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math/0103171 | Let MATH be the lift of MATH to MATH defined in REF . Then MATH . By REF , MATH depends only on MATH; thus, MATH is well defined. Homogeneity of MATH follows from the definition of MATH. To see that MATH is positive, write MATH in the form MATH where MATH are blowup coordinates as in REF. Because MATH is regular on the blowup, MATH is strictly positive. Consequently, MATH where MATH. Finally, observe that MATH is positive because MATH is transverse to MATH. By REF , to conclude the proof we need only show that MATH coincides with the NAME form of MATH. We prove equality via explicit formulas for both forms using blowup coordinates centered at an arbitrary point MATH. Because MATH is arbitrary, we need only verify equality on the fiber MATH. Choose holomorphic coordinates MATH, MATH, centered at MATH as in the proof of REF . In these coordinates, MATH assumes the form MATH and, by REF , the form MATH assumes the form MATH . We next focus on the computation of MATH and MATH. Let MATH denote a tangent vector to MATH at MATH. If the ray generated by MATH is in the coordinate patch of MATH, then MATH. Then by REF , MATH where MATH. The NAME form of MATH is therefore given by MATH . Comparing this formula with REF yields the equality MATH and concludes the proof of the proposition. |
math/0103171 | The proof is a corollary to REF . Because the forms MATH and MATH coincide, the NAME vector field of MATH coincides with the restriction to MATH of the vector field MATH defined by REF . But REF shows that the projection onto MATH of the integral curves of MATH are the intersections of leaves of the NAME foliation with MATH. |
math/0103171 | Choose a point MATH. Because MATH is real analytic, for sufficiently small MATH, it has a holomorphic extension MATH, where MATH. It is easy to check that MATH lifts to a real analytic map MATH which is an extension MATH. By analytic dependence of MATH on MATH and compactness of MATH, there exists a real number MATH such that MATH is defined on MATH for all MATH. We now have a real analytic map MATH which is holomorphic in the second factor. The one-parameter identity MATH then allows us to extend the map to all of MATH as the composition MATH where the integer MATH is chosen so that MATH. REF of MATH follows by construction. To prove REF , first observe that MATH is the identity map on MATH. We, therefore, need only show that the derivative of MATH is injective on all of MATH. It then follows (after shrinking MATH if necessary) that MATH is a diffeomorphism, as claimed. But because MATH is the identity on MATH, it follows that MATH is injective if and only if the vector field MATH is transverse to MATH. It suffices to show that the projection MATH is transverse to MATH. But this is clear, for by construction MATH . This completes the proof of the lemma. |
math/0103173 | Use induction on the length of MATH. If MATH, then MATH by the definition on MATH. Otherwise, MATH and by the induction, MATH. Since MATH is homogeneous, we have MATH, hence MATH. |
math/0103173 | By REF it is enough to prove the lemma for the case when MATH, MATH and MATH are vacuum elements in a lattice vertex algebra MATH for some vectors MATH. Let MATH for some MATH. Using REF together with the formulas MATH for MATH, MATH and MATH, we obtain MATH . Let MATH be the minimal value of MATH such that MATH. We have MATH . It follows that if MATH or if MATH, then MATH; if MATH, then MATH; finally, if MATH, then MATH and the statement follows. |
math/0103173 | CASE: Consider the derivation MATH, MATH, of the associative algebra MATH which acts on a generator MATH by MATH, see REF for the notations. It is easy to see that MATH preserves the ideal MATH generated by the locality relations REF, hence MATH is a derivation of the algebra MATH. Also, we have MATH where MATH is the derivation of MATH defined by MATH. Since MATH can only increase the indices, it preserves the left ideal MATH, therefore MATH acts on the free vertex algebra MATH. CASE: Both MATH and MATH are conformal derivations, therefore they are uniquely determined by their action on the generators MATH. But for every MATH we have MATH and MATH, and the claim follows. |
math/0103173 | Using the dimension argument it is enough to prove only that MATH, though in fact the proof of the other inclusion is no more difficult. Since MATH, it is enough to show that MATH. The NAME algebra MATH is spanned by MATH for MATH. From REF it follows that MATH. Since MATH is a generator of MATH, it follows that each MATH preserves MATH. |
math/0103176 | By construction, MATH is an oriented smooth surface bundle over a surface MATH. The claim about the NAME characteristic of the base is obvious. The claim about the signature is an instance of NAME additivity. |
math/0103176 | The assumption shows that there is a diffeomorphism MATH mapping the sections into each other. Now the statement is obvious - notice that the self-intersection is the difference of the two self-intersections since in the subtracting operation we change the orientation of MATH. |
math/0103176 | The signature is additive when summing along embedded surfaces of self-intersection zero. |
math/0103176 | By the classification of surfaces, there exists a diffeomorphism MATH of MATH such that MATH and MATH. Then MATH . |
math/0103176 | Suppose that the surface of genus MATH with two holes in REF is embedded in MATH. Consider the curves on MATH given in the figure. The sphere MATH with four holes of the lantern relation, see REF , can be embedded in MATH so that the curves MATH, MATH, MATH, MATH, MATH, MATH, MATH become respectively MATH, MATH, MATH, MATH, MATH, MATH, MATH. This gives us the relation MATH . Similarly, two other embeddings of MATH give the relations MATH and MATH . If we multiply both sides of REF by MATH, use REF and cancel MATH, we obtain MATH or, equivalently, MATH . Similarly, the equalities REF yield the equality MATH . Applying REF to REF proves that MATH and MATH are products of two commutators. Any nonseparating simple closed curve is topologically equivalent to MATH. If MATH is a separating simple closed curve on MATH, then the surface of genus MATH on the right hand side of MATH can be embedded in MATH so that MATH is topologically equivalent to MATH. Now, the proof of REF follows from the fact that a conjugate of a commutator is again a commutator. Similarly, two more embeddings of the lantern give the relations MATH . Multiplying REF by MATH from the left and using REF gives MATH . By combining REF, we get MATH . Cancelling MATH and MATH yields MATH . Any simple closed curve on the left hand side is disjoint from each closed curve on the right. Notice also that the complements of MATH, MATH, MATH, MATH, MATH and of MATH are all connected. REF now implies that MATH is a product of three commutators, implying REF . |
math/0103176 | Suppose that the two-holed torus of REF is embedded in MATH in such a way that MATH and MATH are nonseparating on MATH. The curve MATH intersects MATH transversely at one point. Since MATH intersects MATH transversely at one point also and since any two such pairs are topologically equivalent, we can assume that MATH and MATH. By the two-holed torus relation, we have MATH. Let us denote MATH by MATH. Then, we obtain MATH . If MATH and MATH, we have the equality MATH . Now, MATH is a commutator and MATH is a product of two commutators. This observation completes the proof of REF . |
math/0103176 | It is well-known that a NAME twist can be written as a product of two commutators, see CITE. We need to make explicit choices for these commutators. To this end we consider curves MATH, MATH, MATH, MATH, MATH, MATH and MATH on a genus MATH subsurface of MATH as in REF . Further, we add curves according to REF . If we choose the genus MATH subsurface suitably, the vanishing cycle MATH is topologically equivalent to the curve MATH. Define diffeomorphisms MATH and MATH of MATH as follows. If MATH is nonseparating, set MATH and MATH . If MATH is separating, set MATH and MATH . One can check that MATH, MATH and MATH. The lantern relation as in REF implies that MATH . The monodromy representation of the complement of the singular fiber is given by mapping the standard generators of MATH to MATH, MATH, MATH, MATH and MATH respectively, as MATH . Evaluating the signature cocycle, REF shows that the complement of the singular fiber has signature MATH if MATH is nonseparating, and has signature MATH if MATH is separating. Now REF and NAME additivity complete the proof. |
math/0103176 | In all these proofs the signature of the NAME fibration is the same as that of the complement of the singular fibers, because all the vanishing cycles are nonseparating, compare REF . We take curves MATH, MATH, MATH, MATH on a genus MATH subsurface of MATH as in REF . We also add curves MATH, MATH, MATH as in REF , and MATH and MATH as in REF . For each of the Propositions, the vanishing cycles MATH and MATH are topologically equivalent to certain curves MATH and MATH. We fix the latter explicitly, and construct some diffeomorphisms as required by the proofs in REF, so that we can write the monodromy representation of the complement of the singular fibers as a relator in the mapping class group. Then the calculation is done by implementing the formula in REF with the following data. For REF the base is MATH and MATH is taken to be MATH. The relator giving the monodromy representation is MATH with MATH and MATH . For REF the base is MATH and MATH is taken to be MATH again. The relator giving the monodromy representation is MATH with MATH and MATH . For REF consider the curves MATH on a genus MATH subsurface of the fiber as in REF . The base surface is MATH and the curve MATH is taken to be MATH and MATH is taken to be MATH. We first compute the signature corresponding to the relator MATH with MATH which is the monodromy of a fibration MATH with MATH singular fibers. The signature of MATH is equal to MATH. Subtracting off the fibration of REF from MATH gives the claim. |
math/0103176 | We apply the subtraction operation to the NAME fibrations MATH and MATH as in REF , respectively. In MATH we group the singular fibers into two groups each containing four singular fibers with coinciding vanishing cycles; in MATH the singular fibers form one group. Now subtracting two copies of MATH according to the above pattern we get a surface bundle MATH of fiber genus MATH with MATH (compare REF ). Thus MATH, and the claim now follows by pulling MATH back to unramified coverings of MATH of degree MATH. |
math/0103176 | Notice that the relators defining the fibrations we used in the proof of REF represent REF in the mapping class group MATH of a surface with one boundary component. According to REF , this fact shows that the fibrations given by the relators MATH and MATH admit sections with vanishing self-intersection. Since the lifts of the various NAME twists are chosen to be NAME twists in MATH, REF implies that MATH also admits a section with zero self-intersection for all MATH. Now write MATH as MATH where MATH, and apply REF to MATH copies of MATH together with the product MATH. The resulting surface bundle MATH of fiber genus MATH has MATH. |
math/0103176 | First notice that the proof of REF immediately yields MATH for all MATH. (As before, MATH is the mod REF residue of MATH.) Now every surface of odd genus is a covering of a genus MATH surface. It was shown in REF, that after replacing a given surface bundle by a pullback to some covering of the base, the resulting surface bundle admits fiberwise coverings of any given degree. From the multiplicativity of the signature in coverings and the multiplicativity of the NAME characteristic of the fiber in fiberwise coverings, for odd MATH we obtain MATH . For even MATH consider the fibration MATH of fiber genus MATH with signature MATH we got by taking fiberwise coverings. It is easy to see that since MATH admits a section of zero self-intersection, so does MATH. Summing MATH and the product fibration MATH along their sections (as in REF ), we get a fibration over MATH with fiber genus MATH and signature MATH. These examples yield the bound MATH once MATH is even. Consequently the proof of REF is complete. |
math/0103179 | Since MATH by construction we have that MATH is the largest abelian sub-variety of MATH. Moreover, note that we also have induced extensions MATH and MATH yielding, together with REF , the following commutative diagram with exact rows MATH . Since MATH everything follows from a diagram chase. |
math/0103179 | In fact, there is an extension in MATH where MATH has the quotient MATH-mixed structure; as usual, we then get another extension MATH and so on. We therefore get a flasque resolution MATH in MATH: the canonical MATH-mixed flasque resolution. If MATH is flasque then MATH has a canonical MATH-mixed NAME structure as claimed above; note that the filtrations would be given by flasque sub-sheaves. In general, by construction, the cohomology is the homology of the complex of sections in MATH. Thus MATH has a canonical MATH-mixed NAME structure. The same argument applies to the total complex of the double complex of flasque MATH-mixed sheaves given by the canonical resolutions on each component of a simplicial sheaf. Refer to REF and CITE for a construction of canonical NAME resolutions available on any site and compare CITE for the existence of bifiltered resolutions. |
math/0103179 | Follows from construction as sketched above. In fact, all axioms stated in CITE are verified in MATH and the results in CITE can be obtained in MATH. |
math/0103181 | We only give an outline of the proof, and restrict ourselves to the case when each column of MATH has the same number of cells as the corresponding column of MATH. In that case, MATH has the same total MATH-degree as MATH, so that MATH is a polynomial in the MATH variables only. Since the terms of MATH are MATH in view of REF , the terms of MATH are forced to be of the form MATH for tableaux MATH that have the same column set has MATH. On the other hand, since MATH alternates in sign under the action of column fixing permutations, it has to be a multiple of MATH. Hence, both polynomials having the same degree, we must have the equality stated. |
math/0103181 | We first show recursively that MATH is independent, assuming that the statement holds for partitions with at most MATH cells. As before, for MATH, let MATH be the corner MATH, with MATH, and define MATH . In view of REF, for MATH, the dominant monomial of MATH (in reverse lexicographic order) is of the form MATH with MATH. For MATH fixed with MATH, our induction hypothesis gives that the set MATH is independent, since (in reverse lexicographic order) we have the following expansion of MATH where MATH is the restriction of MATH to MATH. Clearly, the sets MATH are mutually independent, so MATH is independent. We will now show that the number of elements of MATH is MATH using a recursive argument, assuming that the statement holds for MATH. By induction, it is clear that MATH so that (in view of REF) MATH . The result follows from the easy observation that MATH since MATH is the length of the row of MATH in which sits the corner MATH. A "geometric" argument, that can be found in CITE, shows that the dimension of MATH is at most MATH. Thus MATH is a basis. |
math/0103181 | In the remainder of this section, we will prove that REF is an independent set, using a downward recursive argument, and that the number of elements of MATH is MATH . To complete the proof of the theorem, we will show in REF that the dimension of MATH is at most equal to MATH, so that MATH has to coincide with the span of MATH. |
math/0103181 | In CITE, the bigraded MATH-modules MATH and MATH are shown to be equivalent, hence their MATH-free components MATH are equivalent. For any injective tableau MATH of shape MATH, with MATH, we have MATH and MATH . If MATH then MATH, so that MATH is equal to MATH . For MATH and MATH two families of pairwise distinct scalars, we construct a set of points MATH in MATH as follow. For every injective tableau MATH of shape MATH, we define the point MATH in MATH, and set MATH . That is MATH when MATH lies in the shadow of MATH in MATH. Note that MATH contain MATH points in correspondence with every injective tableau of shape MATH. We denote by MATH, where MATH is the projection on MATH that keeps only the first MATH entries. We see that the set of tableaux with MATH entries strictly increasing in rows and where MATH lies in a row MATH such that MATH and MATH give all the points of MATH exactly once. One then easily verifies that the cardinality of MATH is precisely MATH . Following CITE we associate to this set MATH, its annulator ideal and define MATH. The dimension of MATH is then MATH as well. Given a polynomial MATH, let MATH denotes its homogeneous component of highest degree. For any polynomial MATH in MATH, let MATH . For any MATH, the two products in the definition of MATH vanish at MATH unless MATH lies in the shadow of MATH in MATH. But if this is the case then MATH vanishes at MATH. This shows that MATH is in MATH, the annulator ideal of MATH. Hence MATH is in MATH, its graded version and MATH. For any injective tableau MATH of shape MATH such that MATH and MATH we have MATH and MATH . Thus MATH is in MATH. We obtain this way that MATH is a subset of MATH for any MATH with the prescribed conditions. The space in REF is thus contained in MATH, which proves the theorem. |
math/0103181 | Each of these assertions can be shown using the basis we have constructed. The third one is just a direct observation. The first one corresponds to a case for which there is just one corner MATH in the shadow of MATH, with MATH, and then REF can be written as MATH . Since, as long as MATH is in MATH, MATH is an isomorphism of representations between the homogeneous MATH-modules MATH and MATH that lowers the degree by REF, we must have MATH where MATH stands for the graded NAME characteristic. We deduce that, in the first case, MATH with MATH. This is clearly equivalent to the statement of the first case. For the second case there are a few subcases, all similarly dealt with, the most interesting one being when MATH and MATH for which the basis can clearly be broken down as MATH and we only need to show that the graded NAME characteristic of the linear span of MATH is given by MATH . Now we clearly have MATH, with MATH being the complement of MATH in MATH. Under the hypothesis of this subcase, the graded NAME characteristic of the span of MATH is thus given by REF. All other subcases are simple to show. |
math/0103183 | Given two trivialisations, expressing one in terms of the other gives a map from MATH to MATH. The homotopy classes of trivialisations are the homotopy classes of such maps. But as MATH is simply connected such maps lift to maps MATH, as do homotopies between them. The homotopy class of a map from MATH to itself is determined by its degree (as orientations have been fixed), making the homotopy classes of trivialisations a torseur of MATH. |
math/0103183 | Suppose MATH is an isomorphism giving a second NAME group structure to MATH. Then we have an induced orientation-preserving diffeomorphism MATH, and we need to show that the pullback of the canonical trivialisation under MATH is homotopic to the canonical trivialisation. But it is well known that any orientation preserving homeomorphism from MATH to itself is isotopic to the identity. Thus, the pullback fixes the homotopy class of any trivialisation, in particular the identity trivialisation. |
math/0103183 | As with MATH, the difference between trivialisations is determined by a map MATH. When MATH, this lifts to a map MATH. It is well known that the homotopy class of such a map is determined by its degree. |
math/0103183 | As above, when MATH, we have a map MATH representing the difference between the MATH. It is easy to see that the map MATH represents the difference between the pullbacks. As MATH has degree MATH and the degree is multiplicative, it follows that MATH is divisible by MATH. In the case when MATH (as also when MATH), there is a map MATH representing the difference between the trivialisations. On composing with the covering map, this gives the a map representing MATH which lifts to MATH. Thus, if MATH is the covering map, we get a commutative diagram MATH . As the degree of maps is multiplicative, and MATH and MATH, we get MATH, or MATH. The result follows . |
math/0103183 | It will be useful to regard MATH as the join MATH, which is embedded in the natural way in the quaternions. Then the action corresponding to the lens space MATH is the join of actions on the circle MATH generated respectively by MATH and MATH. We first find an equivariant trivialisation along the circles MATH and MATH and then extend these to MATH. Henceforth the equivariant trivialisation is given in terms of the map MATH representing the difference with the left-invariant trivialisation. Along the circle MATH, we can take the first vector to point along the circle. An equivariant trivialisation is obtained by choosing the other two vectors so that they rotate MATH times, for some choice in the mod MATH class of MATH. This follows as the arc joining MATH to MATH is a fundamental domain, with the only identification due to the group action being that of the endpoints induced by the element MATH, and this element rotates the normal plane to the circle by MATH. This trivialisation differs from the left-invariant trivialisation by MATH rotations, and if MATH is chosen odd, this gives a comparison map which lifts to MATH. The resulting restriction of MATH is the map MATH of degree MATH from the circle to itself. Likewise, we can trivialise the tangent bundle along the circle MATH. It is convenient to choose the trivialisation so that it differs from the left invariant one at MATH by MATH. Then the map MATH restricted to this circle is the map MATH of degree MATH from the circle to itself as the identification on the boundary of the fundamental domain is induced in this case by MATH. We now extend this map to a map from the disc MATH of the form MATH. Since the only points in the disc which are identified by the group action are on the boundary, and the trivialisation on the boundary is equivariant, we get an equivariant trivialisation of the disc corresponding to this map. We extend this by requiring equivariance to the disc MATH. The map is previously defined on the boundary of this disc, where it is equivariant. Thus, we have a uniquely equivariant extension. The midpoint of MATH is part of the circle MATH, as is an arc MATH joining the midpoints of MATH and MATH. The map MATH as previously defined on MATH, must agree with the definition on MATH at the midpoint as both these have been defined so that the trivialisation is equivariant. The discs MATH and MATH bound a lens MATH, which is a fundamental domain for the group action. The map MATH has already been defined on the boundary discs as well as the arc MATH. This extends to a map on MATH, and hence a trivialisation of the tangent bundle. Up to homotopy, any other choice of MATH can be obtained by replacing MATH in a small neighbourhood of an interior point MATH of the fundamental domain by a degree MATH map from MATH to itself for some MATH. More precisely, MATH is homotopic to a map that is constant on the neighbourhood of the point MATH. Replace this map in this neighbourhood by a map to MATH that maps the boundary of the neighbourhood to a single point and so that the inverse image of a generic point has algebraic multiplicity MATH. We shall call such a transformation `blowing a degree - MATH bubble'. We now extend the trivialisation equivariantly from the fundamental domain to all of the manifold, and define MATH accordingly. This is an extension of the map on MATH and MATH that was previously defined. The resulting map MATH is homotopic to a map obtained from the join of maps MATH defined by taking powers on the unit circle in MATH by blowing a degree - MATH bubble in each of the MATH images of the fundamental domain. Thus, it has degree MATH. This proves our claim. |
math/0103183 | It is well known that MATH as oriented manifolds a homeomorphism is induced by MATH. Conversely, for a fixed MATH, the condition MATH is a quadratic equation in MATH over the field MATH, with roots MATH and MATH. If MATH, then these are two distinct root of the quadratic equation, and hence the only solutions for MATH. If MATH, and MATH is another root, then we also have MATH, otherwise we would have three distinct roots. But this means that MATH and MATH, and we have already seen that MATH. Alternatively, using MATH, the above reduces to a linear equation for MATH that is satisfied by MATH, and hence MATH. |
math/0103183 | As MATH is a homology sphere, the contact-structure is co-orientable. Choose and fix a co-orientation. This induces an orientation on the plane-bundle given by the contact structure, which we identify with MATH. As MATH, the NAME class of MATH is trivial. Hence there is a trivialisation of MATH. Further, two trivialisations differ by a map onto MATH. As MATH, any such map is homotopic to a constant map. Thus, there is a trivialisation MATH of MATH, canonical up to homotopy. This, together with a vector MATH normal to MATH, that is consistent with the co-orientation, gives a framing MATH. The homotopy class of this trivialisation does not depend on the choice of co-orientation since MATH gives a trivialisation corresponding to the opposite co-orientation, and this is clearly homotopic to the trivialisation MATH. As the trivialisation of MATH pulls back to give a trivialisation of the pullback to any cover of MATH, the second claim follows. |
math/0103183 | As MATH is the quotient of MATH by a group acting by right multiplication, it follows that any framing on MATH pulls back to one homotopic to a framing invariant under right NAME multiplication, or one differing from this by MATH units (as MATH). However, if MATH had a universally tight positive contact structure, then the associated framing must pulls back to give the framing associated to left NAME multiplication. This gives the required contradiction. |
math/0103183 | Let MATH be the MATH-cobordism between MATH and MATH, so that MATH. If MATH is a MATH-manifold with boundary MATH, then MATH is a MATH-manifold with boundary MATH with the same signature as MATH. By considering these manifolds, it is immediate that that if MATH is the canonical MATH-framing for MATH, then the canonical framing MATH of MATH is characterised by MATH . The MATH-cobordism MATH lifts to a MATH-cobordism MATH between MATH and MATH, and the NAME class relative to the framings pulled back is the pullback of the NAME class and hence is zero. Let MATH be a MATH-manifold with boundary MATH, and let MATH. Applying REF to MATH and to MATH, the result follows. |
math/0103183 | We use the additivity of the signature and relative NAME classes. The above proof generalises immediately. |
math/0103183 | We proceed as in REF . Given an orientable tangent plane fields MATH on MATH with MATH, fix an orientation on MATH. There is a trivialisation MATH of MATH respecting the orientation. Further, the homotopy classes of such trivialisations are classified by MATH. Hence, as MATH is a rational homology sphere, the trivialisation is canonical up to homotopy. Find a vector field MATH normal to MATH such that MATH respects the orientation on MATH. This is possible as MATH and MATH are orientable. Then MATH is a framing of MATH. As before this does not depend on the choice of orientation. Conversely, given a framing MATH let MATH be the span of MATH and MATH. Evidently these constructions are inverses of each other. |
math/0103183 | If MATH for a properly embedded surface MATH with boundary, then it is easy to see that MATH as MATH splits as the sum of a trivial complex line bundle and MATH. It follows that MATH. Hence MATH is in the image of MATH by the long exact sequence of homology groups. |
math/0103185 | It is known that MATH, the tensor product balanced over MATH, where MATH is the adjoint of MATH. Since MATH, it follows that, as a set, MATH. The multiplication of compact operators is exactly the convolution product on MATH, therefore, as MATH-algebras, MATH. In the same way, using the fact that MATH, we get MATH. |
math/0103185 | REF for the augmented MATH-algebra MATH yields the following six term exact sequence MATH where MATH is the ideal MATH in MATH, and where, recall, MATH. The maps MATH are the boundary maps corresponding to the NAME extension for MATH described in CITE. The map MATH is the map on MATH-theory induced by the inclusion of MATH in MATH, and the two instances of MATH are the maps induced by NAME multiplication by MATH. (Here, MATH is identified with MATH, MATH is identified with MATH, and MATH is identified with MATH.) But MATH, therefore MATH. |
math/0103188 | This follows from REF , and from the fact that NAME liaison is preserved under hypersurface sections REF . Note that the reverse direction is not necessarily true (or in any case is not known to be true): if MATH is glicci, it does not necessarily (or at least immediately) hold that MATH is. See REF for some discussion. |
math/0103188 | The case of NAME lex-segment ideals was observed in CITE. The existence of the decomposition is clear if we choose the ideals MATH as described in REF . Conversely, if we have the decomposition, then we get in REF MATH thus MATH proving REF and the uniqueness of the decomposition. For REF , since MATH is NAME then it contains pure powers of MATH, so these are automatically in MATH, making MATH . NAME. Then the inclusions imply that the other MATH are also NAME. Furthermore, MATH is a minimal generator of MATH, so MATH as claimed. For REF , the hypothesis implies that MATH is a minimal generator of MATH. The fact that MATH is NAME follows immediately from the definition of NAME and the description of MATH in the statement of the Lemma. |
math/0103188 | If MATH then MATH and the claim is clear. If MATH then multiplication by MATH provides the exact sequence MATH . Hence the claim follows by induction on MATH. |
math/0103188 | The proof is by induction on MATH, the codimension. For codimension two it is known that any arithmetically NAME subscheme of projective space is licci, so there is nothing to prove. Hence we assume MATH. By REF we have MATH where MATH and for each MATH, MATH is an NAME ideal in MATH. Notice that the lifting matrix MATH has MATH rows, and if we remove the first row then the remaining matrix MATH can be used to lift the ideals MATH. Let MATH be the ideal obtained by lifting MATH using MATH. Let MATH be the arithmetically NAME subscheme of MATH defined by MATH. Note that MATH has codimension MATH, but the projective space does not change since the linear forms which are the entries of MATH were taken from the ring MATH. Note also that MATH are arithmetically NAME schemes of the same dimension and generically NAME, as in the set-up of REF . Thanks to REF we have MATH . Hence by REF , the scheme MATH obtained from lifting is in fact also obtained by taking the union of the successive hypersurface sections of the MATH. By induction, MATH is glicci. By REF , then, MATH is also glicci. |
math/0103188 | Let MATH be the MATH-st difference of MATH. Let MATH be the NAME lex-segment ideal with NAME function MATH. If MATH then choose a lifting matrix with entries MATH. The lifted ideal MATH defines a glicci subscheme of MATH by REF , and it has NAME function MATH since it is a MATH-lifting. The non-degenerate property comes directly from the lifting, compare CITE. |
math/0103188 | The implications REF are always true. Note that REF implies that MATH in fact contains pure powers of each of the variables MATH, by the NAME property. Then the implication MATH is immediate, since NAME is already assumed. So we have only to prove REF . Since MATH has codimension MATH, it contains a regular sequence of length MATH. By the NAME property we may take this regular sequence to consist of pure powers of variables, and again by the NAME property we can take it to be powers of MATH. Suppose that one of the other variables, say MATH, occurs in one of the minimal generators of MATH to some power MATH. By a standard trick on monomial ideals (compare for instance REF ) we can then decompose MATH as MATH where MATH is again a monomial ideal. But this shows that the primary decomposition of MATH has at least one component of height MATH, contradicting the hypothesis that MATH is equidimensional. |
math/0103188 | Let MATH be a NAME monomial ideal in MATH of height MATH. By REF , we may view MATH as a cone over an NAME ideal in MATH. By REF , MATH where MATH is the initial degree of MATH and the MATH are cones over NAME ideals in MATH satisfying MATH. We can rewrite this as MATH where MATH is a cone over an NAME ideal in MATH, and MATH is a NAME monomial ideal in MATH whose initial degree is one less than that of MATH. Following REF , we can lift MATH to an ideal MATH in MATH; that is, we choose a lifting matrix MATH whose entries are linear forms MATH, MATH. For example, take MATH . We now make some observations. CASE: MATH is also NAME by REF , and it has the same height MATH as MATH since it contains a complete intersection consisting of powers of MATH. CASE: MATH. This follows from the sequence of inclusions on the MATH. CASE: MATH. This follows immediately. For instance, suppose that MATH. Then MATH . By the NAME property of MATH and the fact that MATH, it follows immediately that each term of this polynomial is in MATH. CASE: MATH and MATH are both NAME, and MATH. This follows from the fact that MATH is NAME by REF and that the NAME property and the codimension are preserved under lifting. Let MATH. An analysis similar to REF above shows quickly that MATH. But both are NAME of the same height in MATH, and they have the same NAME function (since the NAME function of MATH is the first difference of that of MATH). Hence we obtain that MATH. Although MATH is not necessarily generically NAME, the lifting results guarantee that MATH is, and we have noted that MATH is NAME. Hence REF says that MATH is NAME to MATH, and in particular MATH is NAME. We have noted that the initial degree of MATH is one less than that of MATH. Hence in a finite (even) number of steps we obtain that MATH is linked to the hyperplane section MATH. Thus it is enough to show that MATH is glicci. Let MATH denote the ideal MATH in MATH. Then MATH is just a cone over MATH. By induction on the height, MATH is glicci in MATH. Then taking cones we get that also MATH is glicci. Hence we have shown that MATH is glicci, as claimed. |
math/0103189 | Let MATH be a legal popping sequence for MATH with MATH, and let MATH be obtained from MATH by transposing MATH and MATH which are not neighbors in MATH. Suppose MATH. Since the edges incident to MATH are disjoint from the edges incident to MATH, we may apply the DSMP to MATH and to MATH and see that MATH is legal as well and maximal if MATH is. If MATH, first apply the DSMP to MATH and then use the same argument. This shows that equivalent sequences are either both maximal popping sequences or neither. (The case of MATH is trivial, since infinite popping sequences are automatically maximal.) To prove the lemma, we induct on MATH, and then deal with the case of MATH. It is clear when MATH. Assuming the lemma for MATH, let MATH with maximal popping sequence MATH. Let MATH be any other maximal popping sequence. If MATH, applying the DSMP and the induction hypothesis completes the induction. If not, then consider the least MATH for which MATH, if any. When we pop MATH, the orientation MATH has a sink at MATH for each MATH, since the sink at MATH exists until one of its edges is popped and no other sink can contain any such edge until MATH is popped. Thus MATH exists and MATH cannot be a neighbor of MATH for MATH. Hence, we can move MATH to the first position by a sequence of adjacent transpositions with non-neighbors in MATH. We have seen that the resulting sequence MATH is a maximal popping sequence. Now apply the DSMP and the induction hypothesis to conclude that MATH is equivalent to MATH, and thus to MATH. All that remains is the case of MATH, that is, the case when all maximal popping sequences are infinite. Given two such sequences MATH and MATH, the argument from the previous paragraph shows that we can transform MATH so that its first element is MATH. Now applying the DSMP shows that we can bring arbitrarily long initial segments into agreement, which is the definition of equivalence. |
math/0103189 | From the previous lemma, we know that MATH, so MATH is independent of the choice rule. Furthermore, since all maximal popping sequences are equivalent, the multisets of popped vertices are the same. The assertion about SFO's follows because the SFO depends only on which vertices were popped. |
math/0103189 | We prove by induction that for any SFO MATH and any finite sequence MATH, the following event has probability MATH, where MATH means the degree not counting self-loops: MATH, and MATH is a legal popping sequence for MATH, and MATH. This is vacuously true when MATH. Now the probability that the singleton MATH is a legal pop is MATH, so applying the DSMP we see that the probability of a maximal popping sequence MATH with MATH is MATH which completes the induction. To find the probability of both MATH and MATH (with no restrictions on the popping sequence), we must sum this probability over all equivalence classes of potential popping sequences of length MATH. We sum over equivalence classes to avoid double counting, since for any given MATH, REF tells us that the set of maximal popping sequences is an equivalence class. Since neither the summand nor the set of sequences depends on MATH, we have proved the lemma. |
math/0103189 | This is proved by stochastic domination: we run sink popping simultaneously on MATH and MATH, using the same stacks for edges common to both graphs. Every legal popping sequence on MATH restricts to a legal popping sequence on MATH as well, so under this coupling MATH always. |
math/0103189 | At any time, some of the arrows point clockwise and others counterclockwise. Let MATH be the number of arrows pointing clockwise at time MATH. Popping at any vertex causes two opposite pointing arrows to be replaced by two random arrows. Thus MATH has the distribution of MATH where MATH and MATH. Therefore MATH is a simple random walk with delay probability of MATH absorbed at MATH and MATH. The expected absorption time from MATH is twice that for simple random walk, and thus is MATH; see REF on page REF. Hence MATH, which is twice the expected number of ordered pairs of edges where the first is initially clockwise and the second initially counterclockwise. There are MATH ordered pairs of distinct edges, each having these orientations with probability MATH, so MATH. |
math/0103189 | Fix MATH and MATH and let MATH be a cycle containing MATH. By monotonicity, MATH which is at most MATH by REF and symmetry. In fact, MATH is strictly less than MATH unless MATH is a MATH-cycle. By the remark following the proof of the monotonicity lemma, the inequality MATH is strict unless no SFO on MATH can require further popping of MATH on MATH. In our case, MATH is a MATH-cycle, and MATH is a MATH-cycle with some chords or self-loops added. Then (assuming MATH is not a MATH-cycle), there is always an SFO on MATH that does not restrict to an SFO on MATH: if MATH has a self-loop, one can choose an SFO on MATH such that MATH has a sink there; if MATH has a chord, one can use the chord to create a short circuit across MATH giving a cycle of length less than MATH, orient this cycle in a loop, and orient the other edges in MATH towards the cycle. If MATH is a sink in the restriction to MATH of such an SFO on MATH, then strict inequality holds for MATH (because with positive probability, sink popping on MATH will produce this SFO, and MATH will still need to be popped in MATH). |
math/0103189 | We induct on MATH. The base step is MATH, which is immediate from the previous corollary. Assume for induction that the conclusion holds for all subgraphs of MATH. There are three cases other than the base step. CASE: MATH is not connected. Then the result follows from the induction hypothesis and the monotonicity lemma applied to the component MATH of MATH containing MATH. Equality never occurs. If MATH is connected and not in MATH, then MATH must contain an isthmus, that is, an edge whose removal disconnects MATH. CASE: Some edge MATH disconnects MATH into two components both in MATH. Again the result follows from the induction hypothesis applied to the component MATH of MATH containing MATH, and equality never occurs. CASE: MATH has an isthmus and removal of any isthmus always leaves a component that is a tree. Then MATH has a leaf MATH. If MATH then the result follows immediately from monotonicity with MATH. If MATH is the only leaf, then let MATH be its neighbor. Choose a popping order that pops MATH whenever possible, and otherwise executes any choice rule for sink popping on MATH. Initially there is a MATH chance that MATH is a sink, in which case it is popped a mean MATH geometric number of times until the edge MATH points to MATH. Then, each time MATH is popped, the probability is MATH that this edge is reversed, in which case it takes another mean MATH geometric number of pops to reverse it again. Thus MATH . By induction, this is at most MATH. Equality occurs for MATH in MATH if and only if it occurs for MATH in MATH, so we see by induction that it holds only at the end of a lollipop. |
math/0103189 | We prove the lemma by induction, following the pattern of the last proof. The base step is MATH, in which case the lemma follows from REF . In REF of the induction, if MATH is disconnected or the union of two graphs in MATH along an added edge, the result is again immediate from the subadditivity of the function MATH and monotonicity. Finally, if MATH has a leaf MATH, we set MATH and observe that the number of pops MATH and MATH on MATH and MATH respectively are related by MATH. Thus MATH . By the previous lemma, the last inequality is strict unless MATH is a MATH-lollipop and its vertex of degree REF is the neighbor of MATH in MATH. This completes the induction. |
math/0103189 | The theorem follows immediately from combining REF , and REF. |
math/0103189 | Number the vertices of the MATH-lollipop MATH with MATH being the leaf. Always pop the sink with lowest number. Let MATH denote the sink popped at time MATH. Clearly MATH is MATH plus a mean MATH geometric random variable, with the proviso that a value of MATH or higher represents the terminal state in which no sink needs to be popped. Let MATH be the MATH-field generated by MATH. We claim that MATH is a time-homogeneous NAME chain with respect to MATH and that from any state MATH its increments are MATH plus a mean MATH geometric, jumping to the terminal state if it reaches MATH or greater, and from state MATH the same thing with MATH replaced by MATH (thus the jump from MATH is resampled if it hits MATH). All that is needed to check this is an inductive verification that the orientations of edges between vertices of higher index than MATH are conditionally i.i.d. fair coin flips given MATH, which is straightforward. Now we show that the running time on a MATH-cycle is also equal to the time for a random walk to hit at least MATH when its increments are MATH plus a mean MATH geometric, resampled if it hits MATH. At time MATH, let MATH denote the least index of a sink when the edge from MATH to MATH is oriented toward MATH and MATH minus the greatest index of a sink when the edge is oriented toward MATH. In other words, this quantity is the distance from the head of the MATH edge to the nearest sink in that direction. We always choose the pop that sink. The only time the MATH edge can change orientations is when MATH, in which case MATH will be MATH plus a mean MATH geometric; when MATH verification of the conditional increment is trivial. The stopping rule is, again, that one must jump to MATH or greater, and MATH has the right distribution for the same reason as before, so the sequence has the same distribution. |
math/0103189 | Number the vertices MATH mod MATH. We first establish that the discrete Laplacian of MATH depends on the initial orientation of MATH via MATH . To see this, choose any popping order and let MATH denote the in-degree of MATH at time MATH, that is, the number of MATH adjacent to MATH for which MATH is oriented toward MATH. Then, conditionally on anything up to time MATH, MATH since any pop at MATH reduces the expected in-degree by mean REF and any pop at a neighbor of MATH increases it by mean MATH. Summing over MATH, conditioning on MATH and using MATH proves REF . From REF we know that MATH where MATH is the number of initial clockwise arrows (edges oriented from MATH to MATH mod MATH for some MATH). This, along with REF , determines MATH, since the difference of any two candidates for this function would be a harmonic function on the cycle, and hence constant. In general, MATH if there are MATH clockwise edges pointing from MATH to MATH for a set MATH (addition taken mod MATH). To prove this formula, we need only prove that the right hand side satisfies the two properties that characterize the left hand side. The sum over all MATH (that is, MATH) is easy, since it equals MATH which does indeed simplify to MATH. To check that the right hand side of REF works in REF, we proceed as follows. Let MATH be the right hand side of REF. Then MATH where MATH is the NAME delta. Hence, MATH as desired. REF makes it easy to see when MATH is maximized: that can occur only when MATH, in which case MATH equals MATH. This quantity is bounded above by MATH, with equality if and only if MATH. |
math/0103189 | Induct again, as in the proof of REF and the main theorem. Simultaneously, we show by induction that MATH, with equality only for the leaf of a MATH-lollipop and initial conditions MATH all pointing toward MATH. Note that the previous proposition proves this bound for all MATH, if we use monotonicity (which also holds conditioned on the initial orientation). For the base case, MATH and REF shows that in fact MATH. There is strict inequality because equality in REF cannot hold simultaneously for all vertices in a cycle. It follows that MATH unless MATH. The cases with MATH are easily dealt with: those with MATH are trivial, and for MATH the worst case contains a MATH-cycle, which can be analyzed using the sharper bounds in the proof of REF . When MATH is not connected or is the union of two graphs in MATH along an added isthmus, the MATH bound is immediate from subadditivity of MATH and monotonicity, and the MATH bound follows from monotonicity. Finally, when MATH has a leaf MATH, set MATH as before. This time, in the worst case we know that MATH is a sink initially, so MATH is bounded by MATH. This verifies the conclusion that MATH, and adding this to MATH gives, by induction, at most MATH, which completes the proof of the upper bound; the conditions for equality are clear from the proof. |
math/0103196 | We have MATH where the third equality follows from REF . Note that MATH yields MATH, or MATH . REF implies that MATH, so that MATH. REF also shows that, given any two points MATH, we can find a (unique) point MATH such that MATH. Therefore, MATH, which shows that the set of linear operators MATH acts transitively on MATH. Hence, MATH is a symmetric cone. For the second assertion, note that if MATH, then MATH . For MATH, we thus have MATH where the second and last equalities follow from REF, respectively. Consequently, MATH is self - scaled under MATH. |
math/0103196 | Let us define MATH. REF implies that for any MATH we have MATH. Expanding both sides of this equation and comparing the MATH terms one gets MATH or MATH. Thus, MATH, and REF follows. |
math/0103196 | Since MATH is self - dual, REF is equivalent to REF . If MATH is self - scaled, then REF follows from REF. Conversely, assume that MATH satisfies REF . Let MATH be arbitrary points. We claim that there exists a point MATH such that MATH. Towards proving the claim, we consider the optimization problem MATH, where MATH. It is well known that the feasible region is bounded, see REF , p. CASE: We have MATH (see REF MATH), and MATH (see REF MATH), which imply MATH. These imply that the objective function of the optimization problem goes to infinity as MATH approaches the boundary of the feasible region, and thus the optimization problem has a minimizer MATH satisfying MATH for some scalar MATH. Since MATH, we have MATH. The point MATH satisfies MATH (see REF MATH), and proves the claim. Next, we claim that MATH . Let MATH be a point satisfying MATH. The fundamental REF is a consequence of REF and gives MATH, or equivalently MATH. From REF, we obtain MATH or MATH. REF also implies that MATH hence proving the claim. Using logarithmic homogeneity alone one can prove that MATH where MATH (see REF MATH). CASE: MATH shows that the mapping MATH is involutive, that is, MATH. These imply MATH. Since MATH by REF MATH, we have MATH which is to say that MATH. This implies MATH where the last equality follows from REF . This concludes the proof. |
math/0103196 | If MATH, then MATH for every vector MATH. That is, MATH, or MATH, see for example REF . This shows that MATH is orthogonal if and only if MATH. The uniqueness of MATH implies that this condition is equivalent to MATH. |
math/0103196 | By virtue of REF , there exists a unique point MATH such that MATH. Then MATH satisfies MATH, which implies that MATH is orthogonal by REF . Since MATH is orthogonal and MATH is symmetric, MATH is indeed a polar decomposition of MATH. Suppose now that MATH where MATH is symmetric and MATH is orthogonal, MATH. Then, MATH is orthogonal, and we have MATH, or MATH. Since MATH and MATH are symmetric, we have MATH and MATH. |
math/0103196 | If MATH, then REF implies that we can write MATH for some MATH and MATH. By the uniqueness of the polar decompositions MATH and MATH it must be true that MATH. Thus, MATH . Conversely, let MATH. By REF , there exists a point MATH such that MATH. But this implies that MATH. By virtue of REF MATH is therefore orthogonal. This means that MATH has the polar decompositions MATH and MATH. The uniqueness part of REF then implies that MATH and MATH. This proves the first statement of the lemma. Now let MATH and define MATH by MATH, see REF . We have MATH where the second equality follows from the fundamental REF . In a similar vein, taking the first part of this lemma into account we obtain MATH . These two equations imply that MATH and MATH . In particular, setting MATH yields MATH, that is, MATH, and MATH. The lemma follows, since this implies that MATH . |
math/0103196 | From REF we see that MATH, implying that MATH for some constant MATH. REF shows that the function MATH satisfies the equation MATH for some constant MATH. These facts combined with REF imply the proposition. |
math/0103196 | We have MATH. Each term MATH, therefore we have MATH and MATH. Since MATH contains no lines it must be true that MATH, that is, MATH, (MATH). |
math/0103196 | Suppose that MATH admits two irreducible decompositions MATH . Note that each non - zero summand in either decomposition of MATH must lie in MATH and that the subspace corresponding to each zero summand must be one - dimensional, for otherwise the summand would be decomposable. This implies that the number of zero summands in both decompositions is MATH. We may thus concentrate our efforts on MATH, that is, we can assume that MATH is solid and all the summands of both decompositions of MATH are non - zero. By REF, each MATH has a unique representation MATH where MATH. Also, REF implies that MATH, and hence MATH. Consequently, every MATH lies in the set MATH. Conversely, we have MATH, implying that MATH. Therefore, we have MATH . We have MATH, MATH, and the intersection of any two distinct summands in the last sum is the trivial subspace MATH. The above decompositions of MATH and MATH are thus direct sums. Since MATH is indecomposable, exactly one of the summands in the decomposition of MATH is non - trivial. Thus, MATH, and hence MATH for some MATH. Arguing symmetrically, we also have MATH for some MATH, implying that MATH. Therefore, MATH for else MATH, contradicting our assumption above. This shows that MATH. The theorem is proved by repeating the above arguments for the cone MATH. |
math/0103196 | Recall that the universal barrier function MATH is given in REF. Changing the inner product used in the definition of the characteristic function MATH changes MATH only by an additive constant, hence we may assume that MATH for the purposes of this definition. Here, MATH is an inner product defined on MATH chosen so that MATH for some elements MATH where MATH denotes the universal barrier function defined on MATH. Then we have MATH for MATH, in full consistency with our previous notation. Moreover, MATH is self - dual under MATH, since MATH. Hence, we may choose the vector MATH specified in REF as the unique element in MATH such that MATH under MATH, see REF . Thus, MATH can be written as the direct sum MATH and MATH has block structure corresponding to the subspaces MATH, MATH where MATH. Consequently, REF implies that MATH also has the same block structure, MATH where MATH with MATH, MATH being the projection at the beginning of this section. So far we know that MATH has block structure corresponding to the direct sum MATH, but it is not a priori clear that MATH is the Hessian of a function defined on MATH. Let the spaces MATH be defined as earlier in this section, and let us consider the vector fields MATH, defined on MATH and taking values in MATH for all MATH, MATH. We claim that MATH depends only on MATH. In fact, for any two vectors MATH such that MATH we have MATH which shows our claim. Hence, the quotient vector fields MATH are well defined and can be identified with vector fields MATH defined on the cones MATH. The direct sum of these vector fields amounts to the gradient field MATH . MATH being conservative, the MATH must be conservative too, implying that these are the gradient vector fields of some functions MATH defined on MATH which are uniquely determined up to additive constants. We may choose these constants so that MATH. Clearly, we have MATH for any MATH. Using REF , it is straightforward to check that the MATH are self-concordant, see CITE. Applying REF MATH and MATH to MATH, using REF and considering variations of MATH only in the part MATH, we obtain that MATH for some number MATH. Moreover, applying REF MATH to MATH and using REF we get MATH for all MATH. Hence, MATH . This shows that the functions MATH are MATH-logarithmically homogeneous. It is a well - known fact that any logarithmically homogeneous self - concordant function is also a barrier function, see for example CITE or CITE. It remains to show that the functions MATH are self - scaled. As previously noted, REF is satisfied, since MATH for all MATH. Finally, REF holds for MATH because we can apply this condition to MATH, choosing MATH and MATH and using the block structures of MATH and MATH. Note that the irreducible components MATH of MATH must be symmetric cones, since the MATH are self - scaled barriers defined on MATH. The symmetry of the MATH can also be directly derived from the block structure of MATH and the fact that the set of cone automorphisms MATH acts transitively on MATH. |
math/0103196 | CITE proves that if MATH is a symmetric cone, then MATH is generated by MATH where MATH is a neighbourhood of the identity MATH, see REF , pp. CASE: NAME 's proof exploits the fact that all derivations of the NAME algebra associated with MATH are inner. An accessible proof for the case where MATH is irreducible is given in REF , pp. REF, based on certain non - trivial results from the theory of NAME algebras. If MATH is orthogonal, it follows from NAME 's result that MATH for some MATH, MATH. Here the second equality follows from REF . Therefore, it follows from REF that MATH . Since MATH, setting MATH above yields MATH and settles the claim of the theorem. |
math/0103196 | Let MATH be any permutation of MATH. REF implies that there exists an orthogonal automorphism MATH such that MATH (MATH). Using REF , we then obtain MATH . Define MATH. Note that MATH is a symmetric function. Consider a point MATH, with arbitrary MATH and MATH. Applying REF repeatedly, we obtain MATH . Using the symmetry of MATH and MATH, the above equation translates into MATH . If MATH above, we have MATH. This gives MATH for all MATH. Consequently, we have MATH . Now, if MATH REF are arbitrary, choose MATH such that MATH. Since MATH is logarithmically homogeneous, we have MATH by the logarithmic homogeneity of MATH. A simple calculation shows that REF holds true for all MATH. The lemma is proved. |
math/0103196 | REF describes the restriction of MATH on MATH. Since the universal barrier function is also self - scaled, the same considerations apply to MATH. Thus, the functions MATH and MATH are homothetic transformations of each other on MATH, that is, there exist MATH, MATH such that MATH . Let MATH be an arbitrary point with the spectral decomposition MATH. REF implies that there exists MATH such that MATH (MATH). We have MATH where MATH. REF gives MATH and MATH, and hence the Identity REF extends to all of MATH. |
math/0103196 | REF imply that there exist constants MATH and MATH such that MATH where MATH. It is known that MATH, where MATH is the rank of the NAME algebra associated with MATH, and MATH is the dimension of the cone MATH, see REF , p. CASE: Finally, REF implies that the function MATH is self - concordant if and only if MATH. |
math/0103197 | Apply REF , using MATH, MATH, etc. and taking the hypersurface sections obtained by suitable multiples of the MATH (starting with MATH and working backwards), working our way down the picture. This corollary can also be obtained using the lifting results of CITE. |
math/0103197 | Consider the complete intersection MATH in codimension REF, where MATH (note that both products start with MATH). Let us denote the scheme defined by MATH by MATH (subscripts here will refer to the codimension). When MATH, the scheme MATH has the following form (where the components of MATH are represented by dots, and the set of all intersection points is MATH): MATH . Using REF , it follows immediately that the MATH-vector of MATH is MATH as claimed. Now consider the case MATH. Let MATH be the subset of MATH consisting of all ``dots" lying on MATH (there is just one). Let MATH be the union of the components of MATH lying on either MATH or MATH, and so on. Clearly we have MATH and all MATH are arithmetically NAME. We will apply REF , so to that end let us set MATH REF and let MATH the hyperplane section of MATH with MATH. Then we have MATH and by REF the union of the MATH is arithmetically NAME of codimension REF with the MATH-vector MATH (use REF ). Clearly we also have MATH . REF quickly shows that each of these is a generalized stick figure. But clearly the union of the MATH has the form described in REF , so this is MATH and we have completed the case MATH. To pass to codimension REF using REF again, we now take MATH to be the codimension REF schemes just produced (that is .set MATH now to be the MATH just produced, MATH to be the MATH just produced, etc), and we take MATH (MATH). As before we have a generalized stick figure in codimension REF which is arithmetically NAME and has maximal NAME function. Note that the components of the arithmetically NAME scheme MATH we have produced have precisely the form described in REF . We continue by induction to finally produce the desired arithmetically NAME scheme of codimension MATH with MATH-vector given by REF . Note that the range for MATH changes: for instance, in codimension REF we have MATH (MATH). It is a generalized stick figure by REF . |
math/0103197 | It is clear from the description that MATH is a reduced ideal. Next we verify that if MATH has the property that each subset of MATH elements generates a complete intersection of codimension MATH then MATH is a generalized stick figure. To prove this, consider first the case MATH even. Suppose we have three components MATH (each taken from MATH, MATH or MATH). In order for MATH and MATH to have MATH entries in common, we must be able to take out an entry from MATH and replace it with a new entry, giving MATH. Putting both of these entries in and removing a third one must give MATH. Because of the rigid form of the components given in the statement of the theorem, one can just check that this is impossible. Next suppose MATH is odd and MATH. Note that in this case MATH is empty. Then the argument is similar to the one above. The case where MATH is odd and MATH is handled similarly. Here is it slightly tricky to prove that we get a generalized stick figure because of the fact that REF allows two consecutive MATH at the end. But the subscript of the second one is bounded below by MATH, and this fact is needed to complete the proof. For example, the linear forms MATH seem at first glance to be a counter-example, but the ``REF" in MATH cannot simultaneously be bounded above by MATH and below by MATH. The fact that MATH follows from the observation that MATH (respectively, MATH) is MATH, thanks to REF . It remains to show that MATH is a NAME ideal with the correct MATH-vector. For this we proceed by induction on MATH. If MATH then MATH is a principal ideal of degree MATH, thus having the claimed properties. If MATH then MATH is a complete intersection by assumption on MATH. It is easy to check its MATH-vector. Now suppose that MATH. We distinguish two cases. CASE: Assume that MATH is odd. We have at our disposal the set MATH where each subset of MATH elements is linearly independent (over MATH). Let us temporarily re-name the linear form MATH with the new name MATH. Hence we now have the set MATH . By induction we can use MATH to get the NAME ideal MATH which defines a generalized stick figure. The configuration MATH is formed using the set MATH. Since MATH, we observe that MATH . Now we re-name MATH back to MATH. The configuration MATH then becomes (with the re-naming) a configuration, which we now call MATH, whose ideal is MATH where MATH where it is understood that MATH if MATH. By REF , this re-naming does not affect any component of MATH, which is defined by MATH. Of course the naming of the linear forms does not affect the properties of the configurations, and so all together we have that MATH is an arithmetically NAME generalized stick figure with the same MATH-vector as MATH and containing MATH. Since MATH is a generalized stick figure containing MATH, it provides a geometric link to the residual MATH . By REF , the ideal MATH given in the statement of the theorem will have the desired properties if we can show that MATH . We have to show CASE: For every choice of a component from MATH (that is, from MATH) and a component from either MATH, MATH or MATH which meet in codimension MATH, their intersection occurs in either MATH, MATH or MATH. CASE: Every component of MATH, MATH, or MATH is the intersection of a component from MATH and a component from either MATH, MATH or MATH. All of these involve an analysis of how many linear forms can be common to two (or three) of the components of our configurations. For REF , in order for the two components in codimension MATH to meet in codimension MATH, their intersection must be defined by only MATH linear forms, so they must have exactly MATH linear forms in common. We have to determine all the ways that this can happen and show that we always get a component of MATH. This analysis then works backwards to show just how to produce any component of MATH as the intersection of two components of MATH, which answers REF . To answer these two questions, we leave it to the reader to verify that in order to meet in codimension MATH, CASE: a component from MATH (that is, MATH) meets a component from MATH either in the form MATH (and all components of MATH arise in this way) or in the form MATH (and all components of MATH arise in this way except those where MATH). CASE: a component from MATH meets a component from MATH in the form MATH (and all components of MATH arise in this way). CASE: a component from MATH meets a component from MATH in the form MATH where MATH, taking care of those components which were ``missing" from the first set above. To do this one checks how it is possible to remove one entry from the first component and one entry from the second and have the remaining entries equal. (There are very few possibilities.) REF Assume that MATH is even. Then MATH. By induction, the arithmetically NAME scheme MATH contains MATH. We put MATH, the residual scheme. The assertion follows because MATH, which can be shown as in REF (and is easier). This completes the construction of the arithmetically NAME generalized stick figure with ``maximal" MATH-vector. |
math/0103197 | The existence of the decomposition is straightforward. Note that the uniqueness comes from the requirement that MATH. It remains to show the inequality. Assume that the regularity of MATH is MATH. This means that MATH but MATH. Hence MATH but MATH. Therefore MATH is also not contained in MATH, that is . MATH. It follows that MATH which concludes the proof. |
math/0103197 | It is a straightforward computation. |
math/0103197 | In view of REF , it suffices to check that MATH is arithmetically NAME, MATH, the MATH-vector of MATH and the last claim. We induct on MATH. Let MATH. By comparison with REF we get MATH because MATH. Now let MATH. Using the notation of REF , MATH is the MATH-vector of MATH, where MATH. Let MATH. Then MATH for all MATH. Moreover, REF shows that MATH. Since MATH, the induction hypothesis provides inclusions of arithmetically NAME configurations MATH . Suppose that MATH is even. Then the configuration MATH is made up using only the linear forms MATH . Hence the assumption on the set MATH and REF imply MATH and hence MATH . Therefore REF gives us that the ideal MATH defines an arithmetically NAME subscheme having MATH as MATH-vector and is in fact the ideal MATH. If MATH is odd we conclude similarly because then MATH . It remains to show that MATH. Let MATH be the decomposition of the lex-segment ideal MATH with MATH-vector MATH. Let MATH be the MATH-vector of MATH. REF implies that MATH for all MATH with MATH. Therefore, the induction hypothesis provides MATH . Then using the definition of the ideals we obtain MATH as desired. Finally, note that if MATH is a ``maximal" MATH-vector then MATH. Hence we conclude that all of the configurations MATH that we obtain for ``smaller" MATH-vectors are contained in MATH as claimed. |
math/0103197 | As a preparatory step we consider the lex-segment ideal MATH. The properties of its decomposition imply MATH . Using REF we obtain MATH . Since MATH, the multiplication by MATH provides the exact sequence MATH where MATH. We claim that this implies for all MATH . Indeed, the long exact MATH-sequence provides maps MATH. If MATH then MATH, and if MATH then MATH. Therefore MATH is the zero map and the MATH-sequence proves our claim. Now we consider MATH. We induct on MATH and MATH. Again, the case MATH is easy. Let MATH. If MATH then MATH contains a linear form which allows us to conclude using induction on the codimension. Let MATH. Assume that MATH is even. Let MATH be defined by MATH . Then we claim that we have an exact sequence MATH . Indeed, the only question is the cokernel. The sequence clearly holds if we write MATH for the cokernel. But MATH is arithmetically NAME by induction, and MATH is arithmetically NAME by REF , so sheafifying and taking cohomology gives that REF is a NAME ideal. But then the generality of MATH gives that the components of the corresponding scheme are defined precisely by the ideal given as the cokernel of REF . By induction on MATH, MATH has the same NAME numbers as MATH and, by induction on MATH, MATH has the same NAME numbers as MATH. Hence we can conclude as above that the NAME numbers of MATH are the sum of the NAME numbers of MATH and MATH and thus equal to the NAME numbers of MATH. The case where MATH is odd is handled similarly. |
math/0103197 | This is a special case of REF. See also CITE. |
math/0103197 | By the definition of MATH, the ideal MATH does not contain exactly the smallest (with respect to MATH) MATH monomials in MATH of degree MATH. Therefore, by REF , MATH contains exactly the smallest (in the ordering MATH) MATH monomials of MATH of degree MATH. Since MATH is an NAME, REF shows that MATH is a LOIM. |
math/0103197 | Let MATH be the lex-segment ideal having MATH as MATH-vector. Let MATH be its decomposition according to REF . Since MATH we have for a monomial MATH with MATH that MATH by REF . The claim then follows. |
math/0103197 | We induct on MATH. First let MATH. Then there is an integer MATH such that MATH and MATH . Thus we have MATH . For MATH we obtain MATH and thus MATH, and MATH. It follows that MATH as claimed. Now let MATH. Let MATH be the decomposition of MATH. Assume that MATH is even. Then we know by REF and the induction hypothesis that MATH . Since MATH, we are done if we can show Claim: Let MATH and let MATH. Then MATH . By REF we know that MATH. Thus we get MATH . Because MATH is generated by MATH and MATH is generated by MATH, it follows that MATH, and so MATH proving the claim. If MATH is odd we conclude similarly. |
math/0103197 | CASE: Comparing REF we observe that MATH is just defined by MATH and MATH respectively, depending on the divisibility of MATH by MATH. CASE: The first two claims follow immediately from the definition of MATH . For REF it suffices to note that MATH implies MATH, and thus MATH and the assertion follows. |
math/0103197 | Notice that MATH is always contained inside MATH, thanks to REF . However, even if MATH is maximal and so MATH (compare REF ), it is not true (if MATH is odd) that MATH, or even that MATH. See also REF . CASE: According to REF , we know that MATH. By REF , MATH is a generalized stick figure containing MATH. Thus MATH is the sum of the geometrically linked ideals MATH and MATH. The fact that MATH is reduced comes from the fact that MATH is a generalized stick figure, using REF . We have MATH . Note that REF's occur in degrees REF and MATH and the last MATH occurs in degree MATH. Note also that MATH. Consider the sequences MATH, MATH, MATH, MATH, where MATH . Note that MATH is the MATH-vector of MATH. By REF we know that MATH has MATH-vector MATH, and again by REF we know that the first difference of the MATH-vector of MATH is given by MATH . We now claim that MATH. CASE: For MATH we have MATH and MATH. CASE: For MATH we have MATH, MATH and MATH. However, in this range we have MATH by the symmetry of MATH. CASE: For MATH we have MATH, MATH. Hence MATH . It follows that MATH has the desired MATH-vector MATH. CASE: It remains to establish the claim on the minimal prime ideals of MATH. Since MATH is a generalized stick figure and MATH is unmixed of codimension MATH, an ideal MATH is a minimal prime of MATH if and only if MATH and MATH for some minimal prime ideal MATH of MATH and some minimal prime ideal MATH of MATH. Thus the assumption on MATH implies that each minimal prime MATH of MATH is generated by a subset of MATH consisting of MATH elements. Suppose that MATH contains another minimal prime MATH of MATH. Then MATH are pairwise distinct minimal prime ideals of MATH such that MATH has codimension MATH. This contradicts the fact that MATH is a generalized stick figure. Thus we have shown the necessity of the conditions for being a minimal prime. CASE: In order to prove sufficiency, let MATH be a minimal prime of MATH, where MATH. We must show that for MATH, if MATH is the only minimal prime of MATH which is contained in MATH then MATH is a minimal prime of MATH. Our proof will make use of the following observation. Given MATH and MATH as in the last paragraph, consider a specific minimal prime ideal MATH of MATH such that MATH. Then we know by REF that MATH is a minimal prime ideal of MATH if and only if MATH is not a minimal prime ideal of MATH (that is, MATH is a minimal prime of MATH). We will distinguish several cases. CASE: Suppose that MATH (see the statement of REF ), that is, MATH . We already know that MATH is a minimal prime ideal of MATH. Write MATH as in REF and denote by MATH the ideal generated by the first MATH linear generators of MATH. Then MATH is a minimal prime ideal of MATH - in fact, it is a primary component of MATH (if MATH is even) or MATH (if MATH is odd). Since MATH, MATH must be a minimal prime of MATH. Thus MATH is a primary component of MATH. CASE: Suppose MATH, that is, MATH for some MATH. As preparation we need a description of the subsets of MATH having the form MATH for some monomial MATH (see REF ). To this end we call a subset MATH of MATH consecutive if MATH for some MATH. If MATH then MATH is called the predecessor of MATH, and if MATH then MATH is called the successor of MATH. Now let MATH be an arbitrary non-empty subset of MATH. It is clear that MATH has a unique decomposition MATH where MATH are consecutive subsets, the successor of MATH is not contained in MATH if MATH, the predecessor of MATH is not contained in MATH if MATH, and every element of MATH is smaller (in the order MATH) than every element of MATH if MATH. Using this notation, the definition of MATH REF implies: Observation: There is a monomial MATH such that MATH if and only if MATH and the cardinality MATH is even for all MATH with MATH. Returning to our previous notation, let MATH again denote a monomial such that MATH. As above, write MATH for the unique decomposition of MATH into consecutive subsets. Since MATH, we obtain MATH, that is, MATH for some MATH (MATH). Write MATH where MATH and MATH are both consecutive unless one of them is empty, and each element of MATH is smaller than each element of MATH. We have to distinguish further cases: CASE: Suppose that MATH is odd and that MATH if MATH. Then MATH has a predecessor, say MATH. Since MATH is even, the observation above implies the existence of a monomial MATH of degree MATH such that MATH . The prime ideal MATH is a minimal prime of MATH, and thus of MATH. Moreover, we have by construction MATH . Hence we are done, as explained at the beginning of REF . CASE: Suppose that MATH is even and that MATH if MATH. Then MATH has a successor, say MATH. Since MATH is even, we can find as above a monomial MATH such that MATH and MATH is a minimal prime of MATH. We conclude as in REF. CASE: Suppose that MATH is odd, MATH and MATH. Then MATH (since if MATH then MATH is not odd) and MATH. CASE:REF Suppose that MATH. Then we form the subset MATH by replacing MATH by MATH and all MATH with MATH by MATH. The observation above shows that there is a monomial MATH of degree MATH such that MATH. Since MATH, we obtain by REF that MATH divides MATH. Using REF again, we conclude that MATH is a minimal prime ideal of MATH. Hence REF shows that MATH is a minimal prime of MATH. It follows that MATH, and thus MATH is a minimal prime ideal of MATH. CASE: Suppose that MATH. Let MATH be the predecessor of MATH. As in REF, construct MATH out of MATH by replacing all MATH by MATH if MATH and by MATH if MATH. Again we get the existence of a monomial MATH such that MATH and MATH is a minimal prime of MATH, and thus MATH is a minimal prime of MATH. CASE: Suppose that MATH is even, MATH and MATH. Then MATH and MATH is odd. (Note that MATH is impossible since otherwise MATH has odd cardinality.) REF: Suppose that MATH. Similarly as above we conclude that there is a monomial MATH such that MATH, and MATH is a minimal prime of MATH, and thus MATH is a minimal prime of MATH. CASE: Suppose that MATH. Let MATH be the successor of MATH. Then there is a monomial MATH such that MATH divides MATH, MATH and MATH is a minimal prime of MATH. Therefore we have shown the claim in all cases, and the proof of REF is complete. |
math/0103197 | Assume that MATH has the weak NAME property with respect to MATH. The fact that we get a perfect ideal comes from the NAME property. Let MATH be the MATH-vector of MATH where MATH and let MATH (so the second MATH occurs in degree MATH). Consider the exact sequence MATH . Since the multiplication by MATH on MATH is injective if MATH, we get MATH where MATH denotes the initial degree. Since MATH we conclude MATH . Conversely, if MATH has the subspace property then the exact sequence REF gives injectivity for the multiplication by MATH on MATH in degrees MATH, and the self-duality of MATH (up to shift) gives the surjectivity for the second half. |
math/0103197 | Consider the commutative diagram MATH . It follows that MATH . |
math/0103197 | It follows from REF . |
math/0103197 | In the proof of REF we use MATH, where MATH. Thus we get for MATH that MATH (this follows from REF ). Put MATH, MATH, MATH. Consider the commutative diagram MATH . Since MATH and MATH are linked by MATH, the ideal MATH is NAME of codimension MATH and has the MATH-vector MATH with the final REF occurring in degree MATH. Hence the rightmost column shows that MATH is a perfect ideal of codimension MATH in MATH and MATH . Therefore MATH has the subspace property with respect to MATH. |
math/0103197 | Following CITE, we define for a NAME graded module MATH, MATH where MATH is the NAME polynomial of MATH. In CITE, MATH is called the index of regularity of MATH. Since MATH is arithmetically NAME by assumption, we get for its index of regularity MATH . Thus we obtain (compare for instance REF ) MATH and the exact sequence MATH (note that MATH is not assumed to be arithmetically NAME), where the vertical sequences denote the corresponding minimal free resolutions. Now we observe MATH and MATH . It follows that MATH . Thus MATH where the last inequality is by assumption. We conclude that MATH for all MATH. On the other hand, it is well-known that the modules MATH are generated in degrees MATH. It follows that the mapping cone procedure provides a free resolution of MATH which is minimal because cancellation cannot occur. This proves the claim. |
math/0103197 | Since MATH is NAME, we get by applying MATH to a minimal free resolution of MATH a minimal free resolution of MATH, that is, MATH . Thus the Lemma implies our assertion. |
math/0103197 | Consider the exact sequence MATH NAME with MATH and taking homology provides MATH for all MATH and MATH . Now we compute MATH using a minimal free resolution of MATH: MATH NAME by MATH gives the complex (with MATH) MATH . Its homology is MATH . Since MATH is a minimal free resolution, the maps in MATH are minimal maps too. Thus we obtain a minimal free resolution of MATH as a MATH-module: MATH . It follows that MATH . The exact sequence MATH implies MATH . Using MATH we obtain the exact sequence MATH . The associated long exact Tor sequence reads as MATH . Taking into account the isomorphisms REF , our claim follows. |
math/0103197 | Using MATH if MATH the claim follows by analyzing the sequence given in REF . |
math/0103197 | We denote by MATH the dualizing functor with respect to MATH. It is exact. Thus, the exact sequence MATH provides the exact sequence MATH . Since MATH is NAME and NAME, we have MATH. Consider the commutative diagram MATH . We conclude that MATH . Put MATH. Since MATH is NAME we get by local duality MATH . It follows that MATH and therefore MATH . In particular, we get for all MATH isomorphisms of MATH-vector spaces MATH . Hence REF proves the claim for all pairs MATH with MATH. Since the minimal free resolution of MATH is self-dual we have MATH . Using what we have already proved we get for MATH concluding the proof. |
math/0103197 | If MATH then there is no integer MATH satisfying MATH. Thus REF proves the claim. |
math/0103197 | We first prove REF . We may assume that MATH is NAME. Then it has the weak NAME property, that is, there is a MATH such that the multiplication map MATH has maximal rank for all MATH. It follows that the NAME function of MATH is MATH. Thus REF gives MATH where MATH. Therefore REF provides for all MATH . The claim then follows because MATH if MATH. For REF , according to REF it suffices to compute the graded NAME numbers of the subscheme MATH constructed in REF . The starting point was an arithmetically NAME subscheme MATH of codimension MATH with MATH-vector MATH which satisfies MATH thanks to REF . Let MATH be the residual to MATH in MATH. Then we have MATH. The graded NAME numbers were computed in REF : MATH . Using again the fact that MATH if MATH, we get the result. |
math/0103197 | We may assume that MATH is NAME and has the weak NAME property with respect to MATH. Thus MATH. Its graded NAME numbers are easily computed. For the reader's convenience we include the argument. Let MATH be the graded homomorphism given by the matrix MATH . The ideal of maximal minors of MATH is MATH and it is resolved by the NAME complex MATH . In particular, we get for all MATH: MATH . If MATH the claim follows by REF because MATH which then forces MATH. If MATH then we apply REF . It proves our assertion since we have for MATH that MATH if MATH. If MATH then REF shows that MATH unless MATH, MATH or MATH with MATH. But then MATH can be computed recursively from the NAME function of MATH. (A similar computation can be found on page REF.) |
math/0103197 | Define the integer MATH by MATH where MATH. We will use the notation of REF . We list the monomials in MATH in order, MATH. Using the bijection MATH, given by MATH, we define MATH as the subset corresponding to MATH. Finally, we put MATH where MATH is defined in REF . This complex is shellable with shelling order MATH according to CITE. As the next step we define the set MATH by MATH with MATH and MATH. We claim that MATH is the NAME ideal of MATH. Indeed, the definitions immediately imply MATH. But REF provides MATH and REF gives MATH . Thus we conclude that MATH. Hence REF shows that MATH is the MATH-vector of MATH, and REF gives MATH. Finally, we get MATH. |
math/0103197 | CASE: It is well known that the boundary complex MATH of MATH is NAME (compare for instance REF ). Moreover, NAME proves in CITE one direction of the MATH-theorem by showing that MATH in fact has the Weak NAME property. Since the NAME ideal MATH has codimension MATH, we conclude by REF . We now turn to REF . We will first describe the boundary complex following CITE and then verify that it has the required properties. Put MATH. We are looking for a polytope with vertex set MATH. Let MATH and MATH. Now we proceed in a manner similar to our proof of REF . We list the monomials in MATH in order, MATH. Using the bijection MATH, which sends MATH, we define MATH as the subset corresponding to MATH. Put MATH and define MATH where again, MATH is defined in REF . It follows immediately that MATH is a pure MATH-dimensional simplicial complex. Let MATH denote the pure MATH-dimensional simplicial complex on the vertex set MATH whose facets are the MATH-dimensional faces of MATH which are contained in exactly one facet of MATH. NAME and NAME have shown in REF (using the construction of REF) that MATH is indeed the boundary complex of a simplicial MATH-polytope. In order to complete the proof we relate the NAME ideals MATH and MATH to the ideals constructed in our REF, respectively. Put MATH and define the sets MATH and MATH by MATH and MATH . Thus, we have MATH and MATH. Using the notation of REF , we see that the ideals MATH and MATH have the same minimal generators. It follows that MATH . Therefore, REF shows that MATH which means in particular that MATH defines a scheme in MATH which is a cone over MATH. Using REF again, we observe that MATH is a minimal prime of MATH if and only if it is generated by MATH of the variables MATH and contains exactly one minimal prime ideal of MATH. Hence REF provides MATH. Since the graded NAME numbers of MATH were computed in REF , the proof is complete. |
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