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cs/9912020
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The proof uses induction. First, the case MATH is just REF . If the equality holds for MATH then the first MATH terms are the same as for MATH and only the last term needs further expansion. In this term each summand is a function of MATH and from REF one obtains: MATH . Replacing the last term in the equation for the case MATH with the right-hand side of this equation leads to the equation for MATH.
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cs/9912020
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The statement that MATH is a projection, that is, MATH, is equivalent to showing that MATH is zero for MATH. This follows directly from the fact that any function MATH for which the function values do not depend on the variable MATH one has MATH. As any MATH consists of a sum of functions which depend only on MATH variables and all the terms in MATH contain MATH there is at least one MATH for which any particular term in the expansion of MATH does not depend on MATH and thus MATH. If MATH is a product distribution as assumed above, then MATH for every projection measure and for any MATH one has MATH . Now as for each MATH-tuple MATH and for each MATH-tuple MATH one has at least one MATH which is different from all the MATH then MATH . This shows that the error term MATH is orthogonal on MATH which implies that MATH is an orthogonal projection into MATH and from this the minimisation characterisation follows.
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cs/9912020
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It is shown the same way as in an earlier theorem that all the terms of the sum defining MATH are orthogonal and so MATH . For simplicity set MATH and application of REF and similar reasoning as for the case MATH gives MATH where MATH . Now one can see that MATH and from this the claimed bound follows.
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cs/9912020
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Because of the equivalence of the measures the error in the original measure is bounded by MATH and REF for the measure MATH then gives MATH.
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cs/9912020
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The lower bound holds by definition. For the upper bound we first use the property that the measures are in the same class to get: MATH . Then we note that MATH is a best approximation with respect to the norm defined by the product distribution by REF . Thus one has for any MATH: MATH and, as the measures are in the same class: MATH and combining these inequalities and taking the minimum over MATH provides the desired estimate.
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math-ph/9912002
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Choose MATH such that MATH and MATH . Note that the latter condition can be achieved for MATH small enough, since MATH. For this choice of MATH, let MATH be the minimal length scale from REF . We can now use REF to find a sequence MATH and a constant MATH such that for every MATH, CASE: MATH, CASE: MATH, CASE: MATH is satisfied. For MATH, denote MATH and MATH . Since MATH and MATH holds, we have MATH . For MATH, denote MATH . Claim. For every MATH, MATH . Proof. We have MATH . Now, for MATH, MATH which gives the assertion. CASE: Denote by MATH the normalized eigenfunctions of MATH, MATH with corresponding eigenvalues MATH. For each MATH, define a center of localization MATH by MATH . Since MATH, such a center always exists. Claim. There is MATH such that for MATH, MATH and MATH, the cube MATH is MATH-bad. Proof. Assume otherwise. Then by (EDI) it follows that MATH . Estimating the number of unit cubes in MATH very roughly by MATH we find that MATH . If MATH is large enough to ensure MATH the inequality above contradicts the choice of MATH. CASE: Let MATH with MATH. Then there exists MATH such that for MATH, MATH and MATH, MATH where MATH is shorthand for MATH. Proof. We divide MATH into angular regions MATH, MATH . We have MATH . By construction of MATH, for every MATH, we find MATH such that MATH and MATH. Since MATH is MATH-bad and MATH, it follows that MATH is MATH-good so that MATH . Since MATH grows only polynomially in MATH, the assertion follows. CASE: There exists MATH such that for MATH, MATH, MATH . Proof. Since MATH is non-decreasing in MATH, and since MATH from REF only depends on MATH, we can restrict ourselves to the case MATH and adapt the constant MATH. We start by observing MATH . We want to show that each of the terms in the sum is at least MATH, thus giving an estimate on the number as asserted. Using REF and suppressing MATH, we have MATH . Plugging this into the above estimate on the trace, we get the claimed bound for the number MATH. CASE: There is MATH such that for MATH, MATH and MATH, MATH . Proof. We have MATH . We now divide the sum according to where the MATH are located: MATH since one of the cubes MATH has to be MATH-good and the number of MATH has been estimated in REF . For MATH large enough, depending only on the indicated parameters, MATH, MATH . We now treat the remaining terms. Note that for MATH and MATH, we find a MATH such that MATH. From REF we know that MATH must be MATH-bad so that MATH has to be MATH-good since MATH. Therefore MATH . Using REF again, we see that MATH if MATH. The latter estimate, together with REF , gives the assertion. CASE: For MATH from REF and MATH, we have MATH . Proof. For MATH, we can estimate the norm by MATH and use REF , while for MATH, we can use REF . Put together, we have MATH . CASE: End of the proof. For compact MATH, we find MATH such that MATH. Then with MATH, we have MATH since MATH and the MATH grow fast enough.
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math-ph/9912002
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It is well known that (INDY), (WEYL), (GRI) and (EDI) are satisfied if we take for MATH the operator MATH restricted to MATH with periodic boundary conditions. Due to CITE we have a NAME estimate of the form MATH where MATH denotes the sidelength of the cube MATH. In particular, MATH is satisfied for a neighborhood MATH of MATH, arbitrarily given MATH and MATH, and MATH large enough. For given MATH, we can start the multi-scale induction with MATH by NAME asymptotics.
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math-ph/9912002
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We have already checked everything except for the initial length scale estimate MATH, and in particular how large MATH can be taken. By an elementary argument, we can take MATH subject to the condition MATH (see CITE), which gives the claimed result.
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math-ph/9912005
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MATH .
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math-ph/9912005
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NAME has shown that MATH is a MATH CITE. If MATH is not empty, then it contains an entire orbit which is dense by minimality.
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math-ph/9912005
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The assertion follows from MATH which is readily verified.
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math-ph/9912007
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The MATH spectral residue vanishes if and only if MATH . If this is the case, then evidently there exists a monic series solution for every possible choice of MATH.
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math-ph/9912007
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Let MATH, MATH and MATH be as above, and let MATH and MATH be the corresponding monic series solutions. From MATH it follows that MATH . From REF it is evident that the residues of MATH at MATH are all zero. The desired conclusion follows immediately.
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math-ph/9912007
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In the non-commutative setting it suffices to take MATH where MATH, MATH are the coefficient series of the operator MATH. In the commutative setting one can use simpler formulas. Indeed, it suffices to introduce an auxiliary function MATH and then set MATH .
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math-ph/9912007
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Taking MATH and MATH, the desired conclusion follows immediately from REF above.
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math-ph/9912007
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We saw above that the eigenfunctions of the NAME Hamiltonian are the product of a power of MATH and of a hypergeometric series. Hence by the gauge-invariance of spectral residues as asserted in REF above, the spectral residues of the NAME Hamiltonian can just as well be calculated by taking residues of the coefficients of the hypergeometric series. These are of course the well-known: MATH where MATH denotes the rising factorial MATH. Calculating the residue of MATH at MATH (recall that MATH), and using the gauge-invariance of MATH we obtain MATH . From REF it follows that MATH . Another calculation will show that if MATH is odd, say MATH, then MATH . This observation explains why the MATH factor in REF can be discarded. The proposition now follows immediately.
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math-ph/9912007
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Let MATH be the unique series solution to MATH . By REF , if MATH, MATH, then MATH is MATH times a MATH degree polynomial of MATH whose coefficients are polynomials of MATH over MATH. Hence for all MATH the residue of MATH at MATH must have MATH as a factor. By the gauge invariance of spectral residues, the same must be true for MATH. The same reasoning also applies to show that MATH is a factor of MATH for all MATH. We have now shown that MATH, as a function of MATH and MATH, is some factor MATH times MATH . Now MATH and hence for all MATH . Using REF, a degree count, and an examination of the coefficient of MATH we see that MATH. The proposition now follows immediately.
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math-ph/9912007
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Upon setting MATH, MATH, the above identity follows from the odd, MATH, case of REF .
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math-ph/9912007
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Upon setting MATH, MATH, the above identity follows from the odd, MATH, case of REF .
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math-ph/9912007
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We proceed in the same way as in the proof of REF . By the gauge-invariance of spectral residues, MATH is equal to the MATH spectral residue of the confluent hypergeometric operator. Taking the residue of MATH at MATH yields MATH . Next, using REF we see that in the first case, where MATH, we have MATH . In the second case, where MATH, MATH we have MATH . The proposition now follows immediately.
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math-ph/9912007
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Note that MATH, or equivalently that MATH. Furthermore, REF imply that MATH . The desired conclusion now follows immediately.
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math-ph/9912025
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CASE: Since the bound REF ensures MATH for MATH-almost all MATH, it follows that for these MATH's the operator MATH is an operator perturbation of MATH with relative operator bound zero. NAME of MATH on MATH is then guaranteed by the NAME theorem. Since operator boundedness with bound zero implies form boundedness with bound zero, the discreteness of the spectrum of MATH follows from that of MATH and the min-max principle, see for example, REF. CASE: It is shown in the proof of REF that REF implies the MATH-almost sure existence of a decomposition MATH with MATH, MATH and MATH for NAME all MATH with some constant MATH. Here we say that a measurable function MATH belongs to the space MATH, if MATH. The claim thus follows from REF, since MATH is a subset of both MATH and the NAME class over MATH, see for example, REF. CASE: This is a consequence of the considerations in REF and of a straightforward generalization to non-zero vector potentials MATH of REF.
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math-ph/9912025
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Concerning REF we refer to REF. The existence and non-randomness of the integrated density of states is shown in REF for the case MATH by using functional-integral techniques for the NAME transforms of the density-of-states measures. Employing the appropriate NAME formula and so-called magnetic translations, these methods generalize in a straightforward manner to the present setting with a constant magnetic field CITE. Note also that pointwise convergence of the NAME transforms of a sequence of measures implies pointwise convergence of the associated distribution functions at all continuity points of the limit. Alternatively, REF may be obtained from a purely functional-analytic argument, which is outlined in CITE. The representation REF claimed in REF is contained in CITE as REF and Prob. II. REF in the case MATH; for the extension to MATH see CITE. The proof of REF uses the resolvent of MATH. In contrast, the proof of REF relies on semigroup techniques. This explains the different assumptions in REF. The assertion on the growth points of MATH follows from REF.
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math-ph/9912025
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A covariance function MATH which is continuous at the origin, where it satisfies MATH, is bounded and uniformly continuous on MATH by the NAME theorem. Consequently, REF implies the existence of a separable version of MATH which is jointly measurable with respect to the sigma-algebra MATH and the NAME sigma-algebra on MATH. From now on it is tacitly assumed that only this version will be dealt with when we refer to a Gaussian random potential. It remains to verify REF. To this end choose an even natural number MATH. Then there exists MATH such that MATH. NAME 's inequality, the homogeneity of MATH and the explicit computation of the arising Gaussian integral now imply MATH .
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math-ph/9912025
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Using MATH, we rewrite the left-hand side of REF as MATH . The first term on the right-hand side of REF follows from a double partial integration and the rest from the reality of MATH and the gauge condition MATH.
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math-ph/9912025
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We infer from REF of MATH and REF that MATH . When applied to MATH, where MATH with MATH, REF yields the claim of the lemma, provided the second term on the right-hand side of REF is shown to be appropriately bounded in terms of MATH. To do so we estimate MATH . Since MATH, we can apply REF to the scalar product. Together with MATH this gives MATH . To derive the first line of the last inequality, we refer to REF. The second line follows from the NAME inequality, REF and MATH. Thus, the claim is obtained by inserting REF into REF and by observing MATH for MATH.
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math-ph/9912025
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We will estimate from above the probability for the event that MATH is not MATH-frame-regular. Introducing the event MATH one gets from elementary set-theoretic algebra MATH . By assumption one has MATH, MATH and MATH for MATH. Thus, the MATH-independence of MATH implies MATH . With the help of this inequality and REF one gets a MATH-independent upper bound for the sums over MATH in the last line of REF . The remaining multiple sum gives a binomial coefficient. Hence MATH if MATH is sufficiently large.
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math-ph/9912025
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REF is proven by induction on MATH. Pick MATH such that MATH is large enough as required for the applicability of REF . Let MATH stand for either MATH or MATH. Due to REF we can apply REF with MATH and MATH. For given MATH and given MATH this yields MATH . Thanks to REF we thus conclude from REF that for all MATH and all MATH . Choosing MATH in REF, we obtain REF for MATH. Choosing MATH in REF, we get for all MATH the substitute for REF on the length scale MATH which allows one to repeat the above procedure inductively.
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math-ph/9912025
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We show that the theorem is a special case of REF. The inequalities MATH and MATH imply the existence of MATH with MATH. Then estimate REF still holds when MATH is replaced by MATH. Now we choose the quantities MATH of CITE according to MATH, MATH, MATH. The sequences MATH, respectively, MATH, are monotone increasing, respectively,decreasing with MATH. Since MATH one has MATH. The sequences MATH, MATH and MATH are summable because MATH and MATH. Hence, all the assumptions of REF are satisfied such that REF follows MATH-almost surely for all MATH. Since the resolvent of MATH is jointly measurable in MATH and MATH, the subset of pairs MATH for which REF holds is measurable with respect to the product-sigma-algebra and has full MATH-measure by NAME 's theorem.
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math-ph/9912025
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Let MATH be a complete orthonormal sequence of vectors in MATH and suppose that MATH for all MATH with some constants MATH and MATH as in REF . It suffices to show that the event MATH occurs with probability one. Here, MATH is the scalar product REF on MATH, and MATH denotes the absolutely continuous component arising in the NAME decomposition of the projection-valued spectral measure of MATH. But this follows from MATH where we have used REF and NAME, see for example, REF, NAME 's Theorem and REF.
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math-ph/9912025
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Let us define the function MATH by setting MATH, if a one-parameter decomposition of MATH exists in the sense of REF , the operator MATH is self-adjoint and its spectrum is pure point in MATH. Otherwise we set MATH. The joint measurability of the random potential MATH implies that the random potential MATH is jointly measurable with respect to the product sigma-algebra MATH, where MATH is the sigma-algebra of the NAME sets in MATH. Accordingly, CITE implies the joint measurability of MATH with respect to the completion MATH of MATH induced by the product measure MATH. We also define the MATH-subsets MATH . Apart from the MATH-null set allowed for in REF , MATH is the set of MATH's for which MATH enjoys the property of being self-adjoint with only pure-point spectrum in MATH. Hence, we know from REF that MATH. Analogously, up to the same MATH-null set, MATH is the set of MATH's such that for NAME all MATH the operator MATH is self-adjoint with only pure-point spectrum in MATH. The MATH-measurability of MATH follows from the completeness of MATH, the joint measurability of MATH and NAME 's theorem. We split the rest of the proof into two parts. In REF we show that MATH implies MATH and in REF that the assumptions of the theorem imply MATH. CASE: Suppose MATH, then we have MATH and NAME 's theorem gives MATH for almost all pairs MATH with respect to the completed measure MATH. It follows the existence of MATH such that MATH, MATH is MATH-measurable and MATH for MATH almost all MATH. Indeed, the level set MATH differs at most by a MATH-null set from a product measurable set MATH. Now define MATH for all MATH and MATH elsewhere. Thus, we conclude from NAME 's theorem that MATH . The second equality in REF follows from a slight generalization of the ``Disintegration REF " in CITE and uses the fact that the MATH-measurable NAME density MATH provides a regular version of the conditional probability measure of MATH for given MATH. Hence, MATH for MATH-almost all MATH. Since MATH this implies MATH. CASE: The aim is to derive MATH from the conclusion of REF . To this end we have to ensure that the assumptions of REF are satisfied for MATH-almost all MATH when setting MATH, MATH and MATH. Recall that by assumption, MATH is a strictly positive, bounded function and that by REF MATH is MATH-almost surely essentially self-adjoint on MATH. Obviously, REF is identical to REF , if we also set MATH. As to REF , suppose that MATH is in the orthogonal complement MATH of the norm-closed subspace MATH of MATH. Since MATH it follows that MATH . But MATH for NAME all MATH implies MATH for NAME all MATH so that MATH and MATH, proving REF. It remains to verify REF , that is, the MATH-almost sure compactness of MATH for all MATH and all MATH. Because of the boundedness of MATH it is sufficient to show that the operator MATH is compact for MATH equal to some fixed MATH for MATH-almost all MATH. Note that compactness of REF for some MATH implies compactness of REF for all MATH. Hence the set of MATH's for which REF is compact does not depend on MATH. By the second resolvent equation MATH it suffices in turn to prove compactness of MATH and of MATH for MATH-almost all MATH. We note that the validity of the resolvent REF itself is ensured by this compactness and the fact that REF obviously holds on the subspace MATH which, according to the Cor. on p. REF is dense in MATH, because MATH is essentially self-adjoint on MATH. To show this compactness put MATH with MATH, MATH, and observe the diamagnetic inequality (see for example, CITE) for the integral kernels of the resolvents of MATH and MATH which are jointly continuous on MATH. Let MATH be the natural number defined in REF . Then, we have for both MATH and MATH the inequality MATH . To obtain the second inequality in REF we have used the iterated NAME inequality in order to employ REF , the convolution property MATH for the MATH-times iterated integral kernel MATH and the change-of-variables MATH, MATH for MATH. The last inequality in REF follows from the iterated NAME inequality and REF. Since MATH and MATH, the upper bound REF is finite because of MATH. Hence, the operator MATH is MATH-almost surely compact for both MATH and MATH by REF on p. REF.
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math-ph/9912025
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We check that the assumptions of REF are fulfilled for MATH. Obviously, REF provides REF. Since REF of a Gaussian random potential implies REF , the requirement REF follows from the NAME estimate in the weakened form REF, viz. MATH for all MATH and all MATH. Thanks to REF there exists a real constant MATH such that MATH. Hence, MATH, MATH and, since MATH according to REF, we obtain the lower bound MATH for the decay exponent MATH.
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math-ph/9912025
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We check the assumptions of REF with MATH. REF is obviously fulfilled for Gaussian random potentials for any MATH. Concerning REF we introduce a one-parameter decomposition of MATH by defining the function MATH for MATH, MATH being a normalization constant, the centred Gaussian random variable MATH with variance MATH and the non-homogeneous Gaussian random field MATH . Due to the Gaussian nature of both MATH and MATH, and due to MATH for all MATH we conclude that the random variable MATH is stochastically independent of MATH. Hence, the conditional probability density MATH of MATH, given MATH, is independent of MATH and has clearly the desired measurability properties. Note that MATH is a strictly positive function because of REF and MATH. Moreover, it follows from the same remark and REF that MATH for all MATH, where MATH and MATH. Hence, MATH for any natural number MATH. Finally REF ensures that MATH is an admissible choice for MATH in REF , and REF is seen to hold by REF and NAME 's theorem.
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math-ph/9912025
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We deduce the lemma from REF. When adapted to a homogeneous random field and a cube MATH with edges of length MATH, this theorem implies that, if MATH then the realizations of MATH are MATH-almost surely continuous functions on MATH, and the inequality MATH holds for all MATH obeying MATH . First, we verify that REF is fulfilled for all MATH as a consequence of REF . To this end we introduce MATH to characterize the size of the neighbourhood referred to in REF by MATH. Without restriction we may assume that MATH. Hence, REF implies MATH for all MATH. By extending the lower limit of the integration in REF to zero, splitting the integral into two parts at MATH and using MATH in the first, respectively REF in the second part, one arrives at the upper bound MATH . Hence, MATH is finite and REF is applicable. In order to reduce the right-hand side of REF to the more explicit (but less sharp) bound claimed in REF, we exploit MATH for MATH, which follows from REF. Using this and MATH, we continue to estimate REF from above by MATH . The term in curly brackets in REF approaches MATH for MATH. Thus one can find a length MATH, depending on MATH and MATH, such that MATH holds for all MATH. Upon inserting this inequality and REF into REF, we deduce REF for all MATH obeying REF . For those MATH not obeying REF is trivially true, because then its right-hand side is bigger than one by REF.
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math-ph/9912025
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Let MATH and set MATH with some MATH fixed, but arbitrary. Let MATH such that the inequality MATH is satisfied. Then, REF and the NAME REF imply the estimate for all MATH . Here, MATH and we have defined a function MATH on MATH by MATH with suitable constants MATH depending only on MATH. Observe that there exists MATH such that MATH satisfies the inequality MATH for all MATH and all MATH. Clearly, REF also holds with MATH, if MATH and MATH, because MATH for all MATH obeying REF. Therefore we have MATH-regularity of MATH under this condition. But according to REF the probability of the event REF is, independently of MATH, at least MATH provided MATH. The proof is completed by setting MATH.
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math-ph/9912025
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Given MATH and MATH, set MATH and MATH as in REF. As in the proof of REF we infer the existence of a finite length MATH, which is independent of MATH, such that REF holds for all MATH. The proof is completed by requiring MATH, that is MATH with MATH .
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math-ph/9912025
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Let MATH and let MATH, MATH, MATH and MATH as in REF . By REF of the strong-mixing coefficient we have for all integers MATH and all MATH, MATH, with MATH and MATH for MATH that MATH . Iterating this inequality and using MATH, where MATH as required in REF , we obtain MATH . The MATH-sum is bounded from above by MATH for MATH and REF yields for all MATH . Since MATH, REF is satisfied for all MATH and REF holds for all MATH.
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math-ph/9912025
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We denote the NAME (-Plancherel) transform of MATH by MATH. The matrix-valued function MATH with matrix elements MATH is continuous in MATH due to MATH, which follows from MATH and thus MATH. Each MATH is a real-valued and even function in MATH, because MATH. Moreover, since each MATH has a non-negative NAME transform which factorizes in MATH and MATH, the NAME theorem implies that MATH is a matrix-valued covariance function. Therefore - see for example, REF or REF to REF - , there exists a sequence MATH of jointly Gaussian homogeneous random fields on some complete probability space MATH with zero mean and covariance MATH. Without loss of generality we assume that the original probability space MATH underlying MATH coincides with MATH. The same arguments as in the proof of REF show that each MATH has a separable and jointly measurable version. We only consider these versions. Since the non-negative function MATH is determined by REF only up to Euclidean translations, we can assume without loss of generality that MATH. Thus we have MATH for the covariance function MATH of MATH for all MATH. Moreover, the sequence MATH is increasing for all MATH. This proves that MATH is a sequence of Gaussian random potentials on MATH in the sense of REF whose covariance functions satisfy REF. In particular, the relation MATH allows one to identify MATH and MATH. As to REF , we remark that for all MATH an indicator function MATH of the box MATH is uniformly NAME continuous with exponent one, MATH as a consequence of the mean-value theorem and the boundedness REF. Note that the constant MATH does not depend on MATH. Therefore we get, uniformly in MATH, MATH for all MATH and all MATH, the neighbourhood of the origin referred to in REF for MATH. Hence, MATH is uniformly NAME continuous for all MATH with the same exponent MATH and the same neighbourhood of the origin as given for MATH. REF follows from MATH for all MATH. Hence, MATH for all MATH by REF, and the Gaussian nature of MATH implies the stochastic independence of events which are at least a distance MATH apart. REF will be deduced from an application of REF to the left-hand side of REF. To this end observe that by construction the difference MATH is also a Gaussian random potential on MATH for all MATH. Its covariance function MATH has REF with the same NAME exponent MATH and the same neighbourhood as given for the covariance function of MATH. This follows from REF and MATH for all MATH. Moreover, since the NAME transformation is an isometry on MATH, we conclude for MATH the inequality MATH with some MATH-independent constant MATH and MATH taken from REF . Hence, REF follows from REF for all MATH with MATH being also independent of MATH and MATH due to the uniform NAME REF of MATH.
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math-ph/9912025
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We show that REF follows from an application of REF . To do so we have to check the assumptions of this theorem. Let MATH and MATH. Choose the exponents MATH and MATH such that MATH in accordance with REF. The precise value of MATH will be fixed later on. In case that MATH has REF - let us call this simply case (PHM) - we set MATH for all MATH. In case that MATH has REF we use the sequence of Gaussian random potentials constructed in REF for the sequence REF of length scales MATH. For the low-energy REF we consider MATH arbitrary, but fixed, and set MATH in REF , MATH being the energy defined in REF and given explicitly below REF. For the weak-disorder REF we set MATH for all MATH, but allow only for disorder strengths MATH on the length scale MATH with MATH given by REF in the proof of REF . For the time being we suppose that the initial length satisfies MATH. The length MATH is to be taken from REF for the low-energy assertion, respectively from REF for the weak-disorder assertion. The length MATH was defined in REF . Hence, considering case (PHM), we conclude that REF , respectively REF , provides the NAME REF for REF , respectively REF , of REF for any MATH and MATH. The same is true for REF , because REF ensure that REF remain true with the same constants MATH, MATH and MATH of case (PHM), if MATH is replaced by MATH. Next we come to REF . First, we consider the case (PHM) and fix MATH as required by REF . Obviously, both assumptions then follow from REF with MATH. Moreover, observing REF and choosing MATH, the left inequality in REF guarantees that REF hold for some MATH . Concerning REF holds for all MATH and all MATH by REF . Thus we may also pick MATH, MATH and MATH as in case (PHM), thereby satisfying REF . Taking into account REF, it follows that a sufficient condition for REF to hold is MATH . We will show below that in case (PHM) REF is satisfied for all MATH which obey MATH . We will also show below that REF is satisfied in REF if in addition to REF the inequality MATH is assumed for the exponent MATH. Since REF can be fulfilled for some MATH and since MATH obeys REF, it follows that REF are compatible. Moreover, since REF requires MATH, REF does not impose a further restriction beyond REF. Up to now we have shown that REF can be satisfied under REF . Thus, in order to complete the proof it remains to show that REF are sufficient conditions for REF to hold. We consider both cases (PHM) and REF simultaneously. Since MATH for all MATH in case (PHM), there is less to prove in case (PHM) than in REF . Let MATH, MATH and MATH or MATH. Then, the inequality MATH is valid for all MATH and all MATH, MATH. It follows from REF and the inequalities MATH where MATH, MATH, MATH. To derive the latter three inequalities REF have been used. Note that in case (PHM) the last two lines of REF have simply to be omitted. Since MATH for MATH, we bound the right-hand side of REF from below by MATH . Choosing MATH with MATH the first probability in REF is seen to be bounded from below by MATH by virtue of REF . The probabilities in the second and third line of REF are estimated from above with the NAME bound REF and the last probability is estimated from above with REF . Thus, REF is bounded from below by MATH . The constant MATH is an upper bound on the arising NAME constants which is uniform in MATH due to REF and because the subset MATH of the real line has a finite supremum. The latter is true because of REF and the estimate MATH where we have used REF of the functional MATH. Again by REF, it follows that MATH . Combining REF , choosing MATH and observing REF, we infer the existence of a finite length MATH such that MATH for all MATH. This establishes REF and thus completes the proof of the theorem.
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math-ph/9912025
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Let MATH. From the NAME representation of the NAME semigroup MATH, MATH, see for example, CITE, and the explicit form of the heat kernel one obtains the inequality MATH . The fact that MATH and MATH are supported in spatially separated regions allows one to bound the right-hand side of REF by MATH . Estimating the scalar product by the NAME inequality and observing MATH, we conclude for the operator norm MATH . Therefore, by NAME transforming MATH with respect to MATH, we get the inequality MATH where MATH, MATH denotes the modified NAME function of the second kind with index MATH, and REF has been used. Upon inserting the series expansion REFEF for MATH, truncating it after the zeroth term and estimating the remainder, we arrive at REF.
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math/9912001
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Fix MATH, MATH, MATH; we may normalize MATH. Let MATH denote the function MATH, and define the quantity MATH by MATH our task is then to show that MATH . Let MATH be the nearest integer to MATH, where MATH is a small constant to be chosen later. The first step in the argument is to construct group elements MATH such that MATH and MATH for all MATH. Intuitively, REF asserts that the MATH are essentially disjoint, while REF asserts that the MATH are similarly disjoint. For future reference, we note that REF and the MATH normalization of MATH implies that MATH . We now construct the desired group elements. We may let MATH be arbitrary since REF are vacuously true for MATH. Now suppose inductively that MATH have already been constructed for some MATH such that REF (and hence REF) holds for all previous values of MATH. We will show that MATH and MATH where MATH is NAME measure on MATH. By NAME 's inequality, this implies that a randomly selected MATH has probability at least REF/REF of obeying REF and probability at least REF/REF of obeying REF, and so there exists a MATH with the desired properties. From NAME 's theorem, REF, and the identity MATH for all MATH, the left-hand side of REF evaluates to MATH . Thus REF holds if MATH is sufficiently small. The left-hand side of REF can similarly be evaluated as MATH . From NAME we have MATH and MATH . On the other hand, from REF and the induction hypothesis we have MATH . Combining all these estimates, we see that MATH . Thus we obtain REF if MATH is sufficiently small. Fix MATH; all constants may now implicitly depend on MATH. By telescoping REF we have MATH . Let MATH be an arbitrary assignment of signs. Then the function MATH has a MATH norm of MATH and is supported on a set of measure MATH. We now apply Let MATH be a function supported on a set MATH. Then MATH . We divide into two cases, MATH and MATH. We normalize MATH in the first case and MATH in the second; in either case our task reduces to showing that MATH . We may restrict ourselves to the set MATH since the contribution outside of MATH is clearly acceptable. In this set MATH may of course be replaced by MATH. The function MATH is increasing for MATH and decreasing for MATH, with a global maximum of MATH. We thus have MATH in the first case and MATH if the second case. In either case the claim follows by multiplying this estimate by MATH and integrating, using the MATH normalization of MATH. From this lemma we obtain MATH . Since MATH is translation invariant and maps MATH to MATH, we thus have MATH . Randomizing the signs MATH and taking expectations using NAME 's inequality, we obtain MATH . In particular, we have MATH . If we integrate the trivial pointwise estimate MATH using REF, we obtain MATH . Telescoping this for all MATH, we obtain MATH . Comparing this with REF we obtain REF as desired.
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math/9912003
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Using REF we calculate the discriminant of MATH then, by NAME inequality CITE, MATH is NAME unstable.
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math/9912003
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Assume that not all weights are equal. By REF , MATH is NAME unstable. We will prove that MATH is not parabolic stable by showing that at least one of the subsheaves of the NAME filtration contradicts the parabolic stability of MATH. Since the parabolic degree of MATH is zero, we have to prove that the parabolic degree of one of the subbundles is non-negative. First we assume that the NAME filtration of MATH has only one term, that is, there is a short exact sequence MATH such that both MATH and MATH are NAME semistable torsion-free sheaves. The objective is to show that the parabolic degree of MATH is non-negative. Let MATH, and MATH. By the formula for the first NAME class of MATH we have MATH, where MATH means numerical equivalence, and MATH. Let MATH, MATH, and analogously for MATH and MATH. Both MATH and MATH are NAME semistable, then by NAME inequality MATH . Using MATH and the formula for MATH in terms of MATH this inequality becomes MATH . Using the formula for MATH in terms of MATH we have MATH. Substituting this into the inequality and simplifying we obtain MATH . Note that because MATH is a NAME filtration we have MATH and then we have MATH . There is an induced parabolic structure on MATH. There is a subset MATH with MATH elements such that the weights of the parabolic structure on MATH induced by MATH are MATH for MATH. Claim. For any subset MATH of cardinality MATH, MATH . Proving this is an easy calculus exercise. Use REF multipliers to minimize MATH subject to the conditions MATH and MATH where MATH is some constant (note that we are minimizing a linear function restricted to a sphere). Then combining REF we get MATH but this is the parabolic degree of MATH, so MATH cannot be parabolic stable. Now we assume that the NAME filtration has length REF, that is, we have MATH . Then MATH (recall that we assume MATH), MATH, and MATH and MATH are torsion free sheaves of rank one. Let MATH, MATH, MATH (since MATH, we can think of the first NAME class as an integer number). By the definition of the NAME filtration we have MATH . Using REF for MATH we have MATH, where MATH is the degree of MATH. Then there exist (rational) numbers MATH, MATH such that MATH . Using the formula for MATH we have MATH where MATH, hence MATH. Combining this with REF we obtain MATH . The number MATH must be real, then MATH. Now we calculate the parabolic degrees of MATH and MATH as functions of MATH (we take the polarization MATH). Using MATH and MATH we obtain MATH . Note that if we fix MATH, MATH is fixed, but MATH could take two values, hence the two signs in the formula. We want to show that for any value of MATH (with MATH), at least one of these is non-negative. REF defines a conic with coordinates MATH in the plane MATH. This conic intersects the axis MATH in the two points MATH and MATH . For MATH, MATH (by REF ). Assume MATH. Then MATH but if MATH is in this interval, using MATH we have MATH (for the equality we used REF , to get the correct sign of the square root), and then REF implies that MATH. Then either MATH or MATH is non-negative.
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math/9912003
|
Let MATH be the corresponding stable parabolic bundle given by REF . By REF the parabolic structure is trivial (that is, all weights are equal), and then MATH is a multiple of identity.
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math/9912003
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Let MATH be a representation, and MATH the associated parabolic bundle. If MATH is irreducible then, by REF , MATH is a multiple of identity. Then the representation MATH induces a representation MATH. If MATH is irreducible then MATH is also irreducible, and we get a contradiction.
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math/9912003
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Given an irreducible local system on MATH, consider its NAME extension MATH CITE. By construction, the real part of the eigenvalues of the residue of MATH are non-negative and less than REF. Define a parabolic structure on MATH, setting the parabolic weights equal to the real parts of the eigenvalues of the residue of MATH. This integrable logarithmic connection satisfies the hypothesis of CITE, and then we obtain a parabolic NAME bundle. It is stable because of the irreducibility hypothesis and because it has a NAME metric. The vanishing of the eigenvalues of the residue of MATH is given in CITE. By CITE, the eigenvalues of the residue of MATH are real, and by CITE they are equal to the parabolic weights of the parabolic NAME bundle, and applying again CITE we obtain the formula for the slopes.
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math/9912003
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Assume that there is an irreducible representation MATH. Let MATH be the stable parabolic NAME bundle associated by REF . We will use the following fact: MATH . The proof was shown to us by NAME. Consider the sequence MATH where the second map is the residue, and note that the image of the constant function REF under the co-boundary map is represented by the NAME cocycle MATH, where MATH is a local equation of MATH on an open set MATH, and this is (up to a non-zero constant) the NAME class of the line bundle MATH (compare CITE). If MATH is a nontrivial line bundle on MATH with MATH, we still have MATH . To see this, note that MATH by NAME vanishing theorem, hence MATH, and then use the sequence REF tensored by MATH. Consider first the case when the parabolic structure of MATH is non-trivial (weights are different). We already know that the eigenvalues of MATH are zero. Since the residue of the NAME field also preserves the parabolic filtration, it yields a map of line bundles MATH. But by REF we have MATH so this map has to be zero. In other words, MATH, and then MATH defines a NAME pair on MATH. Using NAME inequality for NAME bundles (compare CITE) we see that MATH is not stable as a NAME pair. Then there is a short exact sequence MATH where MATH is the ideal sheaf of a zero-dimensional subscheme and MATH and MATH are line bundles with MATH. A calculation similar to the proof of REF (for filtrations of length REF), using formulae REF and the fact that MATH, shows that MATH is not parabolic NAME stable (since MATH is MATH-invariant), contradicting the hypothesis and then finishing the proof in this case. The case when the parabolic structure is trivial is more subtle (it is to cover this case that we need the assumption on the tangent bundle). Since all the weights are equal, MATH is NAME stable iff it is parabolic stable. But if MATH is parabolic stable then there is a corresponding MATH irreducible representation MATH of MATH, and since the weights are equal, MATH is a multiple of the identity. Then there is an induced irreducible MATH representation of MATH, in contradiction with the hypothesis. Then MATH is not NAME stable, and thus there is a sequence as before MATH where MATH and MATH are line bundles, MATH is the ideal sheaf of a zero-dimensional subscheme MATH, and MATH, but now MATH might not be MATH-invariant, so MATH doesn't contradict parabolic NAME stability. The NAME field MATH defines a map MATH. Regard this as a section of MATH. By REF , this is a section of MATH. In other words, MATH. On the other hand, MATH vanishes on MATH, since it factors through MATH. Putting both facts together, MATH can be seen as a section of MATH. If the tangent bundle is globally generated we see that this group is zero (and thus MATH, being invariant under MATH, contradicts stability of the parabolic NAME pair) unless MATH and MATH is empty. Suppose therefore that we have a non-trivial extension MATH . We now produce a ``model" of the extension REF , which is adapted to the divisor MATH. Since the tangent bundle is generated by global sections, the anticanonical line bundle MATH has non-negative degree. Then, since the NAME group has rank one, MATH is ample, and by NAME vanishing theorem MATH. Let MATH denote the normal bundle of MATH. From the sequence MATH it follows that the map MATH is onto. Note that this map is obtained from the natural multiplication map MATH, using the extension class of REF . Then, given a section MATH of MATH, the corresponding extension MATH is given as the kernel in the sequence MATH the map on the right being MATH. From REF we have MATH and since MATH is surjective, there is a section MATH such that MATH. From now on we will use this ``model" of MATH. Note the obvious inclusion MATH. Consider the composition MATH where the last map comes from the left map on REF . We denote this composition by MATH. One can now represent this map as a matrix MATH where MATH are sections of MATH, MATH is a section of MATH, and MATH of MATH. We will use this representation to show that the residue of the NAME field is zero (note that MATH). By REF , MATH. Consider a local holomorphic section MATH of MATH. The residue of MATH sends MATH to MATH hence MATH is an eigenvalue of MATH, and hence of MATH, but those eigenvalues are zero, then MATH and MATH. Since the eigenvalues of MATH are zero, MATH, and then MATH. Finally we will impose the condition that the map MATH comes from the NAME field on MATH, that is, MATH extends to MATH. The exact sequence REF shows (after tensoring with MATH) that a local section of MATH is the same thing as a local section MATH of MATH, with the condition that MATH. We will write local sections of MATH as MATH, where MATH and MATH are respectively holomorphic local sections of MATH and MATH and MATH is a local equation of MATH. Take a local holomorphic section MATH of MATH that doesn't vanish identically on MATH, and let MATH be a local holomorphic section of MATH such that MATH, so that MATH is in fact a local section of MATH. The matrix REF acting on this gives MATH . This should be a local section of MATH. We have seen that MATH is in fact a section of MATH, then MATH is a local section of MATH. We need then MATH to be also a local section of MATH. Recall that MATH is a section of MATH and MATH a local section of MATH (that doesn't vanish identically on MATH). Then for MATH to be a local section of MATH we need the residue of MATH to be zero, that is, MATH. Then we obtain that the residue of REF (and hence of MATH) is zero. Finally, if the residue of MATH is zero, since we have assumed that the weights of the parabolic structure of MATH are equal, we obtain that MATH is a multiple of the identity, then there is an induced MATH representation, and since this is reducible by hypothesis, we conclude that MATH is also reducible.
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math/9912003
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Since MATH (and also the associated NAME connection) is singular on MATH, we cannot directly apply the NAME formula. We will modify the metric to make it smooth, and this will produce the second summand. First note that MATH . This can be proved by first showing that both sides are invariant under change of holomorphic trivialization, and then computing in a holomorphic trivialization where MATH CITE. Note that MATH is a Hermitian metric on MATH, singular on MATH (in fact MATH). Following CITE, choose a section MATH with a zero of order one along MATH. Take any Hermitian metric on the line bundle MATH, and then MATH is a smooth function on MATH, vanishing on MATH. Near MATH, MATH (using an appropriate trivialization of MATH), and then MATH defines a smooth metric on MATH, and we can apply the NAME formula as in CITE. The integral REF can be written, using REF MATH where the second integral is given by the NAME formula. This finishes the proof of the lemma.
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math/9912004
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Since MATH is an isolated singular point, there is a simply connected neighborhood MATH which closure does not contain other critical points of the function and which boundary is a smooth closed curve. Let us assume MATH is a restriction of function MATH to the set MATH . Then the levels of both functions coincide in this set. As MATH is a regular value of the function g, the pre-image MATH is a REF-dimensional submanifold in MATH that is a disjoint union of imbedded circles and open intervals. Let MATH be a connected component of the set MATH . If H is homeomorphic to a circle, then it is the boundary of a closed set MATH. As function MATH is not constant on MATH the maximum or minimum value of it on MATH is distinct from MATH and the appropriate point of maximum or minimum is another, except for MATH critical point in domain MATH . The obtained contradiction proves that MATH is homeomorphic to an open interval. Let us show that all accumulation points of MATH in MATH belong to MATH. By contradiction, let there is such a point MATH that does not belong to MATH. Then, according to the continuity of the function MATH: MATH . As MATH, then MATH belong other component MATH of MATH . Since MATH is a submanifold, there is a neighborhood of a point MATH that does not contain other points from MATH except the points of MATH . It contradicts assumption that MATH is an accumulation point of MATH . Using similar argument to the neighborhood of closure MATH, we obtain that the accumulation points of MATH in MATH belong the same component of MATH as MATH. So in the boundary of MATH there is no more than two accumulation points of interval MATH. Accumulation points of MATH that belong to the boundary of MATH or coincide with MATH we call extremities of the interval. NAME types of the interval MATH are possible: REF both extremities coincide with the point MATH, REF both extremities lay on the boundary of MATH, REF one of extremities coincides with a point MATH and another lays on the boundary U. Let us prove that there doesn't exit a component of the level MATH that form a loop with vertex in the point MATH (that is, both extremities coincide with MATH). By contradiction, let such component exists, then it is a boundary of the two-dimensional disk. In this disk there is a critical point of minimum or maximum that distinct from MATH . It contradicts a choice of a neighborhood MATH . So there doesn't exit an interval of type REF. Note that only the finite number of extremities of intervals can lay on the boundary of MATH. Otherwise, the condition that MATH is a submanifold in a limit point is false. So there are a finite number of intervals of type REF. We contract a neighborhood MATH to such a neighborhood MATH, which closure is not intersected with interval of type REF and such, that the boundary of MATH is smooth and has transversal intersection with MATH. Then MATH consists of finite union of the closed intervals, one of which extremities coincides with a point MATH and another lays on the boundary of MATH. Hence MATH is a cone of the finite number of points. Let us prove that number of extremities laying on the boundary MATH is even. We call domains in MATH with MATH by positive and with MATH by negative. Each interval contains in the boundary both positive and negative domain. These domains alternate. Thus we have the same number of the domains of each type. Hence, total number of the domains, as well as intervals, is even.
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math/9912004
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Let MATH be a neighborhood with the same properties as in the lemma and MATH be the number of arcs, which an extremity is the point MATH . The critical level devides the neighborhood MATH to domains. Let MATH be one of them and such that MATH for MATH . Then in the points of the intersection MATH with MATH the vector field grad-MATH is directed inside of the domain MATH . Let us prove that there is a trajectory that passes through MATH and tends to MATH . Really, let MATH be a map given by the formula MATH . In view of the smoothness of the vector field grad-MATH, the continuity of map MATH follows for points, which image does not coincide with the point MATH . If the set of such points coincides with MATH, then the image of this connected set by the continuous map is connected set. The obtained inconsistency proves the existence of a required trajectory. Let MATH be a neighborhood of a point MATH in MATH. We define a neighborhood MATH in F for MATH by REF . By the same formula, as above we define a map MATH . Let us prove that for any connected neighborhood MATH of the point MATH in MATH such that MATH there exists MATH that MATH . Let MATH . Let MATH be such points that MATH and between points MATH on the trajectory MATH there don't exist points of its intersection with boundary MATH . Let MATH . Denote by MATH union of all arcs of the boundary MATH that their endpoints belong to the set MATH and which do not contain points of MATH. Let MATH . Then it is easy to see that MATH . Let us construct an conjugated homeomorphism from this neighborhood to appropriate neighborhood of the function Re-MATH . We take a set MATH in the capacity of the neighborhood W of the function Re-MATH . Let us select one-to-one the correspondence of domains, into which level lines decompose neighborhoods MATH, so that to adjacent domains corresponded adjacent ones and positive domains to positive ones. We construct an conjugated homeomorphism between appropriate domains. Let for a determination, domain MATH corresponds domain with a polar angle: MATH . Let MATH and MATH be a nearest to MATH on MATH point, for which MATH . We fix a homeomorphism MATH of the curve MATH with endpoints MATH and MATH to MATH so, that MATH . Let us define map MATH from the closed domain D, limited by a curvilinear tetragon with vertex MATH and sides laying on MATH to the rectangle MATH by REF . Thus, the level lines of function and integrated curves are mapped in segments parallel to the sides of a rectangle. Let us prove that the map MATH is a homeomorphism. Since each point lays on an integrated trajectory and the function is monotonic at going on a trajectory in one direction, the uniqueness of map MATH follows. For an arbitrary point MATH and sequence of points MATH, it is necessary to prove that MATH if and only if MATH. By contradiction, let MATH and MATH does not converge to MATH . Then there is such MATH and subsubsequence MATH that MATH or MATH . It is equivalent to that MATH . Thus, points MATH lay out of the neighborhood of a point MATH that are bounded by curves MATH . It contradicts with convergence of the sequence MATH . Let us assume now that the sequence MATH converges to MATH and the sequence MATH does not converge to MATH . Since MATH is the compact set then there is a subsubsequence MATH that converges to MATH . Then MATH . It contradicts that MATH . By analogy with above, we construct a curvilinear tetragon with vertex in points MATH . If MATH as well as above we create, a homeomorphism of this tetragon on a rectangle MATH . If MATH we denote by MATH a curvilinear tetragon with vertex in the points MATH and by MATH a curvilinear triangle with vertex in points MATH . Let us fix homeomorphisms MATH of arcs on level lines of the function between the points MATH and MATH to CITE and [REF/REF], accordingly. Let us set a map MATH by the formula: MATH . As above, this map is a homeomorphism that maps a level lines to horizontal curves. Similarly we build homeomorphisms of domain for function Re-MATH . Then, having taken appropriate compositions of homeomorphisms, we obtain homeomorphisms of domain of function MATH to domain of function Re-MATH . By the construction, these homeomorphisms coincide in the boundaries of domain and thus set a required homeomorphism of the neighborhood MATH .
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math/9912004
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Necessity. If a conjugated homeomorphism of the surface MATH is given, it maps critical levels to critical levels. It sets an isomorphism of the diagrams. Sufficiency. Suppose that the diagrams of two functions are isomorphic. Then there is a homeomorphism MATH of a surface MATH that maps critical levels to critical levels. Let us cut the surface MATH by the critical levels. We obtain surfaces MATH which is homeomorphic to cylinders MATH . Then the homeomorphism MATH induces homeomorphisms of obtained cylinders. We replace these homeomorphisms with homeomorphisms, which map the levels of the function to the levels proportionally to value of function between two critical values and coincide with initial ones on each component of the boundary of the cylinders. As the constructed homeomorphisms coincide on the boundaries, they set an conjugated homeomorphism of the surface MATH .
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math/9912004
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Necessity. If the functions are topologically conjugated, there is an isomorphism of their diagrams. The restriction of this isomorphism to the graphs sets isomorphism of the distinguishing graphs. Sufficiency. Let distinguishing graphs are isomorphic. We construct an isomorphism of the diagrams. By the construction, we have one-to-one correspondence of the cylinders MATH and the pairs of the upper and lower cycles. Thus the correspondence between cylinders is given. Let us construct arbitrarily homeomorphisms between cylinders, so their restrictions on upper and lower foundation coincide with restrictions of isomorphisms of the graphs on appropriate cycles. As these homeomorphisms coincide on the boundaries, they set a homeomorphism of a surface MATH and the isomorphism of the diagrams signifies.
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math/9912004
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The necessity is obvious. Let us prove sufficiency. The conditions of the theorem guarantee that after gluing of cylinders to the graphs we obtain a closed surface. Let us construct the function f. For this purpose we set MATH and for the cylinder MATH with boundary on MATH and MATH: MATH . Thus in vertexes of the graph the cylinders are pasted so, that the function MATH coincides with function MATH in an appropriate coordinate system. The constructed function is desired.
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math/9912005
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Let MATH and let MATH be the multiplication. Let MATH be the open subvariety of MATH of points such that the linear map MATH has maximal rank. Then MATH carries a vector bundle MATH whose fibre above the point MATH is simply the cokernel of MATH. We consider the vector bundle MATH which is an irreducible variety and has a morphism to MATH whose image must contain an open dense subvariety of any left general component since its image contains every point MATH such that the left restriction of MATH is isomorphic to a representation of the form MATH for MATH. Its image is irreducible and consequently must lie in one of these components so it follows that there must be precisely one left general component and the image of MATH must lie in this unique component.
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math/9912005
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The injectivity of this linear map implies that the map considered in the proof of the previous lemma has maximal rank.
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math/9912005
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Let MATH be a MATH-standard family over the algebraic variety MATH. Then for each vertex MATH, MATH is a vector bundle of weight MATH for MATH of rank MATH for a suitable dimension vector MATH. Then by the local isomorphism theorem, REF, there exists an open MATH equivariant subvariety MATH of MATH such that the restriction of MATH to MATH is isomorphic as MATH vector bundle to MATH where MATH acts trivially on MATH and diagonally on MATH. We choose bases for each MATH and a basis of MATH which give bases for each MATH and determines a homomorphism from MATH to each MATH and hence to MATH which we may regard as a diagonal embedding. Each arrow MATH determines a morphism of vector bundles from MATH to MATH and hence determines a morphism of algebraic varieties from MATH to MATH and hence we have a morphism of algebraic varieties from MATH to MATH which is MATH equivariant and has the property that the pullback of the canonical family on the image of the morphism is MATH. It follows that the image actually lies in MATH and in fact must lie in the component MATH. Since the stabiliser in MATH of a point MATH in MATH is isomorphic to the units of MATH and so is the stabiliser in MATH of its image MATH in MATH, we deduce that the morphism from MATH to MATH is injective; indeed that the morphism from MATH to MATH is injective and this latter morphism is also dominant since MATH is birationally representation equivalent to MATH.
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math/9912005
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This argument essentially occurs as a special case of the proof of REF to which the reader should refer for greater detail. We deal only with the first case since the second case has the same proof. Let MATH be the associated left MATH-free family on MATH (after replacing MATH by an open dense subvariety). Let MATH be a MATH-standard family over MATH where MATH is MATH equivariant to a dense open subvariety of MATH for some integer MATH. Let MATH and let MATH be the extension family of MATH on MATH. Then since we assume that MATH is left general with respect to MATH, MATH and MATH are birationally representation equivalent. Let MATH. After shrinking MATH a little we may assume that for all MATH, MATH. Let MATH be a dimension vector for the quiver MATH which has two vertices MATH and MATH, MATH arrows from MATH to itself and MATH arrows from MATH to MATH. Then MATH acts on MATH whilst MATH acts on MATH so that MATH is a MATH vector bundle of weight MATH and MATH is MATH birational to MATH. Since MATH and MATH are birationally representation equivalent, a general representation of type MATH has a MATH-socle which must coincide with its obvious subrepresentation isomorphic to MATH. We therefore pass to a dense open MATH subvariety of MATH where this is true. Then the orbits of MATH correspond to the isomorphism classes of representations in the restriction of MATH to this subvariety. Since a general representation of type MATH and hence of type MATH has trivial endomorphism ring, it follows that MATH has trivial stabilisers generically on MATH and hence MATH is a NAME root for the quiver MATH. Thus the main result of REF , allows us to conclude that MATH is reducible to matrix normal form and hence by REF there exists a MATH equivariant subvariety MATH of MATH where MATH such that the restriction of the canonical family on MATH to MATH is standard and MATH is a dense subvariety of MATH. Since MATH and MATH are MATH birational, there is a corresponding MATH equivariant subvariety MATH of MATH and the restriction of MATH to MATH is what we want.
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math/9912005
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If MATH and MATH are representations for which we have short exact sequences MATH and a short exact sequence MATH then it is clear that the induced extension of the subrepresentation MATH over MATH splits; thus MATH has a subrepresentation MATH isomorphic to MATH such that the factor is isomorphic to MATH and MATH must be the MATH-socle of MATH. Thus by induction all objects in MATH have the required structure. By construction, MATH and clearly MATH; therefore MATH for all MATH in MATH as required. Now let MATH be any object in MATH; so there is a short exact sequence MATH . Then MATH and since MATH, MATH so that MATH is a representation of dimension vector MATH of the MATH-th NAME quiver. In fact, the natural homomorphism from MATH to MATH is surjective with kernel isomorphic to MATH for some integer MATH. Using this short exact sequence it is a simple matter to check that the two functors are mutually inverse and that MATH is preserved by this functor.
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math/9912005
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If MATH is reducible to matrix normal form for the MATH-th NAME quiver we take the family of representations of dimension vector MATH for the MATH-th NAME quiver and apply the functor MATH considered in REF . This gives a family of representations for the quiver with relations MATH that is birationally representation equivalent to MATH since a general representation of type MATH lies in MATH and MATH is left general with respect to MATH and therefore shows that MATH is reducible to matrix normal form of type MATH. In the remaining case where MATH is not reducible to matrix normal form then a general representation of dimension vector MATH for the MATH-th NAME quiver is isomorphic to MATH where MATH and MATH and for positive integers MATH and MATH such that MATH. Using the facts that a general representation of type MATH lies in MATH and that MATH is left general with respect to MATH, we deduce that MATH is birationally constant and the general representation of type MATH is isomorphic to MATH where MATH and the representations MATH and MATH and the integers MATH and MATH satisfy the conditions of the lemma.
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math/9912005
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We shall proceed by induction on MATH. The associated left MATH-free family, MATH, has a NAME reduction of type MATH to the dimension vector MATH for the MATH-th NAME quiver MATH for some integer MATH. Now suppose that the dimension vector MATH is reducible to matrix normal form. Then by REF , MATH, is reducible to matrix normal form of type MATH and so, by REF , MATH is reducible to matrix normal form of type MATH as required. If the dimension vector MATH is not reducible to matrix normal form then, by REF , a general representation of type MATH is isomorphic to MATH where MATH and MATH and MATH and MATH are positive integers such that MATH. Note that MATH unless MATH in which case MATH and MATH. We let MATH for notational convenience. A general representation MATH of type MATH is the middle term of a short exact sequence MATH . Then since MATH and MATH are representations that lie in MATH, it follows that MATH for MATH since MATH for MATH. It also follows that MATH from REF since a general representation of type MATH has trivial endomorphism ring and MATH for the same reason. Since MATH for MATH, it follows that MATH and MATH has a MATH-top. Let MATH be the kernel of the homomorphism from MATH onto MATH. We have a short exact sequence MATH . Therefore the linear map from MATH to MATH is surjective and MATH is a summand of MATH. Therefore, MATH is right general with respect to MATH because MATH is left general with respect to MATH and so an open subvariety of the extensions of MATH on MATH occur in the family MATH. Further, the associated right MATH-free family, MATH has a NAME reduction of type MATH. The only thing remaining to check is that the family MATH is left general with respect to MATH but this follows because MATH is a summand of MATH and the family MATH is left general with respect to MATH. Thus we have shown that the family MATH has a NAME coreduction of type MATH and if MATH then MATH and we are done by induction. If MATH, we noted above that MATH and so when we perform the same argument again for this coreduction the numbers will drop this time so again we are done by induction.
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math/9912005
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Let MATH be the canonical family on the left general component MATH of representations of dimension vector MATH. Let MATH, MATH and MATH. Then MATH has a two-step NAME reduction of type MATH and so by REF , MATH is reducible to matrix normal form.
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math/9912005
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Once stated, this result is also clear. In the derived category MATH, the object MATH is equivalent to the complex MATH which after our identification is the complex of sheaves MATH and this itself is a sheaf if and only if it has homology only at the penultimate term as required.
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math/9912005
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Consider the complex of sheaves on MATH . Let MATH be the support of the homology of this complex except at the penultimate term. Then MATH is closed and so is its image in MATH; therefore the complement of the image of MATH in MATH is open and it is the set of points MATH where MATH is a sheaf. In the second case, the vanishing of MATH is an open condition so the result follows.
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math/9912005
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After tensoring by a suitable line bundle we may assume for a general MATH in the moduli space that MATH is a representation but MATH is not; that is one of MATH and MATH is non-zero. Since MATH has natural cohomology, it follows that MATH; it also follows that MATH since by NAME duality this is dual to MATH and since MATH is a representation any such homomorphism is split surjective. So we consider the exact complex of sheaves on MATH from which we deduce that for a sheaf MATH that is also a representation the groups MATH are the homology of the complex MATH and this complex is exact except at the middle term. In particular, after tensoring with MATH we see that the linear map from MATH to MATH is injective. Let MATH, MATH and MATH. Then MATH considered as a representation has dimension vector MATH and the injectivity of this linear map shows by REF that MATH is a representation in the left general component. Therefore by REF our moduli space of sheaves is birational to an orbit space for MATH on the left general component of representations of dimension vector MATH but we know that the canonical family on this component is reducible to matrix normal form by REF . Therefore our moduli space is birational to a suitable number of MATH by MATH matrices up to simultaneous conjugacy where MATH is the depth of MATH.
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math/9912005
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This follows from the known results on matrices up to simultaneous conjugacy. A good summary of the known results may be found in CITE.
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math/9912006
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We proceed by induction on MATH, the number of components of MATH. The theorem is true for knots in MATH since, by the cyclic surgery theorem in CITE, MATH and MATH surgery on a nontrivial knot cannot both produce a manifold with trivial fundamental group. Now, assuming the theorem is true for all links of MATH or fewer components, we will show the result holds for any link with MATH components. Let MATH be a link in MATH such that MATH for all MATH - tuples MATH . If MATH is reducible, then there is some MATH which does not bound a ball. Since any MATH separates and bounds a ball on each side, the sphere in MATH must split MATH into two nonempty sublinks MATH and MATH. The inductive hypothesis implies that each of these sublinks is the trivial link. Furthermore, they are separated by a sphere, so their union MATH is also the trivial link. In particular, MATH is trivial if some torus MATH on MATH has a compressing disk MATH in MATH because the frontier of a regular neighborhood of MATH would be a reducing sphere. Therefore, we may assume that MATH is irreducible with incompressible boundary. Each component of MATH must be unknotted because performing MATH surgery on any other component results in a link that, by induction, is trivial. Furthermore, each pair of distinct components MATH and MATH have MATH. Suppose that MATH. Let MATH, a knot in MATH. Then MATH which is homeomorphic to MATH by our assumption. See, for example, CITE as a reference for this modification of surgery instructions. However, MATH, so MATH. Therefore, the link resulting from MATH or MATH surgery on any one component is a link in MATH, and the meridian - longitude pairs of the remaining components are the same as the meridian - longitude pairs for the corresponding sublink of MATH. Consider the links MATH and MATH. Each of these are MATH - component links which satisfy the hypothesis of the theorem, so they are trivial by induction. Therefore, the torus MATH becomes compressible after both MATH and MATH . NAME fillings on MATH. Since MATH and since MATH is irreducible with incompressible boundary, by REF (generalized by NAME in CITE) we see that MATH is either MATH or a cable space, a solid torus with a regular neighborhood of a MATH cable knot of the core curve of the torus removed. In either case, MATH is NAME fibered. NAME and NAME produced a complete list in CITE of all links in MATH with NAME fibered complement. Any such link with more than one component has nontrivial linking number between some pair of components. This contradicts that every pair of components of MATH are algebraically unlinked, so MATH must in fact be reducible, proving the theorem.
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math/9912006
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When MATH has one component, this is the well-known result of NAME and NAME. Suppose MATH has MATH components. Because MATH is homologically trivial, its MATH-th meridian has order MATH in MATH. Thus MATH for all MATH. The MATH-th component of MATH is unknotted because MATH is NAME, and so MATH is a link in MATH. The linking number of the MATH-th and MATH-th components of MATH is MATH, where MATH, MATH, and MATH are the linking numbers in MATH. (This holds for any link in which MATH twists are made about an unknotted MATH-th component.) Thus MATH is homologically trivial. Deleting the first component of MATH yields MATH, and since this link may also be obtained by deleting the first component of MATH and then performing NAME surgery on the last component of the resulting link, it is trivial. The same holds for the MATH-th component, MATH, and so MATH is NAME. Thus, by induction, MATH is trivial, and therefore MATH for all integers MATH. NAME and NAME 's result CITE then implies that MATH is a trivial knot for all integers MATH, and therefore MATH for all integers MATH. By REF , MATH is trivial.
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math/9912006
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Let MATH be a homeomorphism. (By a homeomorphism of link complements, we always mean an orientation preserving homeomorphism, as is required for link equivalence.) Since MATH and MATH are homologically trivial, the longitudes of the components of each link are null-homologous in the link complements. These are the only isotopy classes of essential simple closed curves on each boundary component of MATH and MATH that are null-homologous. Since MATH induces isomorphisms on the homology groups, MATH maps longitudes to longitudes. The map induced by MATH on the homology of the torus boundaries of the complement of tubular neighborhoods of the MATH must be invertible, and therefore the MATH-th meridian of MATH must map to a slope MATH of the MATH-th component of MATH. If every meridian of MATH is taken to a meridian of MATH, then it is clear from the definition of NAME surgery that we can extend MATH to a homeomorphism MATH. That is, the two links are equivalent. Assume without loss of generality that the first meridian of MATH is taken to some slope MATH of the first component of MATH. Then we can extend MATH to a homeomorphism from MATH to MATH. The former is trivial because the link MATH is NAME. Thus, the latter is also trivial since it is a link in MATH and the trivial link is determined by its complement. Therefore, MATH so MATH must be the trivial link by REF . As above, the homeomorphism MATH implies that MATH is also the trivial link, proving that MATH and MATH are equivalent.
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math/9912006
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Let MATH be a homeomorphism. Since MATH is homologically trivial, the image of each longitude under MATH must be an essential simple closed curve on the boundary of MATH that is null-homologous in the link complement. Therefore MATH is also homologically trivial, and MATH maps longitudes to longitudes. As in the proof of REF , the MATH-th meridian of MATH must map to a slope MATH of the MATH-th component of MATH. Moreover, the MATH-th meridian of MATH must then map under MATH to the slope MATH of MATH. This implies that MATH is determined by MATH and the slopes MATH, for if there is a homeomorphism MATH, then the composition MATH takes the meridians of MATH to the meridians of MATH, so MATH and MATH are equivalent. Thus the links with complements homeomorphic to that of MATH are parameterized by MATH - tuples of slopes MATH. If MATH for all MATH, then MATH and by REF , MATH is trivial. Therefore, we assume that MATH. Since MATH is NAME, it has a diagram where components MATH are MATH disjoint planar circles. Using this diagram, it is clear that we may perform MATH twists on the MATH-th component for all MATH, and so in this way, we may realize any link MATH whose complement is homeomorphic to MATH.
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math/9912006
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We proceed by induction on the number of components in MATH. For knots, the corollary is true by CITE. Suppose that MATH has more than one component. It must be homologically trivial by the argument in the proof of REF . Performing MATH surgery on any one component results in a trivial link by induction, so MATH is NAME. Since MATH, L is trivial by REF .
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math/9912006
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REF imply that MATH is an HTB link. If REF is true, then MATH so L is trivial by REF .
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math/9912014
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If MATH, then there is some term order MATH such that one of MATH and MATH is a minimal generator of MATH, and the other is standard for MATH. Since MATH is also MATH-graded, it suffices to prove the lemma for monomial MATH-graded ideals, where MATH for all MATH. Suppose there exist a MATH such that MATH is not a Graver binomial. Then there exists MATH with MATH such that MATH and MATH. Since MATH is MATH-graded, one of MATH or MATH is in MATH. If we have MATH then MATH would not be a minimal generator of MATH, and if MATH then MATH would not be standard. Therefore, MATH is a Graver binomial for all MATH.
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math/9912014
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It suffices to prove that MATH is weakly MATH-graded, where MATH is any term order, since MATH has the same NAME series as MATH. If MATH is a Graver binomial, then since there is some MATH with MATH, one of MATH and MATH lies in MATH. Let MATH and MATH be two monomials of degree MATH, and let MATH be a Graver binomial with MATH and MATH. Since one of MATH and MATH lies in MATH, one of MATH and MATH lies in MATH. It thus follows that there is at most one standard monomial of MATH in each degree MATH, and so MATH is weakly MATH-graded.
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math/9912014
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Let MATH be a Graver binomial, with MATH. By REF it suffices to show that either MATH, or MATH. Since MATH, there is some (possibly identical) Graver binomial MATH with MATH a minimal generator of MATH, and MATH, and MATH. If MATH, then MATH, and so MATH. If MATH, then MATH, so MATH.
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math/9912014
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Let MATH be the initial ideal of MATH with respect to MATH. We first show the containment MATH. Let MATH be a minimal generator of MATH. If MATH, or MATH is a minimal generator of MATH other than MATH, then MATH. So we need only consider the case that MATH, where MATH and MATH, as this is the only other form minimal generators of MATH can have. In order to show that MATH is in MATH, it suffices to show that MATH is a Graver binomial, where MATH is the unique standard monomial of MATH of the same MATH-degree as MATH. Suppose MATH is not a Graver binomial, so we can write MATH, MATH, where for each MATH, MATH is a Graver binomial. Since MATH, we must have MATH and MATH for all MATH. If MATH for some MATH, this would mean that MATH, and hence MATH, was in MATH. We can thus reduce to the case where MATH and MATH for all MATH, and so MATH. Now since MATH is a minimal generator of MATH, there must be some minimal generator, MATH, of MATH for which the result of the reduction of the NAME of MATH and MATH is MATH. The only binomial that can be used in the reduction is MATH, and hence there exists MATH such that MATH and MATH is the least common multiple of MATH and MATH. If MATH, then MATH. Since MATH and MATH have no common variables, we get that MATH which contradicts MATH being a minimal generator of MATH. So we must have MATH and MATH is the least common multiple of MATH and MATH. But this implies that MATH is a multiple of MATH and hence in MATH, which is a contradiction. Therefore, this case cannot occur and we conclude that MATH. We now show the reverse inclusion. Suppose MATH is a minimal generator of MATH not equal to MATH, and MATH is the corresponding Graver binomial with MATH. We may assume that MATH is a multiple of MATH, as otherwise it is a generator of MATH, and thus in MATH automatically. Write MATH, where MATH. Suppose that MATH. Then MATH, so we must have MATH and MATH to preserve MATH being a Graver binomial. But then MATH, contradicting MATH being a minimal generator of MATH. Thus MATH, and so there is some MATH with MATH a minimal generator of MATH such that MATH. This means that MATH, and so MATH because MATH. Any monomial in MATH is in MATH, so we conclude that MATH.
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math/9912014
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Since MATH is MATH-homogeneous, MATH is the initial ideal of MATH if and only if MATH is a MATH-graded ideal. But by REF MATH is an initial ideal of MATH, so is MATH-graded exactly when MATH is.
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math/9912014
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Suppose MATH is a flippable binomial for a monomial MATH-graded ideal MATH such that MATH and MATH. If MATH then MATH and if MATH then MATH both of which are contradictions. To see that the converse is false, consider MATH which has REF monomial MATH-graded ideals all of which are coherent. The binomial MATH, but neither MATH nor MATH lies in MATH.
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math/9912014
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The dense torus in MATH is MATH, and the action of this torus on MATH is by coordinate-wise multiplication. It thus follows that if MATH and MATH are in the same torus orbit, they have the same support. Suppose MATH have the same support. If this support is the entire set MATH, then define MATH. Then if MATH is a binomial in MATH, MATH, so MATH is in MATH, and so MATH and MATH are in the same torus orbit. Suppose now that MATH and MATH have the same support MATH. Since MATH and MATH are in MATH, this means that there is no binomial in MATH of the form MATH where MATH and MATH. This is because if such a binomial were in MATH, we will have MATH for MATH, and MATH for MATH, which contradicts MATH. This means that there is no affine dependency between MATH and MATH. But this implies that MATH is a face of MATH, and if MATH, then MATH. This means that MATH and MATH lie in an invariant toric subvariety, and so by a similar argument to above are torus isomorphic.
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math/9912014
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It is straightforward to see that MATH lies in every irreducible component of MATH in which MATH does (this follows from the fact that MATH is a flat MATH module). All points in the MATH-torus orbit of MATH have the same support, and thus lie on the same ambient torus orbit by REF .
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math/9912014
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Let MATH be the ideal corresponding to a vertex MATH of MATH. The orbit of MATH under the ambient torus corresponding to MATH is just the ideal MATH. By REF the MATH-torus orbit of MATH is contained in any ambient torus orbit, so MATH is MATH-torus fixed as well, and thus is a monomial ideal. For the other implication, let MATH be a monomial MATH-graded ideal corresponding to a point MATH in MATH. As a point in MATH, MATH is invariant under any scaling of its coordinates in any fashion, and thus is invariant under any ambient torus action. It thus corresponds to a vertex of MATH.
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math/9912014
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Let MATH and MATH be the two initial ideals of MATH, and let MATH be the universal NAME basis of MATH. The set MATH contains a reduced NAME basis for MATH with respect to a term order for which MATH is the initial ideal, and so there exist binomials MATH with MATH for which MATH is a minimal generator of MATH, MATH. Suppose for all such binomials we have MATH. Then MATH is an inclusion of distinct monomial MATH-graded ideals, which is impossible. So we conclude that there is some binomial MATH with MATH, MATH and MATH. Suppose there is some other binomial MATH with MATH. Without loss of generality we may assume that MATH and MATH. We note that MATH, as by REF the two binomials MATH and MATH are Graver binomials, and they must be distinct since MATH is the universal NAME basis of MATH. We can thus find a supporting hyperplane for MATH, which intersects the cone only at the origin. This implies the existence of a vector MATH which satisfies MATH and MATH. Let MATH. Then MATH, and MATH, so MATH, and MATH. This means that MATH has a third initial ideal, which contradicts our assumption, and so we conclude that MATH is the only binomial in MATH. Pick MATH. Define MATH, and MATH for MATH. Then MATH is in the desired form.
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math/9912014
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Suppose MATH and MATH are connected by an edge MATH. Let MATH be the ideal corresponding to a point MATH in the relative interior of MATH. By REF the MATH-torus closure of MATH is contained in MATH. Thus MATH has at most two initial ideals. If MATH had only one initial ideal, it would be a monomial ideal and thus corresponds to a vertex of MATH, by REF . We thus conclude that MATH has exactly two initial ideals, MATH and MATH, corresponding to MATH and MATH respectively. Now by REF MATH is MATH-torus isomorphic to MATH, where MATH and MATH. Since MATH is MATH-graded, MATH is a Graver binomial. Because MATH has initial ideals MATH and MATH, it is their wall ideal MATH, and so MATH and MATH are connected by a flip over MATH. Conversely, suppose MATH and MATH are connected by a single flip. Then there is an ideal MATH which has as its two initial ideals MATH and MATH. Let MATH be a MATH-graded ideal which is isomorphic to MATH under the ambient torus corresponding to MATH. Let MATH be a minimal generator of MATH, with MATH, and MATH the corresponding Graver binomial with MATH. Then MATH, and thus MATH, as the ambient torus action preserves the monomials in a MATH-graded ideal. So MATH contains all minimal generators of MATH and MATH except MATH and MATH. Suppose MATH has a minimal generator MATH, where MATH is a Graver binomial, MATH, and MATH. Without loss of generality we may assume that MATH. If MATH then MATH by the definition of MATH, and thus also MATH. We thus conclude that MATH. But this means there exist MATH, MATH, such that MATH and MATH are minimal generators of MATH. Since MATH and MATH have disjoint support, we cannot have MATH, so at least one of MATH and MATH is in MATH. But this means at least one of MATH and MATH is in MATH, giving a contradiction. Hence the only binomial minimal generator of MATH is of the form MATH, so as in the proof of REF MATH is MATH-torus isomorphic to MATH. We thus see that all ambient torus closures of MATH are the same as the MATH-torus closure, and so MATH and MATH are connected by an edge.
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math/9912014
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It suffices to show that the reduced scheme MATH is connected if and only if the flip graph of MATH is connected. Since passing to an initial ideal is a flat deformation, each irreducible component contains a monomial MATH-graded ideal. It thus suffices to show that all monomial MATH-graded ideals lie in the same connected component of MATH if and only if the flip graph is connected. The ``if" direction follows from the fact that if MATH and MATH are connected by a single flip, then they are both initial ideals of a single wall ideal MATH, and so lie in the same connected component of MATH. The ``only-if" direction follows from REF , which imply that the flip graph restricted to an irreducible component of MATH is the edge skeleton of a polytope whose vertices are the monomial MATH-graded ideals in that component, and so is connected. As the intersection of two irreducible components of MATH contains a monomial MATH-graded ideal by NAME deformation, this means that if MATH is connected, the flip graph of MATH is connected.
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math/9912014
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By computing the MATH-degree of both terms in each binomial of MATH, it can be seen that MATH is a subset of MATH. It can also be checked that for each binomial in MATH, the positive term is the leading term with respect to MATH. Hence MATH is contained in the initial ideal of MATH with respect to MATH and no generator of MATH is redundant. The monomial ideal MATH will equal MATH if MATH is the reduced NAME basis of MATH with respect to MATH. Consider the elimination order MATH refined by the graded reverse lexicographic order MATH on the first block of variables and the weight vector MATH on the second block of variables. Then the reduced NAME basis of MATH with respect to MATH is the intersection of the reduced NAME basis of MATH with respect to MATH with MATH (see REF). By a laborious check it can be shown that the reduced NAME basis of MATH with respect to MATH is MATH .
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math/9912014
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We will show that the binomials in MATH are flippable for MATH while those in MATH are not. In order to show that a binomial MATH is flippable for MATH we need to show that every MATH-polynomial (monomial in our case) formed from the binomial MATH (with MATH as leading term) and a minimal generator MATH of MATH different from MATH reduces to zero modulo MATH a minimal generator of MATH. We first consider MATH. A binomial MATH in MATH can form a non-trivial MATH-pair (MATH-monomial) with REF MATH, REF MATH, REF MATH, REF MATH, REF MATH, REF MATH, REF MATH and REF a monomial MATH from MATH such that MATH. The remaining generators of MATH (except MATH itself) are relatively prime to MATH and so the MATH-pairs formed reduce to zero by NAME 's first criterion. We consider each case separately. CASE: The MATH-monomials formed from MATH and MATH are MATH, MATH, where MATH if MATH and MATH if MATH. CASE: If MATH, MATH is a multiple of MATH. CASE: If MATH, MATH reduces to zero modulo MATH. CASE: The MATH-monomials formed from MATH are MATH, MATH all of which lie in MATH and hence reduce to zero modulo MATH. CASE: The MATH-monomials between MATH and MATH are MATH for MATH. If MATH, MATH is a multiple of MATH, and if MATH then MATH is divisible by MATH. CASE: The MATH-monomials from MATH are MATH for MATH, all of which reduce to zero as in REF . CASE: The monomial MATH gives MATH for MATH, all of which reduce to zero modulo MATH. CASE: From MATH we get MATH, MATH, all of which are multiples of MATH. CASE: The MATH-monomials from MATH are MATH which are also multiples of MATH for MATH. CASE: For this last case, suppose first that MATH. Then MATH and the MATH-monomial between MATH and MATH is MATH which is a multiple of MATH. If MATH, then the MATH-monomial is MATH which is divisible by MATH since MATH and hence MATH. Similarly, one can check that the binomials in MATH are all flippable for MATH, which shows that MATH has at least MATH flippable binomials. To finish the proof, we argue that no binomial in MATH is flippable for MATH. CASE: The MATH-binomial between MATH and MATH is MATH which is not divisible by any generator of MATH. CASE: The binomials MATH and MATH form the MATH-binomials MATH and MATH respectively with MATH. None of them can be divided by a minimal generator of MATH. CASE: The MATH-binomial of MATH and MATH is MATH which does not lie in MATH. Hence MATH has exactly MATH flippable binomials.
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math/9912014
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The same proof as in REF shows that the generators of MATH form a NAME basis with respect to MATH with initial ideal MATH, for every choice of scalars MATH from the underlying field MATH. REF proved this claim for the case MATH, for a MATH and MATH for all MATH. Since MATH is MATH-graded, the MATH-homogeneous ideal MATH is also MATH-graded for every choice of scalars MATH. Hence there is an injective polynomial map from MATH, such that MATH maps to the point on MATH corresponding (uniquely) to MATH. Since MATH is irreducible, the image of this map lies entirely in one irreducible component of the toric NAME scheme MATH and the dimension of this component is at least MATH.
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math/9912014
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The implication MATH is immediate in the first statement so we need only show that MATH implies MATH. Suppose MATH. Let MATH be a minimal generator of MATH. Then either MATH is a minimal generator of MATH, MATH, or MATH for some MATH. In each case MATH, so MATH. If the containment is proper, REF gives a proper containment of NAME ideals of triangulations of MATH, which is not possible. So we conclude MATH. For the second statement, suppose MATH is not a circuit. Then there exists a circuit MATH with MATH, and MATH where at least one of these inclusions is proper. Since MATH, we must have MATH, and thus MATH. This implies MATH, and so, since we know MATH, MATH. This means MATH, and so MATH. But this means, as above, that MATH, which in turn implies that MATH, contradicting the hypothesis.
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math/9912014
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MATH is the only minimal non-face of MATH, so to show that MATH is a subcomplex of MATH, we need to show that MATH is the only minimal generator of MATH with support in MATH. Suppose MATH is a minimal generator of MATH, with MATH. Then there is some MATH such that MATH. Write MATH, where MATH, and MATH. If MATH, then MATH does not divide MATH and so MATH is in the wall ideal MATH. We can choose MATH with MATH so that MATH for some MATH. Since MATH, it follows that MATH, because MATH. So MATH, and there is thus some MATH such that MATH. This implies that MATH. But, by REF , this means that MATH, which in turn implies that MATH, contradicting our hypothesis. So MATH, and thus MATH. This shows that MATH is the only minimal generator of MATH with support in MATH, as required. From this we conclude that MATH is a subcomplex of MATH. We now show that every maximal simplex MATH has the same link. We do this by showing that any simplex not in the link of one maximal simplex of MATH is not in the link of any other maximal simplex of MATH. Suppose MATH is not a simplex in the link of a maximal simplex MATH of MATH, where MATH for some MATH and MATH. Then MATH, because MATH is not a face of MATH, and so there exists MATH, and MATH a minimal generator of MATH with MATH, such that MATH. Write MATH, where MATH, MATH, and MATH. Choose MATH with MATH such that MATH for some MATH. Then because MATH, we have MATH, and so MATH is in MATH and thus in MATH. So MATH. Let MATH be another maximal simplex of MATH, so MATH for some MATH. Then MATH, and so MATH, and thus MATH is not in the link of MATH in MATH. This shows that every maximal simplex MATH has the same link, as required.
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math/9912014
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If MATH then MATH by REF , and so MATH. Suppose MATH. Then REF implies that MATH is a circuit of MATH. By REF MATH is a subcomplex of MATH with each maximal simplex of MATH having the same link in MATH. It remains to show that MATH is the result of performing a bistellar flip on MATH. Let MATH be the result of performing the bistellar flip on MATH over MATH, and let MATH be the NAME ideal of MATH. We claim that M is the squarefree monomial ideal generated by MATH, all the generators of MATH except MATH, and also all monomials of the form MATH, such that MATH, and MATH is not in the link of the maximal simplices of MATH. Let MATH. Then MATH is a face of MATH exactly when either MATH is a face of MATH and MATH, or MATH, where where MATH, and MATH is in the link of the maximal simplices of MATH. This means that MATH is not a face of MATH exactly when either MATH, or MATH is not a face of MATH and MATH for any MATH and MATH in the link of the maximal simplices of MATH. This proves the claim. We now show that MATH. Let MATH be a minimal generator of MATH such that MATH is a minimal generator of MATH. If MATH is also a minimal generator of MATH, then MATH is in the square free ideal generated by all the generators of MATH except MATH, so MATH. Since MATH, the only case left to consider is MATH for some MATH with MATH. Write MATH, where MATH, MATH, and MATH. Choose MATH so that MATH for some MATH, where MATH for some MATH. Since MATH is a minimal generator of MATH different from MATH, it is in MATH. It thus follows that MATH, and so, because MATH, we have MATH and thus in MATH. Since MATH, MATH and thus MATH is not in the link of the maximal simplices of MATH. Because MATH, this means MATH, and therefore MATH. Now because MATH and MATH are both triangulations of MATH, this inclusion cannot be proper. So MATH, and thus MATH is the result of performing a bistellar flip on MATH.
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math/9912014
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For the matrix MATH, the monomial ideal MATH is MATH-graded. The flippable binomials of MATH are MATH and MATH. In this example, there are REF monomial MATH-graded ideals in total and the flip graph of MATH is connected. Consider the monomial ideal MATH and a degree MATH. All the monomials in MATH of MATH-degree MATH that are divisible by at least one of MATH are in MATH by construction. Among the monomials in MATH of degree MATH (there is at least one such since MATH), there is precisely one that is not in MATH and hence not in MATH and hence MATH is MATH-graded. If MATH is flippable for MATH then MATH minimal generator of MATH, MATH. The only non-trivial MATH-pairs that are produced during this calculation are those between MATH and a monomial minimal generator MATH of MATH. Since the resulting initial ideal equals MATH, it follows that MATH minimal generator of MATH, MATH and hence MATH is flippable for MATH. So MATH must be one of the three flippable binomials of MATH. Additionally, each of the minimal generators MATH of MATH provides a flippable binomial and hence MATH has MATH flippable binomials.
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math/9912015
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One has to show MATH . NAME if the verification is a bit lengthy, we insert it here, for sake of completeness. We have a complex MATH, sum of two complexes MATH, with maps MATH. We denote by MATH. Then we define MATH, with differential MATH, and MATH on MATH given by MATH, that is concretely MATH. Now take local sections MATH and similarly for MATH replaced by MATH. So here, MATH is MATH whereas MATH is the natural embedding. In order to simplify the notation, we omit the MATH from the notations (but not from the computations). One has MATH thus REF is satisfied. One has MATH thus REF is satisfied. One has MATH thus REF is satisfied. One has MATH thus REF is satisfied. One has MATH thus REF is satisfied. One has MATH thus REF is satisfied. One has MATH thus REF is satisfied. One has MATH . The degree of MATH in MATH equals the degree of MATH in the complex MATH plus REF. Thus REF is fulfilled. One has MATH thus REF is satisfied. The homotopy is defined as follows MATH .
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math/9912015
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The NAME sequence can be written MATH . Here MATH is interpreted as infinitesimal symmetries of the bundle MATH. A connection MATH relative to MATH gives rise to an action of MATH, that is, a lifting MATH of the inclusion MATH. The corresponding sequence for infinitesimal symmetries of the projective bundle MATH looks like MATH . Since symmetries of the vector bundle give rise to symmetries of the projective bundle, we get a diagram REF MATH . In particular, the bottom line represents the pushout of the NAME extension to the tracefree endomorphisms. On the other hand the composite of the vertical arrows gives rise by adjunction to a map MATH which is dual to MATH in REF. It follows that the reduction of structure of the tracefree NAME class defined by the connection MATH is given by the bottom line in the diagram MATH . Given the relation between the map labelled MATH above and the map MATH in REF, it is straightforward to check that this bottom line is obtained from REF by dualizing and pushing forward.
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math/9912015
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From the exact sequence MATH the splitting principle on MATH and MATH, and the compatiblity of the products, one just has to see that MATH for MATH. To this end, we will define a quasi-isomorphism MATH . CASE: We write the term in degree MATH in MATH as MATH. CASE: The exact sequence REF defines a decreasing filtration on the exterior powers of MATH, with MATH generated by wedges with at least MATH entries in MATH. MATH. By projecting MATH we find a canonical element MATH. Taking powers gives MATH. Since MATH, multiplication by MATH gives a map as in the statement of the lemma. To see this map is an isomorphism, we argue by induction on MATH. Consider the diagram MATH where the top row comes from the short exact sequence of sheaves MATH . Suppose for a moment we know the lemma is true when MATH. It follows that the maps MATH in the above diagram are isomorphisms in all degrees. It follows that the maps MATH are surjective and the maps MATH are zero. By induction the maps MATH are isomorphisms for all MATH, so it follows that the middle vertical arrow is an isomorphism as desired. It remains to prove the lemma when MATH. One proves by descending induction on MATH that MATH . We have MATH, so the assertion is clear when MATH. The induction step comes from consideration of the long exact sequence MATH noting that MATH when MATH and is zero otherwise. Details are omitted. To finish the proof of REF , it suffices to note that the maps in the lemma are compatible with the NAME differentials, so we get quasi-isomorphisms in the derived category, which gives REF. The desired decomposition REF follows by taking cohomology on MATH.
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math/9912015
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Let MATH, and let MATH. Let MATH and MATH be the natural maps. Define MATH by restricting the cone of complexes of sheaves used to define MATH to MATH. One has a pullback map MATH and MATH. Finally, in degree MATH we have MATH . Compatibility of the connections on MATH and MATH leads to a commutative diagram of sheaves on MATH . Here of course we use the notation MATH, MATH etc to indicate in which projective bundle we consider the construction REF of MATH. Exterior powers of the maps labelled MATH enable one to define MATH . Commutativity of the square labelled MATH above implies, using REF that MATH. Finally, regarding the kernel of MATH in degree MATH, we show here that it is a subgroup of MATH. Once we describe the NAME homomorphism below, it will be clear this kernel is nonzero. Since MATH has codimension MATH, purity results for local cohomology imply MATH . Further we have MATH . It follows that the local cohomology of MATH, which maps onto the kernel of MATH, is a subgroup of MATH as claimed. Let MATH be the inclusion. There is defined a NAME map MATH . The image MATH and MATH maps to the cycle class of MATH in MATH. There is a projection formula MATH . We have a commutative diagram MATH . The left column leads to maps MATH . On the other hand, NAME duality theory, CITE, gives MATH where MATH is the NAME local cohomology sheaf. Composing these arrows, we get MATH . An elaboration on this construction, using the middle column of the previous commutative diagram and the analogous map on MATH-sheaves MATH gives a map of complexes MATH . Finally, using purity, we find MATH . In the special case MATH, we find MATH . Composing with the map MATH, we see that the above map is injective, and in fact is an isomorphism, taking REF to the class of MATH. Finally, the projection formula is a straightforward consequence of the multiplication on the complex MATH, which makes MATH a MATH-module. The proof of the NAME REF is now straightforward. Write MATH for the polynomials associated to MATH and MATH. We know that MATH so necessarily MATH. Also, MATH, so MATH . Since MATH is the unique monic polynomial in MATH of degree MATH vanishing in MATH, we conclude MATH.
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math/9912015
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A connection defines an infinitesimal action of vector fields over MATH on MATH, that is, a splitting of the NAME sequence REF. Thus, a relative connection leads to a diagram MATH . The descent comes by defining MATH .
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math/9912015
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We start with the resolution MATH which we filter MATH . This filtration is compatible with the differential, and (using MATH), we find MATH . The complex MATH is obtained by tensoring the NAME resolution MATH with MATH and is thus exact. It follows that MATH is a resolution of MATH. Taking MATH, we obtain the desired resolution of MATH.
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math/9912015
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Briefly, one has diagrams of complexes of sheaves on MATH and on MATH . Using NAME resolutions and Cousin complexes, one defines a trace map MATH which yields a commutative diagram MATH .
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math/9912021
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It is enough to consider the case when MATH. Suppose first that MATH where each MATH is an integer. Then MATH takes either MATH or MATH. Conversely, suppose that all the MATH with MATH are real. Then MATH equals MATH and MATH for all MATH. This implies that , for each MATH, MATH. Since MATH is semisimple and the MATH forms a basis of MATH, necessarily MATH with MATH and MATH. This proves the first part of the statement in REF . We note that since all the MATH are real then MATH is also real for any root MATH which is not necessarily simple. Therefore each MATH stabilizes all the root spaces of MATH in MATH. Since MATH is also stabilized, we obtain that MATH and thus MATH. Clearly MATH satisfies MATH where MATH. This is because MATH is representable by a diagonal matrix with entries of the form MATH. Moreover, at least one diagonal entry must be equal to MATH.
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math/9912021
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It was shown in REF that the elements MATH with MATH such that MATH is real for all MATH are those for which MATH has the form MATH with MATH as in Notation REF, and MATH are integers. Moreover as in REF we also have that MATH is real for any MATH exactly when MATH for some integers MATH . Therefore all the root spaces of MATH are stabilized under the adjoint action of MATH. Since clearly MATH is also stabilized, then MATH stabilizes all of MATH and this implies that MATH. In fact this shows that MATH if and only if MATH has the form MATH for certain MATH integers. Thus REF and these remarks compute the intersection MATH. From here it is easy to conclude.
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math/9912021
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Let MATH denote the representation of MATH on MATH. Then we have MATH. If MATH is odd, MATH is isomorphic to MATH. Denote MATH with MATH as in REF . we set MATH when MATH is even and MATH when MATH is odd. Let MATH. We have that for each MATH, MATH for MATH. The MATH generate the group MATH with MATH elements where MATH and now the group MATH is a subgroup of MATH and it is isomorphic to MATH for MATH even and to MATH for MATH odd. What remains is to verify that MATH. Let MATH. By the NAME decomposition of MATH and MATH , it suffices to show that MATH, MATH. The right side, MATH is either MATH or MATH according to the parity of MATH. Recall that the maximal compact NAME subgroup MATH of MATH acts transitively on the set MATH of all maximal abelian NAME subalgebras MATH which are contained in the vector space MATH. The action of MATH is also transitive on MATH (by REF) and thus for any MATH there is MATH such that MATH where MATH is the isotropy group (in MATH) of an element in MATH, for instance of the element MATH. However this isotropy group MATH has been computed implicitly in the proof of REF and MATH. Therefore MATH.
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