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math/0005197 | Let MATH be REF generated by operad MATH and let MATH be the universal NAME algebra. According to REF , its internal enveloping algebra MATH exists in category MATH. Let MATH be the NAME extension of MATH obtained by adding kernels of all idempotents in MATH. Since MATH admits infinite direct sums, MATH admits infinite direct sums as well. Therefore, the functor defining the NAME algebra MATH in MATH extends to a functor MATH . Since MATH is defined in MATH by means of split coequalizers, the image MATH represents an internal enveloping algebra for MATH in MATH. Since MATH is a full subcategory in MATH and MATH is isomorphic to the symmetric algebra of MATH in MATH, this proves the existence of the enveloping algebra MATH. The algebra MATH is a split coequalizer of the pair MATH constructed similarly to REF . Let us calculate the algebra MATH. Applying the functor MATH to REF we get MATH . The coequalizer of this pair can be identified with MATH. This proves the second part of the theorem. |
math/0005197 | The category MATH of representations of MATH is a tensor category with direct sums and MATH endowed with the adjoint action is a NAME algebra in MATH. The internal enveloping algebra of MATH in MATH is just MATH endowed with the adjoint action of MATH. Then the external enveloping algebra is precisely MATH . |
math/0005197 | It follows from REF . |
math/0005199 | Since MATH is a union of ``moment-angle" blocks MATH (see REF ), and each MATH is obviously MATH-invariant, the whole MATH is also MATH-invariant. In order to show that MATH is a cellular subcomplex of MATH (with respect to the above constructed cellular decomposition) we just mention that MATH is the closure of cell MATH. |
math/0005199 | See CITE . |
math/0005199 | See CITE . |
math/0005199 | See CITE. |
math/0005199 | It follows directly from the definitions of MATH and MATH that the constructed map is an isomorphism of graded algebras. So, it remains to prove that it commutes with differentials. Let MATH, MATH and MATH denote the differentials in MATH, MATH and MATH respectively. Since MATH, MATH, we need to show that MATH, MATH. We have MATH, MATH. Since any REF-cell of MATH is either MATH or MATH, MATH, it follows that MATH where MATH if MATH and MATH otherwise. Hence, MATH. Further, since any REF-cell of MATH is either MATH or MATH, it follows that MATH . Hence, MATH. |
math/0005199 | We use the cochain complex MATH for calculations. The module MATH has basis consisting of monomials MATH with MATH and MATH. Since MATH, MATH, the bigraded component MATH is generated by monomials MATH with MATH and MATH. In particular, MATH if MATH or MATH, whence REF follows. To prove REF we mention that MATH is generated by REF, while any MATH, MATH, is a coboundary, hence, MATH, MATH. Now we are going to prove REF . Since any MATH has MATH, and any simplex of MATH is at most MATH-dimensional, it follows that MATH for MATH. It follows from REF that MATH for MATH, so it remains to prove that MATH for MATH. The module MATH is generated by monomials MATH, MATH. Since MATH, it follows easily that there no cocycles in MATH. Hence, MATH. To prove REF we mention that MATH and MATH (this follows from REF ). Hence, REF follows from REF . As for REF , it follows from REF that MATH. The module MATH is generated by monomials MATH, MATH. We have MATH and MATH. It follows that MATH is a cocycle if and only if MATH is not a REF-simplex in MATH; in this case two cocycles MATH and MATH are cohomological. REF now follows easily. The remaining REF follows from the fact that the monomial MATH of maximal total degree MATH necessarily has MATH, MATH, MATH. |
math/0005199 | It follows from REF that MATH . Then MATH . Denote MATH. Then it follows from REF that MATH . Substituting above MATH for MATH, we finally obtain from REF MATH which is equivalent to REF . |
math/0005199 | We have MATH . Now the statement follows from REF . |
math/0005199 | This follows from the previous corollary and REF . |
math/0005199 | To prove that MATH is a cellular subcomplex of MATH we just mention that this MATH is the closure of the MATH-cell MATH. So, it remains to prove that MATH is contractible within MATH. To do this we show that the embedding MATH is homotopical to the map to the point MATH. On the first step we note that MATH contains the cell MATH, whose closure contains MATH and is homeomorphic to MATH. Hence, our MATH can be contracted to MATH within MATH. On the second step we note that MATH contains the cell MATH, whose closure contains MATH and is homeomorphic to MATH. Hence, MATH can be contracted to MATH within MATH, and so on. On the MATH-th step we note that MATH contains the cell MATH, whose closure contains MATH and is homeomorphic to MATH. Hence, MATH can be contracted to MATH within MATH. We end up at the point MATH to which the whole torus MATH can be contracted. |
math/0005199 | Indeed, REF-skeleton of our cellular decomposition of MATH is contained in the torus MATH. |
math/0005199 | Since MATH and MATH, we have MATH . It follows from REF that MATH . Hence, MATH by REF . |
math/0005199 | We have MATH. It follows from REF that MATH. By definition, the module MATH is spanned by monomials MATH such that MATH, MATH, MATH, and all these monomials are cocycles. Suppose that MATH are two MATH-simplices of MATH sharing a common MATH-face. Then the corresponding cocycles MATH, MATH, where MATH, MATH, are cohomological (up to sign). Indeed, let MATH, MATH. Since any MATH-face of MATH is contained in exactly two MATH-faces, the identity MATH holds in MATH, hence, MATH and MATH are cohomological. Since MATH is a simplicial sphere, any two MATH-simplices of MATH can be connected by a chain of simplices such that any two successive simplices share a common MATH face. Thus, any two cocycles in MATH are cohomological, and we can take any one as a representative for the fundamental cohomological class of MATH (after a proper choice of sign). |
math/0005199 | To prove the first assertion we construct homotopy equivalence MATH (see REF ) as it is shown on REF . This homotopy equivalence is covered by a MATH-invariant homotopy equivalence MATH, as needed. The second assertion follows easily from the definition of MATH. |
math/0005199 | We need only to check that the differential MATH adds MATH to bidegree. This follows from REF and MATH . |
math/0005199 | REF shows that every moment-angle complex MATH is a cellular subcomplex of MATH, and each cell is the product of cells of REF different types: MATH, MATH, MATH, MATH, and MATH, MATH. These products are encoded by words MATH, where MATH are pairwise disjoint subsets of MATH such that MATH. In the case MATH the definition of MATH (see REF ) shows that the cell MATH belongs to MATH if and only if the following two conditions are satisfied: CASE: The set MATH is a simplex of MATH. CASE: MATH. Let MATH denote the number of cells MATH with MATH, MATH, MATH, MATH, MATH, MATH. It follows that MATH where MATH is the MATH-vector of MATH (we also assume MATH and MATH for MATH or MATH). By REF , MATH . Now we calculate MATH as it is defined by REF , using REF : MATH . Substituting MATH, MATH above, we obtain MATH . Since MATH we obtain MATH . The second sum in the above formula is exactly MATH (see REF ). To calculate the first sum, we mention that MATH (this follows from calculating the coefficient of MATH in both sides of MATH). Hence, MATH since MATH (remember that MATH). Finally, using REF , we calculate MATH . |
math/0005199 | REF shows that MATH and MATH. Moreover, it can be seen in the same way as in REF that relative NAME duality isomorphisms REF regard the bigraded structures in the (co)homology of MATH and MATH. It follows that MATH . Hence, MATH and MATH . Using REF , we calculate MATH . Substituting the formula for MATH from REF and the above expression into REF , we obtain MATH . Calculating the coefficient of MATH in both sides after dividing the above identity by MATH, we obtain MATH, as needed. |
math/0005199 | Since MATH, MATH, we have MATH (the coefficient MATH can be dropped since for odd MATH the left side is zero). |
math/0005207 | Let MATH with MATH be common resolutions and MATH. We choose MATH-exceptional effective MATH-divisors MATH and a MATH-divisor MATH on MATH such that MATH. Then, MATH . |
math/0005207 | The ``only if" part is obvious taking it into account that for any MATH. Actually MATH and MATH. Now we prove the ``if" part. REF means that the coefficient of MATH in MATH is MATH. This says that for any MATH, MATH is the only MATH-exceptional prime divisor on MATH containing MATH and the coefficient of MATH in MATH is MATH. We consider a prime divisor MATH on MATH which is smooth at MATH and define MATH as the largest integer such that MATH. Then MATH is smooth at the generic point of MATH for any MATH, and so we get MATH. Therefore the coefficient of MATH in MATH is MATH. By REF , we can choose MATH such that MATH, that is, MATH because of the above argument. Adding MATH such that MATH, we can take local parameters MATH at MATH. Then MATH equals, as valuations, the exceptional divisor obtained by the weighted blow-up of MATH with its weights MATH, and especially MATH is obtained by the weighted blow-up of MATH with its weights MATH. |
math/0005207 | Let MATH be the greatest common divisor of MATH and MATH, and let MATH . Since MATH is NAME by CITE, so is MATH. Hence MATH and MATH. |
math/0005207 | We consider the exact sequence: MATH . By REF , we get MATH . Since MATH is an integer for any MATH and MATH is positive, we have REF . By REF , MATH . Let MATH. We note that MATH if MATH, and MATH. Because MATH is weak KLT and MATH is MATH-ample for a sufficiently small positive rational number MATH and an integer MATH, using CITE, we have MATH for MATH. So by REF , for any MATH, CASE: MATH CASE: MATH for MATH, and therefore MATH. Putting MATH in REF , we get MATH . Combining this and REF with MATH, we get REF . With REF , we obtain for MATH, MATH . Eliminating MATH with REF , we obtain MATH . Since for MATH, MATH with REF we have MATH . Of course because MATH, combining this with REF , we obtain REF . Putting MATH in REF , we have MATH . Since MATH, we get REF . |
math/0005207 | CASE: By the assumption and MATH, the proof of REF says that we can take local parameters MATH at MATH such that MATH and MATH. Then for MATH, MATH equals, as valuations, the exceptional divisor obtained by the weighted blow-up of MATH with its weights MATH. Hence MATH and so MATH . Here we used REF proved later. CASE: As in the proof of REF , we can take local parameters MATH at MATH such that MATH and MATH. Then for MATH, MATH equals, as valuations, the exceptional divisor obtained by the weighted blow-up of MATH with its weights MATH. We have MATH . But since MATH for any MATH such that MATH we have MATH . Thus MATH and hence MATH . Therefore with REF , MATH . |
math/0005207 | MATH . |
math/0005207 | For MATH is a line in this case and MATH is the discrepancy of MATH with respect to MATH, we get MATH and MATH . By REF , MATH and thus MATH. From REF , we obtain MATH. By REF , MATH . From REF , and REF , for MATH we have MATH . Hence for MATH, MATH where the equality holds if and only if MATH. If there exists a positive integer MATH such that MATH and MATH for any MATH, then by REF , and the condition of the equality in REF , we obtain MATH. Thus with REF , we have MATH, that is, MATH. This contradicts REF and hence we get MATH. |
math/0005207 | In this case we use essentially the same idea as in REF , but it is a little more complicated. By REF , MATH. Thus we have only to consider the two subcases: CASE: MATH CASE: MATH . In REF , we have MATH by REF and thus MATH from REF . But since MATH is a point, we get MATH and MATH. Then choosing local parameters MATH at MATH such that MATH, MATH equals, as valuations, the exceptional divisor obtained by the weighted blow-up of MATH with its weights MATH. So we have only to investigate REF . Recalling that MATH is the discrepancy of MATH with respect to MATH, we have MATH . Calculating with REF we obtain MATH, and thus by REF , MATH . From REF , and REF , for MATH we have MATH . Hence for MATH, MATH where the equality of the first inequality holds if and only if MATH, and the second holds if and only if MATH. MATH. Utilizing REF with MATH, we have MATH . We take local parameters MATH at MATH as in MATH, satisfying MATH if MATH is a line. We have MATH . But since MATH we get MATH and thus MATH . Hence, MATH . From REF , and the condition of the second equality in REF , we have MATH. If there exists a positive integer MATH such that MATH and MATH for any MATH, then by REF , and the condition of the first equality in REF , we obtain MATH. Thus with REF , we have MATH. This contradicts REF and hence we get MATH. |
math/0005209 | These are immediate consequences of the definitions. |
math/0005209 | This follows directly from the observation that for such sequences MATH is nothing but a finite linear combination of the solutions MATH with MATH and MATH of the recursion relation MATH and thus, it is annihilated by MATH. |
math/0005209 | Using the identity MATH that follows from REF , we obtain after some easy algebra MATH . This shows that the right hand side is at least MATH since the denominator is MATH due to MATH. For the MATH variant, the term corresponding to MATH in the numerator is MATH that vanishes for convex MATH. For the MATH variant, that term is MATH that simplifies to MATH for convex MATH. This finishes the proof. |
math/0005209 | Rewriting MATH one may perform the limit for MATH upon using the assumptions according to MATH since MATH. |
math/0005209 | Specializing REF to MATH, it suffices to prove the last assertion. Here, the proof follows from the observation that MATH and MATH imply MATH for large MATH in view of the assumptions. |
math/0005209 | We rewrite MATH as defined in REF in the form MATH for large MATH where we used REF in the numerator, and in the denominator the relation MATH that follows by repeated application of REF . Insertion of REF now yields MATH where REF was used. Also the fact was used that MATH annihilates any linear combination of the solutions MATH with MATH and MATH of the recursion relation REF since each MATH is a zero with multiplicity exceeding MATH. Invoking REF given in REF one obtains MATH . The proof of REF is completed taking leading terms in the sums over MATH. Since MATH where MATH, and MATH, REF is obtained where MATH. |
math/0005209 | The proof proceeds as the proof of REF but in the denominator we use MATH that follows from REF given in REF. |
math/0005209 | We have for fixed MATH whence MATH . If the MATH alternate in sign, we obtain for these limits MATH . This implies REF . |
math/0005209 | Since MATH cannot be a zero of MATH. Then, REF entail that MATH is NAME. Furthermore, REF is applicable and yields MATH. |
math/0005209 | We have for fixed MATH and MATH . Note that the denominators cannot vanish and are bounded from below by MATH. Hence, we have MATH and consequently, MATH since MATH according to REF . |
math/0005209 | Use MATH to obtain MATH . NAME expansion of the polynomial yields due to the zero at MATH . Invoking NAME 's lemma CITE completes the proof. |
math/0005209 | We have MATH for large MATH. Hence, MATH . Since the characteristic polynomial MATH has a zero of order MATH at MATH according to the assumptions, REF follows using NAME expansion. |
math/0005210 | The first step of the proof is to construct a vertex set and an action of MATH. We do not even have a true action of MATH on MATH, however. What we do have is an action of MATH on the space of ends MATH. This can be promoted into an action of MATH on a new vertex space as follows. Bushiness and bounded valence of MATH tells us that the space of ends MATH is a NAME set. Note that each edge of MATH determines a partition of MATH into a pairwise disjoint set of clopens (closed-open subsets). Define a quasi-edge of MATH to be any partition of MATH into a pair of disjoint clopens. Clearly MATH acts on the set of quasi-edges of MATH. If MATH is a clopen then the convex hull MATH is a subtree of MATH. If MATH is a quasi-edge then the intersection of REF-neighborhoods of the convex hulls MATH is bounded; the diameter of this intersection is defined to be the quasi-edge constant of MATH. Note that a REF-quasi-edge is the same thing as (the partition determined by) a true edge of MATH. The action of a MATH quasi-isometry of MATH on a MATH-quasi-edge produces a MATH-quasi-edge, where MATH depends only on MATH. It follows that the MATH-orbit of a single quasi-edge has uniform quasi-edge constant. We therefore define a vertex set MATH to be the MATH-orbit of some arbitrarily chosen quasi-edge of MATH. It is not hard, using uniformity of the quasi-edge constant, to attach edges to MATH in a locally finite, MATH-equivariant manner, thereby producing a locally finite graph MATH on which MATH acts coboundedly, and a coarse conjugation MATH. Unfortunately, MATH may not be a tree, that is, in particular it may not be simply connected. The latter problem can be corrected without too much difficulty by attaching REF-cells to MATH in a locally finite, MATH-equivariant manner, producing a locally finite, simply connected REF-complex MATH on which MATH acts coboundedly; the coarse conjugation MATH extends to a coarse conjugation MATH. The final step is to get from REF-complex MATH back to a tree. This is accomplished by using tracks in the sense of CITE: a track MATH in MATH is a locally separating, connected REF-complex in general position with respect to the skeleta of MATH. Since MATH is simply connected, each track MATH separates MATH into two components, and MATH is essential if each component of MATH is unbounded. Using minimal surface ideas as in CITE, we construct a MATH-equivariant family of pairwise disjoint, essential tracks MATH in MATH. By a NAME finiteness argument as in CITE one shows that a maximal such family MATH exists, and that all components of MATH are bounded. The final tree MATH has vertex set in REF correspondence with the components of MATH, and edge set in REF correspondence with the set MATH. |
math/0005210 | Suppose MATH exists as stated. Choose a free, discrete, cocompact subgroup MATH. We may give MATH a left-invariant metric so that the inclusion of MATH into MATH is a quasi-isometry. Since MATH is quasi-isometric to the free group MATH, it follows that MATH is quasi-isometric to a finite valence tree MATH. The left action of MATH on itself is quasi-conjugate to a cobounded quasi-action of MATH on MATH. Applying REF it follows that MATH has a cobounded quasi-action on a bushy tree MATH of bounded valence. It follows that every virtually free group MATH has a cobounded action on MATH. Pick a prime MATH larger than the maximal valence of a vertex of MATH. The group MATH is virtually free, and therefore has discrete cocompact image in MATH with finite kernel, and so MATH acts coboundedly on MATH. But each of the free factors MATH acts trivially on MATH, making MATH act trivially, contradicting coboundedness. |
math/0005210 | We shall prove: CASE: There exists MATH such that for each MATH and each MATH there is a MATH-raft MATH with MATH. To see why this suffices, consider a raft MATH of MATH and two vertices MATH, and so MATH. Applying MATH we get MATH, MATH for MATH-rafts MATH of MATH, and we want to verify that MATH. Since MATH it follows that the coarse intersection of MATH and MATH coarsely contains a coarse MATH space, namely MATH. If MATH then the coarse intersection of MATH and MATH is coarsely equivalent to some edge space of lower dimension, but a coarse MATH space with MATH cannot contain a uniformly embedded copy of a coarse MATH space. It follows that MATH and MATH. Similarly MATH for some MATH-raft MATH, and so MATH; but this is only possible if MATH. This shows that MATH coarsely respects MATH-rafts. If REF is not true then, taking counterexamples for larger and larger values of MATH, using coboundedness of the isometry group of MATH, and passing to a limit, we obtain a quasi-isometry MATH, a vertex MATH and an edge MATH, such that MATH intersects both components of MATH arbitrarily deeply. It follows that for sufficiently large MATH, the subset MATH coarsely separates MATH into at least two deep components. But the set MATH, with metric restricted from MATH, is uniformly equivalent to a subset of MATH with restricted metric. When MATH has dimension MATH we obtain a contradiction using arguments of coarse algebraic topology: a coarse MATH space cannot be coarsely separated by a subset which is uniformly equivalent to a subset of a coarse MATH space of dimension MATH. If MATH is of dimension MATH then in fact MATH: otherwise, a subset of the coarse MATH space MATH which is not coarsely equivalent to all of MATH would embed uniformly in coarse MATH space MATH, coarsely separating MATH, and that is impossible. We therefore have MATH in MATH. This shows that MATH is a coarse MATH space uniformly embedded in the coarse MATH space MATH, and so MATH has exactly two deep components each coarsely containing MATH, each contained in a separate deep component of MATH. Now the argument breaks into cases. CASE: MATH is contained in a bushy raft MATH. In this case the coarse MATH space MATH coarsely separates MATH, a tree of MATH spaces, and that is clearly impossible. CASE: MATH is a point raft. By hypothesis, the crossing graph of MATH is either connected or empty. CASE: The crossing graph of MATH is connected. Using connectedness of the crossing graph together with some coarse separation arguments, one shows that one of the deep components of MATH coarsely contains the union of all codimension-REF edge spaces inside MATH. But this is absurd, because coboundedness of the MATH stabilizer subgroup of MATH shows that the union of incident codimension-REF edge spaces intersects every deep subset of MATH. CASE: The crossing graph of MATH is empty. Each edge incident to MATH therefore has dimension MATH, and it follows that the inclusion of MATH in MATH has the following ``coarse NAME separation property": for all sufficiently large MATH there are exactly two deep components of MATH which coarsely contain MATH. The inclusion of MATH into MATH therefore has the same property: for all sufficiently large MATH there are exactly two deep components of MATH which coarsely contain MATH. Let MATH be the subtree of MATH spanned by all edges in MATH whose edge space is coarsely equivalent to MATH; we think of MATH as the ``edge raft" containing MATH, although a priori MATH can have MATH-dimensional vertices and edges. But by using the coarse NAME separation property for MATH one shows that MATH contains at most one MATH-dimensional vertex of MATH. Moreover, MATH cannot contain exactly one MATH-dimensional vertex, for then one would be able to find a MATH dimensional valence REF vertex of MATH which would violate irreducibility of the NAME tree MATH. It follows that MATH in fact consists entirely of MATH dimensional vertices and edges, and so is a MATH raft. MATH cannot be a bounded raft, for it has at least one edge, namely MATH, and again that would violate irreducibility. MATH cannot be a line raft, by hypothesis. Finally, it cannot be a bushy raft, because that would violate the coarse NAME separation property for MATH: for larger and larger MATH, the number of deep components of MATH coarsely containing MATH would approach infinity. |
math/0005210 | By passing to asymptotic cones we replace MATH with a bilipschitz homeomorphism taking MATH to MATH for each MATH. Applying the NAME Theorem, at almost any point MATH the derivative MATH gives the desired conclusion. |
math/0005217 | First notice that MATH is actually equal to MATH, and therefore the factor MATH commutes with MATH. By NAME - NAME formula, our generating functions REF are rational functions of MATH and MATH's. Therefore MATH . Now for MATH, we have MATH . It is well known (and follows almost immediately from the definition) that MATH on MATH for any genus. Here the second MATH is the corresponding line bundle on MATH. MATH is the divisor on MATH whose generic elements have the reducible domain curves of two components, with MATH on one component and the rest on the other. By projection formula MATH . From the definition of MATH-theoretic push-forward MATH . Since MATH for all MATH, MATH . Therefore MATH is a vector bundle. We are now reduced to computing MATH. Now by NAME duality MATH, where MATH is the dualizing sheaf. NAME, MATH is the holomorphic differential with poles of order at most MATH at MATH. Thus we have a filtration of degrees of poles at each marked point MATH: MATH and the graded bundles MATH is isomorphic to MATH and MATH. Therefore MATH . Notice that the last equality holds only in MATH-theory (using graded objects). This means MATH which is equivalent to REF . |
math/0005217 | REF follows from the theory of modular forms. The sections of MATH are the modular forms of weight MATH. REF is a rephrase of the classical result which states that the space of modular forms are generated by a weight four and a weight six modular forms. REF follows from REF. Alternatively, MATH can be computed from (generalized version of) NAME 's theorem. |
math/0005217 | Since MATH, this formula follows from REF . |
math/0005217 | Main term: The main contribution comes from the fundamental class of MATH because the generic point in MATH has no automorphism for MATH. Notice that the integrand has the form MATH . But the dimension of MATH is MATH, which implies that the only term contributing to the integration is of the form MATH. Now the main term can be computed by the dilation equation: MATH . NAME terms: A point in MATH is a stable curve of genus one, which consists of one ``elliptic component" MATH with some (chains of) rational curves attached to it. Here MATH could either be a pointed elliptic curve or a loop formed by pointed rational curves. Therefore the NAME stratum will have a form like MATH. The integration over MATH turns out to be zero due to the smoothness of MATH and the following dimension counting: the integrand REF has cohomological degree MATH, the number of special points in a rational component. However, MATH. We conclude from above that the only orbifold contribution MATH come from those NAME stratum which is of the form MATH. It is easy to see that automorphisms occur only in the following two cases: either MATH is a smooth elliptic irreducible component or MATH has only two rational components, each carrying two marked points, as shown in the following graph (call it (MATH)). Let us find all possible genus one stable curves with MATH marked points which has nontrivial automorphisms. They belongs to one of the following REF types. (All marked points must be positioned on the fixed points of MATH.) CASE: The generic element is MATH, a genus one irreducible curve with at most four marked points at four REF-torsion points (MATH symmetry). Note that the corresponding strata contain three (special) moduli points of the form REF with marked points placed at fixed points of MATH action. CASE: MATH is a genus one irreducible curve with MATH such that (at most) three marked points sit at fixed points of MATH symmetry. CASE: MATH is a genus one irreducible curve with MATH with (at most) two marked points at fixed points of MATH symmetry. Now it's obvious that if MATH, MATH. Let us deal with the cases MATH separately. MATH . Only type REF is possible. The only nontrivial contribution of singular stratum comes from type REF . But the singular stratum itself is the moduli space of elliptic curve with all two-torsion points marked, i. CASE: MATH. We can apply NAME 's formula: MATH . We claim that MATH, MATH, MATH. This can be seen by a local computation. First for the generic (smooth) curves the formal cotangent space is represented by MATH, and MATH. Second, the forgetful morphism from MATH to MATH restricted to this singular strata MATH is a MATH-equivariant morphism. Thus MATH on this strata. Third, these bundles are trivial over the NAME open set MATH formed by smooth curves. One can trivialize these bundles on MATH and present bundles on this strata MATH by choosing transition functions from MATH to the neighborhoods of the three degenerate points. A local computation at degenerate curves show that the generic presentation of these bundles on MATH carries through to these three degenerate points. Combining all above we see that MATH. MATH . Type REF and type REF are possible. Since three points out of four two-torsion points are sufficient to determine a basis of the two-torsion points in the Jacobian, the strata for type REF is still MATH. The contribution of type REF to MATH is MATH . The fixed point set of type REF is just one point. The contribution is CASE : MATH . Now all three types can contribute. For type REF the fixed point locus is MATH. The contribution to NAME 's formula is MATH . The type REF contribution is: CASE: MATH . The type REF contribution is CASE : MATH . |
math/0005219 | Define MATH to be the NAME MATH-algebra of MATH. From REF it follows that MATH . Now we define MATH . Because MATH is a multiplicative unitary the linear space MATH is an algebra that acts non-degenerately on MATH. Because MATH is clearly self-adjoint, we get that MATH is a NAME subalgebra of MATH. Working with the NAME algebraic quantum group MATH instead of MATH we obtain that also MATH is a NAME subalgebra of MATH. Observe that it follows from the commutant theorem for the tensor product of NAME algebras that MATH for all MATH. Then we conclude from REF that it is possible to define a one-parameter group MATH of automorphisms of MATH by MATH for all MATH and MATH. It also follows from REF and the fact MATH for all MATH, that we have MATH for all MATH and MATH, which makes sense because MATH. For the same reason we can write MATH and because MATH for all MATH and MATH, we get MATH for all MATH. By the right invariance of MATH it follows that the restriction MATH of MATH to MATH is semifinite. By NAME 's theorem (see for example, CITE) there exists a unique normal faithful conditional expectation MATH from MATH to MATH satisfying MATH for all MATH. From CITE it follows that MATH for all MATH, where MATH denotes the orthogonal projection onto the closure of MATH. So the range of MATH contains MATH for all MATH and MATH. By the right invariant version of REF we get that MATH. So MATH is the identity map and MATH. Working with the NAME algebraic quantum group MATH we obtain MATH. We already proved that MATH is the MATH-strong-MATH closure of MATH and so this concludes the proof of the proposition. |
math/0005219 | Because MATH is manageable and MATH for all MATH, REF imply that MATH is a MATH -algebra, MATH is a non-degenerate MATH-homomorphism from MATH into the multiplier algebra of MATH, such that MATH and both MATH and MATH are dense in MATH. Now define MATH and MATH to be the restriction of MATH and MATH to MATH respectively, giving you two faithful lower semi-continuous weights on MATH. By REF we get that MATH, implying that MATH for all MATH. It follows that MATH. Define MATH to be the restriction of MATH to MATH. Then MATH is a MATH-anti-automorphism of MATH satisfying MATH. It is also clear that MATH. For MATH and MATH, we have that MATH which implies that MATH. We know that MATH and thus MATH for all MATH and MATH. It follows that MATH is densely defined. Define MATH to be the restriction of MATH to MATH. REF guarantees that MATH is dense in MATH. Therefore MATH is a NAME for MATH. REF tells us that MATH implying that MATH for all MATH and MATH. Hence MATH for all MATH. Define the one-parameter group MATH on MATH by setting MATH for all MATH. Notice that REF implies that MATH is norm continuous. Since MATH is a dense subset of MATH, we get that MATH. REF implies for all MATH, MATH and MATH that MATH . Therefore MATH for all MATH and we can define a one parameter group MATH on MATH by setting MATH for all MATH. REF implies that MATH is norm continuous. By now it is clear that MATH is a NAME on MATH (in the MATH -algebraic sense) with MATH as its modular group. Because MATH, we also get that MATH is a NAME on MATH. Take MATH and MATH. Choose MATH. On the MATH -algebra MATH we can make a NAME for the positive functional MATH. This way we obtain a NAME space MATH, a non-degenerate representation MATH of MATH on MATH and a (cyclic) vector MATH such that MATH. By REF we know that MATH belongs to the multiplier algebra of MATH, where MATH denotes the MATH -algebra of compact operators on MATH. Hence the unitary MATH defined by MATH belongs to MATH. Define MATH by setting MATH for all MATH. Then MATH . Therefore the left invariance of MATH implies that MATH belongs to MATH and MATH . Translating REF to the NAME algebra setting, we now conclude that MATH belongs to MATH and therefore to MATH. Taking MATH such that MATH, REF and the left invariance of MATH imply that MATH . So we have proven that MATH is left invariant in the sense of REF. Because MATH and MATH we also get that MATH is right invariant. From all this we conclude that MATH is a reduced MATH -algebraic quantum group. |
math/0005219 | By REF, we know that MATH for all MATH. Therefore the NAME duality theorem guarantees that also MATH for all MATH. Choose MATH. By REF we know that MATH, and so we get that MATH. But NAME theory tells us that MATH belongs to MATH, implying that MATH. NAME duality allows us to conclude that MATH for all MATH and MATH. |
math/0005219 | Take MATH. Then we have for all MATH that MATH . If MATH then the formula above implies for all MATH that MATH and hence MATH. Now suppose that there exists MATH such that MATH. By REF above, we get for all MATH that MATH . Because such elements MATH form a MATH-strong-MATH core for MATH, we get MATH for all MATH. So MATH belongs to MATH. |
math/0005219 | Consider MATH. For every MATH and MATH, we define MATH as MATH . So we have for MATH that MATH and thus MATH from which we conclude that MATH and MATH . It is easy to check that for every MATH we have MATH and MATH. Therefore the closedness of the mapping MATH implies that MATH and MATH . CASE: Let MATH. Then we have for every MATH that MATH. Clearly, MATH converges to MATH. REF implies that MATH converges to MATH. In other words, MATH converges to MATH. Since MATH is dense in MATH and MATH is a core for MATH, we conclude that MATH is dense in MATH and that MATH is a MATH-strong-MATH - norm core for MATH. CASE: Let MATH. Then we have for every MATH that MATH and MATH by REF . So MATH converges to MATH. From this all, we conclude that MATH is dense in MATH. |
math/0005219 | It is clear that MATH is a MATH-subalgebra of MATH. Because MATH is a left ideal in MATH, we get that MATH. Thus in order to prove that MATH is dense in MATH, it is by the previous lemma enough to prove that MATH is dense in MATH. But we have for all MATH with MATH, MATH and MATH that MATH which easily implies that MATH is dense in MATH. Hence MATH is dense in MATH. Since MATH is dense in MATH, MATH belongs to the MATH-strong-MATH closure of MATH. Combining this with the fact that MATH is a MATH-strong-MATH - norm core for MATH and the inclusion MATH, we conclude that MATH is a MATH-strong-MATH - norm core for MATH. |
math/0005219 | Since MATH, the definition of MATH gives clearly that MATH. From REF , we know that MATH is a dense subspace in MATH. We now use the notation MATH as it was introduced in notation REF For every MATH we denote by MATH the element in MATH defined by MATH. Then MATH and MATH for all MATH and MATH. If MATH and MATH, it is not so difficult to check that MATH and MATH. It follows that MATH for all MATH, hence MATH for all MATH. Therefore MATH for all MATH. We conclude from all this that MATH is a core for MATH (see for example, REF) and the lemma follows. |
math/0005219 | Choose MATH. Then MATH . Therefore the definition of MATH and MATH imply that MATH . Thus the previous lemma implies that MATH belongs to MATH and MATH. |
math/0005219 | Using REF and the strong left invariance of MATH (see REF), the previous result implies easily MATH. Define the subspace MATH of MATH as MATH. Let MATH. Remember that MATH for all MATH. Choose MATH. Then MATH and MATH belong to MATH and MATH . Therefore the element MATH belongs to MATH. We conclude that MATH is a dense subspace of MATH, invariant under the family of operators MATH. It follows that MATH is a core for MATH and thus a core for MATH. Combining this with the fact that MATH, we conclude that MATH. Now the uniqueness of the polar decomposition implies that MATH and MATH. |
math/0005219 | Define the anti-unitary MATH such that MATH for MATH. Choose MATH such that MATH and MATH. For MATH, we define MATH such that MATH. Remember that MATH is analytic with respect to MATH and MATH, implying that MATH and MATH . Since MATH we see that MATH for MATH, hence MATH and MATH. By assumption, MATH, so MATH and MATH. Hence MATH is a bounded operator and its closure equals MATH. Define the strongly continuous one-parameter group MATH of isometries of MATH such that MATH for MATH and MATH. The discussion above implies (see for example, REF) that MATH and MATH. By REF, implying that MATH belongs to MATH. So we see that MATH is a bounded operator and that its closure equals MATH. Since MATH, this implies that MATH and MATH . We know that we have for every MATH that MATH and MATH. Since MATH and MATH, we conclude (see for example, REF) that MATH and MATH . Since MATH converges to MATH and MATH converges to MATH, the closedness of MATH implies that MATH and MATH . Consequently MATH . Since such elements MATH form a core for MATH, such elements MATH form a dense subspace of MATH. Therefore MATH and we are done. |
math/0005219 | As already mentioned in REF, MATH is the modular conjugation of MATH in the NAME MATH. Choose MATH. Since MATH for all MATH, we get that MATH and MATH. Hence MATH and MATH. Combining this with the previous proposition and the definition of the modular conjugation, we get that MATH . Therefore MATH. |
math/0005219 | It is easy to check that MATH for all MATH. We already mentioned that MATH for all MATH and by definition we have MATH and MATH for all MATH. Because MATH for all MATH it is easy to verify that MATH for all MATH and by definition we have MATH for all MATH. Using that all three one-parametergroups MATH, MATH and MATH commute and that MATH and MATH for all MATH, it is straightforward to check the first equality in REF by applying the operators to an element MATH with MATH. Using REF and the fact that MATH we can check the equalities MATH and MATH on a vector MATH when MATH. This gives the first equalities of REF , and the rest of REF follows from modular theory, because MATH is the modular conjugation of MATH in the NAME MATH. By the biduality theorem we also get MATH and MATH. Because for all MATH we have MATH, we get MATH. This implies that MATH for all MATH and we can check then the second equality in REF on a vector MATH with MATH. Because MATH for all MATH and MATH, we get the second equality of REF follow because MATH and MATH implement respectively MATH and MATH on MATH. Also MATH implements MATH on MATH and this gives the first equality of REF . Because we already saw that MATH the second equality of REF follows immediately from REF . By the biduality theorem we can indeed perform the operation stated in the proposition, because MATH for all MATH and so in a sense MATH. |
math/0005219 | The first equality follows from REF. Combining REF and our REF we get the second equality. Dualizing this we get MATH and this gives the third equality after applying the first one. The final statement follows from the fact that MATH, the previous formulas and the equalities MATH, MATH, MATH and MATH. The first two of these equalities follow from modular theory and the last two from REF . |
math/0005219 | We will prove the proposition for the dual NAME algebraic quantum group MATH. Because of the biduality theorem this proves the stated result. We also represent MATH on a NAME space MATH and then it is enough to prove the proposition in case MATH. Recall that we introduced the multiplicative unitary MATH in REF . Then define for every MATH the element MATH by the following formula, which makes sense because MATH. MATH . When MATH we denote by MATH the positive rank one operator defined by MATH. Let now MATH and suppose MATH. Choose an orthonormal basis MATH of MATH such that MATH for some MATH. Choose MATH. Then we have MATH . In REF we saw that MATH where MATH. The last equality follows from REF . So it follows that MATH . In REF we saw that for MATH one has MATH if and only if MATH. So it follows that MATH if and only if MATH and in that case MATH . Suppose that MATH is a unitary. And suppose that REF is valid. We claim that MATH and MATH . For this choose MATH and make the following computation: MATH . From this follows our claim. But then we get for every MATH and every unitary MATH that MATH . From this we may conclude that MATH, for all MATH. Let now MATH. Let MATH again be an orthonormal basis for MATH. Then MATH . By lower semicontinuity of MATH we can conclude that MATH. Let now MATH. Then MATH and this clearly belongs to MATH. But it also belongs to MATH by the result in the previous paragraph. Let MATH be the unique spectral decomposition of MATH, considered as an element of MATH. Then MATH because MATH. So take MATH such that MATH and MATH. Then define the element MATH by MATH . Then we get that MATH . Let us now suppose first that MATH. This will prove the special case stated in the proposition. Then MATH and MATH will be a scalar. So we get a MATH such that MATH . Now there are two possibilities. CASE: Either there exists a MATH with MATH such that MATH. Then MATH because MATH . But then also MATH for all MATH, and so MATH for all MATH. Then it follows from REF that MATH and so MATH because of left invariance. CASE: Either we have MATH for all MATH. This means that MATH. Because of left invariance we cannot have MATH and so MATH. Again MATH. In both cases we arrive at MATH. Now we return to the general case. Let MATH and MATH. Then we apply MATH to REF . This gives us MATH . In this computation we used the special case of the proposition proved above. So it follows that MATH and this gives what we wanted to prove. |
math/0005219 | Because MATH, the NAME algebra underlying MATH is given by MATH . The last equality follows from REF . Further we have for every MATH that MATH . Because MATH, this gives MATH . This gives MATH. Applying this last formula to MATH and using the biduality theorem we get MATH. Taking the dual and using once again the biduality theorem this gives our second result MATH. To compute MATH we have to observe once again that the modular conjugation MATH of the left invariant weight MATH on MATH is given by MATH. Then it is clear that MATH. |
math/0005227 | Note that it suffices to show that there exists an equivalence bimodule MATH so that MATH implies MATH, for MATH, MATH (that is, the MATH-action on MATH is strongly non-degenerate and MATH implies MATH for MATH) and MATH. Let MATH be an arbitrary equivalence bimodule. We start by noticing that we can choose it so that the MATH and MATH actions are strongly non-degenerate. Indeed, it follows from the compatibility of MATH and MATH and their fullness properties that MATH. If we denote MATH, then MATH has a natural MATH-bimodule structure and also compatible MATH and MATH valued inner products, defined by simply restricting MATH and MATH to MATH. Note that given MATH, by idempotency we can write MATH, for MATH, and by fullness of MATH, we can write MATH. So MATH, showing that MATH restricted to MATH is still full, and the same holds for MATH. Therefore MATH is a MATH-equivalence bimodule with strongly non-degenerate MATH and MATH actions. We also observe that if MATH is an (arbitrary) equivalence bimodule, then MATH and MATH act on MATH faithfully (that is, MATH for all MATH implies MATH, for all MATH, and the same for MATH). To see that, suppose MATH for all MATH. Then given an arbitrary MATH, it follows that we can write MATH and hence MATH. Therefore MATH by non-degeneracy of MATH and the same argument holds for MATH. It then follows (see the proof of CITE) that MATH and MATH is still a MATH-equivalence bimodule, with the property that MATH implies that MATH, the same holding for MATH. Moreover, if the actions of MATH and MATH were strongly non-degenerate on MATH, this still holds for the quotient MATH. So, we can always find an equivalence bimodule MATH with strongly non-degenerate MATH and MATH actions and inner products satisfying MATH and MATH. We observe that, in this case (see proof of CITE), MATH. Given MATH an equivalence bimodule with these properties, we note that if MATH, then in particular MATH for all MATH. But then MATH for all MATH, and hence MATH. But since MATH is also arbitrary, it follows that MATH. Therefore, MATH and clearly establishes a NAME MATH-equivalence between MATH and MATH. |
math/0005227 | Let MATH be positive in MATH and let MATH be an arbitrary positive linear functional (in the algebraic sense, that is, MATH for all MATH). We must show that MATH. Note that by REF ideals (see CITE), if MATH, then MATH, where MATH denotes the MATH-algebra generated by MATH. Also observe that if MATH, then MATH, and hence MATH, for some MATH. Therefore, MATH. For the converse, note that if MATH is a positive linear functional in MATH, then MATH is positive in MATH and hence if MATH, we immediately have MATH. Hence MATH. |
math/0005227 | We must show that given a complex pre-Hilbert space MATH and a MATH-representation MATH, the formula MATH, for MATH and MATH defines a positive semi-definite inner product on MATH. Note that since MATH extends to a MATH-representation MATH on the NAME space completion of MATH, there is a natural isometric embedding MATH, where the MATH-balanced tensor products are defined using MATH and MATH, respectively. So for MATH and MATH, we have MATH . Therefore, to show the positivity of MATH, it suffices to show that the right hand side of REF is non-negative. This can be done by the usual procedure (see CITE), using the fact that any MATH-representation of MATH on a NAME space can be decomposed into the sum of a trivial representation and a non-degenerate one, and the non-degenerate part itself can be decomposed into the sum (in the topological sense) of cyclic (also in the topological sense) MATH-representations (see CITE). |
math/0005227 | For the first part use the projection MATH to pull-back MATH-representations. For the second observe that if MATH then MATH. |
math/0005227 | The invariance of MATH is obvious and the fact that MATH is strongly non-degenerate follows from the idempotency of MATH. Finally let MATH. Then in particular MATH for all MATH whence MATH follows and thus MATH. The other inclusion is trivial. The strong non-degeneracy of a GNS representation follows directly from the definition. |
math/0005227 | Let us first observe that MATH is a well-defined map. Note that if MATH is a MATH-representation of MATH, then MATH is always norm dense in MATH. To see that, let MATH be an approximate identity in MATH, with MATH. This is possible since MATH is dense in MATH CITE. Then given MATH, we have MATH and MATH. Now let MATH and suppose MATH and MATH are two different MATH-representations of MATH so that MATH. We must show that MATH. Note that if MATH, then we can find MATH with MATH. So MATH, for all MATH, and hence MATH. This shows that MATH and by changing the roles of MATH and MATH we conclude that MATH. This proves well-definedness. It is not hard to check that MATH is a bijection which preserves the order structure of the lattices. This guarantees that MATH is a lattice isomorphism. |
math/0005227 | It remains to show MATH. But this follows by polarization and the fact that MATH is torsion-free. |
math/0005227 | This generalizes REF . Let MATH and MATH such that MATH. Then MATH implies MATH, whence MATH. For MATH, we find by the NAME inequality that MATH and hence MATH. |
math/0005227 | Let MATH. Then MATH for all MATH-representations MATH since otherwise there would exist a MATH such that MATH which contradicts the assumption since MATH is a positive functional. Hence MATH and by REF the equality REF follows. The other statements follow directly from CITE. |
math/0005227 | If MATH is a positive linear functional then MATH is positive, too, whence MATH follows from REF . Then REF is clear. |
math/0005227 | The first part is a simple computation thus assume that the inner product is positive semi-definite and satisfies P. Let MATH be a MATH-representation of MATH on MATH and let MATH be the induced MATH-representation of MATH. Then MATH if and only if MATH for all MATH and MATH since the equivalence classes of those MATH span MATH. But since the inner product of MATH is non-degenerate this is equivalent to MATH for all MATH and MATH. Thus it is equivalent to MATH and hence MATH which implies REF . Hence closed MATH-ideals are mapped to closed MATH-ideals and since MATH and MATH are given by MATH and MATH the proof is completed. |
math/0005227 | By REF if and only if one has MATH for all MATH and MATH. Then it is a simple computation to obtain REF . |
math/0005227 | The fact that MATH and MATH are inverse to each other follows from the above considerations. Hence it remains to show that MATH (and thus MATH) is a lattice homomorphism. But this follows from bijectivity and REF . |
math/0005227 | Let MATH. We may assume, due to REF , that MATH is strongly non-degenerate. Hence, by CITE, we know that MATH is unitarily equivalent to MATH. So they have the same kernel and by REF it follows that MATH. Analogously, one obtains MATH for all closed MATH-ideals of MATH and hence MATH and MATH are inverse to each other. The homomorphism properties follow as in REF . |
math/0005227 | Let MATH be an equivalence bimodule. Note that if MATH is non-degenerate and idempotent, then MATH for all MATH implies that MATH (see the proof of REF ). Hence the conclusion follows. |
math/0005227 | The first statement follows directly from REF by transitivity. It also follows from REF that MATH has trivial minimal ideal, while MATH. So the result follows by REF . |
math/0005227 | Let MATH be a closed set in MATH. Denote by MATH the set of positive compactly supported NAME measures on MATH with support in MATH. By the continuity of functions in MATH, it follows that MATH. Thus MATH is closed. Conversely, if MATH is closed, then it is the intersection of kernels of GNS representations of some positive linear functionals. But the positive linear functionals in MATH are given by positive NAME measures, see for example, CITE. So we can write MATH, for MATH and define MATH. It is then easy to check that MATH. The last assertion follows from the fact that if MATH is compact, then there is a positive NAME measure with support exactly MATH. |
math/0005227 | Due to the nilpotence properties of forms with positive degree, it is easy to see that MATH has to be contained in the minimal ideal. On the other hand, MATH has sufficiently many positive linear functionals and thus the first statement follows. The other statement is clear from the above results and REF. |
math/0005227 | Assuming MATH, the MATH's are just MATH-valued functions on MATH and MATH. Note that if MATH denote the canonical section basis of MATH, then for each MATH one can find a MATH so that MATH (recall that MATH means MATH, where the last expressions is just the usual matrix multiplication). For instance, define MATH to be the matrix valued function with MATH as the first row and zero everywhere else. Now the proof follows the usual trick: MATH. To prove the result in general, let MATH, MATH be an open cover of MATH so that MATH is trivial. Let MATH be a quadratic partition of unity subordinated to this cover. Then observe that for MATH, we have that MATH (since MATH). Hence, MATH . Now note that the last expression is just MATH. But since MATH, and MATH is trivial, the result follows from the previous discussion. |
math/0005228 | By REF, we get MATH for every horizontal vector MATH and for every vertical vector MATH. Therefore MATH for every vertical vector MATH. By REF, we have MATH for every linearly independent horizontal vectors MATH and MATH. |
math/0005228 | Since MATH is geodesically complete, the base space MATH is complete. Let MATH be the Riemannian metric on MATH defined by MATH for every MATH, MATH vector fields on MATH. Since MATH is a horizontally complete Riemannian metric (this means that any maximal horizontal geodesic is defined on the entire real line) and MATH is a complete Riemannian manifold then MATH is an NAME connection for MATH (see REF). By REF, it follows MATH is a locally trivial fibration and we have an exact homotopy sequence MATH . Since MATH is simply connected, we have MATH. If MATH has nonpositive curvature, then MATH by theorem of NAME. It follows MATH. |
math/0005228 | In order to prove that MATH is a locally symmetric space we need to check that MATH. Let MATH, MATH, MATH, MATH be vector fields on MATH and let MATH, MATH, MATH, MATH be the horizontal liftings of MATH, MATH, MATH, MATH respectively. By definition of the covariant derivative we have MATH . In order to prove that the curvature tensor MATH of the base space is parallel, we have to lift all vector fields in relation REF. By REF , the horizontal liftings of MATH, MATH and MATH are MATH, MATH and MATH, respectively. We denote by MATH the horizontal lifting of MATH. The convention for NAME tensor used here is MATH. NAME 's equation MATH gives us the following relation MATH . Using this relation we compute MATH . Since MATH has constant curvature, we have MATH for every vertical vector MATH and for every horizontal vector fields MATH, MATH, MATH. This implies MATH is horizontal and MATH, MATH, MATH are vertical. Hence MATH. Since MATH has constant curvature, we get MATH. So the sum of the first four terms in relation REF is zero. We have MATH. For the case of totally geodesic fibres, NAME 's equation MATH becomes MATH . By REF and by the hypothesis of constant curvature total space we get MATH for every horizontal vector fields MATH, MATH, MATH and for every vertical vector field MATH. It follows MATH and MATH for every horizontal vector fields MATH, MATH, MATH and for every vertical vector field MATH. Therefore MATH and MATH. This implies MATH . By circular permutations of MATH in the last relation we get MATH . So the sum of all terms in relation REF is zero. We proved that MATH for every vector fields MATH, MATH, MATH, MATH, so MATH is a locally symmetric space. By REF , MATH is simply connected and complete. Therefore MATH is a Riemannian symmetric space. By REF , MATH has negative sectional curvature. Hence MATH is a noncompact Riemannian symmetric space of rank one. Let MATH. Since MATH is a totally geodesic submanifold of a geodesically complete manifold, MATH is itself geodesically complete. Since MATH for every MATH, MATH tangent vectors to MATH, we have MATH, by REF . Since MATH is a complete, simply connected semi-Riemannian manifold of dimension MATH and of index MATH and with constant sectional curvature MATH, it follows MATH is isometric to MATH (see REF from page REF). Hence any fibre is diffeomorphic to MATH. We shall prove below that the tangent bundle of any fibre is trivial. From a well known result of NAME it follows that MATH. The tangent bundle of any fibre is trivial. Since MATH, we have that MATH, MATH is an injective map and MATH, if MATH. For any horizontal vector field MATH, we denote by MATH the map given by MATH. By NAME 's equation MATH, we have MATH for every vertical vector fields MATH and MATH. Hence, by REF MATH, we get MATH for every vertical vector field MATH. If MATH anywhere then MATH is surjective and hence MATH. By REF MATH, we have MATH. This implies MATH. Let MATH and MATH with MATH. We denote by MATH the horizontal lifting along the fibre MATH of the vector MATH. Let MATH an arbitrary point in MATH and let MATH be an orthonormal basis of the vector space MATH. Since MATH sends isometrically MATH into MATH we have MATH is a linearly independent system which can be completed to a basis of MATH with a system of vectors MATH. Let MATH be the horizontal liftings along the fibre MATH of MATH respectively. By REF , we have for every MATH and for every MATH . Since the induced metrics on fibre MATH are negative definite, we get MATH. By REF , we have MATH. We proved that MATH. Since MATH sends isometrically MATH into MATH, we get MATH is a basis of the vector space MATH for every point MATH. Let MATH be tangent vector fields to the fibre MATH. We denote by MATH the vector subspace of MATH spanned by MATH. Let MATH be the Riemannian metric on MATH given by MATH for every MATH, MATH vector fields tangent to MATH. Since MATH and MATH, we get MATH is an isometry for every MATH. So MATH is a global frame for the tangent bundle of MATH. It follows the tangent bundle of the fibre MATH is trivial. This ends the proof of REF . |
math/0005228 | Let MATH, MATH be two linear independent horizontal vectors and let MATH, MATH. By REF , the metric induced on fibres are negative definite. This implies MATH. By REF , we get MATH . By NAME inequality applied to the positive definite scalar product induced on MATH, we have MATH . By REF , we get MATH . Thus MATH. Therefore MATH for every orthogonal vectors MATH and MATH. We proved that MATH. We shall prove that if the base space MATH has constant curvature MATH then MATH. It is sufficient to see that for any point MATH there is a MATH-plane MATH such that MATH. We choose MATH where MATH is a horizontal vector and MATH is a vertical vector. By REF , we have MATH . We notice that MATH and MATH are orthogonal, because, by REF , we have MATH. By REF , we have MATH for every horizontal vector MATH and for every vertical vector MATH. By REF , we get MATH. Hence, by polarization, we find MATH for every horizontal vector MATH and for every vertical vector MATH. Therefore the relation REF becomes MATH . Then MATH. Therefore if the base space MATH has constant curvature MATH then MATH. Let MATH be a horizontal vector field. By REF , MATH if and only if MATH . For every MATH, we denote by MATH . With this notation, MATH. Since MATH sends isometrically MATH into MATH, we have MATH . We compute MATH from the geometry of MATH. We have the following possibilities for MATH: MATH. The curvature tensor of hyperbolic space MATH is given by MATH . We have MATH. Hence MATH. It follows MATH. If MATH then MATH is isometric to MATH, which falls in REF , since MATH is isometric to MATH. If MATH then MATH and this is REF. MATH. Let MATH be the natural complex structure on MATH. The curvature tensor of complex hyperbolic space MATH with MATH is given by MATH . We get MATH. So MATH. It follows MATH. MATH. Let MATH be local almost complex structures which rise to the quaternionic structure on MATH. The curvature tensor of the quaternionic hyperbolic space MATH with MATH (see CITE) is given by MATH . It follows that MATH if and only if MATH. Therefore MATH. Hence MATH. We get MATH. MATH. Let MATH be local almost complex structures which rise to the NAME structure on MATH . NAME hyperbolic plane. The curvature tensor of the NAME plane MATH with MATH (see CITE) is given by MATH . We get MATH . So MATH. It follows MATH. |
math/0005228 | The index of the pseudo-hyperbolic space cannot be MATH. Indeed, by REF , for MATH, we get MATH for every horizontal vector MATH and for every vertical vector MATH. But this is not possible. By CITE, any semi-Riemannian submersion with totally geodesic fibres, from a pseudo-hyperbolic space of index MATH onto a Riemannian manifold is equivalent to the canonical semi-Riemannian submersion MATH. It remains to study the case MATH. By REF , any semi-Riemannian submersion with totally geodesic fibres from a pseudo-hyperbolic space of index MATH onto a Riemannian manifold is one of the following types: CASE: MATH, or CASE: MATH, or CASE: MATH. In order to prove that any two semi-Riemannian submersions in one of the categories REF or REF are equivalent we shall modify NAME 's argument (see CITE) to our situation. In the category REF , we shall prove there are no such semi-Riemannian submersions with totally geodesic fibres. First, we shall prove the uniqueness in the case MATH. Let MATH and let MATH the map given by MATH for every MATH and for every MATH. We denote MATH by MATH. It is trivial to see that MATH is skew-symmetric (that is, MATH). The NAME 's equation MATH becomes MATH. This implies MATH. Hence, by polarization in MATH, we have MATH for every MATH. So MATH. Again by polarization we get MATH. Let MATH be the Riemannian metric given by MATH for every MATH, MATH. It follows MATH. This is the condition which allows us to extend MATH to a representation of the NAME algebra MATH of MATH. We also denote by MATH the extension of MATH. Since MATH and MATH is positive definite, MATH has at most two types of irreducible representations. We notice that MATH is a MATH-module which splits in simple modules of dimension MATH. The next step is to show that any two such simple modules in decomposition of MATH are equivalent. Let MATH be an orthonormal basis of MATH. Since the affiliation of a simple MATH-module to one of the two possible types is decided by the action of MATH, it is sufficient to check that MATH. Consider the function MATH defined on the unit sphere in MATH. We have MATH. A straightforward computation shows that MATH is orthogonal to MATH and MATH. Hence MATH is a multiple of MATH. By polarization of the relation MATH, we get MATH for every horizontal vectors MATH and MATH . In particular, we have MATH. Let MATH be the vector subspace of MATH spanned by MATH. By REF , we get MATH for all MATH. By geometry of MATH, there exists a unique totally geodesic hyperbolic line MATH passing through MATH such that MATH. Notice that for every orthonormal vectors MATH, MATH. In particular we have MATH. Hence MATH. It follows that MATH. Hence MATH for all unit vectors MATH. Since the function MATH defined on the unit sphere in MATH is continuous, we get either CASE: MATH for any unit horizontal vector MATH, or CASE: MATH for any unit horizontal vector MATH. We may assume REF holds. If REF is happen, we replace the orthonormal basis MATH of MATH with the orthonormal basis MATH. So for this new basis we are in REF . Since MATH is an isometry, we have MATH for all unit horizontal vectors MATH. So the NAME inequality for the scalar product MATH becomes equality. It follows that MATH for some MATH. Because MATH is an isometry and we assumed REF , it follows MATH. We proved that MATH for all unit horizontal vectors MATH. Obviously, MATH for all MATH. Let MATH be another semi-Riemannian submersion with totally geodesic fibres. For an arbitrary chosen point MATH, we consider horizontal and vertical subspaces MATH and MATH. Let MATH be an orthonormal basis in MATH such that MATH acts on MATH as MATH. Let MATH be the isometry given by MATH for all MATH and let MATH be the extension of MATH to the NAME algebras. The composition MATH makes MATH to be a MATH-module of dimension MATH. Let MATH and MATH be the decomposition of MATH and MATH in simple MATH-modules, respectively. For each MATH there is MATH an equivalence of MATH-modules, which after a rescaling by a constant number is an isometry which preserves the NAME 's integrability tensors. Taking the direct sum of all these isometries, we obtain an isometry MATH which preserves the NAME 's integrability tensors. Therefore MATH is an isometry which maps MATH onto MATH and MATH onto MATH. Since MATH is a simply connected complete symmetric space, there is an isometry MATH such that MATH and MATH (see REF). Therefore, by REF , we get MATH and MATH are equivalent. Now, we shall prove that any two semi-Riemannian submersions MATH with totally geodesic fibres are equivalent. The proof is analogous to the case MATH, but it is easier. Let MATH and let MATH, MATH be the horizontal and vertical subspaces in MATH for MATH, let MATH, MATH be the horizontal and vertical subspaces in MATH for MATH. Let MATH be an orthonormal basis of MATH and MATH be an orthonormal basis of MATH such that MATH and MATH. Since MATH, the irreducible MATH-modules are MATH-dimensional. Since MATH, we get MATH is simple. Because MATH and MATH we get MATH and MATH are MATH-modules equivalent. Analogously to REF , we can construct an isometry MATH, which map MATH onto MATH and MATH onto MATH. This produces an isometry MATH such that MATH and MATH (see REF). Again by REF , we get MATH and MATH are equivalent. Now, we prove that there are no MATH semi-Riemannian submersions with totally geodesic fibres. The proof is analogous to that of NAME (see REF). MATH becomes a MATH-module by considering the extension of the map MATH, MATH to the NAME algebra MATH. Here MATH denotes the NAME algebra of MATH, MATH for every MATH, MATH. Since MATH is positive definite, we have MATH. Hence, MATH splits into two MATH-dimensional irreducible MATH-modules. Since the induced metrics on fibres are negative definite we get MATH is totally geodesic in MATH and isometric to MATH, by REF. Here MATH denotes the NAME hyperbolic line through MATH; We choose MATH be the horizontal space of the restricted submersion MATH. So for every MATH, MATH we find an irreducible MATH-submodule MATH of MATH passing through MATH. Since MATH, we get a contradiction. |
math/0005228 | We denote by MATH the natural almost complex structure on MATH. By REF , we have REF MATH. Hence the fibres have constant holomorphic curvature MATH. CASE: MATH, since the fibres are complex submanifolds. We obtain MATH for every vertical vector field MATH. CASE: MATH, since the induced metrics on fibres are negative definite. By REF , it follows that the fibres are simply connected. Since the fibres are complete, simply connected, complex manifolds with constant holomorphic curvature MATH, we have that the fibres are isometric to MATH. |
math/0005228 | Let MATH be the canonical semi-Riemannian submersion with totally geodesic fibres given in REF (see also CITE or CITE). We have MATH is a semi-Riemannian submersion with totally geodesic fibres, by REF. Since the dimension of fibres of MATH is greater than or equal to REF, we get, by main REF , the following possible situations: CASE: MATH, MATH and MATH is isometric to MATH or CASE: MATH, MATH and MATH is isometric to MATH. First, we shall prove that any two semi-Riemannian submersions MATH with connected, complex, totally geodesic fibres are equivalent. By proof of REF , we have MATH. Let MATH . By proof of the main theorem, this implies MATH. The extension of MATH constructed in proof of the main theorem, to the NAME algebra MATH makes MATH a MATH-module which splits in MATH irreducible modules of dimension MATH. By classification of irreducible representation for case MATH and MATH positive definite, we have any two such irreducible MATH-modules are equivalent. Like in proof of the main theorem, we may construct an isometry MATH, which maps MATH onto MATH and MATH onto MATH. This produces an isometry MATH with MATH and MATH (see REF). Again by REF , we get MATH and MATH are equivalent. For the case MATH we shall obtain that there are no MATH semi-Riemannian submersions with complex, connected, totally geodesic fibres. There are no MATH semi-Riemannian submersions with connected, complex, totally geodesic fibres. The proof is based on NAME 's argument (see proof of main theorem in CITE). Here, we show how to modify NAME 's argument to our different situation. Suppose there is MATH a semi-Riemannian submersion with complex, connected, totally geodesic fibres. By main REF , MATH is equivalent to the canonical semi-Riemannian submersion MATH given by REF . Let MATH be the spin representation of MATH. MATH acts on MATH via double covering map MATH and transitively on MATH. We denote by MATH the even component of NAME algebra MATH. Notice that MATH, MATH and the volume element MATH in MATH satisfies MATH (see CITE). For any MATH, let MATH be the isotropy group of MATH in MATH. If we restrict MATH then MATH breaks MATH into two MATH-spin representations. We will denote them by MATH. Hence MATH. Let MATH. We have MATH, MATH and the following diagram is commutative MATH where all arrows are standard inclusions. Let MATH be an orientated basis of MATH. Then MATH lies in the centre of MATH and MATH acts by MATH on MATH and MATH on MATH. We have MATH on MATH. Since MATH, MATH is invariant under MATH and so is MATH. Here MATH denotes the natural complex structure on MATH. Hence MATH commutes with MATH. Let MATH be the generator of the center of MATH. We have either MATH or MATH. Therefore MATH commutes with MATH for every MATH and hence for every MATH. Consider the linear map MATH given by MATH. It has the following properties: CASE: It factors through MATH CASE: MATH . Hence MATH extends to a homomorphism MATH. But MATH (see CITE). So the above homomorphism is impossible to exist. We get the required contradiction. This ends the proof of REF . |
math/0005234 | Let MATH be a surface homeomorphic to the MATH-times punctured REF-sphere, with a marking MATH, that is, a labelling of the punctures. Let MATH be the space of convex ideal polyhedra in MATH, parametrized by the positions of their vertices on the sphere at infinity of MATH (interpreted as the NAME sphere MATH). NAME of the vertices are fixed at MATH, MATH and MATH, which eliminates the action of the isometry group of MATH. MATH is easily seen to be a MATH dimensional manifold. MATH plays the role of MATH in the invariance of domain principle. We will abuse notation and view a polyhedron MATH both as a geometric object and as a polyhedral isometric embedding of MATH into MATH. MATH will inherit the labelling of vertices from MATH. Let MATH be the set of complete, finite volume hyperbolic structures on MATH. This set is parametrized by shears along the edges of a geodesic triangulation. This parametrization is explained in detail in REF. It will also be shown REF that MATH is a MATH dimensional contractible manifold. Although many of the results of REF are known to NAME theorists, they are so elementary in this particular setting that it was impossible to resist including a full exposition. MATH will play the role of MATH in the invariance of domain principle. The role of the map MATH will be played by the map MATH. MATH is MATH viewed as an abstract Riemannian manifold. The continuity of MATH with respect to the chosen coordinate systems on MATH and MATH is the content of REF . Since MATH and MATH are manifolds of the same dimension and MATH is continuous, REF shows that MATH is an open map. That MATH is closed is the content of REF . |
math/0005234 | Use the upper half-space model of MATH. Recall that the hyperbolic metric is related to the Euclidean metric on the upper half-space by MATH. This means, in particular, that the hyperbolic distance between MATH and MATH is MATH. By a hyperbolic isometry MATH can be sent to MATH, MATH to MATH and MATH to MATH. NAME centered on MATH are then simply horizontal lines, and if MATH, then MATH, since the metric on the horocycle around infinity through MATH is then simply the standard metric on the real line, rescaled by MATH. The assertion of the Lemma now follows. |
math/0005234 | Both MATH and MATH are invariant under the NAME group, and since any three points can be transformed to MATH, MATH, and MATH by a NAME transformation, the Lemma follows from the obvious special case where MATH, MATH, and MATH. |
math/0005234 | All that is necessary to show is that if MATH, MATH, MATH and MATH are cusps and there are non-intersecting curves MATH connecting MATH and MATH and MATH connecting MATH and MATH, then the corresponding geodesics also don't intersect. That this last statement is true can be observed by examining the geometry of universal cover of MATH. The postulated curves MATH and MATH exist if and only if the lifts of MATH and MATH do not separate the lifts of MATH and MATH on the circle at infinity of MATH. In that case, however, it is seen that the corresponding geodesic segments do not intersect either. |
math/0005234 | Pick two distinguished vertices MATH and MATH. For any starting triangulation MATH there is a sequence of NAME moves transforming MATH into a triangulation MATH where MATH is connected to every other vertex. Now consider the complement in MATH of MATH and all the edges incident to it. This will be a triangulation of a MATH-gon, all of whose vertices are those of the MATH-gon. By a similar argument, this can be transformed by a sequence of NAME moves into a triangulation where every vertex is connected to MATH. Thus, it is seen that by a sequence of NAME moves, every triangulation can be transformed to a particular triangulation (where both MATH and MATH have valence MATH), and thus MATH is connected. |
math/0005234 | It is enough to note that the cross ratio of any permutation MATH is a rational function of the cross ratio of MATH (this is most easily seen when MATH). By REF flipping the diagonal of MATH corresponds to permuting the arguments of the cross ratio. Since a permutation of the arguments corresponds to a fractional linear transformation of the cross-ratio itself, REF follows. |
math/0005234 | Consider MATH. REF is equivalent to the observation that the polyhedron MATH is convex REF if and only if all of its edges are convexly bent (Condition MATH). |
math/0005234 | NAME move of the described type corresponds to filling in a missing tetrahedron MATH of a polyhedron lying above MATH. Every time a move as above happens, the edge MATH is buried, never to be seen again. Since the number of possible edges is finite, the result follows. |
math/0005234 | First, observe that the angle between the the faces MATH and MATH is equal to the angle between the circles MATH and MATH. The rest of the proof is contained in REF . |
math/0005234 | This is just a ``bent" three-dimensional version of REF . |
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