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math/0101044 | Let MATH be the sum of positive semidefinite matrices. Then there exist fixed orthogonal matrices MATH and real numbers MATH such that MATH may be written as MATH . Then we have the differentiation formula (see, for example, REF): MATH where MATH is the adjunct matrix of MATH. Now, MATH where MATH is the elementary matrix with MATH in the MATH position and zeros elsewhere. Therefore, by cyclically permuting MATH in the trace above we find that MATH is the MATH the entry of MATH which is nonnegative since MATH is positive semidefinite. Lastly, if MATH is positive definite then MATH is also, which means that MATH is positive. The lemma follows. |
math/0101044 | By definition of MATH there exist MATH with CASE: MATH and MATH. CASE: MATH and MATH. CASE: MATH and MATH. Now MATH so that MATH and we are done. |
math/0101044 | First we show that MATH is itself a subspace hence equal to its span. Recognize that MATH is the union of all tangent spaces to flats which contain MATH. Pick a basis MATH of MATH note that MATH where MATH is the union of all the tangent spaces to flats containing MATH using the notation of CITE. REF states that MATH for some symmetric space of noncompact type and MATH. In particular it is a manifold and the tangent space to it corresponds to MATH, which is a vector space. Then MATH is a vector space. Let MATH be the stabilizer of MATH in MATH. Then MATH where MATH denotes the NAME group (a finite group). Hence MATH. Hence MATH . Since MATH we obtain MATH . Putting this together we obtain, MATH . But REF below gives that this final term is MATH, as desired. |
math/0101044 | For a root MATH in MATH, define MATH, where MATH is the NAME involution at MATH. Then by REF we have that MATH, MATH, and MATH. Note that from the definition of MATH it follows immediately that MATH . Note that in MATH the normalizer mod centralizer is finite for any flat subspace. Therefore for any MATH we may write the NAME algebra MATH of MATH as, MATH . It then follows from the previous statements that, MATH . Consequently, we may assume that MATH in the statement of the lemma is maximally singular: MATH may be written as the intersection of the kernels of the greatest number of roots among all subspaces of dimension MATH. Otherwise MATH is strictly smaller than it would be if MATH were maximally singular. Recall that we have the invariant inner product MATH on MATH and hence on MATH. Let MATH denote the collection of roots. For MATH, let MATH denote the dual root vector (with respect to MATH) corresponding to MATH. For any subset MATH we define the function MATH . Since root vectors lying in a subspace always come in opposing pairs, MATH is a positive integer. Let MATH be any root. Note that if a subspace MATH, then MATH lies in MATH. Therefore the statement of the lemma reduces to showing that MATH or more simply, MATH . Swapping MATH for MATH and vice versa, and using MATH for each MATH, it is sufficient to prove that MATH . Since we are assuming that MATH is simple, we could check this condition by using a classification of root vectors in the simple algebras such as in CITE. However, because this would be tedious we will instead give a synthetic proof. For each MATH, we say that MATH is a maximally rooted subspace of dimension MATH if MATH . In other words, MATH is maximally rooted if MATH is maximally singular. We claim that if MATH are any maximally rooted subspaces of MATH with MATH, then for MATH, MATH . This is true for MATH since MATH is one dimensional it contains a root vector pair and the trivial subspace MATH contains none. By induction, assume the claim holds for all maximally rooted subspace MATH of dimension MATH. In particular, for such a space MATH and for any subspace MATH of codimension one, MATH so MATH . We claim that there exists a root vector MATH which is not in MATH or its perpendicular MATH (with respect to MATH). If not, then every root vector either lies in MATH or MATH which implies the root system is reducible (for example, REF), and hence MATH is reducible, contrary to assumption. Therefore, MATH is a codimension one subspace of MATH and by inductive hypothesis there are at least MATH distinct pairs of root vectors MATH in MATH. For each of these we have MATH. By the standard calculus of roots (for example, REF) this implies that for each MATH either MATH or MATH is a pair of root vectors lying in MATH which does not lie in MATH. Including MATH, these form at least MATH pairs of root vectors which are contained in MATH. Therefore MATH. Since by definition of MATH, MATH, the claim follows. Recursively applying REF shows that for MATH, MATH . Now to prove REF , as noted before we may assume MATH of dimension MATH is maximally rooted, since then MATH is maximally singular. Since MATH is a maximally rooted space, the above expression reads MATH . This is readily seen to be greater that MATH unless MATH and MATH (MATH). However, every irreducible NAME algebras of rank two other than MATH has at least four pairs of roots (see CITE, p. REF ), and hence REF is satisfied in all of the required cases. |
math/0101044 | For any subspace MATH, let MATH be the maximally singular subspaces of dimension MATH which have minimal angle with MATH. Define MATH. If MATH denotes the NAME variety of subspaces in MATH with dimension MATH, then the set of MATH for which MATH is constant has codimension MATH in MATH. For any subspace MATH we define a subspace MATH by MATH where MATH is the subgroup of MATH which stabilizes MATH. By REF , MATH has dimension at least MATH since we always have MATH for some MATH with MATH. If no such constant MATH as in the lemma exists then there is a sequence MATH and MATH with MATH such that MATH . Now since MATH and hence MATH varies upper semicontinuously in MATH (thinking of the map MATH as a self-map of MATH), it follows from the continuity of the MATH function that MATH is lower semicontinuous in MATH. However since both MATH and MATH are compact, for some subsequence of the MATH, the MATH converge to MATH and the MATH converge to a fixed subspace MATH. Furthermore, MATH lies in MATH since MATH must be MATH. It follows that MATH where MATH is the NAME group stabilizing MATH. By construction, MATH and for any MATH, MATH . Therefore, we also have MATH. Continuity of MATH along with the fact that MATH acts isometrically implies that it is sufficient to show that for any fixed subspace MATH the quantity MATH is bounded away from MATH. Note that since this quantity is lower semicontinuous in MATH, and since MATH is compact, it is unnecessary to show that the bound is independent of MATH. First we handle the denominator. Using the bi-invariance of the metric on MATH, the properties of the angle function, and the fact that for all MATH we have MATH, it follows that MATH . So it remains to show that for any sequence MATH in any fixed neighborhood MATH of MATH, that MATH. Furthermore, since MATH for any MATH, we may assume that MATH. By REF, in a sufficiently small neighborhood of MATH we may uniquely write MATH as MATH where MATH and MATH. Furthermore MATH and MATH. NAME of the metric on MATH implies that for MATH . Now MATH is the only subgroup of MATH which both leaves MATH in MATH and also intersects all sufficiently small neighborhoods of the identity. Therefore, in order to show that MATH, we need only show that MATH . Well, the NAME formula implies that MATH . Since the definition of MATH implies that MATH and MATH is perpendicular to MATH, we have MATH . Since we had MATH this finishes the lemma. |
math/0101044 | The first step of the proof is to reduce to the case when MATH is a subspace of MATH, so that REF may be applied. We first observe that the lemma is true if and only if it is true with MATH replaced by MATH for any fixed MATH. Since MATH is compact we may therefore choose MATH among all MATH so that MATH for all MATH. With this assumption, consider the projection MATH of MATH onto MATH. By REF , we obtain a subspace MATH such that MATH for all MATH. Then we let MATH be the projection of MATH onto MATH. By the properties of the angle function (see REF), it follows that MATH . Thus it suffices to bound MATH by a constant times MATH. But MATH and we are done. |
math/0101044 | If MATH then MATH. However, MATH may be expressed as MATH where MATH is the unit vector in MATH pointing to MATH. Then we have MATH . The strictness of the inequality finishes the proof. |
math/0101044 | Since the isometry group of the symmetric space MATH is transitive on MATH and for any isometry MATH, MATH, it is sufficient to prove the lemma for a fixed MATH and all MATH. For now choose MATH. Take a monotone sequence MATH, and any choice MATH for each MATH. It follows that some subsequence of the MATH, which we again denote by MATH, must tend to some point MATH. Let MATH be the weak limit of the measures MATH. From REF , MATH is a probability measure supported on a set MATH satisfying MATH . Therefore we have, MATH . Now whenever MATH then MATH for some unit vector MATH and MATH. Using either case above we may write MATH . So choosing MATH small enough we can guarantee that CASE: any two NAME chambers intersecting MATH for all MATH in the same flat must share a common face of dimension MATH, and CASE: for any MATH, MATH . Let MATH . By the first property used in the choice of MATH above, for any two points MATH, either MATH and MATH are in the boundary of the same NAME chamber, or else there is another point MATH in the intersection of the boundaries at infinity of the closures of the respective NAME chambers. By maximality there is some MATH intersecting the boundary at infinity of the closure of every NAME chamber which intersects MATH for all MATH. Hence, for every MATH, the support of the limit measure MATH satisfies MATH. (While MATH is not necessarily unique, the support MATH of the corresponding limit measure MATH is.) As MATH increases, for any MATH, the measures MATH uniformly become increasingly concentrated on MATH. Then applying the estimate REF to MATH, we may choose MATH sufficiently large so that for all MATH with MATH, MATH . |
math/0101044 | By way of contradiction, let MATH be an unbounded sequence such that MATH lies in a compact set MATH. We may pass to an unbounded subsequence of MATH, which we again denote as MATH, such that the sequence MATH converges within a fundamental domain for MATH in MATH to a point MATH. Since MATH is compact, the set MATH contains an open neighborhood of MATH and MATH. Notice that MATH is itself a cone, being the intersection of cones on a nonempty subset of MATH. We now show that MATH contains the image MATH of increasingly large balls (MATH). However, we observe from the fact that MATH is a cone on an open neighborhood of MATH in MATH that MATH contains balls MATH with MATH. By REF, and hence MATH, is coarsely NAME: MATH for some constants MATH and MATH. Therefore MATH where MATH. In particular MATH. Hence, there exists an unbounded sequence MATH such that MATH. Furthermore, since the NAME curvature is assumed to be bounded from above and the injectivity radius from below, we have that MATH is greater than some constant independent of MATH and hence MATH for some constant MATH. By choice of MATH there is a constant MATH depending only on MATH such that MATH for all MATH. In polar coordinates we may write, MATH . Using this we may estimate, using any MATH, MATH . Therefore for all sufficiently large MATH, MATH . The constant MATH is the constant MATH from REF such that MATH. Set MATH. Recalling that MATH for all MATH, we have that for sufficiently large MATH, MATH but MATH, contradicting the conclusion of REF in light of REF . |
math/0101045 | Let MATH, for an appropriate sequence MATH. Recall that MATH. Then combining REF with REF have that some subsequence of the MATH converges to a MATH-invariant measure MATH supported on MATH. Note that in CITE, the notation MATH refers to a subset of a fundamental set of roots corresponding to the face of a NAME chamber containing MATH in its boundary. If MATH converges then both MATH and MATH exist (note the definition of MATH in CITE). Again in the notation of CITE, MATH is the conjugate subgroup MATH in MATH. Moreover, MATH is the orbit MATH. By REF, any other convergent subsequence of the MATH produces the same measure in the limit, and therefore the sequence MATH itself converges to MATH uniquely. |
math/0101045 | For the strict convexity, it suffices to show that given a geodesic segment MATH between two points MATH, there exists some MATH such that function MATH is strictly convex in MATH, and hence on an open positive MATH-measure set around MATH. We know it is convex by the comment preceding the statement of the proposition. If MATH is constant on some geodesic subsegment of MATH for some MATH, then MATH must lie in some flat MATH such that the geodesic between MATH and MATH (which meets MATH at a right angle) also lies in MATH. On the other hand, MATH is in the direction of the algebraic centroid in a NAME chamber, and MATH is perpendicular to this direction. By the properties of the roots, MATH is a regular geodesic (not contained in the boundary of a NAME chamber). In particular, MATH is contained in exactly one flat MATH. Furthermore, MATH is a finite set (an orbit of the NAME group). As a result, for almost every MATH is strictly convex in MATH. For fixed MATH, by the last property listed in REF , we see that MATH tends to MATH as MATH tends to any boundary point MATH. Then for fixed MATH and MATH, MATH increases to MATH as MATH tends to any boundary point MATH. Hence it has a local minimum in MATH, which by strict convexity must be unique. |
math/0101045 | REF is clear; we prove REF . By the hypothesis on MATH, the group MATH can be written as a product MATH, where each MATH is a simple rank one NAME group. REF states that there exists MATH so that for all MATH, the measure MATH is supported on some MATH-orbit MATH . Hence MATH . Since each MATH has rank one, MATH so that MATH . Let MATH denote the NAME function for the rank one symmetric space MATH with metric MATH. Then for MATH and MATH we have MATH. Since the factors MATH are orthogonal in MATH with respect to the metric MATH, the NAME function of MATH with basepoint MATH at a point MATH is given by MATH . Since MATH, we may verify that MATH . Similarly, MATH . Differentiating again with respect to a fixed orthonormal basis of MATH with respect to MATH we obtain, MATH . Since MATH is supported on MATH, we can use product coordinates on MATH to write the right hand side of REF as, MATH where the superscript MATH means transpose and we have cancelled the factor MATH from the numerator and denominator. Recall the following estimate for positive semi-definite block matrices, MATH . Then since the matrix MATH is positive semi-definite, by iteratively using the above estimate (on sub-blocks) we obtain MATH . Also, we have MATH . Hence we have MATH . Since MATH is the NAME function of rank one symmetric space MATH, it follows that, setting the tensor MATH, we have MATH where the MATH are the matrices representing the underlying complex structure of (the division algebra corresponding to) the symmetric space. REF says that for any MATH matrix MATH with MATH, the following holds: MATH with equality if and only if MATH. Applying this estimate to each term of the above product now gives MATH with equality if and only if MATH. Hence MATH . If equality is attained then we have equality for each term. Let MATH be the subspace which is mapped by MATH onto MATH. So for any MATH and any MATH we have MATH . Recall that MATH . So multiplying each side by MATH we write the above with respect to the metric MATH: MATH . Hence for all MATH we obtain MATH . Now we follow REF. It follows that for any orthonormal basis MATH of MATH we have MATH . Lastly it follows that MATH . The equality implies that MATH . Hence MATH is a homothety of ratio MATH as desired. |
math/0101045 | From its definitions, MATH is continuous in MATH and MATH. Observe that for fixed MATH, MATH. It follows that MATH. This implies the proposition. |
math/0101045 | If MATH then MATH. However, MATH may be expressed as MATH where MATH is the unit vector in MATH pointing to MATH. Then we have MATH . The strictness of the inequality finishes the proof. |
math/0101045 | Since the isometry group of the symmetric space MATH is transitive on MATH and for any isometry MATH, MATH, it is sufficient to prove the lemma for a fixed MATH and all MATH. For now choose MATH. Take a monotone sequence MATH, and any choice MATH for each MATH. It follows that some subsequence of the MATH, which we again denote by MATH, must tend to some point MATH. Let MATH be the weak limit of the measures MATH. From REF , MATH is a probability measure supported on a set MATH satisfying MATH . Therefore we have, MATH . Now whenever MATH then MATH for some unit vector MATH and MATH. Using either case above we may write MATH . So choosing MATH small enough we can guarantee that CASE: any two NAME chambers intersecting MATH for all MATH in the same flat must share a common face of dimension MATH, and CASE: for any MATH, MATH . Let MATH . By the first property used in the choice of MATH above, for any two points MATH, either MATH and MATH are in the boundary of the same NAME chamber, or else there is another point MATH in the intersection of the boundaries at infinity of the closures of the respective NAME chambers. By maximality there is some MATH intersecting the boundary at infinity of the closure of every NAME chamber which intersects MATH for all MATH. Hence, for every MATH, the support of the limit measure MATH satisfies MATH. (While MATH is not necessarily unique, the support MATH of the corresponding limit measure MATH is.) As MATH increases, for any MATH, the measures MATH uniformly become increasingly concentrated on MATH. Then applying the estimate REF to MATH, we may choose MATH sufficiently large so that for all MATH with MATH, MATH . |
math/0101045 | By way of contradiction, let MATH be an unbounded sequence such that MATH lies in a compact set MATH. We may pass to an unbounded subsequence of MATH, which we again denote as MATH, such that the sequence MATH converges within a fundamental domain for MATH in MATH to a point MATH. Since MATH is compact, the set MATH contains an open neighborhood of MATH and MATH. Notice that MATH is itself a cone, being the intersection of cones on a nonempty subset of MATH. We now show that MATH contains the image MATH of increasingly large balls (MATH). However, we observe from the fact that MATH is a cone on an open neighborhood of MATH in MATH that MATH contains balls MATH with MATH. By REF, and hence MATH, is coarsely NAME: MATH for some constants MATH and MATH. Therefore MATH where MATH. In particular MATH. Hence, there exists an unbounded sequence MATH such that MATH. Furthermore, since the NAME curvature is assumed to be bounded from above and the injectivity radius from below, we have that MATH is greater than some constant independent of MATH and hence MATH for some constant MATH. By choice of MATH there is a constant MATH depending only on MATH such that MATH for all MATH. In polar coordinates we may write, MATH . Using this we may estimate, using any MATH, MATH . Therefore for all sufficiently large MATH, MATH . The constant MATH is the constant MATH from REF such that MATH. Set MATH. Recalling that MATH for all MATH, we have that for sufficiently large MATH, MATH but MATH, contradicting the conclusion of REF in light of REF . |
math/0101045 | By the previous lemma, MATH is a contracting NAME map. We note that the local notion of MATH given by MATH is well defined for a.e. MATH. Let MATH be the set of points which have unbounded preimage under MATH. The set MATH is clearly closed and of measure REF, and MATH acts properly on MATH. Since MATH is homotopic to MATH, MATH for a.e. MATH in one connected component MATH of MATH, and MATH a.e. on the other components. We note that REF only require that the injectivity radius be bounded for any finite set of points. The proof of these two lemmas show that for every MATH, MATH. Now we will show that MATH is proper on the closure of MATH. This implies that MATH. If the map MATH were not proper then there would be a sequence MATH tending down an end of MATH such that MATH limits to a point MATH. After pasing to a subsequence (also denoted MATH) we may find compact rectifiable curves of finite length which pass through all of the MATH and MATH. For any such MATH, by continuity, the pre-image MATH is therefore contained in at most MATH curves MATH one of which (say MATH) can be chosen to pass through the MATH. By possibly slightly perturbing the points MATH, NAME 's theorem guarantees that we can choose a curve MATH such that the derivatives of MATH on the pre-image curves MATH are a.e. equal to one. On the other hand, curves MATH are NAME since MATH is and therefore by the fundamental theorem of calculus it must have the same length as MATH which is finite. This contradicts that the MATH are unbounded. |
math/0101046 | It follows from the dynamical twist identity REF that MATH which implies coassociativity of MATH. The counit axiom is obvious. |
math/0101046 | Suppose MATH is not semisimple and let MATH be the maximal non-zero power of its radical. Then MATH is MATH-stable and for all MATH and MATH we have MATH. The last condition is equivalent to MATH for all MATH. In particular, we have MATH but since MATH is invertible in MATH, the elements MATH span MATH, so that MATH, which is a contradiction. |
math/0101046 | Since MATH as a MATH-module, the space of MATH-invariant elements of MATH has dimension MATH. For any orbit of MATH in MATH the corresponding central idempotent of MATH is MATH-invariant, so there is a single orbit. |
math/0101046 | First, we check that MATH is a left MATH-module and right MATH-module. For all MATH we have, using the definition of MATH and dynamical twist REF : MATH . Also, for all MATH we have : MATH . It is immediate from REF that the actions of MATH and MATH commute. To prove that the action of MATH is faithful suppose that MATH for some MATH and all MATH. Then MATH being invertible implies that MATH is identically equal to MATH on MATH, so that MATH. The proof for the other action is completely similar. |
math/0101046 | By REF all simple MATH-modules have the same dimension, which we will denote MATH. Suppose MATH for some MATH. Let MATH be the number of non-isomorphic simple MATH-modules and MATH be the multiplicities with which they occur in the decomposition of MATH. By REF MATH is isomorphic to a subalgebra of the centralizer MATH of MATH in MATH. But MATH, therefore MATH for all MATH. Then we have MATH which is a contradiction. |
math/0101046 | In the proof of REF we must have MATH, that is, every simple MATH-module has the multiplicity equal to its dimension. |
math/0101046 | To show that each MATH is well-defined, we compute MATH for all MATH. The third equality above uses REF . It is clear that MATH is a MATH-module homomorphism. To see that it is invertible we first observe that from REF we have MATH, so that the map in question is between spaces of the same dimension. So it suffices to show that it is surjective. But the range of MATH clearly coincides with the range of the map given by the composition of the right translation by MATH (which is surjective since MATH is invertible) and the usual multiplication map MATH (which is surjective because there are invertible functions in each MATH). |
math/0101046 | From REF we have MATH for any MATH-module MATH (regarded as a MATH-module on which the kernel MATH acts trivially), where MATH is a weight subspace of MATH defined in REF. Taking the trivial MATH-module MATH we get MATH. |
math/0101046 | Observe that a decomposition of MATH into the sum of simple MATH-bimodules establishes an isomorphisms of MATH-sets MATH. But the relative tensor product of bimodules MATH and MATH is isomorphic to MATH by REF , whence the composition rule of MATH's follows. |
math/0101046 | We need to show that the space of MATH-invariant elements of MATH has dimension MATH. Any such an element injectively corresponds to a MATH-invariant element of MATH, but the latter is equivalent to MATH as a left MATH-module, for which the space of invariants is MATH-dimensional. |
math/0101046 | Given a dynamical twist MATH fix a point MATH and a subgroup MATH as above. Consider the corresponding algebras MATH and MATH bimodules MATH and fix simple MATH-modules MATH. Let MATH be the isomorphisms identifying MATH-algebras and choose bimodule isomorphisms MATH. Let MATH denote the action of MATH on MATH by shifts, compare REF . For every MATH choose an irreducible projective representation MATH such that MATH . Then for all MATH we have MATH that is, MATH commutes with the action of MATH on MATH. Therefore, by NAME 's Lemma, MATH for some functions MATH. By REF , MATH as MATH-equivariant bimodules, whence MATH for all MATH. Replacing MATH by a projectively isomorphic representation MATH, where MATH, we obtain a collection of irreducible projective representations MATH of MATH such that MATH is linear and induces MATH, that is, a dynamical datum for MATH. Let us show that gauge equivalent twists define isomorphic dynamical data, for any choice of projective representations MATH. This will imply, in particular, that the isomorphism class of a dynamical datum we have constructed above is well defined. Let MATH be a dynamical twist in MATH gauge equivalent to MATH via a gauge transformation MATH. By REF the map MATH defines an isomorphism between the corresponding MATH-algebras MATH and MATH. This establishes a bijective correspondence between minimal ideals of MATH and MATH of these algebras and allows to identify their stabilizers, so we can assume that both MATH and MATH define the same subgroup MATH. Let MATH be the corresponding MATH bimodules. Fix vector spaces MATH and identify MATH-algebras MATH and MATH. Choose isomorphisms MATH of bimodules, and projective representations MATH such that MATH where MATH denotes the action of MATH on MATH. Arguing as above, we get functions MATH and MATH such that MATH and MATH (where MATH) make MATH linear. Isomorphisms between simple MATH-algebras MATH and MATH yield projective isomorphisms MATH (defined up to scaling) between MATH and MATH, MATH: MATH for some multipliers MATH. Comparing REF and using REF we conclude that MATH . We have MATH, where MATH and MATH so that all multipliers MATH are equal, that is, dynamical data constructed from MATH and MATH are isomorphic. |
math/0101046 | This is clear since MATH is simple if and only if MATH. |
math/0101046 | For all MATH-modules MATH and elements MATH, MATH, and MATH we have : MATH since MATH maps MATH to MATH. |
math/0101046 | We need to show that MATH is surjective for all MATH, where MATH are MATH-modules. Equivalently, we need to prove that the composition map MATH is surjective. But this follows from the fact that MATH as MATH-modules. Indeed, MATH by REF and MATH are mutually non-equivalent by REF . Therefore, a copy of MATH is contained in MATH, and has the same dimension. |
math/0101046 | We have already seen that MATH is an invertible zero weight function, we need to show that it satisfies the twist properties of REF . Let MATH be MATH-modules and MATH. Consider the composition of MATH-module homomorphisms MATH defined by MATH. The associativity law gives two different ways of factorizing it: MATH whence we have MATH, that is, MATH for all MATH. The counital properties of MATH are clear since MATH. In proving the gauge equivalence of dynamical twists coming from isomorphic data we may assume that MATH. Let MATH be an isomorphism between two dynamical data, MATH be associated homomorphisms of modules, and MATH the dynamical twists. Define MATH by the identity MATH . Note that MATH is well defined since all MATH have the same multiplier function, compare REF . Then MATH is a zero weight function taking invertible values and satisfying MATH since MATH. For any MATH we compute MATH therefore MATH, and MATH that is, MATH is a gauge transformation of MATH. |
math/0101046 | Let MATH be a dynamical twist, MATH be (a representative of the class of) dynamical data associated to MATH, and MATH be a dynamical twist coming from the exchange construction for MATH. According to REF (compare REF ) both MATH and MATH are determined by the map MATH corresponding to the composition MATH . Comparing two MATH-module isomorphisms MATH (the one established in REF , corresponding to the construction of MATH from MATH, and the one used in the exchange construction above) we get a zero weight function with invertible values MATH that implements the corresponding MATH-module automorphism MATH for all MATH, and is such that MATH for all MATH. Therefore MATH and MATH are gauge equivalent, that is, MATH, in particular MATH is surjective. Let us show that MATH is also injective. Let MATH and MATH be two sets of dynamical data that produce dynamical twists MATH and MATH gauge equivalent to each other. By REF their MATH-algebras MATH and MATH are isomorphic via MATH . Also, for the corresponding bimodules MATH and MATH it follows from the exchange construction of a dynamical twist in the beginning of this section that there are canonical isomorphisms of MATH-modules MATH where MATH (respectively, MATH) acts by right translations on MATH and in a standard way on the MATH space, whereas MATH acts on MATH by the left translation by MATH and trivially on the first factor. For MATH this implies that all the matrix blocks of MATH (respectively, MATH) are canonicaly isomorphic to MATH (respectively, MATH) and form a MATH-homogeneous space isomorphic to MATH (respectively, MATH). Therefore MATH is conjugate to MATH and we can assume MATH. Thus, we have a canonical MATH-algebra isomorphism MATH and so there is a projective isomorphism MATH such that MATH. Note that MATH is defined up to a scalar, therefore its multiplier function MATH, MATH is uniquely defined. The gauge transformation MATH also defines isomorphisms of simple MATH-equivariant bimodules : MATH where MATH, in particular, MATH. As it was observed above, isomorphisms MATH and MATH of equivariant MATH-bimodules are also canonical. By NAME 's Lemma, for fixed MATH we have MATH for some constant MATH, whence replacing MATH by MATH we get MATH, that is, the system MATH gives an isomorphism between the dynamical data in question. |
math/0101046 | It is clear from the above definitions that MATH are isotropic subspaces for MATH. Let us show that they are maximal. The dimension MATH of a maximal isotropic subspace of MATH is given by MATH where MATH is the orbit of the coadjoint action of MATH containing MATH. We have MATH, where MATH is the stabilizer of MATH. Clearly, MATH. In order to conclude that MATH it suffices to show that MATH. Note that MATH is compatible with the NAME decomposition of MATH, so it is enough to prove that MATH. Let MATH, where MATH with MATH. Let MATH, where MATH is such that MATH. Then MATH, a contradiction that shows MATH. One shows that MATH in a similar way. |
math/0101046 | Let MATH, MATH be a MATH-module, and MATH denote a MATH-dimensional MATH-module determined by MATH, then we have MATH where the second isomorphism is a consequence of the NAME reciprocity and the third uses that MATH (which holds since MATH, MATH, and MATH). Therefore, MATH, as required. |
math/0101048 | According to REF MATH is given by MATH where MATH is a pair of dual bases of MATH and MATH consisting of weight vectors MATH . REF of MATH implies MATH . The map MATH acts on the matrix coefficients of a MATH-module by restricting the module to MATH . Since MATH and MATH . If MATH is a dominant weight, then MATH . Hence MATH is a lowest weight vector for the MATH-submodule of MATH generated by MATH . The corresponding MATH-highest weight vector is MATH and MATH . The lemma now follows from REF. |
math/0101048 | CASE: By a straightforward computation, for all MATH . CASE: Combining REF with the identities REF gives MATH which implies REF in view of REF. |
math/0101048 | The operator MATH in MATH belongs to MATH if and only if MATH because of REF, and the the selfadjointness of MATH . The operator MATH is diagonal in the orthonormal basis REF of MATH and according to REF acts by MATH recall the notation REF. It is clear that the linear operator MATH in MATH is trace class if and only if MATH for MATH which implies the statement because of REF. |
math/0101050 | REF follows easily from REF on p. REF. REF follows readily from REF on p. CASE: In order to prove REF , recall (REF on pp. REF) that each proper projective representation of MATH in characteristic MATH has dimension divisible by MATH where MATH is the exact number of terms in the dyadic expansion of MATH. We have MATH where MATH's are distinct nonnegative integers. In particular, if MATH then MATH and MATH. This proves REF if MATH. If MATH then MATH if MATH. This proves REF in the case of MATH. The remaining case MATH follows from Tables in CITE. Further we assume that MATH. In particular, MATH and MATH is not a power of MATH. Since MATH is even, all MATH and MATH. If MATH then MATH is not divisible by MATH. Therefore if MATH then REF holds. On the other hand, if MATH then MATH, that is, MATH which implies that MATH. We get a contradiction which, in turn, implies that MATH and therefore MATH. This implies easily that MATH. In particular, MATH and MATH. Since MATH is not a power of MATH, there are no proper projective representations of MATH of dimension MATH in characteristic MATH if MATH . Clearly, this inequality holds if MATH. But this inequalty is equivalent to MATH which holds for all MATH. |
math/0101062 | We can assume that MATH and MATH with MATH, MATH. So we have to prove MATH . Let MATH be a symplectic basis of MATH and MATH the corresponding dual basis of MATH, that is, MATH and MATH. It follows that MATH. It is MATH . Since MATH spans MATH and MATH, this proves the proposition. |
math/0101062 | REF should be clear from the definitions. Let us investigate REF a bit more. We can assume that MATH and MATH are NAME diagrams with MATH respectively, MATH univalent vertices and MATH with MATH. So we have to prove MATH since MATH means to remove the components with at least one univalent vertex. Recalling the meaning of MATH, it should be clear that REF follows from the fact that applying MATH on a NAME diagram means to glue MATH pairs of its univalent vertices in all possible ways. |
math/0101062 | First, we show that MATH is well-defined. For every MATH, let MATH be the subspace of MATH spanned by all MATH with MATH. It is enough to show that for all MATH there exists a injective map MATH that fulfills REF. Further, we can restrict ourselves to the case of even MATH due to REF, which says that MATH is homologous to zero for odd MATH. (This follows at once from the anti-symmetry relation.) For any MATH, the map MATH defined in REF can be taken as MATH when restricted to MATH. This is because of REF, where it is shown that MATH evaluated at MATH equals the right hand side of REF. It remains to show that MATH is injective. This can be proven by a dimension argument: By REF the image of MATH under MATH has at least the dimension of MATH, so MATH is injective. |
math/0101062 | In MATH, we calculate MATH . This proves the lemma because there is a well-defined MATH-linear map MATH which is injective. |
math/0101062 | Since MATH is a linear operator of degree MATH on MATH and MATH, we can assume that MATH. Set MATH. We have MATH so we see that MATH is an eigenvector. We just have to calculate the eigenvalue: MATH . Applying the (injectiv) map MATH of REF and using REF yields MATH . Because of the injectivity of MATH, this proves the theorem. |
math/0101062 | As remarked in CITE by NAME, a NAME representative of the NAME class MATH on every NAME manifold MATH is related to the characteristic classes MATH of MATH in the following way: NAME representative of MATH is given by MATH. Here, MATH means taking the MATH-th product of MATH viewed as MATH-form, giving an element of MATH, and then using the associative algebra structure of MATH. Further, MATH means taking the trace on MATH, and MATH is induced by the canonical projection MATH. That this procedure on symplectic NAME manifolds is essentially the same as taking the NAME class of a wheel follows from REF. |
math/0101062 | The statements follow from REF. |
math/0101062 | Due to the irreducibility of MATH, that is, MATH for all MATH, we can write MATH for all MATH. Using this, we have MATH which proves the proposition because of REF . |
math/0101062 | Set MATH in REF. To obtain MATH, compute both sides with MATH, which proves the first equation. Then set MATH in REF. To obtain MATH compute both sides with MATH, which leads to the second equation. |
math/0101062 | Note that MATH. Then MATH . |
math/0101062 | We can assume that MATH. If follows that MATH with MATH. By definition of the NAME polynomials, we have MATH . |
math/0101062 | If MATH, then MATH or MATH. In both cases, MATH due to REF . Because of REF and the usual NAME formula, the assertion follows. |
math/0101062 | This is an immediate application of REF . |
math/0101062 | We can assume that MATH. Since both sides of REF are rational functions in the cohomology class of MATH and the equality holds for every irreducible symplectic complex manifold, by a standard argument of hyperkähler geometry, REF remains true, if we substitute MATH by an arbitrary cohomology class in MATH. Here, we have to substitute MATH by MATH. |
math/0101062 | Using the lemma above the theorem is proven if we show that MATH. By definition of MATH and MATH we get MATH for every MATH. Now HRR in its classical form - using MATH - gives MATH whereas our formula in REF reads MATH . Comparing coefficients, we find MATH . Since both MATH and MATH are positive multiples of MATH and the coefficients of the HRR formula as a polynomial in the quadratic form are independent of the line bundle considered, the theorem follows. |
math/0101062 | Let MATH. One has MATH and since the action of the symmetric group does not flip the factors MATH and MATH, and for an arbitrary line bundle MATH on MATH we have MATH, we get MATH . Now, if MATH is an ample invertible sheaf, MATH is also ample and so is MATH, because MATH is a finite surjective morphism and MATH is ample. Let MATH be large enough that both MATH and MATH have no higher cohomology such that MATH. For the global sections of an symmetrized line bundle MATH, one has the isomorphism MATH . Now replacing MATH by MATH, we find MATH . Since MATH is a polynomial in MATH, evaluation in MATH proves the lemma. |
math/0101062 | The splitting of the sheaf MATH follows from the seesaw principle (compare CITE): For fixed MATH we have seen that the restricted morphism MATH is the isomorphism wich maps MATH to the fiber of the summation morphism MATH over the point MATH. Since MATH is simply connected, its NAME group is discrete and it follows that MATH. Therefore MATH is of the form MATH with MATH, and we can compute the component MATH by considering the restrictions of MATH to MATH. Now complete REF as follows MATH . Here MATH denotes the ``singular NAME, the fiber over REF of the addition morphism MATH, MATH its desingularization, and MATH is defined analogously to MATH. Now consider a point MATH over MATH. Instead of computing the sheaf MATH we equivalently compute MATH. |
math/0101062 | The splitting of MATH is a well known fact, compare REF. The triviality of their first rational NAME classes follows from the relation: MATH . It follows that MATH, which proves the lemma. |
math/0101062 | We start again with a line bundle MATH on MATH that we twist with a sufficiently ample bundle MATH. By abuse of notation we denote the resulting bundle by MATH again. By construction the symmetrized bundle MATH is still ample such that the invertible sheaf MATH on MATH as a pull back along the birational morphism MATH is nef and big. The same argument shows that the line bundle MATH is nef and big: Using the notations of REF the bundle MATH is ample and MATH is big and nef. So by the NAME vanishing theorem (compare CITE), we have MATH . On the one hand, we have due to the NAME formula MATH . On the other hand, since MATH is finite and MATH is a flat morphism, we can compute MATH alternatively MATH . This shows that MATH with MATH. Since the sheaf MATH is still nef and big, REF and NAME implies that MATH equals the NAME characteristic of this line bundle. Therefore, using the classical NAME theorem on MATH we have MATH . Now combining the computations - and noting that MATH - we find MATH . Once again - considering the formula as a polynomial in MATH and evaluating in MATH - the formula holds for a general line bundle MATH. |
math/0101064 | For the details we refer to CITE, we only remark that MATH can be established from the separability of MATH as follows. Apply MATH to REF , for an arbitrary MATH write MATH, and use that MATH is multiplicative to obtain MATH . We also note that in CITE REF are not required for a MATH-bialgebroid. However it can be easily seen that MATH as defined above satisfies REF . |
math/0101064 | CASE: Let MATH be a left MATH-comodule algebra. We will show that MATH is then a left comodule algebra over the MATH-bialgebroid MATH. Note that MATH where MATH is a weak counit, is a well-defined algebra map. Indeed, suppose MATH. This means that MATH, and therefore MATH . Note that the second equality was obtained by using the following observation made in CITE. The unit property of a coaction of a weak comodule algebra in REF is equivalent to MATH . This proves that MATH is well-defined. To prove that MATH is an algebra map we require the following two equalities (compare CITE and CITE respectively). For all MATH and MATH, MATH . Now for all MATH, MATH we have MATH where MATH denotes another copy of MATH, and CITE has been used to derive the second equality and the unit property in REF to obtain the fourth one. This proves that MATH is an algebra map and hence MATH is a MATH-ring. Next, using the canonical projection MATH define a map MATH, MATH, where MATH is the left coaction of a weak NAME algebra. Explicitly, MATH. The map MATH is left MATH-linear since using REF and the fact that MATH is a comodule algebra we have for all MATH and MATH . Then it is clear that MATH. We prove now that MATH. REF implies that for all MATH and MATH . Apply MATH to the last equality to obtain MATH . Using REF which, put together, state that for all MATH, MATH we compute MATH . Hence we have proved that MATH . In particular, MATH as required. It remains to be proven that MATH. Using the unit property of a comodule algebra of a weak NAME algebra in REF we compute MATH . Since MATH is a left MATH-module via MATH we obtain MATH . This completes the proof that MATH is a left MATH-comodule algebra over the MATH-bialgebroid MATH. If MATH is a morphism of left MATH-comodule algebras, then MATH is also a morphism of left MATH-comodule algebras over the MATH-bialgebroid MATH. In view of the definition of MATH, the left MATH-colinearity is obvious. We also know that MATH is an algebra map, so MATH is a map of MATH-rings if MATH. Using the MATH-colinearity of MATH and the fact that MATH, we find MATH REF Conversely, let MATH be a left comodule algebra over the bialgebroid MATH as in REF . We prove that MATH is a weak left MATH-comodule algebra with coaction given by MATH. Explicitly MATH where we used that MATH is a right MATH-module via the target map MATH. The fact that MATH is a left comodule of a weak NAME algebra MATH can easily be established with the help of REF . We prove now that MATH is an algebra map. First, since MATH we have for all MATH and MATH, MATH . Applying the section MATH we obtain MATH . On the other hand, application of MATH to an expression reflecting the fact that MATH is an algebra map leads to equality MATH . Noting that MATH and writing MATH for another copy of MATH we have MATH that is, MATH is multiplicative as required. It remains to prove the unit property of the weak coaction MATH. Since MATH we obtain that MATH hence the unit property of MATH is equivalent to the following equation MATH . It is known, however, that MATH for MATH is a comodule algebra of a MATH-bialgebroid. Application of MATH yields MATH . Next apply MATH to the preceding equality to obtain MATH . Now the required condition follows from these two equations, and REF . Thus we conclude that MATH is a left weak MATH-comodule algebra. Suppose now that MATH is a morphism of left MATH-comodule algebras over the MATH-bialgebroid MATH. Then MATH is an algebra map, and MATH . Applying MATH to both sides, we obtain MATH . MATH is a map of left MATH-modules, so the left hand side equals MATH, and this means that MATH is left MATH-colinear, and MATH is a morphism of left MATH-comodule algebras. CASE: We still need to show that the functors constructed in REF of the proof are inverses to each other. First, let MATH be a left weak MATH-comodule algebra. It is first transformed into a left MATH-comodule algebra MATH over the bialgebroid MATH, and then into a left weak MATH-comodule algebra MATH. We easily compute that MATH as needed. We used REF . Conversely, we start with a left MATH-comodule algebra MATH over the bialgebroid MATH, transform it into a weak left MATH-comodule algebra MATH using REF , and then back into MATH over MATH. We have to show that MATH and MATH. We write MATH, and then easily find that MATH . Using REF , we obtain MATH . |
math/0101064 | CASE: Let MATH be a left weak MATH-module coalgebra and define MATH, MATH and MATH, MATH. We will show that these maps make MATH into a left module coalgebra of the MATH-bialgebroid MATH. Since MATH is left MATH-linear, so is MATH. This implies that MATH is a MATH-bimodule map. Next we prove that MATH is left MATH-linear. First note that setting MATH in REF one immediately obtains MATH for all MATH, MATH. In particular MATH where MATH is another copy of MATH and MATH is the counit of MATH as a MATH-bialgebroid. This implies that MATH is a MATH-bimodule map. Clearly, MATH is a coproduct and MATH is a counit of a MATH-coring MATH. The compatibility of MATH with the left action of MATH on MATH follows immediately from the fact that MATH is a weak module coalgebra. Thus we conclude that MATH is a left module coalgebra over the MATH-bialgebroid MATH as claimed. Let MATH be a morphism of left weak MATH-module coalgebras. Then MATH is left MATH-linear, and it clearly preserves the comultiplication over MATH. Furthermore MATH-and we conclude that MATH is a morphism of left MATH-module coalgebras over the MATH-bialgebroid MATH. CASE: Conversely, assume that MATH is a left module coalgebra over the MATH-bialgebroid MATH, in the sense of REF . We claim that MATH is a left weak MATH-module coalgebra with the coproduct MATH (note that here MATH while MATH), and the counit MATH, MATH, where MATH is the counit of the MATH-coring MATH and MATH is the NAME map MATH, the restriction of the counit of a weak NAME algebra MATH to the NAME algebra MATH. This claim can be proven by a fairly straightforward calculation and hence details of the proof are left to the reader. Let MATH be a morphism of left MATH-module coalgebras over the bialgebroid MATH. Then MATH is left MATH-linear and MATH-and MATH so that MATH is also a morphism in the category of weak left MATH-module coalgebras. CASE: We finally prove that the functors constructed in REF of the proof are inverses to each other. First we take a left weak MATH-module coalgebra MATH, turn it into a left module coalgebra MATH over the bialgebroid MATH, and then back into a left weak MATH-module coalgebra MATH. Using REF , we find MATH-and MATH . Finally take a left module coalgebra MATH over the bialgebroid MATH, make it into a weak left MATH-module coalgebra MATH, and then back into a left module coalgebra MATH over the bialgebroid MATH. Obviously MATH . Proving MATH is slightly more complicated. Recall that MATH is the counit of the weak NAME algebra MATH, and MATH is the counit of the MATH-bialgebroid MATH. Take MATH, and write MATH, for some MATH. Observe that MATH and MATH, and compute MATH this completes the proof. |
math/0101064 | First note that the left action of MATH on MATH is well-defined since the image of the left MATH-coaction of MATH is required to be in MATH. The fact that it is an action indeed follows from the fact that MATH is a left MATH-comodule algebra. Note also that in the definitions of MATH and MATH we used the natural isomorphisms MATH and MATH respectively. Clearly MATH is a right MATH-module map. To prove that it is a left MATH-module map as well take any MATH and MATH and compute MATH where we used that MATH is a left MATH-module coalgebra. On the other hand MATH . This proves that MATH is right MATH-linear, hence it is a MATH-bimodule map as required. Directly from the definition of MATH it follows that it is coassociative. It is clear that MATH is right MATH-linear. To prove that it is also a left MATH-module morphism take any MATH, MATH and compute MATH where we used that MATH is a left MATH-module coalgebra to obtain the second equality, then REF to derive the third one and the fact that the image of the left coaction of MATH on MATH is in MATH to obtain the fourth equality. This proves that MATH is left MATH-linear hence it is MATH-bilinear. The fact that MATH is a counit of MATH follows directly from the definition of MATH. Thus we conclude that MATH is a MATH-coring as stated. To prove the isomorphism of categories, take any left MATH-comodule MATH and view it as a NAME module via the same coaction MATH. Conversely, any NAME module MATH can be viewed as a left MATH comodule via MATH. |
math/0101064 | Consider two maps MATH, MATH, and MATH, MATH. The map MATH is well-defined because using REF we have for all MATH, MATH and MATH, MATH . Clearly, MATH is a right MATH-module map. Recall from CITE that MATH is a left MATH-module via MATH. Now, the fact that MATH is a weak left MATH-comodule algebra implies that MATH is a left MATH-module map. Thus MATH is a MATH-bimodule map. To prove that MATH is a coring map take any MATH, MATH and use the fact that MATH is a MATH-comodule algebra to compute MATH where MATH is another copy of MATH. On the other hand, using the fact that MATH is a left MATH-module coalgebra we have MATH as required. Thus we have proven that MATH is a map of MATH-corings. We now prove that MATH is an inverse of MATH. For a typical element MATH of MATH we have MATH for MATH is a weak MATH-module algebra. On the other hand, since MATH (compare proof of REF ) we have for all MATH, MATH . This completes the proof that MATH is an isomorphism of MATH-corings. |
math/0101075 | First assume MATH for every MATH-thick poset MATH of rank MATH. Let MATH be any graded poset of rank MATH. Since MATH is MATH-thick, MATH . Dividing by MATH gives the desired inequality for all graded posets. Now assume MATH for every graded poset MATH of rank MATH. Let MATH be a MATH-thick poset of rank MATH. For each rank MATH, fix a total order of the elements of MATH of rank MATH. Given an interval MATH of MATH of rank at least REF, let MATH denote the set of the first MATH atoms in MATH. (If MATH covers MATH, set MATH.) The operation MATH satisfies the following: MATH . Let MATH . How many sequences are in the set MATH? Given any MATH-chain of MATH, extend it to sequences in MATH one rank at a time. Having fixed MATH through MATH (MATH), if MATH, then there are exactly MATH choices for MATH. Thus MATH. To each maximal chain MATH: MATH of MATH is assigned an interval system as follows. For MATH, let MATH be the largest MATH such that MATH. Let MATH, and let MATH be the interval system consisting of minimal intervals in MATH. We show MATH belongs to MATH if and only if MATH blocks MATH. Suppose MATH: MATH is in MATH. Then for all MATH, MATH, so by the maximality of MATH, MATH. So for all MATH the interval MATH contains the element MATH of MATH. Thus MATH blocks MATH. For the reverse implication, suppose MATH is a maximal chain of MATH and MATH blocks MATH. Let MATH and MATH. Since MATH blocks MATH, MATH contains an element MATH. So MATH. Apply REF : MATH and MATH, so MATH. Thus MATH is in MATH. Given a system of intervals MATH denote by MATH the number of those maximal chains MATH of MATH for which MATH. (Note that MATH depends not only on MATH but also on the ordering of the elements of each rank.) Then MATH . By REF the sums MATH are all nonnegative, and so MATH . |
math/0101075 | Recall that MATH. Since the interval MATH of MATH is isomorphic to MATH, the result is obtained by substitution in REF . |
math/0101075 | The chain MATH of rank MATH is half-Eulerian. For every positive integer MATH, MATH is a MATH-Eulerian poset. On the other hand, by the definition of the function MATH, for an interval MATH of rank MATH in a MATH-Eulerian poset, MATH . Therefore MATH is the number of elements MATH strictly between MATH and MATH. Thus, if MATH is a MATH-Eulerian poset, then MATH is a positive integer, and MATH is MATH-thick. |
math/0101075 | The fact that the equations (in flag MATH-vector form) hold for all MATH-Eulerian posets follows from REF . Fix a set MATH with gap MATH. For each MATH-chain identify the rank MATH element MATH and rank MATH element MATH, and apply REF to the interval MATH. Sum the resulting equations for all the MATH-chains. Convert the flag MATH-vector equations using REF . Writing MATH and dividing by MATH, the result is MATH . From this we prove by induction that MATH (abbreviated as MATH) for all noneven sets MATH. Let MATH be any noneven set, and let MATH be an odd maximal interval of MATH. REF gives MATH . If MATH is a noneven proper subset of MATH, then by the induction assumption, MATH. So consider an even subset MATH. Since the maximal intervals of MATH contained in MATH are even, MATH, where MATH is odd, MATH is even, and, for MATH, MATH is odd. Thus, for MATH, MATH has the same parity as MATH. The coefficient of MATH in REF is MATH. So REF reduces to MATH. Conversely, suppose MATH for all noneven sets MATH. We show that the equations in REF hold. Let MATH and MATH a maximal interval of MATH. For MATH even, we need to prove REF . (The case of MATH odd is similar, and is omitted.) It suffices to consider the terms MATH with MATH an even set. For such MATH, MATH as above, with MATH odd, and MATH. So the coefficient in REF of MATH is MATH. Thus REF holds. To complete the proof, it suffices to show that the linear span of the MATH-vectors of MATH-Eulerian posets of rank MATH is the subspace of MATH determined by the equations MATH for all subsets MATH that are not even. This can be accomplished by finding a set of linearly independent vectors in the span of the MATH-vectors of MATH-Eulerian posets, one vector for each even subset MATH. In CITE NAME and NAME constructed, for each interval system MATH, a sequence of graded posets MATH. The construction starts with a rank MATH chain, and replicates intervals of ranks in the poset. For an even set MATH, let MATH be the set of maximal intervals in MATH. (For example, for MATH, MATH.) If MATH is an even subset of MATH, then MATH is half-Eulerian for all MATH. Furthermore, the sequence of MATH-vectors of these posets satisfies the following. Here MATH is the number of intervals in MATH. MATH . (See CITE for details.) Using REF , we get for any positive integer MATH, MATH . For fixed MATH the limiting MATH-vectors for each even interval system MATH are linearly independent, since for each even set MATH, the vector formed from the sequence MATH has MATH-entry REF for all MATH not containing MATH. |
math/0101075 | Since every interval of a MATH-Eulerian partially ordered set is MATH-Eulerian, REF gives that MATH for all intervals MATH of positive even rank. Now assume that for every interval MATH of positive even rank, MATH. Then by REF , for every interval MATH and for every MATH that is not even, MATH. For MATH of rank MATH, by REF , MATH . Since MATH is nonzero only if T is an even set, and then MATH is an even number, MATH . The same argument can be repeated for every interval of MATH, showing that it is a MATH-Eulerian poset. |
math/0101075 | Let MATH be a graded poset. By REF MATH is Eulerian if and only if MATH is MATH-Eulerian if and only if for every interval MATH of MATH with MATH even MATH if and only if for every interval MATH of MATH with MATH even MATH if and only if for every interval MATH of MATH with MATH even MATH if and only if for every interval MATH of MATH with MATH even MATH if and only if for every interval MATH of MATH with MATH even MATH. |
math/0101075 | Recall the map MATH of REF ; it maps MATH onto MATH. Clearly it also maps MATH onto MATH. So MATH, which contains the cone of MATH-Eulerian flag vectors. On the other hand, for any half-Eulerian poset MATH, MATH, the flag vector of the MATH-Eulerian poset MATH. If MATH holds, then MATH is the cone of half-Eulerian flag vectors, and its image is contained in the cone of MATH-Eulerian flag vectors. Thus, if MATH holds, then MATH is exactly the closed cone of flag vectors of MATH-Eulerian posets. |
math/0101075 | We first show the inequality is not a convolution of lower rank inequalities. In MATH-vector form the convolution satisfies the rule MATH (see REF ). So the convolution of linear expressions for ranks MATH and MATH with MATH gives a linear combination of MATH involving only subsets MATH not containing MATH. Since each element of MATH occurs in some set MATH in REF , it is not a convolution of lower rank inequalities. We now show that the inequality determines a facet of the cone. NAME and NAME list the facet inequalities for the general graded cone up through rank REF. The inequality of the proposition comes from applying one of the rank REF NAME inequalities to the rank-selected subposet MATH of an arbitrary half-Eulerian poset MATH. To check it is a facet of the half-Eulerian cone, we give twenty linearly independent limiting normalized MATH-vectors of half-Eulerian posets, for which the inequality holds with equality. The first sixteen posets are NAME limit posets determined by interval systems as in the following table. The next three limit posets are obtained from the rank REF Extremes REF by inserting a single new element of rank REF, shifting the old elements up one rank. To describe the last sequence of posets, let us (re)introduce the following generalization of the operator MATH. Given a graded poset MATH of rank MATH denote by MATH the poset obtained from MATH by replacing each MATH satisfying MATH with MATH elements MATH (keep every MATH satisfying MATH unchanged), and by setting the following order relations. The MATH-rank-selected subposet of MATH and of MATH are identical. For MATH satisfying MATH and MATH set MATH or MATH in MATH if and only of the same relation holds between MATH and MATH in MATH. Finally for MATH satisfying MATH set MATH in MATH if and only if MATH and MATH in MATH. For example, REF shows MATH where MATH is a chain of rank MATH. Note that for a graded poset MATH of rank MATH the graded poset MATH is isomorphic to MATH. The same notation is used in CITE. Let MATH be an arbitrary positive integer, and MATH be a chain of rank MATH. Consider now the following four graded posets. MATH . The MATH-rank-selected subposets of MATH and MATH are both isomorphic to MATH, where MATH is a chain of rank MATH; the MATH-rank-selected subposets of MATH, MATH, and MATH are all isomorphic to MATH where MATH is a chain of rank MATH. Let MATH be the graded poset of rank MATH obtained from MATH, MATH, MATH, and MATH by performing the following identifications: -identify the bottom element MATH of all four posets, -identify the top element MATH of all four posets, -identify MATH with MATH, -identify MATH with MATH. REF indicates how the four posets are identified, in a schematic way. Straightforward calculation shows that MATH is a half-Eulerian poset, for each positive MATH. Furthermore the normalized MATH-vectors, MATH, converge. The rows of the matrix below are the normalized MATH-vectors of the twenty limit posets. In the columns are the values of MATH (divided by the appropriate power of MATH), with the sets MATH in the order MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH. It is easy to check the rows are linearly independent. MATH . |
math/0101076 | We first prove that the flag vector of the pyramid is given by REF . For MATH let MATH be the MATH-fold pyramid over the MATH-gon. Let MATH be the MATH-simplex. We prove by induction on MATH that MATH . The formula is valid for MATH, since the sum is empty. We need the MATH case as well. Write MATH. MATH . For the induction step we write MATH in terms of MATH. We need two basic combinatorial facts. For a simplex MATH, MATH . The MATH-faces of the pyramid MATH are all either MATH-fold pyramids over the MATH-gon or simplices. Let MATH be the set of MATH-faces that are simplices. Let MATH be the set of MATH-faces that are pyramids (not simplices); the number of such faces is MATH. Thus, MATH . So the formula for the flag vector of the MATH-fold pyramid over the MATH-gon is valid. Now we show that the same formula holds for the multiplex. This is proved by a recursion different from the one used for the pyramid. Recall the formula for the MATH-vector of the multiplex MATH in REF . MATH . This is easily seen to agree with REF for MATH. Next is the proof for the special case where MATH. For MATH a vertex of MATH, write MATH for the quotient of the polytope MATH by the vertex MATH. This is itself a multiplex of dimension MATH. MATH . This can be shown to agree with REF . We now show that REF holds by induction on MATH. MATH . Algebraic manipulation gives REF . |
math/0101076 | Assume MATH is a MATH-dimensional multiplex with MATH vertices, and suppose MATH is a proper face of some higher dimensional multiplex MATH. Then MATH is a facet of a multiplex of dimension MATH, namely, any MATH-dimensional face of MATH containing MATH. According to the description of the facets in the definition of multiplex, MATH has at most MATH vertices. So MATH. To prove the converse, let MATH be the MATH-dimensional multiplex with MATH vertices. The facets of MATH are MATH for MATH. For MATH, MATH is a facet of MATH with MATH vertices. Thus among the proper faces of MATH are the MATH-multiplexes with MATH vertices for every MATH, MATH. |
math/0101076 | Let MATH be the graph of the multiplex MATH. Let MATH with MATH such that for all MATH, MATH is an edge of MATH. CASE: If MATH, then MATH. Note that MATH is not an edge of MATH, so it is not a subset of MATH. Thus MATH is contained in either the facet MATH or the facet MATH of the multiplex MATH. Since both those facets are simplices, MATH is a simplex that is a face of MATH. If MATH, the argument is similar. CASE: Assume MATH. Note that each vertex MATH of MATH is contained in exactly the facets MATH, MATH, , MATH, MATH, MATH, , MATH. Here we use the conventions that MATH if MATH, and MATH if MATH. (The self-duality of MATH is expressed in the combinatorial description.) Thus the set MATH is contained in the facet MATH if and only if MATH and MATH. (This statement is not true if MATH or MATH.) Let MATH and MATH. Then we can write the set MATH as MATH. The set of vertices in MATH is exactly MATH, that is, MATH. To check this, we need to know that MATH. If MATH, then MATH. Otherwise note that since MATH and MATH are assumed to be adjacent, either MATH (so MATH), or MATH. In the latter case, MATH. Since MATH is not adjacent to MATH, MATH, so MATH. Similarly, MATH. Now suppose MATH; we wish to show that MATH. If MATH, then MATH. If MATH, then MATH. If MATH, then MATH if and only if MATH. Thus if MATH, then MATH, that is, MATH. Now observe that this argument is also valid for every subsequence of MATH. Thus, for every MATH, MATH is a face of MATH, and so also a face of MATH. Therefore MATH is a simplex. |
math/0101076 | NAME the number of complete subgraphs with vertex set MATH, with MATH. MATH . For MATH, the complete subgraphs are obtained by choosing MATH-element sets from MATH not containing the pair MATH, since MATH is the only nonedge using these indices. There are MATH such complete subgraphs. MATH . For MATH, the complete subgraphs are obtained by choosing MATH-elements sets from MATH not containing the pair MATH. There are MATH such complete subgraphs for each MATH, or MATH altogether. MATH . For MATH, the complete subgraphs are obtained by choosing MATH-element sets from MATH. There are MATH such complete subgraphs. MATH . For MATH, the complete subgraphs are obtained by choosing MATH-element sets from MATH. There are MATH such complete subgraphs. Altogether the number of MATH-faces that are simplices is thus MATH . |
math/0101076 | NAME CITE proves that these are indeed two-faces of MATH, but does not show that they are the only nontriangular two-faces. Consider a facet MATH as given in REF , and apply REF (and the subsequent comment) to give all its nontriangular two-faces. Renumber the vertices of the facet MATH as MATH, MATH, . Thus the MATH elements of MATH are numbered MATH through MATH, and MATH. The quadrilateral faces in MATH are of the form MATH. Such a quadrilateral contains no element of MATH, and is of the form MATH, with MATH. Considering all the facets of MATH, MATH can range from REF to MATH. |
math/0101076 | REF implies that for the ordinary polytope MATH, MATH. Write MATH . Then MATH. The ordinary polytope MATH is elementary if and only if MATH. If MATH, then solving MATH for MATH gives MATH. But MATH, so this is not possible. Thus MATH implies MATH, that is, the ordinary polytope is a multiplex. We have already seen that every multiplex is elementary. |
math/0101076 | We consider the pairs not listed in the statement of the theorem. These fall into two categories. First are those pairs MATH with MATH. From REF every facet containing MATH and MATH, with MATH, also contains the nonempty set of vertices MATH. Thus, MATH is not a face (edge) of MATH. The number of these pairs is MATH. The other pairs not listed are MATH with MATH, MATH, and MATH. For such pairs MATH is not an edge of MATH, because it is a diagonal of a two-dimensional face as described in REF . The number of these pairs is MATH. The number of pairs listed in the statement of the theorem is thus MATH, which is the number of edges of MATH, as computed by NAME REF . So all the listed pairs are edges of MATH. |
math/0101076 | If MATH, then MATH is a cyclic polytope, and the graph of MATH is the complete graph on MATH vertices, so its chromatic number is MATH. So assume MATH. Let MATH be the graph of MATH, with vertex set MATH. Assign colors from the set MATH to the vertices of MATH as follows: MATH . Since MATH and MATH are not adjacent in MATH, every edge containing MATH or MATH is assigned two different colors. If MATH and MATH, or MATH, then MATH, so MATH. Thus adjacent vertices have different colors, so MATH gives a proper MATH-coloring of MATH. Now MATH contains a complete subgraph on the vertex set MATH, so the chromatic number of MATH is MATH. |
math/0101076 | If MATH is odd, the vertex sequence, MATH, MATH, MATH, , MATH, MATH, MATH, MATH, , MATH, MATH, MATH, gives a Hamiltonian cycle. If MATH is even, the vertex sequence, MATH, MATH, MATH, , MATH, MATH, MATH, MATH, , MATH, MATH, MATH, gives a Hamiltonian cycle. |
math/0101076 | For MATH, the vertices MATH and MATH are adjacent if MATH or if MATH. In addition MATH and MATH are adjacent, and MATH and MATH are adjacent. So usually MATH, MATH, MATH, , MATH, MATH, with MATH, gives a MATH - MATH path of length MATH. This is valid as long as MATH. If MATH, then MATH, MATH, MATH, , MATH, MATH, MATH, with MATH, gives a path of length MATH. If MATH, then MATH. The remaining cases are the paths MATH, MATH, MATH, , MATH and MATH, MATH, MATH, , MATH, MATH, if MATH, and MATH, MATH, MATH, , MATH, MATH, with MATH, if MATH. These are all of length MATH. Clearly, the MATH - MATH path given is the shortest MATH - MATH path. So the diameter is exactly MATH. |
math/0101082 | CASE: We proceed by induction on MATH. From the short exact sequence MATH we obtain the long exact sequence MATH . From CITE it follows that MATH if MATH, and that MATH is CM of dimension MATH. Thus we get an exact sequence MATH and isomorphisms MATH for all MATH. By REF , the module MATH is sequentially CM and has a CM filtration of length MATH. Hence by induction hypothesis we have MATH for MATH. This implies that MATH, and hence by REF we have MATH, and MATH. Summing up we conclude that MATH and MATH for MATH. Thus the assertion follows from the induction hypothesis and the fact that MATH is CM of dimension MATH. REF follows from REF and CITE since for any NAME MATH of dimension MATH one has MATH. |
math/0101082 | We write MATH, where MATH is a polynomial ring. Let MATH be the graded maximal ideal of MATH, and MATH the graded maximal ideal of MATH. By the Local Duality Theorem (see CITE) we have MATH and MATH where MATH. Since MATH, and since MATH, we see that MATH . Therefore REF imply that MATH and we hence may as well assume that MATH is a polynomial ring. Let MATH be the minimal graded free resolution of MATH. Note that MATH since MATH is a polynomial ring. Then the MATH-dual MATH of MATH is the free complex MATH with MATH, and MATH. Let MATH be the minimal graded free resolution of MATH. Then there exists a comparison map MATH which extends the identity on MATH. Since by REF is CM of dimension MATH, the complex MATH has the same length as MATH, namely MATH. Thus the MATH-dual MATH of MATH induces a natural homomorphism MATH. Here MATH and MATH for all MATH. Since MATH is CM by assumption, the complex MATH is exact, and hence a free resolution of MATH, and so the induced map MATH is given by MATH. It remains to be shown that MATH is a monomorphism. Let MATH be the mapping cone of MATH. Since MATH and MATH are acyclic, it follows that MATH and MATH for MATH. Notice that MATH is an isomorphism, since this is the case for MATH. Hence the chain map MATH is split injective, and so by cancellation we get a new complex of free MATH-modules MATH where MATH. Again we have MATH and MATH for MATH. Now suppose that MATH, and let MATH be a minimal prime ideal of the support of MATH. Since MATH, and since MATH is a NAME of dimension MATH, it follows that MATH is a minimal prime ideal of MATH with MATH. Therefore MATH, is a complex of length MATH with MATH for all MATH, MATH and MATH for MATH. By the NAME lemme d'acyclicité CITE this implies that MATH is acyclic, a contradiction. |
math/0101082 | We proceed by induction on MATH. Let MATH, and let MATH be the image of MATH. By REF, the module MATH is a NAME of dimension MATH. Consider the short exact sequence MATH. As in the proof of REF we get an exact sequence MATH and isomorphisms MATH for all MATH. Since MATH is an isomorphism (see REF), we deduce from the above exact sequence that MATH, and that MATH. Thus we have MATH for MATH, and MATH for MATH. Hence, by induction hypothesis, MATH is sequentially CM, and so is MATH by REF. |
math/0101082 | Let MATH. The first module MATH in the filtration must be the image of MATH. Then one makes use of an induction argument to MATH to obtain the desired result. |
math/0101082 | Since MATH is MATH-regular, the long exact NAME derived from MATH splits into short exact sequences MATH . It follows that MATH is CM of dimension MATH if and only if MATH is CM of dimension MATH. Thus REF implies the assertion. |
math/0101082 | After a generic choice of coordinates we may assume that MATH. The first statement in REF is true for any term order, while the second statement about the depth and REF follow from CITE because we may assume that the sequence MATH is MATH-regular if MATH. By the module version of CITE, and by CITE one has that MATH . On the other hand for a generic choice of coordinates we have MATH. Therefore MATH, which yields the last assertion. |
math/0101082 | Observe that, since we assume MATH, we have MATH where for each MATH, MATH is a strongly stable ideal , compare CITE. Hence MATH, so that, by REF one only has to prove that MATH is sequentially CM for any strongly stable ideal MATH. Recall that a monomial ideal is strongly stable if for all MATH and all MATH such that MATH divides MATH one has MATH for all MATH. Here MATH denotes the unique minimal set of monomial generators of MATH. For a monomial MATH we let MATH, and MATH. Let MATH, and let MATH be the unique monomial ideal such that MATH. It is clear that MATH is a strongly stable ideal in MATH. Thus it follows that MATH. Note that MATH contains MATH properly and is strongly stable. Let MATH. Then MATH, and since the extension MATH is flat, we have MATH. Now MATH is a non-trivial MATH-dimensional CM module over MATH, and therefore MATH is a MATH-dimensional NAME over MATH. Next we observe that MATH and that MATH is strongly stable. Since MATH, the assertion of the theorem follows from REF. |
math/0101082 | MATH : Set MATH and MATH. We proceed by induction on MATH. Suppose MATH, then MATH and MATH. Since MATH, MATH and MATH for MATH, the assertion follows in this case. Now suppose that MATH. Assume first that MATH. We have MATH, and, by REF, MATH. By REF, we also know that MATH is sequentially CM. Thus, if the implication MATH were known for modules of positive depth, it would follow that MATH for all MATH. Notice that MATH and MATH. However, since MATH and MATH, and since MATH and MATH have the same NAME function, we conclude that also MATH and MATH have the same NAME function. These considerations show that we may assume that MATH. Accordingly, MATH by REF, and MATH is sequentially CM by REF. Since MATH and MATH are sequentially CM we have MATH and MATH. We may assume that the coordinates are chosen generically so that MATH and that MATH regular on MATH and regular on MATH. According to REF, MATH is sequentially CM. Therefore our induction hypothesis, together with CITE, implies that the NAME functions of the local cohomology modules of MATH and of MATH are the same. We have short exact sequences of graded MATH-modules MATH and MATH because MATH is regular on MATH and MATH. Therefore applying the induction hypothesis to MATH we get MATH from which we deduce that MATH. MATH : We proceed again by induction on MATH. With the same arguments as in the proof of the first implication, we may assume that MATH. Therefore MATH, too, and since we are working with generic coordinates and MATH is sequentially CM by REF, MATH is MATH-, MATH- and MATH-regular. We shall show that MATH is also MATH-regular. Since MATH is MATH-regular, the long exact cohomology sequence derived from MATH splits into short exact sequences MATH . We show by induction on MATH that the corresponding sequences for MATH are also exact. For MATH the assertion is trivial, since MATH. Now let MATH, and assume that the assertion is true for all MATH. Then MATH is surjective, and we obtain the exact sequence MATH . Suppose multiplication with MATH is not surjective, then there exists a degree MATH such that MATH is not surjective. Using REF , and the hypothesis that the local cohomology modules of MATH and MATH have the same NAME function, one has MATH . This is a contradiction, since MATH and MATH, and consequently MATH. Now it follows that MATH is MATH-regular, and also that the cohomology modules of MATH and MATH have the same NAME functions, whence our induction hypothesis implies that MATH is sequentially CM. Since MATH is MATH-regular, we finalöly deduce that MATH is sequentially CM. |
math/0101082 | REF is proved in CITE. For the proof of REF we shall need the following result which also can be found in CITE: for all MATH and MATH one has MATH . (Observe that MATH for MATH and all MATH, as shown by NAME, see CITE and CITE). Now suppose that MATH is sequentially CM. Then MATH is componentwise linear, and hence MATH by REF of symmetric algebraic shifting. Since MATH, REF shows that MATH for all MATH and MATH, as desired. For the viceversa, let MATH be the MATH-matrix with entries MATH, MATH, MATH the MATH-matrix with entries MATH and MATH the MATH-matrix with entries MATH. Then REF says that MATH. Since MATH is invertible, we see that the numbers MATH are determined by the NAME functions of the local cohomology modules of MATH. Thus, if MATH for all MATH and MATH, then the numbers MATH for MATH and MATH coincide, which in turn implies that their graded NAME numbers are the same (because MATH). By REF of symmetric algebraic shifting this implies that MATH is componentwise linear, and hence MATH is sequentially CM. |
math/0101089 | Let MATH be the rectangle given by the definition of NAME boundary. It is easy to check that MATH has a NAME boundary. The conclusion follows now from the Corollary to REF. |
math/0101089 | Let MATH be a NAME sequence in MATH. We can construct an increasing sequence MATH of connected open sets with NAME boundary such that MATH, and MATH. By REF the functions MATH belong to MATH and MATH q.e. on MATH. As MATH for MATH large enough, by the NAME inequality MATH is a NAME sequence in MATH, and therefore it converges strongly in MATH to a function MATH with MATH q.e. on MATH. It is then easy to construct a function MATH such that MATH q.e. on MATH and MATH strongly in MATH for every MATH. As MATH converges strongly in MATH, we conclude that MATH strongly in MATH. Let MATH be a bounded sequence in MATH. As in the previous part of the proof we deduce that MATH is bounded in MATH for every MATH. By a diagonal argument we can prove that there exist a subsequence, still denoted by MATH, and a function MATH such that MATH weakly in MATH for every MATH. This implies MATH q.e. on MATH for every MATH, hence MATH. As MATH is bounded in MATH, we conclude that MATH weakly in MATH. |
math/0101089 | If MATH is a connected component of MATH whose boundary does not meet MATH, then MATH is simply connected. Indeed, if MATH is a bounded open set with MATH, then MATH, since otherwise MATH and MATH, which contradicts the fact that MATH is connected. Therefore in this case MATH is connected. Let MATH be a connected component of MATH whose boundary meets MATH. If MATH, by applying the previous result to the continuum MATH, we obtain that MATH is simply connected. By our regularity assumptions MATH is a closed simple curve. We want to prove that MATH is connected. If not, there exist MATH such that MATH intersects both connected components of MATH. Let MATH be a simple arc connecting MATH and MATH, whose interior points lie in MATH. As MATH is simply connected, MATH divides MATH into two relatively open connected components, each of which contains one of the connected components of MATH, and hence intersects MATH. As MATH does not meet MATH, this contradicts the fact that MATH is connected. Therefore MATH is connected. If MATH, then MATH has exactly one connected component MATH whose boundary intersects MATH. Let MATH be the connected component of MATH which contains MATH. By our regularity assumption, it is easy to see that MATH is closed and open in MATH, so that MATH coincides with MATH, and hence it is connected. Let us prove that in each connected component MATH of MATH there exists a harmonic function MATH such that MATH. As the differential form MATH is closed, the existence of MATH is trivial if MATH is simply connected. The only non-trivial case is when MATH and MATH is the unique connected component of MATH whose boundary intersects MATH. Let MATH be an open set with smooth boundary such that MATH. Since MATH is a solution of REF , integrating by parts we obtain that MATH. This implies that the integral on MATH of the differential form MATH vanishes, and proves the existence of the harmonic conjugate also in this case. Let MATH and MATH be smooth tangent and normal vectors to MATH. By classical regularity results for solutions to NAME problems on domains with corners (see CITE), we have that MATH has a continuous extension to MATH and MATH on MATH and on MATH. This implies that MATH and MATH have a continuous extension to MATH and MATH on MATH and on MATH, hence MATH is constant on MATH for every connected component MATH of MATH (as MATH is connected, so is MATH, which is obtained by possibly removing some of its end-points). We can choose the integration constant in each connected component so that MATH on MATH, therefore if we define MATH in MATH we obtain that MATH belongs to MATH, is continuous in MATH, and MATH on MATH. Moreover MATH is harmonic in MATH and MATH in MATH. As MATH on MATH, MATH is constant on each connected component of MATH. As MATH on MATH, the continuity of MATH implies that MATH is constant on each connected component of MATH. |
math/0101089 | For every open set MATH we have MATH for MATH large enough. Since the sequence MATH is bounded in MATH, there exists a subsequence, still denoted by MATH, such that MATH converges weakly in MATH to some function MATH. Since the space MATH is closed in MATH, we conclude that there exists MATH such that MATH a.e. in MATH. Repeating the same argument for all MATH, we construct MATH such that MATH a.e. in MATH. Assume now that MATH and let MATH. For every MATH there exists MATH such that MATH for MATH. Let MATH be an open set such that MATH. As MATH for MATH large enough, we have also MATH. Then MATH where MATH is an upper bound for MATH and MATH. From the previous part of the lemma MATH and the conclusion follows from the arbitrariness of MATH. |
math/0101089 | We can express MATH as the union of an increasing sequence MATH of simply connected open sets, with MATH, such that MATH is a finite union of regular open arcs of class MATH that are in one-to-one correspondence with the connected components of MATH, so that corresponding arcs have the same end-points. We can also write MATH as the intersection of a decreasing sequence MATH of open sets with boundary of class MATH such that MATH is connected, MATH, MATH has a finite number of points, and MATH meets MATH forming angles different from MATH and MATH. As MATH, there exists a solution MATH to the problem MATH . Using MATH as test function in REF , we obtain that the norms MATH are uniformly bounded. By REF , there exists MATH such that, up to a subsequence, MATH converges to MATH weakly in MATH. Let us prove that MATH . To this end it is enough to construct a solution MATH of REF such that MATH a.e. in MATH and MATH q.e. on MATH. Let MATH be a connected component of MATH whose boundary meets MATH, and let MATH. Given MATH small enough, let MATH and let MATH be the connected component of MATH containing MATH. Let MATH be the relative interior of MATH in MATH. As MATH meets MATH, for MATH small enough MATH. Using REF we deduce that there exists MATH such that MATH q.e. on MATH and, up to a subsequence, MATH weakly in MATH. Hence MATH a.e. in MATH. Since MATH is arbitrary, we can construct a function MATH such that MATH q.e. on MATH and MATH a.e. in MATH. Let now MATH be a connected component of MATH whose boundary does not meet MATH and let MATH. It is easy to see that MATH is the union of the increasing sequence MATH of the connected components of MATH containing MATH. Since MATH does not meet MATH and MATH is a solution of REF , we conclude that MATH a.e. in MATH, hence MATH a.e. in MATH. We define MATH in these connected components. In this way we have defined a function MATH on each connected component of MATH such that MATH, MATH q.e. on MATH, MATH a.e. in MATH, and hence MATH converges to MATH weakly in MATH. To conclude the proof of REF it is enough to show that MATH is a solution of REF . Let MATH with MATH q.e. on MATH. As MATH and MATH q.e. on MATH, we can use MATH as test function in REF . Then passing to the limit as MATH we obtain REF , and the proof of REF is complete. Let MATH, with MATH q.e. on MATH, be the harmonic conjugate of MATH on MATH given by REF . As MATH a.e. on MATH and MATH a.e. on MATH, we deduce that MATH is bounded in MATH. Since MATH is constant on each connected component of MATH, our hypotheses on the sets MATH imply that this function can be extended to a function, still denoted by MATH, which belongs to MATH and is locally constant on MATH. In particular we have that MATH is constant on each connected component of MATH and MATH is bounded in MATH. Assume that MATH has more than one point. Then MATH; since the sets MATH are connected, we obtain also MATH. As MATH q.e. on MATH, using the NAME inequality (see, for example, CITE) it follows that MATH is bounded in MATH, hence there exists a function MATH such that, up to a subsequence, MATH weakly in MATH. This implies that MATH a.e. in MATH, that MATH q.e. in MATH, and that MATH is constant q.e. on each connected component of MATH. Assume now that MATH has only one point, and let MATH be the mean value of MATH in MATH. By the NAME inequality, the sequence MATH is bounded in MATH, hence a subsequence converges to a function MATH weakly in MATH. It is clear that MATH a.e. in MATH and that MATH is constant q.e. on each connected component of MATH. The condition MATH q.e. on MATH is trivial in this case. |
math/0101089 | Since the differential form MATH is exact in MATH, it follows that there exists a function MATH such that MATH in MATH and MATH in MATH. As MATH, we have also MATH, hence MATH. It remains to prove that MATH is a solution of REF . Let us prove that, if MATH has more than one point and MATH meets the closure MATH of a connected component of MATH, then MATH q.e. on MATH. If this is not true, there exists a constant MATH such that MATH q.e. on MATH. Let us fix MATH. For almost every sufficiently small MATH we have MATH and MATH, hence MATH takes the values MATH and MATH in two distinct points of MATH. This implies MATH which yields MATH, in contradiction with our hypothesis. Therefore, if MATH has more than one point, then MATH q.e. on all connected components of MATH whose closure meets MATH. Since MATH is constant q.e. on the other connected components of MATH, we can apply CITE to a suitable extension of MATH and we can construct a sequence of functions MATH, converging to MATH in MATH, such that each MATH vanishes in a neighbourhood of MATH and is constant in a neighbourhood of each connected component of MATH. Let MATH with MATH q.e. on MATH. As MATH in MATH and MATH near MATH, we have MATH . Passing to the limit as MATH, we obtain MATH showing thus that MATH is a solution of REF . |
math/0101089 | Note that MATH is a minimum point of REF if and only if MATH satisfies REF ; analogously, MATH is a minimum point of REF if and only if MATH satisfies REF with MATH and MATH replaced by MATH and MATH. Taking MATH as test function in the equation satisfied by MATH, we prove that the sequence MATH is bounded in MATH. By REF , there exists a function MATH such that, passing to a subsequence, MATH weakly in MATH. Let MATH be the harmonic conjugate of MATH given by REF . Then MATH is bounded in MATH, MATH q.e. on MATH, and MATH is constant q.e. on each connected component of MATH. If MATH has only one point, using REF , we obtain that MATH weakly in MATH for some MATH, and by the NAME inequality there are constants MATH such that MATH weakly in MATH. This implies that MATH is constant q.e. on each connected component of MATH. Moreover MATH q.e. on MATH since MATH. If MATH has more than one point, we have MATH; since the sets MATH are connected, we obtain also MATH. As MATH q.e. on MATH, using the NAME inequality (see, for example, CITE) it follows that MATH is bounded in MATH. Hence there exists a function MATH such that MATH weakly in MATH. This implies that MATH is constant on each connected component of MATH. Let us fix an open ball MATH containing MATH. Using the same extension operator we can construct extensions of MATH and MATH, still denoted by MATH and MATH, such that MATH, MATH weakly in MATH. Given an open set MATH, any function MATH will be extended to a function MATH by setting MATH q.e. in MATH. By CITE we have MATH . Since the complement of MATH has two connected components, from the results of CITE and CITE we deduce that, for every MATH, the solutions MATH of the NAME problems MATH converge strongly in MATH to the solution MATH of the NAME problem MATH . This implies (see, for example, CITE) that, in the space MATH, the subspaces MATH converge to the subspace MATH in the sense of NAME (see CITE). Since MATH by REF , and MATH weakly in MATH, from the convergence in the sense of NAME we deduce that MATH, hence MATH q.e. on MATH by REF . As MATH is harmonic in MATH and MATH in MATH, we deduce that MATH is harmonic in MATH and MATH in MATH. Let us consider a smooth closed curve MATH in MATH. Since, for MATH large enough, the functions MATH are harmonic in the same neighbourhood of MATH, the weak convergence implies the uniform convergence of MATH to MATH in a neighbourhood of MATH. As MATH, the differential forms MATH are exact, thus their integrals over MATH vanish. It follows that the integral over MATH of the differential form MATH is zero. Then we can apply REF and we obtain that the function MATH is a solution to REF . We now construct a function MATH such that MATH a.e. in MATH and MATH q.e. on MATH. As in the proof of REF , we consider a connected component MATH of MATH whose boundary meets MATH, a point MATH, and the corresponding sets MATH, MATH, and MATH. Since MATH for MATH large enough, we can apply REF and deduce that there exists a function MATH such that MATH q.e. on MATH and, up to a subsequence, MATH weakly in MATH, which implies MATH a.e. in MATH. In this way we construct MATH on all connected components of MATH whose boundary meets MATH, and we define MATH on the other connected components. It is clear that MATH, MATH a.e. in MATH, MATH q.e. on MATH, hence MATH is a solution to REF -REF . By the uniqueness of the gradients of the solutions, we have MATH a.e. in MATH. As the limit does not depend on the subsequence, the whole sequence MATH converges to MATH weakly in MATH. Taking MATH and MATH as test functions in the equations satisfied by MATH and MATH, we obtain MATH . As MATH weakly in MATH and MATH strongly in MATH, from the previous equalities we obtain that MATH converges to MATH, which implies the strong convergence of the gradients in MATH. |
math/0101089 | Given MATH, let MATH. As MATH for MATH large enough, we have MATH. Applying REF with MATH we get MATH . Passing to the limit as MATH we obtain REF . |
math/0101089 | It is enough to prove the lemma when MATH and MATH. Since MATH is separable and locally connected (see, for example, CITE), and MATH is open in MATH, the connected components of MATH are open in MATH and form a finite or countable sequence MATH. Since each MATH is closed in MATH, we have MATH. If MATH, then MATH would contain an open, closed, and non-empty proper subset, which contradicts the fact that MATH is connected. Therefore MATH. As MATH, we conclude that MATH. Therefore for every MATH there exists a point MATH. As MATH in the NAME metric, there exists MATH such that MATH. If there are infinitely many connected components MATH, there exists a sequence of integers MATH tending to MATH such that MATH . If there are MATH connected components MATH, REF is true with MATH for every MATH. Let MATH denote the closed segment with end-points MATH and MATH, and let MATH . Then the sets MATH are continua, contain MATH, and converge to MATH in the NAME metric. As MATH REF , we have MATH which, together with REF , yields MATH . The opposite inequality for the lower limit follows from REF . |
math/0101089 | Since MATH is a solution of REF which satisfies the boundary REF , by linearity for every MATH we have MATH a.e. in MATH, hence MATH where the last equality is deduced from REF . Dividing by MATH and letting MATH tend to MATH we obtain REF . The continuity of MATH implies that MATH is of class MATH. |
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