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hep-th/0101089 | The proof is a direct generalization of that of the analogous statement from CITE. Let us expand MATH as MATH . The solution is constructed iteratively in the form MATH with MATH being MATH . In particular, for the function MATH we have explicitly MATH where MATH denotes terms of higher order in MATH and MATH. |
hep-th/0101089 | Function MATH is a classical BRST observable, because of MATH is obviously a classical BRST charge from REF. In the MATH-th order in MATH . REF implies MATH with MATH given by MATH where we have expanded MATH and MATH in MATH according to MATH . Arguments similar to those of the proof for REF show that MATH provided REF holds for all MATH. Since cohomology of a classical BRST charge vanish in strictly positive ghost number and MATH we see that REF has a solution for MATH. Moreover, MATH can be taken to belong to MATH. The second part of the statement is trivial when MATH carries strictly positive ghost number, since classical BRST cohomology vanishes in this case. Let MATH satisfies REF and MATH. Let us show that there exists MATH such that MATH depends on MATH and MATH only. This is obvious for the zero order term MATH in the expansion MATH of MATH in powers of MATH. Assume that this is the case for all MATH. In the MATH-th order in MATH . REF then takes the form REF. Because MATH is a function of only MATH for MATH then MATH in REF has the form MATH; one can then find MATH such that MATH . Since MATH there exist functions MATH and MATH such that MATH . It follows that MATH is a MATH-th order term in the expansion of MATH in powers of MATH. Proceeding further by induction one can find MATH required in REF. |
hep-th/0101089 | For the degree MATH term one has MATH, since MATH vanishes. Assume that REF holds for MATH. Then MATH is also MATH and MATH independent. Indeed, for MATH one has MATH with MATH given by REF. The space of MATH and MATH independent functions is closed with respect to the quantum multiplication in MATH, since the respective components of the connection MATH and its curvature vanish. In its turn operator MATH also preserves this space. Thus MATH satisfies REF and the statement follows by induction. Up to the terms of degree higher than three MATH has the following form MATH . |
hep-th/0101089 | The proof goes in the same way as that of REF . |
hep-th/0101089 | Let us write down explicitly the following terms: MATH . The first three equalities are trivial. The last one follows from REF . Further, it follows from the explicit form of the matrix REF entering the NAME bracket MATH that each non-vanishing term in MATH is proportional at least to one quantity from REF and thereby vanishes on MATH. |
hep-th/0101089 | Consider the following subset of the constraints REF: MATH . These constraints are also second-class ones. It is easy to write down respective NAME bracket; the non-vanishing basic NAME bracket relations are given by: MATH . It follows from REF that subalgebras MATH and MATH are closed with respect to the NAME bracket REF. Let us introduce the quantum multiplication MATH in MATH build by the restriction of the NAME bracket REF to MATH: MATH where MATH stands for dependence on MATH only. CASE: Let MATH do not depend on MATH and MATH. Then, MATH where MATH in the Right-hand side denotes NAME multiplication REF. CASE: For any MATH its restriction MATH belongs to MATH. Multiplication MATH determines a multiplication in MATH that is identical with MATH: MATH . The only nontrivial is the second statement. It is easy to see that restriction MATH of an arbitrary element MATH belongs to MATH. That MATH restricts to MATH follows from the following properties of MATH: MATH for any MATH. Indeed, MATH (as an algebra with respect to ordinary commutative product) can be identified with the quotient of MATH modulo ideal generated (by ordinary commutative product) by elements MATH. Then, REF implies that this ideal is also an ideal in MATH with respect to the MATH-product. Thus MATH determines a star-product in MATH. One can easily check that in MATH the product coincides with MATH. Let us now rewrite the master REF for MATH in the ``NAME form: MATH where MATH and MATH . Since MATH doesn't depend on MATH the master equation can be equivalently rewritten using multiplication MATH and the NAME connection on MATH: MATH . Taking into account the second item of REF one arrives at MATH where MATH and MATH are restrictions of MATH and MATH defined on MATH to MATH. In the coordinates MATH on MATH one has MATH where MATH are coefficients of a connection MATH on MATH with respect to the coordinates MATH (recall that NAME connection MATH on MATH restricts to MATH). Finally, one can observe that for MATH REF implies: MATH . Similar arguments show the rest of the statement. |
hep-th/0101089 | The statement of the theorem can be explicitly checked for MATH. Further, assuming that REF holds for all MATH one can see that the respective quantities MATH and MATH (see the proof of REF ) do coincide. Since the operators MATH and MATH are precisely the same in both cases one observes that MATH. The statement then follows by induction. |
hep-th/0101107 | See CITE . |
hep-th/0101107 | See CITE . |
hep-th/0101107 | See CITE . |
hep-th/0101107 | See CITE . |
hep-th/0101227 | Let MATH . Corresponding to the decomposition REF of the particle multiplet MATH into particle types MATH there is, for each MATH in MATH a family of linear subspaces MATH satisfying MATH see for example, CITE. Note that the closures of the above vector spaces are independent of MATH by the NAME - NAME property and span MATH if MATH runs through MATH . Similarly, the ``anti - particle" NAME spaces MATH are independent of MATH mutually orthogonal for different MATH and span MATH if MATH runs through MATH (note that MATH). CITE have shown the particle - anti-particle symmetry in this situation: For each MATH and MATH there is a unitary map MATH from the closure MATH of the vector space REF onto the space REF intertwining the respective (irreducible) subrepresentations of MATH. We now recall in detail the relevant result of NAME and NAME. Denote by MATH the set of field operators MATH such that the map MATH is smooth in the norm topology. NAME and NAME consider a special class of spacelike cones: Let MATH be an open, salient cone in the MATH plane of NAME space, with apex at the origin. Then its causal completion MATH is a spacelike cone. Its dual cone MATH is defined as the set MATH . Let MATH where MATH is a spacelike cone as above and MATH . Then MATH is, considered as a function on the mass shell MATH the smooth boundary value of an analytic function MATH on the simply connected subset MATH of the complex mass shell. Further, its boundary value on MATH satisfies MATH where MATH is a complex number of unit modulus which is independent of MATH and MATH, and MATH is the mentioned intertwiner from MATH onto MATH. Note that REF coincides literally with REF. We reformulate this result as follows. Denote by MATH the class of spacelike cones MATH contained in MATH which are of the form MATH as in the lemma and contain the positive MATH-axis. Let further MATH where MATH runs through MATH and MATH . The lemma asserts that on this domain an operator MATH may be defined by MATH . Further, the intertwiners MATH modified by the factors MATH appearing in REF, extend by linearity to a unitary ``charge conjugation" operator MATH on MATH which satisfies REF. Now REF may be rewritten as MATH . This inclusion implies in particular that MATH is closable, its closure satisfying the same relation. But this closure is an extension of the operator MATH as we show in the Appendix REF . Hence we have MATH and it remains to show the opposite inclusion. To this end, we refer to the opposite wedge MATH . Let MATH . We claim that the following sequence of relations holds true: MATH where MATH is the NAME operator. Twisted locality and modular theory imply that MATH . Applying MATH this yields MATH . But MATH because MATH commutes with the modular operators. Using MATH this proves the first inclusion. Since MATH commutes with MATH and both MATH and MATH are involutions, the inclusion REF implies that MATH . Thus, the adjoint of relation REF reads MATH which implies the second of the above inclusions. Finally, the group relations REF and MATH imply that MATH . But the spin-statistics theorem CITE asserts that MATH . (Namely, both operators act on MATH as multiplication by the statistics sign MATH . ) Hence the last equation in REF holds. This completes the proof of REF and hence of the proposition. |
hep-th/0101227 | Consider the set MATH of velocity tupels MATH satisfying the requirements that REF one of the velocities, say MATH has the strictly largest MATH-component: MATH and REF the relative velocities with respect to MATH have different directions: MATH . Given such MATH let MATH . For MATH let MATH be a cone in the MATH plane of MATH containing the ray MATH and with apex at the origin, and let then MATH be its causal closure. Then, having chosen sufficiently small opening angles, the regions MATH are mutually spacelike separated for all MATH and further MATH for MATH . Now MATH exhausts the set of all velocity tupels in MATH except for a set of measure zero. Hence, a scattering state MATH as in REF can be approximated by a sum of scattering states MATH whose localization regions satisfy that MATH for some MATH and MATH for MATH . This is accomplished by a standard argument CITE taking into account the continuous dependence of MATH on the MATH and the NAME theorem. But due to the normal commutation relations obeyed by the scattering states, MATH coincides with MATH and hence is of the form required in the lemma. |
hep-th/0101227 | Let MATH . Considering this operator as an internal symmetry, it should act multiplicatively on the scattering states as shown by NAME in CITE. The complication is that MATH does not act strictly local, but only leaves MATH invariant. We generalize NAME 's argument to this case utilizing the last lemma. By induction over the particle number MATH we show that MATH is the unit operator on each MATH . Let MATH be a scattering state with MATH where the localization regions MATH are as in the above lemma. Since MATH is of fast decrease in MATH while MATH increases at most like MATH one concludes as NAME in CITE: MATH where we have put in the induction hypothesis that MATH acts trivially on MATH . Due to NAME 's result, MATH commutes with the translations, which implies that MATH coincides with MATH . But modular theory and covariance guarantee that MATH is, like MATH in MATH . In addition MATH hence by the standard lemma of scattering theory, REF may be rewritten as MATH . By assumption of the proposition, MATH acts trivially on MATH and hence on the scattering state. By linearity and continuity, the same holds on MATH completing the induction. |
hep-th/0101227 | We proceed by induction along the same lines as in the last proposition. Let MATH be the single particle states appearing in REF , with velocity supports contained in compact sets MATH and with localization regions MATH such that MATH are mutually spacelike separated for suitable MATH . According to REF , we may assume that the localization regions satisfy MATH and MATH . By the same arguments as in the last proof, we have MATH where we have put in the induction hypothesis that MATH acts as in REF on MATH . Now by NAME 's result CITE we know that the commutation relations MATH hold. From these we conclude that the spectral supports of MATH and MATH are related by the transformation MATH and hence their velocity supports are related by MATH where MATH denotes the inversion of the sign of the MATH-coordinate. By virtue of the NAME theorem and the continuity of the scattering states, we may assume that for MATH there are MATH and MATH such that MATH . Further, MATH can be chosen such that MATH which in turn is contained in MATH due to REF . Then MATH can be approximated by an operator MATH localized in the region MATH . These regions are mutually spacelike separated for large positive MATH and hence the incoming MATH particle state in REF may be written as MATH . Similarly, NAME 's commutation relations imply that MATH . Now MATH is in MATH and MATH and therefore the discussion around REF implies that the above operator may be approximated by an operator MATH . Recall that the operators MATH are localized in the regions MATH . For large positive MATH these regions are spacelike to MATH and are hence contained in MATH . Hence the MATH (anti-) commute with MATH for large MATH . Thus the standard arguments of scattering theory CITE apply, yielding that the vector REF may be written as MATH and only depends on the single particle vectors. But these are MATH and MATH for MATH . Hence the limit coincides with the right hand side of REF , completing the induction. |
hep-th/0101227 | CASE: Let MATH have the above representation property on MATH . As is well known CITE, the restriction of MATH to the space of scattering states is equivalent to the second quantization of its restriction to MATH . By virtue of REF , see REF , the assumption thus implies MATH which proves the claim. CASE: By twisted locality and modular theory, one has MATH . Now recall that MATH and that MATH see REF . One therefore obtains, by applying MATH to the inclusion REF and using REF , the opposite inclusion. Hence equality holds in REF. Since every wedge region arises from MATH by a NAME transformation, the claimed REF follows by covariance of the field algebras and the representation REF of MATH . |
hep-th/0101227 | We prove the second of REF for MATH with MATH. The other assertions are shown along the same lines. Recall that the covering homomorphism MATH is characterized by MATH . We have MATH . Using the identity MATH which follows from MATH and the well-known relation MATH one verifies that the argument of MATH in REF equals MATH . By REF , this proves the claim. |
hep-th/0101227 | Recall that for MATH the bounded operator MATH where MATH denotes again the generator of the boosts MATH may be written as MATH . Here MATH is the NAME transform of MATH and the integral is understood in the weak sense. Let now MATH be a smooth function with compact support, and let MATH where MATH for some MATH . Applying the above formula to MATH one finds, using that MATH is analytic and MATH . The one-parameter group MATH extends to an entire analytic function satisfying MATH where MATH acts as multiplication by MATH on the coordinates MATH and MATH and leaves the other coordinates unchanged, and MATH acts as MATH on MATH and as the zero projection on MATH CITE. Note that in particular MATH . Further, one easily verifies that for any MATH the vector MATH is in the dual cone MATH . Hence for all MATH and all MATH the complex vector MATH is in MATH the domain of analyticity of MATH and approaches MATH as MATH . It follows that the integrand in REF is anti - holomorphic in MATH in the strip MATH and that REF coincides with MATH . Here we have used that for all MATH is again in the domain MATH of MATH due to the covariance of the field algebra. This is so because for all MATH there is some MATH such that MATH . Let now MATH be in the (dense) domain of MATH and let MATH . We have shown from REF to REF, that MATH . Let MATH denote the set of finite linear combinations of vectors of the form MATH where MATH and MATH . Then the above equation shows that MATH is in the domain of MATH and that MATH on MATH . But MATH is a core for MATH hence MATH is an extension of MATH . |
math-ph/0101008 | Recall that MATH . Since MATH is independent of MATH, we have MATH unless MATH, in which case we have MATH . Since MATH, we get MATH . Since the support of MATH is contained in MATH modulo REF, and MATH, we get MATH . Hence MATH . |
math-ph/0101008 | For ease of notation we put MATH . By the previous lemma, MATH on using REF. Thus MATH . Applying NAME twice we get that MATH and hence MATH . |
math-ph/0101008 | It is sufficient to show that REF has less than MATH solutions. We argue as in REF, except for a small twist: If MATH corresponds to MATH, then MATH corresponds to MATH since the action is MATH-linear. We may thus make MATH . NAME stable by multiplying by MATH. Since MATH is in MATH, the number of solutions to MATH is bounded by the number of solutions of MATH which in turn is given by the number of solutions to MATH . Let MATH. The number of solutions of REF then equals the product of: CASE: The number of solutions to MATH where MATH ranges over all elements in MATH. CASE: The number of elements MATH that reduce to the same element MATH. From REF, applied with MATH, it follows that the first term is MATH . For the second term, REF gives that the cardinality of the cokernel of the reduction map MATH is uniformly bounded in MATH. Hence there are MATH elements MATH that reduce to MATH modulo MATH. Finally, from REF we have MATH and thus MATH . The number of solutions is therefore bounded by MATH . |
math-ph/0101008 | Let MATH. Since MATH is smooth and compactly supported, MATH for MATH, and MATH for MATH. Trivially, MATH, so MATH . For the sum over small MATH, we have MATH . We note that MATH . Indeed, MATH . Now MATH is bounded by the number of divisors of MATH, and hence is MATH for all MATH. Therefore from REF we get MATH . |
math-ph/0101008 | If MATH is trivial or MATH then we can express the sum as a classical NAME sum, and in this case the result is well known. For MATH nontrivial, we note that the degree of MATH is coprime to MATH, hence we may apply NAME 's bound CITE on exponential sums (see CITE) to obtain MATH . Note that the bound is independent of the order of MATH! |
math-ph/0101008 | MATH since MATH and MATH is unitary. From REF we get MATH since MATH unless MATH. Hence MATH. |
math-ph/0101010 | Suppose that there exists a function MATH such that MATH . Applying REF gives MATH so that MATH or equivalently MATH . Conversely, given a solution MATH of REF , MATH is a solution of REF . |
math-ph/0101010 | Substituting REF gives MATH or alternatively, using REF MATH . Note that, as in REF , the vector part of REF is MATH so that MATH for some function MATH . If MATH this is equivalent to MATH . REF , written in terms of MATH is MATH so that REF is equivalent to MATH . Noting that MATH this equation can be rewritten in the form MATH . |
math-ph/0101010 | REF is equivalent to MATH where MATH . The scalar functions MATH and MATH are two solutions of REF . Substituting the expression MATH for MATH into this equation, where MATH is a scalar function, gives MATH and MATH . Simplifying and using the fact that MATH and MATH-are solutions of REF , the equation is reduced to MATH so that MATH for an arbitrary constant MATH . It remains to show that REF is the gradient of some scalar function MATH . This is the case if the rotational of REF disappears. This is shown as follows, where the identities MATH are used. MATH . |
math-ph/0101035 | Choose an eigenvector MATH of MATH with eigenvalue MATH at some MATH. Now evolve MATH with the differential equation MATH and consider MATH . It follows that MATH and because MATH vanishes at MATH it must vanish everywhere. We conclude that the eigenvalues of MATH are independent of MATH and hence that MATH is independent of MATH. |
math/0101004 | Let MATH be an open set with MATH an isomorphism. MATH meets MATH, hence MATH exists on the complement of a discrete subset MATH. Let MATH be a small disk in MATH, centered at MATH. The projective morphism MATH factors through MATH for some MATH and the composite MATH is given by meromorphic functions. Obviously it extends to a holomorphic map on MATH, and the image of the induced map MATH lies in MATH. |
math/0101004 | CASE: REF is compatible with base change and in REF one enforces the compatibility of REF with base change, as well. CASE: Since MATH has rational NAME singularities, REF for MATH implies that MATH has at most rational NAME singularities. The other properties asked for in REF are obvious. For REF, remark that MATH is normal with rational NAME singularities and hence MATH has the same property. The same argument with MATH replaced by MATH gives REF. The sheaf MATH is torsion free, hence locally free outside of a closed codimension two subvariety MATH of MATH. Since MATH is normal and equidimensional over MATH, for MATH open and for MATH one has MATH and thereby MATH . So MATH coincides with the maximal extension of MATH to MATH. |
math/0101004 | Since MATH are both normal crossing divisors one obtains REF. MATH is normal with rational NAME singularities, hence MATH and MATH . The sheaf on the left hand side is invertible, and the one on the right hand side torsionfree, and both coincide outside of a codimension two subvariety. Hence they are equal and one obtains the first isomorphism in REF. For the second one, one repeats the argument for MATH instead of MATH. By REF, all the three sheaves in REF are reflexive. REF has been shown in CITE see also REF. Let MATH denote the zero divisor of MATH, hence MATH . In order to show that MATH is the pullback of a MATH-divisor on MATH, we have to show, that the multiplicities of two components of MATH coincide, whenever both have the same image in MATH. To this aim, given any component MATH of MATH consider a general curve MATH, which intersects MATH in some point MATH. Replacing MATH by a neighborhood of MATH we will assume that this is the only intersection point. Let us write MATH, where MATH stands for any of the varieties in REF . Similarly, if MATH is any of the morphism in REF , MATH will denote the restriction of MATH to MATH. By REF, the variety MATH is again normal, NAME with at most rational singularities, and for MATH sufficiently general MATH and MATH will be non-singular. Applying REF with MATH replaced by MATH, one obtains a natural inclusion MATH and the zero divisor of MATH is the restriction of MATH to MATH. In order to show REF, we just have to verify that MATH is the pullback of a MATH-divisor on MATH. By CITE there exists a finite morphism MATH, totally ramified in MATH, such that MATH has a semi-stable model MATH. By REF, the pullback of MATH to some non-singular covering of MATH remains normal with rational NAME singularities. By flat base change REF is compatible with further pullbacks. Hence we may as well assume for a moment that MATH factors through MATH. Then MATH are two flat NAME morphism, MATH and MATH are birational, and both are normal with at most rational singularities. Therefore, repeating the argument used to prove REF, one obtains MATH and the divisor MATH is the pullback of a divisor MATH on MATH. Since MATH is totally ramified in MATH, the divisor MATH is itself the pullback of a MATH-divisor on MATH. |
math/0101004 | If MATH is ample with respect to MATH, for all MATH sufficiently large and divisible MATH is globally generated over MATH, as well as its quotient sheaf MATH . We may assume that MATH is very ample, and we obtain the morphism asked for in REF. On the other hand, if there is a morphism MATH surjective over MATH, the sheaf MATH as a quotient of a weakly positive sheaf is weakly positive over MATH. |
math/0101004 | By CITE, the restriction of the sheaf MATH to MATH is locally free. The verification of the weak positivity will be done in several steps. Let us first show: In order to prove REF we are allowed to assume that MATH is ample with respect to MATH. Let MATH be a very ample sheaf on MATH and let MATH be an embedding. For a general choice of the coordinate planes MATH, the intersection MATH is of codimension two in MATH. We choose a codimension two subscheme MATH with MATH, for MATH. By definition, in order to show that MATH is weakly positive over MATH, we may replace MATH by MATH and assume that MATH. Moreover, for MATH large enough MATH will be flat. By the local nature of weak positivity, it is sufficient to show that MATH is weakly positive over MATH . In fact, one can cover MATH by such open sets, for different choices of the coordinate planes. Given MATH we choose MATH, and consider the MATH-th power map MATH . Let MATH be the normalization of MATH, and let MATH be the induced map. For the pullback MATH of MATH to MATH one obtains MATH . Leaving out codimension two subschemes in MATH, not meeting MATH, we may assume that MATH is non-singular. Then MATH is non-singular. In fact, MATH is smooth over MATH and MATH is smooth over MATH . Let us choose MATH and MATH. The sheaf MATH is ample with respect to MATH. Applying REF to MATH instead of MATH one finds MATH to be weakly positive over MATH. By flat base change, this sheaf is isomorphic to MATH . Hence for all MATH sufficiently large and divisible, the sheaf MATH is globally generated over MATH. We obtain a morphisms MATH surjective over MATH, and MATH surjective over MATH. For MATH large enough, MATH is generated by global sections and hence MATH is globally generated over MATH. REF allows to assume that MATH is ample with respect to MATH. Then the sheaf MATH will be globally generated over MATH, for some MATH. Replacing MATH by MATH and MATH by MATH we may as well assume that MATH itself has this property. From now on, this assumption will replace REF. Leaving out a codimension two subset of MATH, not meeting MATH, we will continue to assume that MATH is flat. Let us fix some non-singular compactification MATH of MATH and a very ample invertible sheaf MATH on MATH such that MATH is ample. We write MATH and MATH. MATH is globally generated over MATH. Let us choose a compactification MATH of MATH such that MATH extends to a morphism MATH. Moreover we choose MATH and MATH such that MATH is again globally generated over MATH. Let MATH be a blowing up, such that MATH is a normal crossing divisor and let MATH. REF implies that MATH is an isomorphism over MATH. Hence it is sufficient to show that MATH is globally generated over MATH. Blowing up a bit more, and enlarging MATH by adding components supported in MATH, we can as well assume that MATH is globally generated over MATH. Under this REF has been shown in REF. To finish the proof, we consider for any MATH the MATH-fold product MATH and MATH. Let MATH be a desingularization, MATH, MATH . MATH, MATH satisfies again REF. Moreover we assumed MATH to be globally generated over MATH, hence MATH is globally generated over MATH. REF holds true for MATH by: MATH . The proof, similar to the one of REF, is by induction on MATH. Obviously MATH. Let MATH be the support of the cokernel of the inclusion MATH . Applying CITE, to the MATH-th projection MATH, one finds subschemes MATH of MATH with MATH. Since this holds true for MATH, MATH must be empty. By REF the sheaf MATH is globally generated over MATH. Hence REF follows from the next claim. There exists a morphism MATH surjective over MATH. Proof. The natural morphism MATH induces a morphism MATH which is an isomorphism over MATH. By flat base change, the right hand side is nothing but MATH . |
math/0101004 | Using REF (with MATH), one can copy the arguments presented in the proof of REF to obtain REF as a corollary to REF. We leave this as an exercise since REF has been shown under less restrictive assumptions in CITE, using different (and more complicated) arguments. |
math/0101004 | This has been shown in CITE if the general fibres of MATH are of general type, and in CITE in general (see also CITE or CITE). |
math/0101004 | We use again the notations from REF. By REF, mildness of a morphism is compatible with fibre products, hence MATH is again mild. For the normalization MATH of MATH we choose a desingularization MATH, with centers in the singular locus of MATH, and a non-singular blowing up MATH which dominates both, MATH and MATH. We obtain again a diagram MATH which satisfies the assumptions made in REF. One finds for all integers MATH . In fact, by flat base change and by the projection formula, both sheaves coincide over the largest subvariety of MATH, where MATH is locally free. By definition, the right hand side of REF is the reflexive hull of the tensor product on this subscheme, and by REF the left hand side is reflexive, hence both are equal. REF implies: CASE: MATH is reflexive and there is an isomorphism MATH . CASE: There is an inclusion MATH which is an isomorphism over MATH. REF and the first part of REF is nothing but REF. For the second part of REF allows to replace the left hand side by MATH, the right hand side by MATH, and to apply REF . By construction MATH is smooth over MATH and MATH . Now we play the usual game. For the integer MATH chosen above, and for MATH, there is a natural inclusion MATH which locally splits over the open set MATH, where MATH is locally free, in particular over MATH. By the choice of MATH one obtains an inclusion MATH again locally splitting over MATH. In fact, the splitting inclusions in REF exist over MATH, and since the sheaves on the right hand sides are reflexive they extend to MATH. For MATH and MATH consider MATH. By REF MATH has a section whose zero divisor MATH does not contain a whole fibre over MATH, and MATH . All fibres of MATH are of the form MATH and CITE, implies that MATH . So MATH and for MATH REF holds true. By REF the sheaf MATH is weakly positive over MATH. Since MATH is surjective over MATH we can apply REF (for MATH instead of MATH) and obtain the weak positivity of MATH over MATH. Since MATH is reflexive, one has the multiplication morphism MATH . By REF, the left hand side is MATH whereas the right hand side is MATH, hence the assumption on the surjectivity of the multiplication morphism carries over, and MATH is surjective over MATH. Since MATH is weakly positive over MATH, for all MATH sufficiently large and divisible MATH as well as MATH will be globally generated over MATH. By REF, one has a morphism MATH surjective over MATH. Although the sheaf MATH is not necessarily reflexive, the finiteness of MATH allows to apply the projection formula, and to obtain thereby a morphism MATH surjective over MATH. For MATH large enough, the sheaf on the left hand side will be generated by global sections, hence for those MATH the sheaf on the right hand side is globally generated over MATH. Since we assumed MATH to be surjective over MATH, the same holds true for MATH replaced by MATH, and MATH is globally generated over MATH. |
math/0101004 | By REF we find a smooth birational model MATH of MATH which fits into the diagram in REF. We may replace MATH by MATH, and apply REF. REF obviously hold true. Since we assumed MATH to be ample and MATH, for the invertible MATH and for the divisor MATH one obtains REF and by REF. |
math/0101004 | MATH induces a holomorphic map MATH . Since MATH is holomorphic, MATH is a complex subspace of MATH. Let MATH be the complex subspace of MATH given locally by the following condition. Let MATH be a local equation of MATH on MATH. Then MATH is the analytic subspace of the zero set of MATH, where the multiplicity of MATH is larger than or equal to two. We choose MATH. By CITE, page REF, MATH has a decomposition MATH in irreducible components. The index set MATH is countable, since each point MATH has a small neighborhood MATH such that MATH meets only finitely many of those components. As usual, MATH . MATH. If MATH is not an embedding of a small neighborhood of a point MATH, then MATH consists of all hyperplanes MATH passing through MATH, and its dimension is MATH. The set of those points is discrete. For all other points MATH and for all components MATH of MATH one has MATH . In fact, let MATH denote a sufficiently small neighborhood of MATH. A general MATH does not pass through MATH, and for those who do, the intersection is transversal, except for all MATH in a codimension MATH subset of MATH. If MATH is one of the components of MATH, then for all MATH . In both cases, if MATH is a component of MATH with MATH, we are done. Otherwise choose for MATH two points MATH with MATH . Then MATH is not dense in MATH. Obviously, the dimension of MATH is larger than or equal to MATH, and by NAME 's lemma (CITE, page REF) both must be equal. Hence MATH does not contain an open analytic subset MATH. We will show REF by induction on MATH, just using REF but not the definition of MATH as a dual projective space. If MATH, the set MATH is countable. In general, if MATH we choose a point MATH, such that none of the countably many components of MATH is contained in MATH. Moreover, for each MATH, we choose a point MATH. Let MATH be a hyperplane, passing through MATH but not containing any of the points MATH. Then, for each component MATH, the intersection MATH can not be dense in MATH and MATH . Hence MATH and since MATH is an open analytic subset of MATH, contained in MATH, this contradicts the induction hypotheses. Recall that we assumed that the global sections of MATH generate the invertible subsheaf MATH of MATH. In particular, MATH and for MATH in some NAME open subscheme MATH of MATH the preimage MATH will be non-singular and MATH a normal crossing divisor. By REF we can find points MATH in MATH, and for all of them the properties asked for in REF hold true. |
math/0101004 | All this can be done by blowing up MATH in centers not contained in MATH and replacing MATH by a desingularization of the pullback family. |
math/0101004 | Here we will use the assumption, that the dimension MATH of the fibres of MATH is even. A general curve MATH meets MATH transversally. Replacing MATH by some open subset, we can assume that for a given component MATH of MATH . The restriction MATH of the finite morphism MATH is a cyclic covering of order MATH, totally ramified along MATH. By the definition of MATH and MATH, the fibre MATH has one singular point MATH, and we can choose locally analytic parameters MATH in a neighborhood of MATH and MATH in a neighborhood of MATH such that MATH is the zero-set of MATH near MATH. Then locally near MATH the covering MATH is given by the equation MATH . So MATH has one isolated singularity, a double point of type MATH. As well known (see CITE, p. REF, for example), in even dimension the local monodromy group of a MATH singularity is finite, and as in CITE or CITE, p. REF, one obtains the same for the global monodromy. |
math/0101004 | By CITE and REF, there are natural inclusions MATH and MATH, for MATH. In fact, in CITE this is just stated for MATH, but the general case follows by induction, considering the tautological sequences. Since MATH acts on MATH with MATH as decomposition in sheaves of eigenvectors, one obtains the first decomposition in REF. MATH is totally ramified in MATH. Hence there is an exact sequence MATH and the two sheaves on the right hand side differ only in the MATH invariant part. |
math/0101004 | The proof is similar to the proof of REF. It is well-known, that the bundle in REF is the NAME bundle for the variation of NAME structures on MATH. The condition on the monodromy follows from REF. By REF the sheaf MATH is a direct factor of MATH. The morphism MATH is induced by the natural inclusions MATH . Over MATH the kernel of MATH is a quotient of the sheaf MATH . Since the relative dimension of MATH over MATH is MATH and since MATH is fibrewise ample, the latter is zero by the NAME vanishing theorem. So MATH is injective, as claimed in REF. The injective morphism in REF also exist for MATH replaced by MATH, and the exact sequence REF is a subsequence of MATH . Finally by REF this sequence is obtained by taking the sheaves of eigenvectors in the direct image of the exact sequence REF under MATH. One obtains REF. By REF. Comparing the first NAME classes for the tautological sequence for MATH one finds MATH . Recall that MATH is smooth over MATH, for the divisor MATH considered in REF. Hence MATH and MATH contains MATH. Moreover, by REF, MATH is effective. By definition, MATH, and MATH . Therefore MATH contains MATH, hence the sheaf MATH and REF holds true. For REF, recall that over MATH the sheaf MATH is nothing but MATH . Since MATH, by the projection formula the morphism MATH is the restriction of the edge morphism of the short exact sequence MATH . Since MATH is smooth with MATH-dimensional fibres, the sheaf on the right hand side is MATH and the one on the left hand side is MATH. NAME with MATH and dividing by the kernel of the wedge product MATH on the left hand side, one obtains an exact sequence MATH where MATH is a quotient of MATH. By definition, the restriction to MATH of the morphism considered in REF is the first edge morphism in the long exact sequence, obtained by applying MATH to REF . The wedge product induces a morphism MATH . Since MATH this morphism factors through MATH. Hence the exact sequence REF is isomorphic to the tautological sequence MATH . The edge morphism MATH of REF is the NAME map. Since we assumed MATH to be generically finite over the moduli space, this morphism is injective. |
math/0101004 | All properties hold true, for the NAME bundles constructed above, with MATH semi-ample and big. Choose some MATH such that MATH contains an ample invertible sheaf MATH, and consider the NAME bundles MATH . Again we first consider them on MATH, where MATH is the singular locus of MATH and then we take the maximal extension to MATH. By CITE, page REF, the morphism MATH is given by MATH and similarly for MATH and MATH. The decomposition as a direct sum is MATH with MATH where the direct sums are taken over all MATH with MATH . Again we have morphisms MATH compatible with MATH and MATH. In particular MATH is the MATH-th tensor product of MATH, hence injective. The same holds true for MATH which is the direct sum of morphisms of the form MATH . REF remain true, with MATH and MATH replaced by MATH and MATH, for MATH instead of MATH in REF. In REF one has an injection MATH . The morphism MATH is a direct sum of morphisms of the form MATH hence it induces the diagonal morphism MATH . In particular the injectivity of the morphisms in REF, carries over. As remarked in REF the injectivity of MATH, MATH and of MATH is sufficient to perform the constructions with MATH and MATH instead of MATH and MATH, and to obtain some MATH and the morphisms MATH and MATH satisfying REF, with MATH replaced by MATH. The latter contains the ample sheaf MATH, hence REF holds true. Finally the NAME bundle MATH comes from the locally free extension MATH of MATH. The eigenvalues of the residues of the induced connection lie in MATH, hence MATH is contained in the quasi-canonical extension MATH. Replacing MATH by MATH, we enlarge the sheaves MATH, which is allowed without changing REF. |
math/0101004 | We recall the formula for the curvature calculation of a linebundle with a metric MATH (see for example CITE). Let MATH be a local trivialization of MATH and let MATH be a holomorphic section of MATH which does not vanish in any point of MATH. Then MATH corresponds to a holomorphic function MATH on MATH and the metric MATH is given by MATH . The curvature MATH is given by MATH . Applying this formula (see also CITE, proof of REF), one finds MATH . Rescaling the sections MATH one can assume that the MATH are larger than a large constant MATH, hence that MATH is a continuous and positive definite MATH -form on MATH . Moreover MATH . The MATH form MATH is clearly positive semi-definite on MATH. Assume again that MATH is the zero set of MATH for local coordinates MATH on MATH. Then in a small neighborhood of MATH the form MATH is positive definite on the subspace of MATH spanned by MATH . Near MATH the section MATH can be expressed as MATH where MATH a local basis of MATH and where MATH is a positive Function. So, MATH and MATH . So the leading term in MATH near MATH is MATH . Obviously this form is positive definite on the subspace spanned by MATH . Since we assumed that MATH, MATH . By REF the MATH form MATH is continuous and positive definite on MATH . |
math/0101004 | By REF MATH is a subsheaf of MATH. Using the description of MATH in REF , we choose for a given point MATH a small neighborhood MATH and some MATH with MATH . The number MATH is determined by the monodromy of MATH around the component of MATH containing MATH, hence it is independent of the point MATH. Since the ramification order MATH in REF divides MATH we may choose MATH to be the effective divisor with MATH and MATH . |
math/0101004 | Let MATH denote the current of integration over the divisor MATH. As in CITE, proof of REF, one defines a singular metric MATH on sections of MATH by taking the square of the modulus of MATH viewed as a complex valued function. By the NAME equation MATH is the curvature current of this metric. One finds MATH . By REF, the curvature current of a singular metric on a holomorphic linebundle on a complex manifold is compatible with pullback under holomorphic maps. Hence MATH . By REF the latter is negative semi-definite, hence MATH. Moreover, MATH in the sense of currents, hence MATH . |
math/0101004 | We will use the notations from REF, in particular the metric MATH on MATH. Recall that MATH and that by REF, for all MATH . Hence in order to show REF it remains to verify that for MATH there exists some MATH with MATH . Given a point MATH choose a small polydisk MATH with coordinates MATH, in such a way that the divisors MATH and MATH are defined by the equation MATH . Let MATH be the cover ramified along MATH which we considered in REF . Choosing MATH small enough, we may assume that the connected component MATH are polydisks with coordinates MATH, and that MATH is defined by MATH . Hence for MATH and MATH the restrictions to MATH are the zero sets of MATH respectively. Consider as above the NAME bundle MATH obtained from the canonical extension of MATH along MATH, and let MATH be a basis for MATH. For MATH and MATH sufficiently small, there exist some MATH and a real number MATH with MATH for all MATH. By CITE, MATH can be decomposed into MATH where the open subset MATH depends on the index of the filtration of the mixed NAME structure (see REF), and such that MATH for all MATH where MATH is the multi index of the weight filtration of the mixed NAME structure. Since this index set is finite, there exist some MATH and some MATH such that MATH for all MATH and for all MATH. Hence MATH for all MATH. By the NAME inequality we obtain MATH for all MATH. MATH is compact, hence there is a finite covering MATH of MATH such that for all MATH and each of the finitely many connected components MATH of MATH, REF holds true. We may even assume, that REF remains true, for the same MATH, for all point in a small neighborhood of the closure MATH, not lying on MATH. We choose some MATH such that for all the open sets MATH and for the constant MATH given by REF one has MATH . In order to prove REF it is sufficient to show that on each MATH there is some MATH with MATH . Let us return to REF . As in the beginning of this section, for each component MATH of MATH we consider MATH with the hermitian metric MATH, and MATH with the pullback metric MATH . Let MATH be a section of MATH with zero locus MATH, where we assume that MATH has been rescaled as needed in REF for the constant MATH, chosen above. For the section MATH define MATH and MATH . Obviously one has MATH and MATH. Proof of REF . Let MATH be an orthonormal basis for MATH, with respect to MATH. Then MATH is an orthonormal basis for MATH with respect to MATH and MATH is an orthonormal basis for MATH with respect to MATH . Then, using the morphisms in REF , MATH is an orthonormal basis for MATH with respect to MATH . For the map MATH write MATH . Then MATH . Let MATH be the pullback of the morphism REF . By the commutativity of REF one obtains MATH and MATH . Next we consider the second map MATH and its pullback MATH . For the connected component MATH of MATH let MATH be a local generator of MATH . Then MATH is a basis of MATH and the morphism MATH is given by MATH and one finds MATH . Since the metric MATH on MATH is regular on MATH REF implies that MATH . Here and later we allow ourselves to replace the constant MATH by some larger constant, whenever it is needed. For the ramification order MATH of MATH over MATH, and for some positive function MATH on MATH one has MATH . This description extends to the compactification MATH of MATH. Since MATH is compact, MATH is bounded away from zero, and one finds MATH . On the compact set MATH all MATH are bounded above. Hence, all MATH also are bounded above, and the NAME inequality implies MATH . Since we assumed MATH and MATH, the right hand side in REF is smaller than MATH hence, we obtain the inequality MATH as stated in REF . |
math/0101007 | We first prove the formula for the norm of MATH (MATH). For every MATH such that MATH, MATH acts on the two-dimensional subspace MATH of MATH spanned by MATH and MATH as a self-adjoint operator with eigenvalues MATH and MATH. Since MATH is the NAME sum of the subspaces MATH, we see that MATH extends to MATH as a self-adjoint operator with operator norm equal to MATH. Hence for any MATH, MATH extends as a bounded operator on MATH. Finally notice that MATH, proving the equality MATH. |
math/0101007 | All this can be found in CITE, NAME I, paragraphe REF. In general, MATH is a two-sided ideal of the NAME algebra hull of MATH, but in the presence of the unit MATH the two spaces coincide. In fact, when MATH and MATH is in the strong closure of MATH, it is simple to see that MATH is bounded. |
math/0101007 | A square root MATH is in MATH and is Hermitian (that is, MATH). Then MATH. |
math/0101007 | This is immediate from the definitions, see CITE, NAME REF |
math/0101007 | It is clear that MATH is finitely generated over MATH. When MATH is generated by reflections, it is easy to see that the rank of MATH-adic completion MATH over MATH is exactly MATH (see REF). |
math/0101007 | Elementary and well known. Use the NAME theorem for REF , NAME 's version of NAME 's lemma for REF , and REF . |
math/0101007 | Because MATH is dense, it is clear that a representation of MATH is determined by its restriction to MATH and that (topological) irreducibility is preserved. Hence by the previous Corollary, all irreducible representations of MATH have finite dimension. |
math/0101007 | It is clear that the image is in MATH and that the map is finite (by REF ). Since MATH is NAME and NAME is compact, the map MATH is closed if it is continuous. So it remains to show that MATH is continuous. The closure MATH is a unital commutative MATH-subalgebra of MATH. By the NAME transform it is isomorphic to the algebra of continuous functions MATH on the compact NAME space MATH. Denote by MATH the map MATH defined by the condition MATH. By REF, MATH is surjective. In other words, every primitive ideal MATH of MATH intersects MATH in a maximal ideal MATH of MATH, and all maximal ideals of MATH are of this form. The corresponding surjective map from the set MATH of primitive ideals of MATH to the set of maximal ideals MATH is also denoted by MATH. Next we claim that MATH is continuous. The topologies of MATH and MATH are defined by the NAME topologies on MATH and MATH. This means that MATH is closed if and only if every MATH which contains MATH is in MATH. Let MATH be closed, and put MATH. By the surjectivity of MATH we have MATH. Hence if MATH contains MATH, then MATH contains MATH, implying that MATH. Therefore MATH, proving that MATH is closed as desired. Next, we consider the injective map MATH defined by restriction to MATH. Its image MATH is bounded because for every MATH, MATH, showing that each MATH with MATH has a maximum on MATH. Because MATH, we see that MATH for each MATH and MATH. By the NAME theorem, the restriction to MATH of a continuous function MATH can be uniformly approximated by elements in MATH considered as functions on MATH. In other words, there exists a MATH such that MATH for all MATH. Hence MATH is continuous on MATH for all MATH, showing that MATH is continuous and MATH. Since MATH is compact and MATH is NAME it follows that MATH is a homeomorphism. The proposition now follows from the remark that MATH. |
math/0101007 | A square root MATH, the closure of MATH in MATH, of MATH has obviously the property that MATH. Hence for every MATH, MATH: MATH . The second assertion follows since MATH. |
math/0101007 | CASE: By the above REF , MATH for all Hermitian MATH. In addition, the bitrace MATH is a positive semi-definite Hermitian form, and thus satisfies the NAME inequality. Hence for arbitrary MATH we have MATH, proving the continuity of MATH. CASE: Because every irreducible representation MATH of MATH is finite dimensional REF , it is clear that the character MATH is a well defined positive functional on MATH. It is continuous by REF . CASE: If the character MATH is positive, we have by REF that MATH extends to a finite continuous character of MATH. Because MATH is of finite type I (and thus liminal), the standard construction in CITE, paragraphe REF shows that there is up to equivalence a unique irreducible representation MATH of MATH whose character is MATH. The converse statement follows by REF . |
math/0101007 | CASE: This is trivial. CASE: If there exists a weight MATH of MATH violating the condition, then there exists a MATH such that MATH. We may assume that MATH for all weights MATH of MATH. By REF, the function MATH is not summable on MATH. Hence for all MATH, MATH can not be bounded by a constant times MATH. On the other hand, suppose that MATH is tempered. Since MATH and MATH, we obtain that MATH is bounded by a polynomial in MATH, a contradiction. CASE: Recall that the elements MATH with MATH, MATH span the subspace of MATH with basis MATH, where MATH runs over the double coset MATH (see the proof of REF). It is not difficult to see that in fact we can write, for MATH with MATH, MATH such that the coefficients MATH and MATH are equal if MATH and MATH belong to the same facet of the cone MATH. Moreover, by the length formula REF , we have MATH with MATH. Thus we also have MATH . It therefore suffices to show that the matrix entries of MATH with MATH are polynomially bounded in MATH. Since MATH is a direct sum of generalized MATH-weight spaces MATH, it is enough to consider the matrix coefficients of a generalized MATH-weight space MATH with weight MATH, satisfying REF . Observe finally that it is sufficient to consider the case that MATH, a sublattice in MATH of finite index. By NAME 's Theorem we can put the MATH simultaneously in upper triangular form. Choose MATH-generators MATH for the cone MATH, and a basis MATH for the lattice MATH. The NAME decomposition MATH gives mutually commuting matrices MATH and MATH, with MATH semisimple and MATH unipotent upper triangular. By conjugation in the group of invertible upper triangular matrices we may assume that the commuting semisimple matrices MATH are diagonal. The strictly upper triangular matrices MATH are commuting and satisfy MATH . Hence for MATH, with MATH when MATH, we have MATH . Since MATH by assumption, and the exponential map is polynomial of degree MATH on the space of strictly upper triangular matrices commuting with MATH, we see that the matrix entries are bounded by a polynomial of degree MATH in the coefficients MATH. Observe that MATH when MATH. Since the coefficients MATH are nonnegative this implies that for all MATH, MATH. On the other hand, there exists a constant MATH independent of MATH such that MATH for all MATH. Thus there exists a constant MATH independent of MATH such that MATH for all MATH. This gives us the desired estimate of the matrix entries by a polynomial in MATH, of degree MATH. |
math/0101007 | CASE: Let MATH denote the projector of MATH onto MATH. Then MATH for some idempotent of MATH, and since MATH is open we have MATH. Choose an orthonormal basis MATH of MATH. The corresponding matrix coefficients MATH can be identified with the elements MATH. Conversely, suppose that, given a basis MATH of MATH with dual basis MATH of MATH, there exist elements MATH such that for all MATH, MATH. It follows that for each MATH, the map MATH defines an embedding of MATH as a subrepresentation of MATH. CASE: This is similar to the proof of REF . For the last implication we first remark that REF implies that MATH can not contain MATH for any MATH. Thus MATH in this case, hence MATH. CASE: The number of elements in MATH with length MATH grows polynomially in MATH. Thus, by the exponential decay, it is clear that MATH is convergent in MATH, and moreover MATH. |
math/0101007 | The representation MATH associated with the state MATH is faithful (see Subsection REF). The results thus follow from REF. |
math/0101007 | The set MATH is a finite union of smooth varieties, obviously satisfying the condition of CITE, NAME III, REF for regularity, proving REF . As for REF , first note that the assumption implies that MATH, hence that MATH. Thus the codimension of MATH is positive, and MATH. Clearly MATH since MATH is the unique smallest vector in this affine linear space. Because MATH spans MATH, we can find a MATH such that MATH. This implies the result. |
math/0101007 | The existence is proved by induction on the dimension MATH of MATH, the case of MATH being trivial. Suppose that the result is true for tori of dimension MATH. Choose a smooth path in MATH from MATH to the identity MATH which intersects the real projection MATH of the codimension MATH cosets MATH transversally and in at most one point MATH. We may assume that the intersection points are mutually distinct with possible exception for the cases MATH, that is, when MATH. When we move MATH along the curve towards MATH, then we pick up residues when we pass at a point MATH on the curve. We may assume that the cosets MATH are connected (by factoring the defining equations, and adapting MATH accordingly). Let MATH be such that MATH. For simplicity of notation we write MATH instead of MATH. Recall the decomposition MATH with MATH. Let MATH denote the holomorphic extension to MATH of the normalized NAME measure on MATH, and similarly for MATH on MATH. Let MATH, and let MATH be its order. The product homomorphism MATH has kernel MATH. The residue that is picked up on MATH when we cross at MATH can be written as follows: MATH where MATH denotes a small circle in MATH around MATH. Using the action of MATH and the invariance of MATH and MATH, and in addition using MATH as a base point of MATH, this equals MATH . Let MATH be a generator of MATH. Let MATH be the holomorphic constant vector field on MATH which is dual to MATH. We extend MATH to a constant holomorphic vector field on MATH. We define a MATH-th root of MATH by MATH, so that the pull back of MATH to MATH factors as follows: MATH . With these notations, the above residue contribution is of the form MATH where MATH is a certain polynomial of degree MATH. Note that there may exist other MATH with MATH. We pick up similar residues with respect to these cosets as well when we cross at MATH. The above integral can be rearranged as follows MATH where MATH is itself a rational MATH-form on MATH which is a linear combination MATH on MATH with regular holomorphic coefficients MATH, and MATH-forms MATH which factor as in REF . The forms MATH have poles along the intersections MATH (with MATH) which are of codimension MATH in MATH. A simple computation shows that we can choose this decomposition of MATH such that for every MATH and every connected component MATH of an intersection of cosets of the form MATH, the index MATH of MATH satisfies MATH. It follows that the union over all MATH of the MATH-residual cosets in MATH is contained in the collection of MATH-residual cosets of MATH. Moreover, when we take MATH as a base point of MATH, so that we identify MATH with MATH through the map MATH, then the tempered form of a MATH-residual coset in MATH is equal to its tempered form as a MATH-residual coset in MATH. By the induction hypotheses we can thus rewrite the residue on MATH in the desired form, where the role of the identity element in the coset MATH is now played by MATH. At the identity MATH itself we have to take a boundary value of MATH towards MATH, which defines a distribution on MATH. This proves the existence. The uniqueness is proved as follows. Suppose that we have a collection MATH of distributions such that CASE: MATH. CASE: MATH. We show that MATH by induction on MATH. Choose MATH such that MATH for all MATH with MATH. For each MATH with MATH and MATH we choose a MATH such that MATH does not vanish on MATH REF and we set MATH . It is clear that for sufficiently large MATH, MATH for all MATH. On the other hand, by the theory of NAME series of distributions on MATH, MATH is a dense set of test functions on MATH. Since MATH is nonvanishing on MATH, this function is a unit in the space of test functions in MATH. Thus also MATH is dense in the space of test functions. It follows that MATH. |
math/0101007 | We construct the sequences with induction on the norm MATH. We fix MATH and suppress it from the notation. Let MATH and assume that we have already constructed such sequences MATH satisfying REF for all MATH with MATH. Consider the function MATH constructed in the second part of the proof of REF . By NAME analysis on MATH it is clear that there exists a sequence MATH in MATH such that for each holomorphic constant coefficient differential operator MATH of order at most MATH there exists a constant MATH such that MATH . Applying NAME 's rule to MATH repeatedly we see that this implies that there exists a constant MATH for each holomorphic constant coefficient differential operator MATH, such that MATH . Notice that MATH on all MATH with MATH but MATH. On the other hand, for each holomorphic constant coefficient differential operator MATH the function MATH converges uniformly to zero on each MATH with MATH. Again applying NAME 's rule repeatedly we see that there exist a MATH (depending on MATH) such that the function MATH has the property that MATH where the union is taken over all MATH with MATH. It is clear that the sequence MATH thus constructed satisfies REF . We continue this process until we have only one center MATH left. For this last center we can simply put MATH . This satisfies REF , and forces REF to be valid. |
math/0101007 | Because we are working with distributions on compact spaces, the orders of the distributions are finite. Take MATH larger than the maximum of all orders of the MATH. By REF we can thus express MATH as a sum of derivatives of order at most MATH of measures supported on MATH. The result now follows directly from the defining properties of the sequence MATH. |
math/0101007 | We begin the proof by remarking that REF imply that the functional MATH on MATH indeed defines a distribution on MATH, supported on MATH. Consider for MATH the inner integral MATH . Then MATH is a linear combination of (possibly higher order) partial derivatives MATH of MATH at MATH in the direction of MATH, with coefficients in the ring of meromorphic functions on MATH which are regular outside the codimension MATH intersections MATH: MATH . Hence MATH is equal to the sum of the boundary value distributions MATH of the meromorphic coefficient functions, applied to the corresponding partial derivative MATH of MATH, restricted to MATH: MATH . We see that MATH is a distribution supported in MATH, which only depends on MATH and on the component of MATH in which MATH lies. Hence, by the uniqueness assertion of REF , we conclude that it is sufficient to prove that we can choose MATH in such a way that MATH . In order to prove this it is enough to show that we can choose MATH as in REF for the larger collection MATH of all the connected components of intersections of the MATH (with MATH), such that MATH . Here MATH means that the left hand side and the right hand side are homologous cycles in MATH. The desired result follows from this, since the functional MATH is equal to MATH unless MATH is MATH-residual (because the inner integral REF is identically equal to MATH for non-residual intersections, by an elementary argument which is given in detail in the proof of REF ). Let MATH. Denote by MATH the collection of connected components of intersections of the MATH REF such that MATH. Assume that we already have constructed points MATH satisfying REF for all MATH and in addition, for each MATH with MATH, a finite set of points MATH such that MATH and a cycle MATH for each MATH, such that MATH is homologous to MATH . This equation holds for MATH, with MATH, which is the starting point of the inductive construction to be discussed below. We will construct MATH, MATH and MATH for MATH, and finite sets MATH for MATH, with a cycle MATH for each MATH such that REF holds with MATH replaced by MATH, and MATH by MATH. First of all, notice that we may replace MATH by any MATH in REF , because we can shrink the MATH and MATH within their homology class to fit in the smaller sets MATH. Choose MATH small enough such that for each MATH there exists a point MATH with the property that MATH. The singularities of the inner integral are located at codimension MATH cosets in MATH of the form MATH, where MATH is a connected component of MATH for some MATH. We have MATH, and thus MATH. Choose paths inside MATH from MATH to the point MATH. We choose each path such that it intersects the real cosets MATH transversally and in at most one point, and such that these intersection points are distinct. If MATH is the intersection point with the path MATH from MATH to MATH then MATH is of the form MATH with MATH. Given MATH we denote by MATH the set of all MATH arising in this way, for all the MATH such that MATH, and MATH. Notice that if MATH and MATH for some MATH and MATH, we have that MATH where MATH. Since MATH, this contradicts the assertion that the intersection points of the paths in MATH and the cosets MATH are distinct. We conclude in particular that the compact set MATH is contained in the union of the open sets MATH. We can thus choose MATH small enough such that in fact MATH, as required in REF . Write MATH for the identity component of the MATH-dimensional intersection MATH, and decompose the torus MATH as the product MATH. Let MATH and put MATH for the corresponding intersection point in MATH. Notice that for a codimension MATH coset MATH with MATH we have that MATH where MATH is a coset of the subgroup MATH of the finite group MATH, of the form MATH . The cosets MATH are disjoint. Let MATH be the minimum distance of two points in the union of these cosets, and let MATH denote the minimum of the positive real numbers MATH when we vary over all the MATH and MATH. Choose MATH smaller than the minimum of MATH and MATH. Let MATH be a circle of radius MATH with center MATH in MATH. Next we make MATH sufficiently small so that MATH. For MATH suitably close to MATH with MATH we have in MATH: MATH where the union is over all MATH such that MATH. Define MATH . Observe that MATH is a MATH-fold covering of MATH, and that MATH if MATH. We thus have MATH . By possibly making MATH smaller we get that MATH for all possible choices MATH and MATH. If MATH but MATH, then, since MATH and MATH, we have MATH. If on the other hand MATH then MATH. Finally we put MATH where we take the union over all pairs MATH with MATH such that MATH and MATH such that there is an intersection point MATH with MATH. We have shown that MATH as required in REF . Applying REF for all the intersections of all the paths we chose, we obtain REF with MATH replaced by MATH and MATH by MATH. We thus take MATH for MATH, and for MATH we take MATH and MATH as constructed above. This process continues until we have MATH in REF . In the next step we proceed in the same way. Notice that for MATH, either MATH (if we cross MATH with some curve from MATH to MATH in MATH, for one of the one dimensional residual cosets MATH containing MATH), or else MATH. The process now stops, since also MATH. This proves the desired result, with MATH equal to the MATH obtained in the last step of the inductive construction. |
math/0101007 | CASE: If MATH is moved within a chamber of MATH, the path from MATH to MATH can be chosen equal to the original path up to a path which only crosses codimension one cosets of the form MATH which do not contain MATH. Therefore this does not change MATH. CASE: We may replace MATH by MATH and MATH by MATH and leave MATH is unchanged, because the MATH do not contribute to MATH in the procedure of the proof of REF . By REF we may also replace MATH by MATH without changing MATH. We apply REF in this situation in MATH. Then we intersect with MATH and use the formula MATH. |
math/0101007 | If MATH, then MATH (since then MATH, implying that MATH). Now MATH. |
math/0101007 | By REF it is sufficient to show this in the case where MATH is a residual point. We identify MATH with the real vector space MATH via the map MATH, and we denote by MATH the Euclidean inner product thus obtained on MATH. Notice that the role of the origin is played by MATH. The sets MATH with MATH satisfy MATH, and are equipped with the induced Euclidean inner product. By the assumption and REF we can choose MATH within its chamber such that MATH. Assume by induction that in the MATH-th step of the inductive process of REF we have, MATH with MATH and MATH, that MATH (see REF for the meaning of MATH). Notice in particular that this implies that MATH if MATH. By choosing MATH sufficiently small, we therefore have MATH when MATH. Take the path MATH in MATH from MATH to MATH equal to MATH, where MATH denotes the (geodesic) segment from MATH to MATH in the Euclidean space MATH. Consequently, we have MATH for all MATH. Let MATH with MATH and MATH. If MATH intersects MATH in MATH, then we have MATH. By induction on MATH this proves that we can perform the contour shifts in such a way that REF holds for each MATH. This implies that MATH, and thus that MATH. |
math/0101007 | As above, we may assume that in fact MATH is a point. The only contributions to MATH come from contour shifts inside residual cosets of the configuration MATH as in the proof of REF . Likewise, for the construction of MATH we only need to consider the translated configuration MATH. By the assumption on MATH and REF we can construct MATH by working with MATH and MATH, but with the center MATH of MATH replaced by MATH. We now follow the deformations of the centers MATH with MATH and MATH. The assumption on MATH implies that MATH. This implies we can use MATH also as the cycle associated with MATH relative to the center MATH. |
math/0101007 | These properties are simple consequences of REF. |
math/0101007 | Just notice that for any given collection of approximating sequences MATH and any MATH, MATH is also a collection of approximating sequences for the distributions MATH, and this defines an action of MATH on the set of collections of approximating sequences for the MATH. Hence we can take the average over MATH. |
math/0101007 | We compute MATH, because MATH is also a commutator and MATH is central. |
math/0101007 | CASE: Any MATH can be decomposed as MATH with MATH and MATH, so it suffices to prove the assertion for MATH. Thus by REF it is sufficient to prove the assertion for a positive element MATH. Similarly MATH is a linear combination of positive central elements, so that it is sufficient to show that MATH for each positive central element MATH. By REF this reduces our task to proving that MATH for an arbitrary element MATH. Then MATH . It follows easily that MATH. CASE: This is essentially the same argument that we used to prove REF . Since MATH, we can find an open neighborhood MATH in MATH such that MATH. Let MATH. Then MATH, where MATH. We want to prove that MATH for MATH. By NAME analysis on MATH we can find a sequence MATH such that MATH converges uniformly to MATH on MATH for every holomorphic constant coefficient differential operator MATH on MATH of order at most MATH on MATH. We can then find a sequence MATH of the form MATH such that MATH converges uniformly to MATH on MATH, and to MATH on MATH for every MATH. Hence if we put MATH then for each holomorphic constant coefficient differential operator MATH on MATH of order at most MATH, MATH uniformly on MATH, and MATH uniformly on MATH for MATH. Hence MATH, MATH . If we divide this inequality by MATH and send MATH to MATH, we get that MATH with MATH, MATH . It follows that MATH, MATH. Hence the same is true for arbitrary MATH. |
math/0101007 | It suffices to show that MATH is a positive distribution. Assume that MATH and that MATH. Then the positive square root MATH is also in MATH. Using the approximating sequences as we did before, we can find a sequence MATH such that MATH, uniformly on MATH, and to MATH on MATH for MATH. By REF , the support of MATH is contained in MATH, where MATH. This is itself a regular support for distributions. On MATH, the sequence MATH converges uniformly to MATH up to derivatives of order MATH. Hence MATH . This proves the desired inequality. |
math/0101007 | It is enough to prove this for MATH which are Hermitian, that is, such that MATH. By REF we see that for positive functions MATH, MATH . |
math/0101007 | Everything is clear. REF follows from REF by REF . |
math/0101007 | We know already that MATH is a MATH-invariant measure supported on the union of the tempered quasi residual cosets. We apply REF to the integral MATH (compare REF ). Choose MATH. For a suitably small MATH we can find, for each quasi residual subspace MATH, a MATH in a MATH neighborhood of MATH, and a cycle MATH, where MATH denotes a ball of radius MATH centered around MATH, such that MATH . Here MATH is the holomorphic extension to MATH of MATH, and MATH denotes the NAME measure on MATH, also extended as a holomorphic form on MATH. We assume that MATH is small enough to assure that MATH is well defined on MATH. For the inner integral we use a basis MATH of MATH as coordinates on MATH, shifted so that the coordinates are centered at MATH. We can then write the integration kernel as: MATH where MATH is a rational homogeneous MATH-form (independent of MATH) in the MATH, and MATH is a power series in MATH such that MATH. In fact, the form MATH is easily seen to be (including the factor MATH of REF ) MATH . By REF it follows that the form MATH has homogeneous degree MATH if MATH is residual in the sense of REF . A homogeneous closed rational form of positive homogeneous degree is exact. Hence the inner integral will be nonzero only if MATH is in fact a residual coset. In that case the inner integral will have value MATH with MATH . We note that MATH, since MATH defines a rational cohomology class. Let us therefore assume that MATH is residual from now on. Write MATH. By REF we know that MATH with MATH. When MATH, the expression MATH can be rewritten as MATH . Here we used that if MATH with MATH, we have MATH, MATH, and MATH. By the same argument as was used in REF we see that this expression is real analytic on MATH. This implies that we can in fact take MATH for all residual MATH in REF after we evaluate the inner integrals. This leads to MATH where the sum is taken over residual cosets only. When we combine terms over MATH orbits of residual cosets we find the desired result. Let MATH denote the set of residual cosets in the orbit of MATH. We have to take MATH . When we now define a measure MATH on MATH by MATH then we have the equality MATH (sum over the residual subspaces). We note in addition that MATH , because the cycle MATH is constructed inside MATH, entirely in terms of the root system MATH (see REF ) (the factor MATH comes from the facorization MATH, see REF ). Also, it is clear that MATH is independent of the choice of MATH, because the finite group MATH is contained in the simultaneous kernel of the roots of MATH. Finally, the independence of MATH is clear from REF . When we apply a scaling transformation MATH, the point MATH moves such that the facet of the dual configuration containing MATH does not change. Hence MATH and MATH will be independent of MATH. |
math/0101007 | CASE: We note that for MATH, MATH and MATH are in the same chamber of MATH. Hence, by application of REF , we may replace MATH by MATH. CASE: This is trivial. CASE: The formula MATH was explained in the proof of REF . Let MATH. Let MATH be the stabilizer of MATH in MATH. Observe that MATH and MATH. If we define MATH then MATH is a complementary subgroup of MATH in MATH. Using REF and the remark MATH we see that MATH . Using REF the result follows. CASE: Follows easily from REF . CASE: Since REF involves only roots in MATH, it is sufficient to prove the statement for MATH indecomposable and MATH. Notice that for all MATH, MATH is a root of unity and, by REF , MATH is an integral power of MATH. Looking at the explicit REF , we see that it remains to show that this expression has rational coefficients if MATH is a residual point. Let MATH be the extension of MATH by the values of MATH, where MATH runs over MATH. In the case where MATH is of type MATH it follows by REF that MATH, and we are done. For the other classical cases it follows from the result of CITE that the order of MATH is at most two, and hence that MATH. Next let MATH be of exceptional type, and MATH. Define a character MATH of MATH by MATH. By REF we see that there exists a MATH such that MATH. Moreover, MATH acts as an automorphism and MATH is a MATH-residual point. If MATH contains isomorphic components then these are of type MATH, which has only one real residual point up to the action of MATH. Hence by REF , there exists a MATH such that MATH. Put MATH, so that MATH. By the MATH-equivariance of MATH we see that (with the action of MATH being extended to MATH by its action on the coefficients) MATH, whence the desired rationality. |
math/0101007 | The equality MATH was explained in REF, so it suffices to show that the support of MATH is equal to the union of the tempered residual cosets. By REF we know that MATH is supported on this set, so we need only to show that MATH is contained in the support for each tempered residual coset MATH. By REF this reduces to the case of a residual point MATH. By REF it is enough to show that there exists at least one MATH such that MATH. In other words, using REF we single out the point residue at MATH. In particular, we ignore all residues at residual cosets which do not contain MATH and thus do not contribute to MATH in the argument below. By the MATH-invariance of MATH, we can formulate the problem as follows. Recall from the proof of REF that MATH where MATH is the residue cycle at MATH, which is obtained from REF if we use the MATH-form MATH and a base point MATH such that MATH. By definition, MATH. For MATH we have MATH where MATH is the cycle near MATH which we obtain in REF when we replace MATH by MATH. Hence we have to show that there exists a proper choice for MATH such that when we start the contour shift algorithm from this point, the corresponding point residue at MATH will be nonzero. The problem we have to surmount is possible cancellation of nonzero contributions to MATH. We will do this by showing that there exists at least one chamber such that the residue at MATH consists only of one nonzero contribution. We consider the real arrangement MATH in MATH, and transport the Euclidean structure of MATH to MATH by means of MATH. Then MATH is the lattice of intersections of a central arrangement of hyperplanes with center MATH. We assign indices MATH to the elements of MATH by considering the index of the corresponding complex coset containing MATH, and we note that by REF , MATH. From REF we further obtain the result that there exist full flags of subspaces MATH such that MATH. In particular, there exists at least one line MATH through MATH with MATH. By REF we see that the centers MATH, MATH of two ``residual subspaces" MATH (that is, MATH and MATH) in MATH satisfy MATH unless MATH. Hence MATH (where MATH denotes the distance function), with equality only if MATH. In the case of a residual line MATH, MATH is divided in two half lines by MATH, and MATH lies in one of the two halves (that is, does not coincide with MATH). We want to find a chamber for MATH such that the corresponding point residue MATH at MATH is nonzero. We argue by induction on the rank. If the rank of MATH is MATH, obviously we get MATH if we choose MATH in the half line not containing MATH, because we then have to pass a simple pole of MATH at MATH when moving the contour MATH to MATH (since MATH and MATH are separated by MATH). Assume by induction that for any residual point MATH of a rank MATH root system, we can choose a chamber for MATH such that MATH. Let MATH be a sphere centered at MATH through MATH, and consider the configuration of hyperspheres in MATH. Let us call MATH the north pole of MATH. If MATH with MATH, we denote by MATH the intersection of the half line through MATH beginning in MATH (recall that MATH) and MATH. By the above remarks, MATH is in the northern hemisphere for all residual MATH of dimension MATH. We call this point MATH the center of MATH. In the case when MATH, MATH is disconnected and consists of two antipodal points MATH (north) and MATH (south), its opposite. In this case of residual lines through MATH, both of these antipodal points are considered as centers of MATH. We call MATH the northern center, and its opposite is called the southern center. All centers of MATH lie in the northern hemisphere, with the exception of the southern centers of the residual lines through MATH. Consider a closed (spherical) ball MATH centered at MATH such that MATH contains a southern center MATH in its boundary, but no southern centers in its interior. Since MATH is regular with respect to MATH (a trivial case of REF , as MATH is the center of MATH), we have MATH. We take MATH in MATH, and we apply the algorithm as described in the proof of REF , but now on the sphere MATH, and with respect to the sets MATH and their centers. By the induction hypothesis, we can take MATH close to MATH in a chamber of the configuration MATH which contains MATH in its closure, such that a nonzero residue at MATH is picked up in MATH. Denote by MATH the central subarrangement of elements of MATH containing MATH. Consider any alternative ``identity element" MATH which belongs to the same chamber of the dual configuration of MATH as the real (original) identity element MATH. As was explained in (the proof of) REF , when we apply the contour shifts as in (the proof of) REF to MATH, the residue at MATH only depends on the dual chamber which contains the identity element. In other words, we may use the new identity MATH instead of MATH without changing the residue at MATH. We can and will choose MATH close to MATH, and in the interior of MATH. By REF we can replace the integral over MATH by a sum of integrals over cosets of the form MATH (for some MATH) of the residue kernel MATH (compare REF ) on MATH. As was mentioned above, we are only interested in such contributions when MATH, which means that we may take MATH. The new ``centers" MATH with respect to the new identity element MATH are in the interior of MATH. Next we apply the algorithm of contour shifts as in REF to move the cycles MATH to MATH. Since both the new centers MATH and the original centers MATH belong to the interior of MATH, and since the intersection of MATH with MATH is connected if MATH, we can choose every path in the contour shifting algorithm inside the interior of MATH. Thus, the centers MATH of the residual cosets MATH that arise in addition the one southern center MATH in the above process are in the interior of MATH. In particular, with the exception of MATH, the one dimensional cosets of integration which show up in this way, all have a northern center. Finally, in order to compute the residue MATH at MATH, we now have to move the center MATH of MATH to the corresponding center MATH of MATH, for each residual coset MATH which contains MATH and which contributes to MATH. The only such center of MATH which will cross MATH is the southern center MATH. Since MATH has a simple pole at MATH, we conclude that this gives a nonzero residue at MATH. Hence with the above choice of MATH we get MATH, which is what we wanted to show. |
math/0101007 | This is immediate from REF , except for the last assertion. This fact follows from REF . We know that MATH is regular in MATH by REF . On the other hand, MATH lies in MATH, which is clearly a subset of a chamber of MATH. The anti-dual of the chamber of MATH containing MATH is thus a subset of MATH, with MATH. Thus when MATH is contained in the anti-dual chamber we have MATH as desired. |
math/0101007 | We have MATH . Hence from MATH, MATH and REF we conclude that the generalized weight spaces of MATH indeed satisfy the NAME criterion REF for discrete series. Conversely, if MATH is a discrete series representation, REF implies that MATH. By REF , the central character MATH of MATH is such that MATH. REF implies that such points MATH are necessarily residual. |
math/0101007 | Combine REF , and REF . |
math/0101007 | REF follows by a straightforward computation similar to REF , using REF and the definition of MATH. Since MATH is a positive combination of the irreducible characters of the residual algebra MATH, it follows that the weights MATH of the generalized MATH-eigenspaces of the irreducible characters of MATH all satisfy the condition MATH. This shows, by NAME 's criterion REF , that MATH is a tempered functional on MATH. |
math/0101007 | Both the left and the right hand side are finitely generated modules over MATH, and we have a natural morphism from the right hand side to the left hand side (product map). In order to prove that this map is an isomorphism it suffices to show this in the stalks of the corresponding sheaves at each point of MATH. Let MATH denote the maximal ideal in MATH corresponding to MATH, and let MATH denote the MATH-adic completion. Because MATH is faithfully flat over MATH (the stalk at MATH of the sheaf MATH), it suffices to check that for each MATH, we have MATH where MATH denotes the space of analytic germs at the set MATH. Let MATH denote the maximal ideal of MATH at MATH, and let MATH with MATH. For all MATH we have MATH. Since MATH, the left hand side of REF is equal to MATH, the sum of the completions of MATH with respect to MATH. The right hand side of REF is equal to the completion MATH. By the NAME remainder theorem, MATH, finishing the proof. |
math/0101007 | This is clear from REF by the remark that MATH arises from the MATH-algebra MATH by extension of scalars according to MATH . |
math/0101007 | We only need to show that MATH implies that MATH (the other direction being obvious). Notice that if MATH we have MATH . If MATH then MATH, and thus MATH . By REF , we have MATH. Since MATH is parabolic this implies that either MATH or that MATH. In the first case we are done, so let us assume the second case. By REF it follows that MATH. If MATH we have MATH by definition, contradicting the assumption. If MATH and MATH then, by definition, MATH, contrary to the assumption. If MATH and MATH for some MATH then REF implies MATH, again a contradiction. We conclude that the second case does not arise altogether, and we are done. |
math/0101007 | The difference with NAME 's approach is that he works with the MATH-adic completions of the algebras instead of the localizations to MATH. In addition, we have a different definition of the root system MATH. Using REF we can copy the arguments of CITE, because of REF for MATH and because of REF (which replaces in our situation REF). By this we see that the function MATH is analytic and invertible on MATH for all MATH such that MATH (compare REF ), and this is the crucial point of the construction. |
math/0101007 | This is immediate from the definitions. |
math/0101007 | The fact that MATH is MATH-generic implies that the MATH-equivalence classes of the elements of MATH are equal to the MATH-equivalence classes of these elements. Therefore we have, by the above theorem, MATH, where MATH is the number of equivalence classes MATH in the orbit MATH. And for each MATH, MATH is a union of MATH distinct MATH-equivalence classes. The orbit MATH is the disjoint union of MATH subsets of the form MATH, since the stabilizer of MATH is contained in MATH (because MATH is MATH-generic). Each subset MATH in MATH is partitioned into MATH equivalence classes, and the result follows. |
math/0101007 | We localize both the algebras MATH and MATH and use REF . Using REF we see that the functor MATH is the required equivalence. The relation between the dimensions of MATH and MATH is obvious from this definition. Conversely, again using REF , we have MATH finishing the proof. |
math/0101007 | We have to check that MATH is compatible with the NAME relations. Let MATH with MATH. Then MATH . Since MATH acts trivially on MATH, we have MATH. This implies the result. |
math/0101007 | By REF , the induction functor from representations of MATH to MATH gives rise to an equivalence between the representations of MATH with MATH-regular central character MATH and the representations of MATH with central character MATH. If MATH is a representation of MATH with central character MATH, then it is easy to see that the annihilator of MATH contains the kernel of the homomorphism MATH. Thus MATH is the lift via MATH of a representation MATH of MATH. This gives an equivalence between the category of representations of MATH with central character MATH and the representations of MATH with central character MATH. |
math/0101007 | The above form is clearly Hermitian and positive definite. It remains to show that the inner product satisfies MATH for each MATH. To this end we recall REF. Let MATH denote the minimal principal series representation induced from MATH. Then the nondegenerate sesquilinear pairing defined on MATH by MATH satisfies the property MATH (see REF for the definition of MATH). We have MATH, and the pairing REF on MATH factors as the tensor product of the pairings on MATH and on MATH which are also defined by REF but with MATH both in MATH or both in MATH. We choose MATH such that MATH contains a simultaneous eigenvector MATH for MATH with eigenvalue MATH. Via MATH, the vector MATH has eigenvalue MATH with respect to MATH. Thus there is a surjective morphism of MATH-modules MATH, where MATH is the minimal principal series module MATH for MATH, and MATH is the representation space of MATH. By the above, applied to MATH, we have the adjoint injective morphism MATH, where the action on MATH is via MATH (since MATH, because MATH). By the exactness and the transitivity of induction we get morphisms of MATH-modules MATH and MATH. Notice that MATH and similarly, MATH. By the factorization of the pairing REF we see that MATH and MATH are adjoint with respect to the pairings MATH on MATH and REF on MATH. This, the surjectivity of MATH and REF gives the desired result REF . |
math/0101007 | Recall that we have the identification MATH where MATH denotes the subspace spanned by the elements MATH with MATH. Recall from the proof of REF that we can find a basis MATH of MATH such that MATH acts by upper triangular matrices with respect to this basis. By NAME 's criterion, the diagonal entries are characters MATH of MATH which satisfy MATH for MATH. Let MATH denote an ordering of the set MATH such that the length MATH increases with MATH. We take the tensors MATH, ordered lexicographically, as a basis for the representation space of MATH. From a direct application of the NAME relations we see that the MATH are simultaneously upper triangular in this basis, and that the diagonal entries are the elements MATH. Since MATH, MATH and since the vector part of MATH is an element of the cone generated by the negative roots of MATH, it follows from a well known characterization of MATH that the vector parts of these diagonal entries are in the antidual of the positive chamber. Again using REF we conclude that MATH is tempered. |
math/0101007 | The functional MATH factors to a functional of the finite dimensional MATH-algebra MATH , where MATH is the maximal ideal in MATH corresponding to MATH. We have MATH for MATH, and this defines the extension with the required property uniquely. |
math/0101007 | Because these are both equalities of holomorphic functions of MATH it suffices to check them for MATH regular, and outside the union of the residual cosets (in other words, MATH). CASE: By the defining REF and CITE we need only to show that the left hand side satisfies the properties MATH and MATH. These facts follow from REF . We see that MATH and similarly for MATH. The value at MATH is equal to MATH on both the left and the right hand side. As in the proof of REF , this is enough to prove the desired equality. |
math/0101007 | CASE: This is a direct consequence of REF . Recall the definition of the states MATH. Recall that the support of the measure MATH is the union of the tempered residual cosets. We combine REF , and REF to see that (with MATH the stabilizer of MATH in MATH) MATH for all MATH and MATH. Rewrite the right hand side as MATH where the inner integrals equal, with MATH, MATH where MATH . Hence MATH is a linear combination of (possibly higher order) partial derivatives (in the direction of MATH) of the kernel MATH evaluated at MATH. The MATH-invariant measure on MATH on the left hand side of REF is thus obtained by taking the boundary values MATH of the MATH, and then sum over the NAME group as in REF . Notice that the collection of MATH-generic points in MATH is the complement of a union of algebraic subsets of MATH of codimension MATH. The kernel REF is regular at such points of MATH. The boundary values at MATH-generic points are therefore computed simply by specialization at MATH. We thus have MATH . The expression on the left hand side can be extended uniquely to MATH and MATH. By REF , each summand in the expression on the right hand can also be extended uniquely to such locally defined analytic MATH and MATH. Take MATH. We restrict both sides to MATH. Substitute MATH by MATH with MATH. On the left hand side we get, by REF , MATH where MATH is a central functional on MATH, normalized by MATH. On the other hand, by REF , if MATH with MATH then MATH if MATH, and MATH otherwise. Observe that, because of REF for MATH, MATH implies that there exists a MATH such that MATH. Since MATH is MATH-generic, we see in particular that the stabilizer of MATH is contained in MATH. Thus MATH implies that MATH, and hence that MATH. Therefore the sum at the right hand side of REF reduces, if MATH is of the form MATH with MATH, to MATH . The function MATH is MATH-invariant on MATH, because MATH is MATH-equivariant (that is, MATH when MATH). In other words, this function is in the center of MATH. By REF , and REF applied to MATH therefore, this sum reduces to MATH which we can rewrite as MATH . Comparing this with REF we see that, in view of REF , this implies that for MATH, MATH . Applying REF , and REF we obtain REF . CASE: By REF , MATH is MATH-almost everywhere a positive trace. On the set of MATH-generic points MATH, we have expressed MATH as a positive linear combination of the irreducible induced characters MATH. This gives the decomposition of MATH in terms of irreducible characters of the finite dimensional algebra MATH, where MATH denotes the two-sided ideal of MATH generated by the maximal ideal MATH of the elements of the center MATH which vanish at MATH. On the other hand, we have the decomposition of MATH in irreducible characters of the finite dimensional NAME algebra MATH, as in REF . This algebra is a quotient algebra of MATH. Because MATH is finite dimensional, there is no distinction between topological and algebraic irreducibility. We therefore have two decompositions of MATH in terms of irreducible characters of MATH. The irreducible characters of MATH are linearly independent, and thus the two decompositions are necessarily the same. This implies that the induced characters MATH are characters of the NAME algebra MATH. In particular, the characters are positive traces for all MATH-generic MATH. These induced characters are regular functions of MATH. Hence by continuity, they are positive traces for all MATH. By REF , MATH extends to a continuous trace of MATH for all MATH. According to the construction in CITE, NAME REF this character is the trace of a (obviously finite dimensional) representation of MATH, quasi-equivalent with MATH when restricted to MATH. Hence all subrepresentations of MATH extend to MATH. |
math/0101007 | The first assertion was shown in the proof of REF . If MATH there exists a sequence MATH with MATH and MATH. By REF , the function MATH is a uniform limit of the functions MATH, proving the continuity. |
math/0101007 | The NAME space MATH is the completion of MATH with respect to the positive trace MATH. In REF we have written MATH as a positive superposition of positive traces MATH, with MATH. In REF we established that, outside a set of measure zero, MATH is a finite linear combination of traces of irreducible representations MATH which extend to MATH. Thus REF is a positive decomposition of MATH in terms of traces of elements of MATH. This allows us to apply REFEF and REFEF in order to obtain REF . The statements about the residual algebra of a residual point follow directly from REF . Finally, REF follows from REF in combination with the factorization REF of the inner product MATH on the induced representations. |
math/0101007 | CASE: This follows from the above text, and the fact that MATH is a central functional. CASE: It is clear from NAME 's criteria that MATH indeed defines a bijection between the sets of discrete series representations MATH and MATH. The result follows from REF . |
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