paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9904111 | It follows from REF that MATH is an orthogonal basis of MATH, and that their quadratic norms are given by MATH . The theorem follows now immediately by setting MATH, MATH and MATH and using the explicit REF of the function MATH and the constant MATH. |
math/9904114 | Let MATH be the projection on the MATH-th factor. Let MATH and MATH be the subbundles which are generated by the image and kernel of MATH, respectively. Then MATH and MATH are contained in MATH, MATH and MATH are contained in MATH, and we have sequences of vector bundles MATH which are generically short exact. Hence, MATH and putting MATH and MATH, it follows that MATH . Clearly MATH so that MATH and therefore either MATH or MATH. Provided that MATH and MATH give MATH-subbundles of MATH we can then take MATH to be the MATH-bundle associated to either MATH or MATH. It thus remains to see show that MATH and MATH are MATH-invariant and, therefore, define MATH-subbundles of MATH. First, let MATH. If we write MATH for some MATH in MATH, then MATH . By our assumption on the matrix for MATH, it follows that MATH and MATH. Then MATH and MATH. But MATH because MATH is MATH-invariant, and thus MATH and MATH. Of course, we can repeat the argument with MATH and hence, MATH is MATH-invariant. The proof that MATH is MATH-invariant is similar. Let MATH. By assumption, MATH. But MATH, so MATH as well. It follows that MATH and thus, MATH is MATH-invariant. We have thus seen that MATH and MATH define MATH-subbundles of MATH and this finishes the proof. |
math/9904114 | The only assertion that requires proof is the final one. Suppose that MATH is poly-stable and that MATH is a proper stable NAME subbundle of MATH with MATH. By semi-stability of MATH the bundle called MATH in the proof of REF must then satisfy MATH and, since MATH, it follows by stability of MATH that MATH. But MATH is a MATH-subbundle so this finishes the proof. |
math/9904114 | We give the proof in case of MATH-representations, the MATH case being completely analogous. Let MATH be the poly-stable NAME bundle of the form REF corresponding to MATH, as we already noticed MATH. Without loss of generality we can assume that MATH. In this case MATH, as otherwise MATH would be MATH-invariant, and therefore violate the stability condition. Let MATH be the subbundle of MATH, such that MATH is the vector bundle generated by the image of MATH. Similarly, let MATH be the subbundle, which is generated by the kernel of MATH. Then the bundles MATH and MATH are both MATH-invariant. We therefore get the following inequalities from semi-stability of MATH: MATH . Note that these inequalities also hold in the case when MATH and MATH. Next, we note that MATH induces a non-trivial global section of the linebundle MATH which therefore has positive degree, that is, MATH where MATH is the generic rank of MATH. Combining this with REF , we obtain MATH so MATH as claimed. |
math/9904114 | Let MATH be a NAME bundle of the form REF which represents a critical point of MATH. MATH comes from the standard representation of MATH on MATH and the infinitesimal gauge transformation MATH which produces the decomposition MATH is fibrewise in MATH. Hence each of the bundles MATH is of the form MATH where MATH for MATH. We claim that either MATH or MATH must be zero (unless MATH). To see this, let MATH be the smallest MATH such that MATH and MATH are both non-zero. Then MATH is contained in either MATH or MATH, since the same is true for MATH and MATH interchanges MATH and MATH. Without loss of generality we may suppose that MATH. Then each of the bundles MATH and MATH is MATH-invariant, and so we have a decomposition of MATH as a direct sum of lower rank NAME bundles. This is impossible because MATH is stable. Let MATH be the rank vector whose entries are the ranks of the bundles MATH. We analyze the possibilities for MATH case by case. REFst case: MATH. Note that MATH. In this case we therefore have MATH, where each MATH is a linebundle. Hence the decomposition REF of MATH is of the form MATH where MATH . REF for the NAME index then takes the form MATH . Now we note that MATH is a MATH-invariant subbundle of MATH and thus, by stability, MATH. Combining this with the above result we get MATH . But since MATH interchanges MATH and MATH we must have MATH and MATH or vice-versa. Therefore MATH. Combining this with the above inequality we get MATH where the last inequality comes from the NAME inequality MATH of REF. We conclude that a critical point of this type always has strictly positive NAME index and hence it cannot be a local minimum of MATH. REFnd case: MATH. Again using MATH we see that in this case MATH. We then have MATH and MATH and so, from REF, we get MATH where the second equality is due to the fact that MATH. Since MATH and MATH are MATH-invariant we get from stability that MATH and MATH. Hence MATH, which shows that MATH. Therefore a critical point of this type cannot be a minimum of MATH either. REFrd case: MATH (or MATH). In this case MATH where MATH and MATH (or vice-versa). Since MATH interchanges MATH and MATH, it follows that MATH and so, MATH is reducible. Thus this case cannot occur. The case MATH is analogous. REFth case: MATH. In this case MATH. Then MATH for MATH and hence we see from REF that these critical points are local minima of MATH. Clearly MATH and MATH, or vice-versa. If MATH it would be MATH-invariant and so MATH which is absurd. Thus, in fact, MATH and MATH. In the notation of REF this means that MATH and MATH. This gives the minima with MATH. Finally, note that if MATH then either MATH or MATH is a MATH-invariant subbundle which violates stability. Thus there are no stable NAME bundles with MATH which are local minima of MATH. |
math/9904114 | This is analogous to the proof of REF. However, in this case the infinitesimal gauge transformation MATH which produces the decomposition MATH of the NAME bundle of the form REF belongs to MATH, that is, it is fibrewise in MATH. Thus there are only two possibilities: either MATH and MATH with MATH, that is, MATH here we are using that MATH. These NAME bundles are seen to be minima as before. The other possibility is that MATH and MATH, where MATH. Note that either MATH or MATH. In this case the decomposition REF has the form MATH where MATH . From REF we therefore get the NAME index MATH . Thus we cannot have a minimum unless MATH, and in this case, from REF, we have MATH since otherwise MATH would not be of rank MATH. This gives the second case of the proposition. |
math/9904114 | Let MATH be a reducible NAME bundle of the form REF which is a local minimum of MATH. Consider MATH as the space of solutions MATH to NAME 's equations modulo MATH gauge equivalence. First consider the case MATH. Then, by poly-stability, MATH, and MATH and MATH are poly-stable vector bundles. On the other hand, it is clear that such NAME bundles are, in fact, reducible (absolute) minima of MATH. This gives the first case of the proposition. Suppose now that MATH. The possible reductions of structure group are the following. Reduction to MATH. In this case we have MATH for linebundles MATH and MATH, while MATH. Thus MATH is the direct sum of two NAME bundles MATH and MATH, where MATH, MATH, and the NAME field MATH has zeros along the diagonal. Note also that MATH by poly-stability of MATH. Each of the bundles MATH is a minimum on the moduli space of rank MATH . NAME bundles of this form and hence of the form REF, in other words all components of MATH are zero, except one off-diagonal entry. (compare REF. There are now two cases to consider. The first case is when MATH is zero on one of the bundles MATH and MATH; since MATH we must have MATH. In other words, MATH is of the form MATH with respect to the decomposition MATH. Thus MATH is of the form considered in REF. As in the proof of that proposition one sees that there is no subspace of the NAME tangent space with negative weights and, therefore, these NAME bundles represent local minima of MATH. This case includes the case of one of the MATH being equal to zero. Note that, if MATH then it follows from REF and the remark following it that MATH. This case gives the remaining local minima of the statement of the proposition. The other case is when MATH is non-zero on both MATH and MATH, say that MATH and MATH. By stability, and since MATH, we then have MATH and MATH, and so MATH and MATH. We shall show that in this case MATH is not a local minimum of MATH. Let the infinitesimal gauge transformation MATH which produces the decomposition MATH of REF have weights MATH on MATH, and weights MATH on MATH. We then have the following equations relating these numbers: MATH . From these equations it follows that MATH and hence, either MATH or MATH. For definiteness suppose that MATH (the other case is entirely similar). This means that MATH has weight MATH and that this is a subspace of the highest weight space of MATH. Note that MATH is zero restricted to the highest weight space and so MATH gives a subspace of MATH on which MATH is negative. But since MATH and MATH we have MATH and therefore, from NAME, MATH. It only remains to find a smooth family of NAME bundles in MATH to which an element in MATH is tangent (compare REF. By REF is the direct sum of the stable NAME bundles MATH and MATH. All extensions MATH are parametrized by MATH so, in particular, MATH defines an extension MATH which is non-trivial if MATH. Note that MATH, where MATH is the non-trivial extension MATH defined by MATH. We define a NAME field MATH on MATH of the appropriate form in the following way. To define MATH we use the composition MATH while to define MATH we use the composition MATH . For MATH the NAME bundle MATH is a non-trivial extension of stable NAME bundles and therefore stable. For MATH, the family MATH is thus a smooth family of NAME bundles in MATH to which MATH is tangent. Reduction to MATH. In this case we have a decomposition of MATH as a direct sum of NAME bundles MATH and MATH), where MATH and MATH with MATH and MATH linebundles. Again MATH and MATH) represent local minima on lower rank moduli spaces and so, MATH. If MATH is reducible we are back in one of the previous cases so we may assume that MATH is stable. Since we are at a minimum it must be of the form REF and again there are several possibilities. If MATH is zero on either MATH or MATH then it is zero on either MATH or MATH and MATH is a minimum as above. Thus the only case that remains is when MATH and MATH, and MATH. The weights of the infinitesimal gauge transformation producing this decomposition are MATH, MATH, and MATH on MATH, MATH, and MATH, respectively. As above one sees that MATH gives a subspace of MATH on which MATH is negative and that MATH is tangent to a smooth family of NAME bundles, so that MATH does not represent a local minimum. We omit the details. Reduction to MATH. This case is analogous to the previous one. |
math/9904114 | Immediate from REF . |
math/9904114 | First consider the case MATH. To see that MATH is connected, consider the continuous map MATH where MATH denotes the moduli space of rank MATH poly-stable vector bundles with fixed trivial determinant bundle. From REF we see that this is surjective and, since MATH and MATH are connected, that MATH is connected. Next consider the case MATH (as already noticed, this also takes care of the case MATH). From REF we see that MATH is isomorphic to the moduli space of rank MATH, degree MATH vector bundles with fixed determinant, which is known to be connected. |
math/9904114 | Follows from REF . |
math/9904114 | From REF it follows in particular that MATH is isomorphic to the moduli space of rank MATH, degree MATH poly-stable vector bundles. Since this space is connected, the result is proved. |
math/9904114 | We have to prove that MATH is stable if and only if MATH is. From REF we know that stability of MATH is equivalent to stability of the MATH-bundle MATH. Thus, all we need to prove is that MATH is stable if and only if MATH is. Because stability is unaffected by tensoring with a line bundle, we can equally well prove that MATH is stable. Note, that MATH. Assume MATH is a stable MATH-bundle. Let MATH be a MATH-invariant subbundle. Let MATH be the subbundle such that MATH is generically the image of MATH under MATH. Then MATH maps MATH to MATH, because of the MATH-invariance of MATH. Hence, MATH defines a MATH-subbundle of MATH, and it follows that MATH . But, as MATH is an isomorphism MATH and similarly MATH. Therefore MATH and so, MATH is stable. Conversely, assume that MATH is stable. Let MATH be a MATH-subbundle of MATH. Let MATH be the subbundle which is generically the image of MATH under MATH. Both MATH and MATH are MATH-invariant subbundles of MATH, because MATH is a MATH-subbundle. Hence, MATH and MATH, by stability of MATH. Recalling that MATH and MATH, we get MATH . Note also that MATH because MATH is an isomorphism, and the image of MATH under MATH is contained in MATH, by the assumption that MATH is a MATH-subbundle. Combining REF, and REF, we get: MATH . Of course, MATH and hence the proof is finished. |
math/9904114 | The only thing there is to remark is that any MATH, of the form given above, is stable. But, MATH is the only MATH-invariant subbundle so, this is obvious. |
math/9904114 | Suppose we have a critical point of the type described in MATH of REF, with MATH. Then, MATH is MATH-invariant and therefore, MATH is semi-stable, but not stable. Since we are considering the moduli space of poly-stable NAME bundles, MATH decomposes as a direct sum of rank MATH . NAME bundles of degree MATH. The only subbundles of MATH of rank MATH and degree MATH are MATH and MATH, and MATH is not MATH-invariant so, we conclude that this situation cannot occur. |
math/9904115 | It is straightforward to translate the proof of CITE to this setting. |
math/9904115 | With MATH as in REF , let MATH be the ideal in MATH generated by MATH . Define MATH and MATH, where MATH is the canonical projection. Obviously MATH is a NAME representation which generates MATH, and it is NAME covariant because MATH. If MATH is another NAME covariant NAME representation, then the homomorphism MATH of MATH satisfies MATH whenever MATH, and hence MATH descends to the required homomorphism of MATH (also denoted MATH). |
math/9904115 | Let MATH be universal for NAME representations of MATH, and define MATH and MATH. Since MATH is a NAME representation of MATH, we get a homomorphism MATH. To construct the inverse of MATH, let MATH be the universal NAME representation of MATH in MATH, and fix MATH. By CITE, there is a linear map MATH which satisfies MATH and then MATH is a NAME representation of MATH. Defining MATH thus gives a NAME representation MATH, and it is routine to check that MATH is the inverse of MATH. Now let MATH be universal for NAME covariant NAME representations of MATH. As above, we get a homomorphism MATH. To construct the inverse, we let MATH be universal and define a NAME representation MATH as before; we need to check that MATH is NAME covariant under each of the hypotheses on the left action. By REF is NAME covariant, so we use induction. Assume that MATH is NAME covariant for some MATH, and suppose MATH acts compactly on the left of MATH; that is, MATH. Since the left action is isometric on each fiber, by CITE we have MATH; hence MATH. But CITE gives MATH, so MATH is NAME covariant. Now suppose that MATH acts by compact operators on each fiber. By representing MATH faithfully on a NAME space MATH we can assume that MATH is a NAME representation of MATH on MATH. Assuming again that MATH is NAME covariant for some MATH, CITE gives MATH. Let MATH, and express MATH with MATH and MATH. Since MATH is NAME covariant and MATH, we have MATH . Now MATH can be approximated by a finite sum MATH, hence MATH can be approximated by a finite sum MATH. Thus MATH and MATH is NAME covariant by CITE. |
math/9904115 | Let MATH be an approximate identity for MATH. By REF, MATH is an approximate identity for MATH, and the result follows. |
math/9904115 | Let MATH and MATH. If MATH and MATH, then MATH and hence the product MATH belongs to MATH. Letting MATH vary over an approximate identity for MATH, this product converges in norm to MATH, so the set of products MATH has dense linear span in MATH. Hence to see that multiplication induces an isomorphism MATH, it suffices to check that it preserves the inner product of any pair of elementary tensors: MATH . Multiplication is associative since MATH . If MATH is an approximate identity for MATH, then for each MATH and MATH we have MATH, so each MATH is essential. If MATH, then by writing MATH with MATH, we see that MATH is compact. |
math/9904115 | Since each fiber MATH is essential, any vector MATH can be written in the form MATH with MATH and MATH. But then MATH, and since elements of the form MATH generate MATH as a MATH-algebra, the result follows from the calculations MATH and MATH . |
math/9904115 | Taking MATH in REF, shows that MATH converges strictly to the identity in MATH; that is, MATH is nondegenerate. For REF of a NAME crossed product, suppose MATH is a nondegenerate representation of MATH; we must show that MATH is a NAME covariant representation of MATH. First note that MATH is nondegenerate since MATH and MATH are. REF gives MATH so MATH and applying MATH gives MATH. In particular MATH is a projection, so MATH, and hence MATH, is a partial isometry. To establish REF, take any MATH, write MATH with MATH, and compute: MATH . Taking adjoints, interchanging MATH and MATH, and applying MATH gives MATH. For every MATH we have MATH taking adjoints and applying MATH gives MATH. This completes the proof of REF . For REF , suppose MATH is a NAME covariant representation of MATH on a NAME space MATH. Define MATH by MATH . Since MATH is nondegenerate and MATH for all MATH, MATH is a coisometry. Since MATH, we deduce that MATH. Thus MATH and since we also have MATH is a NAME representation of MATH. Let MATH be the representation MATH. Then MATH and MATH since MATH is nondegenerate, this implies that MATH, as required. For REF , simply note that MATH, and elements of this form generate MATH. We now show that MATH is a crossed product for MATH. Since MATH and MATH are nondegenerate, so is MATH. If MATH is a nondegenerate representation of MATH, then MATH is a nondegenerate representation of MATH. Hence MATH is a NAME covariant representation of MATH. To see that each MATH is an isometry, let MATH. Since MATH is NAME covariant, REF gives MATH . Since MATH is nondegenerate, this implies that MATH, and hence MATH, is an isometry. This gives REF for a crossed product. REF is obvious, so it remains only to verify REF . Suppose MATH is a covariant representation of MATH on a NAME space MATH, and define MATH as before. We have already seen that MATH is a NAME representation of MATH, and it is NAME covariant since, for any MATH, MATH . Defining MATH gives REF . |
math/9904115 | CASE: The uniqueness of MATH is obvious. By CITE, there is a unital homomorphism MATH determined by MATH . Let MATH be the isometry which satisfies MATH (see the proof of CITE), and define MATH . Then MATH is a homomorphism, and MATH vanishes on MATH. If MATH and MATH, then for any MATH we have MATH giving REF. Since MATH, the space MATH reduces MATH; hence for any MATH and MATH, both MATH and MATH vanish on MATH. This and MATH show that MATH. CASE: Let MATH, and suppose MATH and MATH. Vectors of the form MATH have dense linear span in MATH; since MATH is essential, this holds even when MATH (see REF ). Since MATH we deduce that MATH . Once we show that MATH, it follows from the uniqueness of MATH that MATH. From REF we see that MATH. Suppose that MATH; we will show that MATH, which will complete the proof. Since MATH is in the range of MATH, it can be approximated by a finite sum MATH. Then MATH . Now each MATH can be approximated by a finite sum MATH, and then MATH . Thus MATH can be approximated arbitrarily closely by a vector in the range of MATH, and hence MATH. Since each MATH is essential, the assumption that MATH is nondegenerate implies that MATH is a nondegenerate representation of MATH. Since MATH for all MATH and MATH, we have MATH for all MATH. |
math/9904115 | If MATH is a set of pairs MATH consisting of maps MATH and MATH, then MATH is a covariant representation of MATH if and only if each MATH is. The main point here is that the value of MATH on an element of MATH of the form MATH is MATH. Suppose MATH is a nondegenerate covariant representation on a separable NAME space MATH; that is, the MATH-algebra MATH acts nondegenerately on MATH. We will identify the multiplier algebra of MATH with the concrete MATH-algebra MATH . We claim that MATH. For this, it suffices to check that multiplying a generator MATH of MATH on either the left or the right by an operator of the form MATH, MATH, or MATH yields another element of MATH. Certainly MATH and MATH, and since MATH we also have MATH and MATH. Writing MATH with MATH gives MATH . Finally, to see that MATH, we use a trick from CITE. Let MATH be an approximate identity for MATH; we claim that MATH . Since MATH is nondegenerate, MATH converges strongly to MATH. On the other hand, using REF we see that MATH converges in norm (to MATH), and REF follows. Hence MATH and since MATH we deduce that MATH. Since MATH, we have shown in particular that the ranges of MATH and MATH are contained in MATH. Consequently, any decomposition MATH of the identity as a sum of mutually orthogonal projections MATH gives corresponding decompositions MATH and MATH, and by the first paragraph each pair MATH is a covariant representation of MATH. By the usual NAME 's Lemma argument we can choose these projections such that MATH acts cyclically on MATH; since MATH acts cyclically on MATH, this shows that every covariant representation of MATH decomposes as a direct sum of cyclic representations. Let MATH be a set of cyclic covariant representations with the property that every cyclic covariant representation of MATH is unitarily equivalent to an element in MATH. It can be shown that such a set MATH exists by fixing a NAME space MATH of sufficiently large cardinality (depending on the cardinalities of MATH and MATH) and considering only representations on MATH. Note that MATH is nonempty because the system has a covariant representation, which has a cyclic summand. Define MATH and MATH, and let MATH be the MATH-algebra generated by MATH. By the first paragraph, MATH is a covariant representation of MATH, and it is nondegenerate since each MATH is. We deduce that both MATH and MATH map into MATH, and that REF for a crossed product is satisfied by taking MATH to be the identity representation. REF is trivial, and REF holds because every covariant representation decomposes as a direct sum of cyclic representations. We need to show that MATH and MATH are nondegenerate. For this, let MATH and MATH. If MATH is an approximate identity for MATH, then by REF we have MATH and MATH, so MATH is nondegenerate REF . If MATH is an approximate identity for MATH, then MATH, and since MATH is nondegenerate as a representation on NAME space, REF gives MATH. Thus MATH is nondegenerate. For the uniqueness assertion, suppose MATH is another crossed product. REF allows us to assume that MATH and MATH are covariant representations of MATH on NAME spaces MATH and MATH. REF then gives a representation MATH whose image is contained in MATH since MATH. Similarly one obtains a map MATH which is obviously an inverse for MATH. |
math/9904115 | We follow CITE. Let MATH be a faithful nondegenerate representation MATH of MATH such that MATH is a covariant representation of MATH, and let MATH be a unitary representation of MATH whose integrated form MATH is faithful on MATH. We claim that MATH is a covariant representation of MATH. Most of the verifications are routine, so we check only that MATH . For this, we show that MATH satisfies the properties which characterize MATH REF . First, let MATH; we show that REF holds on any vector in the range of MATH: MATH . Next, note that MATH is the projection onto MATH which is precisely the range of MATH. Since MATH vanishes on the range of MATH, REF follows from the uniqueness assertion of REF . Since MATH is covariant, there is a representation MATH of MATH such that MATH . Since MATH and MATH are faithful, MATH is faithful on MATH, and we can define MATH. By checking on generators it is easy to see that MATH satisfies the coaction identity MATH, and MATH is injective since MATH, with MATH the augmentation representation of MATH (that is, MATH for all MATH). When MATH is abelian, MATH is the action canonically associated with MATH. |
math/9904115 | Since MATH is a NAME representation, so is MATH. Let MATH. The range of MATH is MATH . Hence for any MATH, the range of MATH is MATH which is MATH if MATH, and is zero otherwise. If MATH is faithful then so is MATH; since MATH is isometric, this implies that MATH is isometric. |
math/9904115 | Fix MATH. Since MATH is NAME covariant and MATH, CITE gives MATH. But MATH is essential, so the reverse inclusion holds as well, and since MATH is precisely the range of MATH, we deduce that MATH is constant in MATH. Since MATH for all MATH, this implies that MATH is NAME covariant. |
math/9904115 | CASE: See CITE. For the continuity assertion, suppose MATH strictly in MATH, MATH, and MATH. There exists MATH and MATH such that MATH, and then MATH . CASE: Both MATH and MATH are the projection onto MATH. CASE: If MATH and MATH, then MATH since both sides of REF are supported on MATH, this implies that MATH. By REF , MATH commutes with MATH, giving the other half of REF. CASE: When MATH is a projection, MATH is the projection onto MATH, and the result follows from CITE. CASE: See CITE. CASE: MATH. |
math/9904115 | CITE If MATH is NAME covariant, then MATH where the last equality uses REF . Conversely, suppose REF holds for all compact MATH and MATH. If MATH strictly, then MATH where the convergence is in the strong operator topology. Hence MATH for every MATH. Letting MATH strictly shows that MATH is NAME covariant. |
math/9904115 | Suppose MATH, MATH, MATH, and MATH. If MATH is a total order then either MATH or MATH; either way MATH is compact. If the left action of MATH on each fiber MATH is by compact operators, then by CITE, both MATH and MATH are compact. |
math/9904115 | By checking on an operator MATH, one verifies that MATH, and the result follows easily from this. |
math/9904115 | Express MATH with MATH and MATH; similarly, express MATH with MATH and MATH. Since MATH is NAME covariant, MATH is zero if MATH, and otherwise MATH where MATH. Since MATH is compact it can be approximated in norm by a finite sum of operators MATH with MATH, and hence MATH can be approximated by finite sums of the form MATH. But any such MATH can be approximated by finite sums of products MATH with MATH and MATH; similarly, any such MATH can be approximated by finite sums of products MATH with MATH and MATH. Hence MATH can be approximated in norm by finite sums of operators of the form MATH . |
math/9904115 | The proof is formally identical to that of CITE. |
math/9904115 | The proof is formally identical to that of CITE, except that in REF one must also note that MATH since MATH and MATH generates MATH. |
math/9904115 | Let MATH be a nondegenerate representation of MATH on a NAME space MATH, and let MATH be the NAME representation of MATH. By REF , MATH is a NAME representation of MATH. Since MATH is nondegenerate, so is MATH; since MATH is nondegenerate, MATH is as well. The previous Proposition thus gives a covariant representation MATH of MATH. |
math/9904115 | Let MATH be a faithful representation of MATH on a NAME space MATH such that MATH is a covariant representation of MATH. By REF , MATH is a NAME representation of MATH, so we can take MATH to be the restriction of MATH to MATH. Suppose MATH is a (nondegenerate) NAME representation of MATH. REF gives us a covariant representation MATH of MATH, and hence a representation MATH of MATH such that MATH. Restricting MATH to MATH gives the required representation MATH. Uniqueness of MATH follows by the usual argument. Suppose MATH is compactly aligned. Since MATH is the composition of the NAME representation MATH and the homomorphism MATH (restricted to MATH), MATH is NAME covariant by REF . Let MATH, and express MATH for some MATH and MATH. Then MATH, so MATH contains MATH. Obviously MATH is a closed self-adjoint subspace of MATH, and since MATH is compactly aligned, REF shows that MATH is closed under multiplication. This gives REF. Now let MATH. Using REF with MATH, and then applying MATH, gives MATH and MATH REF shows that MATH so MATH. Since MATH is generated by MATH, elements of the form MATH generate MATH as a MATH-algebra; with MATH as above, REF shows that MATH . Hence to establish REF, it remains only to show that MATH is closed under multiplication. But REF shows that the product MATH of two typical generators of MATH is contained in the closed linear span of elements of the form MATH which by REF simplifies to MATH . Suppose the left action of MATH on each MATH is by compact operators; that is, MATH for all MATH. Let MATH and MATH. Since MATH is essential, we can express MATH for some MATH and MATH. With MATH, we then have MATH so MATH. Since elements of the form MATH generate MATH, this gives MATH. If in addition every MATH have a common upper bound, then by REF the universal map MATH is NAME covariant; the integrated form MATH is surjective since it maps generators to generators. |
math/9904115 | Let MATH be a faithful nondegenerate representation of MATH on a NAME space MATH, let MATH be the NAME representation of MATH, and let MATH; by REF , MATH is a NAME representation of MATH on MATH. We claim that MATH is faithful. Since MATH is the orthogonal projection of MATH onto MATH (see the proof of REF ), each projection MATH dominates the projection MATH onto the MATH-invariant subspace MATH. To establish the claim it thus suffices to show that the subrepresentation MATH of MATH is faithful. But MATH decomposes as MATH, so MATH is unitarily equivalent to MATH, and hence faithful. Now suppose that MATH is faithful and MATH. Let MATH . Then MATH giving REF. |
math/9904115 | The proof, based on REF , is identical in form to the proof of CITE. |
math/9904115 | Since MATH is a decomposition of the identity in MATH, and since MATH is a unital representation of MATH, the projections MATH decompose the identity operator. By REF , MATH commutes with each MATH, and thus MATH . Fix an initial segment MATH, and let MATH. By REF , MATH so it suffices to show that MATH . Let MATH . Since MATH satisfies REF, MATH is a faithful representation of MATH. By REF , the representation MATH is thus also faithful. But MATH, and hence REF is satisfied. |
math/9904115 | CASE: Since MATH is compactly aligned, the spanning REF holds. Since MATH is continuous and maps onto MATH, we deduce that finite sums MATH in which MATH for all MATH are dense in MATH. It therefore suffices to fix such a MATH and show that MATH. Let MATH be a faithful nondegenerate representation of MATH such that MATH is a covariant representation of MATH. By REF , MATH is a covariant representation of MATH and MATH. Since MATH is faithful, MATH satisfies REF. Hence with MATH, REF gives MATH . CASE: Since MATH is compactly aligned, finite sums of the form MATH are dense in MATH. We will show that MATH; it follows that MATH is well-defined on operators of the form MATH and extends to the desired linear contraction. Let MATH, and let MATH; by REF, MATH . By REF , there is an initial segment MATH of MATH such that MATH. Let MATH. We will construct a projection MATH such that MATH is faithful, then define MATH, and show that MATH. This will complete the proof, since by REF we then have MATH . For each MATH such that MATH and MATH, define MATH as in CITE: MATH noting in particular that MATH is never the identity in MATH. Let MATH be as in REF, and define MATH . By REF , MATH is faithful. The proof that MATH is exactly as in CITE, so we omit it. |
math/9904115 | Denote by MATH the orthogonal projection of MATH onto MATH. Since the MATH's are mutually orthogonal, the formula MATH defines a completely positive projection of norm one which is faithful on positive operators. We claim that MATH . Since MATH is compactly aligned the spanning REF holds, and hence REF follows from REF. Suppose MATH and MATH. For each MATH, MATH is zero on MATH unless MATH, in which case MATH maps MATH into MATH. Thus if MATH, MATH for every MATH, and MATH. If on the other hand MATH, then MATH for each MATH, and thus MATH . |
math/9904115 | Suppose MATH is faithful. By REF , MATH is faithful on positive elements, hence so is MATH; that is, MATH is amenable. Since MATH satisfies REF (see the proof of necessity of REF), the converse follows from REF . |
math/9904115 | Our proof is essentially that of CITE, suitably modified to handle NAME bimodules. The homomorphism MATH induces a coaction MATH of MATH on MATH, and hence a conditional expectation MATH of MATH onto the fixed-point algebra MATH, such that MATH . Since MATH is amenable, MATH is faithful on positive elements. Let MATH be the NAME representation of MATH, let MATH be a faithful nondegenerate representation of MATH on a NAME space MATH, and let MATH. By REF , for every MATH we have MATH . Since MATH and MATH are faithful on positive elements, to show that MATH is amenable it suffices to show that MATH is faithful on MATH. Let MATH be a faithful representation of MATH such that MATH is a covariant representation of MATH. By REF , MATH is a covariant representation of MATH and MATH. Observe that MATH is isometric since, by REF , MATH . Let MATH be the set of all finite subsets MATH of MATH which are closed under MATH in the sense that MATH whenever MATH and MATH. Exactly as in the proof of CITE, one can use REF to show that, for each MATH, MATH is a MATH-subalgebra of MATH. Applying MATH to both sides of REF gives MATH since MATH is directed under set inclusion (see the proof of CITE), we deduce that MATH . By CITE, to prove that MATH is faithful on MATH it is enough to prove it is faithful on each of the subalgebras MATH. We shall accomplish this by inducting on MATH. First suppose MATH for some MATH. Let MATH be the NAME MATH - MATH bimodule MATH. We claim that, for each NAME representation MATH of MATH on a NAME space MATH, there is a linear map MATH which satisfies MATH, and that MATH is then a NAME representation of MATH. First observe that if MATH satisfy MATH and MATH, then by REF we have MATH, and hence MATH. Now suppose MATH belongs to the algebraic direct sum MATH; such vectors are dense in MATH. Then MATH ensuring the existence of MATH. It is routine to check that MATH is a NAME representation of MATH. Write MATH for the endomorphism of MATH which corresponds to MATH REF , and write MATH for the associated representation of MATH REF . Suppose MATH is a finite sum MATH such that MATH for every MATH; to prove MATH faithful on MATH we will show that MATH. For each MATH, let MATH denote the operator in MATH which is the image of MATH . Define MATH. It is routine to check that MATH and similarly MATH. Since MATH and MATH are faithful representations of MATH, the representations MATH and MATH are isometric, and thus MATH . For the inductive step, suppose MATH and MATH is faithful on MATH whenever MATH and MATH; we aim to prove that MATH is faithful on MATH. Since MATH is finite it has a minimal element; that is, there exists MATH such that MATH for each MATH. As in the proof of CITE we have MATH for each MATH, where MATH denotes the orthogonal projection of MATH onto MATH. On the other hand, we have already demonstrated that MATH maps MATH isometrically into the range of MATH, and an easy calculation shows that MATH, where MATH is the orthogonal projection onto MATH. Since MATH is faithful, by REF the representation MATH is also faithful. Hence the map MATH is faithful. Now suppose MATH and MATH. We will show that MATH, from which the inductive hypothesis implies that MATH. Let MATH be a sequence in MATH which converges in norm to MATH, and express each MATH as a sum MATH, where MATH. For each MATH, MATH and consequently MATH. Thus MATH, which shows that MATH, as claimed. |
math/9904115 | The group MATH is amenable, and by CITE the canonical map MATH satisfies REF. |
math/9904115 | As in the proof of REF , MATH is nondegenerate. We verify the obvious analogues of REF , and REF . For REF , let MATH be a nondegenerate representation of MATH on a NAME space MATH, let MATH, and let MATH; we must show that MATH is a NAME covariant representation of MATH. Exactly as in the proof of REF , MATH is a NAME covariant representation of MATH, so we need to establish REF. Fix MATH. For any MATH and MATH we have MATH and since MATH is nondegenerate this shows that MATH . Since MATH is compactly aligned, MATH is NAME covariant REF , and REF follows. For REF , let MATH be any NAME covariant representation on MATH. As in the proof of REF , MATH defines a nondegenerate NAME covariant representation MATH. To see that it is NAME, let MATH, and note that for any MATH we have MATH . Since MATH is nondegenerate, this implies that MATH, and hence MATH is NAME covariant by REF. Defining MATH gives the desired representation satisfying MATH and MATH. REF is satisfied since MATH, and elements of this form generate MATH. |
math/9904115 | Let MATH be the canonical homomorphism of MATH onto MATH. By CITE, MATH satisfies the hypotheses of REF ; since MATH is compactly aligned, the system MATH is therefore amenable. Identifying MATH with MATH as in the previous Proposition and defining MATH, the initial projection MATH is precisely MATH, and the result follows from REF . |
math/9904115 | Let MATH be a faithful nondegenerate representation of MATH. Then MATH is a partial isometric MATH-representation of MATH which satisfies REF, and applying MATH we see that MATH is as well. Since MATH for every MATH, MATH also satisfies REF, and is hence bicovariant. Since MATH generates MATH linearly and MATH generates MATH as a MATH-algebra, elements of the form MATH are also generating. If MATH is any bicovariant partial isometric MATH-representation of MATH, then by CITE there is a representation MATH of MATH such that MATH for every MATH. For any MATH the product MATH is a partial isometry; hence by CITE the projections MATH and MATH commute, and we deduce that MATH for every MATH and MATH. Further, MATH so MATH for every MATH and MATH. Thus MATH is a NAME covariant representation of MATH. The representation MATH satisfies MATH. |
math/9904115 | MATH is universal if and only if the representation MATH of MATH is faithful. By REF , this occurs if and only if MATH acts faithfully on the range of MATH whenever MATH, and the result follows from CITE. |
math/9904115 | The first statement is obvious. In the left-invariant partial order on MATH, two elements MATH have a common upper bound if and only if one is an initial word of the other, and then the least upper bound is the longer of the two words. We will show that REF holds if and only if MATH of course a similar statement holds for REF using the right-invariant partial order, and together these prove the Proposition. To begin with, REF implies REF since distict generators of MATH are not comparable. For the converse, first suppose MATH; since MATH is a partial isometry, we then have MATH . The case MATH is similar. Finally, suppose MATH and MATH are not comparable. Then there exists MATH such that MATH, MATH, and MATH. REF implies that MATH, and by CITE the range projection of MATH commutes with the initial projection of MATH, so MATH . |
math/9904115 | Suppose each MATH is nonzero. To see that MATH is universal, we apply REF . If MATH, MATH, and MATH, , MATH, MATH, , MATH, then we can choose MATH such that none of the multi-indices MATH begins with MATH, and none of the multi-indices MATH ends with MATH. Then MATH is nonzero since MATH. Hence MATH is universal. Now define MATH by MATH . Then MATH is a bicovariant partial isometric representation of MATH in which each MATH is nonzero. If MATH is universal, then MATH extends to a homomorphism of MATH, and hence each MATH must be nonzero. |
math/9904116 | The Proposition can be proved by modifying the standard argument of, for example, CITE. Briefly, let MATH be a collection of NAME representations on NAME space which are cyclic (that is, generate a MATH-algebra which admits a cyclic vector), and such that every cyclic NAME representation is unitarily equivalent to an element of MATH. Define MATH and MATH. REF holds because every NAME representation of MATH on a NAME space decomposes as the direct sum of a zero representation and a collection of cyclic representations, and the uniqueness assertion follows from a standard argument. |
math/9904116 | CASE: We use REF to calculate MATH noting that MATH since MATH is abelian. CASE: Take MATH in REF to see that the linear span of monomials of the form MATH is closed under multiplication, and hence a MATH-algebra. CASE: For each MATH the map MATH is a NAME representation, and hence integrates to an endomorphism MATH of MATH. Since MATH and MATH are both the identity on MATH, MATH is an automorphism. Obviously MATH is a group homomorphism, and its continuity follows from a straightforward MATH argument. CASE: This is a standard result about automorphic actions of discrete abelian groups. CASE: Since MATH, REF follows from REF and the continuity of MATH. |
math/9904116 | For each MATH, there is a unique homomorphism MATH such that MATH for all MATH. Let MATH be an orthonormal basis for MATH. Since MATH we deduce that MATH for all MATH, and hence there is a homomorphism MATH such that MATH for every MATH. From REF it is obvious that MATH maps MATH onto MATH. If MATH admits a nontrivial NAME representation then MATH is nonzero (see REF ), hence each MATH is nonzero, and finally MATH is nonzero. Since MATH is simple, we deduce that MATH is injective. |
math/9904116 | Define MATH, MATH, and for each MATH let MATH be an orthonormal basis for MATH. Suppose MATH; we will eventually take MATH. By REF we have MATH and by REF MATH so it suffices to find a unit vector MATH such that MATH . Let MATH be an enumeration of these pairs. We claim that for each MATH there is a unit vector MATH such that MATH . Given the claim, MATH satisfies REF, and hence REF, completing the proof. The claim is vacuous when MATH: taking any unit vector MATH works. Suppose inductively that there exists MATH such that REF holds when MATH. Let MATH, MATH, and MATH. Since MATH, by REF, and we can assume without loss of generality that MATH. We have MATH where MATH . Since MATH, there is a unit vector MATH which is orthogonal to each MATH, and taking MATH gives REF for MATH. |
math/9904116 | Suppose MATH is a MATH-algebra and MATH is a nonzero homomorphism. We will show that MATH is injective, thus establishing the simplicity of MATH. It does not harm to assume that MATH is surjective, so that MATH is unital and MATH. Recall from REF that there is a faithful expectation MATH of MATH onto MATH, given by averaging over the orbits of the gauge action. We will show that there is an expectation MATH of MATH onto MATH such that MATH . To see that this implies that MATH is injective, suppose MATH. Then MATH, so MATH. But MATH is simple and contains the identity of MATH, so MATH. Thus MATH, and we deduce from the faithfulness of MATH that MATH. By REF , finite sums of the form MATH are dense in MATH. Hence to prove the existence of an expectation MATH satisfying REF, it suffices to fix such an element MATH and show that MATH. Thus with MATH, we must show that MATH . Let MATH. Since MATH is a NAME representation of MATH, the relations REF allow us to assume that MATH whenever MATH. Since MATH, REF applies and provides a unit vector MATH such that MATH . Let MATH. Then MATH . Since MATH and MATH both extend linearly to nontrivial homomorphisms of the simple algebra MATH, we have MATH . Combining this with REF gives REF, completing the proof of simplicity. Our proof that MATH is purely infinite is an adaptation of the proof of CITE. Let MATH be a hereditary subalgebra of MATH; we will show that MATH has an infinite projection. Fix a positive element MATH, scaled so that MATH. Choose a finite sum MATH such that MATH and MATH. Then MATH is also positive and satisfies MATH. By applying the relations REF we can assume that there exists MATH such that MATH whenever MATH, and such that MATH and MATH for all MATH. Then MATH is a positive element of the algebra MATH and hence its image under the canonical isomorphism MATH has a unit eigenvector MATH with eigenvalue MATH. It follows that MATH, and hence the projection MATH satisfies MATH. By REF , there is a unit vector MATH such that MATH whenever MATH. Let MATH, noting that MATH and MATH. Then MATH and since MATH we thus have MATH. It follows that MATH is invertible in MATH. Let MATH be its inverse, and define MATH. Then MATH is a partial isometry since MATH, and its initial projection belongs to MATH since MATH and MATH is hereditary. Since MATH and MATH contains nonunitary isometries, we deduce that MATH is an infinite projection in MATH. |
math/9904116 | By REF , MATH admits a nontrivial NAME representation, so one direction follows from REF . For the converse, suppose MATH is not injective. Fix MATH such that MATH and MATH. We claim that MATH generates a proper ideal MATH, so that MATH is not simple. Let MATH be the distinguished NAME representation defined in REF . Since MATH we have MATH, and hence MATH is not all of MATH. To see that MATH is nonzero, choose MATH such that MATH, and define MATH for all MATH. Then MATH is a NAME representation of MATH such that MATH and we deduce that MATH, and hence MATH, is nonzero. |
math/9904116 | Fix MATH and MATH. Our first goal is to define MATH. In any expression of MATH as an ordered sum MATH of basis elements we have MATH. For each of these finitely-many ordered MATH-tuples MATH, it is easy to see that there is a unique way of factoring MATH as a product MATH. These factorizations can be obtained from one another by using the commutation relations MATH on adjacent factors a finite number of times, and hence REF implies that the corresponding product MATH is independent of the factorization; we define MATH to be this common element of MATH. It is obvious that MATH, and defining MATH gives the desired representation MATH. If REF holds, then MATH so MATH is a NAME representation. |
math/9904116 | Let MATH and MATH. Since MATH, we can define MATH by MATH . It is easy to see that MATH is a NAME representation of MATH, so there is a homomorphism MATH such that MATH. Note that for any MATH and MATH we have MATH since elements of the form MATH generate MATH, MATH is surjective. To see that MATH is injective, we construct its inverse. First define MATH . Note that the MATH's and MATH's are isometries which generate MATH, that MATH, and that MATH . Now define MATH for MATH and MATH . It routine to check that the MATH's are isometries whose range projections sum to the identity. Suppose MATH, where MATH and MATH. Then MATH so by REF there is a NAME representation MATH such that MATH and MATH. We check that MATH is the identity on MATH, from which it follows that MATH is injective, and hence an isomorphism: if MATH, then MATH if also MATH, then MATH . |
math/9904116 | By REF , MATH is the universal MATH-algebra for collections MATH of isometries satisfying MATH that is, MATH. Suppose MATH. If MATH is irrational, then MATH is injective, and MATH is simple by REF . The existence and uniqueness of MATH is elementary when MATH is rational, and repeated applications of REF give MATH. |
math/9904116 | Let MATH be a twisted unit for MATH. By replacing MATH with MATH, we can assume that each MATH is a unit vector. Then MATH determines a multiplier MATH of MATH. As in REF, let MATH be the inductive limit MATH under the embeddings MATH, and let MATH be the canonical embedding of MATH in MATH. Since tensoring on the left by the rank-one projection MATH commutes with tensoring on the right by the identity, for each MATH there is an endomorphism MATH of MATH which satisfies MATH . Note that MATH is injective. Moreover, for any MATH we have MATH and hence MATH is a semigroup homomorphism. Let MATH be the embedding of REF , and define MATH by MATH. We claim that MATH is a crossed product for the twisted semigroup dynamical system MATH. Each MATH is an isometry, and MATH, so MATH is an isometric MATH-representation of MATH in MATH. If MATH, then MATH and since elements of the form MATH have dense linear span in MATH, we deduce that MATH for all MATH and MATH. Thus MATH is a covariant representation of MATH in MATH. We now verify REF of a crossed product. Suppose MATH is a covariant representation of MATH in a unital MATH-algebra MATH. Define MATH by MATH . We claim that MATH is a NAME representation of MATH in MATH. If MATH, then MATH and MATH . With MATH an orthonormal basis for MATH, REF gives MATH . To see that MATH is multiplicative, suppose MATH and MATH. Then MATH . Thus MATH is a NAME representation of MATH. Let MATH be the integrated form of MATH; that is, MATH satisfies MATH. We claim that MATH and MATH, giving REF of a crossed product. If MATH, then by REF we have MATH and since elements of the form MATH have dense linear span in MATH, we deduce that MATH. For the second part of the claim, we calculate MATH . To verify REF of a crossed product, suppose MATH. Then MATH and since MATH is generated as a MATH-algebra by MATH, this shows that it is also generated by MATH, as required. We have shown that MATH is a crossed product for MATH. Since MATH is the twisted semigroup crossed product of a nuclear MATH-algebra by an abelian semigroup of injective endomorphisms, it is nuclear by CITE. |
math/9904116 | Since MATH is a twisted unit for MATH, REF applies. |
math/9904117 | The theorem follows from the following two facts, applied to MATH and MATH. CASE: Let MATH be a closed sub-manifold and MATH a smooth function. Then there exists a smooth function MATH such that MATH. Moreover, if MATH is bounded from below, MATH can be chosen to be bounded from below too, and if MATH is proper and bounded from below, MATH can be chosen to be proper and bounded from below. CASE: Let MATH be proper and bounded from below. Then the average MATH of MATH by a compact group action is also proper and bounded from below. Let us prove the first fact. Fix a tubular neighborhood MATH of MATH in MATH, and let MATH be a smooth projection which extends to a proper map from the closure MATH to MATH. Let MATH, MATH be a smooth partition of unity subordinate to the covering of MATH by the two open sets MATH and MATH. Pick a smooth function MATH which is proper and bounded from below (see, e.g., REF). Then MATH has the desired properties. To prove the second fact, notice that MATH is contained in MATH, which is the image of the compact set MATH under the continuous action mapping MATH. |
math/9904117 | For any MATH, on every component MATH of MATH, we have MATH . |
math/9904117 | Let MATH be a point in MATH, let MATH be the infinitesimal stabilizer of MATH, and let MATH be the element assigned to the orbit type stratum containing MATH. Let MATH be any element whose projection to MATH is MATH. Pick an open neighborhood MATH of the orbit MATH which equivariantly retracts to the orbit. The constant function MATH is an abstract moment map on MATH whose assignment is MATH. Choose an invariant partition of unity MATH subordinate to the covering of MATH by the open subsets MATH, with the support of MATH contained in the open set MATH. The convex combination MATH is an abstract moment map; this follows from REF , applied to open subsets of the manifold. |
math/9904117 | The function MATH is well defined, proper, and bounded from below. Since MATH is closed, MATH extends to a function MATH that is proper and bounded from below. (See REF in the proof of REF .) For each MATH, let MATH be an element whose projection to MATH is MATH. We choose MATH to meet the following additional requirement: MATH. If MATH, this condition is automatically satisfied, and if MATH, this choice is possible because MATH. Let MATH be a tubular neighborhood of the orbit through MATH which equivariantly retracts to the orbit and on which the function MATH differs from the value MATH by less than MATH. Then the constant function MATH is an abstract moment map on MATH with assignment MATH and whose MATH-component is bounded from below by MATH. Choose an invariant partition of unity MATH subordinate to the covering of MATH by the open subsets MATH, with the support of MATH contained in the open set MATH. Then the convex combination MATH is an abstract moment map; this follows from REF , applied to open subsets of the manifold. Moreover, since the MATH-component of each MATH is bounded from below by MATH, the same holds for MATH. Since MATH, and MATH is proper and bounded from below, MATH is proper and bounded from below. |
math/9904117 | Let MATH be a minimal stratum and let MATH be the connected component of identity of MATH for MATH. Assume MATH. Then the equivariant NAME class MATH of the normal bundle to MATH is a non-zero torsion element in MATH. In fact, MATH is annihilated by the image of MATH. Alternatively, MATH, because MATH. |
math/9904117 | By using a partition of unity on MATH, the corollary can be reduced to the case where MATH is a linear space and MATH. This case follows immediately from REF when MATH is replaced by MATH with the trivial MATH-action on the second factor. |
math/9904117 | Let us first examine the case where the action is locally free near MATH. Fix a basis MATH in MATH. Then the vector fields MATH form a basis in the tangent space to the orbit at every point of a MATH-invariant neighborhood MATH of the orbit through MATH. By setting MATH we thus obtain a form defined along the orbits in MATH. We extend it to a differential form MATH on MATH by taking its composition with an orthogonal projection to the orbit with respect to a MATH-invariant metric. It is easy to see that MATH satisfies the condition MATH. Let us now prove the proposition in the general case. Pick a closed subgroup MATH whose NAME algebra MATH is complementary to MATH in MATH. A small MATH-invariant neighborhood MATH of the orbit MATH through MATH can be identified, by the slice theorem, with a neighborhood of the zero section in the normal bundle MATH to MATH in MATH, with the action induced by that on MATH. We can apply REF to the linear MATH-action on the fibers of MATH, equipped with the abstract moment map MATH induced from MATH. As a result, we get a smooth family MATH of one-forms on the fibers of MATH, such that MATH for all MATH. The MATH-orbits form a foliation which is transverse to the fibration MATH. We extend MATH to a one-form on a whole neighborhood of MATH by making MATH vanish on the vectors tangent to the MATH orbits. The resulting form is a MATH-invariant form MATH on MATH so that MATH for all MATH, and MATH for all MATH. The MATH-action on MATH is locally free. Let MATH be the form defined as above by REF and extended to MATH so that it vanishes on the vectors tangent to the fibers of MATH. Then MATH for all MATH. Since the vector fields MATH for MATH are tangent to the fibers of MATH, we also have MATH for all MATH. The form MATH has the desired property, that MATH for all MATH. |
math/9904117 | By REF , there exists an open covering of MATH by invariant sets MATH, and on each MATH there exists an invariant one-form MATH such that MATH for all MATH. Let MATH be a partition of unity subordinate to this covering, with MATH supported in MATH for each MATH. Define MATH. Then MATH on MATH for all MATH. |
math/9904117 | Consider the new abstract moment map MATH. By REF , there exists an invariant neighborhood MATH of MATH in MATH and a MATH-invariant one-form MATH on MATH such that MATH on MATH. Let MATH; then MATH on MATH. |
math/9904117 | Pick an element MATH whose restriction to MATH is equal to MATH, and apply REF to the abstract moment map MATH. |
math/9904117 | Take a tubular neighborhood of MATH which retracts to MATH, and apply REF to each of its connected components. |
math/9904117 | It is clear by definition that MATH if MATH is Hamiltonian. Conversely, assume that MATH. Then there exists a MATH-equivariant equivariantly closed two-form MATH such that the assignment of MATH is also MATH. The difference MATH is an abstract moment map with the zero assignment. By REF , MATH is exact and therefore Hamiltonian REF . Thus MATH is Hamiltonian as the sum of two Hamiltonian abstract moment maps, MATH and MATH. |
math/9904117 | The exactness of the left column is a particular case of a more general fact, that the cohomology of the basic NAME complex of MATH is equal to MATH, if MATH is compact or, more generally, if the MATH-action is proper, even when the action is not free. This result, due to CITE, is similar to the NAME theorem and can be proved in the same way. An easy proof is as follows. Recall that the sequence of sheaves of singular cochains on MATH (with real coefficients) is a fine resolution of the constant sheaf MATH on MATH. Furthermore, basic forms on MATH-invariant open subsets of MATH form a sheaf on MATH. This sheaf is a resolution of the locally constant sheaf because it is locally acyclic. Indeed, by using the fact that MATH is compact (or that the action is proper) and adapting the proof of the NAME lemma, one can show that the basic cohomology of a neighborhood of an orbit is the same as of the orbit itself, that is, zero in positive degrees. It is easy to see that this sheaf is also fine because it admits partitions of unity. Thus basic forms on MATH provide another fine resolution of the constant sheaf on MATH. Since the cohomology of both resolutions are equal to the NAME cohomology of the constant sheaf on MATH, they are equal to each other. The middle column is exact by the definition of equivariant cohomology via the equivariant NAME complex. (See, for example, CITE and CITE.) Exactness of the right column follows from REF . The fact that the top two rows are exact follows directly from the definitions of the spaces involved. The commutativity of the diagram is clear. Finally, commutativity with the exactness of the columns and the top two rows implies that the bottom row is exact by simple diagram chasing. |
math/9904117 | First note that by adding, if necessary, an additional copy of MATH (with the trivial MATH-action) to MATH, we can always make MATH even. There exists a MATH-invariant complex structure on the vector space MATH; fix one. We obtain a representation MATH of MATH on MATH with weights MATH. The infinitesimal action of the NAME algebra MATH of MATH on MATH is then given by the vector fields MATH . From now on we will forget about the reality conditions and assume that MATH is a complex-valued one-form. When such a form is found, it will suffice to replace it by the real form MATH, which still has the desired properties because MATH are real vector fields. We will also forget about the equivariance conditions. If a form MATH such that MATH for all MATH is constructed, and if MATH is equivariant, we can replace MATH by its average. Any one-form on MATH can be written in the form MATH for some smooth functions MATH and MATH, MATH. For such a one-form, the function MATH defined by MATH, with the vector fields MATH given by REF, is MATH . Conversely, for any function MATH which has the form REF for some smooth functions MATH, MATH, there exists a one-form MATH such that MATH for all MATH. Namely, just take MATH to be given by REF. To prove REF , it is thus enough to prove the following Let MATH be a MATH-valued function on a neighborhood of the origin in MATH, vanishing at the origin and satisfying the second condition of an abstract moment map: for any subgroup MATH, the function MATH is locally constant on the set of MATH-fixed points. Then there exist smooth functions MATH and MATH such that MATH is given by REF on a neighborhood of the origin. The converse is easy: any MATH of the form REF satisfies the second condition of an abstract moment map. Let us start with polynomial functions and one-forms: Let MATH be a polynomial function which vanishes at the origin and which satisfies the second condition of an abstract moment map. Then there exist polynomials MATH and MATH on MATH such that MATH is given by REF. Since MATH is polynomial, we can write it uniquely as a sum of monomials, MATH summing over MATH and MATH in MATH, where the coefficients MATH are in MATH. For every subset MATH, denote by MATH the subset of MATH consisting of all vectors MATH for which MATH if and only if MATH. All the points MATH in MATH have the same stabilizer, MATH, whose NAME algebra is MATH . Since MATH satisfies the second condition of an abstract moment map, MATH is constant on MATH. Since, additionally, MATH is continuous and vanishes at the origin, MATH vanishes on MATH. Let us analyze what this condition tells us about the coefficients MATH. The polynomial MATH vanishes on MATH for all MATH if and only if for each MATH, the summand MATH vanishes on MATH for all MATH. Fix MATH and MATH, and restrict attention to MATH . Since the monomial MATH does not vanish on MATH, its coefficient, MATH, must vanish for all MATH. By REF, a linear functional that vanishes on MATH is a linear combination of MATH, MATH. Therefore, MATH, and MATH . Since for each MATH, either MATH or MATH factors out of the monomial MATH, MATH is of the form REF. We will now show that the theorem we need to prove in the smooth category follows from its polynomial version, which has already been proved. In other words, we will deduce REF from REF . To this end, let us reformulate these propositions as assertions that certain sequences of homomorphisms are exact. Denote by MATH the ring of complex-valued polynomials in MATH and MATH, MATH. Define the modules MATH, MATH, over MATH as follows: CASE: MATH is the space of one-forms MATH with MATH and MATH in MATH. CASE: MATH is the tensor product MATH over MATH. CASE: For each subset MATH, denote by MATH the ring of polynomial functions on MATH. The restriction homomorphism MATH makes MATH into a MATH-module. Set MATH where MATH is the NAME algebra of the stabilizer of MATH, given by REF. Define the sequence of homomorphisms MATH by setting MATH with MATH and MATH to be the homomorphism MATH associated with the restrictions MATH and MATH. In other words, the MATH-th component of MATH is the MATH-component of MATH restricted to MATH. Hence, MATH satisfies the second condition of an abstract moment map if and only if MATH, and MATH is associated with a one-form if and only if it is in the image of MATH. REF is equivalent to the sequence REF being exact. Denote by MATH and MATH, respectively, the algebras of germs of analytic, respectively, smooth, functions on MATH at the origin. Let MATH, where MATH, be the modules defined similar to MATH but in the category of smooth germs at the origin. Note that MATH are modules over MATH. As before, we have a sequence of homomorphisms MATH . To prove the theorem in the smooth category, it suffices to show that this sequence is exact. Note that the inclusions MATH and MATH make MATH and MATH into MATH-modules. The following lemma is obvious: MATH. To finish the proof of REF , we need to recall some facts from commutative algebra. Let MATH be a commutative ring and MATH a sub-ring of MATH. The ring MATH is said to be flat over MATH if for every exact sequence of MATH-modules MATH the sequence MATH is also exact. It is known that MATH is flat over MATH (see CITE, page REF ) and MATH is flat over MATH (see CITE, page REF ). This in turn implies that MATH is flat over MATH. Therefore, the exactness of REF implies the exactness of REF. |
math/9904117 | By REF it suffices to prove REF for the cohomology of the complex MATH. The first part of the theorem follows from the fact that if MATH then MATH. Therefore, the longest possible tuple MATH of distinct strata has MATH. Thus MATH for MATH. When MATH, the maximal stratum MATH is the open dense stratum in MATH, on which MATH. As a result, MATH. The second part of the theorem follows from the fact that if MATH then MATH. The same argument as before shows that MATH whenever MATH. |
math/9904117 | Denote by MATH the tuple MATH in which each MATH occurs MATH times and the strata MATH are ordered and distinct: MATH. Denote by MATH the number of MATH's such that MATH; call this number the fatness of the tuple. As is easy to check, the fatness of a MATH-tuple is no greater than MATH. For each integer MATH let MATH be the space of MATH-cochains that are supported on tuples of fatness MATH. (This is consistent with the previous definition of MATH.) Then MATH as vector spaces. Set MATH . For every nonzero cochain MATH in this space, there exists a unique integer MATH between MATH and MATH and a unique decomposition MATH such that MATH for all MATH and such that MATH. An easy computation shows that MATH for all MATH and that MATH. In particular, MATH and MATH are subcomplexes, and the assignment cohomology splits: MATH . It remains to show that MATH vanishes. Define a linear map MATH by MATH . An explicit computation shows that MATH where MATH . Therefore, MATH is chain homotopic to zero and, as a consequence, induces the zero map MATH in homology. Denote by MATH the natural inclusion. It is clear that MATH is an isomorphism. Thus MATH is an isomorphism. Since MATH, this is possible only when MATH. |
math/9904117 | The theorem is an immediate consequence of the fact that the sequence of complexes MATH is exact. To prove the exactness of REF, note that the space MATH is the kernel of the restriction map MATH by its definition. To see that the restriction map is onto, note that every cochain MATH in MATH can be extended to a cochain MATH in MATH by declaring MATH to be zero whenever not all of MATH's are in MATH. |
math/9904117 | The short exact sequence REF naturally induces a short exact sequence of complexes, MATH and hence the long exact sequence REF in cohomology. |
math/9904118 | Suppose that MATH is a local biholomorphic change of coordinates with MATH. Since REF is independent of the choice of defining function we can choose MATH as a defining function for MATH. By the chain rule we have that MATH . Now the only crucial point is that the matrix on the right hand side has holomorphic entries. Hence, if we compose with MATH, this matrix will be annihilated by the CR - vector fields on MATH. Applying the vector fields MATH as in REF to this equation we conclude that MATH . Evaluating this identity at MATH, we obtain REF. |
math/9904124 | If MATH, this is CITE Hence we will assume that MATH, and fix some MATH for which REF on MATH hold true. Let MATH be a blowing up, such that MATH is non-singular, MATH a normal crossing divisor and such that, for MATH, the image of MATH is an invertible sheaf, isomorphic to MATH over MATH. So MATH, for some divisor MATH, supported in MATH. Let MATH be the induced morphism. MATH is a proper morphism with image MATH. Let MATH be the ideal sheaf of MATH and MATH . Blowing up again, we may assume that MATH for some divisor MATH with MATH. For all MATH the sheaf MATH will be generated by global sections. For some MATH, the number MATH will not divide the multiplicity of any of the components of MATH. Allowing MATH to have multiplicities, we thereby got to the following situation: MATH . We will prove, by induction on MATH, that REF imply MATH . If MATH we may replace MATH by MATH and MATH by MATH for MATH. Thereby we are allowed to assume that MATH is generated by global sections, and that for the zero divisor MATH of a general section of this sheaf MATH is again projective and flat. Moreover MATH is a normal crossing divisor and MATH. Therefore MATH, MATH and MATH satisfy again REF . If MATH, we choose MATH. In both cases, the morphism MATH is affine and MATH is an affine variety. MATH has a section with zero divisor MATH. By REF, MATH has an integrable connection which satisfies the MATH-degeneration. The assumption MATH allows to apply CITE, and to obtain MATH for MATH. If MATH, we are done. If MATH one uses the exact sequence MATH to obtain REF by induction. |
math/9904124 | By REF it is sufficient to show, that for some MATH is generated by global sections, or that MATH is ample. By definition of nef, this hold true, if the sheaf MATH is nef. To this aim see REF we can replace MATH by a covering, unramified in MATH, and MATH by the pullback family. Thereby we may assume that MATH for an invertible sheaf MATH on MATH. For MATH let MATH be the MATH-fold fibre product, let MATH be a desingularization, and MATH . We write MATH. The morphism MATH is NAME and the general fibre is non-singular. Hence there are natural injective maps MATH both isomorphisms on some open dense subset of MATH. REF induces MATH hence a section of MATH with zero-divisor MATH. Blowing up MATH, with centers in a finite number of fibers, we may assume that the image MATH of MATH is invertible. The ideal sheaf MATH is trivial in a neighborhood of the general fibre. By REF MATH is nef, hence MATH is nef, as well. Let us write MATH. Then MATH contains a nef subsheaf, isomorphic to MATH in a neighborhood of the general fibre. Moreover, by REF, MATH . CITE implies that the sheaf MATH is nef. By the definition of MATH and by REF, the natural inclusion MATH is an isomorphism on some open dense subset of MATH. Using REF one obtains a nef subsheaf of MATH of full rank, hence the latter is nef, as well. Then MATH as well as its quotient MATH are nef. |
math/9904124 | If MATH, this holds true with MATH replaced by the smaller sheaf MATH, since MATH . If MATH, that is, if MATH is semi-stable, then MATH. By CITE, one has an inclusion MATH, hence MATH. |
math/9904124 | Choose a non-singular projective compactification MATH of MATH such that MATH extends to MATH. REF hold true for MATH. By REF the degree of MATH is smaller than a constant depending on MATH and MATH. There exists a finite covering MATH and MATH which induces MATH . We may assume, in addition, that MATH has a semi-stable model MATH, with MATH non-singular. By the definition of stable surfaces in CITE (or of stable canonically polarized varieties in CITE), MATH . CITE, gives an injective map MATH which is an isomorphism over MATH. Hence MATH is bounded, as well. |
math/9904127 | Assume first that MATH is invariant under gauge automorphisms MATH. Since the representation MATH of the gauge group on MATH, given by MATH, is unitary, the MATH also form an orthonormal basis in MATH. Since the endomorphism associated with a NAME space of isometries as in REF is independent of the choice of an orthonormal basis in MATH, it follows that, for any field operator MATH, MATH . Conversely, assume that MATH is gauge invariant. Let MATH and let MATH be a gauge transformation. Then one has for any field operator MATH so that MATH. |
math/9904127 | Since MATH commutes with MATH, the implementer MATH is even: MATH . This implies that MATH (see REF) so that, by REF, MATH . |
math/9904127 | If MATH is an antisymmetric operator from MATH to MATH of finite rank, then one readily verifies that MATH (here the expressions of the form MATH are defined by expanding MATH with MATH, and by setting MATH). Approximating MATH by such finite rank operators in the NAME - NAME norm (compare CITE), one finds that MATH because MATH. It follows that MATH . Since MATH commutes with MATH and MATH, it also commutes with all components of MATH and MATH, including the operators MATH etc. The operator MATH is by REF a bounded function of these components, so that MATH . Hence we get MATH as claimed. |
math/9904127 | The subspace MATH is invariant under MATH because MATH commutes with the components of MATH (compare the proof of REF ). Since MATH is an orthonormal basis in MATH, it follows from the canonical anticommutation relations that MATH . Similarly, MATH commutes with the basis projection MATH (compare REF) and leaves MATH invariant, so that MATH is also left invariant. It then follows from the canonical anticommutation relations that MATH transforms like the MATH-fold antisymmetric tensor product of MATH under MATH. Thus we see from REF that the representation of MATH on MATH is unitarily equivalent to MATH, and the same holds true for the representation MATH. |
math/9904127 | Assume that MATH is localized in MATH. Let MATH be the standard basis in MATH, let MATH be a component of the complement of MATH, and let MATH with MATH. Then MATH is gauge invariant, and MATH is an observable localized in MATH. Since MATH is localized in MATH, one has MATH. Since the MATH are linearly independent, it follows that MATH . Now let MATH be the (basis) projection onto MATH, and let MATH be the corresponding NAME state. One has MATH and, since MATH belongs to the annihilator ideal of MATH, MATH . Therefore one gets from REF, in the special case MATH, that MATH. It follows that there exists MATH such that MATH. By the same argument, MATH for some MATH. Then REF yields that MATH. Therefore these phase factors depend only on MATH and not on the functions. Conversely, assume that REF holds. Then MATH acts on fields localized in bounded regions in MATH like a gauge transformation, and therefore like the identity on observables localized in MATH. It follows that MATH is localized in MATH. |
math/9904127 | The rather lengthy estimates of the NAME - NAME norms of these commutators, which are essentially due to NAME, can be found in CITE. |
math/9904127 | Let MATH be the orthogonal projection onto MATH, and let MATH as in REF. Then MATH and MATH commute with MATH because MATH and MATH do so. Therefore the positive part MATH of MATH and the operator MATH defined in REF also commute with MATH. It follows that MATH and MATH commute with MATH as well. Arguing as in the proof of REF , one finds for MATH and finally MATH . |
math/9904127 | MATH is invariant under MATH because MATH is invariant and because MATH commutes with MATH (see the proof of REF ). The assertion hence follows from REF . |
math/9904128 | MATH where MATH ranges over the MATH sub-matrices of MATH of the form MATH for some MATH. Hence MATH . |
math/9904128 | Apply REF to the polynomial MATH from REF . |
math/9904128 | Let MATH and let MATH where MATH is the MATH matrix obtained by deleting the MATH-th row and the MATH-th column of MATH. Then, by multi-linearity of the determinant, MATH hence MATH . Therefore, MATH . |
math/9904128 | Assume without loss of generality that the first MATH lines of MATH are independent. Let MATH, , MATH, , MATH be the sub-matrices obtained from MATH by deleting the last line and the MATH-th column. Then we can scale MATH in such a way that MATH . We have MATH. By reordering rows and columns, we obtain for each MATH that MATH is of the form: MATH where MATH is the sub-matrix of MATH obtained by deleting the last line and the MATH-th column. Set MATH. Now by REF , MATH . We consider now the morphism: MATH . Then MATH and MATH the first inequality because of REF , and the second because of REF . |
math/9904128 | REF implies MATH . We claim that MATH. Indeed, MATH and MATH can be obtained by solving systems of linear equations with coefficients in MATH, thus MATH. Also, MATH so MATH as claimed. By REF. Hence, by REF , MATH . |
math/9904128 | According to REF , MATH . Also, MATH and according to REF MATH and hence MATH. On the other hand, MATH . Hence MATH . |
math/9904128 | Let MATH be the sub-matrix of MATH obtained by deleting the MATH-th row and the MATH-th column. By NAME 's rule, MATH. Therefore we should define the degree MATH morphism: MATH . Then by REF , MATH . |
math/9904128 | We apply REF to the system: MATH with MATH to obtain MATH . We can bound MATH and MATH. We can apply REF to the map MATH to get MATH. REF implies MATH, hence: MATH and by REF , MATH . Thus, we can estimate that MATH . |
math/9904128 | By definition of the norm, MATH. By REF , we have: MATH . Knowing that MATH, we can use REF to deduce that MATH . According to REF , MATH where MATH is a universal constant. Thus, MATH where MATH is a universal constant. |
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