paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0106101 | The proof is by induction on the rank (fibre dimension) of MATH. For MATH the lemma is trivial. Suppose that the assertion of the lemma is true for MATH ; we shall prove that the assertion is true for MATH . Let MATH be the characteristic polynomial of MATH , with respect to MATH (MATH). For MATH , set MATH and set MATH . Because MATH we have that MATH where MATH denotes the closure of the set MATH . To complete the proof, it suffices to prove that each MATH has an open neighbourhood MATH such that MATH where MATH and MATH are complementary orthogonal vector subbundles of MATH of positive rank such that MATH . Let MATH be such that MATH and let MATH . Let MATH be a root of MATH of order at most MATH . Because MATH is not in MATH we have that MATH has order MATH . Then, by the smooth version of the NAME Preparation Lemma, in an open neighbourhood MATH of MATH we have MATH where MATH is a polynomial of degree MATH in MATH such that MATH and MATH. From the fact that MATH and MATH are both polynomials in MATH (with coefficients smooth functions of MATH ), it follows that MATH is also polynomial in MATH . We shall show that there exists an open neighbourhood MATH of MATH such that, for each MATH , MATH has a root of order MATH . Suppose not. Let MATH be such that MATH and, for each MATH , there exists MATH such that MATH with MATH a root of MATH of order less than MATH . Obviously, MATH is also a root of MATH and, because MATH , MATH is a root of order at least MATH of MATH . Hence MATH . Now, by REF , the sequence MATH is bounded and hence, by passing to a subsequence, if necessary, we can suppose that MATH with MATH . Then MATH and also MATH . Because MATH we have that MATH . But this implies that MATH is not a root of order MATH of MATH . It follows that, in an open neighbourhood MATH of MATH , we have that MATH has only roots or order MATH for any MATH. Thus we can write MATH , (MATH), where MATH is the root of MATH so that MATH is smooth. Hence MATH . Moreover, because MATH we can suppose that MATH is non-zero for any MATH . It follows that MATH is an eigenvalue of order MATH for MATH for any MATH . Let MATH be the eigenspace of MATH and let MATH be its orthogonal complement. It is easy to see that MATH , MATH , are smooth subbundles of MATH which have the required properties. The lemma follows. |
math/0106101 | Since any NAME manifold can be given a real-analytic structure (see CITE ), from REF it follows that it is sufficient to prove that one of REF or REF occurs on an open subset of MATH . It is easy to see that a one-dimensional homothetic foliation has integrable orthogonal complement, at least outside the points where the foliation is Riemannian. Thus, if MATH is not Riemannian on MATH , we can find an open subset MATH of MATH on which MATH is not Riemannian and MATH is integrable. Then MATH on MATH , and from REF it follows that MATH for any horizontal vector field MATH where MATH is the dilation of a harmonic morphism produced by MATH . By REF this is equivalent to MATH for any basic vector field MATH . Also, from REF , it follows that MATH is a basic function for any orthogonal horizontal vectors MATH and MATH . Hence, by REF, we have that MATH for any orthogonal horizontal vectors MATH and MATH . Since MATH , it follows easily from REF that, at each point MATH we have that MATH (apply, for example, CITE). Hence (see CITE), REF occurs on MATH and the proof follows from REF . |
math/0106101 | By REF , if the horizontal distribution MATH is integrable on an open subset of MATH , then MATH is integrable on MATH . Thus, by REF , it is sufficient to prove the case when MATH is nowhere integrable. Also, writing MATH , we can suppose that the leaves of MATH are the fibres of a harmonic morphism MATH , where MATH with MATH , and from now on we shall use the notations of REF . Let MATH be the field of self-adjoint negative semi-definite endomorphisms of MATH defined by MATH for horizontal MATH and MATH . By REF , MATH can be consistently diagonalized on a dense open subset of MATH ; let MATH be a point of this subset. There is an open neighbourhood MATH of MATH and an orthonormal frame MATH for MATH over MATH such that MATH for some continuous functions MATH with MATH smooth. Because MATH and MATH are basic we also have that MATH is basic; hence the MATH are basic as well. We can thus suppose that the MATH are basic. From REF we have MATH . We have the following alternative. Either REF there exists MATH and distinct MATH such that MATH , or REF for any MATH there are at most two distinct MATH such that MATH . Suppose that REF holds. By REF we have that MATH is basic for any MATH . Hence MATH is basic, and, because MATH on some open subset of MATH , we have that MATH is basic MATH . Thus, if REF holds, MATH is basic for all MATH , on some open subset of MATH . Then, by REF (see also REF ), MATH is homothetic on MATH and the proof follows from REF . Suppose that REF holds. If MATH for all MATH , then, by REF , MATH has geodesic leaves and the proof of the theorem follows from REF . Therefore we can suppose that, after renumbering if necessary, we have MATH at some point MATH . Then this holds on some open subset of MATH . Then, either MATH for MATH , on some open subset of MATH , or there exists a point MATH such that, after renumbering if necessary, MATH and MATH . In the latter case, because REF holds, we must have that MATH REF on some open subset of MATH . It follows that there exists an open subset MATH of MATH such that MATH , (MATH). From now on we shall work on MATH . By REF we have MATH . From REF we get MATH . From REF it follows that MATH is basic for MATH . Thus, if MATH at some point for some MATH , MATH , then MATH is basic and so MATH is Riemannian on some open subset of MATH ; hence, by REF , MATH is Riemannian on MATH . It remains to consider the case MATH for some function MATH . Now, either, MATH on some open subset, or, we have MATH on a dense open subset. In the former case, by REF, we have that MATH is basic on some open subset. But, by REF, MATH is also basic, and hence MATH , MATH are basic on some open subset. Since MATH for MATH , MATH has basic mean curvature form. Then, by REF , MATH is homothetic on some open subset and hence, by REF , MATH is homothetic on MATH ; the proof of the theorem follows from REF . It remains to consider the case when MATH . Because MATH is skew-symmetric, at each point MATH , for any MATH with MATH there exists MATH , MATH , such that MATH . Hence at each point MATH we have that either MATH and MATH or MATH and MATH . Suppose that MATH ; then this holds at all points of an open subset, and on that subset we must have MATH . NAME, because MATH is skew-symmetric, we must have MATH and so MATH is not identically zero; in particular MATH is even, that is, MATH for some integer MATH . From REF we get MATH hence, MATH is basic. Thus, if MATH , since MATH is basic, we deduce that MATH is also basic and the proof follows as before. There remains the case MATH which we now consider. NAME the previous discussion, we have that MATH , MATH , and we are now on an open subset on which we have the following: MATH . Moreover, we can assume that MATH and MATH are nowhere zero. Furthermore, because MATH we have that MATH . Hence MATH , equivalently MATH . From this and REF it follows that we have for MATH , MATH . Next, we compute MATH . MATH . We can choose a basic orthonormal local frame MATH such that MATH . Then, from the above calculation we have MATH . Recall that MATH for all MATH ; hence MATH . It follows from the last equation that MATH . From REF we get MATH which is equivalent to MATH . From REF with MATH , MATH , we get that MATH and hence on differentiating this with respect to MATH , by REF , we obtain MATH which is equivalent to MATH . From REF we get that MATH which, because MATH is nowhere zero, is equivalent to MATH . From REF it follows that MATH which, if MATH is not Riemannian (equivalently, MATH), implies that MATH . This is impossible if MATH , since MATH , MATH . The proof of the theorem is complete. |
math/0106101 | Let MATH . If MATH then from REF it follows that MATH for any pair of orthogonal vectors MATH and MATH where MATH is the dilation of MATH . Hence MATH for any horizontal vector MATH , which implies that REF holds. From now on we shall assume that MATH is nowhere zero. In particular REF does not hold and hence, by REF , the fibres of MATH form a Riemannian foliation locally generated by Killing vector fields. Therefore MATH is basic and thus both MATH and MATH are pull-backs by MATH of functions on MATH ; we shall denote these two functions by the same letters MATH and MATH . We have MATH , MATH and we shall denote by the same letter MATH both the metric on MATH and the metric on MATH with respect to which MATH is a Riemannian submersion with geodesic fibres. To complete the proof we must show that REF holds on an open subset of MATH . From REF we get MATH for any pair MATH of orthogonal horizontal vectors. As in the proof of REF , there is an open subset MATH of MATH on which we can consistently diagonalise MATH with respect to a basic orthonormal frame MATH of MATH . Then, from REF it follows that MATH for any MATH . It follows that either MATH for all MATH or, after renumbering if necessary, on some open subset we have MATH and MATH is not identically zero. From REF we obtain MATH for any MATH . Hence MATH for some non-negative function MATH with MATH smooth. From REF we also get MATH . Suppose that MATH (equivalently, MATH has geodesic fibres) then MATH is constant and MATH for MATH . Moreover, from REF it follows that MATH constant. Hence MATH must be even and MATH is MATH times the NAME form of an almost Hermitian structure on MATH . Since MATH , this structure is almost NAME giving REF . Suppose instead that MATH . We shall obtain a contradiction. We have MATH and hence, because MATH is skew-symmetric, MATH and MATH . Thus, MATH which implies MATH . Together with MATH , this gives that MATH . Also, REF becomes MATH which can be written as MATH . Since MATH is a harmonic morphism, by the chain rule (see, for example, CITE ), we have that MATH for any smooth function MATH on MATH . From REF we get that MATH . From REF we get that MATH . Because MATH , as before, we have that MATH must be odd. Then we can choose the frame MATH such that, with respect to this frame, MATH is given by REF . In particular, MATH and REF becomes MATH . The relations REF give MATH . Because MATH , relations REF and the fact that MATH imply MATH . Note that REF can be written as follows: MATH . From now on MATH , MATH and MATH , MATH , will be viewed as objects on MATH . From REF and MATH we get that MATH . Let MATH be the foliation formed on MATH by the level hypersurfaces of MATH . Note that MATH are tangent to MATH and MATH is normal to MATH . Also, MATH is constant along the leaves of MATH and, from REF , it follows that MATH restricted to any leaf MATH of MATH is MATH times the NAME form of an almost Hermitian structure MATH on MATH . But MATH and the constancy of MATH along MATH imply that MATH is actually an almost NAME structure on MATH . In particular, MATH which is equivalent to MATH where MATH (as is usual, we denote by the same letters the distributions MATH and MATH and the orthogonal projections onto them). After a short calculation (similar to the one used in the proof of REF to find MATH ), by using the fact that MATH is integrable, we get that MATH . Hence MATH . We claim that MATH and MATH imply that MATH has geodesic fibres. Indeed, we shall see that MATH where MATH is the second fundamental form of MATH . Hence MATH, that is, MATH is geodesic. Next, we prove REF . We have MATH . Obviously, MATH , and MATH implies that MATH for any MATH (recall that MATH is a local frame for MATH). Also MATH and MATH imply that MATH for any MATH . For MATH , we have MATH . Similarly, MATH . Relation REF now follows from REF . From MATH and MATH it follows that MATH . Hence MATH is basic for MATH. It follows also that MATH is basic for MATH , equivalently, MATH is defined by horizontally holomorphic submersions (see the Appendix for the definition of horizontally holomorphic submersion). Moreover, by taking on the codomain of such a submersion the metric induced by any leaf of MATH, from REF of the Appendix we get that MATH is defined by horizontally holomorphic harmonic submersions. It is easy to see that MATH for any MATH . Because both MATH and MATH are basic for MATH we get that MATH is a conformal foliation with dilation MATH . Moreover, because MATH is constant along the leaves of MATH we get that MATH is a homothetic foliation. Thus, MATH is locally defined by horizontally homothetic submersions MATH with geodesic fibres (equivalently, MATH corresponds, locally, to a warped product decomposition of MATH ). Moreover, MATH is an NAME MATH -manifold (apply, for example, CITE ) . From CITE (or CITE ) it follows that MATH satisfies the following equation: MATH where the NAME tensor of MATH is given by MATH . After a straightforward elementary calculation we get that REF are incompatible and the proof follows. |
math/0106101 | As is well-known, by identifying MATH with MATH the MATH -valued one-form induced by MATH becomes the `horizontal' projection MATH . Let MATH be horizontal vector fields. With MATH the connection on MATH we can write MATH . Hence MATH . Let MATH be a local orthonormal frame for MATH . From REF we get MATH . As is well-known (and easy to prove) MATH for any vertical vectors MATH . The proof follows from this fact and REF . |
math/0106103 | Each character MATH is of form MATH with MATH and MATH. |
math/0106103 | For notational convenience, we consider the equation MATH replacing each MATH with MATH. For MATH, we say a tuple MATH to be a MATH-solution and MATH a MATH-admisible sum if MATH with MATH and MATH. We shall show that there are only finitely many MATH-admissible sums. Note that if MATH is contained in MATH or MATH, which we call a degenerate case, then the only MATH-admissible sum is MATH. We show the finiteness by the induction on the size of MATH. If MATH, then it is degenerate. Assume MATH and it is not degenerate. Then the equivalence among MATH-solutions up to scalar multiplications is just the identity. A MATH-admissible sum is either as to a MATH-solution whose non-trivial subsums do not vanish or of the form MATH such that MATH is a MATH-admissible sum and MATH is a MATH-admissible sum for a non-trivial subset MATH of MATH. By the theorem and by the induction hypothesis respectively, there are only finitely many such sums. |
math/0106103 | It is straightforward that MATH in MATH. So the product has at most as strong topology as that determined by MATH. Hence it is sufficient to show that the identity map MATH is continuous. In general, the product topology of topological groups MATH and MATH has following characterization. Let MATH and MATH be natural injections: MATH. Suppose that MATH is a topological group and MATH is a homomorphism between abstract groups. Then MATH is continuous if and only if so are both of MATH and MATH. Due to REF , MATH and MATH are continuous. So we are done. |
math/0106103 | Applying REF to MATH REF , we observe that the product is also sequential. So it is sufficient to show that MATH has no nontrivial convergent sequence. Suppose that MATH is a sequence in MATH converging to MATH. By REF , there exists MATH such that MATH is a sum of less than MATH numbers in MATH and in MATH, respsctively, for sufficiently large MATH. Due to REF , there are only finitely many such sums. Therefore MATH is eventually equal to MATH. |
math/0106103 | We proceed similarly as in the proof of REF . We change the definition of MATH-solutions according to the restriction: they are tuples MATH such that MATH, MATH, MATH and no subsum of MATH or of MATH vanishes. So there is no MATH-solutions in the degenerate case. By the induction on the size of MATH, we prove the finiteness of MATH-solutions. Assume that MATH is not degenerate. Then there are two possibilities for a MATH-solution: either it has no vanishing non-trivial sums or it is the union of a MATH-solution and a MATH-solution for a non-trivial subset MATH of MATH. Hence REF and the induction hypothesis yields the conclusion. |
math/0106105 | We show REF for a product. Suppose that MATH is a family of TA groups. Let MATH be a finite subset of MATH and MATH a neighborhood of MATH in MATH for MATH. We shall prove that the neighborhood MATH generates the whole product MATH. For any MATH, set MATH, that is, MATH with MATH for MATH and MATH for MATH. For each MATH, there exists a finite sequence MATH in MATH such that MATH. We may assume that MATH and MATH for all MATH. Let MATH for MATH denote MATH. Then MATH and MATH. So we are done. |
math/0106105 | CASE: Assuming that MATH is a dense TNA subgroup of MATH, we shall show that MATH is also TNA. For a neighborhood MATH of MATH in MATH, there exists another MATH such that MATH. By the assumption, we find an open subgroup MATH of MATH with MATH. Then the closure MATH in MATH is an open subgroup contained in MATH. |
math/0106105 | Let MATH be an open subgroup of MATH. Then MATH is open in MATH, and hence MATH, that is, MATH. Since MATH is open in MATH, we also have MATH. Therefore MATH. |
math/0106105 | Let MATH be a neighborhood of MATH in MATH. We shall find an open subgroup MATH contained in MATH. We choose neighborhoods MATH, MATH and MATH of MATH in MATH as follows. First let MATH be such that MATH. By the assumption, there is an open subgroup MATH of MATH contained in MATH. Let MATH be open with MATH and MATH. We denote by MATH the natural homomorphism MATH. Since MATH is open in MATH, it contains an open subgroup MATH. We set MATH. We show that MATH. Suppose that MATH. Since MATH, we have MATH. So there is MATH with MATH. Then MATH, and hence MATH. Now let MATH be the subgroup of MATH generated by MATH. Then MATH is open and MATH as desired. |
math/0106105 | CASE: We proceed as in the argument for the universality of MATH among the TNA Polish groups (compare CITE). Since MATH is TNA and of weight MATH, we have that MATH. The implication MATH is straightforward. We show that MATH. Suppose that MATH is a TNA group. Let MATH be a local base for MATH consisting of open subgroups. We may assume that MATH. We denote by MATH the disjoint union of quotient spaces MATH . The natural action of MATH on MATH such that MATH induces an isomorphic embedding MATH. Since MATH for each open subgroup MATH, we get MATH. CASE: The implication MATH follows from the universality of MATH among (discrete) Abelian groups of size MATH (see REF ). Let MATH be an Abelian TNA group. By virtue of CITE, there exists an isomorphic embedding from MATH into MATH . The estimation of the cardinality is similar as in REF . |
math/0106105 | MATH is zero-dimensional: Every open ball is clopen. MATH and MATH have SMOG: Suppose that MATH. Then there is MATH with MATH. The open subgroup MATH excludes MATH. MATH is not TNA: For each MATH, we have that MATH. Accordingly every nontrivial subgroup is unbounded. MATH is zero-dimensional: Since MATH is discrete, zero-dimensionality of MATH is preserved under the quotient. MATH and MATH do not have SMOG: Let MATH denote the MATH-th unit vector MATH. We show that MATH belongs to all open subgroups. Let MATH be an open subgroup. Then for sufficiently large MATH, we have MATH. Accordingly MATH. MATH and MATH are not TA: the subgroup MATH is open and proper. MATH is not zero-dimensional: REF below. MATH is totally-disconnected: Suppose that MATH. Then for MATH sufficiently small, MATH is clopen and does not include MATH, where MATH is the open ball of MATH with center MATH and radius MATH. MATH is not zero-dimensional: NAME is also hereditary to quotients by a discrete subgroup. |
math/0106105 | Assuming that MATH is an open neighborhood of MATH contained in some open ball MATH with center MATH and radius MATH, we show that MATH is not closed in MATH. We define sequences MATH in MATH and MATH and MATH of integers by induction such that MATH. We set MATH and MATH. Since MATH is open and MATH as MATH, there is a natural number MATH with MATH. Due to REF , MATH as well, and hence for sufficiently large MATH, it occurs that MATH by REF , where we use REF freely. So we may find MATH such that MATH . Then we have MATH and MATH for each MATH. We show that the sequence MATH is convergent. Since it is in MATH, the sequence of MATH is bounded due to REF . The latter is increasing as well, and hence it is convergent. Therefore for any MATH, sufficiently large MATH and every MATH satisfy MATH. Since MATH is arbitrary, REF yields that the sequence MATH is NAME, so convergent. Since the sequences MATH in MATH and MATH in MATH converge to the same point in MATH, we conclude that MATH is not clopen in MATH. |
math/0106106 | By REF we may assume without loss of generality that MATH and that MATH is of the form REF . Suppose that MATH is not continuous and let MATH be a sequence of elements in MATH such that MATH . Let MATH be such that MATH. Since MATH is not isolated, then there exists a MATH with MATH. Continuing the process, we can inductively construct a sequence MATH of different points in MATH such that MATH . Let MATH be a sequence of open subsets of MATH and MATH a sequence of continuous bounded real-valued functions on MATH such that for any MATH and MATH . Since MATH we can define a function MATH in MATH by MATH . From REF and since MATH and MATH for MATH we get MATH . Since our space contains only bounded functions, the contradiction shows continuity of MATH. |
math/0106106 | CASE: Assume that the interior of the set of MATH-points of MATH is empty and MATH is a biseparating map. As before by REF we may assume without loss of generality that MATH and that MATH is of the form REF . First we are going to prove that MATH is continuous for every MATH. Assume that there is a MATH such that MATH is discontinuous. If MATH is not a MATH-point we put MATH. If MATH is a MATH-point we put MATH. Since the map MATH sending each MATH into MATH is continuous, all of the sets MATH are neighborhoods of MATH, and so is MATH. Moreover for any MATH the map MATH is discontinuous. Since the interior of the set of MATH-points is empty, the set MATH contains a point which is not a MATH-point. We denote by MATH one of such points. Let MATH be a sequence in MATH such that MATH . Let MATH be the constant function equal to MATH . Since MATH is not a MATH-point there is a sequence of open neighborhoods MATH of MATH with MATH and such that the intersection of all MATH is not a neighborhood of MATH; we may also assume that MATH, for MATH. It follows that MATH belongs to the closure of the following set: MATH . For each MATH, let MATH be such that MATH . Put MATH since MATH and MATH the series is convergent and MATH. For any MATH and any MATH we have MATH . Hence MATH . Since MATH belongs to the closure of MATH we get MATH, and this contradiction proves the continuity of MATH for every MATH. Suppose now that MATH is not continuous. Then there is a compact subset MATH of MATH and a sequence MATH in MATH with MATH and MATH. Let MATH be such that MATH for each MATH. Notice that the set MATH is infinite since otherwise there would be a point MATH with MATH for infinitely many MATH, contrary to the continuity of MATH. Without loss of generality we may assume that MATH when MATH, and take a sequence MATH of pairwise disjoint open subsets of MATH such that MATH for MATH. Let MATH be a sequence in MATH such that MATH and MATH for MATH. It is easy to see that MATH while, by REF, MATH so MATH is not bounded on MATH. Since this is not possible we conclude that MATH must be continuous. CASE: Assume now that the interior MATH of the set of MATH-points of MATH is not empty and let MATH be a continuous real-valued function on MATH such that MATH for MATH, and MATH at some point MATH. Put MATH. By REF MATH is an open subset of MATH, and consequently of MATH; clearly as a zero set MATH is also closed. Let MATH be a discontinuous linear bijection from MATH onto itself and define MATH by MATH . The map MATH is obviously biseparating but we need to check that it is well defined, that is, that MATH is continuous for any continuous MATH. It is of course true on the clopen subset MATH. Let MATH and let MATH. By REF the function MATH is constant on a neighborhood of MATH, hence MATH, and so MATH are both locally constant and consequently continuous. CASE: As above we assume without loss of generality that MATH and that MATH is the identity map. Suppose that MATH and that there is a point MATH which is not a MATH-point. We well see that there exists a linear bijection MATH from MATH onto MATH such that the map MATH is not well defined. Since MATH is not continuous, then there exists a sequence MATH of norm-one elements of MATH such that MATH, for MATH. On the other hand, since MATH is not a MATH-point, there exists a sequence MATH of neighborhoods of MATH such that the intersection of all MATH is not a neighborhood of MATH. We may assume that MATH, for MATH. Consequently MATH belongs to the closure of at least one of the following sets: MATH . Without loss of generality we may assume that MATH. For each MATH, let MATH, MATH, be such that MATH . It is clear that the map MATH belongs to MATH. Also, according to the description of MATH and to the fact that MATH, we deduce that MATH, which is absurd. Consequently, MATH is a MATH-space. To prove the other implication we just repeat the above argument in REF with MATH. |
math/0106107 | We first need to introduce some notation. For MATH we put MATH . For MATH and MATH we denote by MATH the one dimensional operator on MATH defined by MATH . Since the definitions of MATH and MATH involve only the structures that are preserved by MATH, MATH maps these sets on MATH and MATH, respectively. Notice that MATH is trivial if and only if MATH is dense, and MATH consists of operators of dimension one if and only if the closure MATH of the MATH is one-codimensional. Since MATH contains all projections onto closed one-codimensional subspaces, we have MATH . Hence MATH is simply equal to the set of all the one dimensional operators, so MATH maps a one dimensional operator onto a one dimensional operator. Fix a linear continuous functional MATH on MATH. For any MATH we have MATH for some MATH and MATH. If we change the point MATH but keep the same functional MATH, the operator MATH will still belong to the same element of MATH. Hence MATH belongs to the same element of MATH; this means that MATH and the functionals MATH are proportional. So there is a map MATH such that MATH . Since MATH is linear, so must be MATH; since MATH is trivial, the same must be true about MATH. Finally, since MATH is surjective, it maps an element of MATH onto an entire element of MATH so MATH is surjective. Assume MATH is another continuous linear functional on MATH, not proportional to MATH and let MATH be such that MATH . Since MATH then MATH and MATH belong to distinct elements of MATH, for any MATH, so MATH . Suppose the linear maps MATH and MATH are not proportional and let MATH be such that MATH and MATH are linearly independent. Then the operator MATH is one dimensional, while MATH is two dimensional. The contradiction shows that there is a linear bijection MATH and a map MATH such that MATH . As in the case of MATH, since MATH is a linear bijection so must be MATH. We now show that MATH is continuous. For any MATH and MATH we have MATH and by REF MATH . So MATH hence MATH or MATH . Hence MATH has a closed kernel, and consequently is continuous for any MATH. Since MATH is surjective, this means that MATH becomes continuous when composed with any continuous functional. So, by the Closed Graph Theorem MATH is continuous itself. We now can define a map MATH by MATH . By REF MATH and by REF MATH so each MATH is an eigenvector of MATH hence MATH, and MATH . Since MATH is linear it follows that MATH for all finite dimensional operators MATH. Notice that MATH since both MATH and MATH are invertible. Assume that MATH and let MATH be such that MATH and MATH . Put MATH where MATH is such that MATH. We have MATH while by REF MATH which is a contradiction since MATH is biseparating. Hence MATH so MATH . |
math/0106107 | Again we first need to introduce some notation. For MATH we denote MATH either MATH or MATH depending on the domain of the map MATH for MATH we put MATH . Notice that the sets MATH, and MATH have been defined solely using the properties that are preserved by MATH hence MATH . We show that MATH . The implication MATH is obviously true for all functions in MATH. Assume MATH and MATH. Since both MATH and MATH are nonzero maps there is a continuous one dimensional linear map MATH on MATH such that MATH . Put MATH . We have MATH so MATH which concludes the proof of REF . By REF MATH for arbitrary open sets MATH we have MATH if and only if MATH, so we get MATH . Since MATH, for MATH, the above proves that MATH is strictly biseparating: MATH . We need the following result from CITE. Assume MATH are normed spaces, MATH are realcompact spaces, and MATH is a linear bijection satisfying REF . Then there is a bijective homeomorphism MATH and a map MATH from MATH into the set of linear bijection from MATH onto MATH such that MATH . To finish the proof of REF let MATH be such that MATH and denote by MATH the constant functions on MATH equal to MATH, and to MATH, respectively. By REF for any MATH so MATH is separating. By the same arguments applied to MATH we conclude that MATH is biseparating. By REF MATH where MATH. Hence, by REF MATH to check that MATH is a continuous function it is enough to put into the above formula MATH equal, at every point of MATH to the identity map on MATH. |
math/0106112 | We relying on the explicit computations done in CITE. Any spinor field MATH can be written as MATH where MATH is a form of type MATH on MATH and MATH is a holomorphic section of MATH. Then the set of REF is equivalent to MATH for the pair MATH, and, whenever MATH, this is finally equivalent to the NAME Killing spinor equations CITE. Let MATH and MATH on MATH seen as the unit ball. Then, NAME 's CITE shows that, for any choice of multi-index MATH, MATH provides a NAME Killing spinor, as does MATH for any choice of multi-index MATH. In complex dimension MATH (MATH), NAME 's NAME above does not provide a priori NAME Killing spinors (the second set of equations above is not in general equivalent to the NAME Killing condition) but it is easily checked that the spinor fields given a few lines above are indeed NAME Killing spinors. |
math/0106112 | The first step is to prove that there exists a full set of MATH-harmonic spinors on MATH, asymptotic to the model spinors on MATH described in the previous section; once these spinors have been obtained, the proof will go along the arguments of CITE. The key point in this first step is to show that the zeroth order terms in the NAME formula in REF are always nonnegative if the curvature assumptions of the theorem are satisfied. Taking into account MATH, and letting MATH and MATH, the zeroth order term acting on sections of MATH is MATH . A case by case check yields that this is always equal to MATH, hence nonnegative. It then implies that the modified NAME operator MATH is coercive in MATH, so that we may find MATH-harmonic spinors for the MATH - structure on MATH which are MATH perturbations of the model NAME Killing spinors at infinity. As a second step, we apply the NAME formula to any such spinor field MATH and get MATH . Arguing as in CITE, our asymptotic conditions yield that the limit is zero, so that MATH is MATH-parallel. We now write MATH with MATH in MATH. As MATH respects the splitting of MATH into its MATH - components (where it is a modified connection) and the remaining components MATH (where it equals the NAME connection), one gets that MATH and that each component MATH is parallel if MATH. These remaining components are zero since they are in MATH (model NAME Killing spinors live in MATH). We finally get that each solution MATH solves the NAME Killing equation. As model spinors trivialize on MATH, the solutions trivialize MATH as well. |
math/0106112 | Let MATH be any of our special spinors and MATH a tangent vector. We compute in two different ways: MATH . Letting MATH, then MATH and MATH. Since, at any point, either MATH or MATH is non zero, this forces MATH. |
math/0106112 | We define the metric MATH on MATH such that horizontal and vertical spaces given by the NAME connection are orthogonal, the bundle projection MATH becomes a metric-preserving submersion and the vertical vector field MATH (induced from the MATH-action) has norm MATH. From REF it is an NAME metric. We now follow the argument in CITE. We extend the MATH frame bundle MATH to a MATH principal bundle MATH by letting MATH. The bundle MATH is a MATH frame bundle defining a MATH - structure on MATH. Moreover, the auxiliary bundle of this structure is the pull-backed bundle MATH which is trivial. Any choice of a global section then induces a MATH structure on MATH. |
math/0106112 | The first claim follows from straightforward computations, using the NAME and NAME formulas. We must now understand how our NAME Killing spinors transform when pulled back to MATH, then to MATH. We apply the computations of CITE to our case, cum grano salis due to the NAME signature. The pulled-back bundle MATH is made into a NAME bundle over MATH if one fixes MATH (conjugation of spinors on an even-dimensional manifold). The connection MATH induced by the NAME connection on MATH and the pull-backed connection on the auxiliary bundle of the MATH - structure acts then as follows: for any pull-backed spinor field MATH and any horizontal lift MATH on MATH of a vector MATH on MATH, MATH . As the metric MATH is defined from the natural connection on MATH whose curvature is MATH, the usual MATH-term formula yields MATH and finally MATH . Similarly, one computes MATH . Using the canonical global section on MATH over MATH, we can now turn any spinor field for the MATH - structure into a section for the MATH structure whose existence has been already remarked. Hence, if MATH is any spinor field, it may be seen as associated to the MATH - structure (and is acted upon by the connection MATH induced from the NAME connection tensorized with the pull-back connection) or to the MATH - structure (and is then acted upon by the NAME connection MATH which is better seen as NAME tensorized by the trivial connection). But the trivial connection on MATH differs from the pulled back connection by a factor which is exactly the connection MATH-form MATH of MATH and this shows that MATH and MATH. We now apply the preceding remarks to our set of NAME Killing spinors MATH trivializing MATH on MATH. From REF , we get that the pull back spinors MATH solve MATH . And it is now straightforward to show that these can be extended as parallel spinors on the cone MATH living in the desired subbundle. |
math/0106112 | The previous Lemma provides parallel spinors trivializing the component MATH (with respect to the action of the NAME form of MATH) of the spin bundle MATH of MATH. The NAME tensor MATH acts trivially on MATH, that is, for each MATH, MATH in MATH, MATH . Mimicking the representation-theoretic argument in CITE shows that the only MATH-form in MATH whose action on MATH is zero is the zero form itself. This proves MATH is a flat manifold; applying NAME formulas yields that MATH has constant negative holomorphic sectional curvature CITE. |
math/0106112 | We only need to check that MATH is simply-connected. Assume the contrary : let MATH be the universal covering of MATH. If the covering is not trivial, MATH has at least two ends since the unique end of MATH is simply-connected. This is of course a contradiction. |
math/0106114 | This follows because MATH . This has an elementary proof - see for example CITE or CITE. Thus MATH . |
math/0106114 | Let MATH, MATH be MATH-valued independent random variables that are also independent of MATH, where MATH. Applying REF to the sequence MATH we obtain MATH . Next, let MATH denote the MATH-field generated by MATH. Then applying REF , we see that MATH . Thus we also obtain that MATH . The result follows. |
math/0106114 | First, since MATH satisfies the triangle inequality, it follows that MATH. Next, since MATH, where MATH denotes the integer part of MATH, we see that MATH . |
math/0106114 | Let us first show the lower bound. Here the proof is very similar to the proof of CITE. We know that MATH where by convention MATH, and MATH denotes the dual space to MATH. From this, we immediately see that MATH since MATH whenever MATH. To finish the lower bound, we see that MATH and the result follows by REF . Now let us focus on the upper bound. Really the first part of this proof follows by an inequality obtained independently by CITE, CITE, CITE, and CITE. But we shall provide a self contained proof that is essentially a copy of the proof of this same result that may be found in CITE. From REF , it follows that MATH. We have that MATH that is to say, MATH. Now we may apply REF to MATH, where MATH denotes the MATH-th unit vector. In that case we see that MATH, and MATH, and the result follows. |
math/0106114 | Note that because of the normalization on MATH and MATH that MATH. Also while MATH need not be an NAME function, it does satisfy the property that MATH is an increasing function, and hence it is easily seen to be equivalent to the NAME function: MATH. Suppose that MATH, that is MATH. Then in particular MATH. Thus MATH and so MATH, that is MATH. Also, if MATH (so MATH), then MATH that is, MATH. Therefore MATH. Now suppose that MATH. Again we see that MATH, and MATH. Hence MATH . Since MATH, it follows that MATH. |
math/0106114 | The condition on MATH tells us that there exists positive constants MATH and MATH such that MATH for sufficiently large MATH, that is, MATH embeds into MATH. Thus by REF we see that MATH . But MATH . |
math/0106114 | An easy argument shows that if MATH and MATH, then the NAME function MATH constructed in REF is equivalent to the function MATH. Then the function MATH constructed in REF is equivalent to MATH and the rest of the left approximation of REF follows by simple calculations. To see the second approximation of REF , note that for MATH that MATH is equivalent to MATH, and that for MATH that MATH is equivalent to MATH. Then by an argument similar to the proof of REF , we see that MATH . Finally we need to show that MATH . This is essentially a sequential version of results from CITE. Suppose that MATH. Then for any positive integer MATH, MATH that is, MATH. Conversely, if MATH, then MATH that is, MATH. |
math/0106116 | For any MATH, MATH is either REF or REF. Thus, if MATH CAYLEY we may select orthonormal imaginary octonions MATH such that MATH is an oriented basis of MATH. If MATH, MATH, MATH we have MATH, and their corresponding complex structures MATH are associated to REF-planes MATH span-MATH, MATH span-MATH,MATH span-MATH. Since MATH, MATH, MATH, then MATH satisfy MATH, that is, it is a hypercomplex structure on MATH. The converse follows from the aformentioned characterization in CITE, p. REF. |
math/0106116 | The inclusion MATH can be checked either by direct computation, using REF , or by a standard choice of the frame, like MATH, and the observation that MATH is invariant under right multiplication of MATH by any MATH, and hence by MATH (compare CITE, p. REF). It follows also MATH, by the MATH-equivariance of MATH. Conversely, to see that MATH, refer to a standard choice of three vectors to be substituted in the moment map equation MATH, assuming MATH, (compare the similar proof of the MATH-case in CITE). Then the equation MATH and the orthonormality of the frame give MATH. Then it is easy to check that the element MATH of MATH transforms MATH into a NAME REF-frame. |
math/0106117 | It suffices to show the statement when MATH, and we assume this. Recall that the second dual MATH of MATH is naturally regarded as a NAME subalgebra of the second dual MATH of MATH CITE. Let MATH be the normal extension of MATH to the second dual MATH. Then, MATH is a conditional expectation from MATH onto MATH with the same indices as MATH. Let MATH and MATH be the central projections of MATH and MATH corresponding to the discrete parts of MATH and MATH respectively. Then, thanks to CITE, CITE, we have MATH. Let MATH, MATH, and MATH be the restriction of MATH to MATH. Since the direct sum of all irreducible representations of MATH is faithful, it suffices to show the statement for MATH. By REF has no non-zero finite dimensional representation, and MATH is a direct sum of infinite dimensional type I factors. Thus, there exists a system of matrix units MATH in MATH such that MATH is a partition of unity. Let MATH be the restriction of MATH to MATH where MATH. Then, we have MATH . Therefore, MATH and MATH coincide. |
math/0106117 | CASE: Let MATH. MATH implies MATH, and so for MATH there exists MATH such that MATH CITE. Therefore, the statement holds. CASE: This follows from the characterization of the self-adjoint part MATH in MATH as MATH (see CITE for the notation). |
math/0106117 | It suffices to show that for every state MATH, MATH is contained in MATH. Let MATH be the restriction of MATH to MATH. Since MATH is finite, we have MATH . Therefore, MATH is contained in MATH, and we get the result. |
math/0106117 | Assume that MATH holds. Let MATH be an approximate unit of MATH. Since MATH is weakly dense in MATH, MATH strongly converges to REF in MATH, and so MATH strongly converges to MATH in MATH. However, MATH implies that MATH and MATH converge to MATH in norm, which shows that MATH converges to MATH in the strict topology. Assume MATH now. Let MATH and MATH be an approximate unit of MATH. Then, we have MATH . Since MATH and MATH, MATH is a decreasing net in MATH strongly converging to REF in MATH. Thus, for any MATH, MATH . As MATH is compact in weak* topology, the NAME theorem implies that the above convergence is uniform on MATH, which means MATH . Thus, MATH. CASE: We set MATH to be the restriction of MATH to MATH. Then, MATH has the desired properties. CASE: Since MATH is an ideal of MATH and MATH, we have MATH. Thanks to REF , we have MATH, and so MATH. Thus, we get MATH which shows MATH. |
math/0106117 | CASE: This follows from MATH. CASE: This follows from the fact that when MATH is a full corner of MATH, MATH is a full corner of MATH as well, and the central supports of MATH in MATH and MATH are REF. |
math/0106117 | Note that MATH acts on MATH by left multiplication. Since the norm of MATH and MATH are equivalent, the above condition is equivalent to MATH . Therefore we have MATH where convergence is in norm topology. |
math/0106117 | Suppose that MATH is not a scalar. Then, there would exist a positive number MATH and a non-zero central projection MATH such that MATH and MATH. This implies MATH. However, since MATH is simple, every non-zero representation of MATH is faithful. Thus, we would get MATH which is contradiction. Therefore, MATH is a scalar. The rest of the statement follows from the same argument as in the proof of REF using the NAME theorem. |
math/0106117 | We use the same notation as in the proof of REF . Let MATH be an approximate unit of MATH. Then, MATH is an increasing net in MATH converging to MATH in MATH. Thus, REF implies that MATH is a scalar and REF implies that MATH with MATH. |
math/0106117 | CASE: First we assume that MATH is unital. Let MATH be a closed two-sided ideal of MATH, and MATH be an approximate unit of MATH. Since MATH strongly converges to a central projection MATH in MATH CITE, MATH converges to a central element MATH in MATH. Thus, REF implies that MATH is a scalar and MATH converges in norm. However, MATH implies that MATH converges to MATH in norm, and so MATH and MATH. Since the restriction of MATH to MATH is a state satisfying the NAME inequality, MATH is finite dimensional CITE. Thus, MATH is a finite direct sum of simple MATH-algebras. When MATH is non-unital, the same argument as above implies that for every MATH, MATH converges in norm. Thanks to REF , we have MATH. Thus, MATH converges to MATH in strict topology. The same argument as above shows that MATH is finite dimensional, and MATH is a finite direct sum of simple MATH-algebras. CASE: Thanks to REF , we can apply REF to MATH, and so MATH is a finite direct sum of simple MATH-algebras. Since MATH is strongly NAME equivalent to MATH, we get the result. |
math/0106117 | Thanks to REF , MATH is a finite direct sum of simple MATH-algebras. First we assume that MATH is unital. Let MATH be the set of minimal central projections of MATH. Then, MATH is a positive scalar, say MATH. We define a conditional expectation MATH from MATH onto MATH by MATH . Then, MATH. Thus, we can apply REF to MATH and there exist quasi-bases MATH for MATH. MATH is a quasi-basis for MATH with MATH . When MATH is not unital, passing to the second dual, we get the same formula MATH where MATH is defined in the same way using MATH. Thus, REF shows that MATH exists. |
math/0106117 | Note that a MATH-algebra MATH is stable if and only if there exists a system of matrix units MATH in MATH such that MATH holds in strict topology. Thus, REF follow from REF . |
math/0106117 | The following is a modification of NAME 's argument in CITE. We use the same notation as in the proof of REF . Thanks to REF , MATH is stable and MATH-unital. Since MATH is a projection in MATH such that MATH is a stable MATH-unital full corner, NAME 's theorem CITE implies that there exists an isometry MATH satisfying MATH. We set MATH . Note that MATH belongs to MATH thanks to REF . We claim that for every MATH, MATH holds CITE. Indeed, one can easily check it for MATH. Since MATH is weakly dense in MATH, the claim holds. Applying this to MATH, we get MATH . Since MATH is isometry, we have MATH . Applying MATH to the both side, we see that MATH is an isometry. In a similar way, using MATH, we get MATH . Now we show the quasi-basis property. Let MATH. Then REF implies MATH . Applying MATH to the both sides, we get MATH which shows that MATH is a quasi-basis for MATH. |
math/0106117 | We already know that if one of these quantities is finite, so are all. Thus, we assume that they are finite. Thanks to REF , MATH is a finite direct sum of infinite dimensional simple MATH-algebras, and so MATH does not have a non-zero finite dimensional irreducible representation. Therefore, REF implies MATH. On the other hand, in general we have MATH. Note that MATH and MATH do not change if we replace MATH with MATH . Thus, we may assume that MATH is stable. Since MATH is dense in MATH in the strict topology, MATH holds. Therefore, REF implies MATH which finishes the proof. |
math/0106117 | Thanks to REF , we may assume that MATH is simple purely infinite by passing to the dual inclusion if necessary. Furthermore, REF shows that we may assume that MATH and MATH are unital and simple by passing to corners. Let MATH be the conditional expectation from MATH onto MATH naturally induced by MATH. Then, we also have MATH. Since MATH is simple due to REF implies that MATH is a finite direct sum of a simple unital MATH-algebras. Suppose that MATH is not simple. Then, there would exist a central projection MATH with MATH. By the usual perturbation argument, we can take a representing sequence MATH of MATH such that for all MATH, MATH is a projection with MATH. However, since MATH is simple, there exists a sequence MATH such that MATH, MATH, and MATH hold for all MATH. Let MATH. Then, MATH though MATH does not commute with MATH which is contradiction. Thus, MATH is simple and MATH is purely infinite. |
math/0106117 | Since MATH is also unital purely infinite simple being strongly NAME equivalent to MATH, the only point is to show that MATH in MATH. As MATH is a unital inclusion and MATH holds in MATH, we have MATH in MATH. Let MATH be an isomorphism from MATH into the corner MATH defined by MATH. Then, MATH holds as well. Thanks to NAME 's observation in CITE and MATH are equivalent in MATH, and we get a desired isometry MATH. The rest of the argument is the same as the proof of REF . |
math/0106117 | CASE: This immediately follows from REF . We define a projection MATH by MATH where MATH is the implementing unitary representation of MATH in MATH. Since MATH is outer, the crossed product MATH is simple CITE, and so MATH is a full corner of MATH. Thus, MATH is isomorphic to MATH where MATH is the right regular representation of MATH. Therefore, the statement follows from REF . |
math/0106117 | Let MATH be the conditional expectation from MATH onto MATH defined by MATH . Then, we have MATH. As MATH is stable, MATH is stable thanks to REF . Thus, NAME 's theorem CITE implies that there exists such MATH as above. Note that the space MATH is finite dimensional CITE. Let MATH be decomposition of MATH into the simple components, and MATH be a system of matrix units of MATH. For each MATH, we choose an isometry MATH satisfying MATH, and set MATH. Then, MATH satisfy MATH where MATH is defined by MATH . Since MATH is an isomorphism from MATH onto MATH, MATH belongs to MATH. Therefore, we get MATH . It is easy to show that MATH is irreducible and the decomposition is unique. |
math/0106117 | Let MATH and MATH be the isometries defined in REF . Then, we have MATH and MATH. Since MATH and MATH hold, there exists an surjective isomorphism from MATH satisfying MATH for MATH, which implies MATH . Therefore, when MATH is regarded as a map from MATH into itself, MATH. Applying MATH to the both sides of REF for MATH, we get MATH. Note that MATH is given by MATH . Thus, MATH . For MATH, we have MATH which shows MATH. As REF implies MATH, to prove MATH, it suffices to show it for MATH, MATH. Since MATH is a map onto MATH, there exist MATH such that MATH, MATH. Thus, we get MATH which finishes the proof. |
math/0106117 | We apply the previous lemma to MATH and set MATH, MATH, and MATH. Then, these have the desired properties. Note that the conditions in the statement imply that MATH is the basic construction (regardless the particular construction we made). Therefore, we get MATH. Now, we show uniqueness. Assume that MATH satisfies the same property for MATH in the statement. Then, MATH. We show that MATH is a non-zero multiple of a unitary, which will finish the proof. We set MATH, MATH. Then, MATH is a conditional expectation from MATH onto MATH with a quasi-basis MATH. Since MATH and MATH is the minimal expectation, we get MATH. Thus, MATH which shows that MATH is an isometry. In a similar way, we get MATH, MATH. Therefore, MATH . Since MATH is faithful, MATH is a unitary. |
math/0106117 | Note that every type III factor belongs to MATH. REF implies that there exist an endomorphism MATH of MATH and an isometry MATH such that we have MATH . Thanks to NAME 's theorem CITE, every endomorphism of a NAME algebra with a separable predual is automatically normal, and we get the result. |
math/0106117 | CASE: It is straightforward to show that the inverse map is given by MATH which shows the statement. CASE: Since MATH, there exists an isometry MATH from MATH onto MATH such that MATH which shows the first statement. The second one can be shown in the same way using MATH. |
math/0106117 | We denote by MATH and MATH the set of state extensions of MATH to MATH, and that of the pure state extensions of MATH to MATH respectively. MATH is the set of extreme points of MATH, and MATH is the weak-MATH closure of the convex hull of MATH. For a state MATH and a representation MATH of MATH, we denote by MATH and MATH the support of MATH in MATH and the central support of MATH in MATH. We use the same symbols MATH and MATH for their normal extensions to MATH for simplicity. Let MATH. Then, we have MATH . In particular, MATH which proves the second part of the statement. REF implies that there exist finitely many mutually disjoint irreducible representations MATH of MATH such that whenever MATH, MATH is equivalent to MATH for some MATH. We may assume MATH for all MATH. Let MATH be the density matrix for MATH, that is, MATH is a trace class operator in MATH satisfying MATH . Let MATH be the support of MATH. Since MATH is pure, any state MATH satisfying MATH for some number MATH is in MATH. Thus, REF implies that MATH is a finite rank operator. We define a finite dimensional MATH-algebra MATH by MATH . Then, REF also implies that there exists a continuous convex map MATH given by MATH . Since MATH is compact, the image of MATH is compact in weak-MATH topology. Moreover, as the image of MATH contains MATH, MATH is a surjection, which implies MATH . Further decomposing MATH, we get the decomposition of the desired form. |
math/0106117 | CASE: Since we have MATH has a normal extension if and only if MATH does, which is further equivalent to that MATH is quasi-equivalent to MATH. On the other hand, REF implies MATH which shows the statement. CASE: Note that MATH is a self-conjugate sector obeying the fusion rule MATH . We claim that there exist two disjoint irreducible representations MATH, MATH of MATH such that MATH and MATH. We choose an arbitrary irreducible representation MATH. Since MATH, REF implies that MATH is decomposed into at most two irreducible representations, and there are two cases: CASE: MATH is irreducible. CASE: There exist two irreducible representations MATH, MATH such that MATH. We first consider REF . We set MATH, MATH. Then, REF implies MATH which also shows MATH and MATH are disjoint. We consider REF now. REF implies that MATH and MATH contain MATH. On the other hand, as before REF implies MATH. This means that either MATH or MATH is equivalent to MATH and we may assume MATH. Therefore, we get MATH and MATH. MATH and MATH satisfy the claim. Let MATH and MATH. We take a pure state MATH satisfying MATH, and set MATH. Then, since MATH, we have MATH which shows that MATH is pure. We also have MATH . This shows that MATH and MATH are not quasi-equivalent, and MATH does not have a normal extension to MATH. |
math/0106117 | Let MATH be a minimal projection of MATH. If there exists a conditional expectation MATH from MATH onto MATH of a finite index, the restriction of MATH to MATH gives a desired expectation. Therefore, we may assume that MATH, MATH, and MATH are stable from the beginning. Thanks to REF and the argument in the proof of REF , there exists a quasi-basis MATH for the restriction of MATH to MATH. We define a completely positive map MATH from MATH to MATH by MATH which satisfies the following for MATH: MATH . We show that MATH is a conditional expectation from MATH onto MATH. Let MATH be another quasi-basis for MATH. Then, we have MATH . Let MATH be a unitary in MATH. Since MATH is also a quasi basis for MATH, we get MATH, which shows that MATH is a MATH bimodule map. MATH implies that MATH has norm REF. Since MATH is identity on MATH, MATH is a conditional expectation. |
math/0106117 | Let MATH be the right-hand side of the above equality. Then, MATH is obvious. Let MATH be a conditional expectation from MATH onto MATH whose existence is assured by REF . We claim that the restriction of MATH to MATH is the orthogonal projection onto MATH. Indeed, since MATH is a MATH bimodule map, MATH globally preserves MATH. For MATH, we have MATH. Since it is already a scalar, it is equal to MATH which shows the claim. This implies that for every MATH, we get MATH . Therefore, we get the statement. |
math/0106117 | Let MATH be a minimal projection of MATH. If the statement holds for MATH, the restriction of MATH to MATH satisfies the conditions of the statement. Therefore, we may assume that MATH and MATH are stable. The rest of the statement follows from exactly the same arguments as in CITE and CITE. |
math/0106117 | CASE: MATH implies MATH . Since MATH, we get MATH which shows the statement. CASE: Using REF and a similar argument as above, we get MATH . Thanks to the NAME reciprocity, we may replace the orthonormal basis MATH with MATH . Thus, MATH . Since MATH, we have MATH . Thus, MATH which shows the statement. |
math/0106117 | CASE: We show MATH. The general statement can be obtained by applying this and REF to MATH . We choose orthonormal bases MATH . Then, MATH where we use MATH . Since MATH is already a scalar, we get MATH. If MATH, we have MATH. Therefore, MATH . CASE: Using REF , we get MATH . On the other hand, using REF twice, we get MATH . Now, we compare the ``MATH-coefficient" by applying REF , and get MATH which shows the statement. REF . REF implies MATH . Using REF again and applying REF in the same way as above, we get MATH which finishes the proof. |
math/0106117 | Our proof is based on NAME 's MATH-system CITE and the characterization of the MATH-basic construction. We set MATH and MATH . Since MATH is stable, we can choose isometries MATH such that MATH . We define an endomorphism MATH of MATH by MATH . Let MATH and MATH . Then, thanks to REF , MATH and MATH are isometries satisfying MATH . We claim that MATH holds. Indeed, straightforward calculation yields MATH . MATH . Therefore, the claim is equivalent to MATH . This follows from comparison of the MATH-coefficients in two different ways of expansion of MATH. Therefore, the claim is proven. MATH also implies MATH because MATH . We set MATH . Then, MATH is an intermediate MATH-subalgebra between MATH with MATH. For MATH, we define MATH, which is a conditional expectation from MATH onto MATH with a quasi-basis MATH. We claim that MATH is irreducible, and consequently, MATH is simple. Indeed, let MATH be the inclusion map from MATH into MATH. Then, REF implies that MATH. Thus, the NAME reciprocity implies MATH which shows MATH. REF implies that MATH is simple. Now we show MATH. Let MATH which is an isometry satisfying MATH for MATH. We show that MATH satisfies the condition of REF . Direct computation using REF shows MATH . On the other hand, REF implies MATH which shows MATH. Therefore, we get MATH because MATH commutes with MATH and MATH. As we have MATH, to prove MATH, it suffices to show that MATH is equal to MATH which can be shown by direct computation. Since MATH is simple, the map MATH is faithful. Therefore, REF implies that we can identify the basic construction MATH with the MATH-algebra generated by MATH. Obviously, we have an inclusion MATH. On the other hand, MATH which shows MATH. To finish the proof, it suffices to show that MATH and to put MATH. However, this follows immediately from REF . |
math/0106117 | Thanks to NAME 's result CITE, when MATH is an automorphism, the statement holds. Therefore, we may assume that MATH is not an automorphism. Suppose that the statement does not hold. Then, in exactly the same way as in CITE, we can show that there would exist MATH, a non-zero hereditary MATH-subalgebra MATH, and a positive number MATH such that for every unitary MATH and every pure state MATH whose support MATH belongs to MATH, the following holds: MATH . This implies MATH. Let MATH . We claim that the map MATH gives an isometry from MATH onto MATH. Indeed, let MATH be the pure state on MATH given by MATH. Then, MATH is characterized as the set of vectors in MATH inducing a scalar multiple of MATH on MATH, which shows that the above map is an isometry MATH into MATH. On the other hand, for a given MATH, we can construct MATH by setting MATH . Thus, the claim holds. REF imply that MATH (and MATH as well) is finite dimensional. Let MATH be an orthonormal basis of MATH. Then, thanks to the claim, we get MATH which implies MATH . Let MATH be the closure of MATH. Then, REF is equivalent to MATH . We introduce a map MATH by MATH and set MATH . We take a sequence of unit vectors MATH satisfying MATH . We fix a free ultrafilter MATH. For a unit vector MATH, we set MATH . Note that since MATH is bounded on the unit ball of MATH, the above limits exist. For all MATH, we have MATH . This implies MATH . From the latter, we get MATH . Let MATH, and MATH be the projection onto MATH. The above inequality is equivalent to MATH where MATH . Since MATH is a hereditary MATH-subalgebra with MATH, MATH is actually the closure of MATH. Therefore, for any MATH we get MATH . To obtain contradiction applying REF , we need to get rid of MATH. For any orthonormal system MATH in MATH, we have MATH which implies that MATH is a trace class operator. Let MATH be the quotient map from MATH onto the NAME algebra MATH. Then, we get MATH . Let MATH be the finite closed system of irreducible endomorphisms in MATH generated by MATH and MATH, and MATH and MATH be as in REF , where we identify MATH with its image in MATH. Then, thanks to REF , MATH is simple, and so either MATH is trivial or MATH. The latter does not occur because it would imply that MATH would be a proper MATH-subalgebra of MATH acting irreducibly on MATH. Therefore, the restriction of MATH to MATH is faithful, and so we get MATH . Applying MATH to the both sides, we obtain MATH which is contradiction. Therefore, we finally get the statement. |
math/0106117 | It suffices to show the statement for MATH as usual and we may assume that MATH and MATH are stable from the beginning. Then, as in the usual finite group crossed product case, the statement follows from repeated use of REF and the ``crossed product expression" discussed in REF. |
math/0106118 | Since MATH is general with respect to MATH, MATH is MATH-twisted stable. Since MATH is MATH-twisted semi-stable, it is sufficient to show that MATH. Since MATH, we get MATH . |
math/0106118 | Considering NAME filtration of MATH with respect to MATH-stability, we may assume that MATH is MATH-stable. If MATH is locally free, then by REF , MATH is MATH-stable, and hence MATH. Therefore MATH is uniquely determined by MATH. Next we assume that MATH is not locally free. Under the notation of REF, if MATH, then clearly MATH. Hence MATH for all MATH. If MATH, then MATH for MATH. |
math/0106118 | By REF , we have NAME type flips MATH and MATH. By REF , MATH. Therefore we get our claim. |
math/0106118 | By REF , we get the first claim. We next show the second claim. We note that MATH. Hence MATH is a general polarization with respect to MATH (compare CITE). The same is true for MATH, MATH. Hence MATH. In order to prove our claim, it suffices to show that MATH. We first show that MATH. Assume that there is an exact sequence MATH such that MATH is semi-stable with respect to MATH and CASE: MATH or CASE: MATH . Since MATH, REF or REF implies that MATH and MATH are MATH-semi-stable of MATH with respect to MATH. Hence MATH except finite number of points of MATH. If REF holds, then MATH for all MATH, because MATH, MATH is a stable sheaf of MATH with respect to MATH and MATH. Therefore MATH and MATH satisfies MATH and we get an exact sequence MATH . By REF , MATH . Since MATH is semi-stable with respect to MATH and MATH, MATH. Hence we see that MATH . This implies that MATH is not semi-stable with respect to MATH. Therefore REF does not occur. If REF holds, then MATH. By the proof of REF , we get a contradiction. Thus MATH. We next show that MATH. Assume that there is an element MATH. Then we see that there is an exact sequence MATH such that REF MATH, REF MATH, REF MATH and REF MATH is semi-stable with respect to MATH. We set MATH according as MATH is an abelian surface or a KREF surface as in REF. We note that CASE: MATH, MATH is stable of MATH with respect to MATH, CASE: MATH is stable of MATH and MATH with respect to MATH. By REF , we get MATH for all MATH. By REF , we see that MATH for all MATH. Since MATH and MATH are MATH-semi-stable sheaves of degree MATH with respect to MATH, MATH except finite number of points MATH. Hence MATH holds for MATH, MATH and MATH with respect to MATH and we have an exact sequence MATH . In the same way, we see that MATH is MATH-semi-stable with respect to MATH. Since MATH is general, we get MATH. On the other hand, REF implies that MATH . By using REF , we see that MATH which is a contradiction. Therefore our claim holds. |
math/0106118 | By deformation theory and REF , we see that MATH is a locally complete intersection. If MATH or MATH, then MATH. Therefore the singular locus is at least of codimension REF. If MATH and MATH, then a general member of MATH fits in a non-trivial extension MATH where MATH and MATH. Then MATH is simple, which implies that MATH is smooth at MATH. Therefore the singular locus is at least of codimension REF. For the last claim, we use NAME 's criterion. |
math/0106118 | Since MATH is normal, every connected component is an integral scheme. Hence the number of irreducible components of MATH is MATH. By the upper-semicontinuity of MATH, the number of irreducible components of MATH is upper semi-continuous. On the other hand, by NAME 's connectivity theorem, the number of connected components of MATH is lower semi-continuous. Therefore we get our lemma. |
math/0106118 | Let MATH be a smooth curve over MATH and MATH be a family of polarized abelian or KREF surfaces. For a family of NAME vectors MATH, let MATH be the relative moduli space of semi-stable sheaves on MATH, MATH of NAME vector MATH and MATH the open subscheme consisting of stable sheaves. Since MATH is defined over a field of characteristic REF, MATH for MATH, where MATH is the moduli space of semi-stable sheaves on MATH (compare CITE). Since MATH is smooth CITE, it is flat. Assume that MATH is general with respect to MATH for all MATH. By REF , MATH is a dense subscheme of MATH. Since MATH is a smooth curve, MATH is also flat. Therefore MATH is a proper and flat morphism. Then our claim follows from the same argument as in MATH case. |
math/0106118 | If MATH or MATH is effective, then MATH or MATH. If MATH and MATH are not effective, then MATH. Hence we see that MATH. |
math/0106118 | We set MATH. Let MATH be an irreducible component of MATH containing MATH and let MATH be a general point of MATH. We consider the NAME filtration of MATH: MATH . We set MATH. By our assumption, we may assume that MATH consist of curves of genus greater than REF. Hence MATH. Moreover by REF , MATH. Let MATH be the stack of filtrations REF such that MATH for MATH. By CITE, MATH . Since MATH, we get MATH. Hence we see that MATH . Since MATH, we get our claim. |
math/0106118 | We set MATH. Let MATH be an irreducible component of MATH containing MATH and let MATH be a general point of MATH. We consider the NAME filtration of MATH: MATH . We set MATH. Then MATH. It is easy to see that MATH is divisible by MATH and MATH. Then MATH. As in REF , let MATH be the stack of filtrations REF such that MATH for MATH. By CITE, MATH . Therefore we get our claim. |
math/0106118 | We set MATH. Let MATH be an irreducible component of MATH containing MATH and let MATH be a general point of MATH. We consider the NAME filtration of MATH: MATH . We set MATH. Then we see that MATH. As in the proof of REF, we see that MATH. By using CITE again, we see that MATH . Therefore we get our claim. |
math/0106118 | Assume that MATH is not irreducible. Then there is a filtration MATH such that MATH is a pure dimension REF sheaf of MATH and REF MATH is a pure dimension REF sheaf of MATH, where MATH consists of curves of genus greater than REF. We set MATH, MATH. By REF , we may assume that MATH. We first note that MATH for MATH. CASE: We first treat the case where MATH is a KREF surface. Assume that MATH. By REF , we see that MATH . By our assumption, MATH. Hence MATH. By our assumption, MATH. Therefore MATH. We next assume that MATH. Then we see that MATH . Since MATH, MATH. Then MATH, because MATH is not primitive. Therefore we get our claim. CASE: We next treat the case where MATH is an abelian surface. Assume that MATH. Then MATH . Since MATH consists of curves of genus greater that REF, MATH and MATH. Then MATH. If MATH, then MATH . Since MATH is not primitive, MATH. Therefore we get our claim. |
math/0106118 | Let MATH be a germ of a curve intersecting MATH at MATH transversely. Let MATH be the stalk of MATH at MATH. We take a free resolution of MATH: MATH . Then the local equation of MATH at MATH is given by MATH. By restricting the sequence to MATH, we get a free resolution of MATH. Then MATH is given by the local intersection number MATH. Therefore we get our claim. |
math/0106118 | By CITE, MATH is isomorphic to MATH. Since MATH is primitive, CITE implies that it is irreducible. For a smooth curve MATH, the fiber of MATH is MATH. It is easy to see that MATH is a Lagrangian subscheme of MATH. By CITE, every fiber is of dimension MATH. |
math/0106118 | Assume that MATH is not of pure dimension REF and let MATH be REF-dimensional subsheaf of MATH. Then MATH must be injective. Hence it is isomorphic, which implies that the exact sequence split. Therefore MATH is of pure dimension REF. If MATH is not stable, then there is a subsheaf MATH of MATH such that MATH, where MATH. Hence MATH. Since MATH is divisible by MATH and MATH, we get MATH. Thus MATH, which implies that MATH is not stable. Therefore MATH must be stable. |
math/0106118 | By REF , MATH is surjective. Hence we get our claim from REF . |
math/0106118 | By REF , it is sufficient to show that MATH is irreducible. Let MATH be an integral curve. Then MATH is the compactified Jacobian of MATH. By CITE, the compactified Jacobian of MATH is irreducible. Therefore MATH is irreducible. |
math/0106118 | For MATH, we get MATH. If there is a non-zero homomorphism MATH, then the stability implies that it is an isomorphism. Hence MATH. Since MATH, MATH is smooth of MATH. |
math/0106118 | Since MATH is general with respect to MATH, we see that MATH is MATH-twisted stable. Then we see that REF holds. We next show that REF holds. We may assume that MATH is MATH-stable. If MATH, then MATH. If MATH, then MATH for MATH. Thus REF holds. Then the same proof of REF works and we get our claim. |
math/0106118 | If MATH, then NAME index theorem implies that there is a divisor MATH such that MATH and MATH. By NAME theorem, we may assume that MATH is effective. Since MATH is unnodal, MATH is ample. If MATH, then MATH. Hence MATH satisfies assumptions of REF . Then we get an isomorphism MATH where MATH is general with respect to MATH. By REF , we get our claim. |
math/0106118 | We first assume that MATH. We set MATH. Replacing MATH by MATH, MATH, we may assume that MATH is primitive and MATH. Since MATH is primitive, MATH. By REF , we get MATH . Replacing MATH by MATH, we may assume that MATH. By the same argument as above, we may assume that MATH and MATH is primitive. We set MATH, where MATH satisfies that MATH. Then MATH and MATH. Since MATH, MATH. Since MATH, REF implies that our claim holds for this case. We shall next treat the general case. We use induction on MATH. We set MATH. Replacing MATH by MATH, we may assume that MATH. We first assume that MATH. We note that MATH. Replacing MATH by MATH, MATH, we may assume that MATH. Then by REF , MATH for a general MATH. We take an integer MATH such that MATH. Then MATH, where MATH. Since MATH, REF , implies that MATH for a general MATH. By induction hypothesis, we get our claim. If MATH, then we may assume that MATH. If MATH, then we can apply the same argument and get our claim. If MATH, then MATH, so we get our claim. |
math/0106118 | Considering NAME filtration, we may assume that MATH and MATH are MATH-twisted stable. Since MATH is base point free, there is an inclusion MATH. Hence we get an inclusion MATH. If MATH, then MATH, and hence MATH. Assume that MATH. If MATH, then MATH and MATH. If there is a map MATH, then MATH is not an isomorphism, because of MATH. Hence MATH. |
math/0106118 | For the NAME filtration REF implies that MATH for MATH. Moreover we get MATH . Therefore in the same way as in the proof of CITE (see the description of the stack of filtration in CITE), we get this proposition. |
math/0106118 | We shall prove our claim by induction on MATH. By tensoring MATH, we may assume that MATH. If MATH, then MATH. Hence our claim obviously holds. If MATH, then by MATH, we get an isomorphism MATH. By induction hypothesis, MATH. Thus our claim holds. |
math/0106118 | Since MATH. By REF, we get MATH. Since MATH on MATH is obtained by compositions of MATH and MATH from MATH, REF implies our claim. |
math/0106118 | It is sufficient to prove that MATH for some point MATH. We choose a point MATH which is not contained in MATH. Since MATH is of pure dimension REF, we get MATH. |
math/0106118 | By taking account of a NAME filtration, we may assume that MATH is stable. We note that MATH . Since MATH, MATH, by REF holds. So we shall prove REF . Let MATH be the decomposition of the scheme-theoretic support of MATH, where MATH consists of all fiber components and MATH consists of the other components. Then we have an exact sequence MATH where MATH is the torsion submodule of MATH. Then MATH is a pure dimension REF subsheaf of MATH with MATH. By the stability of MATH, we get MATH . Since MATH is sufficiently large (the condition MATH is sufficient, where MATH is ample), we get MATH. Since MATH for all MATH, we shall prove that MATH except finite numbers of points. Proof of the claim: Let MATH be the NAME filtration of MATH with respect to MATH. Then MATH . Since MATH is a subsheaf of MATH, we also have the inequality MATH. If MATH, then MATH and MATH is MATH-equivalent to MATH for some MATH. Hence the choice of MATH is finite. Clearly MATH for MATH. Hence the claim holds. |
math/0106118 | We shall first prove that MATH is MATH-flat. Let MATH be a locally free resolution of MATH on MATH. It is sufficient to prove that MATH is injective for all MATH. We note that MATH and MATH is a torsion sheaf on MATH. Since MATH is torsion free, MATH is injective for all MATH. Thus MATH is a MATH-flat sheaf. Hence we can use base change theorem. Since MATH is relative dimension REF, MATH. Since MATH is semi-stable for general MATH, MATH for a general point MATH of MATH. Thus MATH is a torsion sheaf. By the proof of base change theorem, locally there is a complex of locally free sheaves MATH which is quasi-isomorphic to MATH. Hence MATH, which means that MATH satisfies MATH. Also we get MATH. Hence MATH is of pure dimension REF. |
math/0106118 | We consider the NAME filtration of MATH with respect to MATH, MATH. Applying MATH to this filtration, we get our corollary by REF . |
math/0106118 | Assume that MATH is not semi-stable. Then, there is a stable subsheaf MATH of MATH such that MATH. We set MATH. Applying MATH to the exact sequence MATH we get an exact sequence MATH . By REF , MATH. Since MATH, we also get MATH. Thus MATH and MATH satisfies MATH and MATH is a subsheaf of MATH. We set MATH, MATH. Then MATH. Since MATH is semi-stable with respect to MATH, MATH, CASE: MATH, or CASE: MATH and MATH. On the other hand, MATH where MATH. We note that the choice of MATH is finite. In REF , we shall show that the choice of MATH is also finite. Then there is an integer MATH such that for MATH, CASE: MATH implies MATH and CASE: MATH implies MATH. Then MATH implies that MATH, which is a contradiction. Therefore MATH is a semi-stable sheaf. |
math/0106118 | We fix an ample divisor MATH. Since MATH is a subsheaf of MATH, MATH . Since MATH and MATH, we get our claim. |
math/0106118 | We note that REF imply that MATH satisfies MATH and MATH is torsion free. Assume that MATH is not semi-stable with respect to MATH, MATH. Then there is a destabilizing subsheaf MATH of MATH. We set MATH. It is easy to see that MATH is semi-stable for general MATH. Since MATH is sufficiently large, MATH and MATH are semi-stable vector bundles of degree REF for general MATH. Then REF implies that MATH and MATH satisfy MATH and we get an exact sequence MATH . In the same way as in REF , we get a contradiction. Thus MATH is semi-stable. |
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