paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9912193 | Consider a sequence MATH, MATH. Using Main Lemma construct corresponding sequences of MATH, where MATH and MATH such that MATH is MATH-close to MATH, MATH and MATH close to MATH for every MATH. Then MATH is uniformly NAME. Since MATH is metrizable compactum (actually we assume that MATH is NAME cube), there exists cont... |
math/9912193 | Construct a sequence MATH such that: CASE: MATH. CASE: MATH. CASE: Pair MATH satisfies condition of REF . Let MATH. Check that pair MATH satisfies requirments of lemma. Consider MATH. Let MATH (since MATH is compact, we may choose finite refinment of MATH, but it is not essential for further consideration). Let MATH an... |
math/9912193 | Since MATH is MATH-compactum, for MATH pick MATH such that pair MATH satisfies conditions of REF for space MATH. Further, for MATH choose MATH such that pair MATH meets conditions of REF . Check that pair MATH satisfies condition a. Consider MATH. By our choice of MATH there exists MATH such that MATH. Since MATH there... |
math/9912193 | Let we are given a MATH and MATH. We say, that cover MATH and sequences MATH, MATH form canonical system with respect to MATH and MATH, if the following conditions are satisfied (we use notation of REF ): CASE: MATH. CASE: MATH is canonical refinement of MATH with respect to MATH and MATH. CASE: MATH is canonical refin... |
math/9912199 | The proof is by induction on the number of simplices of MATH. If MATH is a disjoint union of vertices MATH, then MATH is a bouquet of MATH copies of MATH (see REF ). In degree zero MATH is just MATH, while in degrees MATH it is isomorphic to MATH. Therefore, MATH, where MATH is the ideal generated by all square free mo... |
math/9912199 | REF shows that the second assertion follows from the first one. To prove the first assertion we mention that if a point MATH has MATH, then MATH is a simplex of MATH, hence MATH. |
math/9912199 | We construct a retraction MATH that is covered by an equivariant retraction MATH. The latter would be a required homotopy equivalence. The retraction MATH is constructed inductively. We start from the boundary complex of a MATH-simplex and remove simplices of positive dimensions until we obtain MATH. On each step we co... |
math/9912199 | The retraction MATH constructed in the proof of REF is equivariant with respect to the MATH-actions on MATH and MATH. Since MATH, the corollary follows. |
math/9912199 | Let MATH denote the orbit map for the torus action on the moment-angle complex MATH (see REF ). For each subset MATH denote by MATH the following subset of the poly-disk MATH: MATH, where MATH is the disk MATH if MATH, and MATH is the boundary MATH of MATH if MATH. Thus, MATH, where MATH. It is easy to see that if MATH... |
math/9912199 | We have MATH. Now, the corollary follows from REF . |
math/9912199 | Let us consider the commutative diagram MATH where the left vertical arrow is the induced fibre bundle. REF shows that MATH is homotopically equivalent to MATH. From REF we obtain that the cellular cochain algebras MATH and MATH are modules over MATH. It is clear from the proof of REF that MATH and MATH is the quotient... |
math/9912199 | One can make MATH a MATH-module by means of the homomorphism that sends REF to REF and MATH to REF. Let us consider the NAME resolution (see, for example CITE) of MATH regarded as a MATH-module: MATH where the differential MATH is defined as in REF . Since the bigraded torsion product MATH is a symmetric function of it... |
math/9912199 | The spectral sequence under consideration converges to MATH and has MATH . It is easy to see that the differential in the MATH term acts as in REF . Hence, MATH by REF . |
math/9912199 | See CITE. |
math/9912199 | For the proof of REF see CITE. To prove REF we just mention that MATH in MATH (see REF ). |
math/9912200 | If MATH, then MATH for some MATH. In this case we write MATH, where MATH and MATH, MATH, MATH. Then MATH . In both cases MATH. Assume that MATH. Then MATH and MATH . |
math/9912200 | Write MATH, MATH. Let MATH be a coefficient MATH. Then by CITE, MATH where MATH, MATH. Since MATH is lc (see CITE), MATH and we may assume that MATH. Using MATH one can easily show that in REF MATH (see CITE). If MATH for all MATH in REF, then, obviously, MATH. Otherwise MATH for some MATH and MATH for MATH in REF. The... |
math/9912200 | Follows by CITE. |
math/9912200 | Note that MATH is MATH-Cartier and numerically trivial over MATH. Put MATH. Then MATH, MATH and MATH is a MATH-complement for all MATH (by convexity of the lc property see CITE or CITE). Fix an effective NAME divisor MATH on MATH (passing through MATH) and put MATH. For MATH, define a function MATH and put MATH. Fix so... |
math/9912200 | Since MATH is non-exceptional, there is a MATH-complement MATH such that MATH for some MATH. Then one can apply REF . |
math/9912200 | Let MATH be a MATH-complement such that MATH and let MATH be a general hyperplane section of MATH passing through MATH. Since MATH does not contain the center of MATH, MATH and MATH for all MATH. Take MATH so that MATH is maximally lc. Then, as in the proof of REF , MATH for some MATH, a contradiction. |
math/9912200 | Follows by NAME 's lemma (see e. CASE: CITE). |
math/9912200 | First take a log terminal modification MATH of MATH (see CITE, CITE). Write MATH where MATH and MATH are proper transforms of MATH and MATH, respectively, and MATH is exceptional. One can take MATH so that MATH is reduced and MATH (see CITE, CITE). We claim that MATH cannot be nef over MATH. Indeed, write MATH . This g... |
math/9912200 | By REF, MATH is klt and anti-ample over MATH for sufficiently small positive MATH. Take MATH as the crepant pull-back MATH . In other words, MATH . From REF we obtain that MATH is klt CITE, anti-nef and anti-big over MATH. REF holds if MATH, that is, for MATH. |
math/9912200 | By REF such a boundary exists on the first step. If MATH is not nef over MATH, then MATH is also not nef over MATH. Put MATH . By REF this supremum is a maximum and is achieved on some extremal ray. Hence MATH is rational and MATH. Consider the boundary MATH. Then MATH is nef over MATH and MATH. We claim that MATH is a... |
math/9912200 | By REF and the Base Point Free Theorem, MATH is semi-ample. Thus for some MATH the linear system MATH defines a contraction MATH. For any curve MATH in a fiber we have MATH. Since the components of MATH generate MATH (see REF), we have that MATH for some component MATH of MATH. Hence MATH. |
math/9912200 | See the proof of REF. |
math/9912200 | By REF any MATH-complement of MATH can be extended to a MATH-complement of MATH. By REF we can pull-back complements of MATH under divisorial contractions because they are MATH-positive. Finally, note that the proper transform of a MATH-complement under a flip is again a MATH-complement. Indeed, the inequality in REF, ... |
math/9912200 | REF follows by REF . Note that MATH satisfies the conditions of our theorem (see REF ). By inductive hypothesis we may assume that there is a non-klt MATH-complement of MATH for MATH. The rest follows by REF . |
math/9912200 | By the Adjunction CITE it is sufficient to prove that MATH is not plt near MATH. Taking into account discussions above, we see that this is a consequence of REF below. |
math/9912200 | MATH follows by REF . The inverse implication follows by REF . |
math/9912200 | The inequality MATH follows by REF , so we show MATH. Let MATH be a non-klt MATH-complement of MATH. Then MATH. By REF MATH. Consider the crepant pull-back MATH and let MATH be the proper transform of MATH on MATH. Then MATH is a MATH-complement of MATH. |
math/9912200 | Let MATH be a non-klt MATH-complement with MATH. Then MATH (see REF). By definition of exceptional contractions MATH and MATH for all MATH. Hence MATH because MATH is an integer. Since MATH, MATH. Thus we can take MATH. |
math/9912200 | Put MATH and assume that MATH. REF gives us that there is a MATH-complement MATH of MATH, where MATH. Let MATH be the coefficient of MATH in MATH. By REF, MATH. Hence MATH and MATH is lc, a contradiction. |
math/9912200 | We will use notation of the proof of REF . Taking the fiber product with MATH-factorialization we can reduce the situation to the case when MATH is MATH-factorial. Note also that MATH is connected (because MATH is considered as a germ near MATH and MATH is irreducible). Consider the base change MATH where MATH is the N... |
math/9912200 | Notation as in the proof of REF . Take some MATH-complement MATH with MATH. Run MATH-MMP. For each extremal ray MATH we have MATH. Hence MATH is not contracted. At the end we get a model MATH with MATH-nef MATH. Since MATH is numerically trivial, for any divisor MATH of MATH, we have MATH (compare CITE). This shows tha... |
math/9912200 | First, replace MATH with its MATH-factorialization. Then as in REF we take MATH such that MATH is ample (but MATH is not necessarily lc). Next we put MATH, MATH so that MATH is lc but not klt (that is, MATH is the log canonical threshold MATH). Now, the proof is similar to the proof of REF . |
math/9912200 | A boundary MATH such as in REF exists by NAME (see e. CASE: CITE). |
math/9912200 | Let us consider the crepant pull-back MATH, MATH. Write MATH, where MATH is reduced, MATH, MATH have no common components, and MATH. We claim that MATH is a MATH-complement of MATH. The only thing we need to check is that MATH. From REF we have MATH. This gives us that MATH (because MATH is MATH-nef; see CITE). Finally... |
math/9912200 | Let MATH be a log resolution. Write MATH, where MATH is the proper transform of MATH on MATH and MATH. By the Inversion of Adjunction CITE, MATH is normal and MATH is plt. In particular, MATH is a birational contraction. Therefore we have MATH . Note that MATH, because MATH is smooth. It is easy to show (see CITE) that... |
math/9912208 | Let MATH be the ``additive" measure on MATH and let also MATH denote the space of NAME functions on MATH-matrices MATH over MATH. We shall regard MATH as a subspace of MATH. Consider the NAME transform MATH given by MATH . It is easy to see that MATH, where MATH. Let us now show that REF above hold. Without loss of gen... |
math/9912208 | Standard. |
math/9912208 | First, we can make use of NAME 's theorem to construct a proper, integral scheme MATH over MATH such that MATH. By NAME 's theorem we can blow up MATH to build MATH: MATH such that the union of the fiber MATH over MATH and the closure the zero locus of the differential form MATH on MATH is a normal crossing divisor. It... |
math/9912210 | First note that for any complex MATH and any complex MATH, the following Gaussian integral formula holds: MATH where the choice of the integration path MATH, described in the formulation of the theorem, is dictated by the convergence condition of the integral, and the square root is the analytical continuation from pos... |
math/9912210 | The left hand side of REF vanishes at MATH due to the factor MATH. This means that the integral in the right hand side vanishes as well. So, differentiating simultaneously the MATH-function in the left hand side and the integral in the right of REF with respect to MATH, then putting MATH, and rescaling the integration ... |
math/9912210 | At large MATH one can use the steepest descent method for evaluation the integral in REF . The only stationary point at MATH is separated from the integration path by a finite number of poles of the function MATH which are located at MATH, MATH. Thus, taking into account convergence at infinity, we can shift path MATH ... |
math/9912219 | We will proceed in two steps: First, the candidate for a NAME solution will be built up by nets of smooth solutions at fixed MATH. Second, existence and uniqueness in MATH will be proved by verifying additional estimates with respect to the regularization-parameter MATH. To begin with, we consider global solvability of... |
math/9912219 | It will suffice to consider the case of a left-sided mollifier (since a similar argument applies to the case of a right-sided mollifier). Fix some MATH. By our assumption on the support of MATH we have MATH for MATH and MATH sufficiently small (depending on MATH). Also, since MATH is globally bounded MATH so MATH by NA... |
math/9912219 | Assume that MATH is the distributional limit of MATH. Then from MATH it follows that MATH, where MATH. Also, the third line of the regularized system implies MATH, so MATH . Using the coordinate transformation MATH it follows that MATH is of the form MATH (that is, MATH) for some MATH, hence is contained in MATH. By RE... |
math/9912220 | Obviously, the function REF is well defined due to the MATH - boundedness of the contour MATH and since for all MATH there exist a MATH such that the estimate MATH holds. Thus, the proof of this lemma is reduced to the observation that the function MATH is holomorphic for MATH and coincides with MATH for MATH. REF is o... |
math/9912220 | One can prove this theorem making use of NAME 's Fixed Point Theorem (see CITE). |
math/9912220 | First we prove REF . Note that, according to REF , MATH . Thus, in view of the representations REF , the function MATH can be written as MATH which proves REF . The boundeness of the operator MATH for MATH is obvious. Further, we give a sketch of the proof that the factor MATH is a boundedly invertible operator if the ... |
math/9912220 | Let MATH. By REF and MATH . Therefore, the relation REF follows from the factorizations MATH and MATH . By the relation REF MATH belongs to the spectrum of the operator MATH if and only if MATH and MATH. From REF we conclude that MATH if and only if MATH. Again by REF the coincidence of the spectra of MATH and MATH fol... |
math/9912220 | The estimate in REF can be proved by using the relation REF following the proof of the estimate REF . This estimate yields that the sum MATH is a boundedly invertible operator in MATH. To prove REF we recall that due to the factorization REF the following factorization holds for for MATH: MATH where MATH and MATH are h... |
math/9912220 | The proof is carried out in the same way as the proof of the relation REF , only the path of integration is changed. |
math/9912221 | Since wide subcategories are closed under summands, MATH is closed under retracts. It is clear that MATH if and only if MATH. It remains to show that, if we have an exact triangle MATH and MATH, then MATH. We have a short exact sequence MATH where MATH is the cokernel of the map MATH, and MATH is the kernel of the map ... |
math/9912221 | By hypothesis, the complex MATH consisting of MATH concentrated in degree MATH is in MATH. Therefore the thick subcategory MATH generated by MATH is contained in MATH. But MATH is precisely the small objects in MATH. (This is proved in CITE for commutative MATH, but the proof does not require commutativity). Hence MATH... |
math/9912221 | Suppose first that MATH. This means that for every MATH, we have MATH for all MATH. Hence MATH. Thus MATH. Conversely, if MATH, then for every MATH we have MATH for all MATH. Thus MATH. |
math/9912221 | This corollary is true for any adjunction between partially ordered sets. For example, if MATH, then MATH. But MATH, so MATH. Furthermore, combining the counit and unit of the adjunction shows that MATH is contained in and contains MATH. The other half is similar. |
math/9912221 | Suppose first that MATH is the collection of finitely presented modules. Suppose MATH is a finitely generated left ideal of MATH. Then MATH is a finitely presented module, so MATH, as the kernel of the map MATH, is in MATH. Hence MATH is finitely presented, and so MATH is coherent. The collection of finitely presented ... |
math/9912221 | Write MATH as the cokernel of a map MATH. Recall that, given a module MATH, MATH denotes the complex that is MATH concentrated in degree MATH. Define MATH to be the cofiber of the induced map MATH. Then MATH is small and MATH (and MATH is the kernel of MATH). |
math/9912221 | Define an object MATH of MATH to have finite projective dimension if there is a MATH such that MATH for all MATH-modules MATH and all MATH with MATH. Here MATH is the function complex MATH in MATH obtained by replacing MATH by a cofibrant chain complex MATH quasi-isomorphic to it. (In the terminology of CITE, the model... |
math/9912221 | It may be possible to give a direct proof of this, but we prefer to use model categories. REF asserts that any cofibrant complex MATH that is small in the category MATH of chain complexes and chain maps, in the sense that MATH commutes with direct limits, will be small in MATH. Of course, MATH is small in MATH, but it ... |
math/9912221 | We have already seen that MATH. Suppose MATH. Then MATH is small by REF , so clearly MATH. Thus MATH. |
math/9912221 | Because MATH, we can construct the commutative diagram below, MATH where the rows are exact and MATH. The snake lemma then gives us an exact sequence MATH . Since MATH and MATH are in MATH, we find that MATH is in MATH. Similary, we find that MATH is an extension of MATH and MATH, so MATH. |
math/9912221 | Suppose MATH is a map of MATH. Then we have the commutative diagram below, MATH where the rows are exact and MATH. The snake lemma gives an exact sequence MATH . Hence MATH is an extension of MATH and MATH, both of which are in MATH by REF . So MATH. Similarly, MATH, which is in MATH by REF . |
math/9912221 | Note that the union that defines MATH is an increasing one, in the sense that MATH. This makes it clear that MATH is closed under extensions. REF implies that MATH is closed under kernels and cokernels. Therefore MATH is a wide subcategory. Since any wide subcategory containing MATH must contain each MATH, the corollar... |
math/9912221 | It suffices to show that MATH, where MATH is a wide subcategory of MATH. According to REF , MATH, where MATH is the collection of extensions of MATH. We prove by induction on MATH that if MATH acts trivially on some MATH, then in fact MATH. The base case of the induction is clear, since MATH. Now suppose our claim is t... |
math/9912221 | The second statement follows immediately from the first and REF . To prove the first statement, note that, by hypothesis and NAME 's REF , the map MATH is an isomorphism. It suffices to show that MATH is a surjection. So suppose MATH is a wide subcategory of finitely presented MATH-modules. Then MATH for some MATH. Sin... |
math/9912221 | Since MATH is left adjoint to MATH, MATH. But, given MATH, MATH is in MATH, and hence MATH. |
math/9912221 | We have a map MATH, defined as in the statement of the theorem. The composition MATH is proved to be an isomorphism, for MATH . NAME and commutative, in CITE (see also CITE). Since MATH is injective, we conclude that MATH, and hence also MATH, is an isomorphism. |
math/9912224 | For two partitions MATH we say that MATH, if there exists a set MATH of measure MATH, and such that MATH and MATH are either distinct, or coinside, for all MATH. Denote by MATH the partition MATH of MATH. Note that for any partition MATH, there exists a MATH, such that MATH. One can assume, by replacing MATH with MATH,... |
math/9912224 | Write MATH where MATH . The NAME measure on MATH disintegrates as MATH, where MATH is the induced NAME measure on the orbit MATH. For each MATH, let MATH be the set given in REF for MATH. Let MATH . Then MATH. Letting MATH gives the statement. |
math/9912224 | One clearly has MATH. On the other hand, for MATH a polynomial of fixed degree MATH, MATH where MATH denotes the integer part. This is because MATH . It follows that MATH and MATH. It follows after taking limits that MATH which in turn implies that MATH . Hence whenever MATH, we get MATH which in turn gives MATH. |
math/9912224 | This follows from the obvious inclusion MATH where MATH. |
math/9912224 | One has MATH the inequality now follows after taking limits. |
math/9912224 | We shall first prove that MATH. Fix MATH. Let MATH, and consider the set MATH . We claim that MATH. To show this, it is sufficient to verify (by enlarging the set MATH to contain all words in MATH of length at most MATH and also the unit of MATH) that for MATH, MATH, and for MATH and MATH, MATH, MATH or, equivalently, ... |
math/9912224 | Let MATH, MATH be given. Let MATH. Then by REF we have that MATH. Let MATH. Then we have in particular that MATH . Note that MATH. Let MATH, where MATH denotes the integer part of a number. Then for at least MATH numbers MATH in the set MATH, we have MATH. It follows that MATH is contained in the set MATH where MATH wh... |
math/9912224 | We may identify MATH with MATH in such a way that the projections MATH correspond to characteristic functions of the intervals MATH for some points MATH. Fix MATH; we can choose MATH in such a way that MATH for all MATH and all MATH. Choose integers MATH so that MATH are zero starting from some MATH, and and write MATH... |
math/9912224 | Fix MATH . By REF, for all MATH, there exist a subset MATH so that MATH and such that if MATH, then there exist MATH so that MATH and MATH is free up to order MATH and degree MATH in MATH from MATH with amalgamation over MATH. This implies that MATH . Passing to the limit gives MATH . The reverse inequality is obvious. |
math/9912224 | Note first that because of the freeness assumptions, MATH and MATH . The inequality MATH is then clear. Fix MATH. Choose MATH so that for each MATH, MATH . By REF, for all MATH, there exist a subset MATH so that MATH and such that if MATH, then MATH and MATH are free up to order MATH and degree MATH in MATH with amalga... |
math/9912224 | Let MATH be fixed. Then because of REF , we have that, having fixed MATH and MATH, there is a MATH and MATH, MATH for which MATH for all MATH and all MATH. The claimed inequality now follows from the definition of MATH. |
math/9912224 | By REF , given MATH, MATH, for MATH sufficiently large, there exists a permutation MATH, so that MATH where MATH. It follows that MATH, for all MATH. Given MATH, for MATH large enough, there exists a subset MATH, so that MATH, and so that elements of MATH are free from MATH in moments up to length MATH and degree MATH.... |
math/9912224 | Since MATH have f.d.a, given MATH, there are MATH, MATH so that for some MATH there exists an element MATH. By the assumed freeness between MATH and MATH, we find that given MATH, for all MATH and MATH sufficiently large, there is a subset MATH, so that MATH, and so that MATH . It follows that given MATH there exist MA... |
math/9912224 | It is sufficient to prove that, with the same notation as in the definition of MATH, MATH (we caution the reader that the quantity on the left involves entropy in the presence of an infinite number of variables). The inequality MATH is clear. To prove the opposite inequality, fix MATH, and choose MATH so that MATH for ... |
math/9912224 | The first inequality follows from MATH (note that MATH for MATH close to zero). The second inequality follows (under the assumptions of the hypothesis) from MATH and from MATH . (see CITE). The statement for one unitary follows from REF . |
math/9912224 | This follows from REF . |
math/9912224 | It is sufficient to prove the statement for MATH. Write MATH for MATH, MATH for MATH. Denote by MATH the unitary MATH. The We have, using the definition of MATH, REF and subadditivity of entropy that MATH . Since MATH commutes MATH, we have by REF that for some constant MATH independent of MATH, MATH since MATH. It now... |
math/9912224 | It is sufficient to prove the statement for MATH. Henceforth denote MATH by MATH. By REF , we have that MATH. Therefore, under the hypothesis of the Proposition, we have the inequality MATH . Thus, to conclude the proof, it is therefore sufficient to prove that MATH. Since MATH because MATH is free from MATH with amalg... |
math/9912224 | Note that under the orbit-equivalence assumptions, for all MATH, MATH, there exist polynomials MATH with coefficients from MATH having the form MATH, where MATH are orthogonal projections, so that MATH, where MATH and MATH. In particular, MATH commutes with MATH whenever MATH are unitaries and commute with MATH. Take M... |
math/9912224 | Note that the unitary MATH is free from MATH with amalgamation over MATH, and is independent from MATH. In our identification of MATH with MATH the unitary MATH is identified with the matrix MATH . Lastly, if MATH are each MATH-distributed as MATH, are independent from MATH and are MATH-free over MATH, then the matrix ... |
math/9912224 | We first prove the statement for MATH. We must prove that MATH . We shall prove that MATH, which is sufficient, since MATH . Choose cross-sections for the action of MATH on MATH, so that MATH and the action MATH has the form MATH for a cyclic permutation MATH or order MATH acting on MATH. Note that MATH in such a way t... |
math/9912224 | This follows from REF. |
math/9912224 | Denote by MATH the characteristic function of MATH. Then MATH commutes with MATH and MATH. By REF , we obtain that MATH . Note that MATH has the same MATH-valued distribution as MATH, where MATH is independent from MATH and free from MATH with amalgamation over MATH, and has the same MATH-distribution as free Brownian ... |
math/9912224 | We have MATH since MATH are free with amalgamation over MATH. |
math/9912224 | The first and second properties follows from the additivity of MATH for families of unitaries which are MATH-free over MATH, and from REF . The last property follows from the fact that for any finite graphing MATH, denoting by MATH the equivalence relation generated by MATH and by MATH the canonical unitary implementin... |
math/9912224 | We have that for each MATH, MATH because MATH are free from MATH with amalgamation over MATH. The rest follows from REF . |
math/9912224 | Note that if MATH is a generator and MATH, then MATH is also a generator. It is thus sufficient to prove the Proposition for the case that MATH for a single unitary MATH. If we replace MATH with MATH, then MATH, since MATH. Let MATH be fixed. Then there exists a projection MATH, MATH and a unitary MATH, so that MATH an... |
math/9912224 | Let MATH be a family so that MATH. Then MATH because MATH, so that REF applies. Thus, by definition of MATH, we get that MATH . |
math/9912224 | Using the fact that MATH is a generator and REF , we find that for any family MATH, MATH. Combining this with REF gives MATH. It follows that MATH. It follows that MATH since MATH for sufficiently large MATH. |
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