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math/9912224 | Note that MATH is the unitary corresponding to MATH. It follows that MATH together with MATH generates MATH. Hence MATH is a weak generator. For any fixed MATH, we have MATH since the set MATH is free from MATH with amalgamation over MATH (see REF ). On the other hand, MATH by REF . Hence MATH, and thus MATH. Hence MATH is a generator. |
math/9912224 | We have that MATH, since MATH is a weak generator. But MATH, since MATH are free with amalgamation over MATH. Since each MATH implements a free action of the integers, MATH, so that MATH. Thus MATH. |
math/9912225 | This claim hold trivially for MATH. Suppose that it holds for MATH, we prove it for MATH. Let MATH and MATH. By induction the minimum energy contained in the given springs is MATH where we have for convenience let MATH denote MATH and MATH denote MATH. This energy is minimized when MATH at which point the energy takes the value MATH as claimed. |
math/9912226 | CASE: We have : MATH where we used the definition of MATH and REF . The second identity is similar. CASE: Since MATH we can compute : MATH using REF , definition of the source counital subalgebra, and the identity MATH that follows from REF . Observe that MATH is a separability idempotent CITE of MATH. CASE: Using REF and the fact that MATH and MATH commute, we have MATH . |
math/9912226 | First, we need to check that MATH is well defined. For all MATH and MATH we have : MATH where we used definition of the target counital subalgebra, REF , and that MATH for all MATH. Next, we verify that MATH for all MATH. For all MATH we have : MATH using the identity MATH and REF . The following computation shows that MATH commutes with the right action of all MATH : MATH . Finally, MATH for all MATH, therefore, MATH is a homomorphism. |
math/9912226 | We need to check that MATH . For all MATH, and MATH we compute MATH where we used REF and the properties of the element MATH. Also, for every MATH we have : MATH where we used that MATH commutes with the right multiplication by elements from MATH and identities from REF . |
math/9912226 | Follows from REF . |
math/9912226 | We know that MATH, where MATH is the trivial MATH-module algebra, therefore applying REF to MATH we see that MATH is a projective generating MATH-module such that MATH. Therefore, MATH and MATH are NAME equivalent. Since MATH is always semisimple (as a separable algebra), MATH is semisimple. |
math/9912228 | REF implies: MATH . The first part of the statement in the theorem follows directly from REF . If we define MATH then REF is a direct consequence of the equality REF . |
math/9912228 | We only have to prove the existence of the representation of the residues of MATH as stated in the theorem. For MATH and MATH let MATH be the NAME distribution given by the sum of the smooth densities MATH each defined on the connected component MATH of the fixed point set MATH as described in REF and by REF . We have MATH for any MATH, and after a convenient reindexing of the connected components of each fixed point set, we can suppose that MATH as well. Then MATH because, as defined by REF , they are determinants of two linear maps conjugated by a diffeomorphism defined at the tangent space level by MATH. Because the operator MATH is MATH equivariant, a straightforward computation shows that the smooth densities MATH and MATH are conjugated by the the map induced at the cotangent space level by MATH. Then the collection MATH is a MATH . NAME distribution, as described in REF states that the residue of MATH at MATH is equal to MATH which can be rewritten as MATH. |
math/9912228 | Let us fix a MATH invariant metric on MATH. This can be done by averaging any metric over the compact group MATH. Let MATH be a point in the quotient space and MATH such MATH. Let MATH be the isotropy group at MATH. Then MATH acts on the tangent space MATH and keeps invariant the tangent space MATH to the MATH-orbit of MATH. Let MATH be the orthogonal complement of MATH in MATH, MATH. Because the metric on MATH is MATH invariant, MATH will act on MATH by restriction. Also, all its translations MATH will be MATH invariant and MATH. Let MATH be the restriction of the tangent vector bundle to the orbit MATH, and MATH be the subbundle whose fiber above MATH is MATH. This subbundle has a natural MATH vector bundle structure coming from the action of MATH on MATH. Let us consider the principal bundle MATH, where MATH acts by right translations on MATH, and the associated vector bundle MATH. This vector bundle has a natural MATH vector bundle structure, MATH acting by left translations on MATH and MATH. The maps MATH, MATH and MATH, MATH give a MATH-equivariant isomorphism MATH between the two MATH vector bundles considered above. If MATH is the exponential map associated with the MATH invariant metric on MATH, then MATH realizes a MATH equivariant diffeomorphism between a neighborhood MATH of the zero section in MATH and an open tubular neighborhood MATH of the orbit MATH in MATH. Because MATH is compact one can find an open MATH invariant neighborhood MATH of the origin in MATH such that MATH is a MATH equivariant diffeomorphism. We will then take MATH. Passing to the MATH-orbit spaces, we get a homeomorphism MATH, where MATH is an open neighborhood of MATH in MATH. We will construct an orbifold chart over MATH around MATH. The map MATH, MATH gives a homeomorphism when passing to the orbit spaces MATH. Denote by MATH the composition MATH. Then MATH is a linear orbifold chart at MATH, where the action of MATH on MATH is the restriction of the linear representation of MATH on MATH. As shown above, the induced map to the orbit spaces MATH is a homeomorphism. We have to show that any two different charts defined as above are compatible. Let MATH be two orbifold charts around MATH, MATH. Let MATH such that MATH. Then, using the definition of MATH in the construction of the charts MATH, one can find MATH, MATH and MATH, MATH, so that we have MATH. By replacing MATH with MATH we can assume that MATH. Moreover, the map MATH is a MATH equivariant local diffeomorphism, so one gets a local MATH equivariant diffeomorphism MATH between a neighborhood of MATH in MATH and a neighborhood of MATH in MATH. The tangent space to MATH at MATH is equal to MATH and the derivative of MATH will have a block decomposition corresponding to this direct sum as MATH. Because MATH acts by left translations on the first component of MATH and MATH is MATH equivariant, a straightforward computation shows that MATH and MATH, so MATH is a diffeomorphism between MATH and MATH. Then there exist neighborhoods MATH of MATH such that the map MATH is a diffeomorphism, whose derivative at MATH is MATH. Denote this map by MATH. The horizontal dotted lines in the following diagram represent locally defined maps that are local diffeomorphisms: MATH . The map MATH is locally defined in a neighborhood of MATH as MATH. Using the fact that the upper triangle, the left and right quadrilaterals and the large trapezoid are commutative, we get that MATH on a neighborhood near MATH. Because MATH was chosen arbitrarily, we conclude that the orbifold charts MATH and MATH are compatible. |
math/9912228 | Let us fix a MATH invariant Riemannian metric on the base space and a MATH invariant linear connection MATH in the vector bundle. For a fixed MATH with MATH, consider, as described in the proof of REF , the isotropy group MATH and the direct sum decomposition of the tangent space at MATH as MATH modules MATH. Denote by MATH the action of MATH on the vector space MATH. Let MATH be the restriction to MATH of the subbundle of the tangent bundle to MATH whose fiber above MATH is MATH. The map MATH , MATH is MATH equivariant and realizes an isomorphism between the vector bundles MATH and MATH. Let MATH be an open MATH invariant neighborhood of the origin such that the map MATH is a MATH equivariant map and realizes a diffeomorphism onto its image MATH. We showed in the proof of REF that MATH is a linear orbifold chart at MATH for MATH. We will construct a vector orbibundle chart for MATH at MATH. Let MATH be the pull-back of the vector bundle MATH via the natural projection map MATH restricted to MATH. The vector bundle MATH has a MATH vector bundle structure coming from the action of MATH on MATH and MATH; indeed MATH and MATH acts by the diagonal action MATH. The group MATH fixes the point MATH, so it acts on the fiber above MATH by a linear representation MATH. Let MATH act on MATH by the product action MATH. Consider the vector bundle MATH with fiber MATH. The group MATH acts on this bundle by left translations. We already showed that MATH given by MATH is a MATH equivariant diffeomorphism. Let MATH defined as MATH. Then we have the following commutative diagram in which the horizontal maps are MATH equivariant diffeomorphisms MATH . Indeed MATH. MATH is surjective because any MATH with MATH is of the form MATH. Also, if MATH then MATH, so MATH and MATH. But MATH and MATH so MATH, MATH with MATH, MATH. We conclude that MATH, so MATH is injective as well. The pair MATH defines a MATH equivariant isomorphism of MATH vector bundles. In the proof of REF we considered the exponential map MATH with respect to the MATH invariant metric on MATH, which realizes a MATH equivariant diffeomorphism between a neighborhood MATH of the zero section in MATH and MATH - a tubular neighborhood of MATH in MATH. For MATH and MATH, let MATH, MATH be the path that realizes the parallel transport in MATH with respect to the MATH invariant connection MATH above the path MATH, with MATH. Then we define MATH, MATH. The map MATH is MATH equivariant and MATH is a linear isomorphism. Then the restriction of MATH to MATH together with the restriction of MATH to MATH give us a MATH equivariant isomorphism between the restriction of the bundle MATH to MATH and the MATH vector bundle MATH. If we choose a small enough MATH equivariant neighborhood MATH of the origin in MATH as in REF , the pair of maps MATH gives us a MATH equivariant isomorphism between the MATH vector bundles MATH and MATH. Passing to the MATH orbits we get a homeomorphism between MATH and MATH. We will describe a linear vector orbibundle chart around MATH. Let MATH, MATH and MATH be the composition MATH. Also let MATH, MATH and MATH be the composition MATH. Then MATH is a linear vector orbibundle chart around MATH. As in REF , we have to prove that any two vector orbibundle charts defined above are compatible. Let MATH, MATH be two vector orbibundle charts around MATH. Let MATH such that MATH. As described in REF , we can find MATH and MATH so that MATH and MATH. The maps MATH for MATH define MATH equivariant diffeomorphisms. Then one can choose MATH invariant neighborhoods MATH of MATH in MATH so that the composition of the previous diffeomorphisms defines a MATH equivariant diffeomorphism MATH between MATH and MATH. MATH induces a MATH equivariant diffeomorphism MATH between MATH and MATH. Using the fact that MATH is MATH equivariant, one can show, as in the proof of REF , that the composition MATH is a local diffeomorphism which together with the induced local diffeomorphism MATH between neighborhoods of MATH in MATH and MATH in MATH define a vector bundle diffeomorphism MATH which realizes the compatibility of the charts MATH and MATH in a neighborhood of MATH. |
math/9912228 | It is obvious that any MATH invariant section MATH defines a section MATH by MATH. Let MATH and MATH be two MATH invariant sections in MATH which induce equal sections MATH. If MATH then there exists MATH such that MATH. Nevertheless, MATH. Then there exists MATH such that MATH. But MATH so MATH. So MATH. We reached a contradiction so the map MATH is injective. Let MATH be a smooth section. We need to construct a MATH invariant smooth section MATH such that MATH. We will construct MATH from local data. As we showed in REF , the local model of passing from a MATH vector bundle to a vector orbibundle is given by passing from the MATH vector bundle MATH to the vector orbibundle MATH. MATH is a finite group which acts on the left on the open set MATH and on a vector space MATH and by right translations on MATH. The action of MATH on MATH and MATH is given by MATH and MATH. In fact the vector bundle MATH is a MATH vector bundle whose associated vector orbibundle of MATH orbits is the local model MATH, which happens to be a genuine vector bundle because the action of MATH is free. A section MATH is induced by a smooth section MATH. Let MATH be defined by MATH. This is a MATH invariant smooth section which is also MATH invariant and it induces a smooth section MATH. It is obvious that the map MATH induces MATH when passing to the MATH orbit spaces. Because of the first part of the proof such a MATH is unique, so we can glue different sections MATH to get a global MATH invariant smooth section in MATH which will induce MATH. The problem of proving that if MATH is smooth then its image MATH is smooth is a local one. A smooth section MATH is induced by a MATH invariant smooth section MATH. If MATH is MATH invariant then MATH is MATH invariant so there exists a MATH smooth section MATH of the MATH vector bundle MATH such that MATH. The induced section MATH of the vector orbibundle MATH is the same as the one induced by MATH. So MATH is smooth. |
math/9912228 | Let us fix a metric MATH on the base space MATH. Let MATH be the orthogonal group, MATH. We will describe the orthonormal frame bundle MATH. We will show that MATH is a manifold and that the group MATH acts on it with finite isotropy groups. The MATH vector bundle we need to construct will be essentially the pull-back of MATH to MATH via the canonical projection. The description of the space of orthonormal frames uses an atlas MATH of the vector orbibundle MATH. Let MATH be the MATH invariant metric on MATH which induces the metric MATH via the map MATH. Let MATH be the space of orthonormal frames of MATH. Then the orthogonal group MATH acts by right translations on MATH. Because the metric MATH is MATH invariant, the action of MATH on MATH extends to MATH. Observe that this action is free. Indeed, if MATH and MATH such that MATH then MATH and MATH acts on the tangent space MATH as the identity map. But the order of MATH is finite so MATH. Then the space of orbits MATH is a smooth manifold. The local diffeomorphisms between open sets MATH and MATH preserve the metrics so they will induce local diffeomorphisms between MATH and MATH and a diffeomorphism between their images in MATH respectively MATH. The space of orthonormal frames MATH is obtained by gluing the manifolds MATH along these diffeomorphisms and it has a natural structure of smooth manifold. The right action of MATH on MATH commutes with the action of MATH and with the local diffeomorphisms induced between different MATH and MATH so we will get an induced right action of MATH on MATH. Let us consider the MATH vector bundle MATH corresponding to the chart MATH and the pull-back to the space of orthonormal frames MATH together with the induced left action of the group MATH on it and the right action of group MATH. The action of MATH is free and commutes with the action of MATH so if we pass to the spaces of MATH orbits we get a genuine vector bundle MATH endowed with the action of MATH. By gluing these vector bundles together using diffeomorphisms induced by the identifications made on the initial vector orbibundle charts we get a MATH vector bundle MATH. We will show that the action of MATH on MATH has finite isotropy groups and if we pass to the space of MATH orbits we get a vector orbibundle isomorphic with the initial vector orbibundle MATH. Any point in MATH is the MATH orbit of a point MATH for some index MATH. The group MATH acts freely on MATH. If MATH fixes the MATH orbit MATH then there exists MATH such that MATH. This is equivalent to MATH and MATH. The group MATH acts freely and transitively on the frames at MATH. The group MATH acts freely on the frames at MATH as well. Then the above equalities imply that the cardinality of the isotropy group of MATH with respect to the action of MATH is equal to the cardinality of the group MATH, which is finite. Passing to the vector orbibundle of MATH orbit spaces MATH can be realized by first passing to the MATH vector bundles MATH and then gluing together the resulting vector orbibundles of MATH orbits. But the MATH vector bundles MATH and MATH are canonically isomorphic and the resulting vector orbibundles obtained by gluing the vector orbibundles of MATH orbits will be isomorphic. So MATH and MATH are isomorphic. |
math/9912228 | The multiplication operator by MATH is a continuous map on MATH endowed with the NAME norm MATH, being a differential operator of order zero. The the restriction to the MATH invariant smooth functions remains a continuous operator. |
math/9912228 | It is sufficient to prove that any point MATH has a neighborhood MATH such that MATH is a submanifold in MATH. Let MATH be an linear orbifold chart at MATH, MATH. Observe that MATH. Then for any MATH we have MATH. So MATH. The fixed point set MATH is a submanifold given by linear equations on which MATH acts trivially. Then MATH is a smooth submanifold of MATH. |
math/9912228 | We will give the construction of the density MATH on a linear orbifold chart MATH. Let MATH be the MATH . NAME density on MATH that represents MATH. For MATH let MATH. Then MATH is a submanifold of MATH. We have MATH and MATH. We will need the following lemma: The restriction of the projection map MATH is a covering map with MATH sheets. MATH is locally constant on MATH, being either MATH or a conjugate of it by an element in MATH. For each MATH one can choose a small neighborhood MATH of MATH in MATH such that MATH acts trivially on MATH and MATH for MATH. Let MATH and MATH. The map MATH is a homeomorphism. For any other MATH the map MATH is a homeomorphism between a neighborhood of MATH and MATH. Observe also that MATH has exactly MATH points. Let us fix MATH. Then MATH is a smooth submanifold of MATH and MATH is a stratification of MATH. Then MATH is an open and dense set in MATH (MATH is the decomposition in connected components of MATH). Let MATH be the restriction of the density MATH to MATH and MATH . Let MATH. We showed in the previous lemma that MATH realizes a local diffeomorphism between MATH and MATH. Let MATH for some MATH such that MATH. This definition is independent of the choice of MATH and of MATH. Indeed, let MATH be such that MATH is another representative for MATH and MATH. After a reordering of the indices MATH we can suppose that MATH. Because MATH is a MATH . NAME density on MATH we have MATH and because MATH commutes with the action of MATH on MATH we get the independence of MATH of the choices we made. If MATH was not defined on MATH because there were no MATH such that MATH is an open submanifold in MATH, we take MATH. We define MATH on MATH to be the density whose restriction to MATH is equal to MATH. We need to show that MATH . It is sufficient to prove this for MATH. If MATH is such that MATH then MATH which is equal to the right side of REF. In the last step we used the fact that MATH is a covering map with MATH leaves. |
math/9912228 | If we define MATH by MATH then MATH, but MATH will not be, in general, a pseudodifferential operator, unless MATH is smoothing and then the smooth kernel MATH of MATH is defined uniquely by MATH. We will define MATH using a partition of unity of MATH with smooth functions MATH subordinated to a cover of MATH with coordinate chart neighborhoods MATH which are slices with respect to the MATH action on MATH. We will choose MATH so that for MATH at least one of the following conditions holds: CASE: MATH CASE: MATH is included in the same coordinate neighborhood. If MATH then MATH. Denote MATH. We will define MATH and then MATH. If MATH, then MATH is smoothing and we define MATH. Otherwise, choose MATH such that MATH. Because MATH is a slice, its preimage MATH can be identified with MATH endowed with the MATH action by left translations. We have the induced MATH equivariant isomorphism MATH . The operator MATH is localized above MATH and we define MATH to be the block-diagonal operator with diagonal cells equal to MATH, where the block representation is with respect to the decomposition given by the isomorphism REF. Then MATH is a MATH invariant pseudodifferential operator on MATH localized above MATH and MATH. Define MATH to be the sum MATH. Then MATH and MATH is a subset of the MATH equivariant pseudodifferential operators in MATH. So MATH is MATH equivariant. |
math/9912228 | For each MATH such that MATH choose MATH with MATH. Because the orbibundles are isomorphic, there exists a group isomorphism MATH between MATH - the corresponding isotropy groups of MATH in MATH and one can find neighborhoods MATH of MATH in MATH which are MATH invariant and a MATH-equivariant bundle diffeomorphism MATH that makes the following diagram commutative: MATH . Let MATH be a partition of unity of MATH subordinated to the open cover MATH (because MATH are diffeomorphic and the diffeomorphism MATH permutes the two open covers, we will denote the two partition of unity with the same letters, thought we will refer to functions on two different, but diffeomorphic orbifolds). We can choose this partition so that for any MATH at least one of the following conditions holds: CASE: MATH, MATH and MATH CASE: MATH. We will show only one implication of the proposition, the other implication can be proved similarly. Let MATH be a MATH equivariant pseudodifferential operator acting on the smooth sections of MATH whose restriction to the invariant sections is equal to MATH. We will construct a lift MATH as required in the proposition. Because MATH and because a lift for the operator MATH is provided by MATH, we reduced our problem to the operators of the form MATH which we will denote by MATH for simplicity (we denoted by MATH the lift of the smooth function MATH). Denote the lifts MATH by MATH. In the first case the operator MATH is smoothing. In the second case the operator MATH is localized to the open set MATH. We will treat the two cases separately. In the first case define: MATH . Because MATH was smoothing and MATH equivariant, MATH is smoothing and MATH bi-invariant: MATH . Moreover MATH induces the same operator MATH at the orbibundle level because its action on the MATH invariant sections is not changed. If we chooose a MATH invariant metric on MATH then the kernel MATH of the operator MATH is a smooth section in MATH and because of the equalities REF it satisfies the condition MATH for all MATH. This implies that MATH is a MATH invariant smooth section in the MATH vector bundle MATH and so it defines a smooth section in the vector orbibundle MATH. Because the vector orbibundles MATH, MATH, are isomorphic, there exists a unique smooth MATH invariant section MATH in MATH which induces the same section via the isomorphism MATH at the orbibundle level as MATH. The smoothing operator MATH with kernel MATH is MATH bi-invariant, and so it is MATH equivariant. The operator MATH induces at the orbibundle level the operator MATH. The statement of the proposition is proved in REF . In REF MATH is a MATH equivariant operator in MATH. Let us denote MATH. The MATH vector bundle MATH is isomorphic with MATH. Observe that MATH. We will use the observations regarding the operators MATH and MATH made at the beginning of REF. Compare REF , there exists a MATH equivariant pseudodifferential operator MATH which is MATH equivariant as well. An element MATH acts on MATH by right multiplication on MATH by MATH and left multiplication on MATH by MATH. Let MATH be the induced operator to the vector bundle of MATH orbits MATH which is isomorphic to MATH. Because MATH was MATH equivariant, compare REF , the operator MATH is MATH equivariant. We have the natural isomorphisms between the spaces of invariant sections MATH which can be identified with the space of smooth sections in the vector orbibundle MATH above the open set MATH. The operators MATH, MATH and MATH restricted to the above spaces of invariant sections will be equal via the natural isomorphisms. In particular they induce MATH at the level of sections in the associated vector orbibundle MATH. We define the MATH equivariant operator MATH in MATH using the isomorphisms MATH provided in REF . (here, as before, MATH). Let MATH be the MATH equivariant operator which induces MATH at the MATH orbit level, and let MATH be the operator induced at the MATH orbit level. Compare REF , this operator is MATH equivariant. Using the sequence of isomorphisms: MATH observe that the operators MATH and MATH act the same way on the corresponding spaces of invariant sections and induce MATH at the vector orbibundle level. If we extend MATH by MATH outside the neighborhood MATH we obtain a lift of MATH. We proved the proposition in REF as well. |
math/9912228 | It is sufficient to prove that if MATH is a classical pseudodifferential operator that induces the zero operator on MATH, then MATH is smoothing. Let MATH be a point such that its projection MATH onto MATH is a smooth point. We will show that the total symbol of the operator MATH on a neighborhood of MATH is smoothing (though the total symbol of the operator is not well-defined, the property of being a smoothing symbol is independent of the chart around MATH). Let MATH be a slice at MATH with respect to the action of the group MATH. Because the point MATH is smooth, the isotropy group MATH is trivial. Let MATH be a smooth function on MATH with support inside MATH which is equal to REF on a neighborhood of a smaller slice MATH at MATH. Let MATH be a smooth function on MATH which is equal to REF on a neighborhood of MATH and zero outside MATH. MATH is a pseudodifferential operator which is localized on MATH. We will prove that this operator is smoothing. Then MATH will be smoothing above MATH where MATH. Let MATH be the MATH invariant extension of MATH to MATH. Then MATH is a MATH equivariant extension to MATH of the zero operator on MATH. Let MATH be an arbitrary section with support in a neighborhood of MATH that vanishes outside the set MATH. Then MATH is a MATH invariant section that extends MATH. Because MATH restricted to the MATH invariant section is equal to zero, we have: MATH . But MATH has support inside MATH which is disjoint from MATH so MATH and so MATH are smoothing. Because MATH was chosen arbitrary, we conclude that MATH is smoothing so MATH is smoothing above MATH. Then MATH is smoothing on the regular set MATH. If MATH is a singular point then, on an Euclidean chart around MATH the total symbol of MATH will have the asymptotic expansion MATH . But MATH is dense in MATH so MATH on a dense set because MATH is smoothing on MATH. We conclude that MATH on a neighborhood of MATH so MATH is smoothing on that neighborhood and thus on the whole manifold MATH. |
math/9912228 | As described in REF one can find an open cover MATH of the orbifold MATH, a family of finite groups MATH , MATH, and a vector space MATH such that MATH is isomorphic to the associated vector orbibundle of MATH orbits of the MATH vector bundle MATH and MATH can be covered with a family of open sets MATH which are MATH equivariant diffeomorphic to MATH and such that the restriction of the MATH vector bundle MATH to MATH is diffeomorphic with the MATH vector bundle MATH by a MATH equivariant vector bundle diffeomorphism. The cover MATH can be chosen to be finite because MATH is compact. Using a partition of unity, any pseudodifferential operator MATH acting on sections of the vector orbibundle MATH can be written as MATH where MATH is a smoothing operator with smooth kernel which is zero on a neighborhood of the diagonal and MATH are pseudodifferential operators which vanish on sections that are supported outside the open set MATH and take sections with support inside MATH into sections with support inside MATH. For each MATH we will construct a MATH equivariant pseudodifferential operator MATH acting on sections of MATH whose restriction to the MATH invariant sections induces MATH. First let us fix a MATH bi-invariant pseudodifferential operator MATH acting on MATH of order MATH such that MATH on constant functions MATH. In all the situations MATH can be taken to be the MATH power of the Laplacian MATH on MATH with respect to a bi-invariant metric on MATH. Using REF we can find a MATH equivariant pseudodifferential operator MATH acting on the space of sections MATH of the MATH bundle MATH which induces MATH when restricted to MATH invariant sections. The space of smooth sections of the MATH vector bundle MATH is equal to MATH so the pseudodifferential operator MATH acts on this space and it has order MATH. It is MATH equivariant. The finite group MATH acts by a diagonal action on MATH and MATH so we can replace the above pseudodifferential operator with the MATH equivariant pseudodifferential operator MATH. This will induce a pseudodifferential operator on the sections of the vector bundle of MATH orbits MATH. We will transport it using the isomorphism between the previous MATH vector bundle and MATH to a pseudodifferential operator MATH acting on MATH, extending it by MATH outside MATH. We have to compare the actions of MATH and MATH. Let MATH be a MATH invariant smooth section in MATH which induces MATH. It is enough to take MATH with support in MATH. In this case MATH is induced by a MATH invariant section MATH which is MATH invariant as well. Then MATH with MATH, MATH induces MATH and MATH corresponds via the MATH equivariant vector bundle diffeomorphism to MATH and this section induces MATH. We will construct a smoothing operator MATH in MATH whose action on MATH invariant sections induces the smoothing operator MATH. The kernel of MATH is given by a smooth section MATH of the endomorphism bundle MATH which corresponds to a smooth MATH invariant section MATH of the endomorphism bundle MATH. MATH is the kernel of a smoothing operator MATH, which induces MATH when restricted to the MATH invariant sections of MATH. The operator MATH will be the sum of the operators MATH and MATH. If the operator MATH is classical, we can choose MATH to be classical, then MATH, MATH and MATH are classical so MATH is classical. |
math/9912228 | Let MATH and denote by MATH the orthogonal group MATH. Let MATH be the MATH vector bundle, with MATH, such that the vector orbibundle of orbits is canonically isomorphic to MATH, as in REF . Let us fix a MATH bi-invariant metric MATH on MATH so that MATH and a MATH bi-invariant positive selfadjoint pseudodifferential operator MATH of order MATH acting on MATH. As a metric MATH we can choose the left translations of the opposite of the Killing form on the NAME algebra MATH and MATH where MATH is the NAME operator associated with the metric MATH. We will construct a MATH invariant metric on MATH which induces the given metric on MATH, a MATH invariant hermitian structure on MATH which induce the given hermitian metric on MATH and a MATH equivariant elliptic positive selfadjoint pseudodifferential operator MATH acting on MATH whose restriction to the invariant sections is equal to MATH. Let us fix a finite atlas MATH of the vector orbibundle MATH, with MATH. Then MATH forms an atlas for MATH. Let MATH be the MATH invariant metric on MATH which induces MATH on MATH. Then the collection of MATH invariant metrics MATH on MATH induces the MATH invariant metric MATH on MATH. Because MATH is the pull-back of MATH with respect to the canonical map MATH, we define the hermitian structure on MATH to be the pull-back hermitian structure. It will be MATH invariant by construction and induces the initial hermitian structure when passing to the MATH orbit spaces. Consider the MATH invariant scalar product on MATH associated with hermitian structure and the metric MATH constructed above. When restricted to the MATH invariant sections, this scalar product is equal with the scalar product on MATH constructed with the help of the metric MATH and the hermitian structure on MATH. Also MATH. The pseudodifferential operator MATH acting on MATH constructed as in REF will be symmetric with respect to the scalar product on MATH. By construction, the principal symbol of MATH is equal to MATH where MATH is the principal symbol of MATH and MATH is the principal symbol of MATH. Because MATH and MATH are positive symmetric elliptic pseudodifferential operators, the operator MATH is elliptic as well. Because MATH is compact, the operator MATH acting on MATH is essentially selfadjoint. The restriction of MATH to the closed subspace of MATH invariant MATH sections MATH which is equal to MATH is essentially selfadjoint. The spectrum of MATH is real and discrete so the spectrum of MATH is a discrete subset of MATH. |
math/9912228 | The proof is analogous to the proof of REF We construct the operator MATH acting on sections of the MATH vector bundle MATH with the help of a MATH bi-invariant elliptic selfadjoint pseudodifferential operator MATH acting on MATH. The principal symbol of MATH is equal to MATH with MATH the principal symbol of MATH and MATH the principal symbol of MATH. Because MATH is an NAME angle for MATH, the spectrum of MATH is disjoint from the region in the complex plane MATH for a small enough MATH. Because MATH is selfadjoint the spectrum of MATH is disjoint from MATH. We conclude that MATH is an NAME angle for MATH. Then MATH is elliptic and has a discrete spectrum. MATH being the restriction to the MATH invariant sections will have a discrete spectrum as well. The complex powers MATH are well defined and are pseudodifferential operators of order MATH so the restriction to the MATH invariant sections, which are equal to MATH, are pseudodifferential operators of order MATH. |
math/9912228 | Let MATH be a fixed finite atlas of linear orbibundle charts . Using a partition of unity we can decompose the complex powers of the operator MATH as MATH where MATH are pseudodifferential operators of order MATH with support inside MATH and MATH is a smoothing operator. We can arrange for MATH to take smooth sections with support inside MATH into smooth sections with support inside MATH. The trace MATH is defined for MATH and it is a holomorphic function on the complex plane. In order to prove the theorem, we will show that the statement in the theorem holds for the trace of the operators MATH. We will show that MATH is a meromorphic function on MATH with at most simple poles at MATH, MATH, and that the residues are computed as integrals on MATH of NAME densities. MATH can be seen as a holomorphic family of pseudodifferential operators of order MATH acting on the smooth sections of the vector orbibundle MATH. For the sake of simplicity, we will drop the index MATH and denote MATH by MATH. There exists a holomorphic family of MATH equivariant pseudodifferential operators MATH of order MATH acting on the sections of the MATH vector bundle MATH which induce MATH on the MATH invariant sections. This family is unique up to smoothing operators, compare REF . Then the component corresponding to the trivial representation of the trace of MATH and of the residues will be equal to respectively the trace functional of MATH and its residues. REF shows that the component of the trivial representation of trace functional is a meromorphic function with at most simple poles at MATH for MATH. From REF , the component of the residue of the trace functional MATH corresponding to the trivial irreducible representation is: MATH where MATH are MATH . NAME densities on MATH constructed from the total symbol of MATH which on the chosen linear vector orbibundle chart coincides with the total symbol of the pseudodifferential operator MATH on MATH. Though the statements of REF refer to the complex powers of a pseudodifferential operator, their proofs can be slightly changed and adapted to include the case of holomorphic families of pseudodifferential operators and so the conclusions listed above are still true. The MATH . NAME density MATH defines a NAME density on the orbifold MATH which we will denote by MATH. Observe that REF is equivalent to MATH . The NAME density whose integral on MATH is equal to the residues of the zeta function MATH at MATH will be equal to the sum of the NAME densities MATH constructed as above on the charts of the vector orbibundle. |
math/9912228 | For each MATH and MATH, the density MATH can be obtained as follows: let MATH and MATH be a linear vector orbibundle chart centered at MATH. Observe that MATH and the map MATH is a local diffeomorphism. REF gives us NAME densities MATH on MATH whose integrals are equal to the residues of the zeta function and can be computed explicitly in terms of the total symbol of MATH. These densities can be described above MATH as families of densities MATH such that MATH is a smooth density on MATH. Consider the density on MATH given by MATH . Define MATH on MATH to be the density whose restriction to MATH is equal to MATH. As proved in REF , the integral on MATH of the NAME density MATH and the sum of the integrals of the densities MATH on the strata MATH for MATH are equal, so they are equal to the residue of MATH at MATH. |
math/9912228 | Let MATH be a MATH-invariant open neighborhood of MATH in MATH such that MATH for MATH. Let MATH be the linear representation of MATH on the tangent space at MATH generated by the action of MATH on MATH. Choose a MATH invariant metric on MATH (this can always be done by averaging a metric over the finite group MATH). Then the exponential map MATH is MATH equivariant. Indeed, if MATH is a geodesic curve with MATH and MATH then, for MATH, MATH is again a geodesic curve and MATH. So MATH. One can choose a smaller MATH such that the exponential map realizes a diffeomorphism between a MATH invariant neighborhood of MATH in MATH and MATH. If we fix a linear isomorphism MATH and transport the action of MATH on MATH via this isomorphism, we obtain a linear representation MATH and a MATH equivariant diffeomorphism between a neighborhood MATH of the origin and MATH. Using the pull-back via this diffeomorphism we can replace the bundle MATH by a new MATH vector bundle such that the action of MATH on the base will be the restriction of a linear representation to an open neighborhood of the origin. The point MATH corresponds to the origin MATH. For the sake of simplicity we will use the same notation for the new bundle. Consider now a linear connection MATH in the MATH vector bundle MATH. The group MATH acts on the space of connections, and we can make MATH to be invariant with respect to MATH by replacing it with the connection MATH. We will define the bundle map MATH between the trivial MATH bundle MATH and MATH as follows: for a point MATH and a vector MATH, let MATH be the vector in the fiber of MATH above MATH obtained by parallel transport of MATH along the curve MATH. The vector space MATH as in the statement of the proposition will be equal to MATH with the action of MATH induced from the action of MATH on the total space of the bundle. MATH is a vector bundle isomorphism and we have to show that MATH is MATH equivariant. If MATH is the path in MATH that realizes the parallel transport from MATH to MATH above MATH, and MATH, then, because the connection is MATH invariant, the path MATH realizes the parallel transport between MATH and MATH above the curve MATH (we used the fact that the action of MATH on MATH is linear). Then MATH, so MATH is MATH equivariant. For MATH as above the set MATH is the reunion of disjoint open sets MATH with MATH a complete system of left coset representatives for MATH. Then MATH given by MATH is a diffeomorphism, where MATH is the cross-product MATH , with MATH, for any MATH, MATH and MATH. Moreover, MATH is MATH equivariant. The action of MATH on the cross-product MATH is by left translations. The restriction of the vector bundle MATH is trivial and a trivialization is given by MATH . The bundle isomorphism is also MATH equivariant. |
math/9912228 | Let MATH be a finite cover of MATH with open sets as in REF so that the restrictions of the MATH bundle MATH to the subsets MATH are isomorphic to MATH trivial vector bundles for some MATH. Then MATH is a finite open cover with MATH invariant open sets. Consider a partition of unity MATH subordinated to the open cover MATH such that the functions MATH are MATH equivariant. One can choose the open cover and the partition of unity such that for any MATH and MATH either MATH or MATH for some MATH. Then the operators MATH are either smoothing (in the first case) or have the support and range included in the space of sections that vanish outside the open set MATH (in the second case). Consider the trivialization by the MATH vector bundle isomorphism MATH. Let MATH the complete symbol of the operator MATH in this trivialization. Then for a section MATH with compact support in MATH we have: MATH where MATH and the double integral is computed over MATH. To keep our notation simple, we will drop the indexes MATH and MATH and denote MATH by MATH. The action of MATH on MATH is given by MATH. Then MATH . We can choose the trivialization in such a way that MATH. The distributional kernel of MATH will be equal to MATH. For MATH the trace of this operator is equal to the integral MATH . We will choose the open cover MATH so that either MATH has fixed points inside MATH or MATH for all MATH for some fixed MATH. In the second case we will show that the trace function MATH can be extended to the whole complex plane. Indeed, if we denote by MATH the differential operator MATH we have MATH. Then MATH . For a fixed half-plane MATH if we choose a large enough MATH the expression MATH is a symbol in MATH so it is absolutely integrable and gives after integration a holomorphic function in MATH in the half-plane MATH. It follows that on open sets where MATH the trace function MATH has a holomorphic extension to the complex plane MATH. Consider now a set MATH so that MATH has a nonempty fixed point set. The diffeomorphism MATH is given by MATH. Denote MATH. Let MATH such that MATH. As shown in REF , we can choose MATH to be the origin in MATH and MATH act on MATH by linear isometries. Then MATH is the restriction of a linear isometry and its fixed set MATH is the intersection between a linear subspace MATH of MATH and MATH. The dimension of MATH is one of the dimensions MATH of the connected components of the fixed point set MATH. For the sake of simplicity we will denote it by MATH. Let MATH be linear coordinates on MATH so that MATH are coordinates along the fixed-point set MATH and MATH are normal coordinates to MATH. In these new coordinates MATH has the form MATH with MATH a MATH square matrix whose eigenvalues are different from MATH. Let MATH be the coordinates in the cotangent space corresponding to the coordinates MATH. Using the new coordinates in REF MATH will be equal to MATH . Because MATH is a matrix whose eigenvalues are different from MATH, the matrix MATH is non-degenerated and if we use the change of coordinates MATH the above integral becomes MATH where MATH is a classical symbol of order MATH in MATH and with compact support in MATH. For MATH the integrand is absolutely integrable so we can apply NAME 's theorem and get: MATH . We will need the following result: If MATH is a classical matrix valued symbol of complex order MATH then, for MATH, the expression MATH is a classical symbol of order MATH with the asymptotic expansion MATH . If MATH is a holomorphic family of symbols then there exists a holomorphic family of symbols MATH defined for MATH such that MATH is a holomorphic family of smoothing symbols. The asymptotic expansion of MATH is given by REF. We will postpone the proof until the end of the proof of the main theorem. Using the above lemma for MATH, the trace of MATH becomes: MATH where the integration is taken over the cotangent space of the fixed point set MATH. Because MATH is a smoothing symbol, the difference between the integral above and MATH is a whole function. We will show that, as a function in MATH, the above expression has a meromorphic extension to the complex plane. Let us consider the asymptotic expansion as a classical symbol MATH with MATH homogeneous of degree MATH in MATH. Let MATH be a positive, real valued smooth function which vanishes on a neighborhood of the origin and is equal to MATH for MATH. Let us fix a half-plane MATH. For a large enough MATH the difference MATH is a symbol in MATH for any MATH with MATH. Then on this half-plane the difference MATH is holomorphic. We will have to prove the existence of a meromorphic extension and find the poles and residues for each of the functions MATH . Because MATH for MATH and the integral on the compact set MATH yields a holomorphic function in MATH we need to study the meromorphic extension of MATH . Let MATH with MATH and MATH be the decomposition in polar coordinates. The degree of homogeneity of MATH in MATH is equal to MATH. Then MATH and after passing to polar coordinates, REF becomes: MATH . The double integral defines a holomorphic function on MATH, so the above expression has a meromorphic extension to MATH with a simple pole at MATH and residue: MATH . As a consequence, the trace functions MATH and MATH have meromorphic extensions on any half plane MATH and so on MATH. We will proceed with the computation of the poles and residues of MATH. We observed that on the half space MATH the difference MATH is holomorphic. Then MATH has simple poles at MATH for MATH and the residue at MATH is equal to the expression in REF gives an asymptotic expansion of MATH of the form: MATH with MATH . If MATH is the asymptotic expansion in homogeneous terms, with MATH homogeneous of degree MATH, then the homogeneous component of degree MATH of MATH will be equal to MATH with MATH of degree of homogeneity MATH. To conclude the proof of the theorem, consider the fixed point set MATH and a coordinate chart MATH where MATH. Let MATH. We define the densities MATH that compute the residue of the zeta function at MATH as: MATH . |
math/9912228 | We have MATH so for MATH the integral defining MATH is absolutely convergent so one can change the order of integration in REF. Consider the NAME expansion of MATH near MATH . We have MATH the double integral being the composition of the NAME transform in MATH and the inverse NAME transform in MATH evaluated at MATH. Because MATH is a classical symbol of order MATH, the right-hand side term of the above equality is a classical symbol of order MATH. In order to prove that MATH is a symbol with the asymptotic expansion as in REF we need to show that the integral of the remainder MATH is a symbol of arbitrary negative order if MATH is chosen large enough. After changing the order of integration of MATH and MATH we see that it will be sufficient to provide a uniform estimate in MATH for the integrals MATH with MATH. Integrating by parts we get: MATH . Let MATH where MATH is the integral over the set MATH and MATH is the integral over the complement. The volume of MATH is bounded by MATH where MATH doesn't depend on MATH, MATH and MATH. On MATH we have MATH so the integrand in MATH is bounded by MATH. Thus MATH with MATH independent of MATH, MATH and MATH. By using the identity MATH and integrating by parts, we can rewrite MATH as a sum of terms of the form: MATH with MATH. Because MATH, the expression MATH is bounded from above by MATH for MATH and by MATH otherwise. The constant MATH doesn't depend on MATH, MATH, MATH and MATH. For MATH large enough the previous integral is bounded by MATH with MATH arbitrary large. Thus MATH for an arbitrary MATH with the constant MATH independent of MATH. Using Proposition MATH and Theorem MATH in CITE we conclude that MATH is a symbol of order MATH and has the asymptotic expansion given in REF. If MATH is a classical symbol then MATH is classical as well. For the second part of the proof, let us consider the positive, real valued smooth functions MATH, for MATH, such that MATH on the ball of radius MATH and MATH outside the ball of radius MATH. Let MATH . It is obvious that MATH is a smoothing operator. If MATH is a holomorphic family of symbols for MATH then MATH defined as above is a holomorphic family as well. |
math/9912238 | Without loss of generality, MATH is Legendrian and MATH. Given MATH with MATH, let MATH denote the affine MATH-line through MATH and MATH; let MATH denote the contact plane at MATH. Thus MATH is a (round) circle for MATH, and MATH reduces to the singleton MATH. Let MATH denote the (euclidean) diameter of MATH. Because MATH is Legendrian, the limit of MATH as MATH approaches MATH along MATH exists and is MATH. It follows that for all MATH there exists MATH such that if MATH and MATH, then MATH. Consequently, for all sufficiently large MATH, there is a sequence of MATH points MATH on MATH (cyclically ordered identically by their indices and by their position on MATH) and a piecewise-smooth annulus MATH with MATH for all MATH. Denote by MATH the hyperbolic MATH-line MATH. The union MATH (which is, as it were, an ``inscribed hyperbolic MATH-polygon" of MATH) is not a hyperbolic MATH-line arrangement as defined in this paper, since MATH and MATH are parallel; but almost any small perturbation of MATH will be a hyperbolic MATH-line arrangement, and for a suitable such perturbation MATH (easily achieved by moving all the affine MATH-lines MATH slightly closer to the origin) the link at infinity MATH is isotopic to MATH. It remains to be shown that, with MATH and MATH as above, MATH is fibered iff MATH and MATH. This can be done by invoking results about quasipositivity, NAME sums, and fiber surfaces. The annulus MATH is quasipositive (by CITE), as is each MATH, so (by CITE) the plumbed surface MATH is quasipositive and therefore a least-genus surface for its boundary. As is well-known, a least-genus NAME surface bounded by a fibered link must be a fiber surface. By CITE, a NAME sum of NAME surfaces is a fiber surface iff all of the summands are fiber surfaces. Since MATH is indeed a fiber surface, MATH is fibered iff MATH is a fiber surface. The only annuli which are fiber surfaces are MATH and its mirror image MATH. Since MATH (compare CITE), MATH must be MATH. |
math/9912238 | If MATH is not connected, then MATH is a split link and so not fibered. If MATH is connected, then (by REF ) there is a small perturbation MATH of MATH such that MATH and MATH are ambient isotopic and MATH is in general position; of course MATH is also connected, so its closure MATH is a divide. An inspection of NAME 's construction in CITE, translating his symplectic approach into complex language, immediately reveals that MATH (with one of its two natural orientations) and MATH are identical, whence MATH is fibered. The affine case reduces immediately to the hyperbolic case. |
math/9912240 | We begin by introducing some notation and recalling some basic material from CITE and CITE. Let MATH be the trivial bundle over MATH with fiber MATH. Impose the MATH pre-Hilbert space structure on the fibers and consider the smooth subbundle MATH whose fiber above MATH is MATH . The bundle MATH over MATH is MATH-equivariant and hence descends to give a bundle, also denoted by MATH, over MATH, which we regard as the tangent bundle to MATH with the MATH metric as opposed to a NAME metric. Recall the operator MATH on MATH defined by MATH. It can be extended to give an operator MATH by setting MATH . Then MATH is a closed, essentially self-adjoint NAME operator with dense domain, depending smoothly on MATH and MATH. It has discrete spectrum with no accumulation points, and each eigenvalue has finite multiplicity. If MATH is MATH-perturbed flat, then MATH respects the splitting MATH where MATH. Remark. Note that MATH as defined here differs from the operator used in CITE. However, the formula for MATH is the same, because changing the sign of MATH in MATH is equivalent to changing the orientation of REF-manifold, and it is proved in CITE that MATH . We now introduce a closely related operator on MATH. Let MATH be the MATH-orthogonal projection and let MATH be the operator on MATH obtained by restricting MATH. For paths in MATH the spectral flow of MATH and MATH are identical. Let MATH . Choose open neighborhoods MATH of MATH and MATH of MATH small enough so that MATH implies MATH. Over MATH it is possible to decompose MATH into MATH where MATH . Here MATH is an eigenvalue and MATH is its eigenspace. Let MATH be the projection and choose MATH so that MATH for MATH. For MATH define MATH by setting MATH. (Recall that MATH.) A standard argument shows that MATH is a submersion along MATH and so, by the Inverse Function Theorem, for MATH and MATH small enough, MATH is a submanifold of MATH parameterized by a MATH function MATH of the form MATH, where MATH is smooth. Consider part of the parameterized moduli space MATH defined by MATH . Then MATH is the image under MATH of the zero set of the map MATH from MATH to MATH defined by MATH. This zero set is not cut out transversely since MATH is not regular along MATH . We expand MATH about MATH for MATH. For clarity we are using MATH instead of MATH to denote gauge orbits. Since MATH and we have MATH . It then follows that MATH . This last step follows since MATH is the projection onto the kernel of the Hessian of MATH. Thus the function MATH extends to a MATH function MATH defined by MATH . Obviously, for MATH the zero set of MATH coincides with that of MATH. Moreover, the restriction of MATH to MATH is transverse to the zero section of MATH, since by REF is a NAME function. Therefore, for MATH small enough, MATH is a smooth, REF-dimensional submanifold of MATH which intersects MATH transversely and MATH . Following this product cobordism gives a natural bijection MATH. To prove REF , let MATH and denote by MATH a differentiable family of connections representing the path of orbits MATH. Consider the differentiable family of closed, essentially self-adjoint NAME operators MATH. (Here we could equally well work with the path MATH of operators on MATH since we are only concerned with the behavior of the small eigenvalues.) The eigenvalues of MATH of modulus less than MATH vary continuously differentiably in MATH, and their derivatives at MATH are given by the eigenvalues of MATH restricted and projected to MATH (see REF). However, one can see directly that the restriction of MATH to MATH agrees with MATH and this completes the proof. |
math/9912240 | Since the argument is nearly identical to the proof of REF , we only explain the modifications one needs to make. The tangent space to the gauge group MATH at the identity is given by the space of REF completed in the MATH norm. Therefore, the tangent space to the subgroup MATH of based gauge transformations is the subspace MATH consisting of REF vanishing at the basepoint. Consider the bundle MATH whose fiber above MATH is MATH and denote again by MATH the induced bundle on the quotient MATH. Notice that the fiber MATH contains MATH as a subspace of codimension MATH. We regard MATH as the tangent space of MATH at MATH. By restricting and projecting MATH, we obtain an operator MATH on MATH . This operator agrees with MATH on MATH and vanishes on the orthogonal complement to MATH in MATH, which is just the tangent space to the orbit of the residual MATH action. Choose open subsets MATH containing MATH and MATH containing MATH and define MATH as in REF , with MATH replaced by MATH. Over MATH decompose MATH into the two eigenbundles as in REF . Let MATH be the analog of the map MATH from before. The only substantial difference is that now MATH is not NAME but rather equivariantly NAME. This implies that MATH induces NAME functions on MATH and MATH with only finitely many critical points. The argument from REF which produced the map MATH on MATH can also be applied here and results in equivariant maps MATH and MATH whose zero sets together coincide with that of MATH. Reducing modulo MATH, we obtain REF-dimensional (product) cobordisms in MATH and MATH which we follow to define the map MATH. The preimages of the cobordisms under MATH are equivariant product cobordisms in MATH. Nondegeneracy of MATH in the normal direction to MATH guarantees that there are no irreducible orbits in MATH nearby, and the claims about the spectral flow follow as in the previous case. |
math/9912240 | We first show that REF is independent of the choices of MATH and MATH. The argument of REF shows that REF depends only on the gauge orbits MATH and not on their gauge representatives. That argument also shows that REF is independent of the choice of MATH. So, it suffices to show that REF is independent of the choice of MATH and MATH. The NAME group MATH acts smoothly on MATH, and hence REF implies MATH is a smooth submanifold of MATH. Since MATH acts freely on MATH the quotient MATH is also smooth. Thus the dimension of the kernel of MATH is constant as a function of MATH (the tangent space of MATH at MATH can be identified with the space of zero modes of the Hessian). The same is true of the signature operator MATH since it is just the Hessian enlarged by putting MATH and its adjoint MATH in opposite off-diagonal blocks. Given MATH, there is by REF a path in MATH from MATH to MATH which we lift to a path MATH of irreducible connections from MATH to MATH, where MATH. Since none of the eigenvalues of MATH cross zero along MATH, it follows that MATH. This proves REF is independent of the choice of MATH. To prove REF is independent of the choice of MATH choose a lift MATH of MATH and decompose the tangent space of MATH at MATH into the subspaces of vectors tangent to MATH and vectors normal to MATH in MATH. Now MATH connected implies MATH is connected, and hence the dimension of the kernel of MATH is constant as a function of MATH. The same is true for the restriction MATH because its kernel can be identified with the normal bundle of MATH in MATH. Similar statements hold for the signature operator MATH and its restriction MATH (notice that MATH and MATH for MATH). Given MATH there is a path in MATH from MATH to MATH which we lift to a path MATH of reducible connections from MATH to MATH. Since none of the eigenvalues of MATH or its restriction MATH cross zero along MATH, it follows that MATH and MATH. This proves that REF is independent of the choice of MATH. To compute the contribution of MATH to MATH, we choose an admissible function MATH so that MATH is equivariantly NAME and consider the parameterized moduli space MATH for REF-parameter family of perturbations MATH . For MATH small, MATH is a union of connected components corresponding to the connected components of MATH. Let MATH be the component of MATH containing MATH, and let MATH denote the ``t-slice" MATH. Then, by definition, the contribution of MATH to MATH is the sum MATH where MATH is a small positive number and MATH is the decomposition into irreducible and reducible gauge orbits. From REF , it follows that MATH . This uses the previously established fact that MATH for all MATH, together with the observation that the NAME index of MATH at MATH equals that of the induced function MATH on MATH at MATH. Similarly, from REF , it follows that MATH . The second step follows since MATH and MATH are independent of MATH and since MATH is even. The last step is justified by the following lemma. For all MATH and all MATH, MATH. To prove the lemma, suppose MATH is a REF-parameter family in MATH with MATH and MATH for MATH. Then MATH . Indeed, as MATH increases from MATH a pair of eigenvalues of MATH of equal magnitude and opposite sign leave zero. This proves that MATH. It also proves the claim since, as we have already seen, MATH is independent of MATH and MATH is independent of MATH . We now complete the proof of REF . Substituting REF into REF , we see that the contribution of MATH to MATH is given by MATH . Notice that quantity in brackets on the first line of REF is independent of the equivariantly NAME function MATH on MATH. (This follows from an argument similar to but simpler than that given in CITE to show that MATH is independent of perturbation.) Hence we can compute it using any equivariantly NAME function we want. Choosing a function whose Hessian in the normal directions to MATH is positive definite and whose critical values along MATH are all larger than the values along MATH, we see that the quantity in brackets on the first line of REF equals the relative NAME characteristic MATH . A standard argument shows that MATH . This proves REF equals REF and we are done. |
math/9912241 | Assume that MATH. Then MATH is an algebraic fiber space and MATH is an elliptic curve. Let MATH be a general fiber of MATH. Then MATH is a smooth projective surface, so MATH is ample REF . Hence MATH for MATH by NAME Vanishing Theorem (compare REF ) and so MATH for all MATH. Then applying REF , we know that MATH is a MATH-factorial NAME terminal REF-fold. Let MATH be a resolution and put MATH. Claim. MATH for some effective divisor MATH and some MATH. Proof of the claim. Set MATH (respectively, MATH) the fiber of MATH (respectively, MATH) over MATH. For a sufficiently large integer MATH, MATH by the ampleness of MATH. Fix this MATH. We may assume MATH. Let MATH be a sufficiently ample line bundle on MATH. We consider the following exact sequence. MATH . Since MATH, MATH is nef and big. Thus MATH . Therefore looking at the exact sequence above, MATH . Thus MATH by NAME Theorem (compare CITE). Then we can apply REF because of REF . Thus we know that MATH is algebraically equivalent to an effective divisor. The algebraically equivalence of cycles is preserved by proper push-forward (see REF ). Then we obtain the claim. We can write MATH for some MATH and some prime divisors MATH by the claim. Since MATH is nef but not big by MATH, MATH and so MATH for all MATH. Because MATH is NAME by REF, MATH is a NAME surface. Hence MATH is ample for MATH REF , then MATH . The absurdity comes from MATH . |
math/9912241 | Note that MATH is an algebraic fiber space in the case MATH when we regard MATH as a morphism from MATH to MATH. Assume MATH. Denote a terminalization of MATH by MATH (see REF ). Let MATH be a general fiber of MATH. Then MATH is smooth and connected, so MATH is an irreducible reduced NAME surface. Therefore MATH is ample by REF and we obtain MATH for all MATH by relative NAME Vanishing Theorem. Then MATH for all MATH, and so MATH. This lies in contradiction with REF . |
math/9912241 | Assume that MATH. Then we know that MATH because of REF and MATH is an elliptic curve. Let MATH be a MATH-factorialization. Then there exists a sequence MATH such that for MATH each morphism MATH is constructed by REF and MATH has at most MATH-factorial terminal singulalities (if MATH, MATH). Put MATH. Since MATH is not nef, there exists an extremal contraction MATH. Since MATH for MATH as in the proof of REF , we obtain that MATH is NAME REF . Furthermore we know that MATH easily. Let MATH be the NAME morphism. Case: MATH . In this case, MATH is an elliptic curve and so MATH is an isomorphism. REF says that MATH is not MATH-factorial terminal, so there exists an irreducible curve MATH such that MATH is a point. Then MATH and so MATH is a curve. This lies in contradiction with MATH. Case: MATH . When MATH, MATH for some ample divisor MATH. But because MATH is almost nef REF , MATH. Therefore we know MATH. Combining this with MATH for all MATH, we know that MATH is a hyperelliptic surface. If MATH has only terminal singularities, MATH is just a MATH-factorialization so it is small. MATH is not MATH-factorial terminal by REF , then there exists a rational curve MATH on MATH such that MATH, and so MATH is a rational curve on a hyperelliptic surface MATH. This is absurd. Thus MATH, that is to say, MATH is not terminal and there exists a morphism MATH such that MATH is constructed by REF and MATH factors through MATH. Put MATH the exceptional divisor of MATH. If MATH, then MATH is uniruled by the NAME Lemma (REF ! of CITE). But this lies in contradiction with MATH. If MATH, then there exists an irreducible curve MATH on MATH such that MATH is a point. Since MATH, MATH is a smooth rational curve. But this derives contradiction as above. Assume that MATH below. We obtain that MATH for some MATH and some prime divisors MATH as in the proof of REF . Since MATH, MATH and there exists an irreducible curve MATH such that MATH is a point for all MATH. In fact, if not, MATH is strictly nef for some MATH. Moreover we know by REF that MATH for MATH. Then we can derive a contradiction as follows: MATH . Let MATH be the NAME morphism. Note that MATH. MATH . Claim. MATH for some MATH. Proof of the claim. If MATH is a point for all MATH, then there exists an irreducible curve MATH on MATH such that MATH is REF-dimensional and MATH does not meet any MATH. This implies MATH which is contradiction. Therefore we know that MATH for some MATH, then MATH. In fact, assume to the contrary that MATH is a curve. Now that MATH is a curve, MATH. But this is absurd because MATH is a point. Using the claim above, we prove the following. Claim. Assume that MATH for an irreducible curve MATH on MATH. Then MATH is a fiber of MATH. Proof of the claim. If MATH is a curve for all curves MATH, then MATH for all MATH by MATH. From the well known formula MATH (see the proof of REF), we get MATH . Then MATH for all MATH. This lies in contradiction with MATH. Then we know that there exists a curve MATH such that MATH is a point, and so MATH is a fiber of MATH. It follows from REF that MATH is a MATH-bundle over MATH with nef MATH and almost nef MATH. Then we can write MATH for a normalized sheaf MATH on MATH, that is, MATH, but MATH for all line bundles MATH with MATH (see CITE about the treatment and the terminology of geometrically ruled surfaces around here ). Set MATH. We separate into these two cases: MATH or MATH. Let MATH be a fiber of MATH and MATH a canonical section satisfying MATH. In the case MATH, there exists an irreducible curve MATH on MATH such that MATH is numerically proportional to MATH. Since MATH is nef, MATH . Thus MATH . This contradicts the claim above. In the case MATH, MATH is nef. Then we obtain MATH . This also yields contradiction. Case: MATH . Since MATH is NAME, MATH is a divisorial contraction (compare CITE). Let MATH be the exceptional divisor. If MATH is a point, then we have the following equation MATH for some MATH. Thus MATH is nef. Furthermore looking at the description of the normal bundle of MATH in MATH, MATH, in CITE, we get MATH (note that MATH since MATH is a point). Hence MATH by NAME Vanishing Theorem. This derives contradiction. Hence we shall assume that MATH is a curve from now on. We know that MATH is locally a complete intersection and MATH is just the blow-up of MATH along MATH (compare CITE). Then MATH being NAME implies MATH, where MATH denotes the normal bundle of MATH in MATH. Let MATH be the normalization, MATH the normalized sheaf of MATH, that is, MATH, but MATH for all line bundles MATH with MATH. Furthermore let MATH be the projection and MATH its fiber. The tautological line bundle MATH has a canonical section MATH satisfying MATH. Put MATH the genus of MATH. MATH . We have MATH and since MATH where MATH is the base change of MATH by MATH. We can write MATH for some MATH. Then note that MATH. Moreover MATH from MATH . Combining REF , we get MATH and MATH . On the other hand, since MATH is a subsheaf of MATH, the last equation implies MATH . Squaring yields MATH . Note that MATH for any curves MATH in MATH, MATH, and so if MATH, MATH is nef by REF . Furthermore we know MATH by REF . If MATH, we get MATH and MATH . Hence MATH is nef and big, so this contradicts MATH. Consequently we know MATH . So we obtain from REF that MATH. Since MATH, it is enough to consider the following four cases: CASE: MATH, MATH, REF MATH, MATH, REF MATH, MATH, and REF MATH, MATH. Before we consider the four cases above, we prove the following claim. Claim. If MATH, then MATH. Proof of the claim. Assume that MATH. We have MATH . Thus MATH by MATH and MATH. Therefore MATH is MATH and MATH is a point for any curves MATH where MATH is the complete linear system on MATH defined by MATH. Using REF , we obtain that MATH in REF and MATH in REF . MATH . Let MATH be the exceptional divisor of MATH for MATH. Then there exists MATH such that MATH and MATH. We denote the image of MATH on MATH by MATH and the strict transform of MATH on MATH by MATH. We know that MATH is a point and MATH. Thus MATH by MATH. Put MATH for some MATH. Then by REF , MATH . On the other hand, let MATH be a general hyperplane section of MATH. Then the induced morphism MATH is a crepant partial resolution. Now that MATH is a smooth rational curve, MATH . Therefore we obtain MATH. Moreover we get MATH by REF , so MATH, MATH and MATH by the definition of MATH and MATH. Then MATH by REF. This yields a contradiction because MATH for a general fiber MATH of MATH. Now we shall investigate each case. CASE: MATH and MATH. In this case, we know that MATH is normal by the definition of MATH, and so MATH and MATH are isomorphisms. From REF , we obtain MATH . If MATH, then MATH . Since MATH is neither nef or big, MATH . The absurdity comes from MATH . Then the claim above yields MATH . From REF , MATH and MATH. Then MATH . Hence MATH . Therefore REF induces MATH. Hence MATH is a NAME surface of degree REF over a smooth rational curve MATH. But this implies that MATH by REF. CASE: MATH and MATH. Now MATH by REF , and so by the claim above, we get MATH . Thus MATH . Then the absurdity comes from MATH . CASE: MATH and MATH. In this case, MATH by REF . We can derive contradiction in the similar way as REF . CASE: MATH and MATH. In this case, MATH, and MATH and MATH are smooth by the definition of MATH. If MATH, we obtain that MATH is nef and big easily. So we know MATH and MATH . Hence MATH or MATH because MATH is an elliptic ruled surface. Combining this with MATH, we know MATH. Therefore MATH. If MATH, then there exist infinitely many irreducible curves on MATH such that each curve is contracted by MATH. Take such an irreducible curve MATH on MATH. Then by REF , MATH is numerically proportional to MATH in MATH. If MATH, then MATH is a smooth rational curve. But this derives contradiction because MATH is an elliptic curve by MATH. Thus there exists a prime divisor MATH such that MATH. This is also absurd because MATH never contracts infinitely many divisors. Therefore we get that MATH. Let MATH be a general hyperplane section of MATH. Then MATH is an irreducible curve on MATH which dose not intersect MATH. So we know that MATH for some MATH (compare Chap.V, REF). But since MATH is not a point, this lies in contradiction. |
math/9912241 | It is enough to consider the case that MATH is a terminal REF-fold by terminalization. Since a general hyperplane section of a terminal REF-fold is smooth, we can use the proof of CITE. |
math/9912241 | Assume that MATH. Write MATH. Now that the NAME morphism MATH is an algebraic fiber space REF , a general fiber of MATH is MATH. Using REF , we know MATH. Let MATH be a MATH-factorial terminalization as in the proof of REF . Since MATH is not nef, there exists an extremal contraction MATH and we know MATH because REF says in paticular that MATH is surjective. Case: MATH . Now MATH, and so MATH for some ample divisor MATH on MATH. On the other hand, since MATH is almost nef REF , MATH. Thus MATH. So we know that MATH is an abelian surface and hence MATH is an isomorphism where MATH is the NAME morphism. If MATH is not MATH-factorial terminal, there exists a curve MATH on MATH such that MATH is a point. On the other hand, MATH implies MATH is a curve. This contradicts MATH. Hence MATH is a MATH-factorial terminal REF-fold and we can apply REF , so we know that MATH is smooth. Then MATH is ample by REF . The ampleness of MATH yields a cotradiction with MATH. Case: MATH . The following argument is in pages REF. By NAME Theory, there exists a sequence MATH where each morphism MATH is a divisorial contraction or a flip for MATH and MATH is a NAME fiber space to a surface MATH. As in the case MATH above, we know that MATH is an abelian surface, so we get readily that MATH is the NAME morphism of MATH. So by REF , MATH is a smooth morphism and MATH is nef. Thus MATH is divisorial. But this is absurd with REF . |
math/9912241 | Taking a MATH-factorial terminalization, we may assume that MATH has at most MATH-factorial terminal singularities. By NAME theory, MATH is birational to a MATH-factorial terminal REF-fold MATH such that MATH has a NAME fiber space structure MATH, that is, MATH is an extremal contraction with MATH. If MATH, then MATH is a projective surface and MATH. So MATH for some ample divisor MATH. On the other hand, since MATH REF , MATH. This is absurd. |
math/9912241 | Assume that MATH is infinite. Then there exists an infinite tower of normal finite étale NAME covers MATH . We know MATH by MATH, hence MATH for a sufficiently large MATH. Then MATH being strictly nef contradicts REF . |
math/9912241 | Because MATH for a general hyperplane section MATH by NAME REF, page REF, MATH is finite. Assume that MATH below. Write MATH. Let MATH be a terminalization and set MATH. Since MATH is nef and not big, MATH by REF . If MATH, then MATH is ample by REF and so it contradicts MATH. Hence NAME Theorem (compare CITE) says that MATH which implies MATH. Let us consider the exact sequence MATH . Exchange MATH for its multiple if necessary. Then we may assume MATH for MATH. Therefore MATH . Now that MATH (compare REF) and MATH is smooth, we can apply REF and we get MATH for some MATH. Hence MATH . This is a contradiction. So MATH. |
math/9912241 | Let MATH be a sufficiently ample general hyperplane section and MATH. Consider the exact sequence MATH . Therefore MATH. Let MATH be a terminalization. If MATH, then MATH by REF . Furthermore by MATH, MATH . So MATH. This implies MATH is ample by REF . On the other hand, MATH is ample when MATH REF . This completes the proof. |
math/9912241 | Note that MATH by REF . Let MATH be the locus of cDV points of MATH. If MATH is finite, combining REF , we know that MATH is ample because MATH. If MATH is infinite, then there exists an infinite tower of normal finite NAME covers MATH such that MATH is étale where MATH is the morphism from MATH to MATH and MATH. Let MATH be a non-cDV point in MATH and MATH the morphism from MATH to MATH. Since MATH is a finite group REF where MATH is a sufficiently small analytic neighbourhood of MATH and MATH stands for the smooth locus of MATH (we use MATH below as the same way), there exists a positive integer MATH such that MATH is disjoint union of copies of MATH for all MATH. Now that MATH is normal, for all MATH, MATH is also disjoint union of copies of MATH (see NAME REF, page REF) and so MATH is étale over MATH. MATH has only finitely many non-cDV points, so MATH is étale and MATH is infinite for all sufficiently large MATH. Now that MATH is a canonical REF-fold with strictly nef MATH, we derive a contradiction from REF . |
math/9912241 | Assume that MATH. Then we know readily that MATH. Let MATH be a resolution of singularities. Because the NAME morphism of MATH is an algebraic fiber space CITE, so is the one of MATH, MATH and MATH. Consider the case that the equality holds. Then we can regard terminalization of MATH as a resolution of singularities MATH REF . So MATH is an abelian variety by CITE. Hence MATH and MATH are isomorphic. If MATH, it is enough to consider the case that MATH is not big. According to REF , MATH so we get MATH by REF . |
math/9912242 | The distribution of MATH has a support which remains in a fixed compact set, and it converges weakly towards that of MATH. REF follows from this and the fact that the function MATH is a limit of a decreasing sequence of continuous functions. If MATH is large enough, then the union of the supports of the distributions of the MATH is away from MATH, hence the function MATH is continuous there and MATH follows from weak convergence. |
math/9912242 | The function MATH is harmonic in MATH, while MATH is subharmonic there, consequently MATH is a nonnegative superharmonic function on MATH. Since this function attains the value MATH by MATH, it is identically MATH by the minimum principle, therefore MATH is harmonic on MATH, and thus the support of MATH is included in MATH. |
math/9912242 | Let MATH be a symmetry free with MATH, then by CITE MATH and MATH are MATH-free, thus MATH is distributed as MATH. Now using the fact that multiplying with a free NAME unitary MATH does not change the MATH-distribution of MATH, we can replace the latter according to MATH, and get the following equalities of MATH-distributions MATH . |
math/9912245 | By induction on the dimension of simplices, it clearly suffices to show the following. If MATH consists of all simplices MATH such that MATH for each strict subset MATH, then MATH is a direct summand of MATH. As MATH is projective the canonical epimorphism MATH admits a section MATH . The definition of MATH yields for each MATH that MATH, and thus also MATH, is the identity on MATH. Composing the projection MATH with MATH then retracts the natural homomorphism MATH and REF follows. Repeating the argument with respect to degrees instead of simplices yields the second assertion. |
math/9912245 | As said above, MATH is an exact sequence of projective MATH - modules and so splits, whence the induced map MATH is surjective. The first claim follows immediately and the second one is obtained replacing skeleton by degree filtration. For the final assertion, note that the hypothesis yields a natural epimorphism of DG MATH - modules MATH. As MATH is projective it has in particular coherent homogeneous components and so MATH can be generated by a finite number of sections over MATH, which in turn provides an epimorphism MATH. Employing now fully that MATH is projective, this epimorphism splits as a homomorphism of MATH - modules. As MATH is generated in a single degree, the differential necessarily vanishes on the generators and the splitting is already a morphism of DG modules. Finally use that MATH. |
math/9912245 | Consider on each simplex MATH the canonical flabby resolution, say, MATH. As this resolution is functorial, each component of the complex MATH is naturally again a DG MATH - module and the resulting system MATH is a resolution by DG MATH - modules. Now cut this resolution on the simplex MATH at the place MATH to obtain MATH for MATH, MATH for MATH and MATH for MATH. The choice of the cut-off point guarantees first, as MATH, that MATH is MATH-acyclic along with the other components. Secondly, as MATH for MATH, the truncated complexes still form a simplicial system. Finally, MATH is now a complex that locally vanishes above and thus the canonical injection MATH into the DG module obtained as the total complex associated to MATH provides the desired functorial quasiisomorphism into a MATH-acyclic MATH - module. |
math/9912245 | REF can be verified locally and there CITE applies. To prove REF , we may assume by the first part of REF that MATH. In this case the spectral sequence associated to the skeleton filtration on MATH converges and so it is sufficient to show the claim in case that MATH with a projective DG MATH - module MATH. In view of REF this requires to show that the corresponding map MATH is a quasiisomorphism. Because of the second part of REF we may reduce to the situation where MATH is generated in finitely many degrees, in which case the spectral sequence associated to the degree filtration on MATH converges. It remains to deal with the case that MATH is generated in a single degree MATH and then the final part of REF exhibits the map MATH as a direct summand of MATH for some MATH. As MATH are MATH-acyclic, this last map is a quasiisomorphism and the claim follows. |
math/9912245 | It follows from REF that any quasiisomorphism between projective DG MATH - modules that locally vanish above is represented by an actual morphism MATH of DG MATH - modules. Such a morphism is a homotopy equivalence if the identity on the mapping cone MATH is homotopic to REF. With MATH also MATH is projective, thus MATH-acyclic, and locally vanishes above. So the preceding result REF applies to yield that MATH is a quasiisomorphism. The rest follows immediately. |
math/9912245 | If MATH is a projective approximation, then MATH is necessarily bounded above with coherent cohomology as this holds for MATH and MATH is a quasiisomorphism. Conversely, it suffices by REF to show the existence of a projective approximation when MATH is furthermore MATH-acyclic. But if MATH is MATH-acyclic with coherent, thus MATH-acyclic, cohomology, then its submodule of boundaries, MATH is also MATH-acyclic: the exact sequences MATH yield for each MATH and each simplex MATH isomorphisms MATH whence MATH for each MATH. In view of this observation, the classical construction of a projective resolution applies: as MATH is coherent there exists a surjection from a graded free MATH - module MATH with coherent homogeneous components onto MATH such that MATH vanishes in degrees MATH if MATH vanishes for those degrees for all simplices MATH. Due to the acyclicity of the boundaries MATH, this homomorphism lifts from the cohomology to the cycles MATH, and the resulting morphism of DG MATH - modules MATH is surjective in cohomology. Now MATH is projective, thus any submodule is MATH-acyclic, and the usual process, see for example . CITE, produces a complex MATH with projective DG MATH - modules MATH that becomes acyclic when MATH is applied. Moreover, on each simplex MATH the projective modules MATH can be choosen to vanish uniformly above so that the DG MATH - module MATH associated to the total complex of the MATH is again a projective MATH - module. The induced morphism MATH then resolves, compare REF, and constitutes a desired projective approximation. |
math/9912245 | Let MATH and MATH be projective approximations as DG MATH - modules. By REF , MATH is a quasiisomorphism and so MATH . As MATH and MATH is a projective DG MATH - module that locally vanishes above, REF follows from REF. To obtain REF , let MATH be as before and MATH a MATH - acyclic resolution that is a morphism of DG MATH - modules. One has then MATH . For REF , note that MATH on MATH is fully faithful by REF . This functor takes its values in MATH, and to establish it is an equivalence with inverse as indicated, it suffices to remark that for each MATH in MATH the natural morphism MATH obtained from REF is an isomorphism in MATH. Indeed, let MATH be again a MATH - acyclic resolution that is a morphism of DG MATH - modules and choose a morphism of DG MATH - modules MATH that constitutes a projective approximation. As MATH is already a MATH - module, this quasiisomorphism factors as MATH, the morphism MATH is a quasiisomorphism by REF , and thus so is MATH. Now the pair of quasiisomorphisms MATH represents the morphism MATH, whence the latter is an isomorphism in MATH. |
math/9912245 | As mentioned above, the natural morphism of functors MATH is an isomorphism on the level of modules. It induces an isomorphism MATH of the corresponding derived functors on MATH, whence the functor MATH is still fully faithful on MATH. To prove the projection formula, observe first that the natural map MATH is an isomorphism as this holds on each simplex. As MATH is exact, this isomorphism passes to the derived tensor products as soon as those exist. Adjunction then yields a morphism MATH. Now set MATH and compose the corresponding morphism with the one induced by the adjunction map MATH to obtain the morphism in the projection formula. Given the existence of this natural morphism, to establish it as an isomorphism for MATH, we first replace MATH by a projective approximation MATH, then use REF to reduce to the case MATH for some simplex MATH. Now MATH is a finite complex of flat MATH - modules that resolves MATH and MATH resolves MATH. The desired isomorphism in MATH follows thus from the obvious one for MATH - modules, MATH. |
math/9912245 | According to our general assumption on DG algebras, the MATH - module MATH is in MATH, and so REF guarantee a projective approximation of MATH in form of a morphism of DG MATH - modules from a graded free DG MATH - module, say MATH. The induced morphism MATH of DG algebras induces a surjection in cohomology and the structure map MATH is free. Now assume constructed for some integer MATH a morphism of DG MATH - algebras MATH with MATH free over MATH that is surjective in cohomology and such that MATH is an isomorphism for MATH. As the kernel of the surjection MATH is coherent, one may choose a graded free coherent MATH - module MATH that is concentrated in degree MATH and a morphism MATH of graded MATH - modules of degree REF that maps MATH into the cycles of MATH and such that the sequence of MATH - modules MATH is exact. Now set MATH and use MATH to extend the given differential. This DG algebra is free over MATH and the structure map is an isomorphism in degrees greater than MATH. The composed map, say, MATH maps MATH into the boundaries, hence we can find a lifting MATH so that MATH. There is a unique homorphism of MATH-algebras MATH that restricts to MATH on MATH. By construction it is also a morphism of DG algebras, and it is surjective in cohomology with MATH an isomorphism for each MATH. Finally set MATH. |
math/9912245 | In view of REF, the quasiisomorphism REF for MATH shows that MATH and the term on the right is isomorphic to MATH. According to REF, the term on the left is isomorphic to MATH, and so the first isomorphism follows. The second one follows from the fact that MATH is a projective MATH - module. The final assertion is a consequence of the usual universal property of the module of analytic differentials, see REF. |
math/9912245 | As the direct sum of a family of connections is again a connection, we may restrict by REF to the case that MATH, where MATH is a projective MATH - module generated in a single degree, say MATH. If MATH is graded free, then MATH with MATH a finite dimensional vector space over MATH, and the collection of maps MATH defines a connection on MATH. In the general case MATH embeds into a free module such that MATH. If MATH is a connection on MATH, then the composition MATH is easily seen to be a connection on MATH. |
math/9912245 | That MATH is a homomorphism of right MATH - modules is easily verified by explicit calculation. Moreover, MATH whence MATH is a homomorphism of DG modules, thus a cycle. If MATH are connections, then MATH is MATH - linear and so MATH, which means that the cycles MATH are cohomologous. |
math/9912245 | If MATH, the classes in question are equal to MATH itself. If MATH, the map MATH of degree MATH is MATH-linear. As MATH is equal to MATH the claim follows. |
math/9912245 | To compare the NAME classes formed with DG algebra resolutions MATH and MATH of MATH over MATH, note first that by REF there is a free MATH algebra that constitutes a resolvent of MATH. Thus we may suppose that MATH is a free MATH-algebra. In this REF gives the independence from the choice of DG algebra resolution. The construction is as well independent of the embedding MATH. With similar arguments as above it suffices to compare two embeddings MATH and MATH that are related by a smooth map MATH, meaning that MATH is smooth for every MATH. If MATH is a free MATH-algebra forming a resolvent then we can take a free MATH algebra, say, MATH as a resolvent on MATH. Now a projective MATH-resolution MATH of MATH gives a projective MATH-resolution MATH, and REF applies again. Finally, the independence from the choice of locally finite coverings by NAME compact sets is easily seen considering refinements; we leave the simple details to the reader. |
math/9912245 | Consider as in REF a resolvent of MATH over MATH and choose projective approximations MATH and MATH. There is a morphism MATH lifting the given morphism MATH and the assertion follows now easily from REF. |
math/9912245 | By the preceding result and by REF the diagram commutes. |
math/9912245 | Let MATH be a quasiisomorphism into a MATH-acyclic module, MATH a projective approximation, and let MATH be a connection. Using the product rule, the composed map can be extended to a connection MATH on MATH. Hence under the natural map MATH the NAME class MATH maps onto MATH. Since the module on the left is isomorphic to MATH and the module on the right is isomorphic to MATH, the result follows. |
math/9912245 | This is an immediate consequence of REF and the fact that MATH. |
math/9912245 | Let MATH be a resolvent of MATH as in REF. There are projective resolutions MATH, MATH, MATH of MATH, MATH, MATH, respectively that fit into a commutative diagram Consider connections MATH, MATH on MATH, MATH, respectively, and equip MATH with the connection MATH. Using the isomorphisms MATH can be computed as follows. The linear map MATH can be extended to a linear map MATH, and MATH is zero on MATH and so defines a map MATH that represents MATH. Taking as extension the map MATH we get MATH, and so MATH represents MATH. By construction it also represents MATH, proving the second part of the result. The equality MATH follows with a similar argument and is left to the reader. |
math/9912245 | We need to show for MATH that MATH . As the trace map is compatible with taking cup products, the diagram commutes, where as before MATH. Therefore MATH . As MATH is a derivation of degree MATH on MATH and MATH, we obtain that MATH . For homogeneous endomorphisms MATH the trace satisfies MATH, whence taking traces yields MATH . Comparing with REF , the result follows. |
math/9912245 | In the algebraic case this is shown in CITE. In the analytic case we can proceed as follows. Let MATH be a resolvent of MATH over MATH as in REF and let MATH be a projective resolution of MATH as a MATH - module. A MATH-extension MATH of MATH gives rise to an extension MATH of MATH by MATH. Since MATH is smooth over MATH, the embedding MATH can be lifted to a MATH-map MATH, and the surjection of algebras MATH to a map of MATH-algebras MATH. With MATH the differential on MATH, the map MATH is a MATH-derivation of degree REF that represents the class of MATH in MATH . If one equips the trivial extension MATH with the differential MATH, the map MATH becomes a quasiisomorphism of DG algebras that restricts to the identity on MATH. Let now MATH be a connection. Contracting with MATH and projecting onto MATH gives a map MATH of degree REF satisfying the product rule MATH for local sections MATH in MATH and MATH in MATH . The class MATH is represented by the map MATH of degree REF; note that by REF MATH . If the class MATH vanishes then MATH for some MATH-linear map MATH of degree REF. The differential MATH defines then on MATH the structure of a DG module over MATH and we denote this DG module MATH. Using the exact cohomology sequence associated to the exact sequence of DG modules MATH it follows that the MATH module MATH is an extension of MATH by MATH. Gluing yields a MATH modules MATH which is an extension of MATH by MATH. Conversely assume that there exists such an extension MATH of MATH by MATH. As MATH is projective the map MATH can be lifted to a map of MATH - modules MATH. A simple calculation shows that MATH satisfies the product rule MATH for sections MATH in MATH and MATH in MATH. Hence MATH is MATH-linear and satisfies MATH, whence the cohomology class of MATH in MATH vanishes. As this class represents MATH under the isomorphism MATH, the result follows. |
math/9912245 | For the convenience of the reader we include the simple argument. We may assume that MATH is a complex of finite free MATH - modules with MATH for MATH. By REF is a perfect complex and so for MATH the complex MATH has the property that MATH is exact with MATH a free MATH - module. Using induction on MATH and the long exact cohomology sequences associated to the exact sequences of complexes MATH it follows that MATH is exact and that MATH is a free MATH - module. Hence MATH is exact and MATH is free as a MATH - module, proving the lemma. |
math/9912245 | Let MATH be a resolvent for MATH over MATH as in REF and choose a quasiisomorphism MATH into a MATH-acyclic complex of MATH - modules, where MATH and MATH denote the simplicial sheaves on MATH associated to MATH and MATH, respectively. We choose an algebra resolution MATH of the composition MATH so that MATH is a graded free algebra over MATH and MATH is a quasiisomorphism of DG algebras. In particular, MATH is a projective approximation of MATH as a MATH - module, and MATH is a quasiisomorphism by REF. It follows that MATH provides a projective resolution of MATH as MATH - module. Hence MATH for all MATH. Composing the natural inclusions, see REF, MATH gives the desired map in REF . If MATH, then MATH and MATH, and the inclusion of derivations into endomorphisms is a morphism of DG NAME algebras that induces a morphism of graded NAME algebras in cohomology. To show REF , note first that MATH is represented by MATH. Consider a derivation MATH of degree MATH that represents the cohomology class MATH in MATH. Its image in MATH is then represented by MATH. Under the isomorphism in MATH, the image of the latter element in MATH is represented by a MATH-derivation MATH, where MATH is a derivation restricting to MATH on MATH. Let now MATH be a connection on MATH, as exists by REF. If MATH is one of the maps MATH or MATH from MATH to MATH then the product rule MATH is satisfied for local sections MATH in MATH and MATH in MATH. It follows that the difference MATH is MATH-linear and so MATH and MATH represent the same cohomology class in MATH. As MATH by definition, and as MATH is of degree MATH, we have MATH as required. |
math/9912245 | For a manifold, MATH, and according to NAME, CITE, see also CITE, the fundamental class of MATH is given by MATH whence the result follows from REF. |
math/9912245 | Let MATH be a resolvent of MATH and choose a free graded DG MATH - algebra MATH that provides a DG algebra resolution MATH of the composition MATH. By REF, the induced map MATH is a quasiisomorphism and so MATH constitutes a resolvent for MATH over MATH. Thus MATH and moreover MATH where MATH is the NAME functor as in REF. By construction, MATH is a free MATH-algebra and so admits an augmentation, say, MATH. Clearly, the MATH-linear map MATH will not be a morphism of DG algebras in general, but the commutative diagram MATH shows that the MATH-derivation MATH factors through a derivation MATH that in turn represents MATH under the isomorphism MATH, see CITE. The map MATH is induced by the map of complexes MATH and composition with the canonical map MATH yields the map MATH. As this map coincides with the MATH-derivation MATH under the identification MATH the result follows. |
math/9912245 | For suitable representative of the cotangent complexes involved there is a commutative diagram of exact sequences of complexes of MATH - modules by REF and our assumption that MATH is injective on MATH. In MATH this gives rise to a commutative diagram MATH . As well, the commutative diagram of exact sequences of MATH - modules yields in MATH a commutative diagram Combining this square with the one in REF gives the result for MATH. The general case follows from this by taking appropriate exterior powers. |
math/9912245 | Let MATH be the class corresponding to the extension MATH, given by a morphism MATH in the derived category. The result REF induces a commutative diagram where MATH is the NAME connection. The relative partial NAME character MATH is the image of the absolute partial NAME character MATH in MATH. Using REF and the injectivity of MATH in the diagram above, it follows that REF is equivalent to REF. Contracting against MATH sends the absolute partial NAME character to zero, MATH . Now, REF yields a commutative diagram and MATH is injective by REF below. It follows that MATH if and only if REF holds. On the other hand, by REF, the condition MATH is equivalent to REF , so the result follows. |
math/9912245 | Let MATH be the MATH infinitesimal neighbourhood of MATH in MATH so that MATH is an extension of MATH by MATH, where MATH is the maximal ideal. The map MATH is injective and applying REF repeatedly we see that MATH can be lifted to a deformation MATH on MATH for all MATH. Let MATH be a versal deformation of MATH which is a coherent module on MATH where MATH is a complex space germ over MATH. Using versality there are MATH - morphisms MATH with MATH and MATH. Hence MATH admits a formal section, namely MATH. By NAME 's approximation theorem, we can find a convergent section MATH. Now MATH is a coherent NAME module on MATH that induces MATH on the special fibre. The uniqueness of the horizontal lifting gives that MATH as sections in MATH for each MATH. Hence MATH is algebraic for all MATH near MATH and each MATH. |
math/9912245 | For REF observe that the map MATH is surjective due to the versality of MATH. In turn, by REF below, the functor MATH is right exact as MATH is smooth over a subspace of MATH. Thus, MATH is right exact in MATH as well. To show REF , let MATH, MATH be as in REF . By versality, there is a morphism MATH that lies over some MATH - morphism MATH. If now MATH is an extension of MATH over MATH, then the homogeneity condition yields an extension MATH over MATH that satisfies MATH. Thus, there is a natural isomorphism MATH for every artinian MATH - module MATH, whence REF implies REF . In order to show REF consider the MATH infinitesimal neighbourhood MATH of REF in MATH and set MATH. Repeating the argument just given, we have MATH for any MATH. Hence there is a commutative diagram Since MATH is formally semiuniversal, the map MATH is bijective and MATH is surjective. By REF is surjective and so MATH is surjective too. Now the result follows from the Jacobian REF below. It is obvious that REF implies REF . Finally, MATH follows with the same reasoning as above from the smoothness criterion given in REF. |
math/9912245 | The implications REF are obvious. To show REF , note first that MATH as MATH is complete. Hence REF implies that MATH is surjective. This means that there are derivations MATH and elements MATH such that MATH, where MATH. By REF and NAME, see CITE, there is then an isomorphism MATH, where MATH is a MATH - subalgebra of MATH. By construction, MATH, and so MATH must be a quotient of MATH, which gives REF . Finally, assume that MATH and that REF is satisfied. Writing MATH with MATH and MATH, we need to show that MATH. If not, choose a power series MATH of minimal order in MATH. After a linear change of coordinates we may assume that MATH with MATH. Now consider the curvilinear MATH-algebra MATH. The derivation MATH on MATH induces a derivation MATH that by assumption can be lifted to a derivation MATH. Composing with MATH yields a derivation, say, MATH with MATH. Using the product rule and the fact that MATH for MATH, it follows that MATH. On the other hand, MATH, whence MATH and this is a contradiction. |
math/9912245 | As was already observed in the proof of REF, we have MATH, and, as MATH is formally semiuniversal, the map MATH is bijective. Hence we can write MATH with MATH. By definition, MATH, and so the dimension of MATH is just the minimal number of generators of MATH. In the extended NAME sequence MATH the module MATH vanishes by the versality of MATH. Hence MATH injects into MATH and the minimal number of generators of MATH is bounded by MATH. |
math/9912245 | Note that MATH as MATH injects into MATH. By REF, the vector space MATH is dual to MATH, where MATH is as in the proof of REF and MATH is the integral closure of MATH in MATH. Now the result follows from REF. |
math/9912245 | The injective hull MATH of MATH can be obtained as a limit MATH with MATH and each MATH finite artinian. As MATH is the zero map, there is an index MATH so that MATH is already zero. Restricting the map MATH to MATH yields a commutative diagram and REF follows. In order to derive REF , note that the map MATH embeds MATH into MATH, as MATH is formally versal, and under this map MATH becomes thus isomorphic to a subspace of MATH by REF . As MATH is isomorphic to MATH, the space of infinitesimal deformations, the claim follows from REF. |
math/9912245 | Indeed, the functor MATH is left exact on MATH and satisfies the assumptions of REF. |
math/9912245 | Let MATH be a deformation of MATH over an artinian germ MATH, whence MATH is a MATH - module that is flat over MATH and restricts to MATH on MATH. By REF, for every coherent module MATH on MATH there is a semiregularity map MATH and by REF, this map is compatible with base change. According to CITE, the functor MATH is exact; note that using the results of CITE, NAME 's original result extends to the case of compact manifolds that are bimeromorphically equivalent to NAME manifolds with the same proofs as in (loc.cit.). Applying REF follows. Finally assume that MATH is injective. Using induction on the length of MATH it follows easily that MATH is injective for all MATH. In particular, the functor MATH is exact on the left and therefore MATH is exact on the right. Now REF shows that MATH is smooth over a closed subspace of MATH, as claimed. |
math/9912245 | Let MATH be a deformation of MATH over an artinian germ MATH, whence MATH is a MATH - module that is flat over MATH and induces MATH on MATH. The functor MATH is exact. As the semiregularity map is compatible with base change MATH, see REF, the result follows as before from REF. |
math/9912245 | To prove this statement, we use the existence of tangent functors MATH for holomorphic mappings as constructed in CITE, see also CITE. These functors fit into an exact sequence MATH see CITE. The group MATH is canonically isomorphic to the set of all isomorphism classes of extensions of MATH by MATH. Such extension is a holomorphic map MATH, where MATH, MATH are extensions of MATH, MATH by MATH, MATH, respectively, with MATH inducing the map MATH on MATH and the identity on MATH; see CITE for details. Moreover, the map MATH in MATH assigns to MATH the extension MATH. In view of the exactness of the sequence MATH this proves the lemma. |
math/9912245 | Let MATH be a resolvent for MATH over MATH and let MATH be a quasiisomorphism into a MATH-acyclic complex MATH of MATH - modules as in REF. We need to construct a map MATH . Let MATH be the topological preimage of the sheaf MATH and let MATH be the associated simplicial sheaf of rings on MATH. As the topological restriction MATH is a sheaf of abelian groups on MATH, we can form its associated NAME complex and so we can consider the composed map MATH where the first two maps are the natural inclusions and the last one is given by the NAME functor. There is always a natural morphism MATH, thus taking cohomology we obtain a natural map MATH where the final isomorphism results from the fact that the complexes MATH and MATH are quasiisomorphic to MATH, respectively, MATH, see REF. Composing MATH with MATH gives the desired map. If MATH, then replacing MATH by MATH and MATH by MATH, the first inclusion in MATH becomes bijective, and so the argument shows the compatibility with the NAME algebra structure. |
math/9912245 | For every coherent MATH - module MATH the natural map MATH is a quasiisomorphism and so the claim follows from REF. |
math/9912245 | This is a local calculation, so we may suppose MATH . NAME and NAME defined by REF so that the NAME complex MATH is a MATH-resolution of the sheaf MATH. More explicitly, set MATH with the NAME differential given by MATH. Note that MATH is the free graded algebra over MATH with generators MATH, MATH, in (cohomological) degree MATH and that MATH is a derivation of degree REF. In particular, MATH for every coherent MATH-module MATH. Consider on MATH the connection MATH . The NAME class of MATH is now the element of MATH represented by MATH a MATH-linear map of degree REF. Note that MATH as well as MATH are derivations on the ring MATH and that MATH. Thus, the map MATH which is only nonzero on MATH and MATH, is given there by MATH . Consider now MATH and write MATH with sections MATH of MATH. Under the isomorphism MATH, the element MATH corresponds to the derivation MATH with MATH. The composition MATH is a map of degree MATH and is therefore determined by the component MATH . Note that MATH represents MATH. Now we wish to take traces to produce an element in MATH. Let MATH be the Cousin complex, see REF. Let MATH be the basis of MATH dual to MATH and set MATH . With this convention, MATH is the basis element dual to MATH and the differential on MATH is given by multiplying from the left with MATH. The computation above gives that MATH . Using REF below, this is mapped under the trace map to MATH . Comparing with MATH above, the result follows. |
math/9912245 | As MATH is a sheaf of flat MATH - modules, the complex MATH is quasiisomorphic to MATH. Therefore, the canonical projection MATH induces a quasiisomorphism MATH . In a first step, we construct a section of this projection, MATH as follows. The map MATH, where MATH is as above, realizes MATH as a subcomplex of MATH. We define MATH to be the composition MATH where the first map is given by MATH. As the identity corresponds to the element MATH in MATH, the map MATH is given explicitly by MATH . Now we can define the local trace map MATH as the composition of MATH . The image of MATH under these maps is just given by MATH as desired. |
math/9912245 | For REF , let MATH be a curvilinear extension of MATH by MATH and let MATH be a morphism of MATH - algebras lifting the given map MATH, so that MATH corresponds under MATH to the extension MATH. By the valuative criterion of integral dependence mentioned above, MATH is in the kernel of MATH, whence MATH. Thus MATH. To show equality, assume that MATH is a MATH-linear form on MATH that vanishes on MATH. We need to show that MATH vanishes on MATH. Such a linear form can be written as MATH for some MATH. By assumption, for every curvilinear extension MATH of MATH by MATH, the element MATH vanishes. Applying the valuative criterion of integral dependence it follows that MATH and so MATH vanishes on MATH, as desired. In order to deduce REF note that the map MATH assigns to an extension MATH the algebra extension MATH of MATH by MATH that is the quotient of the trivial extension MATH by the ideal MATH, where MATH is the differential. The reader may easily verify that then MATH and that the NAME map MATH becomes the inclusion of MATH into MATH, whence MATH is injective. Conversely, if for an extension MATH the map MATH is injective, then it is easily seen that the construction just described recovers MATH from the extension MATH of MATH by MATH. Finally for REF , if MATH is a curvilinear extension as in the diagram MATH in REF, then the composed map MATH is the map MATH and so is injective. Hence MATH is also injective, proving the inclusion MATH. |
math/9912245 | Replacing MATH by MATH we may assume that MATH. Choose elements MATH that form a basis of the MATH-vector space MATH and consider the natural ring homomorphism MATH given by MATH. In a first step we prove that this map is finite. In fact, the elements MATH generate the ring MATH as an algebra over MATH, and if MATH is an equation of integral dependence for such a MATH over MATH, the coefficients satisfy MATH, whence MATH and finiteness follows. This implies MATH . As MATH appears as the special fibre of MATH we obtain MATH see CITE. Together with REF the result follows. |
math/9912245 | In the absolute case, where MATH, this is a result due to NAME and NAME, see CITE. Alternatively, it follows from the chain of REF where we have applied REF. To deduce the general case, set MATH. The spectral sequence MATH yields MATH . Hence the result follows from the chain of REF . |