paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0102104 | Suppose that MATH is a harmonic MATH - cycle in MATH such that MATH in MATH. In other words, MATH for some MATH - chain MATH in MATH. ( We identify MATH with its image under the inclusion of chains MATH.) The wall MATH divides MATH into two half-spaces; let us call them MATH and MATH. We first claim that we can find a MATH - chain MATH so that MATH and so that MATH is supported on only one half-subspace, say MATH. To see this, first write MATH, where MATH (respectively, MATH) is supported on MATH (respectively, MATH) and therefore, MATH and MATH are both supported on MATH. Then set MATH. Since MATH fixes MATH, MATH . Set MATH. Then MATH so MATH is a MATH - cycle in MATH. Let MATH denote its image in reduced MATH - homology MATH. Consider the NAME - NAME sequence of MATH in unreduced MATH - homology. Let MATH be the connecting homomorphism and let MATH be the induced map of quotients. It follows from the definition of MATH, that MATH in unreduced homology and therefore, MATH, since MATH is harmonic. On the other hand, just as in REF, the map MATH is isomorphic, under NAME duality to MATH. Hence, if MATH is the zero map, then so is MATH. Therefore, our hypothesis implies that MATH and consequently, that MATH is injective. |
math/0102104 | MATH is a MATH - simplex and MATH collapses onto MATH. |
math/0102104 | Let MATH and let MATH denote the inverse image of MATH in MATH. Then MATH (or MATH) can be regarded as a fundamental domain for the MATH - action on MATH. Let MATH denote the result of cutting MATH open along MATH. Take a component of the complement of MATH in MATH and let MATH be its closure. Then MATH is a covering space of MATH with group of covering transformations MATH. Furthermore, MATH is a fundamental domain for the MATH - action on MATH. The only points of MATH that are identified under MATH lie on the common boundary of MATH and MATH. It follows that MATH is a quotient space of MATH by an equivalence relation MATH. Since by hypothesis, D is homeomorphic to MATH, it follows that the quotient space is homeomorphic to MATH, where MATH. The proposition follows. |
math/0102105 | Let MATH be a MATH-stable maximal torus of MATH obtained by twisting MATH by MATH. Let MATH be the image of MATH under the corresponding conjugation map. Then MATH is a maximally split torus in MATH if and only if there is a MATH-stable NAME subgroup of MATH containing MATH, which happens if and only if there is a positive system of MATH such that MATH. |
math/0102105 | From the definition of MATH, MATH . Given a vector MATH with MATH, one can translate it to MATH. The last coordinate of the new vector is MATH and the sum of the coordinates in this new vector is a multiple of MATH. Abusing notation, we call this new vector MATH. Thus MATH . Now let MATH. Then the expression for MATH simplifies to MATH . This proves the first assertion of the theorem. The second assertion follows from the first by viewing partitions diagramatically and taking transposes. The third assertion follows from either the first or second assertions together with the well-known fact that the generating function for partitions with at most MATH parts of size at most MATH is the q-binomial coefficient MATH. The fourth assertion follows from the second and REF . |
math/0102105 | We argue for the case MATH, the case of MATH being similar. Recall that in type MATH the coroot lattice is all vectors with integer components and zero sum with respect to a basis MATH, that MATH for MATH and that MATH. The elements of the coroot lattice contributing to some MATH are: MATH . One observes that the inverses of the permutations in the above card shuffling description for a given MATH contribute to MATH where MATH . The total number of such permutations for a fixed value of MATH is MATH, the number of interleavings of MATH cards with MATH cards preserving the relative orders in each pile. Since MATH, and MATH, the proof is complete. |
math/0102105 | Consider MATH multiplied by the coefficient of the identity in MATH in type MATH. By REF , this is the number of ways of writing MATH mod MATH as the sum of MATH integers from the set MATH. REF states that this is the number of monic degree MATH polynomials over MATH with constant term REF which factor into linear terms. Working in the multiplicative group of MATH, this is clearly the number of ways of writing MATH mod MATH as the sum of MATH integers from the set MATH. |
math/0102105 | This is an elementary NAME inversion running along the lines of a result in CITE. |
math/0102105 | The left hand side is equal to MATH where the last step uses REF on page REF. Note by REF that this expression is precisely the cycle structure generating function under the measure MATH, multiplied by MATH. To complete the proof of the theorem, it must be shown that the right hand side gives the cycle structure generating function for degree MATH polynomials over a field of MATH elements with constant term MATH. Let MATH be a fixed generator of the multiplicative group of the field MATH of MATH elements, and let MATH be a generator of the multiplicative group of the degree MATH extension of MATH, with the property that MATH. Recall NAME 's correspondence CITE between degree MATH polynomials over MATH and size MATH aperiodic necklaces on the symbols MATH. This correspondence goes by taking any root of the polynomial, expressing it as a power of MATH and then writing this power base MATH and forming a necklace out of the coefficients of MATH. It is then easy to see that the norm of the corresponding polynomial is MATH raised to the sum of the necklace entries. The result now follows from REF . Note that there is no MATH term because the polynomial MATH can not divide a polynomial with constant term REF. |
math/0102105 | For the first assertion, from REF, the coefficient of MATH in MATH is MATH . The second assertion is similar and involves two cases. |
math/0102105 | The proof proceeds in several cases, the goal being to show that the inverse of the above processes generate MATH with the probabilities in REF . We give details for one subcase - the others being similar - namely even MATH when MATH satisfies MATH. (The other case for MATH even is MATH). The inverse of the probabilistic description in the theorem is as follows: CASE: Start with an ordered deck of MATH cards face down. Successively and independently, cards are turned face up and dealt into one of MATH uniformly chosen random piles. The odd piles are then flipped over (so that the cards in these piles are face down). CASE: Collect the piles from pile REF to pile MATH, so that pile REF is on top and pile MATH is on the bottom. Consider for instance the permutation MATH given in REF-line form by MATH. Note that this satisfies MATH because the top card has a negative value (that is, is turned face up). It is necessary to count the number of ways that MATH could have arisen from the inverse description. This one does using a bar and stars argument as in CITE. Here the stars represent the MATH cards, and the bars represent the MATH breaks between the different piles. It is easy to see that each descent in MATH forces the position of two bars, except for the first descent which only forces one bar. Then the remaining MATH bars must be placed among the MATH cards as MATH consecutive pairs (since the piles alternate face-up, face-down). This can be done in MATH ways, proving the result. |
math/0102105 | REF shows that the chance that an Affine Type MATH shuffle gives a signed permutation MATH is MATH . Using the fact that these shuffles are a probability measure and dividing both sides of the resulting equation by MATH, it follows that MATH . This can be rewritten as MATH . Since this is true for all MATH, the relation can be inverted to solve for MATH. In the theory riffle shuffles CITE one gets the equation (NAME 's identity) MATH for all MATH. Thus MATH as desired. |
math/0102105 | REF shows that the chance that an Affine type MATH shuffle gives a signed permutation MATH is MATH . Thus the total variation distance is equal to MATH . From CITE, one recognizes this last expression as the total variation distance between a MATH riffle shuffle and uniform. |
math/0102105 | For the first equation, consider the coefficient of MATH on the left hand side. It is the probability that a uniformly chosen unimodal permutation on MATH symbols has MATH cycles of shape MATH. The coefficient on the right hand side is MATH. These are equal by REF . To deduce the second equation, observe that setting all MATH in the first equation gives that MATH . Taking reciprocals and multiplying by the first equation yields the second equation. |
math/0102105 | The result follows from the claim that if MATH has a NAME series around REF and MATH, then the MATH limit of the coefficient of MATH in MATH is MATH. To verify the claim, write the NAME expansion MATH and observe that the coefficient of MATH in MATH. |
math/0102105 | Symmetric function notation from REF is used. Thus MATH are the power sum, complete, elementary, and NAME symmetric functions parameterized by a partition MATH. From REF, the number of MATH cycles with descent set MATH is the inner product of a NAME character MATH and a NAME character MATH. From the proof of REF, MATH where MATH is the number of standard tableaux of shape MATH with descent composition MATH. Thus the sought number is MATH . Expanding these NAME functions using REF on page REF, using the fact that the MATH are an orthogonal basis of the ring of symmetric functions with known normalizing constants (page REF), and using the expansions of MATH and MATH in terms of the MATH's (page REF) it follows that MATH . |
math/0102105 | One argues separately for odd and even characteristic and first for prime powers. Taking the coefficient of MATH on the left hand side of this equation and dividing by MATH gives by REF the probability that MATH chosen according to the MATH probability measure is in a conjugacy class with MATH positive MATH-cycles and MATH negative MATH-cycles for each MATH. By REF , to verify the theorem for even prime powers it is enough to show that the coefficient of MATH on the right hand side of this equation is the number of degree MATH monic self-conjugate polynomials over MATH which factor as MATH (with MATH where MATH) and MATH and MATH. This follows readily from REF . The theorem now follows for arbitrary MATH since two functions analytic in a region and agreeing on a set with an accumulation point REF in that region must be equal. |
math/0102105 | Given REF with MATH, it is enough to define a MATH to MATH map MATH from the MATH type MATH characteristic REF shuffles to unimodal elements of MATH, such that MATH preserves the number of MATH-cycles for each MATH, disregarding signs. To define MATH, recalling REF observe that the MATH shuffles are all ways of cutting a deck of size MATH, then flipping the first pack, and choosing a random interleaving. For instance if one cuts a REF card deck at position REF, such an interleaving could be MATH . Observe that taking the inverse of this permutation and disregarding signs gives MATH . Next one conjugates by the involution transposing each MATH with MATH, thereby obtaining a unimodal permutation. Note that this map preserves cycle structure, and is MATH to MATH because the first symbol (in the example MATH, can always have its sign reversed yielding a possible shuffle). |
math/0102105 | Taking coefficients of MATH on both sides of the equation under question and then setting MATH gives the equation MATH . However this equation follows from REF for odd MATH and REF . |
math/0102105 | The proof proceeds in two steps. REF is to show that there is a bijection between semisimple classes MATH in MATH and pairs MATH such that MATH and MATH. REF is to show that MATH is conjugate to MATH. The theorem then follows from the definition of MATH. CASE: It is known that there is a bijection between simplices of maximal dimension in the NAME complex and elements MATH in the affine NAME group such that MATH. The alcoves MATH are all obtained by first transforming by MATH to give MATH and then translating by MATH for some MATH. Let MATH be the union of the alcoves MATH for MATH; MATH is called the basic star. The sets MATH are called the stars and the centers of the stars are the elements of MATH. Each alcove MATH which lies in MATH lies in some star whose center lies in MATH. Conversely, if MATH we wish to know which alcoves in the star MATH lie in MATH. If MATH does not lie on any boundary wall of MATH all alcoves in MATH lie in MATH. If MATH lies on the boundary wall corresponding to MATH then the alcove MATH lies on the MATH-side of this boundary wall if and only if MATH is a positive root. This can be seen by looking at the star MATH. Thus there is a bijection between semisimple classes MATH in MATH and pairs MATH such that MATH and MATH. CASE: Let MATH be a maximal split torus of MATH and let MATH be its co-character group. Let MATH be a MATH-stable maximal torus of MATH obtained from MATH by twisting with MATH. We have conjugation maps MATH. Under these maps MATH maps to MATH. Let MATH be an element of the affine NAME group such that MATH. From REF, MATH contains a unique MATH satisfying MATH that is, MATH. Let the walls of MATH be MATH. Let MATH. MATH is a proper subset of MATH. The roots MATH form a simple system MATH in a subsystem MATH. From page REF the point MATH maps to an element MATH. Then MATH and MATH lies in the semisimple conjugacy class of MATH corresponding to the point MATH. MATH can be identitied with the root system of the centralizer of the semisimple conjugacy class of MATH corresponding to the point MATH. To complete the proof of REF , it suffices to show (by REF ) that MATH. The construction of the point MATH as the intersection of a sequence of increasingly small alcoves, each obtained from the previous one by a map MATH which preserves the type of the walls, shows that MATH lies in the MATH-face of MATH and of MATH (the MATH-face of MATH is the intersection of the MATH for MATH). For MATH let the wall MATH of MATH coincide with the wall of type MATH of MATH. The root orthogonal to MATH pointing into MATH is MATH. Consider the root orthogonal to the wall of type MATH for MATH pointing into MATH. Since MATH this is the root orthogonal to the wall of type MATH for MATH pointing into MATH. This is the root MATH since the wall of type MATH for MATH is the image under MATH of the wall of type MATH for MATH. Hence MATH. This shows, since MATH, that MATH. Hence MATH, as desired. |
math/0102108 | Let MATH and MATH be two functions on MATH, and denote by MATH the Hamiltonian vector field of MATH. So MATH. As already observed, MATH is the Hamiltonian vector field of MATH. As a consequence, MATH . |
math/0102108 | Since the evaluation map is invariant under the simultaneous action of the inverse of a diffeomorphism on MATH and of the induced diffeomorphism on MATH - that is, MATH - , we obtain, at the infinitesimal level, MATH where MATH and MATH are the lifts of MATH and MATH to MATH. On the other hand, by definition of MATH, we have MATH. Combining these identities, we get MATH . |
math/0102108 | Proceed as in the proof of REF till the last equation which is now replaced by MATH upon using REF. |
math/0102108 | The first statement follows from REF . We have first to check that MATH restricted to MATH is invariant. This is a consequence of the identities MATH . The last equality holds on MATH since MATH for MATH and because the restriction of MATH to MATH vanishes (as one has to set MATH). Finally, MATH. As for the second statement, let us denote by MATH the Hamiltonian vector field of MATH. Observe that MATH has support in MATH, so MATH . However, if we restrict to MATH, then MATH belongs to the zero locus of MATH for all MATH, and MATH is constant on it by assumption. So MATH. Since the vector fields MATH and MATH commute, it follows immediately that the bracket MATH is a constant function, but we want this function to vanish identically. Actually, by REF we have MATH the last equality holding on MATH (and so on MATH) since MATH must belong to the zero locus of MATH for MATH, and MATH vanishes on it by assumption. |
math/0102109 | Assuming that MATH is MATH-quasi-isometric, we fix MATH and put MATH. Let MATH, MATH be the separation and radius constants respectively for MATH involved in the subexponential size estimate. We put MATH and take a maximal separated net MATH. Then MATH is MATH-separated, hence, it is contained in some maximal separated net MATH. Next, we fix MATH and take MATH. If MATH intersects MATH for some MATH then MATH intersects MATH since MATH. Thus MATH intersects MATH and consequently we have MATH for every MATH, MATH. Fix MATH. Then for MATH we obtain MATH . Thus taking MATH we obtain the required estimate MATH for every MATH and MATH. |
math/0102109 | If MATH is quasi-isometric to MATH then it is quasi-isometric to MATH too. Any map MATH parallel to a quasi-isometric one is quasi-isometric, thus there is a continuous quasi-isometry MATH. By REF we have MATH. Hence, the claim. |
math/0102109 | Assume that we have a MATH-quasi-isometric map MATH. We take a maximal MATH-separated net MATH with MATH and note that every ball MATH with MATH contains only a finite number of elements of MATH. This is because MATH is MATH-separated and MATH is finitely compact, thus the ball MATH intersects MATH over a finite set. It follows that the nerve MATH of the covering MATH of MATH is a locally finite simplicial complex. Choosing a continuous partition of unity MATH subordinate to MATH, we obtain a continuous map MATH by MATH where we identify MATH with REF-skeleton of MATH. Note that MATH lies in the simplex MATH spanned by MATH. Next, we extend MATH to a continuous MATH using convexity of MATH and acting by the induction on the dimension of the skeletons. Then MATH and thus MATH is a continuous map parallel to MATH. |
math/0102109 | CASE: By REF , MATH is QPC. Thus MATH by REF , sect. CASE: On the other hand, it follows from CITE that MATH. By REF , MATH. Hence, the claim. CASE: We have MATH by CITE, MATH by REF , and MATH by REF . Hence, the claim. |
math/0102111 | The fact that MATH is positive definite is elementary. Then, after a change of variables, we may assume that MATH is the identity matrix, so that MATH may be written as MATH. The NAME transform of MATH may be written as MATH, with MATH. To prove that MATH implies MATH, it is sufficient to prove that, for MATH, we have the inequality MATH . But this integral is twice the integral on the subset where MATH. So it is bounded by MATH which allows to conclude. It is clear that MATH implies MATH for all functions MATH. Let us now prove that MATH implies MATH. First, writing MATH, it is immediate that the integrability of MATH implies that the homogeneous polynomial MATH is always positive, which implies that MATH. Now, let MATH and MATH be the homogeneous terms of maximal degree of the polynomials MATH and MATH. There exists MATH of norm MATH such that MATH is different from MATH. We call MATH the cone which is obtained as the image of the cone MATH under a rotation which maps the point MATH to MATH. Then, for MATH small enough, there exists a constant MATH such that, for MATH, one has the inequality MATH . The same inequality is valid for MATH and MATH for MATH and MATH large, which implies that MATH . We remark that if MATH then MATH. So, a fortiori, we have that MATH . We know, using NAME 's theorem, that there exists MATH for which MATH which proves that MATH. |
math/0102111 | We may assume that MATH. First step. Both MATH and MATH are in MATH. For almost every MATH, MATH . As MATH, the set of all MATH's such that MATH has positive measure. In particular, there is a basis MATH of MATH such that, for MATH, MATH and MATH . Since, clearly, there exists a constant MATH such that MATH we get MATH. Exchanging the roles of MATH and MATH, we get MATH. MATH . Second step. The function MATH defined by MATH satisfies the following properties (with MATH depending only on MATH) MATH . REF is obvious from the definition of MATH and the fact that MATH is in MATH. As MATH, MATH is bounded thus REF is also obvious. To prove REF , we have MATH with MATH . We claim that MATH which allows to conclude. Indeed, separating the cases of MATH being positive or negative, we get MATH . As MATH, it is enough to get a bound for MATH. Now, fix MATH and write MATH, then MATH . We conclude directly for the first integral. For the second one, it is sufficient to note that, if MATH, then MATH . This completes the proof of REF . Let us finally prove REF . Fix MATH. Then the left hand side is bounded by MATH . Then, if we multiply and divide by MATH in the integral of right side, we get the required REF . This completes the proof of the claim. MATH . Third step. The function MATH admits an holomorphic extension to MATH that is of order MATH. Moreover, there exists a polynomial MATH such that for all MATH, MATH. It follows from REF and NAME inversion that MATH admits an holomorphic extension to MATH which we again denote by MATH. Moreover, MATH with MATH the MATH norm of MATH. It follows that MATH is of order MATH. On the other hand, for all MATH and MATH of modulus MATH, MATH . Let us now define a new function MATH on MATH by : MATH . As MATH is entire of order MATH, so is MATH. By differentiation of MATH, the proof of this step is complete once we show that MATH is a polynomial. To do so, we will use REF and an elementary variant of NAME 's principle which we recall here, and which may be found in CITE : let MATH be an entire function of order REF in the complex plane and let MATH; assume that MATH is bounded by MATH on the boundary of some angular sector MATH. Then the same bound is valid inside the angular sector (when replacing MATH by MATH). Let us fix a vector MATH and define the function MATH on MATH by MATH. Then MATH is an entire function of order REF which has polynomial growth on MATH and on MATH by REF . We cannot directly apply NAME 's principle since we are not allowed to do so on angular sectors of angle MATH. But, to prove that MATH has polynomial growth in the first quadrant, it is sufficient to prove uniform estimates of this type inside all angular sectors MATH. Moreover, it is sufficient to have uniform estimates for the functions MATH, with MATH, and MATH . MATH clearly has polynomial growth on MATH and on MATH, that is MATH . The constant MATH, which comes from the constant in REF , is independent of MATH. The same estimate is valid inside the angular sector by the NAME 's principle, and extends to MATH, which we wanted to prove. Proceeding in an analogous way in the three other quadrants, we prove that MATH is an entire function with polynomial growth of order MATH, so a polynomial of degree MATH. Let us now write MATH . Then MATH shows that MATH is a homogeneous polynomial of degree MATH on MATH. The entire function MATH, which is a polynomial on MATH, is a polynomial. Finally, MATH where MATH is a polynomial and the proof of this step is complete. REF . A lemma about entire functions of several variables. We are now lead to solving REF , where MATH is an entire function of order MATH of MATH variables and MATH is a given polynomial. It is certainly well known that such functions MATH can be written as MATH, with MATH a polynomial of degree at most MATH. Moreover, the equation implies that MATH, so that MATH is homogeneous of degree MATH. So we have completed the proof, up to the study of REF . Since we did not find a simple reference for it, we include the proof of the next lemma, which is a little more general than what we need above. Let MATH be an entire function of order MATH on MATH such that, on every complex line, either MATH is identically MATH or it has at most MATH zeros. Then, there exists a polynomial MATH with degree at most MATH and a polynomial MATH with degree at most MATH such that MATH. Without loss of generality, we may assume that MATH does not vanish. Then, for MATH, MATH is a non-zero entire function of order MATH that has at most MATH zeros. By NAME 's factorisation theorem, for every MATH, there exists a polynomial MATH with MATH and MATH such that : MATH with MATH . From the uniqueness in NAME 's theorem, we easily see that the functions MATH and MATH are homogeneous of degree MATH and MATH respectively, and MATH is of degree MATH. We may assume that MATH is non identically zero. We have MATH . Differentiating REF MATH times with respect to MATH, and then taking MATH, we get that MATH is a homogeneous polynomial of degree MATH. Now, MATH and differentiating MATH times with respect to MATH at MATH, we get that MATH is holomorphic. Thus MATH is also holomorphic, and so a homogeneous polynomial of degree MATH. An analogous proof allows to conclude that MATH is a homogeneous polynomial of degree MATH. Define MATH and MATH. We know that MATH is holomorphic, and we have to prove that it is a polynomial. Then MATH . In particular, MATH is a holomorphic function, thus a homogeneous polynomial of degree MATH. It follows that MATH is a polynomial of degree MATH, which we wanted to prove to conclude for this step. We have also completed the proof of REF . Indeed, MATH has the required form thus so has MATH. |
math/0102111 | It is enough to see that REF are stronger than REF . Thus MATH for some positive definite matrix MATH. A direct computation then shows that the form of the matrix MATH imposed by REF are respectively MATH diagonal and MATH. |
math/0102111 | It is enough to notice that MATH and to apply REF to obtain the second point. The two other cases are easy consequences of REF . |
math/0102111 | For MATH, this is exactly REF . For MATH, we proceed as in CITE to reduce to the one-dimensional case. For almost every MATH, the function MATH defined by MATH is in MATH and has as NAME transform the function MATH . So, for almost every MATH, the function MATH satisfies the assumptions of REF , and MATH is a polynomial of degree at most MATH in the MATH variable. The same is valid in each variable, which allows to conclude. |
math/0102111 | It is sufficient to consider the one-dimensional case: for the first statement we conclude the general case from the one-dimensional one as before, and to find a dense subset of functions we use tensorization. So, let us first assume that MATH is such that MATH for some positive constants MATH and MATH, with MATH not identically zero. Here MATH, and MATH is the conjugate exponent. It follows from the second inequality that MATH extends to an entire function, which satisfies the inequality MATH . Moreover, the same inequality is valid when MATH is replaced by its even part, or its odd part. Such a function may be written as MATH, or MATH, with MATH an entire function. One of them is non zero, and satisfies MATH . In the second inequality, MATH has effectively been replaced by an arbitrarily close smaller constant, which we write MATH again for simplification. We then consider the function MATH . Then MATH is a non zero entire function of order MATH, and, moreover, MATH . Let us consider the indicator function of MATH, given by MATH (see CITE for this topic). Then there exists a positive measure on the unit circle, MATH, such that the MATH-periodic function MATH is the convolution of MATH with the continuous function MATH defined by MATH on MATH. An elementary computation shows that MATH . It implies that MATH and, in particular, MATH. But MATH, while MATH. The inequality MATH follows at once. Let us now prove that, for MATH, there is a dense subset of functions MATH such that MATH . Since this set of functions is stable under multiplications by MATH (changing MATH into an arbitrarily close smaller constant), we see immediately that it is dense, unless it reduces to MATH. Indeed, if MATH is such a non zero function and MATH is orthogonal to all functions MATH, then MATH is identically MATH. Since MATH is analytic, it means that MATH is MATH. Moreover, using a dilation if necessary, we can restrict to the case when MATH and MATH. We shall prove a slightly stronger result, the fact that MATH is not reduced to MATH, where MATH is the class of functions for which, for every MATH, there exists a constant MATH such that MATH . We claim that this follows easily from the following lemma: The function MATH is in MATH if and only if it extends to an entire function of order MATH, and, for every MATH, there exists a constant MATH such that MATH . Indeed, let us take the lemma for granted and finish the proof. We know that there exists non zero entire functions of order MATH which satisfy these two conditions (see CITE or CITE). So, it follows from the lemma that there exists non zero functions in MATH. One implication (not the one that we need) is trivial. So let us assume that MATH is an entire function which satisfies REF . We adapt the proof from CITE with the required modifications to have a precise control of the constants. We first give precise estimates of the function inside each of the quadrants of the complex plane which are delimited by the coordinate axes. It follows from NAME principle and the assumption that, for every MATH there exists a constant MATH such that MATH . From the convexity of the function MATH, we see that MATH. Combining the corresponding inequality for MATH with a simple estimate of MATH in an angular sector around the MATH-axis, and doing the same in the three other quadrants, we deduce the existence of constants MATH and MATH such that, for every MATH, there exists a constant MATH, such that, MATH . Let us now deduce from these last inequalities the required exponential decrease of the NAME transform of MATH. It is sufficient to prove that, for every MATH, there exists a contant MATH such that, for all positive integers MATH and for all MATH, MATH . This is equivalent to proving (possibly by changing the constant MATH), that, for all positive integers MATH and for all MATH, MATH . This in turn is implied by the inequality MATH which we now prove using REF . To estimate MATH, we use the NAME inequality related to the circle centered at MATH and with radius MATH. For MATH, we use the first part of REF , to get the inequality MATH . Then NAME 's formula for the asymptotics of MATH allows to conclude for the integral on the interval MATH. Outside, we get the estimate MATH . We conclude as before using the NAME 's formula. This finishes the proof of REF . |
math/0102111 | This fact is also well known, however to help the reader get familiar with our notation, let us recall the proof of the last assertion (see CITE, CITE). If MATH and MATH are fixed, we will write MATH . The change of variables MATH gives MATH . In particular, for almost every MATH, the integral with respect to MATH is finite, that is, MATH. Noticing that MATH, and using NAME 's formula we obtain MATH which completes the proof. |
math/0102111 | Let us first remark that MATH and MATH. Indeed, it is sufficient to use the triangle inequality MATH . Moreover, MATH and MATH may be replaced by MATH and MATH, where ``for all" has been changed into ``for some". Let us prove that MATH implies MATH. With the notations of REF , NAME identity gives : MATH . So, if MATH holds, for almost every MATH, MATH. As we assumed that MATH, there exists MATH such that MATH, and the first inequality in MATH holds with MATH. Since MATH and MATH play the same role, we conclude for the second part similarly. Conversely, if MATH holds, the right hand side of REF is finite, and MATH holds also. |
math/0102111 | Let us start by proving the inequality. Set MATH. We may assume that MATH so that both factors are finite, and, by homogeneity, that MATH. Moreover, replacing MATH and MATH by translates of these functions (with the same translation), we may assume that MATH . NAME 's REF applied to MATH implies that, for any MATH : MATH . Integrating this inequality with respect to the MATH-variable and appealing to NAME 's inequality, we get : MATH . Let us now transform the first factor on the right hand side of this expression. We write MATH . The second term which appears is, by NAME identity, equal to MATH . The first term is equal to MATH using REF . Finally, including these results in REF , we obtain the desired inequality. Assume now that we have equality. Let us remark that, using REF and MATH of REF , we may as well assume that the constants MATH and MATH are MATH. Moreover, up to a same translations in space and frequency, we may again assume that MATH, and MATH. Let MATH as before. Then, to have equality in REF , we have equality in REF , that is, we have MATH. Similarly, exchanging the roles of the MATH and MATH variables, we also have MATH. We then have equality in NAME 's REF . This implies that MATH are proportional. Further, for almost every MATH, we also have equality in NAME 's REF . From now on, we assume for simplicity that MATH. We then get that, for almost every MATH, MATH is a Gaussian in the MATH variable : MATH where MATH, MATH is not identically MATH and MATH for those MATH for which MATH. Let us first prove that MATH does not depend on MATH. With this expression of MATH, we get that MATH whereas (with NAME identity) MATH . But these two functions are proportional only if MATH is a constant, say MATH. Taking inverse NAME transforms, we get that MATH where MATH is the NAME transform in MATH of MATH. In particular, as MATH is continuous, MATH is also continuous. Further, if one has equality in REF for MATH, one has again this equality if MATH are replaced by their NAME transform, as MATH. So, we have MATH for some function MATH and some MATH. Comparing REF , we get that MATH for some function MATH. Taking the NAME transform in the MATH variable, we get MATH . So, setting again MATH, MATH, we get MATH . This is only possible if MATH and MATH, MATH with MATH. |
math/0102111 | First, it follows from REF that MATH and MATH have moments of order MATH. Let us prove that MATH and MATH are not correlated. Without loss of generality, we may assume that MATH for all MATH, so that MATH and MATH are centered. Let us show that MATH . Using NAME identity , we are led to consider MATH . Writing MATH, and MATH we get eight terms. For four of them we get directly MATH. The sum of the last four may be written, after changes of variables, as MATH . After an integration by parts, the two terms give MATH, so that their difference is MATH. Let us now prove the second assertion. Let MATH be an automorphism of MATH. For a function MATH, we consider MATH the function given by MATH. Then a simple change of variables shows that the probability density of MATH is MATH. Eventually changing MATH and MATH into MATH and MATH we may assume that MATH is the identity matrix. Moreover, we may also assume that MATH is diagonal. Then the inequality follows from REF . Equality holds only if all eigenvalues are equal to MATH, which means that MATH and MATH are Gaussian functions. |
math/0102111 | For MATH, we note MATH. Then MATH . This clearly makes sense for MATH, and defines an entire function. |
math/0102111 | By homogeneity, we may assume that MATH. For each MATH, we consider the function MATH . By REF , the NAME transform of MATH is given by MATH. First step. There exists a polynomial MATH such that MATH . We consider here the function MATH. As MATH is (up to a change of variable) its own NAME transform, by REF , it is sufficient to prove that MATH . It follows immediately from the assumption on MATH, using NAME inequality and the fact that MATH. To complete the proof of the proposition, it is sufficient to prove that MATH extends to an entire function of order MATH. Indeed, REF then implies that MATH where MATH is a polynomial and MATH a polynomial of degree at most MATH. But, as MATH we get MATH for some constants MATH. Next, REF implies that MATH are purely imaginary, MATH with MATH so that MATH with MATH a polynomial. REF allows to conclude. So, we have finished the proof once we have proved the second step. Second step. The function MATH extends to an entire function of order MATH. To prove this, we first show that, for each MATH, the function MATH extends to an entire function of order MATH. Since, up to a change of variable, MATH coincides with its NAME transform, it is sufficient to show that MATH . This last inequality follows from the fact that MATH, and from the assumption on MATH. We get the estimate MATH for all MATH, with a constant MATH which does not depend on MATH. In order to conclude this step, it is sufficient to show that there exists a constant MATH such that, for each MATH, we may find MATH which is of the form required in REF , such that MATH and MATH . By density of the NAME functions we can choose a polynomial MATH such that MATH . We then define MATH by MATH . It follows from the choice of MATH that MATH . Then REF follow from direct computations. Finally, since MATH extends to an entire function for each MATH, and since the second factor is also entire and does not vanish in a neighborhood of MATH, MATH extends also to an entire function. The fact that it is of order MATH follows from REF . We have completed the proof of REF . |
math/0102111 | With the weaker REF , we conclude that REF also holds, using the directional theorem for NAME transforms. We claim that MATH is an analytic function of each of the variables MATH and MATH. Indeed, as before, for every function MATH, the product MATH extends to a holomorphic function of MATH, MATH being fixed, as well as to a holomorphic function of MATH, MATH being fixed. When choosing MATH as before, we conclude that the function extends to an entire function of order MATH in MATH, for fixed MATH. So, for almost every fixed MATH, MATH may be written as MATH, with MATH a polynomial of degree at most MATH. It follows that the continuous function MATH is a polynomial of degree at most MATH in each variable MATH, MATH. So it is a polynomial, and REF is satisfied. We can now use REF to conclude. |
math/0102111 | When MATH, the result follows from REF . Let us now consider the case MATH. We assume that the condition is satisfied for MATH. For MATH and for MATH in MATH, we define MATH. It follows from NAME identity that MATH . Changing variables, and using NAME 's Theorem, it follows that, for almost every MATH and MATH in MATH, MATH . It follows from the one dimensional case that either MATH or MATH is identically MATH. So, for almost every MATH and MATH in MATH, MATH. It follows that MATH. |
math/0102111 | It is sufficient to consider the one dimensional case. Otherwise, the proof works as in REF . Let us prove the first assertion. If we proceed as in the proof of REF , we conclude at once that MATH, using the similar result on NAME transforms. It remains to show that MATH is an analytic function of each variable, which is obtained in the same way as before using an auxiliary function MATH. Let us now prove that for MATH, there exists a non zero function MATH such that the two conditions are satisfied by MATH. Using NAME formula, the first condition may as well be written as MATH . Using the same change of variable as before, and the inequality MATH we see that this integral is bounded by the square of the integral MATH . For the second integral, we use REF to write it in terms of the NAME transform of MATH. We obtain that it is bounded by MATH . The fact that there is a non zero function MATH for which both integrals are finite is an easy consequence of REF and NAME inequality, since MATH. |
math/0102112 | Consider the associated isometric structures MATH to MATH, MATH. Then MATH on MATH is the associated isometric structure MATH to MATH . Let MATH, MATH, and let MATH, where MATH is the rank of MATH. Since MATH and MATH are relatively prime in MATH and since either MATH or MATH is non-singular, MATH and MATH are relatively prime in MATH. Then MATH and MATH are also relatively prime in MATH. For, if MATH is a common factor of MATH and MATH, then MATH, MATH, is a common factor of MATH and MATH. Thus there are polynomials MATH and MATH in MATH and a non-zero integer MATH such that MATH. For MATH, there are MATH and MATH with MATH. As stated right after the definition of the NAME polynomial, each MATH is the characteristic polynomial for MATH, and hence MATH. Using this and MATH, we have MATH . Since MATH is MATH-invariant, MATH. Since MATH is a direct summand of MATH, this implies MATH, and hence MATH. Similarly, MATH. So MATH. This now implies that MATH since MATH and MATH. Since MATH is MATH-invariant, each MATH is MATH-invariant. Since MATH is torsion free, each MATH is a direct summand of MATH. Since the intersection pairing MATH on MATH is unimodular, MATH on MATH. Thus each MATH is a metabolizer for MATH. |
math/0102112 | Let MATH and MATH be NAME surfaces for MATH and MATH. Then a boundary connected sum MATH is a NAME surface for MATH. Let MATH be a metabolizer for the isometric structure on MATH satisfying the conclusion of REF with the exceptional primes MATH, that is, MATH vanishes for all prime powers MATH and all primes MATH except MATH. Then by REF there are metabolizers MATH and MATH for the isometric structures on MATH and MATH, respectively, with MATH. Let MATH be a power of a prime. Let MATH and MATH, where MATH and MATH are the endomorphisms of MATH and MATH, respectively, as denoted in REF. Then since MATH, MATH, and MATH, MATH . Let MATH and MATH denote the MATH-primary components of MATH and MATH, respectively. Let MATH. Then MATH is an element in MATH, where MATH stands for the trivial character in MATH. By the additivity of NAME - NAME invariants CITE, for all primes MATH except MATH, MATH . Also, by CITE MATH is determined by the algebraic concordance class of the knot if the character is trivial. This implies that MATH and hence MATH. Since MATH was chosen arbitrarily, we just have found a metabolizer MATH for the isometric structure on MATH such that MATH vanishes for all prime powers MATH and all primes MATH except MATH, that is, MATH is zero in MATH. Similarly, MATH is zero in MATH. This completes the proof. |
math/0102112 | Define MATH. Then CITE showed that if MATH and MATH, MATH . Note that MATH is nonzero only if MATH for an integer MATH with MATH and MATH. To prove REF , let MATH and MATH. Then MATH is nonzero only when MATH for odd MATH with MATH. Note that MATH. So we have MATH . Write MATH for MATH. Then MATH and MATH. Thus MATH . Since MATH if and only if MATH, we see MATH . To prove REF , note first that MATH since MATH is the mirror image of MATH. Let MATH and MATH. For any integer MATH with MATH, MATH, and MATH, we see MATH and hence MATH . Let MATH be an integer with MATH. Note MATH. We will consider two cases: If MATH, then MATH and MATH and hence MATH. If MATH, then MATH and MATH and hence MATH. Since the sets MATH and MATH have MATH and MATH elements, respectively, MATH changes by MATH over the interval MATH. Incorporating these, we have MATH . Now REF follows. |
math/0102112 | From a basic result of linear algebra, MATH has a basis consisting of eigenvectors of MATH since MATH is invariant under MATH and MATH is diagonalizable over MATH. In particular, MATH, where MATH are the eigenspaces of MATH restricted to MATH. Since the rank of MATH is MATH, one of MATH has rank greater than or equal to MATH. Suppose that MATH has rank MATH. Using the NAME algorithm and rearranging basis elements, we may assume that a basis for MATH consists of vectors of the form MATH, MATH, where MATH are linear combinations of MATH. Adding these basis elements together we see that MATH contains a vector MATH for some MATH. Multiplying the vector by a nonzero integer gives a desired vector. The same argument works if rank-MATH. |
math/0102112 | If MATH then MATH is the unknot and MATH for any MATH. Now suppose that MATH. From REF MATH has local minima MATH at the integers MATH with MATH. Observe that the function MATH is increasing over MATH and has MATH at MATH. Also, a close look at the proof of REF reveals MATH and hence MATH if MATH. Now REF follows. To show REF , let MATH be an integer such that MATH and MATH. Let MATH, MATH a prime power, and MATH a prime dividing MATH. It is easy to see that MATH mod MATH. For simplicity, let MATH and MATH the multiplicative order of MATH mod MATH. Then MATH divides MATH and let MATH. For any integer MATH not divisible by MATH, let MATH, that is, a coset of the multiplicative group MATH modulo MATH. Then there are integers MATH such that MATH and MATH are all disjoint. Let MATH. We will show that there is MATH for which at least half of MATH belong to MATH. Note MATH. Since MATH is a partition of MATH, there is MATH such that MATH. For simplicity, assume MATH. Let MATH. Define a function MATH by MATH mod MATH. Since, for each fixed MATH, MATH, the function MATH is a MATH to MATH map. Since MATH, MATH. For each fixed MATH with MATH, let MATH. Then MATH is a partition of MATH and hence there is an integer MATH such that MATH. Let MATH and, for MATH, let MATH be integers such that MATH and MATH mod MATH. Then MATH and at least half of MATH's belong to MATH. Note that if MATH, then MATH. Let MATH. Then MATH and MATH. By REF , if MATH, MATH . Summing these, we have MATH . This completes the proof. |
math/0102112 | The NAME polynomial of MATH is MATH. Let MATH denote MATH. First, we will show that MATH and MATH are relatively prime for distinct primes MATH and MATH. Suppose to the contrary there is a prime MATH dividing both MATH and MATH. Then, working modulo MATH, for MATH, MATH and MATH have a common root MATH in an extension field of the finite field MATH of MATH elements. We claim that the three polynomial MATH, MATH, and MATH have a common root, MATH, mod MATH. If MATH, this is obvious. Suppose MATH. Then MATH has two distinct roots MATH and MATH mod MATH and hence MATH must be quadratic over MATH. In particular, MATH in MATH. Thus, MATH or MATH. So, MATH mod MATH and MATH is a common root, MATH, mod MATH of the three polynomials. Since MATH and MATH are distinct primes, there are MATH and MATH such that MATH. So MATH mod MATH. This implies that MATH is a root of MATH mod MATH. However, this is a contradiction since MATH mod MATH. Thus there are no primes MATH dividing both MATH and MATH, implying they are relatively prime. It now suffices to show that MATH for any large primes MATH. We will show that MATH as prime MATH. If MATH is a matrix associated with the isometric structure of MATH and MATH is the kernel of MATH as before, then MATH since MATH. NAME matrix for MATH corresponding to the NAME surface in REF is MATH . So MATH . Let MATH, MATH, and MATH . Then MATH and MATH. We now see that, for any odd integer MATH, MATH and hence MATH. Since MATH, MATH, and MATH as prime MATH. In particular, MATH for any large primes MATH. This completes the proof. |
math/0102112 | Let MATH. For MATH, there is MATH satisfying the conclusion of REF . Let MATH. Suppose to the contrary that MATH and MATH is zero in MATH. Then MATH has vanishing NAME slice obstruction. Use the same notation MATH, MATH, MATH, etc. as before for MATH and put a subscript MATH on the objects corresponding to MATH. Let MATH be a metabolizer satisfying the conclusion of REF for the surface MATH. By REF , MATH contains an integral vector MATH that is either MATH or MATH for some MATH and MATH. By REF we can find a prime MATH and an odd prime MATH such that MATH divides MATH, MATH does not divide MATH, and MATH vanishes, where MATH denotes the MATH-primary component subgroup of MATH. Recall that MATH. From the proof of REF , MATH is the identity matrix multiplied by an integer MATH when MATH is odd and MATH. Thus MATH. Since MATH is the direct sum of MATH copies of MATH, we have MATH. Since the prime MATH divides MATH, MATH divides MATH and hence MATH for any MATH. In particular, for the vector MATH chosen above, MATH. Moreover, since MATH, MATH and hence MATH. On the other hand, let MATH be the constant from REF determined by MATH, MATH, and MATH chosen above together with MATH or MATH depending on whether MATH or MATH. Since MATH was chosen not to divide MATH, by multiplying an integer to MATH, we may further assume that MATH mod MATH. For MATH, MATH, let MATH, MATH be integers such that MATH, MATH mod MATH, and MATH mod MATH. Since MATH, we have MATH . By the additivity of MATH, it is equal to MATH which is, by REF , equal to MATH . By REF , we now see that MATH . Thus MATH, which is a contradiction. This proves REF . Next, suppose that MATH and MATH for all MATH. Under the same contradiction hypothesis and notation as above except: let MATH, instead of choosing it from REF . Observe that MATH and hence MATH by REF . We may assume MATH mod MATH as before. Then we have MATH . This is a contradiction, completing the proof of the first part of REF . In addition, suppose that MATH for some MATH. We only need to check the case MATH. By the definition of the averaged signature MATH, there are MATH and MATH such that MATH and MATH for any MATH with MATH. We can take MATH arbitrary large as above so that MATH for some integer MATH. Apply a similar argument as above. The difference from the previous case is that the role of MATH is switched with that of MATH. |
math/0102112 | The NAME polynomial of MATH is MATH. We will see that all MATH are coprime in MATH. Let MATH and let MATH be the greatest common divisor of MATH and MATH in MATH. Then MATH divides MATH that is a unit in MATH. So, MATH and MATH are relatively prime in MATH for any distinct pair of integers MATH and MATH. Let MATH be the set of integers MATH for which MATH has infinite order in MATH. REF implies MATH contains all but finitely many nonnegative integers. Note that under the hypotheses of REF MATH contains all but MATH or only MATH. Suppose that there are distinct MATH such that MATH is zero in MATH for some integers MATH. Since MATH, all MATH are pairwise coprime. Also, note that all MATH have nonsingular NAME forms. Applying REF inductively, each MATH must be zero in MATH and hence each MATH must be MATH. This completes the proof. |
math/0102112 | REF is an immediate consequence of REF since MATH for any MATH if MATH is the unknot. For REF , the torus knots MATH for MATH are such examples with MATH for all MATH and MATH for some MATH. |
math/0102112 | Let MATH, where MATH. Since MATH, it suffices to compute MATH for MATH when MATH. Then MATH. For MATH, let MATH . From REF , we have MATH. Let MATH . Then MATH. Observe that MATH is a quadratic polynomial in MATH with maximum at MATH and that MATH . To prove REF and the first part of REF , assume that MATH. Then MATH and hence MATH if MATH. Now, we will compute MATH when MATH, MATH and MATH. By REF , MATH . So, MATH . Since MATH, MATH can be negative only when MATH, and MATH is the minimum. Thus, MATH. This proves REF . Assuming MATH, we have the first statement of REF . Next, to prove REF and the second statement of REF , we compute MATH when MATH. Let MATH. By REF we have MATH since MATH. Observe that MATH since MATH and so MATH. Since MATH is either an integer or an integer plus MATH, MATH. Then MATH . If MATH, then MATH. If MATH, then MATH. Thus we have MATH . Note MATH or MATH. Thus, if MATH is sufficiently large then so is MATH. This completes the proof. |
math/0102112 | Let MATH. By REF there is MATH such that MATH for any MATH. Let MATH. We will show that MATH is a set satisfying the conclusion of REF . If MATH then MATH has infinite order in MATH by REF . Thus, for any MATH, MATH has no metabolizer for the isometric structure and hence has nonvanishing NAME ribbon obstruction. From now on, assume MATH. Suppose to the contrary that MATH has vanishing NAME ribbon obstruction for a positive integer MATH. Then it satisfies the conclusion of REF for MATH. Let MATH be the NAME surface for MATH as depicted in REF . We take MATH as the boundary connected sum of MATH copies of MATH so that MATH is a NAME surface of MATH. Then there is a metabolizer MATH for the isometric structure on MATH such that MATH vanishes for all primes MATH, where MATH is the MATH-primary component of the kernel, MATH, of MATH. Since MATH is the direct sum of MATH copies of the map MATH corresponding to MATH, MATH is the direct sum of MATH copies of MATH. Note that MATH, where MATH is the matrix of the isometric structure on MATH with respect to the basis MATH as defined in subsection REF. So MATH is generated by an element MATH. Thus every character in MATH is a direct sum of characters of the form MATH and MATH is isomorphic to MATH, where MATH is the maximal power of MATH dividing MATH. CITE showed that MATH and hence MATH. Using this and the NAME algorithm, CITE showed that MATH has an element in MATH having the first MATH entries equal to MATH and all the remaining MATH entries divisible by MATH for some MATH. Let MATH be an integer for which MATH. Multiplying by MATH, we see that MATH has an element MATH of the form MATH, where MATH can be any integers. Thus, MATH by the contradiction hypothesis. On the other hand, by the additivity of MATH, we have MATH . Observe that MATH for any MATH. Now by REF , MATH since MATH. So MATH. This is a contradiction, proving REF . Next, assume MATH and MATH. Let MATH denote the character as given above again. If MATH is a composite number, that is, MATH, then MATH by REF since none of MATH and MATH is MATH. Now assume MATH. Then MATH and MATH, where MATH and MATH are those in REF , and MATH. We now have MATH . This proves the first part of REF . For MATH, an elementary computation shows: MATH . So if MATH for MATH, then MATH are all positive for MATH. Note that MATH. This completes the proof. |
math/0102112 | The exact same proof of REF works here by applying all the counter-parts for the ribbon case: For instance, REF instead of REF . |
math/0102113 | We set MATH. It is enough to check that MATH remains unchanged under the switch of MATH and MATH in the tensor product. Suppose MATH is mapped to MATH under the isomorphism MATH. For MATH define MATH. Note that MATH and MATH satisfies REF if and only if so does MATH. Hence it is enough to show MATH. Set MATH . Then MATH. Since MATH is evident, we are left to show the following: CASE: MATH, CASE: MATH. REF for MATH is trivial. To check REF for MATH use REF , in particular, REF with MATH. For REF notice that MATH. |
math/0102113 | We start from REF . In the limit MATH, one has MATH for all MATH. Thus by setting MATH, the factor MATH gives rise to MATH. The summand MATH contained in the quadratic form REF thereby becomes MATH on account of REF . For a MATH such that MATH (hence MATH and MATH), write MATH in REF as MATH where the MATH-sum actually extends over only the vertices of MATH. To read off the quantum space data MATH for MATH from this, one identifies the first terms here with that for MATH from REF , leading to MATH . Solving this one finds REF . By a direct case check, the other MATH-binomial factors and the remaining terms in the quadratic form REF can be matched with the contributions from the product of MATH's. |
math/0102115 | There are canonical maps MATH and MATH given by MATH . These maps induce a commutative diagram and so, using the universal property of the tensor product, there is a natural map MATH . Conversely, to construct a natural map in the other direction, note that the first graded piece of MATH is just MATH, so there is an induced map MATH . It is easy to check that these maps are inverse to each other. |
math/0102115 | The spaces MATH are the analytic spectra associated to the symmetric algebras MATH, respectively, and MATH is represented by MATH which is the analytic spectrum of MATH. Hence we need to verify that there is a natural isomorphism MATH but this is a consequence of REF . |
math/0102115 | By REF the functors MATH are representable by linear fibre spaces over MATH. There are natural maps MATH and MATH . By definition we have MATH so the result follows from REF . |
math/0102115 | First note that for every complex space MATH we have MATH . Applying this to the case MATH, REF follows immediately from the assumption that MATH is locally free. Moreover, if REF is satisfied we have MATH where for the last isomorphism we have used REF . Thus REF follows. Conversely, assume that REF holds. Using REF we infer from the isomorphism in REF that MATH . Applying this to MATH, MATH a reduced point, it follows that the map MATH is surjective for every point MATH. Using standard arguments (see for example, CITE) we conclude that the functor MATH is exact on the category of coherent MATH-modules whence MATH is locally free, as required. |
math/0102115 | The functor MATH respectively, MATH on the category MATH of all analytic spaces is represented by MATH respectively, MATH. Thus MATH is simple if and only if MATH which is equivalent to the vanishing of MATH. Using NAME 's lemma, the equivalence of REF follows. Finally, the equivalence of REF is immediate from the definition of MATH. |
math/0102115 | The set of points MATH for which MATH is just the complement of the support of MATH and hence NAME in MATH. Using REF we get the desired result. |
math/0102115 | The fact that MATH is locally free over MATH, is well known (see, for example . CITE). Moreover, since MATH is simple we have MATH. As MATH we get MATH so the space MATH represents MATH as desired. |
math/0102115 | As the sheaf MATH is locally free over MATH we may assume that it has constant rank, say, MATH over MATH. By REF the functors MATH are representable by linear fibre spaces MATH respectively, MATH, where MATH and MATH are coherent MATH-modules. If for some space MATH the pairs MATH and MATH are isomorphic, then by REF MATH and MATH are locally free MATH-modules of rank MATH on MATH. Thus applying CITE as in the proof of CITE, we are reduced to the case that MATH and MATH are locally free MATH-modules of rank MATH. Let us consider the pairings MATH these correspond to pairings Using REF it follows as in the proof of CITE that our functor MATH is represented by the open subset MATH . |
math/0102115 | Let MATH be the groupoid where the objects over a space MATH are the coherent MATH-modules that are MATH-flat. As usual, given MATH and MATH, a morphism MATH in MATH consists of a pair MATH, where MATH is a MATH-morphism and MATH is an isomorphism of coherent sheaves. Assigning to a pair MATH the sheaf MATH gives a functor MATH. It is well known that there are semiuniversal deformations in MATH (see CITE or CITE) and that versality is open is MATH (see for example, CITE). The fibre of MATH over a given object MATH is the groupoid MATH as explained in CITE. More concretely, given a space MATH, an object in MATH over MATH is a morphism MATH . As the functor underlying MATH is representable by REF we get that the objects in MATH, MATH, admit semiuniversal deformations and that versality is open in MATH. Applying CITE gives the desired conclusion. |
math/0102115 | We will verify that REF - REF are satisfied. REF hold by REF . Moreover, REF is just REF . Finally, REF holds as simple pairs have no non-trivial automorphism. |
math/0102120 | Since MATH and MATH preserve the equalizer of MATH, we know that it is a MATH - subbicomodule. Analogously, MATH is a MATH - subbicomodule of MATH . In the commutative diagram MATH the second row is exact because MATH preserves the equalizer of MATH. The exactness of the first row is then deduced by using that MATH is assumed to preserve the equalizer of MATH. Now, consider the commutative diagram with exact rows: MATH REF gives the isomorphisms MATH and MATH, which induce the isomorphism MATH. |
math/0102120 | By REF, MATH is natural for every MATH. Thus, we have only to show that MATH is natural for every MATH. We argue first for MATH. Let MATH be a homomorphism in MATH. From the diagram MATH we get that MATH is natural. Now, use a free presentation MATH to obtain that MATH is natural for a general MATH. The rest of the statements are easily derived from this. |
math/0102120 | We need just to prove that REF commutes for MATH. In this case, the diagram can be factored out as MATH . Since all trapezia and triangles commute, the back rectangle does, as desired. |
math/0102120 | The equality will be first proved for MATH. Consider the diagram MATH . The back rectangle is commutative by definition of MATH, while the other two parallelograms are commutative because MATH is natural. Therefore, the right triangle is commutative. The equality is now easily extended for MATH and, by using a free presentation MATH, for any MATH. |
math/0102120 | In order to prove that the coaction MATH is coassociative, let us consider the diagram: MATH . We want to see that the top side is commutative. Since MATH is assumed to be MATH - compatible, we have just to prove that the mentioned side is commutative after composing with the isomorphism MATH. This is deduced by using REF , in conjunction with the naturality of MATH and the very definition of MATH. The counitary property is deduced from the commutative diagram MATH . To prove the second statement, let us consider the diagram MATH . Both triangles commute by definition of MATH and MATH, and the upper trapezium is commutative because MATH is natural. The bottom trapezium commutes by REF . Therefore, the back rectangle is commutative, which just says that MATH is a morphism of left MATH - comodules. This finishes the proof. |
math/0102120 | Let MATH be a right MATH - comodule. We have the following diagram with exact rows MATH where the desired isomorphism is given by the universal property of the kernel. |
math/0102120 | CASE: We need to prove that if MATH, then MATH is MATH - colinear if and only if the image of MATH is included in MATH. But these are straightforward computations in view of the definition of the contensor product. CASE: Since the inclusion MATH splits-off, we get from REF the natural isomorphism MATH. Now, the functor MATH is right adjoint CITE to the forgetful functor MATH and, by hypothesis, MATH is right adjoint to the functor MATH. This implies that MATH is right adjoint to MATH, as desired. |
math/0102120 | Consider the diagram MATH . The pentagon labeled as REF commutes since MATH is a homomorphism of corings. The four-edged REF is commutative since MATH is a left comodule. The commutation of the quadrilateral REF follows easily from the displayed decomposition of MATH. Therefore, the pentagon with bold arrows commutes. Now consider the diagram MATH . We have proved before that REF is commutative. Moreover, REF is obviously commutative and REF commute by REF. It follows that the outer curved diagram commutes, which gives the pseudo-coassocitative property for the coaction MATH. To check the counitary property, let MATH be the canonical isomorphism. We get from the commutative diagram MATH that the diagram MATH commutes, where MATH denotes the canonical isomorphism. Therefore, MATH is a left MATH - comodule structure map. In order to show that the assignment MATH is functorial, we will prove that MATH is a homomorphism of left MATH - comodules for every morphism MATH in MATH. So, we have to show that the outer rectangle in the following diagram is commutative MATH . From the definition of MATH and the fact that MATH is a morphism of MATH - comodules, it follows that the upper trapezium commutes. The lower trapezium is commutative by REF. Since the two triangles commute by definition, we get that the outer rectangle is commutative and MATH is a morphism in MATH. |
math/0102120 | Since the comultiplication MATH is coassociative, we get that the diagram MATH is commutative. This implies that the left trapezium of the following diagram is commutative. MATH . Since we know that the rest of inner diagrams are also commutative, we obtain that the outer diagram commutes, too. This proves that MATH is a MATH - bicomodule. We will see that MATH is a quasi-finite right MATH - comodule, that is, that MATH is left adjoint to MATH. Now, these functors fit in the commutative diagrams MATH where MATH denotes the forgetful functor. Since MATH is left adjoint to MATH and MATH is left adjoint to MATH we get the desired adjunction. The rest of the proposition follows from REF . |
math/0102120 | Assume that MATH is separable. By CITE or CITE, the unit of the adjunction MATH is split-mono, that is, there is a natural transformation MATH such that MATH. By REF the functor MATH is MATH - compatible. By REF , MATH is a MATH - bicomodule map. Obviously, MATH. To prove the converse, we need to construct a natural transformation MATH from the bicomodule map MATH. Given a right MATH - comodule MATH, consider the diagram MATH where the vertical are equalizer diagrams (here, we are using the definition of the cotensor product and the fact that MATH preserves the equalizer of MATH). If we prove that the top rectangle commutes, then there is a unique dotted arrow MATH making the bottom rectangle commute. The identity MATH is obvious; so we need just to prove that MATH which is equivalent to MATH and this last identity is easy to check. Now, consider the diagram MATH . By REF , we have that MATH. Since MATH is natural, and MATH, this implies that the external rectangle commutes. We know that the four trapezia commute, whence the internal rectangle commutes as well. Define MATH, which gives a natural transformation MATH. Moreover, MATH . Therefore, MATH and, by CITE, the functor MATH is separable. |
math/0102120 | Obviously, the forgetful functor coincides with MATH, so that we get from REF a characterization of the separability of this functor in terms of the existence of a MATH - bicomodule map MATH such that MATH. Now, notice that the adjointness isomorphism MATH transfers faithfully the mentioned properties of MATH to the desired properties of MATH. |
math/0102120 | If MATH is separable then, by CITE, there exists a natural transformation MATH such that MATH. In particular, MATH and, by REF , MATH is a bicomodule map (in fact, we easily get that the functor MATH is MATH - compatible from the fact that MATH is a direct summand of MATH as a left MATH - module). The map MATH gives an isomorphism of MATH - bicomodules MATH which implies, after REF, that MATH . Thus, MATH is the desired MATH - bicomodule map. For the converse, assume there is a MATH - bicomodule map MATH such that MATH. For each right MATH - comodule MATH, let us prove that the homomorphism of right MATH - comodules MATH factorizes throughout MATH. Since MATH preserves the equalizer of MATH, we know that MATH . Therefore, we need just to check the equality MATH . This is done by the following computation: MATH . Thus, we have proved that the natural transformation given in REF factorizes throughout a natural transformation MATH . This means that we have a commutative diagram MATH . Finally, we will show that MATH splits off MATH by means of the following computation: MATH . Since MATH is a monomorphism, we get MATH. By CITE, MATH is a separable functor. |
math/0102120 | This follows from REF taking that the MATH - bimodule homomorphisms from MATH to MATH correspond bijectively with the invariants of MATH into account. |
math/0102127 | Since MATH, we have MATH . Thus MATH for MATH. Then REF follows from induction on MATH immediately. Furthermore, REF follows from REF and the fact that MATH . |
math/0102127 | Since MATH for any nonnegative integers MATH, one direction is clear. We shall prove the other direction by using induction on MATH. Suppose that MATH. Let MATH. Then we have MATH for any MATH, so that MATH for MATH. Then MATH, where MATH. Thus it is true for MATH. Now, suppose that it is true for MATH and that a formal series MATH satisfies the relation MATH. Set MATH. Then MATH. By the inductive hypothesis, there are MATH such that MATH . Set MATH . Since MATH for any positive integer MATH, we have MATH . From the base step, we have MATH for some MATH. Then the inductive step follows immediately. This concludes the proof. |
math/0102127 | Suppose that MATH. Then, for any MATH, MATH in MATH. Then from CITE REF we have MATH for any MATH. It follows immediately that MATH for some nonnegative integer MATH if and only if MATH. Similarly, this is true for MATH. |
math/0102127 | Since MATH, for MATH, MATH if and only if MATH for some MATH. Clearly, REF is equivalent to that either MATH, or MATH and MATH. Then MATH if and only if MATH. Thus MATH. From REF , we have MATH assuming that MATH is in the domain of MATH. Then, for MATH, MATH for MATH, that is, MATH. It follows from REF that MATH commutes with MATH for any MATH . This proves REF . REF is immediate from REF by considering the coefficients of MATH . The first part of REF follows from REF by noticing that MATH if MATH and MATH. From REF , we have MATH . From a simple fact in linear algebra, it suffices to prove MATH that is, MATH . Otherwise, there exist MATH such that MATH . Then MATH where MATH and MATH . Compare the highest powers of MATH on both sides of REF . If MATH, we have MATH that is, MATH. This is a contradiction because MATH. Assume MATH. Then MATH if MATH and MATH if MATH. In both cases, we have MATH, noting that MATH. It is also a contradiction. If MATH, using a similar argument we can prove that MATH is a linear isomorphism from MATH onto MATH. REF is obvious. |
math/0102127 | Note that MATH for any MATH. Then in REF we can replace MATH by MATH. For any MATH, by REF we may define MATH by requiring MATH . Recall that MATH and MATH which is isomorphic to MATH linearly. By REF , MATH for MATH or equivalently MATH . |
math/0102127 | It is clear that MATH is an two-sided ideal of MATH, so that MATH is a commutative associative algebra. By REF , MATH is a NAME algebra with the NAME bracket: MATH for MATH. Taking MATH of REF then gives MATH . Then MATH is a NAME ideal of MATH so that MATH is a NAME algebra with the NAME bracket, and the NAME rule is clear. The proof is complete. |
math/0102127 | For MATH, suppose MATH where MATH lies in the space spanned by MATH for MATH and MATH . Then MATH is a MATH-module if and only if MATH . Since MATH is a MATH-module, we have MATH where MATH and MATH is an element of MATH representing MATH. Since MATH is a faithful MATH-module, REF follows from REF . This concludes the proof. |
math/0102127 | Let MATH be any restricted MATH-module. Then MATH is a restricted MATH-module. Set MATH. Then MATH is a local subspace of vertex operators on MATH. Thus MATH generates a canonical vertex algebra MATH such that MATH is a natural MATH-module. In particular, both MATH and MATH are MATH-modules. If we can prove that MATH, then we are done. It is clear that MATH is a MATH-module generated by MATH from MATH by REF . From the axioms of vertex algebra we know that MATH for all MATH and MATH . Thus MATH is a quotient MATH-module of MATH. On the other hand, MATH for all MATH and MATH . It is easy to prove (compare REF ) that MATH for all MATH and MATH . Then by REF , the linear map MATH from MATH to MATH defined by MATH is a MATH-homomorphism. In particular MATH is a MATH-homomorphism. Thus MATH is a MATH-isomorphism. The proof is complete. |
math/0102127 | Since MATH is a local vertex algebra on the base space MATH, MATH is a vertex algebra. Then it is clear that MATH is a quotient vertex algebra of MATH for any complex number MATH. |
math/0102127 | Noticing that MATH for MATH, by using induction on MATH we obtain MATH for MATH and any positive integer MATH. Because of this relation, MATH is linearly spanned by MATH for MATH. Then the linear map MATH from MATH to MATH defined by MATH for MATH is onto. Since MATH is a MATH-module with MATH being represented by MATH for MATH, all MATH for MATH form a NAME subalgebra of MATH. Let MATH be the representation on MATH of MATH. Then MATH is the linear map from MATH to MATH such that MATH. Because MATH for any MATH, MATH is injective, so that MATH is injective. Thus MATH is a linear isomorphism. For MATH, suppose MATH for some MATH. Then MATH . Therefore, MATH. Then the lemma follows immediately. |
math/0102127 | It was proved in REF that if MATH is a simple vertex operator algebra, then MATH. Since MATH is a graded algebra, the center is also graded. Let MATH be any homogeneous vector in the center of MATH. Then MATH for any MATH. By the skew-symmetry we get MATH . Thus MATH for any MATH. Since MATH, we obtain MATH for any MATH, so that MATH if MATH for some MATH. That is, if MATH, then MATH for any MATH. Let MATH be a positive integer such that MATH, so that MATH. Then we have MATH . In particular, MATH. Thus MATH. Then MATH. This concludes the proof. |
math/0102127 | The identity map on MATH induces an onto homomorphism of NAME algebras from MATH to MATH by the universal mapping property of MATH . Since MATH is isomorphic to MATH as a vector space, it follows immediately from PBW theorem and MATH that MATH is, in fact, isomorphic to MATH . |
math/0102127 | Clearly, MATH is a NAME ideal of MATH. Furthermore, by definition we have MATH . Then MATH. |
math/0102127 | MATH is a quotient vertex algebra of MATH modulo the ideal MATH generated by MATH for MATH . Then MATH is isomorphic to MATH by REF . Clearly, MATH is generated by MATH for MATH . |
math/0102127 | First we deal with the case MATH . Then MATH in this case is isomorphic to MATH and MATH is isomorphic to MATH by REF . For general MATH the result follows from REF and the fact that MATH is a quotient vertex algebra of MATH modulo the ideal generated by MATH . |
math/0102127 | From the proofs of REF we see that MATH is isomorphic to MATH . From the structure of MATH (compare REF ) we have MATH, where MATH and MATH is a fixed highest weight vector. Note that MATH is an ideal of the vertex operator algebra MATH. Then MATH by REF and MATH is a NAME ideal of MATH. Since the associative product is defined as MATH, the NAME product is defined as MATH and MATH for MATH, it is clear that MATH is a subset of the NAME ideal generated by MATH of MATH. Thus MATH is the NAME ideal generated by MATH. This concludes the proof. |
math/0102127 | If MATH, we have MATH. Since MATH for MATH, from REF we have MATH . Then the relation REF follows immediately. If MATH, we have MATH . Using REF we get MATH . This proves REF . Using REF again we get MATH . Since MATH, from this we immediately obtain REF . Clearly, REF are special cases of REF with MATH. |
math/0102127 | From the proof of REF we see that MATH contains the following subspaces: MATH for MATH. (Notice that MATH for MATH.) Let MATH be the space of MATH spanned by the above subspaces. Noticing that MATH spans MATH over MATH, we see that MATH as an associative algebra is generated by MATH for MATH with the relations REF . To show that there is no more relation, we prove that MATH. First, notice that the following holds: MATH for MATH. Because of REF is true with mod MATH being replaced by mod MATH. Then because of REF it is clear that MATH for any MATH. Second, since MATH for MATH, using induction we can prove MATH for any MATH. Third, similar to the proof of REF using induction again we can prove that MATH for any MATH. Thus MATH for MATH, so that MATH. Therefore MATH. From this, there exists a subset MATH such that MATH and MATH, where for a set MATH, MATH denotes the free abelian semigroup (with identity) generated by MATH. Then MATH gives rise to a basis of MATH. In view of REF , we have an algebra homomorphism MATH from MATH onto MATH such that MATH . Let MATH be the corresponding subset of MATH with MATH and MATH being replaced by MATH and MATH, respectively. Since MATH in MATH and MATH gives rise to a basis of MATH, MATH is linearly independent. It follows from the relations REF that MATH linearly spans MATH. Then it follows immediately that MATH is a linear isomorphism. Therefore, MATH is isomorphic to MATH as an algebra. |
math/0102127 | It follows from a calculation for the NAME brackets in MATH. |
math/0102129 | first part: Let MATH be an arbitrary decomposition of MATH, MATH the critical point of MATH and let MATH, MATH be the critical points of MATH. Observe that for each MATH there is an unique letter MATH in its itinerary, since MATH has MATH critical values. Hence MATH has itinerary MATH (by REF ). In particular, MATH. Suppose by induction that we know the inner itinerary of MATH, for MATH. Let MATH be the itinerary of MATH and MATH the itinerary of MATH. The itinerary of MATH is equal to the itinerary of MATH, except in two positions: if MATH then clearly MATH and if MATH then MATH. To find the position of the letter MATH, there are two cases: if there is MATH such that MATH, we conclude that the letter MATH happens in the itinerary of MATH in the same position that in the inner itinerary of MATH. If MATH for MATH, then the letter MATH happens in the itinerary of MATH exactly in the position MATH. Now the letter in the MATH-th position in the itinerary of MATH, where MATH, will be MATH if MATH is even and the word MATH has an even number of letters MATH; or MATH is odd and the word MATH has an odd number of letters MATH. Otherwise MATH. second part: We are going to prove that MATH can be deduced of the inner itinerary. We will get a proof by induction in MATH. Suppose that for any multimodal map MATH of type MATH and a decomposition MATH, it is possible obtain MATH of the inner itinerary of the critical points of the associate extended map MATH. Consider a multimodal map MATH of type MATH with decomposition MATH. Then the structure MATH associate to the map MATH can be deduced of the inner itinerary of MATH, by induction hypothesis. Let MATH be the critical points of MATH. By the proof of the first part, the inner itinerary of the points in each interval between the critical points of MATH can be deduced of the structure MATH. Select the interval MATH such that any point in MATH has the same inner itinerary that MATH with respect to MATH. Define MATH and the map MATH by MATH, if MATH and MATH. Then MATH and MATH. |
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