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math/9912111
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Apply REF to MATH and MATH. We get a MATH-complement MATH with MATH. Then MATH is lc and MATH. But if MATH, MATH is klt (because MATH).
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math/9912111
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Assume that MATH is lc for some MATH. By REF there is a regular complement MATH near MATH. Since MATH, MATH. By the definition of complements, MATH. Hence MATH is lc.
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math/9912111
|
Put MATH. Assume that MATH. Clearly, MATH is reduced. Let MATH be an inductive blow up of MATH (see REF). Then we can write MATH, where MATH is the proper transform of MATH. If MATH is not plt, then by REF , MATH is MATH or MATH-complementary. Since MATH, this gives us that MATH, MATH is lc and MATH. Hence, we may assume that MATH is plt. By REF, MATH intersects MATH transversally and MATH . Since MATH, MATH. If MATH, then MATH and in REF, MATH for MATH, a contradiction with MATH. Assume that MATH. Then in REF we have MATH and MATH. Hence MATH and MATH. This yields MATH . Therefore MATH and MATH. Finally, assume that MATH. Similarly, in REF we have MATH. From this MATH and up to permutations MATH is one of the following: MATH, MATH, MATH. Thus MATH, or MATH. In all cases MATH, so MATH.
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math/9912111
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By REF we get that all singular points MATH are cyclic quotients: MATH where the action of MATH on MATH is free outside of MATH. Therefore MATH, MATH, where MATH is negative on MATH. From this it is easy to see that for MATH there are only the possibilities MATH, MATH, MATH, MATH, MATH and MATH. Since MATH is MATH-ample, MATH-complements for MATH can be extended to MATH-complements of MATH. By REF we have the desired MATH-complements. This proves REF .
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math/9912111
|
Let MATH be the ``central" exceptional divisor of the minimal resolution and MATH, MATH, MATH exceptional divisors adjacent to MATH. Write MATH. Intersecting both sides with MATH, we obtain MATH . This yields MATH .
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math/9912111
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One has MATH . This yields MATH and MATH. In particular, MATH is contained in the smooth locus of MATH and MATH. Moreover, it follows from MATH that MATH for any component MATH. Similarly we can write MATH . If MATH is irreducible, then MATH and MATH is a smooth elliptic curve (because MATH is dlt). If MATH, then MATH, MATH and MATH. Since MATH is dlt, MATH intersects MATH transversally at two points. The only possibility is when MATH is a wheel of smooth rational curves.
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math/9912111
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Consider MATH as a small analytic neighborhood of MATH. We calculate the fundamental group of MATH. Denote by MATH the group generated by MATH with relations MATH . MATH . Let MATH be a small neighborhood of MATH and MATH. From REF we have MATH. Denote by MATH the generators of these groups. The set MATH is homotopically equivalent to MATH glued along MATH with sets MATH, MATH, MATH. Denote loops around MATH (with the appropriate orientation) also by MATH. Then MATH. From the description of points REF it follows also that the map MATH is surjective. Now the lemma follows by NAME 's theorem. Now for REF we notice that the groups MATH, MATH, MATH and MATH have finite quotient groups isomorphic to MATH, MATH, MATH and MATH, respectively, such that the images of the elements MATH have orders MATH. This follows from the fact that there exist actions of MATH, MATH, MATH and MATH on MATH with ramification points of orders MATH. Then this finite group determines a finite cover MATH unramified outside of MATH, where MATH is smooth. The NAME factorization gives a contraction MATH of an irreducible curve MATH. If MATH, then this contraction is the minimal resolution of the singularity MATH. Finally, if MATH is ample, then so is MATH. Thus MATH, that is, MATH is a smooth point. But the groups MATH, MATH, MATH and MATH cannot act on MATH freely in codimension one. This proves REF .
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math/9912111
|
By REF , MATH and MATH are MATH-complementary. Therefore there are two curves MATH, MATH such that MATH is lc and linearly trivial over MATH. Moreover, by REF up to analytic isomorphisms we may assume that MATH is a toric pair. For example, assume that MATH has exactly two singular points. Consider the minimal resolution MATH and MATH the composition. It is sufficient to show that the morphism MATH is toric. By REF, in a fiber over MATH we have the following configuration of curves: MATH where the black vertex corresponds to a fiber (and has self-intersection number MATH), white vertices correspond to exceptional divisors and have self-intersection numbers MATH, and the vertices MATH correspond to the curves MATH, MATH. If MATH, MATH is the minimal resolution of a cyclic quotient singularity MATH and in this case the morphism MATH is toric. If MATH, then MATH factors through the minimal resolution MATH of the singularity MATH (which is a toric morphism) and MATH is a composition of blowups with centers at points of intersections of curves. Such blowups preserve the action of the two-dimensional torus, hence MATH is a toric morphism.
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math/9912111
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In REF any nontrivial element MATH have MATH as a fixed point. It can have at most one more fixed point MATH on each component MATH. Moreover, MATH permutes points MATH. Then MATH can be singular only at images of MATH and MATH.
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math/9912111
|
Let MATH be the normalization of MATH. Consider the NAME factorization MATH. Then MATH is flat and a generically MATH-bundle. Therefore for the central fiber MATH one has MATH, where MATH is reduced and it is the preimage of MATH. On the other hand, MATH where MATH is the degree of MATH. Whence MATH, MATH. This proves the assertion.
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math/9912111
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In the construction REF we have REF . Then MATH where MATH, MATH are singular points of MATH and MATH is the index of MATH. By REF and by REF , MATH is MATH-complementary.
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math/9912111
|
If MATH is plt, then by REF we have Case MATH. Thus we may assume that MATH is not plt. We claim that MATH is MATH, MATH or MATH-complementary. Assume that MATH is not MATH-complementary. For some MATH the log divisor MATH is lc, but not plt (so, MATH is maximally lc). Consider a minimal log terminal modification MATH, where MATH is the reduced exceptional divisor, MATH is the proper transform of MATH and MATH is dlt. As in REF, applying the MATH-MMP to MATH at the last step we obtain the blowup MATH with irreducible exceptional divisor MATH. Moreover, MATH is lc, where MATH is the proper transform of MATH and MATH is plt and negative over MATH. Since MATH is antiample for MATH, the curve MATH can be contracted in the appropriate log MMP over MATH and this gives a contraction MATH with purely log terminal MATH. By REF MATH is as in REF . If MATH in nonnegative on MATH, then by REF we can pull back MATH-complements from MATH on MATH and then push-down them on MATH (see REF). Thus we obtain MATH-complement of MATH, a contradiction. From now on we assume that MATH is ample over MATH. Then by REF complements for MATH can be extended on MATH. According to REF, MATH, where for MATH there are the following possibilities: MATH . Further, MATH has exactly two singular points and these are of type MATH and MATH, respectively (see REF ). Since MATH intersects MATH at only one point, this point must be singular and there are two more points with MATH. We get two cases: CASE: MATH, MATH, there is a MATH-complement; CASE: MATH, MATH, there is a MATH-complement. This proves the claim. If MATH is lc (but not plt), then in Construction REF MATH is also lc but not plt (see REF ). Since MATH is a NAME divisor, MATH is canonical. Hence MATH is as in REF . We get the case MATH. To prove that note that MATH and MATH is lc. Hence MATH is not exceptional and MATH is MATH- or MATH-complementary by REF . Assume that MATH is MATH-complementary, but MATH is not lc. Then there exists a reduced divisor MATH such that MATH is lc and linearly trivial. By our assumption and by REF , MATH. Let MATH be a point of index MATH. Then MATH and again by REF there are two components MATH passing through MATH. But since MATH, where MATH is a generic fiber of MATH, MATH, MATH and MATH is the only point of index MATH on MATH. Now assume that MATH is MATH-complementary, but not MATH-complementary and MATH is not lc. Then we are in REF . Therefore MATH . Take the minimal resolution MATH of MATH. Over MATH and MATH we have only single MATH-curves and over MATH we have a chain which must intersect the proper transform of MATH, because MATH passes through MATH. Since the fiber of MATH over MATH is a tree of rational curves, there are no three of them passing through one point. Whence proper transforms of MATH and MATH on MATH are disjoint. Moreover, the proper transform of MATH cannot be a MATH-curve. Indeed, otherwise contracting it we get three components of the fiber over MATH passing through one point. It gives that MATH coincides with the minimal resolution MATH of MATH. Therefore configuration of curves on MATH looks like that in Case MATH. We have to show only that all the curves in the down part have selfintersections MATH. Indeed, contracting MATH-curves over MATH we obtain a MATH-bundle. Each time, we contract a MATH-curve, we have the configuration of the same type. If there is a vertex with selfintersection MATH, then at some step we get the configuration MATH . It is easy to see that this configuration cannot be contracted to a smooth point over MATH, because contraction of the central MATH-curve gives configuration curves which is not a tree. This completes Case MATH. Case MATH is very similar to MATH. We omit it.
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math/9912111
|
First take the maximal MATH such that MATH is lc. Put MATH. Next we consider a minimal log terminal modification MATH of MATH (if MATH is dlt, we put MATH). Thus we can write MATH, where MATH is reduced and nonempty, MATH and MATH is contained in the fiber over MATH. Run MATH-MMP over MATH: MATH . If MATH, then MATH and we can contract a component of MATH. At the end we get the situation when MATH. Taking MATH we see the first part of the proposition. The second part follows by REF and the fact that all contractions MATH are positive with respect to MATH. Finally, assume that MATH is exceptional. Then by REF , there is exactly one nonklt complement MATH (where MATH). Clearly, MATH is irreducible in this case. Contractions MATH and MATH are crepant with respect to MATH. By REF MATH is plt. Assume that there are two dlt models MATH and MATH. Consider the diagram MATH where MATH is the minimal resolution and MATH is a composition of contractions of MATH-curves. Let MATH be a MATH-complement and MATH the crepant pull back, where MATH is the proper transform of MATH and MATH is a boundary. Clearly, MATH for any irreducible component MATH of MATH. Hence MATH is a nonklt MATH-complement, so MATH and MATH. Similarly, we get MATH. By exceptionality, MATH and MATH are irreducible and MATH (as discrete valuations of MATH). Then MATH is an isomorphism in codimension one, hence it is an isomorphism.
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math/9912111
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The implication MATH is obvious (because MATH is reduced, see REF). If MATH is plt, then by REF MATH is the only nonklt complement and MATH is the only divisor with MATH. This shows REF . MATH follows by REF. Let us prove the implication MATH . Assume that MATH is nonexceptional. By REF there are two different divisors MATH, MATH such that MATH. Then in REF we have MATH. Since MATH, MATH, that is, MATH is nonexceptional.
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math/9912111
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Follows by REF.
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math/9912111
|
We are going to apply REF . So we consider a dlt model MATH and MATH the minimal resolution of singularities of MATH. Then we have the following diagram: MATH where MATH is a sequence of contractions of MATH-curves. If MATH, then MATH and MATH is a smooth elliptic curve or a wheel of smooth rational curves. Contracting, if necessary, MATH-curves we obtain case MATH. Further, we assume that MATH. Then MATH is singular, so MATH. Consider the crepant pull back MATH where MATH is the proper transform of MATH, MATH, and MATH. Since MATH, it is easy to see that MATH. It is clear also that the set MATH coincides with the fiber over MATH. By construction, MATH contains no MATH-curves. First we consider the case when MATH also contains no MATH-curves. Then MATH is exactly the minimal resolution of MATH. By REF singular points of MATH are NAME. Cases MATH (MATH), MATH, MATH, MATH, MATH of REF gives cases MATH (with MATH), MATH, MATH, MATH, and MATH, respectively. For example, if MATH is irreducible and there are exactly three singular points of MATH, then similar to REF the graph of the minimal resolution MATH must be as in REF . By REF we have the following possibilities for MATH: MATH . Now, we consider the case when MATH contains a MATH-curve. Since MATH is a minimal resolution, all MATH-curves are contained in MATH, the proper transform of MATH. Using the negative semidefiniteness for the fiber MATH over MATH one can show that the dual graph of MATH cannot contain proper subgraphs of the form MATH . Suppose that MATH is irreducible. Then MATH is plt and MATH is the only a MATH-curve. Thus in the case MATH we obtain the dual graph for a fiber of MATH as below MATH . By the above this is impossible. In other cases we have the dual graphs as in REF . For MATH, MATH and MATH we obtain cases MATH, MATH and MATH, respectively. Similarly Case MATH, MATH of REF gives Case MATH. NAME connected fibers are only of type MATH, so only they can be multiple. This proves the theorem.
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math/9912111
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By REF we may assume that MATH over MATH and a general fiber of MATH is rational. First, as in the proof of REF , we replace the boundary MATH with MATH so that MATH is maximally lc. Replacing MATH with its log terminal modification, we may assume that MATH is smooth and the reduced part MATH of the boundary is nonempty. Next we blow up a sufficiently general point on MATH. We get a new model such that some component MATH of MATH is MATH-curve and it is not contained in MATH. Moreover, MATH is a point which is nonsingular for MATH. Let MATH be a (unique) component passing through MATH. Then the curve MATH can be contracted to a point, say MATH: MATH . The central fiber MATH of MATH is irreducible. Since MATH, the point MATH is lc. Apply REF to the birational contraction MATH. We get a regular MATH-complement MATH in a neighborhood of MATH. We claim that this complement extends to a complement in a neighborhood of the whole fiber MATH. We need to check only that MATH in a neighborhood of MATH. But in our situation the numerical equivalence over MATH coincides with linear one. Therefore the last is equivalent to MATH. Obviously, both sides have the same intersection numbers with all components of MATH different from MATH. For MATH we have MATH, MATH (because the coefficients of MATH in MATH and MATH are equal to MATH). This proves the theorem.
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math/9912111
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By taking a log terminal modification we may assume that MATH is smooth, MATH is dlt and MATH (see REF). Since MATH, MATH. So there is a morphism MATH onto a curve MATH of genus MATH. Let MATH be a contraction from a projective surface onto a curve of genus MATH. Assume that MATH is lc and MATH is nef. Furthermore, assume that the general fiber of MATH is a smooth rational curve. Then no components of MATH are contained in fibers. Let MATH be such a component. Replace MATH with a dlt modification. Then we may assume MATH . If the fiber MATH is reducible, there is its component MATH meeting MATH. Then MATH and MATH for MATH. Thus MATH generates an extremal ray which is negative with respect to MATH for MATH. By Contraction Theorem CITE we can contract MATH over MATH. This contraction preserves all assumptions of the lemma as well as assumption MATH (however, we can lose the dlt property of MATH). Continuing the process, we get the situation when the fiber containing MATH is irreducible. Similarly, components of all reducible fibers can be contracted. We obtain a model MATH such that all fibers are irreducible. Moreover MATH is an extremal MATH-negative contraction. Hence MATH. By our construction, MATH is lc, MATH is MATH-factorial and MATH is a fiber of MATH. Let MATH be an extremal ray on MATH other than that generated by fibers of MATH. Then MATH and MATH. Therefore MATH for MATH. Hence there is a curve MATH on MATH generating MATH (as above, if MATH is not dlt, we can use Contraction Theorem for MATH, see also REF). In this situation, MATH (see REF ). But then the base curve MATH also should be rational, a contradiction with MATH. Notation as in REF . Then the pair MATH has at worst canonical singularities. Replace MATH with a suitable log terminal modification (see REF ) and apply REF . Going back to the proof of REF , denote MATH and MATH. Let MATH be a component of MATH. By REF , MATH. If MATH, there is a curve MATH in a (reducible) fiber such that MATH and MATH. Since MATH, MATH is a MATH-curve and MATH . Hence we can contract MATH and by REF we can pull back complements under this contraction. Repeating the process, we reach the situation when MATH. Thus we may assume that MATH is a (smooth) ruled surface over a nonrational curve MATH, that is, MATH, where MATH is a rank two vector bundle on MATH. Moreover, MATH is dlt (see REF ). Then MATH . This implies MATH . Thus MATH and MATH is a smooth elliptic curve. Hence MATH. We may assume that MATH. Let MATH be a general fiber of MATH. Since MATH, there is exactly one extremal ray MATH on MATH. Assume that MATH. Then MATH is not nef, MATH and MATH. If MATH, then MATH generates an extremal ray (see REF ), so MATH. This contradicts REF. Therefore, MATH. In particular, MATH. Further, if MATH, then MATH is negative with respect to MATH. Since MATH is dlt, the ray MATH must be generated by a rational curve. This implies that MATH is also rational, a contradiction. Finally, we have MATH. Then REF implies MATH (recall that MATH is effective). By the NAME Index Theorem, MATH. Thus, MATH and MATH, where MATH and MATH is an irreducible curve with MATH. If MATH, then MATH is negative with respect to MATH for MATH. Again we have a contradiction. Hence MATH and MATH. As in REF we have MATH . Since MATH, MATH is nef, so MATH. Replacing MATH with MATH we get our assertion. Taking into account the equality MATH, we obtain MATH . Therefore MATH generates an extremal ray MATH on MATH (see REF ). It cannot be MATH-negative (otherwise MATH is generated by a rational curve). Hence MATH and in relations above equalities hold. Thus REF gives that MATH and all the components MATH are numerically proportional (because MATH). In particular, MATH . Let MATH be a general fiber of MATH. Then MATH and MATH. This yields MATH. Let MATH be a ruled surface over an elliptic curve and let MATH be a reduced divisor such that MATH. Then MATH. Moreover, if MATH is reducible, then MATH. Since MATH, MATH for some integral divisor of degree MATH on MATH. First we assume that MATH is irreducible. Then MATH is a smooth elliptic curve and we have MATH. Hence MATH because MATH is of degree two. Now we assume that MATH, where MATH, MATH are sections. Similarly, MATH. Hence MATH because MATH is an isomorphism. Now we finish the proof of REF . If MATH (that is, MATH is a MATH-section of MATH), then MATH, MATH and MATH. Hence MATH. By REF , MATH, that is, we have a MATH or MATH-complement (with MATH). If MATH, then MATH is a section of MATH. Recall that MATH, where MATH is a rank two vector bundle on MATH. By REF, the surface MATH is not such as in REF . On the other hand, by REF MATH and the vector bundle MATH has even degree. From the classification of rank two vector bundles over elliptic curves (see for example, CITE), we obtain that MATH is of splitting type. Hence there is a section MATH such that MATH. Write MATH. Then MATH . From REF we have MATH where MATH, MATH is a pair of disjoint sections. By REF and by the definition of complements, MATH is a MATH-complement of MATH if MATH whenever MATH. By REF this does not hold only if MATH, where MATH is a MATH-section. Consider this case. As in the proof of REF , MATH and MATH. Therefore MATH and MATH. This proves our theorem.
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math/9912111
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Since MATH (see REF ), we have MATH.
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math/9912111
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If MATH is big, then by REF there is a regular complement. Therefore we assume that MATH or MATH. CASE: Applying a minimal log terminal modification MATH we may assume that MATH is dlt and MATH. Take the crepant pull back MATH . By REF, MATH is klt, but it is not necessarily a boundary. Consider the new boundary MATH . Then MATH is klt and MATH is nef and big. Further, by REF we can push-down complements. So we replace MATH, MATH, MATH with MATH, MATH, MATH. Thus we assume now that MATH is dlt, MATH, MATH is nef, and there exists a boundary MATH such that MATH is klt and MATH is nef and big. By REF , MATH is rational. Set MATH and MATH. The following lemma shows that the NAME cone MATH is polyhedral and generated by contractible extremal curves. Let MATH be a log variety such that MATH is klt and MATH is nef and big over MATH. Then there exists a new boundary MATH on MATH such that MATH is again klt and MATH is ample over MATH. Let MATH be a very ample divisor on MATH (over MATH). By NAME 's Lemma, MATH for some MATH (see for example, CITE). Take MATH and put MATH. The lemma shows that we can contract all extremal rays on MATH. Moreover, if an extremal ray MATH on MATH is birational and generated by a curve MATH which is not contained in MATH, then the contraction preserves all assumptions (see REF). If additionally MATH is MATH-trivial, we can pull back regular complements by REF . CASE: By REF , MATH has at most two connected components. As in proofs of REF we distinguish the following cases: MATH is disconnected, MATH is a smooth elliptic curve or a wheel of smooth rational curves, MATH, MATH is a tree of rational curves. Also we should separate cases MATH and MATH. If MATH, then by Base Point Free Theorem the linear system MATH determines a contraction MATH. By REF , MATH is ample for some boundary MATH, hence a general fiber is rational. Then we consider cases when MATH is contained in a fiber of MATH and MATH has a horizontal component. CASE: By REF there exists a contraction MATH onto a curve such that MATH is a pair of two disjoint smooth sections (in particular, MATH is plt). A general fiber MATH of MATH is MATH (see REF), so MATH, MATH and MATH is contained in fibers of MATH. Since MATH is rational, MATH. In our case MATH is numerically trivial on a general fiber of MATH, so it is numerically trivial on all fibers. Contracting curves in fibers we get the situation when MATH. We can pull back all complements by REF . If MATH, then MATH is nef and big for any MATH. By REF there exists a regular complement of MATH. By REF MATH also has a regular complement. Therefore we may assume that MATH and MATH. Then both MATH and MATH generate extremal rays which must coincide because MATH. The ray cannot be birational, so MATH. This shows that there exists a nonbirational contraction MATH such that MATH and MATH are fibers of MATH. If MATH, then again MATH is ample for MATH. As above (by REF ) there is a regular complement. Therefore we may assume that MATH. Now it is sufficient to verify MATH for some MATH. By REF the numerical equivalence in MATH coincides with linear one. Therefore it is sufficient to show only that MATH is NAME for some MATH. Take MATH . Since MATH, MATH. By REF , MATH has a regular MATH-complement MATH near MATH (a fiber of MATH). Then MATH by REF . This yields MATH, MATH is integral and MATH near MATH (because MATH intersects all the components of MATH). Hence MATH. Similarly, we have a regular MATH-complement near MATH and MATH. Let MATH. Then MATH is integral and MATH is NAME near MATH. Let MATH be a fiber of MATH and let MATH. By Adjunction, MATH and MATH . Therefore MATH and MATH is smooth outside of MATH. Further, MATH is NAME everywhere on MATH and it is sufficient to show that MATH. Assume the opposite. Then we have (up to permutations MATH and MATH): MATH, MATH and MATH. Since MATH is integral, MATH. REF gives MATH. By REF, MATH where MATH, MATH are some points. On the other hand, MATH (see CITE) a contradiction. This proves our theorem in the case when MATH is disconnected. CASE: By REF below there is a MATH-complement. Note that in this case the assumption MATH is not needed. First we claim that MATH. Indeed, by Adjunction we have MATH . If MATH, then MATH is contained in fibers of MATH. But MATH has rational fibers (see REF), a contradiction. Let MATH be a rational projective log surface, where MATH is the reduced and MATH is the fractional part of the boundary. Assume that MATH is analytically dlt, MATH, MATH is connected and MATH. Then MATH, MATH, MATH is smooth along MATH and has only NAME singularities outside. By REF , MATH is smooth and MATH near MATH. Replace MATH with a minimal resolution and MATH with the crepant pull back. It is sufficient to show only that MATH and MATH. Now we contract MATH-curves on MATH. Since MATH is not a tree of rational curves, it cannot be contracted. This process preserves all the assumptions, so on each step MATH. Since every MATH-curve MATH has positive intersection number with MATH, we have either MATH, MATH or MATH, MATH. If MATH, then whole MATH also cannot be contracted. At the end we get MATH or MATH (a NAME surface). In the case MATH, MATH is ample. Hence MATH is connected. If MATH we derive a contradiction. Consider the case MATH. Then MATH and the last two terms are nonnegative. Therefore MATH and MATH. On the other hand, for a general fiber MATH of MATH one has MATH . In particular, MATH. Since MATH, MATH is not a section of MATH at a general point. Hence MATH and MATH. Recall that MATH. Since MATH, we have MATH and MATH. We proved that MATH, MATH and MATH in the case MATH. Therefore on our original MATH one also has MATH. By REF we can pull back a MATH-complement of MATH under contractions of MATH-curves. CASE: In this case, MATH is plt. We claim that after a number of birational contractions MATH. Indeed, otherwise there is a component MATH of MATH with coefficient MATH and MATH. If MATH, then as in REF we can reduce MATH a little so that MATH becomes big and obtain a regular complement by REF (note that MATH). This complement is also a complement of our original MATH by REF . If MATH, then we can contract MATH and pull back complements by REF . Consider the case MATH. By Base Point Free Theorem the linear system MATH determines a contraction MATH such that MATH is a fiber of MATH. Contracting curves MATH in reducible fibers we get the situation when MATH, that is, MATH is an extremal contraction. Let MATH be another extremal contraction on MATH and MATH a nontrivial fiber of MATH. Then MATH and MATH (because MATH is not a fiber of MATH). Assume that MATH is nonbirational and MATH is sufficiently general. Then MATH is smooth along MATH, MATH and MATH intersects MATH transversally. Hence MATH and we can write MATH where MATH and MATH. Moreover, the coefficient MATH of MATH at points MATH satisfies MATH. Further, MATH . Easy computations as in REF show that this is impossible. Indeed, MATH . Clearly, MATH (otherwise MATH but MATH). Since MATH, we have MATH. Thus MATH . This gives us MATH and MATH. Hence MATH. It is easy to check that the last inequality has no solutions with MATH and MATH. If MATH is birational and contract MATH (that is, MATH), then MATH (by REF ) and has a coefficient MATH. Moreover, MATH is plt, so MATH is klt. As above we derive a contradiction with MATH. Finally, if MATH is birational and does not contract MATH, we can replace MATH with MATH. We get the situation when MATH and MATH, as above. Thus we may assume now that MATH. By Base Point Free Theorem and assumptions of the theorem, there exists MATH such that MATH. Let MATH be the index of MATH (that is, the minimal positive integer with this property) and MATH the log canonical MATH-cover (see REF). It is sufficient to show that MATH. Write MATH where MATH. Then MATH is linearly trivial and plt (see REF ). By Adjunction every connected component of MATH is a smooth elliptic curve. First we assume that MATH is connected. By construction, MATH and we can identify MATH with MATH. We claim that MATH contains no translations. Indeed, let MATH a translation. Then we put MATH and MATH. By REF , MATH is linearly trivial and plt (because MATH). But then MATH, where MATH is the degree of MATH. By REF is the smallest positive with this property. The contradiction shows that MATH contains no translations. Then MATH is a finite group of order MATH, MATH, MATH, or MATH (see for example, CITE). If MATH is disconnected, then MATH interchange connected components of MATH. By REF there is a contraction MATH onto an elliptic curve with rational fibers such that components of MATH are sections. This contraction must be MATH-equivariant because MATH has a unique structure of a contraction with rational fibers. Set MATH. Since the ramification locus of MATH does not contain components of MATH, MATH. As above we consider MATH and MATH. Then MATH is a smooth elliptic curve, hence MATH by REF . This contradicts our choice of MATH. CASE: By REF , MATH is a chain and MATH has coefficients MATH near MATH. As in REF we claim that after some birational contractions MATH. Let MATH be a component with nonstandard coefficient. If MATH, then we can argue as in REF . The only nontrivial case is MATH and MATH. Then again MATH determines a contraction MATH. Clearly, MATH is a fiber and MATH is contained in a fiber (because MATH). There is an extremal rational curve MATH which is not contained in fibers. Then MATH. If MATH, we contract MATH and replace MATH with a new birational model. If MATH we derive a contradiction computing MATH as in REF of REF . Thus we may assume that MATH. As in REF take the log canonical MATH-cover MATH. It is sufficient to show that MATH. Again we can write MATH where MATH. Obviously, MATH is reducible. We claim that MATH is dlt. By REF MATH is plt outside of MATH. Recall that the ramification divisor of MATH is MATH. Hence none of irreducible components of the ramification divisor intersects MATH. At points MATH the surface MATH is smooth, so MATH is étale over MATH. Therefore MATH is dlt and MATH is smooth at points MATH. Since MATH, MATH is smooth along MATH (see REF) and MATH. By REF , MATH is a wheel of smooth rational curves and by our construction MATH acts on MATH faithfully. Let MATH be the irreducible decomposition and MATH singular points of MATH. If MATH contains an element MATH such that MATH and MATH for all MATH, then MATH is again a wheel of smooth rational curves. As in REF we derive a contradiction. Therefore MATH acts faithfully on the dual graph of MATH and then it is a subgroup of the dihedral group MATH. The same arguments show that every MATH has a fixed point on MATH. This is possible only if MATH. Therefore MATH and MATH is MATH-complementary. CASE: We may contract all components of reducible fibers of MATH which are different from components of MATH. Thus MATH, MATH is a fiber and all other fibers of MATH are irreducible. Again we may assume that MATH-horizontal components of MATH have standard coefficients (otherwise some MATH and MATH is nef and big for MATH, hence we can use REF ). Note that the horizontal part MATH of MATH is nontrivial (because a general fiber of MATH is MATH). Further, there is a regular MATH-semicomplement of MATH (see REF ). For sufficiently small MATH the MATH-divisor MATH is nef and big. Thus we can extend MATH-semicomplements of MATH from MATH by REF . If MATH is not a denominator of coefficients of MATH, then by REF we obtain a regular complement of MATH. If MATH is reducible, then by REF we can take MATH. On the other hand, by REF coefficients of MATH are equal to MATH. Therefore MATH is not a denominator of coefficients of MATH in this case and there is a MATH-semicomplement of MATH. Now we assume that MATH. Then MATH. By REF and by REF we have the following possibilities: MATH . By REF , MATH is not a denominator of coefficients of MATH in all cases. CASE: So, we assume that there is a horizontal component MATH (that is, such that MATH). It is clear that MATH is nef and big for a sufficiently small positive MATH. If MATH (that is, MATH is reducible), then the same trick as in REF (using REF ) gives the existence of regular complements. Thus we may assume that MATH and MATH. In particular, MATH is plt. Contracting curves MATH in fibers we get the situation, when fibers are irreducible, that is, MATH. We can pull back complements by REF. There are two subcases: CASE: MATH is a section of MATH, and REF MATH is a MATH-section of MATH. CASE: If MATH is a section of MATH, then the horizontal part MATH of MATH is nontrivial and either MATH or MATH, where MATH, MATH are sections and MATH is a MATH-section. As above we can take a regular MATH-complement for MATH. If MATH, then again this gives a regular complement of MATH by REF . But on MATH there exists a regular MATH, MATH, MATH, or MATH-complement for the boundary MATH. Indeed, otherwise by REF there is a MATH-complement MATH. By REF , MATH. Therefore MATH and MATH is supported in one or two points (because MATH). Then MATH is also MATH-complement for any MATH. This shows that we have a regular complement in the case when MATH is a section. CASE: Now let MATH be a MATH-section of MATH. Then MATH is contained in the fibers of MATH. Since MATH, the restriction MATH has exactly two ramification points, say MATH, MATH. Put MATH and MATH, MATH. Notation as in REF. Let MATH be a fiber of MATH such that MATH and MATH (where MATH). Then CASE: MATH; CASE: MATH is lc and linearly trivial near MATH; CASE: MATH is plt; CASE: MATH is invariant under the natural NAME action of MATH on MATH. First we show that MATH is lc. Assume the converse and regard MATH as an analytic germ near MATH. Let MATH, MATH be analytic components of MATH. If MATH is not lc near MATH, then MATH is not, either. But in this case, MATH is not connected. This contradicts REF . Now, by Adjunction MATH . On the other hand, MATH . This gives that in REF equalities hold. Hence MATH and MATH near MATH and proves REF . By REF , MATH is NAME near MATH. Since MATH, we have MATH. This proves REF . REF easily follows by REF . Further, MATH . To show REF we just note that MATH is of type MATH of REF near MATH (because MATH is plt). In particular, we have MATH. As in REF using that MATH and MATH, we have the following cases (up to permutations of MATH, MATH): MATH where MATH for some MATH, MATH, MATH and MATH. Our strategy is very simple: we construct a boundary MATH such that MATH is contained in fibers of MATH, MATH is lc and numerically trivial. If MATH, then we can use proved cases with MATH. If MATH, then we show that MATH for some MATH. The numerical equivalence in MATH coincides with linear one (see REF ), it is sufficient to show only that MATH is NAME. Note that MATH is trivial on fibers of MATH (because MATH is contained in fibers). On the other hand, MATH . Since MATH, we have MATH . Let MATH be a germ of a contraction from a surface to a curve, MATH, and MATH a germ of a curve such that MATH is one point. Assume that MATH, MATH is plt and numerically trivial. Then there is a MATH or MATH-complement MATH (with MATH) such that MATH is not plt at MATH. Note that a general fiber of MATH is MATH and MATH is a MATH-section of MATH. Take MATH so that MATH is maximally log canonical (that is, MATH is maximal such that with the log canonical property of MATH, see REF). Then MATH. We claim that MATH is not plt at MATH. Indeed, otherwise MATH is not connected near MATH. This is a contradiction with REF . In particular, MATH is not exceptional. By REF , MATH has a MATH, or MATH-complement MATH which is not plt at MATH. In particular, MATH. Since MATH is a MATH-section of MATH, MATH has no horizontal components. Hence MATH or MATH. Now we consider possibilities of REF step by step. CASE: Then MATH is smooth outside of MATH. By REF there are MATH such that MATH, MATH is lc and not plt at MATH, MATH and MATH near MATH, MATH. By Adjunction (see REF) MATH. By REF, MATH. Moreover, MATH is NAME near MATH and MATH. Therefore MATH is NAME on MATH (because MATH). CASE: Then MATH for some fiber MATH of MATH. By REF MATH is lc. Since MATH, MATH is numerically trivial. By the above cases with MATH there is a regular complement of MATH (actually, MATH and we can use REF to show the existence of a MATH-complement). CASE: By REF there is MATH such that MATH near MATH, MATH is lc and not plt at MATH. Since MATH, MATH. Again by the above cases with MATH there is a regular complement of MATH (more precisely, MATH is not plt, so we can use REF ). CASE: Then we take MATH so that MATH. By REF, MATH. We show that MATH is NAME. First note that MATH (and MATH) is NAME along MATH. Indeed, MATH is smooth and MATH. On the other hand, MATH. Hence the multiplicity MATH of the fiber MATH is at most MATH and MATH near MATH over MATH. By REF there is a MATH-complement MATH near MATH which is not plt at MATH. If MATH, then MATH near MATH. Assume that MATH. Then MATH. Hence MATH and the fiber MATH is not multiple. So MATH is NAME and NAME is smooth along MATH. This yields MATH near MATH. Write MATH and MATH . Then by REF , MATH where MATH. Since MATH, we have MATH. This gives only the following possibilities for MATH: CASE: MATH (that is, MATH is smooth along MATH) and MATH; CASE: MATH (that is, case MATH of REF with MATH) and MATH. It is easy to see that MATH is NAME near MATH in both cases. Now it is sufficient to show only that MATH near MATH. Similarly, we obtain only the following possibilities for MATH: CASE: MATH (that is, MATH is smooth) and MATH; CASE: MATH (that is, MATH is NAME of type MATH) and MATH, Assume that MATH is smooth. Since MATH is plt, MATH intersects MATH transversally (see REF). Hence MATH and the coefficient of MATH in MATH is MATH. We claim that MATH near MATH. Indeed, MATH and MATH implies that the multiplicity of the fiber MATH is at most MATH and MATH over MATH. On the other hand, by REF there is a MATH-complement MATH near MATH which is not plt at MATH. By our assumptions, MATH is plt. Thus MATH and MATH near MATH. This yields MATH near MATH. Now we assume that MATH is NAME of type MATH. Then MATH is not a component of MATH. As above, by REF there is a MATH-complement MATH near MATH which is not plt at MATH. If MATH, then MATH is lc and by REF , MATH. It is easy to see that in this case MATH . Hence we can take MATH. Then near MATH we have MATH . On the other hand, the multiplicity of the fiber MATH divides MATH (because MATH and MATH). This gives as MATH and MATH near MATH. Finally, let MATH. By REF MATH is lc but not plt at MATH. Then MATH . Hence MATH and as above, MATH. Similarly, we obtain MATH and MATH near MATH. This finishes the proof of REF .
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math/9912111
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If MATH is not klt, then there is a regular complement by REF . From now on we assume that MATH is klt. Replacing MATH with suitable MATH we also may assume that MATH is lc (and not klt). First we consider the case when MATH. Let MATH be a component of MATH. Replace MATH with MATH: MATH . If MATH is nef, then we can apply REF (because MATH and MATH). Further, we assume that MATH is not nef. Then there exists a MATH-nonpositive extremal ray MATH such that MATH. If it is birational, then we contract it. Since MATH is nonpositive on MATH, this preserves the lc property of MATH and MATH. We can pull back regular complements of MATH because MATH (now we are looking for regular complements of MATH, see REF ). Note also that MATH. Therefore MATH is not contracted and on each step MATH is not klt. If on some step MATH is nef, we are done. Otherwise continuing the process, we obtain a nonbirational extremal ray MATH on MATH such that MATH. But on the other hand, MATH a contradiction. Consider now the case MATH. Then MATH is lc, but is not plt. Recall that MATH is klt. As in REF we can construct a blowup MATH with an irreducible exceptional divisor MATH such that MATH, the crepant pull back MATH is lc and MATH is plt for any MATH, MATH. Here MATH and MATH are proper transforms of MATH and MATH, respectively. Note also that MATH. Write MATH where MATH. Assume that there exists a curve MATH such that MATH. Then MATH. Therefore MATH is a component of MATH and MATH. Further, MATH. Hence MATH and we can choose MATH so that MATH. Therefore MATH is a MATH-negative extremal curve and its contraction preserves the lc property of MATH. Again we can pull back complements of MATH (see REF ). Repeating the process, we get the situation when MATH is nef. All the steps preserve the nef and big property of MATH. If MATH, then we apply REF to MATH. If MATH, then by the monotonicity, MATH is nef and big. Again apply REF to MATH. This concludes the proof of the corollary.
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math/9912111
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REF is obvious. Assume that MATH. Write MATH where MATH and MATH is the proper transform of MATH. By REF is lc and not klt (see REF). Therefore MATH is not plt and we can take MATH so that MATH is lc but not plt. It is easy to see that MATH is MATH-ample. If MATH, then MATH is a log NAME. By REF (or by REF) there is a regular complement of MATH and by REF it can be extended to a complement of MATH. Similarly, in the case when MATH is a curve, we can use REF .
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math/9912111
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By REF it is sufficient to show the existence of a regular complement. If MATH is not rational, then the assertion follows by REF . Otherwise we can apply REF . The existence of MATH in conditions of the theorem follows by REF below.
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math/9912111
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Replace MATH with its minimal resolution and MATH with its crepant pull back. Then again MATH is a log NAME surface. By REF it is sufficient to construct MATH on this new MATH. Further, there is a sequence of contractions of MATH-curves MATH, where MATH or MATH, MATH, MATH. Put MATH. Then MATH is again a log NAME surface. It is sufficient to construct MATH such that MATH is klt and MATH is nef and big. Indeed, the crepant pull back MATH of MATH satisfies REF . However, MATH is not necessarily a boundary (that is, effective). To avoid this one can take MATH for MATH. Further, if MATH or MATH, then we take MATH. In the case MATH with MATH, we write MATH, where MATH is the negative section of MATH, MATH, MATH, and MATH is not a component of MATH. It is easy to see MATH . Hence, MATH. Thus we can take MATH.
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math/9912111
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Assume that MATH CASE: First we consider the case MATH. CASE: Apply a minimal log terminal modification as in REF. It is easy to see that this preserves the left hand side of REF. Thus we may assume that MATH is dlt. In particular, MATH is klt, MATH is MATH-factorial and MATH. CASE: Write MATH, where MATH and MATH. Then MATH. Hence MATH cannot be nef. Run MATH-MMP, that is, contract birational extremal rays MATH such that MATH. The left hand side of REF does not decrease. Of course, we can lose the dlt property of MATH, but REF - REF are preserved. Moreover, if MATH, then on each step we contract a curve MATH with MATH. In particular, whole MATH is not contracted. At the end we get a nonbirational contraction MATH. CASE: Assume that after REF we get a NAME contraction MATH with MATH. Write MATH, where MATH is the sum of all components such that MATH and MATH is the sum of components which are fibers of MATH. Let MATH be a general fiber of MATH. Then by Adjunction MATH . This gives MATH. In particular, MATH (because MATH). Now, let MATH be the extremal ray of MATH other than MATH. Then MATH. Hence MATH for MATH. By Contraction Theorem there is a contraction MATH of MATH. Assume that MATH. Then, as above, we have MATH (because components of MATH are horizontal with respect to MATH). This yields equalities in REF. If MATH, then MATH is birational. Let MATH be the MATH-exceptional divisor. If MATH is a component of MATH, then again by Adjunction we have MATH . Since any component of MATH meets MATH, by REF we obtain MATH. This yields equalities in REF. Finally, if MATH is not a component of MATH, then we replace MATH with MATH. Note that in this case we get strict inequality MATH in REF and MATH. By the next two steps this is a contradiction. CASE: Assume that MATH is a point (and MATH). Then MATH. We claim that after perturbation of coefficients one can obtain the case when MATH is not klt. Indeed, assume that MATH is klt. Let MATH be the ample generator of MATH (see REF ) and let MATH, MATH. Without loss of generality we may assume that MATH. Take MATH, where MATH and MATH. Clearly, MATH is again nef and MATH is effective. Moreover, for MATH the left hand side of REF remains the same. If MATH is lc but not klt for some for MATH, then MATH gives the required boundary. If MATH is klt for MATH, then we replace MATH with MATH continue the process with another pair MATH. Since the last procedure reduces the number of components of MATH, this process terminates. At the end we get the situation when MATH is not klt. CASE: Now we consider the case when MATH is not klt. Apply REF - REF again. On REF the equality holds. So we assume that MATH and MATH. For any component MATH by Adjunction we have MATH . On the other hand, all components of MATH intersect MATH and MATH (see REF ). Therefore MATH and MATH. This completes the proof in the case MATH. CASE: Consider the case MATH. As in REF we may assume that MATH is dlt. Further, similar to REF , run MATH-MMP. This preserves REF . Let MATH be a birational extremal contraction, MATH, and MATH the exceptional curve. Clearly, MATH, where MATH. Then MATH. Hence MATH. Assume that MATH and MATH a hyperplane section of MATH. Then MATH, a contradiction. Therefore MATH. We can replace MATH with MATH and continue the process. At the end we get a log surface MATH with a nonbirational MATH-negative extremal contraction MATH. In particular MATH. If MATH is a point, then MATH and MATH is ample. Take MATH so that the divisor MATH is integral and very ample. Let MATH be a general member. Then MATH is dlt and numerically trivial. In this case, by the proved REF , MATH, a contradiction. If MATH is a curve, then we can use the arguments of REF . Thus MATH (because we have strict inequality in REF). The last contradiction proves that MATH. Assume that MATH has at least one nonrational singularity MATH. Clearly, MATH is not klt at MATH and MATH. Then by REF MATH is a simple elliptic or cusp singularity. As in REF , let MATH be a minimal log terminal modification. If MATH is connected, then MATH and MATH. By REF MATH and MATH, a contradiction. Therefore MATH is a connected component of MATH. Denote MATH and MATH. Our REF implies that MATH. Then MATH. By REF there is a contraction MATH with rational fibers onto a curve MATH such that MATH and MATH are (smooth) disjoint sections. Then MATH has no horizontal components. Let MATH be a MATH-negative extremal rational curve. Since MATH, MATH cannot be horizontal. On the other hand, MATH (because MATH). Therefore the contraction of MATH is birational. This contraction reduces the left hand side of REF, a contradiction. This proves REF .
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math/9912111
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REF follows by REF . This also shows that MATH. To prove REF we apply REF - REF of the proof of REF to MATH. At the end we obtain one of the following: CASE: MATH and MATH has exactly three components. Clearly they intersect each other and does not pass through one point, so MATH (compare Proof of REF ). By REF , MATH is a MATH-complement. According to REF, MATH on our original MATH is MATH-complementary. CASE: MATH and MATH has exactly four components. Moreover, there is an extremal contraction MATH onto a curve. By discussions in REF of the proof of REF (especially, REF), we have a decomposition MATH such that both MATH and MATH have two irreducible components. Any component of MATH meets all components of MATH. As in REF we have MATH. Finally, as above, MATH is MATH-complementary. By REF , MATH is connected. Since MATH is NAME, MATH (see REF). Thus MATH and MATH. This proves REF . The assertion of REF follows by the lemma below.
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math/9912111
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Let MATH be minimal lt modification of MATH. Write MATH, where MATH is reduced and MATH. The exceptional divisor of MATH is contained in MATH. Hence on MATH all our conditions hold. So it is sufficient to prove our assertion for MATH. By REF , MATH, MATH a wheel of smooth rational curves, MATH is smooth along MATH and has only NAME singularities outside. Run MATH-MMP. By REF, on each step we contracted a component of MATH (which is contained into the smooth locus of MATH). Thus our MMP is a sequence of contractions of MATH-curves. At the end we obtain a NAME contraction MATH, where MATH has only NAME singularities, MATH is lc (in fact it is analytically dlt) and numerically trivial. The sequence of transformations MATH is a sequence of blowups of divisors with discrepancies MATH. They must preserve the action of a two-dimensional torus (if MATH is a toric pair). Thus it is sufficient to show that the pair MATH is toric. If MATH, then MATH has exactly three components which are NAME divisors. Therefore MATH is a log NAME of NAME index MATH (see REF). By REF , MATH and MATH, where the MATH are lines. Obviously, MATH is toric in this case. Finally, assume that MATH. Then MATH has exactly four components and by REF they form a wheel of smooth rational curves. It is an easy exercise to prove that MATH and the fibers of MATH are rational. Therefore MATH, where MATH, MATH are disjoint sections of MATH and MATH, MATH are fibers. We claim that MATH is smooth. Indeed, by construction, MATH is smooth along MATH. Let MATH be a fiber of MATH different from MATH, MATH. Take MATH so that MATH is maximally lc. If MATH, then MATH has three connected components near MATH. This contradicts REF . Hence MATH is lc. By Adjunction MATH. On the other hand, MATH, where MATH. Hence MATH and MATH is smooth along MATH. Thus we have shown that MATH is smooth. Then MATH and MATH is the natural projection MATH. Since MATH, MATH are disjoint sections one of them, say MATH, must be the minimal section MATH. Now, it is easy to show that the pair MATH is toric.
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math/9912111
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By REF the log divisor MATH has a regular MATH-complement MATH near each point MATH. Then MATH is lc and MATH (see REF). Moreover, if MATH, then by definition of complements MATH. Therefore MATH. This gives that MATH is also lc at MATH.
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math/9912111
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Indeed, otherwise MATH and we have an extremal contraction MATH onto a curve, positive with respect to MATH. Let MATH be a general fiber. Then MATH and by REF and by REF, MATH is MATH, MATH, MATH, MATH, or MATH-complementary. But then (as in the proof of REF ) MATH. Hence MATH which is a contradiction.
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math/9912111
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By REF and by REF there exists a regular complement of MATH near MATH. As in the proof of REF , MATH. If MATH is a component of MATH, then MATH, a contradiction.
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math/9912111
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Denote MATH and let MATH be the exceptional divisor. Then we can write MATH, where MATH. Since MATH is MATH-lt, by REF it is sufficient to show only that MATH does not contract components of MATH with coefficients MATH.
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math/9912111
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By REF there is a regular complement MATH. Clearly, MATH. Therefore MATH is lc. By REF MATH has exactly two (analytic) components at MATH and we have an analytic isomorphism MATH . Taking the corresponding weighted blowup, one can compute MATH (see REF ). This yields MATH a contradiction. Hence MATH and MATH is smooth. The rest is obvious.
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math/9912111
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Assume that there is a sequence MATH of boundaries as in our lemma such that MATH for some (fixed) MATH. By REF and because MATH we have MATH, for all MATH. Then MATH for MATH. Denote MATH. By the above, MATH is a MATH-boundary and MATH. Moreover MATH is lc and MATH is nef (see for example, CITE). By the Inductive REF there is a regular MATH-complement MATH of MATH and this is a MATH-complement of MATH if MATH (see REF ). A contradiction with our assumptions.
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math/9912111
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If MATH is rational, then we can omit REF by REF . Otherwise there is a regular complement by REF below.
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math/9912111
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By REF we may assume that MATH is klt. First consider the case when MATH. By REF the pair MATH has at worst canonical singularities. Replace MATH with the minimal resolution and the crepant pull back. It is easy to see that this preserves all the assumptions. Run MATH-MMP. Clearly, whole MATH cannot be contracted. At the end we get an extremal contraction MATH to a smooth curve MATH with MATH. Moreover MATH is smooth, so MATH is a ruled surface. Thus MATH. On each step we can pull back regular complement, so it is sufficient to prove our statement for this new MATH. By REF all the components of MATH are horizontal. Further, if MATH for some component MATH, then MATH is nef and big, a contradiction (see REF ). On the other hand, if MATH, then MATH. Thus MATH is an extremal rational curve, a contradiction with MATH. Therefore MATH for all components MATH. Since MATH, this gives that all the MATH are numerically proportional and MATH for MATH. As in REF we have MATH. Hence MATH and all MATH and MATH are smooth elliptic curves. Moreover, MATH. Since MATH, MATH generate an extremal ray of MATH (see REF ). In particular, MATH contains no curves with negative self-intersections. Restricting MATH to a general fiber MATH of the rulling MATH we get a numerically trivial divisor MATH with MATH. Obviously, MATH and MATH. Thus we can write MATH . There is a structure of an elliptic fibration MATH. All components of MATH are contained in fibers of MATH. If MATH has at least two components MATH and MATH, then MATH is klt and numerically trivial for some small positive MATH. By the Log Abundance Theorem CITE, we have MATH, that is, MATH. This gives an elliptic pencil. Assume that MATH, where MATH is an irreducible smooth elliptic curve. Clearly, MATH. If the rulling MATH corresponds to a vector bundle of splitting type, then there is a section MATH such that MATH. Again we can apply the Log Abundance Theorem to MATH. Finally, consider the case when MATH corresponds to an indecomposable vector bundle MATH. By REF the degree of MATH is odd. Then up multiplication by an invertible sheaf, MATH is a nontrivial extension MATH where MATH is a point on MATH (see for example, CITE). Let MATH be a section corresponding to the above exact sequence. Denote also MATH. It is easy to see that MATH is nef but not big (see CITE). It is sufficient to show the existence of an effective divisor MATH. Indeed, then MATH is irreducible, MATH and the same arguments as above gives an elliptic pencil. To find such MATH we consider the linear system MATH. Since MATH, MATH is ample. By NAME and NAME vanishing, we have MATH . Thus the surjection MATH is surjective. Therefore the linear system MATH is base point free and determines a finite morphism MATH of degree MATH. Since MATH, the images of the fibers MATH form a pencil of curves of degree MATH. Since any pencil of conics contains a degenerate member, we derive that MATH is a line for at least one fiber MATH. Then the residual curve MATH belongs to the linear system MATH. The claim is proved. As above, let MATH, where MATH is an indecomposable vector bundle of odd degree. Describe the structure of elliptic fibration on MATH and multiple fibers explicitly. Going back to the proof of REF , assume that MATH contains also all multiple fibers of the fibration MATH (we alow MATH). By the canonical bundle formula (see for example, CITE) MATH where MATH is a general fiber and MATH is the multiplicity of MATH. This gives us MATH . Since MATH, MATH . Hence the collection MATH is one of the following (see REF): MATH . By REF we see that MATH for MATH, and MATH, respectively. Finally, let MATH. Replace MATH with its minimal resolution and let MATH be the crepant pull back of MATH. If again MATH, then by the classification of smooth surfaces of NAME dimension MATH we have MATH for MATH. If MATH, then we run MATH-MMP (that is, contract MATH-curves step by step). As above, at the end we get a contraction MATH to a smooth curve MATH with MATH. On the other hand, all components of MATH are rational (because so are singularities of our original MATH) and at least one component of MATH is horizontal, a contradiction. The proposition is proved.
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math/9912111
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By REF it is sufficient to check it for two series of nonexceptional klt singularities. Consider, for example, singularities of type MATH, that is, it has the following dual graph of the minimal resolution: MATH . Then the discrepancies MATH can be found from the system of linear equations (see CITE, CITE): MATH . This yields MATH . Finally, by CITE, MATH . We left computations in case MATH (with the dual graph as in REF ) to the reader.
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math/9912111
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Let MATH. From the exact sequence MATH where MATH is a sheaf with MATH, we have MATH . On the other hand, by REF we have MATH. This yields MATH . In particular, MATH. Assume that MATH. Then MATH is a wheel of smooth rational curves and in REF the equality holds. Let MATH be an ample generator of MATH. We have MATH, MATH for some positive rational MATH, MATH, MATH, MATH. Since every MATH intersects MATH transversally at a (unique) nonsingular point, MATH. Hence MATH . This implies MATH . Since MATH is ample, MATH . Therefore MATH is antiample (and lc). We claim that MATH is smooth along MATH. Indeed, otherwise MATH, where MATH. On the other hand, by Adjunction we have MATH . The contradiction shows that MATH is smooth along MATH, and similarly MATH is smooth along MATH and MATH. Thus MATH, MATH, MATH are NAME. In particular, MATH. By REF, MATH and MATH. Since MATH, MATH. The linear subsystem of MATH generated by MATH is base point free and determines a morphism MATH of degree one (see also REF below). Therefore MATH and MATH are lines in the general position. Simple computations show that MATH has no other components. Finally, MATH is a MATH-complements of MATH, a contradiction proves the corollary.
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math/9912111
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Clearly, we may assume that MATH is not plt (otherwise we have REF ). By REF there is a regular complement MATH. Since MATH, MATH. In particular, MATH is lc and MATH has at most two (analytic) components passing through MATH (see REF ). If MATH has exactly two components, then MATH is smooth by REF . Obviously, MATH is analytically dlt at MATH in this case. From now on we assume that MATH is analytically irreducible at MATH. Write MATH, where MATH. Recall that MATH. First we consider the case when MATH is not plt. Then MATH and MATH is such as in REF. In particular, MATH and MATH. Let MATH be an inductive blowup of MATH and MATH the proper transform of MATH. Write MATH where MATH. Here MATH. By Adjunction, MATH is not klt and MATH. Moreover, MATH is not MATH-complementary (because neither is MATH). Therefore we have (compare REF ) MATH for some points MATH and some MATH. From this we have MATH . By Adjunction MATH . Since MATH is a point of type MATH, MATH. This yields MATH . Thus MATH and MATH. On the other hand, MATH is NAME near MATH. Therefore MATH, a contradiction. Now we may assume that MATH is plt. By REF , MATH is MATH-complementary and MATH, so MATH and MATH is integral. We claim that MATH is smooth. Assume the opposite. Then MATH . Consider the weighted blowup with weights MATH. By REF we get the exceptional divisor MATH with MATH where MATH. Since MATH, we have MATH or MATH. But in the second case MATH, a contradiction. Therefore MATH and MATH. Further, MATH . The contradiction shows that MATH is smooth. Now we claim that MATH is a smooth curve. As above, consider the blowup of MATH. For the exceptional divisor MATH, we have MATH where MATH. Hence MATH and MATH is smooth. Finally, MATH is plt for any MATH. By Adjunction, MATH. Hence MATH is reduced. This means that MATH, that is, MATH and MATH have a simple tangency at MATH. The rest is obvious.
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math/9912111
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By NAME vanishing CITE one has MATH for MATH. Therefore by NAME we obtain MATH . This proves REF . Recall (see CITE) that for any polarized variety MATH the following equality holds: MATH . Combining this with REF we obtain REF . Finally, assume MATH. Then by REF , MATH. From REF we have MATH. Moreover, in REF the equality holds. Such polarized varieties (of arbitrary dimension) are classified in CITE. In particular, it is proved that MATH is very ample and MATH are varieties of minimal degree. In the two-dimensional case from CITE we obtain possibilities as in REF .
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math/9912111
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Since MATH and MATH, MATH is ample. Hence MATH is rational. By REF Then MATH has only normal crossings at smooth points of MATH, MATH does not pass MATH and MATH (by REF ). Write MATH . We assume that MATH. Since MATH is nef, MATH . Take MATH so that MATH, that is, MATH . Then MATH . Since MATH is ample, MATH. Recall that MATH is analytically dlt except for REF . In particular, MATH is smooth at points MATH and MATH. By Adjunction, MATH . If MATH, then MATH. This is impossible because MATH. Therefore MATH and MATH. CASE: Then MATH and MATH. From REF we obtain MATH or MATH. On the other hand, MATH is ample. This gives MATH . If MATH, then by REF, MATH, a contradiction. Therefore MATH, MATH are lines on MATH. Then MATH . If MATH, then MATH, MATH and MATH, a contradiction. Hence all the components of MATH are lines. From REF we have only two possibilities: MATH and MATH. These are cases MATH and MATH. From now on we assume that MATH is singular. Since MATH, we have two possibilities: MATH and MATH. CASE: Let MATH. Then MATH . Equality REF gives MATH . Hence MATH . By Inversion of Adjunction, MATH is plt near MATH. Assume that MATH. Then MATH and MATH, a contradiction with MATH. Therefore MATH. Since MATH (see REF ), we have only one possibility: MATH, where MATH is a single point MATH. Moreover, MATH and MATH. If MATH is smooth, then MATH, where MATH is irreducible, MATH and MATH. Thus MATH is NAME (see REF), MATH and MATH. By REF MATH is a cone over a rational normal curve of degree MATH. In this case MATH and MATH. Therefore MATH is a quadratic cone. Further, MATH, so MATH, MATH are hyperplane sections (and they do not pass through the vertex of the cone). Finally, from MATH we see that MATH is a generator of the cone. This is case MATH. Therefore MATH is singular. Then it must be NAME of type MATH. Moreover, MATH and MATH is NAME (but MATH is not, because MATH is smooth at MATH). Hence MATH. Further, MATH. If MATH, then MATH and MATH. By REF and our assumption that MATH is singular, MATH and MATH is a quadratic cone. But then MATH, a contradiction. Hence MATH, MATH. Put MATH. Then MATH, MATH, so MATH, MATH. This gives MATH. On the other hand, by REF , MATH, where MATH. Therefore MATH and MATH. If MATH, then MATH, MATH, MATH. But this contradicts MATH. We obtain MATH, MATH, MATH, MATH, MATH. By REF , MATH. We may assume that MATH and MATH. Then MATH and MATH, MATH. But MATH (otherwise MATH is singular at MATH). Moreover, MATH, because MATH consists of two points. This is case MATH. CASE: By REF we may assume that MATH is a single point, say MATH. As in REF take MATH so that MATH, that is, MATH . Since MATH, we have MATH and MATH, MATH. This yields MATH . Assume that MATH. Then MATH . On the other hand, by REF, MATH where MATH (see REF). Hence MATH . Since MATH, we have only one possibility MATH and MATH. In particular, MATH is NAME (see REF ), so MATH. On the other hand, MATH (see REF). Hence MATH and MATH is not NAME for MATH. This gives us that MATH. Moreover, in REF equalities hold, so MATH and MATH. From REF we have MATH. Hence MATH. Further, MATH, gives MATH. By REF , MATH. We get case MATH. Now assume that MATH. Then MATH is a cone. From MATH we see that MATH and MATH is not NAME. Hence MATH contains the vertex and MATH does not. Thus MATH is NAME. Finally, MATH. Therefore MATH is a generator of the cone and MATH is a smooth hyperplane section. But then MATH is rational, a contradiction. CASE: Then MATH. By REF, MATH. Hence MATH. Using MATH we get the following cases: MATH . By Inversion of Adjunction, MATH is plt near MATH. In particular, either MATH or MATH is NAME (see REF) and MATH has at most two components. Thus MATH or MATH. Note that MATH CASE: It is easy to see MATH, so MATH. We claim that MATH. Indeed, if MATH, then MATH and MATH. Thus MATH is not NAME for MATH. By REF we have that MATH is NAME, MATH and MATH. Therefore MATH and in REF the equality holds. In particular, MATH, MATH. By REF , MATH and MATH. But then MATH is NAME and NAME, a contradiction. Thus MATH and MATH is a cone of degree MATH (see REF). Hence MATH is a smooth hyperplane section and MATH is a generator of the cone (that is, MATH, MATH). Write MATH. (Note that MATH and MATH because MATH in our case). We have MATH . Assume that MATH has a component MATH which does not pass through the vertex. Then MATH, so MATH . This gives MATH or MATH. If MATH, then MATH. From REF we get MATH, that is, case MATH. If MATH, then MATH or MATH. In both cases by REF we have MATH. For MATH we derive a contradiction with the left side of REF. We obtain case MATH. Now we assume that all components of MATH pass through the vertex MATH of the cone (in particular, MATH). Since MATH is plt at MATH (see REF ), there is at most one such a component and MATH. We claim that either MATH or MATH. Indeed, assume that MATH. Then MATH . Since MATH is smooth outside of MATH, MATH. By Adjunction, MATH at MATH. On the other hand, by REF, the coefficient of MATH at MATH is MATH . We obtain MATH a contradiction. Therefore MATH or MATH. But the second case is impossible by the right side of REF. Hence MATH. But this contradicts to the left side of REF. From now on we assume that MATH. If MATH, then MATH is a cone and contains exactly one singular point, say MATH, and MATH. Hence we may assume that MATH and MATH is NAME. Thus we may assume that MATH and MATH, MATH are not NAME. CASE: Then we have MATH . Since MATH, MATH or MATH. On the other hand, MATH. Hence MATH, MATH and MATH. By symmetry, taking into account MATH, one can see that REF holds also for MATH: MATH . From MATH we get MATH. Thus MATH and MATH is NAME by REF . By REF all singular points are contained in MATH. Since MATH is dlt (see REF ), we obtain that MATH has only NAME points of types MATH, MATH. Since MATH, MATH. By REF, MATH and MATH. Now we can use the classification of NAME surfaces with MATH (see for example, CITE). The configuration of singular points on MATH is MATH. We may assume that MATH contains the point of type MATH. Hence MATH (see REF). At least one of points MATH, MATH, MATH, MATH is smooth. Hence MATH and MATH. Thus MATH, where MATH and MATH. This implies MATH. But then MATH, a contradiction. CASE: Since MATH is not NAME, MATH and MATH is singular (of type MATH or MATH). Moreover, no components of MATH pass through MATH. Further, MATH . Since MATH, MATH. On the other hand, MATH. Thus MATH or MATH. Further, by REF , MATH, where MATH and MATH. If MATH, then MATH and MATH . This gives MATH and MATH. By REF , MATH. In particular, MATH. Since MATH is ample, MATH. Therefore MATH and moreover MATH is smooth, MATH and the intersection of MATH and MATH is transverse. Thus MATH and MATH. We may assume that MATH, MATH, and MATH, MATH. But if MATH, then MATH is not lc at MATH. On the other hand, if MATH, then MATH passes through the point MATH, a contradiction. Therefore MATH and we may put MATH. This is case MATH. If MATH, then MATH and MATH . This gives MATH or MATH. But in the first case MATH which is a contradiction with MATH (see REF). Hence MATH and MATH. Since MATH, similar to REF we have MATH. In particular, this means that MATH contains no points of index MATH. But MATH contains such a point (because MATH), a contradiction with REF. CASE: Since MATH is not NAME for MATH, MATH and MATH is a singular point of type MATH or MATH. By REF , MATH is smooth. Thus MATH, where MATH and MATH. Put MATH. Then MATH, MATH. Since MATH, MATH. If MATH, then MATH. Hence MATH. Further, by REF , MATH . On the other hand, MATH is ample, so MATH. This gives MATH . We get the following case: MATH . We claim that MATH is MATH-complementary. Note that MATH is ample. By REF and because MATH, MATH contains exactly one singular point of MATH, say MATH. Therefore MATH is supported at two points MATH and MATH. It is easy to verify that MATH is a MATH-complement. By REF this complement gives a MATH-complement MATH, where MATH is reduced and MATH. By REF , MATH is a toric pair. Such MATH is defined by a fan MATH in MATH. Let MATH, MATH, MATH be generators of one-dimensional cones in MATH. Since MATH is smooth, we may assume that MATH and MATH generate MATH. Thus we can put MATH and MATH. Therefore MATH is a weighted projective space MATH, MATH, MATH and MATH. Since MATH is singular of type MATH, where MATH or MATH and MATH, we can take MATH. Finally, from MATH we obtain MATH. This is case MATH. CASE: Since MATH is not NAME for MATH, MATH and MATH. Hence MATH and points MATH are singular. This contradicts to REF is proved.
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math/9912111
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First we prove REF . We consider only the case of compact MATH. In the case MATH there are stronger results (see REF ). Applying a log terminal modification REF, we may assume that MATH is dlt (and MATH is smooth). Set MATH, MATH. Note that MATH is connected by Connectedness Lemma. Take sufficiently large and divisible MATH and consider the exact sequence MATH . By NAME Vanishing CITE, MATH . Therefore MATH. We claim that MATH. Indeed, if MATH is not a tree of rational curves, then MATH and MATH is either a smooth elliptic curve or a wheel of smooth rational curves (see REF ). Moreover, MATH. But then MATH and MATH in this case. Note also that here we have a MATH-complement by REF . Assume now that MATH is a tree of smooth rational curves. Then MATH is base point free on each component MATH whenever MATH is NAME. Hence so is MATH. This proves our claim. Thus we have shown that MATH. Let MATH be a general member. Then MATH is dlt near MATH (see REF). By Connectedness Lemma, MATH is lc everywhere. Hence MATH is a MATH-complement of MATH. The fact that MATH is free outside of MATH can be proved in a usual way (see for example, CITE, CITE). We omit it. REF is obvious. Let us prove REF . Clearly, we may assume that MATH. It follows by REF that any MATH-negative extremal ray MATH is generated by an irreducible curve MATH. By REF , MATH. Let MATH be the contraction given by the linear system MATH for sufficiently big and divisible MATH. Then MATH. This implies that MATH belongs to a finite number of algebraic families. Thus the cone MATH is polyhedral outside of MATH. Now consider the extremal ray MATH such that MATH. By the NAME Index Theorem, MATH. Thus, by REF MATH is generated by an irreducible curve, say MATH. Since MATH, we have that MATH contracts this curve to a point. Therefore there is a finite number of such curves, so MATH is polyhedral everywhere.
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math/9912111
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We prove only REF . Take a MATH-cycle MATH so that MATH, MATH and MATH a sequence of effective MATH-cycles whose limit is MATH. Write MATH, where MATH are distinct irreducible curves. Since MATH, there is at least one curve MATH such that MATH. Write MATH, MATH. Then MATH . Thus MATH and MATH. Pick MATH. Then MATH is effective for MATH and MATH . Since MATH is an extremal ray, this implies that MATH.
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math/9912111
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Let MATH be the minimal resolution and MATH the crepant pull back. Consider the case MATH. Let MATH be the proper transform of MATH. Then MATH because MATH and MATH is a boundary. Now we assume that MATH. Then MATH is ample (see REF). Thus MATH is a log NAME surface. By REF , MATH is birationally ruled. It is well known, that in this situation MATH is covered by a family of rational curves MATH such that MATH. Take MATH. Then MATH .
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math/9912121
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If MATH and MATH have a same class, then there exists an integer MATH such that MATH. Hence, boxes which have a same class are located on the line parallel to the diagonal of the NAME diagram. For standard tableaux, numbers labeled on boxes strictly increase across each row and each column. Therefore, the box on which MATH is labeled and the box on which MATH is labeled cannot be located on the same line parallel to diagonal.
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math/9912121
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See for example, CITE or CITE .
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math/9912121
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See, for REF or REF .
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math/9912121
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A straightforward computation.
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math/9912121
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Expanding MATH with MATH's, the term which has the longest length is the nonzero scalar multiplication of MATH from the definition of MATH's. Hence MATH's are linearly independent and constitute a basis of MATH.
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math/9912121
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From REF , the multiplication of two monomial elements of even length is expressed by elements whose terms have even length. Hence it is enough to see that MATH(for MATH) are generated by MATH's. If MATH and MATH or MATH and MATH, then MATH is the generator itself. For MATH and MATH, MATH. For MATH, we obtain from REF that MATH . Thus, we complete the proof.
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math/9912121
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Consider all of elements which have even length in the set, MATH . Then they are linearly independent over MATH in MATH. Thus it is sufficient to prove that any element in MATH is expressed as a linear combination of the form, MATH . We will prove this by induction on the length of elements in MATH. Consider the element MATH in MATH. In the exchange and deletion process of expressions of words using relations of REF , leading terms of right hand sides of REF bring us same calculations as in the case of MATH, and remaining terms calculations for the products of at most MATH's. By the induction assumption, remaining terms can be expressed with a linear combination of desired form, hence MATH can be also.
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math/9912121
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Relations REF are obvious. For relation REF, MATH . For relation REF, MATH . Thus we complete the proof.
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math/9912121
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A straightforward computation.
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math/9912121
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For MATH, Let MATH be a monomial in MATH with at least two occurrences of MATH. Displaying two consecutive occurrences of MATH in MATH, we write MATH, where we can assume that MATH is a monomial in MATH, that may be MATH or MATH or MATH. For the first case, MATH. For the second case, from the relation REF , MATH . Hence we get the desired form. For the third case, from the relation REF , MATH . Applying the second case to the term MATH, we get the desired form for the third case. Thus we complete the proof for MATH. Next, let MATH be a monomial in MATH with at least two occurrences of MATH. Displaying two consecutive occurrences of MATH in MATH, we write MATH, where we can assume that MATH is a monomial in MATH. We can assume by induction that MATH contains MATH at most once. If MATH does not contain MATH at all, then by the relation REF , MATH for some MATH. In this case, MATH is vanished. If MATH contains MATH exactly once, we can write MATH with MATH in MATH and then by the relations REF , MATH for some MATH. By the relation REF , MATH reducing the number of occurrence of MATH.
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math/9912121
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The proof will be by induction. For MATH, it is obvious. Let MATH be an element in MATH. If MATH contains no MATH, then MATH. Hence by the induction assumption, MATH is expressed by a linear combination of monomials in normal form in MATH. If MATH contains MATH, then by REF , we can write MATH where MATH. By induction, MATH is a linear combination of the form MATH with MATH for MATH. By the relations REF , we have MATH for some MATH in MATH. By induction again, MATH is a linear combination of monomials of the form MATH with MATH. Thus MATH is a linear combination of monomials MATH as desired.
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math/9912121
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Consider the map MATH such that MATH. MATH defines a homomorphism of algebras. We immediately observe that this map is surjective. Hence it is enough to see that the dimension of MATH is no more than the dimension of MATH, but this was already shown in REF .
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math/9912121
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It is enough to prove for the case MATH. The element MATH acts on MATH in three ways according to where MATH and MATH appear in the standard tableau MATH. Hence, for the action of MATH, we must consider places of MATH, MATH, MATH and MATH. We claim that only two cases are possible for arrangements of MATH and MATH: they appear in the same row, or appear in the same column. If MATH, then there is no interaction between the action of MATH and MATH. But if MATH, then actions of MATH and MATH interact. Hence the action of MATH is more complicate. So, we show equalities of matrix elements as the case may be. The case MATH. CASE: If MATH and MATH appear in the same row(respectively, column) of MATH, and MATH and MATH also appear in the same row(respectively, column) of MATH, then MATH and MATH appear in the same column(respectively, row) of MATH and MATH and MATH also appear in the same column(respectively, row) of MATH. In this case, we easily get MATH and MATH. CASE: If MATH and MATH appear in the same row(respectively, column) of MATH, and MATH and MATH appear in the same column(respectively, row) of MATH, then MATH and MATH appear in the same column(respectively, row) of MATH and MATH and MATH appear in the same row(respectively, column) of MATH. In this case, we easily get MATH, and MATH. CASE: If MATH and MATH appear in the same row(respectively, column) of MATH, and MATH and MATH appear in neither the same row nor the same column of MATH, then MATH and MATH appear in the same column(respectively, row) of MATH. Using the notation of the end of REF, set MATH and we get MATH . The case MATH . By exchanging MATH and MATH, we observe that it is enough to check only in two cases: MATH, MATH and MATH appear in the same row, MATH and MATH appear in the same row and MATH and MATH appear in the same column. CASE: If MATH, MATH and MATH appear in the same row of MATH, then MATH, MATH and MATH appear in the same column of MATH. In this case, we easily get MATH and MATH. CASE: MATH and MATH appear in the same row and MATH and MATH appear in the same column of MATH, then MATH and MATH appear in the same column and MATH and MATH appear in the same row of MATH. We observe that MATH acts on MATH, MATH, MATH and MATH as scalar multiplication. Using the notation of the end of REF, set MATH and we get MATH .
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math/9912121
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From REF , representation matrices of MATH and MATH for generators of MATH are coincide. Therefore, the above statement holds.
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math/9912121
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Let MATH. We set MATH. From REF , we get following calculation for the element MATH of MATH. MATH . Thus, MATH is in MATH. Next, we get following calculation for the element MATH of MATH. MATH . Thus, MATH is in MATH.
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math/9912121
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For MATH, we consider the direct sum of irreducible representations MATH and MATH. For MATH, we have from REF the following. MATH . Thus, MATH is in MATH. The Similar calculation is valid for MATH, hence we omit the calculation for MATH.
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math/9912121
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At first, we will show the semisimplicity of MATH under the assumptions of irreducibilities and mutual inequalities of these representations. We consider the map MATH . Then, by theorems of NAME and NAME, MATH has a quotient MATH isomorphic to the semisimple algebra MATH where MATH runs over irreducible representation spaces listed in the statement of theorem. This semisimple algebra has dimension MATH and we already show that MATH has dimension MATH, thus MATH is isomorphic to MATH and semisimple. Next, we will show the irreducibilities and mutual inequalities of these representations by induction. For MATH, it is obvious. Indeed, MATH is generated by the unit element, and both MATH and MATH are identity maps. For MATH, MATH and MATH are identity map and MATH where, MATH with standard tableau MATH as follows. MATH . We get the following from the proof of REF , MATH . Since every representation has degree MATH, these representations are irreducible. Moreover, since MATH is neither MATH nor MATH-th root of unity with MATH, we immediately have MATH is nonzero. Hence they are mutually inequivalent. Let MATH. By induction assumption, MATH is a semisimple algebra with central primitive idempotents MATH,MATH,MATH, MATH,MATH,MATH,MATH, MATH,MATH with MATH,MATH,MATH and MATH,MATH,MATH are non self-conjugate and MATH,MATH, MATH are self-conjugate. Let MATH be non self-conjugate. Then there is no diagram MATH such that MATH is non self-conjugate and MATH and MATH. Indeed, for such diagram, we immediately obtain, MATH and MATH. Hence, MATH but this is contradiction. When we restrict MATH to MATH-module, we can write(for the reason, see CITE ), MATH where MATH's are all of elements in MATH such that MATH. We suppose that MATH are non self-conjugate and MATH are self-conjugate. We observe that MATH is at most MATH because it is impossible to remove one box and add another box with keeping the self-conjugacy. We can write by induction, MATH or MATH with each subspace is irreducible MATH-module and inequivalent each other. For MATH such that MATH and MATH, there is exactly one MATH such that MATH and MATH. Let MATH be such that the tableau obtained from MATH by removing MATH-th box is shape MATH and the tableau obtained from MATH by removing MATH-th box and MATH-th box is shape MATH. Similarly, let MATH be such that the tableau obtained from MATH by removing MATH-th box is shape MATH and the tableau obtained from MATH by removing MATH-th box and MATH-th box is shape MATH. If MATH is non self-conjugate, then MATH, and we get, MATH . Since MATH and MATH is not a MATH-th root of unity with MATH, MATH is well-defined and nonzero. If MATH is non self-conjugate, then MATH . If MATH is self-conjugate, then as the submodule decomposition MATH we can write MATH as follows MATH where MATH be such that the tableau obtained from MATH by removing MATH-th box is shape MATH and the tableau obtained from MATH by removing MATH-th box and MATH-th box is shape MATH. MATH and MATH. Hence MATH . If MATH is self-conjugate, then other NAME diagrams MATH with MATH are non self-conjugate. We set MATH and MATH where MATH be such that the tableau obtained from MATH by removing MATH-th box is shape MATH and the tableau obtained from MATH by removing MATH-th box and MATH-th box is shape MATH. Then we get, MATH and elements in MATH do not appear in MATH. Hence, MATH is nonzero. The same discussion is valid for MATH and we get, MATH . Because MATH is arbitrary with MATH, every MATH-submodule MATH of MATH includes whole MATH, therefore MATH is irreducible. Let MATH be self-conjugate. When we restrict MATH to MATH-module, we can write(see CITE again), MATH where MATH's are same as the non self-conjugate case. We observe that MATH is at most MATH because it is impossible to remove one box and add another box with keeping the self-conjugacy. We set for MATH, MATH where each direct summand is irreducible MATH-module with MATH,MATH, MATH and (if exist) MATH, MATH are mutually inequivalent each other by induction. Hence, from REF , we can write, MATH . For MATH such that MATH and MATH, there is exactly one MATH such that MATH and MATH. Let MATH be such that the tableau obtained from MATH by removing MATH-th box is shape MATH and the tableau obtained from MATH by removing MATH-th box and MATH-th box is shape MATH. Similarly, let MATH be such that the tableau obtained from MATH by removing MATH-th box is shape MATH and the tableau obtained from MATH by removing MATH-th box and MATH-th box is shape MATH. For MATH, MATH where MATH is a nonzero element in MATH. Hence, MATH . Because MATH is arbitrary with MATH, every MATH-submodule MATH of MATH includes whole MATH, hence MATH is irreducible. The same argument is valid for the case in MATH. Thus the irreducibilities of MATH and MATH were proved. Finally, we show inequivalencies of MATH's as MATH-module. If MATH is self-conjugate, then there is at most one self-conjugate MATH in MATH or MATH such that MATH. In this case, MATH has a direct summand MATH and does not have MATH as MATH nor MATH- module, and MATH has a direct summand MATH and does not have MATH as MATH nor MATH- module On the other hand, If MATH is non self-conjugate, then MATH has both MATH and MATH or has neither MATH nor MATH. Hence, MATH and MATH and MATH are mutually inequivalent. Next, we consider the case that different NAME diagrams MATH and MATH are both self-conjugate or the case that different NAME diagrams MATH and MATH are both non self-conjugate with MATH. In the former case, the set of all MATH such that MATH do not coincide with the set of all MATH such that MATH. Hence, MATH and MATH are inequivalent already as MATH-module. Since MATH as MATH-module and MATH as MATH-module, MATH and MATH are mutually inequivalent already as MATH-module. In the latter case, the set of all MATH such that MATH do not coincide with the set of all MATH such that MATH. Hence, MATH and MATH are mutually inequivalent already as MATH-module.
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math/9912121
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See for example, CITE .
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math/9912121
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Directly from REF we have MATH and the NAME 's Lemma means that its dimension is the multiplicity of MATH in the restriction MATH. Multiplicities of irreducible MATH-modules in the restrictions of irreducible MATH-modules are already shown in REF for the non self-conjugate case and in REF for the self-conjugate case.
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math/9912125
|
We will need the NAME duality isomorphism CITE: MATH where MATH is a compact topological space and MATH its closed subset such that MATH is a smooth orientable MATH-dimensional manifold. Take MATH and MATH. Then REF ensure that the above conditions are satisfied, with MATH. Hence MATH. Since MATH is contractible by REF follows.
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math/9912125
|
The additivity of the NAME characteristic CITE, which applies in view of REF , gives MATH . In the last identity, the left-hand side is equal to REF by REF , while the right-hand side is equal to MATH by REF . Simplifying, we obtain the desired formula.
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math/9912125
|
Since MATH, MATH, and MATH normalizes MATH, it suffices to prove the last statement. It is well known (compare , for example, CITE or CITE) that any MATH is uniquely factored as MATH with MATH and MATH. Hence MATH, as desired.
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math/9912125
|
It is enough to show that MATH implies MATH. Just as in the proof of REF , we can write MATH, where MATH, MATH, and MATH. Then MATH, where the factors belongs to MATH and MATH, respectively. Thus MATH, as desired.
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math/9912125
|
Immediate from CITE.
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math/9912125
|
MATH.
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math/9912125
|
Uniqueness follows from the uniqueness part of REF , together with REF and the fact that MATH is a group. In more detail: assume MATH, where MATH and MATH. Let MATH be as in REF . Then MATH, where MATH, a contradiction. With the notation MATH and MATH, it remains to check that MATH satisfies MATH. Indeed, MATH.
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math/9912125
|
For the type MATH, this is an immediate corollary of CITE. The general case can be deduced from (highly nontrivial) CITE. According to the latter, for any generalized minor MATH and any sequence of indices MATH, the function MATH (compare REF ) is either identically zero, or is a polynomial with positive integer coefficients. (The type MATH version of this statement is well known; see, for example, CITE.) Since MATH does not vanish at some point in MATH, we know that MATH is a nonzero polynomial for any reduced word MATH for MATH. For MATH, any reduced word MATH for MATH contains some reduced word MATH for MATH as a subword (see CITE). Hence MATH is a specialization of MATH, obtained by setting some of the variables equal to MATH. Then MATH implies MATH. On the other hand, MATH is a polynomial with positive coefficients, so MATH for any MATH, or. equivalently, MATH for any MATH.
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math/9912125
|
Follows from REF .
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math/9912125
|
By CITE, the set MATH is defined by several inequalities of the form MATH, where MATH is a generalized minor. Since MATH (by REF ), none of these minors vanishes on MATH - and therefore none vanishes anywhere on MATH, by REF .
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math/9912125
|
Let MATH and MATH, MATH. Then, by REF , MATH as desired.
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math/9912125
|
The inclusion MATH is REF . The opposite inclusion is immediate from REF .
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math/9912125
|
By CITE, for any MATH, we have MATH for some MATH. Moreover, it is clear from the proof of this statement in CITE that MATH, and the lemma follows.
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math/9912125
|
The first statement follows from REF . Proof of the second one: for some MATH and MATH, we have MATH.
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math/9912125
|
Assume MATH, MATH, and MATH. Then MATH (by REF and the definition of MATH), as claimed. Let us prove that the map in question is a surjection. Let MATH, and let MATH and MATH be given by REF ; note that the right-hand sides of these formulas are well defined (by REF ). Thus MATH and MATH, where MATH. Then MATH and MATH (by REF ). Furthermore, MATH (since MATH by REF ) and therefore MATH (by REF ). Let us now prove injectivity. Suppose MATH, MATH, and MATH. We will show that MATH and MATH can be recovered from MATH via REF - REF . Since MATH, the right-hand sides of REF - REF are well defined (by REF ). Then MATH proving REF . Let us prove REF . Denote MATH. We have MATH (by REF ). On the other hand, we already proved that MATH. Thus MATH, that is, MATH, as desired. It remains to prove that MATH is totally nonnegative whenever MATH is. Assume MATH. Consider a path that connects MATH with a point MATH and stays inside MATH (such a path exists since MATH is connected and its boundary contains MATH - see REF ). The image of this path under the projection MATH connects MATH with MATH. Since MATH, REF implies that MATH.
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math/9912125
|
Let MATH, where MATH and MATH, as in REF . Let MATH be such that MATH (such MATH exists and is unique by REF ). Set MATH since both MATH and MATH belong to MATH, the element MATH is well defined in view of REF , and belongs to MATH by REF . Let us prove that the element MATH defined by REF has the desired properties, that is, MATH is well defined and belongs to MATH, as shown below: MATH . Denote MATH. Once again, MATH, and REF implies MATH. Then MATH (because MATH), so MATH is well defined indeed. Furthermore, MATH (by REF ), as desired. Uniqueness is proved by a similar argument. Suppose that MATH is such that MATH. As before, denote MATH. Then MATH and MATH. Since MATH (by REF ) and MATH, it follows from REF that MATH. Hence by REF , MATH, implying REF . It remains to prove the second part of the lemma. In view of REF , a path connecting MATH and MATH within MATH gives a continuous deformation of the identity MATH into MATH within MATH, which gives rise (via REF ) to a continuous deformation of MATH into MATH and, finally, to a path connecting MATH and MATH within the NAME cell containing MATH. Hence MATH is totally nonnegative by REF .
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math/9912125
|
The map MATH establishes a diffeomorphism between MATH and MATH. Let us prove transversality. Consider a point MATH. It will be enough to show that MATH is transversal to the smooth submanifold MATH of dimension MATH in MATH. Assume the contrary, that is, there exists a common tangent vector MATH to MATH and MATH at the point MATH. Let us evaluate the differential MATH of the projection MATH at the vector MATH. On the one hand, the projection is constant on MATH - hence MATH. On the other hand, in view of REF , the restriction of the projection onto MATH is a diffeomorphism - hence MATH, a contradiction. Let us prove the second part of the theorem. From REF , we have MATH, where MATH is given by REF and MATH by REF . The resulting map MATH is rational and therefore differentiable on its domain. Its inverse is again a map of the same kind, with the roles of MATH and MATH reversed. Hence these maps are diffeomorphisms. Furthermore, they preserve the NAME stratification (in view of REF ) and total nonnegativity (by the second part of REF ).
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math/9912125
|
In view of REF , the statement follows from the factorization MATH, where the two factors on the right belong to MATH and MATH, respectively.
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math/9912125
|
First part: MATH. The second part is then a special case of REF .
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math/9912125
|
The equality MATH implies MATH. Then MATH. Since MATH and MATH, we are done.
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math/9912125
|
For a matrix MATH, the matrix element MATH vanishes unless MATH. It follows that MATH unless MATH, proving REF. Let us prove REF . Set MATH, MATH, MATH; thus MATH. In what follows, we denote by MATH the determinant of the submatrix of a matrix MATH formed by the rows MATH and the columns MATH (in this order). Using the definition of MATH and the fact that MATH, we obtain: MATH . Since MATH is totally nonnegative, the sign of MATH is either zero or MATH, where MATH . The sign of MATH can be determined in a similar fashion. Using the notation MATH to indicate that the index MATH is being removed, we obtain: MATH . Hence the sign of MATH is either zero or MATH, where MATH is the same as before, and MATH . Since MATH we conclude that the sign of MATH is either zero or MATH, matching the claim of the lemma.
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math/9912125
|
For MATH, let MATH denote the diagonal matrix whose first MATH diagonal entries are equal to REF, and all other entries vanish. The equality MATH implies MATH where by REF all terms in the first sum are nonnegative while all terms in the second sum are nonpositive. Then MATH, implying MATH. Since MATH, the lemma follows.
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math/9912126
|
REF follows from the Monotonicity Theorem. Assume MATH. Then the MATH-th column (respectively, MATH-th row) does not contain elements of MATH below (respectively, to the left) of MATH. If, in addition, MATH, then MATH and therefore MATH, proving REF . If, on the other hand, MATH (that is, MATH), then MATH is greater than all elements strictly below and to the left of it, so any chain in MATH is extended by MATH. Hence the maximal length, MATH, of a chain is increased by REF. The first row of MATH then contains one more box than the first row of MATH, proving REF . To prove REF , assume MATH. Then MATH contains an element MATH strictly to the left of MATH, as well as an element MATH strictly below MATH. Both MATH and MATH are maximal elements of MATH, and MATH. Let the boxes MATH and MATH be defined by MATH and MATH. REF implies that MATH lies weakly to the left of MATH (see REF a). Now consider the poset MATH on the same ground set as MATH, the difference being that MATH in MATH if and only if MATH and MATH. The chains of MATH are the antichains of MATH. Hence the shape MATH is the transpose of the shape MATH. Notice that in MATH, MATH is maximal while MATH is minimal. REF then implies that MATH lies in either the same row as MATH, or in the row immediately below it (see REF b). We conclude that MATH must lie one row below MATH (see REF c), as desired.
|
math/9912129
|
Let us give two proofs of this statement, both based on a study of the maps MATH defined for MATH by MATH for MATH. By considering a NAME form of MATH, as in the proof of REF, the condition MATH on the eigenvalues of MATH means that there exists a norm on MATH such that MATH in the associated norm on MATH. If MATH, it follows from REF that MATH for MATH, and by iteration, MATH for MATH, MATH. Now, using REF in the form MATH one deduces from REF that MATH . Thus REF follows with MATH .
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math/9912129
|
The identity MATH in conjunction with REF , implies that all monomials MATH, MATH, are contained in the cyclic subspace generated by MATH, and hence this space is dense in MATH. Indeed, for every MATH, there is, by REF , a MATH such that MATH for all MATH such that MATH. Therefore MATH. An application of REF to MATH then yields the desired cyclicity. This cyclicity is the second of the two properties of the subspace MATH in the discussion of REF, that is, REF . The rest (and some more details) follows from the discussion in REF.
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math/9912129
|
Introducing MATH, MATH, the identity REF above reads MATH . Generally for third degree, the correspondence MATH is given by the following functional identity in MATH: MATH and the boundary conditions, MATH, that is, MATH is uniquely determined by these conditions and the normalization MATH. See CITE for details. This applies to both the pair MATH and the pair MATH, so we get MATH . As noted, MATH is the unique normalized MATH-solution to this identity, subject to MATH. But, if MATH is the solution corresponding to MATH, then a direct substitution MATH shows that the mirrored function MATH satisfies REF , and we conclude from the uniqueness that MATH as claimed in the Proposition. The proof of REF is similar, or see REF below. We resume the proof of REF after the following remark.
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math/9912129
|
Let us consider the two operators MATH and MATH in MATH individually, and as part of a pair of MATH-representations. While the two MATH-representations are inequivalent, the two MATH-operators alone are unitarily equivalent. This follows from the general fact that any operator of the form REF coming from a wavelet is unitarily equivalent to the shift of infinite multiplicity by CITE. The explicit intertwiner can also be calculated directly as follows: Let MATH, MATH, and define three operators MATH, MATH, and MATH (acting on MATH) by MATH . Then MATH is a unitary intertwining operator for MATH and MATH, that is, MATH holds, as can be verified by a direct calculation. The minus sign in the second symmetry REF is still reflected in the MATH-representations as follows. Let MATH and MATH, and consider the two MATH-representations MATH, MATH, MATH. We then have MATH, and thus MATH where again MATH. Hence MATH intertwines the MATH-representation MATH with the MATH-representation MATH, modified by the automorphism of MATH induced by MATH; see REF .
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math/9912129
|
From CITE or CITE, we have MATH . Since MATH, we conclude that MATH . This second summation is just one of the MATH residue classes for the full MATH summation in REF . But the formula for MATH yields MATH .
|
math/9912131
|
The function MATH factors as follows. MATH for MATH, with the interpretation that the function MATH is MATH when MATH in MATH.
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math/9912131
|
We first show that the exponentials MATH are mutually orthogonal in MATH where the MATH's are given on MATH by the usual REF from REF. The inner product in MATH of MATH and MATH factors as follows: MATH . If MATH in MATH, then it vanishes since MATH is a spectral pair; and, if MATH but MATH, it vanishes since MATH is one. This proves orthogonality of MATH. To see that it is total, let MATH and suppose MATH is orthogonal to MATH. The inner products (vanishing) are: MATH . If MATH is fixed, and the double integral vanishes for all MATH, then the integral MATH for almost all MATH, by the totality of MATH on MATH. But MATH is arbitrary so the totality of MATH on MATH implies MATH. We conclude, that MATH is total on MATH as claimed.
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math/9912131
|
If MATH is a spectral pair, then so is MATH for any vector MATH. An application of REF completes the proof.
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math/9912131
|
To check orthogonality, let MATH. Then the two points MATH and MATH are in MATH, and the corresponding MATH-inner product is zero. But it is also MATH and since MATH, the orthogonality follows. We now show that MATH is total in MATH if MATH for all MATH, MATH. Let MATH and suppose MATH is orthogonal to all the MATH-exponentials. Let MATH be a general point in MATH. Then the inner product with MATH is MATH . If MATH, then MATH, and the second factor vanishes. If MATH, then the first factor is zero, and we get that MATH is orthogonal in MATH to the MATH-exponentials. They are total, and we conclude that MATH vanishes in MATH. The remaining case is when MATH for some MATH. But it follows that then MATH for all MATH and MATH. In particular, MATH is then not total in MATH.
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math/9912131
|
The assertion in the theorem about MATH-translations tiling the plane with MATH is clear from REF - REF , and it is illustrated graphically in REF . It is immediate from REF that each one of the two REF - REF for MATH make MATH a spectral pair, and the main result is that there are not others. We show this directly by an examination of the possibilities for MATH which are implied by the inclusion MATH where MATH is read off from REF above. Again a translation of MATH by a single vector in the plane will reduce the analysis to the case when MATH is in MATH. Let MATH and suppose MATH. Then either MATH or MATH is not an integer. Suppose MATH is not an integer. Then MATH is a nonzero integer. Let MATH be an arbitrary point in MATH. If MATH is not an integer, then MATH is a nonzero integer, so MATH is not in MATH, contradicting REF . So MATH for any MATH in MATH. To verify MATH is a subset of a set given by REF for MATH, we need only check that if MATH and MATH are different points in MATH, with MATH, then MATH. But recall MATH, so it follows from REF that MATH. Since an orthonormal basis cannot be a strict subset of another orthonormal basis for the same space it follows that MATH is given by REF .
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math/9912131
|
Suppose MATH is a tiling set for MATH. By the tiling property there exist functions MATH so that MATH . Fix MATH then REF (NAME 's theorem) implies MATH is independent of MATH, we will write MATH in place of MATH to indicate this independence. Similarly, MATH and MATH. Considering, for fixed MATH, the intersection of the plane MATH by cubes MATH, MATH it follows that the set MATH is a tiling set for MATH in MATH. Hence, by our MATH result REF , either MATH . It follows that there exist MATH so that MATH, MATH, and MATH . For each MATH is a tiling set for MATH in MATH. Hence, our MATH result implies either REF MATH for all MATH or REF MATH, for all MATH, MATH. Suppose REF , then for each MATH is a tiling set for MATH, hence our MATH result implies that either REF MATH for all MATH or REF MATH for MATH and MATH for MATH. If REF then we are done, so suppose REF : then MATH . Let, if possible, MATH, MATH, MATH be such that MATH and MATH: then MATH does not have any nonzero integer entry, contradicting NAME 's theorem. So, either MATH, MATH, MATH for all MATH, or MATH for all MATH; in the three last cases we are done, so assume MATH. If MATH, MATH, and MATH then MATH does not have any nonzero integer entry, contradicting NAME 's theorem. This contradiction completes the proof of REF . The proof of REF is similar; we leave the details for the reader. Conversely, for every such set MATH, the translates of MATH by the vectors of MATH clearly tile MATH. This completes the description of tilings of MATH by MATH. By REF any set MATH of the form REF is a spectrum for MATH. We sketch a proof of the converse. Apply REF to MATH, to show that if MATH is a spectral pair, then one of the three coordinate intervals may be picked as MATH in REF , that is, MATH, MATH and with MATH satisfying the orthogonality on MATH. By REF , this means MATH . Eventually we show that MATH must be of the form MATH. But to select the one of the three coordinates which has this form, consider the canonical mapping MATH and select the one of the three sets MATH, MATH, of the smallest cardinality. Assume it is MATH for simplicity. We will show that MATH then satisfies REF , so that REF applies. The assertion is that the set MATH is a singleton. The proof is indirect. Suppose ad absurdum that, for some MATH such that MATH, MATH meets both MATH and MATH. We conclude from REF that for MATH in each of the two sets MATH or MATH, the points in MATH must be of the form MATH for MATH and MATH some function, or alternatively MATH, MATH with MATH some possibly different function. Calculating MATH-inner products for associated points MATH with MATH, and MATH (MATH), we get the following possibilities for the respective coordinates in the second and third place: MATH . But if the difference MATH is not in MATH, then one of the corresponding differences in the second place, or the third place, must be in MATH. Making variations, we conclude that then one of the two sets MATH, MATH, must be a singleton. But this contradicts that MATH has two distinct points, and is chosen to have smallest cardinality of the three sets MATH, MATH.
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math/9912131
|
The stated conclusion follows from combining the results in the present section with REF ; for MATH the spectral condition is equivalent to the operator extension property.
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math/9912131
|
If MATH is a MATH-tile then MATH, since any two MATH-tiles have the same volume CITE. Conversely, suppose a measurable set MATH has REF . Let MATH; then MATH is a measure-theoretic partition of MATH. By REF , the sets MATH, MATH, are measure disjoint. Hence, MATH by REF . It follows that MATH, up to sets of measure zero. Hence, up to sets of measure zero, MATH as we needed to show.
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math/9912131
|
CASE: Since the functions MATH are orthogonal it follows that MATH. By CITE MATH has uniform density MATH, hence MATH. CASE: Using MATH we conclude that the functions MATH are orthogonal in MATH. So, since MATH, an application of CITE allows us to conclude MATH is a spectral pair. CASE: That MATH follows from MATH being a MATH-tile, by REF . If MATH in MATH, then MATH has a nonzero integer entry, hence MATH. Hence MATH has volume MATH. By REF is a MATH-tile, so using REF we conclude MATH. CASE: Let MATH, MATH. If MATH or MATH, then MATH so MATH . In particular MATH has volume MATH and MATH for all MATH. Hence MATH is a MATH-tile by REF . Using REF it follows that MATH is a MATH-tile.
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math/9912131
|
Let MATH . We must show that MATH extends, by continuity, to an isometric isomorphism, mapping MATH onto MATH. If MATH and MATH, then MATH by a simple computation using the fact that MATH is an isometric isomorphism. Hence, MATH. Since MATH and MATH both are isometric isomorphisms so is the adjoint MATH.
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