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, for a positive integer MATH, is the matrix of the (positive, one-dimensional) operator on MATH defined by MATH . Notice that MATH for MATH in MATH . We may write MATH in tensor products of MATH matrices and MATH matrices. Now for MATH in MATH , we have MATH whence it follows that MATH . This means that MATH . Once we show that MATH is positive, it will follow by induction that MATH and hence MATH, is positive for all MATH. For MATH in MATH, we have MATH showing MATH as promised. |
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 or inverse generators. |
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 commutant MATH of MATH consists of scalars. Our argument makes essential use of a result of NAME (REF and especially REF) which implies that a nonzero linear combination of elements of MATH must have kernel zero in the left regular representation. CASE: If MATH, then MATH is orthogonal to MATH for each MATH. Proof of REF Since MATH is a MATH-eigenvector for MATH so is MATH . Thus MATH where we have used REF and the identity MATH. This makes MATH . Define MATH on MATH by setting MATH if MATH for some MATH, and MATH for all other MATH in MATH. Then REF for MATH in MATH because MATH is a MATH-eigenvector for MATH; REF for MATH by REF above, because MATH; and REF for all other MATH in MATH because for such MATH all of the values of MATH in the asserted relationship are zero. Because the vectors MATH that appear in the nonzero part of the definition of MATH are mutually orthogonal and norm-bounded, the function MATH belongs to MATH . The claim now follows from REF. CASE: For MATH in MATH and MATH in MATH we have MATH for each MATH, and hence MATH for all MATH in MATH. Proof of REF If furthermore MATH, we see using REF that MATH . Because the claimed relationship is linear in MATH, we can remove the assumption that MATH be selfadjoint. The second assertion follows by induction on the length of MATH. CASE: MATH for MATH in MATH. Proof of REF It will suffice to show that MATH for MATH and MATH in MATH. We proceed more or less as in the proof of REF . Fix MATH in MATH and the index MATH. Define MATH on MATH by setting MATH if MATH for some MATH and MATH for all other MATH in MATH. Thus MATH . As in the proof of REF , we will show that MATH must vanish by showing that it satisfies MATH for all MATH in MATH. This identity holds if MATH or if MATH ends in a positive generator power because all of the values of MATH that appear are zero. It holds if MATH for some MATH because MATH is a MATH-eigenvector for MATH. The only remaining case is MATH for which we must show that MATH . But MATH where we have used REF above, and MATH. It follows from REF that MATH is orthogonal to MATH for every MATH in MATH, and hence MATH by REF , for all MATH in MATH. Because MATH is cyclic for MATH, this proves that MATH is irreducible. |
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 annulus. Let MATH denote the reduced operator norm. Then for every positive integer MATH we have MATH where the lower bound is obvious and the upper bound follows from NAME 's inequality CITE plus the observation that MATH is a linear combination of words of length MATH. Since the coefficient of a MATH-fold product of generators in MATH is the corresponding product of MATH's, we have MATH so MATH . This shows that the reduced spectral radius of MATH is REF-norm of the coefficient vector. It remains to show that we have correctly identified the inner radius MATH which we do by induction on the number of nonzero coefficients. In the case of only one nonzero coefficient, say MATH, the spectrum is the circle about REF of radius MATH which plainly coincides with MATH. Suppose that the proposition gives the correct inner radius when there are MATH nonzero coefficients. Without loss of generality, we may assume MATH where MATH . Let MATH. There are two cases: CASE: MATH; and REF the contrary. In REF , the inner radius of the reduced spectrum of MATH is at most REF by our induction hypothesis (plus the assumption that the MATH's are all at most REF) and the outer radius is at least REF by the spectral radius formula already established. Thus MATH is not invertible in the left regular representation, giving MATH as required in this case. In REF , let MATH . If MATH then the reduced spectral radius of MATH is less than REF, and hence MATH is invertible in the left regular representation. This shows that the reduced spectrum of MATH has inner radius at least MATH. On the other hand, MATH is singular in the left regular representation because MATH has reduced spectral radius precisely REF. This shows that the reduced spectrum of MATH has the correct inner radius in REF , completing the proof. |
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 most only the first two symbols in its argument. CASE: Suppose MATH for some MATH (and hence MATH). Observe that MATH is constant on MATH, in fact on MATH. We have MATH so MATH is constant on MATH, and the remaining factors are constant there for the same reason as in REF . Likewise if MATH (and hence MATH), the two initial factors MATH and MATH are constant on MATH. The remaining factors are constant on MATH for the same reason as in REF . |
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 in the various cases that arise, we obtain MATH . Using MATH, this makes MATH . Since MATH the second term is simply the MATH times the integral of MATH over MATH. We are done with the first part of the proof. CASE: Next we show that MATH if MATH is a reduced word not beginning with MATH (including the possibility MATH). To begin with, MATH . Use REF to find MATH such that MATH for MATH in MATH. Looking up values for MATH in the various cases, we obtain MATH . Evaluating the MATH's gives MATH and we are done with the second part of the proof. CASE: Finally we show that MATH if MATH is either REF or a reduced word ending with MATH for some MATH. Let MATH . Then MATH . By REF , we have a MATH such that MATH in the first two integrals. Thus MATH since MATH . As we observed in REF , and REF above suffice to establish the asserted formula for 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 identification of MATH as MATH above. Finally, it follows from REF that MATH is reduced. |
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 all MATH in MATH, while if MATH ends in MATH for some MATH different from MATH, we have MATH for all MATH in MATH, which finishes the case MATH. |
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 summands above are equal, or the MATH summands are. Hence both are equal, and REF for MATH follows from REF for MATH by multiplying by MATH . |
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. Since MATH this makes MATH that is, MATH vanishes on each MATH. Now fix distinct indices MATH and MATH, and let MATH if MATH in MATH and REF otherwise. As with the MATH's, we have MATH so MATH must be a multiple of MATH, but MATH so MATH . Continuing in this fashion, we see that MATH must be supported on MATH, and it then follows by equating coefficients that MATH and in general MATH . This makes MATH so MATH so MATH is invertible by REF above. If MATH then MATH for some MATH. We have MATH and MATH so we may apply the previous case to the free generators MATH . |
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 that MATH intertwines MATH and MATH. For MATH in MATH we have MATH while MATH . It is readily checked that MATH and MATH are the same for all MATH, namely MATH . (One must check two subcases in each of the two cases above.) The intertwining of each MATH with its inverse, and thus of MATH with MATH now follows. Now let MATH be the closed MATH-invariant subspace of MATH generated by MATH and MATH. REF will both follow once we show that MATH. Since MATH and MATH are linearly independent linear combinations of MATH and MATH, these two indicator functions must both belong to MATH. We have MATH so MATH for each MATH. It will follow that each MATH once we show that the matrix MATH is invertible. Multiplying row MATH by MATH for each MATH and using MATH turns this into MATH whose determinant by REF in the appendix is MATH . We conclude that MATH for each MATH. The unitary MATH preserves MATH and sends MATH to MATH, so MATH as well. Suppose we have shown for some MATH that MATH for all reduced words MATH of length MATH. For such a word MATH not beginning with MATH we have MATH so MATH . Using the unitary MATH, it follows that MATH as well if MATH doesn't begin with MATH. We conclude that the indicator function of every cylinder set belongs to MATH, showing that MATH . This takes care of REF . To prove REF , it will suffice to show that MATH cannot have MATH as an eigenvalue in MATH . We will think of MATH as acting on the NAME space MATH constructed in REF with MATH and MATH as in REF . Thus, MATH is the closed linear span of MATH, while MATH is MATH and so forth. Suppose now that MATH in MATH satisfies MATH . We claim first that MATH must be orthogonal to each subspace MATH. Fix MATH, and MATH in MATH. Define MATH in MATH by MATH . Notice that MATH for every MATH in MATH. We obtain MATH by checking the cases MATH (use MATH), MATH (use MATH and MATH), and MATH separately. This means that MATH . The assumption is ambient that MATH (and hence its adjoint) is not invertible in the left regular representation, so MATH by REF , and thus MATH by REF. We have so far shown that MATH and that MATH (that is, MATH) for each MATH. Take MATH, for instance MATH and MATH. Using REF , we obtain MATH . This forces MATH because MATH cannot be a scalar multiple of MATH. (The scalar in question would have to have modulus REF because these are unit vectors, and would have to be MATH to make the inner product with MATH come out right. However, MATH because MATH is given by the formula in REF .) |
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 REF above that MATH so MATH vanishes identically by CITE. We have shown that MATH must belong to MATH. By REF , then, we have MATH and similarly for MATH in MATH and any index MATH . We can't have MATH because that would force MATH for all MATH in MATH and hence MATH . Hence MATH. We may assume that MATH . The preceding formula becomes MATH whence it readily follows by induction on the length of MATH that MATH for every MATH in MATH, and thus that MATH . |
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 state MATH on the algebra of bounded operators on MATH such that MATH . We will be done once we show that the restriction of MATH to MATH must coincide with MATH. Let MATH be the set of reduced words in MATH beginning with some MATH, and let MATH. Let MATH and MATH be respectively the orthogonal projections of MATH on MATH and MATH, so MATH. Suppose we know that MATH . For a MATH in MATH with MATH (that is, for MATH not in MATH), we have MATH for sufficiently large MATH, and hence MATH. On the other hand, if MATH and MATH is reduced, then MATH because MATH for MATH. Likewise MATH if MATH is reduced. It now follows easily that MATH for MATH in as well as outside of MATH . We show now that MATH must be MATH. Suppose not, that is, suppose MATH . Consider the state MATH defined on bounded operators MATH by MATH so MATH . We have MATH and MATH . Because MATH is in the left kernel of MATH, this makes MATH . The same argument as in the previous paragraph, mirror-imaged by the automorphism MATH of MATH that sends each MATH to MATH, shows that MATH . In particular, we have MATH for MATH . Consider now MATH. We have already observed that MATH so MATH and likewise MATH . Since also MATH it follows that MATH for MATH. All of these quantities are nonnegative, and at least two are positive. This, however, contradicts MATH because the latter forces the sum over unequal MATH and MATH of MATH to vanish. |
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 most one MATH can exceed the sum of the others, all but one of the MATH's must be positive. Assume for definiteness that MATH for MATH . Writing MATH as MATH, we have MATH . Now MATH which is positive for MATH because MATH and MATH . |
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 some MATH, or MATH for some MATH. Suppose that MATH for some MATH. Notice that this forces MATH (eventually along the sequence), else MATH . If MATH then MATH, and we are done. If MATH we may write MATH and conclude that MATH . This makes MATH for every MATH, and thus MATH for every MATH. Since MATH we must have MATH for every MATH. Since MATH it follows that MATH so MATH . The remaining possibility is that MATH for some MATH. It follows that MATH . Since MATH for MATH we also have MATH for MATH. But MATH so MATH which rules MATH out of MATH. |
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 coincide for every MATH except possibly the one for which MATH which means they must coincide for this MATH as well. In order for MATH and MATH to be different, then, we must have MATH . Assume for definiteness that MATH so both sides of REF are greater than zero. Since the MATH's and MATH's both sum to REF, we must have MATH for some MATH. For this MATH, we must have MATH and MATH . Because MATH, it follows that MATH, and hence MATH, is greater than REF. |
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 course MATH so as well MATH . It follows that the images of MATH under MATH and MATH must coincide. (In other words, not surprisingly, the hypersurfaces MATH and MATH must be tangent to one another at MATH.) The observation at the beginning of the proof of REF shows that MATH acts on MATH by multiplying the MATH entry of a vector in MATH by MATH. It follows that if MATH is orthogonal to MATH then MATH is the same for all MATH. Describing the normal to MATH in similar fashion, we deduce from MATH that there is a real number MATH such that MATH for every MATH. Summing on MATH, we obtain MATH . In case MATH this makes MATH and REF simply means that MATH . If we add this to the inequality MATH we obtain MATH . Likewise, MATH . For MATH we have MATH and the linear system REF has the form MATH where MATH is easily seen to be invertible with MATH . Thus MATH and hence in this case as well we have MATH for each MATH. By REF , this contradicts MATH . |
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 the quadratic polynomial defined by MATH . Notice that MATH has no real zeros when MATH . Let MATH be either REF if MATH has no real zeros, or the larger real zero of MATH. That is, MATH . In either case, MATH. (Since MATH makes MATH, the smaller zero of MATH is less than REF in this case.) Define functions MATH and MATH on the interval MATH by MATH . Because the MATH's sum to REF, we must have either MATH or MATH . For MATH the derivative of MATH is MATH . Because MATH and MATH we have MATH . Thus MATH is strictly decreasing on MATH . If MATH the sign choice MATH is ruled out for MATH as well as for the other MATH's, so we must have MATH and the lemma is proved in this case. Assume, then, that MATH . We have MATH all terms of which are negative. Since MATH and MATH is strictly decreasing on MATH, it must be that MATH is strictly increasing on MATH. Since MATH and the two functions in question are strictly monotone in opposite senses, the condition that one or another vanish at MATH determines MATH uniquely. As we saw in the proof of REF , this in turn determines MATH uniquely. |
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 nonempty. Let MATH be the infimum of the numbers MATH that come from pairs in MATH. A sequence in MATH along which the sum of the coordinates in the MATH-slot tends to MATH must by REF have a subsequence converging to a pair MATH, where at least one coordinate in each of MATH and MATH vanishes (and because MATH along the sequence, the zeros occur in same-indexed coordinates). If the number of nonzero coordinates in MATH and MATH is at least two, then the induction hypothesis is contradicted. If this number is one, then REF is contradicted. There must of course be at least one nonzero coordinate because the sum of the coordinates of MATH is MATH . |
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 components of the exceptional divisor of MATH. We will use the notations of CITE for the blowup diagram MATH where MATH is the exceptional divisor of the blowdown morphism. We also have MATH and MATH. Discrepancies of the exceptional divisors of these morphisms are related by MATH where MATH is the multiplicity of MATH along MATH and MATH is the codimension of MATH in MATH. We will use for a while the following technical assumption. MATH . We have the following exact sequences of coherent sheaves on MATH, see REF. MATH . Here MATH is the tautological quotient bundle on MATH. This implies MATH where MATH and MATH. Note also that MATH. Therefore, MATH where MATH. We will now use MATH. We write the NAME expansion MATH of the expression under MATH in the above identity. Observe that MATH is exactly the class in MATH whose integral is MATH, so we need to show that the contribution of the rest of the terms vanishes. Notice that MATH for MATH and MATH where MATH is NAME polynomial of a vector bundle, see CITE. Hence, one needs to calculate MATH . We denote MATH, MATH and use the fact that MATH to rewrite REF as MATH . Here we denote MATH and use MATH. To show that REF is zero, observe that the function whose coefficient at MATH is measured, is elliptic in MATH. Really, MATH obviously keeps it unchanged, and MATH does not change it, because of MATH. We have used here the fact that none of the MATH-s is equal to MATH, which follows from the condition that MATH is log-terminal, see for instance CITE. It remains to show that MATH is the only pole of this function up to the lattice MATH, so the residue is zero. To do so, observe that the normal crossing condition implies MATH, and moreover, whenever MATH the corresponding factor MATH in the denominator of the last product is offset by a factor MATH in the numerator of the second product. We will now get rid of REF . Indeed, it is easy to see that the difference between MATH and MATH can be written as a degree of an element of MATH that is preserved when one deforms MATH to the embedding of MATH into the normal cone, for which REF is satisfied. |
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 implies that MATH . Then elliptic genera defined via MATH and MATH coincide by definition. |
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 to get MATH so that the exceptional sets of MATH and MATH are divisors with simple normal crossings. Then singular elliptic genera of MATH and MATH calculated via MATH are given by the same formula, because the discrepancies coincide. |
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 components of MATH, and is assumed to have normal crossings. Polynomials MATH are defined in terms of mixed NAME structure on the cohomology of MATH, see CITE. Subvariety MATH is MATH, and the sum includes the empty subset MATH. For each MATH where MATH. The adjunction formula for complete intersections yields MATH where MATH is the closed embedding. We then obtain MATH where MATH. When we plug this result into NAME 's formula, we get MATH . |
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 REF . A rather lengthy but straightforward calculation, similar to that of REF, shows that MATH . |
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 therefore has a global section. Global sections of the pluricanonical bundle are birational invariants, so one can consider the divisor MATH of this section on MATH. It is easy to see that for both morphisms MATH and MATH the exceptional divisor is MATH, which we can then assume to have simple normal crossings (perhaps by passing to a new desingularization). Therefore, elliptic genera of MATH are calculated on MATH using the same discrepancies. |
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 theorem. To verify the modular property, we denote MATH where MATH is a character of the subgroup of MATH generated by MATH and MATH. Then MATH where we suppress MATH from the notations for the sake of brevity. We have: MATH and hence MATH, since by REF. It is clear that MATH and hence MATH. We have: MATH and hence MATH since MATH is NAME and NAME. Finally, MATH . Then the NAME transformation properties follows easily from REF, similar to the proof of REF. It is straightforward to see from REF that orbifold elliptic genus is holomorphic and has the NAME expansion with non-negative powers of MATH. |
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 embedded into MATH. The centralizer group is a semidirect product of its normal subgroup MATH which acts by cyclic permutations inside cycles of MATH and the product of symmetric groups MATH that act by permuting cycles of the same length. Our definition of the elliptic genus then gives MATH . The symbol MATH here should be interpreted as the supersymmetric product where the cohomology of MATH is given parity by the sum of the cohomology number and the parity of the exterior algebras. We will now calculate MATH . We denote the MATH-cycle by MATH and observe that MATH . This implies MATH, which allows us to write MATH . Here we have denoted the primitive MATH-th root of unity by MATH. Now REF finishes the proof. |
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 map MATH has ramification if and only if for some one-dimensional rays of MATH points MATH are no longer minimal in the new lattice. NAME divisors on a toric variety correspond to piecewise linear functions on the fan. It is easy to see that the definition of MATH assures that the piece-wise linear linear function that takes values MATH on all MATH gives the divisor MATH. We denote this piece-wise linear function by MATH. One can show that MATH where MATH is the function defined in CITE. More explicitly, MATH where MATH means analytic continuation. The proof of this fact is based on the explicit calculation of the NAME characteristics of the bundles MATH by means of NAME cohomology. The calculation is very similar to that of REF and is left to the reader. We remark that the sum over MATH in REF facilitates the change from MATH to MATH, while taking MATH-invariants is responsible for the switch from MATH to its sublattice MATH. Now let MATH be a toric desingularization of MATH given by the subdivision MATH of MATH. We denote the codimension one strata of MATH by MATH, and the generators of the corresponding one-dimensional cones of MATH by MATH. We also denote the first NAME classes of the corresponding divisors by MATH and get MATH where MATH and MATH. We use MATH to rewrite MATH as MATH which equals MATH by REF. We have used here the fact that MATH does not change when the fan is subdivided. |
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 contributions of points, because it is the NAME characteristics of the bundle on the quotient that is the usual elliptic genus bundle twisted at the ramification points. Since the equality holds in the toric case of the MATH-fold covering of MATH by MATH, which has two points of ramification MATH, the extra terms for two genera coincide, which finishes the proof. |
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 MATH. In addition, one can replace MATH by MATH to get MATH . Let MATH be a refinement of the fan MATH in MATH such that the corresponding toric variety MATH is smooth. Coefficients MATH define a hypersurface MATH in MATH which is a resolution of singularities MATH. We denote the codimension one strata of MATH by MATH, their first NAME classes by MATH and the corresponding generators of one-dimensional cones of MATH by MATH. By REF, we get MATH . We denote MATH and MATH. Because of REF, we get MATH which gives MATH . Observe now that MATH where MATH is the resolution induced by the subdivision of the fan. In addition, MATH is a zero set of a section of MATH. Hence, the adjunction formula gives MATH where MATH is the embedding. The exceptional divisors of MATH are MATH (unless MATH), and their discrepancies are equal to MATH. Then it is easy to see that the above expression is precisely the singular elliptic genus of MATH. |
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 (that is, MATH is a complex subbundle in MATH) satisfying the following normal crossing condition: at each point of MATH the union of (stabilized by adding trivial bundles) tangent spaces MATH is given in the (stabilized) tangent space to MATH by MATH where MATH are linearly independent complex linear forms. A pair MATH is cobordant to zero if there exist a MATH-manifold MATH with a complex structure on the stable tangent bundle and a system of submanifolds MATH such that MATH and MATH. As usual, the disjoint union and product provide the ring structure on cobordism classes. Notice that the numbers MATH where MATH are the classes in MATH dual to submanifolds MATH, MATH is the fundamental class of MATH and MATH are invariants of cobordism of such pairs (indeed, if MATH and MATH then this number is MATH since MATH is homologous to zero in MATH). The lemma therefore will follow if we shall show that for an almost complex null-cobordant MATH-manifold MATH the quotient MATH admits a resolution of singularities MATH, where MATH is the exceptional locus, such that this pair is cobordant to zero. If MATH, where MATH is a MATH-manifold, we can construct resolution of MATH as follows. Let MATH be a subgroup of MATH and MATH. Then MATH are smooth submanifolds of MATH (possibly with boundary) providing a stratification of MATH. Let MATH be the union of subgroups of MATH conjugate to MATH. Then MATH is still a submanifold of MATH and the group MATH acts on MATH so that MATH is an unramified cover (of degree MATH). In particular, MATH is a smooth manifold and these manifolds for all MATH provide a stratification of MATH such that MATH is equisingular along each stratum MATH. A small regular neighborhood of each stratum in MATH is isomorphic to a bundle MATH over MATH with the fiber isomorphic to MATH where MATH is a fiber of the normal bundle to MATH in MATH over a point of MATH (this presentation is independent of a point in MATH and representations at points of MATH and MATH are isomorphic for conjugate MATH and MATH). For each quotient singularity MATH let us fix the universal desingularization constructed by NAME (compare REF). Its universality assures that it is equivariant with respect to the centralizer of MATH in MATH. Hence one can use the transition functions of MATH to construct the fibration MATH with the same base as MATH and having as its fiber the universal resolution of MATH. Moreover, due to universality of canonical resolution (compare REF) this property assures that MATH corresponding to different classes of conjugate subgroups MATH can be glued together yielding an almost complex manifold which boundary is the pair MATH where MATH is the exceptional set of the universal resolution of MATH. This proves the lemma. |
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 MATH evaluated on the fundamental class of MATH this contribution is zero. This yields the lemma. |
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 symmetric and MATH, we have that MATH is an open subgroup of MATH contained in MATH. As for the last assertion, just recall that the unique open subgroup of a connected group is the group itself. |
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 space MATH. Let MATH be the quotient map, and define MATH such that MATH. Since the inclusion MATH given by MATH is continuous, we have that MATH also is. Moreover, if MATH, MATH so MATH is a morphism. We must prove now that the pair MATH has the claimed universal property. Note first that if MATH is a continuous action and MATH is any continuous function, then the map MATH such that MATH is a morphism MATH. Moreover, if MATH is also a morphism, then MATH is compatible with MATH: if MATH in MATH, since MATH, we have: MATH and therefore: MATH. Thus MATH induces a continuous map MATH, such that MATH, MATH, MATH. We have that MATH, and it is also clear that MATH is a morphism, uniquely determined by the relation MATH. Since the pair MATH is characterized by a universal property, it is unique up to isomorphisms (In categorical terms, MATH is a universal from MATH to MATH, where MATH is the forgetful functor from the category of actions to the category of partial actions; see CITE for details). It remains to prove the last three assertions of the statement. The third of them is clear, because MATH, MATH, MATH. As for the first and second, note that MATH is clearly injective, so it suffices to show that it is an open map. Let MATH be an open subset. We have to show that MATH is open in MATH. But MATH, which is open in MATH because MATH is continuous, and hence open in MATH because MATH is open in MATH. |
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 of them containing MATH or MATH, then MATH. Since MATH is a homeomorphism of MATH, by REF we may suppose that MATH. Let MATH, MATH be such that MATH. If every neighborhood of MATH intersects every neighborhood of MATH, then for any pair MATH of neighborhoods in MATH of MATH and MATH respectively, there exists MATH, say MATH, with MATH. Consider the net MATH: it converges to MATH, so MATH, because MATH is closed. Hence MATH, and MATH is NAME. |
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 MATH. Since MATH is a commutative ideal of MATH, MATH is contained in the center MATH of MATH: if MATH, MATH, and MATH is an approximate unit of MATH: MATH . Since MATH is invariant, it follows that MATH, MATH, thus MATH. But then MATH, and therefore MATH, so MATH is abelian. |
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, MATH. Thus, we get a homomorphism MATH, and therefore MATH, MATH. But MATH. Finally, it is clear that MATH is injective if and only if MATH is an essential ideal of MATH. |
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 is an isometry of a *-dense ideal of MATH onto a *-dense ideal of MATH, and therefore it extends uniquely to an isomorphism MATH, which clearly satisfies MATH, MATH, and MATH. Moreover, it is clear that these conditions determine 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 that MATH. Suppose in addition that MATH is a separable MATH-algebra, and let MATH be a countable and dense subset of MATH. Since MATH, whenever MATH, there exists an increasing sequence MATH such that MATH, MATH. It follows that MATH is an AF - ideal that contains any ideal of MATH, and this proves our first assertion. As for the second one, note that, since MATH is separable, then MATH is separable if and only if so is MATH. Now, by the first part, the result is proved as in REF. |
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 subnet that is convergent to some MATH. We may assume without loss of generality that the net itself converges to MATH. But this is a contradiction, because: MATH . The contradiction implies that the Lemma is true. |
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 particular, since MATH, MATH . Thus MATH, and therefore MATH. Moreover, the representation MATH is continuous in the norm MATH: MATH . It follows that MATH, and hence we may extend MATH by continuity to a representation of MATH. This representation is non - degenerate, because MATH is dense in MATH in the inductive limit topology, and therefore also in MATH. |
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 and MATH, we obtain representations of MATH and MATH, which we also call MATH and MATH respectively, and it is clear again that MATH agrees with MATH. Now, by REF (see also the end of REF ), we have isomorphisms MATH, and MATH. Therefore, MATH is an isomorphism. Thus, MATH is naturally identified with MATH in 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 section with MATH, so we have a function MATH defined by MATH, that is supported in MATH, and that is continuous. In fact, let MATH. Since the function MATH that maps MATH into MATH is continuous and equal to zero in every MATH, for MATH, there exist open neighborhoods MATH of MATH, MATH of MATH, such that MATH, MATH, MATH. Since the MATH cover the compact set MATH, there exists a finite subcovering MATH of MATH. Let MATH and pick MATH, MATH. Then either MATH, and then MATH, or MATH, and therefore MATH for some MATH. In this case, MATH, and hence MATH, so MATH. It follows that MATH is continuous, and hence uniformly continuous, because it has compact support. Thus, for any MATH, there exists MATH such that if MATH, then MATH. So we have, for any MATH such that MATH, and for all MATH: MATH . Therefore, since MATH, MATH and MATH, MATH, it follows that MATH in the inductive limit topology, MATH. |
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 and the NAME theorem, MATH . On the other hand, if MATH, MATH, MATH: MATH because MATH, MATH. It follows that MATH, and hence that MATH. Consider now MATH, MATH, MATH. We have: MATH . Let MATH be an approximate unit of MATH as in REF . For MATH, MATH, MATH, define MATH by MATH . Then MATH, and we have: MATH where MATH is such that MATH. Note that MATH is continuous of compact support: MATH. By REF , we see that MATH in the inductive limit topology MATH, and hence also in MATH. So, we have that MATH, where MATH. To see that MATH, it is sufficient to see that MATH is dense in MATH in the inductive limit topology. By II-REF, for this is enough to verify that: CASE: MATH is dense in MATH, MATH, where MATH and REF if MATH is continuous, then MATH, MATH. CASE: We have: MATH. Therefore, MATH by the hypothesis in REF. CASE: Let MATH be a continuous function, MATH, MATH, MATH. Defining MATH as MATH we have: MATH . Since MATH, we conclude that MATH, closing the proof. |
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 REF. |
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 equal to MATH. The amenability of MATH implies that these two norms are equal. Let MATH. Since MATH by assumption, and MATH by REF, we have: MATH and therefore MATH. Then, the completions of MATH with respect to MATH and MATH agree. Let us denote this completion by MATH. We have that MATH is a full NAME module over both MATH and MATH, and hence MATH. This shows that MATH is amenable. |
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 to MATH. Consequently, MATH. CASE: MATH because MATH is an ideal and MATH. Thus, MATH. Thus, we may apply REF , concluding that MATH is a hereditary sub-MATH-algebra of MATH. Suppose now that MATH is the enveloping action of MATH, that is, MATH is dense in MATH. To see that MATH, it is sufficient to show that MATH is dense in MATH, for all MATH. Now: MATH. Therefore MATH. By the NAME -Hewitt theorem, every MATH may be written in the form MATH, for some MATH. So the set above is exactly MATH, which is dense in MATH. |
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, let MATH be defined as MATH, where MATH is the NAME symbol. By REF, MATH defines a unique non - degenerate representation MATH, such that MATH, MATH. In particular, MATH. By REF , we have that MATH is a hereditary sub - MATH-algebra of MATH, which is equal to MATH because of the amenability of MATH. Moreover, MATH is NAME equivalent to MATH. Therefore, there exist a NAME space MATH and a non - degenerate representation MATH, such that MATH is a NAME subspace of MATH and MATH is the compression of MATH to MATH, that is : MATH, MATH, where MATH is the orthogonal projection and MATH is the natural inclusion (CITE, XI-REF). Now, MATH, for some covariant representation MATH of the dynamical system MATH; in particular, MATH is a unitary representation of MATH. Note that MATH is a clopen subset of MATH, so MATH, and we may compute, in MATH: MATH . Therefore: MATH . Observe now that MATH is an orthogonal projection such that MATH, and hence MATH is greater or equal to the orthogonal projection MATH with image MATH. Thus, we have that: MATH. On the other hand, it is clear that MATH, MATH, and consequently: MATH . It follows that 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 contraction. In particular, MATH is continuous, and therefore MATH, MATH, MATH, because this is true for each MATH, a dense subset of MATH. If MATH is surjective, it is clear that so is MATH. Suppose that MATH is injective, and let MATH. Then MATH, MATH, and hence MATH, MATH, and therefore MATH. It follows that MATH, and hence that MATH. Consequently MATH is injective, and therefore isometric. Thus: MATH so MATH is isometric. |
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 follows that the domain of MATH is MATH, and hence that MATH. Therefore, MATH is an extension of MATH. |
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. Then MATH, so MATH, and hence MATH is open. |
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 norms, we have: MATH . Thus, MATH factors through the quotient, where it is a contraction, and hence extends by continuity to a homomorphism of MATH-trings MATH. We have, MATH: MATH, and therefore MATH. Similarly, MATH. Finally, if MATH, MATH are isomorphisms, we apply the first part of the proof to the maps MATH e MATH, and we note that MATH, MATH. |
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 the partial action MATH of MATH on the MATH-tring MATH. Consider the family MATH. Since MATH extends MATH and MATH extends MATH, it follows that MATH extends MATH. It is clear that MATH, MATH. On the other hand: MATH: MATH . Similarly, MATH. Finally, by REF, every MATH is an isomorphism, and MATH, MATH, so MATH is a set theoretic partial action on MATH, and MATH, MATH. It remains to show that MATH is continuous. First, note that the family MATH is continuous by REF, because MATH is a partial action. Let MATH. Note that if MATH, MATH, then MATH such that MATH is a continuous section of the NAME bundle MATH, and that for each MATH, MATH is dense in MATH. On the other hand, the map MATH such that MATH is continuous, MATH, MATH. So we conclude, by CITE, II-REF and II-REF, that the application MATH such that MATH is a continuous homomorphism of NAME bundles, and therefore MATH is a continuous partial action. Thus MATH. |
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 an ideal MATH, and it is easy to see that MATH, MATH. Since MATH, and MATH, MATH, MATH and MATH are continuous families, so it is the family MATH. Let us define now MATH by MATH . Since MATH and MATH, MATH, MATH, MATH, we have that MATH is a partial action of MATH on MATH. We call MATH the linking partial action of MATH. Observe that if MATH is the restriction of MATH to MATH, then MATH and MATH. Similarly, it is enough to restrict MATH to MATH to see that it is also NAME equivalent to MATH. The considerations above show that we may assume that MATH, for some projection MATH, and that MATH and MATH are the restrictions of MATH to MATH and to MATH respectively. Let now MATH and MATH be the Fell bundles corresponding to MATH and MATH respectively, and let MATH. We have that MATH, MATH is a sub - Fell bundle of MATH, and that MATH is a sub - NAME bundle of MATH. On the other hand, if MATH, MATH, and MATH, we have: CASE: MATH. CASE: MATH. CASE: MATH, that belongs to MATH. CASE: MATH. Now, for all MATH, we have that MATH: the left member of this equality contains MATH. Therefore, we may apply REF , from where we conclude that MATH. |
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 is continuous and has compact support, and that MATH, MATH. Thus, we may apply CITE, II-REF, from where we conclude that the map MATH is a continuous section of compact support of MATH. In other words, MATH. As for MATH, we have that MATH is compact, and MATH. Consequently, MATH. Routine computations show that MATH, MATH, so MATH is a *-algebra. It is also immediate that MATH. Finally: 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 such that MATH. Since MATH and the norm on MATH are continuous maps, there exists a precompact open neighborhood MATH of MATH, such that MATH, MATH. The family MATH is an open covering of MATH, so it has a finite subcovering MATH. Let MATH be a partition of unity of MATH subordinated to MATH, and let MATH. We have that MATH and MATH. Therefore, it is enough to show that MATH. For this, it is sufficient to show that MATH, MATH and MATH. But, since MATH is dense in MATH, there exists MATH such that MATH, for a given MATH, and hence MATH. Since MATH, we have that MATH, and therefore MATH. |
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 such that MATH. There also exist sections MATH such that MATH, MATH. Thus we have: MATH. Let MATH, MATH, MATH, MATH, MATH. If MATH, let MATH be given by MATH. Then: MATH and therefore MATH, because MATH, 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 inductive limit topology. Conversely, assume that MATH is not saturated: there exist MATH such that MATH. Let MATH be such that MATH, and let MATH be the distance from MATH to MATH. There exists a continuous section of compact support of MATH that takes the value MATH in MATH. In other words, there exists MATH such that MATH. Now, if MATH, MATH: MATH and therefore MATH, because MATH. |
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 that MATH extends to a bounded operator MATH, and MATH extends by continuity to a bounded linear map MATH. In addition, MATH: if MATH, MATH we have that MATH . As for the involution, we have that MATH. In fact, if MATH, MATH, MATH: MATH . It follows that MATH is a bounded representation. Now, if MATH, MATH, MATH: MATH and hence MATH. Suppose that MATH is faithful. If MATH, for MATH, we would have that MATH. In particular, if MATH is an approximate unit of MATH like in REF, and if MATH, choosing MATH and taking limit with respect to MATH, by REF we have that MATH, and therefore that MATH, MATH. Hence MATH is also faithful. Finally, let us suppose that MATH is non - degenerate, and let MATH such that MATH, MATH, MATH. Then MATH, MATH, MATH. Thus MATH, MATH, MATH, so MATH almost everywhere in MATH, and therefore MATH almost everywhere in MATH, because MATH. In fact, since MATH is non - degenerate, MATH is also non - degenerate, and hence, given MATH, there exist MATH and MATH, such that MATH. Now there exist MATH, MATH such that MATH, and sections MATH, MATH such that MATH, MATH . Define MATH such that MATH. It is clear that MATH. Finally, taking MATH such that MATH we have: MATH . Consequently MATH in MATH, and therefore MATH is non - degenerate. |
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 MATH is determined by the formula MATH. From this we see that MATH, MATH. Indeed, if MATH, MATH: MATH . Therefore, MATH is a homomorphism of positive MATH-trings. |
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 MATH such that MATH is an isomorphism of MATH-trings. Consequently, the application MATH such that MATH is an isomorphism of MATH-algebras. Now, by REF , we have that MATH, and hence MATH. Moreover, if MATH, MATH: MATH . This shows that the map MATH such that MATH, may be extended by continuity to an adjointable operator MATH. This way we have constructed a map MATH. On the other hand, if we define MATH to be the restriction of MATH to MATH, we have that: MATH. Since MATH is injective, it follows that MATH is injective and that the norm of MATH in MATH agrees with MATH. This proves REF. and REF. It remains to prove REF. Let MATH, MATH, MATH: CASE: MATH: MATH . CASE: MATH: MATH . CASE: MATH: MATH . CASE: MATH: MATH . |
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, then MATH. Let MATH be an approximate unit of MATH as in REF , and for each MATH let MATH. Then MATH, and MATH, because if MATH: MATH . Now, for each MATH consider the kernel MATH given by MATH. If MATH: MATH . Thus MATH in the inductive limit topology, and hence MATH. |
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 MATH is an ideal of MATH, it is contractive (by CITE, VI-REF). In consequence we have, MATH: MATH and therefore MATH . |
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 norm on MATH induced by MATH, that is, MATH, where we are considering MATH as an element of MATH. This is a MATH - norm on MATH, because by completing it with respect to this norm, one obtains a NAME module over MATH (see CITE). Let MATH be the norm of MATH as an element of MATH. Recall that MATH, where MATH is the adjointable map given by MATH. So we have that MATH, because MATH. Now, by REF above (with MATH) it follows that MATH. Therefore MATH, and since MATH and MATH are the enveloping actions of MATH and MATH, then MATH, because of the uniqueness of the enveloping action REF . |
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 follows now from REF , |
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 in the inductive limit topology. So REF. is proved. CASE: This is an immediate consequence of REF. CASE: We must show that MATH. Since MATH, we have that MATH. To prove the converse inclusion, by REF it is enough to show that MATH. Let MATH. Then MATH, for some MATH. For a given MATH, let MATH be such that MATH. Then MATH, because MATH is an isometry. Consider now MATH. We have: MATH . By REF., MATH, and therefore MATH from where it follows that MATH. By the continuity of the product, it follows now that MATH, MATH, and this ends the proof. |
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, MATH. On the other hand, if MATH and MATH, we have that MATH. Thus: MATH . It follows that MATH is an isomorphism of MATH-trings, and MATH, which proves REF. Let us see that MATH is a set theoretic partial action: it is clear that MATH and MATH. Suppose now that MATH is such that MATH. Then: MATH . Then MATH is an extension of MATH, and therefore MATH is a set theoretic partial action on MATH. So, REF. is proved, up to the continuity of MATH, that will be proved later. Since MATH by REF, the computation that showed that MATH was an isometry also implies that MATH. To see that MATH, observe that if MATH, MATH, by REF we have: MATH . Hence MATH is an extension of MATH, MATH. By REF , we have that MATH, MATH, from where we conlude that MATH. Then REF. is proved. It remains to show the continuity of MATH. We will prove that if MATH is a net in MATH that converges to MATH, with MATH, MATH, then MATH. Let MATH. By the NAME - NAME theorem, there exist MATH, MATH, such that MATH, and we may suppose that MATH. Since MATH is dense in MATH, there exists MATH such that MATH. Therefore, MATH . Since the family MATH is continuous, there exists MATH such that MATH. Let us define MATH. In particular, MATH. Note that MATH, because MATH, and MATH, MATH. Since the action of MATH on MATH is continuous, and since MATH is continuous, we have that MATH in MATH. Therefore, there exists MATH such that MATH we have: MATH (In fact, we even have that MATH in the inductive limit topology, because MATH). Let MATH. Since MATH we must have: MATH . Then MATH. Hence the map MATH such that MATH is continuous and has compact support. Therefore the map MATH given by MATH is continuous. It follows that MATH when MATH. So we also have that MATH if MATH. Hence there exists MATH such that MATH: MATH . On the other hand, since MATH, there exists MATH such that MATH we have: MATH . Let MATH. Since each MATH is an isometry, if MATH, by REF we have that: MATH . This proves that MATH is continuous, which ends the proof. |
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. The equality follows from the NAME - NAME theorem and from the inclusion MATH . CASE: Let MATH, and consider the dense subalgebra MATH of MATH. If MATH, MATH, MATH, MATH, then: MATH where MATH. Since MATH, MATH, MATH, we see that MATH. Now, let MATH be such that MATH, with MATH and MATH, for some MATH. Let MATH, MATH be such that MATH, MATH, so MATH. Let MATH be a base of neighborhoods of the identity of MATH, and MATH an approximate identity of MATH as in REF. For each MATH consider MATH such that MATH. The map MATH given by MATH is continuous with compact support. Thus, if MATH, there exists a neighborhood MATH of the identity of MATH such that, if MATH, then MATH. Hence, if MATH: MATH . Since MATH, we conclude that MATH is dense in MATH, MATH, and hence that MATH is dense in MATH in the inductive limit topology by REF. This, together with REF., proves that MATH and MATH are NAME equivalent. Let MATH be the natural action of MATH on MATH, and MATH, MATH. Then MATH, and therefore MATH. It follows that MATH, thus MATH. This completes the proof. |
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 that MATH, MATH and MATH satisfy REF . - REF (with MATH in place of MATH and MATH instead of MATH). Now the result follows from REF. |
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 bundle associated with the enveloping action MATH, and MATH the natural action of MATH on MATH. Consider the right ideal MATH of MATH constructed in the proof of REF , that is, MATH. Applying REF , it follows that MATH and MATH are NAME equivalent. Since MATH is a saturated Fell bundle. by REF we have that MATH, and therefore MATH by REF. and REF. In conclusion: MATH. |
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. Thus, MATH is continuous if and only if MATH is continuous; but MATH, and MATH is continuous. |
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 action. It remains to verify that the MATH-orbit of MATH is MATH. Suppose that there exists MATH such that MATH, MATH. Then MATH, MATH, that is, MATH, MATH. But then MATH, MATH, and therefore MATH, because MATH is the enveloping action of MATH. The contradiction implies that every primitive ideal belongs to the MATH-orbit of some element of MATH. As for MATH and MATH, identify MATH with MATH via MATH. Then MATH becomes: MATH, and for MATH, MATH is the class of the unique extension to MATH of the irreducible representation MATH of MATH. From the computations above it follows that MATH. On the other hand, if MATH, then MATH is an extension to MATH of the representation MATH, and therefore MATH agrees with MATH. It remains to show that the MATH-orbit of MATH is all of MATH, and this is similar to which has been done previously for MATH and MATH: if MATH, MATH, then MATH, MATH, and therefore MATH, what is a contradiction. |
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. Now, if MATH, then MATH, where MATH is the unique primitive ideal such that MATH. Since MATH is a lattice isomorphism, it follows that MATH . That is, MATH . Thus, since MATH is a primitive ideal, it must agree with MATH. It follows that MATH is a morphism of set theoretic partial actions. Similarly, its inverse map MATH is a morphism, so MATH is an isomorphism between the set theoretic partial actions MATH and MATH. The proof of the corresponding statement for MATH and MATH is similar and it is left to the reader. |
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 MATH and MATH is an isomorphism of partial actions between MATH and MATH, it follows that MATH is the enveloping action of MATH. The proof of REF is similar. |
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 is no primitive ideal of MATH containing the ideal MATH, and hence MATH, MATH. Therefore MATH is a global action. If MATH is connected, then MATH. |
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, and hence MATH. To prove the converse inclusion, it is enough to show that MATH is dense in MATH in the inductive limit topology. This will follow from REF . For given MATH, MATH, let MATH be such that MATH. It is clear that MATH is continuous and has compact support. Let MATH. Then MATH is dense in MATH by the NAME - NAME theorem. In particular, MATH is a dense subspace MATH in the inductive limit topology. Let us see that MATH: if MATH, MATH, we have: MATH . Now, it is clear that MATH, MATH. This shows that MATH is dense in MATH, and therefore MATH is an isomorphism. Finally, if MATH we have: MATH . On the other hand: MATH so MATH, and since MATH is dense in MATH, it follows that MATH. |
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 MATH's, and by rescaling elements of this basis by a common multiple of the denominators, we can assume the MATH's to be MATH-linear combinations of the MATH's. Hence the image MATH is contained in the MATH-module MATH spanned by the MATH's, which is free by construction. The fact that MATH is torsion is clear because there exist nonzero elements MATH such that MATH. If MATH is a common multiple of the MATH's, then MATH is contained in MATH. On the other hand, the inclusion MATH implies MATH, hence MATH and MATH is torsion. |
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 the simple form REF. We can write MATH, MATH. REF then becomes: MATH . Let MATH be the maximal value of MATH for MATH such that MATH. We want to show that MATH. Among all MATH such that MATH there will be some (nonzero) MATH of maximal degree MATH. Then without loss of generality we can change the basis MATH and assume that the coefficient MATH is nonzero and of degree MATH. If MATH, then no nonzero term in the left-hand side of REF has a third tensor factor of degree MATH or MATH since MATH. Hence, terms from the right-hand side of degree MATH (respectively MATH) in the third tensor factor must cancel against each other. Terms having degree MATH in the third tensor factor cancel, since they give the following sum: MATH which in the third tensor factor has degree MATH and lower. Note also that their coefficients have total degree at most MATH. Terms having third tensor factors of degree MATH, besides REF, arise when we choose MATH. Those with MATH can be expressed in terms of commutators as above, and hence only contribute to lower degree. So, we only need to account for terms with MATH. Let us focus on such terms having a third tensor factor proportional to MATH, whose coefficient must be zero. They occur in REF only when MATH, MATH. When MATH, things cancel as above. The only other nonzero terms are the following: MATH where MATH is the standard basis of MATH. We have seen that all other contributions have coefficients of degree at most MATH, so the sum MATH must lie inside MATH. All MATH are of degree at most MATH and MATH are of degree exactly MATH, hence MATH must lie in MATH too. But this implies that MATH for all MATH, so in particular MATH, which is a contradiction. This proves that MATH. Now if MATH, where MATH, MATH, then we have: MATH and REF becomes MATH . Comparing the terms in MATH, we see that REF is equivalent to the system REF . |
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