanning trees or arborescences in the context of a single cost constraint and uniform $[0,1]$ weights and costs. I.e. the case where $\cS$ is the set of spanning trees of $K_n$ and the case where $\cS$ is the set of spanning arborescences of $\vec{K}_n$. In these cases, they obtain asymptotically optimal estimates for those problems, whereas for the problems in the present paper we have only obtained estimates that are correct to within a constant factor. The first problem we study involves paths, here denoted as minimum weight paths for consistency with the remainder of the paper. Let $\cP(i,j)$ denote the set of paths from vertex $i$ to vertex $j$ in $K_n$. {\bf Constrained Minimum Weight Path (CMWP)}: $Opt(\cP(1,n),\bC)$.\\ Without the constraint $\bc(P)\leq \bC$, there is a beautiful result of Janson \cite{Jan} that gives a precise value for the expected minimum weight of a path, when the $w(e)$'s are independent exponential mean one. With the constraints, we are only able to estimate the expected minimum weight up to a constant (but can do so for a more general class of distributions). Throughout the paper we let \[ \D=\prod_{i=1}^rC_i \] be the product of the cost thresholds. Our results show that for the structures we consider, this product of the cost thresholds controls the dependency of the minimum-weight structure on the vector of cost constraints. In particular, for the minimum weight path problem, we have: \begin{theorem}\label{th1} If $\frac{n{\D^{1/\b}}}{\log^{r/\b}n}\to\infty$ and $C_i\leq 10\log n,\,i=1,2,\ldots,r$ then w.h.p. \[ w^(\bC)=\Theta\bfrac{\log^{r\a/\b+1}n}{n^\a{\D^{\a/\b}}}. \] $A=\Theta(B)$ denotes $A=O(B)$ and $B=O(A)$. And here, the hidden constants depend only on $r,\a,\b$. \end{theorem} For the unconstrained problem, see Hassin and Zemel \cite{HZ}, Janson \cite{Jan} and Bhamidi and van der Hofstad \cite{BH}, \cite{BH1}. Now consider the case of perfect matchings in the complete bipartite graph $K_{n,n}$. Let $\cM_2$ denote the set of perfect matchings in $K_{n,n}$. {\bf Constrained Assigment Problem (CAP)}: $Opt(\cM_2,\bC)$. \begin{theorem}\label{th2} If $\D^{1/\b}\gg n^{r/\b-1}\log n$\footnote{Here $A=A(n)\gg B=B(n)$ if $A/B\to \infty$ as $n\to\infty$.} and $C_i\leq n,\,i=1,2,\ldots,r$ then w.h.p. \[ w^(\bC)=\Theta\bfrac{n^{1+r\a/\b-\a}}{{\D^{\a/\b}}}. \] \end{theorem} We note that requiring a lower bound on $\D$ is necessary in Theorem \ref{th2}. Indeed, if ${\D^{1/\b}}\leq e^{-(r+1)}\b n^{r/\b-1}$ then the optimization problem is infeasible w.h.p. To see this we bound the expected number of feasible solutions as follows: let $Z_1,Z_2,\ldots,Z_r$ be independent sums of $n$ independent copies of $Z_E^{1/\b}$. Then, \[ n!\prod_{i=1}^r\Pr\brac{Z_i\leq C_i}\leq n!\prod_{i=1}^r\frac{C_i^{n/\b}}{\b^nn!n^{n(1/\b-1)}}\leq \bfrac{e^r\D^{1/\b}}{\b n^{r/\b-1}}^n=o(1). \] We use Lemma \ref{lemB} here to bound $\Pr(Z_i\leq C_i)$. We note that a problem similar to this was studied by Arora, Frieze and Kaplan \cite{AFK} with respect to the worst-case. Now consider the case of perfect matchings in the complete graph $K_{n}$. Let $\cM_1$ denote the set of perfect matchings in $K_{n}$. {\bf Constrained Matching Problem (CMP)} $Opt(\cM_1,\bC)$. \begin{theorem}\label{th3} If ${\D^{1/\b}}\gg n^{r-1}\log n$ and $C_i\leq n,\,i=1,2,\ldots,r$ then w.h.p. \[ w^(\bC)=\Theta\bfrac{n^{1+r\a/\b-\a}}{{\D^{\a/\b}}}. \] \end{theorem} Now consider the Travelling Salesperson Problem (TSP). Let $\cT$ denote the set of Hamilton cycles in $K_n$ or the set of directed Hamilton cycles in $\vec{K}_n$., {\bf Constrained Travelling Salesperson Problem (CSTSP)} $Opt(\cT,\bC)$. \begin{theorem}\label{th4} If ${\D^{1/\b}}\gg n^{r-1}\log n$ and and $C_i\leq n,\,i=1,2,\ldots,r$ then w.h.p. \[ w^(\bC)=\Theta\bfrac{n^{1+r\a/\b-\a}}{{\D^{\a/\b}}}. \] \end{theorem} \section{Structure of the paper} We prove the above theorems in their order of statement. The upper bounds are proved as follows: we consider the random graph $G_{n,p}$ (or bipartite graph $G_{n,n,p}$ or digraph $D_{n,p}$) for suitably chosen $p$ associated with the random costs. We then seek minimum weight objects contained in these random graphs. The definition of $p$ is such that objects, if they exist, automatically satisfy the cost constraints. For minimum weight paths we adapt the methodology of \cite{Jan}. For the remaining problems we use theorems in the literature stating the high probability existence of the required objects when each vertex independently chooses a few (close) random neighbors. In Section \ref{mgd} we consider more general distributions. We are able to extend the above theorems under some extra assumptions about the $C_i$. \section{CSP} \subsection{Upper Bound for CSP} In the proof of the upper bound, we first consider weights $\hw(e)$ where the $\hw(e)$ are independent exponential mean one random variables. The costs will remain independent copies of $Z_E^\b$. We will then use Holder's inequality to obtain the final result. \subsection{$\frac{\log^{2+r/\b}n}{n}\leq {\D^{1/\b}}$ and $C_i\leq 10\log n,\,i=1,2,\ldots,r$}\label{upper} Suppose now that we let $L=10\log n$ and \[ E_0=\set{e:c_i(e)\leq \frac{C_i}{L},i=1,2,\ldots.r}. \] The proof in this case goes as follows: \begin{enumerate}[(i)] \item We search for short paths that only use edges in $E_0$ and note that the graph $([n],E_0)$ is distributed as $G_{n,p}$. \item Observe that any path using fewer than $L$ edges of $E_0$ automatically satifies the cost constraints. \item A simple calculation shows that w.h.p. the number of edges between a set $S$ of size $k$ and the remaining vertices is close to the expectation $k(n-k)p$ for all sets of vertices $S$, see \eqref{defES} and \eqref{cE}. \item We run Dijkstra's algorithm for finding shortest (now minimum weight) paths from vertex 1. We use Janson's argument \cite{Jan} to bound the distance to the $m=n/3$ closest vertices $V_1$. We need the claim in item (iii) here. \item We repeat (iv), starting from vertex set $n$, to obtain the $m=n/3$ closest vertices $V_2$. If $V_1\cap V_2\neq \emptyset$ we will have found a path of low enough weight, otherwise we calim that w.h.p. there will be a low enough weight edge joining $V_1,V_2$. \item We then argue that the trees constructed by the Dijkstra algorithm are close to being Random Recursive Trees and we can easily bound their height. Showing that we can use item (ii). \item We finally use Holder's inequality to switch from $\hw$ to $w$. \end{enumerate} We first bound the value of $p$. \beq{defp}{ p=\Pr\brac{e\in E_0}=\prod_{i=1}^r\brac{1-\exp\set{-\bfrac{C_i}{3 L}^{1/\b}}}, } where $e$ is an arbitrary edge. We note that if $0<x\leq 1$ then $x/2\leq 1-e^{-x}\leq x$. This implies that \beq{defpp}{ \frac{\D^{1/\b}}{2^r(3L)^{r/\b}}\leq p \leq \frac{\D^{1/\b}}{(3L)^{r/\b}}. } We consider the random graph $G_{n,p}$ where edges have weight given by $\hw$ and costs $c_i(e)\leq C_i/3L,i=1,2,\ldots,r$. We modify Janson's argument \cite{Jan}. We now deal with item (iii). We observe that w.h.p. for every set $S$ of size $k$, $e(S:\bar{S})\approx k(n-k)p$ where $e(S:T)$ is the number of edges $\set{v,w}$ with one end in $S$ and the other in $T$. We only need to check the claim for $\|S\|\leq n/2$. Let $\e=\frac{1}{\log^{1/3}n}$ and \beq{defES}{ \cE_S=\set{e(S:\bar{S})\notin (1\pm\e)k(n-k)p}\text{ and }\cE=\bigcup_{\|S\|\leq n/2}\cE_S. } Then, using the Chernoff bounds for the binomial distribution, \beq{cE}{ \begin{split} \Pr(\cE)&\leq \sum_{k=1}^{n/2}\binom{n}{k}\Pr(Bin(k(n-k),p)\notin (1\pm\e)k(n-k)p)\\ &\leq 2\sum_{k=1}^{n/2}\bfrac{ne}{k}^ke^{-\e^2k(n-k)p/3}\\ &=2\sum_{k=1}^{n/2}\bfrac{ne^{1-\e^2(n-k)p/3}}{k}^k\\ &\leq 2\sum_{k=1}^{n/2}\bfrac{ne^{-\Omega(\log^{4/3}n)}}{k}=o(1). \end{split} } We now continue with item (iv). We set $S_1=\set{1}$ and $d_1=0$ and consider running Dijkstra's algorithm \cite{Dijk}. At the end of Step $k$ we will have computed $S_k=\set{1=v_1,v_2,\ldots,v_k}$ and $0=d_1,d_2,\ldots,d_k$ where $d_i$ is the minimum weight of a path from 1 to $i,i=1,2,\ldots,k$. Let there be $\n_k$ edges from $S_k$ to $[n]\setminus S_k$. Arguing as in \cite{Jan} we see that $d_{k+1}-d_k=Z_k$ where $Z_k$ is the minimum of $\n_k$ independent exponential mean one random variables. Also, the memoryless property of the exponential distribution implies that $Z_k$ is independent of $d_k$. It follows that for $k<n/2$, \mult{meank0}{ \E(d_k\mid\neg\cE)=\E\brac{\sum_{i=1}^k\frac{1}{\n_i}\bigg\|\neg\cE} =\sum_{i=1}^k\frac{1+o(1)}{i(n-i)p} =\frac{1+o(1)}{np}\sum_{i=1}^k\brac{\frac{1}{i}+\frac{1}{n-i}}\\ =\frac{1+o(1)}{np}\brac{H_k+H_{n-1}-H_{n-k+1}}, } where $H_k=\sum_{i=1}^k\frac{1}i$. By the same token, \beq{vark0}{ \Var(d_k\mid\neg\cE)=\sum_{i=1}^k\Var(Z_i\mid\neg \cE)= \sum_{i=1}^k\frac{1+o(1)}{(i(n-i)p)^2}=O((np)^{-2}). } We only pursue the use of Dijkstra's algoritm from vertex 1 for $m=n/3$ iterations. It follows from \eqref{meank0} and \eqref{vark0} and the Chebyshev inequality that we have w.h.p. \beq{dn}{ d_{m}\approx \frac{\log n}{np}. } We next deal with item (vi). The tree built by Dijkstra's algorithm is (in a weak sense) close in distribution to a random recursive tree i.e. vertex $v_{k+1}$ attaches to a near uniformly random member of $\set{v_1,v_2,\ldots,v_k}$. Indeed, assuming $\cE$ does not occur, \[ \Pr(v_{k+1}\text{ attaches to }v_i)=\frac{e(v_i:\bar{S}_k)}{\n_k}\leq \frac{(1+\e)(n-1)p}{(1-\e)k(n-k)p}. \] Hence, if $T$ is the tree constructed in the first $m$ rounds of Dijkstra's algorithm, then \beq{height}{ \begin{split} \Pr(height(T)\geq L)&\leq \sum_{1<t_1<\cdots<t_L<m}\prod_{i=1}^L\frac{3(1+\e)}{2(1-\e)t_i}\\ &\leq \frac{1}{L!}\bfrac{3(1+\e)}{2(1-\e)}^L\brac{\sum_{i=1}^n\frac{1}{i}}^L\\ &\leq \bfrac{3(\log n+1)e^{1+o(1)}}{2L}^L=o(1). \end{split} } It follows from \eqref{defpp}, \eqref{dn} and \eqref{height} that w.h.p., for every $v\in V_1=S_{m}$, there exists a path $P$ from 1 to $v$ of weight at most \[ \l\approx\l_0=\frac{\log n}{np}\lesssim\frac{30^{r/\b}\log^{r/\b+1}n}{n\Upsilon^{1/\b}} \] and costs $c_i(P)\leq LC_i/3L\leq C_i/3$.\footnote{Here we write $A=A(n)\lesssim B=B(n)$ if $A\leq (1+o(1))B$.} We now deal with item (v). We next consider applying Dijkstra's algorithm to find a minimum weight path from vertex $n$ to other vertices. Using the same argument as above, we see that we can find $m$ vertices $V_2$ that are within distance $\l_0$ of vertex $n$. If $V_1\cap V_2\neq\emptyset$ then we have found a path of weight at most $2\l_0$ between vertex 1 and vertex $n$. If $V_1,V_2$ are disjoint then w.h.p. there is an edge of weight $20/np$ between them. Indeed, \[ \Pr(\exists V_1,V_2\text{ with no such edge})\leq \binom{n}{m}^2(e^{-20/np})^{n^2/9}=o(1). \] This yields a path $P$ with \begin{align} \hw(P)&\leq 2\l_0+\frac{20}{np}\leq\frac{3\cdot 30^{r/\b}\log^{r/\b+1}n}{n\Upsilon^{1/\b}}.\label{eq1}\\ c_i(P)&\leq \frac{2C_i}{3}+\frac{C_i}{3}= C_i,\quad i=1,2,\ldots,r.\label{eq2} \end{align} (Here we have used $C_i\geq {\D^{1/\b}}/L^{r-1}\gg p$.) We now deal with item (vii). We use Holder's inequality to yield \beq{Hold1}{ w(P)=\sum_{e\in P}\hw(e)^\a\leq \brac{\sum_{e\in P}\hw(e)}^\a L^{1-\a}= O\bfrac{\log^{r\a/\b+1}n}{n^\a\D^{\a/\b}}. } This completes the proof of Theorem \ref{th1} for this case. \subsection{$\frac{3\om\log^{r/\b}n}{n}\leq {\D^{1/\b}}\leq \frac{\log^{r/\b+2} n}{n}$ and $C_i\leq 10\log n,\,i=1,2,\ldots,r$}\label{smalldelta} The proof is similar to that of Section \ref{upper}, but requires some changes in places. The problem is that we cannot now assume the non-occurrence of $\cE$. Other than this, the proof will follow the same strategy. Our problem therefore is to argue that w.h.p. $e(S_k:\bar{S}_k)$ is sufficiently large. \begin{enumerate}[(a)] \item We now have to keep track of the size of $e(S_k:\bar{S}_k)$ as a random process. This is equation \eqref{l0}. \item The term $\eta_k$ is the number of edges between $v\notin S_k$ and $S_k$. We don't want this to be large, as it reduces $e(S_{k+1}:\bar{S}_{k+1})$. So, we do not add vertices to $S_k$ if $\eta_k\geq 2np$, which only happends rarely. \item Finally, we have to work harder in the case where $V_1,V_2$ are disjoint. We need to use edges of slightly higher cost in order to get a low weight edge in $e(V_1:V_2)$. \end{enumerate} Let $p$ be as in \eqref{defp} where $L=20\log n$. Note that from \eqref{defpp} we see that \[ p\leq \frac{\log^2n}{n}. \] We again consider the random graph $G_{n,p}$ where edges have weight given by $\hw$ and costs at most $C_i/3L$ and again modify Janson's argument \cite{Jan}. We also restrict our search for paths, avoiding vertices of high degree. We set $S_1=\set{1}$ and $d_1=0$. At the end of Step $k$ we will have computed $S_k=\set{1=v_1,v_2,\ldots,v_k}$ and $0=d_1,d_2,\ldots,d_k$ where $d_i$ is the minimum weight of a path from 1 to $i,i=1,2,\ldots,k$. Let there be $\n_k$ edges from $S_k$ to $[n]\setminus S_k$. We cannot rely on $\cE$ of \eqref{cE} not to occur and so we need to modify the argument here. {\bf Assumption: $1\leq k\leq n_0=1/3p$}\\ {\bf Modification:} if our initial choice $v$ for $v_{k+1}$ satisfies $e(v:\bar{S}_k)\geq 2np$ then we reject $v$ permanently from the construction of paths from vertex 1. The initial aim is roughly the same, we want to show that w.h.p. \beq{l0}{ \sum_{\ell\leq k}\n_\ell\geq (1-o(1))knp. } For $v\notin S_k$, let $\eta_{k,v}=e(S_k:\set{v})$ and $\eta_k=\eta_{k,v_{k+1}}$. Then, w.h.p. \beq{nuk}{ \n_{k+1}\geq \n_k-\eta_{k}+B_k\text{ where }B_k=Bin(n_1,p)1_{Bin(n,p)\leq 2np}, } where $n_1=n-2n_0$. The binomials are independent here. This is because the edges between $v_{k+1}$ and $\bar{S}_k$ have not been exposed by the algorithm to this point. The number of trials $n_1$ comes from the following: we know from the Chernoff bounds that \beq{pq}{ \Pr(Bin(n,p)\geq 2np)\leq e^{-np/3}. } It follows from the Markov inequality that w.h.p. there are at most $ne^{-np/4}$ instances where the modification is invoked. This means that w.h.p. the initial choice for $v_k$ has at least $n-n_0-ne^{-np/4}\geq n_1$ possible neighbors. We now define \[ S_k=\sum_{\ell=1}^kB_k. \] We need a lower bound for $B_k$ and an upper bound for $\eta_k$. We next observe that if \[ \e=(np)^{-1/3} \] then \beq{l1}{ \Pr(B_k\leq (1-\e)np)=\Pr(Bin(n_1,p)\geq 2np)+\Pr(Bin(n_1\leq (1-\e)np))\leq (1+o(1))e^{-\e^2np/3}. } It follows that if $k_0=\min\set{n_0,e^{\e^2np/4}}$ then w.h.p. \beq{l2}{ \Pr(\exists 0\leq k\leq k_0: B_k\leq (1-\e)np)\leq (1+o(1))k_0e^{-\e^2np/3}\leq e^{-\e^2np/12}. } For $k\geq k_0$, we use the fact that $S_k$ is the sum of bounded random variables. Hoeffding's inequality \cite{Hoef} gives that \[ \Pr(S_k\leq \E(S_k)-t)\leq \exp\set{-\frac{2t^2}{4kn^2p^2}}. \] Now $\E(B_k)\geq (1-\e)np$ and so putting $t=k^{2/3}np$ we see that \[ \Pr(S_k\leq (1-\e)knp-k^{2/3}np)\leq e^{-k^{1/3}/2}. \] So \beq{l3}{ \Pr(\exists k\geq k_0:\;S_k\leq (1-\e)knp-k^{2/3}np)\leq \sum_{k\geq k_0}e^{-k^{1/3}/2}=o(1). } We next observe that \begin{align} \Pr(\exists S:\;\|S\|=s\leq 1/3p,e(S:S)\geq s+r)&\leq \sum_{s=1}^{1/3p} \binom{n}{s} \binom{s(s-1)/2}{s+r}p^{s+r}\nonumber\\ &\leq \sum_{s=1}^{1/3p} \bfrac{e^2np}{2}^s \bfrac{sep}{2}^r.\label{Eq2} \end{align} Putting $r=s(np)^{1/2}$, the RHS of \eqref{Eq2} becomes \[ \sum_{s=1}^{1/3p}\brac{\frac{e^2np}{2}\bfrac{sep}{2}^{(np)^{1/2}}}^s\leq \sum_{s=1}^{1/3p}\brac{\frac{e^2np}{2}\bfrac{e}{6}^{(np)^{1/2}}}^s=o(1). \] It follows that w.h.p., \beq{tt}{ \sum_{\ell=1}^k\eta_\ell=e(S_k)\leq 2((np)^{1/2}+1)k. } It then follows from \eqref{nuk} and \eqref{l2} and \eqref{l3} and \eqref{tt} that w.h.p. \beq{case1}{ \n_k\geq (1-o(1))knp-2((np)^{1/2}+1)k\geq (1-o(1))knp. } Arguing as in \cite{Jan} we see that $d_{k+1}-d_k=Z_k$ where $Z_k$ is the minimum of $\n_k$ independent exponential mean one random variables. Also, $Z_k$ is independent of $d_k$. It follows that for $k<n$, \beq{meank}{ \E(d_k)=\E\brac{\sum_{i=1}^k\frac{1}{\n_i}} \leq \sum_{i=1}^k\frac{1+o(1)}{inp} =\frac{1+o(1)}{np}\sum_{i=1}^k\frac{1}{i}=\frac{1+o(1)}{np} H_k, } where $H_k=\sum_{i=1}^k\frac{1}i$. By the same token, \beq{vark}{ \Var(d_k)=\sum_{i=1}^k\Var(Z_i)=\sum_{i=1}^k\frac{1+o(1)}{(inp)^2}=O((np)^{-2}). } It follows from \eqref{meank} and \eqref{vark} and the Chebyshev inequality that w.h.p. we have $d_{n_0}\lesssim\frac{\log n}{np}$. Let $V_1$ denote the $n_0$ vertices at this distance from vertex 1. We next consider applying Dijkstra's algorithm to find a minimum weight path from vertex $n$ to other vertices. Using the same argument as above, we see that we can find $n_0$ vertices $V_2$ that are within distance $\frac{(1+o(1))\log n}{np}$ of vertex $n$. If $V_1\cap V_2\neq\emptyset$ then we have found a path of weight at most $\frac{(2+o(1))\log n}{np}$ between vertex 1 and vertex $n$. If $V_1,V_2$ are disjoint then we will use the edges \[ E_1=\set{e:c_i(e)\in\left[\frac{C_i}{L},\frac{2C_i}{L}\right],i=1,2,\ldots.r}. \] Given $e=\set{x,y}\in V_1:V_2$, then given the history of Dijkstra's algorithm so far, either $e\in E_0$ or we can say that \beq{E1}{ \Pr(e\in E_1\mid e\notin E_0)\geq \Pr(e\in E_1)=(1-e^{-(2^{1/\b}-1)p^{1/\b}})^r. } For the equation in \eqref{E1} we use \mult{p2p}{ \Pr(p\leq Z_E^\b\leq 2p)=\Pr(Z_E^\b\geq p)(1-\Pr(Z_E^\b\geq 2p\mid Z_E\geq p))=\\ e^{-p^{1/\b}}\brac{1-\frac{\Pr(Z_E^\b\geq 2p)}{\Pr(Z_e^\b\geq p)}}=e^{-p^{1/\b}}\brac{1-e^{-(2^{1/\b}-1)p^{1/\b}}}=e^{-p^{1/\b}}-e^{-(2p)^{1/\b}}\geq \frac{(2^{1/\b}-1)p^{1/\b}}{2}. } For the inequality in \eqref{p2p} we use the fact that we now have $p\leq \frac{\log^2n}{n}$. Then we search for an edge in $E_2=\set{e\in E_1:\hw(e)\leq 1/np}$. And, \[ \Pr(E_2\cap(V_1:V_2)=\emptyset)\leq \brac{1-\brac{\frac{(2^{1/\b}-1)p^{1/\b}}{2}}(1-e^{-1/np})}^{1/9p^2}\\ \leq \brac{1-\frac{(2^{1/\b}-1)p^{1/\b}}{2np}}^{1/9p^2}=o(1). \] This yields a path of weight at most $\frac{(2+o(1))\log n}{np}+\frac{1}{np}=\frac{(2+o(1))\log n}{np}$. We deal with the height of the Dijkstra trees. Let $T$ be the tree constructed by Dijkstra's algorithm and let $\xi_i,i\leq k$ denote the number of edges from $v_i$ to $V_1\setminus S_i$. \begin{align} \Pr(height(T)\geq L)&\leq \E\brac{\sum_{1<t_1<\cdots<t_L<n_0}\prod_{i=1}^L\frac{\xi_{t_i}}{\n_{t_{i+1}-1}}}\\ &\leq \E\brac{\sum_{1<t_1<\cdots<t_L<n_0}\prod_{i=1}^L\frac{2np}{\n_{t_{i+1}-1}}}\\ &\leq \E\brac{\frac{1}{L!}\brac{\sum_{i=1}^{n_0}\frac{2np}{\n_i}}^L}\\ &\leq o(1)+\frac{(2enp)^L}{(np)^LL!}\brac{\sum_{i=1}^{n_0}\frac{1+o(1)}{i}}^L\\ \noalign{\text{The first $o(1)$ term here is the probability that there is a small $\n_k$ and this is covered by \eqref{case1}.}}\\ &=o(1), \end{align} since $L\geq 20\log n$. It follows from the above that w.h.p. there exists a path $P$ \beq{final}{ \text{where } \hw(P)\lesssim \frac{2\log n}{np}\text{ and }c_i(P)\leq \frac{(2L+2)C_i}{3L}<C_i,i=1,2,\ldots,r. } Arguing as for \eqref{Hold1} we see that \beq{Hold2}{ w(P)\leq \hw(P)^\a L^{1-\a}= O\bfrac{\log^{r\a/\b+1}n}{n^\a\D^{\a/\b}}. } \subsection{Lower Bound for CSP}\label{lowcsp} This is a straightforward use of the first moment method. Suppose that \[ {\D^{1/\b}}=\frac{\om\log^{r/\b}n}{n},\quad L=\frac{\e\log^{r\a/\b+1}n}{n^\a\D^{\a/\b}}, \] where \[ \e=\brac{\a\b^r e^{-2}\brac{10\brac{\frac{r}{\b}+\frac{1}{\a}}}^{-(r/\b+1/\a)}}^\a, \] then \begin{align} &\Pr\brac{\exists P:w(P)\leq L,c_i(P)\leq C_i,i=1,2,\ldots,r}\nonumber\\ &\leq \sum_{k=1}^{n-2}n^{k-1} \bfrac{L^{k/\a}}{\a^kk!k^{k(1/\a-1)}}\prod_{i=1}^r\frac{C_i^{k/\b}}{\b^kk!k^{k(1/\b-1)}} \label{lowexp}\\ &\leq \frac{1}{n}\sum_{k=1}^{n-1}\brac{n\cdot\frac{e\e^{1/\a}\log^{r/\b+1/\a}n}{\a nk^{1/\a}\D^{1/\b}}\cdot \frac{e\D^{1/\b}}{\b^r k^{r/\b}}}^k\nonumber\\ &=\frac{1}{n}\sum_{k=1}^{n-1}\bfrac{e^2\e^{1/\a}\log^{r/\b+1/\a}n}{\a\b^rk^{r/\b+1/\a}}^k\nonumber\\ &=\frac{1}{n}\sum_{k=1}^{\frac12\log n}\bfrac{e^2\e^{1/\a}\log^{r/\b+1/\a}n}{\a\b^rk^{r/\b+1/\a}}^k+ \frac{1}{n}\sum_{k=\frac12\log n}^{n-1}\bfrac{e^2\e^{1/\a}\log^{r/\b+1/\a}n}{\a\b^rk^{r/\b+1/\a}}^k\nonumber\\ &\leq \frac{1}{n}\sum_{k=1}^{\frac12\log n}\bfrac{\log n}{10(r/\b+1/\a)k}^{(r/\b+1/\a)k}+ \frac{1}{n}\sum_{k=\frac12\log n}^{n-1}10^{-k}\nonumber\\ &=o(1).\nonumber \end{align} {\bf Explanation for \eqref{lowexp}}: we choose a path of length $k$ from 1 to $n$ in at most $n^{k-1}$ ways. Then we use Lemma \ref{lemB} $r+1$ times. Then we use the union bound. \section{Upper Bounds} \subsection{Upper Bound for CAP}\label{CAP} Let $G$ denote the subgraph of $K_{n,n}$ induced by the edges that satisfy $c_i(e)\leq C_i/n$ for $i=1,2,\ldots,r$. Let \[ p=\Pr\brac{c_i(e)\leq \frac{C_i}{n},i=1,2,\ldots,r}=\prod_{i=1}^r\brac{1-\exp\set{-\bfrac{C_i}{n}^{1/\b}}}. \] and note that \[ \frac{\log n}{n}\ll \frac{\D^{1/\b}}{2^rn^{r/\b}}\leq p\leq \frac{\D^{1/\b}}{n^{r/\b}}. \] The approach for this and the remining problems is \begin{enumerate}[(i)] \item Look for a small weight structure in an edge weighted random graph $G$. In this case the random bipartite graph $G_{n,n,p}$. \item Use an idea of Walkup \cite{W1} to construct a random subgraph $H$ of $G$ that only uses edges of low weight. \item Use a result from the literature that states that w.h.p. the edges of $H$ contain a copy of the desired structure. \end{enumerate} $G$ is distributed as $G=G_{n,n,p}$. Note that by construction, a perfect matching $M$ of $G$ satisfies $c_i(M)\leq C_i,i=1,2,\ldots,r$. Let $d=np$ and note that because $dnp\gg\log n$ the Chernoff bounds imply that w.h.p. every vertex has degree $\approx d$. Now each edge of $G$ has a weight uniform in $[0,1]$. Following Walkup \cite{W1} we replace $w(e),e=(x,y)$ by $\min\set{Z_1(e),Z_2(e)}$ where \beq{defZ1}{ \text{$Z_1,Z_2$ are independent copies of $Z_W$ where $\Pr(Z_W\geq x)^2=\Pr(Z_E^\a\geq x)$.} } We assign $Z_1(e)$ to $x$ and $Z_2(e)$ to $y$. Let $X,Y$ denote the bipartition of the vertices of $G$. Now consider the random bipartite graph $H$ where each $x\in X$ is incident to the two $Z_1$-smallest edges incident with $x$. Similarly, $y\in Y$ is incident to the two $Z_2$-smallest edges incident with $y$. Walkup \cite{W2} showed that $H$ has a perfect matching w.h.p. The expected weight of this matching is asymptotically at most \beq{capexp}{ \bfrac{2^\a n}{d^\a}\brac{\G\brac{1+\frac{1}{\a}}+\G\brac{2+\frac{1}{\a}}}\times \frac12= O\bfrac{n^{1+r\a/\b-\a}}{\D^{\a/\b}}. } This follows from (i) the the expression given in Corollary \ref{cor1} for the expected minimum and second minimum of $d$ copies of $Z$ and (ii) the matching promised in \cite{W2} is equally likely to select a minimum or a second minimum weight edge. The selected matching is the sum of independent random variables with exponential tails and so will be concentrated around its mean. \subsection{Upper Bound for CMP} We let $p,d$ be as in Section \ref{CAP}. We replace Walkup's result \cite{W2} by Frieze's result \cite{F2out} that the random graph $G_{2-out}$ contains a perfect matching w.h.p. The random graph $G_{k-out}$ has vertex set $[n]$ and each vertex $v\in [n]$ independently chooses $k$ random edges incident with $v$. We again replace $c(e),e=(x,y)$ by $\min\set{Z_1(e),Z_2(e)}$ where $Z_1,Z_2$ are independent copies of $Z_W$ and associate one copy with each endpoint of the edge. We consider the random graph $H$ where each $v\in [n]$ is incident to the two $Z_W$-smallest edges incident with $x$. This is distributed as $G_{2-out}$ and we obtain an expression similar to that in \eqref{capexp}. We have concentration around the mean as in Section \ref{CAP}. \subsection{Upper bound for CSTSP/CATSP} For the symmetric case we replace $w(e),e=\set{x,y}$ by $\min\set{Z_1(e),Z_2(e)}$ for each edge of $K_n$ and for the asymmetric case we replace $w(e),e=(x,y)$ by $\min\set{Z_1(e),Z_2(e)}$ for each directed edge of $\vec{K}_n$. In both cases we associate one copy of $Z_W$ to each endpoint of $e$. We define $p,d$ as in Section \ref{CAP} and consider either the random graph $G_{n,p}$ or the random digraph $D_{n,p}$. For the symmetric case, we consider the random graph $H$ that includes the 3 cheapest edges associated with each vertex, cheapest with respect to $Z_W(e)$. This will be distributed as $G_{3-out}$ which was shown to be Hamiltonian w.h.p. by Bohman and Frieze \cite{BF}. For the asymmetric case, we consider the random digraph $H$ that includes the 2 cheapest out-edges and the 2 cheapest in edges associated with each vertex, cheapest with respect to $Z_W(e)$. This will be distributed as $D_{2-in,2-out}$ which has vertex set $[n]$ and where each vertex $v$ independently chooses 2 out- and in-neighbors. The random digraph $D_{2-in,2-out}$ was shown to be Hamiltonian w.h.p. by Cooper and Frieze \cite{CF}. The expected weight of the tour promised by \cite{BF} or by \cite{CF} is asymptotically $O(n^{1+r\a/\b-\a}/\D^{\a/\b})$ as in Section \ref{CAP}. We have concentration around the mean as in Section \ref{CAP}. \section{Lower Bounds} We proceed as in Section \ref{lowcsp}. Suppose that $\D=\om n^{r/\b-1}\log n$ and $L=\frac{\e n^{1+r\a/\b-\a}}{\D^{\a/\b}}$ where $\e$ will be a sufficnetly small constant. Let $\Lambda$ denote the relevant structure, matching or cycle. Then, by the union bound and Lemma \ref{lemB}, we have for CAP,CSTSP,CATSP, \multstar{ \Pr\brac{\exists \Lambda : w(\Lambda )\leq L\text{ and }c_i(\Lambda )\leq C_i,i=1,2,\ldots,r}\leq n!\cdot\frac{L^{n/\a}}{\a^nn!n^{n(1/\a-1)}}\cdot \prod_{i=1}^r\frac{C_i^{n/\b}}{\b^nn!n^{n(1/\b-1)}}\\ \leq \brac{\frac{\e^{1/\a}n^{1/\a+r/\b-1}}{\a n^{1/\a-1}\D^{1/\b}}\cdot \frac{e^r\D^{1/\b}}{\b n^{r/\b}}}^n=o(1), } for $\e$ sufficiently small. For CMP, assuming that $n=2m$, \multstar{ \Pr\brac{\exists \Lambda : w(\Lambda )\leq L\text{ and }c_i(\Lambda )\leq C_i,i=1,2,\ldots,r}\leq \frac{n!}{m!2^m}\cdot\frac{L^{m/\a}}{\a^mm!m^{m(1/\a-1)}}\cdot \prod_{i=1}^r\frac{C_i^{m/\b}}{\b^mm!m^{m(1/\b-1)}}\\ \leq \brac{\frac{\e^{1/\a}m^{1/\a+r/\b-1}}{2\a m^{1/\a-1}\D^{1/\b}}\cdot \frac{e^r\D^{1/\b}}{\b m^{r/\b}}}^m=o(1), } for $\e$ sufficiently small. \section{More general distributions}\label{mgd} We follow an argument from Janson \cite{Jan}. We will asssume that $w(e),$ has the distribution function $F_w(t)=\Pr(X\leq t)$, of a random variable $X$, that satisfies $F_w(t)\approx at^{1/\a},\a\leq 1$ as $t\to 0$. For the costs $c_i(e)$ we have $F_c(t)\approx bt^{1/\b},\b\leq 1$. The constants $a,b>0$ can be dealt with by scaling and so we assume that $a=b=1$ here. For a fixed edge and say, $w(e)$, we consider random variables $w_<(e),w_>(e)$ such that $w_<(e)$ is distributed as $Z_E^{\a+\e_n}$ and $w_>(e)$ is distributed as $Z_E^{\a-\e_n}$, where $\e_n=1/10\log n$. (This choice of $\e_n$ means that $n^{\a+\e_n}=e^{1/10}n^\a$.) Then let $U(e)$ be a uniform $[0,1]$ random variable and suppose that $X$ has the distribution $F^{-1}(U)$. We couple $X,w_<,w_>$ by generating $U(e)$ and then $w_<(e)=F_<^{-1}(U)=\log\bfrac{1}{1-u}^{\a-\e_n}$ and $F_>$ is defined similarly. The coupling ensures that $w_<(e)\leq w(e)\leq w_>(e)$ as long as $w(e)\leq\e_n$. Given the above set up, it only remains to show that w.h.p. edges of length $w(e)>\e_n$ or cost $c_i(e)>\e_n$ are not needed for the upper bounds proved above. We can ignore the lower bounds, because they only increase if we exclude long edges. {\bf Assumptions for CMWP.} For the minimum weight path problem we will assume that $\D^{1/\b}\gg \frac{\log^{1+r/\b}n}{n}$, which is a $\log n$ factor larger than required for Theorem \ref{th1}. We will assume that $C_i=o(1)$ and then we only use edges of cost of order $C_i/\log n\ll \e_n$. Observe that the minimum weight of a path from 1 to $n$ is at most $\frac{4\log n}{np}$ w.h.p. and this is less than $\e_n$ because of the assumption $\frac{\log^{1+r/\b}n}{n}\ll {\D^{1/\b}}$ and the definition of $p$ (see \eqref{defpp}). {\bf Assumptions for the other problems,} We deal with costs by assuming that $C_i=o(n/\log n),i=1,2,\ldots r$. It is then a matter of showing that w.h.p. the first few order statistics of $Z_W$ are very unlikely to be greater than $\e_n$. ($Z_W$ is defined in \eqref{defZ1}.) But in all cases this can be bounded as follows: let $W_1,W_2,\ldots,W_m, m\geq n/2$ be independent copies of $Z_W$. Then, \[ \Pr(\|\set{i:W_i\leq \e_n}\|\leq 3)\leq m^3(1-(1-e^{-\e_n^{1/\a}})^{1/2})^{m-3}=m^3e^{-m^{1-o(1)}}. \] This bounds the probability of using a heavy edge at any one vertex and inflating by $n$ gives us the result we need. \section{Conclusion} We have given upper and lower bounds that hold w.h.p. for constrained versions of some classical problems in Combinatorial Optimization. They are within a constant factor of one another, unlike the situation with respect to spanning trees and arborescences, \cite{FT1}, \cite{FT2}, where the upper and lower bounds are asymptotically equal. It is a challenge to find tight bounds for the problems considered in this paper and to allow correlation between length and cost. We have not made any claims about $\E(w^(\bC))$ because there is always the (small) probability that the problem is infeasible. It is not difficiult to similarly bound the expectation conditional on feasibility.	198,621
(I apologize for typos that may or may not occur. This keyboard doesn't like to type well) We had fun last week. The monies paid for the trip to Bacolod and for food stuffs. I miss camping. I'm happy that we're going to the family reunion when I get home. It will be super awesome and stuffs. Elder Sablan and I are doing great. He has really helped me understand the language a lot better and he has taught me better ways to teach the lessons! I don't know if I told you this already but he is a former AP (Assistant to the President) so he is like super super obedient. Apparently like half of what Elder Cano taught me was either wrong or against some rule... But everything is good now! No. I've seen the rubber shoes, and they tear easy. And it's really hard to find my shoe size here (or anything over size like 9 I think, and I'm an 11). My shoes are doing great, but I've started to switch them out every other day because of the rain. So we have been teaching like crazy. We are going to have baptisms every week starting this week! We baptize the RM family next week. We also started doing a sacrament meeting out there, the goal being that there will eventually be a meeting house out there. So technically yes I transferred and what nots, but I still go to the same building, just both sacrament meetings. (and a third later in the day!) Crazy awesome stuffs. Well... Hopefully she wont find out the hard way. I was lucky enough to have someone tell me my first 5 minutes of being there. It's slowly coming back... Just slowly. But other than that I actually GAINED weight (But I'm still getting skinny!) and I'm getting even more tan! :D I will be sending pictures in a bit Sin-cereal-ly, Elder Archuleta Things I forgot: You said to include other things I'd want, and I completely forgot. A black (yes, it has to be black) umbrella. Preferably big (but not too big, I have to carry it in my bag) A blank picture book. Like the one you made me, just with absolutely nothing in it (unless you choose to include pictures) if possible, a blank book. Basically a book that has nothing in it whatsoever. Just blank white pages. If you could get that (or make it) I would love it. With at least 50 sheets (or 100 pages) More deoderant. I'm on my last stick now, and I refuse to use the axe here (it is literally all they have) Other than that, I think that is all I have for now. I will start writing down things as they come to mind. Me :D Me with the ducks :D. (He is wearing one of the UGLY ties we sent...hahahaha!!!) I would have had more time but the comupter is slow so here are some more. Elder Sablan This is how we get to our area. It's through a rice feild then a super super long road that takes like a half hour to walk. Also the binding on my BOM has come unglued... What do I do? Dear Dad, Everything is fine here in the jungle! All we get from the storm is a lot if rain. Even now it is raining like CRAZY outside. We are going to have a lot of baptisms coming up in the next couple of weeks! I would enjoy some pictures of Jake and Jack (when he starts baseball or whatever) at practice. Just so I have more pictures of my family. (Dad included a picture of the Young Men at Folsom Lake) Where is Jake? And seriously the Young Men need to get a tan...they are all white. I will pray that you guys can get some of our rain...we don't really need it! I will be sending the pictures in a second (as stated in Moms email.) Elder Archuleta Glad to hear all is well in the jungle and that you will be having baptisms. I will be sure to get pictures of Jake & Jack. I will take some this week of Jakob at practice. Jake has his first game in a couple of weeks against Elk Grove. Jackson starts baseball in September. But I will get more pictures to add to your family picture book. Do you want them mailed in your next package or can you print them at the computer place? Jake is not in the picture because he was at football practice. Yes the Young Men are all very white boys. Dad What position is he playing this year? I can print pictures here (it costs 7pisos). Elder A. He will be playing guard on the offensive line and defense end on defense. He is one of the bigger boys this year. 7 Pisos is cheap. OK I will be sure to get photos of him at McBean Park during practice this week. Mom says you are working hard with Elder Sablan and he is helping you with the language and teaching approaches. That is good to always learn different ways. One day you will be training someone. Dad How much does Gibby weigh? $5 says he weighs more than I do. Awesome! I really want to train someone, everyone says it is so much fun! Elder A. Jakob weighs 151lbs. Your time will come soon enough to train someone. Mom and I noticed a couple of new blogs on the Bacolod mission that we follow with some Americans coming your direction. I am glad to hear you are having fun and learning everyday and enjoying your experience. Dad Only 151? Dang it. I'll bet you $5 when I come home I'm 170. Elder A.	225,351
The hexadecimal color code #01e6d3 is a shade of cyan. In the RGB color model #01e6d3 is comprised of 0.39% red, 90.2% green and 82.75% blue. In the HSL color space #01e6d3 has a hue of 175° (degrees), 99% saturation and 45% lightness. This color has an approximate wavelength of 495.38 nm. Download: I am seeking. I am striving. I am in it with all my heart. <p style="color: #01e6d3">…</p> Creativity takes courage. <p style="background-color: #01e6d3">…</p> If I could say it in words there would be no reason to paint. <p style="text-shadow: 0.1em 0.1em 0.15em #01e6d3">…</p>	204,593
\begin{document} \title{On the equations and classification of toric quiver varieties} \author{{M. Domokos and D. Jo\'o}\thanks{Partially supported by OTKA NK81203 and K101515. }} \date{} \maketitle {\small \begin{center} R\'enyi Institute of Mathematics, Hungarian Academy of Sciences,\\ Re\'altanoda u. 13-15, 1053 Budapest, Hungary \\ Email: domokos.matyas@renyi.mta.hu \quad joo.daniel@renyi.mta.hu \end{center} } \begin{abstract} Toric quiver varieties (moduli spaces of quiver representations) are studied. Given a quiver and a weight there is an associated quasiprojective toric variety together with a canonical embedding into projective space. It is shown that for a quiver with no oriented cycles the homogeneous ideal of this embedded projective variety is generated by elements of degree at most $3$. In each fixed dimension $d$ up to isomorphism there are only finitely many $d$-dimensional toric quiver varieties. A procedure for their classification is outlined. \end{abstract} \noindent 2010 MSC: 14M25 (Primary) 14L24 (Secondary) 16G20 (Secondary) 52B20 (Secondary) \noindent Keywords: binomial ideal, moduli space of quiver representations, toric varieties \section{Introduction}\label{sec:intro} Geometric invariant theory was applied by King \cite{king} to introduce certain moduli spaces of representations of quivers. In the special case when the dimension vector takes value $1$ on each vertex of the quiver ({\it thin representations}), these moduli spaces are quasi-projective toric varieties; following \cite{altmann-hille} we call them {\it toric quiver varieties}. Toric quiver varieties were studied by Hille \cite{hille:chemnitz}, \cite{hille:canada}, \cite{hille:laa}, Altmann and Hille \cite{altmann-hille}, Altmann and van Straten \cite{altmann-straten}. Further motivation is provided by Craw and Smith \cite{craw-smith}, who showed that every projective toric variety is the fine moduli space for stable thin representations of an appropriate quiver with relations. Another application was introduced very recently by Carroll, Chindris and Lin \cite{carroll-chindris-lin}. From a different perspective, the projective toric quiver varieties are nothing but the toric varieties associated to flow polytopes. Taking this point of departure, Lenz \cite{lenz} investigated toric ideals associated to flow polytopes. These are the homogeneous ideals of the projective toric variety associated to a flow polytope, canonically embedded into projective space. Given a quiver (a finite directed graph) and a weight (an integer valued function on the set of vertices) there is an associated normal lattice polyhedron yielding a quasiprojective toric variety with a canonical embedding into projective space. This variety is projective if and only if the quiver has no oriented cycles. We show in Theorem~\ref{thm:degreethree} that the homogeneous ideal of this embedded projective variety is generated by elements of degree at most $3$. This is deduced from a recent result of Yamaguchi, Ogawa and Takemura \cite{yamaguchi-ogawa-takemura}, for which we give a simplified proof. It follows from work of Altmann and van Straten \cite{altmann-straten} and Altmann, Nill, Schwentner and Wiercinska \cite{altmann-nill-schwentner-wiercinska} that for each positive integer $d$ up to isomorphism there are only finitely many toric quiver varieties (although up to integral-affine equivalence there are infinitely many $d$-dimensional quiver polyhedra). We make this notable observation explicit and provide a self-contained treatment yielding some refinements. Theorem~\ref{thm:directprod} asserts that any toric quiver variety is the product of prime (cf. Definition~\ref{def:prime}) toric quiver varieties, and this deomposition can be read off from the combinatorial structure of the quiver. Moreover, by Theorem~\ref{thm:3regular} any prime (cf. Definition~\ref{def:prime}) $d$-dimensional $(d>1)$ projective toric quiver variety can be obtained from a bipartite quiver with $5(d-1)$ vertices and $6(d-1)$ arrows, whose skeleton (cf. Definition~\ref{def:skeleton}) is $3$-regular. A toric variety associated to a lattice polyhedron is covered by affine open toric subvarieties corresponding to the vertices of the polyhedron. In the case of quiver polyhedra the affine toric varieties arising that way are exactly the affine toric quiver varieties by our Theorem~\ref{thm:vertices} and Theorem~\ref{thm:compactification}. According to Theorem~\ref{thm:compactification} any toric quiver variety can be obtained as the union in a projective toric quiver variety of the affine open subsets corresponding to a set of vertices of the quiver polytope. The paper is organized as follows. In Section~\ref{sec:flowpolytopes} we review flow polytopes, quiver polytopes, quiver polyhedra, and their interrelations. In Section~\ref{sec:repsofquivers} we recall moduli spaces of representations of quivers, including a very explicit realization of a toric quiver variety in Proposition~\ref{prop:realizationofmoduli}. In Section~\ref{sec:reductions} we collect reduction steps for quiver--weight pairs that preserve the associated quiver polyhedron, and can be used to replace a quiver by another one which is simpler or smaller in certain sense. These are used to derive the results concerning the classification of toric quiver varieties. As an illustration the classification of $2$-dimensional toric quiver varieties is recovered in Section~\ref{sec:2-dimensional}. Section~\ref{sec:affine} clarifies the interrelation of affine versus projective toric quiver varieties. Section~\ref{sec:semigroupalgebras} contains some generalities on presentations of semigroup algebras, from which we obtain Corollary~\ref{cor:joo} that provides the technical framework for the proof in Section~\ref{sec:equations} of Theorem~\ref{thm:degreethree} about the equations for the natural embedding of a toric quiver variety into projective space. For sake of completeness of the picture we show in Section~\ref{sec:japanok} how the main result of \cite{yamaguchi-ogawa-takemura} can be derived from the special case Proposition~\ref{prop:theta1} used in the proof of Theorem~\ref{thm:degreethree}. We also point out in Theorem~\ref{thm:affinebound} that the ideal of relations among the minimal generators of the coordinate ring of a $d$-dimensional affine toric quiver variety is generated in degree at most $d-1$, and this bound is sharp. \section{Flow polytopes and their toric varieties} \label{sec:flowpolytopes} By a {\it polyhedron} we mean the intersection of finitely many closed half-spaces in $\mr^n$, and by a {\it polytope} we mean a bounded polyhedron, or equivalently, the convex hull of a finite subset in $\mr^n$ (this conforms the usage of these terms in \cite{cox-little-schenck}). A {\it quiver} is a finite directed graph $\quiver$ with vertex set $\quiver_0$ and arrow set $\quiver_1$. Multiple arrows, oriented cycles, loops are all allowed. For an arrow $a\in \quiver_1$ denote by $a^-$ its starting vertex and by $a^+$ its terminating vertex. Given an integral vector $\theta\in\mz^{\quiver_0}$ and non-negative integral vectors $\mathbf{l},\mathbf{u}\in\mn_0^{\quiver_1}$ consider the polytope \[\polytope=\polytope(Q,\theta,\mathbf{l},\mathbf{u})=\{x\in\mr^{\quiver_1}\mid \mathbf{l}\le x \le\mathbf{u},\quad \forall v\in\quiver_0:\quad \theta(v)=\sum_{a^+=v}x(a)-\sum_{a^-=v}x(a)\}.\] This is called a {\it flow polytope}. According to the generalized Birkhoff-von Neumann Theorem $\polytope$ is a {\it lattice polytope} in $\mr^{\quiver_1}$, that is, its vertices belong to the lattice $\mz^{\quiver_1}\subset\mr^{\quiver_1}$ (see for example Theorem 13.11 in \cite{schrijver}). Denote by $X_{\polytope}$ the {\it projective toric variety} associated to $\polytope$ (cf. Definition 2.3.14 in \cite{cox-little-schenck}). The polytope $\polytope$ is {\it normal} (see Theorem 13.14 in \cite{schrijver}). It follows that the abstract variety $X_{\polytope}$ can be identified with the Zariski-closure of the image of the map \begin{equation}\label{eq:X_P} (\mc^{\times})^{\quiver_1}\to \mathbb{P}^{d-1},\quad t \mapsto (t^{m_1}:\dots:t^{m_d})\end{equation} where $\{m_1,\dots,m_d\}=\polytope \cap \mz^{\quiver_1}$, and for $t$ in the torus $(\mc^{\times})^{\quiver_1}$ and $m\in\mz^{\quiver_1}$ we write $t^m:=\prod_{a\in\quiver_1}t(a)^{m(a)}$. From now on $X_{\polytope}$ will stand for this particular embedding in projective space of our variety, and we denote by $\ideal(X_{\polytope})$ the corresponding vanishing ideal, so $\ideal(X_{\polytope})$ is a homogeneous ideal in $\mc[x_1,\dots,x_d]$ generated by binomials. Normality of $\polytope$ implies that $X_{\polytope}$ is {\it projectively normal}, that is, its affine cone in $\mc^d$ is normal. We shall also use the notation \[\polytope(\quiver,\theta)=\{x\in\mr^{\quiver_1}\mid \mathbf{0}\le x,\quad \forall v\in\quiver_0:\quad \theta(v)=\sum_{a^+=v}x(a)-\sum_{a^-=v}x(a)\}.\] We shall call this a {\it quiver polyhedron}. When $\quiver$ has no oriented cycles, then for ${\mathbf{u}}$ large enough we have $\polytope(\quiver,\theta)=\polytope(\quiver,\theta,\mathbf{0},\mathbf{u})$, so $\polytope(\quiver,\theta)$ is a polytope; these polytopes will be called {\it quiver polytopes}. \begin{definition}\label{def:isomorphicpolytopes} {\rm The lattice polyhedra $\polytope_i\subset V_i$ with lattice $M_i\subset V_i$ $(i=1,2)$ are {\it integral-affinely equivalent} if there exists an affine linear isomorphism $\varphi:\affspan(\polytope_1)\to\affspan(\polytope_2)$ of affine subspaces with the following properties: \begin{itemize} \item[(i)] $\varphi$ maps $\affspan(\polytope_1)\cap M_1$ onto $\affspan(\polytope_2)\cap M_2$; \item[(ii)] $\varphi$ maps $\polytope_1$ \ onto $\polytope_2$. \end{itemize} } \end{definition} The phrase `integral-affinely equivalent' was chosen in accordance with \cite{bruns-gubeladze} (though in \cite{bruns-gubeladze} full dimensional lattice polytopes are considered). Obviously, if $\polytope_1$ and $\polytope_2$ are integral-affinely equivalent lattice polytopes, then the associated projective toric varieties $X_{\polytope_1}$ and $X_{\polytope_2}$ are isomorphic (and in fact they can be identified via their embeddings into projective space given by the $\polytope_i$). As we shall point out in Proposition~\ref{prop:flowpolytope-quiverpolytope} below, up to integral-affine equivalence, the class of flow polytopes coincides with the class of quiver polytopes, so the class of quiver polyhedra is the most general among the above classes. \begin{proposition}\label{prop:flowpolytope-quiverpolytope} For any flow polytope $\polytope(\quiver, \theta,{\mathbf{l}},{\mathbf{u}})$ there exists a quiver $\quiver'$ with no oriented cycles and a weight $\theta'\in\mz^{\quiver'_1}$ such that the polytopes $\polytope(\quiver, \theta,{\mathbf{l}},{\mathbf{u}})$ and $\polytope(\quiver',\theta')$ are integral-affinely equivalent. \end{proposition} \begin{proof} Note that $x\in\mr^{\quiver_1}$ belongs to $\polytope(\quiver,\theta,\mathbf{l},\mathbf{u})$ if and only if $x-\mathbf{l}$ belongs to $\polytope(Q,\theta',\mathbf{0},\mathbf{u}-\mathbf{l})$ where $\theta'$ is the weight given by $\theta'(v)=\theta(v)-\sum_{a^+=v}\mathbf{l}(a)+\sum_{a^-=v}\mathbf{l}(a)$. Consequently $X_{\polytope(\quiver,\theta,\mathbf{l},\mathbf{u})}=X_{\polytope(\quiver,\theta',\mathbf{0},\mathbf{u}-\mathbf{l})}$. Therefore it is sufficient to deal with the flow polytopes $\polytope(\quiver,\theta,\mathbf{0},\mathbf{u})$. Define a new quiver $Q'$ as follows: add to the vertex set of $Q$ two new vertices $v_a$, $w_a$ for each $a\in\quiver_1$, and replace the arrow $a\in Q_1$ by three arrows $a_1,a_2,a_3$, where $a_1$ goes from $a^-$ to $v_a$, $a_2$ goes from $w_a$ to $v_a$, and $a_3$ goes from $w_a$ to $a^+$. Let $\theta'\in\mz^{\quiver'_0}$ be the weight with $\theta'(v_a)=\mathbf{u}(a)=-\theta(w_a)$ for all $a\in\quiver_1$ and $\theta'(v)=\theta(v)$ for all $v\in \quiver_0$. Consider the linear map $\varphi:\mr^{\quiver_1}\to\mr^{\quiver'_1}$, $x\mapsto y$, where $y(a_1):=x(a)$, $y(a_3):=x(a)$, and $y(a_2)=\mathbf{u}(a)-x(a)$ for all $a\in\quiver_1$. It is straightforward to check that $\varphi$ is an affine linear transformation that restricts to an isomorphism $\affspan(\polytope(\quiver,\theta,\mathbf{0},\mathbf{u}))\to \affspan(\polytope(\quiver',\theta'))$ with the properties (i) and (ii) in Definition~\ref{def:isomorphicpolytopes}. \end{proof} \section{Moduli spaces of quiver representations} \label{sec:repsofquivers} A {\it representation} $R$ of $\quiver$ assigns a finite dimensional $\mc$-vector space $R(v)$ to each vertex $v\in\quiver_0$ and a linear map $R(a):R(a^-)\to R(a^+)$ to each arrow $a\in \quiver_1$. A morphism between representations $R$ and $R'$ consists of a collection of linear maps $L(v):R(v)\mapsto R'(v)$ satisfying $R'(a)\circ L(a^-)=L(a^+)\circ R(a)$ for all $a\in\quiver_1$. The {\it dimension vector} of $R$ is $(\dim_{\mc}(R(v))\mid v\in \quiver_0)\in\mn^{\quiver_0}$. For a fixed dimension vector $\alpha\in\mn^{\quiver_0}$, \[\rep(\quiver,\alpha):=\bigoplus_{a\in \quiver_1}\hom_{\mc}(\mc^{\alpha(a^-)},\mc^{\alpha(a^+)})\] is the {\it space of $\alpha$-dimensional representations} of $\quiver$. The product of general linear groups $GL(\alpha):=\prod_{v\in\quiver_0}GL_{\alpha(v)}(\mc)$ acts linearly on $\rep(\quiver,\alpha)$ via \[g\cdot R:=(g(a^+)R(a)g(a^-)^{-1}\mid a\in \quiver_1)\quad (g\in GL(\alpha), R\in\rep(\quiver,\alpha)).\] The $GL(\alpha)$-orbits in $\rep(\quiver,\alpha)$ are in a natural bijection with the isomorphism classes of $\alpha$-dimensional representations of $\quiver$. Given a {\it weight} $\theta\in\mz^{\quiver_0}$, a representation $R$ of $\quiver$ is called {\it $\theta$-semi-stable} if $\sum_{v\in\quiver_0}\theta(v)\dim_{\mc}(R(v))=0$ and $\sum_{v\in\quiver_0}\theta(v)\dim_{\mc}(R'(v))\ge0$ for all subrepresentations $R'$ of $R$. The points in $\rep(\quiver,\alpha)$ corresponding to $\theta$-semi-stable representations constitute a Zariski open subset $\rep(\quiver,\alpha)^{\theta-ss}$ in the representation space, and in \cite{king} Geometric Invariant Theory (cf. \cite{newstead}) is applied to define a variety $\moduli(\quiver,\alpha,\theta)$ and a morphism \begin{equation}\label{eq:quotient} \pi:\rep(\quiver,\alpha)^{\theta-ss}\to\moduli(\quiver,\alpha,\theta)\end{equation} which is a {\it coarse moduli space} for families of $\theta$-semistable $\alpha$-dimensional representations of $\quiver$ up to S-equivalence. A polynomial function $f$ on $\rep(\quiver,\alpha)$ is a {\it relative invariant of weight} $\theta$ if $f(g\cdot R)=(\prod_{v\in \quiver_0}\det(g(v))^{\theta(v)})f(R)$ holds for all $g\in GL(\alpha)$ and $R\in\rep(\quiver,\alpha)$. The relative invariants of weight $\theta$ constitute a subspace $\coord(\rep(\quiver,\alpha))_{\theta}$ in the coordinate ring $\coord(\rep(\quiver,\alpha))$ of the affine space $\rep(\quiver,\alpha)$. In fact $\coord(\rep(\quiver,\alpha))_{\theta}$ is a finitely generated module over the algebra $\coord(\rep(\quiver,\alpha))^{GL(\alpha)}$ of polynomial $GL(\alpha)$-invariants on $\rep(\quiver,\alpha)$ (generators of this latter algebra are described in \cite{lebruyn-procesi}). Now a quasiprojective variety $\moduli(\quiver,\alpha,\theta)$ is defined as the {\it projective spectrum} \[\moduli(\quiver,\alpha,\theta)=\proj(\bigoplus_{n=0}^{\infty}\coord(\rep(\quiver,\alpha))_{n\theta})\] of the graded algebra $\bigoplus_{n=0}^{\infty}\coord(\rep(\quiver,\alpha))_{n\theta}$. A notable special case is that of the zero weight. Then the moduli space $\moduli(\quiver,\alpha,0)$ is the affine variety whose coordinate ring is the subalgebra of $GL(\alpha)$-invariants in $\coord(\rep(\quiver,\alpha))$. This was studied in \cite{lebruyn-procesi} before the introduction of the case of general weights in \cite{king}. Its points are in a natural bijection with the isomorphism classes of semisimple representations of $\quiver$ with dimension vector $\alpha$. For a quiver with no oriented cycles, $\moduli(\quiver,\alpha,0)$ is just a point, and it is more interesting for quivers containing oriented cycles. Let us turn to the special case when $\alpha(v)=1$ for all $v\in\quiver_0$; we simply write $\rep(\quiver)$ and $\moduli(\quiver,\theta)$ instead of $\rep(\quiver,\alpha)$ and $\moduli(\quiver,\alpha,\theta)$. When $\rep(\quiver)^{\theta-ss}$ is non-empty, $\moduli(\quiver,\theta)$ is a quasiprojective toric variety with torus $\pi(\{x\in\rep(\quiver)\mid x(a)\neq 0\ \forall a\in\quiver_1\}) = \pi((\mc^\times)^{\quiver_1})$. On the other hand it is well known (see Proposition~\ref{prop:normal} below) that $\polytope(\quiver,\theta)$ is a lattice polyhedron in the sense of Definition 7.1.3 in \cite{cox-little-schenck}. Denote by $X_{\polytope(\quiver,\theta)}$ the toric variety belonging to the normal fan of $\polytope(\quiver,\theta)$, see for example Theorem 7.1.6 in \cite{cox-little-schenck}. \begin{proposition}\label{prop:hille} We have \[\moduli(\quiver,\theta)\cong X_{\polytope(\quiver,\theta)}.\] \end{proposition} \begin{proof} For quivers with no oriented cycles this is explained in \cite{altmann-hille} using a description of the fan of $\moduli(\quiver,\theta)$ in \cite{hille:canada}. An alternative explanation is the following: the lattice points in $\polytope(\quiver,n\theta)$ correspond bijectively to a $\mc$-basis in $\coord(\rep(\quiver))_{n\theta}$, namely assign to $m\in\polytope(Q,n\theta)\cap\mz^{\quiver_1}$ the function $R\mapsto R^m:=\prod_{a\in\quiver_1} R(a)^{m(a)}$. Now $X_{\polytope(\quiver,\theta)}$ is the projective spectrum of $\bigoplus_{n=0}^\infty\coord(\rep(\quiver))_{n\theta}$ (see Proposition 7.1.13 in \cite{cox-little-schenck}), just like $\moduli(\quiver,\theta)$. \end{proof} A more explicit description of $\moduli(\quiver,\theta)$ is possible thanks to normality of quiver polyhedra: \begin{proposition}\label{prop:normal} (i) Denote by $\quiver^1,\dots,\quiver^t$ the maximal subquivers of $\quiver$ that contain no oriented cycles. Then $\polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ has a Minkowski sum decomposition \begin{equation}\label{eq:recessioncone} \polytope(\quiver,\theta)\cap\mz^{\quiver_1}=\polytope(\quiver,0)\cap\mz^{\quiver_1}+\bigcup_{i=1}^t\polytope(\quiver^i,\theta)\cap\mz^{\quiver_1}.\end{equation} (ii) The quiver polyhedron $\polytope(\quiver,\theta)$ is a normal lattice polyhedron. \end{proposition} \begin{proof} (i) By the {\it support} of $x\in \mr^{\quiver_1}$ we mean the set $\{a\in\quiver_1\mid x(a)\neq 0\}\subseteq \quiver_1$. It is obvious that $\polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ contains the set on the right hand side of \eqref{eq:recessioncone}. To show the reverse inclusion take an $x\in\polytope(\quiver,\theta)\cap \mz^{\quiver_1}$. If its support contains no oriented cycles, then $x\in\polytope(\quiver^i,\theta)$ for some $i$. Otherwise take a minimal oriented cycle $C\subseteq\quiver_1$ in the support of $x$. Denote by $\varepsilon_C\in\mr^{\quiver_1}$ the characteristic function of $C$, and denote by $\lambda$ the minimal coordinate of $x$ along the cycle $C$. Then $\lambda\varepsilon_C\in\polytope(\quiver,0)$ and $y:=x-\lambda\varepsilon_C\in \polytope(\quiver,\theta)$. Moreover, $y$ has strictly smaller support than $x$. By induction on the size of the support we are done. (ii) The same argument as in (i) yields $\polytope(\quiver,\theta)=\polytope(\quiver,0)+\bigcup_{i=1}^t\polytope(\quiver^i,\theta)$. So $\polytope(\quiver,0)$ is the {\it recession cone} of $\polytope(\quiver,\theta)$, and the set of vertices of $\polytope(\quiver,\theta)$ is contained in the union of the vertex sets of $\polytope(\quiver^i,\theta)$. As we pointed out before, the vertices of $\polytope(\quiver^i,\theta)$ belong to $\mz^{\quiver_1}$ by Theorem 13.11 in \cite{schrijver}, whereas the cone $\polytope(\quiver,0)$ is obviously rational and strongly convex. This shows that $\polytope(\quiver,\theta)$ is a lattice polyhedron in the sense of Definition 7.1.3 in \cite{cox-little-schenck}. For normality we need to show that for all positive integers $k$ we have $\polytope(\quiver,k\theta)\cap \mz^{\quiver_1}=k(\polytope(\quiver,\theta)\cap\mz^{\quiver_1})$ (the Minkowski sum of $k$ copies of $\polytope(\quiver,\theta)\cap\mz^{\quiver_1}$), see Definition 7.1.8 in \cite{cox-little-schenck}. Flow polytopes are normal by Theorem 13.14 in \cite{schrijver}, hence the $\polytope(\quiver^i,\theta)$ are normal for $i=1,\dots,t$. So by (i) we have $\polytope(\quiver,k\theta)\cap \mz^{\quiver_1}=\polytope(\quiver,0)\cap\mz^{\quiver_1}+\bigcup_{i=1}^t(\polytope(\quiver^i,k\theta)\cap \mz^{\quiver_1}) =\polytope(\quiver,0)\cap \mz^{\quiver_1}+\bigcup_{i=1}^tk(\polytope(\quiver^i,\theta)\cap \mz^{\quiver_1} ) \subseteq k(\polytope(\quiver,0)+\bigcup_{i=1}^t\polytope(\quiver^i,\theta)\cap\mz^{\quiver1})$. \end{proof} Let $C_1,\dots,C_r$ be the minimal oriented cycles (called also primitive cycles) in $\quiver$. Then their characteristic functions $\varepsilon_{C_1},\dots,\varepsilon_{C_r}$ constitute a Hilbert basis in the monoid $\polytope(\quiver,0)\cap \mz^{\quiver_1}$. Enumerate the elements in $\{m,\varepsilon_{C_j}+m\mid m\in \bigcup_{i=1}^t\polytope(\quiver^i,\theta),j=1,\dots,r\}$ as $m_0,m_1,\dots,m_d$. For a lattice point $m\in \polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ denote by $x^m:\rep(\quiver)\to \mc$ the function $x\mapsto \prod_{a\in\quiver_1} R(a)^{m(a)}$. Consider the map \begin{equation}\label{eq:rho}\rho:\rep(\quiver)^{\theta-ss}\to \mathbb{P}^d,\quad x\mapsto (x^{m_0}:\dots:x^{m_d}).\end{equation} \begin{proposition}\label{prop:realizationofmoduli} $\moduli(\quiver,\theta)$ can be identified with the locally closed subset $\mathrm{Im}(\rho)$ in $\mathbb{P}^d$. \end{proposition} \begin{proof} The morphism $\rho$ is $GL(1,\dots,1)$-invariant, hence it factors through the quotient morphism \eqref{eq:quotient}, so there exists a morphism $\mu:\moduli(\quiver,\theta)\to \mathrm{Im}(\rho)$ with $\mu\circ\pi=\rho$. One can deduce from Proposition~\ref{prop:normal} by the Proj construction of $\moduli(\quiver,\theta)$ that $\mu$ is an isomorphism. \end{proof} This shows also that there is a projective morphism $\moduli(\quiver,\theta)\to \moduli(\quiver,0)$. In particular, $\moduli(\quiver,\theta)$ is a projective variety if and only if $\quiver$ has no oriented cycles, i.e. if $\polytope(\quiver,\theta)$ is a polytope. \section{Contractable arrows}\label{sec:reductions} Throughout this section $\quiver$ stands for a quiver and $\theta\in\mz^{\quiver_0}$ for a weight such that $\polytope(\quiver,\theta)$ is non-empty. For an undirected graph $\Gamma$ we set $\chi(\Gamma):=\|\Gamma_1\|-\|\Gamma_0\|+\chi_0(\Gamma)$, where $\Gamma_0$ is the set of vertices, $\Gamma_1$ is the set of edges in $\Gamma$, and $\chi_0(\Gamma)$ is the number of connected components of $\Gamma$. Define $\chi(\quiver):=\chi(\Gamma)$ and $\chi_0(\quiver):=\chi_0(\Gamma)$ where $\Gamma$ is the underlying graph of $\quiver$, and we say that $\quiver$ is {\it connected} if $\Gamma$ is connected, i.e. if $\chi_0(\quiver)=1$. Denote by $\floweqs:\mr^{\quiver_1}\to\mr^{\quiver_0}$ the map given by \begin{equation}\label{eq:flow} \floweqs(x)(v)=\sum_{a^+=v}x(a)-\sum_{a^-=v}x(a)\qquad (v\in\quiver_0). \end{equation} By definition we have $\polytope(\quiver,\theta)=\floweqs^{-1}(\theta)\cap \mr_{\ge 0}^{\quiver_1}$. It is well known that the codimension in $\mr^{\quiver_0}$ of the image of $\floweqs$ equals $\chi_0(\quiver)$, hence $\dim_{\mr}(\floweqs^{-1}(\theta))=\chi(\quiver)$ for any $\theta\in \floweqs(\mr^{\quiver_1})$, implying that $\dim(\polytope(\quiver,\theta))\le \chi(\quiver)$, where by the {\it dimension of a polyhedron} we mean the dimension of its affine span. We say that we {\it contract an arrow} $a\in\quiver_1$ which is not a loop when we pass to the pair $(\hat\quiver,\hat\theta)$, where $\hat\quiver$ is obtained from $\quiver$ by removing $a$ and glueing its endpoints $a^-,a^+$ to a single vertex $v\in\hat\quiver_0$, and setting $\hat\theta(v):=\theta(a^-)+\theta(a^+)$ whereas $\hat\theta(w)=\theta(w)$ for all vertices $w\in\hat\quiver_0\setminus \{v\}=\quiver_0\setminus\{a^-,a^+\}$. \begin{definition}\label{def:contractable} {\rm Let $\quiver$ be a quiver, $\theta \in \mz^{\quiver_0}$ a weight such that $\polytope(\quiver,\theta)$ is non-empty. \begin{itemize} \item[(i)] An arrow $a\in\quiver_1$ is said to be {\it removable} if $\polytope(\quiver,\theta)$ is integral-affinely equivalent to $\polytope(\quiver',\theta)$, where $\quiver'$ is obtained from $\quiver$ by removing the arrow $a$: $\quiver'_0=\quiver_0$ and $\quiver'_1=\quiver_1\setminus \{a\}$. \item[(ii)] An arrow $a\in\quiver_1$ is said to be {\it contractable} if $\polytope(\quiver,\theta)$ is integral-affinely equivalent to $\polytope(\hat\quiver,\hat\theta)$, where $(\hat\quiver,\hat\theta)$ is obtained from $(\quiver,\theta)$ by contracting the arrow $a$. \item[(iii)] The pair $(\quiver,\theta)$ is called {\it tight} if there is no removable or contractable arrow in $\quiver_1$. \end{itemize} } \end{definition} An immediate corollary of Definition~\ref{def:contractable} is the following statement: \begin{proposition}\label{prop:tightsufficient} Any quiver polyhedron $\polytope(\quiver,\theta)$ is integral-affinely equivalent to some $\polytope(\quiver',\theta')$, where $(\quiver',\theta')$ is tight. Moreover, $(\quiver',\theta')$ is obtained from $(\quiver,\theta)$ by successively removing or contracting arrows. \end{proposition} \begin{remark}\label{remark:altmann-straten-tight} {\rm A pair $(\quiver,\theta)$ is tight if and only if all its connected components are $\theta$-tight in the sense of Definition 12 of \cite{altmann-straten}; this follows from Lemma 7, Corollary 8, and Lemma 13 in \cite{altmann-straten}. These results imply also Corollary~\ref{cor:tightfacets} below, for which we give a direct derivation from Definition~\ref{def:contractable}.} \end{remark} \begin{lemma}\label{lemma:contractable} \begin{itemize} \item[(i)] The arrow $a$ is removable if and only if $x(a)=0$ for all $x\in\polytope(\quiver,\theta)$. \item[(ii)] The arrow $a$ is contractable if and only if in the affine space $\floweqs^{-1}(\theta)$ the halfspace $\{x\in \floweqs^{-1}(\theta)\mid x(a)\geq 0\}$ contains the polyhedron $\{x\in\floweqs^{-1}(\theta)\mid x(b)\geq 0 \quad \forall b\in\quiver_1\setminus\{a\}\}$. \end{itemize} \end{lemma} \begin{proof} (i) is trivial. To prove (ii) denote by $\hat\quiver,\hat\theta$ the quiver and weight obtained by contracting $a$. Since the set of arrows of $\hat\quiver$ can be identified with $\hat\quiver_1=\quiver_1\setminus\{a\}$, we have the projection map $\pi:\floweqs^{-1}(\theta)\to \floweqs'^{-1}(\hat\theta)$ obtained by forgetting the coordinate $x(a)$. The equation \[x(a)=\theta(a^+)-\sum_{b\in\quiver_1\setminus\{a\},b^+=a^+}x(b)+\sum_{b\in\quiver_1\setminus\{a\},b^-=a^+}x(b)\] shows that $\pi$ is injective, hence it gives an affine linear isomorphism $\floweqs^{-1}(\theta)\cap\mz^{\quiver_1}$ and $\floweqs'^{-1}(\hat\theta)\cap\mz^{\hat\quiver_1}$, and maps injectively the lattice polyhedron $\polytope(\quiver,\theta)$ onto an integral-affinely equivalent lattice polyhedron contained in $\polytope(\hat\quiver,\hat\theta)$. Thus $a$ is contractable if and only if on the affine space $\floweqs^{-1}(\theta)$ the inequality $x(a)\ge 0$ is a consequence of the inequalities $x(b)\ge 0$ $(b\in\quiver_1\setminus \{a\}$). \end{proof} For an arrow $a\in\quiver_1$ set $\polytope(\quiver,\theta)_{x(a)=0}:=\{x\in\polytope(\quiver,\theta)\mid x(a)=0\}$. \begin{corollary}\label{cor:tightfacets} (i) The pair $(\quiver,\theta)$ is tight if and only if the assignment $a\mapsto \polytope(\quiver,\theta)_{x(a)=0}$ gives a bijection between $\quiver_1$ and the facets (codimension $1$ faces) of $\polytope(\quiver,\theta)$. (ii) If $(\quiver,\theta)$ is tight, then $\dim(\polytope(\quiver,\theta))=\chi(\quiver)$. \end{corollary} \begin{proof} Lemma~\ref{lemma:contractable} shows that $(\quiver,\theta)$ is tight if and only if $\affspan(\polytope(\quiver,\theta))=\floweqs^{-1}(\theta)$ and $\{x(a)=0\}\cap\floweqs^{-1}(\theta)$ ($a\in\quiver_1$) are distinct supporting hyperplanes of $\polytope(\quiver,\theta)$ in its affine span. \end{proof} The following simple sufficient condition for contractibility of an arrow turns out to be sufficient for our purposes. For a subset $S\subseteq\quiver_0$ set $\theta(S):=\sum_{v\in S}\theta(v)$. By \eqref{eq:flow} for $x\in\floweqs^{-1}(\theta)$ we have \begin{equation}\label{eq:theta(S)}\theta(S)=\sum_{a\in \quiver_1, a^+\in S}x(a)-\sum_{a\in\quiver_1,a^-\in S}x(a)=\sum_{a^+\in S,a^-\notin S} x(a)- \sum_{a^-\in S,a^+\notin S} x(a). \end{equation} \begin{proposition}\label{prop:oneinoneout} Suppose that $S\subset \quiver_0$ has the property that there is at most one arrow $a$ with $a^+\in S$, $a^-\notin S$ and at most one arrow $b$ with $b^+\notin S$ and $b^-\in S$. Then $a$ (if exists) is contractable when $\theta(S)\ge 0$ and $b$ (if exists) is contractable when $\theta(S)\le 0$. \end{proposition} \begin{proof} By \eqref{eq:theta(S)} we have $\theta(S)=x(a)-x(b)$, hence by Lemma~\ref{lemma:contractable} $a$ or $b$ is contractable, depending on the sign of $\theta(S)$. \end{proof} By the {\it valency} of a vertex $v\in\quiver_0$ we mean $\|\{a\in\quiver_1\mid a^-=v\}\|+\|\{a\in\quiver_1\mid a^+=v\}\|$. \begin{corollary}\label{cor:shrinking} (i) Suppose that the vertex $v\in \quiver_0$ has valency $2$, and $a,b\in\quiver_1$ are arrows such that $a^+=b^-=v$. Then the arrow $a$ is contractable when $\theta(v)\ge 0$ and $b$ is contractable when $\theta(v)\le 0$. (ii) Suppose that for some $c\in\quiver_1$, $c^-$ and $c^+$ have valency $2$, and $a,b\in\quiver _1\setminus \{c\}$ with $a^-=c^-$ and $b^+=c^+$. Then $a$ is contractable when $\theta(c^-)+\theta(c^+)\le 0$ and $b$ is contractable when $\theta(c^-)+\theta(c^+)\ge 0$. \end{corollary} \begin{proof} Apply Proposition~\ref{prop:oneinoneout} with $S=\{v\}$ to get (i) and with $S=\{c^-,c^+\}$ to get (ii). \end{proof} \begin{proposition}\label{prop:reflection} Suppose that there are exactly two arrows $a,b\in\quiver_1$ attached to some vertex $v$, and either $a^+=b^+=v$ or $a^-=b^-=v$. Let $\quiver',\theta'$ be the quiver and weight obtained after replacing \begin{tikzpicture}[>=open triangle 45] \node [right] at (2.5,0.5) {by}; \node [below] at (0,0) {$u$}; \node [below] at (2,0) {$w$}; \foreach \x in {(0,0),(1,1),(2,0)} \filldraw \x circle (2pt); \node [above] at (1,1) {$v$}; \node [above, left] at (0.5,0.6) {$a$}; \node[above,right] at (1.5,0.6) {$b$}; \draw [->] (0,0)--(1,1); \draw [<-] (1,1)--(2,0); \end{tikzpicture} \begin{tikzpicture}[>=open triangle 45] \node [right] at (2.5,0.5) {or}; \node [below] at (0,0) {\scriptsize{$\theta(u)+\theta(v)$}}; \node [below] at (2,0) {\scriptsize{$\theta(w)+\theta(v)$}}; \foreach \x in {(0,0),(1,1),(2,0)} \filldraw \x circle (2pt); \node [above] at (1,1) {\scriptsize{$ -\theta(v)$}}; \node [above, left] at (0.5,0.6) {$\hat a$}; \node[above,right] at (1.5,0.6) {$\hat b$}; \draw [<-] (0,0)--(1,1); \draw [->] (1,1)--(2,0); \end{tikzpicture} \quad \begin{tikzpicture}[>=open triangle 45] \node [right] at (2.5,0.5) {by}; \node [below] at (0,0) {$u$}; \node [below] at (2,0) {$w$}; \foreach \x in {(0,0),(1,1),(2,0)} \filldraw \x circle (2pt); \node [above] at (1,1) {$v$}; \node [above, left] at (0.5,0.6) {$a$}; \node[above,right] at (1.5,0.6) {$b$}; \draw [<-] (0,0)--(1,1); \draw [->] (1,1)--(2,0); \end{tikzpicture} \begin{tikzpicture}[>=open triangle 45] \node [right] at (2.5,0.5) {.}; \node [below] at (0,0) {\scriptsize{$\theta(u)+\theta(v)$}}; \node [below] at (2,0) {\scriptsize{$\theta(w)+\theta(v)$}}; \foreach \x in {(0,0),(1,1),(2,0)} \filldraw \x circle (2pt); \node [above] at (1,1) {\scriptsize{$ -\theta(v)$}}; \node [above, left] at (0.5,0.6) {$\hat a$}; \node[above,right] at (1.5,0.6) {$\hat b$}; \draw [->] (0,0)--(1,1); \draw [<-] (1,1)--(2,0); \end{tikzpicture} \quad \noindent That is, replace the arrows $a,b$ by $\hat a$ and $\hat b$ obtained by reversing them, and consider the weight $\theta'\in\mz^{\quiver'_1}$ given by $\theta'(v)=-\theta(v)$, $\theta'(u)=\theta(u)+\theta(v)$ when $u\neq v$ is an endpoint of $a$ or $b$, and $\theta'(w)=\theta(w)$ for all other $w\in\quiver'_0=\quiver_0$. Then the polyhedra $\polytope(\quiver,\theta)$ and $\polytope(\quiver',\theta')$ are integral-affinely equivalent. \end{proposition} \begin{proof} It is straightforward to check that the map $\varphi:\mr^{\quiver_1}\to \mr^{\quiver'_1}$ given by $\varphi(x)(\hat a)=x(b)$, $\varphi(x)(\hat b)=x(a)$, and $\varphi(x)(c)=x(c)$ for all $c\in \quiver'_1\setminus\{\hat a,\hat b\}=\quiver_1\setminus\{a,b\}$ restricts to an isomorphism between $\affspan(\polytope(\quiver,\theta))$ and $\affspan(\polytope(\quiver',\theta'))$ satisfying (i) and (ii) in Definition~\ref{def:isomorphicpolytopes}. \end{proof} \begin{remark}\label{remark:reflection} {\rm Proposition~\ref{prop:reflection} can be interpreted in terms of {\it reflection transformations}: it was shown in Sections 2 and 3 in \cite{kac} (see also Theorem 23 in \cite{skowronski-weyman}) that reflection transformations on representations of quivers induce isomorphisms of algebras of semi-invariants. Now under our assumptions a reflection transformation at vertex $v$ fixes the dimension vector $(1,\dots,1)$. } \end{remark} \begin{proposition}\label{prop:glueing} Suppose that $\quiver$ is the union of its full subquivers $\quiver'$, $\quiver''$ which are either disjoint or have a single common vertex $v$. Identify $\mr^{\quiver'_1}\oplus \mr^{\quiver''_1}=\mr^{\quiver_1}$ in the obvious way, and let $\theta'\in\mz^{\quiver'_0}\subset\mz^{\quiver_0}$, $\theta''\in\mz^{\quiver''_0}\subset\mz^{\quiver_0}$ be the unique weights with $\theta=\theta'+\theta''$ and $\theta'(v)=-\sum_{w\in \quiver'_0\setminus \{v\}}\theta(w)$, $\theta''(v)=-\sum_{w\in \quiver''_0\setminus\{v\}}\theta(w)$ when $\quiver'_0\cap\quiver''_0=\{v\}$. (i) Then the quiver polyhedron $\polytope(\quiver,\theta)$ is the product of the polyhedra $\polytope(\quiver',\theta')$ and $\polytope(\quiver'',\theta'')$. (ii) We have $\moduli(\quiver,\theta)\cong \moduli(\quiver',\theta')\times \moduli(\quiver'',\theta'')$. \end{proposition} \begin{proof} (i) A point $x\in\mr^{\quiver_1}$ uniquely decomposes as $x=x'+x''$, where $x'(a)=0$ for all $a\notin\quiver'_1$ and $x''(a)=0$ for all $a\notin \quiver''_1$. It is obvious by definition of quiver polyhedra that $x\in\polytope(\quiver,\theta)$ if and only if $x'\in\polytope(\quiver',\theta')$ and $x''\in\polytope(\quiver'',\theta'')$. (ii) was observed already in \cite{hille:chemnitz} and follows from (i) by Proposition~\ref{prop:hille}. \end{proof} \begin{definition} \label{def:prime} {\rm \begin{itemize} \item[(i)] We call a connected undirected graph $\Gamma$ (with at least one edge) {\it prime} if it is not the union of full proper subgraphs $\Gamma',\Gamma''$ having only one common vertex (i.e. it is 2-vertex-connected). A quiver $\quiver$ will be called {\it prime} if its underlying graph is prime. \item[(ii)] We call a toric variety {\it prime} if it is not the product of lower dimensional toric varieties. \end{itemize}} \end{definition} Obviously any toric variety is the product of prime toric varieties, and this product decomposition is unique up to the order of the factors (see for example Theorem 2.2 in \cite {hatanaka}). It is not immediate from the definition, but we shall show in Theorem~\ref{thm:directprod} (iii) that the prime factors of a toric quiver variety are quiver varieties as well. Note that a toric quiver variety associated to a non-prime quiver may well be prime, and conversely, a toric quiver variety associated to a prime quiver can be non-prime, as it is shown by the following example: \[\begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(0,0),(1,0),(1,1),(1,-1),(2,0)} \filldraw \x circle (2pt); \foreach \y in {(1,0),(1,1),(1,-1)} \draw [->] (0,0)--\y; \foreach \x in {(1,0),(1,1),(1,-1)} \draw [<-] \x--(2,0); \node [left] at (0,0) {$-2$}; \node [above] at (1,1) {$1$}; \node [above] at (1,0) {$1$}; \node [below] at (1,-1) {$2$}; \node [right] at (2,0) {$-2$}; \end{tikzpicture}\] The quiver in the picture is prime but the moduli space corresponding to this weight is $\mathbb{P}^1\times\mathbb{P}^1$. However, as shown by Theorem~\ref{thm:directprod} below, when the tightness of some $(\quiver, \theta)$ is assumed, decomposing $\quiver$ into its unique maximal prime components gives us the unique decomposition of $\moduli(\quiver,\theta)$ as a product of prime toric varieties. \begin{theorem} \label{thm:directprod} \begin{itemize} \item[(i)] Let $\quiver^i$ $(i=1,\dots,k)$ be the maximal prime full subquivers of $\quiver$, and denote by $\theta^i\in\mz^{\quiver^i_0}$ the unique weights satisfying $\sum_{i=1}^k\theta^i(v)=\theta(v)$ for all $v\in\quiver_0$ and $\sum_{v\in{\quiver_0^i}}\theta^i(v)=0$ for all $i$. Then $\moduli(\quiver,\theta)\cong \prod_{i=1}^k\moduli(\quiver^i,\theta^i)$. Moreover, if $(\quiver,\theta)$ is tight, then the $(\quiver^i,\theta^i)$ are all tight. \item[(ii)] If $(\quiver,\theta)$ is tight then $\moduli(\quiver,\theta)$ is prime if and only if $\quiver$ is prime. \item[(iii)] Any toric quiver variety is the product of prime toric quiver varieties. \end{itemize} \end{theorem} \begin{proof} The isomorphism $\moduli(\quiver,\theta)\cong \prod_{i=1}^k\moduli(\quiver^i,\theta^i)$ follows from Proposition~\ref{prop:glueing}. The second statement in (i) follows from this isomorphism and Corollary~\ref{cor:tightfacets}. Next we turn to the proof of (ii), so suppose that $(\quiver,\theta)$ is tight. If $\quiver$ is not prime, then $\chi(\quiver^i)>0$ for all $i$, hence $\moduli(\quiver,\theta)$ is not prime by (i). To show the reverse implication assume on the contrary that $\quiver$ is prime, and $\moduli(\quiver,\theta)\cong X'\times X''$ where $X',X''$ are positive dimensional toric varieties. Note that then $\quiver_1$ does not contain loops. Let $\{\varepsilon_a\mid a \in \quiver_1\}$ be a $\mz$-basis of $\mz^{\quiver_1}$, and for each vertex $v\in\quiver_0$ let us define $C_v := \sum_{a^+=v}\varepsilon_a - \sum_{a^-=v}\varepsilon_a$. Following the description of the toric fan $\Sigma$ of $\moduli(\quiver,\theta)$ in \cite{hille:canada} we can identify the lattice of one-parameter subgroups $N$ of $\moduli(\quiver,\theta)$ with $\mz^{\quiver_1} / \langle C_v\mid v\in\quiver_0 \rangle$, and the ray generators of the fan with the cosets of the $\varepsilon_a$. Denoting by $\Sigma'$ and $\Sigma''$ the fans of $X'$ and $X''$ respectively, we have $\Sigma = \Sigma' \times \Sigma''=\{\sigma'\times\sigma''\mid\sigma'\in\Sigma',\:\sigma''\in\Sigma''\}$ (see \cite{cox-little-schenck} Proposition 3.1.14). Denote by $\pi':N\rightarrow N'$, $\pi'':N\rightarrow N''$ the natural projections to the sets of one-parameter subgroups of the tori in $X'$ and $X''$. For each ray generator $\varepsilon_a$ we have either $\pi'(\varepsilon_a) = 0$ or $\pi''(\varepsilon_a) = 0$. Since $(\quiver,\theta)$ is tight we obtain a partition of $\quiver_1$ into two disjoint non-empty sets of arrows: $\quiver_1'=\{a\in\quiver_1\mid\pi''(a)=0\}$ and $\quiver_1''=\{a\in\quiver_1\mid\pi'(a)=0\}$. Since $\quiver$ is prime, it is connected, hence there exists a vertex $w$ incident to arrows both from $\quiver'_1$ and $\quiver''_1$. Let $\Pi'$ and $\Pi''$ denote the projections from $\mz^{\quiver_1}$ to $\mz^{\quiver_1'}$ and $\mz^{\quiver_1''}$. By choice of $w$ we have $\Pi'(C_w)\neq 0$ and $\Pi''(C_w)\neq 0$. Writing $\varphi$ for the natural map from $\mz^{\quiver_1}$ to $N \cong \mz^{\quiver_1} / \langle C_v\mid v\in\quiver_0 \rangle$ we have $\varphi \circ \Pi' = \pi' \circ \varphi$ and $\varphi \circ \Pi'' = \pi'' \circ \varphi$, so $\ker(\varphi) = \langle C_v\mid v\in\quiver_0 \rangle$ is closed under $\Pi'$ and $\Pi''$. Taking into account that $\sum_{v\in\quiver_0}C_v=0$ we deduce that $\Pi'(C_w)=\sum_{v\in\quiver_0\setminus\{w\}}\lambda_vC_v$ for some $\lambda_v\in \mz$. Set $S':=\{v\in\quiver_0\mid\lambda_v\ne 0\}$. Since each arrow appears in exactly two of the $C_v$, it follows that $S'$ contains all vertices connected to $w$ by an arrow in $\quiver'_1$, hence $S'$ is non-empty. Moreover, the set of arrows having exactly one endpoint in $S'$ are exactly those arrows in $\quiver'_1$ that are adjacent to $w$. Thus $S'':=\quiver_0\setminus (S'\cup\{w\})$ contains all vertices that are connected to $w$ by an arrow from $\quiver''_1$, hence $S''$ is non-empty. Furthermore, there are no arrows in $\quiver_1$ connecting a vertex from $S'$ to a vertex in $S''$. It follows that $\quiver$ is the union of its full subquivers spanned by the vertex sets $S'\cup\{w\}$ and $S''\cup\{w\}$, having only one common vertex $w$ and no common arrow. This contradicts the assumption that $\quiver$ was prime. Statement (iii) follows from (i), (ii) and Proposition~\ref{prop:tightsufficient}. \end{proof} Note that if $\chi(\Gamma)\ge 2$ and $\Gamma$ is prime, then $\Gamma$ contains no loops (i.e. an edge with identical endpoints), every vertex of $\Gamma$ has valency at least $2$, and $\Gamma$ has at least two vertices with valency at least $3$. \begin{definition}\label{def:skeleton} {\rm For $d=2,3,\dots$ denote by $\sklist_d$ the set of prime graphs $\Gamma$ with $\chi(\Gamma)=d$ in which all vertices have valency at least $3$. Let $\reducedquivers_d$ stand for the set of quivers $\quiver$ with no oriented cycles obtained from a graph $\Gamma\in\sklist_d$ by orienting some of the edges somehow and putting a sink on the remaining edges (that is, we replace an edge by a path of length $2$ in which both edges are pointing towards the new vertex in the middle). We shall call $\Gamma$ the {\it skeleton} $\skeleton(\quiver)$ of $\quiver$; note that $\chi(\quiver)=\chi(\skeleton(\quiver))$. }\end{definition} Starting from $\quiver$, its skeleton $\Gamma=\skeleton(\quiver)$ can be recovered as follows: $\Gamma_0$ is the subset of $\quiver_0$ consisting of the valency $3$ vertices. For each path in the underlying graph of $\quiver$ that connects two vertices in $\Gamma_0$ and whose inner vertices have valency $2$ we put an edge. Clearly, a quiver $\quiver$ with $\chi(\quiver)=d\ge 2$ belongs to $\reducedquivers_d$ if and only if the following conditions hold: \begin{itemize} \item[(i)] $\quiver$ is prime. \item[(ii)] There is no arrow of $\quiver$ connecting valency $2$ vertices. \item[(iii)] Every valency $2$ vertex of $\quiver$ is a sink. \end{itemize} Furthermore, set $\reducedquivers:=\bigsqcup_{d=1}^\infty \reducedquivers_d$ where $\reducedquivers_1$ is the $1$-element set consisting of the Kronecker quiver \begin{tikzpicture}[>=open triangle 45] \draw [->] (0,0) to [out=45, in=135] (1,0); \node [above] at (0,0) {}; \node [above] at (1,0) {}; \draw [->] (0,0) to [out=315, in=225] (1,0) ; \filldraw (0,0) circle (1.5pt) (1,0) circle (1.5pt); \end{tikzpicture}. \begin{remark}\label{remark:combinatorial-tightness} {\rm A purely combinatorial characterization of tightness is given in Lemma 13 of \cite{altmann-straten}. Namely, $(\quiver,\theta)$ is tight if and only if any connected component of $\quiver$ is $\theta$-stable, and any connected component of $\quiver\setminus \{a\}$ for any $a\in\quiver_1$ is $\theta$-stable (see Section~\ref{sec:affine} for the notion of $\theta$-stability). In the same Lemma it is also shown that if $(\quiver,\theta)$ is tight, then $(\quiver,\delta_{\quiver})$ is tight, where $\delta_{\quiver}:=\sum_{a\in\quiver_1}(\varepsilon_{a^+}-\varepsilon_{a^-})$ is the so-called {\it canonical weight} (here $\varepsilon_v$ stands for the characteristic function of $v\in\quiver_0$). It is easy to deduce that for a connected quiver $\quiver$ the pair $(\quiver,\delta_{\quiver})$ is tight if and only if there is no partition $\quiver_0=S\coprod S'$ such that there is at most one arrow from $S$ to $S'$ and there is at most one arrow from $S'$ to $S$. } \end{remark} \begin{proposition}\label{prop:boundonskeletons} For any $d\ge 2$, $\Gamma\in\sklist_d$ and $\quiver\in\reducedquivers_d$ we have the inequalities \[\|\Gamma_0\|\le 2d-2, \quad \|\Gamma_1\|\le 3d-3, \quad \|\quiver_0\|\le 5(d-1), \quad \|\quiver_1\|\le 6(d-1).\] In particular, $\sklist_d$ and $\reducedquivers_d$ are finite for each positive integer $d$. \end{proposition} \begin{proof} Take $\Gamma\in\sklist_d$ where $d\ge 2$. Then $\Gamma$ contains no loops, and denoting by $e$ the number of edges and by $v$ the number of vertices of $\Gamma$, we have the inequality $2e\ge 3v$, since each vertex is adjacent to at least three edges. On the other hand $e=v-1+d$. We conclude that $v\le 2d-2$ and hence $e\le 3d-3$. For $\quiver\in\reducedquivers_d$ with $\skeleton(\quiver)=\Gamma$ we have that $\|\quiver_0\|\le v+e$ and $\|\quiver_1\|\le 2e$. \end{proof} \begin{theorem}\label{thm:finitelymanymoduli} \begin{itemize} \item[(i)] Any $d$-dimensional prime toric quiver variety $\moduli(\quiver,\theta)$ can be realized by a tight pair $(\quiver,\theta)$ where $\quiver\in\reducedquivers_d$ (consequently $\|\quiver_0\|\le 5(d-1)$ and $\|\quiver_1\|\le 6(d-1)$ when $d\ge 2$). \item[(ii)] For each positive integer $d$ up to isomorphism there are only finitely many $d$-dimensional toric quiver varieties. \end{itemize} \end{theorem} \begin{proof} It follows from Propositions~\ref{prop:tightsufficient}, Corollary~\ref{cor:shrinking} and Proposition~\ref{prop:reflection} that any $d$-dimensional prime toric quiver variety can be realized by a tight pair $(\quiver,\theta)$ where $\quiver\in\reducedquivers_d$; the bounds on vertex and arrow sets of the quiver follow by Proposition~\ref{prop:boundonskeletons}. It remains to show (ii). For a given quiver $\quiver$ we say that the weights $\theta$ and $\theta'$ are equivalent if $\rep(\quiver)^{\theta-ss}=\rep(\quiver)^{\theta'-ss}$; this implies that $\moduli(\quiver,\theta)=\moduli(\quiver,\theta')$. For a given representation $R$ of $\quiver$, the set of weights $\theta$ for which $R$ is $\theta$-semistable is determined by the set of dimension vectors of subrepresentations of $R$. Since there are finitely many possibilities for the dimension vectors of a subrepresentation of a representation with dimension vector $(1,\dots,1)$, up to equivalence there are only finitely many different weights, hence there are finitely many possible moduli spaces for a fixed $\quiver$. \end{proof} \begin{remark} {\rm Part (i) of Theorem~\ref{thm:finitelymanymoduli} can be directly obtained from the results in \cite{altmann-nill-schwentner-wiercinska} and \cite{altmann-straten}. From the proof of Theorem 7 in \cite{altmann-nill-schwentner-wiercinska} it follows that the bound on the number of vertices and edges hold whenever the canonical weight is tight for a quiver. While in \cite{altmann-nill-schwentner-wiercinska} it is assumed that $\quiver$ has no oriented cycles, their argument for the bound applies to the general case as well. Moreover Lemma 13 in \cite{altmann-straten} shows that every toric quiver variety can be realized by a pair $(\quiver,\theta)$ where $\quiver$ is tight with the canonical weight. These two results imply part (i) of Theorem~\ref{thm:finitelymanymoduli}. } \end{remark} \begin{remark}\label{remark:fourier-motzkin} {\rm We mention that for a fixed quiver $\quiver$ it is possible to give an algorithm to produce a representative for each of the finitely many equivalence classes of weights. The change of the moduli spaces of a given quiver when we vary the weight is studied in \cite{hille:chemnitz}, \cite{hille:canada}, where the inequalities determining the chamber system were given. To find an explicit weight in each chamber one can use the Fourier-Motzkin algorithm. } \end{remark} Theorem~\ref{thm:finitelymanymoduli} is sharp, and the reductions on the quiver are optimal, in the sense that in general one can not hope for reductions that would yield smaller quivers: \begin{proposition}\label{prop:optimal} For each natural number $d\ge 2$ there exists a $d$-dimensional toric quiver variety $\moduli(\quiver,\theta)$ with $\|\quiver_1\|=6(d-1)$, $\|\quiver_0\|=5(d-1)$, such that for any other quiver and weight $\quiver',\theta'$ with $\moduli(\quiver,\theta)\cong \moduli(\quiver',\theta')$ (isomorphism of toric varieties) we have that $\|\quiver'_1\|\ge \|\quiver_1\|$ and $\|\quiver'_0\|\ge \|\quiver_0\|$. \end{proposition} \begin{proof} The number of inequalities defining $\polytope(\quiver,\theta)$ in its affine span is obviously bounded by the number of arrows of $\quiver$, so the number of facets of a quiver polyhedron $\polytope(\quiver,\theta)$ is bounded by $\|\quiver_1\|$. On the other hand the number of facets is an invariant of the corresponding toric variety, as it equals the number of rays in the toric fan of $\moduli(\quiver,\theta)$. Therefore by Corollary~\ref{cor:tightfacets} it is sufficient to show the existence of a tight $(\quiver,\theta)$ with $\|\quiver_1\|=6(d-1)$ and $\|\quiver_0\|=5(d-1)$. Such a pair $(\quiver,\theta)$ is provided in Example~\ref{example:2sharp}. \end{proof} \begin{example}\label{example:2sharp} {\rm For $d\ge 3$ consider the graph below with $2(d-1)$ vertices. Removing any two edges from this graph we obtain a connected graph. Now let $\quiver$ be the quiver obtained by putting a sink on each of the edges (so the graph below is the skeleton of $\quiver$). Then $(\quiver,\delta_{\quiver})$ is tight by Remark~\ref{remark:combinatorial-tightness} ($\delta_{\quiver}$ takes value $2$ on each sink and value $-3$ on each source). \[\begin{tikzpicture} \draw (0,0)--(3,0)--(3,1)--(0,1)--(0,0) (4,0)--(6,0)--(6,1)--(4,1)--(4,0) (1,0)--(1,1) (2,0)--(2,1) (3,0)--(3,1) (4,0)--(4,1) (5,0)--(5,1); \draw [dotted] (3,0)--(4,0) (3,1)--(4,1); \draw (0,1) to (6,0); \draw (0,0) to (6,1); \filldraw (0,0) circle (1.5pt) (1,0) circle (1.5pt) (2,0) circle (1.5pt) (3,0) circle (1.5pt) (4,0) circle (1.5pt) (5,0) circle (1.5pt) (6,0) circle (1.5pt) (0,1) circle (1.5pt) (1,1) circle (1.5pt) (2,1) circle (1.5pt) (3,1) circle (1.5pt) (4,1) circle (1.5pt) (5,1) circle (1.5pt) (6,1) circle (1.5pt); \end{tikzpicture} \]} \end{example} Relaxing the condition on tightness it is possible to come up with a shorter list of quivers whose moduli spaces exhaust all possible toric quiver varieties. A key role is played by the following statement: \begin{proposition}\label{prop:addanarrow} Suppose that $\quiver$ has no oriented cycles and $a\in\quiver_1$ is an arrow such that contracting it we get a quiver (i.e. the quiver $\hat\quiver$ described in Definition~\ref{def:contractable}) that has no oriented cycles. Then for a sufficiently large integer $d$ we have that $a$ is contractable for the pair $(\quiver,\theta+d(\varepsilon_{a^+}-\varepsilon_{a^-}))$, where $\varepsilon_v\in\mz^{\quiver_0}$ stands for the characteristic function of $v\in\quiver_0$. \end{proposition} \begin{proof} Set $\psi_d = \theta+d(\varepsilon_{a^+}-\varepsilon_{a^-})$, and note that $\hat\psi_d = \hat\theta$ for all d. Considering the embeddings $\pi:\floweqs^{-1}(\psi_d)\to \floweqs'^{-1}(\hat\theta)$ described in the proof of Lemma~\ref{lemma:contractable}, we have that for any $d$, any $y\in\floweqs^{-1}(\psi_d)$ and $b\in\quiver_1\setminus\{a\}$, \[\min\{x(b)\mid x\in\polytope(\hat\quiver,\hat\theta)\}\leq y(b) \leq \max\{x(b)\mid x\in\polytope(\hat\quiver,\hat\theta)\}\] Since we assumed that $\hat\quiver$ has no oriented cycles, the minimum and the maximum in the inequality above are finite. Now considering the arrows incident to $a^-$ we obtain that for any $x\in\floweqs^{-1}(\psi_d)$ we have $x(a) = d-\theta(a^-)+\sum_{b^+ = a^-}x(b) - \sum_{b^- = a^-,b\neq a}x(b)$. Thus for $d \geq \theta(a^-)-\min\{ \sum_{b^+ = a^-}x(b) - \sum_{b^- = a^-,b\neq a}x(b) \mid x\in\floweqs'^{-1}(\hat\theta)\}$ the arrow $a$ is contractable for $(\quiver,\psi_d)$ by Lemma~\ref{lemma:contractable}. \end{proof} For $d\ge 2$ introduce a partial ordering $\ge$ on $\sklist_d$: we set $\Gamma\ge \Gamma'$ if $\Gamma'$ is obtained from $\Gamma$ by contracting an edge, and take the transitive closure of this relation. Now for each positive integer $d\ge 2$ denote by $\sklist'_d\subseteq\sklist_d$ the set of undirected graphs $\Gamma\in \sklist_d$ that are maximal with respect to the relation $\ge$, and set $\sklist'_1:=\sklist_1$. It is easy to see that for $d\ge 2$, $\sklist'_d$ consists of $3$-regular graphs (i.e. graphs in which all vertices have valency $3$). Now denote by $\reducedquivers'_d$ the quivers which are obtained by putting a sink on each edge from a graph from $\sklist'_d$. \begin{theorem}\label{thm:3regular} For $d\ge 2$ any prime $d$-dimensional projective toric quiver variety is isomorphic to $\moduli(\quiver,\theta)$ where $\quiver\in\reducedquivers'_d$. \end{theorem} \begin{proof} This is an immediate consequence of Theorem~\ref{thm:finitelymanymoduli} and Proposition~\ref{prop:addanarrow}. \end{proof} \begin{example}\label{example:threedim} {\rm $\sklist'_3$ consists of two graphs: \begin{tikzpicture} \foreach \x in {(0,0),(1,0),(0,1),(1,1)} \filldraw \x circle (2pt); \draw (0,0)--(1,0)--(1,1)--(0,1)--(0,0); \draw (0,0) to [out=135,in=225] (0,1); \draw (1,0) to [out=45,in=315] (1,1); \end{tikzpicture} \qquad \qquad \begin{tikzpicture} \foreach \x in {(0,0),(1,0),(0,1),(1,1)} \filldraw \x circle (2pt); \draw (0,0)--(1,0)--(1,1)--(0,1)--(0,0); \draw (0,0)--(1,1); \draw (1,0)--(0,1); \end{tikzpicture} Now put a sink on each edge of the above graphs. The first of the two resulting quivers is not tight for the canonical weight. After tightening we obtain the following two quivers among whose moduli spaces all $3$-dimensional prime projective toric quiver varieties occur: \begin{tikzpicture}[>=open triangle 45] \foreach \x in {(0,0),(0,1),(0,2),(-1,1),(0,1),(1,0),(1,2),(1,1),(2,1)} \filldraw \x circle (1.5pt); \draw [->] (0,0) to (1,0); \draw [->] (0,2) to (1,2); \draw [->] (0,0) to (0,1); \draw [->] (0,2) to (0,1); \draw [->] (0,0) to (-1,1); \draw [->] (0,2) to (-1,1); \draw [->] (1,2) to (1,1); \draw [->] (1,0) to (1,1); \draw [->] (1,0) to (2,1); \draw [->] (1,2) to (2,1); \end{tikzpicture} \qquad \qquad \begin{tikzpicture}[>=open triangle 45] \foreach \x in {(0,0),(1,0),(2,0),(1,1),(1,2),(0,1),(2,1),(0.5,0.5),(1.5,0.5),(1,1.5)} \filldraw \x circle (1.5pt); \draw [->] (0,0) to (0,1); \draw [->] (1,2) to (0,1); \draw [->] (0,0) to (1,0); \draw [->] (2,0) to (1,0); \draw [->] (2,0) to (2,1); \draw [->] (1,2) to (2,1); \draw [->] (0,0) to (0.5,0.5); \draw [->] (1,1) to (0.5,0.5); \draw [->] (1,1) to (1,1.5); \draw [->] (1,2) to (1,1.5); \draw [->] (2,0) to (1.5,0.5); \draw [->] (1,1) to (1.5,0.5); \end{tikzpicture} }\end{example} \section{The $2$-dimensional case} \label{sec:2-dimensional} As an illustration of the general classification scheme explained in Section~\ref{sec:reductions}, we quickly reproduce the classification of $2$-dimensional toric quiver varieties (this result is known, see Theorem 5.2 in \cite{hille:chemnitz} and Example 6.14 in \cite{fei}): \begin{proposition}\label{prop:twodim} (i) A $2$-dimensional toric quiver variety is isomorphic to one of the following: \noindent The projective plane $\mathbb{P}^2$, the blow up of $\mathbb{P}^2$ in one, two, or three points in general position, or $\mathbb{P}^1\times\mathbb{P}^1$. (ii) The above varieties are realized (in the order of their listing) by the following quiver-weight pairs: \[ \begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(0,0),(1,0),(1,1),(1,-1),(2,0)} \filldraw \x circle (2pt); \foreach \y in {(1,0),(1,1),(1,-1)} \draw [->] (0,0)--\y; \foreach \x in {(1,0),(1,1),(1,-1)} \draw [<-] \x--(2,0); \node [left] at (0,0) {$-1$}; \node [above] at (1,1) {$1$}; \node [above] at (1,0) {$1$}; \node [below] at (1,-1) {$1$}; \node [right] at (2,0) {$-2$}; \draw [dotted] (3,-1.5)--(3,1.5); \end{tikzpicture} \begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(0,0),(1,0),(1,1),(1,-1),(2,0)} \filldraw \x circle (2pt); \foreach \y in {(1,0),(1,1),(1,-1)} \draw [->] (0,0)--\y; \foreach \x in {(1,0),(1,1),(1,-1)} \draw [<-] \x--(2,0); \node [left] at (0,0) {$-3$}; \node [above] at (1,1) {$2$}; \node [above] at (1,0) {$1$}; \node [below] at (1,-1) {$2$}; \node [right] at (2,0) {$-2$}; \draw [dotted] (3,-1.5)--(3,1.5); \end{tikzpicture} \begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(0,0),(1,0),(1,1),(1,-1),(2,0)} \filldraw \x circle (2pt); \foreach \y in {(1,0),(1,1),(1,-1)} \draw [->] (0,0)--\y; \foreach \x in {(1,0),(1,1),(1,-1)} \draw [<-] \x--(2,0); \node [left] at (0,0) {$-4$}; \node [above] at (1,1) {$3$}; \node [above] at (1,0) {$2$}; \node [below] at (1,-1) {$2$}; \node [right] at (2,0) {$-3$}; \draw [dotted] (3,-1.5)--(3,1.5); \end{tikzpicture} \begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(0,0),(1,0),(1,1),(1,-1),(2,0)} \filldraw \x circle (2pt); \foreach \y in {(1,0),(1,1),(1,-1)} \draw [->] (0,0)--\y; \foreach \x in {(1,0),(1,1),(1,-1)} \draw [<-] \x--(2,0); \node [left] at (0,0) {$-3$}; \node [above] at (1,1) {$2$}; \node [above] at (1,0) {$2$}; \node [below] at (1,-1) {$2$}; \node [right] at (2,0) {$-3$}; \draw [dotted] (3,-1.5)--(3,1.5); \end{tikzpicture} \begin{tikzpicture}[>=open triangle 45] \draw [->] (0,0.8) to [out=45, in=135] (1,0.8); \node [above] at (0,0) {}; \node [above] at (1,0) {}; \draw [->] (0,0.8) to [out=315, in=225] (1,0.8) ; \filldraw (0,0.8) circle (1.5pt) (1,0.8) circle (1.5pt); \draw [->] (0,1.8) to [out=45, in=135] (1,1.8); \draw [->] (0,1.8) to [out=315, in=225] (1,1.8) ; \filldraw (0,1.8) circle (1.5pt) (1,1.8) circle (1.5pt); \node [left] at (0,0.8) {$-1$}; \node [left] at (0,1.8) {$-1$}; \node [right] at (1,0.8) {$1$}; \node [right] at (1,1.8) {$1$}; \end{tikzpicture} \] \end{proposition} \begin{proof} $\reducedquivers_1$ consists only of the Kronecker quiver. The only weights yielding a non-empty moduli space are $(-1,1)$ and its positive integer multiples, hence the corresponding moduli space is $\mathbb{P}^1$. Thus $\mathbb{P}^1\times \mathbb{P}^1$, the product of two projective lines occurs as a $2$-dimensional toric quiver variety, say for the disjoint union of two copies of \begin{tikzpicture}[>=open triangle 45] \draw [->] (0,0) to [out=45, in=135] (1,0); \node [above] at (0,0) {}; \node [above] at (1,0) {}; \draw [->] (0,0) to [out=315, in=225] (1,0) ; \filldraw (0,0) circle (1.5pt) (1,0) circle (1.5pt); \node [left] at (0,0) {$-1$}; \node [right] at (1,0) {$1$}; \ \end{tikzpicture}. $\sklist_2$ consists of the graph with two vertices and three edges connecting them (say by Proposition~\ref{prop:boundonskeletons}). Thus $\reducedquivers'_2$ consists of the following quiver: \[ \begin{tikzpicture}[>=open triangle 45] \foreach \x in {(0,0),(1,0),(1,1),(1,-1),(2,0)} \filldraw \x circle (2pt); \foreach \y in {(1,0),(1,1),(1,-1)} \draw [->] (0,0)--\y; \foreach \x in {(1,0),(1,1),(1,-1)} \draw [<-] \x--(2,0); \node at (0,1) {$A$:}; \end{tikzpicture} \] Choosing a spanning tree $\subtree$ in $\quiver$, the $x(a)$ with $a\in \quiver_1\setminus \subtree_1$ can be used as free coordinates in $\affspan(\polytope(\quiver,\theta))$. For example, take in the quiver $A$ the spanning tree with thick arrows in the following figure: \[ \begin{tikzpicture}[>=open triangle 45,scale=0.8] \node [left] at (1.5,2) {$\affspan(\polytope(A,\theta))$:}; \node [left] at (0,0) {$\theta_1$}; \node [above] at (3,2) {$\theta_2$}; \node [above] at (3,0) {$\theta_3$}; \node [below] at (3,-2) {$\theta_4$}; \node [right] at (6,0) {$\theta_5$}; \node [right] at (8,2) {$ \sum_{i=1}^5\theta_i=0$}; \foreach \x in {(0,0),(3,0),(3,2),(3,-2),(6,0)} \filldraw \x circle (2pt); \foreach \y in {(3,0),(3,2)} \draw [->] (0,0)--\y; \draw [->, very thick] (0,0)--(3,-2); \foreach \x in {(3,0),(3,2),(3,-2)} \draw [<-,very thick] \x--(6,0); \node [above] at (1.5,1) {$x$}; \node [above] at (1.5,0) {$y$}; \node [left] at (1.5,-1) {$-\theta_1-x-y$}; \node [right] at (4.5,1.1) {$\theta_2-x$}; \node [above] at (4.5,0) {$\theta_3-y$}; \node [right] at (4.5,-1.1) {$\theta_4+\theta_1+x+y$}; \end{tikzpicture} \] \begin{figure} \caption{{\it The polytope $\polytope(A,\theta)$}} \begin{tikzpicture}[scale=0.7] \draw [->] (0,0)--(5,0); \draw [->] (0,0)--(0,5); \node [right] at (5,0) {$x$}; \node [above] at (0,5) {$y$}; \draw (-0.3,4.8)--(4.8,-0.3); \node [above right] at (2.1,2.1) {$x+y=-\theta_1$}; \draw (-0.3,4)--(1,4); \node [left] at (-0.3,4) {$y=\theta_3$}; \draw (3.7,1.5)--(3.7,-0.3); \node [below] at (3.7,-0.3) {$x=\theta_2$}; \draw (-0.3,1.3)--(1.3,-0.3); \node [left] at (-0.3,1.3) {$x+y=-\theta_1-\theta_4$}; \draw [pattern=north east lines] (0,1)--(1,0)--(3.7,0)--(3.7,0.8)--(0.5,4)--(0,4)--(0,1); \node at (9,5) {Defining inequalitites:}; \node at (9,4) {$ 0 \le x\le \theta_2$}; \node at (9,3) {$0 \le y\le \theta_3 $}; \node at (9,2) {$-\theta_4-\theta_1\le x+y\le -\theta_1$.}; \end{tikzpicture} \end{figure} Clearly $\polytope(A,\theta)$ is integral-affinely equivalent to the polytope in $\mathbb{R}^2=\{(x,y)\mid x,y\in \mathbb{R}\}$ shown on Figure~1. Depending on the order of $-\theta_1,\theta_3,-\theta_1-\theta_4,\theta_2$, its normal fan is one of the following: \[\begin{tikzpicture} \draw (0,0)--(1.5,0) (-1.5,-1.5)--(0,0) (0,0)--(0,1.5); \node at (0.2,-0.4) {$\sigma_1$}; \node at (1,0.5) {$\sigma_2$}; \node at (-0.4,0.4) {$\sigma_3$}; \end{tikzpicture} \qquad \begin{tikzpicture} \draw (0,0)--(1.5,0) (0,0)--(0,1.5) (-1.5,0)--(0,0) (0,-1.5)--(0,0); \node at (0.8,0.8) {$\sigma_1$}; \node at (0.8,-0.8) {$\sigma_4$}; \node at (-0.8,-0.8) {$\sigma_3$}; \node at (-0.8,0.8) {$\sigma_2$}; \end{tikzpicture} \qquad \begin{tikzpicture} \draw (0,0)--(1.5,0) (0,0)--(0,1.5) (-1.5,-1.5)--(1.5,1.5); \node at (0.2,-0.4) {$\sigma_1$}; \node at (1,0.5) {$\sigma_2$}; \node at (0.6,1) {$\sigma_3$}; \node at (-0.4,0.4) {$\sigma_4$}; \end{tikzpicture} \qquad \] \[\begin{tikzpicture} \draw (-1.5,0)--(1.5,0) (0,0)--(0,1.5) (-1.5,-1.5)--(1.5,1.5); \node at (0.2,-0.4) {$\sigma_1$}; \node at (1,0.5) {$\sigma_2$}; \node at (0.6,1) {$\sigma_3$}; \node at (-0.4,0.4) {$\sigma_4$}; \node at (-1,-0.5) {$\sigma_5$}; \end{tikzpicture} \qquad \begin{tikzpicture} \draw (-1.5,0)--(1.5,0) (0,-1.5)--(0,1.5) (-1.5,-1.5)--(1.5,1.5); \node at (0.4,-0.4) {$\sigma_1$}; \node at (1,0.5) {$\sigma_2$}; \node at (0.6,1) {$\sigma_3$}; \node at (-0.4,0.4) {$\sigma_4$}; \node at (-1,-0.5) {$\sigma_5$}; \node at (-0.6,-1) {$\sigma_6$}; \end{tikzpicture} \qquad \] It is well known that the corresponding toric varieties are the projective plane $\mathbb{P}^2$, $\mathbb{P}^1\times \mathbb{P}^1$ and the projective plane blown up in one, two, or three points in general position, so (i) is proved. Taking into account the explicit inequalities in Figure 1, we see that for the pairs $(A,\theta)$ given in (ii), the variety $X_{\polytope(A,\theta)}=\moduli(A,\theta)$ has the desired isomorphism type. \end{proof} \begin{remark}\label{remark:nosmallerquiver} {\rm (i) Since the toric fan of the blow up of $\mathbb{P}^2$ in three generic points has $6$ rays, to realize it as a toric quiver variety we need a quiver with at least $6$ arrows and hence with at least $5$ vertices (see Proposition~\ref{prop:optimal}). (ii) Comparing Proposition~\ref{prop:twodim} with Section 3.3 in \cite{altmann-nill-schwentner-wiercinska} we conclude that for each isomorphism class of a $2$-dimensional toric quiver variety there is a quiver $\quiver$ such that $\moduli(\quiver,\delta_{\quiver})$ belongs to the given isomorphism class (recall that $\delta_{\quiver}$ is the so-called canonical weight), in particular in dimension $2$ every projective toric quiver variety is Gorenstein Fano.. This is explained by the following two facts: (1) in dimension $2$, a complete fan is determined by the set of rays; (2) if $(\quiver,\theta)$ is tight, then $(\quiver,\delta_{\quiver})$ is tight. Now (1) and (2) imply that if $(\quiver,\theta)$ is tight and $\chi(\quiver)=2$, then $\moduli(\quiver,\theta)\cong\moduli (\quiver,\delta_{\quiver})$. (iii) The above does not hold in dimension three or higher. Consider for example the quiver-weight pairs: \[ \begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(0,0),(2,0),(0,2)} \filldraw \x circle (2pt); \draw [->] (2,0)--(0,0); \draw [->] (2,0) to [out=225,in=315] (0,0); \draw [->] (0,0)--(0,2); \draw [->] (0,0) to [out=135,in=225] (0,2); \draw [->] (2,0)--(0,2); \node [left] at (0,0) {$0$}; \node [above] at (0,2) {$3$}; \node [right] at (2,0) {$-3$}; \end{tikzpicture} \hspace{3em}{\begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(2,0),(4,0),(2,2)} \filldraw \x circle (2pt); \draw [->] (4,0)--(2,0); \draw [->] (4,0) to [out=225,in=315] (2,0); \draw [->] (2,0)--(2,2); \draw [->] (2,0) to [out=135,in=225] (2,2); \draw [->] (4,0)--(2,2); \node [left] at (2,0) {$1$}; \node [above] at (2,2) {$1$}; \node [right] at (4,0) {$-2$}; \end{tikzpicture}} \] The weight on the left is the canonical weight $\delta_{\quiver}$ for this quiver, and it is easy to check that $(\quiver,\delta_{\quiver})$ is tight and $\moduli(\quiver,\delta_{\quiver})$ is a Gorenstein Fano variety with one singular point. The weight on the right is also tight for this quiver, however it gives a smooth moduli space which can not be isomorphic to $\moduli(\quiver,\delta_{\quiver})$, consequently it also can not be Gorenstein Fano since the rays in its fan are the same as those in the fan of $\moduli(\quiver,\delta_{\quiver})$. (iv) It is also notable in dimension $2$ that each toric moduli space can be realized by precisely one quiver from $\reducedquivers_d$. This does not hold in higher dimensions. For example consider the following quivers: \[ \begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(0,0),(0,2),(0,-2),(2,0)} \filldraw \x circle (2pt); \draw [->] (0,2)--(0,0); \draw [->] (0,-2)--(0,0); \draw [->] (0,2)--(2,0); \draw [->] (0,2) to [out=0,in=90] (2,0); \draw [->] (0,-2)--(2,0); \draw [->] (0,-2) to [out=0,in=270] (2,0); \end{tikzpicture} \hspace{3em}{\begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(0,-1),(0,1),(2,-1),(2,1)} \filldraw \x circle (2pt); \draw [->] (0,-1)--(0,1); \draw [->] (2,-1)--(2,1); \draw [->] (0,-1)--(2,-1); \draw [->] (0,-1) to [out=-45,in=235] (2,-1); \draw [->] (0,1)--(2,1); \draw [->] (0,1) to [out=45,in=135] (2,1); \end{tikzpicture}} \] These quivers are both tight with their canonical weights, and give isomorpic moduli, since they are both obtained after tightening: \[ \begin{tikzpicture}[>=open triangle 45,scale=0.8] \foreach \x in {(0,0),(0,1),(0,-1),(2,1),(2,-1)} \filldraw \x circle (2pt); \draw [->] (0,1)--(0,0); \draw [->] (0,-1)--(0,0); \draw [->] (0,1)--(2,1); \draw [->] (0,1) to [out=45,in=135] (2,1); \draw [->] (0,-1)--(2,-1); \draw [->] (0,-1) to [out=315,in=235] (2,-1); \draw [->] (2,-1)--(2,1); \end{tikzpicture} \] } \end{remark} \section{Affine quotients} \label{sec:affine} We need a result concerning representation spaces that we discuss for general dimension vectors. Consider the following situation. Let $\subforest$ be a (not necessarily full) subquiver of $\quiver$ which is the disjoint union of trees $\subforest=\coprod_{i=1}^r\subtree^i$ (where by a {\it tree} we mean a quiver whose underlying graph is a tree). Let $\alpha$ be a dimension vector taking the same value $d_i$ on the vertices of each $\subtree^i$ $(i=1,\dots,r)$. Let $\theta\in\mz^{\quiver_0}$ be a weight such that there exist positive integers $n_a$ $(a\in \subforest_1)$ with $\theta(v)=\sum_{a\in\subforest_1:a^+=v}n_a-\sum_{a\in\subforest_1:a^-=v}n_a$. The representation space $\rep(\quiver,\alpha)$ contains the Zariski dense open subset \[U_{\subforest}:=\{R\in\rep(\quiver,\alpha)\mid \forall a\in\subforest_1:\ \det(R(a))\neq 0\}.\] Note that $U_{\subforest}$ is a principal affine open subset in $\rep(\quiver,\alpha)$ given by the non-vanishing of the relative invariant $f:R\mapsto \prod_{a\in\subforest_1}\det^{n_a}(R(a))$ of weight $\theta$, hence $U_{\subforest}$ is contained in $\rep(\quiver,\alpha)^{\theta-ss}$. Moreover, $U_{\subforest}$ is $\pi$-saturated with respect to the quotient morphism $\pi:\rep(\quiver,\alpha)^{\theta-ss}\to \moduli(\quiver,\alpha,\theta)$, hence $\pi$ maps $U_{\subforest}$ onto an open subset $\pi(U_{\subforest})\cong U_{\subforest}/\!/\! GL(\alpha)$ of $\moduli(\quiver,\alpha,\theta)$ (here for an affine $GL(\alpha)$-variety $X$ we denote by $X/\!/\! GL(\alpha)$ the affine quotient, that is, the variety with coordinate ring the ring of invariants $\coord(X)^{GL(\alpha)}$), see \cite{newstead}. Denote by $\hat\quiver$ the quiver obtained from $\quiver$ by contracting each connected component $\subtree^i$ of $\subforest$ to a single vertex $t_i$ $(i=1,\dots,r)$. So $\hat\quiver_0=\quiver_0\setminus \subforest_0\coprod \{t_1,\dots,t_r\}$ and its arrow set can be identified with $\quiver_1\setminus\subforest_1$, but if an end vertex of an arrow belongs to $\subtree^i$ in $\quiver$ then viewed as an arrow in $\hat\quiver$ the correspoding end vertex is $t_i$ (in particular, an arrow in $\quiver_1\setminus\subforest_1$ connecting two vertices of $\subtree^i$ becomes a loop at vertex $t_i$). Denote by $\hat\alpha$ the dimension vector obtained by contracting $\alpha$ accordingly, so $\hat\alpha(t_i)=d_i$ for $i=1,\dots,r$ and $\hat\alpha(v)=\alpha(v)$ for $v\in\hat\quiver_0\setminus\{t_1,\dots,t_r\}$. Sometimes we shall identify $GL(\hat\alpha)$ with the subgroup of $GL(\alpha)$ consisting of the elements $g\in GL(\alpha)$ with the property that $g(v)=g(w)$ whenever $v,w$ belong to the same component $\subtree^i$ of $\subforest$. We have a $GL(\hat\alpha)$-equivariant embedding \begin{equation}\label{eq:iota}\iota:\rep(\hat\quiver,\hat\alpha)\to \rep(\quiver,\alpha)\end{equation} defined by $\iota(x)(a)=x(a)$ for $a\in\hat\quiver_1$ and $\iota(x)(a)$ the identity matrix for $a\in\quiver_1\setminus \hat\quiver_1$. Clearly $\mathrm{Im}(\iota)\subseteq \rep(\quiver,\alpha)^{\theta-ss}$. \begin{proposition}\label{prop:slices} \begin{itemize} \item[(i)] $U_{\subforest}\cong GL(\alpha)\times_{GL(\hat\alpha)}\rep(\hat\quiver,\hat\alpha)$ as affine $GL(\alpha)$-varieties. \item[(ii)] The map $\iota$ induces an isomorphism $\bar\iota: \moduli(\hat\quiver,\hat\alpha,0)\stackrel{\cong} \longrightarrow \pi(U_{\subforest})\subseteq \moduli(\quiver,\alpha,\theta)$. \end{itemize} \end{proposition} \begin{proof} (i) Set $p:=\iota(0)\in\rep(\quiver,\alpha)$. Clearly $GL(\hat\alpha)$ is the stabilizer of $p$ in $GL(\alpha)$ acting on $\rep(\quiver,\alpha)$, hence the $GL(\alpha)$-orbit $O$ of $p$ is isomorphic to $GL(\alpha)/GL(\hat\alpha)$ via the map sending the coset $gGL(\hat\alpha)$ to $g\cdot p$. On the other hand $O$ is the subset consisting of all those points $R\in\rep(\quiver,\alpha)$ for which $\det(R(a))\neq 0$ for $a\in\subforest_1$ and $R(a)=0$ for all $a\notin \subforest_1$. This can be shown by induction on the number of arrows of $\subforest$, using the assumption that $\subforest$ is the disjoint union of trees. Recall also that the arrow set of $\hat\quiver$ is identified with a subset $\quiver_1\setminus\subforest_1$. This yields an obvious identification $U_{\subforest}=\rep(\hat\quiver,\hat\alpha)\times O$. Projection $\varphi:U_{\subforest}\to O$ onto the second component is $GL(\alpha)$-equivariant by construction. Moreover, the fibre $\varphi^{-1}(p)=\iota(\rep(\hat\quiver,\hat\alpha))\cong \rep(\hat\quiver,\hat\alpha)$ as a $\mathrm{Stab}_{GL(\alpha)}(p)=GL(\hat\alpha)$-varieties. It is well known that this implies the isomorphism $U_{\subforest} \cong GL(\alpha)\times_{GL(\hat\alpha)}\rep(\hat\quiver,\hat\alpha)$, see for example Lemma 5.17 in \cite{bongartz}. (ii) It follows from (i) that $U_{\subforest} /\!/\! GL(\alpha)\cong \rep(\hat\quiver,\hat\alpha)/\!/\! GL(\hat\alpha)=\moduli(\hat\quiver,\hat\alpha,0)$ by standard properties of associated fiber products. Furthermore, taking into account the proof of (i) we see $U_{\subforest}/\!/\! GL(\alpha)=\pi(\varphi^{-1}(p))=\pi(\iota(\rep(\hat\quiver,\hat\alpha))$ where $\pi$ is the quotient morphism \eqref{eq:quotient}. \end{proof} Let us apply Proposition~\ref{prop:slices} in the toric case. It is well known that for a lattice point $m$ in a lattice polyhedron $\polytope$ there is an affine open toric subvariety $U_m$ of $X_{\polytope}$, and $X_{\polytope}$ is covered by these affine open subsets as $m$ ranges over the set of vertices of $\polytope$ (see Section 2.3 in \cite{cox-little-schenck}). For a toric quiver variety realized as $\mathrm{Im}(\rho)$ as in \eqref{eq:rho} and Proposition~\ref{prop:realizationofmoduli}, this can be seen explicitly as follows: $U_{m_i}$ is the complement in $\mathrm{Im}(\rho)$ of the affine hyperplane $\{(x_0:\dots:x_d)\mid x_i=0\}\subseteq \mathbb{P}^d$, and for a general lattice point $m$ in the quiver polyhedron, $U_m$ is the intersection of finitely many $U_{m_i}$. A subset $S$ of $\quiver_0$ is {\it successor closed} if for any $a\in\quiver_1$ with $a^-\in S$ we have $a^+\in S$. A subquiver $\quiver'$ of $\quiver$ is {\it $\theta$-stable} if $\theta(\quiver'_0)=0$ and for any non-empty $S\subsetneq \quiver'_0$ which is successor closed in $\quiver'$ we have that $\theta(S)>0$. The {\it support} of $x\in\mz^{\quiver_1}$ is the quiver with vertex set $\support(x)_0:=\quiver_0$ and arrow set $\support(x)_1:=\{a\in\quiver_1\mid x(a)\neq 0\}$. Now $m\in \polytope(\quiver,\theta)$ is a vertex if and only if the connected components of $\support(m)$ are $\theta$-stable subtrees of $\quiver$. On the other hand for each subquiver $T$ of $\quiver$ that is the disjoint union of $\theta$-stable subtrees and satisfies $T_0=\quiver_0$ there is precisely one vertex $m\in \polytope(\quiver,\theta)$ such that $\support(m)=T$ (see for example Corollary 8 in \cite{altmann-straten}). Given a vertex $m$ of the polyhedron $\polytope(\quiver,\theta)$ denote by $(\quiver^m,\theta^m)$ the quiver and weight obtained by successively contracting the arrows in $\support(m)$. Clearly $\theta^m$ is the zero weight. The following statement can be viewed as a stronger version for the toric case of the results \cite{adriaenssens-lebruyn} on the {\it local quiver settings} of a moduli space of quiver representations: the \'etale morphisms used for general dimension vectors in \cite{adriaenssens-lebruyn} can be replaced by isomorphisms in the toric case. \begin{theorem}\label{thm:vertices} For any vertex $m$ of the quiver polyhedron $\polytope(\quiver,\theta)$ the affine open toric subvariety $U_m$ in $\moduli(\quiver,\theta)$ is isomorphic to $\moduli(\quiver^m,0)$. Moreover, $\iota:\rep(\quiver^m)\to \rep(\quiver)$ defined as in \eqref{eq:iota} induces an isomorphism $\bar\iota:\moduli(\quiver^m,0)\stackrel{\cong}\longrightarrow U_m\subseteq \moduli(\quiver,\theta)$. \end{theorem} \begin{proof} This is a special case of Proposition~\ref{prop:slices} (ii). \end{proof} Conversely, any affine toric quiver variety $\moduli(\quiver',0)$ can be obtained as $U_m\subseteq \moduli(\quiver,\theta)$ for some projective toric quiver variety $\moduli(\quiver,\theta)$ and a vertex $m$ of the quiver polytope $\polytope(\quiver,\theta)$. In fact we have a more general result, which is a refinement for the toric case of Theorem 2.2 in \cite{domokos:gmj}: \begin{theorem}\label{thm:compactification} For any quiver polyhedron $\polytope(\quiver,\theta)$ with $k$ vertices there exists a bipartite quiver $\tilde\quiver$, a weight $\theta' \in\mz^{\tilde\quiver_1}$, and a set $m_1,\dots,m_k$ of vertices of the quiver polytope $\polytope(\tilde\quiver,\theta')$ such that the quasiprojective toric variety $\moduli(\quiver,\theta)$ is isomorphic to the open subvariety $\bigcup_{i=1}^kU_{m_i}$ of the projective toric quiver variety $\moduli(\tilde\quiver,\theta')$. \end{theorem} \begin{proof} Double the quiver $\quiver$ to get a bipartite quiver $\tilde\quiver$ as follows: to each $v\in\quiver_0$ there corresponds a source $v_-$ and a sink $v_+$ in $\tilde\quiver$, for each $a\in\quiver_1$ there is an arrow in $\tilde\quiver$ denoted by the same symbol $a$, such that if $a\in\quiver_1$ goes from $v$ to $w$, then $a\in\tilde\quiver_1$ goes from $v_-$ to $w_+$, and for each $v\in\quiver_0$ there is a new arrow $e_v\in\tilde\quiver_1$ from $v_-$ to $v_+$. Denote by $\tilde\theta\in\mz^{\tilde\quiver_0}$ the weight $\tilde\theta(v_-)=0$, $\tilde\theta(v_+)=\theta(v)$, and set $\kappa\in\mz^{\tilde\quiver_0}$ with $\kappa(v_-)=-1$ and $\kappa(v_+)=1$ for all $v\in\quiver_0$. Suppose that $\subforest$ is a $\theta$-stable subtree in $\quiver$. Denote by $\tilde\subforest$ the subquiver of $\tilde\quiver$ consisting of the arrows with the same label as the arrows of $\subforest$, in addition to the arrows $e_v$ for each $v\in \subforest_0$. It is clear that $\tilde\subforest$ is a subtree of $\tilde\quiver$. We claim that $\tilde\subforest$ is $(\tilde\theta+d\kappa)$-stable for sufficiently large $d$. Obviously $(\tilde\theta+d\kappa)(\tilde\subforest_0)=0$. Let $\tilde S$ be a proper successor closed subset of $\tilde\subforest_0$ in $\tilde\quiver$. Denote by $S\subset \subforest_0$ consisting of $v\in\subforest_0$ with $v_+\in\tilde S$ (note that $v_-\in S$ implies $v_+\in S$, since $e_v\in \tilde\subforest$). We have the equality $(\tilde\theta+d\kappa)(\tilde S)=\theta(S)+\sum_{v^+\in\tilde S,v_-\notin \tilde S}(\theta(v)+d)$. If the second summand is the empty sum (i.e. $v_+\in\tilde S$ implies $v_-\in\tilde S$), then $S$ is successor closed, hence $\theta(S)>0$ by assumption. Otherwise the sum is positive for sufficiently large $d$. This proves the claim. It follows that if $d$ is sufficiently large, then for any vertex $m$ of $\polytope(\quiver,\theta)$, setting $\subforest:=\support(m)$, there exists a vertex $\tilde m$ of $\polytope(\tilde\quiver,\tilde\theta+d\kappa)$ with $\support(\tilde m)=\tilde\subforest$. Denote by $\mu:\rep(\quiver)\to\rep(\tilde\quiver)$ the map defined by $\mu(x)(e_v)=1$ for each $v\in\quiver_0$, and $\mu(x)(a)=x(a)$ for all $a\in\tilde\quiver_1$. This is equivariant, where we identify $(\mc^\times)^{\quiver_0}$ with the stabilizer of $\mu(0)$ in $(\mc^{\times})^{\tilde\quiver_0}$. The above considerations show that $\mu(\rep(\quiver)^{\theta-ss})\subseteq \rep(\quiver)^{(\tilde\theta+d\kappa)-ss}$, whence $\mu$ induces a morphism $\bar\mu:\moduli(\quiver,\theta)\to\moduli(\tilde\quiver,\tilde\theta+d\kappa)$. Restrict $\bar\mu$ to the affine open subset $U_m\subseteq\moduli(\quiver,\theta)$, and compose $\bar\mu\vert_{U_m}$ with the isomorphism $\bar\iota:\moduli(\quiver^m,0)\to U_m\subseteq \moduli(\quiver,\theta)$ from Theorem~\ref{thm:vertices}. By construction we see that $\bar\mu\vert_{U^m}\circ\bar\iota$ is the isomorphism $\moduli(\quiver^m,0)\to U_{\tilde m}$ of Theorem~\ref{thm:vertices}. It follows that $\bar\mu\vert_{U_m}:U_m\to U_{\tilde m}$ is an isomorphism. As $m$ ranges over the vertices of $\polytope(\quiver,\theta)$, these isomorphisms glue together to the isomorphism $\bar\mu:\moduli(\quiver,\theta)\to \bigcup_{\tilde m}U_{\tilde m}\subseteq \moduli(\tilde\quiver,\tilde\theta)$. \end{proof} We note that similarly to Theorem 2.2 in \cite{domokos:gmj}, it is possible to embed $\moduli(\quiver,\theta)$ as an open subvariety into a projective variety $\moduli(\tilde\quiver,\theta')$, such that for any vertex $m'$ of $\polytope(\tilde\quiver,\theta')$ the affine open subvariety $U_{m'}\subseteq \moduli(\tilde\quiver,\theta')$ is isomorphic to $U_m\subseteq \moduli(\quiver,\theta)$ for some vertex $m$ of $\polytope(\quiver,\theta)$ (but of course typically $\polytope(\tilde\quiver,\theta')$ has more vertices than $\polytope(\quiver,\theta)$). In particular, a smooth variety $\moduli(\quiver,\theta)$ can be embedded into a smooth projective toric quiver variety $\polytope(\tilde\quiver,\theta')$, where $\tilde\quiver$ is bipartite. \section{Classifying affine toric quiver varieties}\label{sec:affine-class} In this section we deal with the zero weight. It is well known and easy to see (say by Remark~\ref{remark:combinatorial-tightness}) that $\quiver$ is $0$-stable if and only if $\quiver$ is {\it strongly connected}, that is, for any ordered pair $v,w\in\quiver_0$ there is an oriented path in $\quiver$ from $v$ to $w$. \begin{proposition}\label{prop:valencyfour} Let $\quiver$ be a prime quiver with $\chi(\quiver)\ge 2$, such that $(\quiver,0)$ is tight. \begin{itemize} \item[(i)] For any $v\in \quiver_0$ we have $\|\{a\in\quiver_1\mid a^-=v\}\|\ge 2$ and $\|\{a\in\quiver_1\mid a^+=v\}\|\ge 2$. \item[(ii)] We have $\|\quiver_0\|\le \chi(\quiver)-1$ (and consequently $\|\quiver_1\|=\|\quiver_0\|+\chi(\quiver)-1\le 2(\chi(\quiver)-1)$. \end{itemize} \end{proposition} \begin{proof} (i) Suppose $v\in\quiver_0$ and $a\in\quiver_1$ is the only arrow with $a^-=v$. Then the equations \eqref{eq:flow} imply that for any $x\in\polytope(\quiver,0)$ we have $x(a)=\sum_{b^+=v}x(b)$, so by Lemma~\ref{lemma:contractable} the arrow $a$ is contractable. The case when $a$ is the only arrow with $a^+=v$ is similar. (ii) By (i) the valency of any vertex is at least $4$, hence similar considerations as in the proof of Proposition~\ref{prop:boundonskeletons} yield the desired bound on $\|\quiver_0\|$. \end{proof} Denote by $\reducedquivers''_d$ the set of prime quivers $\quiver$ with $\chi(\quiver)=d$ and $(\quiver,0)$ tight. Then $\reducedquivers''_1$ consists of the one-vertex quiver with a single loop, $\reducedquivers''_2$ is empty, $\reducedquivers''_3$ consists of the quiver with $2$ vertices and $2-2$ arrows in both directons. $\reducedquivers''_4$ consists of three quivers: \begin{tikzpicture}[>=open triangle 45] \foreach \x in {(0,0),(2,0)} \filldraw \x circle (2pt); \draw [->] (0,0) to [out=90,in=90] (2,0); \draw [->] (0,0) to [out=45,in=135] (2,0); \draw [->] (0,0) to (2,0); \draw [<-] (0,0) to [out=270,in=270] (2,0); \draw [<-] (0,0) to [out=315,in=225] (2,0); \end{tikzpicture} \qquad\qquad \begin{tikzpicture}[>=open triangle 45] \foreach \x in {(0,0),(1,1.5),(2,0)} \filldraw \x circle (2pt); \draw[->] (0,0) to (1,1.5); \draw[->] (1,1.5) to (2,0); \draw[->] (2,0) to (0,0); \draw[->] (0,0) to [out=90,in=180] (1,1.5); \draw[->] (1,1.5) to [out=0,in=90] (2,0); \draw [<-] (0,0) to [out=315,in=225] (2,0); \end{tikzpicture} \qquad\qquad \begin{tikzpicture}[>=open triangle 45] \foreach \x in {(0,0),(1,1.5),(2,0)} \filldraw \x circle (2pt); \draw[<-] (0,0) to (1,1.5); \draw[<-] (1,1.5) to (2,0); \draw[<-] (2,0) to (0,0); \draw[->] (0,0) to [out=90,in=180] (1,1.5); \draw[->] (1,1.5) to [out=0,in=90] (2,0); \draw [<-] (0,0) to [out=315,in=225] (2,0); \end{tikzpicture} \begin{example}\label{example:affinedegree} {\rm Consider the quiver $\quiver$ with $d$ vertices and $2d$ arrows $a_1,\dots,a_d,b_1,\dots,b_d$, where $a_1\dots a_d$ is a primitive cycle and $b_i$ is the obtained by reversing $a_i$ for $i=1,\dots,d$. Then $\chi(\quiver)=d+1$, and after the removal of any of the arrows of $\quiver$ we are left with a strongly connected quiver. So $(\quiver,0)$ is tight, showing that the bound in Proposition~\ref{prop:valencyfour} (ii) is sharp. The coordinate ring $\coord(\moduli(\quiver,0))$ is the subalgebra of $\coord(\rep(\quiver))$ generated by $\{x(a_i)x(b_i),x(a_1)\cdots x(a_d), x(b_1)\cdots x(b_d)\mid i=1,\dots,d\}$, so it is the factor ring of the $(d+2)$-variable polynomial ring $\mc[t_1,\dots,t_{d+2}]$ modulo the ideal generated by $t_1\cdots t_d-t_{d+1}t_{d+2}$. } \end{example} \section{Presentations of semigroup algebras}\label{sec:semigroupalgebras} Let $\quiver$ be a quiver with no oriented cycles, $\theta\in\mz^{\quiver_0}$ a weight. For $a\in\quiver_1$ denote by $x(a):R\mapsto R(a)$ the corresponding coordinate function on $\rep(\quiver)$, and for a lattice point $m\in\polytope(\quiver,\theta)$ set $x^m:=\prod_{a\in\quiver_1}x(a)^{m(a)}$. The homogeneous coordinate ring $\homcoord(\quiver,\theta)$ of $\moduli(\quiver,\theta)$ is the subalgebra of $\coord(\rep(\quiver))$ generated by $x^m$ where $m$ ranges over $\polytope(\quiver,\theta)\cap\mz^{\quiver_1}$. In Section~\ref{sec:equations} we shall study {\it the ideal of relations} among the generators $x^m$. This leads us to the context of presentations of polytopal semigroup algebras (cf. section 2.2 in \cite{bruns-gubeladze}), since $\homcoord(\quiver,\theta)$ is naturally identified with the semigroup algebra $\mc[\semigr(\quiver,\theta)]$, where \[\semigr(\quiver,\theta):=\coprod_{k=0}^\infty \polytope(\quiver,k\theta)\cap\mz^{\quiver_1}.\] This is a submonoid of $\mz^{\quiver_1}$. By normality of the polytope $\polytope(\quiver,\theta)$ it is the same as the submonoid of $\mn_0^{\quiver_1}$ generated by $\polytope(\quiver,\theta)\cap\mz^{\quiver_1}$. First we formulate a statement (Lemma~\ref{lemma:joo}; a version of it was introduced in \cite{joo}) in a slightly more general situation than what is needed here. Let $\semigr$ be any finitely generated commutative monoid (written additively) with non-zero generators $s_1,\dots,s_d$, and denote by $\mz[\semigr]$ the corresponding semigroup algebra over $\mz$: its elements are formal integral linear combinations of the symbols $\{x^s\mid s\in\semigr\}$, with multiplication given by $x^s\cdot x^{s'}=x^{s+s'}$. Write $R:=\mz[t_1,\dots,t_d]$ for the $d$-variable polynomial ring over the integers, and $\phi:R \to \mz[\semigr]$ the ring surjection $t_i\mapsto x^{s_i}$. Set $I:=\ker(\phi)$. It is well known and easy to see that \begin{equation}\label{eq:binomspan}I=\ker(\phi)=\mathrm{Span}_{\mz}\{t^a-t^b\mid \sum_{i=1}^da_is_i=\sum_{j=1}^db_js_j\in\semigr\}\end{equation} where for $a=(a_1,\dots,a_d)\in\mn_0^d$ we write $t^a=t_1^{a_1}\dots t_d^{a_d}$. Introduce a binary relation on the set of monomials in $R$: we write $t^a\sim t^b$ if $t^a-t^b\in R_+I$, where $R_+$ is the ideal in $R$ consisting of the polynomials with zero constant term. Obviously $\sim$ is an equivalence relation. Let $\Lambda$ be a complete set of representatives of the equivalence classes. We have $\Lambda=\coprod_{s\in \semigr}\Lambda_s$, where for $s\in\semigr$ set $\Lambda_s:=\{t^a\in\Lambda\mid \sum a_is_i=s\}$. For the $s\in \semigr$ with $\|\Lambda_s\|>1$, set $\mathcal{G}_s:=\{t^{a_1}-t^{a_i}\mid i=2,\dots,p\}$, where $t^{a_1},\dots,t^{a_p}$ is an arbitrarily chosen ordering of the elements of $\Lambda_s$. \begin{lemma}\label{lemma:mingen} Suppose that $\semigr=\coprod_{k=0}^\infty\semigr_k$ is graded (i.e. $\semigr_k\semigr_l\subseteq \semigr_{k+l}$) and $\semigr_0=\{0\}$ (i.e. the generators $s_1,\dots,s_d$ have positive degree). Then $\coprod_{s\in\semigr:\|\Lambda_s\|>1}\mathcal{G}_s$ is a minimal homogeneous generating system of the ideal $I$, where the grading on $\mz[t_1,\dots,t_d]$ is defined by setting the degree of $t_i$ to be equal to the degree of $s_i$. In particular, $I$ is minimally generated by $\sum_{s\in\semigr}(\|\Lambda_s\|-1)$ elements. \end{lemma} \begin{proof} It is easy to see that a $\mz$-module direct complement of $R_+I$ in $R$ is $\sum_{t^a\in\Lambda}\mz t^a$. Thus the statement follows by the graded Nakayama Lemma. \end{proof} Next for a cancellative commutative monoid $\semigr$ we give a more explicit description of the relation $\sim$ (a special case occurs in \cite{joo}). For some elements $s,v\in\semigr$ we say that $s$ {\it divides} $v$ and write $s\mid v$ if there exists an element $w\in\semigr$ with $v=s+w$. For any $s\in\semigr$ introduce a binary relation $\sim_s$ on the subset of $\{s_1,\dots,s_d\}$ consisting of the generators $s_i$ with $s_i\mid s$ as follows: \begin{eqnarray}\label{eq:joo} s_i\sim_s s_j \mbox{ if }i=j\mbox{ or there exist }u_1,\dots,u_k\in\{s_1,\dots,s_d\} \\ \notag \mbox{ with }u_1=s_i,\ u_k=s_j,\ u_lu_{l+1}\mid s\mbox{ for }l=1,\dots,k-1. \end{eqnarray} Obviously $\sim_s$ is an equivalence relation, and $s_i\sim_s s_j$ implies $s_i\sim_t s_j$ for any $s\mid t\in\semigr$. \begin{lemma}\label{lemma:joo} Let $\semigr$ be a cancellative commutative monoid generated by $s_1,\dots,s_d$. Take $t^a-t^b\in I$, so $s:=\sum_{i=1}^da_is_i=\sum_{j=1}^db_js_j\in\semigr$. Then the following are equivalent: \begin{itemize} \item[(i)] $t^a-t^b\in R_+I$; \item[(ii)] For some $t_i\mid t^a$ and $t_j\mid t^b$ we have $s_i\sim_s s_j$; \item[(iii)] For all $t_i\mid t^a$ and $t_j\mid t^b$ we have $s_i\sim_s s_j$. \end{itemize} \end{lemma} \begin{proof} (iii) trivially implies (ii). Moreover, if $t_i,t_j$ are two different variables occuring in $t^a$ with $\sum a_is_i=s$, then $s_is_j\mid s$, hence taking $k=2$ and $u_1=s_1$, $u_2=s_2$ in \eqref{eq:joo} we see that $s_i\sim_s s_j$. This shows that (ii) implies (iii). To show that (ii) implies (i) assume that for some $t_i\mid t^a$ and $t_j\mid t^b$ we have $s_i\sim_s s_j$. If $s_i=s_j$, then $t^a$ and $t^b$ have a common variable, say $t_1$, so $t^a=t_1t^{a'}$ and $t^b=t_1t^{b'}$ for some $a',b'\in\mn_0^d$. We have \[x^{s_1}\phi(t^{a'}-t^{b'})=\phi(t_1(t^{a'}-t^{b'}))=\phi(t^a-t^b)=0\] hence $x^{s_1}\phi(t^{a'})=x^{s_1}\phi(t^{b'})$. Since $\semigr$ is cancellative, we conclude $\phi(t^{a'})=\phi(t^{b'})$, thus $t^{a'}-t^{b'}\in I$, implying in turn that $t^a-t^b=t_1(t^{a'}-t^{b'})\in R_+I$. If $s_i\neq s_j$, then there exist $z_1,\dots,z_k\in\{t_1,\dots,t_d\}$ such that $u_l\in\semigr$ with $\phi(z_l)=x^{u_l}$ satisfy \eqref{eq:joo}. Then there exist monomials (possibly empty) $w_0,\dots,w_k$ in the variables $t_1,\dots.,t_d$ such that \[z_1w_0=t^a,\quad \phi(z_lz_{l+1}w_l)=x^s\quad (l=1,\dots,k-1), \quad z_kw_k=t^b.\] It follows that \begin{equation}\label{eq:3binomspan} t^a-t^b=z_1(w_0-z_2w_1)+\sum_{l=2}^{k-1}z_l(z_{l-1}w_{l-1}-z_{l+1}w_l)+z_k(z_{k-1}w_{k-1}-w_k).\end{equation} Note that $\phi(z_1w_0-z_1z_2w_1)=x^s-x^s=0$. Hence $z_1(w_0-z_2w_1)\in I$. Since $\semigr$ is cancellative, we conclude that $z_1(w_0-z_2w_1)\in R_+I$. Similarly all the other summands on the right hand side of \eqref{eq:3binomspan} belong to $R_+I$, hence $t^a-t^b\in R_+I$. Finally we show that (i) implies (ii). Suppose that $t^a-t^b\in R_+I$. By \eqref{eq:binomspan} we have \begin{equation}\label{eq:2binomspan} t^a-t^b=\sum_{l=1}^k t_{i_l}(t^{a_l}-t^{b_l})\ \mbox{ where }\ t^{a_l}-t^{b_l}\in I\mbox{ and }i_l\in\{1,\dots,d\}\end{equation} After a possible renumbering we may assume that \begin{equation}\label{eq:cancellations}t_{i_1}t^{a_1}=t^a, \ t_{i_l}t^{b_l}=t_{i_{l+1}}t^{a_{l+1}}\ \mbox{ for }\ l=1,\dots,k-1, \ \mbox{ and }\ t_{i_k}t^{b_k}=t^b.\end{equation} Observe that if $t_{i_l}=t_{i_{l+1}}$ for some $l\in\{1,\dots,k-1\}$, then necessarily $t^{b_l}=t^{a_{l+1}}$, hence $t_{i_l}(t^{a_l}-t^{b_l})+t_{i_{l+1}}(t^{a_{l+1}}-t^{b_{l+1}})=t_{i_l}(t^{a_l}-t^{b_{l+1}})$. Thus in \eqref{eq:2binomspan} we may replace the sum of the $l$th and $(l+1)$st terms by a single summand $t_{i_l}(t^{a_l}-t^{b_{l+1}})$. In other words, we may achieve that in \eqref{eq:2binomspan} we have $t_{i_l}\neq t_{i_{l+1}}$ for each $l=1,\dots,k-1$, in addition to \eqref{eq:cancellations}. If $k=1$, then $t^a$ and $t^b$ have a common variable and (ii) obviously holds. From now on assume that $k\ge 2$. From $t_{i_l}t^{a_l}=t_{i_{l+1}}t^{b_l}$ and the fact that $t_{i_l}$ and $t_{i_{l+1}}$ are different variables in $\mz[t_1,\dots,t_d]$ we deduce that $t^{a_l}=t_{i_{l+1}}t^{c_l}$ for some $c_l\in\mn_0^d$, implying that $x^s=\phi(t_{i_l}t^{a_l})=\phi(t_{i_l}t_{i_{l+1}}t^{c_l})=\phi(t_{i_l})\phi(t_{i_{l+1}})\phi(t^{c_l})$. Thus $u_l:=s_{i_l}$ satisfy \eqref{eq:joo} and hence $s_{i_1}\sim_s s_{i_k}$. \end{proof} \begin{corollary}\label{cor:joo} Suppose that $\semigr=\coprod_{k=0}^\infty\semigr_k$ is a finitely generated graded cancellative commutative monoid generated by $\semigr_1=\{s_1,\dots,s_d\}$. The kernel of $\phi:\mz[t_1,\dots,t_d] \to \mz[\semigr]$, $t_i\mapsto x^{s_i}$ is generated by homogeneous elements of degree at most $r$ (with respect to the standard grading on $\mz[t_a,\dots,t_d]$) if and only if for all $k>r$ and $s\in\semigr_k$, the elements in $\semigr_1$ that divide $s$ in the monoid $\semigr$ form a single equivalence class with respect to $\sim_s$. \end{corollary} \begin{proof} This is an immediate consequence of Lemma~\ref{lemma:mingen} and Lemma~\ref{lemma:joo}. \end{proof} \section{Equations of toric quiver varieties}\label{sec:equations} Corollary~\ref{cor:joo} applies for the monoid $\semigr(\quiver,\theta)$, where the grading is given by $\semigr(\quiver,\theta)_k=\polytope(\quiver,k\theta)\cap\mz^{\quiver_1}$. Recall that we may identify the complex semigroup algebra $\mc[\semigr(\quiver,\theta)]$ and the homogeneous coordinate ring $\homcoord(\quiver,\theta)$ by identifying the basis element $x^m$ in the semigroup algebra to the element of $\homcoord(\quiver,\theta)$ denoted by the same symbol $x^m$. Introduce a variable $t_m$ for each $m\in \polytope(\quiver,\theta)\cap\mz^{\quiver_1}$, take the polynomial ring \[F:=\mc[t_m\mid m\in \polytope(\quiver,\theta)\cap\mz^{\quiver_1} ]\] and consider the surjection \begin{equation}\label{eq:varphi} \varphi:F\to\homcoord(\quiver,\theta), \qquad t_m\mapsto x^m.\end{equation} The kernel $\ker(\varphi)$ is a homogeneous ideal in the polynomial ring $F$ (endowed with the standard grading) called {\it the ideal of relations among the $x^m$}, for which Corollary~\ref{cor:joo} applies. Note also that in the monoid $\semigr(\quiver,\theta)$ we have that $m\mid n$ for some $m,n$ if and only if $m\le n$, where the partial ordering $\le$ on $\mz^{\quiver_1}$ is defined by setting $m\le n$ if $m(a)\le n(a)$ for all $a\in\quiver_1$. The following statement is a special case of the main result (Theorem 2.1) of \cite{yamaguchi-ogawa-takemura}: \begin{proposition}\label{prop:theta1} Let $\quiver=K(n,n)$ be the complete bipartite quiver with $n$ sources and $n$ sinks, with a single arrow from each source to each sink. Let $\theta$ be the weight with $\theta(v)=-1$ for each source and $\theta(v)=1$ for each sink, and $\varphi:F \to \homcoord(Q,\theta)$ given in \eqref{eq:varphi}. Then the ideal $\ker(\varphi)$ is generated by elements of degree at most $3$. \end{proposition} For sake of completeness we present a proof. The argument below is based on the key idea of \cite{yamaguchi-ogawa-takemura}, but we use a different language and obtain a very short derivation of the result. For this quiver and weight generators of $\homcoord(\quiver,\theta)$ correspond to perfect matchings of the underlying graph of $K(n,n)$. Recall that a {\it perfect matching} of $K(n,n)$ is a set of arrows $\{a_1,\dots,a_n\}$ such that for each source $v$ there is a unique $i$ such that $a_i^-=v$ and for each sink $w$ there is a unique $j$ such that $a_j^+=w$. Now $\polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ in this case consists of the characteristic functions of perfect matchings of $K(n,n)$. By a \textit{near perfect matching} we mean an incomplete matching that covers all but $2$ vertices ($1$ sink and $1$ source). Abusing language we shall freely identify a (near) perfect matching and its characteristic function (an element of $\mn_0^{\quiver_1}$). First we show the following lemma: \begin{lemma} \label{lemma:nearperfect} Let $\theta$ be the weight for $\quiver=K(n,n)$ as above, and $m_1+\dots+m_k=q_1+\dots+q_k$ for some $k\ge 4$ and $m_i,q_j \in \polytope(\quiver,\theta)\cap\mz^{\quiver_1}$. Furthermore let us assume that for some $0\leq l \leq n-2$ there is a near perfect matching $p$ such that $p \leq m_1 + m_2$ and $p$ contains $l$ arrows from $q_1$. Then there is a $j \geq 3$ and $m'_1,m'_2,m'_j\in \polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ and a near perfect matching $p'$ such that $m_1 + m_2 + m_j=m'_1 + m'_2 + m'_j$, $p' \leq m'_1 + m'_2$ and $p'$ contains $l + 1$ arrows from $q_1$. \end{lemma} \begin{proof} Let $v_1, \dots, v_n$ be the sources and $w_1, \dots, w_n$ the sinks of $\quiver$, and let us assume that $p$ covers all vertices but $v_1$ and $w_1$. Let $a$ be the arrow incident to $v_1$ in $q_1$. If $a$ is contained in $m_1 + m_2$ then pick an arbitrary $j \geq 3$, otherwise take $j$ to be such that $m_j$ contains $a$. We can obtain a near perfect matching $p' < m_1 + m_2 + m_j$ that intersects $q_1$ in $l + 1$ arrows in the following way: if $a$ connects $v_1$ and $w_1$ we add $a$ to $p$ and remove one arrow from it that was not contained in $q_1$ (this is possible due to $l \leq n-2$); if $a$ connects $v_1$ and $w_i$ for some $i \neq 1$ then we add $a$ to $p$ and remove the arrow from $p$ which was incident to $w_i$ (this arrow is not contained in $q_1$). Set $r:=m_1+m_2+m_j-p'\in\mn_0^{\quiver_1}$, and denote by $S$ the subquiver of $K(n,n)$ with $S_0=\quiver_0$ and $S_1=\{c\in\quiver_1\mid r(c)\ne 0\}$. We have $S_0=S_0^-\coprod S_0^+$ where $S_0^-$ denotes the set of sources and $S_0^+$ denotes the set of sinks. For a vertex $v\in S_0$ set $\deg_r(v):= \sum_{c\in S_1} \|r(c)\|$. We have that $\deg_r(v)=3$ for exactly one source and for exactly one sink, and $\deg_r(v)=2$ for all the remaining vertices of $S$. Now let $A$ be an arbitrary subset of $S_0^-$, and denote by $B$ the subset of $S_0^+$ consisting of the sinks that are connected by an arrow in $S$ to a vertex in $A$. We have the inequality $\sum_{v\in A}\deg_r(v)\le \sum_{w\in B} \deg_r(w)$. Since on both sides of this inequality the summands are $2$ or $3$, and $3$ can occur at most once on each side, we conclude that $\|B\|\ge \|A\|$. Applying the K\"onig-Hall Theorem (cf. Theorem 16.7 in \cite{schrijver}) to $S$ we conclude that it contains a perfect matching. Denote the characteristic vector of this perfect matching by $m'_j$. Take perfect matchings $m'_1$ and $m'_2$ of $S$ with $m_1 + m_2 + m_j - m'_j=m_1'+m_2'$ (note that $m_1',m_2'$ exist by normality of the polytope $\polytope(\quiver,\theta)$, which in this case can be seen as an imediate consequence of the K\"onig-Hall Theorem). By construction we have $m_1 + m_2 + m_j=m'_1 + m'_2 + m'_j$, $p' \leq m'_1 + m'_2$, and $p'$ has $l+1$ common arrows with $q_1$. \end{proof} \begin{proofofprop}~\ref{prop:theta1} By Corollary~\ref{cor:joo} it is sufficient to show that if $s=m_1+\cdots+m_k=q_1+\cdots+q_k$ where $m_i,q_j\in \polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ and $k\ge 4$, then the $m_i,q_j$ all belong to the same equivalence class with respect to $\sim_s$. Note that since $k\ge 4$, the elements $m_1',m_2',m_j'$ from the statement of Lemma~\ref{lemma:nearperfect} belong to the same equivalence class with respect to $\sim_s$ as $m_1,\dots,m_k$. Hence repeatedly applying Lemma~\ref{lemma:nearperfect} we may assume that there is a near perfect matching $p\le m_1+m_2$ such that $p$ and $q_1$ have $n - 1$ common arrows. The only arrow of $q_1$ not belonging to $p$ belongs to some $m_j$, hence after a possible renumbering of $m_3,\dots,m_k$ we may assume that $q_1\le m_1+m_2+m_3$. It follows that $q_1\sim_s m_4$, implying in turn that the $m_i,q_j$ all belong to the same quivalence class with respect to $\sim_s$. \end{proofofprop} Now we are in position to state and prove the main result of this section (this was stated in \cite{lenz} as well, but was withdrawn later, see \cite{lenz-withdrawn}): \begin{theorem}\label{thm:degreethree} Let $\quiver$ be a quiver with no oriented cycles, $\theta\in\mz^{\quiver_1}$ a weight, and $\varphi$ the $\mc$-algebra surjection given in \eqref{eq:varphi}. Then the ideal $\ker(\varphi)$ is generated by elements of degree at most $3$. \end{theorem} \begin{proof} By Proposition~\ref{prop:addanarrow} and the double quiver construction (cf. the proof of Theorem~\ref{thm:compactification}) it is sufficient to deal with the case when $\quiver$ is bipartite and $\polytope(\quiver,\theta)$ is non-empty. This implies that $\theta(v)\le 0$ for each source vertex $v$ and $\theta(w)\ge 0$ for each sink vertex $w$. Note that if $\theta(v)=0$ for some vertex $v\in\quiver_0$, then omitting $v$ and the arrows adjacent to $v$ we get a quiver $\quiver'$ such that the lattice polytope $\polytope(\quiver,\theta)$ is integral-affinely equivalent to $\polytope(\quiver',\theta\vert_{\quiver'_0})$, hence we may assume that $\theta(v)\neq 0$ for each $v\in\quiver_0$. We shall apply induction on $\sum_{v\in\quiver_0}(\|\theta(v)\|-1)$. The induction starts with the case when $\sum_{v\in\quiver_0}(\|\theta(v)\|-1)=0$, in other words, $\theta(v)=-1$ for each source $v$ and $\theta(w)=1$ for each sink $w$. This forces that the number of sources equals to the number of sinks in $\quiver$. The case when $\quiver$ is the complete bipartite quiver $K(n,n)$ having $n$ sinks and $n$ sources, and each source is connected to each sink by a single arrow is covered by Proposition~\ref{prop:theta1}. Suppose next that $\quiver$ is a subquiver of $K(n,n)$ having a relative invariant of weight $\theta$ (i.e. $K(n,n)$ has a perfect matching all of whose arrows belong to $\quiver$). The lattice polytope $\polytope(\quiver,\theta)$ can be identified with a subset of $\polytope(K(n,n),\theta)$: think of $m\in\mz^{\quiver_1}$ as $\tilde{m}\in\mz^{K(n,n)_1}$ where $\tilde{m}(a)=0$ for $a\in K(n,n)_1\setminus\quiver_1$ and $\tilde{m}(a)=m(a)$ for $a\in\quiver_1\subseteq K(n,n)_1$. The surjection $\tilde{\varphi}:\mc[t_m\mid m\in\polytope(K(n,n),\theta)]\to\homcoord(K(n,n),\theta)$ restricts to $\varphi:\mc[t_m\mid m\in\polytope(\quiver,\theta)]\to\homcoord(\quiver,\theta)$. Denote by $\pi$ the surjection of polynomial rings that sends to zero the variables $t_m$ with $m\notin \polytope(\quiver,\theta)$. Then $\pi$ maps the ideal $\ker(\tilde{\varphi})$ onto $\ker(\varphi)$, consequently generators of $\ker(\tilde{\varphi})$ are mapped onto generators of $\ker(\varphi)$. Since we know already that the first ideal is generated by elements of degree at most $3$, the same holds for $\ker(\varphi)$. The case when $\quiver$ is an arbitrary bipartite quiver with $n$ sources and $n$ sinks having possibly multiple arrows, and $\theta(v)=-1$ for each source $v$ and $\theta(w)=1$ for each sink $w$ follows from the above case by a repeated application of Proposition~\ref{prop:multiplearrow} below. Assume next that $\sum_{v\in\quiver_0}(\|\theta(v)\|-1)\ge 1$, so there exists a vertex $w\in \quiver_0$ with $\|\theta(w)\|>1$. By symmetry we may assume that $w$ is a sink, so $\theta(w)>1$. Construct a new quiver $\quiver'$ as follows: add a new vertex $w'$ to $\quiver_0$, for each arrow $b$ with $b^+=w$ add an extra arrow $b'$ with $(b')^+=w'$ and $(b')^-=b^-$, and consider the weight $\theta'$ with $\theta'(w')=1$, $\theta'(w)=\theta(w)-1$, and $\theta'(v)=\theta(v)$ for all other vertices $v$. By Corollary~\ref{cor:joo} it is sufficient to show that if \[m_1+\dots+m_k=n_1+\dots+n_k=s\in \semigr:=\semigr(\quiver,\theta)\] for some $k\ge 4$ and $m_1,\dots,m_k,n_1,\dots,n_k \in \polytope(\quiver,\theta)\cap\mz^{\quiver_1}$, then $m_i\sim_s n_j$ for some (and hence all) $i,j$. Set $\semigr':=\semigr(\quiver',\theta')$, and consider the semigroup homomorphism $\pi:\semigr'\to\semigr$ given by \[\pi(m')(a)=\begin{cases} m'(a)+m'(a')&\quad \mbox{ if } a^+=w;\\m'(a)&\quad \mbox{ if }a^+\neq w.\end{cases}\] Take an arrow $\alpha$ with $\alpha^+=w$ and $s(\alpha)>0$. After a possible renumbering we may assume that $m_1(\alpha)>0$ and $n_1(\alpha)>0$. Define $m_1'\in\mn_0^{\quiver'_1}$ as $m_1'(\alpha)=m_1(\alpha)-1$, $m_1'(\alpha')=1$, and $m_1'(a)=m_1(a)$ for all other arrows $a\in\quiver'_1$. Similarly define $n_1'\in\mn_0^{\quiver'_1}$ as $n_1'(\alpha)=n_1(\alpha)-1$, $n_1'(\alpha')=1$, and $n_1'(a)=n_1(a)$ for all other arrows $a\in\quiver'_1$. Clearly $\pi(m_1')=m_1$, $\pi(n_1')=n_1$. Now we construct $s'\in\semigr'$ with $\pi(s')=s$, $s'-m_1'\in\mn_0^{\quiver'_1}$ and $s'-n_1'\in\mn_0^{\quiver'_1}$ (thus $m_1'$ and $n_1'$ divide $s'$ in $\semigr'$). Note that $\sum_{a^+=w}s(a)=k\theta(w)$ and $\sum_{a^+=w} \max\{m_1(a),n_1(a)\}<\sum_{a^+=w}(m_1(a)+n_1(a))=2\theta(w)$ (since $m_1(\alpha)>0$ and $n_1(\alpha)>0$). The inequalities $\theta(w)\ge 2$ and $k\ge 3$ imply that $\sum_{a^+=w}(s(a)-\max\{m_1(a),n_1(a)\})\ge k$. Consequently there exist non-negative integers $\{t(a)\mid a^+=w \}$ such that $\sum_{a^+=w}t(a)=(\sum_{a^+=w}s(a))-k$, $s(a)\ge t(a)\ge \max\{m_1(a),n_1(a)\}$ for all $a\neq \alpha$ with $a^+=w$, and $s(\alpha)-1\ge t(\alpha)\ge \max\{m_1(\alpha),n_1(\alpha)\}-1$. Consider $s'\in \mz^{\quiver'_1}$ given by $s'(a')=s(a)-t(a)$ and $s'(a)=t(a)$ for $a\in \quiver_1$ with $a^+=w$ and $s'(b)=s(b)$ for all other $b\in\quiver'_1$. By construction $s'$ has the desired properties, and so there exist $m_i',n_j'\in\polytope(\quiver',\theta')$ with $s'=m_1'+\dots+m_k'=n_1'+\dots+n_k'$. Since $\sum_{v\in\quiver'_0}(\|\theta'(v)\|-1)$ is one less than $\sum_{v\in\quiver_0}(\|\theta(v)\|-1)$, by the induction hypothesis we have $m_1'\sim_{s'}n_1'$. It is clear that $a\sim_t b$ implies $\pi(a)\sim_{\pi(t)} \pi(b)$, so we deduce $m_1\sim_s n_1$. As we pointed out before, this shows by Corollary~\ref{cor:joo} that $\ker(\varphi)$ is generated by elements of degree at most $3$. \end{proof} The above proof refered to a general recipe to derive a minimal generating system of $\ker(\varphi)$ from a minimal generating system for the quiver obtained by collapsing multiple arrows to a single arrow. Let us consider the following situation: let $\quiver$ be a quiver with no oriented cycles, $\alpha_1,\alpha_2\in\quiver_1$ with $\alpha_1^-=\alpha_2^-$ and $\alpha_1^+=\alpha_2^+$. Denote by $\quiver'$ the quiver obtained from $\quiver$ by collapsing the $\alpha_i$ to a single arrow $\alpha$. Take a weight $\theta\in\mz^{\quiver_0}=\mz^{\quiver'_0}$. The map $\pi:\polytope(\quiver,\theta)\to\polytope(\quiver',\theta)$ mapping $m\mapsto m'$ with $m'(\alpha)=m(\alpha_1)+m(\alpha_2)$ and $m'(\beta)=m(\beta)$ for all $\beta\in\quiver'_1\setminus\{\alpha\}=\quiver_1\setminus\{\alpha_1,\alpha_2\}$ induces a surjection from the monoid $\semigr:=\semigr(\quiver,\theta)$ onto the monoid $\semigr':=\semigr(\quiver',\theta')$. This extends to a surjection of semigroup algebras $\pi:\mc[\semigr]\to\mc[\semigr']$, which are identified with $\homcoord(\quiver,\theta)$ and $\homcoord(\quiver',\theta)$, respectively. Keep the notation $\pi$ for the induced $\mc$-algebra surjection $\homcoord(\quiver,\theta)\to \homcoord(\quiver',\theta)$. We have the commutative diagram of $\mc$-algebra surjections \[\begin{array}{ccc} F=\mc[t_m\mid m\in \polytope(\quiver,\theta)\cap\mz^{\quiver_1}] & \stackrel{\varphi }\longrightarrow & \homcoord(\quiver,\theta) \\ \downarrow{{\scriptstyle{\pi}}} & &\downarrow{{\scriptstyle{\pi}}} \\ F'=\mc[t_{m'}\mid m'\in \polytope(\quiver',\theta)\cap\mz^{\quiver'_1}] & \stackrel{\varphi'}\longrightarrow & \homcoord(\quiver',\theta) \end{array}\] where the left vertical map (denoted also by $\pi$) sends the variable $t_m$ to $t_{\pi(m)}$. For any monomial $u \in F'$ and any $s\in \semigr$ with $\pi(x^s)=\varphi'(u)\in\semigr'$ we choose a monomial $\psi_s(u)\in F$ such that $\pi(\psi_s(u))=u$ and $\varphi(\psi_s(u))=x^s$. This is clearly possible: let $u=t_{m_1}\dots t_{m_r}$, then we take for $\psi_s(u)$ an element $t_{n_1}\dots t_{n_r}$ where $\pi(n_j)=m_j$, such that $(n_1+\dots+n_r)(\alpha_1)=s(\alpha_1)$. Denote by $\varepsilon_i\in\mn_0^{\quiver_1}$ the characteristic function of $\alpha_i\in\quiver_1$ $(i=1,2)$. \begin{proposition}\label{prop:multiplearrow} Let $u_{\lambda}-v_{\lambda}$ $(\lambda\in\Lambda)$ be a set of binomial relations generating the ideal $\ker(\varphi')$. Then $\ker(\varphi)$ is generated by $\mathcal{G}_1\bigcup\mathcal{G}_2$, where \begin{eqnarray}\mathcal{G}_1&:=& \{\psi_s(u_{\lambda})-\psi_s(v_{\lambda})\mid \lambda\in\Lambda, \pi(s)=\varphi'(u_{\lambda})\} \\ \mathcal{G}_2&:=& \{t_mt_n-t_{m+\varepsilon_2-\varepsilon_1}t_{n+\varepsilon_1-\varepsilon_2}\mid m,n\in\polytope(\quiver,\theta)\cap\mz^{\quiver_1}, m(\alpha_1)>0, n(\alpha_2)>0\}. \end{eqnarray} \end{proposition} \begin{proof} Clearly $\mathcal{G}_1$ and $\mathcal{G}_2$ are contained in $\ker(\varphi)$. Denote by $I$ the ideal generated by them in $F$, so $I\subseteq \ker(\varphi)$. In order to show the reverse inclusion, take any binomial relation $u-v\in\ker(\varphi)$, then $\varphi(u)=\varphi(v)=x^s$ for some $s\in\semigr$. It follows that $\pi(u)-\pi(v)\in\ker(\varphi')$, whence there exist monomials $w_i$ such that $\pi(u)-\pi(v)=\sum_{i=1}^kw_i(u_i-v_i)$, where $u_i-v_i\in \{u_{\lambda}-v_{\lambda}, \quad v_{\lambda}-u_{\lambda}\mid \lambda\in\Lambda\}$, $w_1u_1=\pi(u)$, $w_iv_i=w_{i+1}u_{i+1}$ for $i=1,\dots,k-1$ and $w_kv_k=\pi(v)$. Moreover, for each $i$ choose a divisor $s_i\mid s$ such that $\pi(x^{s_i})=\varphi'(u_i)$ (this is clearly possible). Then $I$ contains the element $\sum_{i=1}^k\psi_{s-s_i}(w_i)(\psi_{s_i}(u_i)-\psi_{s_i}(v_i))$, whose $i$th summand we shall denote by $y_i-z_i$ for notational simplicity. Then we have that $\pi(y_1)=\pi(u)$, $\pi(z_k)=\pi(v)$, $\pi(z_i)=\pi(y_{i+1})$ for $i=1,\dots,k-1$, and $x^s=\varphi(y_i)=\varphi(z_i)$. It follows by Lemma~\ref{lemma:quadratic} below $u-y_1$, $v-z_k$, and $y_{i+1}-z_i$ for $i=1,\dots,k-1$ are all contained in the ideal $J$ generated by $\mathcal{G}_2$. Whence $u-v$ is contained in $I$. \end{proof} \begin{lemma}\label{lemma:quadratic} Suppose that for monomials $u,v\in F$ we have $\varphi(u)=\varphi(v)\in \homcoord(\quiver,\theta)$ and $\pi(u)=\pi(v)\in F'$. Then $u-v$ is contained in the ideal $J$ generated by $\mathcal{G}_2$ (with the notation of Proposition~\ref{prop:multiplearrow}). \end{lemma} \begin{proof} If $u$ and $v$ have a common variable $t$, then $u-v=t(u'-v')$, and $u',v'$ satisfy the conditions of the lemma. By induction on the degree we may assume that $u'-v'$ belongs to the ideal $J$. Take $m_1\in\polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ such that $t_{m_1}$ is a variable occurring in $u$. There exists an $m_2\in\polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ such that $t_{m_2}$ occurs in $v$, and $\pi(m_1)=\pi(m_2)$. By symmetry we may assume that $m_1(\alpha_1)\ge m_2(\alpha_1)$, and apply induction on the non-negative difference $m_1(\alpha_1)-m_2(\alpha_1)$. If $m_1(\alpha_1)-m_2(\alpha_1)=0$, then $m_1=m_2$ and we are done by the above considerations. Suppose next that $m_1(\alpha_1)-m_2(\alpha_1)>0$. By $\pi(m_1)=\pi(m_2)$ we have $m_2(\alpha_2)>0$, and the condition $\varphi(u)=\varphi(v)$ implies that there exists an $m_3\in \polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ such that $t_{m_2}t_{m_3}$ divides $v$, and $m_3(\alpha_1)>0$. Denote by $\varepsilon_i\in\mn_0^{\quiver_1}$ the characteristic function of $\alpha_i$, and set $m_2':=m_2+\varepsilon_1-\varepsilon_2$, $m_3':=m_3-\varepsilon_1+\varepsilon_2$. Clearly $m_2',m_3'\in\polytope(\quiver,\theta)\cap\mz^{\quiver_1}$ and $t_{m_2}t_{m_3}-t_{m_2'}t_{m_3'}\in J$. So modulo $J$ we may replace $v$ by $t_{m_2'}t_{m_3'}v'$ where $v=t_{m_2}t_{m_3}v'$. Clearly $0\le m_1(\alpha_1)-m_2'(\alpha_1)<m_1(\alpha_1)-m_2(\alpha_1)$, and by induction we are finished. \end{proof} In the affine case one can also introduce a grading on $\coord(\moduli(\quiver,0))$ by declaring the elements that correspond to primitve cycles of the quiver to be of degree $1$. The ideal of relations can be defined as above, but in this case it is not possible to give a degree bound independently of the dimension. This is illustrated by Example~\ref{example:affinedegree}, providing an instance where a degree $d-1$ element is needed to generate the ideal of relations of a $d$-dimensional affine toric quiver variety. However the following theorem shows that this example is the worst possible from this respect. \begin{theorem}\label{thm:affinebound} Let $\quiver$ be a quiver such that $d := \dim(\moduli(\quiver,0)) >0$. Then the ideal of relations of $\moduli(\quiver,0)$ is generated by elements of degree at most $d-1$. \end{theorem} \begin {proof} Up to dimension $2$ the only affine toric quiver varieties are the affine spaces. Suppose from now on that $d\ge 3$. Clearly it is sufficient to deal with the case when $(\quiver,0)$ is tight and $\quiver$ is prime. Suppose that a degree $k$ element is needed to generate the ideal of relations of $\moduli(\quiver,0)$. In Section 6 of \cite{joo} it is shown that this holds if and only if there is a pair of primitive cycles $c_1$, $c_2$ in $\quiver$ such that the multiset sum of their arrows can also be obtained as the multiset sum of some other $k$ primitive cycles $e_1,\dots,e_k$. Note that each $e_i$ has an arrow contained in $c_1$ but not in $c_2$, and has an arrow contained in $c_2$ but not in $c_1$. It follows that $\mathrm{length}(c_1)+\mathrm{length}(c_2)\ge 2k$, implying that $\quiver$ has at least $k$ vertices. By Proposition~\ref{prop:valencyfour} (ii) we conclude that $d-1=\chi(\quiver)-1\geq \|\quiver_0\|\ge k$. \end{proof} \section{The general case in \cite{yamaguchi-ogawa-takemura}}\label{sec:japanok} \def\osmpoly{OSM(\quiver)} In this section we give a short derivation of the main result of \cite{yamaguchi-ogawa-takemura} from the special case Proposition~\ref{prop:theta1}. To reformulate the result in our context consider a bipartite quiver $\quiver$ with at least as many sinks as sources. By a \textit{one-sided matching} of $\quiver$ we mean an arrow set which has exactly one arrow incident to each source, and at most one arrow incident to each sink. By abuse of language the characteristic vector in $\mz^{\quiver_1}$ of a one-sided matching will also be called a one-sided matching. The convex hull of the one-sided matchings in $\mz^{\quiver_1}$ is a lattice polytope in $\mr^{\quiver_1}$ which we will denote by $\osmpoly$. Clearly the lattice points of $\osmpoly$ are precisely the one-sided matchings. The normality of $\osmpoly$ is explained in section 4.2 of \cite{yamaguchi-ogawa-takemura} or it can be directly shown using the K\"onig-Hall Theorem for regular graphs and an argument similar to that in the proof below. Denote by $\semigr(\osmpoly)$ the submonoid of $\mn_0^{\quiver_1}$ generated by $\osmpoly\cap\mz^{\quiver_1}$. This is graded, the generators have degree $1$. Consider the ideal of relations among the generators $\{x^m\mid m\in\osmpoly\cap\mz^{\quiver_1}\}$ of the semigroup algebra $\mc[\semigr(\osmpoly)]$. Theorem 2.1 from \cite{yamaguchi-ogawa-takemura} can be stated as follows: \begin{theorem} The ideal of relations of $\mc[\semigr(\osmpoly)]$ is generated by binomials of degree at most 3. \end{theorem} \begin{proof} Consider a quiver $\quiver'$ that we obtain by adding enough new sources to $Q$ so that it has the same number of sources and sinks, and adding an arrow from each new source to every sink. Let $\theta$ be the weight of $Q'$ that is $-1$ on each source and $1$ on each sink. Now the natural projection $\pi : \mr^{Q'_1} \rightarrow \mr^{Q_1}$ induces a surjective map from $\polytope(\quiver',\theta)\cap\mz^{Q'_1}$ onto $\osmpoly\cap\mz^{Q_1}$ giving us a degree preserving surjection between the corresponding semigroup algebras. By Corollary~\ref{cor:joo} it is sufficient to prove that for any $k \geq 4$, any degree $k$ element $s\in \semigr(\osmpoly)$, and any $m,n \in \osmpoly \cap\mz^{\quiver_1}$ with $m,n$ dividing $s$ we have $m\sim_s n$. In order to show this we shall construct an $s' \in \polytope(\quiver',k\theta)\cap\mz^{\quiver'_1}$ and $m' , n' \in \polytope(\quiver',\theta)\cap\mz^{\quiver'_1}$ such that $m'\le s'$, $n' \leq s'$, $\pi(m') = m$, $\pi(n') = n$ and $\pi(s') = s$. By Proposition~\ref{prop:theta1} we have $m' \sim_{s'} n'$, hence the surjection $\pi$ yields $m \sim_s n$. The desired $s'$, $m'$, $n'$ can be obtained as follows: think of $s$ as the multiset of arrows from $\quiver$, where the multiplicity of an arrow $a$ is $s(a)$. Pairing off the new sources $\quiver'_0\setminus \quiver_0$ with the sinks in $\quiver$ not covered by $m$ and adding the corresponding arrows to $m$ we get a perfect matching $m'$ of $\quiver'$ with $\pi(m')=m$. Next do the same for $n$, with the extra condition that if none of $n$ and $m$ covers a sink in $\quiver$, then in $n'$ it is connected with the same new source as in $m'$. Let $t\in\mn_0^{\quiver'_1}$ be the multiset of arrows obtained from $s$ by adding once each of the arrows $\quiver'_1\setminus \quiver_1$ occuring in $m'$ or $n'$. For a vertex $v\in\quiver'_1$ set $\deg_t(v):=\sum_{v\in\{c^-,c^+\}} \|t(c)\|$. Observe that $s-m$ and $s-n$ belong to $\semigr(\osmpoly)_{k-1}$, hence $\deg_{s-m}(w)\le k-1$ and $\deg_{s-n}(w)\le k-1$ for any vertex $w$. So if $w$ is a sink not covered by $m$ or $n$, then $\deg_s(w)$ agrees with $\deg_{s-m}(w)$ or $\deg_{s-n}(w)$, thus $\deg_s(w)\le k-1$, and hence $\deg_t(w)\le k$. For the remaining sinks we have $\deg_t(w)=\deg_s(w)\le k$ as well, moreover, $\deg_t(v)=k$ for the sources $v\in\quiver_0\setminus \quiver'_0$, whereas $\deg_t(v)\le 2$ for the new sources $v\in \quiver'_0\setminus \quiver_0$. Consequently successively adding further new arrows from $\quiver'_1\setminus \quiver_1$ to $t$ we obtain $s'\ge t$ with $\deg_{s'}(v)=k$ for all $v\in\quiver'_0$. Moreover, $m'\le t\le s'$, $n'\le t\le s'$, and $\pi(s')=s$, so we are done. \end{proof} \begin{center}Acknowledgement\end{center} We thank Bernd Sturmfels for bringing \cite{yamaguchi-ogawa-takemura} to our attention.	69,848
The Wine and Spirit Trade Association (WSTA) will complete its rebirth' early this year, with Christopher Carson - non-executive chairman and former CEO of Constellation Europe - expected to take the position of chairman, alongside the Association's recently appointed chief executive Jeremy Beadles. Beadles, who formerly held a similar position at the British Retail Consortium, takes up his new position this week. Carson is expected to join him on 6 April after being elected at the AGM. He is the only candidate who has been nominated. Current interim chair Dr Barry Sutton, who was brought in late last year to perform a wholesale review' of the Association's operations, structures and governance', will step aside following Carson's arrival. This is a great nomination for the WSTA, and Carson will make a great chairman. This is the first time in some years that we will have a real trade leader in the role of chairman, and his nomination is the next step on the road to completing the reform of the WSTA. He has the ear of all the important figures across the industry and has knowledge and experience of both spirits and wine.' Carson, who was MD at BRL Hardy Europe before his appointment as head of the merged Constellation Europe group in April 2004, commented: I welcome the opportunity to give continued support to the Association and industry through these challenging times.' Interestingly, Carson was a prominent member of one of the main catalysts for change at the WSTA - the Wine Trade Action Group (WTAG) - which was set up due to dissatisfaction with the Association. In addition to the chairman, WTAG has a very strong presence on the 15-strong executive board. Just over half its members are leading lights in WTAG, including Mike Paul, David Cox and Nick Hyde. Further reforms of the WSTA are expected this year, with Sutton highlighting the role of the WSTA's various committees and subcommittees as an aspect still under review.	143,224
Gov. Jerry Brown’s vain train to Utopia to the tune of $68 billion (with a B), up and down the California’s Central Valley, is an exercise in stupidity with absolutely no redeeming value to it. In the long run, who really wants to travel 200 mph just to get to LaLa Land? Not I, for sure! Don’t even get me started on those two “diversion” water tunnels under the Delta to send more of our precious water down south to their hot tubs, pools and car washes. I do think that we need less-costly imagination and more adult supervision over in Sacramento. I put it to you. Harry M. Short Vacaville The Daily Republic does not necessarily condone the comments here, nor does it review every post. Read our full policy The MisterFebruary 05, 2014 - 7:30 am Oh, there's plenty of reason there, Harry. Underneath the land where that bullet train is going is one of the largest shale oil fields in the world. It would make perfect sense that Jerry Brown will condemn the land under eminent domain and, in compliance with the Kelo decision, will sell the land to big oil interests OR will keep this oil from reaching the market (which is exactly what Bill Clinton did with the worlds second largest clean coal reserve in Utah by making it a "national park"). So, there is plenty of reason... and none of it has to do with you getting from one place to another rather quickly.Reply \| SteveFebruary 05, 2014 - 8:42 am That's a new take on the subject.Reply \| patrickFebruary 05, 2014 - 1:42 pm good pointReply \|	228,812
Hennessy stunned in the images of Boohoo's holiday campaign. 1 This sequin and yellow fur combo is the perfect amount of extra for Hennessy. She lives and breathes fashion. 2 Hennessy loves experimenting with her style and frequently posts killer looks on Instagram. Hennessy is always switching up her hair. 3 From different lengths to bold colors, she'll try it all. Hennessy has major goals she wants to accomplish in the fashion industry. 5 She wants to be a stylist and fashion designer, and she is working towards that by studying at the Fashion Institute of Technology. "My dream customer is every celebrity that’s on the red carpet," she told BET. "Everybody at the Met Gala and on top of that, just everybody and anybody. I would dress everybody." Cardi gave Hennessy a major boost on her road to stardom. 6 Cardi was already wildly popular on Instagram when she joined Love & Hip Hop: New York in 2015, but the reality show helped her reach new levels of fame. Hennessy came along for the ride in 2016, which helped her build a platform of her own. The love goes both ways. 7 She is Cardi's #1 fan, and is always celebrating her on social media. The story behind her name is iconic. 8 Splash Her dad was drunk on Hennessy when she was born and decided to name her after the liquor. Cardi's stage name was inspired by Hennessy. 9 "My sister's name is Hennessy, right?" Cardi, whose birth name is Belcalis Almanzar, told Jimmy Fallon on the Tonight Show last year. "So, everybody used to call me Bacardi. So, I always call myself Bacardi, right? Then it was my Instagram name--like Bacardi, Bacardi B--but for some reason, my Instagram kept getting deleted; I think it was Bacardi that had something to do with it, so I just shortened it to Cardi B." She got involved in Cardi's feud with Nicki Minaj. 11 In the days after Nicki Minaj and Cardi B's massive scuffle at the Harper’s Bazaar ICONS party in September, Hennessy was noticeably silent, but it was for a good reason. When a fan asked why she wasn't clapping back she replied, "I don’t talk, I fights.” The feud escalated when Hennessy accussed Nicki Minaj of leaking Cardi's phone number, which opened her up to attacks from the rapper's fans. Hennessy also got into a war of words with Rah Ali, who Nicki said "really, really beat Cardi’s ass bad” during their New York Fashion Week altercation. Hennessy is just as unfiltered as Cardi. 12 "Some people act a certain way because they are worried about society," she told Paper Magazine. "I act a certain way just because I'm living my life. I don't worry about anybody." She did a reality competition show to help her community. 13 Splash Hennessy told Billboard that she was on the fence about joining MTV's The Challenge: Champs Vs. Stars earlier this year, but she decided to do it for charity. "I chose BronxWorks because I’m from The Bronx, and I got raised in The Bronx, and I just know the struggle and how it is growing up in The Bronx," Hennessy said. "I just did it for The Bronx. ... I wanted to help the kids and everyone there, because there’s not a lot of opportunities out there." Hennessy came out as bisexual on Instagram and is teaching people around her that love is love. 14 Her mom had a hard time accepting her sexuality, but she eventually started to see Hennessy's side of things. "I taught my family and they actually became accepting because I sat them down," she told Paper Magazine. Seeing Hennessy in love with her girlfriend changed them. "We've been believing this is the way things are and it's not--that's just rules that been put on us," she said. "But now I feel free and now I just try to spread the message as much as I possibly can. Love is love no matter what gender." She got her girlfriend by sliding into her DMs. 15 Hennessy is capitalizing on her success in a major way. 16 She is conquering fashion and TV, but she has movies on the way as well. "Everyone thinks I should be an actor," she told Paper Magazine. "Everyone really wants me to act and I'm just like, Sure I'll try anything."	158,324
Betriebswirtschaftliches Forschungszentrum für Fragen der mittelständischen Wirtschaft e. V. (BF/M) an der Universität Bayreuth Betriebswirtschaftliches Forschungszentrum für Fragen der mittelständischen Wirtschaft e. V. (BF/M) an der Universität Bayreuth Founded in 1979, the Bayreuth Research Institute for small and medium sized organizations at the University of Bayreuth (BF/M-B ... - Countries (EU): - Germany - Organisation type: - Research Center - Areas of Interest: - SMEs	303,964
Candidates line up to replace Sen. Margaret Rose Henry Bobby Cummings, former head of the Wilmington Police Department, announced that he's running for the Delaware Senate Friday. 1/5/18 Damian Giletto/The News Journal Several candidates have announced their intention to replace outgoing state Sen. Margaret Rose Henry in the September General Assembly election. Former Wilmington police Chief Bobby Cummings kicked off his campaign for Senate District 2 on Friday. He is joining City Councilman Samuel L. Guy and former Councilman Darius Brown in the race. "If the Delaware Way means trying to preserve the status quo or kick the can down the road at the expense of progress and change, I will be the voice of change," Cummings said to a group of three dozen supporters at the Rose Hill Community Center in New Castle. Henry, 73, is leaving office after more than two decades. She confirmed she didn't plan to run for re-election in September after Guy announced she was retiring and that he would seek her office. District 2 includes northeast Wilmington and the east side as well as parts of New Castle County from Edgemoor to New Castle. RELATED: Sam Guy announces Sen. Margaret Rose Henry to retire, will seek her office Before Henry, the district was represented by Herman Holloway Sr., the first African-American elected to the Delaware Senate. Henry is the first and only black female state senator and the body's only black member. Cummings, 54, retired as chief in April after Mayor Mike Purzycki expressed a desire to hire new leadership. As Mayor Dennis P. Williams' police chief, Cummings oversaw the Police Department during a time of mounting city violence that has since escalated. A Wilmington native, Cummings served in the Police Department for 32 years and said the experience would be an asset to the state Legislature. "Since I have not been a politician, I will look at issues in a new way and seek creative solutions to our state's challenges," he said in his remarks Friday. Guy, 58, has been a controversial figure on the City Council since he was sworn in a year ago. The at-large member calls himself "the watchman" and describes himself as a fighter for the downtrodden and an advocate of transparency, but his colleagues have found him abrasive. Three months into his term, eight of the 13 members voted to censure, or publicly disapprove of, Guy after a pattern of behavior they described as "abusive, berating, degrading and threatening." The measure said that Guy engaged in “baseless allegations” against his colleagues, including claims of racism. This is Guy's second time on the council. The attorney first served from 1997 to 2001 and was later chief of staff to former County Executive Tom Gordon. Guy did not respond to an interview request. MORE: Wilmington’s deadliest year: Mayor withholds 'serious concern' until plan takes hold As a District 3 councilman, Brown, 36, sponsored legislation that provided paid parental leave for city employees, required city contractors to pay workers a minimum of $10.10 an hour and served on a task force that provided a framework for lowering city health care costs. "My record as a council person speaks for itself in terms of creating opportunity in the city of Wilmington," he said. Brown served one term on the council before unsuccessfully running for city treasurer in 2016. If elected to the Senate, he said, he would focus on boosting infrastructure investments that create jobs, improving the school system and creating a "community court" system to address nuisance crimes. "It’s the one seat in the Delaware state Senate for over the last 50 years that has been held by an African-American and is the only seat held by an African-American," he said. "I look forward to representing the issues and concerns of the African-American community in the 2nd senatorial district.".	166,838
The cold breeze is ready to stir you in the magic of winters but are you ready to level up your style game this season? If you thought that styling those mundane jackets is very demanding, then you are wrong. Revamp your wardrobe this season with Wildcraft winter jackets to stay super-warm while earning the styling quotient. These versatile jackets are a great pick this season. So, if you are ready to create some serious fashion goals, then you must look at these Wildcraft winter wear for women to buy online. 8 Best Wildcraft Winter Jackets for Women you can Buy Online 1. Quilted Wildcraft Jacket for Women Quilted jackets are often claimed to be monotonous when it comes to styling. This quilted black Wildcraft jacket offers a sporty look and is great for every woman. Made from polyester, this Wildcraft black lightweight padded jacket looks super appealing with its front zipper, Velcro cuffs and a draw cord hemline. 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Style With: Add some mascara to your eyelashes along with nude pink lip colour. Price: Rs.1491 Onwards 8. Rain Wildcraft Jacket for Women Are you a travel freak? If yes, then you need to add this Wildcraft rain jacket to your wardrobe today. It not only protects you from those chilly winds but also saves you from those ‘uncalled for’ rain showers. The bright orange colour looks ultra-stylish and is perfect for hiking expeditions. This Wildcraft hiking jacket’s light weight feature makes it super-comfortable to wear. It also comes with a storm hood neckline, concealed zipped fastener and functional zipped pockets on the sides. Team it with a pair of simple Bermudas and sneakers to look sporty as ever. Accessorize With: These basic travelling jackets can be accessorized with beanies, backpacks, gloves and a sports watch. Style With: Simply pull up your hair in a bun and just add a dash of lip balm. Price: Rs.1497 Onwards **DISCLAIMER: The mentioned price of the respective product is variable and subject to change.	36,884
North Iceland is home to the beautiful lake region of Myvatn. With an area of 36.5 square kilometers, it is Iceland’s fourth-largest lake, although its size is just one of the qualities that attracts guests throughout the year. It is also home to some incredible geological formations, with a great flora and fauna and it is surrounded by many incredible sites. Travel Hotel in Reykjavik: perfect for the Golden Circle!09/09/2019 Just outside Reykjavik, ready to go on a road adventure in Iceland, Hotel Laxnes welcomes the traveler with typical environments in a truly strategic location. Rent a car in Iceland: all the information to drive in Iceland03/09/2019 3300 km of Icelandic roads. Narrow roads, paved roads, dirt roads, sun, rain, fog, mountain roads, gravel, paved roads, unpaved roads, men at work… At least 30 sheep that crossed the road, thousands at the side of them, numerous birds at risk of being run over. Visiting Santorini with children01/08/2019 The story of Santorini tells about sea and fire, about submerged cities and vanished civilizations, about sunsets and volcanoes. If you plan to visit Greek islands, you will surely have browsed a guide and come across a view of Santorini, with the blue domes, which act as a union between the sky and the sea passing through the white of the houses. Between slides and castles in Aurina Valley10/07/2019 The. London with kids: Covent Garden24/05/2019 Covent Garden is a district in central London that attracts around 30 million tourists a year. It stands out with its wide range of shops, cafes, restaurants, but not only: from Covent Garden you can reach the main London attractions and many small hidden corners.. What to see in Gozo: Malta with children15/04/2019 It’s a perfect destination for Easter and for spring holidays, with a mild climate and easily reachable from many Italian airports with low-cost flights. If Malta surprises you, a long weekend with children in Gozo will give you a unique experience. A museum where you can learn sustainability: Pilke06/02/2019 What do you know about the great forests of the North? Pilke in Rovaniemi, nestled in Finnish Lapland just a stone’s throw from the Arctic Circle, is the right place! Valentine’s Day in Paris in the most romantic location01/02/2019 Paris is one of the most romantic cities in the world: what better way to celebrate Valentine’s Day than to enjoy a weekend in the Ville Lumiere? Snow Weekend: 6 Low cost destinations for mountain travel11/01/2019 If winter for you means snow-covered slopes and you can not wait to wear your boots, it is time to organize a stay in the mountains on the best tracks in Europe. To give you a few days of sport and relax at high altitude, Volagratis.com has selected the best destinations where to spend three days on the snow at convenient prices. Street Art in Valencia with Vingino29/11/2018 Among the many monuments, museums, venues of Valencia, on our last trip to this beautiful Spanish city we have looked for a new itinerary to be discovered by Sissi and her friends. This time we explored Valencia’s street art. Sleep in an igloo in Finnish Lapland: once in your life emotion15/11/2018? How to choose a family hotel in Rome05/11/2018 Rome is the Eternal City, with a thousand monuments, always filled with culture and Italian beauty… but it is so hard to find a good family hotel in Rome!	220,134
GSK and Lyell Immunopharma join forces to develop the next generation of cancer cell therapies Collaboration will combine Lyell’s technologies with GSK’s pipeline of cell therapies and manufacturing capability Issued: London and San Francisco GlaxoSmithKline plc today announced a five-year collaboration with Lyell Immunopharma, a San Francisco biotechnology company,. To date, two cell therapies have been approved for blood-borne cancers[1], but engineered T cells have not yet delivered strong clinical activity in common solid tumours. Improving the “fitness” of T cells and delaying the onset of T cell exhaustion could help engineered T cell therapies become more effective. Combining GSK’s strong cell and gene therapy programmes with Lyell’s technologies may allow the joint research team to maximise the activity and specificity of cell therapies in solid tumour cancers, where there is a high unmet medical need. Dr. Hal Barron, Chief Scientific Officer and President, R&D, GSK said: “We are witnessing significant scientific innovation in cell and gene therapies, transforming the treatment of some blood-borne cancers, but patients with solid tumours are in need of equally effective treatments. Applying Lyell’s novel approach to counter T cell exhaustion and working with world class scientists, such as Rick Klausner and his impressive team, increases our probability of delivering the next generation of cancer cell therapies for patients with solid tumours.” Lyell is exploring several approaches to improving T cell function and increasing T cell “fitness” to enhance initial response rates in solid tumour cancers and to prevent relapses due to loss of T cell functionality. As Lyell addresses inhibition of T cells by the tumour in a fundamental way, there is an opportunity that these technologies can be used as a platform for multiple new cell and gene therapies that can be applied across a broad range of rare and prevalent solid cancers. Dr. Rick Klausner, founder and CEO, Lyell Immunopharma said: “Our approach is to tackle three of the most significant barriers to T cell efficacy in solid tumours. We are redefining the ways we prepare patient cells to be made into therapies, modulating cells’ functionality so that they maintain activity in the tumour microenvironment, and establishing methods of control to achieve specificity and safety for solid tumour-directed cell therapies.” Lyell has a scientific management team with a long history in the field of immune cell therapy. Rick Klausner is the former head of the National Cancer Institute (NCI) and co-founder of Juno Therapeutics, and whose lab discovered the molecular engine behind T cell receptor and CAR signalling; Stan Riddell, co-founder and the Head of R&D, co-founder of Juno whose pioneering work over three decades at Fred Hutchinson Cancer Research Center has helped to define the parameters of successful adoptive cell therapy; Nick Restifo, EVP of Science, whose research over 25 years at the NCI defined the properties of the T cells capable of therapeutic efficacy in cancer; and Margo Roberts, CSO, whose work in adoptive T cell therapy includes serving as CSO of Yescarta® maker Kite Pharma, administering the first CAR T cell into patients in 1993, and demonstrating the role of co-stimulation in T cells for effective CARs. Next generation engineering that leverages Lyell technologies could further enhance the benefit/risk profile of GSK’s lead programme. Lyell co-founders also include Crystal Mackall, M.D., who has pioneered work on T cell exhaustion, and David Baker, Ph.D., Director of the University of Washington Institute for Protein Design, whose novel approaches to protein engineering provide technologies to enable enhanced precision, control and safety in cell-based therapies. About GSK’s cell therapy pipeline GSK has built a pipeline of cell therapy programmes comprising multiple cell platforms and targets including CAR T programmes and an additional TCR T through our existing collaboration with Adaptimmune. While CAR Ts detect cell surface targets, T cell receptor therapies (TCR Ts) detect targets both inside and outside the cancer cells which could expand the number of cancers that could benefit from cell therapies. GSK3377794 is a NY-ESO-1-directed genetically modified autologous T cell immunotherapy and has been granted PRIME designation by the European Medicines Agency and Breakthrough Therapy Designation by the US Food and Drug Administration based on promising activity in synovial sarcoma. NY-ESO-1 is a cancer target expressed in a wide variety of human cancers. GSK3377794 has been administered to over 100 patients across multiple cancer types and has shown encouraging activity in synovial sarcoma with a manageable safety profile. Based on this data, GSK is initiating the IGNYTE-ESO pivotal study of GSK3377794 for patients with relapsed/refractory synovial sarcoma, and concurrently exploring the efficacy of GSK3377794 in other cancer types that also express the NY-ESO-1 target, including non-small cell lung cancer, multiple myeloma, and myxoid/round cell liposarcoma. Lyell Lyell is addressing the unsolved problem of creating reliable, curative adoptive cell therapy for solid tumours. tumour microenvironment, with special attention to preventing, controlling and reversing the differentiation of T cells into dysfunctional states within solid tumours Controlling the specificity and safety of solid tumour. References [1] Yescarta® (axicabtagene ciloleucel, Kite Pharma/Gilead) and Kymriah® (tisagenlecleucel, Novartis) have been approved by US Food Drug Administration and EU authorities as cell-based gene therapies also known as chimeric antigen receptor therapies (CAR Ts) for lymphoma and leukemia.	88,130
2200hp Turbo Cobra Mustang Killing it in Texas! Those of you who follow the drag racing events up in the Chicago area are well familiarized with the Boost 12 Cobra Mustang also seen quite often at Discovery Channel’s Street Outlaws, but Joe and his team showed up down in Texas with a new set up and murdered the competition at Redemption No Prep 4.0. Powered by a 380 cubic inch small block Ford Racing V8 motor turbocharged with a 98mm Precision unit to kick in nearly 2200 horsepower the Boost 12 Cobra Mustang faced some of the toughest cars from the No Prep drag racing world and took home $12,500 for winning the Small Tires Class. See the 1320video from Redemption No Prep 4.0 in Texas. Watch, Enjoy & Share!	349,361
\begin{document} \begin{abstract} Signatures of quadratic forms have been generalized to hermitian forms over algebras with involution. In the literature this is done via Morita theory, which causes sign ambiguities in certain cases. In this paper, a hermitian version of the Knebusch Trace Formula is established and used as a main tool to resolve these ambiguities. \end{abstract} \maketitle \section{Introduction} In this paper we study signatures of hermitian forms over central simple algebras with involution of any kind, defined over formally real fields. These generalize the classical signatures of quadratic forms. Following \cite{BP2} we do this via extension to real closures and Morita equivalence. This leads to the notion of $M$-signature of hermitian forms in Section~\ref{sec:general}. We study its properties, make a detailed analysis of the impact of choosing different real closures and different Morita equivalences and show in particular that sign changes can occur. This motivates the search for a more intrinsic notion of signature, where such sign changes do not occur. In Section~\ref{sec:H} we define such a signature, the $H$-signature, which only depends on the choice of a tuple of hermitian forms, mimicking the fact that in quadratic form theory the form $\<1\>$ always has positive signature. The $H$-signature generalizes the definition of signature in \cite{BP2} and is in particular well-defined when the involution becomes hyperbolic after scalar extension to a real closure of the base field, addressing an issue with the definition proposed in \cite{BP2}. Our main tool is a generalization of the Knebusch Trace Formula to $M$-signatures of hermitian forms, which we establish in Section~\ref{ktf-M}. In Section~\ref{sec:cont} we show that the total $H$-signature of a hermitian form is a continuous map and in Section~\ref{sec:ktf-H} we prove the Knebusch Trace Formula for $H$-signatures. \section{Preliminaries} \subsection{Algebras with Involution} The general reference for this section is \cite[Chapter~I]{BOI}. Let $F$ be a field of characteristic different from $2$. An \emph{$F$-algebra with involution} is a pair $(A,\s)$ consisting of a finite-dimen\-sional $F$-algebra $A$ with centre $Z(A)$ and an $F$-linear map $\s:A\to A$ such that $\s(xy)=\s(y)\s(x)$ for all $x,y\in A$ and $\s^2=\id_A$. The involution $\s$ is either of \emph{the first kind} or of \emph{the second kind}. In the first case $A$ is simple, $Z(A)=F$ and $\s\|_F=\id_F$. In the second case there are two possibilities: either $A$ is simple, $Z(A)=K=F(\sqrt{d})$ for some $d\in F^\x$ and $\s\|_{K}$ is the nontrivial $F$-automorphism of $K$, or $(A,\s)\simeq (B\x B^\op, \wwh)$ with $B$ a simple $F$-algebra, $Z(A)\simeq F\x F$ a double-field and $\wwh$ the exchange involution, defined by $\wh{(x,y^\op)}=(y,x^\op)$ for all $x,y\in B$. We call $(A,\s)$ \emph{degenerate} if it is isomorphic to $(B\x B^\op, \wwh)$. Consider the $F$-subspaces \[\Sym(A,\s)=\{a\in A\mid \s(a)=a\} \text{ and } \Skew(A,\s)=\{a\in A\mid \s(a)=-a\}\] of $A$. Then $A=\Sym(A,\s) \oplus \Skew(A,\s)$. Assume that $\s$ is of the first kind. Then $\dim_F(A)=m^2$ for some positive integer $m$. Furthermore, $\s$ is either \emph{orthogonal} (or, \emph{of type $+1$}) if $\dim_F\Sym(A,\s)=m(m+1)/2$, or \emph{symplectic} (or, \emph{of type $-1$}) if $\dim_F\Sym(A,\s)=m(m-1)/2$. If $\s$ is of the second kind, then $\dim_F(A) = 2m^2$ for some positive integer $m$ and $\dim_F\Sym(A,\s)=\dim_F\Skew(A,\s)=m^2$. Involutions of the second kind are also called \emph{unitary}. Let $\s$ and $\tau$ be two involutions on $A$ that have the same restriction to $Z(A)$. By the Skolem-Noether theorem they differ by an inner automorphism: \[\tau = \Int(u)\circ \s\] for some $u\in A^\x$, uniquely determined up to a factor in $F^\x$, such that $\s(u)=u$ if $\s$ and $\tau$ are both orthogonal, both symplectic or both unitary and $\s(u)=-u$ if one of $\s$, $\tau$ is orthogonal and the other symplectic. Here $\Int(u)(x):=uxu^{-1}$ for $x\in A$. \subsection{$\ve$-Hermitian Spaces and Forms} The general references for this section are \cite[Chapter I]{Knus} and \cite[Chapter~7]{Sch}, both for rings with involution. Treatments of the central simple and division cases can also be found in \cite{GB} and \cite{Le}, respectively. Let $(A,\s)$ be an $F$-algebra with involution. Let $\ve\in \{-1,1\}$. An \emph{$\ve$-hermitian space} over $(A,\s)$ is a pair $(M,h)$, where $M$ is a finitely generated right $A$-module (which is automatically projective since $A$ is semisimple) and $h:M\x M\too A$ is a sesquilinear form such that $h(y,x)=\ve \s(h(x,y))$ for all $x,y \in M$. We call $(M,h)$ a \emph{hermitian space} when $\ve=1$ and a \emph{skew-hermitian space} when $\ve=-1$. If $(A,\s)$ is a field equipped with the identity map, we say \emph{\textup{(}skew-\textup{)} symmetric bilinear} space instead of (skew-) hermitian space. Consider the left $A$-module $M^=\Hom_A(M,A)$ as a right $A$-module via the involution $\s$. The form $h$ induces an $A$-linear map $h^:M\to M^, x\mapsto h(x,\cdot)$. We call $(M,h)$ \emph{nonsingular} if $h^$ is an isomorphism. All spaces occurring in this paper are assumed to be nonsingular. We often simply write $h$ instead of $(M,h)$ and speak of a \emph{form} instead of a space. If $A=D$ is a division algebra (so that $M\simeq D^n$ for some integer $n$) such that $(D,\s,\ve)\not=(F,\id_F, -1)$, then $h$ can be diagonalized: there exist invertible elements $a_1,\ldots, a_n\in \Sym(D,\s)$ such that, after a change of basis, \[h(x,y)=\sum_{i=1}^n \s(x_i) a_i y_i,\ \forall x,y \in D^n.\] In this case we use the shorthand notation \[h=\<a_1,\ldots, a_n\>_\s,\] which resembles the usual notation for diagonal quadratic forms. If $A$ is not a division algebra we can certainly consider diagonal hermitian forms defined on free $A$-modules of finite rank, but some hermitian forms over $(A,\s)$ may not be diagonalizable. Witt cancellation and Witt decomposition hold for $\ve$-hermitian spaces over $(A,\s)$. Let $W_\ve(A,\s)$ denote the Witt group of Witt classes of $\ve$-hermitian spaces over $(A,\s)$. When $\ve=1$ we drop the subscript and simply write $W(A,\s)$. We denote the usual Witt ring of $F$ by $W(F)$. We find it convenient to identify forms over $(A,\s)$ with their classes in $W_\ve(A,\s)$. \begin{lemma} \label{lem:witt-isom} Let $(A,\s)$ be an $F$-algebra with involution. \begin{enumerate}[$(i)$] \item If $\s$ is of the first kind, then $W_{-1}(A,\s)\simeq W(A,\tau)$ for some involution $\tau$ of opposite type to $\s$. \item If $\s$ is of the second kind, then $W_{-1} (A,\s) \simeq W(A,\s)$. \item If $(A,\s)$ is degenerate, then $W(A,\s)=0$. \end{enumerate} \end{lemma} \begin{proof} $(i)$ Let $u\in A^\x$ be such that $\s(u)=-u$ and define $\tau:=\Int(u)\circ \s$. Let $h$ be a skew-hermitian form over $(A,\s)$. Then $uh$ is a hermitian form over $(A,\tau)$. The one-to-one correspondence $h\mapsto uh$ respects isometries, orthogonal sums and hyperbolicity and so induces the indicated isomorphism. $(ii)$ Let $u\in Z(A)$, $u\not= 0$ be such that $\s(u)=-u$. For example, let $u=\sqrt{d}$ if $Z(A)=F(\sqrt{d})$ and let $u=(-1,1)$ if $Z(A)\simeq F\x F$. The one-to-one correspondence $h\mapsto uh$ induces the indicated isomorphism. $(iii)$ Assume that $(A,\s) \simeq (B\x B^\op, \wwh)$ and let $h:M\x M \to B\x B^\op$ be hermitian with respect to $\wwh$. Let $e_1=(1,0)$ and $e_2=(0,1)$. Then $M=M_1\oplus M_2$ with $M_i:=Me_i$ $(i=1,2)$ and $h$ is hyperbolic since $h\|_{M_1\x M_1}=0$ and $M_1^\perp = M_1$, which can be verified by direct computation. \end{proof} In view of Lemma~\ref{lem:witt-isom}$(iii)$ degenerate $F$-algebras with involution are not interesting in the context of this paper. Therefore we redefine \emph{$F$-algebra with involution} to mean non-degenerate $F$-algebra with involution. Observe that such an algebra with involution may become degenerate over a field extension of $F$. \subsection{Adjoint Involutions}\label{adj.invo} The general reference for this section is \cite[4.A]{BOI}. Let $(A,\s)$ be an $F$-algebra with involution. Let $(M,h)$ be an $\ve$-hermitian space over $(A,\s)$. The algebra $\End_A(M)$ is again simple with centre $Z(A)$ since $M$ is finitely generated \cite[1.10]{BOI}. The involution $\ad_h$ on $\End_A(M)$, defined by \[h(x, f(y))=h(\ad_h(f)(x),y),\ \forall x,y \in M, \forall f\in \End_A(M)\] is called the \emph{adjoint involution} of $h$. The involutions $\s$ and $\ad_h$ are of the same kind and $\s(\alpha)=\ad_h(\alpha)$ for all $\alpha\in Z(A)$. In case $\ad_h$ and $\s$ are of the first kind we also have \[\type(\ad_h)=\ve\, \type(\s).\] Furthermore, every involution on $\End_A(M)$ is of the form $\ad_h$ for some $\ve$-hermitian form $h$ over $(A,\s)$ and the correspondence between $\ad_h$ and $h$ is unique up to a multiplicative factor in $F^\x$ in the sense that $\ad_h=\ad_{\lambda h}$ for every $\lambda \in F^\x$. Let $(A,\s)$ be an $F$-algebra with involution. By a theorem of Wedderburn there exists a division algebra $D$ (unique up to isomorphism) with centre $Z(A)$ and a finite-dimensional right $D$-vector space $V$ such that $A \simeq \End_D(V)$. Thus $A \simeq M_m(D)$ for some positive integer $m$. Furthermore there exists an involution $\vt$ on $D$ of the same kind as $\s$ and an $\ve_0$-hermitian form $\vf_0$ over $(D,\vt)$ with $\ve_0\in \{-1,1\}$ such that $(A,\s)$ and $(\End_D(V), \ad_{\vf_0})$ are isomorphic as algebras with involution. In matrix form $\ad_{\vf_0}$ is described as follows: \[\ad_{\vf_0}(X)=\Phi_0\vt^t(X)\Phi_0^{-1},\ \forall X\in M_m(D),\] where $\vt^t(X):=(\vt(x_{ij}))^t$ for $X=(x_{ij})$ and $\Phi_0\in \GL_m(D)$ is the Gram matrix of $\vf_0$. Thus $\vt^t(\Phi_0)=\ve_0 \Phi_0$. \subsection{Hermitian Morita Theory} We refer to \cite[\S1]{BP1}, \cite{FM}, \cite[Chapters 2--3]{GB}, \cite[Chapter~I, \S9]{Knus}, or \cite{Le1} for more details. Let $(M,h)$ be an $\ve$-hermitian space over $(A,\s)$. One can show that the algebras with involution $(\End_A(M),\ad_h)$ and $(A, \s)$ are Morita equivalent: for every $\mu\in \{-1,1\}$ there is an equivalence between the categories $\Herm_\mu(\End_A(M),\ad_h)$ and $\Herm_{\ve \mu}(A,\s)$ of non-singular $\mu$-hermitian forms over $(\End_A(M),\ad_h)$ and non-singular $\ve \mu$-hermitian forms over $(A,\s)$, respectively (where the morphisms are given by isometry), cf. \cite[Chapter~I, Theorem~9.3.5]{Knus}. This equivalence respects isometries, orthogonal sums and hyperbolic forms. It induces a group isomorphism \[W_\mu (\End_A(M), \ad_h) \simeq W_{\ve\mu}(A,\s).\] The Morita equivalence and the isomorphism are not canonical. The algebras with involution $(A,\s)$ and $(D, \vt)$ are also Morita equivalent. An example of such a Morita equivalence is obtained by composing the following three non-canonical equivalences of categories, the last two of which we will call \emph{scaling} and \emph{collapsing}. For computational purposes we describe them in matrix form. We follow the approach of \cite{LU2}: \begin{equation}\label{comp:morita} \Herm_{\ve} (A,\s)\xrightarrow{\phantom{xxx}} \Herm_{\ve} (M_m(D), \ad_{\vf_0}) \xrightarrow{\text{ scaling }} \Herm_{\ve_0 \ve} (M_m(D), \vt^t) \xrightarrow{\text{ collapsing }} \Herm_{\ve_0 \ve} (D, \vt). \end{equation} \textbf{Scaling:} Let $(M,h)$ be an $\ve$-hermitian space over $(M_m(D), \ad_{\vf_0})$. Scaling is given by \begin{equation} (M,h)\mapstoo (M, \Phi_0^{-1}h). \end{equation} Note that $\Phi_0^{-1}$ is only determined up to a scalar factor in $F^\x$ since $\ad_{\vf_0}=\ad_{\lambda \vf_0}$ for any $\lambda\in F^\x$ and that replacing $\Phi_0$ by $\lambda \Phi_0$ results in a different Morita equivalence. \medskip \textbf{Collapsing:} Recall that $M_m(D) \simeq \End_D(D^m)$ and that we always have $M \simeq (D^m)^k \simeq M_{k,\, m}(D)$ for some integer $k$. Let $h: M\x M\too M_m(D)$ be an $\ve_0 \ve$-hermitian form with respect to $\vt^t$. Then \[h(x,y)=\vt^t(x) B y,\ \forall x,y\in M_{k,\, m}(D),\] where $B\in M_k(D)$ satisfies $\vt^t(B)=\ve_0\ve B$, so that $B$ determines an $\ve_0\ve$-hermitian form $b$ over $(D,\vt)$. Collapsing is then given by \[(M,h)\mapstoo (D^k, b).\] \section{Signatures of Hermitian Forms} \subsection{Signatures of Forms: the Real Closed Case}\label{sec:special} Let $R$ be a real closed field, $C=R(\sqrt{-1})$ (which is algebraically closed) and $H=(-1,-1)_R$ Hamilton's quaternion division algebra over $R$. We recall the definitions of signature for the various types of forms (all assumed to be nonsingular) over $R$, $C$, $R\x R$ and $H$. We will use them in the definition of the $M$-signature of a hermitian form over $(A,\s)$ in Section~\ref{sec:general}. \begin{enumerate}[(a)] \item Let $b$ be a symmetric bilinear (or quadratic) form over $R$. Then $b\simeq \< \alpha_1, \ldots, \alpha_n\>$ for some $n\in \N$ and $\alpha_i \in \{-1,1\}$. We let \[\sign b:= \sum_{i=1}^n \alpha_i.\] By Sylvester's Law of Inertia, $\sign b$ is well-defined. \item Let $b$ be a skew-symmetric form over $R$. Then $b$ is hyperbolic and we let \[\sign b:=0.\] \item Let $h$ be a hermitian form over $(C,-)$, where $\bar{\sqrt{-1}}=-\sqrt{-1}$. Then $h\simeq \< \alpha_1, \ldots, \alpha_n\>_{-}$ for some $n\in \N$ and $\alpha_i \in \{-1,1\}$. By a theorem of Jacobson \cite{J}, $h$ is up to isometry uniquely determined by the symmetric bilinear form $b_h:=2\x \< \alpha_1, \ldots, \alpha_n\>$ defined over $R$. We let \[\sign h:= \frac{1}{2} \sign b_h = \sign \< \alpha_1, \ldots, \alpha_n\>.\] \item Let $h$ be a hermitian form over $(R\x R, \wwh)$, where $\wh{(x,y)}=(y,x)$ is the exchange involution. Then $h$ is hyperbolic by Lemma~\ref{lem:witt-isom}$(iii)$ and we let \[\sign h:=0.\] \item Let $h$ be a hermitian form over $(H,-)$, where $-$ denotes quaternion conjugation. Then $h\simeq \< \alpha_1, \ldots, \alpha_n\>_{-}$ for some $n\in \N$ and $\alpha_i \in \{-1,1\}$. By a theorem of Jacobson \cite{J}, $h$ is up to isometry uniquely determined by the symmetric bilinear form $b_h:=4\x \< \alpha_1, \ldots, \alpha_n\>$ defined over $R$. We let \[\sign h:= \frac{1}{4} \sign b_h = \sign \< \alpha_1, \ldots, \alpha_n\>.\] \item Let $h$ be a skew-hermitian form over $(H,-)$, where $-$ denotes quaternion conjugation. Then $h$ is a torsion form \cite[Chapter~10, Theorem~3.7]{Sch} and we let \[\sign h:=0.\] \end{enumerate} \begin{remark}\label{Z-iso} The signature maps defined in (a), (c) and (e) above give rise to unique group isomorphisms \[W(R)\simeq \Z,\ W(C,-) \simeq \Z,\ \text{and } W(H,-)\simeq \Z\] such that $\sign \<1\>=1$, $\sign \<1\>_{-}=1$ and $\sign \<1\>_{-}=1$, respectively. In addition, we have the group isomorphisms \[W_{-1}(R,\id_R)= W(R\x R, \wwh)=0,\ W_{-1}(H,\bbar)\simeq \Z/2\Z.\] See also \cite[Chapter~I, 10.5]{Knus}. \end{remark} \subsection{The $M$-Signature of a Hermitian Form}\label{sec:general} Our approach in this section is inspired by \cite[\S3.3,\S3.4]{BP2}. We only consider hermitian forms over $(A,\s)$, cf. Lemma~\ref{lem:witt-isom}. Let $F$ be a formally real field and let $(A,\s)$ be an $F$-algebra with involution. Consider an ordering $P\in X_F$, the space of orderings of $F$. By a \emph{real closure of $F$ at $P$} we mean a field embedding $\iota : F \rightarrow K$, where $K$ is real closed, $\iota(P) \subseteq K^2$ and $K$ is algebraic over $\iota(F)$. Let $h$ be a hermitian form over $(A,\s)$. Choose a real closure $\iota : F \rightarrow F_P$ of $F$ at $P$, and use it to extend scalars from $F$ to $F_P$: \[W(A,\s) \too W(A\ox_F F_P, \s\ox \id),\ h\mapstoo h\ox F_P:= (\id_A \ox \iota)^(h) ,\] where the tensor product is along $\iota$. The extended algebra with involution $(A\ox_F F_P, \s\ox \id)$ is Morita equivalent to an algebra with involution $(D_P, \vt_P)$, chosen as follows: \begin{enumerate}[$(i)$] \item If $\s$ is of the first kind, $D_P$ is equal to one of $F_P$ or $H_P:=(-1,-1)_{F_P}$. Furthermore, we may choose $(D_P, \vt_P)= (F_P, \id_{F_P})$ in the first case and $(D_P,\vt_P)= (H_P,-) $ in the second case by Morita theory (scaling). \item If $\s$ is of the second kind, recall that $Z(A)=F(\sqrt{d})$. Now if $d<_P0$, then $D_P$ is equal to $F_P(\sqrt{-1})$, whereas if $d>_P 0$, then $D_P$ is equal to $F_P\x F_P$ and $A\ox_F F_P$ is a direct product of two simple algebras. Furthermore, we may choose $(D_P, \vt_P) = (F_P(\sqrt{-1}), -)$ in the first case and $(D_P, \vt_P) = (F_P\x F_P, \wwh)$ in the second case, again by Morita theory (scaling). \end{enumerate} Note that $\vt_P$ is of the same kind as $\s$ in each case. The extended involution $\s\ox \id_{F_P}$ is adjoint to an $\ve_P$-hermitian form $\vf_P$ over $(D_P, \vt_P)$ where $\ve_P=-1$ if one of $\s$ and $\vt_P$ is orthogonal and the other is symplectic, whereas $\ve_P=1$ if $\s$ and $\vt_P$ are of the same type, i.e. both orthogonal, symplectic or unitary. Now choose any Morita equivalence \begin{equation}\label{eq:morita} \M: \Herm(A\ox_F F_P, \s\ox \id) \too \Herm_{\ve_P}(D_P, \vt_P) \end{equation} with $(D_P,\vt_P) \in \{ (F_P, \id_{F_P}), (H_P,-), (F_P(\sqrt{-1}), -), (F_P\x F_P, \wwh)\}$, which exists by the analysis above. This Morita equivalence induces an isomorphism, which we again denote by $\M$, \begin{equation}\label{mor.iso} \M: W(A\ox_F F_P, \s\ox \id) \simtoo W_{\ve_P}(D_P, \vt_P). \end{equation} \begin{defi} \label{def:sign} Let $P \in X_F$. Fix a real closure $\iota : F \rightarrow F_P$ of $F$ at $P$ and a Morita equivalence $\M$ as above. Define the \emph{$M$-signature of $h$ at $(\iota,\M)$}, denoted $\sign_\iota^\nc h$, as follows: \[\sign_\iota^\nc h:= \sign \M(h\ox F_P),\] where $\sign \M(h\ox F_P)$ can be computed as shown in Section~\ref{sec:special}. \end{defi} This definition relies on two choices: firstly the choice of the real closure $\iota: F \rightarrow F_P$ of $F$ at $P$ and secondly the choice of the Morita equivalence $\M$. Note that there is no canonical choice for $\M$. We now study the dependence of the $M$-signature on the choice of $\iota$ and $\M$. Let $\iota_1 : F \rightarrow L_1$ and $\iota_2 : F \rightarrow L_2$ be two real closures of $F$ at $P$, and let $(D_1,\vt_1)$ and $\ve_1$ play the role of $(D_P,\vt_P)$ and $\ve_P$, respectively, obtained above when $\iota$ is replaced by $\iota_1$. Let \[\M_1: \Herm(A \ox_F L_1, \s \ox \id) \too \Herm_{\ve_1}(D_1, \vt_1)\] be a fixed Morita equivalence. By the Artin-Schreier theorem \cite[Chapter~3,Theorem~2.1]{Sch} there is a unique isomorphism $\rho : L_1 \rightarrow L_2$ such that $\rho \circ \iota_1 = \iota_2$. It extends to an isomorphism $\id \ox \rho : (A \ox_F L_1, \s \ox \id) \rightarrow (A \ox_F L_2, \s \ox \id)$. The isomorphism $\rho$ also extends canonically to $D_1 \in\{L_1, (-1,-1)_{L_1}, L_1(\sqrt{-1}), L_1\x L_1 \}$. Consider the $L_2$-algebra with involution $(D_2,\vt_2):=(\rho(D_1), \rho\circ \vt_1\circ \rho^{-1})$. We define $\rho(\M_1)$ to be the Morita equivalence from $(A \ox_F L_2, \s \ox \id)$ to $(D_2, \vt_2)$, described by the following diagram: \begin{equation} \xymatrix{ \Herm(A\ox_F L_1, \s\ox\id) \ar[rr]^--{\M_1} \ar[d]^--{(\id\ox \rho)^}& & \Herm_{\ve_1}(D_1,\vt_1)\ar[d]^--{\rho^}\\ \Herm(A\ox_F L_2, \s\ox\id) \ar[rr]^--{\rho(\M_1)} & &\Herm_{\ve_1}(D_2,\vt_2) \\ } \end{equation} \begin{prop}[Change of Real Closure]\label{prop:iota} With notation as above we have for every $h\in W(A,\s)$, \[\sign \M_1(h\ox L_1) = \sign \rho(\M_1)(h\ox L_2),\] in other words \[\sign_{\iota_1}^{\nc_1} h = \sign_{\iota_2}^{\rho(\nc_1)} h.\] \end{prop} \begin{proof} The statement is trivially true when $\ve_1=-1$, by cases (b), (d) and (f) in Section~\ref{sec:special}, so we may assume that $\ve_1=1$. Consider the diagram \begin{equation} \begin{aligned}[c] \xymatrix{ &W(A\ox_F L_1, \s\ox\id) \ar[rr]^--{\M_1}_--{\sim} \ar[dd]^--{(\id\ox \rho)^}& & W(D_1,\vt_1)\ar[dd]^--{\rho^} \ar[rd]^--{\sign}& \\ W(A,\s)\ar[ur]^--{(\id\ox\iota_1)^} \ar[dr]_--{(\id\ox\iota_2)^} & & & & \Z\\ &W(A\ox_F L_2, \s\ox\id) \ar[rr]^--{\rho(\M_1)}_--{\sim} & &W(D_2,\vt_2) \ar[ur]_--{\sign} & \\ } \end{aligned} \end{equation} The left triangle commutes by the definition of $\rho$. The square commutes by the definition of $\rho(\M_1)$. The right triangle commutes since $\rho^(\<1\>_{\vt_1}) = \<1\>_{\vt_2}$ and by Remark~\ref{Z-iso}. The statement follows. \end{proof} \begin{prop}[Change of Morita Equivalence] \label{prop:morita} Let $\M_1$ and $\M_2$ be two different Morita equivalences as in \eqref{eq:morita}. Then there exists $\delta \in \{-1,1\}$ such that for every $h \in W(A,\s)$, \[ \sign_\iota^{\M_1} h = \delta \sign_\iota^{\M_2} h.\] \end{prop} \begin{proof} Note that $\ve_1=\ve_2$. The statement is trivially true when $\ve_1=-1$, by cases (b), (d) and (f) in Section~\ref{sec:special}, so we may assume that $\ve_1=1$. The two different Morita equivalences give rise to two different group isomorphisms \[m_i: W(A\ox_F F_P, \s\ox \id) \simtoo \Z\qquad (i=1,2),\] by \eqref{mor.iso} and Remark~\ref{Z-iso}. The map $m_1\circ m_2^{-1}$ is an automorphism of $\Z$ and is therefore equal to $\id_\Z$ or $-\id_\Z$. \end{proof} Propositions~\ref{prop:iota} and \ref{prop:morita} immediately imply \begin{cor}\label{iota:morita} Let $\iota_1 : F \rightarrow L_1$ and $\iota_2 : F \rightarrow L_2$ be two real closures of $F$ at $P$ and let $\M_1$ and $\M_2$ be two different Morita equivalences as in \eqref{eq:morita}. Then there exists $\delta \in \{-1,1\}$ such that for every $h \in W(A,\s)$, \[ \sign_{\iota_1}^{\M_1} h = \delta \sign_{\iota_2}^{\M_2} h.\] \end{cor} The following result easily follows from the properties of Morita equivalence: \begin{prop} \label{rem:morsign} \mbox{} \begin{enumerate}[$(i)$] \item Let $h$ be a hyperbolic form over $(A,\s)$, then \[\sign^\nc_\iota h=0.\] \item Let $h_1$ and $h_2$ be hermitian forms over $(A,\sigma)$, then \[\sign^\nc_\iota (h_1\perp h_2)=\sign^\nc_\iota h_1 + \sign^\nc_\iota h_2.\] \item The $M$-signature at $(\iota, \M)$, $\sign^\nc_\iota$, induces a homomorphism of additive groups \[W(A,\s)\too \Z.\] \item Let $h$ be a hermitian form over $(A,\s)$ and $q$ a quadratic form over $F$, then \[ \sign_\iota^\M (q\ox h) = \sign_P q \cdot \sign_\iota^\M h,\] where $\sign_P q$ denotes the usual signature of the quadratic form $q$ at $P$. \end{enumerate} \end{prop} \begin{defi} Let $h$ be a hermitian form over $(A,\s)$. From Definition~\ref{def:sign} and Section~\ref{sec:special} it follows that $\sign^\nc_\iota h$ is automatically zero whenever $P$ belongs to the following subset of $X_F$, which we call set of \emph{nil-orderings}: \[ \Nil[A,\s]:= \begin{cases} \{ P\in X_F \mid D_P= H_P\} & \text{if $\s$ is orthogonal}\\ \{ P\in X_F \mid D_P= F_P\} & \text{if $\s$ is symplectic}\\ \{ P\in X_F \mid D_P= F_P\x F_P\} & \text{if $\s$ is unitary}\\ \end{cases}, \] where the square brackets indicate that $\Nil[A,\s]$ depends only on the Brauer class of $A$ and the type of $\s$. \end{defi} \subsection{The $H$-Signature of a Hermitian Form}\label{sec:H} It follows from Corollary~\ref{iota:morita} that $\sign^\M_\iota$ is uniquely defined up to a choice of sign. We can arbitrarily choose the sign of the signature of a form at each ordering $P$. See for instance Remark~\ref{rem:scal} for a way to change sign using Morita equivalence (scaling). A more intrinsic definition is therefore desirable, in particular when considering the total signature map of a hermitian form $X_F\to\Z$ since such arbitrary changes of sign would prevent it from being continuous. We are thus led to define a signature that is independent of the choice of $\iota$ and $\M$. \begin{lemma}\label{lem:spiff} Let $P\in X_F\sm\Nil[A,\s]$. Let $\iota_1 : F \rightarrow L_1$ and $\iota_2 : F \rightarrow L_2$ be two real closures of $F$ at $P$ and let $\M_1$ and $\M_2$ be two different Morita equivalences as in \eqref{eq:morita}. Let $h_0\in W(A,\s)$ be such that $\sign_{\iota_1}^{\M_1} h_0 \not=0$ and let $\delta_k \in \{-1,1\}$ be the sign of $\sign_{\iota_k}^{\M_k} h_0$ for $k=1,2$. Then \[\delta_1 \sign_{\iota_1}^{\M_1} h = \delta_2 \sign_{\iota_2}^{\M_2} h,\] for all $h\in W(A,\s)$. \end{lemma} \begin{proof} Let $\delta\in\{-1,1\}$ be as in Corollary~\ref{iota:morita}. Then, we have for all $h\in W(A,\s)$ that $\sign_{\iota_2}^{\M_2} h = \delta \sign_{\iota_1}^{\M_1} h$ and in particular that $\sign_{\iota_2}^{\M_2} h_0 = \delta \sign_{\iota_1}^{\M_1} h_0$. It follows that $\delta_1= \delta \delta_2$. Thus $\delta_1 \sign_{\iota_1}^{\M_1} h = \delta \delta_2 \sign_{\iota_1}^{\M_1} h = \delta_2 \sign_{\iota_2}^{\M_2} h$. \end{proof} We will show in Theorem~\ref{thm:main} that there exists a finite tuple $H=(h_1,\ldots, h_s)$ of diagonal hermitian forms of rank one over $(A,\s)$ such that for every $P\in X_F\sm\Nil[A,\s]$ there exists $h_0\in H$ such that $\sign_\iota^\nc h_0\not=0$. \begin{defi}\label{def:s-sign} Let $h\in W(A,\s)$ and let $P\in X_F$. We define the \emph{$H$-signature of $h$ at $P$} as follows: If $P\in \Nil[A,\s]$, define $\sign_P^H h:=0$. If $P\not\in \Nil[A,\s]$, let $i\in \{1,\ldots, s\}$ be the least integer such that $\sign_\iota^\M h_i\not=0$ (for any $\iota$ and $\M$, cf. Corollary~\ref{iota:morita}), let $\delta_{\iota,\M} \in \{-1,1\}$ be the sign of $\sign_\iota^\M h_i$ and define \[\sign_P^H h:= \delta_{\iota,\M} \sign_\iota^\M h. \] \end{defi} Lemma~\ref{lem:spiff} shows that this definition is independent of the choice of $\iota$ and $\M$ (but it does depend on the choice of $H$). A choice of Morita equivalence which is convenient for computations of signatures is given by \eqref{comp:morita} with $(A,\s)$ replaced by $(A\ox_F F_P, \s\ox\id)$. We denote this Morita equivalence by $\CN$ and now describe the induced isomorphisms of Witt groups: \begin{equation}\label{seq5} \begin{split} \xymatrix@R=1ex{ W(A\ox_F F_P, \s\ox \id)\ar[r]^--{\xi_P^}& W(M_m(D_P), \ad_{\vf_P}) \ar[r]^--{\text{scaling}} &W(M_m(D_P), {\vt_P}^t) \ar[r]^--{\text{collapsing}} & W(D_P, \vt_P)\\ h\ox F_P\ar@{\|->}[r] & \xi_P^(h\ox F_P)\ar@{\|->}[r] & \Phi_P^{-1} \xi_P^(h\ox F_P) \ar@{\|->}[r] & \CN(h\ox F_P), } \end{split} \end{equation} where $h$ is a hermitian form over $(A,\s)$, $P\in X_F\sm \Nil[A,\s]$ (so that $\ve_P=1$), $\xi_P^$ is the group isomorphism induced by some fixed isomorphism \[\xi_P: (A\ox_F F_P, \s\ox\id)\simtoo (M_m(D_P), \ad_{\vf_P}),\] and $\Phi_P$ is the Gram matrix of the form $\vf_P$. Observe that $\sign \vf_P$ can be computed as in Section~\ref{sec:special}. \begin{lemma}\label{lem5.5} Let $P \in X_F\sm \Nil[A,\s]$, let $\iota:F \to F_P$ be a real closure of $F$ and let $\CN$ and $\vf_P$ be as above. Then $\sign^\CN_\iota \<1\>_\s =\sign \vf_P$. \end{lemma} \begin{proof} We extend scalars from $F$ to $F_P$ via $\iota$, $\<1\>_\s \mapstoo \<1\>_\s\ox F_P= \<1\ox 1\>_{\s\ox\id}$ and push $ \<1\ox 1\>_{\s\ox\id}$ through the sequence $\eqref{seq5}$, \[\<1\ox 1\>_{\s\ox\id} \mapstoo \xi^_P (\<1\ox 1\>_{\s\ox\id}) = \<\xi_P(1\ox 1)\>_{\ad_{\vf_P}} =\<I_m\>_{\ad_{\vf_P}}\mapstoo \Phi_P^{-1}\<I_m\>_{\ad_{\vf_P}}= \<\Phi_P^{-1}\>_{{\vt_P}^t}.\] (Note that $\xi_P (1\ox 1)=I_m$, the $m\x m$-identity matrix in $M_m(D_P)$ since $\xi_P$ is an algebra homomorphism.) By collapsing, the matrix $\Phi_P^{-1}$ now corresponds to a quadratic form over $F_P$, a hermitian form over $(F_P(\sqrt{-1}),\bbar)$ or a hermitian form over $(H_P, \bbar)$. In either case $\Phi_P^{-1}$ is congruent to $\Phi_P$. Thus $\sign^\CN_\iota \<1\>_\s =\sign \vf_P$. \end{proof} \begin{remark} It follows from Lemma~\ref{lem5.5} that the signature defined in \cite[\S3.3, \S3.4]{BP2} is actually $\sign^{H}_P$ with $H=(\<1\>_\s)$. It is now clear that this signature cannot be computed when $\sign_\iota^\CN \<1\>_\s=0$, i.e. when $\s\ox \id_{F_P}\simeq \ad_{\vf_P}$ is hyperbolic. In contrast, if we take $H=(h_1,\ldots, h_s)$, as described before Definition~\ref{def:s-sign} we are able to compute the signature in all cases. Note that we may choose $h_1=\<1\>_\s$, so that Definition~\ref{def:s-sign} generalizes the definition of signature in \cite[\S3.3, \S3.4]{BP2}. \end{remark} \begin{example} Let $(A,\s) = (M_4(\R), \ad_\vf)$, where $\vf=\<1,-1,1,-1\>$. Then $\sign \vf=0$. Consider the rank one hermitian forms \[ h= \left\< \left(\begin{smallmatrix} 1& & & \\ & 1 & & \\ & & -1 & \\ & & & 1 \end{smallmatrix}\right)\right\>_\s,\ h_1= \left\< \left(\begin{smallmatrix} 1& & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{smallmatrix}\right)\right\>_\s, \text{ and } h_2= \left\< \left(\begin{smallmatrix} 1& & & \\ & -1 & & \\ & & 1 & \\ & & & -1 \end{smallmatrix}\right)\right\>_\s \] over $(A,\s)$. Then $\sign^\CN h=-2$, $\sign^\CN h_1=0$ and $\sign^\CN h_2=4$, where we suppressed the index $\iota$ since $\R$ is real closed. Let $H_1=(h_1)$ and $H=(h_1,h_2)$, then $\sign^{H_1}h$ is not defined, whereas $\sign^H h= -2$. Observe that taking $H=(h_1, -h_2)$ instead would result in $\sign^H h= 2$. \end{example} \begin{remark}\label{rem:scal} Let $a\in A^\x$ be such that $\s(a)=\ve a$ with $\ve\in\{-1,1\}$. The Morita equivalence \emph{scaling by $a$}, \[\Herm(A,\s) \too \Herm_\ve(A, \Int(a)\circ\s) ,\ h \mapstoo ah\] induces an isomorphism \[\zeta_a: W(A,\s) \too W_\ve(A, \Int(a)\circ\s) ,\ h \mapstoo ah.\] It is clear that \[\sign_\iota^\M h = \sign_\iota^{\M\circ (\zeta_a^{-1}\ox \id)} ah.\] Consider the special case where $a\in F^\x$. Thus $\ve=1$ and $\Int(a)\circ \s =\s$. Assume that $a<_P 0$. Then \[\sign_\iota^\M \zeta_a(h) = \sign \M(ah\ox F_P)= \sign \M(-h\ox F_P) = -\sign_\iota^\M h,\] where the last equality follows from Proposition~\ref{rem:morsign}$(iii)$. The same computation shows that $\sign^H_P \zeta_a(h) = - \sign^H_P h$ for any choice of $H$. Thus, scaling by $a$ changes the sign of the signature, which is contrary to what is claimed in \cite[p. 662]{BP2}. \end{remark} \begin{remark}\label{rem:Mchoice} For any choice of $H$, $P$ and $\iota$ as in Definition~\ref{def:s-sign}, there exists a Morita equivalence $ \M'$ such that $\sign_P^H h=\sign_\iota^{\M'} h$ for any $h\in W(A,\s)$ (i.e. such that $\sign_\iota^{\M'} h_i>0$ with $h_i$ as in Definition~\ref{def:s-sign}). Indeed, for $\M$ as in Definition~\ref{def:s-sign}, it suffices to take $\M'=\delta_{\iota,\M} \M$. \end{remark} It remains to be shown that a tuple $H$ as described just before Definition~\ref{def:s-sign} does exist. In order to reach this conclusion we first need to develop more theory in Sections~\ref{sec:inv} and \ref{ktf-M}. \section{Signatures of Involutions}\label{sec:inv} Let $(A,\s)$ be an $F$-algebra with involution. Consider the \emph{involution trace form} \[T_\s: A\x A \too K, (x,y)\mapstoo \Trd_A(\s(x)y),\] where $\Trd_A$ denotes the reduced trace of $A$. If $\s$ is of the first kind, $T_\s$ is a symmetric bilinear form over $F$. If $\s$ is of the second kind, $T_\s$ is a hermitian form over $(K,\s\|_{K})$. Let $P\in X_F$. The \emph{signature of the involution $\s$ at $P$} is defined by \[\sign_P \s:= \sqrt{\sign_P T_\s} \] and is a nonnegative integer, since $\sign_P T_\s$ is always a square; cf. Lewis and Tignol \cite{LT} for involutions of the first kind and Qu\'eguiner \cite{Q} for involutions of the second kind. We call the involution $\s$ \emph{positive at $P$} if $\sign_P \s\not=0$. \begin{example} \mbox{} \label{trace} \begin{enumerate}[$(i)$] \item Let $(A,\s)=(M_n(F), t)$, where $t$ denotes transposition. Then $T_\s\simeq n^2\x \<1\>$. Hence $\sign_P\s=n$ for all $P \in X_F$. \item Let $(A,\s)=((a,b)_F, \bbar)$, where $\bbar$ denotes quaternion conjugation. Then $T_\s \simeq \<2\>\ox \<1,-a,-b,ab\>$. Hence $\sign_P \s=2$ for all $P\in X_F$ such that $a<_P 0, b<_P 0$ and $\sign_P=0$ for all other $P\in X_F$. \item Let $(A,\s)=(F(\sqrt{d}), \bbar)$, where $\bbar$ denotes conjugation. Then $T_\s\simeq \<1\>_\s$. We have $\sign_P \<1\>_\s = \frac{1}{2} \sign_P \<1,-d\>$, cf. \cite[Chapter 10, Examples 1.6(iii)]{Sch}. Hence $\sign_P\s=0$ for all $P \in X_F$ such that $d>_P0$ and $\sign_P\s=1$ for all $P \in X_F$ such that $d<_P0$. \end{enumerate} \end{example} \begin{remark}\label{rem3.8} Let $(A,\s)$ and $(B,\tau)$ be two $F$-algebras with involution. \begin{enumerate}[$(i)$] \item Consider the tensor product $(A\ox_F B, \s\ox\tau)$. Then $T_{\s\ox\tau}\simeq T_\s \ox T_\tau$ and so $\sign_P(\s\ox\tau)=(\sign_P \s)(\sign_P \tau)$ for all $P \in X_F$. \item If $(A,\s) \simeq (B,\tau)$, then $T_\s \simeq T_\tau$ so that $\sign_P\s= \sign_P \tau$ for all $P\in X_F$. \end{enumerate} \end{remark} \begin{remark} \label{rem3.9} Pfister's local-global principle holds for algebras with involution $(A,\s)$ and also for hermitian forms $h$ over such algebras, \cite{LU1}. \end{remark} \begin{remark}\label{rem4.6} The map $\sign \s$ is continuous from $X_F$ (equipped with the Harrison topology, see \cite[Chapter VIII 6]{Lam} for a definition) to $\Z$ (equipped with the discrete topology). Indeed: define the map $\sqrt{\phantom{x}}$ on $\Z$ by setting $\sqrt{k}=-1$ if $k$ is not a square in $\Z$. Since $\Z$ is equipped with the discrete topology, this map is continuous. Since $T_\s$ is a symmetric bilinear form or a hermitian form over $(K,\s\|_K)$, the map $\sign T_\s$ is continuous from $X_F$ to $\Z$ (by \cite[VIII, Proposition 6.6]{Lam} and \cite[Chapter~10, Example~1.6(iii)]{Sch}). Thus, by composition, $\sign \s = \sqrt{\sign T_\s}$ is continuous from $X_F$ to $\Z$. \end{remark} \begin{lemma}\label{lem5.6} Let $P\in X_F$. If $P\in \Nil [A,\s]$, then \[\sign_P\s = \sign \vf_P=0.\] Otherwise, \[\sign_P \s = \lambda_P\, \|\!\sign \vf_P\|,\] where $\lambda_P=1$ if $(D_P,\vt_P) = (F_P,\id_{F_P})$ or $(D_P,\vt_P) = (F_P(\sqrt{-1}), -)$ and $\lambda_P=2$ if $(D_P,\vt_P) = (H_P,\bbar)$. \end{lemma} \begin{proof} This is a reformulation of \cite[Theorem~1]{LT} and part of its proof for involutions of the first kind and \cite[Proposition~3]{Q} for involutions of the second kind. \end{proof} \begin{lemma}\label{h-adh} Let $(M,h)$ be a hermitian space over $(A,\s)$, let $P\in X_F$, let $\iota : F\to F_P$ be a real closure of $F$ at $P$ and let $\M$ be a Morita equivalence as in \eqref{eq:morita}. If $P\in \Nil [A,\s]$, then \[\sign_P\ad_h = \sign^\nc_\iota h=0.\] Otherwise, \[\sign_P \ad_h = \lambda_P\, \|\! \sign_\iota^\nc h\|,\] with $\lambda_P$ as defined in Lemma~\ref{lem5.6}. In particular, \[\sign_\iota^\nc h= 0 \Leftrightarrow \sign_P \ad_h=0.\] \end{lemma} \begin{proof} Assume first that $P\in \Nil[A,\s]$. Then $\sign_\iota^\nc h=0$. Consider the adjoint involution $\ad_h$ on $\End_A(M)$. Since $h$ is hermitian, $\s$ and $\ad_h$ are of the same type. Furthermore, $A$ and $\End_A(M)$ are Brauer equivalent by \cite[1.10]{BOI}. Thus $\Nil[A,\s] = \Nil[\End_A(M), \ad_h]$. By Lemma~\ref{lem5.6} we conclude that $\sign_P \ad_h=0$. Next, assume that $P\in X_F\sm \Nil[A,\s]$. Without loss of generality we may replace $F$ by $F_P$. Consider a Morita equivalence \[\M':\Herm(A,\s) \too \Herm (D,\vt)\] with $(D,\vt) = (F,\id)$, $(D,\vt) = (H, \bbar)$ or $(D,\vt) = (F(\sqrt{-1}),-)$. Let $(N,b)$ be the hermitian space over $(D,\vt)$ corresponding to $(M,h)$ under $\M'$. Then $\sign^{\M'} h=\sign b$. By \cite[Remark 1.4.2]{BP1} we have $(\End_A(M), \ad_h) \simeq (\End_D(N), \ad_b)$ so that $\sign \ad_h=\sign \ad_b$. By \cite[Theorem~1]{LT} and \cite[Proposition~3]{Q} we have $\sign \ad_b=\lambda\, \|\!\sign b\|$ with $\lambda=1$ if $(D,\vt)= (F,\id)$ or $(D,\vt)= (F(\sqrt{-1}),-)$ and $\lambda=2$ if $(D,\vt) = (H, \bbar)$. We conclude that $\sign \ad_h=\lambda\, \|\!\sign^{\M'} h\|=\lambda\, \|\!\sign^{\M} h\|$, where the last equality follows from Corollary~\ref{iota:morita}. \end{proof} \section{The Knebusch Trace Formula for $M$-Signatures}\label{ktf-M} We start with two preliminary sections in order not to overload the proof of Theorem~\ref{ktf} below. \subsection{Hermitian Forms over a Product of Rings with Involution}\label{product-rings} Let \[(A,\s) = (A_1,\s_1) \times \cdots \times (A_t, \s_t),\] where $A,A_1, \ldots, A_t$ are rings and $\s, \s_1, \ldots, \s_t$ are involutions. We write an element $a \in A$ indiscriminately as $(a_1, \ldots, a_t)$ or $a_1 + \cdots + a_t$ with $a_i \in A_i$ for $i=1, \ldots, t$. Writing $1_A = (e_1, \ldots, e_t)$, the elements $e_1, \ldots, e_t$ are central idempotents, and the coordinates of $a\in A$ are given by \[A \too A_1 \x \cdots \x A_t,\ a \mapstoo (ae_1, \ldots, ae_t).\] Note that $e_ie_j=0$ whenever $i\not=j$. We assume that $\s(1)=1$ and thus $\s(e_i)=e_i$ for $i=1, \ldots, t$. Let $M$ be an $A$-module and let $h : M \x M \rightarrow A$ be a hermitian form over $(A,\s)$. Following \cite[Proof of Lemma~1.9]{KRW} we can write \[M = \coprod_{i=1}^t Me_i,\ m = (me_1, \ldots, me_t),\] where $\coprod_{i=1}^t Me_i$ is the $A$-module with set of elements $\prod_{i=1}^t Me_i$, whose sum is defined coordinate by coordinate and whose product is defined by $(m_1e_1, \ldots, m_te_t) a = (m_1e_1a_1, \ldots, m_te_ta_t)$ for $m_1, \ldots, m_t \in M$ and $a \in A$. Define $h_i = h\|_{Me_i}$. Then $h_i(xe_i,ye_i) = \sigma(e_i)h(x,y)e_i = h(x,y)e_i^2 = h(x,y)e_i$ and $h_i : Me_i \x Me_i \rightarrow A_i$ is a hermitian form over $(A_i, \s_i)$. We also have \begin{equation} \begin{split} h(xe_1 + \cdots + xe_t, ye_1 + \cdots + ye_t) &= \sum_{i,j=1}^t h(xe_i,ye_j)e_ie_j \\ &= \sum_{i=1}^t h(xe_i,ye_i)e_i \end{split} \end{equation} which proves that $h = h_1 \perp \ldots \perp h_t$. \subsection{Algebraic Extensions and Real Closures}\label{alg-ext} We essentially follow \cite[Chapter~3, Lemma~2.6, Lemma~2.7, Theorem~4.4]{Sch}. Let $P\in X_F$ and let $F_P$ denote a real closure of $F$ at $P$. Let $L$ be a finite extension of $F$. Writing $L = F[X]/(R)$ for some $R \in F[X]$ and $R = R_1 \cdots R_t$ as a product of pairwise distinct irreducibles in $F_P[X]$ with $\deg R_1 = \cdots = \deg R_r = 1$ and $\deg R_{r+1} = \cdots = \deg R_t = 2$, we obtain canonical $F_P$-isomorphisms $L\ox_F F_P \simeq F_P[X]/(R_1 \cdots R_t)$ and \begin{equation}\label{eq:LFP} L \ox_F F_P \stackrel{\omega}{\too} F_1 \x \cdots \x F_t, \end{equation} where $F_i = F_P[X]/(R_i)$ is a real closed field for $1 \le i \le r$ and is algebraically closed for $r+1 \le i \le t$. We write $1 = (e_1, \ldots, e_t)$ in $F_1 \x \cdots \x F_t$ and define $ \omega_i(x) = \omega(x)e_i$ for $x \in L \ox_F F_P$, the projection of $\omega(x)$ on its $i$-th coordinate. Let $\iota_0 : F_P \rightarrow L \ox_F F_P$ be the canonical inclusion. Then $ \omega_i \circ \iota_0 : F_P \rightarrow F_i$ is an isomorphism of fields and of $F_P$-modules for $1 \le i \le r$. In particular, for $1 \le i \le r$, $F_i$ is naturally an $F_P$-module of dimension one, and it is easily seen that $\Tr_{F_i/F_P} = ( \omega_i \circ \iota_0)^{-1}$. It follows that $\Tr_{F_i/F_P}$ is an isomorphism of fields. Let $\iota_1 : L \rightarrow L \ox_F F_P$ be the canonical inclusion. Then $ \omega_i \circ \iota_1 : L\to F_i$ ($i=1, \ldots, r$) denote the $r$ different embeddings of ordered fields corresponding to the orderings $Q_i$ on $L$ that extend $P$. In other words, if $\{Q_1, \ldots, Q_r\}$ are the different extensions of $P$ to $L$, then for every $1 \le i \le r$, the map \[\xymatrix{ L \ar[r]^--{\iota_1} & L \ox_F F_P \ar[r]^--{\omega_i} & F_i }\] is a real closure of $L$ at $Q_i$. Since $\Tr_{F_i/F_P}$ is an isomorphism of fields, it follows that the map \[\xymatrix{ L \ar[r]^--{\iota_1} & L \ox_F F_P \ar[r]^--{\omega_i} & F_i \ar[r]^--{\Tr_{F_i/F_P}} & F_P }\] is also a real closure of $L$ at $Q_i$. \subsection{The Knebusch Trace Formula} Let $(A,\s)$ be an $F$-algebra with involution. Let $L/F$ be a finite extension. The trace $\Tr_{L/F}: L\to F$ induces an $A$-linear homomorphism \[\Tr_{A\ox_F L} =\id_A \ox \Tr_{L/F} : A\ox_F L \too A\] which induces a group homomorphism (transfer map) \[\trs_{A\ox_F L}: W(A\ox_F L, \s\ox\id) \too W(A,\s),\ (M,h)\mapstoo (M, \Tr_{A\ox_F L} \circ h),\] cf. \cite[p. 362]{B-L}. The following theorem is an extension of a result due to Knebusch \cite[Proposition~5.2]{K}, \cite[Chapter~3, Theorem~4.5]{Sch} to $F$-algebras with involution. The proof follows the general lines of Knebusch's original proof. \begin{theorem}\label{ktf} Let $P\in X_F$. Let $L/F$ be a finite extension of ordered fields and let $h$ be a hermitian form over $(A\ox_F L, \s\ox\id)$. Fix a real closure $\iota : F \rightarrow F_P$ and a Morita equivalence $\nc$ as in \eqref{eq:morita}. Then, with notation as in Section~\ref{alg-ext}, \[\sign_{\iota}^\nc (\trs_{A\ox_F L} h) =\sum_{i=1}^r \sign_{\omega_i \circ \iota_1}^{(\omega_i \circ \iota_0)(\nc)} h.\] \end{theorem} \begin{proof} By definition of signature we have \begin{equation}\label{sign_1} \sign_\iota^\nc (\trs_{A\ox_F L} h) = \sign \nc [ (\trs_{A\ox_F L} h)\ox_F F_P]. \end{equation} Consider the commutative diagram \begin{equation} \xymatrix{ L \ar[rr]^--{\Tr_{L/F} } \ar[d]^--{\ox_F F_P}& & F\ar[d]^--{\ox_F F_P}\\ L\ox_F F_P \ar[rr]^--{\Tr_{L\ox F_P/F_P}} & &F_P\\ } \end{equation} It induces a commutative diagram \begin{equation} \xymatrix{ A\ox_F L \ar[rr]^--{\Tr_{A\ox L} } \ar[d]^--{\ox_F F_P}& & A\ar[d]^--{\ox_F F_P}\\ A\ox_F L\ox_F F_P \ar[rr]^--{\id_A\ox \Tr_{L\ox F_P/F_P}} & &A\ox_F F_P\\ } \end{equation} which in turn induces a commutative diagram of Witt groups \begin{equation} \xymatrix{ W(A\ox_F L,\s\ox\id) \ar[rr]^--{\trs_{A\ox L} } \ar[d]& & \ar[d] W(A,\s) \\ W(A\ox_F L\ox_F F_P, \s\ox\id\ox\id) \ar[rr]^--{(\id_A\ox \Tr_{L\ox F_P/F_P})^* } & &W(A\ox_F F_P, \s\ox\id)\\ } \end{equation} where the vertical arrows are the canonical restriction maps. Thus \begin{equation}\label{sign_2} \sign {\nc} [ (\trs_{A\ox_F L} h)\ox_F F_P]= \sign {\nc} [ (\id_A\ox \Tr_{L\ox F_P/F_P})^ (h\ox_F F_P)]. \end{equation} With reference to Section~\ref{alg-ext} consider the diagram \begin{equation} \xymatrix{ A\ox_F L \ox_F F_P \ar[r]^--{\id_A \ox \omega} \ar[d]^--{\id_A \ox_F \Tr_{L\ox F_P /F_P}}& A \ox_F (F_1\x\cdots \x F_t) \ar[r]^--{\sim} \ar[d]^--{\id_A \ox \Tr_{F_1\x\cdots\x F_t /F_P} } & (A\ox_F F_1)\x \cdots \x (A\ox_F F_t) \ar[d]^--{\sum_{i=1}^t \id_A \ox \Tr_{F_i/F_P} } \\ A\ox_F F_P \ar[r]^--{\id} &A\ox_F F_P \ar[r]^--{\id} &A\ox_F F_P\\ } \end{equation} where commutativity of the first square follows from the isomorphism \eqref{eq:LFP} of $F_P$-algebras, whereas commutativity of the second square follows from \cite[p.~137]{B-alg8}. We push $h\ox F_P$ through the induced commutative diagram of Witt groups: \begin{equation} \xymatrix{ h\ox F_P \ar@{\|->}[r]^--{(\id_A\ox\omega)^} \ar@{\|->}[d] & h' \ar@{\|->}[r] & h'_1 \perp \ldots \perp h'_t \ar@{\|->}[d]\\ (\id_A\ox \Tr_{L\ox F_P / F_P})^* (h\ox F_P) \ar@{\|->}[rr]^--{\id} & & \sum_{i=1}^t (\id_A\ox \Tr_{F_i / F_P})^(h'_i) } \end{equation} where the image of $h'$ equals the orthogonal sum $h'_1\perp \ldots \perp h'_t$ by Section~\ref{product-rings}. Thus \begin{equation}\label{sign_3} \sign {\nc} [ (\id_A\ox \Tr_{L\ox F_P/F_P})^* (h\ox_F F_P)] = \sum_{i=1}^t \sign {\nc}[(\id_A\ox \Tr_{F_i / F_P})^(h'_i)]. \end{equation} We have to consider the following two cases: Case 1: Assume that $1\leq i \leq r$. Observe that $h'_i = (\id_A \ox( \omega_i \circ \iota_1))^(h)$. Then \begin{equation} \sign {\nc}[(\id_A\ox \Tr_{F_i / F_P})^(h'_i)] = \sign {\nc}[(\id_A\ox (\Tr_{F_i / F_P}))^\circ (\id_A \ox (\omega_i \circ \iota_1))^(h)]. \end{equation} The form $(\id_A \ox (\omega_i \circ \iota_1))^(h)$ is defined over $A \ox_F F_i$, and the commutative diagram \[\xymatrix@R=1ex{ & F_P\ar[dd]^{\omega_i \circ \iota_0}\\ F_i \ar[ru]^{\Tr_{F_i / F_P}} \ar[dr]_{\id} & \\ & F_i\\ }\] together with Proposition~\ref{prop:iota} gives \begin{multline}\label{sign_5} \sign {\nc}[(\id_A\ox (\Tr_{F_i / F_P}))^\circ (\id_A \ox (\omega_i \circ \iota_1))^(h)] \\ =\sign (\omega_i \circ \iota_0)(\M)[(\id_A \ox (\omega_i \circ \iota_1))^(h)] \end{multline} Case 2: Assume that $r+1\leq i \leq t$. Since $F_i$ is algebraically closed, it follows from Morita theory that the Witt group $W(A\ox_F F_i, \s\ox\id)$ is torsion and so $h'_i$ is a torsion form. Therefore $(\id_A\ox \Tr_{F_i / F_P})^(h'_i)$ is also a torsion form and thus has signature zero. We conclude that equations \eqref{sign_1}-\eqref{sign_5} yield the Knebusch Trace Formula. \end{proof} \section{Existence of Forms with Nonzero Signature}\label{choose_sign} \begin{theorem}\label{thm2.4} Let $(A,\s)$ be an $F$-algebra with involution and let $P\in X_F\sm \Nil[A,\s]$. Fix a real closure $\iota : F \rightarrow F_P$ and a Morita equivalence $\nc$ as in \eqref{eq:morita}. There exists a hermitian form $h$ over $(A,\s)$ such that $\sign_\iota^\nc h\not=0$. \end{theorem} \begin{proof} Let $P\in X_F\sm \Nil[A,\s]$. Then $A\ox_F F_P \simeq M_m(D_P)$ for some $m\in \N$, where $D_P=F_P$, $H_P$ or $F_P(\sqrt{-1})$ if $\s$ is orthogonal, symplectic or unitary, respectively. In each case there exists a positive involution $\tau$ on $M_m(D_P)$ (namely, transposition, conjugate transposition and quaternion conjugate transposition, respectively, cf. Example~\ref{trace}). By Lemma~\ref{h-adh} the hermitian form $\<1\>_\tau$ over $(M_m(D_P), \tau)$ has nonzero signature since $\tau$ is the adjoint involution of the form $\<1\>_\tau$. After scaling we obtain a rank one hermitian form $h_0$ over $(A\ox_F F_P, \s\ox\id)$ such that $\sign \M(h_0)\not=0$ by Proposition~\ref{prop:morita}. The form $h_0$ is already defined over a finite extension $L$ of $\iota(F)$, contained in $F_P$. Thus we can consider $h_0$ as a form over $(A\ox_F L, \s\ox\id)$ and we have $\sign \M(h_0\ox F_P)\not=0$. In other words, if $Q_1$ is the ordering $L\cap F_P^{\x 2}$ on $L$, then for any real closure $\kappa_1: L\to L_1$ of $L$ at $Q_1$ and any Morita equivalence $\M_1$ as in \eqref{eq:morita}, but starting from $\Herm (A\ox_F L_1, \s\ox\id)$, we have \begin{equation} \sign_{\kappa_1}^{\M_1} h_0\not=0 \end{equation} by Corollary~\ref{iota:morita}. Let $X=\{Q\in X_L \mid P\subset Q\}$. By \cite[Chapter~3, Lemma~2.7]{Sch}, $X$ is finite, say $X=\{Q_1,Q_2,\ldots, Q_r\}$. Thus there exist $a_2,\ldots, a_r\in L^\x$ such that \[\{Q_1\} =H(a_2,\ldots, a_r)\cap X.\] Consider the Pfister form $q:=\<\!\<a_2,\ldots, a_r\>\!\>=\<1,a_2\> \ox \cdots \ox \<1,a_r\>$. Then $\sign_{Q_1} q=2^{r-1}$ and $\sign_{Q_\ell}q=0$ for $\ell\not=1$. It follows that $\sign^{\M_1}_{\kappa_1} (q\ox h_0)=\sign_{Q_1}q\cdot \sign^{\M_1}_{\kappa_1} h_0\not=0$ and $\sign^{\M_\ell}_{\kappa_\ell} (q\ox h_0)=0$ for $\ell\not=1$, where $\kappa_\ell: L\to L_\ell$ is any real closure of $L$ at $Q_\ell$ and $\M_\ell$ is any Morita equivalence as in \eqref{eq:morita}, but starting from $\Herm (A\ox_F L_\ell, \s\ox\id)$. Now $\trs_{A \ox_F L}(q\ox h_0)$ is a hermitian form over $(A,\s)$. By the trace formula, Theorem~\ref{ktf}, we have \[\sign_{\iota}^\nc \trs_{A\ox_F L} (q\ox h_0) =\sum_{i=1}^r \sign_{\omega_i \circ \iota_1}^{(\omega_i \circ \iota_0)(\nc)} (q\ox h_0)=\sign_{\omega_1 \circ \iota_1}^{(\omega_1 \circ \iota_0)(\nc)} (q\ox h_0)\not=0.\] Taking $h:= \trs_{A\ox_F L} (q\ox h_0)$ proves the theorem. \end{proof} \begin{cor} \label{cor4.2} Let $(A_1,\s_1)$ and $(A_2,\s_2)$ be $F$-algebras with involution of the same type. Assume that $A_1$ and $A_2$ are Brauer equivalent. Let $P\in X_F$, let $\iota: F \to F_P$ be a real closure of $F$ at $P$ and let $\M_\ell$ be any Morita equivalence as in \eqref{eq:morita}, but starting from $\Herm (A_\ell\ox_F F_P, \s_\ell\ox\id)$ for $\ell=1,2$. Then the following statements are equivalent: \begin{enumerate}[$(i)$] \item $\sign_\iota^{\M_1} h=0$ for all hermitian forms $h$ over $(A_1,\s_1)$; \item $\sign_\iota^{\M_2} h=0$ for all hermitian forms $h$ over $(A_2,\s_2)$; \item $\sign_P \vt=0$ for all involutions $\vt$ on $A_1$ of the same type as $\s_1$; \item $\sign_P \vt=0$ for all involutions $\vt$ on $A_2$ of the same type as $\s_2$. \end{enumerate} \end{cor} \begin{proof} By Theorem~\ref{thm2.4}, the first two statements are equivalent to $P\in \Nil[A_1,\s_1]=\Nil[A_2,\s_2]$. Thus $(i)\iff (ii)$. $(i)\implies (iii)$ Let $\vt$ be as in $(iii)$. Then $\vt=\ad_{\<1\>_\vt}$ and $\vt = \Int(a)\circ \s_1$ for some invertible $a\in \Sym(A_1,\s_1)$. Thus, with notation as in Remark~\ref{rem:scal} and using Lemma~\ref{h-adh} and Proposition~\ref{prop:morita} we have for any Morita equivalence $\M$, \begin{align} \sign_P \vt&= \lambda_P \| \sign_\iota^\M \<1\>_\vt\| \\ &=\lambda_P \| \sign_\iota^{\M\circ (\zeta_a\ox \id)} \zeta_a^{-1} (\<1\>_\vt)\| \\ &= \lambda_P \| \sign_\iota^{\M_1} \zeta_a^{-1} (\<1\>_\vt)\| \\ &= \lambda_P \| \sign_\iota^{\M_1} \<a^{-1}\>_{\s_1}\| \end{align} which is zero by the assumption. $(ii)\implies (iv)$: This is the same proof as $(i)\implies (iii)$ after replacing $(A_1,\s_1)$ by $(A_2,\s_2)$. For the remainder of the proof we may assume without loss of generality that $A_2$ is a division algebra and that $A_1\simeq M_m(A_2)$ for some $m\in \N$. $(iii)\implies (iv)$: Let $\vt$ be any involution on $A_2$. By the assumption, $\sign_P (t\ox \vt)=0$, where $t$ denotes the transpose involution. Since $t$ is a positive involution, it follows that $\sign_P \vt=0$, cf. Remark~\ref{rem3.8}. $(iv) \implies (ii)$: The assumption implies that $\sign_\iota^{\M_2} h=0$ for every hermitian form $h$ of dimension $1$ over $(A_2,\s_2)$, which implies $(ii)$ since all hermitian forms over $(A_2,\s_2)$ can be diagonalized. \end{proof} \begin{remark}\label{rem:signon} Note that statement $(iii)$ in Corollary~\ref{cor4.2} is equivalent to: $\sign_\iota^{\M_1} h=0$ for all diagonal hermitian forms $h$ of rank one over $(A_1,\s_1)$. \end{remark} \begin{theorem}\label{thm:main} Let $(A,\s)$ be an $F$-algebra with involution. There exists a finite set $H=\{h_1,\ldots, h_s\}$ of diagonal hermitian forms of rank one over $(A,\s)$ such that for every $P\in X_F\sm\Nil[A,\s]$, real closure $\iota : F \rightarrow F_P$ and Morita equivalence $\M$ as in \eqref{eq:morita} there exists $h\in H$ such that $\sign_\iota^{\M} h\not=0$. \end{theorem} \begin{proof} For every $P\in X_F$, choose a real closure $\iota_P : F \rightarrow F_P$ and a Morita equivalence $\M_P$ as in \eqref{eq:morita}. By Corollary~\ref{iota:morita} we may assume without loss of generality and for the sake of simplicity that the map $\iota_P$ is an inclusion. The algebra $A\ox_F F_P$ is isomorphic to a matrix algebra over $D_P$, where $D_P\in \{F_P, H_P, F_P(\sqrt{-1}), F_P\x F_P\}$. There is a finite extension $L_P$ of $F$, $L_P\subset F_P$ such that $A\ox_F L_P$ is isomorphic to a matrix algebra over $E_P$, where $E_P\in \{L_P, (-1,-1)_{L_P}, L_P(\sqrt{-1}), L_P\x L_P\}$, and $P$ extends to $L_P$. Let \[U_P:=\{ Q\in X_F \mid Q \text{ extends to } L_P\}.\] Since $P\in U_P$ we can write \begin{equation} X_F = \bigcup_{P\in X_F} U_P. \end{equation} We know from \cite[Chapter~3, Lemma~2.7, Theorem~2.8]{Sch} that $\sign_{Q} (\trs_{L_P/F} \<1\>)$ is the number of extensions of $Q$ to $L_P$. Thus, \[U_P = \Bigl( \sign (\trs_{L_P/F} \<1\>)\Bigr)^{-1} (\{1,\ldots,k\}),\] where $k$ is the dimension of the quadratic form $\trs_{L_P/F} \<1\>$, and so $U_P$ is clopen in $X_P$ (and in particular compact). Therefore, and since $X_F$ is compact, there exists a finite number of orderings $P_1,\ldots, P_\ell$ in $X_F$ such that \[X_F =\bigcup_{i=1}^\ell U_{P_i}.\] Now let $P\in \{P_1,\ldots, P_\ell\}$ and let $L_P$ be as before. By Theorem~\ref{thm2.4} we have that for every $Q\in U_P \setminus \Nil[A,\s]$ there exists a hermitian form $h_Q$ over $(A,\s)$ such that $\sign_{\iota_Q}^{\M_Q} h_Q \not=0$. By Corollary~\ref{cor4.2} and Remark~\ref{rem:signon} we may assume that $h_Q$ is diagonal of rank one. Consider the total signature map \[\mu_Q: X_F\too \Z,\ P \mapstoo \sign_{\iota_P}^{\M_P} h_Q.\] Then \begin{equation}\label{eq2.4-2} U_P\setminus \Nil[A,\s] = \bigcup_{Q\in U_P\setminus \Nil[A,\s]} \mu_Q^{-1} (\Z\setminus \{0\}). \end{equation} Consider the continuous map \[\lambda_P: X_{L_P}\too X_F, R\mapstoo R\cap F.\] We have $Q\in U_P\setminus \Nil[A,\s]$ if and only if some extension $Q'$ of $Q$ to $L_P$ is in $X_{L_P}\setminus \Nil[A\ox_F L_P, \s\ox\id]$ (this follows from the fact that the ordered fields $(F,Q)$ and $(L_P,Q')$ have a common real closure) if and only if $Q\in \lambda\bigl( X_{L_P}\setminus \Nil[A\ox_F L_P, \s\ox\id]\bigr)$. We observe that $X_{L_P}\setminus \Nil[A\ox_F L_P, \s\ox\id]$ is clopen and compact since $\Nil[A\ox_F L_P, \s\ox\id]$ is either $\varnothing$ or the whole of $X_{L_P}$, which follows from the fact that $A\ox_F L_P$ is a matrix algebra over one of $L_P, (-1,-1)_{L_P}, L_P(\sqrt{-1}), L_P\x L_P$. Hence, \[U_P\setminus \Nil[A,\s] = \lambda\bigl( X_{L_P}\setminus \Nil[A\ox_F L_P, \s\ox\id]\bigr)\] is compact and thus closed. Thus $U_P\cap \Nil[A,\s]$ is open in $U_P$. Using \eqref{eq2.4-2} we can write \[U_P=\bigl(U_P \cap \Nil[A,\s] \bigr) \cup \bigcup_{Q\in U_P\setminus \Nil[A,\s]} \mu_Q^{-1} (\Z\setminus \{0\}).\] Now $\mu_Q^{-1} (\Z\setminus \{0\}) = (\sign \ad_{h_Q})^{-1} (\Z\setminus \{0\})$ by Lemma~\ref{h-adh}, which is open since $\sign \ad_{h_Q}$ is continuous by Remark~\ref{rem4.6}. Thus, since $U_P$ is compact, there exist $Q_1,\ldots, Q_t \in U_P\setminus \Nil[A,\s]$ such that \[U_P=\bigl(U_P \cap \Nil[A,\s] \bigr) \cup \bigcup_{i=1}^t \mu_{Q_i}^{-1} (\Z\setminus \{0\}).\] In other words, for every $Q \in U_P\setminus \Nil[A,\s]$ one of $\sign^{\M_{Q}}_{\iota_{Q}} h_{Q_i}$ $(i=1,\ldots, t)$ is nonzero. Now let $H_P=\{h_{Q_1},\ldots, h_{Q_t}\}$. Letting $H=\bigcup_{i=1}^\ell H_{P_i}$ finishes the proof. \end{proof} \begin{cor} Let $(A,\s)$ be an $F$-algebra with involution. The set $\Nil[A,\s]$ is clopen in $X_F$. \end{cor} \begin{proof} By Theorem~\ref{thm:main} we have $\Nil[A,\s] = \bigcap_{i=1}^s \{P\in X_F \mid \sign^{\M_P}_{\iota_P} h_i=0\}$. The result follows from Lemma~\ref{h-adh} and Remark~\ref{rem4.6}. \end{proof} At this stage we have established all results that are needed for the definition of the $H$-signature in Definion~\ref{def:s-sign}. In the final two sections we show that the total $H$-signature of a hermitian form is continuous and we reformulate the Knebusch Trace Formula in terms of the $H$-signature. \section{Continuity of the Total $H$-Signature Map of a Hermitian Form}\label{sec:cont} Let $h$ be a hermitian form over $(A,\s)$. With reference to Definition~\ref{def:s-sign} we denote by $\sign^H h$ the total $H$-signature map of $h$: \[X_F\too \Z,\ P\mapstoo \sign_P^H h.\] \begin{lemma}\label{lembluefrog} Let $H=(h_1,\ldots, h_s)$ be as in Definition~\ref{def:s-sign}. There is a finite partition of $X_F$ into clopens \[X_F= \Nil[A,\s] \dotcup \bigcupdot_{i=1}^\ell Z_i,\] such that for every $i\in\{1,\ldots, \ell\}$ one of the total $H$-signature maps $\sign^H h_1,\ldots, \sign^H h_s$ is constant non-zero on $Z_i$. \end{lemma} \begin{proof} For $r=1,\ldots, s$, let \[Y_r:=\{ P\in X_F \mid \sign_P^H h_i=0,\ i=1,\ldots, r\}.\] By Lemma~\ref{h-adh} we have \[Y_r= \bigcap_{i=1}^r \{P\in X_F \mid \sign_P \ad_{h_i}=0\}.\] Thus each $Y_r$ is clopen. We have $Y_0:=X_F \supseteq Y_1 \supseteq \cdots \supseteq Y_{s-1} \supseteq Y_s= \Nil[A,\s]$ and therefore, \[X_F=(Y_0\sm Y_1) \dotcup (Y_1\sm Y_2) \dotcup \cdots \dotcup (Y_{s-1}\sm Y_s) \dotcup \Nil[A,\s].\] Let $r\in \{0,\ldots,s-1\}$ and consider $Y_r\sm Y_{r+1}$. By the definition of $Y_1,\ldots, Y_s$ the map $\sign^H h_{r+1}$ is never $0$ on $Y_r\sm Y_{r+1}$. Furthermore, since the rank of $h_{r+1}$ is finite, $\sign^H h_{r+1}$ only takes a finite number of values $k_1,\ldots, k_m$. Now observe that there exists a $\lambda \in \{1,2\}$ such that \[\sign^H h_{r+1}=\frac{1}{\lambda} \sign \ad_{h_{r+1}}\] on $Y_r\sm Y_{r+1}$. This follows from Lemma~\ref{h-adh} and Definition~\ref{def:s-sign} for $P\in Y_r\sm Y_{r+1}$. Therefore, \[ \bigl( \sign^H h_{r+1}\bigr)^{-1} (k_i) \cap (Y_r \sm Y_{r+1})= \bigl(\sign \ad_{h_{r+1}}\bigr)^{-1} (\lambda k_i)\cap (Y_r \sm Y_{r+1}),\] which is clopen by Remark~\ref{rem4.6}. It follows that $Y_r \sm Y_{r+1}$ is covered by finitely many disjoint clopen sets on which the map $\sign^H h_{r+1}$ has constant non-zero value. The result follows since the sets $Y_r \sm Y_{r+1}$ for $r=0,\ldots, s-1$ form a partition of $X_F\sm \Nil [A,\s]$. \end{proof} \begin{theorem} Let $h$ be a hermitian form over $(A,\s)$. The total $H$-signature of $h$, \[\sign^H h : X_F \too \Z,\ P\mapstoo \sign^H_P h\] is continuous. \end{theorem} \begin{proof} We use the notation and the conclusion of Lemma~\ref{lembluefrog}. Since $\Nil[A,\s]$ and the sets $Z_i$ are clopen, it suffices to show that $(\sign^H h)\|_{Z_i}$ is continuous for every $i=1,\ldots, \ell$. Let $i\in \{1,\ldots, \ell\}$, $k_i\in \Z\sm\{0\}$ and $j\in\{1,\ldots, s\}$ be such that $\sign^H h_j=k_i$ on $Z_i$. Let $k\in \Z$. Then \begin{align} \bigl((\sign^H h)\|_{Z_i}\bigr)^{-1} (k) &= \{P \in Z_i \mid \sign^H_P h = k\}\\ &= \{P \in Z_i \mid k_i \sign^H_P h = k_i k\}\\ &= \{P \in Z_i \mid k_i \sign^H_P h = k \sign^H_P h_j \}\\ &= \{P \in Z_i \mid \sign^H_P ( k_i \x h \perp k\x h_j)=0 \}. \end{align} It follows from Lemma~\ref{h-adh} that \[\bigl((\sign^H h)\|_{Z_i}\bigr)^{-1} (k) = \{P \in Z_i \mid \sign_P \ad_{ k_i \x h \perp k\x h_j}=0 \},\] which is clopen by Remark~\ref{rem4.6}. \end{proof} \section{The Knebusch Trace Formula for $H$-Signatures}\label{sec:ktf-H} \begin{theorem} Let $H=(h_1,\ldots, h_s)$ be as in Definition~\ref{def:s-sign}. Let $P\in X_F$. Let $L/F$ be a finite extension of ordered fields and let $h$ be a hermitian form over $(A\ox_F L, \s\ox\id)$. Then, with $H\ox L:= (h_1\ox L,\ldots, h_s\ox L)$, we have \[\sign^{H}_P (\trs_{A\ox_F L} h) =\sum_{P \subseteq Q \in X_L} \sign^{H \ox L}_Q h.\] \end{theorem} \begin{proof} By Theorem \ref{ktf} (and using its notation), we know that \[\sign_{\iota}^\nc (\trs_{A\ox_F L} h) =\sum_{i=1}^r \sign_{\omega_i \circ \iota_1}^{(\omega_i \circ \iota_0)(\nc)} h,\] for any $\iota$ and $\M$. Fix a real closure $\iota : F \rightarrow F_P$ and choose a Morita equivalence $\M$ such that $\sign_{\iota}^\M = \sign^{H}_P$ (cf. Remark~\ref{rem:Mchoice}). We only have to check that $\sign_{\omega_i \circ \iota_1}^{(\omega_i \circ \iota_0)(\nc)} = \sign^{H \ox L}_{Q_i}$ for $i=1, \ldots, r$. By definition of $\M$, there is a $k \in \{1, \ldots, s\}$ such that $\sign_\iota^\M h_j = 0$ for $1 \le j \le k-1$ and $\sign_\iota^\M h_k > 0$. To check that $\sign_{\omega_i \circ \iota_1}^{(\omega_i \circ \iota_0)(\nc)} = \sign^{H \ox L}_{Q_i}$ for $i=1, \ldots, r$, it suffices to check that $\sign_{\omega_i \circ \iota_1}^{(\omega_i \circ \iota_0)(\nc)}(h_j \ox L) = 0$ for $j=1, \ldots, k-1$ and $\sign_{\omega_i \circ \iota_1}^{(\omega_i \circ \iota_0)(\nc)}(h_k \ox L) > 0$. This follows from the fact that \begin{equation} \sign_{\omega_i \circ \iota_1}^{(\omega_i \circ \iota_0)(\nc)} (h_\ell \ox L) = \sign_\iota^\M h_\ell \text{ for every }1 \le \ell \le s, \end{equation} which we verify in the remainder of the proof. By definition, \[\sign_{\omega_i \circ \iota_1}^{(\omega_i \circ \iota_0)(\nc)} (h_\ell \ox L) = \sign (\omega_i \circ \iota_0)(\nc)[(\id_A \ox (\omega_i \circ \iota_1))^(h_\ell \ox L)].\] Consider the commutative diagram \[\xymatrix@R=1ex{ & F_P \ar[r]^--{\iota_0} & L \ox_F F_P \ar[r]^--{\omega_i} & F_i \ar[dd]^{\id} \\ F \ar[ur]^\iota \ar[dr] & & & \\ & L \ar[r]_--{\iota_1} & L \ox_F F_P \ar[r]_--{\omega_i} & F_i }\] Thus, by Proposition \ref{prop:iota}, \[ \sign (\omega_i \circ \iota_0)(\nc)[(\id_A \ox (\omega_i \circ \iota_1))^ (h_\ell \ox L)] = \sign (\omega_i \circ \iota_0)(\nc)[(\id_A \ox (\omega_i \circ \iota_0 \circ \iota))^(h_\ell)].\] Finally the commutative diagram \[\xymatrix@R=1ex{ & F_i \ar[dd]^{(\omega_i \circ \iota_0)^{-1}} \\ F \ar[ur]^{\omega_i \circ \iota_0 \circ \iota} \ar[dr]_{\iota} & \\ & F_P }\] together with Proposition \ref{prop:iota} yields \begin{equation} \sign (\omega_i \circ \iota_0)(\nc)[(\id_A \ox (\omega_i \circ \iota_0 \circ \iota))^(h_\ell)] = \sign \M[(\id_A \ox \iota)^ (h_\ell)] = \sign_\iota^\M h_\ell, \end{equation} which concludes the proof. \end{proof} \section{Acknowledgements} The authors wish to thank the referee, whose comments helped improve the original version of this paper.	52,709
\begin{document} \title{Variational Integrators for Maxwell's Equations with Sources} \maketitle \author {F. M. Lastname} \affiliation {University} \address {} \city {Boston} \postalcode {} \country {USA} \phone {345566} \fax {233445} \email {email@email.com} \misc { } \nomakeauthor \author {F. M. Lastname} \affiliation {University} \address {} \city {Boston} \postalcode {} \country {USA} \phone {345566} \fax {233445} \email {email@email.com} \misc { } \nomakeauthor \begin{authors} {\bf A. Stern}$^{1}$, {\bf Y. Tong}$^{1,2}$, {\bf M. Desbrun}$^{1}$, {\bf and J. E. Marsden}$^{1}$\\ \medskip $^{1}$California Institute of Technology, USA\\ $^{2}$Michigan State University, USA \end{authors} \begin{paper} \begin{piersabstract} In recent years, two important techniques for geometric numerical discretization have been developed. In computational electromagnetics, spatial discretization has been improved by the use of mixed finite elements and discrete differential forms. Simultaneously, the dynamical systems and mechanics communities have developed structure-preserving time integrators, notably variational integrators that are constructed from a Lagrangian action principle. Here, we discuss how to combine these two frameworks to develop {\em variational spacetime integrators} for Maxwell's equations. Extending our previous work, which first introduced this variational perspective for Maxwell's equations without sources, we also show here how to incorporate free sources of charge and current. \end{piersabstract} \psection{Introduction} In computational electromagnetics, as in an increasing number of other fields in applied science and engineering, there is both practical and theoretical interest in developing {\em geometric numerical integrators}. These numerical methods preserve, by construction, various geometric properties and invariants of the continuous physical systems that they approximate. This is particularly important for applications where even high-order methods may fail to capture important features of the underlying dynamics. In this short paper, we show that the traditional Yee scheme and extensions can be derived from the Euler-Lagrange equations of a discrete action, i.e., by designing an electromagnetic \emph{variational integrator}, including free sources of charge and current in non-dissipative media. Furthermore, we present how to use this discrete geometric framework to allow for asynchronous time stepping on unstructured grids, as recently introduced in~\citet{StToDeMa2007}. Variational integrators (not to be confused with variational methods such as finite element schemes) were originally developed for geometric time integration, particularly to simulate dynamical systems in Lagrangian mechanics. The key idea is the following: rather than approximating the equations of motion directly, one discretizes the Lagrangian and its associated action integral (e.g., using a numerical quadrature rule), and then {\em derives} a structure-preserving approximation to the equations of motion by applying Hamilton's principle of stationary action. Since the numerical method is derived from a Lagrangian variational principle, some important results from Lagrangian dynamics carry over to the discretized system, including Noether's theorem relating symmetries to conserved momentum maps, as well as the fact that the Euler-Lagrange flow is a symplectic mapping. (\citet[See][]{MaWe2001,LeMaOrWe2004}.) \paragraph{Overview.} To develop a variational integrator for Maxwell's equations, the discrete Hamilton's principle needs to incorporate more than just the time discretization, as in mechanics; spatial discretization also needs to be handled carefully. Building upon mixed finite elements in space~\citep{Nedelec1980,Bossavit1998,GrKo2004}, we treat the electromagnetic {\em Lagrangian density} as a discrete differential $4$-form in spacetime. Extremizing the integral of this Lagrangian density with fixed boundary conditions directly leads to discrete update rules for the electromagnetic fields, with either uniform or asynchronous time steps across the various spatial elements. \psection{Review of Maxwell's Equations in Spacetime} \paragraph{Electromagnetic Forms.} Let $A$ be a $1$-form on spacetime, called the {\em electromagnetic potential}, and then define the {\em Faraday $2$-form} to be its exterior derivative $ F = d A $. Given a time coordinate $t$, this splits into the components \begin{equation} F = E \wedge d t + B , \end{equation} where $E$ is the electric displacement $1$-form and $B$ is the magnetic flux $2$-form, both defined on the spacelike Cauchy surfaces $\Sigma$ with constant $t$. If $ * $ is the Hodge star associated to the spacetime metric, then we can also split the dual $2$-form \begin{equation} F = \left( _\mu B \right) \wedge d t - _\epsilon E = H \wedge d t - D , \end{equation} where (again, restricted to Cauchy surfaces) $H$ is the magnetic displacement $1$-form, $D$ is the electric flux $2$-form, and $ _\mu $ and $ * _\epsilon $ are respectively the magnetic permeability and electric permittivity. Finally, for systems with free sources, there is a {\em source 3-form} $\mathcal{J}$, satisfying the continuity of charge condition $ d \mathcal{J} = 0 $. In terms of coordinates, this can be split into \begin{equation} \mathcal{J} = J \wedge d t - \rho , \end{equation} where $J$ is the current density $2$-form and $\rho$ is the charge density $3$-form on Cauchy surfaces. \paragraph{Maxwell's Equations.} With the spacetime forms and operators defined above, Maxwell's equations become \begin{equation} d F = 0, \qquad d {F} = \mathcal{J} . \end{equation} Note that the first equation follows automatically from $ F = d A $, since taking the exterior derivative of both sides yields $ d F = dd A = 0 $. The second equation is consistent with the continuity of charge condition, since $ d \mathcal{J} = d d {F} = 0 $. \paragraph{Lagrangian Formulation.} Given the electromagnetic potential $1$-form $A$ and source $3$-form $\mathcal{J}$, we can define the Lagrangian density to be the $4$-form \begin{equation} \mathcal{L} = - \frac{1}{2} d A \wedge d A + A \wedge \mathcal{J} , \end{equation} with the associated action functional $ S[A] = \int _X \mathcal{L} $ taken over the spacetime domain $X$. Suppose that $\alpha$ is a variation of $A$, vanishing on the boundary $\partial X $. Varying the action along $\alpha$ yields \begin{equation} \mathbf{d} S[A] \cdot \alpha = \int _X \left( - d \alpha \wedge d A + \alpha \wedge \mathcal{J} \right) = \int _X \alpha \wedge \left( -d {d} A + \mathcal{J} \right) . \end{equation} Hamilton's principle of stationary action states that this variation must equal zero for any such $\alpha$, implying the Euler-Lagrange equations $d{d}A = \mathcal{J} $. Finally, substituting $ F = d A $ and recalling that $ d F = d d A = 0 $, we see that this is equivalent to Maxwell's equations. \psection{Geometric Properties of Maxwell's Equations} As written in terms of $F$ above, Maxwell's equations have 8 components: 6 dynamical equations, which describe how the fields change in time, and 2 ``divergence constraints'' containing only spatial derivatives. The fact that these constraints are automatically preserved by the dynamical equations (and can therefore effectively be ignored except at the initial time) comes directly from the differential gauge symmetry and Lagrangian variational structure. We discuss these geometric properties here, with a view towards developing numerical methods that preserve them. \paragraph{Reduction by Gauge Fixing.} Maxwell's equations are invariant under gauge transformations $ A \mapsto A + d f $ for any scalar function $f$, since taking the exterior derivative maps $ F \mapsto F + d d f = F $. Therefore, given a time coordinate $t$, we can fix the gauge so that $ A \cdot \frac{\partial}{\partial t} = 0 $, i.e., $A$ has only spacelike components. This partial gauge fixing is known as the {\em Weyl gauge}. Restricted to this subspace of potentials, the Lagrangian then becomes \begin{align} \mathcal{L} &= -\frac{1}{2} \left( d _t A + d _\Sigma A \right) \wedge \left( d _t A + d _\Sigma A \right) + A \wedge \mathcal{J} \\ &= -\frac{1}{2} \left( d _t A \wedge * d _t A + d _\Sigma A \wedge * d _\Sigma A \right) + A \wedge J \wedge d t \end{align} Here, we have adopted the notation $ d _t $ and $ d _\Sigma $ for the exterior derivative taken only in time and in space, respectively; in particular, we then have $ d _t A = E \wedge d t $ and $ d _\Sigma A = B $. Next, varying the action along a restricted variation $\alpha$ that vanishes on $ \partial X $, \begin{align} \mathbf{d} S[A] \cdot \alpha &= \int _X \left( d _t \alpha \wedge D - d _\Sigma \alpha \wedge H \wedge d t + \alpha \wedge J \wedge dt \right) \label{eqn:weylaction}\\ &= \int _X \alpha \wedge \left( d _t D - d _\Sigma H \wedge d t + J \wedge d t \right) .\nonumber \end{align} Setting this equal to zero by Hamilton's principle, one immediately gets Amp\`ere's law as the sole Euler-Lagrange equation. The divergence constraint $ d _\Sigma {D} = \rho $, corresponding to Gauss' law, has been eliminated via the restriction to the Weyl gauge. \paragraph{Noether's Theorem Automatically Preserves Gauss' Law.} There are two ways that one can see why Gauss' law is automatically preserved, even though it has been eliminated from the Euler-Lagrange equations. The first is to take the ``divergence'' $ d _\Sigma $ of Amp\`ere's law, obtaining \begin{equation} 0 = d _\Sigma d _t D - d _\Sigma d _\Sigma H \wedge d t + d _\Sigma J \wedge d t = d _t \left( d _\Sigma D - \rho \right) . \end{equation} Therefore, if this condition holds at the initial time, then it holds for all time. A more ``geometric'' way to obtain this result is to use Noether's theorem, with respect to the remaining gauge symmetry $ A \mapsto A + d _\Sigma f $ for scalar functions $f$ on $\Sigma$. To derive this, let us restrict $A$ to be an Euler-Lagrange solution in the Weyl gauge, but remove the previous requirement that variations $\alpha$ be fixed at the initial time $ t _0 $ and final time $ t _f $. Then, varying the action along this new $\alpha$, the Euler-Lagrange term disappears, but we now pick up an additional boundary term due to integration by parts \begin{equation} \mathbf{d} S [A] \cdot \alpha = \left. \int _\Sigma \alpha \wedge {D} \,\right\| _{t _0 } ^{ t _f }. \end{equation} If we vary along a gauge transformation $ \alpha = d _\Sigma f $, then this becomes \begin{equation} \mathbf{d} S [A] \cdot d _\Sigma f = \left. \int _\Sigma d _\Sigma f \wedge {D} \,\right\| _{t _0 } ^{ t _f } = - \left. \int _\Sigma f \wedge d _\Sigma {D} \,\right\| _{t _0 } ^{ t _f } \end{equation} Alternatively, plugging $ \alpha = d _\Sigma f $ into~\autoref{eqn:weylaction}, we get \begin{equation} \mathbf{d} S [A] \cdot d _\Sigma f = \int _X d _\Sigma f \wedge J \wedge d t = - \int _X f \wedge d _\Sigma J \wedge d t = - \int _X f \wedge d _t \rho = - \left. \int _\Sigma f \wedge \rho \,\right\| _{ t _0 } ^{ t _f } . \end{equation} Since these two expressions are equal, and $f$ is an arbitrary function, it follows that \begin{equation} \left. \left( d _\Sigma D - \rho \right) \right\| _{t_0}^{t_f} = 0. \end{equation} This indicates that $ d _\Sigma D - \rho $ is a conserved quantity, a momentum map, so if Gauss' law holds at the initial time, then it holds for all subsequent times as well. \psection{Geometric Discretization of Maxwell's Equations} Discretizing Maxwell's equations, while preserving the geometric properties mentioned above, can be achieved using \emph{cochains} as discrete substitutes for differential forms, as previously done in, e.g.,~\citet{Bossavit1998}. Therefore, to compute Maxwell's equations, we begin by discretizing the $2$-form $F$ on a spacetime mesh $K$: $F$ assigns a real value to each oriented $2$-face of the mesh. The exterior derivative $d$ is discretized by the coboundary operator, so the equation $ d F = 0 $ states that $ \left\langle d F , \sigma ^3 \right\rangle = \left\langle F, \partial \sigma ^3 \right\rangle = 0 $, where $ \sigma ^3 $ is any oriented $3$-cell in $K$ and $ \partial \sigma ^3 $ is its $2$-chain boundary. Next, given a discrete Hodge star operator~\cite{BoKe1999,TaKeBo1999,Hiptmair2001,AuKu2006}, $ {F} $ is a $2$-form on the dual mesh $ K $, while $\mathcal{J}$ is defined as a discrete dual $3$-form. Then, for every dual $3$-cochain $ \sigma ^1 $ (where $ \sigma ^1 $ is the corresponding primal edge), the equation $ d {F} = \mathcal{J} $ becomes \begin{equation} \left\langle d {* F} , * \sigma ^1 \right\rangle = \left\langle F, \partial { \sigma ^1} \right\rangle = \left\langle \mathcal{J} , * \sigma ^1 \right\rangle . \end{equation} When the cells $ \sigma ^3 $ and $ \sigma ^1 $ are spacetimelike, then these correspond to the dynamical components of Maxwell's equations, and can be used to compute subsequent values of $F$. When the cells are purely spacelike, they correspond to the divergence constraint equations. The exact expression and update of these fields in time now depends on which type of mesh and time stepping method is desired, as described next. \paragraph{Uniform Time Stepping.} For uniform rectangular meshes aligned with the $ (x,y,z,t) $ axes, we can emulate the smooth coordinate expression of $F$ as \begin{align} F &= E _x \,dx \wedge d t + E _y \,dy \wedge d t + E _z \,dz \wedge d t \\ & \quad + B _x \,dy \wedge d z + B _y \,dz \wedge d x + B _z \,dx \wedge d y. \end{align} This suggests the following discretization of $F$, shown in~\autoref{fig:setupEB}: store $ E _x \Delta x \Delta t $ on the $ x t $-faces, $ E _y \Delta y \Delta t $ on the $ y t $-faces, and $ E _z \Delta z \Delta t $ on the $ z t $-faces; likewise for $B$. To discretize $\mathcal{J}$, we can similarly store $ J _x \Delta y \Delta z \Delta t $, $ J _y \Delta z \Delta x \Delta t $, $ J _z \Delta x \Delta y \Delta t $, and $ \rho \Delta x \Delta y \Delta z $ on the corresponding dual $3$-cells. \begin{figure}[ht] \centerline{\includegraphics[width=0.9\linewidth]{figs/setupEB-4inarow.pdf}} \caption{The $2$-form $F = E \wedge d t + B $ can be discretized, on a rectangular spacetime mesh, by storing the components of $E$ and $B$ on 2D faces. The resulting numerical method is Yee's FDTD scheme.} \label{fig:setupEB} \end{figure} If we then enforce the equations $ d F = 0 $ and $ d {F} = \mathcal{J} $ in the discrete sense, the result is precisely the finite-difference time-domain (FDTD) integration scheme of~\citet{Yee1966}. A similar procedure can be applied on unstructured (e.g., simplicial) spatial grids, on which we take uniform time steps $ \Delta t $ (creating prism-shaped spacetime primal elements); in this case, solving the discrete Maxwell's equations recovers the more recent ``Yee-like'' method of~\citet{BoKe2000}. \paragraph{Asynchronous Time Stepping.} As initially mentioned in~\citet{StToDeMa2007}, a more flexible integration scheme can be designed by assigning a different time step per element, in order to focus computational power where needed. A visualization of how to store $F$ on such a mesh structure is shown in~\autoref{fig:setupAVI}. This defines an {\em asynchronous variational integrator} that preserves the numerical properties of the uniform-stepping methods outlined above. The procedure to repeatedly update $E$ and $B$ asynchronously in time is as follows: \begin{enumerate} \item Select the face for which $B$ needs to be updated next. \item $E$ advances $B$, using $ d F = 0 $. \item $B$ advances $E$ on neighboring edges, using $ d { F} = \mathcal{J} $. \end{enumerate} \begin{figure}[ht] \centerline{\includegraphics[width=0.7\linewidth]{figs/setupAVI.pdf}} \caption{Here, $F$ is discretized on an asynchronous time-stepping grid, where each spatial element takes a different-sized time step from its neighbors. This can be done for either a rectangular spatial grid (left) or an unstructured/simplicial spatial mesh (right). \label{fig:setupAVI}} \end{figure} Details of this algorithm, along with initial numerical results, can be found in~\citet{StToDeMa2007}, where only the case $\mathcal{J}=0$ was described. \begin{figure}[ht] \centering \subfigure[regular grid, uniform time steps] { \includegraphics[width=0.45\linewidth]{figs/avi-energy-uniform-steps.pdf} \label{fig:avi-energy-uniform-steps} } \subfigure[random grid, asynchronous time steps] { \includegraphics[width=0.45\linewidth]{figs/avi-energy-random-steps.pdf} \label{fig:avi-energy-random-steps} } \caption{Our geometric integrator robustly maintains near-conservation of energy, even for asynchronous time stepping on a random spatial grid.} \label{fig:avi-energy} \end{figure} One promising result from the initial numerical experiments, shown in~\autoref{fig:avi-energy}, is that this asynchronous integrator matches the FDTD scheme's excellent energy conservation behavior, even on a highly irregular grid, without exhibiting artificial damping or forcing. Future work is expected to include formal accuracy and stability analysis of this asynchronous integrator, focusing both on the choice of Hodge star and on resonance stability criteria for selecting the individual time steps. \ack Our research was partially supported by a Betty and Gordon Moore fellowship at Caltech, NSF grants CCR-0133983 and DMS-0453145 and DOE contract DE-FG02-04ER25657, and by NSF grant CCF-0528101. We gratefully acknowledge these sponsors for their support of this work.	29,846
TA: Ethereum Consolidates, What Could Spark A Fresh Rally Ethereum is consolidating gains near the $3,580 resistance zone against the US Dollar. ETH price could start a fresh rally if it clears the $600 level. - Ethereum remained in a positive zone above the $3,500 and $3,550 resistance levels. - The price is now trading above $3,520 and the 100 hourly simple moving average. - There is a key breakout pattern forming with resistance near $3,580 on the hourly chart of ETH/USD (data feed via Kraken). - The pair could start another increase if it clears the $580 and $600 levels in the near term. Ethereum was able to settle above the main $3,500 resistance zone. ETH even extended its rise above the $3,600 level and settled well above the 100 hourly simple moving average. The price traded as high as $3,674 before there was a downside correction. Ether declined below the $3,600 and $3,550 support levels. However, downsides were limited below $3,500. A low was formed near $3,485 and the price is now rising. It cleared the $3,550 resistance level, and it is now trading above $3,520 and the 100 hourly simple moving average. The price is now testing the 50% Fib retracement level of the recent decline from the $3,674 high to $3,485 low. An immediate resistance on the upside is near the $3,580 level. There is also a key breakout pattern forming with resistance near $3,580 on the hourly chart of ETH/USD.Source: ETHUSD on TradingView.com The main resistance is near the $3,600. It is close to the 61.8% Fib retracement level of the recent decline from the $3,674 high to $3,485 low. A clear break and close above the $3,600 level could start another increase. The next major resistance sits near $3,675. Any more gains could set the pace for a move towards the $3,750 level.Dips Limited in ETH? If ethereum fails to continue higher above the $3,580 and $3,600 resistance levels, it could start another downside correction. An initial support on the downside is near the $3,540 level. The next major support seems to be forming near the $3,520 level and the triangle lower trend line. A downside break below the triangle support could lead ether towards the $3,450 support zone in the near term. Technical Indicators Hourly MACD – The MACD for ETH/USD is slowly gaining pace in the bullish zone. Hourly RSI – The RSI for ETH/USD is still above the 50 level. Major Support Level – $3,520 Major Resistance Level – $3,600	331,595
I Respect Your Email Privacy United States Bankruptcy Court Northern District of Illinois Eastern Division Transmittal Sheet for Opinions Bankruptcy No. 04 B 11045 Adversary No. 04 A 04496 Date of Issuance: October 28, 2005 Judge: Jack B. Schmetterer Appearance of Counsel: Attorney for Movant or Plaintiff: Cameron M. Gulden, Office of U.S. Trustee Attorney for Respondent or Defendant: Jack D. and Lymor K. Wasserman Trustee or Other Attorneys: UNITED STATES BANKRUPTCY COURT NORTHERN DISTRICT OF ILLINOIS EASTERN DIVISION IN RE: ) ) CHAPTER 7 PROCEEDING JACK AND LYMOR WASSERMAN, ) DEBTORS. ) ____________________________________) CASE NO. 04 B 11045 ) IRA BODENSTEIN, ) UNITED STATES TRUSTEE ) ADVERSARY NO. 04 A 04496 PLAINTIFF, ) ) vs. ) ) JACK AND LYMOR WASSERMAN, ) DEFENDANTS. ) ____________________________________) This Adversary Proceeding relates to the Chapter 7 bankruptcy case of Jack and Lymor Wasserman (“Debtors” or “Defendants”). The United States Trustee (“U.S. Trustee”) filed this Adversary Complaint objecting to Defendants/Debtors’ discharge under 11 U.S.C. § 727. The U.S. Trustee contends that Debtors failed to maintain and produce necessary documents and records regarding their income and unsecured claims of at least $313,222.26. He also argues that the Debtors have failed to explain their loss and deficiency of assets. The Debtors contend that all of their business records were seized and disposed on two separate occasions by different landlords, and for that reason they are unable to produce records or account for assets after filing in bankruptcy. Following trial the Court now makes and enters the following Findings of Fact and Conclusions of Law, pursuant to which judgment will separately enter denying each Debtor a bankruptcy discharge under 11 U.S.C. §§ 727(a)(3) and (a)(5). of all monthly bank statements, check registers, and ledgers, beginning with March 1, 2002, through the date of filing. 7. However, the only documents provided were: 1The U.S. Trustee served the discovery request via U.S. Mail, First Class on Debtors former counsel on May 13, 2005. (Trial Tr., 94, Aug. 15, 2005). The Debtors were sent a letter dated July 5, 2005 reminding them of the Request to Admit. (Trial Tr., 96, Aug. 15, 2005). Request to Admit was therefore deemed admitted for the purposes of this Adversary Proceeding. Certain admissions have been supplemented to reflect evidence presented at trial, and the further Findings below come from the Requests thereby admitted and trial evidence. 13. The Debtors have never filed a business tax return of any kind for Full Cup. 22. The Debtors have never filed a business tax return of any kind for Te’Avone. 23. The Debtors have not provided documentation showing what was purchased for the $59,294.73 in credit card charges listed on the Debtors’ Schedule F. CONCLUSIONS OF LAW Jurisdiction Jurisdiction lies under 28 U.S.C. § 1334 and under District Court’s Internal Operating Procedure 15(a). Determination on an objection to discharge is a core proceeding under 28 U.S.C. § 157 (b)(2)(J). Venue is proper under 28 U.S.C. § 1409. The U.S. Trustee has the burden of proof and must prove every element of his objections to discharge by a preponderance of evidence. In re Scott, 172 F.3d 959, 966-67 (7th Cir. 1999); In re Martin, 141 B.R. 986, 992 (Bankr. N.D. Ill. 1992). A Chapter 7 discharge “recognizes the Congressional intent of providing the bankrupt with a ‘fresh start,’” [and] 11 U.S.C. § 727(a) of the Bankruptcy Code must be construed strictly against the objecting [party] and liberally in favor of the debtor.” In re Costello, 299 B.R. 882, 894 (Bankr. N.D. Ill. 2003). However, violation by the debtor of any of the provisions of 11 U.S.C. § 727 completely bars a debtor’s discharge. Id. Section 727(a)(3) provides that a debtor shall be granted a discharge unless the debtor fails to keep or preserve books and records from which the debtor’s financial condition may be ascertained. 11 U.S.C. § 727(a)(3). The purpose of § 727(a)(3) is to make “a privilege of discharge dependent on a true presentation of the debtor’s financial affairs.’” In re Self, 325 B.R. 224, 240 (Bankr. N.D. Ill. 2005) (citing In re Scott, 172 F.3d 959, 969 (7th Cir. 1999)). The language of § 727(a)(3) “places an affirmative duty on the debtor to create books and records accurately documenting his financial affairs.” In re Self, 325 B.R. at 240; In re Hansen, 325 B.R. 746, 761 (Bankr. N.D. Ill. 2005) (noting debtors’ “obligation to maintain, preserve, and produce records of their financial affairs”); In re Costello, 299 B.R. at 897 (noting “affirmative duty to create books and records” imposed on debtors by § 727(a)(3)). In fact, “[s]ubstantially accurate and complete records of a debtor’s financial affairs is a prerequisite to discharge in bankruptcy.” In re Martin, 141 B.R. at 995. The party alleging that a debtor has breached this affirmative duty “has the initial burden of proving that the books and records are inadequate,” and thereafter “the burden of production shifts to the debtor to justify the lack of adequate records under the particular circumstances.” In re Costello, 299 B.R. at 897; In re Self, 325 B.R. at 241. Section 727(a)(3) “insures that creditors receive adequate information so that they can ‘ascertain the debtor’s financial condition and track his financial dealings with substantial completeness and accuracy for a reasonable period past to present.’” In re Costello, 299 B.R. at 897 (citing In re Juzwiak, 89 F.3d 424, 427 ( 7th Cir. 1996) (internal citations omitted)). “The determination of what constitutes a reasonable period prior to filing must be measured on a case-by-case basis, taking into account all of the circumstances of the case.” In re Self, 325 B.R. at 241. Whether a debtor’s books and records are complete and accurate is to be determined on a case-by-case basis, considering the size and complexity of the debtor’s financial situation, as well as “the sophistication of the debtor, his educational background, his business experience and acumen, and his personal financial structure.” In re Self, 325 B.R. at 241; In re Costello, 299 B.R.at 897. Section 727(a)(3) “does not merely require that the debtor not lose any records; it mandates a denial of discharge for the ‘failure to act,” unless such failure to act is justifiable.” In re Self, 325 B.R. at 241. Mere testimony from the debtors regarding their version of their financial transactions is insufficient to satisfy the duty imposed by § 727(a)(3), and neither the U.S. Trustee nor the court is required to “reconstruct a debtor’s financial situation or take a debtor’s word for his financial dealings.” In re Hansen, 325 B.R. at 761; In re Carlson, 231 B.R. 640, 655 (Bankr. N.D. Ill. 1999) (holding that “[c]reditors and the Trustee are not required to accept Debtor’s oral recitations or recollections of his transactions; rather, to qualify for a discharge in bankruptcy, a debtor is required to keep and produce written documentation of all such transactions.”). The Debtors, one with extensive business experience and one with an accounting degree and accounting experience, have failed to produce books and records from which their financial condition may be ascertained. They have not produced any documentation regarding the disposition of the $210,000 they received in starting Full Cup and nothing regarding the income and expenses of the business. Likewise, the Debtors have provided no documentation regarding the $164,000 they received in starting Talk of the Town and nothing regarding the income and expenses of the business. The Debtors have also provided very little documentation regarding the income and expenses of Te’Avone. Despite the fact that much of the Debtors’ $313,222.26 in acknowledged unsecured debt2 appears to relate to these three failed business, their books and records are not only insufficient but are mostly non-existent. The Debtors list $59,294.73 in credit card debt, but have provided insufficient credit card statements explaining the charges for that debt. They have provided only minimal bank statements and no check registers. In addition, neither of the Debtors have filed any tax returns for the past six years (Trial Tr., 72, Aug. 15, 2005), making it impossible to determine even the most elemental aspects of their financial condition. The Debtors’ debt came from a variety of sources. Mrs. Wasserman testified that of the total unsecured debt, less than 1% is from Full Cup Enterprises, 26% is from Talk of the Town, 2The Debtors also listed 49 creditors with “unknown” amounts owed to these creditors, making it likely that the Debtors’ actual unsecured debt is higher than the $313,222.26 listed. 28% is from Te’Avone, 28% is her personal debt, and Mr. Wasserman’s personal debt is 7%. (Trial Tr., 111, Aug. 16, 2005). The scheduled debt comes from different sources, and the Debtors are not necessarily jointly liable for every debt. However, the Debtors jointly wish to discharge over $313,000 of unsecured debt while providing wholly inadequate documentation regarding their books and records. The Debtors have preserved and produced few records. They did supply the court and the U.S. Trustee with random tidbits of their records, but to the extent of the documents that were provided, it is clear that they are in violation of § 727(a)(3). The disorder and nature of the records produced did not allow for a meaningful reconstruction of the Debtors’ financial transactions. Indeed, the documents produced by Debtors did not reconstruct anything with substantial completeness; rather they are snapshots of small parts of the Debtors financial history. Section 727(a)(3) puts an affirmative duty on the Debtors to produce books and records that accurately document their financial affairs. They should not be granted a discharge because, in this case, their failure to do so was not justifiable. “[C]ourts and creditors should not be required to speculate as to the financial history or condition of the debtor, nor should they be compelled to reconstruct the debtor’s affairs.” In re Juzwiak, 89 F.3d at 428 (citations omitted). The Debtors have not offered an explanation which justifies their lack of adequate books and records under the particular circumstances of this case. As to the failure to provide any documentation whatsoever regarding the startup funds and income they received from their businesses, the Debtors’ Answer to the Complaint asserted only that all of their financial records were seized and destroyed by different landlords. However, they failed to provide any evidence to corroborate their stories about two previous evictions in which the substantial part of their books and records were supposedly seized and destroyed. Under these circumstances, such explanations do not constitute an appropriate justification for their failure to keep adequate records. Even if the asserted seizures by landlords were true, the Debtors are not excused from making a good-faith effort to reconstruct those records, which they simply have not done. Accordingly, pursuant to 11 U.S.C. § 727(a)(3) the U.S. Trustee has proven by a preponderance of the evidence that the Debtors have each failed to keep or preserve books and records from which the their financial condition may be ascertained, and the Debtors’ discharges are therefore denied on that basis. Section 727(a)(5) provides that a discharge will be granted unless the debtor has failed to satisfactorily explain any loss of assets or deficiency of assets to meet the debtor’s liabilities. 11 U.S.C. § 727(a)(5). By penalizing a debtor who is insufficiently forthcoming about what happened to his assets, § 727(a)(5) is one of several Code provisions meant to “relieve [ ] creditors and courts of the full burden of reconstructing the debtor’s financial history and condition, placing it instead upon the debtor.” In re Olbur, 314 B.R. 732, 740 (Bankr. N.D. Ill. 2004) (quoting In re Hermanson, 273 B.R. 538, 545 (Bankr. N.D. Ill. 2002)). Under § 727(a)(5), the bankruptcy judge has “broad power to decline to grant a discharge . . . where the debtor does not adequately explain a shortage, loss, or disappearance of assets.” In re Olbur, 314 B.R. at 740 (quoting In re D’Agnese, 86 F.3d 732, 734 (7th Cir. 1996)); In re Martin, 698 F.2d at 886. Proof under § 727(a)(5) comes in two stages. Initially, the objecting party bears the burden of showing that the debtor “at one time owned substantial and identifiable assets that are no longer available to his creditors.” In re Olbur, 314 B.R. at 740 (quoting In re Bostrom, 286 B.R. 352, 359 (Bankr. N.D. Ill. 2002) and In re Hermanson, 273 B.R. at 545). If this showing is made, the burden then falls on the debtor to offer a “satisfactory explanation” for the unavailability of those assets. In re Olbur, 314 B.R. at 740-41 (quoting In re Hermanson, 273 B.R. at 545 and In re Bostrom, 286 B.R. at 364). Whether an explanation is satisfactory or not rests in the discretion of the court. Id. at 741 (citing In re Bostrom, 286 B.R. at 364). “The debtor’s account need not be ‘far-reaching and comprehensive,’ but it must amount to more than a ‘vague, indefinite, and uncorroborated hodgepodge of financial transactions.’” Id. (quoting In re Costello, 299 B.R. at 901). The explanation must meet two criteria in order to be deemed “satisfactory.” In re Mantra, 314 B.R. 723, 730 (Bankr. N.D. Ill. 2004) (citing In re Bryson, 187 B.R. 939, 955 (Bankr. N.D. Ill. 1995)). First, it must be supported by documentation. In re Costello, 299 B.R. at 901. Second, this documentation must be sufficient to “eliminate the need for the Court to speculate as to what happened to all the assets.” Id. (citing In re Bostrom, 286 B.R. at 364-65); see also In re Hansen, 325 B.R. at 763 (noting the intertwined relationship between the documentation required by § 727(a)(3) and the adequacy of the explanation under § 727(a)(5)). The debtor must explain in good faith “what really happened to the assets in question.” In re Olbur, 314 B.R. at 740 (quoting In re Bailey, 145 B.R. 919, 925 (Bankr. N.D. Ill. 1992)). A debtor “cannot abuse the bankruptcy process by obfuscating the true nature of his affairs and then refusing to provide a credible explanation.” In re Mantra, 314 B.R. at 730 (citing In re Johnson, 98 B.R. at 366). Even though a satisfactory explanation must be convincing about lack of records or assets, the focus of the inquiry is not exclusively on the subjective nature or honesty of the debtor’s explanation, but is also on the objective adequacy of such explanation. Id. (citing In re D’Agnese, 86 F.3d at 734-35). A debtor’s failure to satisfactorily justify a substantial loss of assets, however, need not be the product of fraudulent intent. In re Mantra, 314 B.R. at 730 (citing In re Gannon, 173 B.R. 313, 317 (Bankr. S.D.N.Y. 1994)). In this case, the U.S. Trustee has met his burden of showing that the Debtors at one time owned substantial and identifiable assets that are no longer available to their creditors. Specifically, the U.S. Trustee established that the Debtors owned the following identifiable assets: The U.S. Trustee has also met his burden of showing that these substantial and identifiable assets are no longer available to the Debtors’ creditors. Indeed, the Schedules admit, under penalty of perjury, that the Debtors own no real property and only $6,550 in personal property. The Debtors failed to meet their burden to offer a satisfactory explanation for the unavailability of the substantial and identifiable assets. The Debtors produced only a small assortment of documentation, none of which explained what happened to the significant assets mentioned above. Although the Debtors stated that the funds were spent on their various businesses, nothing corroborates this explanation. Records were not produced to document how the Debtors’ spent the substantial funds they received in starting Full Cup or Talk of the Town. There are no ledgers, no invoices, no receipts, few bank statements, and no tax returns or business computer records showing how a total of at least $374,000 was spent and is no longer available for creditors. The Debtors’ vague, indefinite and uncorroborated explanation can only allow the Court to speculate as to what actually happened to the Debtors’ assets, speculation which § 727(a)(5) does not allow or require. Based on the Debtors’ testimony, on two completely separate occasions and in two different buildings with two different landlords they were supposedly evicted, and those two different landlords allegedly destroyed all of their documents without their permission. Even the computer back-ups that the Debtors allegedly created were also seized and destroyed on the two different occasions, they say. (Trial Tr., 47, Aug. 16, 2005). No corroborating witnesses or documents were presented. But even if the Debtors were truthful, they failed to reconstruct their records. With the minimum documents provided, it is not possible to trace the money spent without resorting to speculation, and no way to determine where the money was spent. Regarding Talk of the Town, Mrs. Wasserman testified that to start the business, people were hired to tear down a wall and install flooring. (Trial Tr., 38, Aug. 16, 2005). In addition, Mrs. Wasserman testified that she leased and bought some restaurant equipment. (Trial Tr., 38, Aug. 16, 2005). While she answered in the affirmative that she specifically knew who she paid to tear down the wall, and lease equipment from, she did not subpoena anyone that testified to that effect. (Trial Tr., 40, Aug. 16, 2005). Despite the fact that Mrs. Wasserman testified that she was aware of the names of certain vendors, knew the bank who held certain funds, and knew the lender of loans used to start the business, no witnesses were subpoenaed or produced to corroborate her testimony. During closing arguments, the Debtors expressed their dissatisfaction with their former counsel. Mrs. Wasserman stated that their former counsel did not feel the necessity to subpoena people that could testify to receiving money from the Debtors, subpoena the landlords, or obtain copies of the eviction records. (Trial Tr., 40, Aug. 16, 2005). Mr. Wasserman also expressed his disappointment and anger with their former counsel. (Trial Tr., 52, Aug. 16, 2005). It is possible that with the aid of skilled legal help the Debtors might have been able to produce additional records and proof, but one cannot speculate as to what might have been. Furthermore, even if some such records were produced, the Debtors were still unable to trace their expenditures. Despite having extensive business experience, education, and running three businesses and taking in tens of thousands of dollars each month, they have provided only a small assortment of documents covering only select random periods of time that prevent adequate tracing of their assets. Under such conditions, Debtors are not entitled to a discharge. The Debtors have emphasized the fact that they have not been accused of fraud or intent to deceive. As Mr. Wasserman stated in his closing argument, “And nobody, nobody here has said anything about us committing fraud. Nobody has said that we burned documents; that we threw away documents; that we withheld documents” (Trial Tr., 54-55, Aug. 16, 2005). In this case, there were no allegations by the U.S. Trustee that the Debtors acted with the intent to deceive or committed fraud in connection with their bankruptcy filing. However, the intent to deceive is not a requisite element for denying a discharge under § 727(a)(3) and § 727(a)(5). 11 U.S.C. § 727(a)(3); 11 U.S.C. § 727(a)(5); In re Juzwiak, 89 F.3d at 430 (“Although the ultimate goal of § 727(a)(3) is to distinguish between the honest debtor, who should be granted the privilege of discharge, and the abusive or unworthy debtor, who does not deserve such a benefit (citations omitted) creditors do not need to prove that the debtor intended to defraud them in order to demonstrate a § 727(a)(3) violation.”). Accordingly, pursuant to 11 U.S.C. § 727(a)(3) and 727(a)(5) the U.S. Trustee has proven by a preponderance of evidence that Debtors and each of them have failed to explain loss or deficiency of assets to meet their liabilities and lack of adequate records. The Debtors’ discharges must therefore be denied. Judgment will separately enter denying both Debtors a bankruptcy discharge. ENTER: Jack B. Schmetterer United States Bankruptcy Judge Entered this 28th day of October 2005 Back to Bankruptcy Articles	282,267
>> Private Banks Soon to Get Shot at Bailout Money Topics:Economy (U.S.) \| Economy (Global) \| Banking Sectors:Financial Services \| Banks The U.S. Treasury Department said Friday it will soon allow privately held banks to apply for government capital injections under a $250 billion bank rescue program. The Treasury said it will post application forms and term sheets for private banks, which are generally smaller institutions. This month, it began injecting capital in larger publicly traded institutions by buying senior preferred shares. It said it will set a "reasonable" deadline for private institutions that will be after the Nov. 14 deadline for publicly traded institutions. MORE FROM CNBC TOPIC : Economy (U.S.) TOPIC : Economy (Global) SECTOR : Financial Services	399,936
Heath Ledger Gallery Back to Heath Ledger Source: Heath Ledger on Freebase , licensed under CC-BY Other content from Wikipedia, licensed under the GFDL Other content from Wikipedia, licensed under the GFDL Heath Ledger is ranked on: - #369of 973 Hottest CelebritiesCrowdRanked Icon CrowdRanked List views: 721388 May 18 2013 This list of hot celebrities is ranked by pop culture junkies worldwide, making it the best place to find...more - #39of 191 The Best Actors in Film HistoryCrowdRanked Icon CrowdRanked List views: 24470 May 18 2013 The best actors ever are ranked here in this list of the best actors in film history. These are the great...more - #33of 348 The Greatest Film Actors & Actresses of All TimeCrowdRanked Icon CrowdRanked List views: 78299 May 15 2013 These are the best actors and actresses ever to grace the silver screen, ranked by film fans worldwide. T...more around the web	314,528
Donate Why should I donate? The programs you find on this website have been developed in my spare time. All programs are released as freeware. Thus using my programs is free of charge and there are no functional restrictions at all. You can show your appreciation for oddgravity and support future development by donating! Thank you very much for your support and generosity! How can I donate? You can donate via Bitcoin, Litecoin or via PayPal. I much prefer receiving donations via Bitcoin or Litecoin because PayPal charges rather high fees. Don't have a PayPal account? Sign up for a PayPal account which is completely free! Last 20 donors The following list shows the last 20 generous donors. You must donate at least 1,00 USD to be listed here. - Class On Glass Designs - Tracy Kreckman - Entcho Anguelov - S Stuart Morrison - Robert Burling - Ricky Townsend - Peter Sharkey - Maurice Read - wgo@juno.com - Matthias Hilberer - Robert Coffin - Steve Shergold - Boris Scharinger - Jaime Baraqui - Joyce Ohta - William Garcia - Petr Jochec - William Krauss - Darío Calvo Sánchez - Yves Devillers Mail me if you don't want to be listed here or if you want the entry to be modified. Keep on donating! Your ongoing donations keep this private site running smoothly!	397,576
I begin looking over my recipe and thinking about how this will be good for my daughter's skin. But, wait, what did I read about chamomile being good for eczema? It's supposed to have calming properties. Well, I do have some chamomile tea and I did read where instead of water you could add tea to the lye (or lye to the tea - safety first). Why don't I just do that? So, I made very strong chamomile tea. While I wait for the tea to finish brewing, I get my mold ready, get all my ingredients together and then my mind begins to review what I have seen in the many videos I've watched. I saw an interesting segment on about.com showing how to add goats milk into the soap. Well, goats milk is good for the skin! I should do that! Since I had some evaporated goats milk on hand I was set! All I had to do was use half the water(or tea in this case)with the lye and mix equal parts goats milk and the other half of the water to add later. This would make my lye solution stronger, but, hey, I can handle it! Reviewing my recipe and everything I had learned in my mind I remember that the lye couldn't be hotter than the oils....or was it the other way around? Up the stairs I go to the computer to read. If you haven't found this website yet it is a wonderful resource! Millers Homemade Soap Pages. This has pages and pages of information regarding sources, recipes, how tos, trouble shooting, etc. (No, I did not have this resource before I went on the hunt for lye - see chapter 1). This is where I found the information. Okay, the oils can't be hotter than the lye. Good! Off downstairs to check my lye temp. 145 degrees. Sheesh! I go sit and watch tv for 10 minutes or so......check temp again. 140~ oh boy. Now, keep in mind it is recommended that the temp be between 100 or 110 degrees. This is going to take forever! I check my temp on the oils and they were about 110 so we were good there, but waiting on the lye. boy that olive oil stinks! Off to watch tv again...... well about an hour and a half later the lye is cool enough. It is time to mix. I, once again don my gloves and safety glasses, grab my stick blender and get ready to pour the lye. (I had to slightly heat the oils again b/c they were too cool. Everything now is about 110 degrees.) First I pour in the goats milk. Interesting combination oil and goats milk. All separate as you can image. No worries, I pour in the lye solution and begin mixing. Now this is beginning to be a weird orange color. Really weird orange. But, maybe it'll be good once it gets to trace. I mix and stir, mix and stir (alternating running the stick blender on low and stirring with it). It takes about 15 minutes to get to trace (olive oil soaps are notoriously slow to trace - so I've been told). But, I finally got to trace. It is a thin trace, not what I have seen on the videos (I realized that AFTER when I re-watched the videos) but it'll do. Now, I'm looking for a pure soap so I didn't plan on adding any scent. I don't know how my daughter's skin will react so it's best just to leave it out. I do, however, add some dried chamomile tea to the mix, which adds pretty little specs in the soap. ~I wish I could add a scent, because this really is a little smelly!~ And, what's with the orange color ? Did the chamomile do that? I pour the mixture into my molds which, since I'm not that concerned with appearance, just quality of soap, are plastic glad containers. ~Man, what an awful smell, though~. I keep going anyway and set them on a towel and then wrap them up in about four towels. They have to be kept warm in order to set up right. A couple of hours later I go to peek at how they are doing I am so happy they are getting to a nice gel stage. Apparently the reaction is still going on and it heats all up again and liquefies somewhat jello like and then re-solidifies. "saponification" class, now everyone say it with me. Anyway, it needs to stay warm so I cover it up again. Later I have to peek again because, hey, I'm nosy that way. Well, the tops are all wrinkly! Hmm, I've seen that before on the trouble shooting page of the miller website. So I cover it all back up and run upstairs again to read the website. "Too much insulation during gel phase". Oh boy! I run downstairs and take off the towels. I had thought, "if one towel is good, four must be better". Well no... but no major damage was done, I'm just wanting a quality soap. But, boy, what is with the smell??? and the weird orange color? 2 comments: Looks like you did a great job. I've never made soap..looks like fun to try. anji Mia--I've only made 3 batches of soap so far (I love it!), but I've read that goat's milk can turn orange with the heat from the lye. Do some more computer research about goat's milk soaps and read what others do to try to keep it from turning orange. I don't think it hurts the soap other than the color is less desirable. As for the smell, I'm not so sure. What kind of olive oil did you use? Was it extra virgin? If so, that might be the smell. Keep after the soap making! I love how less drying it is on my skin. My mother says it's less "scummy" than store bought "soaps." Good luck!	101,360
TITLE: Minimum alpha for the following inequality QUESTION [0 upvotes]: The following formula is given: $(a+b)^\alpha \leq a^\alpha+b^\alpha$ What is the minimum $0<\alpha<1$ somehow the above inequality is being satisfied for all $a,b > 0$? Intuitively as if $\alpha \rightarrow 0$, the above inequality is satisfied, for the set of solutions of the question, there is no any infimum. Is it correct? Also, I've heard that the above inequality is true based on the above constraints, but I am seeking for a formal proof. REPLY [3 votes]: Define $f(a) = a ^ \alpha + b ^ \alpha - (a+b)^\alpha$. So, $$ f'(a) = \alpha a ^{\alpha - 1} - \alpha (a+b)^{\alpha-1} = \alpha (a ^{\alpha - 1} - (a+b)^{\alpha-1})= \alpha (\frac{1}{a ^{1-\alpha}} - \frac{1}{(a+b)^{1-\alpha}}) > 0$$ so, $f$ is increasing for $a$ and $b$. As $a,b > 0$ and for $a = 0$ and $b = 0$, we have $a ^ \alpha + b ^ \alpha - (a+b)^\alpha = 0$, so the phrase is a non-negative function and the proof is completed.	94,077
. ____________________________________ Keinlife: Hi, Jenée, thanks for doing this! How would you respond to people who claim that we're past significant racial issues and tensions in the U.S. as evidenced by our electing a black president (also up for discussion)? The more practical the response the better. Jenée Desmond-Harris: Hi! Thanks for your question. I'd probably point out that we have one black president and tens of thousands of other people in the country. Focusing on the tens of thousands of others reveals huge and well-documented racial disparities in everything from education to criminal justice to access to health care. If that's not enough proof, tell them to search for the n-word on Twitter and report back on what they find. ____________________________________ Tracylc: Hey! I have a question. So my nephew is 5 and very, very fair-skinned. At some point in a conversation the issue of my nephew being black came up, and my nephew says, "I'm not black, I'm yellow!" presumably because his dad calls him "lil yellow boy" from time to time. I thought briefly about trying to explain race and all that to him in a way that he'd understand, but stopped because I didn't know how, firstly, or if it was time for such things. My question: When do you begin to talk with kids about race, in your opinion? How young is too young? Jenée Desmond-Harris: That's funny. I definitely used to think I was "pink." And I just heard a friend's story about how she thought her (black, fair-skinned) mom was white until she was 10 years old. The kid just learned what colors were a few years ago, so I can definitely see how this could be confusing. How about a focus on ancestry/history (which is the more meaningful part of race) instead? As in "Our family is African-American. That means our ancestors came from Africa. African-American people come in all different colors, including brown and yellow. We use the word 'black,' but that's silly because no one is actually the color black." I think a 5-year-old could start to get that. Jenée Desmond-Harris: OK, have to wrap this up. Let's do it again. In the meantime, send questions to racemanners@theroot.com. Bye! Stay black! Or yellow -- or pink. Like The Root on Facebook. Follow us on Twitter. Get what we're watching delivered to your inbox everyday! Sign up for our daily video newsletter below. Page 1 of 3 NEXT PAGE > Powered by Livefyre 0	65,011
What Einstein Told His Barber: More Scientific Answers to Everyday Questions - Robert L. Wolke (2000) Chapter 4. The Earth Beneath Our Feet There are other planets in the universe, but we have a firm attachment to our very own Mother Earth. It's called gravity.2 Gravity not only limits our golf drives and makes our body parts sag with age, but serves a number of useful functions, not the least of which is keeping the atmosphere from flying off our spinning Earth like spit from a roller coaster. Gravity makes dust settle and hot air rise. It does innumerable big and little jobs for us, such as keeping the moon up and skirts down. It even allows us to make electricity from water. Gravity is ubiquitous. Even astronauts don't leave home without it. This chapter will attempt to tell you how this most wide-ranging of all forces—its effects are felt across the breadth of the universe—operates, even though we can't yet explain what makes it tick, or should I say “stick”? Earth is, of course, spinning at more than 1,000 miles per hour (1,600 kilometers per hour) as it sails around the sun at more than 10,000 miles per hour (16,000 kilometers per hour). And we're not even dizzy (most of us). But in spite of the fact that we are totally oblivious to them (and I'll tell you why), these motions have crucial consequences in our daily lives. They affect hurricanes, ocean currents and ocean tides. They affect—no, they cause— every day, night and season of our lives. In examining the Earth beneath our feet, we'll visit the center of the planet, the North and South Poles, Mount Kilimanjaro in Tanzania, a swirling hurricane and a toilet bowl the size of North America. And finally, lest we overlook the fact that living things constitute a rather important component of our planet, we'll see how we use radiocarbon dating to explore the past lives of plants, animals and humans. A Matter of Some Gravity Why does gravity try to attract all things to the center of Earth? Why to the center? Why not to Mecca, or Disney World? Because. NITPICKER'S CORNER.) A Lotta Hot Air Everybody says that heat rises. But for heaven's sake, why? Why. NITPICKER'S CORNER When hot air rises through the atmosphere, the very act of rising cools it off somewhat. I know this sounds paradoxical, but don't turn the page quite yet. When a warm mass of air rises, it, of course, gains altitude. Masses of air can gain altitude, even if they're not warm, perhaps by drifting up against a mountain and being forced to swoop upward along its slope. Whatever the reason for air masses' moving upward, there must always be equal masses of air moving downward to replace them. The result is that there are rising and falling masses of air all over the world. Let's see what happens to a particular passel of rising air as it gains altitude. At higher altitudes, the atmosphere is thinner. That's because there's less atmosphere above it, so it's not under as much compression by gravity. (Gravity pulls air down just as it does everything else; air may be light, but it still has weight.) In other words, at higher altitudes there is less pressure from the atmosphere, and that allows our rising passel of air to expand. But in order to expand, the passel's molecules have to elbow aside the air molecules that are already occupying that space. And that uses up some of the passel's own energy. What kind of energy? The only energy the air has is the constant flitting-around motion of its molecules. So in elbowing aside the other molecules, the expanding air's own molecules will be slowed down. And slower molecules are cooler molecules, because heat itself is nothing more than moving molecules. (The faster its molecules are moving the hotter any stuff is, and the slower they're moving the cooler it is.) Therefore, as our passel of air rises and expands, it gets cooler. The higher a mass of warm air rises through the thinner and thinner atmosphere, the more it expands and the more it cools. This is one reason that it's colder up on a mountain than down in the valley. (But see p. 112 for the main reason it's colder at higher altitudes.) You have undoubtedly experienced the automatic cooling of an expanding gas, whether you paid attention to it or not, because there isn't anybody who hasn't used an aerosol spray can for paint, hair spray, deodorant or whatever. Grab the nearest one, and try this: TRY IT Point an aerosol spray can in a harmless direction and spray for three or four seconds. Notice that the can gets cold. It contains a compressed gas—usually propane, now that chlorofluorocarbons (CFCs) have been banished because they chew up the ozone layer. When you press the valve to spray the liquid, the gas is allowed to expand and push the liquid out the nozzle. During that expansion, the gas becomes cooler. You Didn't Ask, but … Is there any way to tell whether a barbecue grill's propane is going to run out in the middle of a cooking session? It's pretty hard to look inside that steel tank and see how much propane is left before you fire up the grill, isn't it? Not all grills have pressure gauges. But hardware stores sell an ingenious little indicator that looks like a strip of plastic because it is. You stick it onto the outside of the tank and, by changing color, it shows you exactly where the propane level is inside the tank. It works by detecting the cooling of the propane gas as it flows out through the valve during use. The propane inside the tank is under pressure, so it is actually mostly in the form of liquid, with some gas above it. (You can hear the liquid sloshing around if you jostle the tank.) While burning your hamburgers, you are tapping off some of the gas, and more liquid evaporates to replace it. This evaporation cools the gas, so you have a layer of cool gas above a layer of warmer liquid. The strip of plastic contains liquid crystals, which have different optical properties at different temperatures. What it shows you, then, is one color above the liquid's surface, reflecting the temperature of the cool gas, and a different color below the surface, reflecting the temperature of the warmer liquid. The borderline between the colors is where the liquid's surface lies within the tank. You'll find that the gauge works only while you're bleeding off gas. After you shut down the tank and it warms up, there is no temperature difference inside, and there are no different colors on the gauge. Where It's Hilly, It's Chilly How does a mountaintop, even in the tropics, stay covered with snow all year ‘round? Obviously, because it's always colder up there. But why is it always colder up in the mountains than down at the seashore? After all, doesn't hot air rise? Shouldn't it therefore be hotter up there? There's certainly plenty of hot air in equatorial Tanzania, but Kilimanjaro, which thrusts its peak 19, 340 feet (5,895 meters) into the tropical atmosphere, is always capped with snow. It all starts with the sun. And what doesn't? With the sole exception of nuclear energy, the sun is the source of all the heat and all other forms of energy on Earth. As the sun shines down on Earth, its light passes quite transparently through the atmosphere, as you must have concluded from the fact that you can see the sun. Not much happens to the light until it strikes the planet's surface. Then, the various types of surfaces—oceans, forests, deserts, car roofs, George Hamilton—absorb the sunlight and are warmed (and in some cases tanned) by it. This makes the entire surface of Earth a giant, warm radiator, and anything nearby—such as the air above it—will also be warmed, just as you are warmed when you stand near the radiator in an old house. (A radiator, not surprisingly, is something that radiates heat radiation..) It stands to reason, then, that the closer you are to the heat-radiating surface of Earth, the more heat you will be getting from it, just as if you were standing closer to a house's radiator. So the air nearest Earth's surface is warmed the most, and the higher you go away from it, the colder the air will be—cold enough above about 10,000 feet (3,000 meters) that all precipitation will be in the form of snow and it will almost never melt. (A lesser reason why it's cold in the mountains is that as air masses sweep up the mountainside, they expand because of the lower atmospheric pressure, and when gases expand they get cooler.) Exactly how does Earth's surface, once warmed by the sun, transmit its heat to the air above it? Mostly by radiation— the same way a radiator warms you. But radiation isn't the only way that heat can be transmitted from a warm substance to a cooler one. It can also move by conduction and by convection. Let's take a quick look at each mechanism. Conduction: When you grab a hot frying pan handle (DO NOT TRY THIS AT HOME!), the heat travels into your hand by conduction. The heat energy is being conducted, or transmitted, by direct molecule-to-molecule contact. Hot frying pan molecules knock up against your skin molecules and pass their heat energy directly to them. Yelping and releasing your grip breaks this molecule-to-molecule contact. (Actually, the yelp doesn't accomplish much.) Unfortunately, the heat will already be in your skin, continuing to do its damage and replacing your yelp with a more leisurely string of expletives. (Tip: That heat will stay in your skin, continuing to hurt for a much longer time than you might expect, because flesh is a poor conductor of heat. For a minor burn, get that heat out as quickly as possible by holding it under the cold water faucet.) Convection: When you open your oven door quickly to peek in at your turkey and you feel a blast of hot air on your face, it's the air that is carrying the heat to you. That's convection: heat being carried on the wings of a moving fluid, such as air or water. In this case, the heat is moving by hitchhiking on the air. When hot air rises, the heat is moving upward by convection. So-called convection ovens are ordinary ovens with fans in them that circulate the hot air around, which speeds up the cooking. Radiation: The next time you're in a blacksmith's shop (okay, so imagine it) notice that you can feel the heat of his red-hot furnace on your face clear across the room. You're not touching anything hot, so it's not conduction. And there's no moving air, so it's not convection. The heat is reaching you by radiation: infrared radiation. Infrared is a type of electromagnetic radiation, like visible light except that it has a longer wavelength and human eyes can't see it. What's unique about it is that it is of just the right wavelength that most substances can absorb it, “swallowing” its energy and becoming warmed by it. The infrared radiation isn't heat per se, in spite of what many books may tell you; I call it “heat in transit.” It is emitted by hot objects and travels through space at the speed of light, but it doesn't actually turn into heat until it strikes some substance and is absorbed by it. Only a substance can be hot, because heat is the movement of molecules, and only substances—not radiations—have molecules. You Didn't Ask, but … Does the air keep getting colder and colder without limit as we go higher and higher in altitude? No, but it. You Didn't Ask, but … Which pole is colder, the North or the South? The. BAR BET It's much warmer at the North Pole than at the South Pole. Wheeeeee! If the whole Earth is spinning at 1,000 miles per hour (1,600 kilometers per hour), why don't we get dizzy, feel the wind or somehow notice the motion? Is it just because we're used to it? No, it's because Earth's rotation is a uniform, unvarying motion, and we can feel only changes in motion (Techspeak: acceleration). Any time a moving object is diverted from its motion, either by a change in its direction or a change in its speed, we say that it has experienced an acceleration. Acceleration doesn't just mean going faster. Say you're a passenger in a car that's moving in a straight line and is operating on cruise control—that automatic speed governor that keeps the car moving at a constant speed. You don't feel any forces pushing your body around, do you? But as soon as the road changes from straight to curved your body becomes aware of it, because you are thrust slightly toward the outside of the curve. Or if the driver suddenly steps on the gas (the “accelerator”), your body becomes aware of it because you are thrust against the back of the seat. Or if the driver suddenly hits the brakes (another accelerator, but a slowing-down one instead of a speeding-up one), your body becomes aware of it because you are thrust slightly toward the front of the car. But as long as the car doesn't speed up or slow down or go around a curve (Techspeak: angular acceleration), your body feels no forces trying to push it around. In effect, your body doesn't know it's moving, even if your brain does. Well, your brain knows that Earth is spinning, but your body doesn't because the motion is smooth, uniform and continuous. As Isaac Newton put it in his First Law of Motion, a body (including yours) that is moving at a constant speed in a straight line will continue moving that way unless some outside force acts on it. Without such an outside force, the body (including yours) doesn't even realize it's moving. But, you protest, we're certainly being carried around a curve, aren't we? We're following the curvature of Earth's surface. It may be a constant speed, but it isn't a straight line. So why aren't we being thrust outward? Well, we are. But the curvature is so gradual—Earth is so big—that the circular path is virtually a straight line, so that the outward force is minuscule. When you think about it, even your car on that perfectly straight road was going around the same big curve: the curvature of Earth. If you continued in that “straight line” long enough, you'd get right back to where you started. This is all very discouraging to the diabolical designers of amusement parks (I call them abusement parks), who want us to experience motion to the max. They deliberately make us feel unbalanced, unstable, precarious, disoriented, pushed around and insecure. That's why nothing in the whole place moves at a constant speed in a single direction, except perhaps the outward flow of money from your wallet. Every ride either spins you around, hurls you first up and then down or slings you through some crazy combination of up, down and around at the same time. The best (?) roller coasters are those that combine ups and downs with speedups, slowdowns, twists and curves. These changes of motion, which we certainly can feel, all fall into the category of accelerations. Even the merry-go-round is accelerating you, because it is continually diverting you from a straight line, forcing you to turn in a circle. Oh, you asked why we don't feel the wind as the cosmic merry-go-round named Earth spins us around? It's because the air is being carried around at the same 1,000-mile-an-hour (1,600-kilometer-per-hour) speed as ourselves, our cars, our houses and even our airplanes. So there is no relative motion between us and the air. This Dizzy World If Earth is rotating at around 1,000 miles per hour (1,600 kilometers per hour), why can't I see it moving beneath me when I'm in an airplane that's going a lot slower? Because”? NITPICKER'S CORNER. You Didn't Ask, but … If I can't see Earth turning from an airplane, can the astronauts see it turning when they look down from their orbiting shuttle? No,. How to Lose Weight I'm not sure if this is science or a riddle, but my ten-year-old daughter asked me if a polar bear would weigh less at the equator than it does at the South Pole. It's both. The. You Didn't Ask, but … Would I weigh less at the bottom of a deep mine shaft than I do on the surface? Boy,. Flushed with Knowledge Do toilets really flush counterclockwise in the northern hemisphere and clockwise in the southern hemisphere? No.? TRY IT. NITPICKER'S CORNER? BAR BET If mechanical clocks had been invented in Australia, they'd all be running counterclockwise. The Infernal Equinox Is it true, as some people claim, that during the vernal equinox it is possible to stand an egg on end? Absolutely. And during the autumnal equinox as well. And on Tuesdays in February, and anytime during the fourth game of the World Series when the count is three and two on a left-handed batter. Get the picture? The point, of course, is that equinoxes have nothing whatsoever to do with balancing eggs. But old superstitions never die, especially when perpetuated year after year by kooks who like to chant and perform pixie dances in the meadows on the day of the vernal equinox. You can balance an egg on end anytime you feel like it. TRY IT Take a close look at an egg. It isn't glassy smooth, is it? It has little bumps on it. Go through a dozen and you're sure to find several that are quite bumpy on their wide ends. Now find a tabletop or some such surface that is relatively smooth, but not glassy smooth. With a steady hand and a bit of patience, you'll be able to accomplish this miraculous astronomical (more appropriately, astrological) feat without any contribution from Mother Earth, except for supplying the gravity that makes the task challenging. If the balancing surface is rather rough, like a concrete sidewalk, a textured tablecloth or a low-pile rug, for example, it's a piece of cake. An old after-dinner trick—on any day of the year—was to conceal a wedding ring under the tablecloth and, with feigned difficulty, “balance” the egg on it. So much for the old egg game. But what is an equinox, anyway? Picture Earth, circling the sun at the rate of 1 revolution per year. The circle made by Earth's orbit around the sun lies in a plane, just as a circle drawn on paper lies in the plane of the paper; it's called the plane of the ecliptic. Now Mother Earth wears another circle around her middle; it's called the equator, and it also lies in a plane, called the equatorial plane. We can imagine the equatorial plane being extended beyond Earth, way out toward the sun. Funny thing, though: It misses the sun. You usually won't find the sun anywhere in the equatorial plane. That's because Earth is tilted, so its equatorial plane passes above or below the position of the sun. (The equatorial plane is tilted from the plane of the ecliptic by 23½ degrees.) As the tilted Earth moves around the sun, there will be two times in the year when the two planes intersect—that is, two times when the sun, in its ecliptic plane, is also in the equatorial plane, meaning that it is directly over the equator. For half of the year, the sun is north of the equator and the northern hemisphere has spring and summer; for the other half of the year the sun is south of the equator and the southern hemisphere has spring and summer. The two “crossover” instants usually occur on March 21 and September 23. Those two instants are how we define the beginnings of spring and fall in the northern hemisphere; they are called the vernal (spring) equinox and the autumnal (autumn) equinox. The word equinox comes from the Latin meaning equal night, because at those instants the periods of daylight (the days) and darkness (the nights) are of equal duration all over the world. You can see that from the fact that the sun is directly over the equator, favoring neither more daylight in the north nor more daylight in the south. Without knowing all of this, primitive people found the equal-light-and-darkness dates to have special significance, ushering in, as they do, seasons of warmth and growth or cold and barrenness. So all sorts of superstitions grew up around these dates. You can see, though, that there is no “alignment of the planets” or any other possible gravitational effects of the equinoxes that would make eggs do anything weird. The only things that are weird are the nuts who still claim that these dates have magical powers. Oh, yes, then there's the matter of the solstices. They occur halfway between the equinoxes. The summer and winter solstices are the instants at which the sun gets as far north or south of the equator as it ever gets during the year. For northern hemispherians, the summer solstice falls on June 21 or 22 and the hours of daylight are longest; you might call it “maximum summer” or “midsummer.” It has no more mystical power over eggs than the equinoxes do, although in Scandinavia, where the winters are long and dark and Midsummer Day is an excuse for great revelry, it does seem to have a mysterious effect on alcohol consumption. O, Solar Mio When the world runs out of coal and petroleum, could we get all our power from solar energy, which is inexhaustible? Probably not, if you mean making electricity from solar panels. There certainly is lots of sunshine, but capturing it and converting it efficiently is the problem. Let's do the arithmetic. Every day, the sun shines down upon the surface of Earth an amount of energy equal to three times the world's annual energy consumption. That means that to keep up with consumption we would have to capture and convert all the sunlight falling on about one-tenth of a percent of Earth's surface. That may not sound like much, but it's about 180,000 square miles (470,000 square kilometers) of solar panels, or about the size of Spain. Double that to take care of the inescapable fact that it's always nighttime in half the world. And oh, yes: There are clouds. But if you think about it, all of our energy sources today come from the sun, with only one exception: nuclear energy, which we discovered how to make about sixty years ago. Nuclear energy, in the form of nuclear fusion, is where the sun gets its energy in the first place. So speaking cosmically, there is really only one source of energy in the universe, and it's nuclear. Even Earth's internal heat, the source of volcanos and hot springs, is fed by nuclear energy from radioactive minerals. But until we learned how to make some of our own nuclear energy down here on Earth, we had to procure our share of cosmic nuclear energy through a go-between: Old Sol. The sun converts its own nuclear energy into heat and light for us, and all of our current energy sources come from that heat and light. They are therefore solar energy in a real sense. Let's look at our “solar energy” sources one at a time. Fossil fuels: Coal, natural gas and petroleum are the remnants of plants and animals that lived millions of years ago. But what created those plants and animals? The sun. Plants used the energy of sunlight to grow by photosynthesis and the animals came along and ate the plants (and, alas, one another). All life on Earth owes its existence to the sun and so, therefore, does the energy we get today from fossil fuels. Water power: Hydroelectric power plants suck the gravitational energy out of falling water by enticing it into plummeting down onto the blades of turbines, our modern version of the waterwheel. Instead of your having to have a waterwheel or turbine in the kitchen to grind your coffee beans, the turbine-driven generators convert the water's gravitational energy into electrical energy, which is then piped to your wall outlet through copper wires. The water cascades down Niagara Falls or spills over Hoover Dam because in deference to Sir Isaac Newton (the falling-apple guy), it is trying to get closer to the center of Earth. Then isn't water power really the power of gravitational attraction? Isn't it Earth-provoked, rather than sun-provoked? Yes, but hold your horsepower. How did that water get so high in the first place that it can then fall down under the influence of gravity? It's the sun again. The sun beats down on the oceans, evaporating water into the air, where it is blown around by the winds, forms clouds and eventually rains or snows back down. So without the sun's water-lifting power, we wouldn't have water-falling power. We wouldn't have waterfalls or rivers, because without being replenished from above by sun-raised rain and snow, they'd all run dry. Wind power: Windmills capture energy from moving air. But what makes the air move? You guessed it: the sun. The sun's rays shine down upon Earth's surface, a little stronger here and a little weaker there, depending on the seasons, the latitudes, cloud cover and a number of other things. But the land is warmed up by the sun's rays much more than the oceans are, and that creates unevenly heated air masses around the globe. As the warmer masses rise and the cooler masses rush in at ground level to replace them, the air flows, producing everything from balmy breezes to monsoons. Because all of these winds are ultimately traceable back to the sun's heat, wind power is truly sun-provoked. NITPICKER'S CORNER All right, so all of our winds aren't caused by the sun. Some of them are caused by Earth, without any outside help. Earth is rotating, and as it rotates it carries along a thin surface layer of gas—the atmosphere. Now gases and liquids are what we call fluids, substances that flow easily, unlike solids. (Most people use the word “fluid” to mean only liquids, but gases also flow, and are therefore fluids.) Any fluid will have a tough time staying in place when the solid body it's trying to hang on to is moving. In an airplane, for example, the coffee in your cup slops around when the plane hits bumpy air the moment after the flight attendant pours it. In the same way, the rotational movement of Earth makes the air slop around to a certain extent, like the coffee in the cup. And what is air that's slopping around? Wind. So some of our winds are Earth-provoked, rather than sun-provoked. One way in which Earth's rotation affects air movements is quite interesting. It's called the Coriolis effect. How to Date a Mummy Can radiocarbon dating tell us how old anything is? It won't help you to determine the age of anything that is still alive, such as a twelve-year-old posing as a twenty-five-year-old in an Internet chat room. Radiocarbon dating is useful for determining the ages of plant or animal matter that died anywhere from around five hundred to fifty thousand years ago. Ever since its invention by University of Chicago chemistry professor Willard F. Libby (1908–1980) in the 1950s (he received a Nobel Prize for it in 1960), the radiocarbon dating technique has been an extremely powerful research tool in archaeology, oceanography and several other branches of science. In order for radiocarbon dating to tell us how old an object is, the object must contain some organic carbon, meaning carbon that was once part of a living plant or animal. The radiocarbon dating method tells us how long ago it lived, or more precisely (as we'll see), how long ago it died. Radiocarbon tests can be done on such materials as wood, bone, charcoal from an ancient campfire or even the linen used to wrap a mummy, because linen is made from fibers of the flax plant. Carbon is the one chemical element that every living thing contains in its assortment of biochemicals—in its proteins, carbohydrates, lipids, hormones, enzymes and so on. In fact, the chemistry of carbon-based chemicals is called “organic chemistry” because it was at one time believed that the only place that these chemicals existed was in living organisms. Today, we know that we can make all sorts of carbon-based “organic” chemicals from petroleum without having to get them from plants or animals. But the carbon in living things does differ in one important way from the carbon in nonliving materials such as coal, petroleum and minerals. “Living” carbon contains a small amount of a certain kind of carbon atom known as carbon-14, whereas “dead” carbon contains only carbon-12 and carbon-13 atoms. The three different kinds of carbon atoms are called isotopes of carbon; they all behave the same chemically, but they have slightly different weights, or, properly speaking, different masses. What's unique about the carbon-14 atoms, besides their mass, is that they are radioactive. That is, they are unstable and tend to disintegrate—break down—by shooting out subatomic particles: so-called beta particles. All living things are therefore slightly radioactive, owing to their content of carbon-14. Yes, including you and me; we're all radioactive. A typical 150-pound person contains a million billion carbon-14 atoms that are shooting off 200,000 beta particles every minute! BAR BET You are radioactive. If the world's nonliving carbon isn't radioactive, where do living organisms get their carbon-14? And what happens to it when the organisms die? The answers to those questions is where the radiocarbon story really gets exciting. Professor Libby, working right down the hall from my laboratory at the University of Chicago, was able to recognize the relationships among a series of seemingly unconnected natural phenomena that, when put together, gave us an ingenious method for looking into our ancient past and into the history of our entire planet. Follow this sequence of events. (1) Carbon-14 is continuously being manufactured in the atmosphere by cosmic rays, those high-energy subatomic particles that are shooting through our solar system in all directions at virtually the speed of light. (Some of them come from the sun, but the rest come from outer space.) When these cosmic particles hit Earth's atmosphere, some of them crash into nitrogen atoms, converting them into atoms of carbon-14. The carbon-14 atoms join with oxygen to become carbon dioxide gas, which mixes thoroughly around in the atmosphere because of winds. So the entire atmosphere has a certain amount of carbon-14 in it, in the chemical form of carbon dioxide. This process has been going on for eons, and the carbon-14 in the atmosphere has settled into a fixed amount. (2) The radioactive carbon dioxide is breathed in by plants on Earth's surface and used to manufacture their own plant chemicals. (You know, of course, that plants take in carbon dioxide to use in photosynthesis.) All plants on Earth therefore contain carbon-14. They all wind up with about 1 atom of carbon-14 for every 750 billion atoms of carbon that they contain. (3) For as long as a plant is alive, it continues the process of taking in atmospheric carbon dioxide, thus maintaining its 1-in-750-billion atom ratio of carbon-14. (4) As soon as the plant dies it stops breathing in carbon dioxide and its accumulation of carbon-14 atoms, no longer being replenished by the atmosphere, begins to diminish by radioactive disintegration. As time goes by, then, there are fewer and fewer carbon-14 atoms remaining in the dead plant material. (5) We know the exact rate at which a number of carbon-14 atoms will diminish by radioactive disintegration (visit the Nitpicker's Corner). So if we count how many of them are left in some old plant material, we can calculate how much time has gone by since it had its full complement of 1 in 750 billion and, hence, how long ago the plant died. In the case of a piece of wood, for example, we will know when the tree was cut down; in the case of a mummy, we can measure its linen wrapping and calculate when the flax plant was harvested to make the linen, and so on. Neat, huh? But what about animal relics such as bones and leather? How can we tell when an animal lived and died? Well, animals eat plants. Or else they eat animals that have eaten plants. Or in the case of human animals, both. So the carbon atoms that animals eat and from which they manufacture their own life chemicals have the same ratio of carbon-14 atoms as the plants do: 1 out of every 750 billion. When the animal dies it stops eating and exchanging carbon atoms with its surroundings, and its load of carbon-14 begins to diminish in its precisely known way. By measuring how much carbon-14 remains today, we can calculate how much time has elapsed since the relic was part of a living animal. There have been several spectacular applications of radio-carbon dating in the past few decades. One of these was the dating of the Dead Sea Scrolls, a collection of some eight hundred manuscripts that were hidden in a series of caves on the coast of the Red Sea, ten miles east of Jerusalem, by Essene Jews around 68 B.C. They were discovered by Bedouin Arabs between 1947 and 1956. The linen-wrapped leather scrolls contain authentic, handwritten portions of the Old Testament that were determined by radiocarbon dating to have been written around 100 B.C. Another triumph of radiocarbon dating was the finding that the Shroud of Turin, believed by some to be the burial cloth of Jesus, is a medieval fake concocted sometime between 1260 and 1390 A.D., which is very A.D. indeed. This unambiguous scientific result, obtained independently in 1988 by three laboratories in Zurich, Oxford and Arizona, continues to be rejected by those who prefer to believe what they prefer to believe. NITPICKER'S CORNER How do we know precisely at what rate the amount of carbon-14 diminishes? Every. 2 The name of the gravitational force is gravitation, not gravity; gravity simply means heaviness. But everybody outside of the Physicists' Club calls the force gravity, and whenever I feel like it throughout this book, so do I.	228,682
Pumpkin Seeds ₹385.00 – ₹699.00 Select options Super Healthy Super 5 Seeds Mix – Roasted Mixed Seeds \| Crunchy Snack \| Seed Mix for Eating \| 5+ Varieties Like Pumpkin, Sunflower, Watermelon, Flax, Chia (Mini Pack – 200g) 5+ VARIETIES: An assorted combo of seeds for healthy snacking. PREMIUM QUALITY: This mix constitutes world-class internationally sourced organic seeds. USAGE: Recommended 1 spoon daily to achieve a long-term healthy lifestyle for any age group. ALL-DAY SNACK: An healthy evening snack or combo with your breakfast cereal or use it to make your own granola bars – customize as you wish! Reviews There are no reviews yet.	17,397
TITLE: associated $\mathbb{C}[t]$- module is cyclic iff cyclic vector exists QUESTION [1 upvotes]: I'm stuck on a part of a question: if $T : V \rightarrow V$ is a linear endomorphism of a $\mathbb{C}$-vector space $V$, then the associated $\mathbb{C}[t]$- module is cyclic (that is $V =\frac{k[t]}{<g>}$ for some monic polynomial $g \in k[t]$) if and only if there is a vector $v$ such that $V = span\{T^{i}(v)\}$. I really can't even see where to start with this! Any hints greatly appreciated. REPLY [0 votes]: I think you want to translate the condition of being cyclic into the condition that there exists $v\in V$ for which $$\varphi:k[t]\rightarrow V$$ $$p(t)\mapsto p(T)v$$ is a surjective map of $k[t]$-modules. In this case, the kernel has a monic generator.	200,095
Search Royalty Free Globe Stock Clipart Illustrations 3d Puzzle Globe with a Missing Piece Blue Jigsaw Puzzle Earth with One Missing Piece, on a Bursting Blue Background and Grid Surface Single Blue Jigsaw Puzzle Globe with One Missing Piece, on a Blue Bursting Background Blue Jigsaw Puzzle Piece Globe with One Missing Piece, on a Blue Wave Background Blue Jigsaw Puzzle Globe with a Single Missing Piece, on a Blue Background with Waves Yellow Jigsaw Puzzle Globe with a Single Missing Piece, on a Yellow Wave Background Globe of Planet Earth Being Separated onto Different Pieces Laptop Computer with a Jigsaw Puzzle on the Screen, Beside a Globe on a Blue Background 3d Golden Wire Globe in the Center of a Blue Labyrinth Digital Set of 3d Logo Icons Professor Owl Holding a Desk Globe Happy Boy Doing a Globe Puzzle Jigsaw Puzzle Pieces over a Globe Open Globe with Puzzle Pieces Inside Digital Set of 16 Colorful Abstract Web Design Elements or Logos	384,108
pls help me am new how can i set session and cow can i retrive from it Discussion in 'Ruby' started by damod.php Can I retrive a session object by a session ID?=?Utf-8?B?VG90bw==?=, Jun 9, 2006, in forum: ASP .Net - Replies: - 4 - Views: - 416 - =?Utf-8?B?VG90bw==?= - Jun 10, 2006 Q: Scheduling in scipy.cowMathias, Dec 28, 2004, in forum: Python - Replies: - 2 - Views: - 289 - Fernando Perez - Dec 29, 2004 " They say why buy the cOw when the milk is fOr frEE "U S Contractors Offering Service A Non-profit, Nov 30, 2006, in forum: C Programming - Replies: - 0 - Views: - 610 - U S Contractors Offering Service A Non-profit - Nov 30, 2006 Differences between Copy on Write (COW) and counted referencemosfet, May 10, 2007, in forum: C++ - Replies: - 3 - Views: - 363 - Chris Thomasson - May 11, 2007 using Command to set Parameters and Recordset to retrive the QueryBruno Alexandre, Mar 3, 2004, in forum: ASP General - Replies: - 5 - Views: - 166 - Bob Barrows - Mar 3, 2004	213,050
. This Winnie the Pooh Embroidered Nursery Sheet Set is the perfect way to customise babies and toddlers nursery's. Comprising of a green fitted sheet, top sheet and pillow case, this set features finely embroidered pictures of Winnie the Pooh and friends. Pale green with… more ABN 64 096 509 134 \| PO Box 1012 Ingleburn NSW 1890 Australia \| Site Map \| Popular Searches \| Contact Us \| Mobile site	138,573
49 Denver Furniture Repair and Upholstery Services Home furniture repair companies in Den Denver Furniture Repair and Upholstery Services Sara Upholstery & Cushions LLC Denver Furniture Repair and Upholstery Services “Sara Upholstery & Cushions did awesome job with re-cushioning our kitchen chairs. The price is cheap and work is better than any other upholstery company. I would highly recommend them.” Loft Design Group Denver Furniture Repair and Upholstery Services _2<< DC Upholstery & Design Workshop Denver Furniture Repair and Upholstery Services Denver Furniture Repair and Upholstery Services “I had a loveseat pull out bed recovered. Beautiful job!! it is just like new and they finished it at the quoted price. I highly recommend them.” Cushion World Denver Furniture Repair and Upholstery Services !” Heartistech, LLC Denver Furniture Repair and Upholstery Services !” Black Canyon Design Group Denver Furniture Repair and Upholstery Services .” List your business here for free - Learn More	24,116
Rodial’s sister brand Nip + Fab have launched a new range inspired by the popular Rodial Dragon’s Blood range that I for sure am a fan of. The Nip + Fab FIX range has the same end game, to plump and hydrate the skin and uses similar ingredients including hyaluronic acid and the all important dragon’s blood. I’m thinking this post might have been a little more relevant a few days ago for Halloween but never fear the dragon’s blood is actually a powder that is taken from a tree called the Croton Lechleri tree, so no hocus pocus here. So lets take a look at the new range! First up we have the Dragon’s Blood Fix Cleansing Pads £9.95. I am a big fan of the Rodial Dragon’s Blood cleansing water so I was looking forward to trying these. They are similar in the ingredients, but obviously one being a cleansing water and the other cleansing pads. Another notable difference is the rose-water in the water, which for me works so well on my skin. With the cleansing pads they have Dragon’s Blood, Salicylic Acid, Hyaluronic Acid and Witch Hazel. The hyaluronic acid plumps and hydrates skin whilst the salicylic acid cleanses pores and the dragon’s blood comforts and restores skins hydration. The witch hazel tightens pores and reduces redness. I use the Nip + Fab glycolic pads on a regular basis and these were exactly the same size, however they are much wetter and definitely feel soaked with the ingredients, like you would find of a cleansing wipe. The one thing you will notice is the smell, they have such a strong smell from them, but it does fade once you have used the pad. These are great for really targeting red areas and spots with the witch hazel. But for me, I felt like they were quite harsh on my skin, or maybe I just love the Rodial water much more for cleansing. Dragon’s Blood Fix Plumping Mask £14.95. This is a hydrating gel mask that provides hydration to the skin and a brighter complexion. The mask itself smells the same as the cleansing pads and again is packed with hyaluronic acid, dragon’s blood and amino acid. With this you apply it all over the face and neck and leave on or ten minutes, then rinse with warm water. It had a nice feeling on and didn’t nip or hurt in any way, or cause any redness when I removed it. My skin felt lovely and revitalised and soft to touch. This is definitely a great weekly treat. The Dragon’s Blood Plumping Serum £19.95 is quite like the plumping mask yet more of a daily product. You can use this one in the morning and evening after cleansing and before moisturiser. The addition of velvet flower enhances moisturisation to make your skin really hydrated and it also includes Gransil EP-9 which absorbs any excess oil and smoothes skin. The serum is lightweight, not heavy in the slightest and dries in pretty quickly. Have you tried this range yet? It’s available separately or as a set for £34.45 from Boots & Superdrug. 1 Comment BlondeofcarbsNovember 3, 2014 at 22:02 The mask is my absolute favourite, even though I have acne it doesn’t flare up my skin and makes it feel amazing- so rare when a product is suitable for all skin types! Must try more from this range xx	270,673
\begin{chapter}{\label{cha:stochastic_app}Fast Bayesian parameter estimation for stochastic logistic growth models} \section{\label{sto:intro}Introduction} In this Chapter, fast approximations to the stochastic logistic growth model (SLGM) \citep{capo_slgm} (see Section~\ref{int:stochastic_logistic_gro}) are presented. The SLGM is given by the following diffusion equation: \begin{align} \label{eq_det_sde} dX_t&=rX_t\left(1-\frac{X_t}{K}\right)dt+\sigma X_t dW_t, \end{align} where $X_{t_0}=P$ and is independent of $W_t$, $t\geq t_0$. A deterministic logistic growth model (see Section\ref{int:logistic_gro}) is unable to describe intrinsic error within stochastic logistic growth time course data. Consequently a deterministic model may lead to less accurate estimates of logistic growth parameters than a SDE, which can describe intrinsic noise. So that random fluctuations present within observed yeast QFA data (\ref{int:QFA}) can accounted for as intrinsic noise instead of being confounded within our measurement error we are interested in using the SLGM in (\ref{eq_det_sde}), instead of its deterministic counterpart (\ref{eq_det}). Alternative stochastic logistic growth equations exist (see Section~\ref{int:stochastic_logistic_gro}) but we find (\ref{eq_det_sde}) to be the most appropriate as intrinsic noise does not tend to zero with larger population sizes. The SLGM (\ref{eq_det_sde}) is analytically intractable and therefore inference requires relatively slow numerical simulation. Where fast inference is of importance such as real-time analysis or big data problems, we can use model approximations which do have analytically tractable densities, enabling fast inference. For large hierarchical Bayesian models (see Chapter~\ref{cha:modelling_den_int}), computational time for inference is typically long, ranging from one to two weeks using a deterministic logistic growth model. Inference for large hierarchical Bayesian models using the SLGM would increase computational time considerably (computational time is roughly proportional to the number of time points longer) with relatively slow numerical simulation approaches, therefore we may be interested in using approximate models that will allow us to carry out fast inference. First an approximate model developed by \citet{roman} is introduced. Two new approximate models are then presented using the linear noise approximation (LNA) \citep{LNA,komorowski} of the SLGM. The model proposed by \citet{roman} is found to be a zero-order noise approximation. The approximate models considered are compared against each other for both simulated and observed logistic growth data. Finally, the approximate models are compared to ``exact'' approaches. \input{sections_SDE/roman} \input{sections_SDE/LNAM} \input{sections_SDE/LNAA} \input{sections_SDE/application} \end{chapter}	178,347
TITLE: Entourage asymmetry and Cauchy nets: $x_a$ is $V$-close to $x_b$ does not imply $x_b$ is $V$-close to $x_a$ QUESTION [2 upvotes]: I have a few quibbles about the nature of the ordering $(a,b)$ versus $(b,a)$ when it comes to membership of an entourage in a uniform space. The Wikipedia article on uniform spaces nowhere asserts that $(a,b)\in V\iff(b,a)\in V$, but rather states that only some entourages will have this property, and they are called symmetric entourages. We can see that for any $V$ an entourage, two symmetric entourages may be generated; $V\cup V^{-1}$ and $V\cap V^{-1}$ are both entourages and are both symmetric, the issue being that although Wikipedia acknowledges asymmetry, I note that it is not mentioned anywhere else that I could find in a problematic way. First issue: A net $x_{\bullet}=(x_\alpha)_{\alpha\in A}$ whose domain is $X$ is termed a Cauchy net if $X$ is a uniform space, with uniformity $\Phi$, and the net has the property that $\forall V\in\Phi,\,\exists\gamma\in A:\forall\alpha,\beta\ge\gamma,\,(x_{\alpha},x_{\beta})\in V$. This definition is natural, and innocuous seeming, but there is a fundamental issue (as far as I can tell) in the fact that $(x_\alpha,x_\beta)\in V$ does not imply $(x_\beta,x_\alpha)\in V$. This means that Cauchy nets will have to have the unspoken restriction that their image always lies in a symmetric subset of any entourage. I will comment more about resolutions to this after I raise the next issue. Second issue: Judging by the ordering of pairs in this article relating the uniform convergence topologies and the compact-open topology, Wikipedia believes $x$ to be $V$-close to $y$ if $(x,y)\in V$, where $V$ is an entourage on a uniform space which contains $x,y$. The composite $W\circ W$ is defined to be $\{(x,z):\exists y\in X,\,(x,y)\in W\wedge(y,z)\in W\}$. Notice the structure $(x,y),(y,z)$. The article at first adheres to this structure by saying that $(f(x_j),\color{red}{f(y)})\in W\wedge(\color{red}{f(y)},g(y))\in W\implies(f(x_j),g(y))\in W\circ W$, in their first proof. However, in the second proof they break this structure - they say that $(f(x_j),f(x))\in W\wedge(f(x_j),g(x))\in W\implies(f(x),g(x))\in W\circ W$ despite the fact that we need an alternating placement of the shared term - it surely should have been $(f(x),f(x_j))$. This is a technicality which really cannot be avoided in their proof, as they implicitly define $b$ is $V$-close to $a$, in the notation $b\in V[a]$, when $(a,b)\in V$. It is not the case that $a$ is $V$-close to $b$, so this damages their proof. We actually only have that $(f(x),g(x))\in W\circ W^{-1}$. I have come up with potential resolutions to these problems: When Wikipedia writes $(x,y)\in V$, they really mean to say that either $(x,y)\in V$ or $(x,y)\in V^{-1}$, so that $(x,y)\in V$ becomes an abbreviated notation for "$x$ and $y$ are $V$-close to each other". An entourage $V$ is really taken to be either $V\cup V^{-1}$ or $V\cap V^{-1}$ (which?) for these purposes A Cauchy net is really defined by $\forall\alpha\ge\beta\ge\gamma,\,(x_\alpha,x_\beta)\in V$, where I have ordered $\alpha\ge\beta$ to avoid the issue of $(x_\alpha,x_\beta)\in V$ not implying $(x_\beta,x_\alpha)\in V$ The restriction that Cauchy nets must actually converge in only symmetric subsets of an entourage is OK, since $V\cap V^{-1}\subseteq V$ is always a symmetric sub-entourage that the Cauchy net would be required to converge in anyway, so the issue I raise is perhaps not a problem Side-note: in the uniform space article, they put: $V\circ U:=\{(x,y):\exists y\in X,\,(x,y)\in U\wedge(y,z)\in V\}$. Is this the conventional order as accepted by most authors? I ask this because I note Wikipedia (potentially) was sloppy with ordering as shown above, and the order does seem to matter; $V\circ U\neq U\circ V$ in general. Question: Are the issues I'm raising actually issues at all? If so, are the potential resolutions right? How am I to interpret the Wikipedia definition of a Cauchy net (and the ordering issue in their second proof!) in light of these issues? REPLY [0 votes]: As is completely standard in such proofs (but not always explicitly written) we take $W$ to be a symmetric entourage such that $W\circ W \subseteq V$. The entourage axioms ensure that this can be done for any entourage $V$. It’s the analog of taking $\frac{\varepsilon}{2}$ when we’re given $\varepsilon >0$ in a metric proof. If we do that everything in the linked proof works as advertised.	126,352
Noel Asmar Oxford Show Shirt - Grey £50.00 A colorful spin on a classic equestrian show shirt design - wear it on its own or under a stock tie for a secret pop of color! A show shirt you can wear anywhere. Each style fatures a magnetic collar closure and white collar and cuffs. The wide box pleat in the back and 4 way stretch body mapping on the sides create a flattering fit that can be worn tucked into breeches or out with jeans. Composition : 100% Cotton, Trim 84% Nylon, 16% Spandex Care Instructions : Machine wash cold, no bleach, tumble dry low or hang to dry.	167,416

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