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math/9907073 | The space MATH is a partial MATH-monoid by MATH. |
math/9907073 | NAME has shown in CITE that if MATH is the fan associated to the variety MATH CITE then the union of some components of MATH is homeomorphic to MATH. The corollary follows from the theorem and from REF . |
math/9907073 | We need to exhibit a natural transformation MATH such that MATH is the identity and the diagram MATH commutes. The morphism MATH is obtained by grafting of trees. |
math/9907073 | The proof is similar to that of REF. In this case we use for each MATH a MATH-equivariant retraction MATH such that MATH preserves MATH, where MATH is the blowdown. |
math/9907073 | NAME has shown in CITE that MATH is homotopy equivalent to MATH. |
math/9907073 | The space of open configurations MATH retracts onto MATH, considered as space of configurations of a single labelled point in MATH. The retraction is achieved CITE by pushing radially the particles away onto the boundary. But the inclusion MATH is a weak equivalence by REF. |
math/9907073 | It is sufficient to carry out the discussion in the non-symmetric case: in fact MATH, where MATH compactifies the space of strictly ordered maps from MATH to MATH. Let MATH be the closure of the subspace of maps MATH such that MATH, MATH. Its elements are described by appropriate trees as in REF. For MATH, we have homeomorphisms MATH, where MATH is the space of configurations in MATH macroscopically concentrated at the point MATH. If MATH, MATH, then MATH. If MATH, MATH, then MATH. We have seen in REF that MATH is the associahedron. Under the identification MATH the NAME space MATH is defined to be the quotient of MATH, seen as space of forests labelled by MATH, under the following steps: CASE: We replace a tree on MATH twigs by a point having as label the action of the tree on its twigs via MATH. CASE: We can cut twigs labelled by MATH. CASE: We identify any two labelled forests coinciding outside REF. But this quotient is exactly MATH. In a similar way one shows that MATH is homeomorphic to MATH. In this case in REF we identify forests coinciding outside REF. |
math/9907073 | We follow the proof of REF. The space MATH has a filtration by MATH. There is a homeomorphism MATH such that MATH is the projection onto the factor MATH. Choose a collared neighbourhood MATH of MATH in MATH and a smooth isotopy retraction MATH such that MATH. For each MATH there is an open neighbourhood MATH of MATH in MATH such that MATH induces a smooth isotopy retraction MATH, and a smooth isotopy retraction MATH covering MATH. For any point MATH we need to show that the restriction MATH of MATH is a weak homotopy equivalence. If we identify domain and range of MATH to MATH by MATH, then MATH pushes the labelled particles away from MATH, and adds a finite set of trees MATH in proximity to MATH. But if the pair MATH is connected, then the trees in MATH can be moved continuously to MATH, where they vanish, and MATH is homotopic to a homeomorphism. On the other hand, if MATH has a homotopy inverse, then MATH has a homotopy inverse that pushes the particles away from MATH and adds some homotopy inverses of the trees in MATH in proximity to MATH. |
math/9907073 | Note that MATH is homotopic to the framed MATH-monoid completion of MATH. For each MATH the base point of MATH is the empty configuration. The translation MATH of the first coordinate by MATH induces a map MATH, composite of the induced map MATH and the projection MATH. Then the `scanning' map MATH is defined for MATH by MATH. For MATH the translation of the MATH-th coordinate by MATH induces similarly a map MATH, and MATH is given by MATH. We define MATH, MATH, and we identify MATH to MATH via MATH. We consider for MATH a commutative diagram MATH . The top row is a quasifibration and the bottom row a fibration. The scanning map MATH is defined on the total space MATH by MATH and is consistent with MATH. Now the space MATH is contractible. In fact by excision MATH, with MATH and MATH. Moreover there is a smooth isotopy MATH, such that MATH is the inclusion and MATH. For example define MATH as the dilatation by MATH centered in MATH, with MATH at the MATH-st position. We conclude by comparing the long exact sequences in homotopy and by induction on MATH. |
math/9907073 | Consider MATH as a partial MATH-monoid as in REF . Now MATH by the same argument of REF . Moreover MATH by REF, MATH by REF and MATH by REF. Now we can apply the theorem. |
math/9907073 | Apply REF , and restrict to the relevant components. |
math/9907073 | We follow the proof of REF. There is a finite handle decomposition of MATH with no handles of index MATH. If MATH is obtained from MATH by attaching a handle MATH of index MATH , then we apply REF and we obtain a quasifibration MATH. On the other hand we have a fibration MATH. But MATH, and MATH. We compare the two sequences by the scanning maps and we conclude by REF and induction on the number of handles. In the case of MATH we have even a handle of index MATH and we apply the second part of REF . |
math/9907073 | We consider MATH as partial framed MATH-monoid as in REF . By REF MATH. We apply the second part of the theorem, and note that MATH because MATH is parallelizable. |
math/9907074 | Choose a complete intersection MATH. Then we can continue as in the proof of REF . |
math/9907074 | REF is REF is REF. Note that the latter can fail if MATH is a finite field. |
math/9907074 | We will show the existence of the elements MATH successively. Suppose we have found minimal generators MATH of MATH where MATH such that MATH is unmixed for all MATH. Choose elements MATH (of degree MATH in the graded case) such that MATH. Put MATH and MATH. Since MATH is unmixed for all MATH, MATH is not just a subsystem of parameters for MATH but even a MATH-filter regular sequence. Hence REF yields for all integers MATH . Since MATH is a parameter ideal for all MATH with MATH we conclude by local duality that we have for all MATH where MATH: MATH . Here the latter estimate is due to REF . Now we consider the set MATH . We have MATH and, for all MATH, MATH because MATH. Thus (using arguments as in REF ) there is an element MATH (homogeneous of degree MATH in the graded case) which is a parameter for MATH and all MATH with MATH. Since MATH is unmixed the module MATH has finite length. Thus the exact sequence MATH implies MATH . Multiplication by MATH provides the exact sequence MATH . Using the isomorphisms above we see that it induces exact sequences MATH . Let MATH. Since MATH is unmixed we have MATH and MATH where for the latter estimate equality holds if and only if MATH. But then MATH is a parameter for MATH due to our choice of MATH. Thus in either case we have MATH. Therefore, using the exact sequence above we obtain MATH . Applying the isomorphisms MATH we get that MATH . According to REF we conclude that MATH is an unmixed module. Thus we may put MATH and we are done. |
math/9907074 | Let MATH be integers and let MATH for MATH. We define for homogeneous elements MATH where MATH: MATH . It is clear that this induces a well-defined MATH-homomorphism MATH of degree MATH. Hence it provides a well-defined MATH-linear map MATH which preserves degrees. Straightforward computations show that MATH is even a natural MATH-homomorphism. |
math/9907074 | First, we assume that the modules MATH are free. Thus, since the map MATH is natural we can even assume that MATH with MATH for MATH. Then the claim is easy to check. Second, we consider the general case. Let MATH be free graded presentations. Put MATH for MATH. We we have exact sequences MATH and MATH . Since the functor MATH is exact they induce the following commutative diagram with exact rows MATH where MATH are the corresponding natural homomorphisms. These maps are even isomorphisms by the first part of the proof. Thus MATH is an isomorphism as well. |
math/9907074 | According to results of CITE we know that MATH is a direct sum of injective hulls MATH where MATH and MATH. Since the tensor product and local cohomology commute with direct sums we may assume that MATH. If MATH for MATH then we have MATH. This implies MATH for all MATH. Otherwise we can find homogeneous elements MATH such that MATH has degree MATH, that is, MATH. Due to our choice of the elements MATH the multiplication gives isomorphisms MATH. Therefore the multiplication MATH where MATH is an isomorphism as well. Thus we obtain for all MATH isomorphisms MATH. Since MATH it follows that MATH for all MATH. |
math/9907074 | First we prove REF . Let MATH, MATH, be injective resolutions of MATH. Put MATH. For the (co)homology modules of these complexes we have by REF MATH . According to REF we have natural homomorphisms of complexes of MATH-modules MATH . Since MATH is an acyclic right complex with MATH there are homomorphisms MATH (compare , for example, REF ). Now composition of the maps described above gives the desired homomorphisms. Second we prove REF . Let MATH be an integer such that the ideals MATH and MATH can be generated by less than MATH elements. Then we have for all positive integers MATH canonical epimorphisms MATH . They induce the homomorphisms MATH . Taking the direct limit over all MATH we obtain MATH . Now we use REF . It provides natural homomorphisms MATH . Taking again the direct limit and using the isomorphism above we get natural homomorphisms MATH . According to REF holds, if MATH is an isomorphism for all modules MATH and MATH and MATH is an isomorphism for all injective modules MATH and MATH. The second statement is true by REF . Thus it remains to show that MATH is an isomorphism. Due to REF the maps MATH are isomorphisms. Thus taking the direct limit shows that MATH is an isomorphism. |
math/9907074 | Write MATH where MATH is a polynomial ring over MATH and MATH is a homogeneous ideal. Let MATH denote the minimal free resolution of MATH as MATH-module. Then MATH is a minimal free resolution of MATH as MATH-module. In conjunction with the NAME formula REF follows. Moreover, we see that the NAME type of MATH is the product of the NAME types of MATH and MATH. Hence REF follows from the fact that MATH is NAME if and only if its NAME type is one. |
math/9907074 | CASE: Since MATH is NAME by the previous lemma we can apply REF . Now suppose that MATH is not unmixed. Then there is a MATH such that MATH. Let MATH. Then we have MATH and thus MATH . Hence REF implies MATH . Since MATH, REF shows that MATH is not unmixed. Conversely, if MATH and MATH are unmixed then MATH is unmixed too using similar arguments as above. REF follows from the cohomological characterization of depth and the NAME formula. CASE: According to REF it suffices to show that REF implies REF . Suppose that MATH is not NAME. Then there is a MATH such that MATH. Let MATH. Since MATH is not finitely generated, the module MATH is not a finitely generated MATH-module and, in particular, is not of finite length. Hence MATH cannot be equidimensional and locally NAME which completes the proof. |
math/9907074 | CASE: The assumption implies MATH . Hence MATH proves the claim. CASE: Since MATH and MATH meet properly we have MATH . If MATH then one of the factors of MATH has finite length because of the assumptions on MATH and MATH. It follows that MATH . This shows that MATH and MATH intersect very properly. |
math/9907074 | The exact sequence MATH provides for the NAME polynomials MATH . Since MATH if and only if MATH for all MATH where MATH the claim follows by comparing the coefficients of the polynomials above. |
math/9907074 | Let MATH. Then we have by REF the isomorphisms MATH . Since MATH has MATH minimal generators and MATH and MATH intersect very properly we see that MATH is a parameter ideal for all MATH and for MATH. Therefore REF shows that MATH is an unmixed module proving the first claim. In order to show the claim on the degrees we show firstly MATH . Let MATH and MATH denote the dimension of MATH and MATH, respectively. For all integers MATH we have the following relation of NAME functions: MATH . Let MATH be an integer such that the NAME functions of MATH in degree MATH are given by their respective NAME polynomials. Then we obtain for MATH: MATH . It follows that MATH as claimed. Above we have already applied REF . Now we use its full strength. It provides a minimal basis MATH of MATH such that MATH is an unmixed module for all MATH. This implies MATH according to our assumptions and the previous lemma. |
math/9907074 | In the special case where MATH is a ring our claim follows from REF . The proof in the general case is similar to those of NAME and NAME. Since it is short we give it for the reader's convenience. By the definition of a slight modification we have an exact sequence MATH which implies MATH for all MATH. In particular, it follows MATH. Let MATH denote the degree of MATH. Then the exact sequence MATH provides the exact cohomology sequence MATH . According to REF the module MATH is finitely generated. Since MATH, NAME 's lemma yields MATH. It follows that MATH which implies our claims. |
math/9907074 | Since MATH and MATH intersect very properly we can apply REF as in the proof of REF . Thus there is a minimal basis MATH of the diagonal ideal MATH consisting of linear forms such that MATH is an unmixed module for all MATH. Moreover we have MATH . Now suppose that REF is fulfilled. Then MATH is a module of dimension one and degree MATH. Since the module MATH has the same degree we conclude by REF that the irrelevant maximal ideal is not an associated prime ideal of MATH, that is, MATH is NAME. It follows that MATH is NAME. Next, assume that REF is satisfied. Then we can apply REF successively. It follows that MATH is NAME. Due to REF we know that MATH is an unmixed module, thus MATH. Therefore we have seen that any of REF implies that MATH is a NAME module. Thus REF proves our assertion. |
math/9907074 | With regard to the previous theorem it suffices to note that the NAME type of MATH is the product of the types of MATH and MATH. |
math/9907074 | REF is a special case of REF because a maximal MATH-module is torsion-free if and only if it is unmixed. In order to prove REF we recall that a maximal MATH-module MATH is reflexive if and only if MATH for all MATH (compare REF ). Thus, the NAME formulas provide that the join MATH is a reflexive MATH-module. Since MATH and MATH intersect very properly the diagonal MATH is a parameter ideal for MATH and all MATH where MATH. Thus, arguing as in the proof of REF we obtain MATH . Hence MATH is a reflexive MATH-module. |
math/9907074 | Multiplication by MATH provides the exact cohomology sequence MATH . Since we have MATH by assumption on the depth MATH the multiplication by MATH is surjective on MATH. But MATH is a finitely generated MATH-module because MATH is reflexive. Hence NAME 's lemma implies MATH. Furthermore, the reflexivity of MATH provides MATH. Thus, considering the exact sequence above we conclude that MATH . It follows that MATH and we are done. |
math/9907074 | CASE: The assumption and REF provide that the tensor product of MATH and MATH is a free MATH-module. Therefore, due to REF , MATH and MATH are NAME, thus free. CASE: Put again MATH. Our assumption implies MATH. Since MATH we conclude by REF that MATH and in particular MATH. Let us denote the MATH-dual MATH of a graded MATH-module MATH by MATH. The NAME formulas and local duality provide MATH . Since each direct summand must be trivial we conclude that MATH for all MATH, that is, MATH is a maximal NAME module and thus free. |
math/9907075 | REF is obvious. For REF , since MATH is dense in MATH, we see that MATH is dense in MATH. But MATH is finite dimensional and therefore closed, hence MATH is finite dimensional and the result is proven. |
math/9907075 | Since MATH is a nonzero divisor in MATH, it is invertible in MATH and it follows that MATH. Also MATH is a nonzero divisor in MATH, and we deduce from this that MATH is dense in MATH. |
math/9907075 | For each MATH, let MATH, a right ideal of MATH. We first show that MATH for some MATH. Since MATH is a NAME algebra, there is a unique projection MATH such that MATH. Let MATH denote the trace map, as described for example in CITE. If we write an element MATH of MATH in the form MATH where MATH, then MATH. Also if MATH is a nonzero projection in MATH, then MATH. Note that the sum MATH is direct, so by using CITE, we see that MATH for any finite subset MATH of MATH. It follows that the number of MATH for which MATH is countable, and we deduce that there exists MATH (in fact uncountably many such MATH) such that MATH. For this MATH, we have MATH and MATH is a nonzero divisor in MATH. Since every element of MATH can be written in the form MATH and also MATH with MATH, we conclude that MATH is a nonzero divisor in MATH. But every element of MATH is either a zero divisor or invertible, and the result follows. |
math/9907075 | Suppose MATH. Then we need to show that MATH and MATH have finite rank. There are nonzero divisors MATH such that MATH. Then MATH and MATH, hence MATH and we see that MATH has finite rank. Using REF , we deduce that MATH has finite rank. Similarly MATH has finite rank. If MATH, then in a similar fashion we can show that MATH and MATH have finite rank. This establishes the result. |
math/9907075 | Write MATH and MATH, where MATH and MATH are nonzero divisors. Then MATH, MATH, MATH, MATH have finite rank and MATH, consequently MATH has finite rank. Similarly MATH has finite rank. Using REF , we deduce that MATH. |
math/9907075 | By REF , there exist MATH such that MATH, MATH are invertible in MATH, so using REF , we may assume that MATH are invertible in MATH. This means when we write MATH with MATH, not only MATH but also MATH are nonzero divisors in MATH. Write MATH where MATH are nonzero divisors in MATH, and then write MATH where MATH are nonzero divisors in MATH. Then MATH and MATH. Thus we may write MATH, MATH where MATH are nonzero divisors in MATH, and similarly we may write MATH, MATH where MATH are nonzero divisors in MATH. Then MATH. Since MATH, we have MATH have finite rank and hence MATH has finite rank. Similarly MATH has finite rank and an application of REF completes the proof. |
math/9907075 | Write MATH where MATH, all nonzero divisors because MATH is invertible in MATH. Then MATH. Since MATH, we know that MATH have finite rank. Therefore MATH have finite rank. The result now follows from REF . |
math/9907075 | Write MATH, where MATH and MATH are nonzero divisors. Then MATH. Since MATH, we know that MATH and MATH have finite rank. If MATH is a bounded linear map between NAME spaces with finite rank, then MATH also has finite rank. Furthermore MATH because MATH is a unitary operator. Therefore MATH and MATH have finite rank. The result now follows from REF . |
math/9907075 | We have already shown that MATH, and we see from CITE and REF that MATH. Now let MATH and set MATH and MATH. Then MATH are well defined elements of MATH and MATH is a nonzero divisor, by CITE. Since MATH is a subring of MATH closed under taking inverses and adjoints, we see that MATH. The result follows. |
math/9907075 | First suppose MATH. Then by REF , there exist MATH such that MATH. Then MATH and MATH. Since MATH and MATH have finite rank, there are subspaces MATH and MATH of finite codimension in MATH such that MATH is zero on MATH and MATH is zero on MATH. Then MATH, hence there are subspaces MATH of finite codimension in MATH such that MATH and MATH. Now choose subspaces MATH of finite codimension in MATH such that MATH and MATH, and set MATH. Suppose MATH. Then MATH for some MATH and so MATH. Using the property that MATH is zero on MATH and MATH is zero on MATH, we see that MATH and we deduce that MATH. Conversely if MATH, then MATH for some MATH. Using the property that MATH is zero on MATH, we see that MATH. But MATH, so MATH and we deduce that MATH. Therefore MATH. Finally for MATH, we have MATH where MATH and MATH where MATH. Thus MATH because MATH. Since MATH by REF , we see that MATH and hence MATH. Also MATH because MATH. Therefore MATH and since MATH by REF , we deduce that MATH. We conclude that MATH on MATH. Conversely suppose there is a subspace MATH of finite codimension in MATH such that MATH on MATH. Write MATH where MATH and let MATH be a subspace of finite codimension in MATH such that MATH. Then MATH and MATH on MATH, so MATH for all MATH and we deduce that MATH has finite rank. Furthermore MATH on MATH, a subspace of finite codimension in MATH, so by a similar argument we see that MATH also has finite rank. Therefore MATH and we conclude from REF that MATH, as required. |
math/9907079 | Fix any MATH, and let MATH denote the NAME algebra for MATH with respect to MATH. Since MATH is semisimple, there exists a positive integer MATH and irreducible MATH-modules MATH, MATH,.,MATH such that MATH . For each integer MATH, MATH, let MATH (respectively, MATH) denote the dual endpoint (respectively, dual diameter) of MATH. Now fix any nonnegative integer MATH. Then for any MATH, MATH, MATH . So we can now argue that, since MATH is dual thin, MATH . In other words, MATH. Similarly, MATH . This yields MATH. |
math/9907082 | For MATH choose an element in MATH as MATH, choosing MATH whenever possible. These choices induce a groupoid morphism MATH, and MATH. Therefore, by the previous Theorem, the elements MATH for MATH are a complete set of generators for MATH. |
math/9907084 | This follows immediately from considering the components of the product MATH for MATH. |
math/9907084 | Fix MATH. There exists MATH, MATH, such that MATH. For all MATH, MATH . Thus MATH. Similar calculations show the other inclusions. |
math/9907084 | The necessity of the stated condition is REF , while the sufficiency is clear from REF . |
math/9907084 | CASE: This is clear from REF . CASE: If MATH, then MATH and thus MATH. For MATH, define mappings MATH and MATH by MATH and MATH . Then MATH, MATH, and MATH for each MATH, which proves the desired result and also shows that MATH and MATH are the projectors onto MATH and MATH, respectively. CASE: For MATH, MATH, we have MATH . Summing up entries shows MATH. CASE: Fix MATH, and let MATH denote the difference between the first and second entries of the product MATH. Then MATH using anticommutativity of the NAME bracket. Adding the quantity MATH, we obtain MATH since MATH. Similarly, we can show that the difference between the second and third entries of MATH is MATH. Thus all three entries are equal, which proves that MATH. |
math/9907084 | Assume MATH is simple and let MATH be a nonzero ideal of MATH. By REF , MATH is an ideal of MATH. Thus MATH, which implies MATH. Conversely, assume MATH is simple, let MATH be an ideal of MATH, and let MATH. By REF , MATH for MATH. By REF , MATH is an ideal. If MATH, then MATH, and thus we may assume MATH since MATH is simple. Now for MATH, MATH, the NAME identity gives MATH. Therefore MATH. But MATH since MATH is simple, and thus MATH. Hence each MATH is contained in the center of MATH, and thus each MATH. Therefore MATH, which shows that MATH is simple. |
math/9907084 | Let MATH be semisimple with each MATH simple. By REF , each MATH is an ideal of MATH and a simple algebra. Since MATH, a direct sum, MATH is semisimple. Conversely, assume MATH is semisimple with each MATH simple. Let MATH be a solvable ideal of MATH. Then MATH for some MATH, where MATH , MATH for MATH, and each containment MATH is proper. From REF , MATH is an ideal of MATH. Now MATH is a proper containment, contradicting REF . Thus MATH has no nonzero solvable ideals, which implies that MATH is semisimple. |
math/9907084 | First observe that MATH by a standard isomorphism theorem for vector spaces. Thus MATH is semisimple by REF . Now suppose MATH is an ideal of MATH such that MATH is semisimple. Then MATH is an ideal of MATH. Now MATH is a solvable ideal of MATH since MATH is a solvable ideal of MATH. This implies MATH is a solvable ideal of MATH. Since MATH is semisimple, we must have MATH. This completes the proof. |
math/9907084 | Let MATH be a NAME decomposition with MATH a semisimple NAME factor. Then MATH. By REF , MATH is semisimple. By REF , MATH. This completes the proof. |
math/9907084 | Compute MATH using REF and take the trace. |
math/9907084 | Let MATH be given. Identify indices modulo MATH: MATH, etc. Using REF and the invariance of the trace form MATH, we compute MATH . Since we are summing over all terms, we may reindex and combine them to obtain MATH . This completes the proof. |
math/9907084 | Fix MATH. For all MATH, MATH which shows that MATH, and similar computations show MATH. The reverse inclusion is clear. |
math/9907084 | This is immediate from REF . |
math/9907084 | We will show that MATH is a derivation. That MATH and MATH are derivations follow similarly. For MATH, we compute MATH . On the other hand, we have MATH . Comparing REF , and using the skew-symmetry of the NAME bracket, the result follows. |
math/9907084 | The identity MATH is an easy consequence of the NAME identity in MATH. |
math/9907084 | For MATH, let MATH. Multiplying matrices, we find that the equation MATH is equivalent to the equations MATH . If we similarly let MATH and MATH, and consider the corresponding matrix equations, then by matching matrix entries, we finally obtain the following system of equations in MATH: MATH for MATH. By NAME 's Lemma, REF implies MATH for MATH. Identifying indices modulo MATH, REF implies MATH for MATH. By NAME 's Lemma, for MATH, there exists MATH such that MATH. But then REF implies that MATH. This completes the proof. |
math/9907084 | For MATH, we compute MATH using the invariance and bilinearity of the standard form, REF , and MATH. Since MATH is nondegenerate REF , MATH for all MATH. On the other hand, since MATH is a derivation, MATH for all MATH. Adding REF , we obtain MATH for all MATH. By REF , there exists MATH such that MATH. |
math/9907084 | Since MATH is a derivation, REF are equal. Take MATH, MATH in REF and match the entries. This gives MATH and MATH for all MATH. By similar arguments, we obtain MATH for all MATH where MATH. Linearizing REF and rearranging, we have MATH for all MATH where MATH. Take MATH in REF , simplify using REF , and match the first entries. This gives MATH for all MATH. Successively taking MATH and MATH give the equations MATH for all MATH. Taken together, REF imply MATH for all MATH, that is, MATH is a derivation of MATH. Since MATH, MATH is also a derivation. |
math/9907084 | Using REF , the equality of REF simplifies to MATH for all MATH. Set MATH in REF and match entries. Using REF , this gives MATH for all MATH. Similar calculations give MATH for all MATH where MATH. Iterating REF , we have MATH for all MATH. Thus MATH for all MATH. Reversing the roles of MATH and MATH in REF and subtracting the resulting equation from REF , we obtain MATH for all MATH since MATH is a representation. Since MATH is simple, we have MATH and thus REF implies MATH for all MATH. By NAME 's Lemma, there exists MATH REF such that MATH. If we set MATH for MATH, then the action of MATH on MATH is given by the action of the matrix MATH. What remains is to show that MATH, and for this purpose we will use REF . Let MATH be given. Using REF , we compute MATH where MATH is the usual transpose of the matrix MATH. Since MATH is nondegenerate, the action of MATH on MATH is given by the action of the matrix MATH. By REF , we have MATH for some MATH. But the diagonal entries of MATH, and hence MATH, are all zero, and thus MATH. It follows that MATH. This completes the proof. |
math/9907084 | The equality of REF gives the following system of REF for all MATH. Set MATH and MATH, and add REF to obtain MATH for all MATH. Iterating REF , we obtain MATH for all MATH. Thus MATH for all MATH. Exchanging MATH and MATH in REF and subtracting the resulting equation from REF gives MATH for all MATH since MATH is a representation. Now MATH because MATH is simple, and thus MATH for all MATH. By NAME 's Lemma, there exists MATH such that MATH. Applying this to REF , we obtain MATH for all MATH. Thus MATH, and hence MATH. Similar arguments, mutatis mutandis, show that MATH. Now REF , say, shows that MATH is a derivation of MATH. Since MATH is simple, there exists MATH such that MATH. It follows that MATH as claimed. |
math/9907084 | Let MATH be given. Define MATH by MATH for all MATH. Then MATH is an automorphism of MATH. Now MATH using REF and the fact that MATH CITE. We have MATH. Thus there exists MATH such that MATH for all MATH. Since MATH is simple, MATH. Thus define MATH by MATH for all MATH. Then MATH and hence MATH for all MATH. Next, there exists MATH such that MATH for all MATH. Putting this together, we find that MATH for all MATH, where MATH. By REF , it follows that MATH for some MATH. Since the left side is an automorphism, MATH, and thus MATH. This completes the proof. |
math/9907085 | CASE: For MATH, MATH if and only if MATH if and only if MATH if and only if MATH. CASE: For all MATH, a computation using REF gives MATH . From this, it is clear that MATH is a pseudo-automorphism with companion MATH if and only if MATH. CASE: This follows from taking MATH in REF and using REF . |
math/9907085 | CASE: If MATH, then by REF , MATH for all MATH. CASE: This follows from REF . This follows from REF . |
math/9907085 | From REF , MATH is a two-sided identity. For MATH, we have MATH and MATH . Thus each left translation MATH has an inverse given by MATH. |
math/9907085 | The equivalence of REF is obvious. Since MATH is a transversal, MATH is a transversal if and only if, for every MATH, there exists a unique MATH such that MATH. This is equivalent to MATH, or MATH, which establishes the equivalence of REF . |
math/9907085 | CASE: For MATH, MATH, we have by REF , MATH . In particular, MATH, and thus MATH. CASE: For MATH, MATH, we have by REF , MATH and thus MATH as claimed. CASE: For MATH, REF imply MATH, which gives REF . For MATH, we use REF , and REF to compute MATH . This establishes REF . CASE: If MATH, then for all MATH, MATH, that is, for all MATH, MATH. This is equivalent to MATH for all MATH, or MATH, as claimed. |
math/9907085 | Since MATH, REF follows. By REF follows. Finally, for MATH, MATH, we use REF , and REF to compute MATH . This establishes REF . |
math/9907085 | This follows from REF , and REF . |
math/9907085 | This is a direct computation. |
math/9907085 | CASE: Applying both sides of REF to MATH, we have MATH. Cancelling, we obtain MATH for all MATH, or equivalently, MATH for all MATH. If LIP holds, then MATH for all MATH, which is AIP. Conversely, if AIP holds, then MATH. Cancelling MATH, we obtain LIP. CASE: See REF . Using LAP, REF , and LAP again, we have MATH for all MATH. From this the equivalence is clear. CASE: From the remarks following REF , we have that in a NAME loop satisfying AIP, each left inner mapping MATH is an automorphism, that is, the MATH property holds. By REF holds. The converse follows from REF , since every NAME loop has LIP. |
math/9907085 | CASE: For all MATH, we have MATH. By REF , MATH has LIP if and only if, for each MATH, MATH, or equivalently, MATH. This is equivalent to MATH. Since MATH, LIP holds if and only if, for each MATH, MATH, which is (G-LIP), or equivalently, MATH, which is (G/N-LIP). CASE: For all MATH, we have MATH. By REF , MATH has LAP if and only if, for each MATH, MATH, or equivalently, MATH, or MATH. Since MATH, LAP holds if and only if, for each MATH, MATH, which is (G-LAP), or equivalently, MATH, which is (G/N-LAP). CASE: For all MATH, we have MATH. By REF , MATH is a NAME loop if and only if, for every MATH, MATH, or equivalently, MATH, or MATH. Since MATH, MATH is a NAME loop if and only if, for every MATH, MATH, which is REF , or equivalently, for every MATH, MATH, which is (G/NAME). CASE: The proof is similar to that of REF , mutatis mutandis. CASE: For all MATH, we have MATH. Thus MATH satisfies REF if and only if, for every MATH, MATH, or MATH, which is REF , or equivalently, MATH, which is (G/NAME). |
math/9907085 | CASE: For MATH, we have MATH . Thus the equivalence of REF is clear. For MATH, we have MATH and MATH . Thus MATH is a pseudo-automorphism with companion MATH if and only if, for all MATH, MATH, or equivalently, MATH. This establishes the equivalence of REF . CASE: This follows from REF with MATH (using REF for REF ). |
math/9907085 | CASE: (G-PsA-MATH) holds if and only if, for every MATH, MATH, there exists MATH such that MATH, or MATH . This is (G/N-PsA-MATH). CASE: This follows from applying REF to MATH. CASE: The proof is similar to that of REF , mutatis mutandis. CASE: This follows from applying REF to MATH. |
math/9907085 | CASE: Assume MATH is injective and MATH for some MATH, MATH. Then MATH, using REF . Matching MATH and MATH components, this is equivalent to MATH and MATH. Now MATH implies MATH, and cancelling MATH, we have MATH. Thus MATH is injective. Conversely, assume MATH is injective and MATH. Then using REF again, MATH . Thus MATH. Matching components, MATH. CASE: MATH is surjective if and only if, for each MATH, there exists MATH and MATH such that MATH, that is, MATH. This implies MATH, and conversely, if MATH, then MATH for some MATH. This establishes the desired equivalence. CASE: This follows from REF . CASE: Using REF , we have for all MATH, MATH . Now MATH . If MATH, then matching components gives MATH (that is, MATH is a two-sided inverse of MATH), and MATH. Conversely, if MATH for all MATH, then MATH. |
math/9907085 | From REF , we have that MATH for all MATH. This implies REF , and the result follows from REF . |
math/9907085 | CASE: For MATH, MATH, we compute MATH and MATH . Thus MATH is an automorphism if and only if MATH for all MATH, MATH. Taking MATH gives REF . Taking MATH gives REF . Conversely, REF clearly imply REF . CASE: If MATH is corefree and REF holds, then REF holds, so the result follows from REF . CASE: Assume MATH. We will show that REF implies REF . First note that REF implies MATH, or MATH for all MATH. Fix MATH. Then MATH for some MATH, where MATH, MATH. Using REF , we have for MATH, MATH where MATH if MATH and MATH if MATH. Thus MATH . This completes the proof. |
math/9907085 | CASE: This follows from REF , and REF . This follows from REF . |
math/9907086 | If MATH, then we can take MATH. If MATH lies at a point of MATH with one preimages, take MATH. Otherwise, let MATH be a neighbourhood of MATH with such that MATH consists of two disjoint copies of MATH. Let MATH be a semicircular neighbourhood of MATH such that MATH maps MATH homeomorphically onto MATH, a subset of MATH. Let MATH. MATH is connected, so MATH is connected, and since MATH, MATH is connected, so lies in one of the components of MATH. Take MATH to be the preimage of MATH under MATH in this component. Clearly the map so defined is continuous at MATH, and MATH . |
math/9907086 | Let MATH be the quotient space obtained by collapsing each component of MATH to a point, and MATH the quotient map. Clearly MATH is exact, and since neighbourhoods of MATH are topological discs, MATH has a homotopy inverse MATH. Further, if MATH consists of cross-cuts, this homotopy inverse can be made an embedding, as shown in REF Choose a simplicial subdivision of MATH, such that no simplex contains more than one point of MATH. Since MATH is the quotient of a surface by the curves MATH, each REF-simplex of MATH is contained in no more than two REF-simplexes of MATH. Then any two vertices lying in the same component of MATH can be joined by an edge-path which does not touch MATH. Let MATH be a minimal REF-complex with the property that any two vertices in the same component of MATH lie in the same component of MATH. By the minimality of MATH, each component of MATH is contractible, so MATH. Hence there exists an edge MATH such that MATH and MATH is an edge of exactly one REF-simplex MATH of MATH. Let MATH be the simplicial complex formed by removing MATH and MATH from MATH. There is a strong deformation retract MATH such that MATH, and both MATH and the corresponding inclusion MATH are exact. By iterating this procedure to remove one simple at a time, we obtain the graph MATH. Since the homotopy inverse for MATH can be made an exact embedding if MATH consists of cross cuts, and each inclusion is an exact embedding, we obtain the required homotopy inverse in the case where MATH consists of cross-cuts. |
math/9907086 | Suppose MATH are NAME, and MATH is a relating family for MATH. Suppose MATH. Let MATH and MATH . Since MATH, MATH. Let MATH be given by MATH. Then MATH is a relating family for MATH, and further, there are regions MATH such that MATH. If MATH, then we let MATH, so again there are regions MATH such that MATH. By homotoping if necessary to remove any folds, we can assume that all curves MATH are locally injective. Since MATH is MATH-tight, MATH is locally injective, so, up to parameterisation, MATH . Hence MATH so MATH. Thus MATH , and so all points of MATH are fixed by MATH. If MATH is an isolated repelling fixed point of a graph map MATH and MATH does not lie on a vertex of MATH, then MATH. |
math/9907086 | Since MATH is the inverse image under the glueing map of a submanifold of stable manifold for MATH, MATH has a neighbourhood MATH for which every point of MATH eventually leaves MATH. Since MATH is a union of disjoint copies of an interval with endpoints in MATH, we can find neighbourhoods MATH, MATH and MATH of MATH each of which deformation retract onto MATH such that MATH, MATH, MATH, and every point of MATH eventually leaves MATH . Choose an open cover MATH containing the components of MATH and MATH, and such that for all other MATH, MATH and MATH intersects at most one component of MATH (This is where we need MATH). Let MATH. We claim that MATH and MATH are the required open cover and neighbourhood of MATH. First notice that if MATH and MATH lie in the same component of MATH, but different components of MATH (equivalently, every path from MATH to MATH in MATH crosses MATH), then MATH and MATH lie in different components of MATH. Suppose MATH, and MATH and MATH are MATH-close for all MATH. Then there exists least MATH such that either MATH or MATH are not in MATH. By minimality of MATH, MATH. Since MATH and MATH are MATH-close, they must lie in the same component of MATH. This means that MATH and MATH lie in the same component of MATH, and since components of MATH are simply connected, every path in MATH from MATH to MATH is homotopic to one which does not intersect MATH. |
math/9907086 | MATH since the gluing map is a finite-to-one surjective semiconjugacy, and MATH by REF. |
math/9907088 | The lemma is clearly true if there are no critical points of the height function between the points MATH and MATH where we insert humps. In case there is one critical point between MATH and MATH the lemma follows from the argument on REF . This also proves the lemma in the general case. |
math/9907088 | Let MATH where MATH and MATH be a homotopy between MATH and MATH, that is, for each MATH the map MATH defines a long knot and MATH and MATH. In MATH consider the subset MATH of pairs MATH such that MATH . Without loss of generality we can assume that MATH is a union of smooth compact non-singular curves whose boundary is either empty or belongs to MATH and that there are only a finite number of tangencies of MATH with horizontal lines of the form MATH. In addition we require these tangencies to take place at different values of the parameter MATH; see REF . These assumptions imply, in particular, that for all but a finite number of values of MATH the knot MATH is NAME and that the perestroikas at the bifurcation values of MATH are generic, that is, are insertions (or removals) of humps. If there are no points of tangency of MATH with horizontal lines the knots MATH and MATH are NAME equivalent and MATH. Otherwise, choose the point of tangency of MATH with a horizontal line which corresponds to the insertion of a hump with the smallest value of MATH. It is clear that we can connect it with the lower boundary line MATH by a segment MATH of a curve which is disjoint from MATH and whose tangent is nowhere horizontal, see REF . In the neighbourhood of each point of MATH we can modify the knots MATH by inserting humps, this changes MATH as shown on REF . Notice that the number of points where MATH has a horizontal tangent has decreased by one and the knot MATH was changed by an insertion of a hump. Thus, proceeding inductively, we eliminate all insertions of humps. In the same way we eliminate the removals of humps with the only difference that we connect them to the upper boundary line and proceed from the bifurcation with the largest value of MATH downwards. The result is that we construct a NAME equivalence between MATH, possibly with several humps inserted, and MATH, also with some extra humps. However, from REF we know that MATH and MATH with humps inserted are NAME equivalent to MATH and MATH respectively (here MATH is the number of maxima of the modified knots) and this proves the lemma. |
math/9907096 | Consider the following diagram MATH where MATH. Since the right square is a fibered square it follows from the properties of the MATH morphism that MATH. Therefore it remains to show that MATH. It follows from the construction in CITE that the MATH morphism is determined by a NAME open subset whose intersection with each fiber is dense. In other words, MATH coincides with MATH . We have used the fact that MATH since MATH is proper. |
math/9907096 | Let MATH. Denote by MATH the graph obtained from MATH by attaching a tail labeled MATH to the vertex MATH of MATH. For each MATH the graph MATH determines a substack of MATH, and there are natural morphisms MATH . Consider the following commutative diagram MATH . Note that the left square is a fibered square, the right square is close to a fibered square in the sense of REF , and all morphisms MATH, MATH are representable and flat. Therefore, one can apply REF . Also note that for each MATH one has MATH . Now MATH . |
math/9907096 | It is clear that the morphisms MATH are MATH-equivariant. In order to prove the restriction properties fix MATH, MATH such that MATH, and a stable graph MATH of genus MATH with MATH tails. Let MATH be the set of all MATH decorated graphs such that the underlying graph without decoration is MATH. Let MATH and let MATH be the product of the corresponding virtual fundamental classes. Consider the following commutative diagram: MATH . We want to see how the MATH summand of MATH restricts to MATH. In the sequence of equations below we will use the following properties. The right square of the above diagram is a fibered square. All vertical morphisms MATH are proper. If MATH is invariant under the action of MATH, then MATH, where MATH. In addition, we use the following result of CITE: MATH where MATH is the diagonal morphism. If MATH, and MATH, then we denote by MATH the tensor product of the corresponding MATH's on MATH, and by MATH the formal sum on MATH. For the sake of brevity we will write MATH for MATH, MATH for MATH, and MATH for MATH. The sums below are always taken over MATH. MATH . Summing over all MATH gives the statement of the theorem taking into account that MATH is the cap-product with the NAME dual of the diagonal in MATH. |
math/9907096 | The morphisms MATH and MATH are flat as compositions of flat morphisms. Let MATH be MATH, and MATH be MATH. Then the commutative diagram above is close to a fibered square in the sense of REF (compare CITE). Therefore one has MATH and MATH. Iterating, one obtains the statement of the lemma. |
math/9907096 | Consider the following commutative diagram close to a fibered square: MATH where MATH. Let MATH associated to the MATH marked point. One has MATH . |
math/9907096 | We proceed by induction. When MATH the statement of the lemma is trivial. Assume that the statement is true for MATH and prove it for MATH. We denote by MATH the universal curve MATH. One has MATH . |
math/9907096 | We proceed by induction. Let MATH, and MATH. Then, using REF , one gets MATH . This proves the statement of the lemma when MATH. Now assume that the statement is true for MATH and prove it for MATH. Denote by MATH the natural morphism MATH. MATH . The rest follows applying the induction hypothesis to the product MATH and using REF . |
math/9907096 | (of REF .) We assume that MATH is determined by REF. It follows that MATH, and MATH . It follows that for each MATH, MATH, and MATH such that MATH one has MATH and MATH if MATH. Consider the function MATH, where MATH is a constant. Differentiating it with respect to MATH provides using REF MATH . Therefore MATH does not depend on MATH. It follows that for all values MATH and MATH one has MATH . In particular, MATH for every MATH. |
math/9907096 | Applying the chain rule one gets: MATH . The second equation uses that MATH satisfies the topological recursion relations. Similarly, MATH . REF implies that MATH, and the proposition follows. |
math/9907096 | We use the chain rule, REF, and the topological recursion relations for MATH: MATH . In the second equation we used the divisor equation for MATH CITE. |
math/9907102 | Substituting in the integral MATH, we obtain MATH . For MATH we use MATH . As MATH, the left-hand and right-hand side of this inequality are integrable functions over MATH, and we obtain MATH for some MATH. For MATH we have MATH, and therefore MATH . Now we substitute MATH and get MATH . Again we split up MATH and use an estimate of the form MATH for the second integral. NAME in this way, we receive MATH . For the last integral we use MATH . As MATH, the left-hand and the right-hand side of this inequality are integrable functions on MATH. Therefore MATH . |
math/9907102 | Instead of MATH we use the right-hand side of REF . From REF with MATH we obtain (see REF ) that MATH where MATH is chosen according to REF . From REF applied to MATH we see that the sides of the NAME polygon corresponding to the weight function REF are given by MATH with MATH. But this means that the NAME polygon for MATH is constructed from MATH by a shift of MATH to the left, that is, we have MATH. |
math/9907102 | From REF we trivially obtain REF and, setting MATH, REF . Taking MATH and dividing REF by MATH, we receive MATH . Taking the limit for MATH, we obtain REF . Now let REF - REF be fulfilled. For MATH we write MATH in the form MATH with MATH . The coefficients of MATH (considered as a polynomial in MATH) are homogeneous functions in MATH, and therefore MATH is a homogeneous function in MATH of degree MATH. From this and from REF it follows that MATH . Multiplying these estimates, we see that MATH is MATH-elliptic with parameter in MATH. |
math/9907102 | CASE: We write MATH with MATH and set MATH. After division of MATH by MATH we obtain the equation MATH . First we fix MATH with MATH. Let MATH be a zero of MATH of multiplicity MATH. Then there exists a MATH such that MATH holds for all MATH. Therefore, for every MATH REF has exactly MATH roots in MATH which we denote by MATH. NAME in this way for all zeros of MATH, we obtain the set MATH of zeros of MATH. Now we assume that the statement in REF is false. Then there exists a sequence MATH with MATH and a constant MATH such that MATH for all MATH where we have set MATH. Due to compactness, we may assume that MATH converges to MATH. As the zeros of MATH depend continuously on MATH, we obtain for large MATH that MATH . But from the same considerations as above we see that for every sufficiently small MATH there exists a MATH and a MATH such that MATH has exactly MATH roots in MATH for all MATH and MATH. Taking MATH, we obtain a contradiction to REF . CASE: We set MATH and MATH and obtain the equation MATH with MATH. First we fix MATH with MATH. We write MATH . Let MATH be a zero of MATH of multiplicity MATH. Then we know from the theory of algebraic functions that there exist MATH roots MATH of MATH for which we have an expansion (NAME series) of the form MATH (compare , for example, REF). In REF we have to take the MATH different branches of the function MATH to obtain the zeros MATH. The series on the right-hand side is a holomorphic function in MATH for MATH for some MATH. From the construction of the NAME series (compare REF) we know that the coefficients MATH in the series REF depend continuously on the coefficients of the polynomial MATH and therefore on MATH. Thus there exists a MATH, independent of MATH, such that the right-hand side of REF is a holomorphic function in MATH for MATH. As the function MATH is continuous in MATH and MATH for MATH and MATH, it is bounded by some constant MATH, independent of MATH and MATH, which finishes the proof of REF . |
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