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math/0104187 | For simplicity we will only prove that MATH is reduced and that it is the closure of the locus of those smooth pointed curves MATH for which MATH and MATH are in different fibers of the same MATH. Then by iteration we will get a similar statement for MATH. Let MATH with MATH be a general point in a component of MATH. A standard dimension count shows that MATH must be smooth. There exists a limit MATH on MATH, say MATH, together with sections MATH and MATH, such that MATH, MATH, MATH and moreover MATH (apply REF ). Clearly MATH, hence MATH. The contraction map MATH collapses MATH and identifies MATH and MATH, so the second part of the claim follows. To conclude that MATH is also reduced we use that both MATH and MATH are reduced and that they meet transversally. This is because the limit MATH we found on MATH is smoothable in such a way that all ramification is kept away from the nodes (compare REF ), hence the tangent spaces to MATH and MATH at the intersection point MATH cannot be equal. |
math/0104187 | Once again, let MATH be a point in MATH, with MATH and MATH. Then there exists a limit MATH, say MATH on MATH together with sections MATH and MATH such that MATH and moreover MATH. The NAME formula on MATH and the condition defining a limit linear series give that MATH. On the other hand, since MATH moves in a family of dimension MATH it follows that MATH also moves in a family of dimension MATH in MATH (that is, codimension MATH). Since according to REF , the locus of pointed curves MATH carrying a MATH having MATH has codimension MATH, we get MATH. There are two possibilities: CASE: MATH. Let us denote MATH, hence MATH and MATH for MATH. Therefore MATH . Moreover, since MATH, we obtain that MATH, hence MATH, that is, MATH is a MATH on MATH with MATH in a fiber, or equivalently MATH, where MATH. To see that conversely MATH we pick a general pointed curve MATH having a MATH with MATH in a fiber and we construct a NAME admissible covering MATH of degree MATH, where MATH is a curve semistably equivalent to MATH defined as above, and MATH is the transversal union of two lines (see REF ): we take MATH to be the degree MATH covering such that MATH, while MATH is the degree MATH map containing MATH and MATH in different fibers and with MATH in the fiber over MATH. It is clear that there is a unique such MATH on MATH. Furthermore, at MATH we insert a rational curve MATH mapping isomorphically onto MATH and at the remaining MATH points in MATH we insert rational curves mapping with degree MATH onto MATH while at the MATH points in MATH we insert copies of MATH mapping isomorphically onto MATH. We denote the resulting curve by MATH. If MATH, then MATH is stably equivalent to MATH and MATH and MATH appear in distinct fibers of the MATH-sheeted map MATH. Thus we get that MATH. CASE: MATH. We denote MATH. Since MATH we get that MATH. Now MATH is already a codimension MATH condition on MATH, so it follows that MATH, hence MATH. This yields MATH and MATH. We thus get that MATH. Conversely, given MATH together with a MATH on MATH with MATH in a fiber, we construct a degree MATH admissible covering MATH, which will prove that MATH: we first take MATH of degree MATH with MATH. Then MATH is of degree MATH, completely ramified at MATH and with MATH. At MATH we insert a rational curve MATH which we map MATH to MATH such that we have total ramification both at MATH and at the point MATH characterized by MATH. Finally, at each of the points in MATH we insert a MATH which we map isomorphically onto MATH. Thus we have proved that MATH. The conclusion now follows if we notice that MATH is reduced and all admissible coverings we constructed are smoothable, hence MATH is reduced too. |
math/0104187 | We will compute the class of MATH when MATH. The class of MATH is computed similarly. Let us write the following relation in MATH: MATH . We use the method of test curves to determine the coefficients MATH and MATH, that is, we intersect the classes appearing on both sides of the previous relation with curves inside MATH . By computing intersection numbers we obtain linear relations between the coefficients MATH. By REF we have that MATH . Using the same reasoning as in REF , we obtain that MATH is the closure in MATH of the locus of curves MATH carrying a MATH with MATH in a fiber. In order to determine the coefficient MATH we intersect both sides of MATH with a general fiber MATH of the map MATH: we get that MATH (compare REF ). To determine MATH and MATH we use two test curves in MATH: first, we fix a general curve MATH of genus MATH and we obtain a family MATH, by fixing a general point MATH and letting MATH vary on MATH. From MATH, clearly MATH. On the other hand, according to REF MATH. For a new relation between MATH and MATH we use the test curve MATH in MATH, where this time MATH is a fixed general point while MATH varies on MATH. We have the equation MATH, and since MATH is already known we get in this way both MATH and MATH. We are only left with the computation of MATH. From REF we know that the class of MATH is a linear combination of the NAME class and of the class of the divisor of NAME points, that is, MATH, where MATH . We already know that MATH. To determine MATH we use the following test curve in MATH: we take a general curve MATH of genus MATH and a general MATH-pointed elliptic curve MATH. We consider the family MATH obtained by identifying the variable point MATH with the fixed point MATH. We easily get MATH, MATH, while MATH vanishes on all the other boundaries. On the other hand MATH is the number of limit MATH's on the curves MATH having vanishing MATH at the fixed point MATH. If MATH is such a linear series, then using again the additivity of the NAME numbers (compare REF ) and the assumption that MATH is not torsion, we obtain that MATH, so either MATH or MATH. Thus MATH and we can write a new relation enabling us to compute MATH. |
math/0104187 | Let us write MATH, where MATH and MATH. As noticed before, the MATH-part of MATH and the MATH-part of MATH coincide, hence using REF MATH . Finally, to determine MATH one has to compute the MATH coefficient of the divisor MATH on MATH. Arguing in a way that is entirely similar to REF we obtain that MATH. |
math/0104193 | MATH is slc and MATH is MATH-Cartier by NAME REF . Thus MATH is slc, and MATH at the strictly slc singularities of MATH. Now MATH and moreover MATH if MATH, so the bound is clear at stricly slc points of MATH . Since MATH has a MATH-Gorenstein smoothing, the slt singularities are of REF types (compare REF ): CASE: MATH where MATH. CASE: MATH where MATH. CASE: MATH. We have MATH, MATH, and MATH in REF . We now use the two properties MATH is slc and MATH is NAME to bound MATH and MATH in REF . In REF , we first remark that MATH and MATH by REF , using the fact that MATH smoothes to MATH. Consider the local smooth cover MATH of MATH at MATH, let MATH denote the inverse image of MATH, say MATH. Then MATH since MATH is log canonical at MATH. Let MATH be the local equation of MATH at MATH and MATH be a monomial of minimal multiplicity. Then MATH, and MATH, using MATH . NAME and NAME. Thus in particular MATH. Suppose MATH, then MATH, and MATH, MATH. It follows that if MATH we have MATH, and if MATH we have MATH. This proves the bounds in REF . In REF , let MATH be the local analytic component MATH of MATH at MATH, let MATH be the double curve, MATH. Then MATH is log canonical at MATH. Let MATH be the smooth cover MATH of MATH, let MATH denote the inverse images of MATH. By adjunction MATH is log canonical, equivalently MATH is reduced. Write MATH. Then MATH where MATH. We have MATH (respectively MATH) if MATH (respectively MATH), using MATH . NAME (respectively MATH . NAME). It follows that MATH if MATH, MATH if MATH. |
math/0104193 | We use the valuative criterion. So, suppose we are given MATH and MATH in MATH such that MATH. We need to show that MATH. Note that we may assume that MATH and MATH is smooth for MATH and MATH, since the open subset of MATH of pairs MATH where MATH and MATH is smooth is dense. After base change, we may assume that MATH. Possibly after base change, we can construct a common semistable log resolution MATH of MATH and MATH that is an isomorphism over MATH. We now claim that we can reconstruct the two families as the MATH canonical model of MATH for MATH. So MATH and MATH as required. To prove the claim, we just need to verify that MATH is log canonical and relatively ample for MATH. We can check ampleness on the central fibre. We have MATH by adjunction. This is ample since MATH is ample for some MATH and MATH. Now, MATH is slc for some MATH by the definition of MATH, and the same is true for MATH. It follows that MATH is log canonical by REF . This completes the proof. |
math/0104193 | If MATH is log canonical, then trivially MATH is slc by adjunction. We now prove the converse. We may work locally at MATH. Take a semistable log resolution MATH of MATH. Let MATH be the MATH canonical model over MATH. Then MATH is an isomorphism over the locus where MATH is log canonical. MATH is log canonical at any codimension MATH point of MATH. For the implication MATH slc MATH MATH log canonical is easy in the case MATH. So no exceptional divisor of MATH has centre a codimension MATH point of MATH. Now let MATH be the normalisation of a component of MATH, MATH the inverse image of the double curve of MATH on MATH, and MATH. Let MATH be the normalisation of the strict transform of this component in MATH, MATH the inverse image of the double curve of MATH on MATH, MATH and MATH. Write MATH for the induced map. We have MATH where the primes denote strict transforms and the MATH are the MATH-exceptional divisors. Then MATH for all MATH because MATH is slc. Now MATH since there are no MATH-exceptional divisors with centre a codimension MATH point of MATH. So, we have MATH where MATH for all MATH since MATH is relatively ample. If MATH for some MATH, we have MATH, a contradiction. So, in particular, for each component MATH of MATH, with strict transform MATH in MATH, the map MATH does not contract any component of the double curve of MATH on MATH. It follows that there are no MATH-exceptional divisors over MATH (recall that every such divisor has centre MATH). Moreover, for each component MATH of MATH, MATH. Hence MATH, so MATH has no exceptional divisors. Then MATH and MATH-ample implies MATH is an isomorphism, so MATH is log canonical as required. |
math/0104193 | First complete MATH to a family MATH. Possibly after base change (which we suppress in our notation) we can take a semistable log-resolution MATH of MATH which is an isomorphism over MATH. Then MATH is dlt. Run a MATH MMP over MATH. Let MATH be the end product. Then MATH must be relatively nef, since it is zero on the generic fibre. Now it follows that MATH. For we can write MATH, where the MATH are the components of MATH, since MATH is zero on the generic fibre. Without loss of generality we may assume MATH and MATH for MATH, using the relation MATH. Restricting to MATH, we see that MATH relatively nef implies that MATH for all MATH such that MATH. Since MATH is connected, repeating the argument we obtain MATH for all MATH, so MATH as claimed. Compare REF . We have MATH dlt and MATH is MATH-factorial, so MATH is dlt. We now run a MATH MMP over MATH and let MATH be the end product. Then MATH is a NAME fibration, since the generic fibre is a NAME surface (namely MATH). Now, MATH and MATH-factorial gives MATH irreducible - for we have an exact sequence MATH by REF , where the MATH are the irreducible components of MATH. The dlt property of MATH now gives that MATH is normal and log terminal. Since MATH is dlt and MATH, we have MATH log canonical and MATH. We obtain MATH is log canonical and MATH by adjunction. The sharpening in the case MATH is clear. Finally, since MATH is MATH-Cartier and MATH, MATH is MATH-Cartier. |
math/0104193 | First complete MATH to a family MATH. By REF and its proof, we may assume that MATH is log canonical, and MATH. After base change, take a semistable log resolution MATH which is an isomorphism over MATH. Now run a MATH MMP over MATH. Here MATH is chosen such that for MATH an exceptional divisor of MATH, if MATH then MATH. Let MATH be the end product. By the choice of MATH, we have MATH. Now take the MATH canonical model over MATH, denote this MATH. Then MATH is log canonical and relatively ample over MATH, and MATH since MATH. We obtain MATH is slc and ample and MATH by adjunction. The sharpening in the case MATH is obvious. Finally since MATH and MATH is MATH-Cartier, MATH and MATH are MATH-Cartier. |
math/0104193 | We use the valuative criterion of properness. Taking MATH, we need to show that, possibly after base change, there exists an extension MATH. We may assume that MATH and MATH is smooth as in the proof of separatedness above. Applying REF we obtain our result. |
math/0104193 | MATH is ample by REF . There exists a MATH-Gorenstein smoothing MATH of MATH by REF . Finally, MATH is slc, and MATH is MATH-Cartier, so MATH is slc. |
math/0104193 | First, by REF , either MATH and there is at most one strictly log canonical singularity on MATH, or MATH, MATH is connected, and MATH is log terminal away from MATH. Now let MATH be a component of MATH. We have MATH is ample, so MATH. Expanding the left hand side, MATH . Now MATH and MATH, thus MATH, MATH, and MATH. We know that MATH is log canonical, so we have the classification given in REF of the singularities of MATH at MATH. We calculate that each point of type REF on MATH contributes MATH, MATH, MATH to MATH respectively. The theorem now follows easily. |
math/0104193 | The classification in the Theorem is immediate from the classification of the components MATH in REF using the classification of slc singularities in REF . |
math/0104193 | REF is the well known classification of smoothable log terminal surface singularities (CITE, p. CASE: NAME REF). REF follow from REF . |
math/0104193 | Write MATH where the MATH are the exceptional divisors of MATH, and MATH is the strict transform of MATH. Then MATH. Here MATH for all MATH since MATH is log canonical. MATH is minimal, so MATH is MATH-nef, hence also MATH is MATH-nef. It follows that MATH for all MATH. This proves REF . REF is immediate since MATH by definition and MATH is ample. REF now follows since MATH . |
math/0104193 | First, we claim MATH is birationally ruled. For MATH is nef and big, so MATH for all MATH. MATH is effective, so MATH for all MATH. It follows that MATH is birationally ruled. We may assume that MATH is not rational. Then MATH is ruled over a curve of positive genus, let MATH be a ruling and let MATH denote the composite MATH. Suppose that there is no horizontal component of MATH. Thus MATH is a sum of fibres. Now MATH nef and big implies that MATH is nef and big. Then MATH by NAME duality and NAME vanishing, a contradiction. Thus there is an irrational component of MATH. By REF we know that MATH has only rational components, so by the classification of log canonical singularities it follows that MATH has a simple elliptic singularity. Let MATH denote the MATH-exceptional elliptic curve on MATH. Then MATH has multiplicity MATH in MATH and MATH is horizontal. Now MATH is big so MATH for MATH a fibre of the ruling, thus MATH, MATH is a section. Next we claim that MATH is in fact ruled. Suppose not, then there exists a degenerate fibre. Let MATH be a component meeting MATH. Then MATH is not contained in MATH. We have MATH, with equality iff MATH is contracted by MATH. But also MATH with equality only if MATH is a MATH curve. Thus MATH is a MATH curve and is contracted by MATH, a contradiction since MATH is minimal. So MATH is ruled over an elliptic curve. MATH is obtained from MATH by contracting the negative section and so MATH is an elliptic cone. Finally MATH by NAME 's connectedness theorem (compare REF ) so MATH. |
math/0104193 | CASE: Certainly MATH is exact, so we just need to show the first map is injective. Suppose MATH, then using MATH, we can replace this with a relation MATH, where, without loss of generality, MATH and MATH for MATH. Then MATH is effective and linearly equivalent to MATH. So MATH since MATH is proper. Thus MATH for all MATH such that MATH. Since MATH is connected, repeating the argument we obtain MATH for all MATH, as required. CASE: Suppose MATH is a NAME divisor which is numerically equivalent to MATH. Then MATH, thus MATH, some MATH. Now a similar argument to the above shows MATH as required. |
math/0104193 | CASE: First we show that MATH is an isomorphism. From the exponential sequence for MATH we obtain the long exact cohomology sequence MATH . We have MATH by NAME duality. Thus MATH since MATH is ample. Now MATH gives MATH. So we obtain that MATH is an isomorphism. Now, since MATH, cohomology and base change (CITE, p. REF ) gives MATH, thus MATH since MATH is affine. Using the exponential sequence for MATH we obtain that MATH is an isomorphism. Finally, MATH is an isomorphism since MATH is a homotopy retract of MATH, so MATH is an isomorphism as claimed. Now, since MATH is surjective, MATH is surjective and MATH is injective. But MATH is clearly surjective since MATH is the only closed fibre of MATH. Hence MATH and MATH are isomorphisms. CASE: First, we show that we have isomorphisms MATH, MATH and dually MATH, MATH. Well, first from the proof of REF above we have MATH, and by REF we have MATH, so we obtain MATH. This gives MATH. For MATH, let MATH be a component, let MATH be a resolution. We may assume that MATH is not normal, otherwise we are done by the above. Then MATH is rational by REF . Thus MATH, it follows that MATH. We have MATH. For MATH since MATH is ample, and MATH since MATH is rational (using NAME). So MATH using the exponential sequence. Thus MATH as desired. It remains to show that MATH is exact, where the map MATH is as in the statement of the lemma, and we work with MATH coefficients. We use the NAME sequence inductively here, separating off one component at a time. First we separate the double curves MATH where the two branches of MATH at MATH belong to the same component. In this case, MATH is homotopy equivalent to MATH, where MATH is MATH with the two branches at MATH separated, MATH is MATH, and MATH and MATH are glued along MATH, where MATH is obtained using REF-to-REF map MATH. Note that MATH is either REF MATH, REF-to-REF, or REF MATH where the two components are joined at a node and each maps isomorphically onto MATH. Then MATH is exact, using MATH. In REF we have MATH thus MATH is an isomorphism. In REF , writing MATH, we have MATH thus MATH is exact. Next we separate the components of MATH. Let MATH be a component (assumed normal), and write MATH, MATH. Then MATH is exact, using MATH (recall MATH is either MATH or MATH). Now, using these steps repeatedly, we obtain our result. |
math/0104193 | CASE: We have MATH exact, where MATH maps to zero if MATH is obtained by folding MATH. Now MATH, MATH and MATH. Thus MATH, with equality iff MATH is MATH-factorial. So, using the exact sequence above, we obtain MATH, equal only if MATH is MATH-factorial. CASE: The maps MATH and MATH are dual. Let MATH be the component of MATH meeting MATH, let MATH denote the intersection. By the exact sequence of REF , we find that MATH is injective (recall that MATH). For, if MATH is in the kernel, MATH for some MATH. Intersecting with a relatively ample divisor on MATH, we find that MATH, so MATH. Now using the description of MATH in REF we see that MATH. For locally at MATH, MATH is generated by MATH and MATH (where MATH is a divisor flat over MATH, of degree MATH on MATH). So we obtain MATH as desired. CASE: Similarly, in this case we find that MATH, and MATH, so we obtain MATH. |
math/0104193 | All this follows from REF except for the claim that MATH for MATH a middle component of a surface of type MATH. To prove this, observe that in this case MATH is injective. For write MATH, where MATH is glued to MATH by glueing MATH to MATH. Suppose MATH is in the kernel of MATH. By the exact sequence of REF , we have MATH for some MATH. Thus MATH. Working inductively, we obtain MATH and MATH, so MATH as claimed. We now conclude as in the proof of REF . |
math/0104193 | Suppose MATH is a counter example, let MATH be a MATH-Gorenstein smoothing with MATH. We have MATH by REF . We have MATH since MATH is MATH-Cartier. Thus MATH, so MATH, using MATH ample and MATH. So MATH. Let MATH be the minimal resolution of MATH. Then MATH is rational by REF . Applying NAME 's formula we obtain MATH. So the resolution MATH has strictly decreased MATH. However, we calculate below that the only possible singularities on MATH will increase MATH when we take the minimal resolution, so we have a contradiction. MATH has singularities of type MATH and MATH, with the latter cases occurring in pairs MATH and MATH, by REF . Now, given a cyclic quotient singularity MATH, let MATH be the expansion of MATH as a NAME continued fraction. The geometric interpretation of this is that the minimal resolution of the singularity has exceptional locus a chain of smooth rational curves with self-intersections MATH. Then on taking the minimal resolution of the singularity, the change in MATH is given by MATH where MATH denotes the inverse of MATH modulo MATH (CITE, p. REF). We have MATH in the case MATH (CITE, p. REF). Finally, we calcuate MATH in the case of a pair of singularities MATH. To see this, note that if we write MATH and MATH, we have MATH. |
math/0104193 | First, we know that every normal log terminal singularity of MATH is of the form MATH, since we assume there exists a MATH-Gorenstein smoothing (compare REF ). We need to show that MATH. We sketch the proof here, for details see CITE, p. REF, NAME REF . We compute that the NAME fibre MATH of a smoothing of a singularity of type MATH has MATH and negative definite intersection product. Now, since MATH has positive definite intersection product, it follows that MATH, so MATH as required. Finally, we show MATH. Let MATH be a MATH-Gorenstein smoothing of MATH to MATH. Let MATH. Then locally at MATH, we have MATH, generated by MATH (see CITE, p. REF, NAME REF, and CITE, p. REF, NAME REFEF). But MATH locally at MATH, where MATH is a divisor, flat over MATH, that restricts to a hyperplane section on MATH. Thus MATH. For another proof using the NAME fibre, see CITE, p. REF . |
math/0104193 | Suppose given a normal log canonical NAME surface MATH which admits a MATH-Gorenstein smoothing to MATH. We may assume MATH is strictly log canonical, otherwise MATH is a NAME surface. Let MATH be the minimal resolution of MATH. If MATH is not rational, then MATH is an elliptic cone by REF . Then MATH gives that MATH is an elliptic cone of degree MATH. So we may assume MATH is rational. The NAME spectral sequence gives an exact sequence MATH . Now MATH since MATH is ample, and MATH since MATH is rational. So MATH, MATH has rational singularities. We can now use a result of NAME (CITE, p. REF , and p. REF ): Let MATH be a birational morphism, with MATH maximal (so MATH is an isomorphism over the negative section MATH of MATH). Let MATH denote the birational ruling so obtained. Then the exceptional locus of MATH is the strict transform MATH of MATH together with the irreducible components of the degenerate fibres of MATH of self-intersection MATH. In particular, MATH. Moreover every degenerate fibre contains a unique MATH curve. We quickly sketch the proof of this. First, since MATH smoothes to MATH, we have MATH, and MATH since MATH is minimal. NAME deduces there is no horizontal curve MATH on MATH with MATH except possibly MATH. Now since MATH, it follows that the exceptional locus of MATH is MATH together with all the components of the degenerate fibres of self intersection MATH, and every degenerate fibre has a unique MATH curve. There are two types of rational strictly log canonical surface singularities - namely a MATH quotient of a cusp and a quotient of a simple elliptic singularity. Consider the minimal resolutions of these singularities. In each case the exceptional locus is a union of smooth rational curves. For a MATH quotient of a cusp, the exceptional locus consists of a chain of curves with two MATH curves off each end component of the chain. For a quotient of a simple elliptic singularity, the exceptional locus consists of a curve with three chains of curves off it. We now analyse how these could possibly fit into the minimal resolution MATH of MATH. Consider the minimal model program yielding MATH in the neighbourhood of a given fibre of MATH. At each stage we contract a MATH curve, meeting at most MATH components of the fibre, and disjoint from MATH (since MATH is an isomorphism over MATH). We know that MATH . This set decomposes into the exceptional loci of the minimal resolutions of one log canonical rational singularity and some MATH singularities. First concentrate on the log canonical singularity; let MATH denote the exceptional locus of its minimal resolution. Then MATH contains a curve MATH which meets MATH other components of MATH - we call such a curve a fork of MATH. Suppose MATH is a degenerate fibre of MATH containing a fork MATH of MATH. Let MATH denote the component meeting MATH (this is the strict transform of the corresponding fibre MATH of MATH). Then we have a decomposition MATH, where CASE: MATH is the unique -REF curve in MATH. CASE: MATH is a string of curves, with one end component meeting MATH, MATH contracts to a MATH singularity (or is empty). CASE: MATH,MATH and MATH are nonempty configurations of curves meeting MATH and MATH, MATH, and MATH and MATH respectively. Then in the MMP MATH, we contract MATH, MATH, MATH and MATH in that order. (Note: MATH and MATH are nonempty because MATH is a fork. One might think MATH could be empty since MATH contains MATH, but in that case MATH and must be contracted before we can contract MATH, a contradiction). Next suppose that MATH is a degenerate fibre of MATH that does not contain a fork of MATH. Then we have a decomposition MATH, where CASE: MATH is the unique -REF curve in MATH. CASE: MATH is a string of curves, with one end component meeting MATH, MATH contracts to a MATH singularity (or is empty). CASE: MATH is a non-empty string of curves meeting MATH and with one end component meeting MATH. We now analyse these two cases for each of the two types of singularity. We call them fibre types I and II. First suppose MATH has a MATH quotient of a cusp singularity. So MATH where MATH is a chain of MATH's, and MATH (respectively MATH) are MATH-curves meeting MATH (respectively MATH). Suppose MATH is contained in a degenerate fibre MATH. Then as above we can write MATH where without loss of generality MATH, MATH, MATH and MATH for some MATH. Note that MATH cannot contain the other fork MATH of MATH, since then MATH contradicting the description above. We contract MATH first. We deduce that the curves in the string MATH have self-intersections MATH. Thus MATH contracts to a MATH singularity, where MATH is the length of the string. But this is never a MATH singularity (since MATH), a contradiction. So MATH is empty. We can now calculate that the curves in the string MATH have self-intersections MATH if MATH. Then MATH, a contradiction. Hence MATH. Next suppose MATH is not contained in a degenerate fibre. Then MATH is horizontal, hence MATH. Then it follows that we have a fibre of type II with MATH, a -REF curve. We deduce that MATH is a single -REF curve. But then MATH contracts to a MATH singularity, which is not MATH, a contradiction. Combining, we deduce that MATH, and we have two fibres of the form MATH as above, and MATH. There are no further degenerate fibres. It only remains to calculate MATH. We use MATH to deduce MATH. I claim that the surface MATH constructed above does not admit a MATH-Gorenstein smoothing. Let MATH be the index one cover of MATH at the singular point. Then MATH has a cusp singularity and the exceptional locus of the minimal resolution MATH is a cycle of rational curves of self-intersections MATH. Suppose MATH has a MATH-Gorenstein smoothing, then, taking the canonical cover of the smoothing at the singular point we obtain a smoothing of MATH. Let MATH denote the NAME fibre of the smoothing of MATH. Consider the intersection product on MATH, write MATH, where MATH, MATH and MATH are the number of zero, positive and negative eigenvalues of the intersection form. Since MATH is normal and NAME, we have CITE MATH . In our case we calculate MATH, a contradiction. So MATH is not smoothable, hence MATH does not have a MATH-Gorenstein smoothing (not even locally). Now suppose MATH has a quotient of a simple elliptic singularity. So MATH where MATH, MATH is a chain of smooth rational curves and MATH meets MATH, for MATH and MATH. We first give a partial classification of these singularities. We can contract the chains MATH to obtain a partial resolution MATH. Write MATH for the image of MATH under MATH. Then the chains MATH contract to singularities of type MATH on MATH. Let MATH be the indices of these singularities, then MATH (because MATH is assumed to be strictly log canonical - the condition is equivalent to MATH). Thus MATH or MATH after reordering. In particular, we see that each chain MATH is either a single MATH of self-intersection MATH, or a chain of MATH's of self-intersection MATH. We claim that the fork MATH of MATH cannot be contained in a fibre MATH. By the classification above, its enough to show that MATH, since this then forces MATH, a contradiction. Write MATH. Let MATH. Then MATH . Since MATH is an isomorphism over MATH, we have MATH because MATH is MATH-exceptional. So MATH, writing MATH for a fibre of MATH. Here MATH where MATH is the multiplicity of MATH in MATH. Now MATH since MATH is log canonical and MATH is minimal. Hence MATH so MATH as required. Thus MATH is horizontal, MATH and we have REF degenerate fibres of type II. In each case MATH is a single curve of self-intersection MATH or a string of MATH-curves. If the fibre MATH is a string, we deduce that MATH is a string of MATH-curves or a single curve of self-intersection MATH respectively. Now, since MATH contracts to a MATH singularity, we deduce MATH is a single MATH-curve and MATH. If MATH is not a string, we find that MATH is a string of three MATH-curves, MATH meets the middle component, and MATH is empty, hence again MATH. So MATH for all MATH, contradicting the classification above. It remains to show that an elliptic cone of degree MATH admits a MATH-Gorenstein smoothing to MATH. We prove this in REF below. |
math/0104193 | Given an elliptic cone MATH of degree MATH, write MATH for the minimal resolution of MATH. Then MATH is a ruled surface over an elliptic curve MATH, that is, MATH, where MATH is a line bundle on MATH of degree MATH. We claim that MATH is determined up to isomorphism by its section MATH. For MATH acts transitively on MATH, and given a line bundle MATH of degree MATH we have MATH for some MATH (compare CITE, p. REF ), thus MATH acts transitively on the line bundles of degree MATH. Our claim follows. Let MATH be the spectrum of a DVR, and write MATH. Let MATH be an elliptic curve in the special fibre. Let MATH be the blowup of MATH in MATH. Then the special fibre MATH consists of the strict transform MATH of MATH together with a ruled surface MATH of degree MATH over the elliptic curve MATH. We contract MATH to obtain a family MATH which is a smoothing of an elliptic cone of degree MATH over MATH to MATH. |
math/0104193 | CASE: Write MATH. We work locally at a closed point MATH, say MATH. Then MATH is a closed subscheme with support MATH. The sheaf MATH is MATH by assumption, so there exists a regular sequence MATH for MATH at MATH. Replacing MATH by MATH if necessary, we may assume that MATH. Now lift MATH to MATH, then MATH is a regular sequence for MATH at MATH (CITE, p. REF, Corollary of REF ). Equivalently, we have an exact sequence MATH . Consider the natural map MATH, write MATH for the kernel and MATH for the cokernel. MATH and MATH have support contained in the set MATH, so any given element of MATH or MATH is annihilated by some power of MATH. So, if MATH, there exists MATH such that MATH, so in particular MATH, contradicting the exact sequence above. NAME if MATH, there exists MATH such that MATH. Again using the exact sequence above, since MATH we obtain MATH for some MATH, it follows that MATH, a contradiction. Thus MATH, so the map MATH is an isomorphism as claimed. CASE: If MATH is invertible at MATH then, working locally at MATH, lifting an isomorphism MATH we obtain a surjection MATH, by NAME 's Lemma. Now, by flatness of MATH, it follows that MATH is an isomorphism, so MATH is invertible at MATH. |
math/0104193 | We may assume that MATH and MATH are affine, write MATH, MATH. We have a commutative diagram: MATH . Thus MATH iff MATH is surjective, by the snake lemma. This proves the first part using REF . Moreover, we see that in this case MATH for all MATH, hence MATH is flat over MATH as required. |
math/0104193 | If MATH commutes with base change then MATH is an isomorphism, and MATH is invertible by assumption. Since MATH is coherent, it follows that MATH is invertible by NAME 's Lemma and the flatness of MATH (using REF ). Conversely, suppose MATH is NAME. By REF , we need only show that MATH is an isomorphism. Both sides are invertible by assumption and the map is an isomorphism in codimension MATH, thus it is an isomorphism since MATH is MATH. |
math/0104193 | This is a straightforward generalisation of the usual cyclic covering trick CITE. |
math/0104193 | Let MATH, MATH, then MATH . NAME gives MATH . NAME, so MATH. We claim that MATH. We may work locally at MATH. First suppose that MATH where MATH is a local NAME MATH-algebra. Let MATH be a cyclic cover defined by MATH. Let MATH be the locus where MATH is NAME (then MATH is the open subscheme of MATH with underlying space MATH) and let MATH be the inverse image in MATH. We have MATH. Thus MATH connected implies MATH is connected. Hence MATH is connected, since MATH and MATH have the same underlying space. But MATH thus the cover MATH of MATH is connected, it follows that the index MATH of MATH is equal to MATH. The general case now follows using REF - we have MATH . NAME for every NAME subscheme MATH, hence MATH is NAME, so MATH. |
math/0104193 | This follows from CITE, p. REF and p. REF . |
math/0104193 | See CITE, REF . |
math/0104193 | For the local case, we use CITE, p. REF , and p. REFEF (which shows that MATH for each MATH). The global case follows formally. |
math/0104193 | We are given a family MATH such that MATH commutes with base change for all MATH. Then in particular MATH is MATH-Cartier and MATH by REF . Let MATH be an index one cover. Then MATH is flat over MATH, because MATH and MATH is flat over MATH for each MATH by REF . Given MATH in MATH, the isomorphisms MATH show that MATH is an index one cover of MATH. Given MATH a MATH invariant extension of MATH we write MATH, then MATH is a flat family extending MATH. We claim that the quotient MATH is an index one cover of MATH. First, MATH is invertible, and MATH commutes with base change, thus MATH is invertible. Thus MATH is NAME. In particular MATH is MATH-Cartier, so MATH by REF . Let MATH denote the open subscheme of MATH where MATH is invertible, and write MATH for the corresponding open subscheme of MATH. Then MATH is an etale MATH quotient (because the map is etale over MATH). Finally MATH by REF . So MATH is an index one cover, by REF . Moreover, we claim MATH commutes with base change for all MATH. First MATH commutes with base change since it is invertible, using REF . Also, by the above, MATH is an index one cover of MATH, and MATH is an index one cover MATH of MATH by assumption. It follows that MATH is an isomorphism for MATH. Now MATH commutes with base change for all MATH by assumption, thus MATH commutes with base change for MATH using REF . So MATH commutes with base change for all MATH by REF . Finally, if MATH is a infinitesimal extension, there is a unique index one cover of MATH extending MATH - for such an index one cover MATH is determined by the choice of an isomorphism MATH extending a given isomorphism MATH and thus by a unit MATH. Moreover multiplying MATH by MATH for some MATH does not change the isomorphism type of the extension MATH of MATH. But we can always take MATH-th roots in MATH since MATH is infinitesimal, thus MATH is uniquely determined as claimed. The last part of the Proposition follows. |
math/0104193 | We may work locally at MATH. Let MATH be an index one cover. It is enough to show that the map MATH of the special fibres is an index one cover of MATH, by REF . Now MATH extends to an index one cover MATH, we need to show that MATH is MATH to deduce MATH. First, MATH slc and MATH canonical implies that MATH is canonical. For there exists a finite base change MATH such that MATH admits a semistable resolution, then MATH is canonical by REF . Using CITE, p. REF we deduce that MATH is canonical. Now MATH canonical implies that MATH is canonical, and canonical singularities are rational so in particular CM. Hence MATH is CM since MATH (where MATH is a uniformising parameter). This completes the proof. |
math/0104193 | Let MATH be the inclusion of the locus where MATH is invertible. Let MATH be a local index one cover, a MATH quotient say, and write MATH for the inverse image of MATH, then MATH is etale. We have MATH and MATH using REF , thus MATH. NAME, MATH, so, using MATH, we find MATH. But MATH and MATH agree on MATH, so we obtain our result. |
math/0104193 | For the local case we use REF together with REF . We may assume MATH is a local ring. Given a MATH-Gorenstein family of slc surfaces MATH extending MATH, take an index one cover MATH (a MATH quotient, say). This is flat over MATH and restricts to an index one cover MATH by REF . Now, by REF , there is a canonical element MATH such that MATH iff MATH. By functoriality of the obstruction map, we have MATH using REF . We define MATH. In order to show REF holds, we just need to verify that if MATH then MATH, for by REF we have a bijection MATH - we use REF below. REF follows immediately from REF since MATH. Finally, REF is a special case of REF . The global case now follows formally exactly as for ordinary deformation theory using the spectral sequence of REF (we use REF to identify MATH with MATH, the sheaf of infinitesimal automorphisms). |
math/0104193 | Write MATH, MATH, and MATH. Let MATH be an ideal defining a closed subscheme MATH (not necessarily flat over MATH) extending MATH. Then MATH is flat iff MATH for some liftings MATH of MATH and every relation MATH, MATH, between the MATH lifts to a relation MATH, MATH, between the MATH (compare CITE). In our case, pick generators MATH of MATH which are MATH eigenfunctions (using CITE, p. REF ). Pick MATH eigenfunctions MATH lifting MATH. Let MATH be exact. We define a MATH action on MATH (in the obvious way) such that MATH is MATH equivariant, and hence obtain a MATH action on the module of relations MATH. We have a well defined map MATH where MATH are some lifts of the MATH. Here we first regard MATH as an element of MATH and then take its image in MATH. Then MATH iff MATH is flat over MATH, by the criterion above. Thus there exists a flat extension MATH of MATH iff MATH - for this is the condition that we can change the MATH so that MATH becomes the zero map. Now, by construction, MATH is a MATH invariant element of MATH, so equivalently we require that MATH . In this case we can replace the MATH by MATH eigenfunctions MATH to obtain a MATH invariant flat extension MATH. This completes the proof. |
math/0104193 | We just need to show that the smoothability property for a stable pair MATH of degree MATH gives MATH if MATH, MATH, and MATH. First note that trivially MATH if MATH admits a smoothing to MATH. If MATH the only possible singularities of MATH are of local analytic types MATH where MATH and MATH (using REF ). In the second case MATH shows MATH. Thus MATH. We now prove MATH. First note that MATH by NAME duality, and MATH is ample. We have an exact sequence MATH where MATH is the normalisation of MATH, and MATH is the normalisation of the double curve MATH of MATH. We obtain a short exact sequence MATH where MATH. In particular, since MATH is ample, we have MATH. Thus the long exact sequence of cohomology associated to the short exact sequence above gives MATH. If MATH is klt, then MATH by NAME vanishing. Otherwise, MATH is an elliptic cone by REF , and MATH, MATH. An easy calculation shows that MATH in this case. |
math/0104193 | This is immediate from REF - MATH is the subgroupoid of smoothable families. |
math/0104193 | The 'only if' part follows from REF . So, suppose given a family MATH of quasistable pairs such that MATH is canonical and MATH and MATH are MATH-Cartier. Then MATH is a MATH-Gorenstein family by REF . Let MATH be an index one cover, with special fibre MATH an index one cover of MATH. Since MATH is MATH-Cartier, and MATH is NAME (using MATH and MATH), we have MATH . NAME, using REF . It follows that MATH locally, some MATH, hence MATH commutes with base change for all MATH. Thus MATH is allowable as required. |
math/0104193 | Given MATH, MATH is flat over MATH and MATH for all MATH by REF . The other conditions are satisfied by the definition of MATH. Conversely let MATH satisfy REF to REF . We claim that MATH. Note that MATH is a relative NAME divisor (compare REF ). It is enough to show that MATH commutes with base change for all MATH. By REF , we may assume that MATH is a local NAME ring. So, by induction, it is enough to show the following: Let MATH be a local ring with residue field MATH and MATH a small extension (that is, the kernel MATH of MATH is annihilated by the maximal ideal of MATH). Suppose given MATH and MATH extending MATH which satisfies REF to REF . Then MATH. We work locally on MATH. Let MATH be an index one cover (a MATH quotient, say), MATH the index one cover obtained by restriction to MATH, and MATH. Given a MATH invariant extension MATH of MATH we obtain an extension MATH of MATH. We claim that every extension MATH of MATH occurs in this way. Let MATH be the special fibre of MATH. Then MATH is an index one cover, and since MATH and MATH we have MATH . NAME. Thus MATH has unobstructed embedded deformations (locally) and the extensions MATH of MATH form a prinicipal homogeneous space under MATH (using REF ). There exists a MATH invariant extension by REF , hence also MATH has unobstructed embedded deformations (locally) and the extensions MATH of MATH form a prinicipal homogeneous space under MATH. To prove our claim, we just need to identify the sheaves MATH and MATH. We have the exact sequence MATH applying the functor MATH we obtain MATH . Now MATH, since MATH is invertible in codimension MATH and MATH. Hence we have a short exact sequence MATH and similiarly MATH . Applying the exact functor MATH to the last exact sequence we obtain MATH thus MATH as required. So, given an extension MATH of MATH, it is obtained from an extension MATH of MATH by taking the MATH-quotient. In particular, MATH is NAME. Thus locally MATH for some MATH, hence MATH commutes with base change for all MATH as required. |
math/0104193 | Given an infinitesimal extension MATH in MATH and a MATH-Gorenstein family MATH extending MATH, we need to show that there exists an extension MATH of MATH (using REF ). By the proof of REF there are no local obstructions. We may assume that MATH is local and MATH is a small extension. Let MATH have residue field MATH and write MATH for the special fibre of MATH. Then the obstruction to extending MATH to some MATH lies in MATH by REF . Moreover, since there are no local obstructions, using the local-to-global spectral sequence for MATH we see that the obstruction lies in MATH. We have an exact sequence MATH (see the proof of REF ). Now MATH by assumption, and MATH since MATH is ample, thus MATH. So the obstruction is zero, and there is an extension MATH of MATH as required. |
math/0104193 | For REF see CITE, p. REF (in fact only the case MATH, MATH local is treated in CITE, but the same argument proves the general case). We now prove REF . If MATH, it is a principal homogeneous space under MATH by REF . Now MATH acts on MATH, so we obtain a homomorphism MATH . In the case MATH, MATH, MATH we obtain the map MATH above. We have a natural isomorphism MATH by REF , which is compatible with the maps to MATH. The result now follows. |
math/0104193 | We have MATH and MATH, so we have a natural map MATH with kernel MATH and cokernel MATH (compare REF ). In particular, MATH is a MATH-module, moreover MATH is a MATH-module since the functor MATH satisfies NAME 's REF . Now MATH and MATH so we have a natural map MATH with kernel MATH by REF . Finally, MATH is surjective by REF . |
math/0104193 | REF is immediate from REF - because the forgetful map MATH is smooth, the obstruction theory for MATH gives an obstruction theory for MATH. We have a natural map MATH which is surjective by REF . Assuming MATH, MATH is a principal homogeneous space under MATH by REF . Moreover, given MATH, MATH is the set MATH. This is a principal homogeneous space under MATH, where MATH as in REF . Now, the set MATH has a natural action of MATH, since the functor MATH satisfies NAME 's REF . This action is compatible with the actions just described via the exact sequence of REF , it follows that MATH is a principal homogeneous space under MATH as required. Finally, REF is a special case of REF . |
math/0104193 | The theorem is obtained by applying REF to our functor MATH. We verify the conditions of the theorem in REF , and REF below. |
math/0104193 | We have that MATH is MATH-Cartier and relatively ample using REF and the base change property for MATH, MATH. So, taking a sufficiently large and divisible multiple of MATH, we obtain a projective embedding of MATH. |
math/0104193 | We have a canonical polarisations MATH and MATH on MATH and MATH for some MATH, and so by CITE we know that the functor MATH is represented by a quasiprojective scheme MATH. It is then easy to construct MATH as a locally closed subscheme of MATH. |
math/0104193 | CASE: The locus where the fibres MATH are CM is open by CITE, p. REF, Corollary to REF . MATH is reduced iff it is regular in codimension MATH - this is an open condition. Since MATH commutes with base change, the requirement that MATH is NAME in codimension MATH is open. The condition MATH is equivalent to requiring that the sheaf MATH is MATH, which is an open condition, again by CITE (note that MATH is flat over MATH). Assuming this is satisfied, the natural map MATH is an isomorphism, thus the condition MATH . NAME in codimension MATH is open. Hence MATH is open as required. Using REF we deduce that MATH and MATH are relative NAME divisors. CASE: It is enough to show the following: Let MATH be the spectrum of a DVR with generic point MATH and closed point MATH. Let MATH be a family of pairs such that MATH and MATH commute with base change for all MATH, and such that the special fibre MATH is quasistable of degree MATH. Then MATH is quasistable of degree MATH. Clearly MATH is ample, and it is also slc by REF . We are given MATH , we claim that this implies MATH - the essential point here is that MATH, so MATH is discrete. To prove the claim, observe that MATH is NAME by REF , and the restriction map MATH is an isomorphism (compare Proof of REF ). Thus MATH, and restricting to the generic fibre we obtain our result. We similiarly obtain MATH in the case MATH. If MATH then we are given MATH, now MATH using REF , it follows that MATH. Given MATH, it follows that MATH by semicontinuity (using the base change property for MATH, note that MATH is flat over MATH by REF ). Finally, MATH is trivially an open condition. This completes the proof. |
math/0104193 | Let MATH be a direct system in MATH such that the limit MATH lies in MATH. Suppose given MATH, we need to show that this is obtained from some MATH by pullback. We know that MATH is projective, fix an embedding MATH. Then MATH is obtained by pullback from some projective flat family MATH for some MATH (since MATH is of finite type, where MATH denotes the NAME polynomial of the fibres of MATH). By REF we have that MATH is a codimension MATH closed subscheme which is flat over MATH. It follows (since MATH is of finite type, where MATH denotes the NAME polynomial of the fibres of MATH) that there exists MATH such that MATH, and MATH, such that MATH is obtained from MATH by pullback to MATH. By REF , we may also assume that each fibre MATH is CM, reduced and NAME in codimension MATH, and that MATH and MATH are relative NAME divisors. We now analyse the push forward and base change conditions. Let MATH be a coherent sheaf on an open subset MATH, flat over MATH, and MATH, MATH the corresponding objects obtained by pullback to MATH. We are interested in the cases CASE: MATH where MATH is the locus where MATH is NAME. CASE: MATH, where MATH is the locus where MATH is NAME. Assume that the push forward of MATH commutes with base change, and that MATH is coherent for each MATH. We work locally at MATH, say MATH. The natural map MATH is surjective, so pick a finite set of elements of MATH which generate MATH over MATH (recall MATH is assumed to be coherent). Since MATH, these are defined over some MATH where MATH. Write MATH, MATH for the objects obtained by pullback from MATH to MATH, and say MATH. Then, by construction, we have that MATH is surjective, hence by REF the push forward of MATH commutes with base change in a neighbourhood of MATH, which we may assume is MATH. As in REF , we only need to consider a finite number of sheaves MATH. Thus there exists MATH such that MATH and, writing MATH for the pullback of MATH, the sheaves MATH and MATH commute with base change for all MATH. Finally, by REF we may assume that every geometric fibre of MATH is quasistable. Then MATH is an element of MATH. This completes the proof of the lemma. |
math/0104193 | Let MATH be a diagram in MATH as in the statement of REF . Write MATH, then MATH is an infinitesimal extension of MATH. We need to show that MATH is surjective for MATH. So, let MATH and MATH be families which extend some MATH. Define MATH as follows: let MATH ( where MATH denotes the underlying topological spaces), and MATH, MATH. Then MATH is a flat family of pairs extending MATH and MATH (to prove flatness, use CITE p. REF ). Thus, by REF , we only need to verify that MATH is a MATH-Gorenstein family to obtain MATH as required. To see this, take an index one cover MATH, this gives an index one cover MATH on restriction to MATH by REF , extend this to an index one cover MATH. Define MATH by MATH as above, then MATH is an index one cover extending MATH and thus MATH is a MATH-Gorenstein family by REF as required. Suppose given MATH, an extension MATH in MATH, a finite MATH-module MATH, and MATH. REF states that the natural map MATH is an isomorphism. By REF , each side is naturally identified with MATH, so the map is an isomorphism as required. Finally, the finiteness REF for MATH is obvious from the construction of the module MATH using the properness of MATH. |
math/0104193 | Write MATH. An element of MATH is a sequence of compatible families MATH. This defines a pair of formal schemes MATH, where MATH denotes the formal spectrum. We have a line bundle MATH on MATH such that the restriction to the special fibre is ample, for example, MATH where MATH. By NAME 's Existence Theorem (CITE, III. REFEF), MATH is the completion of a proper pair MATH along the fibre MATH. We claim that MATH. First observe that MATH and MATH commute with base change for all MATH by REF . Then, by REF , the set of points MATH is open. But it contains the closed point by assumption, hence it is the whole of MATH. Thus MATH as claimed, so the map MATH is surjective. Now suppose given MATH and MATH which give the same element of MATH, that is, we have compatible isomorphisms MATH for each MATH. Equivalently, using REF , we have compatible maps MATH and thus a map MATH so there is an isomorphism MATH over MATH extending the MATH. Thus the map MATH is injective. |
math/0104193 | CASE: Let MATH in MATH be etale (in fact we only require MATH flat). Given a flat family of schemes MATH and a coherent sheaf MATH on MATH we have natural isomorphisms MATH by CITE, p. REFEF, using MATH flat. Given a MATH-Gorenstein family of slc surfaces MATH and a coherent sheaf MATH on MATH, let MATH be a local index one cover (a MATH quotient, say), then MATH by REF . Thus we obtain MATH by applying MATH to the natural isomorphism above with MATH. By CITE, p. REF we have natural isomorphisms MATH for any quasi-coherent sheaf MATH since MATH is flat. Thus MATH for each MATH, using the spectral sequence of REF and flatness of MATH. Also, given coherent sheaves MATH, MATH on MATH, we have MATH since MATH is flat (CITE, p. REF ). Now suppose given MATH an extension in MATH, MATH, MATH and MATH a finite MATH-module. Since MATH we see that MATH commutes with etale localisation by REF above. Using the exact sequence of REF together with REF above we find that MATH commutes with etale localisation. CASE: We need to show that, given MATH, MATH a finite MATH-module, and MATH a maximal ideal, we have MATH . By REF , we may assume that MATH, then we need to show MATH that is, MATH . This is a result in the style of REF , and is proved in the same way. CASE: We may assume that MATH is an integral domain, since we are given that MATH is reduced, and MATH and MATH commute with etale localisation. We first show that, for MATH integral, and MATH a family of schemes over MATH, there is an open affine subset MATH such that if MATH is a morphism in MATH which factors through MATH, then the natural maps MATH are isomorphisms. Recall the construction of the MATH: MATH where MATH is a cotangent complex for MATH, and MATH is obtained via the spectral sequence MATH . If MATH is a cotangent complex for MATH, then given MATH, MATH is a cotangent complex for MATH (CITE, p. REFEF ). Recall that, for a coherent sheaf MATH on MATH (of finite type), where MATH integral, there is a non-empty open affine subset MATH of MATH such that MATH is flat over MATH (CITE, NAME REF, p. REF). It follows that, given coherent sheaves MATH on MATH, there exists an open affine subset MATH such that the natural map MATH is an isomorphism whenever MATH factors through MATH (compare the proof of CITE, p. REF ). Thus, in our case, there is an open affine subset MATH such that the natural maps MATH are isomorphisms for all MATH and all MATH. NAME, shrinking MATH if necessary, we may assume the natural maps MATH are isomorphisms for all MATH and all MATH. So combining, we have isomorphisms MATH for all MATH and all MATH. Next, we may assume that the natural maps MATH are isomorphisms for all MATH, MATH and MATH, by the theory of cohomology and base change for projective morphisms (see for example, CITE, III. REF, in particular p. REF ), thus MATH . Finally, we may assume that we obtain MATH using the local-to-global spectral sequence, by imposing certain flatness requirements. The same argument shows that the MATH commute with base changes MATH for some MATH, where MATH is a family of slc surfaces. We use the local identification MATH where MATH is a local index one cover which is a MATH quotient. We use the exact sequence of REF to show that MATH commutes with base changes MATH for some MATH. We may assume that the modules MATH, MATH and MATH commute with base changes MATH factoring through some MATH by the above, and if we assume some flatness conditions we obtain that MATH commutes with base changes MATH using REF-lemma. Thus, given MATH integral, MATH a finite MATH-module, and MATH an extension, MATH and MATH commute with base changes MATH for some open affine subset MATH. This completes the proof. |
math/0104193 | First, it is clear that MATH defines a stack. The only non-trivial point is that etale descent data for MATH are effective - this follows since for each MATH and MATH we have a canonical polarisation MATH for some MATH. By REF , we can construct a scheme MATH locally of finite type over MATH and MATH such that the morphism of stacks MATH is smooth and surjective. The construction is as follows: For every MATH, let MATH be an algebraic ring with a closed point MATH and MATH, formally smooth over MATH, such that MATH. Let MATH be the disjoint union of the schemes MATH, and MATH the union of the families MATH. We have that the diagonal MATH is representable, quasicompact and separated, that is, for every MATH and every pair of elements MATH, MATH, the functor MATH is represented by a scheme which is quasicompact and separated over MATH. For MATH is represented by a scheme MATH, quasiprojective over MATH, by REF . Thus MATH is an algebraic stack, locally of finite type over MATH. Finally, we claim that MATH is a NAME algebraic stack, that is, there exists a scheme MATH locally of finite type over MATH and MATH such that the morphism of stacks MATH is MATH and surjective. By CITE, p. REF , it is enough to show that the diagonal map MATH is unramified, that is, for every MATH and MATH, MATH, the scheme MATH is unramified over MATH. We may assume that MATH, MATH algebraically closed. We know that MATH is finite for MATH by CITE, since MATH is slc and ample. It follows that MATH has no infinitesimal automorphisms (since MATH) and hence MATH is unramified as required. |
math/0104193 | It is immediate that MATH is a NAME algebraic stack, locally of finite type over MATH by REF . It remains to show that it is of finite type. By REF there exists MATH such that MATH is NAME for every stable pair MATH of degree MATH. By smoothability, the NAME polynomials of MATH and MATH with respect to the polarisation MATH are the same as the NAME polynomials of MATH and MATH a curve of degree MATH with respect to the polarisation MATH. Also, by REF , there exists MATH such that MATH is very ample and has no higher cohomology for all MATH. Thus there is a NAME scheme MATH of pairs MATH such that every stable pair of degree MATH occurs in the universal family. Now, let MATH be a local affine patch of MATH with universal family MATH. Pick an projective embedding MATH defined by MATH. We obtain a map MATH, and moreover taking all possible such embeddings we obtain a map MATH . We perform this construction everywhere locally on MATH. The (set-theoretic) image of the maps MATH is the set MATH of points MATH of MATH such that MATH is a stable pair of degree MATH embedded by MATH. We claim that MATH is a locally closed subset of MATH and moreover that the images of the maps MATH give an open cover of MATH. Assuming this for the moment, we deduce that there exists a finite open subcover of MATH, and hence a finite affine open cover of MATH. Thus MATH is of finite type as required. We now prove our claim above. We first apply REF to deduce that there exists an open locus MATH containing MATH such that for all MATH, MATH is CM, reduced and NAME in codimension MATH, and MATH is NAME in codimension MATH. Moreover, writing MATH for the universal family, MATH and MATH are relative NAME divisors on MATH. Next, we apply REF , to deduce that there exists a locally closed subset MATH containing MATH such that, writing MATH for the universal family, MATH and MATH commute with base changes MATH with MATH reduced, for all MATH. There now exists an open subset MATH containing MATH such that MATH is quasistable for all MATH by REF . Let MATH be the closure in MATH of the locus where MATH, then MATH contains MATH and MATH is a stable pair of degree MATH for all MATH. Finally, MATH is the locus where MATH is defined by MATH - this is an locally closed condition. Thus MATH is a locally closed subset of MATH. Now let MATH be a local patch of MATH with universal family MATH and MATH a map as constructed above, we claim that the image of MATH is an open subset of MATH. Let MATH be the spectrum of a DVR with closed point MATH and generic point MATH, and MATH a morphism (where MATH is given its reduced structure) such that MATH lies in the image of MATH. It is enough to show that MATH also lies in the image of MATH. Writing MATH for the pullback of the universal family of MATH to MATH, we have MATH (note that the base change conditions are satisfied by construction). We deduce our result by the versality of MATH. |
math/0104193 | All this follows from REF apart from the exact sequence of REF - see CITE, p. REFEF. |
math/0104193 | Let MATH, we first work locally analytically at MATH. Suppose that MATH is normal crossing at MATH, that is, MATH . Then, using the exact sequence MATH, we deduce that MATH is a line bundle on MATH. Moreover we obtain a natural isomorphism MATH - for we have MATH . Suppose now that MATH has a singularity MATH at MATH. Let MATH be the canonical cover, then MATH is normal crossing. Write MATH, MATH for the inverse images of MATH, MATH, and MATH, MATH, MATH for the inverse images of MATH, MATH, MATH on MATH, MATH, MATH. We have a natural isomorphism MATH, and MATH. Patching together this local information, we obtain that globally MATH, where MATH is defined as detailed above. |
math/0104193 | Let MATH for MATH and MATH be the minimal resolutions of the components of MATH. Then MATH for MATH and MATH by NAME 's formula, and MATH where the MATH are the indices of the non-Gorenstein singularities of MATH at MATH (see the proof of REF ). Thus MATH . We also have MATH and MATH for MATH and MATH. Solving for MATH we obtain MATH. |
math/0104193 | We have an exact sequence MATH . Apply the functor MATH - we obtain an exact sequence MATH . Thus we have an inclusion MATH with cokernel supported on MATH. It follows that MATH is surjective. So it is enough to show that MATH for MATH. Write MATH for a component MATH. By NAME duality, MATH . We claim that MATH has a nonzero global section. Assuming this, we have MATH . Now, letting MATH be the minimal resolution, we have MATH since MATH has only quotient singularities REF . Thus MATH, since MATH is rational by REF . So MATH as required. It remains to show that MATH has a nonzero global section. We have an exact sequence MATH . Now MATH by NAME duality and NAME vanishing (recall that MATH is log terminal and MATH is ample). So we are done if MATH has a nonzero global section. A local calculation shows that MATH, where MATH is the sum of the singular points of MATH lying on MATH. Now MATH is isomorphic to MATH, and there are at most MATH singular points of MATH on MATH by assumption, thus MATH and MATH has a nonzero global section as required. |
math/0104193 | For a smoothable surface MATH of type B, the non-Gorenstein singularities at MATH are of type MATH. Moreover, if we have MATH such singularities with indices MATH then MATH, in particular MATH. We show that if MATH smoothes to MATH, then MATH. We assume MATH and obtain a contradiction. Let MATH be a MATH-Gorenstein smoothing of MATH over the spectrum MATH of a DVR with generic point MATH, such that MATH. Write MATH, then there are two cases : CASE: MATH and MATH, MATH, MATH is MATH-factorial. CASE: MATH, MATH, MATH is not MATH-factorial. Here we use REF , also MATH by REF and we can calculate MATH using REF . Suppose first MATH is of type REF . We claim that we can contract the divisor MATH to obtain a relative minimal model MATH. We have MATH canonical by REF , so we can use the relative MMP theory. It's enough to show that the double curve MATH of MATH generates an extremal ray in MATH (note that MATH is relatively ample so certainly MATH). Let MATH have singularity MATH generically along MATH. Then MATH . Now MATH since MATH, thus MATH and MATH generates an extremal ray on MATH. By the exact sequence MATH (compare REF ), we see that MATH generates an extremal ray on MATH. Hence we can contract MATH to obtain a MATH-factorial family of surfaces MATH with generic fibre MATH. The special fibre MATH is obtained from MATH by contracting the double curve MATH. Thus MATH has a log-terminal singularity whose minimal resolution has exceptional locus a tree of rational curves with one fork. But this singularity is not smoothable, a contradiction. Now suppose that MATH is of type REF . Then MATH is not MATH-factorial. Thus we have a point MATH such that, locally analytically at MATH, MATH where MATH, MATH. Working locally at MATH, we construct a small partial resolution MATH. Assume first that MATH. Let MATH be the blowup of MATH. Writing MATH and MATH, we have two affine pieces of MATH as follows: MATH . Thus MATH is normal crossing at the strict transform MATH of MATH and MATH has a MATH singularity along MATH. The only possible singularity of MATH away from MATH is a MATH singularity at MATH. Write MATH, MATH, and MATH, MATH for the strict transforms. Then MATH is an isomorphism, and MATH contracts a MATH to the smooth point MATH. In the case MATH we obtain MATH as the MATH quotient of the construction above. Now work globally on MATH. Note that MATH is slt, MATH is canonical, and MATH. We first claim that MATH is projective. For it is easy to see that the special fibre MATH is projective, and the restriction map MATH is an isomorphism (using MATH nef and big, compare proof of REF ). Moreover MATH is MATH factorial (compare REF ). We now obtain a contradiction as for REF above. |
math/0104193 | First suppose that MATH is a surface of type B that admits a MATH-Gorenstein smoothing to MATH. REF on the singularities of MATH is satisified by REF . Let MATH be a MATH-Gorenstein smoothing of MATH, where MATH is the spectrum of a DVR with generic point MATH, such that MATH. We have MATH since MATH is MATH-Cartier, so REF holds. Finally, REF is satisfied by REF . Thus the conditions are necessary. Now let MATH be a surface of type B which satisfies REF . We first construct a first order MATH-Gorenstein deformation MATH of MATH, where MATH. We then show that this extends to a MATH-Gorenstein smoothing of MATH over a DVR. Recall that there is a natural isomorphism MATH. Now, by the local-to-global spectral sequence, we have an exact sequence MATH . By REF we have MATH, so MATH is exact. We next specify an element of MATH, and describe the local first order deformations of MATH it determines. We can then lift this to an element of MATH defining a global MATH-Gorenstein first order deformation MATH of MATH. The sheaf MATH is supported on the singular locus of MATH. So, to define a global section of MATH, we have to define a section in a neighbourhood of each connected component of the singular locus of MATH. An isolated singularity MATH of MATH is of the form MATH by assumption. Write MATH for the local index one cover of MATH, then MATH . We define a local first order MATH-Gorenstein deformation of MATH as follows: MATH . This corresponds to a local section of MATH. Now consider MATH, the locus of non-isolated singularities of MATH. In a neighbourhood of MATH, MATH is a line bundle on MATH of degree MATH by REF . Moreover MATH and MATH by REF . So there exists a section MATH of MATH at MATH, with reduced divisor of zeroes missing the non-Gorenstein points of MATH, that is, MATH where MATH are distinct points of index one on MATH. Then MATH defines local first order deformations of MATH at the points MATH of MATH of the following forms. If MATH and MATH at MATH, some MATH, then we have MATH for some unit MATH. If MATH for some MATH then MATH at MATH and we have MATH for some unit MATH. Observe that MATH has unobstructed MATH-Gorenstein deformations. For the local index one covers of MATH are local complete intersections, so MATH (thus MATH has unobstructed MATH-Gorenstein deformations locally). Also, we have MATH, since MATH is an invertible sheaf on MATH of non-negative degree, and MATH by REF . Thus MATH by the local-to-global spectral sequence, so MATH has unobstructed MATH-Gorenstein deformations as claimed. Hence we can extend the deformation MATH to a sequence of compatible MATH-Gorenstein deformations MATH, where MATH. They determine a formal scheme MATH, where MATH, MATH is the completion of MATH, and MATH denotes the formal spectrum. By the NAME Existence Theorem, MATH is the completion of a projective MATH-Gorenstein family MATH (compare Proof of REF ). We claim that this is a smoothing of MATH. For, from the explicit descriptions of the local first order deformations above, we see that MATH has only singularities of types MATH and MATH, hence MATH is smooth over the generic point MATH of MATH as required. The geometric generic fibre MATH is a smooth NAME surface with MATH, hence MATH. Thus MATH is a MATH-Gorenstein smoothing of MATH to MATH as required. |
math/0104193 | MATH is either a NAME surface or of type B. We proved the result in the type B case in the proof of REF . If MATH is a NAME surface then the local index one covers of MATH are local complete intersections so MATH, MATH since MATH has MATH dimensional support, and MATH by CITE, p. REF, Proof of REF . Thus MATH, so MATH has unobstructed MATH-Gorenstein deformations as required. |
math/0104193 | Write MATH for a component of MATH, and let MATH denote MATH. We need to show that MATH is log terminal. First we show that there are no singularities of types MATH or MATH. For suppose MATH is a component of MATH which passes through such a point. Then there is at most one other singularity of MATH on MATH, of type MATH. We calculate MATH (we allow MATH - this is the case where there are no other singularities on MATH). Now MATH, so MATH. But MATH misses the strictly log canonical point of MATH, since MATH is slc, thus MATH is NAME near MATH. So MATH, a contradiction. It only remains to show that no normal strictly log canonical singularities can occur. Otherwise, MATH is a normal surface, so is an elliptic cone by REF . Let MATH be a ruling, MATH the minimal resolution of MATH and MATH the exceptional curve. Thus MATH is a ruled surface over an elliptic curve, MATH is the negative section, and the strict transform MATH of MATH is a ruling. We calculate MATH. Now MATH implies MATH. But MATH misses the singularity of MATH since MATH is log canonical, so MATH is NAME. Thus MATH, a contradiction. |
math/0104193 | First, by REF above, for MATH, MATH has unobstructed MATH-Gorenstein deformations. Second, by REF , the map of functors MATH is smooth. Note that MATH in this case, that is, every MATH-Gorenstein deformation of MATH is smoothable. Thus MATH is smooth as required. |
math/0104193 | We have MATH with MATH, and MATH by definition (compare the proof of REF ). We need to show MATH for all MATH. If MATH for some MATH, then MATH for all MATH in the same connected component of MATH using MATH-nef. It follows that this connected component contracts to a canonical singularity of MATH. This must be a NAME singularity of MATH with MATH locally. But then we have a normal log-terminal singularity on MATH which is not of type MATH, a contradiction. |
math/0104193 | See the Proof of REF - a finer analysis in the cases MATH and MATH gives our result. |
math/0104193 | X is slt for MATH by REF , thus MATH is log terminal. Recall that the log terminal singularities of MATH are of types MATH and MATH. REF states which values of the indices MATH and MATH are possible. It only remains to determine what combinations of singularities can lie on MATH for MATH. Suppose MATH, and let MATH have singularities of indices MATH at MATH. Then MATH and MATH (compare REF ). Moreover, MATH is REF-divisible by REF below. Using the restrictions on the indices in REF , we obtain the solutions MATH, MATH, or MATH, MATH, MATH for MATH, and MATH, MATH, or MATH, MATH for MATH. Combining our results we obtain the lists of possible connected components of MATH above. |
math/0104193 | We use the relation MATH. Thus MATH, so it's enough to show that the index of MATH is not REF-divisible. See the proof of REF . |
math/0104193 | We know MATH is not MATH-exceptional by REF , hence MATH. But MATH. Hence MATH and MATH as claimed. |
math/0104193 | This is immediate from the index calculations of REF together with REF . |
math/0104193 | Let MATH be a stable pair of degree MATH, where MATH or MATH. Let MATH be a component of MATH. Then MATH is rational since MATH. If MATH then MATH, since MATH is ample and REF-divisible (again using MATH). So assume MATH, and choose MATH as in Notation REF. We classify the pairs MATH, and hence the pairs MATH, and finally glue these together to form the surfaces MATH. Each degenerate fibre of the birational ruling MATH consists of some irreducible components of MATH and some MATH-curves. We have a list of possible connected components of MATH in REF , and REF describes how MATH-curves and irreducible components of MATH may intersect. Note also that two MATH-curves MATH in a degenerate fibre MATH do not intersect unless MATH. We classify the MATH by calculating all the ways we can piece together connected components of MATH and MATH-curves to form a net of curves which is a union of fibres and horizontal curves on some surface MATH with a birational ruling MATH. So, suppose given some MATH, and assume MATH (otherwise MATH is smooth, so MATH). Let MATH be a connected component of MATH. We consider the cases of REF . We first treat the case MATH. In REF MATH cannot intersect a MATH-curve. Thus if MATH intersects a degenerate fibre, it must contain the whole degenerate fibre. But by inspection MATH cannot contain a degenerate fibre, so MATH does not intersect any degenerate fibre. It follows that there are no degenerate fibres. But we know there are at least two curves of negative self-intersection, a contradiction. So REF does not occur. It now follows that REF cannot occur. For a MATH with singularities at MATH corresponding to REF must be glued to a MATH with singularities at MATH corresponding to REF in order to form a smoothable surface MATH. In REF , let MATH be the components of MATH, ordered as in REF . Only MATH can intersect a MATH-curve and MATH cannot contain a degenerate fibre. Suppose that a component of MATH is contained in a degenerate fibre MATH. Then MATH consists of MATH together with some MATH-curves meeting MATH, and possibly MATH or MATH. The only possibility is MATH where MATH are MATH-curves. Then MATH and MATH must be horizontal, but MATH cannot intersect MATH, a contradiction. So every component of MATH is horizontal. In particular there can be no degenerate fibres (because MATH cannot intersect a MATH-curve), a contradiction. Thus REF does not occur. In REF MATH cannot intersect a MATH-curve. So there are no degenerate fibres (compare REF above). Thus MATH, so MATH. In REF again MATH cannot intersect a MATH-curve. Thus MATH is ruled, MATH, and so MATH. Finally MATH, since MATH is ample and REF-divisible. This completes the classification of the MATH for MATH. Now let MATH. In REF MATH cannot intersect a MATH-curve. We obtain a contradiction as for MATH, REF . So REF does not occur. It follows that REF cannot occur, as for MATH, REF . For a MATH with singularities at MATH corresponding to REF must be glued to a MATH with singularities at MATH corresponding to REF to form a smoothable surface MATH. In REF , let MATH be the components of MATH, ordered as above. Only MATH can intersect a MATH-curve, and a MATH-curve intersecting MATH does not intersect any other irreducible components of MATH. We proceed as in MATH, REF to obtain a contradiction, thus REF does not occur. In REF cannot intersect a MATH-curve. Thus MATH is ruled, MATH and MATH. The curve MATH cannot be horizontal using MATH ample and REF-divisible, thus MATH is a fibre and MATH. In REF , write MATH for the components of MATH, ordered as above. Then only MATH and MATH can intersect a MATH-curve. Now MATH cannot contain a fibre, thus we see that MATH has components in at most one fibre, and then this fibre contains MATH or MATH. If every component of MATH were horizontal, then MATH would intersect a MATH-curve in a degenerate fibre, a contradiction. Thus MATH has components in a unique degenerate fibre MATH. We classify the possible fibres MATH above. We have MATH because MATH intersects MATH, it follows that MATH. If MATH, then MATH consists of MATH, some MATH-curves (meeting MATH) and possibly MATH - but such a configuration is never a fibre, a contradiction. Hence MATH. If there is a MATH-curve MATH meeting MATH, then MATH. Otherwise, we have some MATH-curves meeting MATH. If MATH, MATH are two such curves then MATH, a contradiction. Thus there is exactly one MATH-curve, MATH say, meeting MATH. Then MATH intersects another irreducible component MATH of MATH which has multiplicity MATH or MATH (by REF ) and hence self-intersection MATH or MATH by REF . We have MATH. If MATH we find MATH, a contradiction. Thus MATH, we deduce MATH. Note that MATH is a connected component of MATH of type REF . So, we have two possible types of degenerate fibre MATH - either MATH where MATH meets MATH, or MATH, a chain of curves, where MATH. In each case MATH is horizontal - it follows that there are no more degenerate fibres, since MATH cannot intersect a MATH-curve. We see that MATH is obtained from MATH by a sequence of blowups, and MATH. Also MATH so MATH. A graded ring calculation shows that MATH in the first case. In the second case we see MATH by toric methods. REF is very similiar: writing MATH for the components of MATH, we obtain the same possible degenerate fibres MATH as above, MATH horizontal, and MATH a fibre. MATH is obtained from MATH by a sequence of blowups, and MATH. A graded ring calculation shows that MATH in the first case, where MATH. In the second case we have MATH, where MATH. Finally, for REF , we may assume that MATH has index MATH, because I have already classified the MATH with an index MATH singularity above. Then the same calculation as for MATH gives MATH or MATH. This completes the classification of the MATH for MATH. We calculate that each surface MATH has MATH, and MATH is a log NAME surface. We now glue the MATH together to obtain the surfaces MATH. The normal surfaces MATH are the surfaces MATH with MATH. The non-normal surfaces MATH are obtained by glueing components MATH, MATH along MATH,MATH so that each MATH singularity on MATH is glued to a MATH singularity on MATH to give a singularity of type MATH on MATH. We also require that MATH, equivalently MATH (in fact this is automatic in our cases MATH and MATH). Finally, we conclude that each surface MATH constructed as above occurs in a stable pair MATH of degree MATH or MATH. We need to show that there exists a divisor MATH on MATH such that MATH, MATH is slc - this is easy to check. Moreover, we require that MATH admits a MATH-Gorenstein smoothing MATH such that MATH and MATH is MATH-Cartier - this follows from REF , using the existence of a MATH-Gorenstein smoothing MATH REF . |
math/0104193 | Let MATH be a component of MATH. Then there is a unique toric blowup MATH which blows up the node of MATH such that the strict transforms of the components of MATH give fibres of a fibration MATH. For, if MATH is given by the fan defined by the vectors MATH, where MATH and MATH correspond to the components of MATH, the map MATH corresponds to the subdivision of the fan obtained by adding the vector MATH. Write MATH for the strict transform of MATH. Glueing the MATH together so that the negative sections form a cycle, we obtain a partial resolution MATH. We claim that, for a suitable choice of MATH and MATH in Construction REF, we can recover MATH as the special fibre of MATH. Hence we obtain a smoothing of MATH as above. To prove the claim, note that MATH is uniquely determined by the following data - the self-intersections MATH of the negative sections of the components MATH, and the MATH singularities at the double curves. Let MATH be any smoothing of MATH. Possibly after base change, we can choose MATH locally at the nodes of MATH to obtain the required singularities (compare Construction REF). These local choices extend to a global sheaf MATH. We require that MATH in order to obtain the correct self-intersections of the negative sections - we can achieve this by twisting MATH by a suitable line bundle. |
math/0104198 | Given a non-empty subset MATH of MATH let MATH be a MATH-minimal element of MATH. Then MATH, that is, MATH is isolated in MATH. Thus MATH is scattered. Next we prove that every MATH is compact by well-founded induction on MATH. Assume that MATH is compact for each MATH. By NAME 's subbase lemma it is enough to prove that any cover of MATH with subbase elements contains a finite subcover. So let MATH be such that MATH . If MATH for some MATH then MATH and so MATH. Hence we can assume that MATH, that is, MATH for some MATH. If MATH then MATH, and we are clearly done. So we can assume that MATH, and consequently MATH. Then MATH. Since MATH the set MATH is compact by the induction hypothesis, hence MATH is covered by a finite subfamily MATH of MATH. Therefore MATH is a finite cover of MATH. Consequently MATH is compact. |
math/0104198 | We prove this by induction on MATH. Assume that MATH for all MATH. If MATH then MATH and so MATH is an isolated point of MATH, that is, MATH. Thus we have MATH. Now assume that MATH. Then by our above remark MATH, moreover there are MATH such that MATH . Let MATH. Then MATH, moreover we have MATH by REF , contradicting MATH. Note that this argument is valid for MATH as well. Indeed, in this case we have MATH, moreover MATH. Thus we have concluded that MATH. |
math/0104198 | Let MATH. Then it is clear from coherence that MATH hence MATH. Next let MATH. If MATH then MATH . Now assume that MATH. Then, again by coherence, MATH and we have MATH . In both cases MATH is MATH-open. Similarly we can see that MATH is MATH-open, hence the topologies MATH and MATH coincide. To show that MATH is well-founded, assume that MATH and MATH. If MATH then MATH because MATH. Thus there is MATH with MATH for each MATH because MATH is well-founded. Finally, that MATH is MATH-closed is an easy consequence of coherence and the MATH-closednessof the families MATH. |
math/0104198 | Let MATH and MATH, where MATH. Then MATH by the choice of MATH and so MATH because MATH is MATH-closed. Consequently we have MATH as required by the definition of coherence. |
math/0104198 | The system MATH is coherent by REF , thus REF holds by REF . Consequently we have MATH . Now if MATH and MATH then MATH by REF and MATH by REF , hence MATH . This, by MATH, proves REF . |
math/0104198 | It is obvious that MATH is tree-like. To show that MATH is chain-closed, let MATH be ordered by MATH . If MATH then MATH. If, however, MATH then put MATH. Since MATH is tree-like, MATH is also ordered by MATH and clearly MATH. So MATH and MATH, which was to be shown. To show that MATH is well-founded assume that MATH, where each MATH. If MATH for some MATH, then we are done. Otherwise for each MATH we have MATH, hence as MATH is MATH-closed we can assume that MATH. Since MATH is well-founded, there is MATH such that MATH, and so MATH as well, for each MATH. |
math/0104198 | Given MATH, the family MATH is ordered by MATH because MATH is tree-like. Thus either MATH and so MATH, or if MATH then MATH, for MATH is chain-closed. Assume now that MATH and let MATH. Clearly MATH. Since MATH is tree-like, MATH is ordered by MATH , so it has a MATH-least element, say MATH, because MATH is also well-founded. Pick MATH and let MATH. We claim that MATH. Clearly MATH because MATH and MATH. On the other hand, if MATH, MATH and MATH then either MATH or MATH because MATH is tree-like. But MATH implies that MATH, that is, MATH can not hold. Thus MATH and so MATH is proved. |
math/0104198 | Assume that MATH is a limit ordinal and MATH for all MATH. We want to show that MATH. Since MATH and MATH, moreover MATH, it is enough to show that MATH. So assume that MATH with MATH and verify that then MATH. Fix MATH. For each MATH with MATH let MATH be the MATH-minimal element of MATH with MATH. Then MATH is a chain because MATH for MATH by the minimality of MATH and because MATH is MATH-closed. Thus MATH and clearly MATH. |
math/0104198 | We shall write MATH for MATH, and MATH for MATH. First observe that because MATH is a closed subspace of MATH, hence REF holds. Now let MATH. Then there are MATH such that MATH . Since here MATH and MATH, we can fix MATH such that MATH for every MATH. Now consider the basic neighbourhood MATH of MATH in MATH. We claim that MATH. The inclusion MATH is clear from the choice of MATH. On the other hand, if MATH with MATH and MATH, then MATH hence MATH, so as MATH is MATH-closed we can assume that MATH. If we had MATH then MATH would imply MATH for some MATH, hence MATH and so MATH, a contradiction, thus we must have MATH. Moreover, since MATH, we must also have MATH. By REF we have MATH. Moreover, the topologies MATH and MATH coincide because the above argument also shows that for each MATH and MATH we have MATH . Hence MATH is a clopen subspace of MATH and so MATH, what proves REF . REF follows immediately from REF . Finally, MATH, as follows immediately from REF , proving REF . |
math/0104198 | For each MATH consider the well-founded, MATH-closed, MATH-good family MATH constructed in REF : MATH . Fix a bijection MATH, and let MATH, that is, MATH is simply an isomorphic copy of MATH on the underlying set MATH. As MATH is also chain-closed and tree-like, hence so is MATH. We shall now show that the *-modified families MATH satisfy REF . Since MATH it follows that MATH and so REF is true. For MATH, the height of MATH is MATH and the height of MATH is MATH, hence MATH and MATH are not isomorphic. Thus MATH and so MATH because MATH and MATH. Hence MATH is a successor ordinal by REF , that is, REF is satisfied. REF holds by REF. CASE: To show REF , let us fix MATH. Then MATH where MATH for all MATH by REF , consequently MATH. Thus we may apply REF to the family MATH and conclude that the space MATH is LCS, MATH, moreover since for every MATH the space MATH is an open subspace of MATH, we have MATH, consequently MATH. |
math/0104198 | We do induction on MATH. If MATH then we let MATH be REF-point compactification of the disjoint topological sum of countably many copies of MATH. If MATH is limit then we first fix an almost disjoint family MATH, for MATH. Applying the inductive hypothesis for each MATH we also fix a locally compact scattered space MATH of height MATH such that MATH and MATH for MATH. Now amalgamate the spaces MATH as follows: consider the topological space MATH where MATH is the topology generated by MATH. Since MATH is a finite and open subspace of both MATH and MATH it follows that each MATH is an open subspace of MATH. Consequently, MATH is LCS with countably many isolated points, and MATH. |
math/0104198 | If MATH, then REF gives such a space. If MATH then MATH and so according to REF for each MATH there is locally compact, scattered space of height MATH and countably many isolated points. |
math/0104199 | We will use the standard notation that MATH denotes the MATH . NAME space over MATH with MATH derivatives. We will let MATH denote the MATH . NAME space over MATH with MATH derivatives. We first recall the energy inequality obtained by pairing REF with MATH . From this, we obtain by integrating over time (and observing that the MATH norm is always positive), that if the solution MATH remains smooth up to time MATH, we have the estimate MATH . We now pair REF with MATH in order to estimate MATH. We obtain MATH . Clearly, we must estimate the nonlinear problem term MATH . Since MATH is divergence free, we can bound the absolute value of the above by MATH with MATH. Now since we are assuming MATH, we obtain by the NAME imbedding theorem, MATH since we must spend MATH derivatives to get all three terms in MATH and we have spent MATH. Applying NAME, we get immediately MATH . Combining this with REF, we get MATH . In turn, combining this with REF and with NAME 's inequality gives global solvability. |
math/0104199 | Since the NAME dimension of E is the infimum of the set of all MATH for which MATH, it is enough to prove that MATH for all MATH. Pick MATH such that MATH. We can cover E by the MATH. Now MATH which limits to zero as MATH goes to MATH whenever MATH. |
math/0104199 | Let MATH. Then (since we are working in MATH) we have MATH . Now MATH so by NAME 's inequality, we have MATH which is finite since MATH is a NAME function. But by the relation between MATH and MATH, we have MATH which proves the proposition. |
math/0104199 | Let us define MATH where MATH. We have MATH . By our estimates on the derivatives of MATH, we can get MATH while, because of the NAME transform supports, MATH . Thus the proposition is proved. |
math/0104199 | We neglect negligible terms. Then we have MATH, by REF . We estimate MATH by REF . |
math/0104199 | MATH is a composition of type MATH pseudodifferential operators whose symbols have disjoint support. Thus it is smoothing. |
math/0104199 | We simply observe that MATH . |
math/0104199 | Note that MATH . Note that by REF , we have MATH . To estimate MATH, we observe that MATH is of order MATH since MATH is of type MATH. Further by REF , we have MATH . NAME further, applying REF , we get MATH . Now applying the mapping properties of operators of order MATH and NAME, we get for any number MATH, MATH . Now applying REF , we get the desired result. |
math/0104199 | Simply sum REF over MATH. |
math/0104199 | By conservation of energy, for every MATH and every MATH we have, MATH . Since our constants can depend on MATH this means MATH . Now suppose the lemma is false. Applying the opposite of REF, for MATH we get MATH which implies that REF holds for MATH. |
math/0104199 | We consider for MATH, the relevant expression MATH . We divide this into two cases which are MATH and MATH. In the second case, we use the idea of REF (we just have to use a higher degree of smoothing than is used in that proposition) to observe MATH . (This is because MATH acts on MATH only where it has a negligible effect on the whole quantity by REF .) Now we simply observe by REF that MATH and by REF , the proof of REF (to control the action of MATH) and REF that MATH . On the other hand for the first case, there is no point in localizing to MATH because MATH is too small and so the error is too big. Thus in in this case, we simply estimate MATH . Summing these estimates gives the desired bound for MATH. The bound for MATH proceeds likewise (and gives a better estimate since the derivative falls on the level MATH term.) |
math/0104199 | From the definition of MATH we have: MATH . Since the above sum has only MATH many terms by applying REF , we observe that for some particular values of MATH . Now we apply REF together with NAME to obtain MATH . Using REF , we observe that MATH . Finally direct calculation shows MATH where the first inequality comes from the fact that MATH on the support of MATH and the second inequality comes again from REF . Combining all these estimates proves the lemma. |
math/0104199 | We begin similarly to before by estimating MATH . With the other terms one can proceed likewise. Now, similarly to what we have already done, we observe that we can write this as, MATH . Now we estimate MATH . Combining these estimates gives the desired inequality. |
math/0104199 | Let MATH be the collection of cubes MATH where MATH is a bad cube at level MATH. From REF we know that there are disjoint cubes MATH such that the collection consisting of MATH covers the set MATH. Any cube of sidelength MATH can be covered by REF cubes of sidelength MATH (and REF is a constant.) We will define MATH to be the covering formed by the union of the thousandths of the elements of MATH . In order to count cubes in MATH it is enough to count the disjoint cubes used in the construction of MATH. However we know since the MATH's are disjoint that MATH by conservation of energy, while we know MATH by the badness of the cubes. Combining REF implies the claim. |
math/0104199 | We proceed by contradiction. Suppose the theorem is false. We let MATH be the first time and MATH be the largest cube so that MATH . Now since our initial data is smooth, at the initial time, (since MATH is chosen sufficiently large), we have MATH . Therefore, it must be the case that MATH . By using REF , we can replace MATH by an extended nuclear family MATH, with MATH for which MATH . For the current theorem, this is all we need. Since MATH also begins MATH, we must have MATH where we define MATH . Since there are only MATH cubes in MATH, for this to be the case, there must be a MATH so that MATH . We contradict this by using REF , and REF to estimate MATH. By our definition of MATH, the cube MATH is contained in no larger bad squares. Each estimate contains a factor of MATH which we take out using the estimate MATH, which we get from the definition of MATH. First let MATH be a nuclear family member of MATH and MATH be a distant ancestor at level MATH. Suppose MATH, we see that MATH. Thus we need not worry about this term. Suppose MATH. We must estimate MATH . Note that both MATH and MATH are good squares. Thus we have the estimates MATH and MATH . Applying NAME, we get (using MATH) MATH . Summing over (there are only MATH terms) provides the desired estimate on MATH. We can estimate MATH in the same way (by allowing MATH as large as MATH.) We are left to estimate MATH. We fix a scale MATH and pick MATH within REF of MATH. Now we are left to estimate MATH . However since MATH is a good square, we have MATH . Thus MATH which is an estimate that decays geometrically in MATH when MATH. By using the similar estimate for MATH, applying NAME and summing over MATH, we get the desired result. |
math/0104199 | We refer to the elements of MATH as the bad cubes. We say that a cube of sidelength MATH is very bad if either it intersects a bad cube of the same length or it intersects more than MATH elements of MATH for some MATH with MATH a small constant to be specified later. Let MATH be the union of all very bad cubes of length MATH , then by the estimates on the cardinality of the MATH's and by the NAME lemma, we can see that that MATH can be covered by MATH cubes of length MATH. We refer to these cubes as MATH. Now we need only prove REF. Let MATH be a cube of length MATH which does not intersect MATH. Let MATH be the set of elements of MATH which intersect MATH. Then we have the estimate MATH . Let MATH be the function defined from MATH to MATH which counts how many elements of MATH intersect MATH. For each MATH define MATH to be that number so that the center of MATH lies on MATH. Then MATH . Thus MATH . Since MATH this estimate decays geometrically with MATH. By choosing MATH sufficiently small, we may arrange that MATH . Thus by NAME 's inequality, there must be a value of MATH between MATH and MATH so that MATH for all MATH. This is the value of MATH that we choose. |
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