paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/9912101 | By definition of MATH and MATH, MATH, so that, by definition of the torsion: MATH . Taking the scalar product with MATH and using the fact that the MATH are parametrized at unit speed shows that: MATH . Therefore: MATH and we obtain the first equation. Coming back to REF , we see that: MATH and the second equations fol... |
math/9912101 | Integrating REF shows that, for MATH: MATH so that: MATH . Let: MATH . For MATH: MATH . Thus there exists MATH such that, if MATH, then: MATH which contradicts the definition of MATH. So MATH, and REF follows. REF is a direct consequence using REF . |
math/9912101 | Let: MATH . REF shows that: MATH while it is easy to check that: MATH because this corresponds to a change in the parametrization of the geodesic starting at MATH in the direction of MATH. |
math/9912101 | Let MATH be the inverse image of MATH by the restriction of MATH to the ball of radius MATH. By the previous corollary and the local convexity of MATH, the restriction of MATH is a diffeomorphism onto its image. |
math/9912101 | REF is a simple consequence of REF , and REF then follows from REF . |
math/9912101 | Let MATH be a small real number; we will see later how small MATH has to be. For MATH, let: MATH . Then define: MATH . For MATH, apply the NAME theorem to an infinitesimal strip bounded by MATH, MATH, MATH and MATH. This shows that: MATH so that: MATH where MATH if MATH is small enough (this last step uses REF applied ... |
math/9912101 | If MATH or MATH is compact, the result is obvious, so we suppose here that neither MATH nor MATH is compact. The proof is by contradiction, so we suppose that MATH. First note that a rather direct smoothing argument shows that, for any MATH, there are smooth curves MATH such that: CASE: MATH bounds a connected closed s... |
math/9912101 | Let MATH and MATH. We can suppose that the integral curve of MATH starting at MATH meets the integral curve of MATH starting from MATH: otherwise, the proposition would fail slightly before the first value of MATH such that the intersection does not exist, because then the length of both MATH and MATH would go to infin... |
math/9912101 | The intersections between those integral curves remain at bounded distance as long as the lengths of the integral curves of MATH and MATH going from MATH to MATH and to MATH remain below MATH, and the integral curves of MATH and of MATH do not meet MATH because MATH is complete (compare REF). |
math/9912101 | Let MATH be the vector fields and MATH be the functions on MATH such that: MATH . Call MATH and MATH. Then: MATH . But MATH, so that: MATH and therefore: MATH . Now, by definition of MATH: MATH so that: MATH and therefore: MATH . Using this and REF shows that: MATH . Now the definition of a convex map and the bounds on... |
math/9912101 | Let MATH be a MATH-quasi-geodesic, with either MATH or MATH and MATH. Note MATH the vector field on MATH defined by: MATH . Let MATH be the angle between MATH and MATH, and MATH the angle between MATH and the parallel transport at MATH of MATH along MATH. By REF : MATH while the definition of a MATH-concave map indicat... |
math/9912101 | The proof is similar to that of REF ; we call MATH, and we suppose that MATH. Then: MATH . It is the clear that MATH will soon become positive; moreover, if we let MATH and MATH, then: MATH . The second equation indicates that MATH remains positive while MATH, and both equations taken together again show that MATH inte... |
math/9912101 | Note that, if MATH is small enough, then, for any MATH, if MATH is close enough to MATH, then MATH. This comes from REF and from REF . This immediately implies that MATH is a decreasing family of subsets of MATH. |
math/9912101 | Fix MATH. For MATH, let MATH be the maximal integral curve of MATH in MATH containing MATH. We consider two cases. CASE: There exists MATH and a sequence MATH such that, for each MATH, MATH has one end on MATH. If there exists MATH such that MATH, then the proposition is proved. Otherwise, call MATH the connected compo... |
math/9912101 | This is again a consequence of REF , along with REF , which bounds the rate of variation of the direction of MATH along MATH. |
math/9912101 | Since the integral of MATH on MATH is finite, REF shows that: MATH . One can therefore define a parallel vector field on MATH as: MATH and, by REF : MATH . The same works for MATH because MATH; set: MATH and then: MATH . Let MATH; we suppose (without loss of generality) that MATH. The proof will proceed differently acc... |
math/9912101 | By linearity, it is enough to prove this proposition when MATH is positive. Let MATH be a one parameter family of curves such that MATH and that: MATH . To simplify somewhat the notations, we call MATH the speed of MATH, that is: MATH . We also call MATH the unit vector along MATH, and MATH. Therefore: MATH for some fu... |
math/9912101 | For any MATH, we have: MATH . Moreover, MATH and MATH, so that: MATH and the corollary follows. |
math/9912101 | Let MATH. The relation REF becomes: MATH . Let: MATH . The relation now is: MATH so, for MATH: MATH with: MATH . Now the eigenvalues of MATH are the roots of: MATH . From the NAME theorem, MATH, so MATH, and the eigenvalues are MATH, where: MATH . The associated eigenvectors are MATH. Suppose now that MATH. Then MATH, ... |
math/9912101 | Let MATH, so that MATH by REF . By REF , if MATH is small enough, there exists a function MATH such that, on MATH: MATH so that: MATH . For MATH small enough and MATH, set: MATH where MATH is the unit normal to MATH at MATH towards the convex side of the complement. This defines, for MATH small enough, a smooth curve M... |
math/9912101 | Apply REF recursively to obtain a MATH-concave map MATH such that MATH is a segment of MATH. |
math/9912101 | Let MATH. First apply REF after translating the origin to MATH in MATH, so as to obtain MATH and a solution MATH of REF with: MATH . Apply REF again, now after a translation of the origin to MATH; this provides us with MATH and with a solution MATH of REF with: MATH . Repeat this procedure to find a sequence MATH with ... |
math/9912101 | Let MATH be a sequence of smooth curves, MATH, such that: CASE: MATH; CASE: for MATH, MATH lies entirely on the concave side of MATH; CASE: the curvature of MATH is at least MATH, where MATH. The existence of such an approximating sequence is not too difficult to prove. The MATH are not parametrized at speed one. We su... |
math/9912101 | The underlying idea is to apply REF recursively, to obtain the existence of such a deformation for MATH for some MATH. The formal proof, however, has to be done in a slightly different way. Suppose that such a deformation can not exist for all MATH. Let MATH be the set of couples MATH, where MATH and MATH satisfies the... |
math/9912101 | It is similar to the proof of REF . First choose a sequence of smooth curves MATH converging to MATH, such that MATH is in the interior of MATH for MATH and that the curvature MATH of MATH is at least MATH, with MATH. For MATH and MATH, let: MATH where we suppose again that MATH is towards the non-compact side of MATH.... |
math/9912101 | Like the proof of REF from REF . |
math/9912101 | First note that, by a direct approximation argument, it is enough to prove the result when MATH is smooth, so we suppose that is the case. Let MATH. By REF , there exists MATH (depending only on MATH and MATH), such that MATH is a diffeomorphism from the subset of the ball of radius MATH where it is defined onto its im... |
math/9912102 | The natural homomorphism of algebraic groups MATH induces a morphism at the level of NAME (and Spin)-bundles. By REF semistability is preserved under this operation. Thus, taking into account that the spinor norm is preserved under the above group homomorphism and that MATH is a coarse moduli space, we get the above cl... |
math/9912102 | First let us prove REF. By REF (MATH), the multiplication map MATH is surjective. Consider the character space decomposition REF of the two spaces. Since the above tensor product is compatible with the linear action of the group MATH, we get a surjective map between the two character spaces of REF corresponding to the ... |
math/9912102 | We will show that the duality between the above vector spaces is given by a reduced NAME divisor, whose set-theoretical support equals MATH . Indeed, as shown in REF of REF, the family MATH of rank MATH vector bundles MATH parametrized by MATH is equipped with a MATH-valued quadratic form. Hence, by REF, we can conside... |
math/9912102 | Since MATH, we have an isomorphism MATH, which gives rise to a homomorphism MATH. Since both MATH and MATH have degree MATH and since MATH is poly-stable, it follows that MATH is a subbundle of MATH. It remains to determine its supplement bundle. There is a natural homomorphism of MATH into MATH. Using the isomorphism ... |
math/9912102 | We know that the orthogonal bundle MATH is the tensor product MATH. Let us do the computations for MATH. MATH where we used the fact that MATH is self-dual and the projection formula for the map MATH. Now by example REF, MATH is a semistable orthogonal bundle over MATH, hence this bundle splits MATH. The computations f... |
math/9912102 | It is enough to prove linear independence, which is done by induction on MATH. For the first cases we refer to REF (MATH), REF (MATH), REF (MATH). The statement for MATH will follow by pulling back MATH under the map MATH. We observe that MATH. Since the MATH form a basis, we can conclude. Let us assume that the MATH's... |
math/9912103 | We need to count the number of integer points MATH inside the region MATH which consists of the points in the cube MATH which lie between the hyper-planes MATH . Note that the region MATH is convex and contained in a ball around the origin of radius MATH. By the NAME principle (see CITE) we know that MATH . The distanc... |
math/9912103 | We first remark that since MATH and the MATH are distinct, the hyper-planes REF with MATH replaced replaced by MATH are not parallel to the hyper-plane given by REF . Moreover, the fact that our sequence is lacunary insures that the angle between these hyper-planes is not small. Thus when we solve for MATH in REF and i... |
math/9912103 | Our proof is by induction on MATH. The case MATH is clear, the number of solutions in this case being zero. Let us assume that the statement holds true for MATH and prove it for MATH. Let MATH be a solution to the system REF. If there exists MATH such that MATH then MATH will be a solution for the same system with MATH... |
math/9912103 | Each solution MATH to REF produces a partition of the set MATH as a disjoint union of subsets MATH, where MATH are the above sets MATH without repetitions. Let us count the number of non-degenerate solutions to REF which correspond to a given partition MATH of the set MATH. For MATH denote MATH, MATH for MATH, then wri... |
math/9912103 | We write MATH where MATH we will show that MATH and thus prove our lemma. Fix MATH, and let MATH, MATH. Since MATH, MATH for large MATH. Now divide the range of summation in REF into MATH and MATH: MATH . The second sum is bounded by MATH by our choice of MATH and MATH. As for the first sum, it is bounded by the number... |
math/9912103 | By REF we have MATH . Moreover, MATH . Now summing over all MATH we get MATH . Fix MATH, and choose MATH and MATH sufficiently large in terms of MATH and MATH, say MATH. Also set MATH. We have MATH for large MATH. In REF we break up the sum over MATH into ranges MATH and MATH, and likewise for the sum over MATH. In the... |
math/9912103 | Let MATH. For MATH we write MATH in the form MATH with MATH and MATH. From MATH, with MATH and MATH it follows that MATH and since MATH we get MATH for MATH . For any MATH let MATH . Then for any MATH . In particular one has: MATH . It remains to bound MATH. In order to do this we produce for any MATH an upper bound fo... |
math/9912103 | Given MATH we choose a positive integer MATH, depending on MATH only, whose precise value will be given later. Let MATH and MATH. There exists MATH such that the set MATH has more than MATH elements. Arrange the elements of MATH in increasing order: MATH and pick from this set the first element MATH, then ignore the ne... |
math/9912103 | We use the representation of MATH as in REF: MATH where MATH. Note that MATH so we may assume MATH. Now fix MATH, and set MATH; then for MATH to lie in the support of MATH, we need MATH. By REF , for almost all MATH there are at most MATH integers MATH satisfying this. Similarly, we need MATH for all MATH which forces ... |
math/9912105 | Induction in MATH. For MATH the identity MATH is true. Now let MATH. Using the commutation of MATH with each of MATH we rewrite the left hand side of REF as: MATH . Applying the to above expression the basic relation REF written in the form MATH, we see that the left hand side of REF equals: MATH . Finally, applying th... |
math/9912105 | Denote MATH. Note that MATH. Then the identity REF can be rewritten as MATH . This identity is true because the morphism MATH is right MATH-equivariant. Lemma is proved. |
math/9912105 | It suffices to take MATH. In this REF implies that for MATH we have: MATH where MATH. Lemma is proved. |
math/9912105 | Clear. |
math/9912105 | It suffices to prove the proposition in the assumption that MATH, that is, when MATH is semisimple of types MATH and MATH respectively. REF follows. Let us prove REF . It suffices to analyze the case when MATH. Due to REF it suffices to prove the statement only for MATH, MATH. In this case MATH, and we set MATH in the ... |
math/9912105 | It is easy to see that MATH satisfies REF . Since the centralizer in MATH of any element MATH is trivial we see that MATH does not depend on a choice of REF . Lemma is proved. |
math/9912105 | It suffices to show that MATH. By definition, we have: MATH for MATH. Lemma is is proved. |
math/9912105 | Note that MATH for all MATH. So we prove the lemma in the assumption that MATH. Let MATH. Express MATH as MATH, where MATH, MATH. Then MATH because MATH. Furthermore, for each MATH we have: MATH, where MATH. Finally, MATH . Lemma is proved. |
math/9912105 | Throughout the proof we denote for shortness MATH. In this case we have MATH, MATH (where MATH) which implies that MATH. The following result is obvious. Let MATH. then for any MATH and MATH we have: MATH . Furthermore, let MATH, MATH, MATH. Then MATH . Applying REF twice, we obtain MATH . By by REF , we have MATH, and... |
math/9912105 | Taking into the account that MATH, we rewrite REF: MATH . REF is proved. |
math/9912105 | It is easy to see that for any MATH and MATH we have MATH. It is also clear that for any MATH, MATH we have: MATH where MATH as in REF. Furthermore, by definition of MATH, one has MATH . Composing MATH with MATH, we obtain for MATH, MATH: MATH . Lemma is proved. |
math/9912105 | We proceed by the induction in MATH. If MATH, that is, MATH for some MATH then MATH . Let us now assume that MATH. Using REF and the fact that MATH is a group homomorphism, we obtain for any MATH: MATH REF is proved. |
math/9912105 | It follows from REF that for any MATH satisfying MATH we have: MATH . This proves REF . Let us prove REF now. Denote MATH. We will proceed by induction in MATH. If MATH, the statement is obvious. Let now MATH. Let us express MATH, where MATH, MATH and MATH. Thus, the inductive hypothesis implies that MATH is dense in M... |
math/9912105 | Prove REF . The multiplication in MATH induces the open inclusions MATH, MATH. Furthermore, MATH . Therefore, it follows from REF that MATH . REF is proved. Let us prove REF now. Recall that MATH is MATH-equivariant, and MATH. Note that MATH intersects MATH non-trivially. Thus MATH . Recall that the action MATH is defi... |
math/9912105 | Prove REF . Indeed, MATH. Thus MATH is a section of MATH because MATH is MATH-equivariant. Prove REF . It is easy to see that MATH. Thus, for MATH and any MATH, MATH we have MATH. On the other hand, MATH. This implies that MATH. Prove REF . We proceed by the contradiction. Assume that MATH is not injective. That is, we... |
math/9912105 | Follows from REF . |
math/9912105 | Let us describe the sub-torus MATH of MATH. We have MATH . That is, MATH can be described in MATH by the equations CASE : MATH . On the other hand, for any MATH one has MATH . REF is proved. |
math/9912105 | For any MATH, MATH let MATH be the representative of MATH defined by MATH for any MATH such that MATH. We need the following obvious refinement of REF . MATH. For any MATH choose an element MATH such that MATH. By REF , we have: MATH where MATH. Clearly, MATH, and MATH. It is also clear that MATH commutes with each MAT... |
math/9912106 | It suffices to prove that the following diagram commutes. MATH . Recall that MATH, where MATH and MATH. Recall further that the minimality condition on MATH implies that the linear part of its differential vanishes. The linear part of MATH is the linear map MATH defined by the condition MATH. Recall that MATH and MATH ... |
math/9912106 | It suffices to prove the dual statement, namely that MATH factors over the surjection MATH to induce a differential in MATH. But MATH consists of products along with elements of the form MATH for MATH, MATH. Since MATH is a MATH-derivation, MATH is a product. It follows that MATH, completing the proof. |
math/9912106 | We proceed by induction. For MATH, let MATH. Since MATH, MATH, establishing the first statement. For the second statement we may take MATH to be the identity map. Let MATH be the homotopy NAME algebra of MATH. Apply REF to the minimal algebra MATH to get a graded NAME algebra isomorphism MATH. Since MATH in MATH, MATH ... |
math/9912106 | Let MATH be a minimal model. Recall that the underlying algebra of MATH is MATH, where MATH. Let MATH and MATH be constructible acyclic closures for MATH and MATH, respectively. The model MATH determines a MATH-morphism MATH where MATH is an isomorphism. The composition MATH induces an isomorphism of NAME spectral sequ... |
math/9912106 | NAME in CITE proves that there is a dgl MATH and a dgh quasi-isomorphism MATH. Thus as NAME algebras, for MATH, MATH and MATH. The result follows by applying REF to the dgl MATH. |
math/9912106 | Observe that, for MATH, the sequence of functors MATH identifies MATH as the natural dual of the free MATH-algebra MATH, with MATH naturally dual to MATH, and the inclusion MATH naturally dual to the projection MATH. Therefore if MATH, then MATH is a MATH-morphism. Conversely, if MATH is a MATH-morphism, then MATH, so ... |
math/9912106 | Straightforward. |
math/9912109 | The first part contains standard facts in the theory of toric varieties. The only point to be made is that, since the vectors MATH belong to all maximal simplices in the triangulation MATH no monomial in MATH corresponding to MATH will contain any of the elements MATH . The second part is a direct consequence of the wa... |
math/9912109 | Indeed, for any MATH we have that MATH for some holomorphic function MATH (we have used the classical identity MATH). REF is then an immediate consequence of the relations REF in the definition of the NAME - Reisner ring MATH . To check REF ., assume that for some MATH we have that MATH is a simplex in MATH . Then ther... |
math/9912109 | The proof is by induction. For MATH the operator MATH is just the identity operator. REF follow directly from REF . To prove the induction step, we consider two adjacent cones MATH and MATH and let MATH and MATH be the corresponding regular triangulations. Assume that the analytic continuation operator MATH correspondi... |
math/9912109 | We make repeated use of the NAME - NAME - NAME formula and of the properties of the NAME map in cohomology. For MATH we have that MATH is given by MATH . The last expression is exactly MATH which proves the lemma. |
math/9912109 | According to the previous lemma, the induced action on MATH of the NAME - NAME functor determined by the complex MATH is identical to the action NAME - NAME functor determined the same complex viewed as a complex of sheaves on MATH . The lemma shows then that, for MATH this action is given by MATH where the sign is det... |
math/9912109 | According to the last part of REF . we have MATH . We will need the following lemma. For any cohomology class MATH with the function MATH defined by MATH and the contours MATH give the residues at MATH and MATH respectively. (of the lemma) For any series MATH around the origin MATH we denote by MATH the coefficient of ... |
math/9912109 | The NAME divisors (spanned by global sections) MATH come from a nef - partition of the vectors in MATH . As a consequence of the second part of REF , for any divisor MATH with MATH we have that MATH . This shows that any curve that is contracted under the map MATH does not intersect the generic divisors MATH for MATH .... |
math/9912109 | The fan defining the weighted projective space MATH has the property that the multiplicity index in MATH of any maximal cone generated by the vectors MATH is equal to MATH . Since the vectors MATH do not belong to any maximal cone of the fan MATH we have that MATH . For each MATH the closed toric orbits MATH given by t... |
math/9912109 | Indeed, we have that MATH so MATH . |
math/9912109 | According to REF , the automorphism MATH acts on MATH by MATH with the canonical maps (see REF ) MATH . REF shows that MATH where, for MATH we denote by MATH the pull-back hypersurfaces MATH (see REF ). It follows that MATH . According to the construction of REF , the subvarieties MATH and MATH are complete intersectio... |
math/9912109 | The proof reduces to a calculation very similar to the one performed in the beginning of REF and it is designed to match the formulae of REF . Recall REF that the solutions to the GKZ system are given by the series MATH where the parameter MATH satisfies REF , and MATH is the dual of the cone MATH given by MATH . As ex... |
math/9912111 | It is sufficient to show only the implications MATH . Let MATH be a log resolution of MATH, MATH exceptional divisors and MATH, MATH proper transforms of MATH and MATH, respectively. By NAME 's theorem, MATH is a simple normal crossing divisor, so MATH is also a log resolution of MATH. We can choose MATH so that MATH d... |
math/9912111 | Let us prove, for example, the first inequality. Write MATH, where each MATH is the proper transform of MATH and MATH is the (reduced) exceptional divisor. Then MATH. From the exact sequence MATH (compare CITE) we have MATH . |
math/9912111 | We have MATH . From this by NAME Vanishing Theorem CITE, MATH . Applying MATH to an exact sequence MATH we get the surjectivity of the map MATH . Let MATH be a component MATH. Then either MATH is MATH-exceptional or MATH is the proper transform of some MATH whose coefficient MATH. Thus MATH is MATH-exceptional and MATH... |
math/9912111 | Consider a log resolution MATH. We have MATH where MATH is the proper transform MATH on MATH and MATH, MATH are effective exceptional MATH-divisors without common components. Then MATH is a boundary and MATH is dlt. Apply log MMP to MATH over MATH. We get a birational contraction MATH from a normal MATH-factorial varie... |
math/9912111 | Take a sufficiently small MATH and put MATH . Then MATH . Next, run MATH-MMP over MATH. By REF, each extremal ray is negative with respect to the proper transform of MATH, an effective divisor. Such a divisor can be nef only if it is trivial. Hence the process terminates when the proper transform of MATH becomes zero. |
math/9912111 | As a first approximation to MATH we take a minimal log terminal modification MATH as in REF. Then MATH, where MATH is an integral reduced divisor and MATH is the proper transform of MATH. In particular, MATH is the number of components of MATH. Since MATH has only klt singularities, MATH, where the MATH are components ... |
math/9912111 | The relation in REF follows from the equality MATH . REF can be proved similarly. |
math/9912111 | Let MATH be a minimal log terminal modification (see REF). Since the fibers of MATH are connected and MATH, it is sufficient to prove the assertion for MATH. Let MATH be the composition map. Set MATH and MATH. Then MATH. Assume that MATH is disconnected. Run MATH-MMP over MATH. If the fiber MATH is reducible, then ther... |
math/9912111 | As in the proof of REF, MATH a minimal log terminal modification. Again set MATH and MATH. Assume that MATH is disconnected. Run MATH-MMP. All intermediate contractions is MATH-nonpositive. Therefore the log canonical property of MATH is preserved (see REF). Since at each step MATH is klt, MATH is klt outside of MATH a... |
math/9912111 | Let us proof, for example, REF . Write MATH, where MATH and MATH, MATH, MATH. Then MATH . In both cases MATH. |
math/9912111 | Let MATH. Then MATH for some MATH. This yields MATH and MATH. Hence MATH. |
math/9912111 | REF is trivial. As for REF we notice that MATH. Hence by REF , MATH and we may assume that MATH and MATH. Put MATH. It is sufficient to show that there exists MATH such that MATH . This is equivalent to MATH . Taking into account that MATH, we have MATH for some MATH. So there exists MATH such that MATH . This proves t... |
math/9912111 | Take MATH and apply REF. |
math/9912111 | Consider the crepant pull back MATH . Write MATH, where MATH is reduced, MATH, MATH have no common components, and MATH. We claim that MATH is a MATH-complement of MATH. The only thing we need to check is that MATH. From REF we have MATH. This gives that MATH (because MATH is MATH-nef; see CITE or CITE). Finally, by Mo... |
math/9912111 | Let MATH be a log resolution. Write MATH, where MATH is the proper transform of MATH on MATH and MATH. By Inversion of Adjunction, MATH is normal and MATH is plt. In particular, MATH is a birational contraction. Therefore we have MATH . Note that MATH, because MATH is smooth. By REF we see that MATH. So we can apply RE... |
math/9912111 | Similar to the proof of REF. By REF , the curve MATH is nodal. Further, we can take a log resolution MATH so that MATH. |
math/9912111 | By REF the number of singular points is MATH. Assume that MATH has exactly six singular points MATH. Then by REF we have MATH. This means that MATH are ordinary double points. In particular, MATH is NAME. Applying NAME 's formula to the minimal resolution MATH of MATH, we obtain MATH. Let MATH be a MATH-curve and MATH ... |
math/9912111 | Replacing MATH with a log terminal modification, we may assume that MATH is dlt. Then MATH. In this situation we can apply REF . The last statement follows by Connectedness Lemma. |
math/9912111 | If MATH is not klt, the assertion follows by REF . Thus we may assume that MATH is klt (in particular, MATH). Let MATH . As in REF put MATH . Note that MATH (because MATH, see REF). If MATH is lc, then this is a regular complement. Assume that MATH is not lc. Take MATH so that MATH is maximally lc. It is clear that MAT... |
math/9912111 | NAME gives that MATH is sufficiently large, where MATH is divisible enough and MATH. Then standard arguments show that MATH is not klt for some MATH and MATH (see for example, CITE or the proof of REF ). |
math/9912111 | First note that MATH, because MATH is not klt and MATH is connected by REF. REF follows by REF . To prove REF we note that there exists an extremal MATH-negative contraction MATH. By REF , MATH is not birational. Hence MATH is a curve of genus MATH and fibers of MATH are irreducible. If MATH is contained in fibers of M... |
math/9912111 | Let MATH be an effective NAME divisor on MATH containing MATH and let MATH. First we take the MATH such that MATH is maximally lc (see REF) and replace MATH with MATH. This gives that MATH. Next we replace MATH with a log terminal modification. So we may assume that MATH is dlt and MATH. Then REF give us that there exi... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.