paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0105017 | One reduces to the two rectangle case by restriction to the subinterval MATH. Recalling that the tensor product of two rectangles is multiplicity-free and using the duality symmetry of tensor product multiplicities, one has MATH from which the result follows. |
math/0105017 | This holds by REF . |
math/0105017 | This follows from REF , and REF. |
math/0105017 | Note that MATH. Since MATH, MATH and MATH it follows that MATH so that indeed MATH. The other assertions follow easily from the definitions. |
math/0105017 | Let MATH, MATH, and MATH. The vacancy numbers change by MATH . Let MATH (respectively, MATH) be the column index of the box MATH (respectively, MATH). Then MATH. Since MATH it follows from REF that there are no singular strings of length MATH with MATH in MATH. Hence MATH. By induction on MATH it follows that MATH for MATH. Since by REF the strings in the range MATH are nonsingular we must have MATH. |
math/0105017 | By REF , one may reduce to the case that MATH is a sequence of single-rowed shapes. Recall that MATH. By REF it is obvious that MATH commutes with the duality map on rigged configurations. Thus it suffices to prove the theorem where MATH replaces MATH. We proceed by induction on MATH. The theorem is true for MATH. Let MATH and MATH be the column indices of the letters MATH in MATH. Then in MATH the column indices of the letters MATH are in the alphabet MATH with letters MATH omitted. Call them MATH for MATH. Let MATH be the tableau obtained from MATH by removing all letters MATH and MATH. By induction MATH. Hence it suffices to show that the addition of the letters MATH to MATH under MATH is (up to reversal of the order of all partitions) equal to the addition of the letters MATH to MATH under MATH. Set MATH and let MATH be the rigged configuration corresponding to MATH with letters MATH added in columns MATH. Adding the letter MATH in column MATH has the effect on the rigged configurations of selecting singular strings in MATH for MATH of length MATH maximal such that MATH, adding a box to the selected strings and making them singular again. By REF , MATH. It follows that MATH with MATH and MATH is uniquely defined to be the length of the maximal singular string in MATH such that MATH where MATH if MATH and MATH if MATH or MATH. Define a matrix MATH with MATH rows and MATH columns with entries MATH. The entries are weakly increasing along a row and weakly decreasing along a column. Some of the entries may be MATH. Now do the analogous construction for MATH. Call the selected strings under MATH, MATH for MATH and MATH. By the same arguments as before they are uniquely defined as the lengths of the maximal singular string in MATH such that MATH where MATH if MATH and MATH if MATH or MATH. This yields a matrix MATH with entries MATH. It is clear from REF that the entries in MATH can either be defined inductively row by row, top to bottom, right to left, or column by column, right to left, top to bottom. The same is true for MATH. Since by induction MATH for MATH it follows that MATH or equivalently MATH. This implies MATH as desired. |
math/0105017 | The proof follows from the Evacuation Theorem CITE, the Transpose Theorem CITE and MATH for MATH which is true by REF . |
math/0105017 | All the properties are immediate except for REF. Since MATH and MATH (respectively, MATH and MATH) commute for MATH, by the definitions we have for MATH . Here MATH denotes the largest integer smaller or equal to MATH. First let us verify REF for MATH. We have that MATH is a multiple of MATH by REF applied to MATH and REF for MATH viewed as an element of MATH. For MATH we have MATH by REF, and REF. A similar calculation establishes REF for MATH. Let MATH. Observe that MATH using REF for MATH and REF. Suppose first that MATH. Then MATH and one has MATH . If MATH then MATH and MATH . In either REF follows by REF. |
math/0105017 | Let MATH be aligned virtual MATH-crystals for MATH. Since they are aligned, it follows that for all MATH and all MATH, MATH and MATH either act entirely on the left factor or on the right factor of MATH. Therefore the operators MATH and MATH coincide with those given by the MATH-crystal structure on the tensor product MATH defined by REF, which define a tensor category. Finally, it is easy to verify that MATH is aligned. |
math/0105017 | MATH is an aligned simple virtual MATH-crystal, by REF , generated from the extremal vector MATH which holds by REF of simple virtual crystal MATH. Similarly MATH is an aligned simple virtual MATH-crystal with generator MATH. Now MATH is a MATH-crystal isomorphism such that MATH. As such it intertwines with the virtual operators MATH and MATH for MATH. Therefore MATH is a MATH-crystal isomorphism from the MATH-component of MATH to that of MATH, that is, from MATH to MATH. But there is a unique such map, namely, MATH. This proves REF. By abuse of notation, let MATH be defined by REF. It suffices to show that MATH satisfies the defining properties of the energy function given in REF . This is entirely straightforward except for REF. Let MATH for MATH and write MATH. Suppose first that MATH and MATH. Since MATH is aligned for MATH, this means MATH and MATH by REF. Applying MATH to MATH times and using REF, one has MATH . Dividing by MATH one obtains REF for MATH in the first case. The other cases are similar. |
math/0105017 | First let MATH. Since MATH is multiplicity-free as a MATH-crystal, it suffices to show that both MATH and MATH are MATH-highest weigh vectors of the same weight. Since MATH is a MATH-crystal isomorphism, MATH is a MATH-highest weight vector of weight MATH. MATH is a MATH-highest weight vector of weight MATH by REF. Since MATH it follows that MATH by REF. So REF holds. Now suppose MATH satisfies REF and MATH for some MATH. It will be shown that MATH also satisfies REF. This, together with a similar proof with MATH replacing MATH, suffices. Let MATH. We have MATH by REF , and the fact that MATH is a MATH-crystal isomorphism. REF holds for MATH and MATH by applying REF with MATH and MATH respectively. REF holds for MATH for MATH by observing that MATH and MATH commute, and applying REF for MATH and then for MATH. |
math/0105017 | Let MATH. Since MATH is an isomorphism of MATH-crystals we have MATH by REF . The proof for MATH is similar. |
math/0105017 | Let MATH. Suppose MATH for some MATH. Since MATH is an aligned virtual MATH-crystal, it follows that MATH where MATH for MATH and MATH for MATH. Replacing MATH with MATH and continuing in this manner, eventually one has MATH for MATH. |
math/0105017 | The proof that elements of MATH satisfy REF is similar to the proof of REF and is omitted. Suppose MATH. Let MATH be such that REF holds. Let MATH be the virtual MATH-crystal generated by MATH. It suffices to show that MATH. By the proof of REF holds for any element in MATH. But REF implies that MATH is MATH-aligned for MATH, in a manner similar to the proof of REF . By REF MATH is MATH-aligned. Therefore MATH is an aligned MATH-crystal. By REF MATH. |
math/0105017 | The following are equivalent: CASE: MATH does not satisfy REF. CASE: There is an index MATH such that MATH, MATH, and MATH. CASE: There is an index MATH such that MATH. CASE: MATH is not a tableau. Each of the above conditions is obviously equivalent to the next except for REF implies REF. For that case just take MATH to be minimal. If MATH or MATH then the index MATH satisfies REF, contradicting minimality. |
math/0105017 | The construction in the definition of MATH and MATH is equivalent to the algorithm of REF used to compute MATH. |
math/0105017 | For existence, define MATH and MATH by MATH . By REF and MATH are column words of the correct length. Applying the map MATH to REF, by REF and the fact that MATH is an involution, one has MATH . By REF and MATH, so that MATH and MATH are column words of the same length and MATH and MATH are column words of the same length. Thus to prove REF it is enough to show that MATH. Since the shapes of both MATH and MATH have two columns, it suffices to show that CASE: MATH for MATH and for MATH. CASE: MATH and MATH have the same shape. Now MATH and MATH. We have MATH by REF and MATH by REF. This proves REF. REF is equivalent to: MATH and MATH have the same shape. But MATH by REF with MATH. So it is enough to show that MATH. But this holds since MATH has at most two rows and exactly MATH cells (since MATH is taken within the MATH rectangle). For uniqueness, observe that MATH is a tableau with MATH ones and MATH twos, and MATH is a tableau of the same shape with MATH ones and MATH twos. The second tableau is uniquely specified by this property. |
math/0105017 | REF implies REF: Let MATH under the MATH-crystal isomorphism MATH given by REF. Then MATH, MATH, MATH, and MATH with MATH defined by REF . By REF MATH is a tableau, with MATH. REF implies REF: Let MATH be such that MATH and MATH. By REF MATH. By REF it follows that MATH under the isomorphism MATH. For the equivalence of REF, suppose that MATH and MATH are such that REF holds. Then the following are equivalent: CASE: MATH is a tableau. CASE: MATH is the column-reading word of a skew two-column tableau with unique southeast corner. CASE: MATH is a tableau. CASE: MATH is a tableau. Moreover it is easily seen that MATH. Now suppose REF holds. Write MATH in the form MATH. By the proof of REF it follows that REF holds. Thus REF follows. The proof that REF implies REF is entirely similar. |
math/0105017 | Let MATH be any weight such that MATH occurs in the MATH-crystal decomposition of MATH. MATH occurs in MATH with multiplicity one, and on this component MATH has value MATH if MATH has one tensor factor and value MATH otherwise CITE. Let MATH be the unique MATH-highest weight vector in MATH with MATH. There is an embedding of virtual MATH-crystals MATH given by MATH where MATH. The map is a morphism of virtual MATH-crystals since MATH is a morphism of MATH-crystals. Now let MATH be a weight appearing in REF. Then MATH occurs in MATH. By the argument in the previous paragraph it suffices to show that MATH. It can be shown that there is an explicit sequence of virtual operators going from MATH to MATH. Moreover one has MATH if MATH has a single tensor factor, and otherwise MATH . |
math/0105017 | The forward direction holds by REF . For the reverse direction, suppose MATH satisfies REF. Let MATH be the virtual MATH-crystal generated by MATH. It follows from the proof of REF that every element of MATH satisfies REF. But MATH is automatically aligned by REF . By REF there is a MATH-highest weight vector MATH that is also a MATH-highest weight vector. Since MATH satisfies REF, MATH. Since MATH it follows that MATH for some MATH appearing in REF. By REF MATH. It follows that MATH. |
math/0105017 | This follows from the proof of REF . |
math/0105017 | The lemma follows directly from MATH, which holds by REF for MATH, REF. |
math/0105017 | By REF MATH and the embedding is inclusion. To show equality it suffices to show that MATH is closed under MATH and MATH. To check this we use the explicit computation of the MATH-string in MATH-crystals given in REF. Let MATH such that MATH. Write MATH. Then MATH and MATH both contain MATH and do not contain MATH. Computing MATH on MATH, in all cases MATH and MATH are obtained from MATH and MATH respectively by removing MATH from the bottom and adding MATH at the top. Thus MATH is a tableau. MATH satisfies REF by REF . Let MATH and MATH be defined as MATH for MATH. Then MATH is a tableau, being obtained from MATH by removing MATH from the bottom of each column. From this one also sees that MATH. By REF , MATH. In an entirely similar manner, one may show that MATH is closed under MATH. By REF it remains to show that the elements of MATH are MATH-aligned. By REF , we need only check the case that MATH and MATH occur in MATH once each. Since MATH is a tableau, MATH. If MATH then MATH and MATH is MATH-aligned. So assume MATH. If MATH, then MATH has two tensor factors, and by REF MATH is MATH-aligned. MATH is MATH-aligned. So let MATH. Write MATH and MATH. We will show that MATH, as the proof of MATH is similar. MATH by REF. Thus it is enough to show that MATH. Since MATH has a single MATH and a single MATH, this holds if MATH is in the right hand column of MATH. But MATH by REF. So it suffices to show that MATH remains in the right hand column in passing from MATH to MATH. But there is only one letter that moves from the right hand column to the left under MATH, and it must be in MATH (and therefore is not MATH) since MATH is a tableau with columns of equal size. |
math/0105017 | The following are equivalent: CASE: MATH satisfies REF. CASE: MATH. CASE: MATH satisfies REF. The equivalence of the first two items follows from the fact that MATH is multiplicity-free as a MATH-crystal and MATH is a MATH-crystal morphism. The equivalence of the second and third items is a consequence of REF , and the definition of MATH. Write MATH where MATH and MATH are column words. Since MATH and MATH are column words of lengths MATH and MATH respectively, it follows that MATH and MATH have lengths MATH and MATH for some MATH. Moreover one can pass between MATH and MATH using a two column jeu de taquin. Let MATH and MATH be defined in a manner similar to MATH for MATH. Suppose first that MATH. Then MATH for some MATH. By REF , MATH satisfies REF. By the above argument MATH satisfies REF. By REF MATH is a tableau and MATH. To finish the forward direction it is enough to show that MATH and MATH. In passing from MATH to MATH, some letters go from the left column to the right. Since all letters in MATH are blocking letters of MATH, the only letters that can block MATH from moving to the right are those in MATH. But by REF , MATH, so there are MATH letters in MATH available to move to the right. The smallest MATH of these, actually do move. In particular MATH and MATH. For the reverse direction, suppose MATH satisfies the three properties. Since MATH is a tableau, it suffices to show that MATH satisfies the properties in REF for MATH. Since MATH satisfies REF, MATH satisfies REF, arguing as above. This time, passing from MATH to MATH, we assume that MATH is a tableau. That means that all the letters of MATH cannot move to the left, and so stay in their column. The left column only gets larger, and can only get larger by letters in the interval MATH. Therefore MATH and MATH again. |
math/0105017 | Observe that alignedness follows from the above MATH-crystal decomposition, since MATH for MATH and the virtual MATH-crystals MATH are MATH-aligned for MATH. The map MATH coincides with MATH by REF , since both are MATH-crystal embeddings MATH. Let MATH be as in REF , which asserts that MATH. To show equality it suffices to show that MATH is closed under MATH and MATH. Recall that for MATH, MATH and MATH. Let MATH be such that MATH. Write MATH. Let MATH and MATH be defined as MATH is for MATH. It will be shown that MATH satisfies the three properties in REF . Since MATH, MATH satisfies REF by REF . Suppose first that MATH. Then MATH and MATH. It is easily seen that REF hold for MATH, so that MATH. Otherwise suppose MATH. Then MATH, MATH, MATH, and MATH is obtained from MATH by removing MATH from the bottom and putting MATH at the top. In particular MATH and MATH. The only way that MATH fails to be a tableau is if MATH, when the first column is shorter than the second. Since MATH is a tableau with columns of equal length, MATH. By REF there are no ones present. But then MATH, contrary to assumption. Therefore MATH is a tableau. To check REF it is enough to show MATH. Suppose not, that is, equality holds. Following the proof of REF , write MATH and define MATH and MATH. Then MATH. As in the aforementioned proof it can be shown that MATH for some MATH and that MATH and MATH. By our assumption it follows that MATH and hence that MATH and MATH. In particular MATH is a tableau. By REF MATH or MATH but not both. But MATH is not possible since MATH is a tableau and MATH. And if MATH then MATH, contrary to our assumption. The proof that MATH is closed under MATH, is similar. |
math/0105017 | This follows from REF . |
math/0105017 | It suffices to show that MATH is MATH-aligned and closed under MATH and MATH. Let MATH. By REF , MATH is a tableau and MATH where MATH is the number of occurrences of MATH in MATH for MATH. Write MATH where MATH and MATH are column words of type MATH. Suppose first that MATH. Then MATH and MATH, so that MATH is MATH-aligned. Writing MATH the column words MATH and MATH are obtained from MATH and MATH by removing MATH from the top and putting MATH at the bottom. Clearly MATH is a tableau. Since MATH it satisfies REF by REF and is therefore in MATH by REF . An entirely similar argument applies for the case MATH. Finally, suppose MATH. Since MATH is a tableau, MATH. Regardless of where the single symbol MATH appears in MATH, one has MATH, so that MATH is again MATH-aligned. |
math/0105017 | MATH satisfies the conditions of REF for all elements of MATH; this follows immediately from the known properties of MATH recorded in REF . To see that MATH stabilizes MATH, by REF it is enough to show that MATH. But this element is of extremal weight and is easily seen to be in the MATH-component of MATH by direct computation. |
math/0105017 | Let MATH. Write MATH and MATH where MATH and MATH and similarly for MATH. By REF MATH is a tableau and MATH. Suppose first that MATH. By direct computation MATH where MATH means to add or subtract MATH from each entry in the tableau MATH. Applying REF again, it follows that MATH since MATH has the same above properties that MATH does. Otherwise let MATH. Then MATH for MATH. Now MATH which has two columns of equal length. This means that MATH has two columns of equal length, so that it is equal to its MATH tableau. But MATH. Therefore the columns of MATH are given by MATH and MATH, and the previous argument goes through. |
math/0105017 | By CITE it is known that there is an irreducible integrable MATH-module MATH with crystal basis which is isomorphic to MATH as a MATH-crystal. Since the MATH-highest weight vectors in MATH have distinct weights, it follows by REF that the map MATH of REF for the crystal basis of the module MATH is uniquely determined by REF, and REF where only MATH are used. By REF , the involution MATH of MATH restricts to an involution on MATH. As such it satisfies the properties of REF , by REF . |
math/0105017 | In CITE a MATH-module MATH was constructed. Its crystal basis MATH was shown to be isomorphic to MATH as a MATH-crystal. This agrees with the decomposition specified by REF. Combining REF and the explicit action of MATH and MATH computed in CITE, it follows that MATH as MATH-crystals. |
math/0105017 | The MATH-module MATH is also the MATH-module MATH. Therefore their crystal bases MATH and MATH are equal as sets (but not as weighted crystals). By abuse of notation we write MATH for either crystal, using MATH to distinguish the two structures on this set. The isomorphism of NAME diagrams implies that MATH on MATH. Therefore there is an automorphism MATH such that MATH for all MATH. We claim that MATH as MATH-crystals. In light of the above facts and the MATH-crystal isomorphism MATH, it is enough to show that there is a unique bijection MATH such that REF holds for all MATH. But the map MATH is unique since MATH is connected as a virtual MATH-crystal and there is no choice for MATH by weight considerations. Recall that MATH. It is easily verified that the automorphism MATH of the MATH-crystal MATH satisfies REF. |
math/0105017 | This follows directly from REF . |
math/0105017 | We prove REF by induction on increasing MATH and decreasing MATH. Suppose that for given MATH and MATH is not a nut-bolt pair for any MATH. We want to show that then MATH is a nut-bolt pair. First assume that MATH is maximal, that is MATH. Note that from the duality and the tableau condition of MATH as stated in REF it follows that MATH and MATH for MATH. Since we know already that the final rigged configuration has to be symmetric, that is, satisfies REF , and since MATH for all MATH by REF we therefore must have MATH so that MATH forms a nut-bolt pair. Now assume that MATH. Then by induction REF must hold for MATH. Assume that MATH is a nut-bolt pair. If MATH is also a nut-bolt pair then by symmetry MATH so that MATH indeed forms a nut-bolt pair. If MATH is a nut-bolt pair for some MATH or MATH then MATH since otherwise MATH would be a nut-bolt pair which contradicts our assumptions. Then again by symmetry MATH and MATH forms a nut-bolt pair. Otherwise assume that MATH is a nut-bolt pair for some MATH. If MATH is also a nut-bolt pair then MATH cannot be a nut-bolt pair for any MATH since otherwise MATH would be a nut-bolt pair which contradicts our assumptions. Hence MATH must be a nut-bolt pair. By the change of vacancy numbers there are no singular strings of length MATH in the MATH-th rigged partition after the addition of all letters up to MATH. Since MATH this shows by symmetry that MATH and hence MATH forms a nut-bolt pair. If MATH is not a nut-bolt pair then MATH by REF (note that MATH since otherwise MATH would be a nut-bolt pair which contradicts our assumptions). By the same arguments as before there are no singular strings of length MATH in the MATH-th rigged partition after the addition of all letters up to MATH. Using MATH REF follows again. REF are proven in a similar fashion. |
math/0105017 | The lower bound follows directly from REF. The upper bound is proven by increasing induction on MATH and decreasing induction on MATH. Fix MATH. Let MATH be maximal, that is MATH. Assume that MATH is a nut-bolt pair for some MATH. The conditions of the Lemma require that MATH. By REF is the length of the largest singular string in MATH subject to the condition MATH. There is a singular string of length MATH in MATH. Hence by symmetry MATH must form a nut-bolt pair so that MATH. This contradicts MATH. Therefore, by REF, either MATH or MATH is a nut-bolt pair. By the definition of MATH there is no singular string of length MATH in MATH and hence by symmetry in MATH. Using that MATH and taking into account the change in vacancy numbers this implies that MATH. If MATH then by REF and the crossing conditions on MATH-slides exactly one of the following three must be a nut-bolt pair for MATH: CASE: MATH; CASE: MATH; CASE: MATH. We are going to treat each case separately. Assume that REF holds. If MATH, then by the definition of MATH there are no singular strings of length MATH in MATH. By symmetry there are no singular strings of length MATH in MATH. Taking into account the change of vacancy number, there are still no singular strings of length MATH in the MATH-th rigged partition after the addition of all letters up to MATH. Since MATH by REF it follows that MATH. If MATH, then MATH must be a nut-bolt pair. By the same arguments as above there are no singular strings of length MATH in the MATH-th rigged partition after the addition of all letters MATH. Since MATH by REF this implies MATH. Assume that REF holds. Then the conditions of REF are satisfied for MATH replaced by MATH and by induction MATH. Note that MATH in REF . Since MATH it follows that MATH. Hence by REF it follows that MATH. Assume that REF holds. Let us first show that MATH. If MATH is not a nut-bolt pair then by induction MATH (note that MATH since otherwise MATH would be a nut-bolt pair which contradicts our assumptions). Since MATH by REF it follows that MATH. If MATH is a nut-bolt pair then REF MATH or REF MATH is a nut-bolt pair. In REF MATH so that the assertion holds. In REF the assumptions of the Lemma hold for MATH replaced by MATH so that by induction MATH. Since MATH by REF the assertion holds. There are no singular strings of length MATH in the MATH-th rigged partition after the addition of all letters up to MATH because of the change of the vacancy numbers. Furthermore there are no singular strings of length MATH in MATH by the definition of MATH. Hence by symmetry and change in vacancy numbers, there are no singular strings of length MATH in the MATH-th rigged partition after the addition of all letters up to MATH. Now assume that MATH. Then these conditions and the fact that MATH is a nut-bolt pair if MATH imply that MATH. Hence the only case left to consider is MATH. By the above arguments this case can only occur when MATH and MATH are nut-bolt pairs with MATH. We will show that MATH cannot occur unless MATH in which REF obviously holds. Assume that MATH. By the definition of MATH and symmetry there is no singular string of length MATH available for MATH unless REF MATH or REF MATH. Assume REF holds. By REF either MATH or MATH has to be a nut-bolt pair. However, the first case yields a contradiction since MATH so that MATH must be a nut-bolt pair. Assume that REF holds. Then MATH must be a nut-bolt pair. By REF this implies that MATH and MATH must be a nut-bolt pair. Hence in both REF MATH must be a nut-bolt pair and MATH. Now, there is no singular string of length MATH for MATH unless MATH. But then MATH must be a nut-bolt pair since otherwise MATH is a nut-bolt pair which contradicts our assumptions. This in turn implies that MATH. By REF either MATH or MATH must be a nut-bolt pair. However, the first case contradicts MATH since MATH. Hence MATH must be a nut-bolt pair. By induction MATH. In summary both MATH and MATH are nut-bolt pairs with MATH. Now repeat the entire argument which shows that both MATH and MATH are nut-bolt pairs with MATH etc. Since there are only finitely many MATH-slides these conditions must eventually break down which shows that MATH cannot occur when MATH. This concludes the proof of REF . |
math/0105017 | We prove REF by induction on MATH. CASE: If MATH then by REF also MATH and MATH. This implies that MATH is the largest singular string in MATH and MATH is the largest singular string in the central rigged partition after the addition of MATH. Since MATH is singular it follows that MATH. After the addition of MATH the vacancy numbers of the strings of length MATH in the central rigged partition decrease by one. However, since all labels are even no string of length MATH becomes singular. Hence MATH. Now assume that MATH. If MATH is a nut-bolt pair then by REF MATH. Furthermore MATH since otherwise the folding of the MATH-th MATH-slide would not cross the MATH-th MATH-slide in a bolt. This implies that MATH. Also MATH since otherwise MATH was not a nut-bolt pair. Since there is a singular string of length MATH this implies that MATH. After the addition of MATH, the vacancy numbers corresponding to the strings of length MATH in the central rigged partition have decreased by REF. Again, since there are no singular strings of length MATH in MATH by the definition of MATH and all labels are even, it follows that there are no singular strings of length MATH after the addition of MATH. Hence MATH. If MATH is a nut-bolt pair then by REF we have MATH and MATH. This implies MATH. After the addition of MATH the vacancy number in the central rigged partition corresponding to the strings of length MATH decreases by one and of length MATH increases by one. Hence there are no singular strings of length MATH and by the now familiar arguments also not for MATH since all labels are even. This proves MATH. Finally let MATH be a nut-bolt pair. Note that this case can only happen if MATH. We will show that MATH. Since by REF MATH it then follows that MATH. If MATH then MATH for MATH to hold. The length of the longest singular string in MATH is MATH and hence by symmetry also in MATH. This forces MATH. Hence assume MATH. If MATH is a nut-bolt pair then MATH holds by REF . If MATH is a nut-bolt pair then the vacancy numbers in the MATH-st rigged partition corresponding to strings of length MATH are increased by one after the addition of MATH so that there are no singular strings of this length. By REF MATH (note that the case MATH is included here). Hence using MATH and REF this implies MATH. By the definition of MATH there are no singular strings in MATH of length MATH and hence by symmetry also not in MATH. After the addition of MATH there is a singular string of length MATH so that MATH. To conclude assume that MATH is a nut-bolt pair. Then by REF MATH. By the same arguments as in the previous case there are no singular strings of length MATH in the MATH-st rigged partition after the addition of MATH so that MATH as asserted. CASE: Assume that MATH is a nut-bolt pair. Then MATH, MATH and MATH. Since there is a singular string of length MATH after the addition of MATH and the letters MATH it follows that MATH by REF. By the definition of MATH there are no singular strings of length MATH in MATH. The vacancy numbers of the central rigged partition corresponding to the strings of length MATH decrease by one. Hence due to the even labels there are no singular strings of length MATH in the central rigged partition after the addition of all letters up to MATH. This proves REF. Now assume that MATH is a nut-bolt pair. By induction MATH. The vacancy numbers of the central rigged partition after the addition of all letter up to MATH for the strings of length MATH increase by one and for the strings of length MATH decrease by one. Since there are no singular strings of length MATH in MATH by the definition of MATH there are no singular strings of length MATH in the central rigged partition after the addition of all letters up to MATH due to the even labels. Also MATH. Hence REF holds unless MATH and MATH. We will show that in the present case this cannot occur. Assume that MATH and MATH. This requires MATH and MATH by REF. By REF either REF MATH or REF MATH forms a nut-bolt pair. Since MATH is a nut-bolt pair by assumption, MATH in REF must be MATH. Hence if REF holds there is no singular string of length MATH in the central rigged partition after the addition of all letter up to MATH by the change in vacancy number. Hence MATH cannot hold. By the same arguments there is no singular string of length MATH if MATH. Hence either REF MATH or REF MATH. Since MATH REF requires REF which we already argued yields a contradiction. Hence assume REF holds. But then by REF MATH must be a nut-bolt pair which contradicts MATH. This concludes the proof of REF when MATH is a nut-bolt pair. Finally assume that MATH is a nut-bolt pair. If MATH then MATH and MATH, MATH and MATH. By symmetry MATH forms a nut-bolt pair which contradicts our assumption. Hence we may assume from now on that MATH. We will show that MATH. Then by REF it follows that MATH. If MATH is a nut-bolt pair then MATH since otherwise the MATH-st and the MATH-th MATH-slide would not cross in a bolt. Hence by the definition of MATH there are no singular strings of length MATH in MATH. By symmetry there are no singular strings of length MATH in MATH. Since MATH by REF it follows that MATH. If MATH is a nut-bolt pair then MATH follows from REF . The last case to consider is that MATH is a nut-bolt pair. Note that due to the change of vacancy numbers there are no singular strings of length MATH in the MATH-st rigged partition after the addition of all letters up to MATH. Also by the definition of MATH and symmetry there are no singular strings of length MATH in the same rigged partition. Hence, if MATH then MATH as asserted. We may therefore restrict our attention to the case MATH with MATH (note that MATH would imply the assertion). We will show that these conditions cannot hold simultaneously. By REF MATH can only happen if REF MATH is a nut-bolt pair or REF MATH. Note that REF implies REF since otherwise MATH is a nut-bolt pair which contradicts our assumptions. REF implies by induction that MATH. In summary MATH is a nut-bolt pair with MATH. There are no singular strings of length MATH available in the MATH-st rigged partition for MATH unless REF MATH or REF MATH. In REF by REF either MATH or MATH forms a nut-bolt pair. The first case contradicts MATH. In REF MATH so that MATH must form a nut-bolt pair. Hence in both REF MATH forms a nut-bolt pair and MATH. By the change in vacancy numbers there is no singular string of length MATH available for MATH unless MATH. Then MATH must be a nut-bolt pair since otherwise MATH forms a nut-bolt pair which contradicts our assumptions. By induction MATH. Repeating the same arguments shows that MATH forms a nut-bolt pair with MATH etc. Since there are only finitely many MATH-slides these conditions must be eventually broken which shows that MATH is not possible unless MATH. This concludes the proof of REF. |
math/0105018 | Remember that MATH and MATH are isotopic iff there is a (differentiable) homotopy MATH between them such that for fixed MATH, MATH is a MATH-diffeomorphism MATH. Consider then the identity cobordisms MATH and MATH. The map MATH, where MATH is the projection MATH, is clearly a MATH-diffeomorphism. Since when restricted to the boundary it is the disjoint union of MATH and MATH, relation REF gives the desired result. |
math/0105018 | Let MATH be an homotopy between MATH and MATH. Then we have a MATH-cobordism MATH. On the other hand if we define MATH by MATH we have another cobordism MATH. What we are going to show is that they are inverse to each other. Of course we can "shift" this cobordism to MATH with MATH now defined by MATH, by the diffeomorphism MATH. Now we perform the gluing of these two MATH-cobordisms to obtain MATH and we will prove that it is the identity by first deforming MATH to a "constant" map and then using a contracting diffeomorphism MATH and relation REF. A homotopy MATH is a map MATH. Define MATH by MATH . Note that MATH and MATH. The fact that it is a well-defined and continuous map and also a homotopy relative to the boundary are also easy-to-establish facts that we leave to the reader. This implies that MATH. Now define the diffeomorphism MATH by MATH . This extends to a MATH-diffeomorphism MATH which is the identity on the boundary, so, by applying relation REF we have finally MATH. This gives MATH. To get the equality in the other direction first shift by MATH and rerun the above proof with suitable changes. |
math/0105018 | If MATH is a MATH-diffeomorphism MATH, then by relation REF and relation REF we have MATH and the term on the right is clearly a cylinder cobordism. For the second statement, note that if MATH is a cobordism with a NAME function MATH that has no critical values then by the cylinder recognition theorem (see the book CITE, page REF) there is a diffeomorphism MATH such that the triangle in REF is commutative. Since MATH is the identity in MATH, relation REF implies the equality MATH . The right hand side is an isomorphism by REF , and pushing MATH, which is a diffeomorphism, to the right side we have the result. |
math/0105018 | Note that MATH is the gluing of four cylinders, each copy with only trivial homotopy information on it. So we can first "straighten" out the gluing by a convenient diffeomorphism to conclude the equality with MATH and then squeeze MATH to the unit interval MATH to get the identity. The other triangular identity is obtained in the same way. Now let us prove that MATH is just MATH but as a morphism MATH. By definition, MATH is obtained from MATH by gluing four cylinders, two at the beginning and the other two at the end. So, once again, by "straightening" out the gluing, we get MATH . Squashing MATH to the unit interval MATH we get the identity on both sides, and the equality is proved. Now let MATH be a MATH-diffeomorphism, then by REF we have the equality MATH . Applying the above calculation to the cobordism on the right we have MATH . Now, a gluing made along MATH is clearly diffeomorphic to a gluing made along MATH, so we have MATH . From these calculations it is clear that the duality is strict. REF is also obvious from the above computation of duals. The equality REF follows from exactly the same type of arguments used up to now and is left to the reader. |
math/0105018 | By REF the naturality question is reduced to proving the commutativity of the square in REF for a general MATH-cobordism MATH. But since MATH is an isomorphism, commutativity of REF is the same as the equality MATH . The right term consists of two cylinders glued at the ends of MATH and the map into MATH is defined by MATH. The proof will be accomplished if we can deform this map into MATH by a homotopy that is fixed on the boundary. In MATH we define this homotopy simply by MATH . Note that if MATH then this reduces to MATH and if MATH then this reduces to MATH. For MATH we define the homotopy by MATH . This is a continuous function because for MATH the first branch is MATH. For MATH this is just MATH and therefore continuity of the whole homotopy is ensured. For MATH this is MATH and therefore the homotopy remains fixed in this component of the boundary. We leave to the reader the easy task of supplying the definition of the homotopy for MATH as well as making the necessary checks. As for being a monoidal transformation, since the induced functors are strict monoidal, this is equivalent to ask that MATH . But this equality follows from the diffeomorphism MATH and relation REF. |
math/0105018 | Since our manifolds are differentiable, they admit (at least) one structure of NAME with cells up to the dimension of the manifold. Standard obstruction theory says that the map MATH induces the isomorphisms MATH . It is clear that MATH is a MATH-diffeomorphism iff MATH is a MATH-diffeomorphism, and this fact, together with an application of the second isomorphism for MATH-cobordisms, implies that MATH is full and faithful. The first isomorphism coupled with an application of REF implies that MATH is isomorphism-dense. |
math/0105018 | The first step is to use the collapsing construction and its universal property depicted by the diagram in REF where the map of pairs MATH is the collapsing map. This universal property just says that any map of pairs MATH factorizes through a unique map (drawn in dashed lines) MATH. In terms of isomorphisms it entails MATH but as MATH we have MATH . Now use the exact sequence of a pair to get, for MATH, MATH . Finally, with MATH and the fact that the inclusion MATH, being a closed cofibration, implies the natural isomorphism MATH we get the promised isomorphism of the proposition. |
math/0105018 | Part of the proof was done above when we have showed that any object is isomorphic to a disjoint union of copies of MATH and MATH and shown that every MATH-diffeomorphism is either a symmetry or can be turned into a cylinder cobordism. Note that by the classification theorems of MATH-manifolds, plus the gluing REF and the above commutative diagrams in REF , for any MATH-manifold that has a non-trivial homology class MATH in it, we can always push the MATH class to a cylinder. Also by the classification theorems of MATH-manifolds, every such manifold with a given orientation can be decomposed into a gluing and disjoint union of the basic morphisms listed above. On the other hand every such manifold, viewed as a cobordism in some particular way, can be decomposed in terms of these basic morphisms and their duals, that is, compositions of the basic morphisms with the pairings MATH and MATH. Of course, for a given cobordism there may be more than one way of writing it as a composition of the basic cobordisms so we have to ensure that all the relations are covered by the algebraic relations of a NAME object and the relations of a symmetric monoidal category. But this is a purely topological fact pertaining to NAME theory and that the reader can see in CITE. |
math/0105018 | We need to prove that the relation on arrows is in fact a congruence. So, let MATH and MATH, then we have the commutative squares in REF . Since MATH is thin we have MATH. So, compose the two squares in REF top to bottom to obtain the desired MATH. To see that MATH is an equivalence note that if MATH and MATH in REF are both in MATH then because MATH is thin we have MATH and MATH but this immediately implies that the map MATH is bijective. Finally, the fact that MATH is surjective on objects implies that it is an equivalence. |
math/0105018 | Just define the functor, called MATH, say, as MATH and MATH. The above conditions on MATH guarantee that this is a sound definition and the rest follows. |
math/0105018 | The above step-by-step construction provides a map MATH . It is not difficult to prove that it is indeed a functor and that it has an inverse follows from the (trivial) reconstruction of an HQFT from the respective functor MATH. The fact that it is a natural isomorphism is also easy to prove and we leave to the reader the task of providing the specific details. |
math/0105020 | This is REF, in the case MATH. Our element MATH is NAME 's MATH, and his MATH is our MATH. There is a slight misprint in the statement: the element MATH is MATH, not MATH. Geometrically, the square comes from the NAME diagram of cofibrations, as studied in CITE. In this context, MATH is identified with MATH, and MATH with MATH. |
math/0105020 | This is standard, but we give a proof for completeness. First, it is clear that there is a commutative square as shown. Thus, if we let MATH denote the pullback of MATH and MATH, we get a map MATH, and we must show that it is an isomorphism. Suppose that MATH has image zero in both MATH and in MATH. As MATH in MATH we have MATH for some MATH. As MATH in MATH we see that MATH in MATH, so MATH say. It follows that MATH but MATH so MATH or in other words MATH. This means that the map MATH is injective. Now suppose we have an element MATH, so MATH and MATH, and MATH and MATH have the same image in MATH. We then have MATH for some MATH and MATH, and MATH in MATH for some MATH. Next, we can choose a sequence MATH in MATH such that MATH and MATH in MATH. It follows that the element MATH maps to zero in MATH, say MATH. We may replace MATH by MATH and thus assume that MATH for all MATH. Now let MATH be such that MATH. As MATH, we see that MATH. As MATH for all MATH, we have MATH. When MATH this gives MATH and thus MATH. We put MATH. It is clear that MATH in MATH. Moreover, the equation MATH shows that MATH in MATH, so MATH. Thus, the map MATH is also surjective, as required. |
math/0105020 | First, note that the definitions of MATH and MATH are simply the appropriate specialisations of the definition of MATH so they can essentially be ignored. To see that MATH respects the relations, note that MATH . It is trivial to check that MATH. There is an evident map MATH, which induces a map MATH. As MATH is already complete at MATH, we also get a map MATH. It is easy to see that the composite MATH is just the usual quotient map. Next, one checks by induction on MATH that in MATH we have MATH . In each case the right hand side converges MATH - adically to MATH, so in MATH we have MATH . Using this, we see that MATH is surjective. Moreover, as a special case of the first equation, we have MATH. Putting this into the second equation gives MATH. In particular, we have MATH, so MATH factors through a map MATH. As MATH is surjective and MATH we see that MATH is an isomorphism. |
math/0105020 | First observe that MATH so the notation is self-consistent. To see that MATH respects the relations, note that MATH . It follows that we have a ring homomorphism as described. It induces a map MATH, which we again call MATH. On the other hand, it is clear that there is a unique map MATH sending MATH to MATH, and that MATH. As MATH in MATH, we see inductively that MATH lies in the image of MATH for all MATH and MATH. A similar argument shows that MATH lies in the image of MATH, so MATH is surjective. As MATH we deduce that MATH and MATH give an isomorphism MATH. |
math/0105020 | First, recall that MATH is the coefficient of MATH in the series MATH, so MATH. Next, we have MATH. By putting MATH we can rewrite this as MATH . It follows that MATH. By inspecting REF , we see that this is the same as MATH. Next, recall that MATH, so MATH (by putting MATH and MATH and MATH and noting that MATH). Next observe MATH . It follows that MATH . On the other hand, we have MATH . This is the same up to reindexing, as required. |
math/0105020 | For MATH we put MATH . We define a polynomial MATH by MATH . We then put MATH. One checks that MATH, which is a regular element in MATH. It follows easily that MATH is regular in MATH, and thus that MATH is regular in MATH. Next, we define a map MATH by MATH . In the case MATH of the second equation, the term MATH is to be interpreted as MATH. It is easy to see that MATH and MATH and MATH, so the notation is consistent. Moreover, we see that MATH. Using this, we see that MATH induces an isomorphism MATH, where MATH . It is clear from this that MATH. As MATH is regular in MATH for all MATH, it must be regular in MATH as well. |
math/0105020 | Suppose that MATH and MATH for some MATH. It is clear from the proposition that the image of MATH in MATH must be zero, so MATH, so MATH. |
math/0105020 | Combine REF , and REF . |
math/0105021 | CASE: Suppose that MATH for all positive integers MATH. It suffices to prove REF for MATH-homogeneous elements MATH, MATH, where MATH. Note first that MATH and let MATH, MATH be fixed. Then REF imply that MATH . On the other hand MATH by REF, and then REF follows by comparing the coefficients of MATH in the right-hand sides of REF respectively. Assume now that MATH is a faithful weak MATH-twisted MATH-module and that REF holds, and let MATH be such that MATH for all MATH. Then REF yield MATH . Thus MATH by REF. Using CITE one gets that MATH for MATH. Therefore MATH for MATH as well (since MATH is faithful), which proves REF. CASE: We first prove the second statement. Suppose that MATH is irreducible. If MATH were a REF-dimensional trivial MATH-module, then MATH would be a nonzero MATH-submodule of MATH for each nonzero MATH and MATH satisfying MATH (such elements exist since MATH is faithful), hence a contradiction. Suppose now that MATH is a nonzero proper MATH-submodule of MATH. Then there exists MATH for some MATH, where MATH and MATH are the MATH-gradations of MATH and MATH respectively. Note that MATH are MATH-submodules of MATH by REF, and thus MATH since MATH is irreducible. It follows from REF that there exist MATH and MATH such that MATH and MATH. Then REF implies that MATH for all MATH, where MATH. Consequently, MATH is a (trivial) MATH-submodule of MATH by REF. Clearly, MATH, so that MATH and thus MATH. Hence MATH (again by the faithfulness of MATH), and one gets that MATH, which is a contradiction. MATH must therefore be an irreducible MATH-module. Conversely, suppose that MATH is a nontrivial irreducible MATH-module. Then MATH is MATH-homogeneous of some conformal weight MATH. Recall REF and notice that MATH . Let MATH be a nonzero MATH-submodule of MATH. Then MATH (MATH being MATH-stable), where MATH. By REF there exist MATH and MATH such that MATH, and it follows from REF that MATH for every MATH and MATH. Since MATH is irreducible, one has MATH, and then by iterating REF one gets MATH for all MATH. Therefore MATH and the proof is complete. |
math/0105021 | Let MATH. By the twisted associator REF and induction, it suffices to prove that MATH. Recall that MATH and notice that MATH in REF , while MATH in REF . Let MATH. Then we can embed MATH into MATH with canonical central element MATH in REF . Since MATH is a level MATH integrable MATH-module, it is à fortiori an integrable MATH-module of level MATH. As such, MATH becomes by REF an (untwisted) weak module for the VOA MATH via the restricted vertex map. It then follows from REF that MATH. Since MATH, one gets from REF that MATH in REF , as needed. Note that in REF we can embed MATH into MATH with canonical central element MATH and that the restriction of MATH to MATH coincides with the inner automorphism MATH. As above, MATH is an integrable MATH-module of level MATH and thus MATH becomes by REF a weak MATH-twisted MATH-module via the restricted vertex map. Using REF with MATH and MATH, one gets that MATH is an untwisted weak MATH-module. Consequently, MATH acquires a structure of level MATH integrable MATH-module via the vertex map MATH. Moreover, it follows from MATH for MATH that MATH, which combined with REF gives MATH . REF again implies that MATH, and then REF yields MATH. The proof is thereby complete. |
math/0105021 | Let MATH be the associative algebra automorphism of MATH induced by the NAME algebra involution of MATH defined by MATH, MATH, MATH. Let MATH be such that MATH and MATH, where MATH is arbitrarily fixed. An easy induction gives MATH for MATH, so that if MATH it follows that MATH . Suppose now that MATH. Using REF and induction one gets that MATH whenever MATH, and therefore MATH . It follows from REF that in both cases MATH . Consequently MATH by using REF and the action of MATH. Then REF follows by induction on MATH from MATH together with REF. |
math/0105021 | Using the explicit form of the root systems MATH, it is readily checked that in REF the MATH's are sums of at most three commuting root vectors of MATH. Therefore MATH for some MATH by REF. The same argument implies that in REF MATH while obviously MATH. Applying now REF with MATH, MATH, and MATH one gets MATH, so that the proposition is true in all cases. |
math/0105021 | Let MATH be a highest weight vector of MATH and set MATH. By REF it suffices to prove that MATH, which in turn reduces to showing that MATH by REF. According to REF , MATH for some MATH. It then follows from REF and induction that MATH and consequently MATH (since MATH for MATH). Hence MATH if MATH, and MATH where MATH according to REF . Since MATH and MATH, the representation theory of MATH together with REF imply that MATH as required. |
math/0105021 | Note first that by REF there exist indeed MATH and MATH as specified in the assumptions. Set MATH, MATH, MATH, MATH, and recall the automorphism MATH of order MATH of the NAME diagram of MATH. Then MATH is a MATH of MATH and MATH is the diagram automorphism of MATH induced by the automorphism MATH with respect to MATH. Let MATH be the MATH-decomposition of MATH and notice that MATH and that MATH. Moreover, MATH is MATH-invariant and it obviously satisfies MATH for all integers MATH. Then REF follows by combining REF with REF applied to the data MATH. |
math/0105021 | By REF it suffices to consider only the case when MATH and MATH. Let MATH be the highest weight of M and let MATH be a highest weight vector. If MATH, then one can argue as in the proof of REF to show that for MATH one has MATH for some MATH, so that MATH is necessarily integrable. The converse is given by REF . |
math/0105022 | A base MATH contributes to MATH if and only if MATH, namely if and only if there exists a number MATH palindromic in base MATH of length MATH, that is such that MATH and MATH for MATH, and MATH. Then MATH, and hence MATH contributes to MATH if and only if MATH. We consider separately the cases MATH and MATH. In the first case, the largest MATH-palindrome in base MATH is MATH and so MATH. Now for the second case we let MATH be the largest integer such that MATH, that is MATH. Then MATH and so every MATH-palindromic number in base MATH begins with MATH. Hence MATH . Set MATH . We split the sum MATH with MATH, MATH, and MATH. From the previous considerations we deduce that MATH . On the other hand MATH . Finally MATH and thus MATH . |
math/0105022 | Proof. - First note that the hypothesis that the MATH-expansion of MATH has length REF is equivalent to the fact that MATH. Now MATH is MATH-palindromic in base MATH if and only if there exists MATH and MATH such that MATH . Solving the associated Diophantine linear equation MATH with respect to MATH we see that the above representation is equivalent to the existence of MATH satisfying MATH . The second pair of inequalities is equivalent to MATH, and so it implies that MATH. Then this condition is necessary for MATH to be REF-palindromic in base MATH. Let's check that it is also sufficient: the integer MATH satisfies the second pair of inequalities. Then it only remains to prove that it also satisfies the first pair, which is equivalent to MATH. This follows from the inequalities MATH which are in term a consequence of the hypothesis MATH. |
math/0105022 | Proof. - Take MATH and assume that MATH for all MATH, so that MATH . We will see in a minute that this is contradictory: Set MATH and MATH. For MATH we let MATH be the largest integer such that MATH . Then MATH, and also every REF-palindromic number MATH with MATH is less or equal than MATH. Hence MATH which implies that MATH as MATH. We have that MATH . We have that MATH and MATH and thus we conclude MATH, which contradicts REF for MATH large enough. It follows that for each (sufficiently large) MATH there exists MATH such that MATH . The fact that MATH implies that the set of such MATH's is infinite. |
math/0105027 | Since MATH, no reducible solutions will arise in a generic path of metrics. In a neighborhood of a given path MATH in MATH, there is a uniform lower bound on the scalar curvature and the norm of MATH. By a basic argument in NAME - NAME theory, this implies that there are a finite number of MATH structures for which the parameterized moduli space has dimension MATH. But for generic paths, a moduli space with negative dimension will be empty. |
math/0105027 | The proof rests on the isomorphism of moduli spaces MATH given by pulling back the spin bundle, spinors, and connection. This isomorphism preserves orientation precisely when MATH. Note that MATH. Thus, if MATH is either a generic element of MATH or a generic path in MATH, we have a diffeomorphism MATH and the lemma follows by summing up over MATH. |
math/0105027 | As remarked before, each term in MATH depends only on the endpoints MATH of the path. Start by choosing a generic path MATH from MATH to MATH, and note that MATH is necessarily generic for MATH. Now take generic paths MATH and MATH, and define MATH and MATH. Thus we have a well-defined loop in the contractible space MATH, which may be filled in with a generic MATH - parameter family MATH. Consider the MATH - parameter moduli space MATH which is a smooth compact MATH - manifold with boundary. (For compactness, we need to ensure that there are no reducibles in a MATH - parameter family, which is why we require that MATH.) Any boundary point lies in the interior of one of the sides of the rectangle indicated in REF . Now form the union MATH where the union is taken over the same set of MATH as is the sum defining MATH. Note that by construction the moduli space corresponding to the left side of the MATH square (that is, the MATH - parameter moduli space corresponding to MATH) matches up with the right side of the MATH square (corresponding to MATH). The argument diverges slightly according to whether MATH is finite or infinite. In the former case, assume that the orbit has exactly MATH elements. Hence we have a diffeomorphism from the right hand boundary of the MATH square MATH to the left hand boundary of the first square. Note that the last isomorphism, because it is defined up to gauge equivalence, is independent of the choice of isomorphism between MATH and MATH. Moreover, by the assumption that MATH, this isomorphism is orientation preserving. All of the other contributions from the side boundary components cancel, according to the observation in the last paragraph. So, counting with signs, the number of points on the top boundary of MATH is the same as the number of points on the bottom part of the boundary. It follows that the conclusion of REF holds in the case that MATH is finite. When MATH is infinite, we use the principle REF that for MATH some MATH, the moduli spaces MATH are all empty. Thus the union of all of the MATH - parameter moduli spaces provides a compact cobordism between the union of MATH - parameter moduli spaces corresponding to the initial point MATH and that corresponding to MATH. |
math/0105027 | Since MATH, there are no reducible solutions to the NAME - NAME equations for generic metrics or paths of metrics. Because MATH is open in the space of metrics, any path may be perturbed to be generic while staying in MATH. Since MATH, there are no reducible solutions to the NAME - NAME equations for generic metrics or paths of metrics. Now if MATH is a path from MATH to MATH, then all of the paths MATH lie in MATH. By the standard vanishing theorem, all of the moduli spaces MATH are empty, and so the invariant MATH must vanish. |
math/0105027 | In the proof we use the notation MATH for the composition MATH of functions, and MATH for the composition of paths (the path MATH followed by the path MATH.) We will use an isotopy between MATH and MATH to construct a diffeomorphism between the various MATH - parameter moduli spaces, taking advantage of the fact that any with the correct endpoints may be used to calculate MATH. First choose an isotopy MATH such that MATH and MATH. Let MATH be generic, and choose a generic path MATH from MATH to MATH. By definition, MATH where MATH. We use the path MATH to calculate the MATH term in the sum. Let MATH which gives an isotopy of MATH to MATH. We calculate MATH using MATH for the initial point. The path MATH has endpoints MATH for MATH and MATH for MATH. So the MATH - parameter moduli spaces MATH may be used to calculate MATH. But MATH is homotopic to the identity, and so acts as the identity on the set of MATH structures on MATH. Hence there is an orientation preserving diffeomorphism MATH which is the moduli space used to calculate MATH. (Note that the gen-ericity of the path MATH follows from this diffeomorphism.) Therefore-MATH. |
math/0105027 | Choose a generic MATH and path MATH with which to define MATH. If MATH, consider the concatenation of paths MATH, with endpoints MATH and MATH. The union of the moduli spaces MATH corresponding to these paths may be used to define both MATH and MATH, and so these invariants are equal. The same argument works if the orbit of MATH is finite, except that each moduli space MATH is counted MATH times when one takes the union of the moduli spaces MATH which defines MATH. |
math/0105027 | If MATH were an isometry of a generic metric MATH, then this would be clear because one could choose a constant path from MATH to MATH. Since MATH might not be generic, we make use instead of REF : the isometry group of MATH is compact, so MATH is in the connected component of the identity for some MATH. But this implies that MATH is isotopic to the identity, and so MATH. By REF , this implies that MATH as well. |
math/0105027 | To prove these statements, we describe the MATH - parameter NAME - NAME moduli space on MATH in terms of the moduli spaces on MATH and MATH. The idea is basically the same as the `fundamental lemma' in CITE; details of a very similar argument (without the extra parameter) are given in the proof of REF. As a first step, note that the path MATH induces a path MATH on MATH and a constant path MATH on MATH. The assumption that MATH is generic implies that MATH is generic. Moreover, the moduli space MATH is diffeomorphic to the cylindrical end moduli space associated to MATH. To establish the first part of the proposition, note that if MATH is disjoint from MATH, then the same will be true for MATH. Since the finite-energy NAME - NAME moduli space on MATH has formal dimension MATH and contains no reducibles, the parametrized moduli space MATH will be empty. Because the gluing map is a diffeomorphism, the parameterized moduli space on MATH will also be empty. The second part is a little more complicated; we need to work out the NAME model for the MATH - parameter gluing problem for the NAME - NAME equation for a path of perturbations for which the equations on MATH admit a reducible solution MATH at MATH. As above, this path gives rise to a path of perturbations on MATH. Using the relation between wall-crossing on MATH and MATH, there will be exactly one small MATH for which the NAME - NAME equations with perturbation MATH admit a reducible solution MATH. To simplify notation, we will assume MATH. Suppose that MATH is a smooth irreducible solution to the NAME - NAME equations on MATH. As remarked above, it determines a unique finite-energy solution on MATH. For small MATH, say in an interval MATH, we have the constant configuration MATH on MATH. Note that this will be a solution to the NAME - NAME with perturbation MATH only when MATH. Consider the MATH - parameter family of configurations on MATH gotten by gluing MATH to MATH, and the problem of deforming this family to give elements in the MATH - parameter moduli space on MATH parameterized by MATH. Adapting CITE to this context, the portion of the parameterized moduli space close to the glued-up path near MATH is described (for MATH sufficiently large) in terms of the NAME model. In other words, there is a MATH - equivariant map MATH such that the parameterized moduli space on MATH is diffeomorphic to MATH. The assumption that the path MATH is generic means that MATH vanishes. From the NAME - NAME - NAME index theorem, the index of MATH is zero, so that MATH . As in CITE, the derivative MATH is given by MATH where MATH is defined in REF . Hence MATH consists of exactly one point; the orientation of this point is the same as the orientation of the corresponding point in MATH. Each point of the moduli space MATH contributes therefore MATH to MATH. By REF , the rest of the path (where MATH misses MATH) doesn't contribute at all, and the result follows. |
math/0105027 | Let MATH, where MATH and MATH are generic, and choose a generic path MATH from MATH to MATH. This glues up to give a path MATH which we can use to compute MATH. According to REF , we need to know the intersection number of the wall MATH with the path MATH for each MATH. Now MATH, viewed as a transformation of the hyperbolic space MATH, is a parabolic element, with fixed point MATH in the unit disc model of MATH. In this model, MATH is the geodesic meeting the boundary of the unit disc orthogonally at MATH and MATH. Hence for any generic starting point MATH, there is a unique MATH for which MATH is on the left side of MATH and MATH is on the right side. It follows that MATH for MATH, and that MATH. Hence MATH. |
math/0105027 | By REF , the invariants MATH are all non - zero, as long as MATH. Suppose that MATH and MATH are in the same path component of MATH. Hence they are joined by a path MATH. As in the proof of REF , use the translates of MATH by MATH to show that MATH. But this implies that MATH. For the second part, suppose to the contrary that MATH and MATH are connected by a path of positive scalar curvature metrics for all MATH. Then by the usual cobordism argument we would have MATH . (The assumption that the metrics are connected in MATH provides a MATH - parameter family in which the MATH - parameter moduli spaces corresponding to the `sides' are empty.) This would imply that the corresponding MATH invariants are equal, contradicting our assumption. |
math/0105027 | Let MATH denote the elliptic surface with MATH, blown up at MATH points, and let MATH as in REF . According to CITE, MATH decomposes as a connected sum as in the statement of the corollary. Therefore CITE MATH admits a metric MATH of positive scalar curvature. By REF , MATH supports a diffeomorphism with MATH (for an appropriate choice of MATH) and so by REF , MATH has infinitely many components. |
math/0105027 | For any MATH, we start with two simply-connected MATH - manifolds MATH satisfying CASE: MATH. CASE: MATH, inducing a correspondence between MATH and MATH. CASE: For corresponding MATH structures MATH and MATH we have that MATH. CASE: MATH and MATH decompose as a connected sum of MATH's and MATH's. There are plenty of sources of such manifolds. One could, for example take MATH to be a (blown-up) elliptic surface, and build the other MATH to be the corresponding decomposable manifold. An infinite family MATH of such manifolds arises via the NAME - CITE knot surgery construction on MATH. The decomposition REF of the stabilized manifold MATH follows readily from the analogous fact for MATH. Given MATH and MATH, we get, as above, diffeomorphisms MATH of a single manifold MATH with the property that for an appropriate MATH structure on MATH, the invariants MATH and MATH are distinct. This MATH structure has the property that MATH is infinite for all MATH. It is explained in CITE and CITE how to arrange the identification of the stabilized manifolds so that MATH and MATH are homotopic to one another. Use diffeomorphisms MATH to pull back a positive scalar curvature metric MATH on MATH to get metrics MATH. A result of CITE implies that MATH is concordant (or pseudoisotopic) to MATH. In other words, there is a diffeomorphism MATH of the MATH - manifold MATH, which is the MATH in a collar of MATH and MATH in a collar of MATH. Pulling back the product metric MATH via MATH gives the required concordance between MATH and MATH. But by the second part of REF , these metrics are not isotopic. |
math/0105027 | By construction, MATH is homotopic to MATH. If MATH were isotopic to the identity, then MATH would be isotopic to MATH. But, using REF , this would imply that MATH which is a contradiction. |
math/0105027 | The main topological ingredient is the computation of the group MATH of bordisms of diffeomorphisms. Recall that MATH consists of pairs MATH where MATH is a diffeomorphism of MATH, modulo cobordisms over which the diffeomorphism extends. (All manifolds are oriented, and cobordisms and diffeomorphisms are to respect the orientations.) The group was computed by CITE and CITE; we will recall NAME 's computation of MATH shortly. To make use of this computation, we need the following observation. Let MATH be a non-spin, simply-connected MATH - manifold, with a metric MATH of positive scalar curvature. Suppose that MATH is a diffeomorphism which is bordant to MATH. Then MATH is concordant to MATH. By definition, there is a manifold MATH with two boundary components MATH, and a diffeomorphism MATH with MATH . According to CITE we can (and will) assume that MATH is simply connected. Now there is a cobordism MATH, relative to the boundary of MATH, to the product MATH. Again, by preliminary surgery on circles if necessary, we can assume that MATH is simply connected. By surgery on MATH - spheres in MATH, we can also assume that MATH . (It is at this point that we used the assumption that MATH is not spin, in order to be able to choose a basis of MATH - spheres with trivial normal bundle.) As a consequence, we can find a handle decomposition for MATH, relative to the boundary, with only MATH - handles. In other words, MATH is obtained from MATH by surgery on MATH - spheres, and vice-versa. Now MATH has the product metric MATH, which can be pushed across the cobordism MATH to give a metric of positive scalar curvature on MATH which is a product near the boundaries. Use the diffeomorphism MATH to pull this metric back to give a new positive scalar curvature metric on MATH which is MATH near MATH and MATH near MATH. Push this new metric back across MATH to give a positive scalar curvature metric on MATH; since the metric is not changed near the boundary components, this is a concordance. (Compare CITE for an argument of this sort.) To make use of the claim, we need to show that the diffeomorphism MATH on MATH constructed in REF is bordant to the identity. By the basic theorem of CITE, this is determined by the action of MATH on MATH, as follows. Let MATH denote the intersection form on MATH. Then MATH is bordant to MATH if (and only if) there is a MATH - dimensional subspace MATH which is MATH invariant and on which the form MATH vanishes. (MATH is called a metabolizer for the isometric structure MATH . ) This will clearly hold if there is a metabolizer for the action of MATH on the summand MATH of MATH. Now a straightforward calculation shows that on MATH, with respect to the basis MATH the action of MATH is given by the matrix: MATH . The reader can check that with respect to the basis given above, the vectors MATH and MATH span a metabolizer in MATH. It follows that MATH is bordant to the identity. To conclude the proof, let MATH be a representative of any concordance class in MATH. (Because of the decomposition of MATH as a connected sum, this space is non-empty.) Then all of the metrics MATH are concordant, by the claim and the preceding calculation. On the other hand, these metrics are mutually non-isotopic, because of REF . |
math/0105027 | There are two framings for the surgery on MATH, which differ by twisting by the non-trivial element in MATH. But since MATH has a circle action with a circle's worth of fixed points, this rotation can be undone via an isotopy of MATH with support in a neighborhood of a copy of MATH of MATH. Thus the two surgeries give the same manifold. The key observation for showing that MATH does not depend on MATH is that it has a very simple structure as a double of a manifold with boundary. Here is a more precise description: First note that splitting the MATH fibers into pieces with MATH or MATH exhibits MATH as the double of the non-orientable MATH - bundle MATH. MATH be MATH minus an open disc, and consider the restriction MATH of this MATH - bundle. Then MATH is the double of MATH. To see this, note that the curve MATH along which the surgery is done is the union of two MATH fibres whose endpoints are identified when MATH is doubled. MATH minus a neighborhood of MATH is clearly gotten by doubling MATH along the part of its boundary lying in the boundary of MATH. When doing the surgery, one glues in a copy of MATH for each circle in the boundary of a neighborhood of MATH. This has the effect of the doubling along the rest of the boundary of MATH. The rest of the argument is handlebody theory, and can be expressed in either MATH or MATH - dimensional terms. Both versions start with the observation that MATH has a handlebody decomposition with a MATH - handle, a MATH - handle, and a single MATH - handle. MATH, being the double of MATH, is the boundary of the MATH - manifold MATH, which has a handlebody decomposition with handles which are thickenings of those of MATH and therefore have the same indices. But the isotopy class of the attaching map of the MATH - handle, being an embedding of a circle in the MATH - manifold MATH, depends only on its homotopy class and its framing. By considering the MATH - dimensional picture, the framing is also seen to be independent of MATH. But this homotopy class is MATH times a generator, and is therefore independent of MATH. By considering the MATH - dimensional picture, the framing is also seen to be independent of MATH. It follows that MATH for any MATH. The MATH - dimensional version is a little more subtle; the fact that MATH is a double implies that it has a (MATH - dimensional) handle decomposition with a (non-orientable) MATH - handle, two MATH - handles, a MATH - handle, and a MATH - handle. As usual, we can ignore the handles of index MATH. The MATH and MATH - handles, plus the first MATH - handle, give a handlebody picture of MATH as in CITE. If one draws this first MATH - handle as in REF below, then it has framing MATH, because the framing is the same as the normal framing of the MATH torus knot which is the attaching region in the usual NAME splitting of MATH. The MATH - handle is of course non-orientable; this means that in the diagram, a curve going into the front of one of the attaching spheres of the MATH - handle comes out the back of the other one. The second MATH - handle is attached, with framing MATH, along a meridian to the first MATH - handle. By sliding the first MATH - handle over the second, one can change crossings; however this is not quite enough to change the diagram for MATH into the one for MATH. The reason for this is that in the diagram for MATH the MATH - handle is drawn as a tangle of MATH arcs going from one attaching sphere of the MATH - handle to the other, and changing crossings does not change the endpoints of this tangle. Associated to the tangle is a permutation of the MATH points where the MATH - handle crosses the attaching sphere of the MATH - handle. For differing MATH, these may not be the same permutation. However they are both MATH - cycles and are therefore conjugate. Choose a permutation conjugating one to the other, and a MATH - string braid MATH inducing this permutation on the endpoints of the strings. Without changing MATH, we can alter the MATH - handle by inserting MATH on one end and MATH on the other end, where MATH is MATH with all of the crossings reversed. (Note that MATH becomes MATH when pushed over the MATH - handle.) Now we can use the second MATH - handle to change crossings until the diagrams for the two manifolds are identical. Since MATH and MATH are both even, we can arrange by further handle slides to make the framings the same. |
math/0105027 | According to REF , if MATH and MATH are in the same component of MATH, the corresponding MATH - invariants must coincide. We will show that this implies the stated equality of MATH and MATH. First we must compute the MATH - invariants. For at least one choice of framing, the cobordism MATH is a MATH cobordism from MATH to MATH. The obstruction to the extending a MATH structure over the cobordism is a relative MATH, lying in MATH. This is the same as the obstruction to extending the restriction of the MATH structure to a neighborhood MATH of the attaching curve MATH over the MATH - handle. There are two MATH structures on MATH, which differ by twisting by the non-trivial element of MATH; one of them extends over the MATH - handle. If, with respect to one framing, the MATH structure does not extend, then it will if the handle is attached by the other framing, which also differs by twisting by the non-trivial element of MATH. Fix a MATH structure on MATH, and choose the framing of the MATH - handle so that MATH gives a MATH cobordism to MATH. Let MATH, MATH be the MATH representations of MATH. By extending over the cobordism MATH, these give rise to MATH representations of MATH, for which we will use the same name. Each of these representations gives rise to a twisted MATH structure on MATH. According to REF : MATH A MATH structure twisted by a MATH representation is in a natural way a MATH structure, and so the MATH invariants in REF may be calculated as in CITE: MATH where the reduced MATH - invariant MATH (in the original notation of CITE, but not of CITE) is given by: MATH . Note that with the given metrics, these formulas are valid over MATH, not just reduced modulo MATH. This follows, for example, from CITE where MATH is computed via a spectral decomposition for the relevant NAME operators in terms of spherical harmonics. Thus: MATH . Dividing the MATH roots of unity into those for which MATH, this may be rewritten as: MATH . When MATH is odd, we can substitute MATH for MATH to get MATH where we have also used the fact that MATH is odd. Suppose now that MATH and MATH are connected in MATH. Taking into account that a diffeomorphism might permute the MATH structures on MATH, this means that for some MATH, we must have MATH . Under the assumption that MATH is odd, we will show that REF implies that either MATH with MATH, or MATH with MATH. Since MATH is odd, and we have assumed that MATH are odd as well, it suffices to show that one of these congruences holds mod MATH. Let MATH be a primitive MATH root of unity, and note that for any MATH we have MATH . Suppose that the metrics MATH and MATH are connected in MATH, so that the MATH - invariants correspond as in REF . By repeating these same calculations for MATH, we get that MATH where MATH. We now use NAME 's independence lemma CITE which states that there are no non-trivial multiplicative relations amongst the algebraic numbers MATH, that is, the terms on either side of this equality must match up in pairs. Eliminating some trivial possibilities we find that either MATH (implying MATH) or MATH which implies MATH. REF follows. |
math/0105028 | This follows from standard transversality arguments; see for example CITE and CITE. |
math/0105028 | MATH-links are defined modulo inner automorphisms of the free group. One need only observe that inner automorphisms follow from the basing relation. |
math/0105028 | Consider MATH so that MATH and consider the corresponding element MATH satisfying REF . We start by presenting MATH as a result of a contraction. Inserting MATH and MATH to MATH, we have that MATH where MATH . NAME, we have: MATH . Now, we bring MATH to MATH, at the cost of pushing MATH hair on the MATH-colored legs. Indeed, we have: MATH . Remembering that we need to contract MATH legs and MATH legs, we can push the MATH-hair on MATH. Let MATH denote the result of pushing MATH hair on each MATH-colored leg of MATH. Then, we have: MATH . Now, we may get MATH from MATH by attaching MATH-rooted labeled forests MATH whose root is colored by MATH; see CITE. Since MATH it follows that MATH . Letting MATH, it follows that MATH . In other words, MATH. The converse follows by reversing the steps in the above proof. |
math/0105028 | For REF , it is easy to see that MATH satisfies the above properties, by looking at the MATH-degree MATH and MATH part in MATH. This determines MATH uniquely on MATH. By the universal property, we know that MATH extends over MATH; a priori the extension need not be unique. Since however, MATH extends uniquely to matrices over MATH, and since every element of MATH has a matrix presentation (see the discussion in the beginning of the section), it follows that MATH is characterized by the derivation properties. REF follows easily from REF . |
math/0105028 | It is identical to the proof of REF and is omitted. |
math/0105028 | See REF. |
math/0105028 | Our first task is to define MATH. In order to do so, we need to introduce some more notation. Given a word MATH in the letters MATH and MATH, let CASE: MATH denote the corresponding skeleton. CASE: MATH denote the identity element. CASE: MATH denote the gluing word at the MATH-th gluing site, that is, MATH with a label of MATH on each component, in the sense described by the following example: if MATH, then MATH. Given a word MATH in the symbols MATH and MATH, let MATH denote MATH with the crosses forgotten. Let MATH be MATH with every crossed strand broken. For example, MATH . Take a MATH-indexed element MATH. CASE: Select MATH-tangle lifts of each tangle in the sequence, MATH, such that CASE: the strands of each MATH-tangle which lie on boundary lines of the following form MATH are bracketed MATH, where MATH (respectively, MATH) corresponds to the strands going to the left (respectively, right) of the removed disc, and these bracketings match up, CASE: boundary words at boundaries of the form MATH are given the canonical left bracketing. CASE: Compose the (usual) NAME invariants of these MATH-tangles in the following way: CASE: if two consecutive tangles are as follows, MATH with a word MATH in the symbols MATH describing their matching boundary word, then they are to be composed MATH CASE: if, on the other hand, two consecutive tangles meet as follows, MATH with their matching boundary word at the MATH-th gluing site factored as MATH, then they are to be composed MATH . This completes the definition of MATH. A normalized version MATH (useful for invariance under NAME moves) is defined by connect-summing a copy of MATH into each component of the crossed link associated to MATH. Observe that the result does not depend on the choice of a place to connect-sum to. The proof of REF follows from REF below. |
math/0105028 | For the first part we only need to show that the function does not depend on the choice of bracketings. This follows because of the following, where MATH are some expressions built from the associator. MATH where the first step follows from REF . The invariance of MATH under the MATH equivalence relation follows the same way. |
math/0105028 | It suffices to show that MATH respects basing relations. In other words, it suffices to show that MATH implies that MATH for MATH. In the first case, consider two diagrams MATH such that MATH, and let MATH denote the diagram in MATH such that MATH is obtained by forgetting some cross, and MATH is obtained by forgetting the other cross on that component. Enlarge MATH to MATH, the set of MATH-indexed sliced crossed diagrams on MATH whose presented links in MATH have at least one cross on every component, thus MATH. The function MATH immediately extends to MATH. The key point is the fact that the product of labels around each component of the skeleton equals to MATH; this is true since the locations of the two crosses cut the component into two arcs, both of which have trivial intersection word with the gluing sites. In other words, we have: MATH . It follows by construction that MATH . Thus, MATH. In the second case, consider MATH. Then, for some appropriate expression MATH built from the associator, we have that: MATH . Thus, MATH. |
math/0105028 | Consider MATH such that MATH or MATH (based on some element MATH). Using REF , we need to show that MATH . The case of MATH. Consider the decomposition of MATH . Suppose that an element MATH is pushed onto the legs labeled by MATH, to form MATH. Now, let MATH be MATH, where MATH is in the MATH-th entry, and let MATH denote MATH with a MATH pushed onto every leg marked MATH. Then, MATH because all the beads labeled MATH and MATH match up and cancel, after we pairwise glue the MATH-colored legs. The case of MATH. Consider MATH such that MATH . Notice that if MATH are MATH-integrable, so is MATH. Furthermore, MATH commutes with MATH. The result follows from REF below. |
math/0105028 | The idea of this lemma is the same as ``in the case MATH", but instead of pushing an element of the free group onto legs labeled by some element, we push an exponential of hair. The substitution MATH defines a map MATH with the property that MATH is MATH-substantial for all MATH. Assume that the canonical decomposition of MATH is given by REF . Let us define MATH where MATH is in the MATH-th entry. Then, REF from REF implies that MATH where the last equation follows from REF . |
math/0105028 | We prove this in two steps, which usually go by the names of the ``lucky" and ``unlucky" cases. The lucky case: MATH. In this case, we can use the Iterated Integration REF . MATH . Now, using the Integration by REF , it follows that MATH because MATH . The unlucky case: MATH. This pathology is treated by deformation. Namely, we ``deform" MATH by multiplying it by MATH for MATH a ``small real," commuting with MATH. Formally, we let MATH be a variable and consider an integration theory based on MATH and its NAME localization MATH. We denote by MATH, the subring of rational functions in MATH non-singular at zero, and we denote by MATH the corresponding noncommutative localization of MATH. Letting MATH it follows from the argument of the ``lucky case" that MATH . There is a ring homomorphism MATH defined by ``setting MATH to zero". Since MATH the result follows. |
math/0105028 | Consider two pairs MATH related by a wrapping move, as in REF . Observe that a wrapping move preserves the covariance matrix. Using REF, it follows that if MATH is the common covariance matrix of MATH and MATH, then MATH . |
math/0105028 | We will show that for MATH, we have MATH . Given sliced crossed links MATH such that MATH, consider an accompanying link MATH as in the previous section. MATH may be presented so that all the strands going through the MATH-th meridianal disc proceed, parallel to each other, through the MATH-th meridianal disc, and then onto the MATH-st meridianal disc. That is, we can find a sequence of tangles, containing some arc MATH (not exactly the arc MATH of REF), such that the link MATH is presented by taking a parallel the arc MATH according to some word (in the example at hand, MATH). Let us define MATH by MATH and MATH . Continuing to work with the example of the above paragraph, then the sequence will look as follows (where the MATH-th, the MATH-th and then the MATH-st meridianal discs are displayed): MATH . Now, the link MATH (respectively, MATH) is presented by the diagrams recovered by gluing back in the MATH-st (respectively, MATH-th) removed disc. So, on the one hand we have: MATH A MATH-move has been used here to move the gluing word MATH up to (just underneath) the gluing word arising from the MATH-th meridional slice. This is legitimate because the meaning of the MATH-move is that such an element (if it involves every appearance of MATH) can be slid along a paralleled arc (here arising from the component MATH). On the other hand, we have: MATH . Observe that this line results from the line ending the previous paragraph of equations by the algebraic action of MATH, as is required for the proof. |
math/0105028 | Let the canonical decomposition of MATH with respect to MATH be MATH for some matrix MATH. Then, the canonical decomposition of MATH with respect to MATH is MATH . Observe that MATH which implies that MATH is MATH-integrable. Furthermore, we have that MATH . Note that we could have used REF to deduce the above equality. |
math/0105028 | Recall that MATH is the subgroup of MATH that consists of inner automorphisms of the free group MATH. For the first part, an inner automorphism by an element MATH maps a bead MATH to MATH. The relations of REF cancel the MATH at every trivalent vertex of a diagram. The remaining vertices change a diagram by a basing move MATH. The second part follows from the fact that MATH for MATH, see CITE. |
math/0105028 | REF implies that the horizontal maps are well-defined. It is easy to show that the MATH-flavored group-like and infinitesimal basing relations (defined in REF) are mapped by the MATH-map to MATH-flavored infinitesimal basing relations. It is also easy to see that the MATH and MATH relations are mapped to MATH-flavored basing relations. |
math/0105028 | It follows from REF. |
math/0105028 | The invariant MATH that satisfies REF was defined in REF. Let us show REF . We begin by observing that the following diagram commutes, by the definition of the invariant MATH, using the work of REF: MATH . We now claim that the following diagram commutes: MATH . The above two diagrams, together with REF . In order to show the commutativity of REF, we will assemble several commutative diagrams in one with two layers. The top layer consists of the rational invariants MATH and its integration theory MATH; the bottom layer consists of the NAME integral MATH and its integration theory MATH. MATH REF imply that all faces of the above cube except the side right one commute. Since the map MATH is onto, it follows that the remaining face of the cube commutes. Using the identification MATH (see REF ) it follows that REF commutes. This concludes the proof REF follows from CITE; see also the discussion of REF. |
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