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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 continuous MATH. Obviously, MATH is selection of MATH. |
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 and MATH. Consider MATH, which is canonical refinement of MATH in the sense of REF . Let MATH. Using REF of canonical refinement, for each MATH find MATH such that MATH. Notice that generally speaking, MATH and MATH may coincide for MATH. Finally, consider related to MATH sets MATH and MATH, introduced in REF. We are ready now to construct map MATH. Using induction by MATH construct a sequence of map MATH such that: CASE: MATH. CASE: MATH. CASE: MATH is MATH-close to MATH. CASE: MATH for each MATH. For MATH define MATH letting MATH. Observe, that MATH is defined correctly and continuous on MATH (see properties G REF - G REF on the page REF). By our choice of MATH, MATH satisfies REF g - REF g. Assuming that MATH has been already constructed, let us construct MATH. To accomplish this it is enough (see G REF on the page REF) to define MATH on each non-emty MATH for each MATH such that MATH. Fix MATH such that MATH. Consider MATH. Let MATH. Obviously, MATH is closed subset of MATH. Since MATH for MATH (recall that we consider MATH) by our choice of MATH we have MATH. Therefore, by REF g we conclude that MATH. Hence by our choice of MATH, map MATH has an extension MATH such that MATH. Let MATH. Observe, that MATH is continuous (see property G REF on the page REF). Check that MATH satisfies REF - REF . Indeed, REFg and REFg are met by construction. Further, we have MATH, therefore, since MATH for every MATH, REF is also met. Finally, our choice of MATH and MATH coupled with REF for MATH and just checked REF for MATH (as well as the choice of MATH) yields REF for MATH. Since MATH (see property F REF on the page REF) we complete the proof letting MATH. |
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 exists extension MATH. Therefore, by our choice of MATH, there exists MATH such that MATH. The last fact implies that MATH. CASE: Consider MATH. For MATH pick MATH such that pair MATH satisfies conditions of REF for space MATH. Let MATH. Observe, that MATH. Further, let MATH. For MATH find MATH as in REF , applied to space MATH. We may assume that MATH. For MATH pick MATH as in REF . Check, that MATH and MATH satisfy our requirments. Consider MATH such that MATH. By the choice of MATH there exists MATH, which is MATH-close to MATH and hence MATH-close to MATH. This fact implies, that MATH, which, in turn, implies by the choice of MATH, that MATH has an extension MATH such that MATH. Since MATH and MATH are MATH-close, by our choice of MATH we may now conclude that MATH has an extension MATH such that MATH is MATH-close to MATH. Finally, by the choice of MATH, MATH and MATH. |
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 refinement of MATH. CASE: MATH such that MATH. CASE: MATH such that MATH. Note, that canonical system exists for each MATH. Note also, that some of MATH may coincide. Finally, observe, that since MATH is MATH collection, we may assume without loss of generality that MATH and MATH which REF provide us with for every MATH do not depend on MATH. CASE: Fix MATH. Construct sequence MATH such that MATH and for each MATH pair MATH satisfies conditions of REF for any MATH. Let MATH. In addition, we may assume that MATH for every MATH. Consider also a cover MATH, where MATH. Let MATH, MATH, MATH be canonical system for MATH and MATH. For each MATH pick MATH. Finally, consider sets MATH and MATH, constructed with respect to MATH (see REF). Now we construct map MATH. Using induction by MATH construct a sequence of maps MATH such that: CASE: MATH. CASE: MATH. CASE: MATH is MATH-close to MATH. CASE: For each MATH such that MATH there exists MATH having property MATH. For MATH define MATH letting MATH. Observe, that MATH is defined correctly and continuous on MATH (see properties G REF - G REF on the page REF). By REF - REF of canonical system and by the choice of MATH, MATH satisfies requirements REF - REF. Assuming that MATH has been already constructed, let us construct MATH. To accomplish this it is enough (see G REF on the page REF) to define MATH on each non-empty MATH for each MATH such that MATH. Fix MATH such that MATH. Consider MATH. Let MATH. Obviously, MATH is closed and non-empty subset of MATH. The idea is to define MATH on MATH extending MATH from MATH over MATH. For each MATH consider MATH which exists by propery REF for map MATH. Consider MATH. By REF of canonical system we have: MATH and MATH . Since MATH from MATH we can conclude that there exists MATH with the following property: MATH . Define MATH. REF coupled with REF implies that for any MATH we have MATH. Last inclusion and REF of MATH (as well as our choice of the sequence MATH) yields the following chain of inclusions for each MATH, MATH : MATH. Hence (see property G REF on the page REF) MATH. The last fact and our choice of sequence MATH allow us to extend MATH to MATH over MATH such that MATH . Observe, that MATH is correctly defined and continuous on MATH by the properties G REF and G REF on the page REF. Let us check that MATH satisfies REF - REF . Indeed, REF are met by construction. Further, since MATH, REF follows from MATH. Finally, since MATH, from REF applied to MATH we have MATH, which shows that REF is also met. Since MATH (see property F REF on the page REF) we complete the proof letting MATH. CASE: Fix MATH. Construct sequences MATH, MATH such that MATH, MATH and for each MATH we have MATH, MATH in the sense of REF applied to MATH (for any MATH). Let MATH and MATH. In addition, we may assume that MATH and MATH for every MATH. Suppose that we are given a map MATH such that MATH is MATH-close to MATH. Consider a cover MATH, where MATH. Let MATH, MATH, MATH be canonical system for MATH and MATH. Since MATH, for each MATH we can pick MATH such that MATH. Finally, consider sets MATH and MATH, constructed with respect to MATH (see REF). Now we construct map MATH. As in REF, using induction by MATH construct a sequence of maps MATH such that: CASE: MATH. CASE: MATH. CASE: MATH is MATH-close to MATH. CASE: For each MATH such that MATH there exists MATH having property MATH. CASE: For each MATH, MATH . For MATH define MATH letting MATH. Observe, that MATH is defined correctly and continuous on MATH (see properties G REF - G REF on the page REF). By REF - REF of canonical system and by the choice of MATH, MATH satisfies requirements REF - REF. Assuming that MATH has been already constructed, let us construct MATH. As before in the proof of a, to accomplish this it is enough (see G REF on the page REF) to define MATH on each non-empty MATH for each MATH such that MATH. Fix MATH such that MATH. Consider MATH. Let MATH. Obviously, MATH is closed and non-empty subset of MATH. Again, the idea is to define MATH on MATH extending MATH from MATH over MATH. Using the same arguments as in proof of a, one can show, that MATH . Let us show, that MATH . Indeed, since MATH, we have MATH. Therefore MATH for any MATH. Further, by construction we have MATH for any MATH. These inequalities coupled with REF and the fact that MATH yield MATH. REF is checked. From properties MATH and MATH we can conclude according to our choice of sequences MATH and MATH that MATH can be extended over MATH to MATH such that MATH . Using the same arguments as in proof of a one can show that MATH is continuous map satisfying REF - REF . Let us check that REF is also met. Since MATH, for each MATH, from REF applied to map MATH we have: MATH and REF is checked. Finally, let MATH. Obviously, MATH is MATH-close to MATH. Check, that MATH is MATH-close to MATH. REF of MATH implies that MATH. Since MATH, we have MATH. These inclusions coupled with inequality MATH imply that MATH for each MATH and consequently MATH is MATH-close to MATH. |
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 monomials of degree MATH, and MATH is the projection onto the quotient ring. Thus, the lemma holds for MATH. Now suppose that the simplicial complex MATH is obtained from the simplicial complex MATH by adding one MATH-dimensional simplex MATH. By the inductive hypothesis, the lemma holds for MATH, that is, MATH. By REF , MATH is obtained from MATH by adding the subcomplex MATH. Then MATH, where MATH is generated by MATH and MATH. |
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 construct a retraction, and the composite map would be required retraction MATH. If MATH is the boundary complex of a MATH-simplex, then MATH and the retraction MATH is shown on REF . Now suppose that the simplicial complex MATH is obtained by removing one MATH-dimensional simplex MATH from simplicial complex MATH. By the inductive hypothesis, the lemma holds for MATH, that is, there is a retraction MATH with the required properties. Let us consider the face MATH (see REF ). Since MATH is not a simplex of MATH, the point MATH having coordinates MATH, MATH, MATH, do not belong to MATH. Hence, we may apply the retraction from REF on the face MATH, starting from the point MATH. Denote this retraction by MATH. Now take MATH. It is easy to see that this MATH is exactly the required retraction. |
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 is a face of cubical complex MATH (see REF ) then MATH. Since for MATH the MATH is canonically identified with a subset of MATH, we see that those MATH for which MATH is a simplex of MATH fit together to yield MATH. (This idea can be used to prove that MATH is a smooth manifold provided that MATH is the dual to the boundary complex of a simple polytope, see CITE.) For any simplex MATH the subset MATH is invariant with respect to the MATH-action on MATH. Hence, the NAME construction MATH is patched from NAME constructions MATH (compare this with the local construction of MATH from CITE). The latter can be factorized as MATH, which is homotopically equivalent to MATH. Hence, the restriction of the projection MATH to MATH is homotopically equivalent to the embedding MATH. These homotopy equivalences for all simplices MATH fit together to yield a required homotopy equivalence between MATH and 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 epimorphism. Since MATH is contractible, we have a chain equivalence MATH. Therefore, there is an isomorphism MATH . The NAME - NAME spectral sequence (see CITE) of commutative square REF has the MATH-term MATH and converges to MATH. Since MATH it follows from REF that the spectral sequence collapses at the MATH term, that is, MATH. Now, REF shows that the module MATH is an algebra isomorphic to MATH, which concludes the proof. |
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 its arguments, one has MATH where we denoted MATH. Since MATH by REF , we obtain the required isomorphism . |
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. Then MATH. If MATH, then MATH, MATH and MATH. If MATH, then MATH. In both cases MATH. |
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 some log resolution of MATH and let MATH be the union of the exceptional divisor and the proper transform of MATH. Then MATH can be computed as MATH . (see e. CASE: CITE). In particular, MATH. Hence MATH is a MATH-complement. By the above, MATH can be computed from linear inequalities MATH, where MATH runs through a finite number of prime divisors MATH. Therefore the function MATH is piecewise linear and continuous in MATH and so are the coefficients of MATH. By construction, MATH is not klt for all MATH. We claim that MATH. Indeed, MATH. Thus MATH. Since MATH is lc, MATH. Now, take MATH . By the above discussions MATH is rational (and MATH). If MATH, then we put MATH and MATH. Otherwise, for any MATH, MATH. Hence there is a divisor MATH of MATH such that MATH. Again we can take MATH to be a component of MATH. Thus MATH does not depend on MATH if MATH. Obviously, MATH and we can put MATH. |
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 give us MATH, so MATH where MATH is effective, exceptional and MATH. This divisor cannot be MATH-nef (see e. CASE: CITE). Now, run MATH-MMP over MATH. At the last step we get a birational contraction MATH which satisfies REF - REF. |
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 also big over MATH. Assume the opposite. By the Base Point Free Theorem, MATH is semi-ample over MATH and defines a contraction MATH onto a lower-dimensional variety. Let MATH be a general curve in a fiber. Then MATH, so MATH. Since MATH is nef, MATH and MATH. By REF, MATH, a contradiction. Further, MATH is polyhedral, so there is an extremal ray MATH such that MATH and MATH. Hence MATH. Let MATH be the contraction of MATH. Put MATH. Then MATH. Therefore MATH is MATH-Cartier, klt and MATH is nef and big over MATH. If MATH is divisorial, we can continue the process replacing MATH with MATH and MATH with MATH. Assume that MATH is a flipping contraction and let MATH be the flip. Take MATH. Again we have that MATH is nef and big over MATH. Thus we can continue the process replacing MATH with MATH. |
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, obviously, is preserved under any birational map which is an isomorphism in codimension one. The log canonical property is preserved by CITE. |
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 NAME factorization. Then MATH is a contraction and MATH is a finite morphism. Define MATH and MATH by MATH (see CITE). This means that, for example, the coefficient of a component MATH of MATH in MATH is as follows MATH where MATH is the ramification index at the generic point of MATH. Then by REF, MATH (and MATH). Obviously, MATH is ample over MATH. First we consider REF (that is, when MATH is plt, MATH, MATH, MATH). We put MATH. By REF , MATH is compact and MATH. Applying CITE (or CITE) we get that MATH is plt. By the Connectedness Lemma CITE and the Adjunction CITE, MATH is connected, irreducible and normal. Define MATH from MATH, where MATH. Let MATH be the normalization of MATH in the function field of MATH. There is the commutative diagram MATH where MATH is a finite morphism and MATH and MATH are birational maps such that both MATH and MATH do not contract divisors. Hence MATH has the ramification divisor only over MATH and the ramification index of MATH at the generic point of a component over MATH is equal the ramification index of MATH at the generic point of the corresponding component over MATH. Applying MATH and MATH to REF, we obtain MATH where MATH and MATH. Recall that MATH is irreducible. Now, REF yields MATH where MATH and MATH. By CITE and REF (see also CITE), MATH is plt. Moreover, REF give us that MATH is nef and big over MATH. It is sufficient to prove the boundedness of the degree of the restriction MATH. Indeed, MATH, where MATH is the ramification index over MATH. By REF, MATH is bounded. Now, we consider log pairs MATH and MATH. Restricting REF on MATH, we obtain MATH . In particular, MATH. Both sides of this equality are positive by REF . By the proof of REF , there is a MATH-complement MATH of MATH with MATH. Define MATH from MATH and put MATH and MATH. Then MATH and MATH are MATH-complements. Since MATH is klt (see REF ), we have MATH . Similarly, MATH . By REF, MATH and MATH belongs to a finite number of algebraic families. Taking into account that MATH (see REF ) and MATH, MATH, we see that so are MATH and MATH. This gives us that MATH is bounded. Now, we consider REF . Let MATH be the normalization of a dominant component of MATH and let MATH be the proper transform of MATH on MATH. We claim that MATH is a plt blow-up of MATH. Consider the base change MATH . It is clear that MATH is finite and its ramification divisor can be supported only in MATH. Then MATH is the exceptional divisor of the blow-up MATH. We have MATH where MATH is a boundary. This divisor is plt CITE, CITE and anti-ample over MATH. By the Adjunction CITE, MATH is normal. On the other hand, MATH is connected near the fiber over MATH. Indeed, MATH is nef and big over MATH, by REF. Since MATH, it is connected by the Connectedness Lemma CITE. This proves our claim. Now, as in REF we consider the commutative diagram MATH . Similar to REF are bounded. Hence we may assume that MATH is bounded, where MATH. It remains to show that the ramification index MATH of MATH at the generic point of MATH is bounded. Clearly, MATH is equal to the ramification index of MATH at the generic point of MATH. Similar to REF write MATH . Then MATH (see CITE or CITE). We claim that MATH belong to a finite number of algebraic families. Note that we cannot apply REF directly because MATH is not necessarily nef. As in REF , take MATH-complement MATH with MATH. Similar to REF define MATH and MATH (see REF): MATH . Then MATH and by REF, MATH is anti-ample. Hence MATH is ample for any MATH. Note that MATH . Thus we can apply Conjecture REF to MATH for small positive MATH. We obtain that MATH is bounded. Now, as in REF we see that so is MATH. Take a sufficiently general curve MATH in a general fiber of MATH. From REF we have MATH . Clearly, MATH depends only on MATH, but not on MATH. So we assume that MATH is fixed. Recall that the coefficients of MATH are standard (see CITE, CITE), so we can write MATH, where MATH, MATH. Put MATH. By CITE both MATH and MATH are NAME along MATH. So REF can be rewritten as MATH, where MATH is a fixed natural number and MATH is also natural. Thus by REF MATH. This gives us that MATH is bounded and proves the theorem. |
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 that MATH is plt. Further, by REF , MATH. Since MATH is nef over MATH, we see that MATH coincides with the fiber over MATH. By construction, MATH and MATH is klt (by the Adjunction CITE). Therefore MATH . Obviously, MATH. By CITE, MATH belongs to a finite number of algebraic families. Thus we may assume that MATH is bounded by MATH. Now, consider the exact sequence MATH . By NAME vanishing MATH, MATH. Hence, MATH. Similarly, MATH. Since MATH is a topological retract of MATH, MATH. Hence MATH is bounded, and so is MATH (because MATH is a sequence of flips). This shows REF. To prove REF one can use that MATH is equal to the number of components of MATH (by the same arguments as above, see CITE). |
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, since MATH is an integral divisor, we have MATH . |
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 the coefficients of MATH satisfy REF . So we can apply REF to MATH. We get a MATH-complement MATH of MATH. In particular, by REF, there exists MATH such that MATH . By NAME Vanishing, MATH . From the exact sequence MATH we get surjectivity of the restriction map MATH . Therefore there exists a divisor MATH such that MATH. Set MATH . Then MATH and MATH. Note that we cannot apply the Inversion of Adjunction on MATH because MATH can have negative coefficients. So we put MATH. Again we have MATH and MATH. We have to show only that MATH is lc. Assume that MATH is not lc. Then MATH is also not lc for some MATH. It is clear that MATH is nef and big over MATH. By the Inversion of Adjunction CITE, MATH is plt near MATH. Hence MATH near MATH. On the other hand, by the Connectedness Lemma CITE, MATH is connected near MATH. Thus MATH is plt. This contradiction proves the proposition. |
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 generality we may assume that MATH. Then MATH. Since MATH it follows that MATH. The verification of the second conditions is left to the reader. |
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 follows that for any MATH-point MATH of MATH there exist local coordinates MATH on a neighborhood MATH of MATH satisfying the following conditions: CASE: MATH, for some function MATH on MATH with MATH and MATH ii)MATH on MATH for some function MATH on MATH with MATH and MATH. It is clear that for a point MATH sufficiently close to MATH the fibers MATH and MATH have the same measure. Hence we can replace MATH by MATH. By the implicit function theorem we can choose a neighborhood MATH of MATH in p-adic topology: MATH such that the coordinates MATH give an isomorphism (of sets) MATH, where MATH is a positive integer. Moreover, if we choose MATH sufficiently large the function MATH is well defined on MATH, hence, changing MATH for MATH we can suppose that on MATH one has MATH. Without loss of generality we can assume that MATH is constant on MATH. Since MATH is compact it can be covered by finitely many open sets MATH satisfying the properties stated above. In fact one can choose them to be disjoint. To prove it we need the following lemma. Let MATH be an open, compact subset of the affine space MATH. We claim that MATH can be covered by finitely many disjoint balls contained in MATH. (A ball is a subset of the form MATH. We apply the lemma to the sets MATH. (By construction each of them is identified with an open compact subset of MATH). Since MATH is compact we can cover it by finitely many balls contained in MATH. Let MATH be the smallest radius of these balls. Pick an integer MATH such that MATH. Consider the projection MATH. The pre-image of each element of Im-MATH is a ball. These balls constitute the desired covering. Now the proposition follows from the following simple lemma Let MATH stand for differential form MATH where MATH and MATH . Then MATH for any MATH, MATH. |
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 positive values of MATH. Now, starting from the right hand side of REF , collect the terms, containing the summation variable MATH, into a complete square: MATH - now REF can be applied to the MATH-dependent exponential - MATH - with subsequent evaluation of the summations - MATH - the exponential MATH in the integrand, being multiplied by an odd function of MATH (with respect to MATH), can by replaced by it's odd part - MATH - reversing the previous argument, replace MATH by an exponential - MATH - and rescale the integration variable (MATH) - MATH with notation REF being used. |
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 variable by MATH, we rewrite the result in the form of REF . |
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 by imaginary unit and add integration along a closed contour encircling points MATH in the counterclockwise direction. The integration along the shifted path MATH can be transformed by the change of the integration variable MATH: MATH where in the last line we have used the (quasi) periodicity property of the MATH-function, the fact that MATH and MATH are mutually prime, and disregarded the odd terms with respect to the sign change MATH. Now, the obtained formula straightforwardly leads to the asymptotic power series MATH in REF through the NAME series expansion of the function MATH at MATH, and evaluation of the Gaussian integrals. The other terms in REF come from the evaluation of the contour integral by the residue method. |
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 the corresponding integro-differential equations where we will suppress all references to the index MATH. REF : integrating with respect to MATH and inserting initial conditions we get an equivalent system of integral equations MATH . Writing MATH instead of MATH and denoting MATH and MATH, which is globally bounded by assumption, we consider the right hand side as an operator MATH on the set MATH . Using MATH, MATH for MATH, and NAME 's inequality for the convolutions, one derives from REF MATH which shows that MATH for MATH small enough. To see that MATH is actually a contraction on MATH for small MATH we can estimate the terms occurring in MATH for MATH, MATH after applying NAME 's inequality and/or taking MATH for all integrands. First, in the integral involving convolutions the non-trivial factors are handled as follows: by the mean value theorem applied to the function MATH and MATH we have MATH; since also MATH is globally bounded by MATH and by definition of MATH we further estimate MATH . For dominating the MATH of the integrands in the second integral we simply observe MATH to get terms we already estimated above. Altogether for MATH small, depending on MATH, MATH, and MATH the map MATH is a contraction. Therefore a solution of the fixed point equation exists for small MATH. Assuming that MATH and MATH are two fixed points for MATH arbitrary one can estimate their difference similar to the above except that only MATH is to be considered inside integrals with respect to MATH. A standard NAME argument then yields global uniqueness of the solution. An a priori estimate for a solution MATH in MATH is derived by using the fixed point REF , MATH, taking MATH inside the integrals, and again applying NAME 's lemma: MATH . Therefore existence of a global solution follows. Clearly this solution is bounded on MATH by definition of MATH. Finally to get unique solutions in MATH from initial data in MATH we proceed inductively. Set MATH with MATH; then for MATH we define MATH where MATH . Then estimates can be done essentially by the same arguments as above because the nonlinear combinations of unknowns remain of the same type (for example, all derivatives of the function MATH are again bounded and therefore application of NAME 's inequality, mean value theorem, and NAME 's lemma are still possible). The weight factor MATH for derivatives of order MATH in the above norms serves to control inner derivatives and assures again for example, the contraction property for small MATH. This concludes the purely classical base for the next step. CASE: Existence and uniqueness of solutions in MATH. In REF all involved functions except MATH now carry an index MATH and MATH is to be replaced by MATH. We assume MATH. In order to establish existence of a solution we will show that the net MATH belongs to MATH. From the integral equations for MATH and using NAME arguments as above we conclude MATH which (by MATH) gives the MATH-estimates of order MATH. The MATH-estimates for the MATH-derivatives of higher order follow inductively by similar arguments if in each step we write the MATH-derivative of any convolution product of MATH with any function MATH as MATH and use MATH. The MATH-estimates for the MATH-derivatives follow directly by induction from the differential equations. Concerning uniqueness, suppose that MATH is another solution in MATH. This means that there exists some MATH with MATH . From this by arguments similar to those used above we inductively derive MATH-estimates for each derivative of MATH. Thus MATH in MATH. |
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 NAME 's inequality which (since MATH was arbitrary) establishes the claim for MATH. Then since MATH it follows that MATH. Finally, MATH . From this it follows that MATH for MATH. But then since MATH in this region we obtain MATH . Thus also MATH and therefore MATH for MATH. |
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 REF the supports of MATH and MATH are contained in MATH and from the initial conditions we get MATH. Thus MATH, which contradicts MATH. |
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 obtained from REF using the Residue Theorem. |
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 condition MATH holds. The formula MATH the definitions of MATH and MATH and REF imply that MATH . Using again REF we find MATH . The inequality MATH yields MATH and one obtains for MATH that MATH . Hence, for MATH, MATH . The last inequality is a direct consequence of the second assumption in REF . We conclude that MATH is invertible and that REF holds. |
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 follows. |
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 holomorphic functions with values in MATH. By the resolvent equation and REF the product MATH can be written as MATH where MATH and MATH . Further, REF yields that MATH . The function MATH is holomorphic inside the contour MATH, MATH, since the argument MATH of the integrand in REF belongs to MATH and thereby MATH. Thus the term MATH does not contribute to the integral MATH while the resolvent MATH gives the identity MATH which proves REF . Regarding REF we obtain MATH . The first integral vanishes whereas the second integral equals MATH. The second equation of REF can be checked in the same way. |
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. Hence MATH and MATH are in MATH, and so MATH as well. Thus MATH. |
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 for all small objects MATH and all MATH. It remains to prove that MATH for all but finitely many MATH if MATH is small. This is proved analogously; the collection of all such MATH is a thick subcategory containing 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 modules over any ring is clearly closed under cokernels and is also closed under extensions. Indeed, suppose we have a short exact sequence MATH where MATH and MATH are finitely presented (in fact, we need only assume MATH is finitely generated). Choose a finitely generated projective MATH and a surjection MATH. We can lift this to a map MATH. Then we get a surjection MATH, as is well-known. Furthermore, the kernel of this surjection is the same as the kernel of MATH, which is finitely generated since MATH is finitely presented. Hence MATH is finitely presented. Now suppose that MATH is coherent. We show that the kernel of a map MATH of finitely presented modules is finitely presented. The point is that the image of MATH is a finitely generated submodule of the finitely presented module MATH. Because the ring is coherent, this means that the image of MATH is finitely presented. The kernel of MATH is therefore finitely generated, but it is a submodule of the finitely presented module MATH, so it is finitely presented, using coherence again. |
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 category MATH of chain complexes over MATH with the projective model structure is a MATH-model category, and we are using that structure). If MATH, then MATH, so MATH has finite projective dimension if and only if MATH does. It is easy to see that complexes with finite projective dimension form a thick subcategory containing MATH. Therefore every small object of MATH has finite projective dimension. |
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 will not be cofibrant. To make it cofibrant, we need to replace MATH by a projective resolution. Since MATH is finitely presented and the ring MATH is coherent, each term MATH in a projective resolution for MATH will be finitely generated. Since MATH has finite projective dimension, the resolution MATH is finite. Hence MATH is small in MATH, and so also in MATH. Since MATH is isomorphic to MATH in MATH, the result follows. |
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 corollary follows. |
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 true for MATH, and MATH acts trivially on MATH. Write MATH as an extension MATH where MATH. Then MATH acts trivially on MATH and MATH, so MATH. Furthermore, this is an extension of MATH-modules; since MATH is a wide subcategory of MATH-modules, MATH. The induction is complete, and we find that MATH. |
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. Since MATH for any MATH-module MATH such that MATH and any MATH not containing MATH, we have MATH for a unique MATH. One can easily see that MATH, but we claim that in fact MATH. Indeed, MATH, so this claim boils down to showing that the inclusion MATH is a proper map. This is well-known; the inclusion of any closed subset in any topological space is proper. Naturally, we claim that MATH. We have MATH, since MATH. To show the converse, it suffices to show that MATH, since MATH is an embedding by REF . But any module MATH in MATH has MATH for all MATH and for all MATH. Therefore MATH. |
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, that MATH; one can also assume that MATH is an integer. Let MATH be the partition of MATH given by MATH . It follows that MATH. For MATH sufficiently large, MATH and MATH; hence there exists a step-function MATH for which MATH; we may also require there exists a step-function MATH, for which MATH. There is a permutation MATH, so that MATH; this is because the measure of MATH and MATH is the same, for a set MATH of measure MATH. It follows that MATH. Finally, we get that 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, using independence of MATH and MATH, MATH for all MATH. Writing MATH, we see that the equation above is satisfied if MATH for all MATH, MATH and MATH. But this is precisely the condition that MATH. It follows that MATH since MATH, we get that MATH which implies MATH. To prove the opposite inequality, let now MATH be such that MATH . Then MATH for all MATH. Hence by REF , MATH . Since the unit of MATH occurs among MATH, this implies that any MATH satisfy MATH for all MATH, MATH, MATH. Let MATH with MATH. Notice that MATH is a diagonal matrix, whose MATH-th diagonal entry is MATH . If MATH is so large that MATH for some MATH, it follows that for each MATH there exists a subset MATH of MATH with MATH, and so that for all MATH, MATH . Hence MATH where MATH if MATH and MATH if MATH. It follows that MATH . Taking the limit MATH and using REF gives MATH . Since MATH is chosen so that MATH, taking the limit as MATH and infimum over MATH and MATH gives the desired inequality. The proof for MATH is identical. |
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 when MATH and MATH is the ball MATH for MATH (the MATH norm is with respect to the normalized trace MATH on MATH). It follows that MATH . The limit as MATH of MATH is zero. Hence we get MATH . As MATH and MATH, we get as estimate MATH for some universal constant MATH. The desired estimate now follows from the definition of MATH. |
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 (note that MATH are zero for sufficiently large MATH). Then choosing MATH and letting MATH, we have that MATH provided that MATH, where MATH is the NAME measure of the symmetric difference of MATH and MATH. Hence for MATH sufficiently large, we can choose MATH for MATH, MATH to be the first MATH for which MATH is zero and MATH, and have that: MATH . This implies that MATH . Taking the limit MATH and noticing that in this case each MATH, since MATH, and MATH, gives that MATH . Note that we have the same inequality for MATH and MATH replaced by MATH. For the opposite inequality, we may assume that MATH, with MATH, MATH and MATH. Moreover, assume that for some MATH, MATH . Since MATH, it follows that, given MATH we can find MATH and MATH (independent of MATH and MATH), for which, after letting MATH, one has MATH . Let now MATH . Let MATH. Denote by MATH the diagonal matrix having all entries zero, except that the MATH-th entries for MATH are equal to MATH. Then for a subset MATH of size at most MATH, we have that MATH . Therefore, one has MATH . Taking the limits MATH (so that MATH) gives finally MATH . Note that the same argument gives the same inequality for MATH instead of 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 amalgamation over MATH. It follows that MATH which implies the proposition after taking limits. |
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. It follows that the set MATH. The claimed inequality now follows from the definition of 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 MATH for which the image of the map MATH lies in MATH. It follows after taking limits that MATH . By the independence and freeness assumptions on MATH we finally get MATH which is the desired estimate. |
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 some non-commutative polynomial MATH with coefficients from MATH. Then one has the inclusion MATH for all MATH. Taking limits gives the opposite inequality, and hence implies REF . |
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 follows that MATH . Since MATH was arbitrary, this implies the desired inequality. The proof for MATH, MATH, etc. is the same. |
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 amalgamation over MATH, and MATH, it follows that we must prove MATH . Thus it would be sufficient to prove the inequality MATH . Given MATH, there exists a NAME map MATH, assuming a finite number of values, from the set MATH to the set MATH having the form MATH so that MATH for all MATH. Moreover, since MATH is free with amalgamation over MATH from MATH, there exists a subset MATH in MATH, so that MATH and so that for all MATH, we have that MATH . In particular, for MATH, the values of the map MATH lie in the set MATH . Since this map preserves NAME measure on the unitary group, we conclude, after passing to limits, that MATH thus finishing the proof. |
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 MATH to be free Brownian motion, as in the definition of the free dimension MATH. Since MATH, MATH normalize MATH and define the same automorphisms of MATH as MATH, it follows that there exists unitaries MATH, MATH, normalizing MATH, so that MATH (one can simply choose any extension of the isometry MATH to a unitary normalizing MATH). Therefore, since MATH (compare CITE), MATH hence MATH. It follows that the hypothesis of REF is satisfied, and hence MATH. By Proposition, REF we get also that MATH. Since MATH are arbitrary, we get that MATH. Reversing the roles of MATH and MATH gives finally that MATH . |
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 is independent from MATH, is MATH-distributed in the same way as MATH and is MATH-free with amalgamation over MATH from the MATH-algebra generated by MATH and MATH. It follows that the MATH-valued distribution of MATH is the same as the MATH-valued distribution of MATH. Since MATH is free from MATH over MATH, and because MATH is a NAME unitary, independent from MATH, it follows that MATH is free from MATH over MATH, and has the same MATH-valued distribution as MATH. Write MATH. It remains to show that MATH and MATH have the same MATH-valued MATH-distributions. Indeed, that would imply that the MATH-valued distribution of MATH is the same as that of MATH (hence the same as MATH), and also that MATH is MATH-free from MATH over MATH, since MATH is MATH-free from MATH over MATH. As a matrix, MATH . To prove that the MATH-valued MATH-distributions of MATH and MATH are the same, it is sufficient to prove that the families of their entries have the same joint MATH-distributions; that is, that the joint MATH-distribution of family MATH is the same as that of MATH. Write MATH. Hence it is sufficient to prove that: CASE: MATH are NAME unitaries, independent from MATH and MATH-free with amalgamation over MATH; CASE: MATH is MATH-free from MATH over MATH and REF MATH has the same MATH-valued MATH-distribution as MATH. To prove REF , notice that we can, by replacing each MATH with MATH where MATH are NAME unitaries, independent from MATH and MATH-free from each other and from MATH with amalgamation over MATH, without changing the joint MATH-valued MATH-distribution of the family, replace MATH by MATH. Since the unitaries MATH are MATH-free over MATH, it follows that MATH are MATH-free over MATH. Clearly, each MATH is independent from MATH; and each MATH is a NAME unitary (note that we can always replace, say, MATH by MATH for arbitrary MATH, without changing the joint distribution of MATH). For the second claim, we have MATH. Notice that MATH is MATH-free over MATH from MATH (which are all MATH-free over MATH among each other) and hence from MATH. Hence MATH is MATH-free over MATH from MATH. Lastly, MATH is clearly independent from MATH, and has the same MATH-distribution as MATH. Since MATH are MATH-free and form a multiplicative free Brownian motion, the MATH-distribution of MATH is the same as that of MATH. |
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 that identifies MATH with diagonal matrices in MATH with values from MATH, and MATH with the permutation matrix MATH. Let MATH be unitaries, independent from MATH, and free from each other over MATH, and so that each MATH is a NAME unitary, and MATH is MATH-distributed as a free multiplicative Brownian motion started at identity and evaluated at time MATH. Let MATH be fixed, and let MATH . Set MATH and MATH. Let MATH, MATH, MATH, MATH be given. For MATH sufficiently large, we can write MATH, where MATH and MATH. Then there exist MATH, MATH, MATH, MATH and MATH for which the map MATH for a chosen matrix MATH has values in MATH, and is injective. The union of its images over possible different MATH has the same volume as MATH, and is a subset of MATH. It follows that MATH from which, after taking limits we get MATH . Consider now the set MATH. We may assume, for MATH large enough, that MATH, MATH, and MATH has the form MATH, where MATH has the form MATH, with MATH a cyclic permutation of order MATH. Let MATH be given. We may furthermore assume by REF that for this choice of MATH, there exist MATH for which MATH is within MATH of MATH. Note that each element of MATH lies in MATH and can be represented as a matrix MATH in which MATH and each MATH. By REF , for large enough MATH, there is a subset MATH of MATH, with MATH, so that each MATH is MATH free from the algebra MATH, and in particular, from MATH. It follows that given MATH, MATH, MATH, MATH, there exist MATH, MATH, MATH and MATH, so that if the matrix MATH above lies in MATH, then MATH. It follows that MATH . Hence MATH for MATH arbitrarily small. Combining this with REF gives, in view of independence and freeness assumptions: MATH as we claimed. The same proof can be modified to work for MATH instead; we point out the necessary changes. We claim that MATH where MATH are MATH-free NAME unitaries and MATH are MATH-free unitaries, MATH-free from MATH, and each MATH has the same distribution as free multiplicative Brownian motion started from MATH and evaluated at time MATH. The map MATH which sends MATH to the pair of matrices MATH has values in MATH . This gives, just like in the first part of the proof, the inequality MATH . Conversely, we can assume that there is a subset MATH of MATH, so that MATH and so that for all MATH there exists a matrix MATH, commuting with MATH, for which MATH and MATH are MATH-free from MATH with amalgamation over MATH. Then the map sending such a pair MATH, MATH to MATH is valued in MATH . To see this, observe that the family MATH has the same joint MATH-valued MATH-distribution as MATH . Next, observe that (as in the proof of REF ) that the family of matrices MATH are MATH-free with amalgamation over MATH from the permutation matrix MATH since they can be written as MATH and MATH, where MATH . From this it follows that MATH can be embedded into MATH arguing as in the first part of the proof now gives MATH and hence, MATH . To finish the proof, we must compute MATH where we use freeness of MATH with amalgamation over MATH, and REF . |
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 motion started at identity and evaluated at time MATH. It follows from REF that MATH, and from REF that MATH. Hence MATH. By REF , we get that MATH for MATH. Hence MATH. This is the same as the cost of MATH, see CITE. |
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 implementing MATH, we have: MATH . Choose now a measure-preserving automorphism MATH of MATH, so that MATH implements a free and ergodic action of MATH, and the induced equivalence relation MATH is free from MATH. Let MATH. Then MATH for any finite graphing MATH of MATH. Let MATH be a graphing of MATH. Then MATH is a graphing of MATH. If MATH, then there exists a finite graphing MATH of MATH, with the same cost as MATH (indeed, given MATH so that MATH, one can find integers MATH and MATH so that the domains and ranges of MATH, MATH are disjoint, and hence replace MATH by a single automorphism, keeping the cost of the graphing the same). It follows that MATH since MATH. Hence MATH. |
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 and MATH. Write MATH. Then MATH . Since MATH, we get MATH . Since MATH is arbitrary, the conclusion follows. |
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. |