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math/0009011 | Note that MATH, as there are no further differentials in this part of the spectral sequence. By REF , MATH, which is MATH by REF . |
math/0009011 | We need the fact that an extension MATH is NAME if and only if MATH. This follows from REF . We have shown that MATH is obtained from MATH by adjoining the square roots of elements in MATH, while MATH is obtained by adjoining the square roots of elements in MATH. Therefore we have MATH if and only if MATH. To show that MATH implies MATH we note that for any element MATH, the orbit of MATH under the action of MATH is finite. This is because we may choose a representative of MATH, which is an element of MATH, hence algebraic over MATH, so that any image of this element under the action of the NAME group satisfies the same minimal polynomial. Therefore, if MATH, the MATH submodule generated by MATH is finite and has a fixed point by the usual counting argument. Thus if MATH, MATH, which is the contrapositive of the desired statement. The converse is trivial. |
math/0009011 | Since MATH is not a NAME, by CITE there exists an anisotropic binary form MATH over MATH such that the set MATH of values of MATH, regarded as a subgroup of MATH, has at least three elements. (As MATH is an elementary abelian MATH-group, so are any of its subgroups, so ``MATH has at least three elements" immediately implies ``MATH has at least four elements". The fact that MATH is a subgroup follows from the identity MATH.) We may assume that MATH for some MATH by an appropriate transformation. Because MATH is anisotropic, we know that MATH in MATH. Since MATH, there exist elements MATH, MATH in MATH which are linearly independent over MATH. The statement ``MATH is a value of MATH in MATH" means that there exist MATH, MATH, and MATH in MATH such that MATH. By REF we see that MATH, and similarly that MATH. This information allows us to construct directly a nonzero element in the spectral sequence for MATH. Because the relations MATH and MATH exist in MATH but not in MATH, there exist MATH, MATH in MATH such that MATH and MATH. Then, setting MATH, we can compute MATH . Thus MATH survives to MATH and hence to MATH by its position in the spectral sequence. |
math/0009011 | Recall that we have the exact sequence MATH arising from the long exact sequence in cohomology associated to the short exact sequence of coefficients MATH. Note that since the action of MATH on the coefficient group MATH is trivial, we may identify MATH with MATH. In other words MATH. We shall make this isomorphism explicit. Let MATH. Then we can associate to this MATH the function MATH which is given by the formula: MATH . It follows from the definition of socle series that MATH. The connecting homomorphism MATH is given by MATH. In other words, we have a one-to-one correspondence MATH given by MATH. Using the basis MATH introduced earlier each function MATH can be written as MATH simply because MATH. Observe that our function MATH has the values: MATH and MATH . (Recall that each continuous homomorphism MATH has only finitely many values by definition of the NAME topology on MATH.) Combining the above with the fact that REF and the definition of MATH completes the proof. |
math/0009011 | We have shown that for MATH-fields, MATH, so that MATH and MATH. So we must show that MATH. From the isomorphism between NAME cohomology and the NAME ring given by the NAME conjecture, it is enough to show that MATH and that MATH, where MATH is the MATH-th power of the fundamental ideal in the NAME ring. Ware CITE has shown for fields satisfying our hypotheses that MATH, while it follows from a result of CITE and the basic theory of NAME forms that MATH in this case. |
math/0009011 | Suppose that MATH. Let MATH. Then MATH is an open subgroup of MATH and MATH. Therefore from the above we obtain MATH. However from NAME theory we know that MATH. |
math/0009011 | From local class field theory (see CITE) we know that there is a natural isomorphism MATH for any intermediate extension MATH. If MATH we obtain the first statement. For MATH we observe (see CITE, page REF) that for any proper quadratic extension MATH of MATH, MATH; however MATH. We conclude that MATH cannot be contained in MATH. |
math/0009011 | Let MATH denote an extension of MATH such that MATH. Consider MATH; then by restriction MATH will also be a module over this group. Denote the usual module-theoretic norm map by MATH. From the definition of the socle series for MATH it follows that MATH, where the intersection is taken over all extensions as above, of co-degree MATH. Let MATH denote a proper quadratic extension of MATH. The norm map MATH is induced from the composition of the field-theoretic norm map MATH with the maps MATH and MATH (the latter induced by the inclusion MATH). Hence if MATH, we see that MATH if and only if MATH. However from NAME theory (see CITE, page REF ) we have that MATH, and so MATH if and only if MATH. However, from REF , pp. REF, we know that MATH belongs to MATH if and only if MATH. Therefore, from the transitivity property of the norm we see that MATH if and only if MATH. |
math/0009011 | From our description of MATH and the triviality of the map induced by the norm, we infer that MATH, whence MATH. Taking a quadratic extension MATH we see that in the socle series for MATH as a MATH-module, MATH. Hence MATH. |
math/0009011 | Let MATH denote a basis for the subgroup MATH. For each MATH one can construct a REF - dimensional real representation of the NAME subgroup of MATH which restricts non-trivially to MATH. This can then be induced to yield a MATH-dimensional representation MATH for MATH. Doing the analogous construction for MATH and pulling back to MATH, we obtain the representaions MATH. It is elementary to verify that every involution in MATH must acts freely on the product of associated spheres; indeed the unique maximal elementary abelian subgroup satisfies this by construction. |
math/0009011 | Suppose that MATH is as in the statement of our Proposition. In order to show that MATH it is enough to observe that for each monomial MATH that can occur in the expression of the left hand side as an element of MATH, the corresponding coefficient is zero. This can be proved using a case by case analysis, in which division into cases depends on the multiplicities of the MATH which appear in the monomial. We limit ourselves to providing a complete analysis for the case where all the multiplicities are MATH; the other cases are very similar and left to the reader. Assume that CITE then the term MATH can occur as the summand of the following terms of our sum MATH: MATH . From REF we see that when this term does occur, one of our triple commutators MATH, MATH, or MATH will be MATH and not the identity. However from the identity REF: MATH and the relation MATH we see that the term MATH occurs in our sum MATH either zero or two times. This completes the analysis of the case when all the multiplicities are MATH; an analysis of the other cases proves that all terms MATH occur an even number of times in the sum and thus we have MATH as desired. |
math/0009015 | We have to show that MATH respects the relations REF , in other words, MATH maps equivalent sums of triples to equivalent ones. It is trivial with REF . To check REF , let us recall REF . Consider a sum of triples MATH belonging to REF , that is MATH, and MATH on the smooth part of MATH. Since the irreducible components of MATH can be treated separately it is natural to consider only the case when all the triples have the same support, MATH. Let MATH be the divisor of poles of MATH and let MATH. We want to prove that MATH on the smooth part of MATH, where MATH for each MATH is the restriction of the map MATH. Suppose first that there exists a smooth point of MATH which is smooth also in MATH. Then REF applied in a neighborhood of that point gives us the desired vanishing MATH, as a consequence of the equality MATH. This is however not enough for our proof since some components of MATH may lie entirely in the set of singular points of MATH. To overcome this problem we apply the NAME theorem replacing MATH with a smooth variety MATH, a blow-up of MATH, and correspondingly blowing up all MATH, so that the following diagram is commutative: MATH . Then we apply REF on the blown up side. We must recall now that the divisor MATH could have components that were mapped by MATH to subvarieties of dimension less than MATH; hence, we conclude that we have just proved the following statement (symbolically): if REF then REF + REF . Now, it remains to prove the compatibility of MATH with REF . Let MATH be a degenerate triple described in REF , that is, MATH. We shall show that REF + REF in this case. The polar divisor MATH, is, by assumptions of REF , a normal crossing divisor in MATH. Let us split the components of MATH into two parts: non-degenerate and degenerate ones. That is MATH where MATH and MATH. According to this splitting, MATH is represented as a sum of two terms corresponding to MATH and MATH. The second term belongs to REF and we have to show only that the first one belongs to REF , that is, that MATH, where MATH. Recall that we suppose that MATH. If it happens that MATH, we have MATH and there is nothing to prove. Therefore we may assume that MATH and, by irreducibility of MATH, MATH. Then, for a generic smooth point MATH, its preimage in MATH, MATH, is a smooth projective curve. This curve intersects with MATH over the set MATH and we may suppose that the latter consists of a finite number of points MATH which are smooth in MATH and that the intersections are transverse there. Let MATH denote the values of MATH at the points MATH and pick up a non-vanishing MATH-form MATH at MATH (recall that MATH). Let us show that MATH (this would mean that MATH on the smooth part of MATH - the required result). To prove this, let us notice that there exists a meromorphic REF-differential MATH on MATH such that MATH . (MATH is obtained by dividing MATH by the non-vanishing form MATH.) This equality is understood in the sense of the natural isomorphism MATH where MATH is the smooth part of MATH. It is easy to see now that for MATH we have MATH . The latter equality follows from the observation that MATH are the only points on MATH where MATH has poles. Indeed, the poles of MATH are located on MATH. One part of this gives us the points MATH, while the rest, MATH, corresponding to the ``degenerate" part MATH of MATH can be assumed to be empty, MATH. Indeed, we could have assumed from the very beginning that MATH does not meet MATH because MATH and we might suppose that MATH. |
math/0009015 | We need to prove this for triples MATH, that is, for forms MATH with normal crossing divisors of poles. The repeated residue at pairwise intersections differs by a sign according to the order in which the residues are taken, see REF. Thus the contributions to the repeated residue from different components cancel out (or, the residue of a residue is zero). |
math/0009015 | By the NAME duality, MATH is the map MATH and it is sufficient to verify that the pairing REF vanishes if MATH and MATH, or if MATH and MATH. This follows immediately from the NAME - NAME REF : MATH that is MATH. |
math/0009015 | As we have already mentioned, the homomorphism MATH of REF can be conveniently described in terms of the following natural map of polar chains: MATH where MATH is the space of currents of degree MATH (that is, a space of linear functionals on smooth MATH-forms, see CITE). As a matter of fact, this map is described by the integral REF . For a MATH-dimensional submanifold MATH, let the current MATH denote the linear functional on MATH-forms corresponding to the integration over MATH. The current MATH is supported on MATH. Therefore, for a MATH-form MATH defined on MATH, the product MATH makes sense and defines a current in MATH. Recalling the isomorphism of the cohomology of currents with the cohomology of smooth forms, MATH we can use MATH and MATH in place of MATH and MATH in the integral REF . Thus, for a transverse intersection of smooth polar cycles we derive that MATH . The second equality can be checked in local coordinates. This proves the theorem, since MATH is supported on MATH. |
math/0009015 | A straightforward verification. |
math/0009015 | This will be similar to the proof of REF and will use the same notations. We first represent the polar cycles MATH and MATH by the currents MATH and MATH respectively, then MATH where MATH on the right is understood as taking the MATH-cohomology class. On the other hand, for MATH introduced in REF , the current representing MATH is MATH and it is easy to show that MATH what implies the statement of the Theorem. The last equality is easily checked by noticing that MATH is a MATH-form (in fact, a current) conormal to MATH and similarly for MATH, while MATH. This is to be compared to MATH and MATH in REF . One has to note only that, for example, MATH is conormal to MATH over MATH (that is in the sense of MATH-forms) while MATH is conormal to it over MATH (that is in the sense of MATH-forms). |
math/0009023 | This follows from REF . Write the orthogonal projection of MATH onto the span of MATH as the linear combination MATH. Then MATH, which proves REF . Similarly, MATH. This proves REF with MATH. If MATH are linearly independent then the representation MATH is unique. |
math/0009023 | Since MATH are bounded random variables, to prove REF we need only to verify that for arbitrary polynomials MATH, MATH . This is equivalent to MATH see REF . The latter follows from REF , proving REF . To prove REF , we verify that for arbitrary polynomials MATH we have MATH . By REF , this is equivalent to MATH . It suffices to show that REF holds true when MATH and MATH are the MATH-Hermite polynomials defined by REF . REF implies that the left-hand side of REF is given by MATH and the right-hand side becomes MATH . Using formulas from REF we can see that both sides are zero, except when MATH or MATH. We now consider these three cases separately. CASE: Using REF simplifies to MATH . This equation is satisfied when coefficients MATH satisfy the equation MATH . CASE: Using REF simplifies to MATH . This equation is satisfied whenever MATH . CASE: We use again REF . On both sides of REF we factor out MATH, and equate the remaining coefficients. (This is allowed since we are after sufficient conditions only!) We get MATH . Now we use MATH. Suppressing the correction to the constant term (that is, the term free of MATH), we get MATH where MATH denotes the suppressed constant term corrections. This equation holds true when the coefficients at MATH match, which gives MATH and the constant terms match: MATH. The latter holds true when the expectations are equal (MATH), and hence this condition is equivalent to REF . The remaining three REF , and REF have a unique solution given by REF . |
math/0009023 | Using the definition of vacuum expectation state, REF we get MATH. Therefore REF , and REF imply MATH . The latter is zero, except when MATH or MATH. We will consider these two cases separately. If MATH, by orthogonality we have MATH . Clearly, MATH; this can be seen either from REF , or directly from REF . By REF the second term splits into the sum over permutations MATH such that MATH and the sum over the permutations such that MATH. This gives MATH . Elementary algebra now yields REF for MATH. If MATH, then the right-hand side of REF consists of only one term we get MATH . Since MATH is given by the same expression with the roles of MATH switched around, this ends the proof of REF . The remaining expectations match the corresponding commutative values, and can also be evaluated using recurrence REF , and REF . To prove REF notice that since MATH and MATH commute, using REF we get MATH . The only non-zero values are when MATH, or MATH. Using REF again, and then REF we get REF . Since by REF we have MATH recurrence REF used twice proves REF . REF is an immediate consequence of REF . |
math/0009023 | Since the case MATH is well known, we restrict our attention to the case MATH. Our starting point is REF . A computation shows that the conditional variance MATH is as follows. MATH . The right-hand side of this expression must be non-negative over the support of MATH. It is known, see CITE or CITE, that MATH have the joint probability density function MATH with respect to the product of marginals MATH. Moreover, MATH is defined for all MATH and from its explicit product expansion we can see that MATH is strictly positive. In particular, the right-hand side of REF must be non-negative when evaluated at MATH. Using REF with the above values of MATH we get the rational expression for the conditional variance which can be written as follows. MATH . Therefore MATH . Since the assumptions imply that MATH, this implies MATH, proving REF . |
math/0009024 | We follow the proof of REF. Let MATH be the canonical decomposition of the restriction of MATH to MATH into a direct sum of isotypic representations. Since the MATH-module MATH is simple, MATH permutes the MATH transitively. If MATH is some MATH, then the MATH-module MATH is isotypic and REF holds. Assume from now on that REF does not hold. Let MATH . Then MATH. Since MATH contains MATH, it is normal in MATH. Thus for every MATH, MATH . Every MATH is a simple MATH-module, because MATH is a simple MATH-module. The kernels of the natural maps MATH have trivial intersection and are conjugate in MATH. It follows that if MATH is a non-identity element of MATH, then it has a conjugate which does not lie in the kernel of MATH. Since MATH divides MATH, it is a MATH-power, and therefore odd. Thus MATH is odd, so at least one of the simple MATH-modules MATH is self-dual. This implies easily that all the MATH are self-dual. Suppose MATH is not symplectic. Then none of the MATH are symplectic. Since MATH is a simple MATH-module, every MATH-invariant alternating bilinear form on MATH is zero. Since the MATH are mutually non-isomorphic simple MATH-modules, every MATH-invariant bilinear pairing beween MATH and MATH for MATH induces the zero map MATH, and therefore is zero. Therefore, every MATH-invariant alternating bilinear form on MATH is zero, contradicting that MATH is symplectic. Thus MATH is symplectic. |
math/0009024 | The MATH-module MATH splits into a direct sum of MATH-modules MATH such that the restriction of MATH to MATH is non-degenerate, and either MATH is simple or MATH where MATH is simple. In the latter case choose a MATH-stable MATH-lattice MATH in MATH and apply REF . |
math/0009024 | Let MATH be the non-zero pairing induced by MATH. Its kernel is a MATH-submodule of MATH, so is zero by the MATH-simplicity of MATH; that is, MATH is non-degenerate. By NAME 's Lemma, MATH is perfect. |
math/0009024 | Let MATH choose a section MATH, and let MATH . Note that MATH is independent of the choice of section MATH. Define a homomorphism MATH by MATH for MATH, MATH, MATH. Then the desired conditions are all satisfied. |
math/0009024 | By REF there is an injective homomorphism MATH. Now apply REF to MATH. |
math/0009026 | The proof is by induction on MATH. CASE: MATH. By REF , MATH and MATH are adjacent regions. Let MATH be the common facet of the closures of MATH and MATH and let MATH be the affine span of MATH. Since functions MATH are continuous, MATH for all MATH and therefore for all MATH. We may assume that MATH (the other case is treated similarly). Then MATH, and MATH satisfies conditions of the lemma. CASE: MATH. By REF , there is a region MATH adjacent to MATH such that MATH. By the induction hypothesis, there is MATH such that MATH . If MATH, then MATH satisfies conditions of the lemma. Otherwise, we have MATH. By REF , the unique hyperplane MATH that separates MATH and MATH also separates MATH and MATH. The same argument as in REF shows that MATH for all MATH. Since MATH and MATH, we have MATH for all MATH. Consider function MATH. It is zero on the hyperplane MATH and positive on the full - dimensional region MATH. Thus it is positive on the open halfspace containing MATH. Hence, it must be negative on the open halfspace containing MATH and MATH. We conclude that MATH. Since MATH, MATH satisfies conditions of the lemma. |
math/0009026 | Let MATH be a central point in MATH. For MATH, let MATH be the unique intersection point of the ray from MATH through MATH with MATH. We define MATH . Clearly, MATH is a piecewise linear function on MATH and MATH. Thus MATH admits representation REF . |
math/0009027 | Choose a regular value MATH such that the symplectic reduction at MATH is defined. Define induced horizontal and vertical spaces at MATH by the intersections with MATH: MATH . By definition, MATH. As MATH is Abelian, MATH and MATH, so that MATH. Using the following set-theoretical identity MATH if MATH, we obtain MATH . Hence, MATH. The corresponding connection one-form MATH is defined by the horizontal space via MATH. The collection of these MATH then define a reconstruction connection MATH as defined in Section MATH. It is MATH-invariant because MATH is MATH-invariant for Abelian groups. Finally, for the connections on the symplectic stratification of MATH determined by connections on NAME and symplectic bundles, that is, by MATH and MATH, respectively, we have at MATH: MATH and MATH where MATH with MATH and we have used that MATH. Comparing the last two expressions we conclude that MATH. |
math/0009027 | The proof readily follows from the fact that the momentum map factors through the quotient map, so that MATH, and the definition of the map MATH for any MATH given by REF , where MATH is any vector in MATH that satisfies MATH, with MATH and MATH: MATH . |
math/0009027 | For any tangent vector MATH and any NAME algebra element MATH : MATH where MATH for some MATH is a vertical (fiber) component, MATH is a horizontal component, and MATH by definition. By definition of the momentum map MATH, so that MATH where MATH is thought of as a MATH-valued one-form on MATH and we have used the fact that the pairing between MATH and its dual is independent of MATH. Thus, MATH and the result follows from the non-degeneracy of the pairing. To verify that MATH indeed defines a connection we check that it satisfies MATH and is MATH-equivariant. Consider the pairing of MATH with an arbitrary element from the NAME algebra MATH and use above definitions of the connection and the abstract locked inertia tensor: MATH . From the non-degeneracy of the pairing it follows that MATH. The MATH-equivariance means MATH and follows from equivariance of the momentum map and equivariance of the abstract locked inertia tensor in the sense of a map MATH (see REF ). |
math/0009027 | Using REF , MATH and omitting MATH for simplicity we have MATH where for the last equality we used the definition of a symplectic form and considered the one-form MATH as a tangent map MATH acting on vectors in MATH. |
math/0009027 | The proof follows readily from REF of the previous corollary and the MATH-orthogonality of MATH and MATH: MATH where we used that MATH for a subspace MATH. |
math/0009027 | Using REF we obtain MATH: MATH . |
math/0009027 | Consider any vectors MATH and MATH. By definition of the induced metric MATH where MATH are horizontal components of the pre-images: MATH, and MATH. From MATH it follows that MATH. But MATH, so that by REF MATH . For the vector MATH it holds MATH and, hence, by the commutativity of the diagram in REF , MATH for any of its pre-images. In particular, for the horizontal pre-image MATH we have MATH . From the expressions for MATH and MATH it follows that MATH. Finally, REF follows from MATH and REF . |
math/0009027 | By the definition of the momentum map, MATH is a Hamiltonian for the vector field MATH of the infinitesimal transformations, that is, for any vector MATH . The one-form MATH can be thought of as the tangent map MATH acting on vectors in MATH and paired with MATH. Take MATH to be MATH for some infinitesimal generator MATH corresponding to MATH. Then, MATH where we used the definition of the map MATH given by REF . On the other hand, MATH . Alternatively, this expression can be obtained from REF using an explicit form of the connection one-form given by REF . From the last two expressions and the non-degeneracy of the pairing we conclude that MATH, then from REF it follows that MATH . Notice that MATH is an isomorphism as both MATH and MATH are. |
math/0009027 | The proof is quite straightforward and relies on the constructions discussed in this section. Using the definition of the transverse derivative, REF and MATH-invariance of the abstract locked inertia tensor and the almost complex structure we obtain MATH where the last equality follows from REF . As it was pointed out earlier, both MATH and MATH are MATH-invariant and, hence, can be dropped to the quotient MATH, so that REF can be computed at any MATH. |
math/0009027 | We start with the definition of the two-form MATH above and shall demonstrate that the following three special cases hold for any MATH : CASE: MATH, here MATH is the characteristic distribution and MATH, CASE: MATH, CASE: MATH for any MATH and MATH, which all together prove the statement of the theorem, according to the definition of the assembled form REF . CASE: From the definition of the reduced symplectic form in the NAME reduction it follows that MATH where MATH and MATH, that is, the pre-images lie in the horizontal space of the reconstruction connection MATH. Recall that in our case, MATH denotes the metric orthogonal to the group orbit within the kernel of MATH: MATH . From REF and the fact that MATH-orbits of any point MATH are isotropic for an Abelian group it follows that MATH . Hence, using the definition on the two-form MATH above, for any vectors MATH lying in the characteristic distribution at MATH, their pre-images MATH satisfy MATH so that MATH . CASE: Let MATH, then MATH, by the definition of the reconstruction connection. But MATH, so that MATH . Using the modularity property and the fact that MATH is isotropic, that is, MATH, and, hence, MATH we obtain that MATH but this space is also isotropic, that is, it is contained in its symplectic orthogonal because of REF : MATH . Thus, MATH. CASE: Finally, combining previous arguments, for any MATH and MATH, MATH and MATH. But, MATH so that MATH and MATH. |
math/0009034 | REF is trivial. If MATH is MATH-absorbing, there exists MATH such that MATH. Then, of course MATH. Thus MATH . Since MATH is convex we obtain MATH . But this is true for every MATH. Hence we can take limits: MATH which gives MATH and by REF , MATH is norming. That REF is trivial. To see that REF does not imply REF does not imply REF take MATH, the unit vectors in MATH. We now prove that REF implies REF for convex sets. Let MATH. Since MATH is MATH-almost absorbing and convex MATH for every natural number MATH. Since MATH, MATH is eventually less than MATH. To see that REF does not imply REF take a dense subset of the unit ball in a separable NAME space. That REF for closed convex and symmetric sets follows from REF . To end the proof, by REF , it is enough to construct a closed, convex, symmetric set MATH such that MATH is fundamental but not norming. For this, let MATH . It is easy to check that MATH has the desired properties. |
math/0009034 | If MATH is norming for MATH, MATH for some MATH. Thus MATH and MATH is invertible. If MATH and MATH are isomorphic by MATH then, for some MATH, MATH . But, by the construction of MATH, MATH for all MATH. We use that MATH inductively for a constant MATH: MATH . Continuing this way gives after MATH steps MATH . This is true for any MATH. Now choose MATH so big that MATH. Then, by letting MATH, we obtain MATH . This proves that MATH is norming. |
math/0009034 | REF clearly implies REF . Suppose MATH is thin. Since MATH is thin we can pick a countable, increasing covering, MATH of MATH, consisting of sets which are non-norming for MATH, and a sequence MATH such that MATH but MATH. Let MATH be an arbitrary element of MATH. Then there is a natural number MATH such that MATH. Thus, since MATH is increasing, MATH while MATH . Thus REF implies REF . To show that REF implies REF , suppose MATH is pointwise bounded on MATH, that is, MATH . Put MATH . Then MATH is an increasing family of sets which covers MATH. Since MATH is thick, some MATH is norming. Then, using REF , there exists a MATH such that MATH . But then, for arbitrary MATH, MATH . Thus MATH and the theorem is proved. |
math/0009034 | Suppose MATH is covered by an increasing family MATH. Since MATH is of second category some MATH, contains a ball. Then MATH contains a ball centered at the origin, and hence MATH is norming. Since MATH was arbitrary, MATH must be thick. |
math/0009034 | REF implies of course REF . The proof that REF implies REF is completely analog to the corresponding part of the proof of REF . The proof that REF implies REF is also very similar to the corresponding part of the proof of REF . Just put MATH . |
math/0009034 | Of course REF implies REF implies REF . To show that REF implies REF suppose REF is not true, that is, MATH is thin. We will construct a NAME injection which is onto MATH but not onto all of MATH. Let MATH be an increasing family of subsets of MATH such that MATH. Since MATH, we may assume each MATH to be contained in MATH. Put MATH and MATH. Define MATH . Then MATH is closed, bounded, convex and symmetric. We now show that MATH is non-norming for MATH. To do this, let MATH and take MATH such that MATH. Since MATH is not a norming set, there is a functional MATH such that MATH. But then, by the definition of MATH, MATH . Hence, by REF , the NAME construction will produce a NAME space MATH and an operator MATH with the desired properties, that is, it is injective, NAME, onto MATH but not onto all of MATH. It remains to show that REF implies REF . To do this, let MATH be any bounded, linear operator into MATH and onto MATH. Put MATH, where MATH is the domain space of MATH. Since MATH is onto MATH, MATH, an increasing covering of MATH. Since MATH is thick some MATH is norming for MATH. By REF , there exists a MATH such that MATH . Hence MATH and, by for example, CITE, MATH is onto. |
math/0009034 | Only the implication MATH needs proof since the remaining implications are trivial. Suppose MATH is thin. Write MATH an increasing union of sets which are non-norming for MATH. Then MATH, an increasing union of sets. Since MATH is thick, there exists a number MATH and a MATH such that MATH. Thus MATH . Thus MATH . This shows that MATH is non-norming. |
math/0009034 | That REF implies REF is trivial. To show that REF implies REF we make necessary adjustments in the corresponding proof of REF . First substitute MATH's with MATH's. Then define MATH by MATH-closure. Note that MATH is now a non-MATH-norming set. Define MATH to be the MATH-closure of MATH. Two cases must be considered: CASE: MATH CASE: MATH . In the first case let MATH be the embedding of MATH into MATH. Then, since MATH is MATH-closed, MATH, where MATH is the annihilator of MATH in MATH. Moreover, MATH is the adjoint of the quotient map MATH. In the second case MATH separates points on MATH and the set MATH can be used to define a new norm MATH on MATH by the formula MATH . Then, by the definition of MATH, MATH is strictly weaker than the original norm. Let MATH be the completion of MATH in this weaker norm, let MATH be the embedding of MATH into MATH. The adjoint MATH of MATH is then injective (since MATH is dense). Moreover, MATH is by definition onto MATH and hence onto MATH. Thus REF implies REF . To show that REF implies REF , mimicking the corresponding proof of REF gives the existence of a natural number MATH such that MATH . Now we use that MATH is an adjoint operator. This gives us that the set MATH is MATH-compact, and hence MATH which concludes the proof. |
math/0009034 | We will use REF . Let MATH be a sequence in MATH such that MATH for all MATH and all MATH. Since MATH is MATH- thick we conclude that MATH for all MATH. Thus MATH for all MATH and every MATH. Since MATH is MATH- thick, MATH for all MATH and the result follows since MATH is thick. |
math/0009034 | CASE: Note that the restriction to a subspace MATH of a NAME boundary is a NAME boundary. CASE: Put MATH. By NAME 's generalization of the NAME lemma, there is a sequence MATH on MATH which converges weakly to MATH. By the NAME selection principle MATH can be assumed to be a basic sequence. Let MATH. We look for MATH inside MATH. CASE: Let MATH be the natural embedding of MATH in MATH. Put MATH. Then we can show that MATH is a NAME boundary for MATH. CASE: Show that each MATH is relatively norm-compact as done on page REF. Thus MATH has a MATH-compact NAME boundary MATH. CASE: Use REF to renorm MATH equivalently to have a countable NAME boundary MATH. CASE: Follow the proof of CITE to construct a copy of MATH inside a once more equivalently renormed version of MATH. This copy is also a copy in MATH. |
math/0009034 | Since MATH and MATH does not contain a copy of MATH the sets MATH and MATH are both MATH-thick. Hence, by REF , MATH is MATH-thick. |
math/0009034 | Since MATH is a separable dual it has the RNP and thus doesn't contain a copy of MATH. By the main result from CITE the sets MATH and MATH are both MATH-thick. Hence, by REF MATH is MATH-thick. But by CITE MATH is exactly the set of exposed points of MATH. |
math/0009034 | Use that MATH. |
math/0009048 | If a simply degenerate honeycomb has a loop, we can breathe it in and out; one direction will increase the value of any (generic) superharmonic functional. A largest lift is by assumption already at the maximum value of the functional, so there can be no loops. |
math/0009051 | It is straightforward to see that the intersection of any number of the MATH is smooth. This implies that MATH is smooth, and it is clearly MATH-equivariant. It will now suffice to prove that MATH. We claim that every MATH for MATH is a proper preimage of some MATH for MATH that is not contained in MATH and vice versa. Let MATH be a proper subvariety of MATH. Denote the image of MATH under the blowdown map by MATH, and consider MATH. If MATH, there is nothing to prove. If MATH, notice that MATH is a subvariety of MATH, so the tangent space MATH is a MATH-submodule of MATH . If the connected component of MATH passing through MATH equals MATH, then the fiber of the blowup over MATH, which equals MATH, has no MATH-fixed points. Therefore the tangent space to MATH is a proper submodule of MATH, which we assume to equal MATH. The point MATH then corresponds to a one-dimensional submodule of MATH, which we will identify with MATH. The tangent space MATH is isomorphic as a MATH-module to MATH . Therefore, the tangent space to MATH at MATH is MATH which is the tangent space of the proper preimage of MATH. The statement in the opposite direction is proved similarly. To show that MATH, suppose that MATH is contained in MATH different proper preimages of the MATH. Then MATH corresponds to a normal direction to MATH which is contained in all these MATH's, which leads to a contradiction. |
math/0009051 | First, it is easy to see that the MATH stratification is conical. Moreover, this can be said about any stratification induced by some connected components of fixed point sets of some subgroups of MATH. Indeed, for any MATH there exists a MATH-equivariant isomorphism of a neighborhood MATH and a neighborhood of the origin in MATH. Under this isomorphism all strata map to linear subspaces, so the stratification is conical. Second, the maximal wonderful blowup is defined by successively blowing up proper preimages of all strata, starting with the strata of smallest dimension. This is equivalent to blowing up strata in any order that is compatible with the partial ordering on the strata, and NAME 's algorithm clearly provides that. |
math/0009051 | The statement is clearly local in MATH, so it could be assumed that MATH is a neighborhood of the origin in a MATH-vector space MATH. Denote by MATH and MATH the sets of strata in the MATH and stabilizer stratifications respectively. According to CITE the maximal blowups MATH and MATH can be described as closures of the images of an open subset of MATH under the map to MATH . Here MATH is the tangent space to the stratum, and MATH is either MATH or MATH. The functoriality is then obvious. |
math/0009051 | Combine REF . |
math/0009051 | Suppose MATH has points with non-solvable stabilizers. Consider the point MATH with a non-solvable stabilizer MATH that lies in the minimum number of exceptional divisors of MATH. If this minimum number is zero, then MATH has a point with non-solvable stabilizer that lies outside the image of the exceptional divisors. Otherwise, let MATH be an exceptional divisor of MATH that passes through MATH. The tangent space MATH splits as a MATH-module into MATH and a one-dimensional module. This one-dimensional module corresponds to a character MATH. The kernel of this character is again non-solvable; however its fixed point set contains points in the neighborhood of MATH not lying in MATH. This contradicts the minimality of MATH and proves the ``if" part. The ``only if" part is obvious. |
math/0009051 | In view of REF , one needs to show that every point MATH with non-solvable stabilizer is contained in an irreducible stratum of codimension bigger than one. Since all MATH have codimension at least two, it remains to observe that each point MATH with a non-solvable stabilizers is contained in MATH. |
math/0009055 | The ``only if" direction is clear. For the ``if" direction choose disjoint neighborhoods MATH, each with a Riemannian metric coming from REF for every critical point of MATH. Now choose finitely many contractible open sets MATH with MATH-critical points-MATH that together with the MATH's cover MATH. Using REF., it is easy to find a Riemannian metric on each MATH that turns MATH into a gradient of MATH. Now the required Riemannian metric is obtained by using a partition of unity. |
math/0009055 | It is to be shown that MATH vanishes on boundaries. So let MATH satisfy MATH, then MATH since MATH and MATH is a homomorphism. |
math/0009055 | CASE: We need to show that MATH is a well defined matrix over MATH. Note that MATH. We will look at terms of the form MATH and get an estimate for MATH. The idea is to write MATH as a word MATH where the length of the word MATH is smaller than MATH and the words MATH are of the form MATH. So assume that MATH where MATH. Let MATH be the first index whose value appears more than once. Since these numbers are between MATH and MATH we have MATH. Let MATH be the largest number such that MATH, then MATH where MATH, MATH, MATH, MATH and MATH. Notice that MATH. Now look at MATH; among the indices MATH,MATH are at most MATH numbers; the numbers MATH,MATH, which are all different, do not appear. If MATH, one of these numbers will appear more than once. Let MATH be the first such index and MATH the last index equal to MATH. Again we get MATH, hence MATH. As above we can write MATH . We continue this process until we get MATH with MATH. Since MATH consists of MATH letters we get MATH. Let MATH be a number such that MATH for all MATH. Then MATH. Since MATH as MATH, MATH is a well defined element of MATH. CASE: The argument is the same as in CITE. Using elementary row reductions we obtain a matrix of the form MATH where MATH is a MATH matrix which is again MATH-regular with the same MATH. Induction gives the result. |
math/0009055 | As before denote the image of MATH in MATH by MATH. Then MATH . It is sufficient to show that MATH . Both sides are clearly MATH for MATH with MATH. Call the left side of REF MATH and let MATH with MATH. Then MATH . Let MATH act on MATH by the cycle MATH and on MATH by rotation. For MATH denote by MATH the orbit of MATH and by MATH the orbit set. We get MATH . Now MATH since the orbit is obtained by shifting MATH, so MATH . |
math/0009055 | Let MATH be a triangulation of MATH which has MATH as a subcomplex MATH. This induces a triangulation MATH of MATH which has two copies of MATH as subcomplexes. Denote the one corresponding to MATH by MATH and the one corresponding to MATH by MATH. Assume MATH has the following properties: CASE: MATH is adjusted to MATH. CASE: MATH is adjusted to MATH. CASE: There is a MATH such that if MATH is a MATH-simplex in MATH, then MATH. CASE: There is a MATH such that if MATH and the trajectory of MATH starting at MATH ends in MATH, then MATH. The existence of such a triangulation is shown in REF gives a simple chain homotopy equivalence MATH . Let us define a chain map MATH. If MATH is a simplex in MATH, look at the liftings MATH and MATH used for the basis of MATH. There is exactly one MATH such that MATH in MATH. Set MATH. Look at the diagram MATH . Because of REF . above, this diagram commutes. Therefore the map MATH is a simple homotopy equivalence between the mapping cones MATH and MATH, where MATH represents inclusion. But by REF and CITE, MATH is chain homotopy equivalent to MATH, in fact simple homotopy equivalent by CITE (the corresponding matrix term is just MATH). But MATH is easily seen to be MATH. After tensoring with the NAME ring we have the following sequence of chain homotopy equivalences MATH and all except the last one are simple. The first map is described in the proof of CITE. Because of the special form of the vector field MATH the NAME complex MATH can be identified with MATH, see CITE and CITE. We claim that this composition is exactly MATH. Denote the composition by MATH. We denote MATH. Let MATH, lift it to MATH (if MATH is a cell in MATH, lift it to MATH). Then MATH where MATH and MATH are defined to give MATH. Now MATH. MATH represents the part of MATH that flows into critical points of index MATH in MATH under MATH while MATH represents the part that flows into MATH. Now MATH . With MATH we thus get MATH . But MATH represents the part of MATH that flows into critical points of index MATH in MATH under MATH. Therefore MATH and MATH is a chain homotopy equivalence whose torsion is given by REF . |
math/0009055 | We have chosen a basis of MATH by choosing MATH-cells in MATH, call these cells MATH. If MATH supp MATH, then there exist negative flowlines from MATH to MATH by the construction of MATH. We need to show that there exists a MATH with the property that given MATH and indices MATH and MATH supp MATH, MATH supp MATH supp MATH we have MATH. Now we have to recall the proof of the Main Theorem in CITE. Every cell MATH defines a thickened sphere in MATH that we denote by MATH, also let MATH. The spaces MATH and MATH are defined in CITE. Since MATH supp MATH we have that MATH induces a homologically nontrivial map from MATH to MATH. Similarly every MATH supp MATH gives rise to a homologically nontrivial map from MATH to MATH. The composition of all these maps plus MATH is homologically nontrivial and hence has a fixed point other than the base point, compare the proof of REF. Notice that the existence of this fixed point does not require the condition that closed orbits of MATH are nondegenerate. This fixed point corresponds to a flow line MATH of MATH with MATH and MATH which passes through the cells MATH. We need the following There exists a MATH such that for every flowline MATH of MATH that starts in MATH and ends in MATH we have MATH. We have MATH and MATH with MATH as in REF Since MATH is rational and MATH has no critical points in MATH there is a MATH such that MATH also contains no critical points. So if MATH is a flowline of MATH with MATH, there is a MATH which depends smoothly on MATH such that MATH. Now MATH by REF . Since MATH acts cocompactly on MATH and the value of the integral depends smoothly on MATH there is MATH such that MATH . This MATH now works for the Lemma, since integrating over a longer flowline will just make the integral smaller. Conclusion of the proof of REF: our flowline MATH is the concatenation of flowlines MATH to which REF applies. Let MATH satisfy MATH. Then we get MATH . Therefore MATH for all MATH supp MATH supp MATH which implies that MATH is MATH-regular. |
math/0009055 | Since MATH we can form the MATH complex MATH from the proof of REF which is simple homotopy equivalent to MATH. The matrices MATH are MATH-regular by REF , so the projection of MATH coker MATH is a chain homotopy equivalence. We have already seen that the boundary homomorphisms of MATH and MATH are the same when viewed as matrices over MATH. The same holds for coker MATH and coker MATH. But since we identified coker MATH with MATH we now get that coker MATH is the same complex as MATH. Also, if we use the triangulation from the proof of REF we get that MATH factors as MATH, where MATH is simple. Therefore MATH . By REF we have MATH . By the proof of the Main Theorem in REF the right hand side is exactly MATH. Of course, CITE only shows this in the rational case, but MATH is independent of the homomorphism MATH when viewed as an element of MATH. |
math/0009055 | There exists a filtration MATH of MATH such that MATH is a compact cobordism containing all critical points of index MATH and such that MATH and the boundary homomorphism comes from the long exact sequence of the triple MATH, see CITE. Also, MATH, where MATH denotes the MATH-skeleton of the triangulation. A simplex MATH is represented by a map MATH. Let MATH be induced by the flow of MATH, where a flowline is supposed to stop once it hits the boundary. For MATH let MATH. Since MATH is adjusted to MATH there is a MATH such that MATH maps MATH to MATH and MATH to MATH. It follows from intersection theory that MATH . Furthermore this does not depend on MATH as long as MATH is large enough. A diagram chase gives that MATH is a chain map. |
math/0009055 | Let MATH be a homotopy between MATH and MATH. As above we can change MATH to a homotopy MATH between MATH and MATH such that MATH intersects MATH transversely for all critical points MATH with MATH, where MATH is a MATH-simplex. Describe the NAME complex as in the proof of REF . Then we define MATH by MATH. Here MATH is so large that MATH, MATH and MATH is a lifting of MATH. Use MATH and MATH to identify the triangulated chain complexes. Then MATH is the desired chain homotopy. |
math/0009055 | Let MATH be a subdivision of MATH. If MATH is adjusted to MATH, so is MATH. Moreover, the diagram MATH commutes, where sd is subdivision, a simple homotopy equivalence. By CITE it is good enough to show the theorem for a special smooth triangulation. As in the proof of REF we have the filtration MATH where MATH is a compact cobordism containing the critical points of index MATH. Choose a triangulation such that each MATH is a subcomplex for all MATH and so that for each critical point MATH of index MATH the disc MATH is a subcomplex. We set for MATH. The complex MATH is given by MATH . The chain map MATH induces maps MATH and MATH. Since the diagram MATH commutes, it suffices to show that each MATH is a simple homotopy equivalence to finish the proof. Clearly MATH induces an isomorphism in homology, so it remains to show that it is simple. We set MATH. Then the inclusion MATH is the inclusion of the core of the handles into the handles, hence a simple homotopy equivalence. Now MATH is a simple homotopy equivalence by CITE, since we can choose the lifts of MATH so that the matrices representing MATH and the boundary operators have only integer values. Therefore MATH is a simple homotopy equivalence. |
math/0009055 | For MATH choose triangulations MATH of MATH adjusted to MATH and let MATH be the flow of MATH. Then there is a MATH such that REF. is satisfied for MATH. Extend MATH to a triangulation MATH of MATH. Choose a diffeomorphism MATH so close to the identity that MATH is adjusted to MATH and REF. and REF. still hold. Modify this triangulation again so that if MATH is a MATH-simplex and MATH flows to MATH, then MATH for critical points MATH of MATH with MATH. Notice that MATH is at most MATH dimensional. By making only very small changes we now have a triangulation satisfying REF.-REF. and by compactness the condition of REF for some MATH, but not necessarily for MATH. Rename this triangulation MATH and the two subcomplexes of the boundary MATH and MATH. Notice that REF . already holds for MATH and MATH. This is trivial for MATH and for MATH it follows from REF. By continuity it also holds in a small collar neighborhood of the boundary. We can think of this collar as MATH for some MATH and assume that there are no REF-simplices in MATH. Let MATH be a smooth function which is REF outside of the collar and REF in a smaller collar. Let MATH be the flow of the vector field MATH. There is a MATH such that if MATH is a MATH-simplex in MATH which does not meet MATH, then MATH. Furthermore MATH satisfies REF.-REF. and the same form of REF as MATH does. Let MATH for MATH and MATH be diffeomorphic to MATH with MATH corresponding to MATH and MATH lying on the same trajectory of MATH. Let MATH, then MATH is a compact cobordism. By changing MATH if necessary we can assume that MATH is a NAME function on this cobordism. The NAME function MATH is ordered so we can assume that MATH gives a filtration with MATH for MATH. Let MATH be a smooth function with MATH and MATH and define MATH and let MATH be the flow of MATH. Define MATH as follows. Let MATH. If MATH, let MATH. If not, the trajectory of MATH reaches MATH. Let MATH. Now MATH flows along MATH all the way back to MATH. For if not, we have MATH. By MATH we have MATH. But then MATH because MATH and then MATH flows all the way back to MATH. This and the implicit function theorem give that MATH is smooth on MATH. Similarly, a smooth homotopy MATH between MATH and MATH with MATH and MATH defines a smooth map MATH such that all MATH and MATH define an isotopy MATH with the following property : if MATH with MATH, then there is a MATH such that MATH and MATH for all MATH. By compactness there is a MATH such that MATH satisfies the Lemma. |
math/0009061 | The second statement follows immediately from the first, since MATH is a domain. To prove that MATH, we first show that MATH is a binomial ideal. We can factor the map MATH as follows. Let MATH and MATH be defined like MATH, except that if MATH, then it appears as an exponent of the variable MATH instead of MATH. Then MATH is equal to the composition of MATH followed by the projection MATH . It is straightforward to see that MATH. Let MATH . Then it follows immediately from CITE that MATH. We obtain a generating set for MATH by computing a reduced NAME basis for MATH using an elimination ordering with MATH for all MATH, and then intersecting it with MATH. Since MATH is generated by binomials, any reduced NAME basis also consists of binomials. This shows that MATH is a binomial ideal. Now, let MATH be a binomial in MATH. We may assume that the two monomials have no common factors, that is, MATH. Then MATH . Thus, MATH if and only if MATH and MATH . Since MATH, we obtain the following facts from the first condition in REF . If, for some MATH, MATH, then MATH, so that MATH. Now suppose that MATH for some MATH. Then MATH. But then MATH, otherwise MATH, which cannot be since MATH for all MATH. And this, in turn, implies that MATH. Hence MATH . It follows from the last two equations in REF that MATH. This completes the proof of the theorem. |
math/0009061 | The proof is similar to that of CITE. It was shown in the proof of the previous theorem that any binomial in MATH is of the form MATH for some MATH. Hence the NAME basis MATH of MATH is of the form MATH . We first show that the set MATH is a generating set for MATH. Suppose not, so there exists MATH which is not a MATH-linear combination of elements in MATH. We can choose MATH so that MATH is minimal with respect to the chosen term ordering of MATH. Since MATH, we have MATH, so that MATH. Hence the leading term of this binomial is divisible by a binomial in MATH. That is, there exist MATH such that MATH . Thus MATH. But MATH is a prime ideal, and it is immediate from the definition of MATH that it contains no monomials, so MATH. Moreover, MATH. Hence, by the choice of MATH, MATH is a linear combination of elements of MATH, which implies that MATH is also a linear combination of elements of MATH. This is a contradiction to our assumption on MATH. Thus, MATH is a generating set for MATH. To show that MATH is minimal, suppose that some MATH or MATH is a MATH-linear combination of elements in MATH. Since the NAME basis MATH is reduced, the linear combination cannot contain any summands coming from the MATH. But observe that all the vectors MATH are symmetric, so that MATH . A similar argument disposes of the other cases. Thus, we have shown that MATH is the NAME basis of MATH. CASE: MATH. |
math/0009061 | According to CITE the dimension of MATH is equal to the number of linearly independent column vectors in the matrix MATH . The theorem now follows. |
math/0009061 | Obviously, every solution of REF is also a solution of REF . Conversely, let MATH be a solution of REF . Then MATH and MATH for MATH . Taking into account that MATH is a linear operator, we conclude that MATH on the right-hand side of REF is non-negative. |
math/0009061 | It is enough to show that the above REF form a NAME basis of the ideal MATH. To compute MATH, one can use the Symmetry Component Algorithm, with any computer algebra system. Here we are using the specialized system CITE which performs NAME basis calculations substantially faster than most general purpose symbolic calculation packages. REF shows the NAME session used to compute MATH for system REF . To simplify notation, we renamed the variables MATH and MATH as follows: MATH . |
math/0009061 | Similar to the proof of REF . To simplify notation we rename the variables MATH and MATH: MATH . REF is the NAME session used to find MATH. |
math/0009061 | Similar to the proof of REF . |
math/0009061 | It is sufficient to show that the ideal MATH is prime in MATH. Consider the ring homomorphism MATH defined by MATH. It is sufficient to prove that MATH . It is clear that MATH. We will show the other inclusion by induction. Let us suppose that MATH and MATH is linear in MATH, that is, MATH . Then MATH where MATH. Therefore MATH . Hence, MATH. Assume now that for all polynomials of degree MATH in MATH and MATH, we have MATH. Let MATH be of degree MATH in MATH. We can write MATH in the form MATH (here every term of MATH contains MATH). Consider the polynomial MATH . Then MATH where MATH . Since MATH we get that MATH. Then, by the induction hypothesis, MATH and, hence, MATH. |
math/0009061 | It is easy to check (using, for example, the radical membership test, see for example, CITE) that MATH . Thus we only have to show that MATH for all MATH and MATH . Indeed, the systems corresponding to the points of MATH are Hamiltonian. For the variety MATH in the case MATH one can easily find an invariant conic MATH and invariant cubic MATH and check that the system has the first integral MATH defined on MATH, where MATH . Computing the quotient of the ideals (for example, by means of the algorithm from CITE) we get MATH. Therefore, MATH . This implies that the system REF has a center on the whole component MATH . Finally, according to REF , MATH is the symmetry component of the center variety. The irreducibility of the components MATH and MATH is obvious, and MATH is irreducible because due to REF the ideal MATH is prime. |
math/0009061 | Computing the intersection of the ideals MATH we find MATH . Hence MATH is radical because due to REF MATH are prime, and according to REF MATH is prime as well. Similarly, for quadratic system we easily check that, for MATH, the ideals MATH are prime and MATH . This yields that the ideal of focus quantities of quadratic system is a radical ideal. |
math/0009061 | CASE: MATH REF and the ideal MATH REF is a radical ideal. Therefore MATH . |
math/0009066 | If MATH has all MATH non-negative and some MATH, then MATH meets the conditions for the Descent Axiom to apply, so we have MATH . But by REF , we have MATH . |
math/0009066 | This follows from repeated application of the Descent Axiom and REF . |
math/0009066 | The convexity axiom of a MATH-spin virtual class (compare CITE) gives that MATH where MATH and MATH is the MATH-th root sheaf of the universal MATH-spin structure. The MATH-th root sheaf MATH of the universal MATH-spin structure of type MATH is isomorphic to MATH under the identification MATH . Since MATH is a subsheaf of MATH, it is also convex and MATH . Therefore, MATH where, in the second equality, we have used the following simple lemma. |
math/0009066 | Consider the exact sequence of sheaves on MATH . Since MATH and MATH, the corresponding long exact sequence for the functor MATH gives a short exact sequence of bundles on MATH . Now the statement of the lemma follows from multiplicativity of the total NAME class. |
math/0009079 | Let MATH. For a condition MATH, let MATH be the set of all accumulation points of MATH. Clearly, MATH . Let MATH be a generic filter. Then MATH is generic in MATH. If MATH then there is some MATH so that MATH. So MATH and MATH. Thus MATH is closed under finite intersections. Clearly, MATH is upwards closed. Thus MATH is a filter. Suppose that MATH is dense and downwards closed. Let MATH be arbitrary, and consider MATH. Let MATH be chosen in MATH. For MATH define MATH, and let MATH. MATH so there is some MATH with MATH. By this Lemma it follows that MATH . We aim now to show that MATH . First, we shall see that MATH. Let MATH be arbitrary. Let MATH and let MATH be the increasing enumeration of MATH. A condition MATH is normal if it satisfies MATH . The set of normal conditions is dense in MATH. Given a condition MATH, let MATH be the least so that MATH and choose MATH of order type MATH. Let MATH. Thus MATH, and is of the form REF above. Let MATH be a fixed elementary submodel of MATH for a sufficiently large regular cardinal MATH so that MATH, MATH and the cardinality of MATH is MATH. Let MATH. Clearly, MATH. MATH is dense in MATH. Let MATH be a condition in MATH and assume, without loss of generality, that it is normal. Let MATH. For every MATH, there exists a closed subset of MATH of order type MATH which belongs to MATH. Let MATH. In MATH, fix an increasing and continuous function MATH with MATH. Let MATH. Thus MATH is a club in MATH. The club MATH itself may not belong to MATH (because MATH may not belong to MATH). But since MATH and both (regular) cardinals belong to MATH, MATH contains some club guessing sequence MATH by the club guessing REF above. Thus there is some MATH so that MATH. Clearly, MATH. Since MATH, also MATH belongs to MATH, and is a closed subset of MATH of order type MATH. Using the claim, choose, for every MATH, a closed set MATH so that MATH, MATH and MATH. Since MATH is closed under countable sequences, MATH, and clearly MATH is a normal condition in MATH. This has established that MATH. We need the following simple fact about MATH and MATH: MATH is MATH-complete and MATH is MATH-complete. Suppose that MATH is a decreasing sequence of conditions in MATH. By induction on MATH, let MATH be chosen so that MATH, and choose a closed subset MATH of MATH with MATH. The condition MATH belongs to MATH and MATH for all MATH. If each MATH belongs to MATH then the sequence itself belongs to MATH because MATH is closed under taking MATH-sequences, and hence some MATH which satisfies MATH for all MATH belongs to MATH, by elementarity. (Alternatively, one can do the induction for proving completeness of MATH inside MATH). Thus, MATH contains a MATH-complete dense set of size MATH. To prove REF from REF it remains to show that MATH is MATH-nowhere distributive. For this purpose we inspect the generic cut which MATH creates in MATH, where MATH is the generic ultrafilter over MATH which forcing with MATH introduces. Suppose MATH is a generic filter. Then MATH is an ultrafilter over MATH. If MATH are normal conditions, then MATH is finite. Thus MATH is an order preserving map onto MATH. Furthermore, if MATH, then there is some MATH such that MATH. Therefore the image of a generic MATH under MATH is an ultrafilter over MATH. Given a normal condition MATH, define the following two functions on MATH by letting, for each MATH, MATH . The set of REF for which MATH is clearly dense in MATH, so we always assume that MATH. Since MATH for every MATH, and MATH", it follows that MATH". Let MATH be a generic filter. Define MATH . Now for each normal MATH, MATH". Let MATH . Suppose that MATH and MATH is a normal condition. Then there exists a normal condition MATH so that MATH . If for some infinite set MATH, MATH then MATH is a normal condition and for all MATH it holds that MATH. Since MATH, the second alternative holds for MATH. If MATH is finite, let MATH be fixed so that for every MATH it holds that MATH and let, for MATH, MATH be the initial segment of MATH whose order type is MATH. Now MATH is a normal condition and MATH. Since MATH is MATH-complete, no new members are added to MATH after forcing with MATH. Therefore, by REF , it holds that MATH is a lower half of a NAME cut in MATH whose upper half is MATH; that MATH has no last element and that MATH has no first element. By the definition of MATH, it is clear that MATH is cofinal in MATH. Furthermore, if MATH is a set of functions, MATH and for all MATH it holds that MATH then by iterated use of REF and MATH-completeness there exists MATH so that MATH. As a consequence, the cofinality of MATH is uncountable. We shall need the following strengthening of REF , which says that the generic cut MATH is not trapped by any product of countable sets. Suppose MATH is a countable set for each MATH, and MATH is a normal condition. Then there is a condition MATH in MATH so that for every MATH it holds that MATH. Let MATH be the order type of MATH and let MATH be the increasing enumeration of MATH. Partition MATH to the intervals MATH, MATH and MATH. For every MATH, choose an interval MATH in the partition of MATH so that MATH, and let MATH be closed of order type MATH. Let MATH. The cofinality of MATH is MATH. We have seen that MATH. Suppose now, to the contrary, that MATH is regular and that MATH is MATH increasing and cofinal in MATH". The Trichotomy Theorem applies to MATH, but: The third condition ( ``NAME) cannot hold , since MATH is an ultrafilter. The first condition (``NAME") cannot hold, because in MATH there is no first element. Let us see now that the second condition (``NAME) cannot hold either. Suppose that MATH, witnesses `NAME for MATH". Then, MATH . By MATH-completeness, we may assume that MATH and each MATH belong to the ground model. By REF there is a condition MATH so that for all MATH, MATH . Since MATH forces that MATH is cofinal in MATH and MATH, there is some MATH and MATH so that MATH . By strengthening MATH we may assume that for some MATH and MATH, MATH . So there is some MATH (in fact, infinitely many) so that MATH . This is a contradiction to REF , since MATH. Thus, the cofinality of MATH is at least MATH and no more than MATH; so it is exactly MATH. Since MATH and MATH is cofinal in MATH, using MATH-completeness of MATH it is easy to find a sequence of REF such that MATH and MATH is (MATH-increasing and) cofinal in MATH. Fix a MATH-name MATH for such a sequence. Observe that if MATH are incompatible, then MATH, since any two conditions in MATH are compatible. For every MATH there is a set MATH of pairwise incompatible conditions below MATH, so that for each MATH there is MATH and MATH so that MATH, and MATH are pairwise incompatible. Let MATH be a normal condition, and let MATH. MATH is a closed set of order type MATH. For each MATH let MATH be the initial segment of MATH of order type MATH, the lexicographic ordering of all sequences MATH in the product MATH. Identify each member in MATH with the sequence in MATH it corresponds to via the order isomorphism, and define a projection MATH for MATH by mapping a sequence of length MATH to its initial segment of length MATH. The inverse limit of this system is the set of all functions MATH with the property that for all MATH, MATH. Denote this set of functions by MATH. Choose a set of MATH different functions MATH and for each MATH let MATH. Let MATH. Thus each MATH is a condition below MATH. Furthermore, if MATH then from some point MATH on, either MATH or MATH. Thus MATH is a set of pairwise incompatible conditions below MATH. For each MATH, MATH . Fix MATH so that for some MATH and MATH, MATH. For MATH, since MATH, MATH and MATH are incompatible, MATH is oncompatible with MATH and MATH is incompatible with MATH. Fix, for each MATH, a maximal antichain MATH of conditions that decides MATH. For every condition MATH there exists some MATH so that MATH is compatible with MATH members of MATH. Let MATH be an arbitrary condition. By REF there are MATH pairwise incompatible conditions MATH below MATH, each of which is forced to be MATH for some MATH, by some extension MATH, and MATH are pairwise incompatible extensions of MATH. Since MATH, there is necessarily some fixed MATH so that MATH. Since different MATH in this set force different values for MATH, they cannot be compatible with the same member of MATH. Thus MATH is compatible with MATH members of MATH. The last claim established MATH-nowhere distributivity of MATH. By REF MATH, and since MATH, the proof is complete. |
math/0009079 | Since MATH is a complete subalgebra of MATH, the universe MATH is contained in MATH. Therefore, there is an onto function MATH in MATH. Since MATH is MATH-complete, MATH is preserved in MATH. Thus, the cardinality of MATH in MATH is MATH. |
math/0009079 | By REF there exists a dense subset of MATH which is isomorphic to the dense subset MATH of MATH, namely, there are conditions MATH so that MATH. Let MATH. Define MATH. It should be clear that MATH is a maximal antichain in MATH. Let MATH be arbitrary. By density of MATH there exists some MATH with domain MATH so that MATH. By REF, for each MATH there exists some MATH so that MATH. Thus, for every MATH there exists some MATH so that MATH is compatible in MATH with one member from each MATH. Now clearly MATH is dense open in MATH for each MATH. Also, MATH. Thus MATH is coverable by MATH nowhere-dense sets. Since it is known CITE that MATH cannot be covered by fewer than MATH nowhere-dense sets, its NAME number is equal to MATH. |
math/0009083 | Recall from CITE that the map MATH defines a group morphism MATH . It follows that MATH if and only if MATH, that is, if MATH is a MATH-th root of unity. |
math/0009083 | As a first step, we will prove that MATH. In order to see this, recall from deformation theory that - after shrinking MATH, if necessary - MATH is of the form MATH where MATH is a function MATH. In particular, if MATH is the non-normal locus and MATH its preimage in the normalization, then MATH is a unit disc and MATH either a unit disk or a union of irreducible components which are each isomorphic to MATH and meet in a single point. In this setup, we may use the NAME sequence for reduced cohomology to calculate: MATH . See CITE for more information about the sequence. NAME pointed out that MATH can also be shown by deforming MATH into a bundle of cuspidal plane cubics where the claim is obvious. Now choose a section MATH which is entirely supported on the smooth locus. After shrinking MATH, this will always be possible. Consider the exponential sequence MATH . The element MATH satisfies MATH and is therefore contained in MATH. Let MATH be a preimage and note, that, since MATH is a MATH-vector space, we can find an element MATH such that MATH . We may therefore finish by setting MATH. |
math/0009083 | Let MATH be a line bundle such that MATH; the existence of MATH is guaranteed by REF . Note that MATH is locally free of rank one. Thus, after shrinking MATH if necessary, a section MATH exists whose restriction to any fiber of MATH is not identically zero. But since the relative degree of MATH is one, the restriction of MATH to a fiber is a section which vanishes at exactly one smooth point of the fiber. This point must therefore be MATH-osculating. Thus, the divisor MATH associated with the section MATH contains only smooth MATH-osculating points and maps bijectively onto the base. |
math/0009083 | Let MATH be any point and MATH a small unit disk centered about MATH. The preimage MATH will then be isomorphic to MATH. Let MATH be the MATH-osculating section whose existence is guaranteed by REF and find an isomorphism MATH such that MATH. Apply REF to see that MATH is then given as MATH . Hence the claim. |
math/0009083 | Choose a unit disk MATH centered about MATH and equip MATH with a coordinate MATH. Then MATH is a cuspidal curve. After shrinking MATH we may assume that MATH is the only point in MATH whose preimage is cuspidal. By REF , we can find an index MATH such that MATH. Therefore we can choose a bundle coordinate on MATH so that we can write MATH for an integer MATH. For a point MATH, MATH, the map MATH with MATH parameterizes the smooth part of MATH. Apply REF with MATH and write MATH . The claim follows. |
math/0009083 | It follows immediately from the construction of the elementary transformation that the intersection number between the strict transforms of MATH and MATH drops exactly by one with each transformation in the sequence MATH. In particular, the strict transforms MATH, MATH of MATH and MATH are disjoint. Likewise, it follows from REF that the strict transforms MATH of the MATH are sections which are mutually disjoint and disjoint from both MATH and MATH. It follows that MATH must be the trivial bundle MATH and that MATH, MATH and MATH are fibers of the projection MATH. It remains to find the right coordinate on MATH. To accomplish this, choose a general point MATH. The fiber MATH will then be a nodal curve and there exists a point MATH such that MATH. It follows directly from REF that we find coordinates on MATH such that MATH corresponds to MATH, MATH corresponds to MATH, and the MATH-osculating points correspond to MATH where MATH are roots of unity. Note that MATH is isomorphic in a neighborhood of MATH and use the coordinates on MATH to obtain a global bundle coordinate on MATH. This coordinate will have the desired properties, and the proof is finished. |
math/0009085 | MATH for MATH by the naturality property of the first obstruction. |
math/0009085 | Since MATH is a proper oriented submanifold MATH has a NAME dual MATH in MATH. Therefore MATH. On the other hand MATH (if follows from the analogous statement for non equivariant cohomology see for example, in CITE). |
math/0009085 | Let MATH denote a tubular neighborhood of MATH in MATH. Replacing MATH with its maximal compact subgroup doesn't change the equivariant cohomology groups, so we can assume that MATH is a MATH-equivariant subset of MATH. Now looking at the MATH-equivariant NAME sequence of MATH: MATH we can see that the Lemma is equivalent to the statement that MATH. Considering now the relative exact sequence of the pair MATH and comparing with the NAME sequence of the normal bundle of MATH we get the commutative diagram: MATH . The map MATH is injective since MATH is not a zero divisor. This implies that MATH is the zero map. But MATH factors through MATH so must be zero, too. |
math/0009085 | Let MATH. By REF is equivalent with the statement that MATH is equal to MATH where MATH . So it is enough to show that the inclusion MATH induces an injection in degree MATH. From the relative cohomology exact sequence MATH we can see that it is enough to show that MATH. By excision MATH where MATH and by REF (and by looking at the relative cohomology exact sequence of MATH) we get that MATH for MATH. |
math/0009085 | Easy computation shows that the maximal compact symmetry group of MATH is MATH, with cohomology ring MATH (MATH are the NAME classes of rank MATH and MATH are the NAME classes of rank MATH). In other words this ring is MATH, MATH with the ``substitutions" MATH . Let us call these substitutions MATH. With this notation the map between cohomology groups induced by the inclusion MATH is MATH and the NAME class of MATH is MATH . To prove the theorem let us compute MATH. It is MATH . The MATH's in the last sum corresponding to MATH are all zero, since the MATH-th and the MATH-th rows are identical for the greatest MATH such that MATH. The term corresponding to MATH is MATH if MATH, and has only REF's in the first row if MATH. These prove the principal and the homogeneous equations, so the theorem. |
math/0009085 | The orbits of MATH are described by the NAME normal forms. A generic MATH has MATH different eigenvalues and the stabilizer group MATH of the orbit MATH is isomorphic to MATH. The inclusion MATH induces a map MATH. This map is injective. (This fact is usually called the splitting lemma. For semisimple NAME groups the corresponding map is still injective rationally by REF.) In other words MATH. It implies that for any invariant subset MATH for which MATH the avoiding ideal MATH so MATH. And if MATH contains every generic orbit then MATH and MATH. |
math/0009085 | Since MATH the restriction map MATH factors through MATH so the homogeneous equations contain no extra information. |
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