paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0007106 | Notice that MATH . For the inductive step, regard MATH as the disjoint union of MATH copies of MATH by writing MATH . We can then write MATH in terms of MATH as a MATH matrix of MATH matrices. For MATH in MATH we have MATH . Thus the MATH entry of MATH, viewed in this way, is MATH if MATH and MATH if MATH where MATH, f... |
math/0007106 | CASE: Both statements follow from the observation that MATH for all MATH in MATH. CASE: Notice that MATH for all MATH in MATH because there is no cancellation in REF The MATH's with MATH as in REF span a dense subspace of MATH, so this follows from REF . |
math/0007106 | This is obvious if MATH the case in which MATH . By what we have observed just above, MATH makes MATH and MATH . For MATH in MATH and MATH we also have MATH and furthermore MATH while MATH when MATH . And so forth - the asserted form plainly remains intact under further noncancelling left multiplication by generators o... |
math/0007106 | Write MATH as in REF . If MATH does not begin with MATH then MATH and so MATH . Furthermore, for MATH we have MATH and MATH . If MATH begins with MATH, then MATH, and for any MATH . Notice also that MATH (because MATH is real). It now follows from REF that MATH coincides with MATH . |
math/0007106 | We begin by noticing that MATH is cyclic for MATH. Indeed, by REF , the linear span of MATH contains MATH for each MATH in MATH and each MATH. It follows that the closed linear span contains each summand MATH. Of course, it also contains MATH because MATH for MATH in MATH. We will show in several steps that the commuta... |
math/0007106 | The spectrum is connected because the reduced MATH-algebra of MATH contains no nontrivial idempotents CITE. It is rotationally invariant because there is an automorphism of this MATH-algebra that multiplies each MATH by a given scalar of modulus one. Thus, the spectrum must be either a closed disc about REF or a closed... |
math/0007106 | Since MATH this amounts to showing that MATH for all MATH . By construction, MATH for each MATH. If MATH for some MATH, we have MATH . Otherwise, MATH for some MATH, and we have MATH . |
math/0007106 | CASE: Write MATH and MATH . Observe that MATH is constant on MATH, in fact on MATH . Whether MATH is MATH for some MATH or MATH, the factor MATH in the formula is constant on MATH; notice that MATH. The remaining factors in the formula for MATH are constant on MATH because for MATH in MATH, the function MATH reads at m... |
math/0007106 | The calculation is in three parts. CASE: We show first that if MATH, and MATH is a reduced word not beginning with MATH (including the possibility that MATH), then MATH. We have MATH . By REF , there is a number MATH such that MATH for all MATH in MATH . Taking this into account, and looking up values for MATH and MATH... |
math/0007106 | See REF below for the existence and uniqueness of the MATH's. The formula for MATH follows from REF above and the observation that the MATH's and MATH's there satisfy MATH for MATH. (Notice as well that MATH.) REF implies that MATH is a MATH-eigenstate for MATH. That MATH is pure follows from results in REF and our ide... |
math/0007106 | When MATH, this is immediate from the descriptions in the vicinity of REF above for MATH . Suppose MATH begins with MATH, so there is no canceling in MATH for MATH in MATH . If MATH doesn't end in MATH, then MATH for all MATH in MATH. This takes care of the case MATH. If MATH ends in MATH for some MATH, then MATH for a... |
math/0007106 | We must show that MATH for all MATH in MATH. This is immediate when MATH since MATH for each MATH. Suppose that REF holds for some given MATH in MATH and that MATH in MATH is such that MATH is reduced. We have MATH because MATH is the integral of MATH over MATH. It follows from REF and MATH that either the MATH summand... |
math/0007106 | Write MATH where without loss of generality the MATH's are all nonzero. Since MATH at least one of MATH must be different from zero. Suppose that MATH. For each index MATH, let MATH be MATH times the indicator function of MATH (the set of reduced words ending in MATH). One checks readily that MATH . Thus, MATH CITE. Si... |
math/0007106 | Consider the symmetry MATH taking each string in MATH to the string obtained by inverting each symbol. It is immediate that MATH . Since MATH for a nonempty reduced word MATH with the MATH's in MATH, it follows that MATH so MATH . The operator MATH on MATH defined by MATH is unitary and takes MATH to MATH. We claim tha... |
math/0007106 | We use the notation of the proof of REF of the previous proposition. Suppose MATH for some complex number MATH and some nonzero MATH in MATH. Fix an index MATH, pick MATH in MATH and consider MATH in MATH defined by MATH . One checks readily that MATH . Since MATH belongs to the reduced spectrum of MATH it follows from... |
math/0007106 | If only one of the coefficients is nonzero, this is REF. Assume therefore that at least two coefficients are nonzero. The argument from CITE for the case MATH also works here with just a few cosmetic changes. Let MATH thought of as an operator on MATH . Because MATH belongs to the reduced spectrum of MATH there is a st... |
math/0007106 | Subtract the first column of MATH from the other columns to obtain MATH then put MATH for MATH in the first term. |
math/0007106 | Notice that MATH . Let MATH . Apply the lemma above to MATH and then divide by MATH to write MATH . Write the first term as MATH and add the MATH-th term of this sum to the MATH-th of the remaining terms of MATH to obtain MATH . If all of the MATH's are positive, we conclude immediately that MATH . Otherwise, since at ... |
math/0007106 | We have seen by now that MATH is an open map of MATH into MATH . It will suffice to show that if MATH is the limit of a sequence MATH, where MATH is a sequence in MATH with no limit points in MATH, then MATH. Omit the superscript MATH and write MATH . After passing to a subsequence, we may assume that either MATH for s... |
math/0007106 | The partial derivative with respect to MATH of the sum of the entries of MATH is MATH . |
math/0007106 | Let MATH (as usual) be the sum of the entries of MATH, and let MATH be the same for MATH. Further let MATH and MATH . It follows easily that MATH for each MATH, and further manipulation shows that MATH . Since MATH there is at most index MATH such that MATH . If we had MATH, then by REF MATH and MATH would have to coin... |
math/0007106 | Suppose the pair REF minimizes MATH subject to MATH, and MATH. Write MATH. Let MATH be a local inverse for MATH such that MATH. (The existence of MATH follows from REF .) Write MATH and let MATH be the hyperplane MATH . For any MATH in MATH, the smooth real function MATH is minimized by MATH and hence MATH . But of cou... |
math/0007106 | Assume that MATH . Let MATH in MATH be such that MATH, and write MATH . In light of REF , we may assume that MATH . As in the proof of that lemma, we have MATH so MATH for every MATH. For MATH the sign choice must be MATH rather than MATH, because MATH for such MATH and the choice of MATH would make MATH . Let MATH be ... |
math/0007106 | We proceed by induction on MATH. The case MATH is straightforward; one checks easily that the map from MATH to MATH defined by MATH gives the identity map on MATH when preceded by MATH. Suppose now that MATH and that the assertion is true in all dimensions less than MATH. Suppose that for some MATH, the set MATH is non... |
math/0007108 | Because of the Weak Factorization Theorem of CITE it suffices to show that MATH when MATH is obtained from MATH by a blowup along a nonsingular subvariety MATH. We remark that the algorithm of CITE is compatible with the normal crossing condition (compare REF ), so we may assume that MATH has normal crossings with the ... |
math/0007108 | Any two resolutions of singularities of MATH can be connected by a sequence of blowups and blowdowns. Let MATH and MATH be two resolutions of MATH, such that MATH is the blowup of MATH as in the proof of REF . For any MATH and MATH big enough to assure that all discrepancies are not equal to MATH, the proof of REF impl... |
math/0007108 | We will show that elliptic genus of a crepant resolution MATH of a variety MATH equals to the singular elliptic genus of MATH. If the exceptional set of the morphism MATH is a divisor with simple normal crossings, then it is enough to observe that in REF the second product is trivial. In general, we can further blow up... |
math/0007108 | To avoid confusion, we immediately remark that the second arguments in singular elliptic genus and in NAME 's MATH-function have drastically different meanings. The definition of MATH in CITE could be stated as MATH where MATH is the exceptional divisor of a resolution MATH together with proper preimages of the compone... |
math/0007108 | Transformation properties of MATH under MATH and MATH together with NAME condition MATH assure that MATH . We needed here that MATH. Similarly, the transformation properties of MATH under MATH show that MATH . It remains to investigate what happens under MATH. For this, one considers the change MATH in the formula of R... |
math/0007108 | Let MATH and MATH be two birationally equivalent NAME manifolds or their generalizations above. Let MATH be a desingularization of the closure of the graph of the birational equivalence, so that MATH are regular birational morphisms. Let MATH be the smallest integer so that MATH is rationally equivalent to zero, and th... |
math/0007108 | We replace the contribution of each conjugacy class by an average contribution of its elements to obtain MATH . Using holomorphic NAME theorem, we obtain: MATH where MATH is a the normal bundle to MATH in MATH. An explicit calculation of the NAME and NAME classes then yields MATH which proves the first part of the theo... |
math/0007108 | It is easy to see that it is enough to check the lemma for a one-dimensional space MATH. If MATH is even, then MATH . If MATH is odd, then MATH . |
math/0007108 | We observe that for a fixed MATH the conjugacy classes of MATH are indexed by the numbers MATH of cycles of length MATH in the permutation. For each MATH the fixed point set MATH consists of the Cartesian products of several copies of MATH, one for each cycle. For a cycle of length MATH the corresponding MATH is embedd... |
math/0007108 | At MATH the function MATH specializes to MATH of CITE. Then the result of CITE allows one to rewrite it in terms of MATH, and REF finishes the proof. |
math/0007108 | Let MATH be the defining cone of MATH in the lattice MATH, see for example CITE. Let MATH be the generators of one-dimensional cones of MATH. The group MATH can be identified with MATH where MATH is a suplattice of MATH of finite coindex. Then the variety MATH is given by the same cone MATH in the new lattice MATH. The... |
math/0007108 | Expanding MATH functions as (linear) polynomials in cohomology classes, one obtains that singular genus is equal to MATH plus sum of contributions of singular points that depend on the ramification numbers only. Here MATH is the genus of MATH. For the orbifold genus, one needs to notice that MATH term gives MATH plus c... |
math/0007108 | Combine REF. |
math/0007108 | First of all, observe that MATH due to the product formulas for MATH and MATH, see CITE. Therefore, we only need to show that MATH . Denote by MATH the piece-wise linear function on MATH whose value on the generators of the one-dimensional cones of MATH is MATH. Notice that MATH consists of all points MATH such that MA... |
math/0007108 | We shall consider cobordisms of pairs MATH CITE where MATH is a stably almost complex manifold (that is, MATH manifold such that a direct sum of a trivial bundle MATH with the differentiable tangent bundle MATH admits a complex structure) and MATH is a finite union of codimension one stably almost complex submanifolds ... |
math/0007108 | Let MATH be a null-cobordant MATH-manifold. Then for each MATH the pair MATH where MATH is the normal bundle of the fixed point set MATH in MATH is cobordant to zero as well. Since the contribution of the term in MATH corresponding to a conjugacy class MATH is a combination of the products of NAME classes of MATH and M... |
math/0007108 | This follows from the result of NAME CITE describing generators of MATH-cobordisms. If MATH, then additive generators of cobordism group in any dimension are toric varieties with group being a subgroup of the big torus. Hence REF yields the claim. |
math/0007109 | Let MATH, and MATH. It is clear that MATH and MATH whenever MATH, MATH; that is, MATH is a submonoid of MATH. For every MATH there exist open neighborhoods MATH of MATH and MATH of MATH such that MATH, and MATH. Since MATH is compact, there exist MATH such that MATH. Consider now the neighborhood MATH. Since MATH is sy... |
math/0007109 | Let us consider the action MATH such that MATH, MATH, MATH. We endow MATH with the product topology, so MATH is a continuous action. Moreover, MATH is compatible with respect to the equivalence relation MATH on MATH given by: MATH and MATH. Thus MATH induces a continuous action MATH of MATH on the quotient topological ... |
math/0007109 | Let us suppose that MATH is a NAME space, and let MATH. In particular, MATH. Since MATH is continuous, MATH, and it must be MATH because of the uniqueness of limits in NAME spaces. Conversely, assume that MATH is closed in MATH, and let MATH. We want to show that if there does not exist disjoint open sets in MATH, each... |
math/0007109 | It is clear that REF. implies REF., and REF. and REF. are equivalent by REF , because of the categorical equivalence between locally compact NAME spaces and commutative MATH-algebras. To see that REF. implies REF, let us suppose that MATH is an enveloping action of the partial action MATH on an abelian MATH-algebra MAT... |
math/0007109 | Let MATH, where MATH, MATH, and consider MATH. Then MATH is a MATH-algebra with MATH, and MATH is an ideal of MATH. In particular, MATH may be considered as a right NAME MATH-module with the inner product: MATH, so there is a homomorphism MATH given by MATH. On the other hand, we have an inclusion MATH given by MATH, M... |
math/0007109 | Let MATH be an approximate unit of MATH. Then MATH, MATH, and since MATH is both the restriction of MATH and MATH to MATH, we have: MATH . |
math/0007109 | For MATH, let MATH and MATH be given by MATH and MATH. By REF , MATH and MATH are MATH-seminorms, and MATH, MATH. Let MATH and MATH. We want to show that MATH. For this, it is enough to prove that MATH. Let MATH and MATH. By REF we have: MATH . It follows that MATH, MATH, and hence that MATH. Thus, MATH such that MATH ... |
math/0007109 | Note that MATH is MATH - invariant, and since MATH, we have that MATH. From this, the result follows immediately. |
math/0007109 | Let MATH be the set of AF - ideals of a MATH-algebra MATH. Since MATH, we have that MATH. Let MATH, and consider the following exact sequence of MATH-algebras: MATH where MATH is the inclusion and MATH is the quotient map. Since the class of AF - MATH-algebras is closed by ideals, quotients and extensions, it follows t... |
math/0007109 | Suppose that there exists MATH such that for any open neighborhood MATH of MATH there exist MATH and MATH such that MATH, that is, MATH. Note that MATH, MATH. Since MATH, then MATH. Thus there exist a compact set MATH and a neighborhood MATH of MATH such that MATH, MATH. Then the net MATH, and hence it must have a subn... |
math/0007109 | Let us fix MATH, and let MATH in MATH. Given MATH, let MATH such that MATH, and let MATH such that MATH, MATH and some constant MATH. Then, if MATH: MATH . It follows that MATH, MATH, and therefore MATH in MATH. |
math/0007109 | REF tells us that MATH is a NAME representation, in the sense of VIII-REF So we may apply VIII-REF, and conclude that MATH is integrable. That is, there exists a representation MATH such that MATH . Moreover, MATH is unique. We set MATH. We want to see that MATH, it is MATH. Now, for MATH, we have that MATH. In particu... |
math/0007109 | Let MATH be a representation of MATH on the NAME space MATH, such that MATH is faithful. Then MATH is a representation of MATH, such that MATH is faithful. Let MATH such that MATH, where MATH is the left regular representation of MATH; that is: MATH, MATH. Similarly, define MATH. It is clear that MATH. Integrating MATH... |
math/0007109 | The function MATH such that MATH is continuous and MATH, MATH. So by CITE, II-REF, the function MATH given by MATH is continuous. In particular, MATH is continuous, and thus REF. is proved. As for REF., let MATH, MATH be compact sets such that MATH. Since the function MATH is continuous, each MATH is a continuous secti... |
math/0007109 | It is straightforward to check that MATH and MATH, and from these facts the two first inclusions follow easily. Let us prove the third assertion. Since MATH, we have isometric inclusions MATH. MATH is a sub-*-Banach algebra with approximate unit of MATH, and it is contained in the right ideal MATH of MATH. Thus, by REF... |
math/0007109 | By REF, MATH is naturally isomorphic to the closure of MATH in MATH. By REF., MATH is a right ideal of MATH, and hence its closure MATH in MATH is a right ideal of MATH. Now, it follows from REF that MATH, and therefore MATH is a hereditary sub-MATH-algebra of MATH. Finally, the last assertion follows immediately from ... |
math/0007109 | Suppose that MATH is amenable, and let MATH be the norm on MATH and MATH the norm on MATH. The closure of MATH in MATH is MATH, by REF . We also have that the closure of MATH in MATH is MATH, because any representation of MATH induces a representation of MATH by restriction, and therefore the norm of MATH is greater or... |
math/0007109 | Let MATH be the Fell bundle associated with MATH, MATH the Fell bundle associated with MATH, and MATH, where MATH. It is clear that MATH as NAME bundles, and that MATH is a sub - Fell bundle of MATH. Moreover, if MATH, MATH, MATH, MATH, we have: CASE: MATH, because MATH. Therefore: MATH CASE: MATH, because MATH belongs... |
math/0007109 | It follows immediately from REF. |
math/0007109 | Let MATH be the partial action described before. Since MATH is amenable, we have that MATH. First of all, note that MATH has an enveloping action MATH acting on MATH, and given by the same formula as MATH. Let MATH be the Fell bundle over MATH associated with MATH, MATH the characteristic function of MATH and, if MATH,... |
math/0007109 | If MATH exists, it must be MATH, MATH. Therefore we must see that MATH implies that MATH. Now, if MATH: MATH . Thus, we have a homomorphism of *-algebras MATH. Since MATH is a dense *-ideal of MATH, then MATH has a unique extension to a homomorphism MATH (CITE, VI-REF). Now, if MATH: MATH, and hence MATH is a contracti... |
math/0007109 | By REF, MATH, MATH is an isomorphism, MATH, and MATH. It follows from REF that, since MATH is an extension of MATH, then MATH is an extension of MATH. Now, the domain of MATH is MATH, and MATH is an isomorphism. By REF, the domain of MATH is MATH. Since MATH is a partial action, we also have by REF that MATH. It follow... |
math/0007109 | Let MATH be an open set, MATH, and MATH. Consider MATH. By NAME - NAME, MATH, for some MATH, and MATH. Since the action of MATH on MATH is continuous, there exist open sets MATH and MATH, such that MATH, MATH, and MATH. Now, MATH, and since MATH is continuous, the set MATH is open and contains MATH. For each MATH take ... |
math/0007109 | Let MATH be the unique linear transformation such that MATH, MATH, MATH; it is a homomorphism of * - ternary rings. Let MATH and MATH be the corresponding MATH - pre - inner products on MATH and MATH respectively. Pick MATH, MATH, MATH. Since MATH, MATH, and MATH, we conclude that MATH. By taking MATH and computing nor... |
math/0007109 | The reflexive and symmetric properties are immediately verified (see REF ). Suppose now that MATH is a partial action of MATH on MATH, MATH is a partial action of MATH on MATH, and MATH is a partial action of MATH on MATH, such that MATH through the partial action MATH of MATH on the MATH-tring MATH, and MATH through t... |
math/0007109 | Suppose that MATH and MATH are NAME equivalent through a partial action MATH; say that MATH, and MATH. Consider the full left NAME MATH - module MATH, where MATH. Then MATH establishes a NAME equivalence between MATH and MATH, the linking algebra of MATH (see CITE for instance). By REF, every ideal MATH corresponds to ... |
math/0007109 | Since NAME equivalence of MATH-algebras is transitive, the proof follows immediately by combining REF above with REF . |
math/0007109 | Let MATH be such that MATH, and let MATH be the retraction of MATH with respect to MATH (CITE, II-REF). Then MATH is a NAME bundle over MATH, and the fiber of MATH over MATH is MATH, which we may naturally identify with MATH. Therefore, MATH as a linear space. Consider now the map MATH given by MATH. We have that MATH ... |
math/0007109 | All these properties are easy to verify. As an example we prove REF., and leave REF. - REF. to the reader. If MATH, MATH, MATH, MATH . |
math/0007109 | For MATH, the map MATH such that MATH is a continuous linear map in the inductive limit topology . Therefore MATH. Conversely, suppose that MATH is such that MATH, MATH. Since MATH is locally compact, there exists a compact subset MATH of MATH whose interior contains MATH. Now, given MATH and MATH, there exists MATH su... |
math/0007109 | Recall that MATH, where MATH is the retraction of MATH with respect to the map MATH such that MATH. By REF , it suffices to show that, if MATH is the linear MATH - orbit of MATH, then: CASE: MATH is dense in MATH, MATH and REF. MATH, and MATH, the function MATH belongs to MATH. If MATH, by NAME - NAME there exist MATH,... |
math/0007109 | First Suppose that MATH is saturated. If MATH, MATH, MATH, MATH, then MATH. On the other hand, since MATH is saturated, given MATH and MATH, there exist MATH, MATH, such that MATH. Moreover, there exist sections MATH such that MATH and MATH, MATH. Therefore: MATH. It follows from REF that MATH is dense in MATH in the i... |
math/0007109 | Since MATH and the map MATH such that MATH is continuous (CITE, VIII-REF), we have that the map MATH such that MATH is continuous, and since it has compact support, we see that MATH is a continuous map of compact support from MATH to MATH, so in particular it belongs to MATH. On the other hand: MATH so MATH. It follows... |
math/0007109 | Let MATH and MATH; then: MATH . It follows that MATH. On the other hand, since MATH is a positive operator, the equality above implies that MATH. Thus, MATH, and MATH if MATH is faithful. It follows that we may extend MATH to MATH, and this extension is an isometry if MATH is faithful. An easy computation shows that MA... |
math/0007109 | Let us suppose that REF. is true, and let MATH be a representation such that MATH is faithful. Then the representation MATH extends uniquely to a faithful representation MATH, where MATH is the completion of MATH with respect to the norm MATH. On the other hand, MATH is an injective homomorphism of MATH-trings. Thus MA... |
math/0007109 | Both parts are direct consequences of REF. |
math/0007109 | Let MATH. Since MATH is a non - degenerate right NAME module over MATH, the NAME - NAME theorem implies that there exist MATH, and MATH, such that MATH. Let MATH such that MATH. The function MATH such that MATH is continuous and vanishes at MATH. Thus, given MATH, there exists a neighborhood MATH of MATH, such that if ... |
math/0007109 | Let MATH be the completion of MATH with respect to MATH, and MATH the corresponding right inner product. Let MATH, MATH, and assume that MATH. Then, MATH: MATH and hence MATH, because MATH is dense in MATH. Thus we may define MATH such that MATH. It is straightforward to check that MATH is a * - homomorphism. Since MAT... |
math/0007109 | Let MATH be the set defined in REF, that is, MATH is the closed cone of MATH generated by the elements of the form MATH, with MATH. Since the convergence in the inductive limit topology implies the convergence in MATH, and therefore in MATH, we have that MATH. Then, by REF, MATH is an inner product. Let MATH be the nor... |
math/0007109 | Since MATH and MATH are NAME equivalent, MATH has property MATH if and only if MATH has property MATH. On the other hand, by REF, MATH has property MATH if and only if MATH has property MATH. |
math/0007109 | Let MATH. There exist MATH, MATH, such that MATH. In particular, MATH, MATH. Now, if MATH, then MATH, and therefore MATH, MATH, that is, MATH, MATH. Equivalently, MATH, MATH. Let MATH, which is a dense subalgebra of MATH in the inductive limit topology. If MATH, MATH, MATH, MATH, then MATH, and hence MATH. The result f... |
math/0007109 | To prove REF., recall that MATH, MATH. Then MATH REF. MATH. CASE: Let MATH, MATH, MATH and MATH, MATH. We have: MATH . Hence, if MATH, MATH, then MATH . Since MATH is a closed linear space, REF shows that the closure of MATH in the inductive limit topology is MATH, and by REF, this set agrees with the closure of MATH i... |
math/0007109 | Let MATH, MATH. Identifying MATH with the unit fiber of MATH over the unit element of MATH, and the kernel MATH with MATH such that MATH, we have the identities: MATH . If MATH, MATH, by REF we have that: MATH . Then MATH is an isometry with dense range, and hence it extends to a bijective isometry MATH, such that MATH... |
math/0007109 | As we have already remarked at the end of REF, if MATH, then MATH. Let MATH, MATH, MATH, MATH, MATH, MATH. Then: CASE: MATH. Therefore, MATH, and hence MATH. CASE: Note that the second assertion is a consequence of the first one. On the other hand: MATH, which proves the first assertion of REF. CASE: MATH. Thus, MATH e... |
math/0007109 | Suppose that MATH is a partial action that implements the equivalence between MATH and MATH. Since the NAME equivalence of partial actions is transitive, we may replace MATH by the linking partial action of MATH (see the proof of REF in page REF). In the proof of REF , we constructed a right ideal MATH of MATH such tha... |
math/0007109 | Since the NAME equivalence of partial actions is transitive, by REF above it is enough to show that if MATH is an enveloping action of the partial action MATH and MATH is the natural action on MATH, where MATH is the associated Fell bundle of MATH, then MATH and MATH are NAME equivalent. Now, let MATH be the Fell bundl... |
math/0007109 | This is a direct consequence of REF. |
math/0007109 | A proof of REF may be found in CITE. To prove REF., note that the following diagram commutes: MATH . Since MATH has the initial topology induced by MATH, then MATH is precisely the topological quotient space of MATH with respect to MATH, and therefore MATH is the topological quotient space of MATH with respect to MATH.... |
math/0007109 | We identify MATH with MATH through the homeomorphism MATH. Thus MATH becomes: MATH, and if MATH, then MATH is the unique primitive ideal of MATH such that MATH. Let us see first that MATH: MATH . Now, if MATH, we have that MATH, and hence MATH. But MATH. Then MATH agrees with MATH; in particular MATH is a partial actio... |
math/0007109 | It is well known that if MATH, then MATH, and also that MATH is a homeomorphism (see for instance REF). Since MATH, it follows that MATH. Let MATH. Then MATH is an ideal of MATH, and therefore there exists a unique MATH such that MATH. We claim that MATH. Indeed, MATH, so MATH. Similarly, if MATH, we have that MATH. No... |
math/0007109 | Since MATH is the NAME enveloping action of MATH, this one is NAME equivalent to MATH, for some ideal MATH of MATH. By REF, MATH is a partial action on MATH. On the other hand, MATH is a homeomorphism, and therefore MATH is continuous, that is, MATH is a partial action on MATH. Since MATH is the enveloping action of MA... |
math/0007109 | By REF we have that MATH is the enveloping action of MATH and that MATH is the enveloping action of MATH. Suppose that MATH. Since the enveloping action is unique, we have that MATH. The converse is clear. |
math/0007109 | If MATH is unital, then MATH is compact by CITE, VII-REF Now, if MATH is compact, by REF there exists an open subgroup MATH of MATH such that MATH restricted to MATH is a global action on MATH, and therefore every primitive ideal of MATH is in the domain of MATH, MATH. By the definition of MATH, this implies that there... |
math/0007109 | We may assume, without loss of generality, that MATH non - degenerately, for some NAME space MATH. Therefore also MATH, and MATH. On the other hand, by REF, the inclusion MATH defines a faithful representation MATH such that, if MATH, MATH, then MATH. Now, if MATH, MATH, and MATH, MATH, we have: MATH where MATH. Thus, ... |
math/0007121 | The kernel of the natural map MATH is the torsion of MATH. The image of MATH under this map is contained inside a free MATH-submodule of MATH. In order to see this, let us consider a set of MATH-generators MATH of MATH, and a MATH-basis MATH of MATH. We can express the elements MATH as MATH-linear combinations of the M... |
math/0007121 | For MATH, we have: MATH which proves the first identity. The second one is proved in the same way. |
math/0007121 | For MATH we have: MATH . This proves REF . To prove REF, it is enough to show that MATH. This follows from the above equation and the fact that MATH. |
math/0007121 | Follows immediately from definitions. Indeed, if MATH, then: MATH . |
math/0007121 | Using an argument similar to that of CITE, we will show that if MATH satisfies REF then MATH. Then REF will imply the first claim, that MATH. Let MATH be a basis of MATH, and let us consider the corresponding NAME - NAME - NAME basis of MATH given by elements MATH, where MATH. In this basis the comultiplication takes t... |
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