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math/9906083 | To see that MATH one may follow the proof in CITE, except for one detail. By the argument of REF , it is enough to show that MATH is isomorphic to REF corner MATH of MATH. As in REF , there is a REF continuous map MATH, which we need to show is onto. However if it were not onto, then there would exist a MATH such that MATH. Then MATH, where MATH is the compact preimage in MATH of MATH under the quotient map. If MATH is not in the closed convex hull of MATH, then the rest of NAME 's proof works to give a contradiction. However if MATH is in this hull, then by elementary NAME theory there is a probability measure MATH supported on MATH (and therefore also on MATH) which represents MATH. Since MATH is an extreme point, MATH, which forces MATH, which is a contradiction. The last two assertions now follows from REF; clearly REF holds since the conditions there hold for the collection of functions MATH for all MATH. |
math/9906083 | If MATH, then it follows from the definition of the inner product on MATH, that MATH for all MATH. By REF , MATH. We also get MATH. Conversely, if MATH, then by REF, we see that MATH, with norm in that latter space MATH. This proves that MATH isometrically. By REF this will be a complete isometry (since both are MIN spaces). To see the last assertion, suppose that MATH, with MATH. We let MATH be the weak*-closure of MATH, with MATH removed, as in REF. There exist MATH such that MATH, for any MATH. Thus MATH . This gives the complete isometry needed. |
math/9906083 | We observed above REF that MATH is e.n.v. iff MATH is compact, that is, iff MATH is unital. However MATH by REF . |
math/9906083 | Suppose that MATH is an oplication, with `identity' MATH. If one looks at NAME 's proof of the BRS theorem (CITE) it is clear that the same idea works in our case to show that there are NAME spaces MATH and MATH, a linear complete isometry MATH, and a linear complete contraction MATH, such that MATH for all MATH, and such that MATH. Indeed to get this one need only use part of NAME 's argument (see the first proof in REF for details if the reader needs them). By REF , for any MATH, the map MATH on MATH is in MATH, with multiplier norm dominated by MATH. Thus MATH is a linear unital contractive map MATH. That MATH is completely contractive follows easily from the fact that MATH. For if MATH then MATH may be identified, by the last isomorphism, with a MATH. For MATH as above, let MATH be the usual amplification. Then MATH, and via these identifications, we have MATH for any MATH. Hence the MATH norm of MATH is dominated by MATH. Thus MATH is completely contractive. The uniqueness of MATH is clear. If MATH is an algebra, then MATH is an algebra homomorphism if and only if MATH, for all MATH, which is obviously equivalent to MATH being a module action. Now we appeal to REF , to obtain all the results of the theorem except for those in the last line. To see these, we note that if MATH is an operator system, then since MATH is unital and completely contractive, it is completely positive. Hence MATH maps into MATH. |
math/9906083 | We will assume MATH is unital, and leave the general case to the reader, who may want to use REF and the first term principle too. CASE: This follows from REF, since if MATH is a left operator MATH-module, then there is an associated unital homomorphism MATH. Hence MATH is a left operator MATH-module, and a MATH-rigged MATH-module, containing MATH as a MATH-submodule. CASE: The MATH direction is clear (see REF). The converse follows from the proof of REF , for if MATH is a left operator MATH-module, then by that proof, the associated homomorphism maps MATH. |
math/9906083 | . Here are the one-line proofs: To obtain BRS, take MATH; MATH is a complete isometric homomorphism into the operator algebra MATH, since MATH. The strengthening of BRS referred to above is obtained by the same idea: take MATH again, apply REF , and take the third operator space structure on MATH to be MATH. The NAME result is evident, as is its nonselfadjoint version. It is well known that the characterization of operator bimodules in CITE follows in a few lines from the NAME result, by a well known MATH matrix trick. Finally, the NAME hypothesis implies that MATH is a MATH-operator module, so that by REF , there is an isometric homomorphism MATH of MATH into MATH; and by REF (or our REF ), MATH is a commutative MATH-algebra. |
math/9906083 | We define MATH, and identify MATH. Define matrix norms on MATH by MATH . It is easy to check that this makes MATH an operator space. Extend MATH to MATH, by letting MATH act as the identity on MATH. The first conclusion will follow from the `unital case' of the theorem, if we can show that the extended oplication satisfies REF. To that end, let MATH, and MATH be given. Then MATH . The last two assertions follow from the `first term principle' as before, and the fact that a contractive representation of a MATH-algebra is a *-homomorphism. |
math/9906083 | (Sketch.) CASE: This can be seen by following the idea of proof of REF , but working on the NAME space of the universal representation of the linking algebra MATH of MATH. The key first step in his proof is to carefully compute the commutants MATH and MATH. One obtains an explicit formula for the weak*-closure MATH of MATH in MATH, and also that MATH is a self-dual MATH-algebra NAME equivalence bimodule over MATH. The latter follows from REF (see also REF ). Then one needs to check that via the usual identification of MATH and MATH, we have MATH, and also that MATH, ``as bimodules", and it is easy to show that MATH is self-dual as a MATH-module. Similar assertions follow by symmetry for the left action. CASE: This may be proved using basic facts about self-dual modules from CITE analogously to the proof of the stable isomorphism theorem, or NAME 's stabilization theorem (see for example REF , or the proof of REF ). Indeed this is one way to show the folklore fact that MATH-Morita equivalent MATH-algebras are w*-stably isomorphic. CASE: Follows from REF. |
math/9906083 | CASE: For any such MATH, and for MATH and MATH, we have: MATH . Taking the supremum over such MATH gives MATH. CASE: In this case the canonical MATH is a *-homomorphism into MATH. It is easy to see that MATH is faithful if and only if MATH is REF, and it is well known that the latter happens for a *-homomorphism if and only if it is completely isometric. |
math/9906083 | Suppose that MATH is a state on MATH with MATH . Then we have that MATH for all MATH. By the polarization identity we see that MATH for all MATH. Thus MATH. Suppose that MATH is a GNS representation of the state. Then if MATH are each of the form MATH , for some MATH, then MATH from which it is clear that MATH. |
math/9906083 | That REF implies REF follows from REF . If REF holds, then we have the following completely isometric MATH-isomorphisms: MATH . If the generation is algebraic, then the closures in this string of equalities are unnecessary, and thus we see that MATH is closed. In any case if MATH has c.a.i. MATH then MATH, which implies that MATH. Hence MATH is unital in the sense of the introduction to REF. Notice that MATH is also a right MATH-module, and MATH is an algebraic single bigenerator. Thus we have REF . Notice that MATH, so that MATH. If MATH is the `right' polar decomposition of MATH, then MATH. The map MATH is a completely isometric MATH-module map from MATH onto MATH, hence it is a `imprimitivity bimodule isomorphism' or `triple isomorphism'. Thus we may take MATH to be the `noncommutative NAME boundary' of MATH. That is, we may replace the `NAME representation' MATH by MATH where MATH. However MATH, for any MATH, and hence MATH. It is easy to see from this, that MATH, and MATH. Since MATH is a unital MATH-algebra, MATH is e.n.v. By symmetry, MATH is right-nonvanishing. The last assertion of the theorem is clear: if MATH is norm closed then MATH, so MATH is a.s.g. The converse assertion was noted earlier. |
math/9906083 | Both follow from the last assertion of the theorem. Also REF uses the easy fact that for a MATH-faithful operator module, the MATH is bicontinuous and consequently has closed range; whereas for MATH is a MATH-algebra we have the fact that the range of a *-homomorphism is closed. As for REF , if the MATH-action is faithful, then MATH is REF and onto MATH, so that MATH is unital. |
math/9906083 | The MATH direction was proved in REF . If MATH is as above, and if MATH, then the subsets MATH and MATH of MATH contain MATH, which implies by spectral theory (since we can uniformly approximate the functions MATH and MATH on any closed interval not containing MATH by polynomials with no constant term) that the MATH-subalgebra of MATH generated by the copy of MATH in the MATH-corner, is unital. Hence MATH is e.n.v., since MATH is a quotient of this unital MATH-subalgebra. Moreover MATH is a right MATH-module, and MATH is a bigenerator for MATH considered as a bimodule. |
math/9906083 | Suppose that MATH is the linear completely isometric isomorphism. By the above, there exist completely isometric unital surjective homomorphisms MATH and MATH. Define MATH. Then MATH is a completely isometric unital surjective homomorphism MATH by REF, and MATH for all MATH. So MATH is a MATH-isomorphism from MATH onto MATH. The last assertion follows since MATH via the map MATH. |
math/9906083 | Taking MATH shows that MATH. Conversely, the oplication theorem shows that MATH. REF shows that MATH. Finally, if we have REF , and if MATH is as in REF , then MATH is a strictly positive element of MATH, and MATH, where MATH is regarded as an element of MATH. This implies that MATH for every state on MATH. Since MATH is strictly positive, we see that MATH on MATH, for every state MATH on MATH. Hence, MATH for every such MATH and MATH, which implies that MATH for every MATH. Thus MATH. To see the last part, note that if MATH then MATH for some MATH. Hence MATH by REF . |
math/9906083 | By the oplication theorem, there exists a completely contractive linear map MATH, such that MATH for all MATH. Moreover MATH is isometric: indeed we have for any MATH that MATH . Any MATH has MATH for some MATH, so that MATH. Thus MATH. For MATH we have MATH. By REF of the previous lemma, we have that MATH. Hence MATH is associative. The rest is clear. |
math/9906083 | CASE: This is easy, extend MATH to a map MATH, and use rigidity of MATH. CASE: Similar to REF - extend MATH to a map MATH, and use the essentiality of MATH. CASE: Suppose that MATH is any NAME MATH-extension of MATH. Inside MATH consider the operator system MATH given by the image of MATH in MATH, together with the two idempotents on the diagonal of MATH. So MATH generates MATH as a MATH-algebra. By NAME 's lemma, the canonical complete isometry MATH is the MATH-corner of a complete order isomorphism MATH. By the NAME theorem (REF above) MATH extends to a surjective *-homomorphism MATH. Let MATH be the restriction of MATH to MATH. Let MATH (respectively, MATH) be the `MATH corner' (respectively, MATH-corner) of MATH. If MATH is an essential operator superspace of MATH, we see that MATH is a complete isometry onto MATH. We claim that MATH is a NAME MATH-isomorphism of MATH onto MATH; that is, MATH is REF. To see that MATH is REF, suppose that MATH. Without loss of generality, MATH. We can then approximate MATH by sums of the form MATH with MATH. We have MATH, since MATH is a complete isometry. Thus MATH. Similarly MATH, and consequently MATH, is REF on MATH. This gives the result. |
math/9906083 | Clearly MATH is an essential NAME MATH-extension of itself. |
math/9906083 | Clearly MATH is an essential NAME MATH-extension of MATH. The proof of REF produces a NAME MATH-isomorphism MATH. Since MATH, we have MATH. |
math/9906083 | By NAME 's factorization theorem, if necessary, we may suppose without loss of generality that MATH is a full MATH-module over each MATH. By the previous corollary, MATH may be supplemented by two *-isomorphisms MATH and MATH, becoming an imprimitivity bimodule isomorphism. Hence it is easy to check from the definition of the latter term that `rank one' operators on MATH with respect to the one inner product, are rank one with respect to the other inner product. Hence MATH. The rest is fairly clear. For example, if you takes the definition of adjointability with respect to the MATH-action, and hit this equation with MATH, then one sees the MATH-adjointability. Finally note that MATH iff MATH. |
math/9906083 | By CITE, MATH is a NAME MATH-module. By the above correspondences, MATH generates MATH. |
math/9906083 | MATH is an ideal in MATH, and therefore also in MATH. We view MATH, so that there is a canonical completely contractive map MATH . By NAME 's lemma, this map is a complete order isomorphism, if and only if the restriction of this map to the MATH-corner MATH, is a complete isometry. However this restriction maps into MATH. Since MATH, so we see that it is necessary and sufficient that the map MATH be a complete isometry. |
math/9906083 | This follows as in CITE, from the Lemma, and so there is no advantage in rewriting it here. Alternatively, one can use the last proposition and the corresponding result in CITE. |
math/9906083 | CASE: It is enough to prove the first assertion; for then by symmetry there is a matching assertion for MATH, and then one can use the relation MATH. We will use REF . Let MATH be the MATH-th inclusion map, which is an isometric MATH-module map. Suppose that MATH is a boundary submodule for MATH in MATH. Thus MATH is a MATH-submodule of MATH, and the canonical map MATH is a complete isometry. Clearly MATH is also a MATH-submodule of MATH. Letting MATH, we see that this may be identified as a MATH-submodule MATH of MATH. If we can show that the canonical map MATH is a complete isometry then we would be done, since MATH has no nontrivial boundary submodules. To this end notice firstly that MATH completely isometrically; and secondly, notice that the map MATH is a complete isometry from MATH, which maps into MATH. A similar but easier argument proves REF . Namely, any boundary submodule for MATH in MATH, may be written as MATH, where MATH are submodules of MATH and MATH respectively. It is clear that MATH is a boundary submodule for MATH in MATH, and is consequently trivial, and so on. Another way to prove this result is first to prove that MATH, which is quite easy to see. One way to see the failure of the second tensor product relation, is to set MATH. Then MATH, which has dimension at least MATH, whereas MATH has dimension MATH. If there was some quotient of the latter isomorphic to the former, then MATH would have dimension MATH. This implies that MATH, and that MATH is a finite dimensional MATH-module, which implies that MATH is injective, since it is a complemented summand of MATH, and every finite dimensional MATH-module is clearly injective. However there are only three distinct two dimensional injective operator spaces, as is clear from a result of NAME 's from REF (written up recently in CITE), and none of these is MATH. To see the failure of the first tensor product relation, take MATH. In this case MATH, and therefore one can identify MATH and MATH from CITE (see also REF above). |
math/9906083 | In the case that MATH and MATH are unital MATH-algebras this is probably well known (compare CITE). We will sketch a quick proof of this case for the readers interest, using REF . In this case, MATH is an `imprimitivity bimodule isomorphism' or `triple isomorphism'. If MATH, and MATH, then MATH . Similarly MATH. So MATH is unitary. Then MATH is a unital isometric homomorphism, since MATH, for MATH. It is therefore a *-homomorphism (or this may be proved directly from the definition of triple morphism: MATH). In the more general case of operator algebras with c.a.i., extend MATH to a completely isometric linear surjection MATH (which is clearly possible by injectivity and rigidity). Then by the first part, there exists a faithful *-isomorphism MATH and a unitary MATH such that MATH for all MATH. Inside MATH, we have that MATH is an operator algebra with c.a.i. We have MATH by NAME 's factorization theorem. Thus MATH. Also, MATH, by NAME 's theorem again. Thus by REF , MATH. Thus MATH and MATH. Hence MATH by the characterization of this space above the statement of REF. A similar argument applied to MATH, shows that MATH. So MATH. |
math/9906089 | MATH . By REF MATH, we just need to show that MATH is log canonical in some neighborhood of MATH if MATH. Suffices to show that MATH for all MATH. Assume by contradiction that MATH. Let MATH be a prime divisor on an extraction MATH such that MATH and MATH. Since MATH is a proper point, there exists a proper point MATH such that MATH. From REF MATH, MATH, hence MATH. Contradiction! MATH . Follows from REF . |
math/9906089 | MATH . Let MATH be a prime divisor on an extraction MATH such that MATH and MATH. Since the induced map MATH has generic fibers of positive dimension, there exists a proper point MATH such that MATH. Thus MATH, hence suffices to check MATH. MATH . We may assume that MATH has log nonsingular support. Let MATH be the exceptional divisor on the blow-up in MATH, and let MATH be a component of MATH dominating MATH. Inductively, let MATH be the exceptional divisor on the blow-up in MATH, and let MATH be a component of MATH dominating MATH. A simple computation gives MATH . In particular, MATH. |
math/9906089 | MATH . We first check that MATH. Indeed, MATH CITE, so MATH if MATH. The equality is attained on the generic point of MATH. MATH . Assume MATH is the generic point of a connected component of MATH. Blowing up MATH in MATH we have again a log nonsingular pair, and the new divisor MATH has log discrepancy MATH. From the previous case, we infer MATH. MATH . Otherwise, shrinking MATH, we may assume that MATH, and there exist divisors MATH such that MATH has log nonsingular support and MATH is a connected component of MATH. Set MATH for all MATH. From REF , MATH. |
math/9906089 | Indeed, let MATH with MATH. There is nothing to prove if MATH, so we may assume that MATH has only log canonical singularites in MATH. Then MATH where MATH. |
math/9906089 | For good resolutions that compute minimal log discrepancies, MATH is a divisor with normal crossings, and MATH . For simplicity, we may assume MATH (the other cases are similar). Therefore MATH near MATH and MATH respectively, and REF gives MATH . |
math/9906089 | Suffices to prove that MATH takes a finite number of values and its fibers are constructible subsets, for every closed subset MATH. There is nothing to prove if MATH, so let MATH. Let MATH be an irreducible components of MATH. From REF MATH and REF , there exists an open subset MATH such that MATH, MATH is constant, and MATH does not intersect the other irreducible components of MATH. Thus MATH and we are done by NAME induction. |
math/9906089 | Let MATH be a log resolution with a normal crossing divisor MATH on MATH supporting MATH and the divisor MATH. Shrinking MATH near MATH, we may assume MATH for some subset MATH, and MATH for every MATH. We may assume that MATH has only log canonical singularities, and MATH. Note that MATH. Removing from MATH all components of MATH that do not contain MATH, we may assume that MATH, or MATH for every (connected) component MATH of MATH. We call relevant those components MATH with MATH. The following hold: CASE: MATH and MATH for every relevant MATH. CASE: If MATH, then MATH, the unique component of MATH containing MATH, is relevant. Since the generic fibers of the morphisms MATH have expected dimension, there exists an open subset MATH such that MATH and MATH for every relevant MATH and for every closed point MATH. Let MATH and MATH. Then MATH, hence MATH . But MATH and all MATH's are non-negative numbers, hence MATH. Thus MATH. Taking infimum after all MATH's as above, we obtain MATH . Finally, let MATH be an index such that MATH. Let MATH be the generic point of an irreducible component of MATH of maximal dimension. Since MATH is relevant, MATH. Moreover, MATH since MATH. Therefore MATH, and the above inequality is in fact an equality. |
math/9906089 | Assume REF is valid, and let MATH be a closed point. Using REF , we may shrink MATH such that MATH for every irreducible component MATH of the fibers of the map MATH. For MATH, there exists a MATH such that MATH. Since MATH, we infer that MATH. But MATH, so we are done. Assume REF is valid. According to REF , we may assume that MATH is a closed point and MATH. Let MATH be a neighborhood of MATH such that MATH for all MATH. Then MATH is an open dense subset. From REF , there exists some MATH such that MATH. Therefore MATH. |
math/9906089 | Since MATH if MATH, we may assume MATH. By assumption, MATH, hence REF implies that MATH is nonsingular in MATH and MATH . Therefore MATH. If MATH, decompose MATH, with MATH and MATH. Then MATH. |
math/9906089 | We may assume MATH and MATH is a closed point. MATH . By REF , MATH for every curve passing through MATH. From the codimension MATH case, MATH has only terminal singularities. MATH has MATH-factorial singularities. Indeed, from LMMP we can find a MATH-factorialization MATH, where MATH is a log variety again. If MATH, there exists MATH with MATH, hence MATH from the codimension MATH cases. Then MATH. Contradiction! Otherwise, MATH. NAME 's Main Theorem implies that MATH is an isomorphism over a neighborhood of MATH, hence MATH is MATH-factorial. MATH . Assume by contradiction that MATH is a singular point. Then it must be an isolated terminal point. From REF , MATH, where MATH is the index of MATH at MATH. Contradiction! |
math/9906089 | We may assume MATH and MATH. MATH . Assume MATH and MATH. By easy divisorial adjunction, MATH, where MATH is the different of MATH on the normalization MATH of MATH. The log variety MATH has dimension MATH, so MATH. MATH . Assume MATH and MATH. From LMMP, there exists a crepant extraction MATH such that MATH is effective and there exists a prime divisor MATH on MATH with MATH and MATH. Let MATH be the generic point of a curve in the fiber of MATH over MATH. From the codimension MATH case, MATH. But MATH, so we are done. |
math/9906089 | We may assume that MATH is a closed point MATH on the MATH-fold MATH. MATH . Assume MATH is a curve MATH passing through MATH. From REF , we may assume that MATH. Then we may also assume MATH, hence MATH is nonsingular in both MATH and MATH. By REF , MATH and MATH. Therefore MATH . Assume MATH is a surface MATH passing through MATH. Let MATH be a curve. Then MATH from the codimension MATH case. From the previous step we get MATH, thus MATH. |
math/9906089 | The second part follows from REF , so we just have to prove MATH. We may assume MATH and MATH is a closed singular point. MATH has only canonical singularities. Indeed, MATH for every curve passing through MATH. From the codimension MATH case, MATH has only canonical singularities on MATH. But MATH, hence we are done. MATH . Assume that MATH is MATH-Cartier. Then MATH is MATH-Cartier, and let MATH be the index of MATH at MATH. Since MATH, we infer that MATH near MATH and MATH. We just have to prove that MATH, since then MATH has only canonical NAME singularities, and therefore MATH is a cDV point due to REF MATH. Note that if MATH is a terminal point, then MATH by NAME, hence MATH. MATH admits a terminal crepant extraction by CITE, that is, there exists an extraction MATH such that MATH has only terminal singularities and MATH. Note that MATH for every closed point MATH. Thus the terminal subcase implies that MATH is NAME. In particular, MATH is NAME near MATH, that is, MATH. MATH . Assume that MATH is not MATH-Cartier at MATH. We have to show that this is impossible. From LMMP we can find a small extraction MATH such that MATH is MATH-factorial. Let MATH be the proper transform of MATH. In particular, MATH. If MATH, then MATH . We have MATH. Otherwise, NAME 's Main Theorem would imply that MATH is an isomorphism over a neighborhood of MATH. Thus MATH is MATH-Cartier, contradicting our assumption. Therefore MATH is a connected union of curves and MATH has only cDV isolated singularities in MATH from REF . Moreover, MATH intersects MATH in a finite set of points. Otherwise, if some curve MATH over MATH is included in MATH, then MATH, a contradiction. We arrive at the final contradiction with the following argument, kindly suggested by NAME: MATH is MATH-nef, but not MATH-trivial, since MATH intersects the fiber MATH. However, MATH admits no flipping contraction since its difficulty CITE is MATH. Contradiction! |
math/9906089 | MATH . The equality holds for the generic closed point MATH from REF . This extends to all the points in MATH since MATH acts transitively on orbits and leaves the boundary fixed. MATH . Let MATH be a proper face of MATH and let MATH for some MATH. We can find primitive vectors MATH on the MATH-dimensional faces of MATH such that MATH . Therefore MATH. |
math/9906089 | We use induction on MATH. If MATH, there is nothing to prove, so let MATH. By REF MATH, every proper face MATH has the same property with respect to MATH. By induction, all proper faces of MATH are nonsingular cones. MATH . Assume MATH is a simplicial cone, that is, MATH. It is known that MATH is nonsingular iff MATH . Assume MATH. Since all proper faces are non-singular cones, MATH. Therefore there exists MATH. Then MATH for every MATH, hence MATH. Therefore MATH. This implies MATH, a contradiction. Therefore MATH, hence MATH is a nonsingular cone. MATH . If MATH is a face of codimension MATH and MATH, then MATH is a nonsingular cone of dimension MATH. Indeed, let MATH. By assumption, MATH is a simplicial cone of dimension MATH. This also implies that MATH, hence MATH has the same property with respect to MATH. Therefore MATH is nonsingular from REF . MATH . We may assume that MATH. Indeed, if MATH we are done from REF . Otherwise, MATH, and we show that this leads to contradiction. Let MATH be a cone of dimension MATH generated by MATH of the vectors MATH's. Then MATH has the same property with respect to MATH, since MATH. Therefore suffices to show that the case MATH is impossible. MATH . Assume MATH is a face of codimension MATH and MATH. Then MATH. Indeed, let MATH be the generators of MATH, which also form a basis of the lattice MATH. From REF , MATH and MATH are both basis for the lattice MATH. The transition matrix has determinant MATH, hence the statement. MATH . Let MATH be generated by MATH. By REF , we may assume that MATH is a basis of the lattice MATH, hence MATH . We show that MATH for every MATH. At least one MATH is positive (negative). Assume MATH. Then MATH generates a codimension MATH face, hence MATH . On the other hand, MATH, hence MATH. Therefore MATH. Assume MATH. If MATH, then MATH generates a codimension MATH face, so MATH . Since MATH, we deduce that MATH. Therefore MATH . But MATH, thus MATH. Therefore MATH. MATH . Let MATH, where MATH and MATH. One can easily check that MATH . Therefore MATH. Contradiction! |
math/9906092 | Take the conjugate NAME pair REF and replace MATH by MATH. Then multiply both MATH and MATH by MATH and sum MATH over the integers. Using REF this transforms MATH and MATH of REF into those of REF. |
math/9906092 | The terms within the curly braces can be combined to MATH. Using this as well as MATH and MATH, REF can be written as MATH . We now use the MATH-binomial sum CITE MATH as well as the limiting case MATH to express the left-hand side of REF as the following quadruple sum, MATH . After shifting MATH, MATH and MATH this becomes MATH . The sum over MATH can readily be performed thanks to CITE MATH leading to MATH . The sum over MATH yields MATH by the MATH-Chu - NAME sum CITE MATH with MATH, MATH and MATH, where the following standard notation for basic hypergeometric series is employed MATH . As a result we are left with MATH . This corresponds to the right-hand side of REF as MATH . |
math/9906092 | Substitute the conjugate NAME pair REF into the defining relation REF. After the shift MATH this becomes MATH where we have set MATH and MATH. To obtain this identity we take REF and choose MATH, MATH and perform a few trivial operations. |
math/9906092 | We only present the proof of the second identity. The other three identities can be proven in a similar fashion. (The second rather than the first identity is chosen because all equations are more compact in this case.) We start with MATH and interchange the sums over MATH and MATH and shift MATH. Then we again swap the order of summation yielding MATH . Now consider the first double sum denoted by MATH and write this as MATH . Using the MATH-Kummer - NAME - NAME formula CITE MATH with MATH, MATH, MATH and MATH this can be put in the form MATH . Once more the order of summation is reversed, then MATH is replaced by MATH and the summation order is again changed. Thus, MATH . Next we deal with MATH, given by the second double sum in REF. Shifting MATH gives MATH . By REF with MATH, MATH, MATH and MATH this is equal to MATH . By an interchange of sums followed by the successive transformations MATH and MATH this becomes MATH . Computing MATH results in the claim of the proposition. |
math/9906092 | Inserting the polynomial identities REF into REF one can apply the MATH transformation REF with MATH, MATH, MATH, MATH, and MATH, MATH, MATH, MATH, respectively. This yields identities for MATH and MATH which immediately imply the expressions of the proposition. |
math/9906092 | After inserting the definition of MATH in the above equation make the variable changes MATH in the first term and MATH in the second term of the left-hand side. Then, by the symmetry REF can be rewritten as MATH where MATH. When MATH this follows directly from the symmetries REF, and in the remainder we assume that MATH and MATH. The complication is now that we no longer have MATH. In view of this let us first investigate the origin of this difficulty. Consider REF of the MATH string functions. The summand has two different terms corresponding to the two terms within the curly braces. In the first term make the variable change MATH, MATH and in the second term make the change MATH, MATH. The result of these changes is exactly the same expression as before except that MATH has been replaced by MATH and that the sum over MATH now runs over all integers greater than MATH. We may therefore conclude that MATH where MATH . By a shift MATH in the second term of the summand this becomes MATH which shows that the MATH term in the summand vanishes and hence that MATH for MATH integer. Inserting REF into REF with MATH we are done with the lemma if we prove that MATH . Using the explicit form for MATH, this is equivalent to showing that MATH . After the shift MATH in the second term in the sum over MATH we are done. |
math/9906092 | First observe that MATH . To prove this shift MATH in the first term in the curly braces and successively MATH and MATH in the second term in the curly braces. Using the symmetry of MATH the resulting terms can be combined to MATH where the middle term follows by application of the MATH-binomial REF and the last term by MATH for MATH. With this result we can write the sum over MATH in the right-hand side of REF as a sum over MATH and MATH. Then using the symmetry of MATH the right-hand side becomes MATH . Comparing with the left-hand side of REF we are done since MATH. |
math/9906092 | Inserting REF into REF gives the identity MATH for MATH even and MATH. Applying REF this can be simplified to MATH . Now observe that for MATH the only contribution to MATH comes from the MATH term in the summand of REF. Therefore, MATH . By induction on MATH this implies REF . |
math/9906093 | The proof of the Lemma is a straightforward calculation, which is given in the Appendix. |
math/9906093 | We shall make use of REF and sum up the NAME series. Introduce the functions MATH with the NAME coefficients MATH so that by REF MATH . Recall that all the characters of MATH are real and have the property MATH, so we can just write MATH. Further, the character MATH is given at MATH by the NAME character formula, MATH . Substituting this into REF and putting together the NAME series, we get MATH . We now gather the components MATH of the fixed point set into pairs MATH with MATH by means of the NAME element MATH. For MATH with MATH we let MATH. If MATH or MATH, then MATH, and we denote MATH. Then MATH . Use the change of variables MATH to relate the functions MATH and MATH. For MATH we have MATH, so the action of MATH at MATH corresponds to the action of MATH at MATH. The form MATH is invariant, then MATH . By the equivariance of the moment map, MATH. Now pick together the terms with MATH from MATH and those with MATH from MATH, and vice versa. Substituting MATH and MATH, we get MATH . (Here we assume that MATH; otherwise a factor MATH must be added). We assume that the moment map has regular points. It follows that the codimension of MATH in MATH must be at least MATH. Then the rational function MATH is of order of at least MATH. It means that REF consists of two series to which the Lemma applies. If MATH, both series sum up by means of REF , and we get MATH which is exactly REF . If MATH, we use both REF , and obtain REF . To get the formulas for MATH with MATH we must just add a factor MATH into REF . If MATH or MATH are the regular values of the moment map, we can consider the limits MATH and MATH to obtain REF , respectively. |
math/9906093 | We have MATH, so it suffices to prove REF for MATH, MATH. Consider a function MATH . Its poles are the real integer numbers, and MATH for MATH, MATH. Integrate MATH over the contour MATH . By the NAME theorem MATH and we have only to prove that MATH as MATH. While MATH, MATH. Then MATH which tends to MATH since MATH. For MATH we get MATH which tends to MATH because MATH. Now cut MATH into MATH . If MATH, then MATH. It follows that MATH . If MATH, we refine the estimate. We have MATH for MATH, and MATH for MATH. Therefore MATH . It follows that MATH. The integral over MATH is treated in the same way. |
math/9906095 | Use the equality MATH and note that MATH when MATH and MATH. |
math/9906095 | From REF , write the MATH of MATH as MATH . Change variables with MATH and MATH . Note that MATH with the absolute value of the inverse Jacobian MATH . Thus MATH with MATH and MATH . |
math/9906096 | When MATH, the lemma is easily verified. Suppose now MATH and the lemma holds for MATH. Then we have MATH . Now by induction hypothesis, MATH, this implies MATH . Hence we have MATH . But the left hand side lies in MATH while the right hand side lies in MATH, therefore, they must both vanish modulo MATH. |
math/9906097 | We prove by induction on MATH. The claim is trivial for MATH. Now suppose that MATH, and the claim has been proven for MATH. Define an element MATH to be popular if MATH and define MATH to be those elements MATH such that MATH is popular. Since each unpopular element of MATH contributes at most MATH elements to MATH, we see that MATH so MATH . By applying the induction hypothesis to MATH we have MATH . Since MATH is popular, we thus have MATH . From REF we thus have MATH . To eliminate the factor of MATH, we let MATH be a large integer, and apply REF with MATH, MATH replaced by MATH, MATH, and MATH replaced with the function MATH defined by MATH to obtain MATH . The claim then follows by letting MATH (compare CITE). |
math/9906109 | We first handle the case MATH. Let MATH be the map MATH, let MATH be an element with MATH, and let MATH be the lattice generated by MATH and MATH. The map MATH identifies MATH with MATH, and MATH with the local system defined by the homomorphism MATH, MATH. There is a unique cocycle MATH in MATH with MATH, MATH; let MATH be the corresponding holomorphic line bundle on MATH. Computing MATH by using the exponential sequence, we find that MATH. By NAME, we have MATH; let MATH be the corresponding global holomorphic function on MATH, that is, MATH and the divisor of MATH on MATH is MATH, with MATH. Take MATH with MATH and MATH. Let MATH. Then MATH and MATH. Thus, multiplication by MATH defines an isomorphism MATH . The proof for MATH is essentially the same, where we replace MATH with the rational function MATH. |
math/9906109 | Let MATH be a quasi-projective surface over a field MATH. By CITE, there is an isomorphism MATH. The product MATH gives the cup product MATH . In addition, let MATH, MATH be NAME divisors which intersect properly on MATH, and suppose that MATH. Then MATH where MATH is the intersection product and MATH denotes the class in MATH. Since MATH, REF implies MATH so it suffices to show that MATH in MATH. Write MATH for MATH. Let MATH be the image of MATH in the constant sheaf MATH. By NAME 's conjecture, the surjection MATH is an isomorphism at each regular point of MATH, hence MATH induces an isomorphism on MATH. Let MATH be the normalization, giving the normalization MATH. Let MATH be the inclusion of the singular point. We have the exact sequence of sheaves on MATH and the exact sequence of sheaves on MATH: MATH with augmentations MATH, MATH. The various cup products in MATH-theory give the map of complexes MATH over the cup product MATH . The augmentation MATH is an isomorphism. The augmentation MATH is an injection, and the cokernel is supported on MATH, so MATH induces an isomorphism on MATH. Thus, the complexes REF give rise to maps MATH . The compatibility of REF with REF yields the commutativity of the diagram MATH . Since MATH for each MATH, we have MATH . |
math/9906109 | We first consider the case in which both MATH and MATH are smooth elliptic curves, MATH, MATH, where MATH and MATH are in MATH and MATH, MATH. We have the maps MATH which are group homomorphisms with MATH, MATH. Suppose that MATH is contained in an algebraic curve MATH. For each MATH, MATH is a finite set (possibly empty), hence, for each MATH, the set of points of MATH of the form MATH has finite image in MATH. Thus, for each MATH, there are integers MATH, MATH and MATH, depending on MATH, such that MATH and MATH . Since there are uncountably many MATH, there is a single choice of MATH, MATH and MATH for which REF holds for uncountably many MATH. But then MATH . If MATH, then MATH, contradicting the condition MATH. If MATH, then we can solve REF for MATH, so REF only holds for this single MATH, a contradiction. If say MATH, then MATH is injective, and we have the infinite set of points MATH in the image of MATH, all lying over the single point MATH. |
math/9906109 | We first give the proof in case MATH and MATH are both non-singular. For a quasi-projective MATH-scheme MATH, we let MATH denote the MATH-th symmetric power of MATH. For MATH smooth, we have the map MATH . For each integer MATH, we have the morphism MATH . By CITE, MATH is a countable union of NAME closed subsets of MATH. On the other hand, since MATH, the NAME kernel MATH is ``infinite dimensional" CITE; in particular, MATH. Since MATH is generated by cycles of the form MATH, it follows that MATH is a countable union of proper closed subsets of MATH. If MATH is a proper algebraic subset of MATH, then, by REF , MATH is a proper analytic subset of MATH, hence MATH is countable. Thus, the set of MATH such that MATH is torsion is countable, which completes the proof in case both MATH and MATH are non-singular. If say MATH, we use essentially the same proof. We let MATH be the open subscheme MATH of MATH. We have the map MATH defined as above. By CITE, MATH is a countable union of closed subsets MATH of MATH. By CITE, we have the similar infinite dimensionality result for MATH as in the smooth case, from which it follows that each MATH is a proper closed subset of MATH. Thus, the closure of each MATH in MATH is a proper algebraic subset of MATH. The same argument as in the smooth case finishes the proof. |
math/9906109 | We give the proof in case both MATH and MATH are non-singular; the singular case is similar, but easier, and is left to the reader. Since MATH it follows from REF that we need to show that MATH. The class MATH is the image of MATH under the map of sheaves MATH, and similarly for MATH and MATH. Thus, by REF, it suffices to see that MATH vanishes in MATH. The MATH-covers MATH, MATH give natural maps MATH . Similarly, the MATH-cover MATH gives the natural map MATH . Letting MATH denote the natural inclusion, the maps above are compatible with the respective cup products: MATH . Each MATH gives the corresponding homomorphism MATH, MATH. Since MATH is MATH and MATH is MATH, it suffices to show that MATH in MATH, where MATH are the respective homomorphisms MATH, and MATH. We have the spectral sequence MATH . Since MATH has cohomological dimension two, and since MATH for MATH, it follows that the natural map MATH is an isomorphism. Since MATH, we need to show that the image of MATH in MATH is zero. By definition of the cup product in group cohomology, we have MATH which clearly vanishes in MATH. |
math/9906109 | The map MATH is obviously surjective, and by the NAME resolution on the smooth points of MATH, the kernel is supported in codimension REF. Thus MATH induces an isomorphism on MATH. On the other hand, MATH . The term MATH maps to MATH via the difference of the restrictions to MATH and MATH, while MATH restricts to REF to either MATH or MATH. This shows the long exact sequence associated to the short one defining MATH. Finally, the value MATH of the map is given by the boundary morphism MATH induced by the normalization sequence MATH on the right argument MATH. The formula for MATH thus follows from REF . |
math/9906109 | By the definition given in REF, MATH is generated by the expressions MATH, with MATH and MATH. By the formula for MATH given in REF , this expression is MATH, after identifying MATH with MATH. Clearly MATH is generated by the elements of MATH of the form MATH, whence the lemma. |
math/9906109 | We just have to verify REF. From the normalization sequence MATH one has a natural map MATH which obviously fulfills REF. |
math/9906109 | The kernel of the composition MATH is the composition MATH hence the MATH map sends MATH to the subgroup MATH of MATH. Take MATH, MATH, and let MATH. Then MATH . Since MATH is just the absolute MATH map, we see that MATH lands in the image of the cup product map MATH which is MATH. Since MATH is algebraically closed, the cup product MATH is surjective, from which one sees that the cup product maps MATH onto MATH. Combining this with the computation above completes the proof. |
math/9906109 | By the remark above, we may replace MATH with MATH. Let MATH be a model for MATH, with equation MATH defined over a number field MATH. Let MATH be the standard global one-form on MATH. Choosing an isomorphism MATH defines the period lattice MATH for MATH. Choose a basis for MATH of the form MATH, and let MATH. Let MATH be the NAME MATH-function for the lattice MATH. The map MATH gives rise to the isomorphism of NAME surfaces MATH making the diagram MATH commute, that is, MATH . We let MATH be the corresponding isomorphism of algebraic elliptic curves over MATH. By CITE, théorème REF, MATH has transcendence degree REF over MATH for all MATH, MATH. (We thank NAME for giving us this reference). Fix a MATH, let MATH be the algebraic closure of the field MATH, and let MATH be the point MATH. Then MATH is a generic point of MATH over MATH. We take MATH . By construction, MATH, where MATH is the element MATH. Here MATH denotes the class in MATH, and MATH denotes the class in MATH. Since MATH has transcendence degree one over MATH, the class of MATH in the absolute NAME cohomology of MATH vanishes, by REF . By REF , MATH dies in the analytic motivic cohomology of MATH as well. It remains to show that MATH is a non-torsion element of MATH. We give an analytic proof of this using the regulator map with values in NAME cohomology. Let MATH be a smooth projective surface over MATH, and let MATH denote the NAME group of divisors modulo homological equivalence. Then NAME theory implies that MATH and that MATH . We note that the map MATH induced by the cup product in NAME cohomology factors through MATH, and that the induced map MATH is injective. Indeed, MATH . Now take MATH, and let MATH be the complement of a non-empty finite set MATH of points of MATH. Let MATH be the class of MATH in MATH, and let MATH be the map MATH. Let MATH be the composition of MATH with the restriction map MATH. We claim that the sequence MATH is exact. Indeed, we have the localization sequence MATH the isomorphism MATH and the identity MATH which proves our claim. In particular, let MATH, where MATH is the diagonal, MATH is an element of MATH which is not a root of unity, and MATH is the class in MATH. Since MATH is not torsion in MATH, we see that MATH has non-torsion image MATH in MATH where the limit is over non-empty NAME open subsets MATH of MATH. Let MATH be the image of MATH in MATH. Then MATH is the image of MATH under the regulator map MATH. Similarly, letting MATH be the pull-back of MATH to MATH, MATH is the image of MATH under the regulator map MATH. Thus, MATH is a non-torsion element of MATH for each non-torsion element MATH. Let MATH be the diagonal in MATH, let MATH be the image of MATH in MATH, and let MATH be the image of MATH in MATH. Clearly, after choosing a complex embedding MATH, MATH (for MATH) is the image of MATH under the extension of scalars MATH, hence MATH is a non-torsion element of MATH. Since MATH is a geometric generic point of MATH over MATH, there is an embedding MATH such that MATH is the composition MATH. Thus, MATH is the image of MATH under MATH, and hence MATH is the image of MATH under the map MATH induced by the extension of scalars MATH. Since the kernel of MATH is torsion, it follows that MATH is a non-torsion element of MATH, as desired. |
math/9906111 | Recall that MATH is the category of functors from discrete MATH-algebras to sets. Define MATH by MATH and MATH. It is clear that MATH is a MATH-semiring object in MATH, and MATH is a MATH-ring object. There is an evident inclusion MATH. As MATH is a commutative monoid scheme, the set MATH is a commutative monoid for all MATH. As MATH is the free commutative monoid generated by the set MATH, there is a unique homomorphism MATH extending MATH. These maps are natural in MATH so we get a map MATH in MATH. If we interpret the colimits in MATH then we have MATH; this translates to the statement that MATH is the initial example of a formal scheme in MATH equipped with a map MATH in MATH. By similar arguments, we find that MATH is the initial example of a formal scheme over MATH with a map MATH in MATH. Moreover, one can check that the schemes MATH and MATH enjoy the evident analogous universal properties for all MATH. It now follows that there is a unique map MATH making the following diagram commute: MATH . Similarly, all the other structure maps for the MATH-semiring structure on MATH induce operations on MATH, and one checks easily that this makes MATH into a MATH-semiring scheme. A similar argument works for MATH. It is clear that there is a map MATH as described. |
math/9906111 | Our coordinate MATH gives an isomorphism MATH (the set of nilpotents in MATH). As MATH lies in the maximal ideal of MATH and MATH is a discrete MATH-algebra, we see that MATH for some MATH, and thus MATH is divisible by MATH in MATH. It follows that MATH is divisible by MATH. For any MATH we have MATH for large MATH, so MATH for large MATH, so MATH for large MATH as required. |
math/9906111 | First suppose that MATH. We can then choose a faithfully flat map MATH such that the image of MATH in MATH has the form MATH. The map MATH is automatically injective, so it suffices to show that for large MATH we have MATH, which is immediate from the previous lemma. Now suppose that MATH. We can then write MATH in the form MATH for some MATH and we reduce easily to the previous case. Finally, consider a general divisor MATH, which need not have constant dimension. Instead, we have a splitting MATH giving a bijection MATH under which MATH becomes a MATH-tuple MATH with MATH for some integers MATH. This means that MATH becomes MATH under the bijection MATH. The cases considered previously imply that MATH when MATH, as required. |
math/9906111 | Let the exponent of MATH be MATH, where MATH is coprime to MATH. The map MATH is an isomorphism and fixes MATH, and MATH so it suffices to show that MATH. To see this note that MATH and MATH for all MATH, so MATH is the trivial representation of rank MATH. As MATH is a ring map, we have MATH, as required. This implies that MATH for MATH but REF says that MATH for MATH, so MATH, so MATH as claimed. |
math/9906111 | Let MATH be the irreducible representations of MATH, and let MATH be their degrees. We assume that these are ordered so that MATH is the trivial representation of rank one. There are then natural numbers MATH and MATH for MATH and MATH such that MATH (Here MATH means the direct sum of MATH copies of MATH.) To give a homomorphism MATH is the same as to give divisors MATH for MATH such that MATH . This exhibits MATH as the equaliser of a pair of maps from MATH to MATH . In particular, this is a pair of maps between formal schemes over MATH, so the equaliser is a formal scheme over MATH. More explicitly, we have MATH, where MATH is defined as follows. We start with MATH and adjoin power series variables MATH for MATH and MATH, and put MATH. We then put MATH and impose the relations obtained by equating coefficients in the following identities between polynomials: MATH . The resulting quotient ring is MATH. |
math/9906111 | We can write MATH for some MATH and MATH. Put MATH. We have MATH so MATH so MATH as claimed. |
math/9906111 | Clearly MATH for MATH so MATH for MATH, and MATH by REF so MATH by the proposition. |
math/9906111 | We will assume that MATH, so MATH. Suppose that MATH and MATH. Put MATH so MATH, and put MATH and MATH. Then MATH and MATH so MATH, so MATH. Thus MATH for MATH. If MATH is in MATH we must have MATH. Note also that MATH, which is a unit multiple of MATH, so the condition is equivalent to MATH. The universal example for MATH is MATH (where MATH). Clearly in this case we have MATH so MATH. |
math/9906111 | Put MATH, so MATH, and let MATH be a MATH-algebra. If MATH is a MATH-semiring homomorphism, then MATH induces a group homomorphism MATH. Conversely, given a group homomorphism MATH we get a map MATH of MATH-semirings. We can compose this with the map MATH in the proof of REF to get a map MATH, or in other words a point of MATH. One checks that these constructions give a bijection MATH, or equivalently an isomorphism MATH. There is also an isomorphism MATH (see CITE), so we have an isomorphism MATH. A straightforward comparison of definitions shows that this is the same as MATH. |
math/9906111 | We may assume that MATH for some elements MATH. For any MATH with MATH we put MATH. We then have MATH and thus MATH . |
math/9906111 | As usual, we let the exponent of MATH be MATH, where MATH is coprime to MATH. If MATH and MATH has MATH-power order then MATH is conjugate to an element of MATH and thus MATH. If MATH is an arbitrary element of MATH then the order of MATH divides MATH so the order of MATH divides MATH, so MATH. This proves that MATH, so MATH in MATH. However, the action of MATH on MATH is induced by the action of MATH on MATH, which is invertible because MATH is coprime to MATH. This implies that MATH as claimed. |
math/9906111 | We already know that MATH is a free Abelian group, so we just need to determine its rank, so it is enough to show that MATH. Let MATH be the set of conjugacy classes of MATH-power order, and let MATH be the set of all other conjugacy classes. Let MATH be the ideal of virtual representations MATH whose character is zero on MATH. It is well-known that MATH. This isomorphism carries MATH to MATH and MATH to MATH so the map MATH is an isomorphism. Clearly MATH, and the claim follows. |
math/9906111 | Let MATH be the usual MATH-spectrum for equivariant MATH-theory, and let MATH be its MATH-completion. It is well-known that MATH, which is free Abelian of finite rank, and it follows that MATH. We give this ring and all its quotients the MATH-adic topology, or equivalently the profinite topology, which is compact. The argument of the NAME completion theorem shows that MATH is the completion of MATH at the augmentation ideal MATH. By a compactness argument, we deduce that the map MATH is surjective; the kernel is MATH. Now let MATH be a NAME MATH-subgroup, so by the same arguments MATH. It is well-known that MATH for MATH (use the fact that MATH for all MATH, for example) and it follows that MATH, so MATH. We now have a diagram as follows. MATH . We have seen that the columns are short exact. As MATH is flat over MATH and MATH, MATH and MATH are finitely generated Abelian groups, we see that the middle row is left exact. The bottom horizontal map is injective by a transfer argument. By a diagram chase we deduce that MATH, so MATH as claimed. |
math/9906111 | First suppose that MATH is a complete regular local ring. (In the topological context, this occurs when MATH is NAME exact.) Consider the following diagram: MATH . In the right hand square, all the corresponding rings are complete regular local rings. A finite injective map of such rings always makes the target into a free module over the source CITE. The maps MATH and MATH are finite injective maps of degrees MATH and MATH. It follows that MATH is finite and injective, and thus (as MATH in this context) that MATH is flat of degree MATH. The left hand square is a pullback by definition, and it follows that MATH is flat of degree MATH over MATH. Using CITE, it is not hard to deduce that this result remains true even if MATH is not regular. Next, let MATH be a coordinate on MATH. Then MATH is a basis for MATH over MATH, and MATH is a basis for MATH. Using this we obtain bases for the rings MATH and MATH that are permuted by MATH, and the orbit sums give bases for the rings MATH and MATH . Using these, it is easy to see that the map MATH is surjective, so the map MATH is a closed inclusion. A similar argument shows that MATH is a closed subscheme of MATH. Next, put MATH . This is clearly a nilpotent ideal, and MATH which is a nilpotent ideal in MATH. Thus our map MATH is an infinitesimal thickening. It is clear that MATH is contained in MATH. Next, let MATH be the preimage of MATH in MATH, or equivalently the scheme of MATH-tuples MATH such that MATH. If MATH then for each MATH we have MATH so MATH. This means that MATH and thus MATH, so the map MATH factors through MATH. As MATH is faithfully flat, it follows that MATH as claimed. We now have maps MATH . We know that MATH, MATH and MATH are closed inclusions and that MATH is an infinitesimal thickening. It follows easily that MATH, MATH and MATH are closed inclusions and MATH and MATH are infinitesimal thickenings. Now suppose that MATH is a field of characteristic MATH. We then have an iterated NAME map MATH corresponding to the ring map MATH and thus a formal group MATH over MATH. The map MATH gives rise to a map MATH. As MATH has height MATH, the map MATH factors as MATH, where MATH is an isomorphism. This is just the geometric statement of the fact that MATH for some invertible power series MATH. By REF is the fibre of the map MATH induced by MATH, and it follows easily that it is also the fibre of the map MATH induced by MATH. It is easy to identify this with the map MATH. If we use the usual generators for the coordinate rings of MATH and MATH then the corresponding ring map sends MATH to MATH, so MATH. |
math/9906111 | Let MATH be the irreducible representations of MATH, and let MATH be their degrees. As in the proof of REF , we see that MATH is a closed subscheme of MATH. Now let MATH be the MATH-part of the exponent of MATH. We see from REF that MATH is actually contained in MATH, which is finite over MATH by the theorem. It follows that MATH itself is finite, as claimed. |
math/9906111 | We can reduce easily to the case where MATH for some MATH. REF tells us that MATH for some MATH, so MATH. Next note that MATH is a discrete MATH-algebra so MATH is nilpotent in MATH, say MATH. It follows that MATH for some power series MATH with MATH, and thus that MATH is divisible by MATH. Thus, for large MATH we have MATH, so MATH, so MATH. If we put MATH this tells us that MATH, and the action of MATH on MATH clearly depends only on the congruence class of MATH mod MATH, as required. |
math/9906111 | For brevity we will write MATH and MATH. This is a slight abuse because these functors do not arise from a spectrum MATH. We also let MATH be any integer greater than or equal to the MATH-adic valuation of the exponent of MATH. Any homomorphism MATH factors through MATH and thus is automatically continuous (for the discrete topology on MATH). It thus gives a positive homomorphism MATH, and it is well-known that this depends only on the conjugacy class of MATH, so this construction gives a natural map MATH. It is easy to see using NAME operations that any positive homomorphism MATH actually lands in the subring MATH. Suppose we have a level structure MATH. As MATH is a finite Abelian group, this gives rise as in REF to a positive homomorphism MATH, which we can compose with MATH to get a positive homomorphism MATH, or in other words a point of MATH, which we call MATH. This construction produces a map MATH of formal schemes over MATH, corresponding to a map MATH. After tensoring by MATH we obtain the required map MATH. We next recall the definition of MATH. Suppose that MATH and MATH. We then have a point MATH. This construction gives a map MATH and thus a map MATH. After tensoring by MATH we obtain the required map MATH. One can check from the definitions that the following diagram commutes: MATH . It follows easily that the diagram in the statement of the theorem commutes. To understand MATH more explicitly, let MATH be the universal example of a level structure. For any element MATH we then have a point MATH and thus an element MATH. These elements satisfy MATH and MATH as divisors, or equivalently MATH is a unit multiple of MATH in MATH. It is also known that MATH is invertible in MATH whenever MATH (because it is a unit multiple of MATH, which divides MATH). Let the representations MATH and the elements MATH be as in the proof of REF . If MATH and MATH then MATH is the MATH'th symmetric function in the variables MATH, and this characterises MATH. We next show that MATH is surjective. For any MATH we define MATH by MATH, and we put MATH. By the NAME Remainder Theorem, it will suffice to show that MATH whenever MATH. If MATH we can choose MATH such that MATH. If MATH and MATH then we must have MATH for some MATH, and without loss we may assume MATH. Define MATH . We can write MATH in the form MATH with MATH for all MATH. We also write MATH, so MATH is invertible in MATH. On the other hand, the representation MATH gives rise in a tautological way to a divisor MATH with equation MATH, say. We have MATH and MATH. The polynomial MATH is monic and thus is not a zero-divisor, so MATH. We evidently have MATH so MATH. As MATH is invertible in MATH, we deduce that MATH as required. Finally, we must show that the kernel of MATH is nilpotent. This kernel is the intersection of the ideals MATH, so by well-known arguments it suffices to show that every prime ideal in MATH contains MATH for some MATH. To see this, put MATH and let MATH be a prime ideal. If MATH is a divisor satisfying MATH, then REF implies that MATH and thus that MATH as divisors, or equivalently MATH divides MATH, which is a unit multiple in MATH of MATH. Now let MATH be the field of fractions of MATH, and note that MATH is invertible in MATH and thus in MATH when MATH. As MATH is a unique factorisation domain, we see that MATH in MATH for a unique system of integers MATH. We define MATH. In particular, if MATH we can let MATH be the tautologically associated divisor over MATH and put MATH. One can check that this gives a homomorphism MATH of MATH-semirings, or in other words an element MATH. From the construction it is automatic that MATH. |
math/9906111 | There are evident natural maps MATH . We have also already constructed a map MATH. If MATH are pointwise-conjugate then the induced maps from class functions on MATH to class functions on MATH are evidently the same, so the induced maps MATH are the same, so MATH. This shows that MATH factors through the projection MATH. We next define a map MATH. Suppose that MATH and MATH. Then MATH extends in a natural way to give a MATH-algebra map MATH and thus a ring map MATH. Using the fact that MATH is the set of MATH-valued class functions on MATH, we see that MATH can be identified with the set of conjugacy classes in MATH. Thus there exists MATH (unique up to conjugation) such that MATH for all MATH. We can choose MATH so that MATH for all MATH, and then we have MATH for all MATH, so MATH. This means that the conjugacy class MATH lies in MATH, so we can define MATH. We leave it to the reader to check that this gives a map MATH as claimed. The maps MATH (as MATH runs over MATH) are jointly injective, an it follows that MATH is injective. One can also check that the composite MATH is just the obvious inclusion, which implies that the map MATH is injective. |
math/9906111 | Put MATH and MATH. Here MATH is the MATH'th elementary symmetric function, and in particular MATH. Put MATH. If we regard MATH as a map MATH then MATH, so we see that MATH and MATH. Next, observe that the inclusion MATH induces an inclusion MATH. We have MATH so MATH. It follows easily that MATH and thus that MATH. |
math/9906111 | As MATH is invertible in MATH, multiplication by MATH is an automorphism of MATH. Define maps MATH by MATH and MATH, and then define MATH by MATH. Clearly MATH is inverse to MATH, so MATH is an isomorphism, giving an isomorphism MATH of rings. If we let MATH act trivially on MATH then everything is equivariant, so we have MATH, so MATH. |
math/9906111 | Given a positive homomorphism MATH, let MATH be the element such that MATH and put MATH. We know from REF that MATH and it follows easily from our description of MATH that MATH and MATH have the properties listed. Conversely, given MATH and MATH as described, we can define a homomorphism MATH of additive groups by MATH . It is straightforward to check that this gives a homomorphism of MATH-rings, and that these constructions give the required bijection. The argument for MATH is essentially the same. |
math/9906111 | Define MATH . Recall from REF that MATH. An easy case-by-case check shows that the sets MATH are disjoint and that the maps MATH are bijections. It will thus be enough to show that the union of the sets MATH is the whole of MATH. Suppose we have an element MATH, with MATH and MATH say. Let MATH, MATH and MATH be the cyclic subgroups generated by MATH, MATH and MATH respectively. Put MATH, and recall that this determines MATH, because MATH generates MATH as a MATH-ring. As MATH we have MATH. By REF , we have MATH . Suppose that MATH. Without loss of generality we may assume that MATH so MATH. The third equation implies that MATH, so we may assume that MATH. As MATH we must have MATH. Recall also that MATH so MATH. Putting all this in the last equation and cancelling MATH gives MATH . Note that MATH and MATH have order MATH, but MATH, MATH, MATH and MATH do not. It follows that we must have MATH and thus MATH. We conclude that MATH. We may thus assume that MATH, so MATH. Suppose that MATH. As MATH we see that MATH is a subgroup of MATH, of order MATH say (so MATH). This implies that MATH, so MATH. We may thus assume that MATH and MATH. The equation MATH then reduces to MATH . It follows that MATH and without loss we may assume that MATH. Note that MATH (because MATH). If MATH this gives MATH so MATH, so MATH. The same argument works if MATH, so we reduce to the case where MATH and MATH and MATH are also nonzero. We then have MATH so MATH. |
math/9906111 | Put MATH and MATH . Put MATH, so MATH is free of rank MATH over MATH. Put MATH and MATH . For any element MATH we write MATH, so MATH is a ring map and MATH for MATH. Put MATH and MATH. The claim is that the ideal in MATH generated by MATH is free of rank one over MATH, and that MATH, so that MATH. We will think of MATH as being embedded in MATH by the map MATH, so MATH . There is a faithfully flat map MATH sending MATH to MATH, and clearly MATH so MATH. It follows that MATH, and thus that MATH as claimed. This implies that we must have MATH and MATH for some MATH. Now work in MATH. We have MATH . We also find that the ideal MATH maps to zero in this ring. Using this, we find that MATH and MATH, so MATH and MATH. It follows that MATH as claimed. |
math/9906111 | Recall that MATH, so MATH is a basis for MATH over MATH. As MATH, it is easy to see that MATH and that these elements are linearly independent in MATH. Moreover, we have MATH . It is easy to check that MATH does not lie in the span of these elements, and to deduce that MATH and MATH are jointly injective as claimed. Thus, to show that MATH we need only check that MATH and MATH, which is a straightforward computation. |
math/9906111 | First recall that MATH and MATH so MATH for any MATH. Next note that MATH, so MATH is the MATH'th elementary symmetric function of MATH, for example MATH . By similar computations, our other two equations also become true when we apply MATH. In the same way, we have MATH, so MATH is the MATH'th elementary symmetric function of the list MATH, or equivalently the list MATH . We thus have MATH . By similar computations, our other two equations also become true when we apply MATH. As MATH and MATH are jointly injective, it follows that our equations hold in MATH as claimed. |
math/9906111 | REF is equivalent to the statement that MATH . This means that MATH is the largest quotient of MATH over which we have MATH, where MATH . Here we write MATH and similarly for our other divisors. As usual we write MATH for MATH, and we recall from REF that MATH. We also recall that MATH, so that MATH . The polynomial MATH can be read off from REF . Putting all this together and expanding it out, we find that MATH, where MATH . We thus have MATH . As MATH is invertible, we can replace MATH by MATH which is one of the relations in the statement of the theorem. As MATH we have MATH and MATH, so the relations MATH and MATH are redundant. Similarly, we can replace MATH by the relation MATH which is another of the relations in the statement of the theorem. One can check that MATH so MATH is redundant. We deduce that MATH as claimed. We next show that the MATH monomials listed form a basis for this quotient ring. We order the set of monomials in MATH, MATH and MATH by saying that MATH iff MATH or (MATH and MATH) or (MATH and MATH and MATH). We claim that our relations form a NAME basis for MATH with respect to this ordering. We first recall briefly what this means. The list of leading terms of our relations is MATH. A polynomial is said to be top-reducible if any of its monomials is divisible by one of these leading terms; if so, we can subtract off a multiple of the corresponding relation to cancel the monomial, a process called top-reduction. Clearly, if a polynomial can be reduced to zero by iterated top-reduction then it must lie in MATH, but the converse need not hold for an arbitrary list of generators of an arbitrary ideal. Let MATH and MATH be any two of our relations, let MATH and MATH be their leading terms, and let MATH be the greatest common divisor of MATH and MATH. The corresponding syzygy is the element MATH. To say that our relations form a NAME basis means precisely that all these syzygies can be reduced to zero by iterated top-reduction. This can be checked by direct computation. For example, the syzygy of MATH and MATH is the element MATH. The first monomial is divisible by the leading term of MATH, so we can top-reduce by subtracting MATH to get MATH. We can then do two more top-reductions by subtracting MATH times the relation MATH and MATH times the relation MATH to get MATH, as required. Now observe that the MATH monomials listed in the statement of the theorem are precisely those that are not top-reducible. It follows from the theory of NAME bases that they form a basis for MATH, as claimed. |
math/9906111 | One sees easily from the relations listed that MATH, MATH and MATH annihilate MATH, so MATH lies in the socle. Now let MATH be an arbitrary element of the socle. It will be convenient to put MATH (so that MATH) and to use the basis given in the Proposition but with MATH replaced by MATH. Using the equation MATH we see immediately that MATH lies in the span of MATH. Using the equation MATH and the fact that MATH we find that the coefficient of MATH is zero, so MATH say. One can check that MATH and MATH, so MATH so MATH, so MATH. This shows that the socle is one-dimensional, so the ring is NAME as claimed. |
math/9906111 | Note that every nontrivial ideal in MATH contains the socle, so it will suffice to show that the socle is not contained in MATH, or equivalently that MATH in MATH. Let MATH be the NAME subgroup in MATH; it will be enough to show that MATH has nontrivial image in MATH. Put MATH; one can check that this consists of the identity and the three transposition pairs, so it is isomorphic to MATH. Recall that the series MATH is defined to be MATH, which in our case is just MATH. As MATH is the NAME class of MATH and MATH, standard arguments show that MATH. This means that MATH. To see that this is nonzero, we use the canonical bilinear form on MATH defined in CITE. This satisfies NAME reciprocity, so MATH. If we let MATH and MATH be the NAME classes of two of the nontrivial characters of MATH, then MATH and the NAME class of the third character is MATH. One checks that the restriction of MATH to MATH is the regular representation minus the trivial representation, which is the sum of the three nontrivial characters. This implies that the restriction of MATH to MATH is MATH. Using CITE we see that MATH is MATH if MATH and MATH otherwise, so MATH. As MATH we see that MATH, as claimed. |
math/9906111 | Choose a maximal isotropic subspace MATH, so MATH. Put MATH, which is isomorphic to MATH as a group because MATH is isotropic. Let MATH be the projection and put MATH. We claim that MATH. To see this, first note that MATH is normal in MATH, so MATH for MATH. Next, suppose that MATH, say MATH with MATH. Let MATH be such that MATH, so MATH and MATH. From the definitions we see that MATH. The map MATH is a surjection from MATH to MATH, each of whose fibres has the same order, and MATH is a nontrivial homomorphism; it follows easily that MATH, as required. Finally, suppose that MATH, say MATH. Then MATH so MATH. This shows that MATH as claimed, so MATH is a character. One checks easily that MATH, so MATH is irreducible. As MATH runs over MATH this gives MATH distinct irreducibles of degree MATH, and MATH gives a further MATH distinct irreducibles of degree MATH. We have seen that MATH has MATH conjugacy classes and thus MATH irreducible characters, so our list is complete. It follows that MATH as claimed. |
math/9906111 | Let MATH be a generator of MATH, so MATH and MATH. We have MATH and so MATH. If MATH then MATH is also a generator, and it follows that MATH is an integer multiple of MATH. On the other hand, it is easy to check that MATH. If we put MATH then MATH is a subring of MATH (with MATH) and MATH so MATH also lies in MATH, say MATH. Moreover, if we work mod MATH we have MATH so MATH. Thus, if MATH divides MATH then MATH, and if MATH does not divide MATH then MATH. Moreover, by counting dimensions we see that MATH for all MATH. The lemma now follows easily. |
math/9906111 | Everything except for MATH can be done by easy manipulation of characters. For the remaining case, it suffices to check that the claimed equations hold when restricted to any cyclic subgroup MATH. First consider the case MATH, so MATH restricts on MATH to the trivial representation of degree MATH. Then MATH becomes MATH, so MATH becomes MATH. Using this, it is easy to check that the equations hold when restricted to MATH. Now suppose instead that MATH is a cyclic group not contained in MATH (which implies that MATH). Then MATH and MATH for all MATH. Using REF we deduce that our equations for MATH are correct when restricted to MATH, as required. |
math/9906111 | Put MATH so that MATH. Suppose MATH and that MATH. Put MATH, so MATH and MATH. Thus MATH also, so MATH, so we can write MATH for suitable natural numbers MATH. By looking at the multiplicity of MATH in the equation MATH we see that MATH. On the other hand, as MATH we have MATH. It follows that MATH so MATH for all MATH. If we now let MATH be the coset MATH we find that MATH. Conversely, it is trivial to check that any element of this form lies in MATH. |
math/9906111 | Let MATH be the usual character MATH of MATH. Given MATH it is clear that the restriction of MATH to MATH gives a homomorphism MATH, and we put MATH. As MATH is a MATH-ring homomorphism we have MATH so MATH. Conversely, suppose we start with MATH. Let MATH and MATH be as in REF . We define a homomorphism MATH of additive groups by MATH for MATH and MATH for MATH. It is easy to check that this is a ring homomorphism that sends MATH to MATH and commutes with the NAME operations. As MATH is torsion free it follows that MATH commutes with MATH-operations as well, so MATH. Clearly these constructions give the required bijection MATH. Now suppose we have a homomorphism MATH. Then MATH for some functions MATH and MATH. As MATH and the projection MATH are homomorphisms we see tht MATH is a homomorphism. Let MATH be the image of MATH, and put MATH. As the image of MATH must be commutative, it is not hard to see that MATH is isotropic, so MATH. As MATH as groups, we see that MATH is also a homomorphism. If we conjugate MATH by MATH we get the homomorphism MATH where MATH. As MATH is a perfect pairing, MATH can be any map MATH that factors through MATH, so MATH is conjugate to MATH if and only if MATH. Now let MATH be the dual of MATH and put MATH, so MATH. We also have a map MATH and thus a point MATH. In MATH we have MATH and it follows that MATH. Thus, if we write MATH for the conjugacy class of MATH then MATH. It follows that MATH determines MATH, and it also determines MATH modulo MATH, so it determines MATH modulo MATH, so it determines the conjugacy class MATH. This proves that MATH is injective as claimed. We leave it to the reader to check that the image is as described. |
math/9906111 | The statement can easily be translated as follows: Let MATH and MATH be finite-dimensional vector spaces over MATH equipped with actions of MATH and invariant Hermitian inner products. Then if there exists an equivariant isomorphism MATH, then MATH can be chosen to preserve the inner products. To see this, we first recall some facts about invariant Hermitian products. For any complex vector space MATH we let MATH be the same set with the conjugate action of MATH, and let MATH be the dual of MATH. The set of Hermitian products MATH on MATH bijects with the set of isomorphisms MATH satisfying certain symmetry and positivity conditions. For any representation MATH one can always choose an invariant Hermitian product so MATH is equivariantly isomorphic to MATH. For each irreducible representation MATH we fix a Hermitian product MATH on MATH; NAME 's lemma implies that MATH and that any other invariant Hermitian product is a positive scalar multiple of MATH. Now let MATH be a Hermitian product on MATH and suppose that MATH and MATH. Then MATH and MATH so the equivariant isomorphism MATH must have the form MATH for some MATH. This implies that MATH and MATH are orthogonal with respect to MATH. Now let MATH be the distinct irreducible representations that occur in MATH. Then there is a unique decomposition MATH, where MATH for some MATH and thus MATH when MATH. By the previous paragraph, the subspaces MATH are orthogonal to each other. As MATH, we find that the restriction of MATH to MATH has the form MATH for some Hermitian product MATH on MATH. By NAME, the space MATH is isomorphic to MATH with its usual Hermitian product, so MATH is equivariantly and isometrically isomorphic to the orthogonal direct sum of MATH copies of MATH. This means that the numbers MATH determine the isometric isomorphism type of MATH, and the lemma follows immediately. |
math/9906111 | It is easy to reduce to the case where the list MATH has length MATH, say MATH. As any representation of MATH admits a Hermitian inner product, we see that the map MATH is surjective. It is also injective by REF . Similarly, by considering invariant Hermitian products we see that if MATH is compact and MATH then MATH is conjugate to a homomorphism MATH. By applying this to the inclusion map, we see that any compact subgroup of MATH is conjugate to a subgroup of MATH. It follows that any two homomorphisms MATH are identified in MATH iff their restrictions to MATH are conjugate, so we have a well-defined and injective restriction map MATH. It is an easy consequence of the theory of roots and so on that any representation of MATH extends uniquely to a complex-analytic representation of MATH, so our restriction map is also surjective. |
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