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math/9906205 | The corresponding assertions for the uncompleted tensor product are easy. It is clear that the bornologies on MATH and MATH coincide with the bornology generated by the trismall sets MATH. Thus the uncompleted bornological tensor product is associative and the MATH- fold uncompleted tensor product is universal for bounded MATH- linear maps. To carry this over to the completed tensor product, we have to extend bounded multi-linear maps to the completions. REF implies that MATH for all convex bornological vector spaces because both sides are complete bornological vector spaces universal for bounded bilinear maps MATH with complete range MATH. Especially, MATH. REF follows. |
math/9906205 | View the multiplications in MATH and MATH as bounded linear maps MATH, MATH. Then we can tensor them to get a bounded linear map MATH. Up to a flip of the tensor factors, this is the linearized version of the natural multiplication in the tensor product. Hence the natural multiplication is bounded. |
math/9906205 | An equibounded family of bounded linear maps MATH corresponds to an equibounded family of bounded bilinear maps MATH. By definition, boundedness for bilinear maps MATH means that if we fix MATH and let MATH vary in MATH, then the family of linear maps MATH, MATH maps any small set MATH into a small set in MATH. This is precisely the condition for a subset of MATH to be equibounded. |
math/9906205 | It is clear that the differential and the grading are bounded operators on MATH. It suffices to verify that the multiplication is bounded on MATH because bounded bilinear maps extend to the completion by REF . Of course, the multiplication is the usual multiplication of non-commutative differential forms CITE. Since MATH is bounded, MATH for all MATH. Thus MATH for all MATH. That is, the multiplication on MATH is bounded. |
math/9906205 | By REF , any small set MATH is contained in a set MATH with MATH. The boundedness of the multiplication in MATH implies that MATH. Thus MATH is small in MATH. |
math/9906205 | Since MATH has positive radius of convergence, we have MATH for some MATH. Let MATH be the disked hull of MATH. Since MATH is a-nilpotent, there is a completant small disk MATH containing MATH. Thus MATH for all MATH and hence MATH . Hence MATH is an absolutely convergent series in the NAME space MATH. |
math/9906205 | This follows immediately from the definitions. |
math/9906205 | Let MATH. We have to show that MATH is small. Let MATH. This set is small in MATH by a-nilpotence. Since MATH is a quotient map, there is MATH with MATH. Let MATH . Then MATH because for all MATH, there is MATH with MATH. That is, MATH and thus MATH by definition of MATH. Let MATH . We claim that MATH . To see this, observe that MATH implies MATH . Thus for all MATH, there is MATH with MATH and therefore MATH and MATH by definition of MATH. Let MATH . Since MATH is an ideal, MATH. The sets MATH are therefore small because MATH is a-nilpotent. REF implies MATH, MATH, and MATH. Thus MATH. This implies MATH. Moreover, MATH because MATH. Thus MATH is small. Consequently, MATH is a-nilpotent. |
math/9906205 | Let MATH be a-nilpotent. By REF , it suffices to show that the uncompleted bornological tensor product MATH is a-nilpotent. Any small subset of MATH is contained in a set of the form MATH with MATH, MATH. Choose MATH such that MATH is small. Since MATH the set MATH is small, and hence so is MATH. Thus MATH is a-nilpotent. |
math/9906205 | Let MATH be a bounded subset. Rescale MATH such that MATH for all MATH, MATH. We claim that MATH is bounded in MATH. This implies that MATH is tensoring by REF . We have to prove that MATH is bounded in the algebras MATH of MATH times continuously differentiable functions for all MATH. The NAME transform for these commutative NAME algebras is the inclusion into MATH. Therefore, the spectral radius in MATH is equal to the norm in MATH and thus bounded above on MATH by MATH. By definition of the spectral radius, the set MATH is norm-bounded in MATH. The condition of REF is trivially verified for the NAME algebra MATH. Thus MATH is tensoring. |
math/9906205 | By the universal property of completions, bounded homomorphisms MATH are in bijection with homomorphisms MATH that map MATH to a small set in MATH for all MATH. The universal REF of MATH asserts that bounded homomorphisms MATH are in bijection with bounded linear maps MATH. In addition, the homomorphism MATH corresponding to a linear map MATH satisfies MATH . The sets MATH are small for all MATH iff MATH is a lanilcur. Consequently, we can extend MATH to a bounded homomorphism MATH iff MATH is a lanilcur, and this extension is necessarily unique. |
math/9906205 | By REF , it suffices to show that MATH is a-nilpotent because MATH is the completion of this convex bornological algebra. Thus we have to verify that MATH for all MATH. There is MATH such that MATH is contained in the disked hull of MATH. Hence it suffices to prove that MATH. We claim that there is MATH such that MATH and MATH (the letter MATH stands for ``invariant"). Since multiplication is bilinear, it follows by induction that MATH for all MATH and hence MATH. Thus MATH as desired. Let MATH and MATH . Clearly, MATH. Pick MATH and MATH and compute MATH . This is a sum of (at most) MATH terms in MATH and thus lies in the disked hull MATH of MATH as desired. |
math/9906205 | The curvature of the identity map MATH is MATH for all MATH. If this map is a lanilcur, then MATH for all MATH. Hence MATH. That is, MATH is a-nilpotent. |
math/9906205 | If the curvature of MATH factors as in REF, then MATH for all MATH because MATH is a homomorphism and MATH is a-nilpotent. That is, MATH is a lanilcur. |
math/9906205 | If each map into MATH is a lanilcur, then the identity map MATH is a lanilcur, so that MATH is a-nilpotent by REF . Conversely, if MATH is a-nilpotent, then we can factor the curvature of any bounded linear map MATH through the a-nilpotent algebra MATH by writing simply MATH. Thus any bounded linear map into MATH is a lanilcur by REF . |
math/9906205 | By REF , MATH is a lanilcur if its curvature can be factored as in REF. Conversely, if MATH is a lanilcur, then we have MATH with a bounded homomorphism MATH by the universal REF of MATH. Hence the curvature of MATH can be factored as MATH. The range of MATH is contained in MATH because MATH is a homomorphism. Thus we can view MATH as a bounded linear map MATH. By REF , MATH is a-nilpotent. Hence MATH factors through an a-nilpotent algebra as asserted. |
math/9906205 | We have MATH and MATH because MATH and MATH are multiplicative. Use REF to factor the curvature of MATH through an a-nilpotent algebra MATH as MATH with a bounded homomorphism MATH and a bounded linear map MATH. Then MATH and MATH factor through MATH as well. Thus MATH and MATH are lanilcurs by REF . In particular, if MATH is a bounded homomorphism, then MATH is a lanilcur. By the universal REF , there is a unique bounded homomorphism MATH with MATH. The symmetry of REF implies that MATH is functorial. |
math/9906205 | REF implies REF . Let MATH. Since MATH is split surjective with bounded linear section MATH, we have an allowable extension MATH. By REF , MATH and MATH are a-nilpotent. Thus MATH is a-nilpotent by REF . Since MATH is a homomorphism, the curvature of MATH takes values in MATH. That is, it can be factored through the inclusion MATH. Since MATH is a-nilpotent, MATH is a lanilcur by REF . Let MATH and MATH be lanilcurs. By the universal REF of MATH, we can factor MATH and MATH with bounded homomorphisms MATH and MATH. By REF , we can replace MATH by MATH. Thus MATH . Since MATH is a bounded homomorphism and MATH is a lanilcur, MATH is a lanilcur by REF . Consequently, lanilcurs form a category. REF implies REF . Since MATH is a lanilcur, the universal REF of MATH yields a bounded homomorphism MATH. By construction, MATH . The uniqueness part of REF implies MATH. Hence MATH is a section for MATH. Thus MATH is a-quasi-free in the sense of REF implies REF . Assume that there is a bounded homomorphism MATH satisfying MATH. The universal REF of MATH implies that MATH is a lanilcur. It is unclear whether the quasi-freeness of MATH alone, that is, the existence of any bounded splitting homomorphism MATH for MATH, suffices to prove REF . REF implies REF . Let MATH be an allowable extension of complete bornological algebras in which MATH and MATH are a-nilpotent. We choose a bounded linear section MATH of our allowable extension. To verify that MATH is a-nilpotent, we have to verify that any bounded linear map MATH is a lanilcur by REF . Since MATH is a-nilpotent, MATH is a lanilcur by REF . By the universal property of MATH we can extend it to a bounded homomorphism MATH. We consider the bounded linear map MATH defined by MATH . By adding MATH, we achieve that MATH. Since MATH, the composition MATH is a homomorphism. Hence the curvature of MATH factors through the inclusion MATH. By assumption, MATH is a-nilpotent and hence MATH is a lanilcur by REF . Since MATH is a lanilcur as well, the composition MATH is a lanilcur by REF . Thus REF follows from REF . |
math/9906205 | REF implies REF . Let MATH be a lanilcur. By REF , we can factor the curvature of MATH through an a-nilpotent algebra MATH. That is, MATH with a bounded linear map MATH and a bounded homomorphism MATH. Let MATH be the multiplication map. We have MATH and thus MATH. Hence the curvature of MATH factors through MATH. Since MATH is a-nilpotent by REF , MATH is a lanilcur by REF implies REF . The linear map MATH is a lanilcur. Hence MATH is a lanilcur by REF . By REF , the composition MATH is a lanilcur. The universal REF of MATH yields that there is a unique bounded homomorphism MATH making REF commute. REF implies REF . If we apply REF to the bounded homomorphism MATH, we obtain that MATH is a lanilcur because it can be extended to a bounded homomorphism MATH. A lanilcur MATH can be extended to a bounded homomorphism MATH by the universal REF of MATH. NAME with MATH, we get a bounded homomorphism MATH satisfying MATH. Since MATH is a lanilcur, so is its composition with the bounded homomorphism MATH by REF . Thus MATH is a lanilcur as desired. REF implies REF . Let MATH be an a-nilpotent complete bornological algebra. We have to show that MATH is a-nilpotent. By REF , the identity map MATH is a lanilcur. Thus the identity map MATH is a lanilcur as well by REF . By REF , this implies that MATH is a-nilpotent. |
math/9906205 | Let MATH be a splitting homomorphism for MATH. Let MATH be a bounded linear section for MATH. Since MATH is a homomorphism, the curvature of the bounded linear map MATH has values in MATH. Since MATH is a-nilpotent by assumption, MATH is a lanilcur by REF . The universal REF of MATH yields an associated bounded homomorphism MATH. Define MATH. By construction, MATH . The uniqueness part of REF yields MATH and hence MATH as desired. Thus MATH, so that there is a unique MATH making MATH a morphism of extensions. The uniqueness of the lifting MATH up to smooth homotopy relative to MATH follows from the more general statement about the lifting of homotopies. We use the bounded homomorphism MATH and the bounded linear map MATH defined similarly. Since MATH for MATH, there is a bounded linear map MATH satisfying MATH for MATH and MATH. For instance, we can take MATH . Since MATH is a homomorphism, the curvature of MATH has values in MATH. This algebra is a-nilpotent by the Homotopy REF . Thus MATH is a lanilcur by REF . Let MATH be the associated homomorphism by the universal REF . Define MATH. By construction, MATH and thus MATH by the uniqueness assertion of REF . Therefore, MATH. A similar computation shows that MATH for MATH. The above argument used no other information about smooth homotopies than the Homotopy Axiom, stating that MATH is tensoring. Since MATH is tensoring as well, the same argument shows that we can lift absolutely continuous homotopies to absolutely continuous homotopies between morphisms of extensions. |
math/9906205 | The algebra MATH is a-nilpotent by REF and MATH is a-quasi-free by the Extension REF . Thus MATH is a universal a-nilpotent extension. Let MATH, MATH, be two universal a-nilpotent extensions of MATH (we write MATH as an abbreviation for these extensions). By the existence part of the Universal Extension REF , the identity map MATH can be lifted to morphisms of extensions MATH and MATH. The composition MATH is a lifting of the identity map MATH to an automorphism of MATH. Another such lifting is the automorphism MATH. By the uniqueness part of the Universal Extension REF , these two liftings are smoothly homotopic relative to MATH. An analogous argument shows that the composition MATH is smoothly homotopic to the identity automorphism relative to MATH. This means that MATH and MATH are smoothly homotopy equivalent relative to MATH. Especially, MATH and MATH are smoothly homotopy equivalent algebras and MATH and MATH are smoothly homotopy equivalent algebras. In addition, any morphism of extensions between MATH and MATH of the form MATH is a smooth homotopy equivalence. In fact, by the uniqueness part of the Universal Extension REF , all morphisms of this form are smoothly homotopic relative to MATH. |
math/9906205 | REF is the definition of analytic quasi-freeness. If REF holds, then MATH is a universal a-nilpotent extension of MATH because the zero algebra is certainly a-nilpotent. If we apply the Universal Extension REF to this universal extension, we get REF with the additional information that the lifting MATH is unique up to smooth homotopy. If we apply REF to the extension MATH, we get REF . If we apply REF to the a-nilpotent extension MATH, then we get REF . Thus REF - REF are equivalent. Finally, if MATH is a-quasi-free, then the Uniqueness REF implies that the two universal a-nilpotent extensions MATH and MATH are smoothly homotopy equivalent relative to MATH. Thus MATH is a smooth deformation retract of MATH and MATH is smoothly contractible. In addition, the Uniqueness Theorem asserts that all bounded splitting homomorphisms MATH are smoothly homotopic. |
math/9906205 | Let MATH be a-quasi-free. By REF , all a-nilpotent extensions of MATH split by a bounded homomorphism. Since MATH- nilpotent algebras are a-nilpotent by REF , MATH satisfies REF . That is, MATH is quasi-free. |
math/9906205 | Since MATH is a free left MATH- module, the left MATH- module homomorphisms MATH are those linear maps of the form MATH with arbitrary bounded linear maps MATH. The map MATH is even a bimodule homomorphism iff MATH for all MATH, that is, MATH is a derivation. The bounded linear sections MATH are those maps of the form MATH with an arbitrary bounded linear map MATH. The multiplication in MATH is defined so that MATH is multiplicative iff MATH is a derivation. |
math/9906205 | Let MATH be a MATH- bimodule. We claim that composition with MATH gives rise to a bijection between MATH- bimodule homomorphisms MATH and bounded linear maps MATH. This is precisely the universal property of a free module. Since universal objects are uniquely determined, it follows that MATH is a bornological isomorphism. REF yields a bijection between bimodule homomorphisms MATH and bounded splitting homomorphisms MATH for the natural projection MATH. The bimodule homomorphism MATH corresponds to the homomorphism MATH. By the universal REF of MATH, composition with MATH gives rise to a bijection between bounded homomorphisms MATH and lanilcurs MATH. Furthermore, the homomorphism associated to a lanilcur MATH is a section for MATH iff MATH. Thus we get a bijection between bimodule homomorphisms MATH and bounded linear maps MATH with the additional property that MATH is a lanilcur. To a bimodule homomorphism MATH, this bijection associates the bounded linear map MATH. To finish the proof of REF, it remains to show that each map MATH of the form MATH with a bounded linear map MATH is a lanilcur. We have MATH. Since this is a homomorphism, the curvature of MATH factors through MATH. Since MATH is split surjective, we have an allowable extension MATH. Since MATH and MATH are a-nilpotent by REF , the Extension REF implies that MATH is a-nilpotent. Thus MATH is a lanilcur by REF . The proof of REF is complete. To obtain the isomorphisms REF, we consider the natural allowable extension MATH (see REF below) and replace MATH by MATH. This extension splits naturally as an extension of left MATH- modules. Hence if MATH is a left MATH- module, then MATH is an allowable extension of right MATH- modules. We apply this to MATH with the zero module structure. That is, MATH for all MATH. We get an extension MATH. The map MATH is the usual one with kernel MATH and the map MATH is induced by the bilinear map MATH. Thus REF is an isomorphism. The isomorphism in REF is obtained similarly by applying the functor MATH to MATH. |
math/9906205 | Since MATH is a lanilcur, the map MATH is a lanilcur as a composition of two lanilcurs. Thus MATH is a well-defined bounded homomorphism by the universal REF of MATH. In fact, the uniqueness part of the universal property asserts that MATH is the only bounded homomorphism satisfying MATH, that is, making the above diagram commute. The composition MATH is a bounded homomorphism with MATH. Hence MATH. |
math/9906205 | Copy the proof of REF of the Tensor REF , using that lanilcurs form a category instead of REF . Use that MATH and MATH are tensoring. |
math/9906205 | Since MATH is a-nilpotent, so is MATH by the Homotopy REF . Hence the bounded linear map MATH defined by MATH is a lanilcur. Thus MATH is smoothly contractible in the lanilcur category. Let MATH be a bounded linear section. The curvature of MATH has values in MATH. Since MATH is a-nilpotent, MATH is a lanilcur. By definition, MATH. Define MATH by MATH, that is, MATH. Clearly, MATH (that is, MATH for all MATH). Thus the curvature of MATH factors through the a-nilpotent algebra MATH, so that MATH is a lanilcur by REF . The lanilcur MATH is a smooth homotopy between MATH and MATH. Thus MATH is a smooth deformation retract of MATH in the lanilcur category. Upon applying the smooth homotopy functor MATH, we get that MATH is smoothly contractible and that MATH is a smooth deformation retract of MATH in the homomorphism category. |
math/9906205 | Since MATH and MATH are lanilcurs, it is evident that MATH is well-defined. Furthermore, MATH because MATH and MATH by the definition of the functoriality for lanilcurs in REF . The proof of the Homotopy REF carries over to MATH without change. Thus MATH is tensoring. This implies that MATH is a smooth map. Details are left to the interested reader. |
math/9906205 | Let MATH, MATH, be NAME homomorphisms. REF implies that the homomorphisms MATH are NAME as well. By the Extension REF , the algebra MATH is a-quasi-free and a fortiori quasi-free by REF therefore implies that the induced maps MATH are chain homotopic. Thus MATH in MATH as asserted. Since the functoriality of the analytic cyclic theories is given by composition with the class MATH, it follows that NAME homomorphisms induce the same map on all analytic cyclic (co)homology groups. |
math/9906205 | By the Uniqueness REF , the extension MATH is smoothly homotopy equivalent to MATH. In particular, MATH and MATH are smoothly homotopy equivalent. That is, there are homomorphisms MATH and MATH such that MATH and MATH are smoothly homotopic to the identity. Furthermore, MATH and MATH are unique up to smooth homotopy. Since MATH and MATH are a-quasi-free, they are quasi-free by REF . The homotopy invariance of the NAME REF therefore applies. It shows that MATH, MATH, and that MATH is independent of the choice of MATH. If MATH is a-quasi-free, then we can apply this to the universal a-nilpotent extension MATH. This yields that MATH and MATH are inverses of each other up to chain homotopy. |
math/9906205 | We abbreviate MATH. We construct a chain map MATH that is a candidate for the inverse of MATH. Since MATH is tensoring, the Tensor REF yields a natural bounded homomorphism MATH that is determined by MATH. This implies, by the way, that MATH for MATH, MATH. Thus we get a natural chain map MATH. If MATH is any complete bornological algebra, we define MATH by MATH . The tracial property of MATH easily implies that MATH is a well-defined chain map. For MATH, this yields a natural bounded chain map MATH. We claim that MATH is inverse to the homology class MATH of the chain map MATH. Since MATH is a homomorphism, one easily checks that MATH. Since MATH for any rank one idempotent, it follows that MATH. That is, MATH in MATH. It remains to show that MATH in MATH. There are two standard injections MATH, namely MATH . We have MATH and MATH. The latter equation follows formally from the naturality of MATH. We will prove that MATH and MATH are smoothly homotopic maps MATH. It follows that MATH and MATH are smoothly homotopic for any MATH. Thus MATH by REF . Hence MATH as desired. It remains to prove that MATH and MATH are smoothly homotopic. We define a multiplier MATH of MATH such that conjugation MATH with MATH implements the coordinate flip MATH sending MATH. Recall that MATH as a complete bornological vector space. We define left and right multiplication by MATH by MATH . Left and right multiplication by MATH extend to bounded linear maps MATH. Moreover, MATH, MATH, and MATH for all MATH. That is, MATH is a multiplier of MATH. In the case of MATH, we have MATH and MATH is the coordinate flip MATH viewed as a multiplier of MATH. We have MATH, that is, MATH and MATH for all MATH. Therefore MATH. There is a natural path of invertible multipliers joining MATH to the identity, defined by MATH and MATH for MATH. A computation using MATH shows that MATH for all MATH. Moreover, MATH and MATH. The bounded homomorphisms MATH provide a smooth homotopy between MATH and MATH as desired. |
math/9906205 | Let MATH be the unit element. A splitting homomorphism MATH is equivalent to a lifting of MATH to an idempotent MATH in MATH. That is, MATH and MATH. Our construction of MATH follows CITE. Adjoin units to MATH and MATH and consider the involution MATH. It suffices to lift MATH to an involution MATH in MATH. That is, MATH and MATH. We make the NAME MATH with MATH. This is always a lifting for MATH and satisfies MATH because MATH is commutative. Hence MATH is an involution iff MATH. The function MATH is holomorphic in a neighbourhood of the origin and satisfies MATH. Since MATH is a-nilpotent, MATH is well-defined by REF and solves our problem. Thus MATH is a-quasi-free. Explicitly, the power series of MATH is MATH . Thus MATH . By the way, the binomial coefficients MATH are always even. Thus the coefficients in REF are integers. |
math/9906205 | Let MATH be a-quasi-free. We have to construct a bounded splitting homomorphism MATH. Since the algebra MATH is a-quasi-free by REF , we can lift the homomorphism MATH sending MATH to a homomorphism MATH by REF . That is, we can lift MATH to an idempotent MATH. Thus MATH. We cut down MATH by this idempotent. Since MATH, we get an allowable extension of complete bornological algebras MATH . A bounded linear section for this extension is MATH. Since MATH is a-quasi-free, another application of REF allows us to lift the natural homomorphism MATH to a bounded homomorphism MATH. We extend MATH to a bounded linear section MATH by putting MATH and MATH for MATH. This map MATH is a homomorphism because MATH for all MATH. Thus MATH is a-quasi-free. Conversely, assume that MATH is a-quasi-free. We have to construct a bounded splitting homomorphism MATH. REF provides a bounded splitting homomorphism MATH for the a-nilpotent extension MATH . Since MATH for all MATH, the restriction of MATH to MATH has values in MATH. Therefore, MATH is a bounded splitting homomorphism MATH and MATH is a-quasi-free. |
math/9906205 | By REF , the extension MATH is a universal a-nilpotent extension of MATH. The Uniqueness REF implies that MATH and MATH are naturally smoothly homotopy equivalent. REF implies that MATH is naturally chain homotopic to MATH. We claim that MATH for any complete bornological algebra MATH. Clearly, MATH. Let MATH be the unit and let MATH be the linear span of terms MATH and MATH with MATH. We will prove that MATH. Since MATH, this implies that MATH. The claim MATH follows. Since MATH for all MATH, we have MATH. Computing the commutators MATH, MATH, MATH, MATH, we find that MATH. Thus MATH. The isomorphisms of the associated (co) homology groups listed above follow immediately. |
math/9906205 | By the universal property of MATH, a lifting homomorphism MATH is determined by liftings MATH of MATH and MATH (that is, MATH, MATH) satisfying MATH. Lift MATH to MATH. Then MATH. Similarly, MATH. Since MATH is a-nilpotent, both MATH and MATH are invertible. Thus MATH is invertible with inverse MATH . Thus MATH defines a lifting homomorphism MATH. It is clear that MATH. We claim that the odd part MATH is isomorphic to MATH via the map MATH. Since MATH is commutative, the derivation rule for MATH implies that MATH for all MATH, MATH. It follows that MATH for all MATH and MATH. Moreover, MATH for all MATH. Thus MATH is a well-defined map MATH. It is bijective. The boundary MATH is the zero map, whereas the boundary MATH is the map MATH. The map MATH is injective because MATH implies that MATH is constant and thus MATH. The range of MATH is the linear span of MATH, MATH. Hence MATH is the sum of a contractible complex and the complex MATH. The copy of MATH is generated by MATH. |
math/9906205 | The proof follows CITE. We only have to verify that certain maps are bounded. This is straightforward if we write down explicit formulas for these maps. Some details are relegated to REF. Let MATH be the boundary that corresponds to MATH under the isomorphism MATH. That is, MATH. Using the definition MATH, we easily compute that MATH . We have to show that the complexes MATH and MATH are chain homotopic. The operator MATH satisfies a polynomial identity MATH on MATH for all MATH. Hence MATH satisfies such an identity, too. It follows that the restriction of MATH to MATH has discrete spectrum. One of the eigenvalues is MATH. This eigenvalue has multiplicity MATH, that is, the corresponding eigenspace is equal to the kernel of MATH. Let MATH be the projection onto the MATH- eigenspace annihilating all other eigenspaces. Let MATH be the operator that is zero on the MATH- eigenspace and the inverse of MATH on all other eigenspaces. Explicit formulas for MATH and MATH show that they are bounded with respect to MATH (see REF). Any operator commuting with MATH commutes with MATH and MATH. Thus MATH and MATH are chain maps with respect to MATH and MATH. The restriction of MATH to the MATH- eigenspace of MATH is equal to MATH. If the eigenvalue MATH had no multiplicity, we could derive this immediately by replacing MATH with MATH in REF. Since MATH on MATH, we have to be a bit more careful. See REF or the proof of CITE. The proof will therefore be finished if we show that MATH is chain homotopic to the identity with respect to the boundaries MATH and MATH. Since MATH and MATH is a chain map, this follows if MATH is chain homotopic to MATH. We write MATH for chain homotopy. REF implies that MATH because MATH is homogeneous of degree MATH and therefore commutes with MATH. Thus MATH with respect to the boundary MATH. We have MATH by REF. However, MATH implies easily that MATH. For example, if we evaluate MATH on MATH, then we get MATH. Thus MATH with respect to the boundary MATH. |
math/9906205 | This follows from REF . We only have to examine the dual complex MATH of bounded linear maps MATH. By the universal property of completions, bounded linear maps MATH correspond to bounded linear maps MATH. Since the sets MATH generate the bornology MATH, the bounded linear functionals on MATH are those linear maps MATH that remain bounded on all sets of the form MATH. Identifying MATH and MATH, we find that MATH can be identified with the space of families MATH of MATH- linear maps MATH satisfying the ``entire growth condition" MATH for all MATH. Here MATH if MATH or MATH and MATH is a constant depending on MATH but not on MATH. The boundary on MATH is simply composition with MATH. Thus MATH is the complex used by CITE to define entire cyclic cohomology for non-unital REF algebras. NAME 's original definition CITE of entire cyclic cohomology is restricted to unital algebras and uses a slightly different complex. However, for unital algebras the complex used by NAME is homotopy equivalent to the complex used by NAME. The proof is written down in CITE for NAME algebras, but carries over to locally convex algebras without difficulty. |
math/9906205 | Since the operator MATH is bounded, we have MATH for all MATH. Thus MATH is bounded. Since MATH, it is a bornological isomorphism. It is trivial to verify that MATH is multiplicative, when considered as a map MATH. |
math/9906205 | It suffices to prove that MATH is a bornological isomorphism. We obtain the other map MATH by exchanging left and right as in REF : MATH. We write down explicitly the inverse of MATH and verify that it is bounded. We will need some detailed estimates about this inverse for the proof that MATH is a-quasi-free. Let MATH be a right polarized monomial. We write MATH and thus derive a formula for the inverse MATH. If MATH for all MATH, then MATH. In particular, this applies to MATH. Otherwise, some MATH is in MATH. Pick the first MATH with MATH. Hence MATH for MATH. If MATH is even, then MATH . If MATH is odd and MATH, then MATH . Finally, if MATH, then MATH . Of course, we define MATH to be the pre-image of MATH described above. Writing elements of MATH as linear combinations of right polarized monomials, we obtain a linear map MATH. By construction, this map satisfies MATH. It is easy to verify that MATH. REF - REF imply that MATH maps right polarized small sets to small sets. Hence MATH extends to a bounded linear map MATH. The equalities MATH and MATH carry over to the completions. Thus MATH is the inverse of MATH as asserted. For later application, we describe the image of a right polarized subset of MATH under MATH more explicitly. Let MATH, MATH, MATH. We write MATH to denote that MATH and get MATH . The constant factor MATH converts sums of two terms into convex combinations. It is important that in the very first tensor factor, we get nothing bigger than MATH. |
math/9906205 | REF and MATH imply that MATH is a bornological isomorphism. Since MATH is a free right MATH- bimodule by REF, it follows that MATH is a free MATH- bimodule. We claim that MATH . It is evident that MATH. Moreover, MATH . Thus MATH. Conversely, MATH . This implies MATH by REF and finishes the proof of REF. The inverse MATH of the isomorphism MATH in REF maps MATH onto MATH and maps MATH onto MATH. Thus MATH is a bornological isomorphism. Since MATH is a free left MATH- module by REF and MATH is a free right MATH- module by REF, MATH is a free MATH- bimodule. Thus MATH is an allowable free MATH- bimodule resolution of MATH. MATH is quasi-free because it satisfies REF . The inclusion MATH is a MATH- bimodule homomorphism and therefore descends to a chain map MATH. Using REF, we compute that MATH is an isomorphism MATH: MATH . REF imply that MATH is an isomorphism MATH: MATH . Thus MATH is an isomorphism of chain complexes MATH as asserted. The standard resolution MATH is a subcomplex of MATH. Let MATH be the inclusion. The comparison theorem implies that MATH is a homotopy equivalence. Let us make this more explicit. Let MATH and define MATH by MATH . It is evident that MATH is a bornological isomorphism and that MATH. Thus MATH. Let MATH and let the boundary MATH be the identity map. Then MATH is a contractible complex of MATH- bimodules and MATH. Taking commutators quotients, we obtain MATH . The copies of MATH match and can be left out. The map MATH in this isomorphism is equal to MATH. Thus MATH is split injective. Let MATH be the image of MATH in MATH. We find that MATH is the range of the map MATH, MATH and that MATH is the range of the map MATH, MATH. Thus we get the complex MATH described in the theorem. Evidently, MATH is a bornological isomorphism. Thus MATH vanishes on MATH, so that MATH is a subcomplex of MATH as well. Since MATH is compatible with MATH, we get a direct sum decomposition MATH. Since MATH is contractible, the chain map MATH is a homotopy equivalence. |
math/9906205 | Since MATH is a lanilcur, it can be extended to a bounded homomorphism MATH. Extend this further to a unital bounded homomorphism MATH. By REF there is a bornological isomorphism MATH. Thus MATH is a well-defined bounded linear map. It remains to prove that MATH is multiplicative and that MATH. It is clear from REF that MATH is invariant under MATH and thus under MATH. The induced left MATH- action on MATH coincides with the usual left MATH- multiplication by REF. Thus MATH for all MATH, MATH. Consequently, MATH. To show that MATH is a splitting homomorphism, we prove first that MATH is a left multiplier for all MATH. The bornological isomorphism REF applied to MATH implies that MATH as a right MATH- module. Thus in order to show that MATH is a left multiplier, it suffices to prove that MATH for all MATH, MATH. This follows if MATH. Since MATH is a graded derivation, REF implies MATH . It follows that MATH as desired. Thus MATH is a left multiplier. In particular, REF implies MATH . Hence MATH for all MATH, MATH. Since MATH is a left multiplier for all MATH, it follows that MATH is a left multiplier for all MATH because the left multipliers form a bornologically closed unital subalgebra of MATH. Therefore, MATH for all MATH, MATH. Hence MATH is multiplicative. |
math/9906205 | If MATH is small, that is, equibounded, then MATH by equiboundedness. This small set is MATH- invariant by definition. Conversely, if MATH is MATH- invariant, then MATH. Hence if MATH is small, so is MATH. This means that MATH is equibounded. |
math/9906205 | By definition, MATH. Since MATH anti-commutes with MATH and commutes with MATH, we have MATH for all homogeneous MATH. In addition, REF implies that MATH. Consequently, MATH commutes with MATH for all MATH. Let MATH, MATH, then REF implies MATH . Thus MATH is a closed graded trace. Since MATH and MATH are graded commutators by definition, it follows that MATH and MATH. |
math/9906205 | If MATH and MATH, then MATH because MATH and MATH commute. This implies MATH. The boundary MATH from even to odd degrees is given by REF. Since MATH, the desired equation MATH is equivalent to MATH and MATH. The first equality is the definition of MATH. For the second equation, let MATH, MATH and compute MATH . Decorated with appropriate normalization constants, this means MATH. |
math/9906205 | REF shows that MATH. REF implies that MATH for all MATH, MATH. Since either MATH or MATH is even, the graded commutator of MATH and MATH is MATH. Hence MATH vanishes on all graded commutators. |
math/9906205 | If MATH and MATH, then MATH because MATH and MATH anti-commute. This implies MATH. By definition, MATH. Therefore, MATH. Moreover, MATH . Thus MATH and consequently MATH. |
math/9906205 | Let MATH be precompact. Then there are compact disks MATH, MATH such that MATH by NAME 's REF . We have MATH. Since MATH converges to MATH uniformly on compact subsets, it follows that MATH for MATH. Consequently, MATH as asserted. Let MATH be precompact. Let MATH and MATH be the unit balls. We claim that MATH is precompact. Given any MATH, we have MATH for a suitable finite set MATH. By definition, a compact operator maps MATH to a precompact set. Thus MATH for a suitable finite set MATH. Furthermore, MATH maps MATH into MATH. Consequently, MATH. Since this holds for all MATH, MATH is precompact. The sequence of operators MATH converges to MATH uniformly on the precompact subset MATH. Hence MATH for MATH. Thus MATH for MATH. |
math/9906205 | We only consider the odd case. The even case is similar. Let MATH be the range of the idempotent MATH. Then MATH. Write MATH as a MATH- matrix MATH with respect to this decomposition. Computing commutators with MATH shows that MATH iff the off-diagonal terms MATH and MATH are compact, and MATH iff MATH and MATH are nuclear. Since MATH has NAME 's approximation property, MATH is dense in MATH. Thus MATH is dense in MATH. |
math/9906205 | We will give the details of the proof in the odd case. The even case is similar but slightly more complicated because the relevant part of the boundary MATH involves MATH. Since we no longer have MATH, this gives rise to complicated expressions. Let MATH. Thus MATH is precompact. Since MATH and MATH are equibounded, there is a constant MATH such that MATH for all MATH. REF implies that MATH for MATH. Since MATH commutes with MATH, we have MATH for all MATH. Thus MATH if MATH. Since MATH for sufficiently large MATH, the set of numbers MATH is bounded. Thus MATH is bounded on the set MATH. By the universal property of the completion, MATH extends uniquely to a bounded operator MATH. The same argument works for MATH instead of MATH. To estimate the growth of MATH, we want to use the equation MATH proved in REF . Define a linear map MATH by MATH for MATH and MATH (in the even case, it is better to take MATH for MATH and MATH). By convention, MATH. Since MATH, we get MATH . Evidently, MATH and hence MATH . In the even case, we find similarly that MATH with MATH on MATH. Since MATH does not commute with MATH, this gives rise to complicated sums. These can be handled. It is, however, more convenient to replace MATH by the boundary MATH given by MATH on MATH and MATH on MATH. The boundaries MATH and MATH give rise to chain homotopic complexes. Hence it is admissible to replace MATH by MATH. The summands MATH or MATH are harmless. In fact, MATH is zero because we declared MATH. In the even case MATH is not zero unless MATH. NAME if MATH, it is easy to prove that MATH is bounded. Hence we concentrate on the other two summands in REF. Let MATH and MATH. It suffices to verify that MATH and MATH remain bounded on MATH because MATH. The set MATH is precompact. Let MATH . The sequence MATH is decreasing by definition and converges towards MATH for MATH by REF . Fix MATH. We compute MATH on MATH. Since MATH maps closed forms again to closed forms, MATH annihilates closed forms MATH. We compute MATH . We used that MATH commutes with MATH, so that MATH. If MATH, we can estimate MATH . Since MATH is a cut-off sequence MATH has at most exponential growth. Since MATH, the term MATH is MATH for any MATH. Hence it decays fast enough to compensate the at most exponential growth of MATH. Therefore, MATH for all MATH. In particular, MATH remains bounded on MATH. Next, we compute MATH on MATH: MATH . If MATH, we can estimate MATH . This is again MATH for any MATH for MATH because the super-exponential decay of MATH overcompensates the at most exponential growth of MATH. The estimation on MATH is similar. The only difference is that the terms in the sum with MATH and MATH drop out. Consequently, MATH remains bounded on sets of the form MATH with MATH and MATH. It remains to verify that the choice of the cut-off sequence does not matter. Assume that MATH is produced by another cut-off sequence MATH. Then MATH extends to a bounded linear functional on MATH. This is verified in the same way as the boundedness of MATH. Therefore, MATH and MATH are cohomologous, the difference being the coboundary MATH. |
math/9906205 | Let MATH. This is a closed unital subalgebra of MATH and thus a unital NAME algebra. We can replace the path MATH by the path of idempotents MATH in MATH. Standard techniques of MATH- theory CITE imply that there is a continuous path of invertibles MATH in MATH such that MATH and MATH. It is well-known that MATH is dense and closed under holomorphic functional calculus in MATH. Thus we can replace the path MATH by a nearby smooth path of invertible elements MATH. In addition, we can arrange that MATH and MATH. |
math/9906205 | An element of MATH is represented by an even MATH- NAME module. The NAME character of such a module was defined in the previous section. We have already observed that it is invariant under operator homotopies, additive, and vanishes on degenerate modules. Hence it descends to a map MATH. The same applies in the odd case. It remains to compare the pairings MATH and MATH for MATH, MATH, MATH. We consider the even case. Represent MATH by an even NAME module MATH. The NAME character is defined by MATH. Since the NAME character in MATH- theory is natural, we have MATH . Comparing with the index pairing, we see that it suffices to show that the functional MATH sending MATH is equal to MATH. Since MATH is generated by MATH, it suffices to compute MATH. Since MATH, we can replace MATH by MATH. We have already checked MATH in REF. Similarly, in the odd case the compatibility with the index pairing can be reduced to the computation MATH that was done in REF. |
math/9906205 | Of course, REF hold if MATH is tensoring and REF implies REF . Furthermore, REF implies that MATH is tensoring by REF . Therefore, we will be finished if we show that REF implies REF and that REF are equivalent. Since MATH, there is MATH. Let MATH, then MATH. Since MATH is a-nilpotent, the set MATH is small. That is, it is contained in MATH for suitable small disks MATH and MATH. Since the map MATH is injective, this implies that MATH for all MATH (Use that the projective tensor product norm on a NAME space satisfies MATH.) Let MATH, then MATH. That is, MATH. Thus REF implies REF . Let MATH be an inductive system of NAME algebras, let MATH. If MATH, then MATH is contained in the image of a bounded subset of MATH for some MATH. Hence MATH is contained in the image of the unit ball of MATH for suitable MATH. It follows that MATH is small. Thus MATH satisfies REF . Conversely, if MATH satisfies REF , then the sets MATH with MATH for some MATH are cofinal. If MATH, then MATH is a NAME algebra with respect to some norm. We have MATH by REF . We can replace MATH by any cofinal subset without changing the limit. Thus we can write MATH as an inductive limit of NAME algebras. |
math/9906205 | We only prove the assertion about extensions, the other assertions are trivial. Let MATH be an extension, possibly without a bounded linear section. Since MATH is nuclear, tensoring with MATH preserves extensions. Thus MATH is again an extension. (If we had bounded linear sections, we would not need the nuclearity of MATH.) If MATH and MATH are tensoring, then MATH and MATH are a-nilpotent. The Extension REF implies that MATH is a-nilpotent. Thus MATH is tensoring by REF . |
math/9906205 | If MATH is tensoring, then so is MATH as an extension of MATH by the a-nilpotent and thus tensoring algebra MATH. Conversely, if MATH is tensoring, so is its quotient MATH. |
math/9906205 | In CITE, NAME proves the assertion if MATH and MATH are NAME algebras with the bounded bornology. This is the special case of entire cyclic cohomology for NAME algebras. Furthermore, the chain maps between MATH and MATH and the homotopy operators joining the compositions to the identity are natural. We can reduce the case of tensoring algebras to this special case by writing a tensoring algebra as an inductive limit of NAME algebras. The functors MATH and MATH commute with inductive limits. That is, if MATH for MATH, then MATH . Thus if we have natural maps between these complexes for NAME algebras, we get induced natural maps for arbitrary tensoring algebras. |
math/9906205 | Any pro-linear map MATH with constant range factors through MATH for suitable MATH by the concrete construction of projective limits. The MATH- fold multiplication in the pro-algebra MATH is zero because a product of MATH forms of positive even degree has degree at least MATH and is therefore zero in MATH. Thus MATH. |
math/9906205 | The curvature of the identity map MATH is equal to MATH. If MATH is a lonilcur, then for any pro-linear map MATH with constant range there is MATH such that MATH, that is, MATH. Thus MATH is locally nilpotent. |
math/9906205 | Assume that the pro-linear map MATH factors as described above. Let MATH be a pro-linear map with constant range. Then MATH for some MATH because MATH is locally nilpotent. Since MATH is a homomorphism, we have MATH. Hence MATH. This means that MATH is a lonilcur. |
math/9906205 | Let MATH be a lonilcur. We have to construct a homomorphism of pro-algebras MATH satisfying MATH. Let MATH, where MATH is the multiplication in MATH and MATH and MATH is the unital extension of MATH. The map MATH describes the map MATH in the situation of pro-algebras. Let MATH be a pro-linear map with constant range. We claim that there is some MATH such that MATH for all MATH. The pro-linear map MATH factors as MATH with pro-linear maps MATH, MATH with constant range and a bounded linear map MATH. Since MATH is a lonilcur, there is some MATH such that MATH. Hence MATH . To construct MATH, write MATH as a projective system. For each MATH, we have the pro-linear map MATH with constant range and have found above a number MATH such that MATH for all MATH. We define pro-linear maps MATH by MATH . The pro-linear maps MATH form a morphism of projective systems from MATH to MATH. This follows from MATH for all MATH. This morphism of projective systems induces a pro-linear map MATH. The pro-linear map MATH is defined by the same formula that occurs in the universal property of the algebraic tensor algebra MATH of an algebra in REF. Since that formula defines a homomorphism, the map MATH is a homomorphism of pro-algebras. Details are left to the reader. Moreover, MATH by construction, and MATH is the only homomorphism MATH with that property. It remains to show that any linear map of the form MATH with a homomorphism of pro-algebras MATH is a lonilcur. If MATH is a pro-linear map with constant range, then MATH factors through MATH for some MATH. Hence MATH factors through MATH. Since the latter pro-algebra is MATH- nilpotent, it follows that MATH. |
math/9906205 | Let MATH be a pro-linear map with constant range. It factors as MATH for pro-linear maps with constant range MATH, MATH, and a bounded linear map MATH. Since MATH is locally nilpotent, there is MATH with MATH. The MATH- fold multiplication in MATH is MATH up to a coordinate flip MATH. Hence MATH. Thus MATH is locally nilpotent. |
math/9906205 | By definition, an extension of pro-algebras is a projective limit of extensions. That is, we have extensions MATH of complete bornological vector spaces indexed by a directed set MATH and morphisms MATH for MATH between these extensions such that MATH, MATH, MATH, MATH, MATH. We can describe the MATH- fold multiplication MATH in MATH by maps MATH. Since the map MATH is multiplicative, we can achieve (by increasing MATH if necessary) that MATH maps the kernel of MATH into MATH. The maps on the quotients MATH induced by MATH describe the MATH- fold multiplication in MATH. The restrictions to MATH describe the MATH- fold multiplication in MATH. We write MATH for all MATH. We have to show that for all MATH, there are MATH and MATH such that MATH. Since MATH is locally nilpotent, there are MATH and MATH, MATH, such that MATH annihilates MATH. Since MATH is locally nilpotent, there are MATH and MATH, MATH, such that MATH is the zero map. That is, MATH is mapped into MATH. Thus MATH. Hence MATH for all MATH with MATH and MATH. Thus MATH is locally nilpotent. |
math/9906205 | We may assume that MATH is a homogeneous monomial and that MATH. More general monomials MATH can be treated by induction on MATH. Thus we have to verify MATH for all MATH and MATH for all MATH. These assertions follow easily by inspecting the summands in REF and estimating their degrees. The crucial observation is that the map MATH may decrease the internal degree by at most MATH. That is, MATH is a linear combination of homogeneous monomials MATH with MATH for all MATH. This follows from the construction of MATH in the proof of REF . To conclude that MATH for all MATH, it suffices to consider the case MATH. This implies the assertion for monomials of higher degree: MATH . It is easy to compute MATH and MATH. These terms have modified total degree at least MATH. |
math/9906205 | We have the following easy recipe to compute MATH. If MATH, we subtract from MATH an element of MATH such that the result is in MATH. If MATH, we subtract from MATH elements of MATH and MATH until the result is in MATH. On the even part MATH, we have MATH . This suffices to determine the even part of MATH by REF. We decompose the odd part MATH into the direct sum MATH. The summand MATH is easy to handle. We have MATH . The isomorphism REF shows that this determines MATH on MATH. The summand MATH requires two steps. As above, we have MATH. Bringing MATH into standard form, we get terms in MATH and terms in MATH involving the curvature of MATH. Since MATH vanishes on MATH, the summands in MATH do not contribute. We apply MATH to each curvature term to get an element of MATH. Then MATH . This procedure determines MATH on MATH because of REF. Thus to obtain MATH, we have to apply the isomorphism MATH at most once on MATH and twice on MATH. Application of MATH either leaves the internal degree unchanged or looses one internal degree. In the latter case, we split a monomial MATH into MATH, producing the kind of expression that occurs in the definition of MATH. The assertion of the lemma follows in a straightforward way. To prove the assertion about MATH, we also have to consider expressions of the form MATH with either MATH or MATH REF . The details are rather boring and therefore omitted. |
math/9906205 | The even part of MATH maps MATH to MATH by REF . This space is mapped by MATH to MATH by REF . Since MATH and MATH is integer valued, we get only terms with MATH. These are mapped by MATH into MATH because MATH. |
math/9906205 | We get as above that MATH maps MATH with MATH and MATH to MATH . (We first take the completant linear hull, then map to the commutator quotient.) This is contained in MATH by REF . |
math/9906205 | The terms in MATH are treated by REF and the estimate MATH. To handle the summand MATH we need the extension of MATH to MATH. Pick a generator MATH of MATH with MATH. We compute MATH . REF implies that MATH and MATH have total degree MATH. Hence the first two summands above are of the required form. Write MATH as a sum of homogeneous monomials MATH. Since MATH, we have MATH or MATH and MATH. If MATH, then MATH has total degree MATH. The second summand in REF is there to accommodate MATH with MATH and MATH. We even have MATH or MATH or MATH with MATH. These estimates are needed to handle MATH. The first two cases are harmless because MATH decreases total degrees MATH by at most MATH and does not decrease the external degree. That is, MATH implies MATH and MATH, and MATH implies MATH and MATH. The definition of MATH in REF implies MATH . This tackles the third case because MATH. |
math/9906205 | We have MATH . Rewriting MATH according to REF, we only get terms with at least MATH's. Moreover, MATH . Both summands on the right hand side are in MATH by REF. |
math/9906205 | Up to commutators, we have MATH . Thus MATH. A similar computation proves the claim. |
math/9906205 | Since the MATH in REF are NAME spaces, we have bornological isomorphisms MATH . This follows directly from the definitions of bounded linear maps. REF follows if we plug this into the universal property of inductive limits REF. Let MATH be a complete bornological vector space. The universal property of the inductive limit gives rise to a natural bounded linear map MATH extending the natural inclusions MATH. The concrete construction of inductive limits shows that this map is a bornological isomorphism. If we apply REF to MATH and MATH, we get REF. Thus the functor MATH is fully faithful. REF can be proved similarly. By REF, we can reduce to the case where MATH are all primitive spaces. If MATH are primitive, the definition of an equibounded family of MATH- linear maps implies immediately that MATH . Let MATH be an injective system of NAME spaces and let MATH be its inductive limit. Let MATH be the image of the unit ball of MATH under the structure map MATH. By the concrete definition of the inductive limit bornology, a subset of MATH is small iff it is contained in MATH for some MATH for some constant MATH. It follows that MATH is equivalent to the inductive system MATH. Since MATH is injective, the structure maps MATH are all injective. Hence we have natural isomorphisms MATH for all MATH. It follows that MATH is equivalent to MATH. Conversely, inductive systems of the form MATH are injective by construction. Thus MATH is an equivalence between the category of complete bornological vector spaces and the category of injective inductive systems of NAME spaces. Its inverse is the inductive limit functor MATH. Of course, we still have MATH if MATH is only essentially injective. Therefore, if MATH is essentially injective, then REF implies MATH . If we consider separated convex bornological vector spaces instead, we have to replace MATH by MATH. The spaces MATH for MATH are then only normed spaces. Otherwise, nothing changes. If we drop the separation assumption, then the spaces MATH are only semi-normed. We have to take the non-separated inductive limit instead of the separated inductive limit to use REF. |
math/9906205 | The sets MATH are completant disks by construction. Each set of MATH is contained in a completant small disk and MATH for all MATH. Thus MATH is a completant bornology on MATH. For example, REF for a bornology follows because MATH. By construction, MATH for all MATH. Hence we can view MATH as a bounded linear map MATH. It remains to verify that any bounded linear map MATH with complete range MATH can be factored as MATH with a bounded linear map MATH. Let MATH, then MATH is contained in a small completant disk MATH because MATH is complete. We claim that MATH can be extended to a bounded linear map MATH. It is clear that MATH can be extended to a bounded linear map on the completion MATH of MATH. The problem is that the natural map MATH may fail to be injective. An element in the kernel is the limit MATH of a (bornological) NAME sequence MATH in MATH for which MATH. By assumption, since MATH is a bornological null sequence, MATH is a bornological null sequence in MATH and hence MATH because MATH is bounded. Thus the kernel of the map MATH is annihilated by the extension MATH. Consequently, MATH descends to a bounded linear map MATH. Finally, we observe that the extensions MATH, MATH piece together to a bounded linear map MATH. Thus MATH is isomorphic to the completion of MATH. |
math/9906205 | Recall NAME 's fundamental theorem about compact subsets of the projective tensor product of two NAME spaces. Let MATH and MATH be two NAME spaces. Let MATH be a compact subset. Then there are null-sequences MATH and MATH in MATH and MATH, respectively, and a compact subset MATH of the unit ball of the space MATH of absolutely summable sequences such that MATH . In particular, it follows that each point of MATH is contained in a set of the form MATH with bounded MATH and MATH. Thus MATH is a bornology on MATH. Since MATH is topologically complete, the bornologies MATH and MATH are completant. The natural bilinear map MATH is bounded. Hence there is a bounded linear map MATH. REF implies that each precompact subset of MATH is contained in a set of the form MATH with MATH, MATH, and null-sequences MATH. Since the points of a null-sequence form a precompact set, MATH and MATH are precompact. Thus MATH is a quotient map. Similarly, the bilinear map MATH is bounded and induces a bounded linear map MATH. By definition of MATH, each MATH is of the form MATH for a small set MATH. That is, MATH is a quotient map. It remains to show that MATH and MATH are injective. For an injective quotient map is automatically a bornological isomorphism. By the concrete definition of the completion, MATH is the separated inductive limit of the spaces MATH with MATH running through MATH, MATH. Pick any MATH with MATH. We have to show that MATH. There are completant precompact disks MATH, MATH, such that MATH. We view MATH as an element in the kernel of the natural map MATH. This natural map can indeed fail to be injective because there are NAME spaces that do not have NAME 's approximation property. However, for MATH to describe the zero element of MATH, it suffices that there are completant precompact disks MATH, MATH, such that MATH is annihilated by the natural map MATH. This follows from the following corollary of REF also due to NAME. Let MATH and MATH be NAME spaces. Let MATH be a null-sequence in MATH. Then there are null-sequences MATH in MATH, MATH, and a null-sequence MATH in MATH such that MATH for all MATH. There is a NAME sequence MATH in the uncompleted tensor product MATH with MATH. The sequence MATH converges bornologically towards MATH. We apply the proposition to this null-sequence. Let MATH be the disked hull of MATH. Thus MATH is a precompact subset of MATH and MATH is a null-sequence in MATH. Hence the image of MATH in MATH is zero. Thus MATH in MATH as desired. The proof of REF is almost finished. It remains to compare MATH and MATH. It is trivial that MATH if both MATH and MATH are NAME spaces. For MATH and nuclear spaces, see CITE CITE. |
math/9906205 | Write MATH and MATH as inductive limits of nuclear NAME spaces, MATH, MATH. The inductive tensor product MATH is isomorphic to MATH (see CITE). Since all MATH are nuclear, the maps MATH are topological embeddings for all MATH. Hence MATH is a nuclear NAME. We claim that MATH . The first and the last isomorphism follow from REF . The second isomorphism follows from REF . The third isomorphism follows because completed bornological tensor products commute with inductive limits. |
math/9906205 | REF implies REF . Let MATH be a null-sequence in MATH. Since MATH is NAME, any topological null-sequence is a bornological null-sequence. That is, there is a null-sequence of positive real numbers MATH such that MATH is precompact. By REF , there is MATH such that MATH is bornopotent. The circled hull of MATH is still bornopotent. There is MATH such that MATH for all MATH. Thus for MATH, MATH is contained in the circled hull of MATH. It follows that the set MATH is bornopotent. REF implies REF . Assume that REF is false. Choose a countable basis MATH of neighborhoods of the origin in MATH. Since REF is false, there are precompact sets MATH for all MATH such that MATH not precompact. Since MATH is a NAME space, there is a null-sequence MATH such that MATH is contained in the completant disked hull of MATH and MATH for all MATH. Rearrange the numbers MATH in a sequence. We claim that this sequence is a null-sequence. Fix a neighborhood of the origin MATH. Since MATH is a null-sequence for each MATH, there is MATH such that MATH for all MATH. Since MATH for some MATH, MATH for all MATH. Thus all but finitely many of the MATH are in MATH. That is, MATH is a null-sequence for any ordering of the indices. Let MATH be finite and let MATH. There is MATH for which MATH. Thus MATH is contained in the disked hull of MATH. Therefore, MATH is not precompact. Consequently, the null-sequence MATH violates REF . Thus if REF is false, then REF is false. REF implies REF is evident because neighborhoods of the origin absorb all bounded sets. |
math/9906205 | Since the extension is allowable, MATH as graded bornological vector spaces. Thus MATH and MATH. Consequently, MATH are short exact sequences of complexes of vector spaces. These induce the long exact homology sequences REF as usual. |
math/9906205 | If MATH is projective, then REF splits by a MATH- bimodule homomorphism. Conversely, if REF splits by a MATH- bimodule homomorphism, then MATH is projective as a direct summand of the free MATH- bimodule MATH. Thus REF are equivalent. The extension REF splits iff there is a MATH- bimodule homomorphism MATH satisfying MATH. The bimodule homomorphisms MATH correspond to bounded linear maps MATH via MATH. One verifies easily that MATH is equivalent to REF. Thus REF is equivalent to REF . Since MATH is the free left MATH- module on MATH, left MATH- module homomorphisms MATH correspond to bounded linear maps MATH via MATH. One verifies easily that MATH satisfies REF iff MATH satisfies REF. Thus REF are equivalent. If MATH is a projective MATH- bimodule, then MATH has an allowable projective MATH- bimodule resolution of length MATH, namely MATH. Thus REF implies REF . The converse direction is usually proved using the Ext functors. If we cared to define and study these functors, we could follow that line. Instead, we use the following REF of NAME. |
math/9906205 | Let MATH and let MATH be the map MATH. The claim is that both MATH and MATH are isomorphic as modules to the kernel of MATH. It suffices to prove this for MATH. As vector spaces, MATH and MATH and therefore MATH and MATH. Thus we have an allowable extension of modules MATH. Since MATH is projective, this extension splits. That is, MATH as modules. Since a module is projective iff it is a direct summand of a projective module, it follows that MATH is projective iff MATH is projective. Since we can get long allowable resolutions by concatenating allowable extensions, it suffices to prove the second assertion for the case MATH. Adding the allowable resolutions MATH and MATH, we can reduce to the special case MATH. In that special case we have MATH. The second assertion follows. |
math/9906205 | It is clear that MATH above defines a bounded linear map MATH with MATH. Furthermore, it is the only map that has a chance to be multiplicative: If MATH is a homomorphism with MATH, then necessarily MATH and thus MATH . It remains to verify that MATH is indeed multiplicative. Let MATH be the set of all MATH satisfying MATH for all MATH. Evidently, MATH is a bornologically closed subalgebra of MATH. It is easy to verify that MATH for all MATH and MATH of the form MATH. It follows that MATH for all MATH, MATH, that is, MATH. Consequently, MATH and therefore MATH for all MATH. Hence MATH. That is, MATH is multiplicative. |
math/9906205 | The equivalence of REF - REF is the special case MATH of REF is equivalent to REF . Identify MATH as a bornological vector space. The multiplication is the NAME product, where terms of degree MATH or higher are omitted. The bounded linear sections for the natural projection MATH are of the form MATH with bounded linear maps MATH. The section MATH is multiplicative iff MATH satisfies REF because MATH up to terms of degree MATH and higher. Thus REF are equivalent. REF implies REF . Let MATH be a bounded splitting homomorphism. Let MATH, MATH be as in REF. Consider the bounded linear map MATH. Its curvature is MATH because MATH is a homomorphism. Since MATH is a homomorphism, it follows that MATH maps MATH into MATH. Thus MATH for all MATH. Consequently, the bounded homomorphism MATH of REF annihilates MATH. Therefore, it descends to a bounded homomorphism MATH. Furthermore, MATH with MATH the natural projection. The composition MATH is a bounded homomorphism satisfying MATH. Thus REF implies REF. Of course, REF is the special case MATH and MATH of REF. Thus REF implies REF. Since MATH is an allowable MATH- nilpotent extension, REF is a special case of REF. Therefore, REF - REF are equivalent. REF implies REF . The homomorphisms MATH are constructed by induction, starting with MATH. In each induction step we apply REF to the MATH- nilpotent extension MATH to lift MATH to a bounded homomorphism MATH. Thus REF implies REF. The implication MATH is proved as the implication MATH. Since REF is a special case of REF - REF are equivalent. |
math/9906205 | Since MATH, the operators MATH and MATH agree up to the perturbation MATH, which produces terms of higher order. Hence it makes no difference whether we add MATH or MATH to MATH. |
math/9906205 | We can replace MATH by MATH without changing the image under the map to MATH. This follows from the derivation rule for MATH as in REF. We compute MATH . Hence the map MATH sending MATH to MATH is equal to MATH up to perturbations of higher degree. Thus it makes no difference whether we add MATH or MATH to MATH. This proves the assertion about the odd part of MATH in the absolute case. The assertion about the even part is trivial. The relative case is handled similarly. |
math/9906205 | Let MATH, MATH. For MATH, REF implies MATH . Thus MATH on MATH. Since MATH commutes with MATH, this also holds on MATH. If MATH, then MATH on MATH because MATH on MATH and MATH. We have MATH on MATH because MATH vanishes on MATH. Hence MATH on MATH. Thus MATH is idempotent. In addition, the range of MATH is contained in MATH. The range of MATH is equal to MATH because MATH on MATH. Thus MATH is a direct complement for MATH. It is invariant under MATH because MATH commutes with MATH. |
math/9906205 | We compute MATH on MATH for MATH: MATH . MATH . Since the range of MATH is MATH, we can omit MATH. On MATH we have MATH as desired. |
math/9906206 | The existence of such a phase function can be established using CITE and the transversality condition (MATH to MATH). Indeed, this proposition shows that there exists a splitting MATH and a corresponding splitting MATH such that MATH give coordinates on MATH. It also shows that there exists a splitting MATH such that MATH give coordinates on the fibers of MATH. Taking into account the transversality condition, we deduce that MATH give coordinates on MATH. We write MATH . Here the special form of MATH and MATH at MATH follows from the fibration requirement. We also write MATH, and similarly for MATH. Then the argument of CITE shows that MATH gives a desired parametrization. Indeed, the proof of that proposition guarantees that MATH parametrizes MATH in the interior of MATH but near MATH, and the fact that MATH is transversal to MATH, hence it is the closure of its intersection with MATH proves that MATH gives a desired parametrization of the NAME submanifold MATH of MATH. In particular, MATH . |
math/9906206 | It is easiest to parametrize the pair using the total boundary defining function MATH for which MATH is parametrized by MATH, as above, and then change back to the original coordinates. Using the parametrization argument of CITE for the conic pairs of NAME submanifolds MATH, we see that there are coordinates MATH and a splitting MATH, which we can take to be MATH, MATH, of these such that after the span of MATH is blown up inside MATH the image of the base point MATH is MATH, MATH, MATH, MATH where MATH. Then MATH are coordinates on MATH, and therefore, on MATH for MATH in a neighbourhood of MATH. Using the transversality, we conclude that MATH give coordinates on MATH. Then, following CITE, MATH where on MATH (so MATH, MATH), gives a parametrization in our sense. Notice that, taking into account MATH, this is of the same form as REF for MATH (with no MATH variables). |
math/9906206 | The heart of the proof is an adaptation of NAME 's proof of equivalence of phase functions REF to this setting. If MATH lies above an interior point of mf, then the result is proved in CITE, so we only need to consider MATH lying above a point on the boundary of mf. Using coordinates as in REF, we will prove the following Lemma. Assume that the phase functions MATH and MATH parametrize a neighbourhood MATH of MATH containing MATH . If MATH, MATH, and the signatures of MATH and MATH at MATH and MATH are equal to the corresponding signatures for MATH, then the two phase functions are equivalent near MATH. That is, there is a coordinate change MATH and MATH mapping MATH to MATH such that MATH in a neighbourhood of MATH. To prove this we follow the structure of NAME 's proof. The first step is to show that MATH is equivalent to a phase function MATH that agrees with MATH at the set MATH, given by REF to second order. This is accomplished by pulling MATH back with a diffeomorphism of the form MATH that restricts to the canonical diffeomorphism from MATH to MATH. This diffeomorphism is given by the identifications of MATH and MATH with MATH. The mapping MATH may be constructed as in REF 's proof. In the second step, we first note that by the equivalence result for NAME on manifolds with boundary REF , there is a change of variables MATH near MATH such that MATH . Therefore, we may assume that MATH. If that is so, then MATH is MATH, in addition to vanishing to second order at MATH. Therefore, one can express MATH for some smooth functions MATH of all variables. We look for a change of variables of the form MATH . Using NAME 's theorem, we may write MATH for some smooth functions MATH. We wish to equate MATH and MATH. Substituting REF into REF and equating with REF gives an expression which is quadratic in the MATH and MATH. Matching coefficients gives the matrix equation MATH to be solved, where MATH and similarly for the others. The expression MATH is a polynomial in the entries which is homogeneous of degree two in MATH. By the inverse function theorem, this has a solution MATH if MATH is sufficiently small. The proof is completed by showing that there is a path MATH of phase functions parametrizing MATH that interpolate between MATH and MATH. We define MATH by considering a path of MATH's and using REF. Recall that the nondegeneracy hypothesis is that MATH . At this point, by REF, MATH regardless of the value of MATH, so the first condition is automatically satisfied. The other derivatives are MATH where MATH is evaluated at MATH. Thus the second nondegeneracy condition at MATH is that MATH is nondegenerate. As shown in REF 's proof, the condition that the signatures of MATH and MATH are equal is precisely the condition that interpolation between MATH and MATH is possible by nondegenerate phase functions. This completes the proof of the lemma. Returning to the proof of the proposition, if the dimension and signature hypotheses of the lemma are satisfied, then the phases are equivalent in some neighbourhood of MATH and thus, by a change of variables, the function MATH can be written with respect to MATH. In the general case, one modifies MATH and MATH by adding a nondegenerate quadratic form MATH in extra variables to MATH, and a similar expression to MATH, so as to satisfy the hypotheses of the Lemma. This requires a compatibility mod REF between the difference of the dimension of MATH and MATH and the signature of the corresponding Hessian, but this is always satisfied by REF. The function MATH can be written with respect to the modified MATH since the effect of the quadratic term is just to multiply by a constant times powers of MATH and MATH. By the Lemma, MATH can be written with respect to the modified MATH, and therefore, with respect to MATH itself. |
math/9906206 | Again we follow the structure of NAME 's proof. First, we may assume that MATH agrees with MATH at the set MATH, given by REF to second order. As before, this is possible if we can find a diffeomorphism of the form MATH that restricts to the canonical diffeomorphism from MATH to MATH. This is simply a version of the result from REF , with MATH an extra parameter. In the second step, we fix MATH and MATH and use the result of NAME and NAME for Legendrian pairs with conic points, which says that there is a coordinate change mapping MATH to MATH. We can deduce the same result, with the same proof, letting MATH vary parametrically. Thus we can assume that MATH at MATH. Consequently, MATH. Thus, we may write MATH . We look for a change of variables of the form MATH . Using NAME 's theorem, we may write MATH for some smooth functions MATH. Equating MATH and MATH gives the matrix equation MATH where MATH is homogeneous of degree two in MATH. This always has a solution for small MATH by the inverse function theorem. This completes the proof of the Lemma. |
math/9906206 | This result follows readily from CITE (see also CITE for more details) by noting that the estimates of REF , which are uniform scattering wavefront set estimates on MATH as MATH, MATH, can also be shown to be uniform on a small MATH-interval not containing zero. Then the argument in the proof of REF extends to show joint continuity of MATH in both MATH and MATH. |
math/9906206 | Regarding the spectral projection MATH as a map MATH, and using REF we may take the pointwise limit as MATH in the integral REF to obtain MATH . Thus it remains to show that MATH as operators MATH. To derive REF, first recall that MATH, MATH, has an asymptotic expansion similar to REF. Namely, MATH . Thus, MATH since both sides are generalized eigenfunctions of MATH, have an expansion as in REF and the leading coefficients of the MATH term agree, namely they are MATH. Thus, we only have to determine MATH in REF. For this purpose, recall the boundary pairing formula from NAME 's paper CITE. Thus, suppose that MATH and MATH. Let MATH. Then MATH . We apply this result with MATH, MATH, MATH. Thus, MATH, MATH, MATH, MATH and MATH with the notation of REF. We conclude that MATH . Thus, MATH. Combining this with REF, we deduce the first equality in REF. The second equality can be proved similarly. |
math/9906206 | First we show near MATH that MATH and MATH satisfy the conditions of REF . This is clear for MATH, which projects under MATH to MATH (see REF) at the corner with fibres MATH. To see this for MATH, notice that at the corner, that is, when MATH, then by REF we have MATH which also projects to MATH, with fibre above MATH equal to MATH. It is shown in CITE that MATH form an intersecting pair with conic points. Since MATH is invariant under the transformation MATH, MATH, MATH, MATH, MATH, this is also the case near MATH. It remains to show that in the interior of MATH that they form an intersecting pair with conic points. To see this, notice that using the description REF of MATH, near MATH, MATH is near MATH and MATH is near MATH there, so we may write, using REF, MATH, MATH, MATH (MATH takes values in MATH as usual), MATH, and then near MATH, MATH is given by MATH that is, the MATH variables MATH give coordinates on MATH near MATH. This means that MATH is the image of a smooth manifold in MATH-polar coordinates. However, notice also that MATH, that is, away from MATH it is a bounded multiple of MATH, and similarly MATH is a bounded multiple of MATH, so MATH is the image of a smooth manifold in MATH-polar coordinates. A similar argument works near MATH with the role of MATH and MATH interchanged, so we can conclude that MATH is the image of a closed embedded submanifold MATH of MATH under the blow-down map. Also MATH is transversal to the front face of the blow-up (since away from MATH respectively, MATH, MATH respectively, MATH is a coordinate on MATH). These, together with corresponding arguments near MATH, establish the result. |
math/9906206 | We prove the result by microlocalizing in each factor of the composition MATH and showing quite explicitly that we get a NAME conic pair associated to MATH. Thus, let MATH and MATH be pseudodifferential operators on MATH, MATH, associated to a microlocal partition of unity, such that each MATH, MATH, MATH and MATH (which are all Legendrian conic pairs associated to MATH or MATH) can be parametrized by a single phase function. Then we write MATH and analyze each term separately. There are three qualitatively different types of parametrizations that can occur: REF if MATH is microlocalized near MATH then MATH has a phase function of the form MATH; REF if MATH is microlocalized where MATH then MATH has phase function MATH parametrizing MATH; and REF if MATH is localized near MATH then MATH has phase function MATH parametrizing MATH. Similarly for MATH, MATH, etc. A subcase of the last case is MATH, that is, the phase function parametrizes only MATH. Let us denote by MATH the case that MATH falls into case MATH and MATH into case MATH. Then there are nine possibilities in all, but we can eliminate several of them. First, writing MATH as MATH this has the effect of `switching MATH and MATH' so that case MATH is the same as case MATH with the sign of MATH reversed. Thus we can neglect cases MATH. Similarly, case MATH is the same as MATH. Thus we have only to consider cases MATH, MATH, MATH, MATH and MATH. Also, since MATH is formally self-adjoint in the sense that MATH we may restrict attention to a neighbourhood of lb and the interior of bf. First consider case MATH. In this case, the composition is an integral of the form MATH where MATH and MATH and MATH both parametrize MATH in region two. (We will omit the MATH-half-density factors for notational simplicity.) The integral is rapidly decreasing in MATH unless MATH. This implies that MATH but in case MATH, MATH is never zero, so the integral is rapidly decreasing at lb and rb. Thus we only need to analyze REF in the interior of bf. Let us show that MATH is a nondegenerate phase function in the interior of bf, with phase variables MATH. Consider the phase function MATH. With MATH held fixed, MATH parametrizes MATH, the Legendrian defined in REF. (It is a non-degenerate parametrization since the coordinate functions MATH are linearly independent on MATH, hence on MATH, MATH and MATH are independent. That is equivalent to independence of MATH when MATH, which is to say that MATH with MATH held fixed parametrizes MATH nondegenerately.) Note that on MATH, the rank of the MATH differentials MATH is everywhere at least MATH, by REF. Therefore, when MATH, the rank of the MATH differentials MATH and MATH is at least MATH. Next, note that MATH and when MATH, we have MATH, so this is MATH . Finally, by the same reasoning as above, the differentials MATH are independent when MATH. Putting these facts together we have shown that MATH is a nondegenerate phase function. It is not hard to see that the Legendrian parametrized by MATH is MATH. Setting MATH we see that MATH is on the Legendrian only if there are points MATH and MATH in MATH with MATH and MATH on the same geodesic, with MATH not the middle point, with MATH. With MATH and MATH this is the same as REF. Thus REF is in MATH with MATH by REF. Next consider case MATH. In fact we can treat MATH and MATH simultaneously, by considering the integral of the form MATH where MATH and MATH is supported where MATH is close to MATH. In the interior of bf the phase function MATH is non-degenerate since the differentials MATH are linearly independent when MATH is close to MATH, and MATH are linearly independent as above. The same reasoning as above shows that MATH parametrizes MATH in the interior of bf. Near lb the phase is of the form MATH, which has the form REF with MATH and MATH. It is a non-degenerate parametrization of MATH near the corner since MATH and MATH are independent. Since MATH parametrizes MATH and for fixed MATH, MATH parametrizes MATH, we see that REF is in MATH. Near the interior of lb, we have an integral expression MATH . The second factor is a smooth function in this region since MATH lies in the interior of MATH. The phase is nondegenerate at MATH, so performing stationary phase we get MATH which is an expression of the form REF with MATH. The case MATH is easily treated, because in this case we have an integral MATH where MATH whenever MATH. In this case, we get nontrivial contributions when MATH, which implies that MATH there. But this means that MATH, so we only get contributions at the interior of bf (since we are neglecting a neighbourhood of rb), and in this region the argument is similar to that of case MATH. Finally we have case MATH. There the composite operator takes the form MATH where MATH is supported where MATH is small. When MATH, the Hessian of the phase function in MATH is nondegenerate, with a critical point at MATH, so for small MATH we get a critical point MATH. When MATH the critical point is also MATH so the critical value takes the form MATH. Using the stationary phase lemma we may rewrite the integral as MATH where MATH. This is a nondegenerate parametrization of MATH both near the interior of bf and near the corner MATH with orders MATH at MATH, MATH at MATH and MATH at lb, so the integral is in MATH. The argument in the interior of lb is the same as in case MATH. This completes the proof of the theorem. |
math/9906206 | Clearly MATH, so we prove the reverse inclusion. Setting MATH in REF and using REF we see that MATH so MATH on the first piece of MATH in REF and then we have MATH, MATH, MATH. It is easy to check on the second and third pieces using REF. Since MATH is contained in the characteristic variety of MATH (see REF) which is MATH, it is tangent to the bicharacteristic flow of MATH. Under this flow MATH evolves by MATH, so on MATH, MATH and hence MATH. |
math/9906206 | We restrict our attention to MATH for the sake of definiteness. Similar results hold for MATH with some changes of signs; in particular, MATH changes to MATH in the statement above. We start by defining MATH. Let MATH be identically MATH near MATH and supported in a small neighborhood of MATH, and let MATH . Then MATH is the distributional limit of MATH where MATH, so MATH is a `classical' symbol for all MATH, as MATH. Thus, by the symbolic functional calculus of CITE MATH for all MATH; this means that its kernel is conormal to MATH on MATH, polyhomogeneous at MATH, and vanishes to infinite order on all other boundary hypersurfaces. Since MATH, we have MATH as bounded operators on MATH, hence certainly as distributions on MATH. Hence, MATH is a scattering pseudodifferential operator. Note that if MATH is another cutoff function with the same properties and MATH is defined similarly to MATH then MATH, so MATH by the symbolic functional calculus, that is, its kernel is polyhomogeneous on MATH (has no singularity at MATH) and vanishes to infinite order an all other boundary hypersurfaces. Thus, the splitting MATH is in fact well defined modulo such terms, and it separates the diagonal singularity (which is really a high energy phenomenon) represented by MATH and the singularity due to the existence of generalized eigenfunctions of energy MATH, represented by MATH. To analyze MATH, note that the kernel of MATH is the distributional limit of MATH which is a function of MATH supported on the positive real axis; hence we can use our knowledge of the spectral projections for MATH to compute this term as MATH . The term MATH can be expressed as MATH (presently we will change variable to MATH). We have shown in the proof of REF that the kernel of MATH is a sum of expressions of the form MATH where MATH parametrizes some part of MATH and MATH is smooth on MATH and in the phase variables. We need to prove smoothness of MATH in MATH as well; for ease of exposition this is left until the end of the section. We may suppose that we have broken up MATH in such a way that on each local expression, MATH is either strictly less than zero, strictly greater than zero or takes values in MATH for some small MATH. Thus to understand MATH we must understand the expressions MATH as MATH. (Note that MATH will also depend on a MATH variable if it is parametrizing MATH near at a conic point; this is not indicated in notation for simplicity.) We define the phase function MATH. It is inconvenient that REF is not compactly supported in MATH, so first we insert MATH into the integral, where MATH is identically equal to one on MATH and is compactly supported. The phase function MATH is stationary in MATH when MATH which is always less than MATH. We may also produce any negative power MATH in the integrand by writing MATH and repeatedly integrating the MATH derivative by parts. Thus, by stationary phase, when MATH is inserted the integral is smooth on MATH and rapidly decreasing as MATH, uniformly as MATH. Hence we may ignore this term and insert the factor MATH into REF. Let us consider REF at MATH with the term MATH inserted, that is, MATH when the phase MATH is negative. In this case there are no critical points of MATH since MATH requires MATH and MATH is nonnegative. Thus in this case, again, the expression is smooth and rapidly decreasing as MATH uniformly in MATH. When the phase is positive, we get critical points when MATH parametrizes MATH, MATH and MATH. We can ignore the fact that the range of integration of MATH has a boundary since MATH there. The phase function MATH is nondegenerate in MATH and exactly quadratic, and moreover the rest of the expression is independent of MATH. It follows that the first term of the stationary phase expansion is accurate to all orders, so that up to a smooth function which is rapidly decreasing in MATH, REF is equal to MATH which is MATH times the corresponding piece of MATH. We take MATH to be the sum of the contributions from all terms with MATH in our decomposition. Since MATH when MATH, we see that MATH with MATH as in REF , and is (up to a numerical factor) microlocally identical to MATH there. When the phase changes sign, then we have an intersecting Legendrian distribution associated to MATH. To see this, note that when MATH, MATH implies MATH, MATH implies MATH and MATH implies that MATH parametrizes MATH. Thus we get the Legendrian MATH for MATH, subject to the restriction MATH which gives us MATH. The phase is non-degenerate since MATH is nondegenerate and the Hessian of MATH in MATH is nondegenerate. In addition, MATH at MATH on MATH. If MATH denotes MATH restricted to MATH, then MATH implies that MATH and MATH implies that MATH parametrizes MATH. The Legendrian parametrized is then MATH which is MATH. Thus in this REF is an intersecting Legendrian distribution when MATH. We define MATH to the sum of all the contributions when MATH changes sign. It is easy to see that as MATH we get convergence in distributions to MATH, since by multiplying by a suitable power of MATH, we have convergence in MATH by the Dominated Convergence Theorem; therefore the expression converges in some weighted MATH space. We have now written MATH as a sum MATH, where MATH have the required properties and MATH is a smooth kernel vanishing with all derivatives at the boundary, hence trivially fits any of the descriptions of the MATH. Thus, together with the proof of smoothness of REF in MATH, the proof of the theorem is complete. |
math/9906206 | Smoothness in MATH and MATH is proved in CITE, so it remains to prove smoothness in MATH. Let MATH. One can show, as in CITE, proof of REF , that for MATH, MATH has a limit in MATH as MATH approaches the real axis from MATH, and the limit is continuous in MATH. (Indeed, this is completely analogous to NAME 's proof of the corresponding statement in the Euclidean setting, see CITE.) Moreover, it is the unique solution to MATH satisfying the radiation condition MATH on MATH. Let us denote the limit by MATH. Since MATH, we see that both MATH and MATH have limits in MATH as MATH. Hence, we have MATH . Since MATH, we see that the MATH derivative of MATH exists in MATH. If we calculate MATH we see that MATH . We can find a solution to MATH of the form MATH, with MATH and varying continuously with MATH (since MATH depends on MATH only through taking MATH and MATH derivatives, not MATH derivatives). Therefore, MATH with the right hand side continuous in MATH. It follows from REF, and REF that MATH is a smooth function on MATH which is continuous in MATH. This implies that actually the right hand side of REF is MATH in MATH. Thus, we can repeat the argument with MATH in place of MATH. Inductively we see that MATH is MATH in MATH for any MATH, proving the lemma. |
math/9906206 | By REF , for MATH, and a dense set of MATH, namely, MATH, the function MATH is continuous on MATH and MATH. It follows that for such MATH, and MATH, MATH . Sending MATH and MATH, we see that MATH for a dense set of MATH. Since MATH, this shows that MATH extends to an isometry from MATH to MATH. The same reasoning shows that MATH also extends as an isometry. It is easy to see that the adjoint of MATH is MATH. The intertwining property follows from MATH by multiplying on the left and right by MATH. |
math/9906206 | Multiplying both sides of the equation on the left by MATH, it suffices to establish MATH. Since each operator acts fibrewise in MATH, it is sufficient to establish MATH. Recall that for MATH, MATH is the solution to MATH with incoming boundary data MATH. Denote by MATH the outgoing boundary data of MATH, that is, MATH. Then MATH is also equal to MATH. But this proves MATH, so the proof is complete. |
math/9906206 | We write MATH as in the proof of REF . Then MATH is a scattering pseudodifferential operator, so MATH. Thus we may restrict attention to MATH. The action of MATH on MATH may be represented by pulling MATH up to MATH from the left factor, multiplying by the kernel of MATH and then pushing forward to the right factor of MATH (see figure). We will apply the pushforward theorem to this pushforward to obtain scattering wavefront set bounds. Recall that the pushforward theorem from CITE applies in the situation when we have a map MATH between manifolds with boundary, which is locally of the form MATH in local coordinates where MATH is a boundary defining function for MATH, and MATH is one for MATH. Corresponding coordinates on the two scattering cotangent bundles over the boundary are then MATH and MATH. Then the pushforward theorem states that for densities MATH, MATH . Unfortunately there is no pushforward theorem currently available on manifolds with corners. Since MATH has codimension REF corners, we cannot apply the theorem directly, but must cut off the resolvent kernel away from the corners and treat the part near the corners by a different argument. It turns out that nothing special happens at the corners and the bound on the wavefront set is just what one would expect from a consideration of the boundary contributions alone. To localize the kernel of MATH, choose a partition of unity MATH, where MATH is zero in a neighbourhood of lb and rb, MATH is supported near MATH, MATH is zero on bf and rb and MATH is supported close to rb. Correspondingly decompose MATH . We investigate each piece separately. The easiest piece to deal with is MATH. The product of the two distributions on MATH is localized near lf, and MATH is smooth there. We know that MATH is a Legendrian associated to MATH and is smooth in MATH (where MATH is a set of coordinates in the interior of the right factor of MATH). Thus if MATH is a rescaled dual variable to MATH, the wavefront set of the product is contained in MATH and by REF, the result has scattering wavefront set in MATH. Next we treat MATH. This is localized near the interior of bf. We will see that the contribution here is the part that propagates inside the characteristic variety of MATH. To apply REF, we must first choose compatible coordinates on MATH and MATH. Let us take MATH, MATH, MATH and MATH near the interior of bf, and let MATH, MATH, MATH and MATH be the corresponding rescaled cotangent coordinates. They are related to coordinates REF by MATH . The scattering wavefront set of the lift of MATH to MATH is MATH . The wavefront set of MATH is contained in the union of MATH and MATH. In our present coordinates this is the union of MATH corresponding to MATH, and MATH by REF (the condition MATH, and the restrictions on MATH in the last two lines, gives the correct `half' of the Legendrian for MATH since this is precisely the part of REF where MATH). It is not hard to see that REF, associated to MATH, gives us MATH back again, so consider the second part. The first piece can be re-expressed as MATH . The scattering wavefront set of the product is in each fibre the setwise sum of the scattering wavefront sets of the factors, that is, MATH . By REF, we only get a contribution to the scattering wavefront set of the pushforward when MATH, that is, when MATH that is, when the corresponding point of MATH is in the characteristic variety, and then we get points MATH . The first set is precisely the bicharacteristic ray from MATH in the direction in which MATH decreases, and the second set is the intersection of the scattering wavefront set of MATH with MATH. This shows that the contribution from the interior of bf is contained in the set specified in the theorem. Next we consider the corner contributions. We use a rather crude method here to bound the scattering wavefront set contributions near the corner; it is not optimal, but as the support of MATH shrinks to the corner, we recover the optimal result. First we treat MATH. Notice that MATH at the corner, so, by taking the support of MATH sufficiently small, we may suppose that the phase function, MATH, of the kernel of MATH takes values arbitrarily close to MATH. If we localize in the coordinate MATH, then via a diffeomorphism that takes a neighbourhood of the boundary of MATH to a neighbourhood to a boundary point of MATH (the radial compactification of MATH) we may suppose that MATH is a function on MATH. Then we may exploit the following characterization of the scattering wavefront set on MATH (see CITE): MATH . Here, MATH is related to the MATH coordinates by MATH and MATH, that is, MATH and MATH are the components of MATH with respect to MATH. Thus, we need to investigate where MATH is smooth in MATH. Here MATH is a function that localizes the integral in MATH. We have multiplied by a large negative power of MATH to make the integral convergent; this does not affect the scattering wavefront set. Note that MATH has stable regularity under repeated application of the vector fields MATH, since MATH is independent of MATH and MATH are symbolic in MATH. Integrating by parts, and expressing MATH as MATH, we find that the integral has stable regularity under the application of MATH . But MATH is close to MATH, and MATH is close to zero, so for some small MATH, if MATH then the factor multiplying the vector field is always nonzero, showing smoothness in MATH. Thus, MATH . As the support of MATH shrinks to the corner, MATH shrinks to zero, so we see that the contribution of the corner to the scattering wavefront set is contained in MATH, which corresponds under REF to MATH. This is just what one would expect by taking the result for the left boundary or the b-face and assuming that it were valid uniformly to the corner. Finally we come to MATH. To deal with this, we use a duality argument and exploit the fact that the adjoint of MATH is MATH for MATH. We use the characterization Let MATH and MATH. Then MATH if there exists MATH, elliptic at MATH, and a sequence MATH converging to MATH in MATH such that MATH converges for all MATH. To prove the lemma, note that the condition on the sequence MATH implies that MATH converges weakly in every weighted NAME space, hence by compact embedding theorems, strongly in every weighted NAME space, and thus in MATH. Also, we know that MATH in MATH, so this implies that MATH. Since MATH is elliptic at MATH this shows that MATH. Now given MATH that is not in the set REF, choose a self-adjoint pseudodifferential operator MATH elliptic at MATH and with MATH disjoint from the set REF. Since REF is invariant under the forward bicharacteristic flow of MATH, one may assume that MATH is invariant under the backward bicharacteristic flow. Also, since by REF does not meet MATH, we may assume that MATH is elliptic on MATH. Choose a zeroth order self-adjoint pseudodifferential operator MATH with is microlocally trivial on REF and microlocally equal to the identity on MATH. Since MATH, it is possible to choose MATH converging to MATH in MATH, such that MATH in MATH. Consider the distributional pairing MATH for MATH. This is equal to MATH . Note that MATH is defined on any distribution since the kernel is smooth in the interior and vanishes at rb (hence there is no integral down to any boundary hypersurface which would lead to convergence problems). Since MATH converges in MATH, the first term converges. As for the second term, the scattering wavefront set of MATH is contained in MATH which is invariant under the backward bicharacteristic flow and contains MATH, so taking into account that MATH is essentially the term MATH for the incoming resolvent MATH, the above results show that the scattering wavefront set of MATH is contained in MATH and is disjoint from the essential support of MATH. Thus MATH, and so the second term converges. This establishes that MATH by the Lemma above, and completes the proof of the proposition. |
math/9906209 | Since MATH, we can write MATH, where MATH is an effective divisor, that is, a curve, of degree MATH, which can be chosen independently of MATH,MATH. So MATH where MATH is a projective space of dimension MATH parametrizing curves of degree MATH. Thus we reduce to considering MATH. This is the non-trivial part, which was proved by REF . The main ideas of their proof are as follows. We regard MATH as a closed subscheme of MATH, where MATH is the scheme of zero dimensional closed subschemes of MATH of length MATH, and MATH is a projective space of dimension MATH parametrizing curves of degree MATH in MATH. Let MATH and MATH denote the universal families over MATH and MATH respectively. It is REF , that MATH is smooth irreducible of dimension MATH. NAME and NAME first show that MATH is the zero scheme of a section of the rank MATH vector bundle MATH on MATH, and so has codimension at most MATH at every point. Thus the dimension of MATH is at least MATH at every point. Next, observe that MATH contains an open subset MATH corresponding to divisors of degree MATH on smooth curves, that is smooth irreducible of dimension MATH. To complete the proof, they use REF that the punctual NAME scheme of zero dimensional schemes of length MATH at a point has dimension MATH. This shows that the fibre of MATH over any curve MATH has total dimension at most MATH. Thus the dimension of MATH must be strictly less than MATH, and hence it is contained in the closure of MATH. |
math/9906209 | By REF , MATH is irreducible of dimension MATH. To see that MATH is an irreducible component, it is enough to observe that its generic point is not contained in a different MATH. Now this is clear, since the support of the generic point of MATH is the union of two smooth plane curves of degrees MATH and MATH respectively. By REF , as a topological space MATH is the disjoint union of its subschemes MATH, where MATH,MATH,MATH are nonnegative integers satisfying MATH, MATH, MATH if MATH, and MATH . It remains to determine the triples MATH satisfying these conditions. We must have MATH . We now impose the conditions on MATH, MATH and MATH. The conditions MATH and MATH translate into MATH and MATH. As MATH and MATH must be nonnegative, we have MATH with equality if and only if MATH or MATH, MATH and MATH. In all other cases we have MATH and either MATH, or MATH,MATH and MATH, hence MATH . |
math/9906209 | We use the fact proven in REF that the sheaf of MATH-modules MATH, defined as the quotient MATH has depth MATH at closed points. It follows from this and the exact sequence MATH that MATH is locally free of rank two provided MATH is locally free of rank two . But this is clear as we have an exact sequence MATH . Finally, from the exact sequence MATH we see that the NAME module of MATH is annihilated by the equation of MATH, hence it is isomorphic to MATH as a modules over MATH. |
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