paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9912021 | This follows from the expression of the NAME group action on elements in MATH given by: MATH. If this expression is applied to MATH it gives MATH. On the level of root characters this becomes, by exponentiation of the previous identity, MATH . Now recall that MATH is just MATH evaluated at MATH. Also MATH will be MATH ... |
math/9912021 | REF follows from REF but with the observation that in, this case, MATH may fail to be in MATH even if MATH. This happens exactly when MATH and MATH with MATH odd (either MATH or MATH). Under these circumstances MATH (rather than one as required in the definition of MATH). We can fix this problem by factoring MATH as a ... |
math/9912021 | This follows from REF and the correspondence MATH giving a bijection between MATH and MATH. |
math/9912021 | The two sets MATH and MATH are clearly in bijective correspondence and an element MATH corresponds to the coset MATH. By REF , Notation REF , the actions on these two sets agree. They are both given by REF . |
math/9912021 | We first point out that the exponential map in the NAME groups MATH, MATH gives a diffeomorphism between MATH and the corresponding connected NAME group. Thus the NAME groups MATH, MATH are isomorphic. This takes care of REF on the level of the connected component of the identity. Also the image under MATH of the set M... |
math/9912021 | Recall that MATH is a NAME group, REF and it thus has defining relations MATH and MATH where MATH is MATH depending on the number of lines joining MATH and MATH in the NAME diagram. The case MATH occurring when MATH and MATH are not connected in the NAME diagram. The only relevant cases then are when MATH, MATH are bot... |
math/9912021 | The groups in REF are isomorphic by REF . We can write MATH where MATH is defined by MATH if and only if MATH for all MATH. Now the group MATH is the center of MATH. Intersecting with MATH we obtain the center of MATH. To find this intersection with MATH, we must find all MATH such that MATH is real for all MATH (see t... |
math/9912021 | We set MATH for simplicity. By REF , we have that each MATH can be written as a union over MATH of sets of the form MATH and thus by REF of MATH, we conclude the statement. |
math/9912021 | We start with the last statement, that MATH is a bijection; we have that MATH is injective because the scalars MATH determine all the root characters MATH, MATH for MATH and these scalars determine MATH in the adjoint group REF . From REF it follows that we can regard MATH as MATH (see REF ). We prove surjectivity by p... |
math/9912021 | This follows from REF and the fact that MATH for MATH. |
math/9912021 | This follows from REF but is better understood in REF . We omit details. |
math/9912021 | The manifold MATH is the finite union of the chambers as in REF . Since the MATH action on MATH is by continuous transformations, it then suffices to observe that the antidominant chamber is compact. The antidominant chamber MATH of REF can be seen to be compact by describing explicitly its image under MATH. This image... |
math/9912021 | By REF , MATH is regular. Then as in REF, MATH must be conjugate, under an element in MATH, to an element MATH in MATH. Using REF MATH is connected. Since MATH is also a regular element it follows that MATH is a NAME subalgebra and MATH is a NAME subgroup of MATH. Using conjugation by an element in MATH we may conjugat... |
math/9912021 | The map MATH is the restriction to the NAME group MATH of the diffeomorphism of complex analytic manifolds MATH. We obtain that MATH must be an injective map. We show surjectivity. If MATH then by surjectivity of MATH there is MATH such that MATH and MATH. Thus MATH with MATH and MATH. Therefore MATH. Since MATH we obt... |
math/9912021 | First we point out that we already have a smooth manifold MATH and the assumption MATH will simply ensure that we can define the map to the flag manifold. We first show the continuity of MATH: We use the local coordinates, MATH for MATH given in REF . In these local coordinates, we assume that for each MATH, MATH (whic... |
math/9912021 | We use REF . The manifold MATH has a MATH action and the only fixed points are the MATH with MATH. These get mapped to the fixed points of the MATH action in MATH, the cosets MATH, MATH. Thus MATH. This also follows directly if if we consider REF with MATH and we just let each MATH go to zero and obtain MATH. |
math/9912021 | This is just REF and the definition of MATH. The two conditions in REF are satisfied by REF (Subsection REF) and by REF in p. REF as noted above in REF . |
math/9912042 | Thanks to CITE MATH is a projective MATH-module of rank MATH. By CITE the NAME group of projective modules over MATH is trivial, in other words MATH . In particular, if MATH is a projective MATH-module whose rank is greater than the NAME dimension of MATH, then MATH is necessarily free, CITE. Since MATH we have MATH, s... |
math/9912042 | Let MATH. In these circumstances CITE states that the NAME locus MATH of MATH coincides with the smooth locus of MATH if MATH is NAME in codimension one, that is the set of points of MATH which are not annihilators of simple MATH-modules of maximal dimension has codimension at least two, see CITE. Since MATH is the cen... |
math/9912042 | CASE: We overline to indicate images modulo MATH in a MATH-module. Now MATH and MATH, with MATH, a unit in MATH when MATH. Thus MATH and MATH, as claimed. CASE: That MATH acts transitively on the simple MATH-modules follows from CITE. Let MATH, and write MATH for MATH. We write MATH for a MATH-module MATH. Since MATH, ... |
math/9912042 | It is clear that MATH factorises as MATH where MATH. It is therefore enough to prove this for the case MATH. By REF we can consider MATH as an affine variety embedded in MATH (with co-ordinate functions MATH). Under this identification MATH takes MATH to MATH. Note that MATH. Indeed the equations MATH show that MATH as... |
math/9912042 | Using REF it is enough to show that any maximal ideal of MATH lying over MATH is singular. Denote MATH by MATH, the fibre product MATH. Claim. Let MATH be the projection map. If MATH is singular then any point of MATH is singular in MATH. Proof of claim. For ease of notation let MATH, MATH and MATH. Thus we are conside... |
math/9912042 | Let's write MATH for MATH. The equivalence in the first sentence is a consequence of REF . By REF there is an algebra isomorphism MATH . We have isomorphisms MATH . Here MATH is, in the notation of REF , specified by MATH and MATH. Recalling our decomposition in REF , MATH we see that MATH is the tensor product of ring... |
math/9912042 | For MATH this follows from REF . For MATH with MATH the only point in CITE where the bound MATH was required was to deduce that the algebra MATH is NAME, that is its projective indecomposable modules are uniserial. If we knew this to be so without the bound then the lemma would follow. Let MATH and MATH be unipotent an... |
math/9912042 | Suppose we have an inclusion of algebras MATH. Suppose further that MATH has a MATH-bimodule decomposition MATH. Then, by CITE, the representation type of MATH is a lower bound for the representation type of MATH (where finite is less than tame is less than wild). It's clear that MATH is a bimodule direct summand of MA... |
math/9912042 | The proof is based on the fact that the truncated polynomial algebra MATH has wild representation type if MATH. This is a consequence of CITE. In order that MATH we must have one of the following for MATH and MATH: CASE: MATH where MATH (and the symmetric case obtained by exchanging the roles of MATH and MATH); CASE: M... |
math/9912042 | By REF MATH is a free MATH-module of rank MATH. Let MATH be a basis for this module and define MATH for MATH by the following equations, MATH . Then for any MATH the structure constants of MATH with respect to the basis MATH are given by MATH. As a result the map MATH defined by MATH, is a morphism of varieties. Let MA... |
math/9912042 | The first part follows directly from REF . Indeed, if MATH then the algebra MATH is semisimple whilst if MATH then MATH is NAME equivalent to a direct sum of truncated polynomial algebras in one variable, hence NAME. Let MATH denote the NAME order on MATH. For the second part note that if MATH then there exists MATH su... |
math/9912042 | The first part follows from REF and the observation that MATH has finite representation type by REF . For REF , arguing as in the proof of REF we see that it is sufficient to show that MATH is wild in the case that MATH, that is MATH for MATH. Let MATH be such that MATH where MATH and MATH are unipotent. By REF we have... |
math/9912042 | The second claim is obvious. For the first we can assume that MATH is concentrated in the MATH position. Then MATH where the right hand side is concentrated in the MATH position. On the other hand, since MATH is non-zero only in the MATH position, we have MATH as required. |
math/9912042 | Let MATH send MATH to MATH. We must check this is well-defined. First note that, for MATH. Thus we have, for MATH and MATH, MATH . Moreover MATH is a MATH-module isomorphism, since, for MATH, MATH . Since MATH is natural in MATH it follows that MATH is naturally isomorphic to the identity functor. One shows similarly t... |
math/9912042 | The discussion above shows that MATH is a skew group ring MATH, and that we may reduce to the case where MATH is a basic algebra. By CITE MATH. By Lying Over for MATH, CITE, we deduce that MATH is a basic algebra. Thus MATH is a finite direct sum of copies of MATH. Commutativity of MATH forces the action of MATH on MAT... |
math/9912042 | CASE: Since MATH by CITE, MATH, so that MATH is basic and has quiver with vertices labelled by MATH. By definition (see for example . CITE), to find the arrows of the quiver of MATH we may assume without loss that MATH. The orthogonality relations for MATH show that the primitive idempotents of MATH (and hence of MATH)... |
math/9912042 | CASE: By REF the winding automorphisms afforded by MATH permute the irreducible MATH-modules transitively. Thanks to the definition of the coproduct and of the winding automorphisms (see REF ), the latter act trivially on MATH. Hence, if a simple MATH-module does not feature in a component of any one simple MATH-module... |
math/9912042 | See CITE. |
math/9912042 | It is clear that MATH since by construction MATH is a combination of the simple roots MATH for MATH. We prove the opposite inclusion by induction on MATH, the case MATH being trivial. Let MATH so that MATH, and let MATH be the set of positive roots corresponding to MATH. We have MATH and MATH for MATH and MATH. Therefo... |
math/9912042 | Only REF and the final sentence of REF are not immediate from REF . By REF , MATH where the sum is taken over the set MATH of all MATH such that MATH does not appear in a reduced expression of MATH. Thus each such MATH is fixed by MATH and we have MATH. The same argument shows that MATH. That MATH is clear from REF . F... |
math/9912042 | Recall that MATH is the minimal length of MATH when written as a product of arbitrary reflections. Suppose that MATH. Then MATH so MATH. Conversely suppose that MATH for all MATH. We prove that MATH by induction on MATH, the case of MATH being clear. Let MATH so that MATH. Suppose that MATH. Since a reduced expression ... |
math/9912042 | By REF , under the hypothesis of the corollary MATH has cardinality MATH. By REF , MATH, proving the corollary, in the light of REF . |
math/9912042 | CASE: By CITE the set MATH is closed in MATH. Thus, by REF the algebra MATH has no more blocks than MATH. CASE: For MATH in the non-empty open set MATH of MATH, MATH is semisimple with MATH simple modules, by REF . The first part of the claim now follows from REF . CASE: MATH . Since MATH is integrally closed, MATH if ... |
math/9912042 | We freely use the notation of the proof of REF . By REF it is sufficient to show that MATH has exactly MATH maximal ideals. By the proof of REF MATH where MATH is as in REF . Let MATH and MATH. We have already seen in REF that if MATH then MATH is isomorphic to MATH, whilst if MATH and MATH then MATH is isomorphic to M... |
math/9912042 | Let's write MATH for the regular function MATH, and similarly MATH for MATH. Let MATH be an arbitrary element of MATH. We have MATH for MATH since MATH is a highest weight vector. Hence MATH is not identically zero if and only if MATH is a lowest weight vector, and in this case it is non-zero for all MATH. Note that th... |
math/9912042 | By REF different blocks arise from the different maximal ideals of MATH lying over MATH, so we need to see how these are permuted by the right winding automorphisms. By REF such maximal ideals are determined by the central elements MATH for MATH. Hence we need only study the action of the winding automorphisms on MATH ... |
math/9912043 | Using REF , the set of isomorphism classes of holomorphic principal MATH-bundles is equal to the set of isomorphism classes of the category MATH. Two isomorphic extensions as in REF give isomorphic pairs MATH and MATH, because REF implies that the image of MATH in MATH under the map induced by MATH is MATH. Conversely,... |
math/9912043 | MATH . This follows from holonomy and REF . MATH . Take and object in MATH, that is, an extension MATH with a flat connection MATH preserving MATH. Take a MATH splitting MATH. This is given by a smooth morphism MATH (that is, a smooth section of MATH) with MATH. The fact that MATH preserves MATH and that it induces the... |
math/9912043 | Since MATH, MATH and then MATH . |
math/9912043 | MATH . This is given by holonomy. MATH . Follows from CITE (see the remarks after NAME 's proof). MATH . Note that since the flat filtration induces semisimple flat connections on the quotients MATH, we get polystable NAME bundles MATH, and as we have already discussed, the flat connection MATH gives a well defined ele... |
math/9912048 | Let MATH be a stable set of MATH such that MATH, and MATH. If MATH, then MATH is adjacent to some MATH, otherwise MATH is a stable set larger than MATH, which contradicts the maximality of MATH. Hence, MATH, and MATH. Therefore, using this exchange procedure, after a finite number of steps, we have to obtain a maximum ... |
math/9912048 | Suppose, on the contrary, that MATH. Hence, any MATH has MATH. Since MATH is a tree and MATH is a stable set of size MATH, it yields the following contradiction: MATH . Consequently, we infer that MATH is not empty. We can assert now that there exists some MATH, such that MATH. CASE: There exist two vertices from MATH ... |
math/9912048 | Let MATH be a bipartition of MATH. Both MATH and MATH are maximum stable sets, because MATH are stable and MATH. Hence, REF implies that both MATH-and-MATH, that supports the conclusion. |
math/9912048 | CASE: If MATH, then REF implies MATH. The equivalence REF is clear. CASE: Without loss of generality, we may suppose that MATH. Since MATH is also a stable set, REF ensures that MATH, and, consequently, MATH. Let MATH and assume that MATH. Then MATH and, hence, MATH. Since no vertex in MATH is adjacent to any vertex in... |
math/9912048 | We prove that if a stable set MATH of MATH satisfies MATH, then MATH. If MATH and MATH, then MATH is the bipartition of the forest MATH, while MATH is the bipartition of the forest MATH. Since MATH and MATH, it follows that MATH. Consequently, by REF , every connected component of MATH or MATH, which is different from ... |
math/9912048 | CASE: Let MATH. Then MATH is stable in MATH, and, therefore, it follows that MATH, for each MATH. Hence, we get that MATH . CASE: There are MATH, such that MATH. Then MATH is stable in MATH and MATH and this implies that MATH. CASE: There are MATH, such that MATH, that is, MATH, (or MATH). Hence, MATH is stable in MATH... |
math/9912048 | According to REF , we infer that MATH. Since MATH is a tree, it follows that MATH. To prove the theorem we use induction on MATH. The result is clearly true for MATH. Let MATH be a tree with MATH, and suppose that the assertion is valid for any tree with fewer vertices. If MATH, the result is clear. If MATH, let MATH a... |
math/9912048 | Let us partition MATH into MATH subsets MATH, each one having at least two vertices. Then we can write MATH where MATH is the subtree of MATH containing MATH as the neighborhood of MATH. Since MATH is pendant in no MATH, we get MATH. By REF , it follows that MATH. According to REF , the vertex MATH, and, consequently, ... |
math/9912048 | Let MATH be the bipartition of MATH. Notice that the distance between two vertices is even if and only if both of them belong to one set of the bipartition. To prove the theorem we use induction on MATH. If MATH, then MATH and the assertion is true. Let now MATH be a tree with MATH vertices. According to REF , MATH yie... |
math/9912051 | REF is CITE. MATH follows from the fact that a very ample divisor plus a divisor generated by its sections is a very ample divisor. Given any divisor MATH and a very ample divisor MATH, choose MATH such that MATH is generated by global sections for MATH. Then MATH is very ample for MATH. REF is trivial. CASE: For any a... |
math/9912051 | Let MATH be right MATH-ample. Then for any coherent sheaf MATH, there exists a MATH such that MATH for MATH and MATH. Since cohomology is preserved under automorphisms, pulling back by MATH, we have MATH for MATH and MATH. So MATH is left MATH-ample. |
math/9912051 | The first claim is CITE. For the second claim, the property of MATH not being quasi-unipotent is reduced to saying MATH has an eigenvalue of absolute value not MATH. Since MATH has determinant MATH, MATH has an eigenvalue of absolute value greater than MATH. |
math/9912051 | Let MATH be the cone generated by numerically effective divisors in MATH. In the terminology of CITE, MATH is a solid cone since it has a non-empty interior CITE. Since MATH maps MATH to MATH, the spectral radius MATH is an eigenvalue of MATH and MATH has an eigenvector MATH CITE. Since MATH, there exists a curve MATH ... |
math/9912051 | Suppose that MATH is a right MATH-ample divisor. Let MATH. By REF , we may replace MATH with MATH and MATH with MATH and assume that MATH is ample. Let MATH be the action of MATH on MATH. Suppose MATH is non-quasi-unipotent with spectral radius MATH and choose an integral curve MATH as in REF . Let MATH be the ideal sh... |
math/9912051 | Again choose a curve MATH as in REF with ideal sheaf MATH. By REF , there exists MATH such that for all MATH and MATH, MATH for any ample divisor MATH. In particular, the above cohomologies vanishes for MATH where MATH. Then repeating the last paragraph of the proof of REF shows that MATH has exponential growth. |
math/9912051 | CITE constructs a family of MATH surfaces whose general member MATH has MATH . (That is, MATH is the free product of two cyclic groups of order REF.) The ample generators MATH and MATH of MATH have intersection numbers MATH . MATH has two generators MATH whose actions on MATH can be represented as two quasi-unipotent m... |
math/9912051 | Since MATH, there exists a divisor MATH and curve MATH such that MATH. Choose MATH so that MATH is ample. By REF , the intersection numbers MATH are given by a polynomial in MATH with leading coefficient MATH. Since this polynomial must have positive values for all MATH, we must have MATH. |
math/9912051 | Choose MATH such that MATH is ample. Let MATH . Then MATH is ample and thus MATH is ample. |
math/9912051 | Let MATH be the span of MATH. MATH is a MATH-dimensional vector space by REF . By REF , it contains the real cone MATH generated by MATH. Using REF, any element of MATH can be written as a linear combination of MATH elements of MATH with non-negative real coefficients. Thus for all MATH, MATH where MATH and MATH. Expan... |
math/9912051 | Let MATH be right MATH-ample. By REF , MATH is quasi-unipotent and MATH is ample for some MATH. Then MATH is quasi-unipotent and MATH is ample. Applying the theorem again, we have that MATH is right MATH-ample. Thus MATH is left MATH-ample by REF . The same lemma gives the second statement of the corollary. |
math/9912051 | In the first case, for MATH to be ample or minus-ample, it must be a NAME divisor. Thus the intersection numbers MATH are defined, where MATH is a curve. Since MATH must be fixed by MATH, some power of MATH must be numerically equivalent to the identity by REF . In the second case, the action of MATH itself must be num... |
math/9912051 | Take MATH such that MATH is ample and MATH is respectively ample, generated by global sections, or numerically effective. Then MATH is ample and we again apply the main theorem. |
math/9912051 | As MATH and ampleness depends only on the numerical equivalence class of a divisor, the corollary follows from our main theorem. |
math/9912051 | Since MATH is unipotent and intersection numbers only depend on numerical equivalence classes, we may replace MATH by the divisor on the right hand side of REF . As noted below that equation, it is not a problem to treat the MATH as divisors. Since the intersection form is multilinear and integer valued on divisors, MA... |
math/9912051 | If the left hand side is larger than the right hand side, then by the discussion before the lemma, there is a subvariety MATH with MATH where MATH is an irreducible component properly containing MATH. This cannot happen by REF . On the other hand, if the right hand side is larger, then there exists a subvariety MATH wi... |
math/9912051 | By REF we may pick an irreducible component MATH of MATH with MATH . Then MATH has an irreducible component MATH with MATH. Combining REF , the claim is proven. |
math/9912051 | The first claim is obvious since the intersection numbers in REF are numerical polynomials, as noted in REF . The independence of the degree comes from REF . If MATH is numerically equivalent to the identity, then MATH. So MATH has degree MATH. |
math/9912051 | Given an ample divisor MATH, one has MATH by REF . So there exists a curve MATH such that MATH. Since MATH for all MATH and in particular for MATH, MATH. However, if MATH is odd, REF implies that the leading term of MATH is MATH where MATH. Then MATH for large MATH, which cannot occur. For the lower bound, note MATH. C... |
math/9912051 | If MATH the lemma is trivial. So assume that MATH. Let MATH. Expanding MATH gives terms of the form MATH where MATH and MATH. We will show that if MATH then MATH. Order MATH in the following way: MATH if the right-most non-zero entry of MATH is positive. We proceed by descending induction on this ordering. The largest ... |
math/9912053 | There is a standard bijection between lozenge tilings and families of nonintersecting lattice paths. This bijection is explained in REF (see in particular REF ). Thus, the problem of enumerating lozenge tilings is converted to the problem of counting certain families of nonintersecting lattice paths. By the NAME - NAME... |
math/9912053 | The first steps are identical with those in the preceding proof: the lozenge tilings are converted into nonintersecting lattice paths, in the way that is described in REF. Therefore, REF yields a determinant for the MATH-enumeration that we are interested in. Unlike in the previous proof, this provides a determinant fo... |
math/9912053 | Again, we use the strategy from the proof of REF . We convert the lozenge tilings into families of nonintersecting lattice paths as described in REF. The starting and ending points are slightly different from the ones used before. They are given in REF yields a determinant for the number we are interested in for even M... |
math/9912053 | We proceed analogously to the proof of REF . The lozenge tilings are converted into nonintersecting lattice paths, in the way that is described in REF. Therefore, REF yields a determinant for the MATH-enumeration in the case of odd MATH (see REF ). Again, the considerations in REF show that this number is a polynomial ... |
math/9912053 | We follow the arguments of the proof of REF , as given in CITE. Suppose we are given a cyclically symmetric lozenge tiling MATH of our cored hexagon MATH. It is completely determined by its restriction to a fundamental region, the lower-left fundamental region, say. (In the example in REF , the lower-left fundamental r... |
math/9912053 | We adapt the arguments used in the proof of REF . (Clearly, here we want to count the same objects, but with respect to a different weight.) So, again, we draw paths that connect the lozenges which are cut in two by the borders of the fundamental region. This time, we choose the top-right region as the fundamental regi... |
math/9912053 | Let us denote the determinant in REF by MATH. We proceed in several steps. An outline is as follows. The determinant MATH is obviously a polynomial in MATH and MATH. In REF - REF we show that the right-hand side of REF divides MATH as a polynomial in MATH and MATH. In REF we show that the degree of MATH as a polynomial... |
math/9912053 | We proceed analogously to the proof of REF . The only difference is the parity of MATH, so we have to read through the proof of REF and find the places where we used the fact that MATH is even. As it turns out, the arguments in REF - REF in the proof of REF can be used here, practically without change, to establish tha... |
math/9912053 | We proceed analogously to the proof of REF . The only difference is the parity of MATH, so we have to check the places in the proof of REF where we used the fact that MATH is even. Again, REF - REF can be reused verbatim, except that the products corresponding to the individual subcases are slightly different, and in R... |
math/9912053 | Analogously to the previous cases, we can show that the product MATH divides the determinant as a polynomial in MATH and MATH. Although not completely obvious, this is implied by the linear combinations of REF - REF . The degree in MATH of this product is MATH which is larger than the maximal degree MATH of the determi... |
math/9912053 | Let us denote the determinant in REF by MATH. We will again proceed in the spirit of the proof of REF . That is, we first show, in REF - REF below, that the right-hand side of REF divides MATH as a polynomial in MATH and MATH. Then, in REF , we show that the degree of MATH as a polynomial in MATH is at most MATH, the s... |
math/9912053 | We proceed analogously to the proof of REF . The only difference is the parity of MATH, so we have to check the places in the proof of REF where we used the fact that MATH is even. REF - REF can be reused verbatim, but the corresponding products are slightly different. In REF we are now interested in the factors corres... |
math/9912053 | We proceed analogously to the proof of REF . The only difference is the parity of MATH, so we have to check the places in the proof of REF where we used the fact that MATH is even. REF - REF can be reused verbatim, but the corresponding products are slightly different. In REF we are now interested in the factors corres... |
math/9912053 | We proceed analogously to the proof of REF . The parameters MATH and MATH are odd, so we have to check the places in the proof of REF where we used the fact that MATH or MATH is even. REF - REF can be reused verbatim, but the corresponding products are slightly different. The computation of the multiplicative constant ... |
math/9912053 | As in the proof of REF, define matrices MATH, MATH and MATH and MATH, MATH . Thus, MATH equals MATH. Now, as in CITE, multiply MATH on the left by MATH and on the right by MATH. Subsequently do the manipulations given in CITE (which amount to applying the NAME - NAME summation formula several times). The result is that... |
math/9912053 | Now choose MATH, MATH, MATH in REF . Then all the sums appearing in REF can be evaluated by means of the NAME - NAME summation REF. The result is MATH . Both determinants on the right-hand side of this identity can be evaluated by means of REF, which reads MATH . This completes the proof of the theorem. |
math/9912059 | By the following short exact sequence of chain complexes MATH the assumption MATH for all MATH is equivalent to the acyclicity of the chain complex MATH (notice that MATH). Now if Conjecture REF holds, then take an element MATH which is a cycle. Then MATH where MATH and MATH. Then MATH is a cycle in MATH and a linear c... |
math/9912059 | We give only a sketch of proof. By definition, MATH for MATH. Because of the hypothesis on MATH, every element MATH of MATH is determined by the values of the MATH where MATH runs over the set of words on the alphabet MATH. It turns out that there is a bijective correspondence between MATH and the word of length MATH o... |
math/9912059 | By REF or by REF , the simplicial nerve of the MATH-category of any composable pasting scheme is contractible. |
math/9912059 | It is obvious for MATH and for MATH, MATH. But MATH, therefore it suffices to notice that the MATH-simplex is contractible. |
math/9912059 | It suffices to calculate the simplicial homology of a simplicial set homotopic to a MATH-sphere. |
math/9912059 | It is clear that MATH. For MATH, MATH by REF . But MATH is the free MATH-category generated by the permutohedron, and with REF , one gets MATH for MATH. |
math/9912059 | The induction equations define a fillable MATH-shell (see REF ). |
math/9912059 | Obvious for MATH and MATH. Recall that MATH is defined by MATH . Let us suppose that MATH and let us proceed by induction on MATH. Since MATH by the previous proposition, then the evaluation map MATH from MATH to MATH is surjective. Now let us prove that MATH and MATH and MATH imply MATH. Since MATH and MATH are in MAT... |
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