paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0011238 | We can write MATH. Assume that the statement is false, and let MATH be a sequence of examples with all entries of MATH bounded by MATH and some entries of MATH or MATH going to infinity. By passing to a subsequence we may assume that all entries converge in MATH. Call a position of a matrix `small' if the corresponding entries stay bounded and otherwise call it `large'. For example, all positions of MATH are small. Note that by NAME 's Rule REF all positions of MATH are also small. For simplicity, we drop the subscripts from MATH. Consider the lowest nonzero off-diagonal entry MATH of MATH or MATH, that is, MATH or MATH such that MATH for all MATH. There are two cases: CASE: The position of MATH is small. We can replace MATH by REF without affecting other entries of the same matrix by multiplying on the right by the elementary lower-triangular matrix that has entry MATH in the position MATH. We can compensate this operation by either multiplying MATH on the right by the same matrix (in case that the entry MATH belongs to MATH) or by multiplying MATH on the left by the inverse of this matrix (in case MATH belongs to MATH). All entries of the new matrix MATH are still small (though not bounded by MATH perhaps). Now proceed to the next lowermost off-diagonal entry of MATH or MATH. CASE: The position of MATH is large. Say MATH belongs to MATH (the other case is analogous). We write MATH . The right hand side is obtained from the matrix MATH with small entries by applying elementary row operations in which a multiple of the MATH row is added to the row below for certain MATH. Therefore all rows bellow row MATH of the matrix MATH are equal to the corresponding entries of the small matrix MATH. The left hand side is obtained from MATH by applying elementary row operations in which a multiple of a row is added to some row above. It follows that the entry of MATH in position MATH is MATH, which is large. Contradiction . |
math/0011238 | Let MATH and MATH be two matrices representing points in cones on disjoint simplices of MATH. Assuming that the two points are far away from the cone point, we have to show that the distance between MATH and MATH is large. If one of MATH has a large entry, this follows from REF . If both MATH and MATH have small entries, then MATH is close to MATH, while MATH is close to MATH, so we must show that MATH and MATH are far apart. In this case MATH and MATH will have large entries, so the statement follows from REF . |
math/0011238 | MATH is a lattice in MATH by CITE. It is a cocompact lattice since the MATH-rank of MATH is REF. The manifold MATH can be taken to be the product of the symmetric space of the semi-simple factor of MATH and of the Euclidean factor corresponding to MATH (real points in the torus modulo maximal compact subgroup). See REF . |
math/0011238 | Note that MATH intersects MATH in a finite subgroup. After passing to a finite cover, we have a split exact sequence MATH and MATH is a quotient of MATH with finite kernel. The image of MATH is a lattice in MATH. A maximal compact subgroup MATH of MATH lifts to a maximal compact subgroup of MATH. We set MATH and apply REF to MATH. To prove the last statement, recall CITE that MATH acts cocompactly and properly discontinuously on a contractible manifold with corners whose interior can be identified with MATH and MATH is the stabilizer of a lowest dimensional stratum, which is a contractible manifold and has codimension MATH (and is really a copy of MATH). |
math/0011238 | Both statements are well-known over MATH (see for example, CITE for REF and CITE for REF ). REF is proved the same way over MATH, and REF over MATH follows by decomposing each root space into REF-dimensional subspaces which are root spaces with respect to a maximal torus that contains MATH (but is split only over MATH, not over MATH). |
math/0011238 | We have MATH where MATH is the sphere of dimension MATH and MATH is the disjoint union of a sphere of dimension MATH and a point. Thus MATH is a MATH-obstructor complex and MATH is a MATH-obstructor complex. The Join Lemma implies that MATH is a MATH-obstructor complex for MATH where the last two equalities follow from REF and the observation that MATH. |
math/0011238 | Every root can be written as a linear combination of simple roots. The key to this proof is the fact that the coefficients in such linear combinations are explicitly known (and can be found for example, in CITE). We first note that if MATH is not irreducible, then the statement follows immediately from the corresponding statements for the irreducible components. The reason for this is that MATH is then the disjoint union of its irreducible components, and any ordering of MATH that restricts correctly to each component will work. Moreover, if there are elements of MATH that are not labeled at all (which is the case when MATH is not the highest node) then we may restrict our consideration to the root system MATH generated by the labeled nodes (and in fact to the irreducible component of MATH that contains MATH). Here is one situation when we can take MATH to be the negative of the largest positive root: MATH and if we write MATH as a linear combination of simple roots, then all other roots in MATH have their MATH-coefficient MATH. These situations, together with the coefficients, are pictured below, with the node MATH circled. In the case of MATH (that is, MATH) we can take MATH and MATH. This leaves us with type MATH. The ordering is the usual linear ordering of the nodes. Say MATH is node MATH in this ordering and let a labeling by ``U"'s and ``D"'s of nodes MATH be given as in the lemma. Let MATH be such that the label of node MATH is ``D" but the label of node MATH is not ``D" (or MATH). Define MATH as the negative of the sum of the nodes MATH. In general, to define the ordering on MATH, we follow this procedure: Work separately on components. If a component is of type MATH order the nodes linearly. Otherwise, define the highest root in the ordering to be the circled node in the corresponding figure above. Then pass to the subdiagram consisting of unlabeled nodes and repeat the procedure. |
math/0011239 | It is clear from REF that MATH for all MATH and that MATH for all MATH. Suppose that MATH and that MATH for some MATH. Let MATH and MATH, and note that MATH. For each MATH, we have MATH for some MATH such that MATH. For MATH, let MATH and let MATH . Note that MATH . Thus, MATH . Hence MATH . Thus, MATH is approximately convex with respect to MATH. |
math/0011239 | Without loss of generality we may assume that MATH. We shall prove that the result holds for all MATH by induction on MATH. Note that the result is vacuously true if MATH and is trivial if MATH. So suppose that MATH and that MATH, so that MATH. By inductive hypothesis, MATH is the disjoint union of sets MATH such that MATH for MATH. Since MATH, and since MATH, there exists MATH such that MATH . Put MATH and MATH for MATH to complete the induction. |
math/0011239 | Let MATH, so that MATH. The proof is by induction on MATH. If MATH then MATH for some MATH, so that MATH . As inductive hypothesis, we suppose that MATH whenever MATH. Now suppose that MATH and that MATH. Without loss of generality we may assume that MATH, so that MATH, where MATH. Note that each MATH since MATH. If MATH, let MATH, MATH, and MATH for MATH. Note that MATH for MATH and that MATH. On the other hand, if MATH, then applying REF with MATH, we can write MATH as the disjoint union of sets MATH such that MATH for each MATH. Note that this implies that MATH for MATH. If MATH, let MATH, where MATH. If MATH, let MATH and let MATH. Thus MATH, where MATH and MATH. Note that MATH . If MATH, then MATH for all MATH, and MATH. Since MATH, our inductive hypothesis implies that MATH. Finally, MATH . This completes the induction. |
math/0011239 | Let MATH. A standard NAME multiplier calculation yields MATH where MATH. Using REF in place of REF, minor changes in the proof of REF show that MATH is approximately convex with respect to MATH. Suppose that MATH for some MATH satisfying MATH. Let MATH. Then MATH, and so MATH . On the other hand, since MATH is approximately convex with respect to MATH, it follows from REF that MATH. |
math/0011239 | To prove that MATH is piecewise linear it is enough to show that MATH is piecewise linear on the interior MATH of MATH. For then by an induction on MATH we will have that MATH is piecewise linear on MATH and the induction hypothesis implies that it is piecewise linear when restricted to any of the facets of MATH, which implies that MATH is piecewise linear on MATH. For fixed MATH and MATH let MATH be the set of feasible MATH-tuples. For MATH let MATH be the linear function MATH so that MATH is given by MATH . Let MATH be the set of extreme MATH-tuples. Then MATH and therefore showing that MATH is piecewise linear is equivalent to showing that MATH is finite. Let MATH and MATH with MATH for MATH. Then MATH. For if not then there is an index MATH with MATH. As all the components of MATH are positive on MATH this implies that on MATH . This contradicts that for MATH there is a MATH with MATH. Let MATH be the group of permutations of MATH. Then it is easily checked that MATH is invariant under the action of MATH given by MATH. Therefore if MATH is the set of monotone decreasing elements of MATH, that is MATH then MATH and to show that MATH is finite it is enough to show that MATH is finite. Suppose that MATH. Let MATH be a non-increasing sequence of MATH positive integers, and let MATH be a positive real number such that MATH and such that if MATH are any positive integers with MATH for MATH, then MATH implies that MATH. (We will say that MATH is extreme for MATH.) Let MATH . Then MATH. (The explicit value of MATH is MATH.) From the definition of MATH we have MATH and MATH which is equivalent to MATH . Assume, toward a contradiction, that MATH. Then MATH . This can be rearranged to give MATH . This contradicts that MATH is MATH extreme and completes the proof. We now prove MATH is finite. First some notation. For positive integers MATH let MATH. If MATH then by REF (and with the terminology of REF ) for each MATH with MATH the tuple MATH is MATH extreme, and MATH itself is MATH extreme. Therefore, by REF , MATH, whence there are only a finite number of possible choices for MATH. For each of these choices of MATH we can use REF again to get MATH, and so there are only finitely many choices for the ordered pair MATH. And for each of these pairs MATH we have that so there are only finitely many possibilities for MATH. Continuing in this manner it follows that MATH is finite. This completes the proof that MATH is piecewise linear and thus point REF of NAME REF To prove point REF let MATH be a nonempty subset of MATH. In proving point REF we have seen that there is a finite collection MATH of linear mappings MATH, each one of the form MATH for some nonnegative integers MATH, MATH, with MATH, such that MATH for all MATH such that MATH. Clearly, we may also assume that MATH whenever MATH. Suppose that MATH and that MATH as MATH. Note that MATH for all sufficiently large MATH, so that MATH for all sufficiently large MATH. Thus, MATH . Thus, MATH is lower semi-continuous. Finally we prove point REF . It follows from REF that the restriction of MATH to the interior of any facet is the minimum of a finite collection of linear functions, and hence is continuous and concave. The lower semi-continuity of MATH forces MATH to be concave on all of MATH. |
math/0011239 | MATH is a symmetric function of MATH and MATH is also concave. Thus MATH achieves its maximum at the barycenter MATH. So there exist nonnegative integers MATH REF such that MATH and MATH. We may also assume that MATH have been chosen to minimize MATH among all possible choices of MATH. Suppose that there exist MATH and MATH such that MATH. Note that MATH . Thus replacing MATH by MATH and replacing MATH by MATH leaves MATH unchanged while it reduces MATH, which contradicts the choice of MATH. Thus MATH for all MATH. It follows that there exist integers MATH and MATH such that MATH and MATH . Moreover, it is clear from REF that MATH is the least nonnegative integer satsifying REF for some MATH, that is, MATH . For this value of MATH it is clear from REF that MATH is the smallest integer in the range MATH satisfying REF, that is, MATH . Substituting these values for MATH and MATH into REF gives REF. |
math/0011239 | By replacing MATH by MATH, we may assume that MATH. Set MATH and define MATH by MATH . Clearly MATH. Suppose that MATH. By NAME 's Theorem (see for example, CITE) there exist MATH points MATH such that MATH. Let MATH. Then MATH lies on the boundary of MATH and so it is a convex combination of MATH of the points MATH. Without loss of generality, MATH for some MATH. Note that MATH is approximately convex with respect to MATH and satisfies MATH for MATH. By REF , MATH. Thus MATH . Taking the infimum over all MATH yields MATH, that is, MATH. Finally, set MATH. The fact that MATH is the best constant follows by taking MATH to be MATH, where MATH is the extremal approximately convex function (with respect to MATH) with domain MATH. |
math/0011243 | First we construct the MATH-module MATH, such that MATH. The anti-involution MATH acts also on the MATH-module MATH, generated by the series MATH, see REF. (In fact MATH has a structure of an associative conformal algebra, see CITE). We let MATH to be the MATH-submodule of MATH consisting of those elements of MATH which change sign under the action of MATH. We have MATH. The NAME algebra MATH is spanned by the coefficients MATH, see REF. Therefore the space MATH is spanned by the elements MATH for all MATH. Then the series MATH span MATH over MATH. Since MATH, we have MATH therefore, if MATH is even, then we get MATH . The remaining series MATH form a basis of MATH over MATH. Some simple calculations show that the algebra MATH of inner derivations of MATH is equal to the whole MATH. So MATH is a module over MATH, where the action is given by MATH for MATH and MATH. This action induces the following action of MATH on MATH: MATH for all MATH. So we obtain TKK conformal algebra MATH, where a MATH-basis of MATH is given by MATH for MATH, a MATH-basis of MATH is given by MATH for MATH, and MATH is spaned over MATH by the basis MATH for MATH. Here MATH is the involution identifying MATH with MATH. Since the coefficients of these series span MATH and are linearly independent, we have MATH. REF shows that MATH and MATH in MATH are local, hence so are MATH and MATH. The series MATH and MATH are local as well because the product MATH is just the the associative product in MATH if we identify MATH and MATH as linear spaces. |
math/0011243 | Express the linear generators of MATH in terms of elementary matrices MATH . Then using REF we obtain by some calculations that MATH . |
math/0011243 | Using REF we get: MATH . |
math/0011243 | Recall that MATH is a module over the NAME algebra MATH, which is a central extension of the NAME algebra MATH of infinite matrices with finitely many non-zero diagonals, and that MATH is embedded in MATH by REF . Let MATH be subalgebra of diagonal matrices of MATH. Since MATH, we get that MATH is a subalgebra of MATH. Any functional from MATH takes values on MATH as well, so we have MATH. Let MATH be the subalgebra of MATH consisting of all elements of degree REF. We have MATH under the mapping REF. To prove REF it is enough to show that any two different weights MATH remain different after restriction to MATH. But this follows from the fact that MATH for MATH are linearly independent on MATH because MATH. For the proof of REF we note that by REF every element MATH is an infinite linear combination of MATH's such that either MATH or MATH. Therefore, by REF we get that if MATH then for any MATH such that either MATH or MATH the weight MATH appears in MATH. Every weight MATH such that MATH could be obtained by a successive application of this operation but no weight of length more than MATH can be obtained. |
math/0011243 | Let MATH. We have to prove that the conformal algebra generated by MATH and MATH is the whole MATH. By REF we can assume that MATH is homogeneous of some weight MATH such that MATH. The only weight of degree REF which is not in MATH is MATH which is of length REF, hence by REF, MATH. Therefore it is enough to prove that MATH is generated as a conformal algebra by MATH and any element MATH. Take MATH. Let MATH be the conformal algebra generated by MATH and MATH. Assume on the contrary that MATH. Then REF implies that there is an integer MATH such that MATH does not contain weights of length MATH. Let MATH be the minimal possible among such integers. Let MATH, and let MATH. Let MATH so that MATH and hence MATH. We prove that the projection of MATH onto MATH is non-zero, therefore MATH so we arrive to a contradiction. Using that MATH and REF, we get MATH . The only pairs of MATH's in the above expression that do not kill MATH and move MATH to MATH are MATH for MATH, MATH for MATH, and MATH for either MATH or MATH. Let MATH. Using REF we get that MATH for all MATH and MATH for all MATH. Also, MATH. Summing up, we get MATH . |
math/0011243 | Both statements are proved by a standard argument. Let us prove REF , the prove of REF being identical. Assume on the contrary that MATH for every MATH. Let MATH be such that MATH. Then MATH could be expressed as a linear combination MATH of elements MATH with integer coefficients. Since MATH we can assume that all MATH. Assume that the above is a combination of this kind with the minimal possible value of MATH. Then since MATH for all MATH and MATH we must have MATH for some MATH, but then MATH and we could make the expression for MATH with a smaller sum of coefficients. |
math/0011243 | First we show that if MATH or MATH the conformal algebra MATH generated by MATH and MATH is as claimed. As it was remarked at the end of REF, all calculations in vertex algebra MATH are very explicit. So we just have to read off the defining relations of conformal superalgebras from REF for the products in MATH. Of course, the case when MATH is REF or REF is well-known. Let us do the most difficult case when MATH. In this case we can identify MATH with MATH by letting MATH. By REF the elements MATH and MATH span a copy of MATH over MATH so that the products are given by REF. Set MATH. Recall then there is an involution MATH induced by the involution MATH of the lattice MATH, so that we have MATH. We have to show that REF - REF hold for MATH, MATH and MATH and also MATH for all integer MATH. Let us for example check REF. First we note that MATH . Using this we calculate MATH . The first sum here is REF because MATH, hence MATH commutes with MATH and MATH. The second sum gives MATH and REF follows. The other formulas are checked in the same way. Instead of checking REF directly, we note that by REF the MATH-modules MATH and MATH are both irreducible highest weight MATH-modules corresponding to the same highest weight, hence they must be isomorphic. The verification of the conformal cocycle REF is done in the same way. We are left to show that if MATH, then MATH. Recall from REF that we have an embedding MATH given by REF such that MATH for MATH so that MATH for MATH. Simple calculations show that MATH for MATH and also MATH. Since MATH is generated by MATH, we have that MATH. However, MATH, therefore, since MATH is a maximal conformal subalgebra in MATH by REF , we must have MATH. |
math/0011243 | By REF the square lengths of MATH and MATH could be only REF or REF. Also, MATH, because otherwise we would have MATH. So we have only finitely many possibilities. One can easily check using REF that every choice of vectors MATH and MATH, not listed in REF - REF will give an infinite root system. So it remains to show that in each of REF the conformal algebra MATH generated by MATH will be in fact graded by the corresponding root system MATH. Let us check this for the most difficult REF . In this case MATH is a direct sum of two copies of MATH, corresponding to vectors MATH and MATH, so that the subalgebras MATH and MATH of MATH are isomorphic to the conformal algebra MATH constructed in REF. So the claim follows from REF . |
math/0011243 | Let MATH be a pair of roots. The root system which they generate must be of one of the types REF - REF from REF . It follows that the NAME number MATH is an integer, hence MATH must lay inside a NAME root system MATH. Moreover it follows from the structure of root systems of rank REF - REF that MATH cannot be of types MATH and MATH, and the condition for the square lengths of the root vectors hold. Next we want to prove that if MATH is a finite NAME root system of the type other than MATH and MATH and the length of roots are as prescribed by the theorem, then MATH is indeed the root system of the conformal superalgebra MATH generated by the set MATH. If MATH is a simply-laced root system of types MATH-MATH then MATH is the affine NAME conformal algebra, as it is well-known. If MATH is of type MATH then it is equally easy to check that the space MATH will be closed under the products REF, hence it is the desired subalgebra. Let MATH be of type MATH. Let MATH be the basis of MATH consisting of pairwise orthogonal extra-long roots. Take MATH, where MATH is the NAME conformal algebra spanned by MATH, see REF . Take MATH to be equal to the conformal NAME triple system, constructed in REF, so that the subalgebra MATH is isomorphic to the conformal algebra MATH. If MATH is a short root then take MATH. Using calculations of REF it is easy to see that MATH is closed under the products REF. Finally, if MATH is of type MATH then the corresponding conformal superalgebra MATH is easily obtained by combining the conformal superalgebras corresponding to the subsystems of MATH of types MATH and MATH. So let MATH be a finite root system, and let MATH be a minimal NAME root system containing MATH. We prove that if MATH is either simply-laced or is of type MATH or MATH then MATH. Assume first that MATH is simply-laced. Let MATH. Since MATH spans MATH over MATH we can write MATH as a linear combination of elements of MATH with integer coefficients. Let MATH, be such a linear combination of the minimal length. Since MATH we can assume that all MATH. But then since MATH we must have MATH for some pair MATH, hence MATH and we can make the combination shorter, contrary to our assumption. Hence MATH. Same argument shows that if MATH is of type MATH or MATH then MATH contains all short roots, and if MATH is of type MATH then MATH contains all long roots. Let MATH be of type MATH. Then MATH contains all long roots of MATH. Since MATH is a minimal NAME root system containing MATH, the latter must contain at least one extra-long root MATH. Let MATH be a long root such that MATH. Then by REF MATH contains the whole root system of type MATH generated by MATH and MATH, therefore the extra-long root MATH also belongs to MATH. Continuing this argument we get that all extra-long roots lay in MATH, hence MATH. Assume now that MATH is of type MATH. When MATH or MATH we refer to REF , so assume that MATH. Let MATH be a basis of MATH consisting of pairwise orthogonal short roots. Let MATH be the set of all long roots in MATH. They form a simply-laced NAME root system of type MATH (MATH if MATH). The root system MATH must contain some long roots too. Let MATH be the set of long roots of MATH. They must form a root system as well. It is not too difficult to see that for MATH to be indecomposable the root system MATH must be either equal to the whole MATH or to MATH, in which case MATH is of type MATH. This choice of MATH is unique up to the action of the NAME group. Therefore MATH is either equal to MATH or is of type MATH. Finally, let MATH be of type MATH. Then by the result of the previous paragraph the set MATH of long roots must at least contain the set MATH. Also, MATH must contain at least one extra-long root. So, as in the case of type MATH, using REF we obtain that MATH. |
math/0011243 | The root system MATH must contain some isotropic roots, otherwise it would be positive definite. At least some isotropic root MATH must be of the form MATH, where MATH and MATH are real roots. If MATH is not of type MATH then MATH and MATH have square lengths more than REF and hence we are in the situation of REF or REF , so MATH for all integer MATH and MATH is infinite. If MATH is of type MATH, then it might happen that MATH and MATH have length REF. If this is the case, let MATH and MATH be real roots such that MATH and MATH. Then by REF , MATH is an isotropic root such that MATH is also a root for all integer MATH. |
math/0011243 | Let MATH be the conformal superalgebra generated by the set MATH. We claim that for any MATH we have MATH. Let us first show that the theorem follows from this claim. Let MATH and MATH. If MATH, then using REF, we get MATH hence MATH. If MATH, take some MATH and then using as before, REF, we get MATH hence MATH. Let us now prove the claim. REF assures that any isotropic root MATH is obtained as a sum MATH of two real roots MATH. Since MATH does not have any short roots, the pair MATH must generate a root system of type either REF or REF . So REF implies that MATH. Assume now that MATH is such that MATH and MATH satisfies MATH. Then we have MATH hence MATH. Therefore the real roots MATH and MATH form a root system of type REF or REF , so we get that MATH. It follows that for every real root MATH which is not orthogonal to either MATH or MATH we have MATH, therefore, since MATH is indecomposable, MATH for all MATH. REF implies that MATH, and the claim follows. |
math/0011244 | CASE: We have seen that, in the above discussion that MATH . If MATH, MATH . REF gives log terminality. If MATH is irreducible and so by the result of NAME and NAME MATH is log terminal. So in either case MATH is log terminal. CASE: Again we write MATH with MATH log terminal. So in particular, for all points MATH, MATH . Suppose now that MATH. Then MATH, MATH and MATH. Therefore by the result of NAME and NAME, MATH must split as a product of (principally polarized) elliptic curves. |
math/0011244 | CASE: By MATH-cohomology and MATH-cohomology coincide. CASE: The usual proof of the MATH-lemma goes through since we have the NAME 's operator, harmonic decomposition, etc. as usual and MATH . |
math/0011244 | REF are obvious. CASE: Since MATH we have that MATH is MATH-closed and MATH-exact and so by the NAME MATH-lemma in REF there exist MATH such that MATH . Thus MATH . Clearly MATH . CASE: The form MATH is MATH-exact and MATH-closed and so by the NAME MATH-lemma REF MATH for some MATH . |
math/0011244 | CASE: By MATH . We filter MATH by subcomplexes MATH and form the associated spectral sequence with MATH and MATH given by MATH . By REF i-ii, REF of the theorem will follow from the fact that this spectral sequence degenerates at MATH, that is, MATH for all MATH. To compute MATH let MATH and let MATH denote the intersection MATH . By REF iii-iv we can define a MATH-linear function MATH and the natural map MATH is surjective. So we can compute MATH on an element MATH by applying the operator MATH to a preimage MATH. So we can compute MATH on an element MATH by applying the operator MATH to a preimage MATH. But using REF repeatedly we can recursively pick our representative MATH for the equivalence class MATH such that MATH in MATH for all values of MATH. Thus MATH for all MATH. CASE: We wish to construct a mapping MATH whose composition with the natural map MATH is the identity. The mapping MATH above can be chosen in such a way as to do the job. Namely, if MATH is in the kernel of the composition of MATH and the projection MATH then MATH where MATH . So we can map MATH to MATH which lies in MATH and is zero in MATH . |
math/0011244 | Restrict MATH to a one-dimensional linear subspace MATH of MATH . If MATH to first-order at MATH, then the same is true to all orders on an analytic neighborhood of MATH. |
math/0011244 | Harmonic decomposition of relative forms over MATH and all parts of REF and of REF can be carried out over MATH. |
math/0011244 | Let MATH. By REF we have MATH . But the left-hand expression is isomorphic to MATH . And the right-hand expression is isomorphic to MATH . Comparing eigenspaces, the theorem now follows. |
math/0011244 | When MATH this is REF . The proof of the general case follows immediately from REF . |
math/0011244 | CASE: Let MATH be a general point of MATH. Then by REF we have, for MATH the inequality MATH . Suppose MATH is the harmonic representative for a non-zero element of MATH . For general MATH, let MATH . Now the codifferential of the NAME map MATH is given by MATH so that, for a general point MATH, MATH has rank MATH . So MATH has dimension MATH . Let MATH be a basis of MATH, then we can assume that, for example, MATH is a partial basis for the image of MATH in MATH. Writing the non-zero form MATH in terms of a completion of this partial basis to a full basis of MATH, the condition MATH implies that MATH for some form MATH. Thus MATH, that is, MATH . So MATH . CASE: Suppose MATH . Then MATH so that there exists a non-trivial holomorphic MATH-form MATH with coefficients in MATH. If MATH there is a MATH such that MATH at the chain level and so also at the level of cohomology since holomorphic forms are automatically harmonic. So MATH and therefore MATH . |
math/0011244 | Throughout this proof let MATH . Let MATH be a general point of MATH. Let MATH be an appropriate neighborhood of MATH such that MATH is smooth and defined on MATH by the equation MATH. Let MATH be such that MATH . Since MATH, it follows that MATH . By REF , the map MATH is surjective. So, by the MATH-antilinear isomorphism REF , the map MATH is surjective. The assertion now follows from a local computation. Let MATH, we may assume that MATH is also smooth and defined by the equation MATH. Choosing local parameters MATH for MATH, we may write a section MATH locally in the form MATH where MATH for MATH and MATH. Therefore, evaluating at MATH, since MATH (where MATH is the index of ramification of MATH along MATH), we have MATH . Since MATH is general, it follows that MATH. |
math/0011244 | Following CITE, consider a log resolution MATH of the pair MATH, that is, MATH is obtained from MATH by a sequence of blowing-ups with smooth centers lying over MATH and MATH is a divisor with simple normal crossings. Notice that MATH and MATH and, for MATH we have MATH. We wish to define the multiplier ideal sheaf MATH associated to the MATH-divisor MATH. Define MATH . Notice that if MATH for all MATH, then MATH . Let MATH . So we must show that for all MATH, MATH REF MATH. Then MATH is an ample MATH-divisor. Therefore by CITE we have that MATH for all MATH and MATH. The quantity MATH is constant. It is impossible that MATH, since then MATH for all MATH and MATH so that, by CITE, MATH would have to be the zero sheaf. CASE: MATH . We have MATH. Let MATH . We wish to apply the results of REF where MATH so that MATH and therefore MATH . So, by REF and the NAME spectral sequence for MATH, MATH . So it follows from CITE or REF that MATH . If MATH then MATH for all MATH and MATH and we have a contradiction as above. So, let MATH be an irreducible component of MATH, and denote by MATH the corresponding map of abelian varieties. We complete the proof by showing that MATH . To prove MATH suppose that MATH . Then MATH is the union of two disjoint subsets MATH where MATH consists of those MATH such that MATH . Notice that MATH since MATH is not ample. Now by REF so that it is impossible that MATH that is, that MATH . Thus by REF for each MATH and MATH all sections of MATH vanish on MATH. Thus we have that MATH for each MATH. So computing with numerical equivalence classes on MATH we have MATH . Let MATH be the NAME class of an ample divisor on MATH. Then from MATH have on MATH that MATH . By MATH the expression before the equals sign in MATH is non-negative. On the other hand the expression after the equals sign is negative since since MATH and MATH for every MATH. |
math/0011244 | If MATH, the assertion is clear. We may therefore assume that MATH. We may also assume that MATH, since if this is not the case, it suffices to replace MATH by MATH, MATH by MATH and MATH by MATH. Let MATH. Define MATH . Therefore, we have that MATH. Let MATH and MATH. It follows for MATH that MATH . Now there is an abelian quotient MATH of the abelian variety MATH such that MATH is the pull-back of an ample divisor MATH and MATH is the pull-back of a divisor MATH on MATH. Let MATH denote the image of MATH in MATH . Now one can apply REF with MATH to obtain a section of MATH vanishing on MATH. Since MATH contains MATH, the theorem follows. |
math/0011244 | By REF (with MATH), for all MATH, one has MATH. It follows that MATH vanishes on all the translates of the cosupport of MATH. The only way this can occur is if the cosupport of MATH is empty, that is, if MATH. |
math/0011244 | NAME 's celebrated result MATH says that the NAME spectral sequence for MATH degenerates at MATH. But the filtration in the NAME spectral sequence is compatible with that of the spectral sequence defined in the proof of REF. |
math/0011244 | Let MATH be a covering of MATH by coordinate disks and MATH the restriction of this covering to MATH. We construct a MATH partition-of-unity MATH subordinate to the induced covering of MATH. Recall that MATH is given with respect to the trivialization MATH by holomorphic local patching data MATH where MATH and MATH . Notice that, if MATH and MATH are three open sets of the cover which have non-empty intersection, then, for all MATH, MATH . Define the mapping MATH over MATH by MATH . This map is well defined since, over MATH we have MATH and so MATH . |
math/0011245 | Let MATH be a facet in the boundary of MATH such that MATH, and let MATH be an alcove in MATH. We claim that MATH. There exists an alcove MATH, MATH which has a wall in a boundary hyperplane of MATH. Let MATH be the simple reflection corresponding to this wall. Then MATH (see for example, REF , or REF); but then also MATH, which shows the claim for MATH. The claim will follow inductively if we can prove it for any alcove MATH which shares a wall with an alcove MATH for which it has already been established. Let MATH be the simple reflection corresponding to such a wall, and observe that MATH. Then MATH, from which one easily deduces the claim for MATH. Assume now that MATH. Then necessarily any alcove MATH has to be in MATH. If MATH, we get MATH from REF and the definitions of MATH and of MATH, see REF. If MATH for some MATH, then MATH using REF. To prove the second claim, observe that because of MATH we get MATH where the first equality follows from the definition of MATH REF , the second equality from REF , and the last equality from REF for MATH and REF for MATH. |
math/0011245 | To prove REF , observe that MATH is equal to the NAME element MATH of the longest element MATH in MATH. In particular, it is self-dual and hence so is MATH. Hence this product can be written as a MATH linear combination of self-dual elements MATH; in fact the same statement is true for MATH for any MATH. We will show the second statement of the claim more generally for MATH for any MATH by induction on the length MATH of MATH. If MATH, MATH is a simple reflection of MATH. In this case the claim follows from the original proof by NAME and NAME: in fact, MATH is a linear combination of MATH with MATH majorized by the larger of MATH or MATH. Either of those elements contains MATH in its boundary, and hence is majorized by MATH. For the induction step, let MATH, with MATH and MATH. Then the claim follows by applying the induction assumption twice for MATH. Now let MATH denote MATH. Let MATH be the set of simple reflections MATH. By definition of MATH and MATH, we have MATH for MATH. But then MATH and MATH for all MATH, using basic properties of NAME elements (see REF ). But then also MATH for all MATH from which one easily concludes MATH. Let MATH be an alcove, and let MATH be the facet in the boundary of MATH which is conjugate to MATH. Then it follows from REF that MATH is equal to a multiple of MATH, and hence also MATH is a linear combination of MATH's. On the other hand, we have MATH by the previous paragraph of this proof. Hence already MATH itself is a linear combination of MATH's. From this follows REF , as the coefficient of MATH is equal to MATH. Using REF , we get MATH where MATH is the unique facet conjugate to MATH which is contained in the closure of MATH. This proves REF , and REF follows from MATH (see REF). |
math/0011245 | Let MATH . , and let MATH and MATH be the facets obtained by applying MATH to MATH and MATH. Observe that MATH and that the line segment from MATH to MATH is contained in MATH. Moreover MATH and MATH have right stabilizers MATH and MATH respectively. It follows from REF that MATH where the summation goes over all facets MATH conjugate to MATH which contain MATH in their boundary. Using REF correspondence between the facets MATH in the sum above, and the points MATH in the orbit of MATH under the right action of MATH, we obtain similarly MATH . The claim about the coefficient of MATH follows from the last two equations, and the definition of MATH, see REF. As MATH is self-dual, so is MATH. From this follows that MATH is self-dual, by the last two equations and the definition of duality on the module MATH (see end of REF). |
math/0011245 | It was shown in REF that MATH is a MATH- linear combination of MATH-s. Moreover, using the fact that MATH is in MATH, we check easily that MATH has coefficient REF in MATH. The claim follows from this and REF. |
math/0011247 | One can symbolically express MATH . Let MATH satisfy REF : MATH . We have MATH by REF . This implies that each MATH is MATH-invariant because of REF : MATH . Let MATH. Since MATH each MATH belongs to MATH. Write MATH for some MATH. Let MATH be the orthogonal reflection about MATH. Then MATH . Therefore one has MATH . Hence MATH is divisible by MATH and thus MATH. This implies MATH because MATH was arbitrary. Recall NAME 's REF . Since MATH, we can conclude that MATH form a basis for the MATH-module MATH. We also have MATH . |
math/0011247 | CASE: Let MATH. By REF , one has MATH . Thus the matrix MATH is MATH-invariant. Next we will show that MATH is independent of choice of the basis MATH for MATH. Suppose that a new basis MATH for MATH is connected with the old basis MATH through an invertible matrix MATH as REF : MATH . The new objects, which are defined using the new basis, will be denoted by MATH etc. By REF , one has MATH . Therefore MATH is independent of choice of MATH. Let MATH. Then we may choose an orthonormal basis MATH for MATH without affecting MATH. Then MATH is the identity matrix. By REF , the assertion MATH holds. Thus each entry of the first row of MATH is divisible by MATH. Thus, outside the first column, each entry of MATH is divisible by MATH. Since det MATH each entry of MATH outside the first column, has no pole along MATH. The MATH-entry of MATH is equal to MATH by REF . Thus each entry, except the first, of MATH has no pole along MATH. Note that the first entry of MATH has pole of order at most MATH along MATH. It follows that each entry of MATH, outside the first column, has no pole along MATH. Each entry of the first column of MATH has pole of order at most MATH along MATH. Since each entry of the first row of MATH is divisible by MATH, each entry of the product MATH has pole of order at most one along MATH. So MATH has no pole along MATH. Since MATH was arbitrarily chosen one can conclude that MATH. For any MATH, we have MATH . Thus MATH is an anti-invariant CITE and hence lies in MATH. Therefore MATH. This proves REF . CASE: By REF , we have MATH . |
math/0011247 | CASE: The MATH-entry of MATH is equal to: MATH which is equal to the MATH-entry of MATH. REF follows from: MATH . CASE: Let MATH in REF to get MATH. Apply MATH . CASE: By REF , and REF , one has MATH . |
math/0011247 | Since MATH and MATH by REF , each entry of MATH lies in MATH. Thus MATH by REF . Multiply MATH on the right side and we get MATH . Since MATH by REF , we finally have MATH . |
math/0011247 | Using REF repeatedly we have REF . Write MATH for simplicity. The shape of MATH is known CITE as follows: CASE: If MATH is not of type MATH with MATH even then MATH where MATH . CASE: If MATH is of type MATH with MATH then the MATH block in rows and columns MATH of the matrix REF - the center of the matrix - is to be replaced by a MATH block MATH with constant entries, where MATH and MATH . REF still holds true outside the MATH block MATH. Therefore, by REF , we can verify the following statements: CASE: If MATH is not of type MATH with MATH even then MATH . CASE: If MATH is of type MATH with MATH then the MATH block MATH in rows and columns MATH of the matrix MATH is MATH with constant entries, where MATH. REF holds true outside MATH. In both cases we conclude MATH. |
math/0011247 | Only the third formula needs to be verified. Let MATH. Compute MATH . |
math/0011248 | We check only the commutativity of the cyclic operators and the cylindrical condition. CASE: MATH . CASE: MATH . |
math/0011248 | We show that MATH commutes with the cyclic and simplicial operators: CASE: MATH . CASE: MATH . With the same argument one can check that MATH where MATH and MATH . |
math/0011248 | One can check that MATH is a cyclic map and MATH , where MATH is defined by MATH . Precisely, MATH and MATH . If MATH is invertible then we can see that MATH and MATH will be represented as MATH . |
math/0011248 | We only check that MATH, the other identities are similar to check. We see that MATH . Therefore we have, MATH . |
math/0011248 | We can see that these operators are directly well defined on the equivariant space, that is, they commute with the action defined in REF . |
math/0011248 | Since MATH is semisimple, we have MATH for MATH and the spectral sequence collapses. The first row of MATH is exactly MATH . |
math/0011249 | . We say that an element of MATH is primitive if MATH for all MATH and MATH. For every MATH there is a primitive MATH such that MATH . The proof makes use of the two following claims: CASE: Assume that MATH is a subgroup of MATH, such that MATH and the bilinear form restricted to MATH is not trivial. Then there are primitive elements MATH such that MATH and MATH. CASE: Let MATH, MATH and MATH with MATH. Then MATH . Using induction and REF it is easy to prove: Let MATH be a subgroup of MATH. Then there is MATH, (where MATH may be REF), such that MATH generate MATH, MATH is linear independent and MATH, MATH . From this fact and induction the Theorem follows. |
math/0011249 | Let us consider all systems MATH, MATH, such that MATH, MATH. Between them we choose a system MATH with maximal MATH . Then MATH is a basis with the conditions that we need. |
math/0011249 | It is a consequence of REF . |
math/0011249 | Let MATH and MATH denote MATH and MATH respectively. Assume that MATH and MATH are strongly equivalent, then according to REF there exists a homeomorphism MATH, which induces an isomorphism MATH such that MATH. Since MATH is induced by a homeomorphism then MATH preserves the intersection form and induces an isomorphism MATH such that MATH and MATH. Hence, for MATH we have MATH . Assume now MATH . We consider some isomorphisms MATH and MATH, such that MATH and MATH, for any MATH and MATH. We note MATH, and MATH . Let MATH be the isomorphism given by MATH. Then, for every MATH, we have MATH . From REF follows that there is MATH such that MATH restricted to MATH is MATH. Consider now MATH . Since MATH then MATH comes from an isomorphism MATH sending the intersection form of MATH to the intersection form of MATH REF MATH . Then by a classical result of NAME in REF (see CITE, pg. REF), there is some homeomorphism MATH inducing MATH and MATH, and by construction MATH . Then by REF the actions MATH and MATH are strongly equivalent. |
math/0011249 | First we construct the action from the form and the numerical conditions. Applying REF we have a basis of MATH, MATH, such that MATH, MATH and MATH. Let MATH be the dual basis of the above one. Now consider a surface MATH of genus MATH and a basis of MATH. Then we construct the epimorphism MATH, defined by MATH if MATH if MATH and MATH if MATH if MATH . Then the epimorphism MATH defines a regular covering MATH with automorphism group MATH and the action of MATH on MATH satisfies MATH. Conversely if there is an action MATH such that MATH it is obvious that MATH. |
math/0011249 | Let us call MATH and MATH and MATH. Let MATH and MATH be the epimorphisms defined by the two actions, MATH be the image of MATH in MATH by MATH and MATH be the image of MATH in MATH by MATH. Assume that MATH and MATH. Since MATH then there exists an isomorphism MATH such that MATH . Then, using REF , and that MATH, there is an isomorphism MATH giving by restriction MATH and sending the intersection form of MATH to the intersection form of MATH. By [MKS, pag REF], there exists a homeomorphism MATH inducing MATH on cohomology. Then by REF , the actions MATH and MATH are strongly equivalent. The isomorphism MATH, defines an automorphism of MATH giving the weak equivalence between MATH and MATH . |
math/0011249 | Using NAME twists along curves around the branched points (see REF , pg. REF, move REF ) it is possible to obtain a basis MATH of MATH such that MATH and MATH . In the same way, we can construct a basis MATH of MATH such that MATH and MATH remark that by REF , MATH and MATH . By REF , then the fixed point free action of MATH on MATH given by MATH and the fixed point free action of MATH on MATH given by MATH are strongly equivalent. Then there exists a homeomorphism, preserving the orientations, MATH inducing on homology an isomorphism MATH and by the proof of REF we can construct MATH such that MATH, and MATH. We now consider a disc MATH on MATH containing MATH and a disc MATH on MATH containing MATH. Then we can modify MATH by composing with an isotopy in MATH in order that MATH, and MATH where MATH is a permuntation of MATH such that MATH . Now MATH defines an isomorphim MATH such that MATH, MATH then MATH. Hence the actions MATH and MATH are strongly equivalent. |
math/0011249 | It is similar to the proof of REF but using REF . |
math/0011251 | By symmetry, it is enough to show that MATH has a bigraded linear resolution. The residue class field MATH has a MATH-linear minimal free MATH-resolution MATH because MATH is NAME. Let MATH be the MATH-diagonal. Applying the functor MATH we get the exact complex MATH. By REF , the MATH module MATH is a direct sum of copies of MATH shifted by MATH. Thus MATH is a standard bigraded NAME algebra. Let MATH be the projection map and MATH the inclusion. Note that both maps MATH and MATH are bigraded homomorphisms. Since MATH is a ring epimorphism and since MATH, the map MATH is a bigraded algebra retract. We may apply a result from CITE to the bigraded situation. It yields that MATH where MATH is considered as a bigraded MATH-module. Since MATH and MATH are NAME, the equality of bigraded NAME series implies that MATH has a bigraded MATH-linear MATH-resolution. This concludes the proof. |
math/0011251 | Since the proofs of REF are similar, we only consider REF . Moreover, is enough to show that all modules MATH with MATH have linear resolutions. Let MATH be the bigraded minimal free MATH-resolution of MATH. Since MATH has a bigraded MATH-linear resolution, every free module MATH is of the form MATH with non-negative integers MATH. Observe that for MATH and MATH we have MATH. Applying the functor MATH, we obtain an acyclic complex MATH where MATH . By REF , all occurring shifts MATH are at most MATH. Similarly, let MATH be the minimal free resolution of MATH. Then for MATH and MATH we observe that MATH and the shifts in MATH are bounded by MATH. To conclude the proof we show by induction that MATH for all MATH and MATH. First we use induction on MATH. The modules MATH are generated in degree MATH, thus MATH. Let now MATH. We apply induction on MATH where MATH. For MATH it follows that MATH and MATH and therefore trivially MATH. Let now MATH. Then MATH or MATH. We discuss the case MATH first. In order to apply REF to the exact complex MATH we show that MATH for all MATH and MATH. Observe that MATH is a direct sum of MATH copies of MATH. Since MATH, the induction hypothesis on MATH implies that MATH for all MATH. For MATH and MATH, we have MATH where the first inequality holds because MATH for all occurring MATH and the second inequality holds by induction on MATH. Now REF implies that MATH. If MATH and MATH the argument above similarly applies to the complex MATH. |
math/0011251 | Let MATH be the minimal free MATH-resolution of MATH. Since MATH we have MATH for some non-negative integers MATH. For the proof of REF we restrict to the case MATH. The other cases follow similarly. By REF we observe that MATH . Now the claim follows from REF . For REF we also restrict to the case MATH and MATH. Use REF to observe that MATH . The claim follows from a case by case computation using MATH and REF . Then the upper bound for MATH follows from the fact that MATH decomposes into the finite sum MATH . |
math/0011251 | Note that a graded MATH-algebra MATH is NAME if and only if MATH. Since MATH and MATH, the claim follows from REF . |
math/0011251 | Let MATH be the MATH-diagonal. Then MATH, and MATH. Thus MATH is NAME and, by REF , the module MATH has a linear MATH-resolution. |
math/0011251 | We use some results about MATH. It is known that MATH is a NAME ideal and that MATH (see CITE). Thus MATH is quadratic. Since MATH we obtain that MATH is stable (see CITE) and therefore satisfies condition MATH. |
math/0011251 | With the condition MATH it is easy to see that MATH for all monomials MATH of MATH with MATH. |
math/0011251 | By REF , we may assume that the defining ideal MATH of MATH has a quadratic NAME basis MATH with respect to the reverse lexicographic term order induced by MATH and that MATH satisfies REF MATH. It is easy to see that MATH has a presentation MATH where MATH. Let MATH denote the reverse lexicographic term order on MATH induced by MATH. We will show that the set MATH is a NAME basis for MATH with respect to MATH which concludes the proof of the theorem. Let MATH be a bihomogeneous polynomial of degree MATH. Then MATH. We show that MATH is divided by some MATH with MATH. Let MATH where MATH and MATH . If there exist indices MATH such that MATH then MATH divides MATH which is the claim. Otherwise we have MATH. Let MATH denote the homomorphism from REF . Since MATH, it follows that MATH. By REF , we have MATH where MATH. Since MATH is a NAME basis for MATH, there exists a MATH such that MATH divides MATH. By REF MATH, we may assume that MATH, if MATH, or MATH, if MATH. Now MATH or MATH divides MATH. |
math/0011251 | Let MATH the standard graded polynomial ring and MATH. We may assume that the defining ideal MATH of MATH does not contain linear forms. Then MATH is generated in degree MATH. Let MATH be the graded maximal ideal of MATH. We denote by MATH the NAME complex of the sequence MATH. Let MATH be the first homology group of this complex. Recall that MATH for some bihomogeneous ideal MATH and that MATH is generated by the residue classes of all monomials in degree MATH. We consider MATH as a MATH-module generated in degree MATH. For MATH, there exists the downgrading homomorphism MATH mapping a residue class of MATH to the residue class of MATH (see CITE). Note that it does not matter which of the factors MATH is replaced by MATH. To show that MATH for all MATH we use induction on MATH. For MATH, we have MATH which has a linear resolution because MATH is NAME. Let now MATH. We have the short exact sequence MATH where MATH. By CITE, MATH is a subquotient of the module MATH for some integer MATH. The module MATH is annihilated by MATH. Since MATH is generated in degree MATH, it follows that MATH is generated in degree MATH. Therefore, MATH for some integer MATH and MATH has a MATH-linear MATH-resolution because MATH is NAME. By induction hypothesis, MATH has MATH-linear MATH-resolution. Thus, by CITE, the module MATH has a MATH-linear MATH-resolution. The assertion follows when we apply the long exact sequence of the functor MATH to the sequence MATH. |
math/0011251 | Let MATH be the MATH-diagonal. Then MATH and MATH . Thus, by REF , the ideal MATH has a linear MATH resolution. |
math/0011251 | Let MATH (respectively, MATH) be the minimal free resolution of MATH over MATH (respectively, MATH). Then the tensor product MATH gives a minimal free resolution of MATH over MATH. Thus if MATH and MATH are NAME, MATH is NAME. Now the NAME product MATH is the MATH-diagonal MATH which is NAME by REF . Let MATH be a positively graded NAME algebra. Consider MATH as a standard bigraded algebra where all generators have degree MATH. Then MATH is a diagonal of MATH and, by REF , MATH is NAME. |
math/0011251 | Consider MATH as a bigraded algebra generated in degree MATH. Let MATH be the MATH-diagonal of MATH. Then MATH and, as a MATH-module, we have MATH. By REF , the claim follows. |
math/0011252 | Let MATH be an element of MATH for some object MATH. Then, MATH if and only if MATH. Let us assume that MATH. Since MATH contains the face opposite REF, we must have that MATH. Then MATH has the form MATH and so MATH is the image under MATH of a unique element of MATH, that is, MATH. Thus, MATH is the pushout claimed. |
math/0011252 | This follows from REF on letting MATH be the face opposite REF (MATH if MATH). |
math/0011252 | Letting MATH in REF (notice that this requires MATH), we have the following pushout diagram. MATH . But then standard results about simplicial sets say that MATH is a weak equivalence. |
math/0011252 | In fact, we shall show that, for every object MATH of MATH, the functor MATH is a deformation retraction. Let MATH. Tracing through the definitions, we see that an object of MATH is a pair MATH where MATH in MATH and MATH. A map MATH is given by a pair MATH where MATH with MATH and MATH. With this notation, MATH is given by MATH and MATH. Now define MATH by MATH on objects and MATH on maps. We have MATH, and we can define a natural transformation MATH by MATH. Thus, MATH is the inclusion of a deformation retract and MATH is a weak equivalence. |
math/0011252 | This statement is part of CITE, but we reprove it here. Let MATH be an object of MATH. We begin by describing the simplicial set MATH. We have the following pullback diagram. MATH . Thus, a MATH-simplex of MATH is given by two compatible maps MATH and MATH. The map MATH is determined by a sequence MATH where MATH is the composite map MATH. The map MATH is then an element of MATH. By REF we can describe such an element by a sequence MATH where MATH and MATH for MATH. In summary, we can write a MATH-simplex of MATH as MATH with the compatibility conditions above. The map MATH is then given by MATH . We define MATH on a MATH-simplex MATH by MATH where the maps MATH are all the identity. Clearly MATH, and it is straightforward to define a simplicial homotopy from MATH to the identity (the construction is very similar to standard arguments involving the bar construction). Thus MATH is a weak homotopy equivalence. |
math/0011252 | We first check the special case MATH. In this case, it is easy to see that MATH is the inclusion MATH, which is a right anodyne extension since MATH. If we let MATH and let MATH denote the set that results when we apply MATH to each map in MATH, then we have shown that every map in MATH is a MATH-cofibration. To get to the general case we note first that, by adjunction, liftings in a diagram as on the left below are in one-to-one correspondence with liftings in the diagram on the right. MATH . In particular, if MATH has the RLP with respect to MATH, then MATH has the RLP with respect to MATH (so is a NAME fibration). Since every map in MATH is a MATH-cofibration, every map in MATH has the LLP with respect to MATH-inj, the set of MATH-injectives (the right fibrations). By adjunction, every map in MATH has the LLP with respect to MATH. It follows that every MATH-cofibration has the LLP with respect to MATH, hence, by adjunction, that every map in MATH has the LLP with respect to MATH-inj. But this says that, if MATH is an anodyne extension, then MATH is a right anodyne extension. |
math/0011252 | We first define MATH. Recall that MATH has MATH top-dimensional nondegenerate simplices MATH through MATH, where MATH is given by the sequences of vertices MATH, MATH, , MATH, MATH, , MATH. Corresponding to these are MATH top-dimensional simplices of MATH we shall call MATH through MATH. We define MATH to be MATH. We leave it to the reader to verify that this does define a simplicial map MATH with the stated properties, and that MATH is natural on MATH. We then extend MATH to all of MATH using the coend: MATH . |
math/0011252 | We show that MATH has the RLP with respect to every map MATH. Consider the following lifting problem: MATH . That this lifting problem can be solved for MATH is just the statement that the fibers, being acyclic, are nonempty. Let us now assume that MATH and think of MATH as MATH. We shall solve the lifting problem by first pulling it into a fiber, where the problem is solvable by assumption. To do this, we consider first the following lifting problem: MATH . The map MATH is the restriction of the composite of MATH with the contraction MATH of REF . We claim that the map on the left is a right anodyne extension. We see this by writing it as the composite MATH . The first of these maps is obtained by pushing out along MATH, which is a right anodyne extension by REF . The second map is obtained by pushing out along MATH which is a right anodyne extension by REF . Therefore, the composite is a right anodyne extension as claimed, and so we can find the map MATH in the diagram above. Now note that the image of MATH lies entirely in a fiber of MATH. By assumption, then, we can extend MATH over MATH. We now consider the following lifting problem: MATH . The map on the left is a right anodyne extension by REF , so we can find the lift MATH. The restriction of MATH to MATH is then the solution to our original lifting problem. |
math/0011252 | Since right lifting properties are preserved by pullback and MATH is a right fibration, so is MATH. Let MATH be the deformation constructed in REF and consider the following lifting problem. MATH . Since MATH is an inclusion, it follows from REF that the map on the left is a right anodyne extension, hence we can solve this lifting problem. The lifted map is the desired deformation. |
math/0011252 | We shall use the ``recognition principle" CITE for cofibrantly generated categories. NAME gives six criteria to check to verify that there is a cofibrantly generated model category structure with the stated weak equivalences, with the stated set MATH as a set of generating cofibrations and the stated set MATH as a set of generating acyclic cofibrations. CASE: The collection of weak equivalences has the two-out-of-three property and is closed under retracts. This is straightforward to check given that it is true for ordinary weak equivalences of simplicial sets. CASE: The domains of MATH and MATH are small. This follows from the fact that all simplicial sets are small. CASE: Relative MATH-cell complexes are acyclic MATH-cofibrations. Every map in MATH is an inclusion, from which it follows that relative MATH-cell complexes are MATH-cofibrations. It suffices, then, to check that every map in MATH is a weak equivalence over MATH. But this is the content of REF . (The case MATH is easy.) REF MATH-injectives are weak equivalences. Let MATH and MATH, and let MATH be a MATH-injective over MATH. Then MATH is an acyclic NAME fibration. If MATH is any vertex in MATH, a diagram chase shows that the following is a pullback diagram. MATH . It follows that MATH is an acyclic NAME fibration for every MATH, hence that MATH is a weak equivalence over MATH. CASE: Acyclic MATH-injectives are MATH-injectives. Let MATH and MATH, and let MATH be an acyclic MATH-injective over MATH. By the inclusion in MATH of the maps MATH, it follows that MATH is a NAME fibration for every vertex MATH in MATH. On the other hand, we have the following commutative diagram, in which the horizontal maps are inclusions of deformation retracts (by REF ) and the map on the right is a weak equivalence by assumption. MATH . It follows that MATH is a weak equivalence, hence an acyclic NAME fibration, for each MATH. From this we can conclude that, for each vertex MATH of MATH, MATH is an acyclic NAME complex. We can now appeal to REF to conclude that MATH is an acyclic NAME fibration, hence is a MATH-injective. Thus, by CITE, MATH is a cofibrantly generated model category with the stated weak equivalences and generating cofibrations. What remains to show is the characterizations of cofibrations and fibrations over MATH. But, it is obvious that the MATH-cofibrations are just the usual cofibrations of simplicial sets, that is, inclusions. It is also clear that the MATH-injectives are precisely the right fibrations for which each map of fibers MATH is a NAME fibration. |
math/0011252 | We show first that MATH preserves cofibrations and acyclic cofibrations. To show that MATH preserves cofibrations, it suffices to show that it takes each generating cofibration to a cofibration. But, REF shows that, if MATH and MATH is the composite MATH, we have the following pushout diagram. MATH . From this it is clear that MATH is a cofibration, so MATH preserves cofibrations. On the other hand, MATH preserves weak equivalences by definition, so MATH also preserves acyclic cofibrations. Therefore, MATH is a NAME adjunction. To show that MATH is a NAME equivalence, we need to show that, if MATH is cofibrant (which is always true) and MATH is fibrant, then a map MATH is a weak equivalence if and only if its adjoint MATH is. Since MATH is a weak equivalence if and only if MATH is, it suffices to show that the counit MATH is a weak equivalence, but this was done in REF . |
math/0011254 | This is just an adaptation of the proof of MATH in CITE. Assume by contradiction that MATH. A fortiori MATH, so that the integral in REF would actually converge in MATH. Now let MATH with MATH. If MATH is the order of the pole, then we have MATH for some MATH. On the other hand the little MATH condition implies there is a MATH such that MATH for MATH, so that splitting the right-hand side integral in REF as MATH we obtain MATH which contradicts REF . |
math/0011254 | It is obviously enough to deal with only one of the above relations. So assume that REF is false. Then for some MATH there is a MATH such that MATH . Therefore MATH is not singular at MATH, but clearly MATH is also the abcissa of convergence of the left-hand side integral above, which contradicts the theorem that the NAME transform of a positive measure has a singularity on the real axis at the abcissa of convergence REF . |
math/0011254 | It is obviously enough to consider that MATH. By hypothesis there exists some MATH, and an unbounded set MATH such that MATH . Now take an arbitrary MATH. It is easy to see that there exists MATH such that MATH and MATH so that MATH . But NAME 's inequality gives MATH which, introduced in REF , yields MATH . |
math/0011254 | As in NAME 's monograph CITE we write for MATH . This proves REF . Proceed likewise with MATH to obtain REF . Now using the relation REF between MATH, MATH and MATH subtract the preceding two NAME transforms to get REF . Finally, from REF and the trivial MATH we deduce MATH. Next note that MATH is continuous and piecewise differentiable, which justifies the following integration by parts at least for MATH . Now apply REF to arrive at REF . |
math/0011254 | The prime number theorem and REF yield REF . Decomposing the integral in REF in the intervals MATH one gets REF . Letting MATH in REF yields REF . The absolute convergence follows from the elementary estimate MATH. |
math/0011254 | Take MATH. Then MATH by REF , so the divergent integral REF yields MATH . |
math/0011254 | An extension of NAME 's well-known criterion for the NAME hypothesis is that REF is equivalent to MATH (see REF ), so choose MATH with MATH and it is obvious how REF implies REF . Now we prove that not REF implies not REF . So assume there is a MATH with MATH and MATH. Then by REF MATH but (probability space) MATH since MATH. |
math/0011254 | Let MATH and MATH, then splitting the range of integration at MATH in REF we get MATH . The result now emerges from the well-known, elementary NAME inequalities (see REF ). |
math/0011254 | Fix MATH. For any bounded interval MATH is a proper NAME integral for each MATH. Let MATH, and MATH . The NAME sums MATH and MATH-for each MATH. Furthermore it is trivial to see that MATH for all MATH, whereas MATH when MATH, so that MATH . Hence MATH. By REF we conclude MATH which the time honored density argument and the continuity of MATH convert into REF , and, a fortiori, MATH. To finish the proof of REF we need to show that each function MATH, MATH is in MATH. This is achieved as follows: For MATH take MATH. Clearly MATH . By REF MATH, and the above inequalities show it converges in MATH-norm to the function MATH. If MATH the modification to the above proof is obvious. |
math/0011254 | Here we denote MATH if the statement MATH is true, otherwise MATH. We start from the well-known elementary identity MATH which we multiply by MATH and integrate thus: MATH . |
math/0011254 | The upper limit of integration in REF can trivially be substituted by MATH, so we get MATH from REF and the (absolute) convergence of the last integral, again due to MATH. Now make the change of variables MATH, and in the formula obtained substitute MATH. |
math/0011254 | If MATH, then MATH by REF , and REF . |
math/0011254 | If MATH, then MATH as remarked in REF . Then by the easy sufficiency part of the NAME REF MATH for MATH, and this implies by REF that MATH for all MATH. |
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