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math/9905061 | It is enough to decode the approximate formulas corresponding to MATH . Since MATH, then MATH and MATH. It follows that MATH. Note that for every integer MATH, the constant function MATH belongs to MATH. Hence it follows from the Uniformity Theorem REF that there exists an integer MATH such that: MATH which decoded says MATH which is MATH . Invoking now REF we get that MATH. |
math/9905063 | This statement is certainly known. We reproduce a proof for the convenience of the reader. As a first step, the assumption implies that MATH is an open subgroup in MATH. Admit this for the moment. The MATH-rational maximal tori in MATH are given by maximal commutative semisimple subalgebras MATH in MATH which are stable under the involution MATH. For any such subalgebra MATH, the MATH-subspace of elements fixed under MATH has dimension MATH. Among these maximal tori, the elliptic ones correspond to subfields of dimension MATH over MATH, and they are known to exist. For any element MATH, the neutral component of the NAME closure of the subgroup generated by MATH is an elliptic maximal torus if and only if the algebra MATH is a field. Let MATH be the subset of MATH consisting of all elements of MATH such that MATH is a field of degree MATH over MATH. By NAME 's lemma CITE, MATH is an open subset of MATH. More concretely, if MATH is a field of degree MATH over MATH, then for all MATH sufficiently close to MATH in the MATH-adic topology, the characteristic polynomial and the eigenvalues of MATH are close to those for MATH. So MATH is a field of degree at least MATH, therefore the degree is exactly MATH. It is known that MATH is not empty; so there exists an open normal subgroup MATH such that MATH is a union of cosets for MATH. Recall that NAME 's density theorem states that for any finite étale NAME covering MATH between normal integral schemes of finite type over MATH, the NAME conjugacy classes in the covering group attached to the closed points of MATH is equidistributed, see CITE. Applying this density theorem to the finite étale covering of MATH corresponding to the finite quotient MATH of the monodromy group MATH, we see that there exist infinitely many NAME tori which are elliptic maximal tori in MATH over MATH. This finishes the proof of REF , except for the fact that MATH is open in MATH. This is done in the next two paragraphs. The reader may wish to skip them because it is a little technical and not central to this note; compare REF . Both MATH and MATH are closed subgroups of MATH; hence they are analytic NAME subgroups of MATH. Let MATH (respectively, MATH) be the NAME algebra of MATH (respectively, MATH), and let MATH (respectively, MATH) be the NAME algebra of MATH (respectively, MATH). The assumption of the theorem means that MATH is equal to the NAME algebra MATH of MATH. Therefore MATH is equal to the NAME algebra MATH of MATH. By CITE, we have MATH. We reproduce a proof here. The algebraic subgroup of MATH consisting of all elements MATH such that MATH clearly contains MATH, hence MATH. Repeating this argument again, one concludes that MATH. From MATH, we deduce that MATH and MATH. Especially MATH is an open subgroup of MATH. |
math/9905063 | Suppose that MATH is a closed point such that the NAME torus MATH is an elliptic maximal torus in MATH over MATH. We know that each elliptic maximal torus MATH in MATH is given by a subfield MATH in MATH of degree MATH stable under MATH, hence MATH acts irreducibly on MATH. So the action of any open subgroup of MATH on the MATH-adic NAME module of MATH is irreducible over MATH. This implies that MATH is not isogenous to a non-trivial product over any finite extension of the residue field MATH of MATH. |
math/9905063 | Since the hypothesis does not change if one extends the base field MATH to a bigger number field MATH, it suffices to prove that the abelian variety MATH is simple over MATH. Recall that whether the abelian variety MATH is simple or not is a question on the existence of an endomorphism MATH of MATH such that MATH for some non-zero integer MATH, while MATH is not finite. Since every endomorphism of MATH over MATH extends to an endomorphism of the abelian scheme MATH over MATH which extends MATH, we are done. |
math/9905064 | REF is a result of REF follows from REF. |
math/9905064 | Using REF for MATH and for any MATH proves MATH . REF with MATH shows MATH . Finally by REF , we have MATH . |
math/9905064 | By REF the composition of the linear map MATH and the canonical map MATH which sends MATH to MATH is a surjective homomorphism. Then REF shows that MATH is generated by MATH and MATH for MATH. |
math/9905064 | Note that MATH is spanned by MATH where MATH and MATH are integers. We can assume that MATH is a monomial MATH in MATH . For such MATH we define the length MATH to be MATH . As usual we define MATH. We prove by induction on the length of monomial MATH that MATH . If MATH, then MATH and it is clear. Suppose the lemma is true for all monomials with lengths strictly less than MATH. Let MATH with the length MATH. Recall that the component operator MATH of MATH is defined by MATH . It follows from REF that MATH for MATH . Now we compute the MATH product MATH . Note that if MATH and MATH, then MATH. Therefore, if MATH, then MATH and if MATH, then we see either MATH or MATH, namely, MATH . Hence we have MATH where MATH and where MATH. Then by the induction hypothesis MATH . As a result, we have MATH. |
math/9905064 | It is enough to prove MATH. Recall the vector MATH from the proof of REF . We also use induction on the length of the monomial MATH. If MATH, it is clear. Let MATH be a positive integer and suppose that the claim is true for all monomials MATH with lengths strictly less than MATH. Now, consider MATH for MATH with MATH. REF shows MATH . From the proof of REF we see that MATH where MATH and MATH. Thus each MATH is a linear combination of MATH's where MATH and where MATH are monomials with lengths less than MATH . By the induction hypothesis each MATH lies in MATH . Thus MATH and MATH . |
math/9905064 | Recall the definition of the circle operation MATH . Also recall that MATH and note that if MATH then either MATH or MATH in the sum. This immediately gives MATH for MATH. Thus we have MATH . It is easy to see that MATH . So MATH . Similarly, when considering MATH and MATH respectively, we obtain MATH and MATH . Add REF together and use REF to yield MATH . Consequently we have MATH. Since MATH is invariant under the permutations of MATH, we can apply the same result to indices MATH and MATH and finish the proof of the lemma. |
math/9905064 | We prove the lemma by induction on MATH. When MATH, it is nothing but REF . Let us suppose MATH and decompose MATH as MATH where MATH and MATH with MATH and MATH for MATH . Note that MATH . By the induction hypothesis, MATH . So we have MATH . The proof is complete by the fact that MATH which follows from either REF or the induction hypothesis. |
math/9905064 | The proof is similar to that of REF . For any MATH and positive integers MATH set MATH . Then MATH is spanned by all possible MATH . We again use induction on MATH for a monomial MATH to show that MATH lies in MATH and MATH which are the left and right MATH-modules generated by MATH respectively. If MATH then by REF , MATH lies in MATH and MATH . If MATH it is clear that MATH where MATH and MATH. Thus by the induction hypothesis, MATH lies in both MATH and MATH . REF shows MATH is an element of MATH . It remains to show that MATH . As in the proof of REF we have MATH . Note that MATH . Since MATH there is at least one MATH positive. Thus MATH is a linear combination of vectors like MATH for negative MATH and a monomial MATH whose length is less than the length of MATH . By the induction hypothesis, MATH . Thus MATH as required. |
math/9905064 | REF shows MATH . |
math/9905064 | Set MATH. Since MATH we have MATH . Note that MATH, that is, MATH . Combining REF together proves MATH . Finally, using REF we get the desired result. |
math/9905064 | We prove the lemma by induction on MATH. When MATH, it is clear. Let MATH and MATH with MATH . Then the relation MATH gives MATH which implies MATH . Thus MATH . This shows that MATH . The induction hypothesis then yields that MATH is spanned by MATH . |
math/9905064 | We first prove MATH. It is enough to show that MATH for MATH is expressed as a linear combination of MATH. Note from REF that MATH for any positive integer MATH where MATH . Since MATH is a linear combination of MATH by REF , the element MATH is also a linear combination of MATH for MATH . We also see from REF that the space spanned by MATH is closed under the bracket MATH . We now prove by induction on positive integer MATH greater than or equal to MATH that MATH is a linear combination of MATH . Suppose that this is true for MATH. Then since MATH is a linear combination of MATH . From the discussion in the previous paragraph, MATH is also a linear combination of MATH and so is MATH . It remains to show that MATH are linearly independent. Suppose that there exist scalars MATH such that MATH . We have to prove that MATH . This is achieved by the following evaluation method. We evaluate REF on the top levels of irreducible modules MATH, MATH and MATH. Consider the basis MATH for MATH and the basis MATH for MATH. Let MATH be the corresponding matrix element with respect to the basis. Also write MATH. Then the action of MATH on the top levels of MATH, MATH and MATH are listed in REF ; MATH acts trivially on the top levels of the modules MATH and MATH. CASE: Then we have the linear system MATH . The only solution to this linear system is MATH as expected. |
math/9905064 | One could prove the lemma using the fact that MATH is a linear combination of the elements MATH's with a lengthy computation. Here we give a short proof by evaluation method discussed in the proof of REF . Consider MATH or MATH or MATH. Clearly MATH is a linear combination of MATH for MATH modulo MATH and acts trivially on top levels of irreducible modules in REF . The proof of REF shows that MATH. |
math/9905064 | Write MATH as a linear combination of the elements MATH and MATH. Then the result is immediate. |
math/9905064 | Since all equalities in the lemma hold on any top levels of the modules MATH and MATH, it suffices to prove by REF that the left hand sides of the equalities belong to MATH. First we consider the case that MATH and MATH are all distinct. Then by REF , we know that MATH is expressed as a linear combination of the elements MATH and MATH and MATH modulo MATH for some scalars MATH . Evaluation of REF on the top level of MATH shows MATH which implies MATH. Therefore MATH is an element of MATH. Further, since MATH, we have MATH. Next we consider the case either MATH or MATH. REF show that MATH is a linear combination of MATH and MATH modulo MATH with some scalars MATH. The same evaluation in the previous paragraph shows MATH, that is, MATH. Note that MATH and MATH. Since the space spanned by MATH for MATH is invariant under MATH for MATH lies in MATH . The same evaluation then concludes that MATH. |
math/9905064 | We only consider the case that MATH or MATH here. The case that MATH and MATH are all distinct is treated in Subsection REF; see REF . We first prove that if MATH then MATH is a linear combination of MATH modulo MATH . By REF , it is enough to show that MATH is a linear combination of MATH modulo MATH . Recall that MATH where we set MATH. The space MATH is a NAME module with respect to a NAME element MATH and decomposes into the direct sum of irreducible modules MATH where MATH denotes an irreducible highest weight module with highest weight MATH and central charge MATH for the NAME element MATH (compare CITE). Since MATH we see MATH for MATH. Therefore, MATH is a linear combination of the elements MATH . It is enough to show that MATH are linear combinations of the elements MATH modulo MATH . By REF we can assume that MATH. Set MATH . Then by REF , MATH . Note that MATH . Induction on MATH gives the desired result. Since MATH the same argument above shows that MATH and thus MATH are also linear combinations of MATH modulo MATH . If MATH by REF , the span of MATH for MATH is the same as the span MATH for MATH . This implies that both MATH and MATH are linear combinations of MATH modulo MATH . Then by REF , MATH and MATH are linear combinations of MATH and MATH modulo MATH and so are MATH and MATH. Now, let MATH modulo MATH with some scalars MATH and MATH. Then the evaluation of these expressions on the top level of MATH shows MATH and then MATH. Similarly we can prove MATH. Namely, MATH is closed under both right and left multiplications by MATH. |
math/9905064 | Let MATH and MATH. REF shows that there exists a scalar MATH such that MATH . If MATH, the left hand side of REF acts trivially on the top level of the module MATH. So the evaluation of REF on this top level shows MATH, which proves MATH. Similarly, MATH. Next, the evaluation of MATH on the top level of the module MATH leads MATH as MATH . |
math/9905064 | In the case MATH, MATH, it is clear. Suppose MATH or MATH, then since MATH by REF , the commutator belongs to MATH. Since MATH acts trivially on the top levels of the known irreducible modules by REF asserts that MATH . The relation MATH is proved by the same argument in the proof of REF . That is, we express MATH as a linear combination of MATH and MATH and then evaluate the expression on the top levels of the known irreducible modules to show that MATH . |
math/9905064 | We first consider REF . Let MATH and MATH. We can assume MATH without loss of generality. Then by REF we have that MATH with some scalars MATH and MATH. Note that in REF case, element MATH acts trivially on the top levels of the modules MATH, MATH and MATH . This implies that all MATH and MATH are zero. Let us now prove REF . Using the same argument in the previous paragraph shows MATH if MATH and MATH if MATH as MATH acts trivially on the top level of MATH. Then the result follows from REF . Finally the evaluation of the equality MATH on the top level of the modules MATH and MATH (see REF ) gives the linear system MATH . Thus we have proved MATH . Then substituting MATH (see REF ) into REF and using REF gives REF . |
math/9905064 | By REF , MATH . We have already proved in REF that MATH . Using REF again shows that MATH . Similarly, MATH. Since MATH and MATH act trivially on the top levels of MATH and MATH, the corollary follows. |
math/9905064 | Since MATH for all MATH we see that MATH is a polynomial in MATH and MATH (compare REF ). In particular, MATH commutes with MATH. Then by REF , we have MATH . Similarly, MATH. Since MATH commutes with MATH by REF , we also have MATH. Thus the same argument in the previous paragraph shows that MATH. The rest of the equalities are proved similarly. |
math/9905064 | Suppose MATH. Then REF or REF shows MATH. If MATH, then MATH. Since MATH lies in the subalgebra generated by MATH, we use REF (also see REF ) to show MATH. Then the evaluation method proves MATH. The rest of the equalities in REF - REF are proved similarly. |
math/9905064 | Since MATH by REF , we have MATH and MATH by REF . |
math/9905064 | Using REF , we see MATH. Similarly we prove MATH by REF . |
math/9905064 | It suffice to prove that MATH and MATH are ideals. But this is clear from REF . |
math/9905064 | First, we see from REF that MATH . Note that MATH and MATH or MATH is nonnegative if MATH. Thus MATH . We now deal with MATH . From REF , MATH . Since the weights of MATH and MATH are less than or equal to REF, we see that MATH with some scalars MATH and polynomials MATH and MATH . Evaluating REF on the top level of the module MATH and noting that MATH (see REF ) yields MATH for all MATH, which implies MATH. Therefore, we have MATH . Further the evaluation of REF on the top level of the module MATH shows that MATH and MATH . Similarly, compute MATH in two different ways gives MATH for some scalars MATH and polynomials MATH and MATH . Again the evaluation on MATH shows that MATH . That is MATH . Now the evaluation of REF on the top level of the module MATH shows MATH the evaluation of REF on the top level of the module MATH yields MATH . Solving the linear system REF we find MATH . |
math/9905064 | Recall that MATH . So MATH . Now using MATH (compare REF ), we have MATH and MATH . Then REF shows MATH with some scalars MATH and polynomials MATH and MATH. Now the evaluation of REF on the top level of the module MATH shows MATH . This implies MATH and MATH. Thus we have MATH . Finally the evaluation of REF on the top levels of the modules MATH and MATH respectively gives MATH . Then solving the linear system REF gives MATH and MATH. For the second relation, note from REF that MATH and MATH . Thus MATH . On the other hand, the evaluation method shows MATH and then we have MATH . |
math/9905070 | Let MATH satisfy REF. Then, MATH . Let MATH and note that MATH . Moreover, note that MATH . Now suppose that MATH for MATH, MATH. If MATH, then MATH. Thus by REF we obtain, MATH by which we obtain the first part of the lemma. However, if MATH then by REF there is a MATH such that MATH . For this case, by REF we obtain MATH which implies that MATH. |
math/9905070 | Except for the conditions given in REF, the claims made in the statement of this lemma are an immediate consequence of REF . To verify the conditions given in REF, we begin by letting MATH where MATH and MATH are defined in REF, and where MATH. Now, let MATH in REF and note that MATH . Moreover, if MATH, then also note that MATH . If MATH, then by REF there is a matrix MATH such that MATH. Thus, MATH and REF follows immediately from REF when MATH. On the other hand, if MATH, then REF implies that MATH. As a result, MATH and again REF follows from REF, and REF when MATH. |
math/9905070 | Differentiating REF with respect to MATH and inserting MATH immediately yields REF. |
math/9905070 | With MATH and with MATH defined in REF, MATH satisfies REF. If MATH, then MATH. MATH has rank MATH, hence MATH is nonsingular. As a result, MATH . However, if MATH, then MATH . By REF, MATH; thus, by REF there is a MATH such that MATH. From this, we obtain MATH . Hence for MATH which satisfy REF, MATH . For all MATH and MATH, note that MATH . Together, REF imply that MATH . If for some MATH there is a MATH, MATH, such that MATH, then MATH . Together, REF imply that MATH, and hence that MATH. Given the uniqueness of solutions for system REF, we conclude that MATH; thereby producing a contradiction which completes the proof. |
math/9905070 | With MATH and MATH, we see that MATH by REF. With MATH defined in REF, then by REF , we can define MATH for MATH. As defined, MATH satisfies REF. Then for MATH, REF implies that MATH . This is equivalent to MATH . Given the invertibility of MATH shown in REF , we infer that MATH satisfies REF. The proof of this theorem will be completed upon showing that MATH satisfies REF. First, observe that MATH and hence that MATH . Upon differentiating REF and using the fact that MATH satisfies REF, we obtain MATH . By REF, it follows that MATH which is precisely REF. |
math/9905070 | Let MATH be the greatest value such that REF holds for MATH. We show that REF must hold for some MATH with MATH, thus proving MATH. The solution of REF, MATH, presumed to be defined on MATH, can be continued onto some MATH with MATH; MATH then satisfies MATH for MATH and for some MATH. For brevity, let MATH denote REF with MATH, and let MATH denote REF in the following. Integrating REF, we obtain MATH . Note that MATH and by REF, as MATH and for MATH, that MATH and by REF, and REF, that MATH . Together with REF, we observe that as MATH REF is equivalent to MATH uniformly so, for MATH. Since MATH and MATH for MATH, MATH . Thus REF yields MATH . In light of REF, an application of NAME 's inequality to REF yields MATH uniformly for MATH. For MATH sufficiently large and MATH, MATH satisfies REF with MATH, and MATH satisfies REF, hence MATH satisfies REF for MATH, where MATH. |
math/9905070 | Utilizing the standard connection between the explicit exponential solutions of the second-order NAME REF and the NAME REF (in the special case MATH, compare REF ), performing a conformal map of the type REF, and the variable transformations REF then yields the following solution for REF, MATH associated with the general initial condition MATH for some MATH with MATH, MATH. Since by REF, the exponential terms in REF enforce MATH unless MATH implying REF. |
math/9905070 | Though stated for elements of the NAME disk MATH, the proof of REF shows that the asymptotic expansion given in REF holds, uniformly with respect to MATH for MATH in MATH, for all elements of the NAME disk MATH as noted in REF. Note that the system REF is independent of the reference point MATH. Recall that MATH, defined in REF, is determined by our apriori choice of the sequence MATH subject only to MATH being in MATH (compare REF). Next note that MATH, which is defined as a limit in REF, which is described explicitly in REF , and which gives solutions of REF which satisfy REF for MATH, is also independent of the reference point MATH. Thus, had we chosen MATH as our reference point at the start, the asymptotic analysis begun in REF and continued in REF - REF would remain the same after the variable change in REF except for the integral expression present in REF in which MATH would be replaced by MATH. However, given the local integrability assumption on MATH present in REF , we see that the integral expression in REF is uniformly continuous for MATH in a compact subset of MATH. Thus REF and consequently REF are uniform for MATH, and for MATH, in compact subsets of MATH. As a consequence, we see that REF holds for elements of the NAME disk MATH, that this asymptotic expansion is uniform with respect to MATH for MATH in MATH and that it is uniform in MATH as long as MATH varies in compact subsets of MATH. The asymptotic expansion described in REF then follows by REF . |
math/9905070 | In the following let MATH, MATH, and MATH. The existence of an expansion of the type REF is shown as follows. First one considers a matrix NAME integral equation of the type (compare CITE, CITE, CITE, CITE, CITE) MATH and observes that the solution MATH of REF satisfies MATH in accordance with REF. Next, introducing MATH one rewrites REF in the form MATH and thus infers, MATH . Iterating REF then yields MATH for some MATH depending on MATH. Finally, we need one more ingredient, proven in CITE using appropriate maximal functions. Let MATH, MATH, for some MATH, and suppose MATH is a right NAME point of MATH. Then MATH . An alternative proof of REF follows from CITE, which implies MATH for any right NAME point MATH of MATH. Given these facts, one iterates REF and its MATH-derivative, integrates by parts, applies REF to MATH for all MATH, and estimates MATH by REF. Inserting the expansions for MATH and MATH-into REF (using a geometric series expansion for MATH) then yields the existence of an expansion of the type REF. The actual expansion coefficients and the associated recursion relation REF then follow upon inserting expansion REF into the NAME REF . The assertion following REF is an immediate consequence of REF and its derivative with respect to MATH, REF, and the NAME lemma. |
math/9905070 | Clearly REF satisfies REF since MATH . Conversely, let MATH be a solution of REF and MATH a fundamental matrix of solutions of REF. Define MATH . Then MATH implies MATH . Thus, there exists a constant MATH matrix MATH (possibly singular), such that MATH with MATH a fundamental matrix of solutions of REF. Hence, MATH. |
math/9905070 | This is obvious from REF. |
math/9905070 | Define MATH by MATH then MATH due to the uniform nature of the asymptotic expansion REF for MATH varying in compact intervals. Next, introduce MATH then MATH . Using MATH a standard NAME iteration argument in REF then yields MATH and hence REF. |
math/9905070 | Define for MATH, MATH . By REF , MATH and hence by REF , MATH where MATH and MATH are fundamental matrices of MATH respectively, with MATH . By REF , MATH as MATH, MATH. Thus, as MATH, MATH, MATH for some constant MATH by REF, and REF. |
math/9905070 | Define MATH and apply REF with MATH, MATH. Then (in obvious notation) MATH as MATH, MATH, and hence the asymptotic expansion REF for MATH in REF coincides with that of MATH (this also applies to the case MATH). |
math/9905075 | Since MATH, MATH, MATH, MATH, we have MATH where MATH . Since the summation MATH is the same as MATH, we have MATH with MATH. Replacing MATH with MATH, the summation turns out be MATH and we have MATH where MATH . Note that from REF , we have MATH . As easily seen from REF , MATH and MATH vanishes if MATH and if MATH respectively. We will use this fact repeatedly from now on and divide the proof into some cases according to the order of MATH. First we divide the proof into two cases; MATH and MATH. REF (MATH). In this case MATH and so from REF, MATH if MATH. If MATH, we see that MATH . Therefore MATH with MATH. From REF we have MATH . We divide the case into two subcases; MATH and MATH. CASE:REF (MATH). Since MATH we have MATH and so MATH where the second equality follows from MATH. Noting that MATH, we see that MATH . Therefore we have MATH if MATH and zero otherwise. Thus the proof is complete for the case where MATH. (Note that MATH if MATH.) REF (MATH). In this case we have MATH if MATH and MATH otherwise. Therefore MATH if MATH. If MATH, we have MATH . This vanishes if MATH and equals MATH otherwise, completing the proof for the case MATH and MATH since MATH if MATH. REF (MATH). First note from REF, MATH if MATH. If MATH, we have MATH and so MATH . Since MATH, MATH vanishes and so does MATH if MATH. Therefore we assume that MATH. In this case since MATH we have MATH . Therefore MATH . There are two subcases; MATH and MATH. CASE:REF (MATH). Since MATH we have MATH and the proof for the case where MATH and MATH is complete. (Note that MATH vanishes unless MATH.) REF (MATH). In this case we only have to consider the case where MATH for otherwise MATH. Now MATH and so we have MATH . The proof for the case where MATH and MATH is now complete. (Note again that MATH unless MATH.) |
math/9905075 | Since MATH, MATH, MATH and MATH, we see that MATH vanishes except for the following four cases, which have already appeared in REF : CASE: MATH, REF MATH, REF MATH and REF MATH. We will only prove the first case because the other cases are similar. Noting that MATH we have MATH since MATH, MATH, MATH and MATH. Now since MATH, we see that MATH . Therefore MATH as required. |
math/9905075 | From REF , we only have to check that MATH. We have MATH but this coincides with MATH as shown below. We have MATH . On the other hand from CITE, we have MATH . Divided by MATH and taking the limit MATH, we have MATH . Therefore we have MATH completing the proof. |
math/9905075 | It is sufficient to show that MATH. Since MATH, MATH and MATH . But these two coincide since MATH vanishes unless MATH (the charge conservation law), completing the proof. |
math/9905075 | From REF , we have MATH . From the NAME - NAME equation for MATH these two coincide, completing the proof. |
math/9905075 | Since MATH, MATH, MATH, MATH and MATH, we have MATH completing the proof. |
math/9905075 | Noting that MATH and MATH commutes since they are diagonal, the first equality follows immediately from that in REF . The second equality follows from MATH completing the proof. Note that the lemma can also be proved by using CITE. |
math/9905075 | In fact we can show that MATH is of the form MATH by using REF repeatedly to `push' MATH and MATH from left to right (See the proof of REF ). Details are omitted. |
math/9905075 | This lemma comes from REF and the following three facts CITE (see also CITE). CASE: the NAME polynomial of a non-trivial torus knot is not trivial. CASE: the NAME polynomial is multiplicative under the connect sum. Therefore if MATH and MATH are non-trivial, then MATH is also non-trivial. CASE: if MATH is a knot obtained from MATH by a cabling operation, then MATH is MATH with some NAME polynomial MATH. Hence if MATH is non-trivial, so is MATH . |
math/9905075 | First note that every coefficient of both colored NAME polynomial and NAME polynomial as a power series in MATH is a NAME invariant. So a knot MATH with every NAME invariant trivial has the trivial colored NAME polynomial for any color and the trivial NAME polynomial. In particular MATH for any MATH. Therefore assuming the volume conjecture, MATH vanishes. From REF , MATH should be trivial, completing the proof. |
math/9905075 | We only prove the equality for MATH since the other case is similar. We use the following quantized NAME relation. MATH . Then since MATH we have the following recursive formula. MATH . Now the required formula follows since MATH for any integer MATH. |
math/9905076 | Suppose that MATH with MATH, or REF. We begin with MATH. Suppose that MATH splits off REF times from MATH. Then the residual system MATH. MATH is disjoint from MATH only if MATH, and so MATH implies that MATH. Thus we have the class MATH with virtual dimension MATH, but dimension MATH. Suppose that MATH splits off REF times from MATH. Then MATH. We require that MATH so that MATH. Then MATH implies that MATH. There are four possibilities for MATH. MATH: MATH implies that MATH. Then MATH implies that MATH. This, however, makes MATH . MATH: MATH implies that MATH. But MATH and MATH, so MATH and MATH. Again this makes MATH. MATH: MATH implies that MATH. But MATH and MATH, so MATH and MATH. This makes MATH. MATH: MATH implies that MATH. This gives the system MATH, for all MATH. Suppose that MATH splits off REF times from MATH. Then MATH. We require that MATH so that MATH. Now MATH whenever MATH. This gives the system MATH whenever MATH. This completes the MATH analysis. Suppose that MATH splits off REF times from MATH. Then MATH, and MATH. This can never be zero, so this case cannot occur. Suppose that MATH splits off REF times from MATH. Then MATH. We require that MATH . This forces MATH. Then MATH requires MATH, and there are five possibilities for MATH. MATH: MATH implies that MATH, but MATH. This case cannot occur. MATH: MATH implies that MATH, but MATH. This case cannot occur. MATH: MATH implies that MATH and MATH. In this case MATH and MATH forces MATH to be REF or REF. These are the systems MATH and MATH. MATH: MATH, so MATH and MATH. Now MATH has virtual dimension REF for all MATH. Therefore we have the (-REF) special system MATH for all MATH. MATH: MATH, so MATH and MATH. Now MATH has virtual dimension REF for all MATH. This gives the system MATH for all MATH. Suppose that MATH splits off REF times from MATH. Then MATH. We require that MATH so that MATH. Now MATH if MATH. This gives the systems MATH for every MATH. This completes the MATH analysis. Suppose that MATH splits off twice from MATH. Then MATH, and MATH, so MATH. MATH implies that MATH. This gives the systems MATH, MATH, and MATH. Suppose that MATH splits off twice from MATH. Then MATH and MATH. This forces MATH. We also have MATH so that MATH. There are six possibilities for MATH. MATH: MATH, so MATH and MATH. But MATH, so this case cannot occur. MATH: MATH, so MATH and MATH. Now MATH implies that MATH is REF or REF. This leads to the systems MATH and MATH. MATH: MATH, so MATH and MATH. Then MATH implies that MATH is REF, or REF. These systems are MATH, MATH, MATH, and MATH. MATH: MATH, so MATH and MATH. In this case MATH for all MATH, giving the systems MATH for all MATH. MATH: MATH, so MATH and MATH. This time MATH for all MATH, giving the systems MATH for all MATH. MATH: MATH, so MATH and MATH, but MATH for all MATH. This means that MATH cannot split off by itself in this case. We will see later that this case does occur when MATH and MATH both split. Suppose that MATH splits off twice from MATH. Then MATH and MATH. This forces MATH. Now MATH occurs whenever MATH. This gives a (-REF) special system MATH whenever MATH. Suppose that MATH splits off twice from MATH. Then MATH and MATH. Now MATH only if MATH, but MATH since MATH splits off twice. Therefore MATH and the system is MATH . Suppose that MATH splits off twice from MATH. Then MATH implies that MATH, and MATH implies that MATH, giving the system MATH. This completes the analysis of the case where one curve splits off twice. The final possibility is that MATH, where MATH and MATH are disjoint (-REF) curves or configurations. As mentioned previously, MATH and MATH are the only possibilities for MATH and MATH. In this situation, the residual system is MATH. MATH and MATH intersect MATH as MATH . MATH . From the first equation we get that MATH. From the second we then have MATH. Now MATH implies that MATH. This leads to the (-REF) special systems MATH for every MATH. |
math/9905076 | If either MATH or MATH is empty then MATH is empty as well since the kernel systems are empty. If MATH and MATH are not empty, then MATH . The first equality follows from REF . The second is true because the systems are non-special and not empty. The third equality is REF a. The final inequality holds by REF. Therefore REF applies and MATH. Now MATH must be empty since MATH . |
math/9905076 | The proof relies on the identities from REF . We claim that with the given hypotheses, MATH. There are three possibilities. If both MATH and MATH are empty, then MATH since MATH. If both systems are non-empty and non special, then MATH and MATH. Then using the three identities we get MATH . The inequality holds since MATH. If one of the systems is empty and the other is not, identities REF give MATH or MATH . The inequalities follow from REF . Now, MATH . We apply REF to get MATH, by REF a. Finally, we have MATH. Therefore MATH and MATH is non special. |
math/9905076 | We may assume that MATH and that MATH. The proof is by induction on MATH. Assume the theorem is true for smaller values of MATH. Then assume that MATH is not (-REF) special, and prove that it is non-special. Begin with the case MATH. We perform a MATH degeneration with MATH. This gives the following relevant systems. MATH . We wish to find a MATH so that both MATH and MATH are empty, and all four systems are non-special. We pick MATH so that MATH is not (-REF) special, and thus non-special by REF. Then MATH, which makes MATH. Therefore MATH is empty. If, in addition, MATH, then MATH since MATH and MATH. Now if MATH is not (-REF) special (and non-special by induction), it will be empty. If MATH this can be ensured by choosing MATH so that MATH is odd. This is also enough to conclude that MATH is not (-REF) special (and non-special by induction). In order for MATH to be not (-REF) special (and thus non-special by REF), we choose MATH. An integer MATH, satisfying the inequalities MATH and such that MATH is odd, may be found when MATH is REF, or MATH. This choice of MATH makes all of the systems involved non-special and the kernel systems empty. We apply REF . This proves that MATH is empty, provided MATH, MATH and the conjecture holds for lower MATH. If MATH, MATH might be (-REF) special even when MATH is odd, but only if MATH. If MATH is not (-REF) special, REF can be used to conclude that MATH is empty. We would like MATH . For a given MATH and MATH, we only need to prove the theorem for the smallest MATH which makes MATH negative. For MATH, this value is the smallest integer MATH such that MATH. It is clear that we need to choose MATH as small as possible. We will pick MATH such that MATH is odd. Then MATH. Now MATH whenever MATH. This guarantees that MATH is not (-REF) special when MATH and MATH. For values of MATH between REF, one may directly check that this choice of MATH makes MATH not (-REF) special for the smallest value of MATH. The same is true for MATH and REF. We are assuming the theorem for smaller values of MATH, so REF applies and these systems are empty. The only outstanding case with MATH is MATH and MATH. For MATH and MATH, the smallest MATH which makes MATH negative is MATH. We choose MATH and MATH. This makes MATH non-special and empty by REF and MATH non-special and empty by induction. MATH is not (-REF) special since MATH and non-special by REF. MATH is not (-REF) special since MATH is odd for MATH and non-special by induction. Now all the systems are non-special and the kernel systems are empty so we use REF to conclude that this system is empty. We have proven the theorem for all MATH provided it is true for smaller values of MATH. The theorem will be proved case by case for smaller values of MATH. If MATH, we perform a MATH degeneration giving the systems MATH . We are assuming that MATH. If all these systems are non-special and the kernel systems are empty then we use REF to conclude that MATH is empty. The kernel systems have negative virtual dimension, so are empty if they are non-special. The systems MATH and MATH are not special by REF. The system MATH is (-REF) special only if it has the form MATH or MATH with MATH. The first comes from the system MATH which has virtual dimension MATH, contrary to hypothesis. The second type comes from a system MATH with virtual dimension MATH. Again, MATH is positive (since MATH), contrary to hypothesis. The system MATH is (-REF) special only if it has the form MATH. This comes from the system MATH with virtual dimension MATH. The system MATH is special, but NAME reduces to the class of a line, and so has dimension REF. MATH is non-special with virtual dimension REF. So in this case, MATH and we may appeal directly to REF to conclude that MATH is empty. The next open case is MATH. (The cases MATH and MATH were mentioned earlier.) We perform a MATH degeneration giving the systems MATH . The kernel systems have negative virtual dimension, hence they are empty if they are non-special. MATH and MATH are non-special by REF. If MATH and MATH are non- special we apply REF to conclude that MATH is empty. MATH is only (-REF) special if it of the form MATH or MATH where MATH. Both of these come from systems with positive virtual dimension. MATH is (-REF) special only if it is MATH or MATH. In these cases we appeal directly to REF . In both cases MATH is non-special of dimension REF. MATH . NAME reduces to the zero dimensional space of constant polynomials. Now MATH, so we use REF to conclude that MATH is empty. In the other case, we notice that MATH . NAME reduces to REF dimensional space of quadratics. This means that MATH, and REF tells us that MATH is empty. When MATH we pick MATH as before (MATH and MATH odd). This allows us to apply REF unless MATH. It is enough to prove MATH is empty. We perform a MATH degeneration in this case and apply REF . For MATH, a MATH degeneration produces systems which are not (-REF) special and kernel systems which are empty, unless MATH or MATH with MATH. The later does not have negative virtual dimension. The former degenerates as MATH is non-special of dimension MATH. MATH . NAME reduces to the zero dimensional class of a quadruple line. In this case, MATH, and REF tells us that MATH is empty. If MATH, we perform a MATH degeneration. This produces systems which are not (-REF) special and kernel systems which are empty unless MATH or MATH, where MATH. These exceptions all come from systems with positive virtual dimension, therefore REF handles all cases with MATH. When MATH we pick MATH as before (MATH and MATH odd). This satisfies the conditions of REF unless MATH. It is enough to show that MATH is empty. A MATH degeneration allows us to use REF in this instance as well. For MATH, a MATH degeneration satisfies the conditions of REF in all but five cases, only two of which have negative virtual dimension. In these cases we use NAME reduction to find the dimension of MATH and apply REF . The MATH case has one exception when a MATH degeneration is used, and it is handled in the same way. A MATH degeneration also suffices to prove the theorem in case MATH and REF. When MATH, a MATH degeneration works unless MATH is MATH, MATH, or MATH. In the first two of these exceptions MATH is special and we proceed as above using NAME reduction to find the dimension. In the final case, it is MATH which is special and we must find another approach. A MATH degeneration allows us to use REF to conclude that MATH is empty in this case as well. If MATH, we use a MATH degeneration. This lets us apply REF in all but two cases. In these cases, we use NAME reduction and REF . For MATH, a MATH degeneration allows us to apply REF unless MATH. This system NAME reduces to the system MATH. The theorem is true for MATH CITE, and this system is not (-REF) special, so this system is empty. Therefore MATH is empty, as well. When MATH, we use a MATH degeneration. We may appeal to REF in all but a finite number of cases. In all but one of these cases we use NAME reduction to determine the dimension of MATH and apply REF . In case MATH, we use a MATH degeneration and REF . If MATH, a MATH degeneration succeeds. If MATH there are eleven possibilities for MATH. In each case, it is enough to prove the theorem for the smallest MATH which makes MATH negative and the system not (-REF) special. We have the following systems MATH . The first system NAME reduces to a homogeneous system with MATH, which is empty CITE. A MATH degeneration is used for all but the third and ninth systems. For these degenerations, MATH is special but we use NAME transformations to find its dimension and apply REF . The remaining systems yield to a MATH degeneration and REF . For MATH or REF, a MATH degeneration works. A MATH degeneration also works when MATH in all but one case. The case MATH did not yield to any of the approaches used so far. To see that this system is empty, Maple was used to analyze the MATH matrix constructed using the points MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, and MATH. The matrix was found to have full rank, therefore there are no thirteenth degree polynomials passing through these points with multiplicities. This implies that MATH is empty for points in general position. In the MATH case, we again list the systems corresponding to possible values of MATH and the critical MATH for each. We have the following seven systems. MATH . It is sufficient to show that the second, sixth, and seventh systems are empty, as these imply the rest are empty. The second system NAME reduces to MATH, which is non-special and empty by CITE. To show the sixth is empty, we employ a MATH degeneration and REF . Finally, we use a MATH degeneration with the seventh system. MATH . NAME reduces to a quadruple line, and so has dimension REF. We apply REF . The MATH case proceeds similarly. There are six possibilities MATH, and we only need to check the smallest MATH in each. We only need to show that MATH and MATH are empty since these imply the other four are empty. In both of these cases we use a MATH degeneration. MATH is special, but NAME reduces to the two dimensional class of a line. In both cases MATH is not (-REF) special with virtual dimension REF. We apply REF . In the MATH case, for similar reasons, we only need to show that MATH and MATH are empty. The first NAME reduces to the empty system of constant polynomials with a double point. The second reduces to the empty system of quadratics with a triple point. When MATH, it suffices to show that MATH and MATH are empty. The former reduces to the empty system of lines through three general points. The latter reduces to the empty system of non-zero constant polynomials passing through three points. For MATH, we need to show that MATH and MATH are empty. The first reduces to the empty system of non-zero constant polynomials passing through a point. The second we may conclude is empty by applying a MATH degeneration and REF . Finally, we must show that MATH and MATH are empty. MATH reduces to the empty system of quadratics with a quadruple point. MATH reduces to the empty system of constant polynomials with a quadruple point. Now we consider the MATH case. Again, we assume that MATH and that MATH. We assume that MATH is not (-REF) special with MATH and prove that it is non special. For each MATH and MATH it is enough to prove the theorem for the largest MATH which makes the virtual dimension MATH. If MATH is non empty and non special, then MATH will be non-empty and non-special for MATH. This is because the conditions imposed on curves of degree MATH in MATH are a subset of the independent conditions for the system MATH. For MATH, MATH when MATH. For lower MATH, the largest value of MATH which makes MATH must be at least as big as in the MATH case. Therefore, we will assume that MATH. We use a MATH degeneration where MATH is chosen to satisfy the hypotheses of REF . MATH is not (-REF) special when MATH. MATH is not (-REF) special if MATH. MATH is not (-REF) special when MATH is odd. MATH is not (-REF) special if MATH is odd and MATH. Now MATH and MATH are non-special and MATH and MATH are non special by the inductive hypothesis. We need MATH to make MATH. To force MATH, the second identity in the proof of REF gives us that it is enough to have that MATH and MATH. The former is true by hypothesis and the later is true when MATH. All of this may be achieved (when MATH) by choosing MATH so that MATH is odd. As before, we can do this for MATH, REF, or whenever MATH. Recall that MATH, so that this choice makes MATH for all MATH. When MATH, MATH will be (-REF) special if MATH. For MATH the largest MATH making MATH is the integer MATH. We need MATH. Putting these last two inequalities together, we see that we should choose MATH. This guarantees that MATH is non-special even when MATH. If we select MATH such that MATH is odd, then we may appeal to REF to conclude that MATH is non-special. This may be done for MATH. For MATH, the hypotheses of REF are satisfied (for MATH and the largest MATH which makes MATH) if we choose MATH, for all but MATH and MATH. For the system MATH, we perform a MATH degeneration. MATH is empty, MATH is non-special of dimension REF, MATH is non-special (by induction) of dimension REF, and MATH is expected to be empty, but is (-REF) special. MATH . NAME reduces to the zero dimensional space of constants. We compute MATH . Therefore, MATH by REF , and this is also the virtual dimension. The MATH case follows in the same fashion. Use a MATH degeneration. MATH is special, but NAME reduces to the zero dimensional space of constants. The hypothesis of REF are satisfied and MATH since MATH and MATH. The theorem is proved for MATH, provided it is true for smaller values of MATH. We now prove the theorem for MATH case by case. In each case we choose MATH so that MATH and MATH are at least -REF, then check that all the systems are not (-REF) special. If they are all not (-REF) special we may apply REF . When MATH we use a MATH degeneration. We analyzed this degeneration already in the MATH case. If the four main systems are all non-special and MATH and MATH have dimension at least -REF we apply REF to conclude that MATH has the expected dimension. Now, MATH was chosen so that MATH and MATH have virtual dimension at least -REF, so it is enough that the systems are all non-special. We saw earlier that this only fails (when MATH) if MATH or if MATH is of the form MATH where MATH. In the second case it is enough to prove the theorem for the largest MATH making MATH, so we may assume MATH. In both cases MATH is special, but we may use NAME transformations to find its dimension. Then directly apply REF . For MATH, we may choose MATH to be REF or REF to make MATH odd (for the largest MATH making MATH at least REF) and use REF . If MATH, this MATH also works unless MATH. In this case we appeal to REF (using NAME reduction to find the dimension of MATH). When MATH we use a MATH degeneration and REF unless MATH or is of the form MATH where MATH. In the first case we appeal to REF . In the second case, note that we only care about the largest MATH (the largest MATH). This is the system MATH, but MATH is non-empty and non special (by REF ) so MATH is as well. When MATH, we may choose MATH or REF so that MATH is odd for the largest MATH making MATH at least REF and use REF . For MATH a MATH degeneration allows us to apply REF . A MATH degeneration also allows us to apply REF in all but three cases. For these exceptions, we use NAME reduction to find the dimension of the special systems and use REF . If MATH, we let MATH be either REF or REF so that MATH is odd and use REF unless MATH. In this case we use a MATH degeneration and REF . For every MATH from REF to REF the process is the same. For each MATH and the corresponding largest MATH, there is a MATH (between MATH and MATH so that MATH and MATH are at least -REF) such that either a MATH degeneration satisfies the hypothesis of REF , or so that we may use NAME reduction to find the dimension of the special systems and apply REF . When MATH, we may use a MATH degeneration and REF in all but two cases. In these cases (MATH or MATH) a MATH degeneration satisfies the requirements of REF . For MATH or REF, a MATH degeneration works. If MATH there are eight cases to check. For MATH or REF the systems (with the largest MATH) NAME reduce to a known case and are non-special. Either a MATH or a MATH degeneration works for the rest of the cases except MATH. To prove the theorem for the system MATH a MATH degeneration satisfies the hypotheses of REF . When MATH there are six cases to check. The top two (MATH or REF) NAME reduce to a known case and are non-special. A MATH or a MATH degeneration works for the rest. For MATH, MATH, and the corresponding largest MATH making MATH at least REF and the system non-special, the linear systems all NAME reduce to known cases and are non-special. Therefore, all linear systems MATH which are not (-REF) special are non- special. In other words the only special quasi-homogeneous linear systems with MATH are the (-REF) special systems listed in REF . |
math/9905079 | Write each of the terms in REF as a multiple of MATH to get the equation MATH where MATH . It suffices to show that MATH. But this follows from the standard NAME identities MATH and MATH. |
math/9905079 | First we show that MATH is an integer. We use the well known fact that if MATH is even and MATH is odd, then MATH is even. If MATH is even, then obviously MATH is an integer, so assume that MATH is odd. Now if MATH is also odd, then MATH is even, so we may assume that MATH is even. Now one of MATH and MATH is even. REF below shows that that MATH is the inverse of the matrix MATH. |
math/9905079 | Let MATH so that MATH is the MATH-entry of MATH. Then MATH satisfies the recurrence MATH . The preceeding recurrence was found by NAME. The theorem will follow if we can establish the correct values of MATH, MATH, and MATH. CASE: Maple computes MATH, and it computes MATH . Now MATH, and with MATH as input, the function sumrecursion gives the recurrence MATH. Maple gives the initial value MATH, for MATH. CASE: Maple computes MATH . Similarly to the previous case, sumrecursion gives the recurrence MATH, and obviously MATH. CASE: We need to do something different in this case. First, we show that our conjectured inverse is symmetric. Let MATH so that MATH. Now zeil produces the recurrence MATH, where MATH . This implies that MATH. Now Maple tells us that MATH. This means that MATH. Maple tells us that MATH which implies MATH. Since MATH and MATH are symmetric, the MATH entry of MATH equals the MATH entry of MATH. The former is MATH, where MATH . The function zeil produces MATH which satisfies MATH . Thus we have MATH and Maple tells us that MATH. All that remains is to check the initial value MATH. Maple tells us that MATH which implies that MATH when MATH. |
math/9905079 | First we show that each summand of the sum which defines each entry is an integer. It is well known that if MATH, MATH, MATH, then MATH, MATH, and MATH are all divisible by REF. Using this fact, we find that one of the terms MATH, MATH, or MATH is divisible by REF unless MATH and MATH. But now MATH, MATH, and MATH. Thus REF divides one of the terms MATH or MATH. The proof that MATH is the inverse of MATH is similar to the proof of REF . Let MATH, so that MATH is the MATH entry of MATH. Then MATH satisfies the recurrence MATH . Now the proof proceeds similarly to the proof of REF , except that we don't have to do the difficult initial value MATH. |
math/9905080 | The proof consists to establish the differentiability of the following MATH-linear map MATH uniquely defined by: MATH . The map MATH is a formal sum of linear maps MATH with MATH being the identity map on MATH. We will show that the MATH's are differential operators. By REF for MATH and MATH. It is easy to see from REF that the MATH's satisfy the following recurrence relation for MATH: MATH (the MATH's are REF-cochains of the star-product). For MATH the sum on the right-hand side is omitted and MATH is the NAME differential. Before going further we need a lemma. Let MATH be a MATH-linear map such that MATH, for MATH, and let MATH be a bidifferential operator vanishing on constants. If MATH satisfies MATH then there exists a differential operator MATH on MATH such that MATH. For two functions MATH, let MATH be the expression of MATH in local coordinates, where MATH and MATH are multi-indices and MATH is a smooth function vanishing for MATH or MATH greater than some integer (for MATH, MATH denotes its length MATH). In REF , only first derivatives can be applied to the first argument of MATH and one can check the following series of equalities: MATH where MATH. Now REF can be written as MATH and by observing that MATH vanishes on MATH and MATH we get that MATH on MATH. This shows the lemma. The term of order MATH in REF yields MATH. We have that MATH for some differentiable MATH-cochain MATH, which can be chosen such that MATH for MATH by adding an appropriate NAME MATH-cocycle (that is, a vector field). Then as before MATH gives us MATH on MATH showing that MATH is a differential operator. With the help of REF , a simple recurrence on MATH in REF shows that for each MATH, MATH coincides with the restriction of a differential operator to MATH. Clearly the map MATH can be naturally extended to a MATH-linear map on MATH. We still denote this extension by MATH. The map MATH is invertible as MATH is the identity map and we can use it to define an equivalent star-product MATH to MATH by: MATH . Notice that MATH for MATH and MATH, therefore MATH is a NAME star-product. |
math/9905080 | Any star-product MATH on MATH is determined by the quantities MATH, MATH. Suppose that MATH is a MATH-covariant NAME star-product, then the star-exponential of MATH defined by: MATH coincides with the usual exponential MATH. The covariance property of MATH allows to use the NAME formula: MATH where MATH is the usual NAME series with respect to the bracket MATH. For MATH, MATH is an element of MATH. We still have MATH for MATH. It follows from REF that there is at most one MATH-covariant NAME star-product on MATH, that is, NAME star-product. The second statement of the lemma follows from REF and from the fact that the equivalence operator MATH preserves the covariance property, that is, MATH for MATH (compare REF ). |
math/9905080 | We just need to see what kind of graphs contribute to MATH, MATH. The graphs for MATH must be such that the vertices MATH and MATH receive only one edge, respectively. For MATH, we simply have the NAME bracket MATH. If MATH, we need to draw MATH edges in such a way that each vertex MATH, MATH receives at most one edge (since the NAME bracket MATH is linear in the coordinates) and this is possible only if MATH, that is, MATH. For MATH, the only graph contributing (up to symmetry factors) is the graph MATH in REF , whose associated bidifferential operator MATH is symmetric. Thus we have MATH. |
math/9905080 | As the vertex MATH can receive at most one edge, we distinguish two cases. CASE: The vertex MATH receives no edge. We will see that the vertex MATH must receive exactly MATH edges. If there are strictly more than MATH edges ending at vertex MATH, then there must be a vertex MATH, MATH, such that the edges MATH and MATH are ending at vertex MATH. This is excluded by definition of MATH. If there are strictly less than MATH edges ending at vertex MATH, then at least one of the vertices MATH must receive two or more edges and the bidifferential operator associated to such a graph is vanishing since the NAME bracket is a linear function of the coordinates. We are left with the case where exactly MATH edges are ending at vertex MATH. Then every vertex in MATH must receive exactly one edge and, up to an isomorphism, there is precisely one such a graph, that is, graph MATH in REF . CASE: The vertex MATH receives one edge. For this case, the vertex MATH receives MATH edges. By relabeling the vertices and the edges, we may suppose that the edge ending at vertex MATH is MATH. Then the second edge starting at the vertex MATH, that is, MATH, cannot end at vertex MATH because, by skew-symmetry of the NAME bracket, the associated bidifferential operator is vanishing on MATH. Hence we may suppose that the edge MATH is ending at vertex MATH. We still have MATH edges starting from the vertices MATH to draw. Let MATH be the number of edges ending at vertex MATH and let MATH be the total number of edges ending at the vertices MATH. We have MATH. Since each vertex in MATH can receive at most one edge, we have that MATH and it follows that MATH. If MATH, it means that there are parallel multiple edges between at least one of the vertices MATH and the vertex MATH. Hence the vertex MATH must receive MATH edges. Clearly every such edge must start at one the vertices MATH. The other MATH edges must end at the vertices MATH. Thus, up to an isomorphism, we find that there is only the graph MATH in REF for this case. |
math/9905080 | The form MATH, where MATH, is MATH. This easily follows from a simple recurrence using explicit expressions for the forms MATH. |
math/9905080 | The first statement follows directly from REF , and REF. The bidifferential operator for the graph MATH is MATH clearly it has constant coefficients and using REF for MATH, we see that the previous equation can be written as a trace of adjoint maps. |
math/9905080 | Recall that MATH is defined by MATH, MATH, MATH. It was shown that the MATH's in MATH are differential operators. Here we have only to solve the recurrence relation for MATH appearing in the proof of REF : MATH where MATH. According to REF , there exist differential operators MATH such that MATH. From REF it follows that MATH . For each MATH, MATH is a differential operator with constant coefficients and is homogeneous of degree MATH in the derivatives. To a differential operator MATH on MATH with constant coefficients we can associate a polynomial MATH on MATH. Here we have MATH and one can check that MATH . The preceding implies that the MATH's, MATH, have constant coefficients and are homogeneous of degree MATH. We have MATH and a recurrence on MATH in REF shows the property for all of the MATH's. Using REF we can express REF in terms of the polynomials MATH and MATH and find that: MATH . By defining MATH to be identically equal to zero and MATH to be MATH, we can rewrite the previous equation as MATH then by considering the formal series MATH and MATH (recall that MATH), we see that REF simply states that MATH where the prime denotes formal derivative with respect to MATH. Thus MATH and REF follows. |
math/9905081 | Write MATH. Then MATH, so we have surjective maps MATH . The first map is MATH and the composition is MATH, proving the first inclusion. A similar argument works for MATH and MATH. |
math/9905081 | If MATH is a connected subgroup of the upper triangular matrices then MATH with MATH a torus and MATH unipotent CITE. An argument similar to that used in CITE implies that MATH so we assume MATH is a torus. Let MATH and MATH. We will show that the filtrations of MATH given by MATH and MATH give the same topology. CASE: For any MATH we must show there exists MATH such that MATH. Since multiplication by MATH maps MATH to MATH, if MATH the desired inclusion holds. CASE: For any MATH we must show there exists MATH such that MATH. CITE has shown that MATH is generated as a MATH-module by fundamental classes of MATH-invariant subvarieties. Any such fundamental class lies in MATH for some MATH, where MATH. Hence if MATH, then MATH. Since MATH is a torus MATH is generated as in algebra in degree REF. Thus, MATH and we can conclude that if MATH, we have MATH. |
math/9905081 | The NAME group of MATH is MATH over any field, so by CITE MATH. In this case it is well known that MATH and MATH where MATH is the MATH-th elementary symmetric polynomial in the variables MATH. This fact obviously implies REF . REF follows from CITE applied to the maximal compact subgroups MATH and MATH. |
math/9905081 | By induction it suffices to prove the lemma when MATH. Let MATH be the disjoint union of MATH and MATH. Then MATH. The finite surjective map MATH gives a map of localization exact sequences MATH A diagram-chase shows that the map MATH is surjective. |
math/9905081 | Let MATH and MATH be the projections. By the homotopy property of equivariant MATH-theory the smooth pullbacks MATH and MATH are isomorphisms of MATH-modules. Since MATH is a regular embedding and MATH we have, by the compatibility of flat and l.c.i. pullbacks CITE, that MATH. Thus, the pullback MATH is an isomorphism of MATH-modules. Hence it suffices to show that MATH. By the projection formula MATH acts by multiplication by MATH. Since MATH is an invariant linear subspace, MATH which is in MATH. |
math/9905081 | Let MATH be a separated algebraic space. Following CITE we define a NAME envelope MATH to be a proper morphism from a quasi-projective scheme MATH, such that for every integral subspace MATH, there is a subvariety MATH of MATH such that MATH maps MATH birationally to MATH. Using NAME 's lemma for algebraic spaces CITE, the argument of CITE shows that every algebraic space has a NAME envelope (of the same dimension as MATH), and that the proper push-forward MATH is surjective. Since MATH is quasi-projective, MATH CITE where MATH is the MATH-th level of the MATH-filtration. Since MATH preserves the rank of a locally free sheaf, MATH. By definition, MATH, so MATH. Hence, by the projection formula, MATH. Since MATH is surjective, the lemma follows. NAME if MATH is not separated it has an open set MATH which is a separated scheme. Then MATH for MATH. Using the NAME induction and the localization sequence we see that MATH for MATH. |
math/9905081 | Let MATH be the quotient algebraic space. The proof of CITE extends to algebraic spaces and shows that MATH is a MATH-principal bundle. Thus, there is an equivalence between the categories of coherent sheaves on MATH and MATH-equivariant sheaves on MATH; that is, MATH. Under this isomorphism, MATH. The lemma now follows from REF . |
math/9905081 | We prove only REF . The corresponding result about NAME groups has essentially the same proof. To show that the filtrations of MATH by the submodules MATH and by powers of the ideal MATH generate the same topology, there are two steps. CASE: We must show that given any pair MATH, there is an integer MATH such that MATH, or in other words, that MATH. Since MATH acts freely on MATH this follows from REF . We must show that given a positive integer MATH, there is a pair MATH such that MATH. Let MATH be any good pair and set MATH. Then MATH. By definition, MATH where MATH is contained in an invariant linear subspace MATH of MATH, and by REF , MATH. By REF , MATH is generated by the images of MATH, so MATH. The group MATH acts freely on the open set MATH, and by induction, MATH . Thus, MATH is the desired good pair. |
math/9905081 | It suffices to show that given MATH and MATH with MATH and MATH greater than MATH the construction using MATH and MATH agrees with that using MATH and an open subset MATH of MATH. We can choose the open subset of MATH arbitrarily, provided the codimension is sufficiently large, so we take the open subset MATH. The constructions agree because the following diagram commutes. MATH . Here the vertical arrows are flat pullback, and commutativity follows from CITE, using the fact that the relative tangent bundle of the morphism MATH is the bundle MATH . |
math/9905081 | REF follow from the definition of MATH and the non-equivariant NAME theorem of CITE. REF follows because, as in REF , the diagram MATH commutes. If MATH is the augmentation ideal of MATH, then MATH. Thus, by REF , MATH . Thus, MATH restricts to a map MATH . Taking the limit as MATH gives the desired factorization MATH proving REF . The proof of REF uses CITE but requires some care, particularly in the category of algebraic spaces. Suppose that MATH is quasi-projective. Let MATH be an open set in a representation on which MATH acts freely. Denote the mixed spaces MATH and MATH by MATH and MATH. Using descent CITE as in CITE, we see that the induced map MATH is smooth and quasi-projective. Let MATH be a NAME envelope for MATH. Then we have a Cartesian diagram MATH where, by base change, MATH is smooth and quasi-projective. Since MATH is a quasi-projective scheme and MATH is a quasi-projective morphism, it follows from CITE that MATH embeds in projective space as well. Hence CITE applies to the morphism MATH. If MATH then MATH for some MATH. Since flat pullback commutes with proper pushforward, MATH. Thus, Taking the limit over all pairs MATH in the construction of MATH gives REF . To prove REF argue as follows: Suppose that MATH and MATH where MATH and MATH are smooth MATH-schemes. The requirement that MATH is special or connected ensures that we can choose open sets MATH so that MATH and MATH are smooth schemes CITE. Thus the mixed space MATH and MATH are embeddable in smooth schemes, and again by descent CITE, the induced map MATH is l.c.i., so REF follows again from CITE. Finally, suppose that MATH is another map with REF . Suppose MATH is given and denote by MATH (respectively, MATH) the term of MATH (respectively, MATH) in MATH. Let MATH be a representation and open set on which MATH acts freely such that MATH. Let MATH be the composition of projection with open inclusion. Since MATH we may identify MATH. The morphism MATH is smooth and quasi-projective so by REF MATH . Since MATH acts freely, MATH and MATH coincide on on MATH. Thus MATH . |
math/9905081 | If MATH is a model for calculating MATH, then the model MATH for calculating MATH is a MATH bundle over MATH. Since MATH, the projection formula implies that MATH. Now if MATH is in MATH, then we can write (using the relation MATH) MATH, so MATH which is the desired formula. |
math/9905081 | Let MATH be a representation of MATH and let MATH be an open set on which MATH acts freely. Then MATH is a MATH-equivariant vector bundle on MATH. The group MATH acts freely on the open set MATH which surjects onto MATH. Identifying MATH with MATH we see that the maps MATH and MATH are the same. Using pairs of the form MATH we define MATH as MATH, as before. On the other hand, by REF we can use vector bundles and open sets of the form MATH to define MATH as MATH. Hence the maps coincide. |
math/9905081 | We will only prove REF for NAME groups; the arguments for NAME groups are similar. Let MATH be a MATH-space. Define a MATH-action on MATH by: MATH . We also define a MATH-action on MATH by: MATH . The projection MATH is MATH-equivariant, and moreover a MATH-principal bundle (MATH-torsor). The projection MATH is MATH-equivariant and a MATH-torsor. Hence we have equivalences of categories between: MATH-equivariant coherent sheaves on MATH, MATH-equivariant coherent sheaves on MATH, and MATH-equivariant coherent sheaves on MATH. (This uses the general fact CITE that if MATH is a principal bundle, then there is an equivalence of categories between coherent sheaves on MATH and equivariant coherent sheaves on MATH.) Hence MATH. Under these equivalences of categories, the MATH-equivariant vector bundle MATH corresponds to the MATH-equivariant vector bundle MATH. This translates into the fact that the isomorphism is an isomorphism of MATH-modules. This proves REF in MATH-theory. REF follows from REF . We prove REF . We now assume that the MATH-action on MATH is the restriction of a MATH-action. We can then define another MATH-action on MATH, this time by: MATH . Let MATH be a MATH-equivariant sheaf on MATH. Then MATH is naturally a MATH-equivariant sheaf, with respect to either action of MATH. As noted above, the projection MATH is MATH-equivariant with respect to action REF on MATH, so MATH is a MATH-equivariant sheaf on MATH. Next, the action map MATH is MATH-equivariant with respect to action REF on MATH. Therefore, MATH is also MATH-equivariant. In particular, REF give two MATH-actions on MATH. Since MATH is a MATH-equivariant sheaf on MATH, by definition there is an isomorphism of coherent sheaves MATH satisfying the cocycle condition (see CITE). The cocycle condition implies: If MATH is a MATH-equivariant sheaf on MATH, then the map MATH is an isomorphism of MATH equivariant sheaves on MATH. Consider the diagram MATH . The left two vertical maps are the forgetful maps, and the left square commutes. The map MATH comes from taking the quotient by MATH, and using the identification MATH with MATH taking MATH to MATH. Likewise, we identify MATH with MATH. Tracing through the definitions, the latter identification is given by MATH. By REF the composition along the top row is the identity. The right vertical arrow is flat pullback, and the reader can verify that the right square also commutes. This proves REF . |
math/9905081 | This follows from the argument of CITE, using the fact that MATH is a tower of projective bundles and the projective bundle theorem of CITE. |
math/9905081 | In general, if MATH is a smooth proper morphism with MATH, then the NAME theorem implies that MATH for all MATH. (Proof: By the projection formula we may assume MATH. Then by NAME MATH.) Applying this to the case where MATH and MATH are the mixed spaces MATH and MATH yields the result. |
math/9905081 | In view of the previous REF follows from REF from REF |
math/9905081 | The rings MATH are local, so MATH. Thus, MATH . Hence MATH. Since MATH, MATH is MATH-adically complete, proving the result. |
math/9905081 | The proof proceeds as in previous proofs, by building up from a torus to MATH to a general group. CASE: MATH is a torus. By NAME 's generic slice theorem CITE, there is a MATH-equivariant open subset MATH such that MATH is equivariantly isomorphic to MATH where MATH for a diagonalizable subgroup MATH, and MATH acts trivially on MATH. If MATH is any representation of MATH which is trivial on MATH, then the vector bundle MATH is equivariantly trivial, so MATH. Hence, under the map MATH, the ideal MATH annihilates MATH. Note that MATH. Since the stabilizers are finite MATH is finite. Let MATH be the character group of MATH, and let MATH be the character group of MATH. Then we can choose a basis MATH of MATH such that MATH is a basis for MATH, where the MATH are positive integers. Using this basis, we have MATH and MATH . Hence MATH, so MATH. By NAME induction we may assume that MATH for some sufficiently large MATH. Then by the localization exact sequence, MATH. Note that the above discussion and NAME induction also show that there is a non-zero ideal MATH such MATH. CASE: MATH. Let MATH be the group of upper triangular matrices in MATH and let MATH be the group of diagonal matrices. Restriction from MATH to MATH induces isomorphisms MATH and MATH (see CITE). Claim: There is an ideal MATH with the support of MATH finite, such that MATH. For, MATH is finite, so if MATH is the inverse image of MATH under this map, then the support of MATH is finite. Since MATH is a MATH-submodule of MATH (see the proof of REF ), we have MATH, proving the claim. Because MATH has finite support and MATH is contained in the maximal ideal MATH, it follows that MATH is MATH-primary, hence contains MATH for some MATH (note that MATH is NAME). Then MATH as desired. CASE: The general case. Embed MATH. In this case, MATH. Thus, by REF , MATH for some positive integer MATH. Equivalently, given MATH and MATH, there exists MATH such that MATH. Now, the action of MATH on MATH factors through MATH. Also, the augmentation map MATH factors through MATH, so if MATH then MATH. REF implies that for some MATH, MATH. This implies that MATH. Indeed, if MATH and MATH, write MATH for MATH, choose MATH as in the preceding paragraph; then MATH and MATH. This proves the result. |
math/9905081 | By the generalization of CITE to algebraic spaces there is a finite cover MATH on which MATH acts freely. Since MATH acts freely, MATH is generated by invariant cycles CITE. On the other hand, the proper pushforward MATH is surjective because MATH is finite and surjective. Therefore, MATH is generated by invariant cycles. |
math/9905081 | The kernel is just the kernel of the localization map MATH. |
math/9905081 | By the localization theorem for diagonalizable group schemes MATH CITE. Since MATH acts trivially on MATH we have CITE MATH . Let MATH be the augmentation ideal of MATH. Since MATH, MATH . Thus, by the NAME isomorphism of REF we have MATH . (Here the NAME character MATH makes MATH into a MATH-module.) The first statement follows. As noted in REF there is an ideal MATH such that MATH is supported at a finite number of points and MATH. Hence, MATH. Then MATH with MATH a MATH-primary ideal. By the NAME remainder theorem MATH so MATH and the second statement follows. |
math/9905081 | CASE: To show that the filtrations induced by powers of the ideals MATH and MATH induce the same topology, we must check two things. First, we must show that for any MATH, there exists a MATH such that MATH. This is clear because under the map MATH, the image of MATH is contained in MATH, so we can take MATH. Second, we must show that for any MATH, there exists a MATH such that MATH. As above we will do this in steps: first for MATH the group of upper triangular matrices, then for MATH, and finally for arbitrary MATH. Suppose then that MATH is the group of upper triangular matrices. We use the notation of the proof of REF . We know there exists MATH such that MATH. Since MATH and MATH generate the same topology on MATH, we can assume MATH. So we must show that there exists MATH such that MATH, that is, such that MATH. Since MATH acts freely on MATH, we have MATH. Under this isomorphism, MATH where MATH denotes is the augmentation ideal of MATH. By REF , MATH for MATH. The analogous statement for NAME groups, that MATH for MATH, also holds. Assume now that MATH. Then we have MATH where the two maps are MATH and MATH. We have proved that there exists MATH such that MATH . Hence MATH where the last equality follows by the projection formula. Now, we also have MATH. Combining these facts, we see that MATH . Since MATH is the identity MATH. Finally consider the case where MATH is any subgroup. By REF , there is an isomorphism or MATH-modules MATH . Under this isomorphism MATH corresponds to MATH. Moreover, MATH corresponds to MATH since the equivalence of categories, obtained from descent of MATH-equivariant sheaves on MATH, takes locally free sheaves to locally free sheaves of the same rank CITE.] Because of these correspondences, the theorem follows from the case where MATH. |
math/9905081 | By embedding MATH it suffices to prove the result for MATH. The corollary now follows by applying the theorem when MATH. |
math/9905081 | The action of MATH on MATH factors through the map MATH. Since MATH, REF implies that the ideals MATH and the MATH generate the same topology. Thus, MATH proving REF . By the projection formula applied to the commutative triangle MATH. Hence, MATH is continuous with respect to the MATH-adic topology, proving REF . |
math/9905083 | The cases MATH, MATH, and MATH are given in CITE; more precisely, MATH is given there as REF , while MATH and MATH are given in REF . (Note that if MATH, then MATH is a fixed-point-free involution with decreasing subsequences corresponding to the increasing subsequence of MATH.) We also give new, elementary, proofs below (REF , and REF). It remains to consider MATH and MATH. Via the NAME correspondence (see REF, for an excellent introduction), we can associate a pair MATH of NAME tableaux of the same shape to MATH, satisfying the relations MATH, MATH, where MATH is the duality operation (``evacuation") of NAME (ibid.). But there is a bijective correspondence between self-dual tableaux and domino tableaux (see, for example, CITE), and further from domino tableaux to pairs of ordinary tableaux with disjoint content, and with shape determined only by the shape of the domino tableau CITE. Thus we have associated four NAME tableaux MATH, where MATH and MATH have the same shape, MATH, MATH have the same shape, MATH and MATH have disjoint content, and MATH and MATH have disjoint content. This corresponds to a choice of MATH, independent choices of two subsets of size MATH of MATH, and independent choices of MATH and MATH of length MATH and MATH. Furthermore, the longest increasing subsequence of MATH has length at most MATH precisely when the longest increasing subsequences of MATH and MATH are of length at most MATH. Putting this together, we find MATH . On the other hand, the integral formula simplifies as follows: MATH . Similarly, an element MATH corresponds to a pair of NAME tableaux of the same shape with disjoint content, and thus MATH . On the other hand, MATH . |
math/9905083 | For MATH, we have MATH . But MATH for MATH odd, so this is MATH as required. The calculations for MATH and MATH are analogous. |
math/9905083 | Using the NAME integration formula for the unitary group, we have MATH . The result follows from the standard theory of NAME determinants, or by the classic formula for the integral of a product of two (generalized) NAME determinants (see, for instance, CITE): MATH for any measure MATH on any set MATH. |
math/9905083 | As observed in REF, integrals over the orthogonal and symplectic groups can be expressed as NAME determinants; thus, for instance, MATH for any polynomials MATH with MATH. In particular, this must be true when MATH (NAME polynomials). In that case, noting that MATH, the MATH coefficient of the determinant is MATH . The constant of proportionality can be determined by comparing the two sides when MATH, and thus MATH. For the other cases, we take MATH to be MATH as appropriate. |
math/9905083 | Use MATH which follows from REF below. |
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