paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006131 | Any sublattice MATH of MATH must be interval-connected. Assume by induction that every interval MATH is rank-connected. We will show that MATH is rank-connected. Let MATH be the subposet of MATH that contains all vertices except MATH. Let MATH be the subgraph of the NAME diagram induced by MATH and MATH. Since MATH is interval-connected, there must be some path between any two vertices in MATH in MATH. Suppose that MATH is not connected. Let MATH be elements of MATH such that MATH and MATH are not in the same connected component of MATH and such that MATH is the shortest possible path between any two elements in different connected components of MATH that is contained in MATH. Let MATH be the set of vertices in MATH. Then MATH does not intersect the vertices in MATH, so MATH both have rank MATH or both have rank MATH, and the rank of all vertices in MATH must be less than MATH or greater than MATH, respectively. Without loss of generality, assume that MATH have rank MATH and let MATH be the last element of MATH that is strictly less than MATH. We observe that if MATH, then MATH is smaller than MATH, and therefore is rank-connected by induction. This means there is a path from MATH to MATH in MATH, which is clearly a path in MATH. Therefore we may assume that MATH. Similarly, we can assume MATH. Now MATH and MATH since MATH. Let MATH be the next element after MATH in MATH. Then MATH and MATH are comparable, hence MATH, since MATH. If MATH, then MATH, a contradiction. So MATH is incomparable to both MATH and MATH. There exists a path from MATH to MATH in MATH that is strictly rank increasing. Let MATH be such a path. Now MATH. Let MATH be the element on MATH that has rank equal to MATH. Then MATH and MATH imply that MATH. Therefore, MATH and MATH must be in the same connected component of MATH. If MATH is in the same connected component of MATH as MATH, then by transitivity MATH and MATH are in the same component. If MATH and MATH are in different components, then we replace the pair MATH and MATH by the pair MATH and MATH and take a path from MATH to MATH that consists of starting at MATH and following a strictly rank decreasing path to MATH and then following the portion of MATH from MATH to MATH. This must be a shorter path than MATH between a rank MATH element and a rank MATH element in different connected components of MATH, because the portion of MATH from MATH to MATH goes through MATH and is therefore longer than the strictly rank decreasing path from MATH to MATH. We bypass MATH, and MATH has lower rank than MATH. This contradicts the selection of MATH as the shortest possible path. |
math/0006138 | The proof is similar to that in REF and uses REF together with uniform local finiteness of MATH. More precisely we use the fact that the valency of any vertex in MATH is uniformly bounded, say MATH. A fortiori the same is true (with the same constant MATH) for MATH for all MATH. We now estimate the MATH norm of MATH for a function MATH . Now MATH so that MATH where we have used REF and NAME inequality. In the last sum above, for every vertex MATH, MATH appears at most MATH times. This proves that MATH. Identical estimate holds (with the same proof) for MATH which yields the lemma if we set MATH. |
math/0006138 | Observe that MATH is in the kernel of the DML MATH if and only if it is in the kernel of the twisted coboundary operator MATH. That is, MATH for all edges MATH. It follows that MATH for all edges MATH. Since MATH is connected, it follows that MATH is a constant function on MATH. Since MATH is infinite, it follows that MATH is identically zero. |
math/0006138 | First observe that if MATH is such that MATH, then MATH . By REF MATH . Therefore we see that MATH . Using REF , we see that there is a positive constant MATH such that MATH . The proof of the lemma is completed by taking the limit as MATH, as the MATH form a regular exhaustion of MATH. |
math/0006138 | See REF. |
math/0006138 | Suppose MATH is not an eigenvalue of any of the MATH, that is MATH. Let MATH be the projection onto the eigenspace for MATH of MATH. Let MATH be the outer MATH-boundary of MATH and let MATH and MATH be the projections onto the functions with support in MATH and MATH respectively. One has MATH . Note that by REF , MATH differs from MATH only on MATH. One can write MATH where MATH encodes this difference. Consider MATH, MATH, MATH an eigenfunction of MATH for MATH. From MATH we have MATH and so MATH . Consider two solutions MATH and MATH of this equation, with MATH. Then MATH and we have MATH . As MATH, this implies that MATH. So the value of MATH on MATH uniquely determines MATH, giving MATH and thus MATH demonstrating that MATH is not in the point spectrum of MATH. As the restricted DMLs MATH are all finite operators with matrix elements belonging to the set of algebraic numbers, the union of their spectra must in turn be a subset of the algebraic numbers; this result therefore implies that any MATH in the point spectrum of the DML must be algebraic. |
math/0006138 | Fix MATH and define for MATH a continuous function MATH by MATH . Then clearly MATH and MATH as MATH for all MATH. For each MATH, choose a polynomial MATH such that MATH holds for all MATH. We can always find such a polynomial by a sufficiently close approximation of MATH. Hence MATH and MATH for all MATH. Recall that MATH denotes the number of eigenvalues MATH of MATH satisfying MATH and counted with multiplicity. Note that MATH by REF . MATH . Hence, we see that MATH . In addition, MATH since MATH for a positive constant MATH independent of MATH, MATH being the size of the fundamental domain MATH. It follows that MATH . Taking the limit inferior in REF and the limit superior in REF , as MATH, we get that MATH . Taking the limit as MATH in REF and using REF , we see that MATH . For all MATH we have MATH . Since MATH is right continuous, we see that MATH proving REF . Next we assume that MATH be a rational weight function, that is MATH for some positive integer MATH. In the notation of REF , let MATH be a non-negative algebraic number, and MATH be the characteristic polynomial of MATH, and MATH with MATH. Then MATH, MATH and MATH being constants independent of MATH. By REF and the remarks following it, for MATH one has, MATH . That is, MATH . Taking limit inferior in REF as MATH yields MATH . Passing to the limit as MATH, we obtain MATH. A similar argument establishes that MATH. By REF , one has MATH. Therefore we see that MATH which proves REF . |
math/0006138 | Let MATH be points of continuity of the spectral density function MATH of the DML MATH with MATH. Then by REF , one has MATH . We also notice that MATH and MATH . This immediately implies the corollary. |
math/0006138 | Notice that the interval MATH is in a gap in the spectrum of the DML MATH if and only if MATH. Then by REF , one has MATH . The converse is proved similarly to the proof above. |
math/0006138 | This corollary follows immediately from REF , together with REF . |
math/0006138 | CASE: In this instance, the operator MATH is invertible, and positivity of the NAME determinant follows immediately. CASE: The argument for the positivity of MATH exactly parallels that of REF, using the lower bound on the modified determinant MATH determined in REF . This case calls for a slightly modified version of the argument found in REF, but using the results in REF instead. For notational convenience, let MATH be the DML offset by MATH, MATH, with NAME spectral density function MATH where MATH is the spectral density function of MATH. Let MATH be the restricted DML similarly displaced by MATH, MATH. Recall that the normalized spectral density functions of MATH are right continuous, and note that the corresponding normalized spectral density functions of MATH will be given simply by MATH . Observe that MATH are step functions. Recall from REF that there exists some MATH such that MATH, and MATH for all MATH. In the following, let MATH be an algebraic number greater than MATH, so that MATH and MATH. Denote by MATH the modified determinant of MATH, that is the absolute value of the product of all nonzero eigenvalues of MATH. Choose positive MATH and MATH such that MATH is less than the least absolute value of any non-zero eigenvalue of MATH, and MATH is greater than the largest absolute value of any eigenvalue of MATH. Then MATH . Integration by parts transforms this NAME integral as follows. MATH . Integrating REF by parts, one obtains MATH . Using the fact that MATH one sees that MATH . We now complete the proof by estimating a lower bound of MATH from a lower bound on MATH. MATH is the absolute value of the product of all the non-zero eigenvalues of MATH, and hence for a rational weight function MATH by REF (and using the notation described there) MATH and thence MATH for some positive constants MATH and MATH independent of MATH. The following estimate is proved exactly as in REF, so we will omit its proof here. MATH . Combining REF with REF we obtain MATH . By the condition under regard, MATH, giving MATH and so one has MATH . From REF , we conclude that MATH . Now MATH and by REF MATH (and similarly for MATH), hence MATH that is, MATH . |
math/0006139 | An element MATH maps the generating monomials MATH to some MATH with MATH. The condition that MATH yields that MATH unless MATH and MATH. On the other hand, if MATH, then MATH, and the value MATH does not matter at all. Hence, we may restrict the knowledge of MATH to MATH. The linearity of MATH translates into the last condition in REF . Eventually, the trivial deformations are spanned by MATH. CASE: One obtains the description of MATH with the same arguments. We should only remark that it is the NAME relations MATH that are responsible for the vanishing of MATH in case of MATH. Afterwards, to get MATH, one needs to divide out the canonical generators MATH (MATH). They have degree MATH, and applied to MATH, they yield non-trivial values MATH and MATH only if MATH or MATH, respectively. Hence, in degree MATH, the map MATH yields MATH with MATH denoting the characteristic function of MATH. On the other hand, this contribution requires MATH. |
math/0006139 | CASE: If MATH, then MATH, hence MATH. On the other hand, let MATH with MATH for some MATH. It follows that MATH and MATH. With MATH, this implies MATH; with MATH being a proper subset of MATH, we obtain the MATH statement. The claims in REF are obvious. |
math/0006139 | Denote, just for this proof, the spaces given by REF of the previous proposition by MATH and MATH, respectively. Then we have to ascertain that pulling back via MATH induces isomorphisms MATH and MATH with MATH and MATH being the corresponding spaces from REF . The maps MATH are correctly defined: This is clear for the Hom case. For MATH, we set MATH, and the only non-trivial task is to check the two conditions that should lead to the vanishing of MATH. If MATH, then both MATH and MATH, hence MATH by REF , hence MATH. On the other hand, if MATH, then we also obtain MATH and MATH: Otherwise, if MATH, it would follow that MATH and MATH, thus, MATH, but this contradicts MATH. CASE: The maps MATH are injective: The Hom case follows from the surjectivity of MATH. For MATH, assume that MATH. In particular, if MATH belong to a common fiber MATH, then MATH. Hence, MATH only depend on MATH and MATH. CASE: The maps MATH are surjective: Let MATH represent an element of MATH. Then, the property that MATH for MATH implies that MATH only depends on MATH. In particular, MATH. To check the MATH case, we would like to proceed similarily with elements MATH. However, this requires a correction by coboundaries: By the cocycle property of the MATH's, we have to find MATH such that MATH vanishes if MATH belong to a common fiber MATH. Using the cocycle property again, we see that MATH will almost do the job for any fixed MATH; but we also have to ensure that MATH whenever MATH. This is done by proving the following Claim: Let MATH with MATH, MATH, and MATH. Then MATH. With MATH, we have that MATH. Thus, MATH (or the NAME condition in REF immediatly implies MATH). In particular, MATH, and since MATH and MATH, we obtain that MATH . Eventually, it is possible to define MATH, and it remains to show its vanishing for MATH. Since, by REF , MATH, we may assume that MATH. Now everything follows from applying the previous claim again. |
math/0006139 | If MATH, then we may identify MATH with the union of all MATH in MATH where MATH and MATH. Thus, MATH as subset of MATH is the union of such MATH with MATH. For such a MATH let MATH be the maximal subflag consisting only of faces in MATH. Now we can continuously retract MATH onto MATH and this can be done simultaneously for all MATH belonging to the above union. |
math/0006139 | We will prove the dual statement in homology using the method of acyclic models (see for example, CITE). If MATH has property U, consider the simplicial complex MATH where the vertices are the elements of MATH and a set of vertices MATH is a face if MATH. If MATH is the chain complex of MATH, then our MATH is the dual of MATH, and we are finished if we can prove that MATH is chain equivalent to MATH. To this end, consider the set MATH of U subsets of MATH as a category with inclusions as morphisms. Let the models in MATH be MATH. Finally, define the two functors from MATH to chain complexes by MATH and MATH. We must show that both MATH and MATH are free and acyclic with respect to these models. Now a basis element of MATH, MATH, comes from MATH and a basis element MATH of MATH comes from MATH, so both functors are free. If MATH, one may check that MATH is a simplex and MATH is a cone over the vertex MATH. Thus, in both cases the chain complexes are acyclic. |
math/0006139 | MATH is the kernel of MATH. Hence, since MATH, the previous theorem implies that an element MATH belongs to MATH if and only if MATH, that is, iff MATH. On the other hand, this means that, for every MATH, the conditions MATH and MATH imply MATH. Since the assumption MATH can be omitted, this translates into MATH. Finally, MATH is generated, as a module, by MATH if and only if MATH cannot happen for faces MATH with MATH. But this is equivalent to the condition formulated in the corollary. |
math/0006139 | First assume that there is a vertex MATH. If MATH, then MATH (otherwise MATH and MATH). Thus, MATH unless MATH. If MATH, then MATH and MATH. Thus, MATH may be contracted to MATH and its reduced cohomology is trivial, so MATH by REF . It is a simple matter to check that the map between the MATH sets is a bijection (with inverse MATH) and that it restricts to a bijection of the MATH subsets. Since it clearly preserves inclusions, it induces a simplicial isomorphism on the complexes MATH. From REF , it follows that we get an isomorphism in the relative cohomology. |
math/0006139 | If MATH is in MATH (respectively MATH), then MATH is in MATH (respectively MATH). Now, if MATH for all MATH, then there is a standard retraction taking MATH to a MATH which fits together to make MATH a strong deformation retract of MATH. (See for example, CITE.) If MATH is a face, then MATH is impossible, since then MATH. If MATH is a non-face and MATH, that is, MATH, then all four spaces are cones and there is nothing to prove. If MATH is a non-face and MATH, then both MATH and MATH are cones, so MATH and MATH and the above retraction works again. |
math/0006139 | It is straightforward to check that MATH is well defined and has the necessary properties to induce MATH by REF . Moreover, it is clear that this means exactly dehomogenization with respect to MATH, hence MATH coincides with the localization map. Finally, to prove that MATH is an isomorphism in case of MATH, we use that MATH and apply REF to both MATH and MATH. |
math/0006139 | If a graded piece MATH meets the image of MATH, then its pre-image is a unique summand MATH. Indeed, by REF , the only possibility to get degree MATH is MATH, MATH if MATH and MATH, MATH if MATH. This means that images of elements with different multidegree MATH cannot cancel each other, and it remains to consider the multigraded pieces MATH with MATH. Since, by REF , every summand MATH with MATH is an isomorphism, we obtain the injectivity of the above map whenever MATH has vertices at all. On the other hand, since MATH, the face MATH cannot be empty. |
math/0006139 | CASE: We know that MATH if MATH. This would not be the case if MATH. REF is a straightforward calculation. For REF , we must first recall what we did in REF of the proof of REF . Antisymmetric functions MATH on the MATH level had been turned into antisymmetric functions MATH on the MATH level via MATH with certain elements MATH. Hence, setting MATH and MATH, we may compute the value of MATH by applying REF on the relation MATH described in the beginning of REF. Using MATH, we obtain MATH . Plugging this into REF, we get MATH . To finish the proof, it is still necessary to take a closer look at the occurring arguments, that is, to calculate MATH and MATH . One can check that all the arguments of MATH and MATH are in MATH if they are in MATH. |
math/0006139 | Via the cup product, we see that each part MATH is responsible for MATH in the quadratic part of the obstruction equations. Moreover, since MATH is concentrated in degree MATH, no higher order obstructions involving only degree MATH deformations can appear, that is, MATH is described by the desired equations. Thus, in any flat deformation of degree MATH, every other parameter must vanish. One directly checks that this means that any fiber is singular, in fact reducible. |
math/0006140 | Assume that there is such a diophantine model MATH, with MATH. Then there is an affine variety MATH over MATH admitting a finite morphism MATH defined over MATH such that MATH. If MATH is discrete (that is, infinite and totally disconnected in the real topology), the traditional proof applies: the real topological closure of MATH in MATH is also mapped to MATH by MATH, and hence it has infinitely many components. If MATH is not discrete (which seems to be the case for the typical infinite diophantine set in MATH, say, the set of squares), then we show that one can select (in a computable way) a discrete subset MATH of MATH. Then the above proof, applied to MATH, gives the result. Here are the details of the construction of MATH. We only have to treat the case where the real topological closure MATH of MATH has only finitely many connected components. Since MATH is continuous, the mean value theorem implies that MATH is the union of finitely many closed subsets in MATH. In particular, the topological closure MATH of MATH contains finitely many closed subsets, and since MATH is infinite, one of these subsets, say, MATH, is not a point. By composing MATH with a suitable MATH-rational projection MATH which does not map MATH to a point, we may assume MATH. By composing with a fractional linear transformation defined over MATH, we may assume MATH to be the unit interval MATH. Let MATH be the element of MATH corresponding to MATH. Let us consider the set MATH . The set MATH is NAME computable (since MATH is a listable subset of MATH, it is easy to write a NAME program to check the inequalities), hence it is recursively enumerable (by NAME 's normal form theorem, compare CITEEF), so by CITE, it is diophantine in MATH. Also, MATH is infinite, since MATH is dense in MATH. We now set MATH . By construction, the set MATH is diophantine in MATH, and hence a fortiori in MATH. So there exists a variety MATH over MATH and a MATH-morphism MATH such that MATH. However, the real closure of the set MATH has infinitely many connected components in the real topology by construction. Hence the same holds for MATH, contradicting NAME 's conjecture. |
math/0006140 | In CITE REF NAME proves that, for MATH, projection onto the MATH-coordinate of the MATH-rational points of the space curve MATH given by MATH gives the set MATH . For MATH, NAME CITE proved that the set MATH is the projection onto the MATH-coordinate of MATH . Already the sets MATH for MATH are disjoint, since their MATH-coordinates are separated (MATH for all MATH). This gives a negative answer to question REF. |
math/0006140 | The proof is a bit indirect: we show that the polynomial ring has a diophantine model in the positive rational integers, and the latter has a diophantine model in the field of rational functions. More precisely, MATH is a recursive ring (compare CITE), because MATH is recursive (since finite), and hence the same holds for the polynomial ring over MATH (compare CITE). So there exists an injective map MATH such that the graphs of addition and multiplication are recursive on MATH, and hence MATH is a diophantine model of MATH in MATH. For the second step, we first recall a construction of NAME and NAME (CITE, CITE). Let MATH denote the MATH-valuation on MATH, that is, MATH is the order of MATH at zero. For any MATH, let MATH denote the equivalence class of elements MATH with MATH. For positive integers MATH and MATH, let MATH denote the relation MATH. Consider the structure MATH. Firstly, multiplication is diophantine in MATH (CITE, corollary on p. REF). Secondly, the set of equivalence classes MATH as above is a model for MATH in which the relations in MATH can be defined by diophantine formulas in MATH between arbitrary representatives of the equivalence classes in MATH. We conclude that for arbitrary elements MATH the relations MATH and MATH are diophantine in MATH. The problem with this encoding is that we do not know the existence of a diophantine set in MATH which contains exactly one representative for each such equivalence class. We fix this problem as follows. We know from the proof of REF that the set MATH is diophantine in MATH, and this will be our model. To define addition and multiplication on elements of this set, we introduce the following switching between MATH and MATH: the set MATH is recursively enumerable in MATH, so by NAME 's theorem, it is diophantine in MATH. Then, by the aforementioned results, the set MATH is diophantine over MATH. For the function symbols MATH on MATH, we let the corresponding symbol MATH for MATH be defined by MATH . For MATH, the righthand side of the equivalence is diophantine in MATH by what we have said before and the fact that for any two elements MATH, the statement MATH is equivalent with MATH, which is diophantine by CITE and CITE. Finally, MATH is a diophantine model of MATH in MATH. This finishes the proof of the theorem. |
math/0006143 | We have MATH with MATH, MATH and MATH. We omit here the isomorphisms between MATH and MATH. Then MATH . The result follows from the corresponding property in the category MATH. |
math/0006143 | By the induction hypothesis we can apply REF to NAME diagrams of size MATH. The statement MATH follows then from REF applied to MATH and MATH with MATH. The square of MATH is equal to MATH times the skein element represented by the following tangle. MATH . Let us consider the intermediate morphism MATH (the subscript in MATH and MATH indicate which isomorphism in MATH is used). The minimality of the idempotent MATH implies that this morphism is equal to MATH, up to a coefficient which is obtained by considering the trace (we use the absorbing REF ). MATH . Statement MATH follows. |
math/0006143 | We want to show that MATH for any MATH. We write MATH . By the induction hypothesis, we have the result if MATH is in MATH. So it is enough to consider the case where MATH for MATH an element of the matrix units basis described in REF, that is, for MATH, where MATH and MATH are standard tableaus with the same shape, whose size is MATH. If MATH, then we have MATH and the result follows. It remains to check the case where MATH. MATH . The result MATH follows; MATH can be obtained similarly. |
math/0006143 | From REF , we can see that the NAME skein relation is respected, whence we have that MATH is well defined. We have that MATH. This implies that MATH. The computation below shows the multiplicativity. MATH . |
math/0006143 | The recursive construction of the preceding section can be done. The only point to check is that at each step the quantum dimensions MATH are not zero. By the induction hypothesis, we have non trivial minimal idempotents MATH for every standard tableau of size MATH. If the shape of MATH is MATH, then we have that MATH is a non zero integer. |
math/0006143 | From REF if MATH, and from REF if MATH, we get MATH . This implies that MATH if MATH. Using REF , we show that, if MATH is the length of MATH, we have MATH . Hence we have that MATH . The independence follows. To show that the family MATH generate MATH, we proceed recursively on MATH. Using the map MATH, we see from the known result in MATH that the MATH, where MATH and MATH is the same NAME diagram with MATH cells, generate the NAME part MATH of MATH. It remains to consider MATH. From the induction hypothesis, we get that MATH is generated by the following elements MATH . These are zero if MATH, and else are equal to MATH where MATH. |
math/0006143 | The coordinates of MATH in the standard basis are obtained by computing MATH. The result is zero unless MATH and MATH have the same shape MATH and MATH; and in the latter case the result is MATH. |
math/0006143 | We can decompose MATH by using the minimal central idempotents in MATH. We get MATH . In the above, only those up and down tableaus MATH with MATH contribute. |
math/0006143 | REF follow from the corresponding in the NAME algebra (see CITE). Using the definition of the idempotent MATH, we can bring REF to the form MATH where MATH is depicted below. MATH . By the NAME lemma, MATH with MATH. Taking the quantum trace of this morphism, we get MATH . The first equality is due to the NAME skein relations and REF above. |
math/0006143 | Let MATH be the element of the algebra MATH defined below. MATH . For MATH the equation in MATH, MATH defines a central element MATH in MATH. We consider the formal power series in MATH, MATH . We have that MATH . We denote by MATH the series given by the action of MATH on the simple component of MATH indexed by MATH. A canonical basis of MATH is given in REF . Let MATH be an up and down tableau, whose length is MATH, and such that MATH . Write the products MATH and MATH in the canonical basis. MATH . Let MATH be the set of up and down tableaus MATH such that MATH for every MATH. By considering MATH (respectively, MATH), we get MATH . Using the three lemmas below the proof can be accomplished as follows. From REF we have MATH . The required formula follows now from REF . |
math/0006143 | Let MATH be as usual obtained by removing the last term in the sequence MATH. We have MATH . We will obtain the diagonal term MATH by computing the quantum trace of MATH. MATH where MATH is represented in the picture below. MATH . Clearly, MATH and we get the required result for MATH; MATH is calculated similarly. |
math/0006143 | We have that MATH . Multiplying on the left by MATH, we can express the above formula in the canonical basis. We denote by MATH the diagonal term of index MATH. From the left hand side, we get MATH . Let us compute the coefficient from the right hand side. MATH . In the the above sum, the term of indices MATH and MATH is zero unless MATH, and in this case the action of MATH multiplies by the coefficient among MATH corresponding to the NAME diagram MATH. These coefficients are distinct, and we know that MATH is not zero. This implies that MATH is a rational function in MATH, whose residue at MATH is equal to MATH. |
math/0006143 | In the following computations, we will drop some MATH. This means that by drawing the figures corresponding to these computation, one may have to add a vertical string on the right in order to get coherent equalities. For example, we write MATH . From the above, using the skein relation, we obtain MATH . This implies the following equalities for the formal series MATH . We also have the formula symmetric to REF MATH . We note that MATH and MATH commute, and that MATH . Multiplying REF on the left by MATH, we get MATH . Using REF and the skein relations, we get MATH . We multiply on each side by MATH, and use the relations MATH . MATH . The recursive REF can be deduced. It can be written MATH . Hence we have that MATH . Recall that MATH is the eigenvalue of MATH corresponding to MATH. REF is then established recursively. Note that MATH . Whence we have the formula for MATH. |
math/0006143 | We will prove REF . We first write the recursive REF in a more convenient form. We denote by MATH the unique cell in the skew diagram MATH. MATH . Here the first big product gives the contribution of the coefficients MATH corresponding to cells in the rows MATH to MATH. Note that some factors cancel if two among these rows have equal length. We can write MATH, where MATH and MATH satisfy the following recursive formulas. MATH . By induction, we can obtain the general formulas for MATH and MATH. MATH . Whence we get REF . |
math/0006143 | Suppose that MATH, and that MATH is obtained from the NAME diagram MATH by removing one cell from the MATH-th row, then we can write the recursive REF as follows. If MATH and MATH, then MATH else MATH . In REF case, MATH, if MATH and MATH, we get MATH else MATH . The announced result follows. In REF case, MATH, if MATH and MATH, we get MATH else MATH . The formulas in REF follow. In REF case, MATH, if MATH and MATH, we get MATH else MATH . The formula follows. |
math/0006145 | We argue as in our discussion of the walk on MATH in REF. Assume first that MATH is central, so that MATH has an identity. The lattice associated with MATH is MATH, so REF gives us an eigenvalue MATH as above for each MATH, with multiplicities MATH characterized by MATH for each MATH, where MATH is the number of chambers in MATH. We wish to show that MATH for MATH and that MATH for MATH. This will follow from REF if we show MATH for each MATH. Now CITE counted the number of regions obtained when an open convex set is cut by hyperplanes (see his REF and the comments at the bottom of p. REF). His result, in our notation, is MATH . This is the case MATH of REF for arbitrary MATH can be obtained by applying REF with MATH replaced by the set of hyperplanes MATH that contain MATH. REF is now proved if MATH is central. The noncentral case is treated by adjoining an identity to MATH, as in REF. The essential point is that REF still holds, and this implies that the ``extra" eigenvalue MATH has multiplicity REF. |
math/0006145 | Let MATH be an irreducible MATH-module. Then MATH is a submodule of MATH, so it is either MATH or REF. (Here MATH is the set of finite sums MATH with MATH and MATH.) It cannot be MATH, because then we would have MATH for all MATH, contradicting the fact that MATH is nilpotent and MATH. So MATH, and the action of MATH on MATH factors through the quotient MATH. Now consider the action on MATH of the standard basis vectors MATH of MATH. Each MATH is a submodule (because MATH is commutative), so it is either MATH or REF. There cannot be more than one MATH with MATH because MATH for MATH. Since MATH, it follows that exactly one MATH is nonzero on MATH, and it acts as the identity. Hence every MATH acts on MATH as multiplication by the scalar MATH, and irreducibility now implies that MATH is REF-dimensional. |
math/0006145 | Recall from REF that the eigenvalues of MATH are the same as the eigenvalues of MATH acting by left multiplication on the ideal MATH. For each MATH, let MATH be the number of composition factors of MATH given by the character MATH. Then the discussion at the end of REF shows that MATH has eigenvalues MATH with multiplicity MATH. Now MATH so the proof will be complete if we show that MATH for all MATH. Consider an arbitrary MATH. It acts on MATH as an idempotent operator, projecting MATH onto the linear span of the chambers in MATH. The rank MATH of this projection is therefore the number MATH defined in REF, where MATH. On the other hand, the rank of a projection is the multiplicity of REF as an eigenvalue, so MATH . Equating the two expressions for MATH gives MATH, as required. |
math/0006145 | Consider the NAME decomposition MATH; here the MATH are pairwise orthogonal idempotents summing to MATH, the MATH are nilpotent, and MATH. If MATH is diagonalizable, then each MATH and we have MATH as required. If MATH is not diagonalizable, then for some eigenvalue MATH we have MATH. Since MATH can be computed in each NAME block separately, we may assume that MATH, where MATH but MATH for some MATH. Then MATH . Thus MATH is rational and has a pole of order MATH at MATH. |
math/0006145 | Suppose MATH is split semisimple with primitive idempotents MATH, and write MATH. Then MATH, MATH, and REF for MATH follows at once. Conversely, suppose MATH is not split semisimple. Assume first that the minimal polynomial MATH of MATH splits into linear factors in MATH, say MATH, where the MATH are distinct. By the NAME remainder theorem, MATH and the assumption that MATH is not split semisimple implies that some MATH. Since MATH can be computed componentwise with respect to the decomposition REF of MATH, we may assume that there is only one factor, that is, that MATH for some MATH, where MATH. Then MATH, where MATH but MATH; hence MATH . Thus MATH has a pole of order MATH at MATH and hence does not have the form REF. If MATH does not split into linear factors, extend scalars to a splitting field MATH of MATH and apply the results above to MATH. Then MATH, viewed now as a function MATH, has poles at the roots of MATH, at least one of which is not in MATH. Once again, MATH does not have the form REF. |
math/0006145 | Let MATH be independent picks from the probability measure MATH. We will get a formula for MATH by computing the distribution of the reduced word MATH; for we have MATH . Given a reduced word MATH with associated chain MATH, we compute the probability in REF as follows. Let MATH, and let MATH. In order to have MATH, the MATH-tuple MATH must consist of MATH elements of MATH, then MATH, then MATH elements of MATH, then MATH, and so on, ending with MATH elements of MATH, where MATH and MATH. The probability of this, for fixed MATH, is MATH. Summing over all possible MATH, we see that the probability in question is MATH, whence REF. |
math/0006145 | Fix a reduced word MATH of length MATH, and let MATH be as above. Then MATH . Setting MATH and multiplying by MATH, we obtain MATH; REF now follows from REF. |
math/0006145 | Suppose MATH. Given MATH, let MATH be the face of MATH of type MATH and let MATH; see REF . We know that MATH and MATH are on opposite sides of MATH, so MATH must be strictly on the MATH-side of MATH, hence MATH and MATH. This proves MATH. Conversely, suppose MATH. To show MATH, it suffices to show that MATH and MATH are on the same side of every hyperplane MATH, where MATH is a codimension REF face of MATH; see CITE. This is automatic if MATH and MATH are on the same side of MATH, so assume they are not. Writing MATH, we then have MATH, hence MATH. Then MATH, so MATH is strictly on the MATH-side of MATH and therefore MATH is on the MATH-side of MATH. |
math/0006145 | Let MATH be the set of simplices of type MATH. There is a REF map MATH, given by MATH, where MATH is the fundamental chamber. It is REF because we can recover MATH from MATH as the face of type MATH. Its image, according to REF , is the set of chambers with descent set contained in MATH. Hence MATH . The proposition now follows from REF. |
math/0006145 | Let MATH and MATH be the MATH-bases of MATH introduced in REF. We have MATH . As we noted in the proof of REF , the chambers MATH that occur in the sum are those with descent set contained in MATH. Under our bijection between MATH and MATH, a given chamber MATH corresponds to the element MATH such that MATH. So we can write MATH where MATH. Our characterization REF of MATH therefore yields MATH . Let MATH and let MATH. Then MATH, so MATH for all MATH, hence MATH. Since the MATH are clearly linearly independent, it follows that MATH is injective and hence gives an anti-isomorphism of MATH onto a subalgebra of MATH. It remains to show that MATH is the descent algebra. For each MATH, let MATH be the reflection with respect to the face of MATH of type MATH. Then our set of generators of MATH is MATH. Moreover, for any MATH the stabilizer of the face of MATH of type MATH is the subgroup MATH generated by MATH; for example, the face of type MATH has stabilizer of order REF, generated by MATH. Finally, our definition of descent sets has the following translation: MATH if and only if MATH, where MATH is the length function on MATH with respect to the generating set MATH. Using these remarks, the reader can easily check that our basis vector MATH for MATH coincides with NAME 's MATH, where MATH. Hence MATH is equal to the descent algebra. |
math/0006145 | A minimal gallery from MATH to MATH crosses only the hyperplanes that separate MATH from MATH. This shows that REF implies REF . For the converse, it suffices to show that if REF holds and MATH is a chamber not in MATH, then there is a hyperplane MATH separating MATH from MATH. Choose a gallery MATH of minimal length, starting in MATH and ending at MATH. By minimality, we have MATH. Let MATH be the (unique) hyperplane in MATH separating MATH from MATH. Then MATH also separates MATH from MATH. For any MATH, we have MATH, where the sign depends on which MATH-halfspace contains MATH. The sign cannot be MATH, because then we could construct a minimal gallery from MATH to MATH passing through MATH, contradicting REF . So MATH, which means that MATH and MATH are on the same side of MATH. Thus MATH separates MATH from MATH, as required. |
math/0006145 | For the ``if" part see the beginning of REF. To prove the converse, we may assume that MATH is an idempotent semigroup for which the relation is antisymmetric, and we must prove REF. Note that MATH, so MATH. On the other hand, MATH, so MATH. Thus antisymmetry implies that MATH, as required. |
math/0006145 | The construction of MATH is forced on us by REF: Define a relation MATH on MATH by MATH. This is transitive and reflexive, but not necessarily antisymmetric. We therefore obtain a poset MATH by identifying MATH and MATH if MATH and MATH. If we denote by MATH the quotient map, then REF holds by definition. To see that MATH is order-preserving, suppose that MATH, that is, MATH. Multiplying on the right by MATH and using REF, we conclude that MATH; hence MATH and MATH, that is, MATH. It remains to show that MATH is the least upper bound of MATH and MATH in MATH. It is an upper bound because the equations MATH and MATH show that MATH and MATH. And it is the least upper bound, because if MATH and MATH, then MATH and MATH, whence MATH, so that MATH. |
math/0006145 | We have MATH for all MATH. In view of REF, this holds if and only if MATH, so REF are equivalent. If REF holds then MATH is maximal, because MATH. For the converse, note that MATH for all MATH; so if MATH is maximal then REF holds. |
math/0006145 | The inequality is immediate by induction on the size of MATH. We have already observed that equality holds if MATH has exactly one atom. If MATH has no atoms, then MATH and MATH. If MATH has more than one atom, then consideration of the term MATH in REF shows that MATH. |
math/0006145 | Let us temporarily take the statement of the proposition as a new definition of MATH. It then suffices to show that, with this definition, REF holds. We may assume that MATH and that REF holds for smaller posets. Then MATH . So we are done if we can show MATH, that is, MATH . The sum on the right counts all chains MATH in MATH, where MATH. The sum on the left counts all such chains of length MATH. Adding REF counts the chain of length REF, so the equation holds. |
math/0006145 | Let MATH denote the right-hand side of REF if MATH, and let MATH. It suffices to show that MATH satisfies the recurrence REF, that is, MATH where MATH is the set of elements of MATH of rank MATH. We may assume MATH. Group the terms on the right-hand side of REF in pairs, where MATH is paired with MATH. This leaves one term unpaired: If MATH is even, we have MATH but MATH, while the reverse is true if MATH is odd. In both cases we obtain MATH . Two simple observations now complete the proof of REF. The first is that MATH by REF with MATH. The second observation is that MATH for MATH. This is proved by expanding both terms on the left-hand side by REF, noting that many terms cancel, and applying the following fact to the remaining terms: MATH for MATH. |
math/0006145 | Take MATH to be the subspace lattice of MATH, so that MATH. It is known CITE that MATH . The corollary now follows at once from the proposition. |
math/0006147 | All the preceding computations amount to show that MATH . Then MATH represents a class since the double complex MATH computes the hypercohomology. |
math/0006147 | It is sufficient to show that changing the definition of the various logarithm branches in MATH amounts to change it by a coboundary. First, we change these branches, MATH where MATH. The effect of these changes on the representatives of the NAME classes of MATH and MATH is MATH . While the term MATH is obviously invariant under these changes, MATH and MATH, by descent theory, transform as follows MATH where MATH and MATH. Note that if MATH for any MATH, then MATH, where MATH, and we are done. To prove that MATH, we actually compute the shift for MATH. First, we explicitly determine MATH . Next, we explicitly compute the shift of the total cocycle representing MATH. This is a straightforward calculation, using relations REF, with the result: MATH . Similar formulas are valid for the shift of MATH. Putting everything together, we get MATH . |
math/0006147 | The procedure is to compute the variation of the various components of MATH by applying MATH to the descent equations. Start with MATH, that can be computed in two different ways: from REF, and from the variational relation MATH. Since MATH, we have MATH and we deduce, using NAME Lemma, that MATH for MATH. An explicit calculation using REF confirms this relation with MATH . The last term in this formula is obtained by varying the difference MATH, that enters REF . Clearly, the variation of MATH is zero and for the variation of MATH we have MATH since MATH (see REF). Computing the coboundary of REF yields MATH . On the other hand, the variation of REF gives MATH . Using the first equation in REF: MATH we get MATH . Finally, putting it all together, we obtain MATH as wanted. |
math/0006147 | The MATH-form MATH in the relation MATH is a source form CITE, hence it is uniquely determined by the NAME class of MATH. Moreover, given a specific MATH, the form MATH is also determined (so MATH is determined up to an exact form). Since MATH, we must have MATH as both MATH and MATH are source forms for the same Lagrangian problem. Here the requirement that the variation be vertical is crucial in order to ensure that MATH glue as a geometric object - a vector field on MATH. Therefore, from MATH, we get MATH by NAME lemma. Proceeding in the same fashion we also get MATH . Now, both MATH and MATH are forms of degree one in the field direction, that is, they contain one variation. NAME 's acyclicity theorem CITE asserts the variational bicomplex is acyclic in all degrees except the top one in the NAME direction, provided the degree in the variational direction is at least one. Hence, MATH and we reach the same conclusion as in the previous proof. |
math/0006147 | Indeed, the push-forward of MATH is MATH, where MATH. It is a projective connection on MATH because of the transformation law MATH . The NAME equation is equivalent to the equation MATH, which is precisely the condition that projective connection MATH is holomorphic on MATH. |
math/0006147 | Let MATH be a local section of MATH. As a result of a direct calculation we have (omitting the coordinate index MATH) MATH where MATH is the NAME derivative operator on MATH. Thus the ``if" part is clear. For the ``only if" part, assume the Right-hand side of REF is zero for all MATH. Therefore, if we consider MATH for any local MATH, then we must have MATH, implying REF. |
math/0006147 | Consider the cone of MATH: MATH . Its cohomology sheaf complex equals MATH, thus by standard homological algebra arguments (see, for example, CITE) one has MATH and from the canonical sequence MATH one gets REF. On the other hand, the Right-hand side of REF is the first cohomology group of the complex MATH which is equal to the first term MATH of the spectral sequence computing MATH. |
math/0006147 | We just repeat the computation of the vertical variation with additional term MATH, where MATH. Since the main term, given by MATH vanishes "on shell", this proves the result. |
math/0006147 | For MATH we have MATH where MATH is the function MATH . Using the infinitesimal action, MATH where MATH is the NAME derivative on MATH, and the NAME bracket in MATH is the usual vector field NAME bracket: MATH. Using the identity MATH we are left with MATH because of the commutativity condition and the skew-symmetry of the operator MATH. |
math/0006153 | A simple proof of this Lemma can be constructed using induction on MATH. |
math/0006153 | MATH . The double sum over permutations, MATH and MATH is equivalent to summing each MATH independently over MATH (terms for which two or more components of MATH are equal make zero contribution) and the first result follows. The second result follows in the same way by interchanging the roles of MATH and MATH. |
math/0006153 | (of REF MATH) REF We first obtain the cyclic REF as follows. MATH . The critical step, and the whole reason for introducing the NAME, is to enable one to go from the restricted sums of REF to the unrestricted sums in REF. This is justified for two reasons, CASE: since MATH is a determinant, if any of the MATH's are equal then MATH - this allows the restriction MATH on the sum to be relaxed to MATH CASE: the MATH are in strictly increasing order combined with the fact that the matrix elements of MATH, are only non-zero if MATH allows the restriction on the sum to be removed altogether. The second part of REF follows mutatis mutandis. CASE: The vector MATH is a right eigenvector of MATH with eigenvalue MATH, since MATH which follows from REF. Similarly MATH is a right eigenvector of MATH with eigenvalue MATH. CASE: Let us start by deriving the first result of REF , namely the orthogonality and normalisation condition. Using REF, for MATH since the components of MATH and MATH are in the same order only the identity permutation gives a non-zero delta function product. Thus MATH . Our derivation of the second part of REF , namely the ``completeness condition", closely parallels the derivation of the orthogonality and normalisation condition. Using REF, for MATH so MATH . Notice that MATH which is the row (and column) space dimension, as it should be for completeness. CASE: Using basic linear algebra gives MATH and MATH . |
math/0006153 | If the conditions of REF hold then we have that REF are valid. Using REF it follows that MATH . Also, using REF it follows that MATH . Substituting these into REF and using REF gives the result immediately. |
math/0006153 | Using REF (which follows from REF by REF ) MATH which is an expansion of the required determinant. |
math/0006155 | If MATH is right-orderable (bi-orderable), just take MATH, the set of positive elements. Conversely, if there exists MATH verifying the required hypothesis, then define the order MATH by: MATH if and only if MATH. |
math/0006155 | Notice that, under the action of such a MATH, the degree and the lexicographical order on the monomials are preserved. Hence, MATH preserves the order we defined on the monomials, thus the order MATH on MATH. Therefore, the NAME order on MATH is also preserved. |
math/0006155 | We argue by induction on MATH. If MATH, then MATH is a free group (of infinite rank), so it is bi-orderable. Suppose then that MATH, and that MATH is bi-orderable. By REF , we have an exact sequence MATH where MATH. By definition of bi-order, conjugation by an element of MATH is an automorphism of MATH which preserves the NAME order. We also know, by REF , that conjugation by an element of MATH is an automorphism of MATH which is trivial on MATH. Hence, by REF , it also preserves the NAME order on MATH. Therefore, conjugation by an element of MATH preserves the NAME order of MATH and thus, by REF , MATH is bi-orderable. |
math/0006155 | Let MATH. For all MATH, we write MATH, where MATH is a path on MATH. We denote by MATH the permutation induced by MATH on MATH. Consider the braid MATH. For all MATH, one has MATH . Now consider the isotopy MATH given by MATH . Then MATH and MATH. |
math/0006156 | Consider the following cartesian diagram where MATH is a stable map to MATH, and MATH is the forgetting map, that is, MATH is a prestable curve: MATH . Remember that if we have two morphisms of schemes (or stacks) MATH we get a distinguished triangle of cotangent complexes: MATH . Moreover MATH naturally maps to MATH, so we get the following diagram: MATH . This diagram is in fact commutative since MATH is flat, and so by CITE we have MATH and the morphisms in the diagram above are just the morphism induced by the distinguished triangle MATH . Applying the cut - off functor MATH to MATH and taking the mapping cone yields the following diagram in the derived category: MATH . The projection formula yields the desired morphism MATH. |
math/0006156 | First we will construct a two - term resolution of MATH that is MATH - equivariant if such an action exists on MATH. We will use similar arguments as NAME does for MATH (compare CITE). Let MATH be an ample invertible sheaf on MATH and let MATH. Then by CITE, for MATH sufficiently large and MATH a vector bundle on MATH we have that CASE: MATH is surjective, CASE: MATH, CASE: for all MATH we have that MATH. Let us set MATH and MATH, and consider the complexes (compare CITE) indexed at MATH and MATH where the morphism within the complex MATH is induced from the composition map MATH. Hence there are morphisms MATH where MATH is surjective, by REF and duality. As before we also have MATH so MATH. Observe that the complex MATH fits into the short exact sequence MATH therefore we get a corresponding long exact sequence of higher direct image sheaves: MATH . Hence MATH for MATH. Moreover, since MATH for MATH, we also get MATH for MATH. Now note that these two complexes fit into the following short exact sequence: MATH yielding the long exact sequence MATH . Thus we have found a two - term resolution of MATH by locally free sheaves: MATH . Moreover, the entire construction is MATH - equivariant, so we actually have found a MATH - equivariant resolution of MATH, if such an action exists on MATH. Finally, we observe that MATH in the derived category. Then by using the fact that MATH is an obstruction theory for the relative problem, and by applying the five lemma we get that MATH is an isomorphism and that MATH is surjective: MATH . Hence MATH is indeed a (MATH - equivariant) perfect obstruction theory for the moduli stack of stable maps MATH. |
math/0006156 | The equality of the two virtual fundamental classes can easily be seen by looking at the complex MATH dual to MATH, MATH . Here we have used that by CITE there exists a morphism of functors MATH and that this morphism MATH is an isomorphism. For convenience we also use the notation MATH . Therefore, the MATH's fit into an exact sequence MATH where the sheaves MATH are given by taking cohomology of MATH: MATH . |
math/0006156 | We will prove the lemma for the obstruction complex MATH, the arguments for the dual complex MATH are similar. There is a natural morphism MATH and we have to show that this morphism is a isomorphism in the derived category, that is, a quasi - isomorphism between complexes. Let MATH, indexed at MATH and MATH. We then have to show that MATH . Now MATH fits into a short exact sequence of complexes MATH such that MATH and MATH are locally free and MATH (see above). Since MATH is a proper flat morphism, we have by NAME 's continuity theorem (see for example CITE) that MATH . This yields the same property for the complex MATH. |
math/0006156 | For a stable map MATH to be a fixed point of the MATH - action on MATH means that for any element MATH in the torus MATH, the stable map MATH is isomorphic to the original curve MATH, that is, that there exists a morphism MATH such that the following diagram is commutative (compare REF ): MATH . Now, it is obvious that a curve MATH satisfying the three conditions stated in the lemma is isomorphic to MATH for any MATH, taking for MATH the morphism defined on the irreducible and reduced components MATH by MATH . On the other hand, let MATH be a fixed point of the MATH - action on MATH. We thus have to show that MATH satisfies the three conditions of the lemma. Let MATH be a marked point of the curve MATH. Then it is obvious that MATH has to be mapped to a fixed point in MATH: since MATH has to be constant on the marked points, we have MATH . Now, assume that MATH is a special point of MATH that is not mapped to a fixed point in MATH. Then the orbit of MATH under the MATH - action contains certainly a subspace isomorphic to MATH. On the other hand, the image of the special points of MATH by MATH is a finite set. Hence we obtain a contradiction, since the image of a special point under any MATH is always again a special point. So if MATH is an irreducible and reduced component of MATH that is mapped to a point by MATH, it has to contain at least three special points by the stability condition, and thus is mapped to a fixed point in MATH as well. Similarly, if MATH is an irreducible and reduced component of MATH that is not mapped to a point by MATH, and the image of which is not contained in the closure of a one - dimensional MATH - orbit MATH, then MATH contains a point whose MATH - orbit is at least two - dimensional. On the other hand, MATH always has to be contained in the image MATH of MATH by MATH that is one - dimensional, hence a contradiction. |
math/0006156 | First we will describe the family of curves MATH. For each edge MATH, let us number its vertices: MATH. For each such edge MATH we also fix a map of degree one MATH, such that MATH and MATH. We set MATH to obtain a map MATH of degree MATH. Note that up to parametrization such a map is a unique. Also set MATH. Let MATH be the function assigning to each vertex the number of outgoing edges: MATH . For each vertex MATH, chose an ordering of the set MATH of the outgoing edges: MATH . For convenience, let us number the vertices in the graph MATH: MATH, where MATH is the number of vertices of the graph MATH. We will now glue together the stable map MATH corresponding to a point MATH in the product MATH. The curve MATH is the union MATH where the different parts are glued together along ordinary nodes according to the following rule: CASE: If MATH, MATH then MATH for some MATH, and we will glue the MATH - th marked point MATH of MATH to MATH of MATH, if MATH, or to MATH of MATH otherwise. The ordered set of marked points MATH is constituted of the ``unused" marked points of the curves MATH, that is, the marked points at which we have not glued curves. The function MATH is defined as follows: MATH . Let us explain this construction in plain English. Remember first, that the moduli spaces of stable maps to a point are isomorphic to NAME - NAME spaces: MATH . The points in MATH therefore encode the part of the fixed stable map that is send to fixed points in MATH. The parts of the fixed map that are send to REF - dimensional invariant subspaces are in fact rigid modulo parametrization, and their ``topology" is encoded in the graph MATH. Let us make a remark to the above construction for vertices MATH with MATH: these curves MATH are points that are glued to other points, that is, they do not contribute irreducible components to the constructed curve MATH. Also, if MATH and MATH (otherwise we must have MATH for MATH!), the remaining marked point of MATH is identified with MATH or MATH of MATH, respectively. The proof of the lemma is now straightforward. The family MATH is the space constructed above - essentially it is the product of the universal curves over the NAME - NAME spaces modulo the constant curves corresponding to the edges of the graph. Since the image of a fixed stable map is rigid, the constructed family maps onto a fixed point component of MATH. Also, all fixed point components can be obtained this way. |
math/0006156 | The MATH - dimensional cone MATH gives a local chart MATH of our toric variety MATH, and the coordinates on MATH are given by (compare REF ): MATH . The MATH - dimensional submanifold corresponding to MATH is given by the equations MATH. In the coordinates of MATH, these equations are equivalent to MATH . Hence we have to look at the MATH - action on the MATH co - ordinate. Thus the action of MATH on MATH is given by MATH using multi - index notation in the last line. Hence the weight of the action on MATH is indeed MATH in the chart MATH. |
math/0006156 | The action of MATH on MATH is just the pull back by MATH of the action on MATH. Since MATH has multiplicity MATH, the formula follows immediately from REF . |
math/0006156 | Let MATH be the complex MATH indexed at MATH and MATH. It fits into the following short exact sequence: MATH . The corresponding long exact sequence of higher direct image sheaves corresponding to MATH is then REF where exactness on the left, that is, the injectivity of the map MATH induced by the natural map MATH is equivalent to the stability of the map MATH (compare REF and the remark following the lemma). Exactness on the right of the long exact sequence follows from the fact of the fibers of MATH being curves. |
math/0006156 | The bundle MATH parameterizes infinitesimal automorphisms of the pointed domain. The induced MATH - action on MATH is obviously trivial on all automorphism MATH of MATH that restrict to the identity MATH on all irreducible components MATH corresponding to edges MATH in the graph MATH. Thus, the moving part of MATH splits into MATH . Note in particular, that the bundle MATH is topologically trivial on MATH since it only depends on the irreducible components that are not mapped to a point, that is, that are rigid in MATH. The MATH obviously depend on the special points of the irreducible component MATH, that is, on the way it is ``glued" at the nodes corresponding to the vertices MATH and MATH of the edge MATH. Since we only look at moduli stacks of stable maps with at least three marked points, we can exclude the two special cases (compare REF ) where both vertices are in MATH, or where one is in MATH and the other in MATH. We are left with two cases: CASE: One edge in MATH, the other in MATH - Without loss of generality we can assume that MATH. In this case, MATH corresponds to a non - contracted MATH attached to another non - contracted (or, in the MATH - case, contracted) component (see REF ). Therefore we have to look at NAME transformations that fix one point, infinity say: MATH . Let MATH be the flag corresponding to MATH. We have seen above that the induced MATH - action on MATH is given by (using again multi - index notation): MATH since the co - ordinate MATH corresponds to the chart of the flag MATH while MATH corresponds to the chart around infinity, that is, at the vertex MATH. To determine the MATH - action on the group MATH of infinitesimal automorphisms of MATH, we have to compute: MATH hence the MATH - action on an automorphism of MATH is given by: MATH . The moving part of MATH is thus spanned by the second co - ordinate, and the weight of the action there is MATH - in this case we have: MATH . CASE: Neither vertex is in MATH - In this case, any automorphism of MATH restricts to an automorphism on MATH that fixes the two points corresponding to the special points of the vertices MATH and MATH. Any such automorphism on MATH (w. l. o. g. we take the two points to be zero and infinity) has to look like MATH where MATH is a non-negative integer. With the same analysis as above of the MATH - action on such automorphism MATH, we see that the MATH - action on MATH is trivial, that is, MATH . |
math/0006156 | Just apply NAME duality: MATH where MATH is the dualizing bundle of MATH. |
math/0006156 | Just assemble the long sequence REF , and REF . To obtain the exponent of MATH just observe that there MATH flags at the vertex MATH. |
math/0006156 | First we observe that formally MATH . Since MATH is the product of NAME - NAME spaces of stable curves MATH corresponding each to some vertex MATH of the graph MATH, the integral above is the product of integrals over the NAME - NAME spaces MATH of the classes corresponding to their associated vertex. So let MATH be a vertex with at least three special points, MATH be the number of flags at MATH, and MATH be the flags of MATH. Then MATH . The dimension of the moduli space MATH corresponding to the vertex MATH is equal to MATH, hence MATH . |
math/0006156 | Remember that by REF we have MATH . Hence, by the binomial formula, we obtain MATH . |
math/0006156 | By REF , the formula is obviously true for the parts coming from vertices with at least three flags (that is, special points). Thus it remains to show that MATH . The first term on the right hand side of the equation obviously coincides with the first term on the left hand side. The third factor on the right hand side is trivial, while the second term on the right hand side is equal to the second term on the left, since MATH . |
math/0006156 | Given a graph type MATH, a graph MATH in this graph type is given by the positions MATH of the MATH marked points (where the classes MATH are attached), and vice versa any such tuple MATH of vertices of MATH yields a graph MATH. Hence, up to automorphisms, the first part of the lemma is obvious. For the automorphism groups, first remark that MATH injects into MATH. The cokernel consists of automorphisms of MATH that change the location of the marked points on the graph MATH and hence reflect multiply counted instances of the same graph in MATH. To prove the second part, we observe that MATH yielding the desired expression by applying the binomial formula. |
math/0006156 | Let MATH be a graph type of MATH. Remember, that the weights MATH in the sum of the right hand side of REF are given by MATH where MATH are the flags at MATH corresponding to its MATH edges. Hence we can write MATH . Note that as the notation suggests, MATH only depends on the class MATH and the vertex MATH of the polytope MATH, hence it is well defined for the image flag MATH on MATH. For a flag MATH, let MATH and MATH. Then the weight MATH is defined by MATH hence is given by the two vertices MATH of the polytope MATH, and the multiplicity of the edge MATH. Accordingly, for a flag MATH in a topological graph type MATH corresponding to an edge MATH with multiplicity MATH and ends MATH, let us define the weight MATH . Then for a graph type MATH such that MATH one easily sees that MATH . Hence we obtain that indeed MATH only depends on the topological graph type MATH. |
math/0006158 | This can be seen by pulling back the extension MATH along MATH and then applying REF . |
math/0006158 | The first statement follows from the previous proposition as the mapping MATH is MATH-equivariant. The second follows from the first by taking MATH to be the adjoint representation. The third statement is clear. Set MATH . The inclusion of this into MATH induces a surjection on MATH as MATH has all negative weights. Since MATH and MATH are both nilpotent, this implies that they are equal. This proves the fourth statement. The exact sequence MATH is an exact sequence of graded MATH-modules. Since MATH is central in MATH, it follows that MATH. Since MATH has only negative weights, it follows that the projection MATH induces an isomorphism MATH and that MATH whenever MATH. |
math/0006158 | Since MATH and since the action MATH is compatible with the gradings, each term MATH of the weight filtration is a MATH-module. Since MATH, the image of the action MATH is contained in MATH, from which the result follows. |
math/0006158 | If MATH is reductive, this follows directly from the classical NAME decomposition REF . When MATH connected, the existence of the splitting follows from the existence of a splitting of MATH as MATH is simply connected. The existence of such a NAME algebra splitting follows by first taking a lifting MATH of MATH, and then decomposing MATH under the restriction of the adjoint action to MATH via MATH. The summand of weight zero (the invariant part) is a NAME algebra (as the bracket preserves weights) and projects isomorphically to MATH. First we will prove that any two liftings have to be conjugate by an element of MATH. The intersection of MATH with the centralizer MATH of a lift MATH of MATH is just the identity as the NAME algebra of MATH has strictly negative weights with respect to MATH under the adjoint action. It follows that if MATH is a splitting and MATH, then MATH. Since the lifts MATH are unique up to conjugation by an element of MATH, any two splitting of MATH have to be conjugate by an element of MATH. To prove the existence of the splitting in general, first take a NAME decomposition MATH of MATH, where MATH is reductive and MATH unipotent. This also induces a NAME decomposition MATH of MATH, the connected component of the identity. By the connected case, we can lift MATH to MATH. In particular, we can lift MATH to MATH. Since MATH and MATH are unipotent, MATH is the reductive quotient of MATH. By the classical NAME decomposition, we have a splitting of MATH. The subtlety is that we have to choose this splitting so that its image normalizes the lift of MATH. The splittings MATH and MATH determine lifts MATH and MATH of MATH. By conjugating one of the splittings by an element of MATH, we may assume they are the same. But then the images of both splittings lie in the centralizer MATH. It follows that the image projection MATH is surjective, and therefore an isomorphism. Inverting this isomorphism provides the splitting. |
math/0006158 | Suppose that MATH is another lift of MATH. Then, by REF , there is MATH such that MATH. Write MATH where MATH is the subspace of MATH where MATH acts with weight MATH via MATH. Then MATH. Since each term of the weight filtrations associated to MATH and MATH are MATH-modules, it follows that, for each MATH, MATH . The reverse inclusion follows by reversing the roles of MATH and MATH. The equality of the weight filtrations follows. To prove the second assertion, we choose a splitting MATH of the canonical projection MATH, whose existence is assured by REF . Choose the lift MATH to be the composite of MATH with the splitting MATH. Let MATH be the corresponding weight decomposition of MATH. Since MATH is central in MATH, it follows that the image of MATH preserves this decomposition and that each MATH is a MATH-module. The result follows from REF . The last assertion is clear. |
math/0006158 | We use the notation of Paragraph REF. Since MATH and MATH are finite dimensional, we may assume that MATH and MATH are algebraic. To prove that the weight filtration is preserved by the morphism MATH, choose a lift MATH, and define MATH to be MATH. Then both MATH and MATH become MATH-modules, and MATH is MATH-equivariant. The first assertion follows. Strictness follows similarly as MATH becomes a map of graded vector spaces after choosing the lift MATH. The exactness of MATH follows for similar reasons. If MATH is an exact sequence of weighted modules, one can choose compatible liftings of the central cocharacters, so that the exact sequence becomes an exact sequence of MATH-modules. The sequence MATH is thus exact for all MATH. The exactness of MATH follows as the MATH-th weight graded quotient of a module is naturally isomorphic to its weight MATH part. |
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