paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9912111 | 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). |
math/9912111 | 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. |
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 assu... |
math/9912111 | 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-ampl... |
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 . |
math/9912111 | 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/9912111 | 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 homotopical... |
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 resoluti... |
math/9912111 | 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. |
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 ... |
math/9912111 | 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. |
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,... |
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 an... |
math/9912111 | 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/9912111 | Follows by REF. |
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... |
math/9912111 | 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 nonempt... |
math/9912111 | 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, ass... |
math/9912111 | Since MATH (see REF ), we have MATH. |
math/9912111 | 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 .... |
math/9912111 | 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 ap... |
math/9912111 | 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 ... |
math/9912111 | 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. |
math/9912111 | 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 surfa... |
math/9912111 | 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. The... |
math/9912111 | 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 ... |
math/9912111 | 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 ... |
math/9912111 | 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. |
math/9912111 | 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. |
math/9912111 | 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. |
math/9912111 | 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. |
math/9912111 | 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.... |
math/9912111 | 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 ... |
math/9912111 | If MATH is rational, then we can omit REF by REF . Otherwise there is a regular complement by REF below. |
math/9912111 | 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 t... |
math/9912111 | 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... |
math/9912111 | 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, MAT... |
math/9912111 | 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... |
math/9912111 | 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 equa... |
math/9912111 | 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 t... |
math/9912111 | 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/9912111 | 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 e... |
math/9912111 | 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 know... |
math/9912121 | 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/9912121 | See for example, CITE or CITE . |
math/9912121 | See, for REF or REF . |
math/9912121 | A straightforward computation. |
math/9912121 | 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. |
math/9912121 | 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 ... |
math/9912121 | 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 elemen... |
math/9912121 | Relations REF are obvious. For relation REF, MATH . For relation REF, MATH . Thus we complete the proof. |
math/9912121 | A straightforward computation. |
math/9912121 | 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 . Henc... |
math/9912121 | 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/9912121 | 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 . |
math/9912121 | 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: the... |
math/9912121 | From REF , representation matrices of MATH and MATH for generators of MATH are coincide. Therefore, the above statement holds. |
math/9912121 | 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. |
math/9912121 | 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. |
math/9912121 | 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 ... |
math/9912121 | See for example, CITE . |
math/9912121 | 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. |
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... |
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. |
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. |
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. |
math/9912125 | Immediate from CITE. |
math/9912125 | MATH. |
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. |
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 coeffi... |
math/9912125 | Follows from REF . |
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 . |
math/9912125 | Let MATH and MATH, MATH. Then, by REF , MATH as desired. |
math/9912125 | The inclusion MATH is REF . The opposite inclusion is immediate from REF . |
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. |
math/9912125 | The first statement follows from REF . Proof of the second one: for some MATH and MATH, we have MATH. |
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... |
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, t... |
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 th... |
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. |
math/9912125 | First part: MATH. The second part is then a special case of REF . |
math/9912125 | The equality MATH implies MATH. Then MATH. Since MATH and MATH, we are done. |
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... |
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 follow... |
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 th... |
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 ... |
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 th... |
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 a... |
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 u... |
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. |
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 o... |
math/9912131 | If MATH is a spectral pair, then so is MATH for any vector MATH. An application of REF completes the proof. |
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-exp... |
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 ... |
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 t... |
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. |
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 zer... |
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 bein... |
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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