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math/0004098
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It is given that both MATH and MATH satisfy REF relative to MATH, and further that MATH. Equivalently, by REF , MATH. Using MATH, we get MATH . Now multiplying through from the right with MATH on both sides in REF , the conclusion of the lemma follows. To see this, notice first from REF - REF that MATH . The proof of REF is based on the observation that the representation MATH REF is permutative, see CITE. Specifically, MATH if MATH, and MATH, and MATH. The desired REF now follows from this and MATH, since MATH is relatively co-invariant for the representation MATH by assumption.
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math/0004098
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Let MATH. Suppose MATH satisfies MATH, or equivalently MATH. Then by REF , we have MATH . The MATH-indices are in MATH. For the matrices MATH, we have the identities MATH . See REF above. The terms on the left-hand side in REF are MATH again with the convention that the terms are defined to be zero when the subscript indices are not in the prescribed range. If MATH, consider the lexicographic order on the subscript indices of the corresponding matrix entries MATH (in REF ). The range on both indices MATH is MATH where MATH is determined as in REF , see also REF . Then pick the last (relative to lexicographic order) nonzero MATH, that is, MATH are determined such that MATH . It follows that there are only the following possibilities for this MATH: MATH . If MATH, then, using REF , we arrive at the matrix identity MATH and therefore MATH where MATH is the first canonical basis vector in MATH. By REF , and REF , we conclude that MATH for all MATH, and therefore MATH for some MATH and MATH; see REF . If MATH, then, using again REF , we arrive at the matrix identity MATH and therefore MATH. Using again REF , and REF , we conclude that MATH for all MATH, and therefore MATH for some MATH and MATH. The reason for why the matrix of MATH has diagonal form relative to the natural NAME basis is as follows: Let MATH, for simplicity. (The argument is the same, mutatis mutandis, in the general case.) Then pick the last term MATH, MATH, with MATH, where again ``last" refers to the lexicographic order of the matrix-entry indices, see REF . We then get, using REF , the following matrix-identity (where we specialize to MATH): MATH and MATH, as mentioned. This forces MATH, which is impossible by REF , since MATH is positive and MATH is not. Let MATH, and suppose MATH, that is, assume that MATH, and MATH if MATH or MATH, referring to the lexicographic order. Then by the same argument which we used in the earlier cases, MATH . But all the double indices MATH of the matrix on the right are strictly bigger than MATH in the lexicographic order, and we conclude that MATH . The formula for MATH then yields MATH. Moreover, the additional restrictions are: MATH as in REF , and MATH . Now substituting MATH, we arrive at MATH and MATH . Since MATH is a projection, and MATH, we must have MATH and MATH and MATH. The conclusion of the theorem can then be checked case by case, using REF . In general, let MATH be the last (in lexicographic order) nonzero term, and assume MATH. Then by REF - REF , we get MATH, and therefore MATH as before. Using this, the equation for the MATH term is then MATH . If MATH, then all the entries in the matrix on the right must vanish, and we get MATH. If not, we proceed as in REF . If MATH, we go to the MATH term, viz., MATH . Eventually the matrix on the right will have nonzero terms, starting with MATH, and terms before that in the lexicographic order. Suppose, for example, that the matrix on the right in REF has nonzero entries. Then the equation for the MATH term is MATH and the argument is done by a case-by-case check, using that the coordinates MATH are in MATH while MATH. There is a similar argument, based on the reversed lexicographic order, starting with MATH, which will account for a possible lower right matrix corner of diagonal form. This completes the proof of the theorem.
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math/0004098
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To show that a subspace MATH is minimal in the sense specified in the lemma, we must check that whenever MATH is a subspace satisfying MATH then MATH cannot be cyclic for the representation MATH of MATH, that is, it generates a cyclic subspace which is a proper subspace of MATH . The cyclic subspace generated by MATH is the closed subspace spanned by MATH and all multi-indices MATH. This follows from REF , and we will denote this space MATH . We will prove the assertion by checking that if REF - REF hold, then there is a MATH such that MATH is contained in the closed span of MATH. Note that this integer MATH might be negative, and also that MATH might well be a proper subspace. Now let MATH be given subject to REF - REF , and suppose (as in the lemma) that MATH. Let MATH, MATH. Then by REF there are MATH with MATH or equivalently, MATH . Using CITE, CITE we conclude that MATH after adjusting with a constant multiple, and moreover that MATH . Setting MATH, this may be rewritten as MATH . Now pick MATH such that MATH, and apply the operators MATH, MATH, to both sides in REF . It follows that there is some MATH such that MATH . Since MATH, the first part of REF implies equality in a NAME inequality. Then REF yields MATH, or equivalently, MATH . Using the formula in REF for the coefficients in MATH, and REF , we note that REF implies MATH and in particular MATH, where MATH is as in REF . For the convenience of the reader, we sketch the argument, but only in the simplest case MATH. With the notation of REF , we get MATH where MATH is the (fixed) number determined from REF as described. Introducing the projections MATH from REF , the first part of REF then reads MATH and since MATH . Hence MATH is in the range of MATH, and MATH. We get MATH and by induction, MATH and therefore MATH, and MATH. This proves REF . Since we wish to prove that MATH is contained in MATH for some MATH, where MATH is the NAME space in MATH, we may assume that MATH in REF is taken as MATH. Since the invariant subspaces for a representation MATH of MATH are the same as for MATH where MATH, we may replace MATH with MATH, or equivalently, reduce to the special case MATH of REF . Since the matrix of the basis permutation MATH is in MATH, the same argument leaves us with the simpler case MATH, or equivalently, MATH . Then CITE implies MATH, or equivalently MATH in REF , and so MATH. Since MATH, we have proved that MATH is not cyclic, that is, the representation on the single vector MATH does not generate all of MATH. This concludes the proof of the lemma.
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math/0004098
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The two vectors MATH and MATH, corresponding to the endpoints in MATH, are special in that MATH and when MATH divides MATH, MATH . It follows that, when REF holds, then MATH where MATH the one-dimensional space of the constants, and neither of the two subspaces in this orthogonal sum is zero. Hence REF implies that MATH is not cyclic. The same argument proves that MATH is not cyclic if REF holds. We also note that REF holds if and only if MATH or equivalently, MATH . (The issue is resumed in REF below.) We now turn to the converse implications in the corollary, doing the details only for MATH . If REF does not hold, then MATH satisfies MATH . Using the following formula, MATH we therefore have the conclusion: For each MATH, MATH is in MATH, and it splits according to the sum MATH as follows: MATH where MATH is computed according to REF . Applying MATH to both sides in REF , we conclude that MATH or equivalently, MATH . Since in the second part of the corollary, MATH by assumption, we conclude from REF that MATH, and the inclusion MATH follows. Finally, we get MATH which proves that the reduced space MATH is then cyclic, and the corollary is established.
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math/0004098
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Once MATH has been chosen as in REF, the three REF - REF follow from REF . The significance of REF - REF is that they imply that if MATH then the fixed-point set MATH is in fact an algebra. This is a result of NAMECITE. Using REF , we conclude that the projections in MATH are characterized by the condition of REF . Now, by CITE, there are projections MATH such that, for each MATH, we have the covariance properties MATH where MATH, or equivalently, MATH and in addition, we have MATH and each subspace MATH irreducible, in the sense that each MATH reduces the representation MATH to one which is irreducible on the subspace. It follows from REF - REF that the complementary projection MATH then also satisfies REF , and so in particular MATH must commute with each MATH (MATH). Then by REF , we conclude that each MATH has a matrix which is diagonal with respect to the NAME basis MATH. Since the loop MATH is picked to be purely non-diagonal, we finally conclude that the decomposition MATH of REF can only have one term, and the proof is concluded.
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math/0004098
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We will restrict to the case MATH, although for MATH, we cover arbitrary MATH in CITE. (If MATH, then MATH. Setting MATH, we showed in CITE that the density matrix MATH given by MATH satisfies MATH, where MATH and MATH, and where MATH is the adjoint of MATH with respect to the trace inner product. Defining the state MATH on MATH by MATH we check that MATH satisfies REF . We know from CITE that MATH and MATH have the same dimension. But MATH is irreducible by REF when MATH. Hence MATH is one-dimensional by REF , and there are therefore no other states MATH satisfying REF . But the state MATH in REF is clearly not faithful, and the proof is complete, in the special case MATH.) We now turn to the details for MATH, MATH, and it will be clear that they generalize to arbitrary MATH. If MATH, MATH, we get MATH, and MATH may easily be computed; see, for example, the details in REF below, especially REF - REF . As a result, we get MATH, and therefore MATH where the MATH summations are both over MATH. In addition, by REF , MATH where MATH. So the complement of MATH in MATH is invariant under MATH, and the element MATH which is fixed by MATH must be diagonal in the NAME basis, by REF . Using REF and the argument from the previous step, we then check that a density matrix MATH may be found in the form MATH such that the state MATH on MATH will satisfy REF . But if MATH, the wavelet representation MATH is irreducible, and so REF has no other state solutions. Finally, it is clear from REF that MATH is not faithful.
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math/0004098
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REF : If MATH, then MATH, and therefore MATH that is, MATH. This means that the coefficient matrices in the expansion MATH satisfy the conditions in REF ; and, if we define matrices MATH by REF above, then it follows that MATH where MATH . Hence REF holds. CASE: This is clear from reading REF in reverse. The equivalence MATH follows from the observation that the following sum representation MATH holds for some MATH and MATH if and only if MATH . The conclusion can therefore be read off from the following general fact: MATH . REF : If MATH, then MATH for MATH, and conversely. This follows from the identity MATH which is part of the defining axiom system for MATH. Hence, the result follows from CITE, once we note that the only polynomials MATH such that MATH for all MATH are the monomials; see also CITE. CASE: We already showed in REF that REF implies the first of the conditions in REF . The second one then follows from REF , that is, the second line in REF follows from the first one. CASE: This follows from REF above. For if MATH, then it follows from REF that MATH and so MATH is not in the subspace MATH and the same argument, based on REF , shows that MATH is not in MATH concluding the proof of REF . CASE: We already saw that if MATH, then the conditions in REF hold if and only if MATH is not in the spectrum of the matrix from REF . Having now MATH from REF above, we can use the identity REF relating the MATH-numbers to the MATH-numbers. But MATH is in the spectrum of the matrix MATH if and only if MATH . The matrix on the right-hand side in REF is of this form, and MATH by NAME 's inequality. Here MATH . But using REF , we also get MATH. Now REF - REF yield MATH, and therefore MATH. Substituting this back into REF then yields MATH, which amounts to ``equality" in NAME 's REF ; and so the corresponding vectors REF are proportional. We already noted the equivalence of REF ; and we just established that the negation of REF amounts to linear dependence of the vectors in REF . So REF is equivalent to the linear independence, as claimed in REF . This completes the proof of the proposition.
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math/0004098
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The details are somewhat technical, and it seems more practical to first do them for the special case when MATH, and then comment at the end on the (relatively minor) modifications needed in the proof for the case when MATH is arbitrary MATH. Using the terminology of REF , we then get MATH where MATH and MATH, MATH, and MATH are MATH-MATH complex matrices satisfying MATH . When MATH is given, we denote that corresponding wavelet representation by MATH, or just MATH for simplicity. Recall MATH or simply MATH where MATH . As we saw in REF , the subspace MATH is then MATH-invariant, and also cyclic for the representation. But the issue is when MATH is minimal with respect to these two properties. The minimality of some subspace MATH then means that MATH is MATH-invariant, and cyclic, and that no proper MATH-invariant subspace of MATH is cyclic. In working out details on MATH, we use REF - REF in conjunction with REF - REF , and it is more helpful to work with the complex conjugates MATH and so MATH consists of polynomials of degree at most MATH. It follows from REF - REF that MATH is then spanned by the functions (polynomials) in the following list: MATH . Hence, by REF , MATH consists of the space spanned by the complex conjugates of these functions. By REF it is clear that MATH.
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math/0004098
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Now, the functions MATH in REF are the matrix elements of the loop MATH, and so it follows from REF that each of them is a polynomial of degree at most MATH, say MATH (since MATH itself has degree MATH when MATH). Hence, MATH or equivalently, MATH . Now set MATH where MATH is as in REF . From REF , we then get MATH or equivalently, MATH using REF . In view of REF , we need then only to compute the following: MATH or equivalently, MATH . Now putting the formulas together, we get the value of MATH on each of the functions REF which go into the definition of MATH, and the conclusion of the lemma follows.
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math/0004098
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Since both MATH and MATH are MATH-invariant, the conclusion will follow if we check the inclusion MATH . For the space on the right-hand side in REF , we shall use the terminology-MATH, and similarly, the space spanned by all the spaces MATH will be denoted MATH. In REF , we vary the multi-index MATH over all the MATH possibilities. It follows from the MATH-invariance of MATH (in REF ) and MATH (in REF ) that we get different families of nested finite-dimensional subspaces: MATH and a similar sequence for MATH. Since MATH is cyclic, we have MATH . But MATH, for some MATH, so MATH is also cyclic. The conclusion of the lemma follows from REF , once we check that MATH and so MATH works, and-MATH for all MATH. Turning now to the details: Since MATH is spanned by MATH, MATH, we must check that each of these basis functions has the representation MATH for MATH, where we refer to REF (see also REF ) for the characterization of the space MATH, or rather MATH. But REF is equivalent to the assertion that MATH for all MATH and MATH; and REF can be checked by a direct calculation, which is very similar to the one going into the proof of REF . Specifically, using REF - REF we get the following: MATH . Recall that the complex conjugates of the functions on the right-hand side in this list are precisely the ones from REF , or equivalently, REF . This proves REF , and therefore the cyclicity of MATH, which was claimed in the lemma. As a bonus, we get from REF - REF that the inclusion MATH holds if and only if MATH. To see this, use the fact (for MATH) that MATH .
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math/0004098
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We will establish the conclusion by proving that if MATH is any subspace of MATH which is both MATH-invariant and cyclic, then MATH. So in particular, MATH does not contain a proper subspace which is both MATH-invariant and cyclic. Now suppose that some space MATH has the stated properties. Since it is cyclic, we must have MATH satisfied for some MATH. As noted in the proof of REF , this is equivalent to MATH for all MATH, and all MATH. But we also saw in the proof of REF that the functions on the left-hand side in REF are precisely those which are listed in REF . Note that the functions in REF , or REF , are those given by MATH . But MATH by assumption, so for some MATH, we have MATH, and the calculation in the proof of REF , and in the previous two sections, then shows that the families of functions in REF are the same, that is, we get the same functions in REF for MATH as the ones which are already obtained for MATH in REF . This is the step which uses the assumption MATH. Since MATH is spanned by the vectors in REF , the desired inclusion MATH follows. More details are worked out in REF below.
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math/0004098
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The result in the theorem is now immediate from the three lemmas, and we need only comment on the size of the genus MATH. We argued the case MATH; but, for the general case, MATH is spanned by MATH, MATH, and the functions from the list REF , or equivalently REF , will then be MATH . Otherwise, all the arguments from the proofs of the lemmas carry over. See REF for more details.
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math/0004098
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The proof comes down to a dimension count. Since MATH is of dimension MATH, we just need to check that the space MATH (MATH), spanned by the MATH functions in REF , is of dimension MATH for a generic set of loops MATH in MATH, and that can be checked by a determinant argument based on the conditions for the matrices MATH defining MATH; see REF - REF above. The above-mentioned dimension count is based on the following consideration (which we only sketch in rough outline). A possible linear relation among the functions from REF takes the form MATH where the MATH summation indices are MATH, and the MATH summation is over MATH. As a result, we get the following system of relations: MATH for all MATH, and all MATH. Note that REF is a matrix multiplication. Using finally MATH we see that the dimension of the space spanned by MATH is MATH, as claimed. See REF for details.
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math/0004098
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For if not, the greatest common divisor MATH of the family REF would have a root MATH, that is, MATH. By REF , we would then have MATH where MATH is the originally given loop. Recall that, by REF , we may view MATH as an entire analytic matrix function, that is, an entire analytic function, MATH, whose restriction to MATH takes values in MATH. (These are also called inner matrix functions CITE.) But REF is impossible (for MATH) in view of REF and its corollary. We will give the details for MATH, but they apply with the obvious modifications to the general case of MATH. If REF holds, then by REF , MATH where we use the projections MATH in MATH from REF . Setting MATH, MATH, REF then yields the following estimate: MATH where the order MATH is that of positive operators on MATH. But this is impossible, since MATH. The latter is from the assumption MATH.
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math/0004098
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By REF , we have the estimate MATH . If MATH, then by REF , we get MATH . If MATH, then the value MATH is not a common root of the two complex polynomials MATH, MATH, and so by REF the two polynomials MATH and MATH have no common roots at all, by REF . Therefore, when REF are combined, we get MATH. Hence MATH, and the proof is completed.
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math/0004098
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CASE: In this example, MATH and MATH . Let MATH be the usual NAME in MATH, viewed as a column vector. By REF , we have MATH where MATH, MATH and MATH are projections, and as before, MATH. Here MATH and MATH, MATH are the MATH, MATH projections specified in REF (see also REF below). We need only check that MATH for some number MATH. But MATH . Substitution of MATH yields MATH . Hence MATH for some number MATH, and therefore, with MATH we have MATH . The result REF follows. REF follows upon solving MATH for some number MATH, in addition to the conditions in REF . But we only need to work out the next derivative, using REF to eliminate the cosine terms where possible: MATH where we used MATH . Noting again that the right-hand side of REF takes the form MATH, we arrive at MATH which directly gives the second part of REF . The specific solution is found by transforming REF into MATH . Squaring and using REF to eliminate MATH yields MATH which gives MATH by the quadratic formula, using MATH to choose the positive radical. Working this back through REF yields MATH and substitution of REF into REF then shows that MATH and MATH must have the same sign, so that the solutions stated in the proposition are the only ones possible, the numerically exhibited pair being those for which MATH and MATH are both positive.
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math/0004098
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If MATH is divisible by MATH, then its six coefficients MATH must be of the form MATH, and the associated loop MATH, MATH . The corresponding MATH-conditions then yield: MATH . Substitution of the second into the first yields MATH. Hence, of the three numbers MATH, at most one, and therefore precisely one, can be nonzero. But each of the three cases is determined up to scale, and REF decides the scale. We are therefore led to the three loops in REF , and REF then gives the three scaling functions MATH which are listed in REF . The cycle of the last line in REF is of order MATH. Let MATH (MATH). Then the cycle is MATH, and MATH, which is the MATH for the last line of REF . (It is from a root of MATH of order MATH REF for MATH.) The other length-MATH loops which would be possible are, as noted, MATH and MATH, with MATH. We claim that they do not in fact occur. If one of them did occur, then the corresponding MATH would be divisible by either MATH, or by MATH. But MATH, so the factorization would be MATH where MATH. (We have picked the normalization of MATH given by MATH for convenience.) From REF we note that the coefficients MATH are real. Divisibility by MATH means that MATH, MATH, MATH, and MATH are roots of MATH. So the complex conjugates MATH, MATH, MATH are also roots. But that would give us all seven points, MATH, as distinct roots of MATH, which is impossible since MATH is of degree at most MATH.
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math/0004100
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Let MATH be a NAME basis of the ideal MATH and MATH be a polynomial set which computed by the algorithm. Apparently, MATH. Consider MATH. If MATH is MATH-autoreduced, then MATH initiated as MATH in line REF does not change in the process of the algorithm, and, hence, MATH. Otherwise, consider the output polynomial sets MATH. Denote MATH by MATH. Then, by construction, MATH. Consider a monomial MATH and show that MATH. Let MATH be a MATH-divisor of MATH, that is, MATH, and MATH be such that MATH. If MATH, by REF of NAME division in REF , we are done. Let now MATH. The while-loop provides that MATH. If MATH with MATH, then from REF it follows that MATH. We have to prove that any variable MATH satisfy MATH. Consider two alternative cases: MATH and MATH. In the first case, by REF , both MATH belong to the same group MATH of monomials in MATH. It follows that MATH. In the second case, by REF , MATH and REF we find again that MATH is MATH-multiplicative for MATH as an element in MATH. Therefore, MATH is a NAME monomial basis of MATH. Thus, by REF , MATH is a MATH-basis. In so doing, every MATH-normal form computed in line REF apparently vanishes. This implies the inclusion MATH.
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math/0004100
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Let MATH be a minimal NAME basis. From REF it follows that MATH is MATH-autoreduced. Thus, MATH is MATH-autoreduced.
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math/0004100
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MATH: Suppose MATH which, by REF , is also a MATH-basis of MATH, is not its MATH-basis. Our goal is to prove that a NAME basis of MATH is an infinite polynomial set. From REF it follows that MATH . Among nonmultiplicative prolongations MATH satisfying REF choose one with the lowest MATH with respect to the pure lexicographical ordering MATH generated by REF . If there are several such prolongations choose that with the lexicographically lowest MATH, that is, with the lexicographically highest MATH. This choice is unique since MATH is MATH-autoreduced. We claim that MATH. Assume for a contradiction that MATH. Then from NAME involutivity REF we obtain MATH . In accordance with REF MATH contains MATH-nonmultiplicative variables for MATH and from REF it follows CITE that MATH. If MATH, then MATH, and both MATH and MATH belong to the same monomial group MATH appearing in REF . Hence, MATH in contradiction to REF . Therefore, MATH and REF can be rewritten as MATH with MATH. Show that MATH. If MATH we are done. Otherwise there is MATH such that MATH. Since MATH, our choice of MATH and MATH implies the existence MATH such that MATH and MATH. If MATH we are done. Otherwise we select again a MATH-nonmultiplicative variable for MATH occurring in MATH and rewrite the corresponding prolongation in terms of its MATH-divisor MATH. Continuity of NAME division CITE provides termination of the rewriting process with an element in MATH such that MATH is a MATH-divisor of MATH. Because MATH is MATH-autoreduced, by REF MATH is also a MATH-divisor of MATH. By this means there are two different NAME divisors MATH and MATH of the same monomial that contradicts REF and proves the claim. Let now MATH be a MATH-basis of MATH. Denote MATH by MATH and show that MATH. Suppose there is an element MATH, such that MATH with MATH. Then MATH and there is MATH satisfying MATH where MATH. Thus, there is a MATH-nonmultiplicative prolongation MATH with MATH such that MATH and MATH that contradicts the above choice of MATH and MATH. Now consider monomial set MATH. By REF , MATH and MATH is obviously the lexicographically lowest MATH-nonmultiplicative prolongation of elements in MATH. It is easy to see that MATH. Indeed, since MATH, it follows that MATH, and a MATH-divisor of MATH would also MATH-divide MATH that is impossible as MATH is MATH-autoreduced. Thus, we find that MATH. By sequential repetition of this reasoning for MATH we deduce that every such monomial is an element in MATH, and, therefore, MATH is infinite. MATH: If MATH is a minimal NAME basis of MATH, then REF implies MATH, and the above arguments show that either MATH is also a MATH-basis or the latter is infinite. MATH: This implication is obvious.
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math/0004100
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This follows from the above proof of REF .
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math/0004100
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MATH has a finite NAME basis CITE. If MATH is homogeneous, then its NAME basis is the reduced NAME basis CITE.
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math/0004103
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Recall the following formula: if MATH is any REF-form on a manifold, and MATH and MATH are vector fields, then MATH . See for example, CITE for an explanation. We choose local bundle coordinates on MATH and apply REF with MATH, MATH and MATH. Since MATH and MATH commute and MATH, we obtain MATH . Note that MATH and evaluate at MATH, MATH.
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math/0004103
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Assume to the contrary, that is, assume that for a general point MATH there exists a singular line MATH passing through MATH. Recall that the rational curve MATH can always be dominated by an integral singular plane cubic, that is, by a rational curve with a single node or cusp. We will reach a contradiction by constructing a section of the pull-back of MATH to the plane cubic which vanishes at a prescribed generically chosen point. This section will be constructed by a deformation of the singular curves. Because MATH was chosen to be general, there exists a singular (that is, cuspidal or nodal) plane cubic MATH and an irreducible component MATH such that the universal morphism MATH is dominant and such that for all MATH we have MATH. Fix a general morphism MATH and note that there is an open set MATH such that for all MATH, the tangent map of the restricted morphism MATH has maximal rank at MATH: MATH . Recall from CITE that the smooth points of MATH are in REF:REF-correspondence with line bundles of degree one, and fix a point MATH such that MATH. Next, let MATH be a Legendrian submanifold of MATH which contains MATH and is transversal to MATH at MATH. By REF , these exist in abundance. Since MATH has maximal rank, we can find a section MATH over MATH, that is, a submanifold MATH such that MATH is an isomorphism. By construction, MATH has dimension MATH and cannot be Legendrian. It follows that there exists a unit disc MATH with coordinate MATH centered about MATH such that MATH . For this, recall that the tangent vector MATH is canonically identified with an element in MATH. By choice of MATH, we have MATH. By choice of MATH, this is impossible, a contradiction.
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math/0004103
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It follows from the definition of the contact structure that MATH. Since MATH, and since vector bundles on MATH always decompose into sums of line bundles we may therefore write MATH where MATH. Thus, the splitting of MATH has exactly MATH positive entries. It follows that the splitting of MATH has at most MATH positive entries. By CITE, MATH is nef, and since MATH, the claim follows.
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math/0004103
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The fact that MATH and MATH are immersive follows from CITE and REF . It follows from an argument of NAME that MATH is birational because all lines through MATH are smooth. For this, see CITE.
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math/0004103
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Since MATH, it follows from the estimate REF for the dimension of the parameter space that MATH . By NAME 's bend-and-break argument CITE, for a given point MATH, there are at most finitely many lines containing both MATH and MATH. It follows that MATH . CASE: The subvariety MATH is MATH-integral where it is smooth. CASE: It follows immediately from NAME 's theorem and from the non-degeneracy of the contact distribution that MATH, and we are done. CASE: Let MATH be a general (smooth) point, MATH a curve which contains MATH and MATH and is smooth at MATH. By general choice of MATH, such a curve can always be found. Let MATH be a birational morphism with MATH and MATH. If MATH is an irreducible component of the reduced NAME which contains MATH, then we have that MATH where MATH is identified with MATH and the map MATH is an application of the tangent map MATH and evaluation at MATH. In other words, the tangent space MATH at MATH is spanned by the tangent space to the curve MATH and by sections of MATH which vanish at MATH. We refer to CITE for a proof of REF . In REF we have already seen that MATH so that it suffices to prove that MATH . In other words, we have to show that if MATH is any unit disk centered about MATH with coordinate MATH, then the section MATH associated with MATH is contained in MATH. For this, note that REF asserts that the section MATH has a zero at MATH whose order is at least two. But since MATH, this implies that MATH vanishes identically. In particular, MATH. This proves that MATH. Since MATH was chosen generically, MATH is MATH-integral where it is smooth, and we are done.
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math/0004103
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We have already seen in REF that MATH is equidimensional of relative dimension MATH. Thus, MATH is a well-defined family of algebraic cycles over MATH in the sense of CITE and the universal property of the NAME yields a map MATH such that MATH is the pull-back of the universal family over MATH. Because MATH, it is clear that the image of MATH is not a point. Use the assumption that MATH to obtain that MATH is actually a finite morphism. Because two reduced algebraic cycles are equal if and only if their supports agree, it follows that for a given point MATH, there are at most finitely many points MATH such that MATH . In particular, if MATH is an open set such that MATH is an embedding, then MATH has dimension MATH. Hence the claim.
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math/0004103
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Our argument involves an analysis of the deformations of MATH which are obtained by varying the base point MATH. We shall argue by contradiction and assume that the assertion of the proposition is wrong. With this assumption we will construct a family of morphisms MATH which contradicts REF , and we are done. We will now produce a map MATH to which REF can be applied. Assuming that the statement of the proposition is wrong, we can find an analytic open neighborhood MATH and a subbundle MATH such that CASE: For all MATH, the vector spaces MATH and MATH are perpendicular with respect to the non-degenerate form MATH which comes with the contact structure. CASE: All lines through MATH are smooth. After shrinking MATH, if necessary, let MATH be a nowhere-vanishing vector field. Thus, if MATH is any point and MATH is any line through MATH, then MATH where MATH again means: perpendicular with respect to the non-degenerate form on MATH. Let MATH be a unit disc with coordinate MATH and MATH be an integral curve of MATH with MATH. Now let MATH be the family of morphisms parameterizing the curves associated with MATH. Set MATH . If MATH is the universal morphism, then it follows by construction that MATH . In particular, since MATH is not MATH-integral, for a general point MATH there exists a tangent vector MATH such that the image of the tangent map is not in MATH: MATH . Decompose MATH, where MATH and MATH. Then, since MATH is MATH-integral, it follows that MATH and therefore MATH . As a next step, choose an immersion MATH such that MATH and such that MATH . In particular, if MATH is the section associated with MATH, and MATH, then the following holds: CASE: it follows from REF and from CITE that MATH is not identically zero. CASE: at MATH, the section MATH satisfies MATH. In particular, MATH. CASE: If MATH is a local coordinate on MATH about MATH, then it follows from REF that MATH and MATH are perpendicular with respect the the non-degenerate form. REF ensure that we can apply REF to the family MATH. Since the section MATH does not vanish completely, the proposition states that MATH has a zero of order at least two at MATH. But MATH is an element of MATH, and MATH is a line bundle of degree one. We have thus reached a contradiction, and the proof of REF is therefore finished.
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math/0004103
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By CITE, the canonical bundle MATH is not nef. But then it has already been shown in CITE that MATH is automatically of type MATH if MATH. We may therefore assume without loss of generality that MATH is NAME and that MATH. Let MATH be a general point, and MATH a minimal rational curve through MATH. It follows from the classical argument of NAME that MATH. Since MATH is a contact manifold, MATH so that MATH . If MATH . , then MATH. If MATH . , then REF shows that the contact distribution MATH is canonically defined, hence unique.
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math/0004104
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Consider the subset MATH which is dense in MATH. To see that MATH is a closed subspace of MATH, it will suffice to show that MATH is a subspace. Let MATH and MATH. Since MATH and MATH, it is clear that MATH and MATH. Moreover, since MATH it is clear that MATH, and hence also MATH, is invariant for MATH. To show that MATH it will be enough to show that MATH whenever MATH is a unitary operator on MATH such that MATH. Moreover, MATH will follow once we show that MATH for every MATH. But this holds because MATH .
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math/0004104
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If MATH then MATH so MATH while MATH. This shows MATH. Suppose MATH and MATH. Then MATH. As an operator on MATH, MATH is invertible and its inverse has spectral radius MATH. Therefore MATH and hence MATH, so MATH. This shows that MATH, and therefore that MATH.
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math/0004104
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Let MATH be given by REF and let MATH . Letting MATH denote the algebraic tensor product of vector spaces, we clearly have MATH, so the inequality MATH holds in REF. Given a unit vector MATH let MATH be MATH. Then MATH, so if MATH then MATH for every MATH. Therefore MATH and MATH holds in REF.
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math/0004104
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Let MATH be a NAME space and MATH a normal, faithful MATH - representation. Using CITE, one finds a NAME space MATH such that the representations MATH and MATH are unitarily equivalent, via a unitary MATH. Now applying REF twice, we have MATH . Hence MATH.
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math/0004104
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Write MATH . Let us first fix MATH, MATH and let us denote MATH simply by MATH, MATH by MATH, MATH by MATH and the diagonal entries of MATH by MATH, (MATH). Now each word in MATH and MATH is a sum of words in MATH, MATH, MATH and MATH; hence we will investigate words of the form MATH for arbitrary MATH . Now, using the independence of T and D, we see that MATH where we have used the convention MATH, where MATH is defined by MATH and where MATH . Using that the joint distribution of MATH is invariant under permutation of the MATH variables, we see that each moment MATH appearing in REF is equal to a moment MATH where MATH and by further permutation the moment REF corresponds to a unique moment of the form REF if we make the additional stipulation that MATH . Hence, rearranging the sum in REF we get MATH where the first sum is over all MATH satisfying REF , and the second sum is over all MATH such that there is a permutation, MATH, of MATH for which MATH . Let MATH REF denote the word on the right - hand - side of REF where MATH is taken to be MATH. Then MATH . Now in order to prove the lemma it will suffice to show that MATH. Since by REF , the difference of moments of MATH and MATH tends to zero as MATH, it will suffice to show that for every MATH and every MATH, the quantity MATH remains bounded as MATH, where the sum is over all MATH such that there is a permutation, MATH, of MATH for which REF holds. But this follows from the sort of counting arguments used by NAME in CITE. Indeed, for any MATH and for any MATH, MATH implies MATH. Taken together with the independence assumtion on the entries of MATH, this shows that for any MATH, a necessary condition so that MATH is that there be a bijection, MATH, from MATH to itself, without fixed points, such that MATH and MATH . Moreover, there is a constant, MATH, depending only on MATH, such that for all MATH and all choices of MATH, MATH . If MATH is a bijection of MATH, let MATH be the number of choices of MATH such that REF holds. There are only finitely many bijections, MATH, of MATH. Hence, in light of the bound REF , in order to show that REF is bounded as MATH, it will suffice to show that for each bijection MATH of MATH without fixed points such that MATH, the quantity MATH remains bounded as MATH. However, MATH, where MATH is the number of vertices in the quotient graph, MATH, which is obtained from the MATH - gon graph by identifying the MATH-th and MATH-th edges with opposite orientations, for every MATH. The graph MATH has MATH edges, hence at most MATH vertices, which shows that MATH and hence that REF remains bounded as MATH.
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math/0004104
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If MATH then let MATH. If MATH but MATH then let MATH. If MATH and MATH then let MATH .
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math/0004104
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We may without loss of generality assume that MATH is a transposition of neighbors: MATH . We will use REF to show that there is a unitary random matrix, MATH, such that MATH has the same distribution as MATH, and this will prove the theorem. Recall that MATH is the usual probability space underlying our random matrices MATH. For MATH let MATH be the block diagonal matrix MATH . By this we mean that MATH has MATH ones down the diagonal, then a MATH block that is the matrix MATH from REF , then MATH ones. Let MATH denote the MATH-th entry of the random matrix MATH. If we write MATH then MATH . Let MATH . If MATH then MATH and if MATH then MATH . In order to show that MATH and MATH have the same distribution, it is thus enough to show CASE: MATH is an independent family of complex MATH - Gaussian random variables; CASE: MATH and MATH are independent sets of random variables. From the facts that MATH are independent sets of random variables, each MATH is complex MATH - Gaussian and MATH is everywhere unitary and is independent from the right - hand set in REF, we see that MATH is an independent family of complex MATH - Gaussian random variables. Moreover, the joint distribution of the family REF is not changed by conditioning on the values of MATH. Hence MATH are independent sets of random variables. But this implies that REF hold.
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math/0004104
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We may introduce a uniformly distributed MATH - valued random variable, MATH, that is independent from MATH. Then MATH has the same distribution as the random matrix, MATH, which at a point MATH takes the value MATH . Now each MATH - moment of MATH is the average over MATH of the corresponding MATH - moments of the matrices MATH described in REF . By REF , each of these is in turn equal to the corresponding MATH - moment of MATH.
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math/0004104
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We may without loss of generality suppose MATH and MATH. That MATH follows from the last paragraph of CITE and the fact that the limit moments of each MATH are those of a centered semicircle law with second moment MATH. To show MATH we will use CITE and an idea from the proof of CITE. Let MATH be a nontrivial ultrafilter on MATH. On the algebra, MATH, of polynomials in noncommuting variables MATH and MATH, let MATH be the tracial linear functional defined by MATH, where MATH is the algebra homomorphism defined by MATH and MATH. Let MATH denote the subalgebra of MATH generated by MATH and MATH. We will check that the conditions MATH and MATH of CITE hold for the sequence MATH and the subalgebra MATH with respect to MATH. Note that every element of MATH is a linear combination of words of the form MATH and that MATH for some MATH. Moreover, MATH. Therefore MATH and hence condition MATH of CITE follows from the boundedness of REF as MATH. To see that REF-a of CITE holds, it suffices to see that if MATH and if MATH is such that MATH has only one element and if MATH then MATH . But this follows from the independence of MATH from all the other matrices appearing in REF and the fact that all entries of MATH have expectation zero. Now conditions MATH-b and MATH-c of CITE follow from REF . Therefore, by CITE, given an injection MATH and defining MATH the family of sets of noncommutative random variables, MATH is asymptotically free with respect to MATH as MATH. However, using the NAME of the entries of the MATH, we see that for every MATH, MATH has the same moments as MATH. Hence MATH is free with respect to MATH. Since MATH was arbitrary, and since each MATH converges in moments as MATH, this implies that MATH is asymptotically free as MATH.
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math/0004104
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We will apply REF . Let us first show that the quantity REF remains bounded as MATH. Write MATH for the MATH-th entry of MATH. We have MATH where MATH. Using the generalized NAME inequality, we have MATH . But by REF , there is MATH such that MATH . Now consider MATH . From the nature of the entries MATH, we see that the quantity REF can be nonzero only if there is a bijection MATH such that MATH, MATH has no fixed points and MATH . One also calculates MATH . Let us call a bijection, MATH, of MATH without fixed points and such that MATH, a pairing of MATH and for every pairing MATH let MATH . From the above estimates we obtain MATH where the sum is over all pairings, MATH of MATH. To each pairing we associate the quotient graph, MATH, of the clockwise oriented MATH - gon graph obtained by identifying with opposite orientation each pair of MATH-th and MATH-th edges. The resulting graph has MATH edges, hence at most MATH vertices. Consequently MATH, and the quantity in REF is bounded by MATH times the number of pairings, which is finite and independent of MATH. This shows that the quantities REF remain bounded as MATH. We now show that REF are satisfied. Let MATH be as described there. Then MATH where we let MATH. Consider the clockwise oriented MATH - gon graph, label the edges consecutively MATH and the vertices MATH so that the vertices of the edge MATH are MATH and MATH, (mod MATH). Let MATH be the quotient of the MATH - gon graph obtained by identifying edges MATH and MATH with opposite orientation, (MATH). The resulting identification of vertices of the MATH - gon graph gives an equivalence relation MATH on MATH whose equivalence classes MATH are precisely the lists of indices labeling the MATH vertices of MATH. The expression MATH in REF is nonzero if and only if whenever MATH and MATH then MATH, and then the value of REF is MATH. For each equivalence class MATH of MATH there is MATH so that MATH; for MATH let MATH be the MATH-th diagonal entry of MATH. Thus MATH . Using the NAME inequality estimate REF, we see that the terms MATH in REF are uniformly bounded in modulus. Moreover, because MATH has MATH edges, it follows that MATH. If MATH satisfies the hypothesis in REF , then every vertex of the MATH - gon graph is equivalent to at least one other vertex, so MATH and the quantity REF tends to zero as MATH, as required. We have proved that REF is satisfied. (In fact, a similar analysis shows that the limit moment is zero unless the pairing of MATH given by MATH is non - crossing - this was examined in a slightly different context in CITE, but unfortunately without the benefit of the idea of a non - crossing pairing.) Suppose that MATH satisfies the hypothesis in REF , namely that MATH. We are interested in the limit of the moment REF as MATH. As the number of terms in the sum REF where MATH for some MATH becomes negligably small compared to MATH as MATH, we may in REF sum over only all distinct choices of MATH. The assumption MATH implies that MATH is not equivalent to any other vertex under MATH. Therefore, renumbering if necessary, we may take MATH and hence MATH. By REF , we have that MATH is independent of the choice of distinct MATH and that MATH as MATH. Moreover, an analysis of the quotient graph MATH similar to that used to obtain REF, and keeping the same notation as in REF, shows that MATH . But then MATH . Taking the limit as MATH, we can at will require MATH's to be distinct and then relax this requirement; using that MATH, we obtain the conclusion of REF .
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math/0004104
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We shall use REF and adapt the proof of CITE to our situation. The family MATH of sets of random variables is asymptotically free as MATH. With MATH each of MATH and MATH is in MATH and MATH is an independent family of sets of matrix - valued random variables. Thus, by REF , each MATH and each MATH converges in moments to a semicircular element and the family REF is asymptotically free as MATH. This proves REF . Now take MATH so that MATH is an independent family of sets of matrix - valued random variables. By REF , MATH is asymptotically free as MATH. If MATH is the polar decomposition of MATH, then MATH is almost everywhere a unitary, and is distributed according to NAME measure on the group of MATH unitaries. Therefore, letting MATH be the polar part, MATH, of MATH, these random unitary matrices satisfy the hypotheses of the theorem. We will follow the proof of CITE to show the asymptotic freeness of REF. For MATH let MATH. For every MATH, the family MATH is asymptotically free as MATH. Given MATH and MATH, let MATH; moreover, let MATH be the essential supremum of the operator norm of MATH evaluated at points of the underlying probability space. Let MATH be a noncommutative monomial in MATH variables (for nonegative integers MATH), with coefficient equal to MATH. Let MATH. By REF of the proof of CITE, letting MATH be the function MATH, there is a polynomial MATH such that, letting MATH we have MATH . Because MATH, it follows that MATH. REF on the entries of MATH implies that for all MATH and for all MATH, MATH. Moreover, the convergence in MATH - moments as MATH of MATH implies that MATH whenvever MATH is an even integer; however, as MATH is increasing in MATH, this holds for all MATH. Fix MATH, MATH, MATH and let MATH . We may chose a constant MATH indepent of MATH and large enough so that MATH . Using NAME 's inequality we find MATH and therefore MATH . The asymptotic freeness of MATH follows from that of REF; this together with REF implies the asymptotic freeness of REF, and REF is proved. REF of the proof of CITE shows that for every MATH there is MATH such that MATH whenever MATH. Let again MATH be a noncommutative monomial having coefficient equal to MATH and with degree MATH, and let MATH, MATH, MATH. Let MATH . Letting MATH be a constant so that REF hold, we easily see using REF and NAME 's inequality that if MATH then MATH . Therefore MATH . This, together with REF shows that the family REF is asympototically free as MATH and finishes the proof of the theorem.
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math/0004104
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From MATH - freeness of MATH and MATH we get MATH for every MATH. Let MATH and let MATH be so that MATH for every MATH. For every MATH let MATH be such that MATH. In order for freeness of REF to hold, it will suffice that MATH . But the above equality follows directly from freeness of REF.
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math/0004104
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For brevity we shall prove REF simultaneously; while proving REF , we may from the outset assume that the stronger hypotheses of REF hold, because if we require MATH to be a single element then they will in any case be satisfied. We may write MATH where MATH is a random unitary matrix and MATH. For every MATH let MATH be a random unitary matrix so that MATH is diagonal, so that the joint distribution of the diagonal entries of MATH is invariant under all permutations of the MATH variables and so that MATH is a mutually independent family of sets of matrix - valued random variables. Let MATH be so that MATH is a mutually independent family of sets of matrix - valued random variables. It follows from the hypotheses of REF that MATH has the same joint MATH - moments as the family REF. We have MATH . Let MATH . Then MATH and MATH is an independent family of matrix - valued random variables. Let MATH . By hypothesis, each MATH converges in moments as MATH. Since MATH forms a commuting family of self - adjoint random matrices, and since the family MATH is independent, it follows that MATH converges in moments as MATH; moreover, the subfamily REF converges in MATH - moments to a family MATH is some MATH - noncommutative probability space MATH, where MATH is positive and has the same moments as the measure MATH, and where for distinct MATH and any MATH, MATH . We shall show that the entries of the set MATH of diagonal random matrices satisfy REF in the statement of REF . Let MATH denote the MATH-th diagonal entry of MATH. Note that MATH stays bounded (in fact converges) as MATH, for every MATH and MATH. This, together with the independence of the family REF, implies REF follows from the independence of REF and the fact that the joint distribution of the diagonal entries of each MATH is invariant under permutations of the MATH variables. Because MATH converges to a product measure, we have for every MATH, MATH, MATH and every MATH - tuple MATH of distinct, positive integers, that MATH . This together with the independence of REF implies REF . Hence we conclude from REF that MATH is asymptotically free as MATH. Therefore the family REF converges in MATH - moments to a family MATH in some MATH - noncommutative probability space, where the joint MATH - moments of MATH are as described above, where each MATH and each MATH is a NAME unitary and where MATH is free. Therefore, the family REF converges in MATH - moments as MATH to the family MATH . It is clear that MATH has the same moments as MATH, namely the same moments as the measure MATH. From the freeness of REF , it follows that the family MATH is MATH - free, and the theorem is proved.
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math/0004104
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Since any noncountable standard NAME space is NAME isomorphic to the unit interval, and since MATH is NAME isomorphic to the one - point compactification of MATH, it is no loss of generality to assume that MATH is a separable compact NAME space and MATH is the NAME MATH - algebra associated to this topology. For any compact set MATH, let MATH denote the set of NAME probability measures on MATH. Consider the folowing subsets of MATH: MATH . Clearly MATH. By CITE, every MATH has a representation MATH for a unique MATH. In fact, (see CITE), MATH is the set of extreme points of the compact simplex MATH. Now let MATH be as in the formulation of the theorem. Using REF we have MATH for a unique MATH. In particular MATH . By the assumption on MATH, the space MATH of complex valued continuous functions on MATH is a separable NAME space (in the uniform norm), so we may let MATH be a countable dense subset of MATH. Given MATH and MATH let us write MATH . With this notation, we have for all MATH, MATH . But REF shows that the above quantity is zero. Hence MATH for all MATH, for MATH - almost all MATH. Hence MATH for MATH - almost all MATH, which implies MATH, the NAME measure at the point MATH. Therefore MATH.
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math/0004104
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It will be more convenient to consider the measure MATH whose density with respect to NAME measure on MATH is MATH for some constant MATH; thus MATH is the push forward measure of MATH under the transformation MATH. We will find the limit as MATH of the marginal distributions, MATH, of MATH corresponding to the variables MATH. Let MATH and let MATH be the polynomials obtained via NAME - NAME orthonormalizaton of MATH in MATH. Thus MATH, where MATH are the (generalized) NAME polynomials. Using the NAME determinant we have MATH for some constant MATH. Therefore, the density of MATH with respect to NAME measure on MATH is CITE MATH for a constant MATH. Writing MATH and noting that as MATH ranges over the permutation group MATH these form an orthonormal family with respect to the measure MATH on MATH, we find MATH. Moreover, the density with respect to NAME measure on MATH of the marginal distribution MATH is MATH . But then the treatment in REF shows that MATH converges in the weak-MATH topology and in moments as MATH to MATH. The density with respect to NAME measure on MATH of the marginal distribution MATH is MATH . As elements of MATH, we thus have MATH . Since we know that MATH converges in the weak-MATH topology as MATH to MATH, it follows from REF that MATH converges in weak-MATH topology as MATH to MATH. Consider the measures MATH on MATH and let MATH be a MATH cluster point in MATH of these. Let MATH be the marginal distribution of MATH corresponding to the first MATH coordinates of MATH. Then from what we have proved above we have CASE: MATH is invariant under finite permutations of the coordinates in MATH; CASE: MATH; CASE: MATH. Hence, by REF , MATH. Since MATH was an arbitrary cluster point of MATH it follows that MATH converges in weak-MATH topology to MATH as MATH. Therefore, for all MATH, the marginal distribution MATH converges in weak-MATH topology to the measure MATH as MATH. It remains to show that REF holds whenever MATH is of polynomial growth. Let MATH and let MATH be a positive continuous function on MATH. Then MATH. Choose MATH, MATH, so that MATH increases pointwise to MATH as MATH. Then for all MATH, MATH . But MATH is supported on MATH; therefore MATH, and the claim is proved. Let MATH and suppose MATH and MATH are continuous functions on MATH satisfying MATH and MATH, and suppose that MATH. Then MATH. Applying REF to MATH gives MATH, while applying REF to MATH yields MATH; the claim is proved. In order to finish the proof of the lemma, it will suffice to show MATH for every MATH and all integers MATH. Letting MATH, we have MATH. Moreover, because MATH converges in moments to MATH, we have MATH . Now REF follows from REF , and the lemma is proved.
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math/0004104
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Clearly for every non - random MATH unitary matrix MATH, the distribution of MATH is equal to the distribution of MATH. Let MATH be the symmetrized joint distribution of the eigenvalues of MATH and let MATH be the symmetrized joint distribution of the eigenvalues of MATH. For MATH let MATH, respectively MATH, be the marginal distribution of MATH, respectively MATH, corresponding to the first MATH variables. Given MATH, MATH . By REF , it follows that MATH where MATH is the free NAME distribution of parameter MATH. Therefore MATH converges in moments to MATH, where MATH has density MATH with MATH and MATH. Now REF applies and finishes the proof.
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math/0004104
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Let MATH be the manifold of matrices in MATH having MATH distinct eigenvalues. Then MATH has full NAME measure in MATH. Let MATH be the NAME group of MATH unitary matrices, and let MATH be the manifold of all upper triangular MATH complex matrices, no two of whose diagonal elements are the same. Let MATH be given by MATH. NAME proved his result by evaluating the Jacobian of MATH (after throwing away the directions in MATH) and thereby finding the measure MATH on MATH such that letting MATH be NAME measure on MATH, the push - forward measure MATH on MATH has density MATH with respect to NAME measure on MATH, that is, the density of a random matrix MATH. This measure MATH was found to have density MATH with respect to NAME measure on MATH, where for a matrix MATH, MATH is the MATH-th diagonal entry of MATH; this density REF is that of the matrix MATH in REF . The matrix MATH in the corollary has density MATH with respect to NAME measure on MATH; since MATH, and building on NAME 's calculation, it follows that the random matrix MATH of REF has density REF with respect to NAME measure on MATH, as required. An application of REF shows that MATH, and hence also MATH, converges in MATH - moments as MATH to a circular free NAME element.
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math/0004104
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This is quite similar to the proof of REF . Let MATH. Consider first the case MATH. Let MATH be the polynomials obtained via NAME - NAME orthonormalization of the sequence MATH in MATH. Then MATH . Using the NAME determinant we have MATH for some constant MATH. Therefore MATH . Writing MATH and noting that as MATH ranges over the permutation group MATH these form an orthonormal family with respect to the measure MATH on MATH, we find MATH. Moreover, the density of MATH with respect to NAME measure on MATH is MATH . We shall show that MATH converges in MATH - moments to MATH. Clearly if MATH and if MATH then MATH . Hence we need only show MATH for all MATH. We have MATH . Writing MATH we have MATH . Hence MATH converges in MATH - moments to MATH as MATH; since MATH is compactly supported it follows that MATH converges in the weak-MATH topology to MATH. The density of MATH with respect to NAME measure is MATH . Hence as a linear functional on MATH, the norm of MATH is bounded above by MATH. Therefore MATH converges in the weak-MATH topology as MATH to MATH. Arguing as in the proof of REF and using REF , we conclude that for every MATH, MATH converges in weak-MATH topology to MATH, which we will denote by MATH. It remains to show that MATH converges to MATH in MATH - moments, namely that MATH for every MATH. Exactly as in the proof of REF , one shows that if MATH is a positive continuous function on MATH then MATH . Then, considering the real and imaginary parts separately and arguing as in the proof of REF , one shows that if MATH and MATH are continuous functions on MATH, if MATH, if MATH and if MATH then MATH. But letting MATH, we have MATH . Moreover, because MATH converges in MATH - moments to MATH, we have MATH . Hence we have REF and the lemma is proved.
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math/0004104
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Using the NAME determinant we find MATH . Averaging REF over the action of MATH gives MATH . From this one easily sees that-MATH and that MATH is obtained from MATH by averaging. In order to show that MATH is obtained from MATH by averaging, it suffices to note that for any measure MATH on MATH, the average over the action of MATH on the marginal distribution, MATH, corresponding to the first MATH variables, is equal to the marginal distribution of the average over the action of MATH on MATH. Since, by REF , MATH converges in MATH - moments and in weak-MATH topology as MATH to MATH, which is invariant under the action of MATH, it follows that MATH converges in MATH - moments and in weak-MATH topology to MATH.
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math/0004104
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For every MATH let MATH be a MATH random matrix whose distribution has density with respect to NAME measure MATH . Then by REF , MATH converges in MATH - moments as MATH to a circular free NAME element of parameter MATH. By REF , each MATH has the same MATH - moments as MATH, where MATH, MATH is a diagonal MATH random matrix, the distribution of whose diagonal entries has density MATH with respect to NAME measure on MATH, and where MATH and MATH are independent. We will use previous results to show that also each MATH, (MATH), converges in MATH - moments as MATH to a circular free NAME element of parameter MATH, where MATH is a diagonal random matrix such that MATH and MATH are independent and where the joint distributions of the diagonal entries of MATH have the following densities with respect to NAME measure on MATH: MATH . The proof that MATH converges in MATH - distribution to a circular free NAME element relies for MATH on REF , (see REF ), and REF ; for MATH we use REF ; the density for MATH is just a rewriting of that for MATH; for MATH we use again REF ; for MATH we use again REF , and REF . We may characterize the above successive transformations as follows: from REF to MATH is decoupling; from MATH to MATH is desymmetrization; from MATH to MATH is regrouping; from MATH to MATH is partial resymmetrization; from MATH to MATH is partial recoupling. Taking blocks of consequetive rows and columns to write MATH as a MATH matrix of MATH random matrices, we have MATH where MATH is an independent family of matrix - valued random variables, where MATH for every MATH and where MATH with MATH, with MATH a diagonal random matrix, the joint distribution of whose diagonal entries has density with respect to NAME measure MATH and with MATH and MATH independent. Let MATH, (MATH), be such that MATH is an independent family of matrix - valued random variables. By conjugating the matrix MATH with MATH and by using REF and the fact that the class MATH is invariant under left and right multiplication by independent unitaries, it follows that MATH has the same MATH - moments, as MATH where MATH is an independent family of matrix - valued random variables, where MATH for every MATH and where the distribution of MATH has density MATH with respect to NAME measure on MATH. If MATH, MATH, MATH, MATH, are non - random MATH unitary matrices, then MATH continues to be an independent family of matrix - valued random variables, MATH has the same distribution as MATH and MATH has the same distribution as MATH. Therefore, the family REF has the same joint MATH - moments as the family REF. Taking into account also REF (as in the proof of REF ), we see that the conditions of REF are fulfilled, allowing us to conclude that the family REF is asymptotically MATH - free as MATH. Moreover, (by REF ), each MATH converges in MATH - moments to a circular free NAME element of parameter MATH, while MATH converges in MATH - moments to a circular element. Therefore, the entries of the matrix MATH model as MATH the entries of the matrix MATH in the statement of the theorem. As MATH converges in MATH - moments to a circular free NAME element of parameter MATH, the theorem is proved.
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math/0004104
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We may without loss of generality assume that MATH, which implies MATH and MATH is a trace. Let MATH and let MATH be the circular free NAME element of parameter MATH as in REF , where we take MATH to be a MATH - noncommutative probability space. Thus MATH is a circular free NAME element of parameter MATH, each MATH is a circular element and the collection of all MATH and MATH is MATH - free. For MATH, let MATH . Another application of REF shows that in the MATH - noncommutative probability space MATH the element MATH is a circular free NAME element of parameter MATH. Hence by CITE, MATH has spectrum MATH . Similarly, if MATH then denoting by MATH the identity element of MATH, we find that in the MATH - noncommutative probability space MATH the element MATH is a circular free NAME element of parameter MATH. Hence MATH has spectrum MATH . Therefore, by REF , if MATH and if MATH then MATH and consequently MATH. Letting MATH grow without bound and choosing MATH appropriately implies REF.
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math/0004104
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The positive part MATH of MATH has the same moments as the measure MATH on MATH whose density with respect to NAME measure is MATH with MATH and MATH. Since MATH has the polar decomposition MATH where MATH is a NAME unitary and where MATH and MATH are MATH - free, the MATH - moments of MATH can be expressed as certain polynomials in the moments of MATH. Clearly, the MATH-th moment of MATH converges to the MATH-th moment of MATH as MATH.
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math/0004104
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We know from general principles that MATH if MATH, and from REF we have that MATH; as MATH is faithful we conclude REF . Let us now prove REF . Arguing as in the proof of REF , we have MATH whenever MATH. We may take the MATH-noncommutative probability space MATH so that MATH is a faithful trace, in which case, since MATH, it follows that MATH . Thus MATH is the limit in strong-MATH topology of MATH as MATH tends to MATH from above. Using REF , it follows that MATH. For REF , let us show that MATH is circular free NAME of the desired parameter, first in the case when MATH is rational. We may take MATH to be MATH and MATH to be equal to the MATH matrix MATH as in REF . By REF , the noncommutative probability space REF is MATH and MATH where MATH is circular free NAME of parameter MATH. Applying again REF , we obtain that MATH is circular free NAME of parameter MATH. When MATH is such that MATH is irrational, then using REF we have that MATH is the strong-MATH limit of MATH as MATH tends to MATH through rational numbers. Hence by REF and the continuity in MATH of of MATH implied by REF , it follows that MATH is circular free NAME of parameter MATH. The statement about the spectrum follows from the result of CITE that we've been using repeatedly. REF is proved similarly. When MATH is rational then we get MATH where MATH is circular free NAME of parameter MATH. The remaining part of the argument is like for REF above.
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math/0004104
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Note that REF implies MATH. Let MATH be normally and faithfully represented on a NAME space MATH. For MATH we have MATH where the last inequality is because the spectral radius of MATH is MATH. Hence MATH. In order to prove the reverse inequality, it will suffice to show MATH for all MATH, because MATH is strong-MATH - continuous by REF . Let MATH, MATH and let MATH, that is, MATH . Set MATH. Then MATH and, because the spectral radius of MATH is MATH, we have MATH. Since MATH is an invariant subspace for MATH, we have MATH. Therefore MATH, so using REF and NAME 's inequality we have MATH, which shows that MATH. Hence MATH and therefore MATH.
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math/0004105
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CASE: Let MATH be a smooth point. It is clear. CASE: Let MATH be a NAME terminal singular point. Since we have MATH, then MATH or MATH. Hence, MATH . MATH . CASE: Let MATH be a terminal quotient singular point of type MATH with MATH . Let MATH for MATH. For MATH, we have MATH . Then by REF , MATH .
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math/0004105
|
(compare [KMMT REF ]) By [MM REF ], there is a covering family of rational curves MATH such that MATH. If MATH has a fixed point MATH, then REF , we have MATH. We have MATH. By CITE, we have MATH. Hence MATH in this case. If MATH has a fixed point MATH, the proof is the same as the one of [KMMREFa, Theorem.]. By [KMMT, NAME REF ], there is a covering family of rational curves MATH with a fixed point MATH such that MATH. Hence by REF , MATH in this case.
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math/0004105
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The proof is the same as the one of [KMMT REF ] except that we can use REF instead of [KMMT REF ].
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math/0004105
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We have MATH and MATH. The rest of the proof is the same as the one of [K, REF .].
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math/0004106
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In the sequel the admissibility of every block basis of MATH will always be considered with respect to MATH. Given MATH, we set MATH . In the above, the admissibility of MATH is measured with respect to MATH, the sequence of functionals biorthogonal to MATH. Because MATH, we have that MATH. Indeed, suppose that MATH and let MATH, MATH. Put MATH, MATH. Since MATH is bimonotone, MATH. Furthermore, MATH is MATH admissible. Hence, MATH and the assertion follows. We define an equivalent norm MATH on MATH in the following manner: MATH . Let MATH be a normalized block basis of MATH, and let MATH. Let MATH be the block subspace of MATH generated by MATH. Since MATH, there exists a normalized block basis MATH of MATH in MATH having no subsequence which is a MATH spreading model. It follows, by the main result of CITE combined with REF, that there exists a subsequence MATH of MATH such that for every MATH, the block basis MATH, is MATH admissible. We next choose MATH, a generic MATH average of MATH. It is easily seen that for some MATH we have that MATH. Therefore, MATH. On the other hand, MATH is MATH admissible, for every MATH and MATH. It follows that MATH. We let MATH and observe that MATH. Let now MATH. Arguing similarly, we can find a normalized block basis MATH of MATH and a generic MATH average MATH of MATH such that MATH and MATH. We let MATH. We are going to show that MATH. Suppose that MATH, and let MATH denote the collection of the ranges of the MATH's. Let MATH. Observe that by the choice of MATH we have that MATH is MATH admissible. On the other hand MATH is MATH admissible and thus MATH is MATH admissible. Because MATH, we obtain the estimate MATH. Hence, MATH, as claimed. Finally, MATH. The proof is now complete since MATH was arbitrary.
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math/0004106
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Suppose the assertion is false. A standard perturbation argument yields a normalized block basis MATH of MATH equivalent to a block basis MATH of MATH. Let MATH be an isomorphism from MATH onto MATH such that MATH, for all MATH. We can choose MATH such that MATH, for every MATH. Our assumptions allow us to choose MATH such that MATH. Let MATH be a normalized block basis of MATH having no subsequence which is a MATH spreading model. But since MATH is a block basis of MATH, it follows that for every MATH and all choices of scalars MATH . Hence, MATH, for every MATH and all choices of scalars MATH. However, MATH, and therefore MATH. Thus, MATH is a MATH spreading model contrary to our assumptions.
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math/0004106
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We first choose MATH such that every MATH is MATH-good. To see that such a MATH exists, set MATH . We can easily verify that MATH is closed in the topology of pointwise convergence in MATH, and therefore it is a NAME set. Because MATH, for every MATH, the infinite NAME theorem yields MATH such that MATH, as claimed. It is a well known fact that MATH endowed with the topology of pointwise convergence is a perfect Polish space. We let MATH and set MATH . A straightforward application of the NAME category theorem yields that MATH is a dense MATH subset of MATH. By a result of CITE and CITE (compare CITE, p. REF , or REF), there exists MATH homeomorphic to the NAME set such that MATH, whenever MATH, MATH are distinct elements of MATH. We can now apply REF to obtain a family MATH of reflexive H.I. spaces such that for every MATH, MATH satisfies the MATH distortion property, where MATH, MATH and MATH are as in the statement of REF . Since MATH whenever MATH and MATH are distinct elements of MATH, REF implies that MATH and MATH are totally incomparable. The proof of the theorem is now complete.
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math/0004106
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Let MATH be an infinite block basis of MATH, and let MATH. Set MATH, where MATH. We can find MATH such that MATH, for every MATH. Let MATH. Choose MATH in MATH, according to the hypothesis. There exists MATH such that MATH. Put MATH, MATH, and note that MATH is non-increasing. We now have that MATH . On the other hand, MATH as MATH is non-increasing. Hence, MATH . Since MATH was arbitrary, MATH is not unconditional. The moreover statement is immediate.
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math/0004106
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By induction on MATH. If MATH the assertion of the lemma is trivial. Assuming the assertion true when MATH, MATH, let MATH with MATH. If MATH there is nothing to prove. So assume MATH. Let MATH be the root of MATH and let MATH for some MATH. We denote by MATH the set of immediate successors of MATH in MATH. Given MATH let MATH. Because MATH we can apply the induction hypothesis on MATH and the set MATH to deduce that the collection MATH is MATH-admissible, where MATH. Since MATH and MATH whenever MATH, we obtain that MATH is MATH-admissible, for every MATH. But also, MATH is MATH-admissible whence MATH is MATH-admissible.
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math/0004106
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Let MATH denote the set of all branches of MATH (a branch is a maximal well ordered subset of MATH). If MATH the assertion is trivial. So assume that MATH for some MATH. Given MATH set MATH . Note that MATH is well defined and that MATH since MATH (MATH being the root of MATH). Let us say that MATH is of type MATH if MATH is terminal in MATH. If MATH is not of type MATH then it is of type MATH (respectively, MATH), if the last MATH-entry of MATH is greater than or equal (respectively, smaller than) MATH. We then denote by MATH the immediate successor of MATH in MATH. We let MATH, MATH and MATH. Observe that the following properties hold: CASE: If MATH then all MATH-entries of MATH are smaller than MATH, MATH, yet MATH. CASE: If MATH, then MATH is non-terminal, all MATH-entries in MATH are smaller than MATH, the last MATH-entry of MATH is greater than or equal to MATH and MATH. CASE: If MATH then MATH is terminal, all MATH-entries in MATH are smaller than MATH and MATH. It is not hard to check now that MATH consists of pairwise incomparable nodes of MATH and hence MATH consists of successive subsets of MATH. Moreover, MATH. Because MATH and all MATH-entries of MATH are smaller than MATH whenever MATH, we obtain that MATH for all MATH. REF now yields that MATH is MATH-admissible. Finally, we let MATH. Since MATH, for all MATH, we set MATH for MATH. We can easily verify that the desired properties hold.
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math/0004106
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Observe that MATH and hence MATH, for all MATH. Let MATH, where MATH. Set MATH and suppose that MATH, for some MATH. Let MATH denote the collection of the ranges of the MATH's, and let MATH denote the collection of those MATH's whose support intersects at least one member of MATH. Put MATH, MATH. Because MATH is MATH-admissible and MATH, we obtain that MATH. On the other hand we clearly have that MATH. Thus, MATH.
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math/0004106
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Let MATH, where MATH for MATH. We can assume without loss of generality that MATH for every MATH where MATH. We are going to show that there exists a normalized block basis of MATH admitting a generic MATH average of norm at least MATH. Suppose instead that this were false. Then it is easy to construct for every MATH, a block basis MATH of MATH so that letting MATH and MATH the following are satisfied: CASE: MATH is a block basis of MATH. CASE: MATH, for all MATH. CASE: MATH, for all MATH. CASE: For every MATH, if MATH with MATH for MATH, then MATH and MATH is MATH-admissible. The construction is easily done by induction. Taking MATH we see from REF that MATH. On the other hand REF. implies that MATH as MATH. Thus, MATH contradicting the choice of MATH.
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math/0004106
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Note first that REF guarantees the existence of the block basis MATH. Let MATH, where MATH. By passing to a subsequence of MATH, if necessary, we can assume that the union of any MATH subsets of MATH belongs to MATH. Choose MATH, MATH, such that MATH and MATH, for every MATH (if MATH, then MATH). Let MATH. Suppose first that MATH. We show that in this case MATH. Indeed, suppose first that MATH. REF yields that MATH for all MATH whence MATH. If MATH choose MATH so that MATH. Observe that if MATH then MATH . When MATH, REF yields MATH. Hence MATH and so our claim holds. The final case to consider is that of MATH. Clearly, MATH, if MATH. We employ the decomposition REF to find a MATH admissible subset MATH of MATH and scalars MATH satisfying the conclusion of REF . Let MATH denote the collection of the ranges of the MATH's (MATH). Our previous work implies that MATH is MATH admissible. But also, MATH is MATH admissible since MATH is MATH admissible. It follows that MATH is MATH admissible.
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math/0004106
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Let MATH be a block subspace of MATH spanned by the normalized block basis MATH of MATH. Let MATH and choose a block basis MATH of MATH satisfying the conclusion of REF . Applying REF (compare also REF), we obtain that for every subsequence of MATH which is a MATH spreading model, it must be the case that MATH and thus MATH.
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math/0004106
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We choose MATH and MATH as we did in the proof of REF . Suppose first that MATH. Because MATH, the argument in the proof of REF shows that MATH. When MATH, we apply the decomposition REF to find a MATH admissible subset MATH of MATH and scalars MATH satisfying the conclusion of REF . Note that if MATH and MATH, then MATH and thus for all MATH, MATH, by REF . Using the splitting argument of REF , we conclude that MATH is MATH-admissible.
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math/0004106
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Set MATH. Let MATH, MATH and MATH. There exists MATH so that MATH, where MATH and MATH. Note that MATH. Applying a splitting argument similar to that of REF and taking in account REF , we obtain that MATH is MATH-admissible. The assertion follows from REF and the fact that MATH.
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math/0004106
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It follows from REF and our preceding remarks that MATH satisfies the MATH distortion property. We show that MATH is H.I. This is accomplished through REF . Let MATH be a normalized block basis of MATH and let MATH. Set MATH, where MATH. We can assume that the union of any MATH subsets of MATH belongs to MATH. Successive applications of REF yield a normalized block basis MATH of MATH, MATH in MATH, and integers MATH with MATH, satisfying the following: CASE: MATH is a normalized MATH average of MATH resulting from REF . CASE: MATH, MATH and MATH, for all MATH. CASE: MATH, for all MATH. CASE: MATH is maximally MATH-admissible. Put MATH, MATH, and note that MATH, MATH. NAME show that MATH satisfies REF . and REF , with MATH, MATH and MATH. REF . is immediate since MATH is MATH-dependent. REF . is achieved by establishing the following Claim: Given MATH, there exist intervals MATH in MATH so that CASE: MATH is MATH-admissible. CASE: MATH is constant for all MATH. CASE: MATH, for all MATH. To prove the claim suppose first that MATH. REF yields that MATH, for at most one MATH, and thus the claim holds in this case. Next assume that MATH. Without loss of generality, there exist a MATH-dependent sequence MATH in MATH and an interval MATH so that MATH. Let MATH be the largest MATH for which MATH is an element of MATH, or let MATH, if no such MATH exists. The injectivity of MATH and REF imply that if MATH, or if MATH, then MATH, for all MATH. If MATH, then the injectivity of MATH yields MATH, MATH for MATH yet MATH. It follows now by REF , that MATH, for all MATH. We also observe that there exists MATH such that MATH, if MATH, while MATH if MATH. Concluding, there exist four intervals MATH in MATH, some of which may possibly be empty, such that MATH is constant for every MATH, while MATH, for each MATH. Finally, assume MATH. If MATH, the claim trivially holds so suppose that MATH. Choose MATH-admissible and scalars MATH according to the decomposition REF . By splitting the MATH's into two sets, those whose support intersects at least two of the ranges of the MATH's, and those whose support intersects at most one, we deduce from our previous work that there exist intervals MATH in MATH so that MATH is MATH-admissible, MATH is constant for all MATH, and MATH, for all MATH. Thus the claim holds and the proof is complete.
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math/0004109
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Apply REF to MATH.
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math/0004109
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By REF , MATH contains a primitive set. Let MATH be the primitive curve class corresponding to this primitive set. Write MATH. Now MATH, so we are done by induction on the degree of MATH (with respect to a fixed projective embedding of MATH).
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math/0004109
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Let MATH be a primitive curve class, and write MATH with MATH and MATH very effective. Then MATH and REF follows since MATH is a non zero divisor in MATH. If MATH is NAME, then a presentation for MATH is obtained by starting with a presentation for MATH in terms of generators and relations, and replacing each relation by a MATH-deformed relation which holds in MATH (CITE, or compare CITE). The presentation REF is of this form.
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math/0004109
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For REF , we induct on the dimension MATH. The case MATH is trivial, and the base case MATH is easily verified. For the inductive step, let us suppose MATH satisfies REF , but that REF fails to hold. Then there is a primitive set MATH whose associated primitive relation REF satisfies MATH. Let MATH be a maximal cone containing MATH, MATH, MATH, and let us denote the remaining generators of MATH by MATH, MATH, MATH, MATH, MATH, MATH (suitably rearranging indices). We insist that the sets MATH and MATH be disjoint. Now MATH is the cone spanned by MATH . Let MATH be the point corresponding to MATH (so MATH for all MATH). We have MATH. Since MATH is NAME, we have MATH for every ray generator MATH, with equality if and only if MATH. So, for MATH we have MATH, for some nonnegative integer MATH. Now MATH . In particular, MATH, and so MATH. Consider the fan MATH in MATH. Let us give MATH coordinates by identifying the elements of MATH (in the order listed in REF ) with the standard basis elements. Then MATH consists of all cones of MATH containing MATH, projected by forgetting the first coordinate. The divisors associated to the projections of MATH, MATH, MATH form a primitive set for MATH. Note that MATH has first coordinate equal to zero; so, if we define MATH by MATH for all MATH, then we have MATH. We are assuming every toric subvariety of MATH is NAME. The induction hypothesis applies to the toric subvariety MATH implies MATH, so we have a contradiction. For REF , we let MATH be a maximal cone, and we give MATH the coordinates thus dictated. Suppose some ray generator MATH, when written in coordinates as MATH, satisfies MATH. If the MATH on MATH corresponding to the MATH-dimensional cone MATH, has fixed points MATH and MATH, then in the coordinate system of MATH, we find MATH has first coordinate MATH. So, if REF fails then, for some MATH and MATH, the coordinates MATH for MATH satisfy MATH or MATH (after shuffling indices). Among all such pairs MATH and MATH we may assume MATH is as large as possible. Now, MATH, MATH, MATH, MATH fail to generate a cone, hence by REF the sum MATH of MATH and some nonempty subset of MATH is also a ray generator. But MATH must have either some coordinate MATH or at least two coordinates MATH, and the sum of the coordinates of MATH is strictly larger than MATH. This is a contradiction. REF implies MATH is NAME, and for any cone MATH, REF for MATH implies REF for MATH, and hence that the toric subvariety MATH is NAME. So every toric subvariety of MATH is NAME, and we have REF .
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math/0004109
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Induct on MATH, and apply REF .
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math/0004109
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We induct on MATH. Suppose MATH. Then MATH must contain a primitive set. The set MATH itself cannot be a primitive set, since MATH is not a ray generator in MATH. So, we may suppose MATH is primitive, with MATH. Then we have MATH for some ray generator MATH, and now MATH, with MATH, MATH, MATH, MATH linearly independent. This contradicts the induction hypothesis.
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math/0004109
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We need to show that for all MATH, MATH, and every cone MATH with MATH, MATH . Suppose REF fails for MATH. We may suppose MATH, and in fact, MATH is a primitive set, with MATH. Hence MATH for some MATH. Now MATH, MATH, MATH, MATH are linearly independent and MATH. So, we have a contradiction to REF . Suppose REF fails for MATH. That is, we have MATH but MATH. Then (rearranging indices further) there is a primitive set, composed of MATH, some subset of MATH, and without loss of generality, all of MATH, with MATH positive. So, MATH for some MATH and some MATH. We now have MATH . This contradicts REF .
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math/0004109
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Suppose not: MATH, say. In the case MATH, then we find MATH, a contradiction. In the case MATH, then by REF , the fact that MATH implies that MATH and MATH are two sets of cone generators. Now MATH and we have a contradiction.
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math/0004109
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Since a NAME toric variety is determined uniquely by the set of ray generators we have REF , and REF is clear. We obtain REF from the characterization of how primitive relations behave under blow-down. By CITE, if MATH is the blow-down corresponding to the primitive relation MATH, then the primitive sets of MATH are precisely the primitive sets of MATH not containing MATH (other than MATH), plus the sets MATH (disjoint union, by REF ) for some (though perhaps not all) primitive sets MATH containing MATH. For such MATH and MATH (primitive sets for MATH and MATH, respectively), the respective primitive relations have the same right-hand sides. Given REF , then, every blow-down of an exceptional divisor is a toric variety which satisfies REF and, additionally, REF of this proposition, and hence by induction on the number of toric divisors is a NAME toric variety all of whose toric subvarieties and toric blow-downs along divisors are NAME.
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math/0004109
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For a maximal cone MATH, let MATH denote the affine span of the generators of MATH, and let MATH denote (signed) integer distance to MATH in MATH. Then the quantity MATH appearing in the statement is MATH. We prove the statement by induction on the degree MATH of a tree of MATH's. The induction hypothesis is, first, that given any tree MATH of MATH's of total degree MATH, meeting MATH, the toric point MATH lies in MATH only if MATH for any maximal cone MATH. Second, if MATH and MATH then the homology class of MATH is that indicated in REF . Third, for any maximal cone MATH with MATH, there exists a tree of MATH's joining the corresponding toric point to a point of MATH, having degree equal to MATH. Let MATH be a tree of MATH's, of total degree MATH, joining MATH to a point of MATH. It suffices to assume MATH, where MATH is a toric MATH joining MATH to MATH, for some toric point MATH, and MATH is a tree of MATH's joining MATH to a point of MATH. Shuffling coordinates, we may suppose MATH, where MATH. Denote the additional generator of the maximal cone MATH corresponding to MATH by MATH, that is, MATH, and let us write MATH in coordinates. Then MATH has intersection numbers MATH with MATH and with MATH, and MATH with MATH for MATH. So, MATH. We claim MATH with equality if and only if MATH. This is a computation: MATH, so the right-hand side minus left-hand side of REF equals MATH and by REF we have MATH. By the induction hypothesis, then, we have MATH, with equality only if MATH and the homology class MATH satisfies satisfies MATH if MATH, otherwise MATH, MATH, MATH for MATH, and MATH for all other MATH. So MATH satisfies REF . For the existence portion of the inductive step, if MATH then MATH must have some coordinate equal to MATH, so without loss of generality we have MATH. We can now take MATH to be the union of MATH (as defined in the previous paragraph) and a tree MATH of MATH's joining MATH to a point of MATH satisfying MATH (the existence of such MATH follows from the induction hypothesis).
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math/0004109
|
Let MATH, and let MATH be a maximal cone containing MATH, MATH, MATH, with MATH. For each ray generator MATH, let MATH be a tree of MATH's joining MATH to a point of MATH, with MATH. For each MATH, let MATH; we have MATH for all MATH. Now the sum over all MATH of MATH copies of MATH has homology class MATH.
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math/0004109
|
Suppose not. Since MATH and MATH, it follows that MATH is special for MATH. By REF , then, if MATH denotes the unique element of MATH not in MATH, then MATH for every MATH. So MATH, and hence some MATH has intersection number MATH with MATH. Without loss of generality, then, MATH. Then MATH is special exceptional, say with MATH the unique element of the associated exceptional set not in MATH. As before, MATH for every MATH, and we may iterate this process. We eventually reach a contradiction to the existence of only finitely many exceptional classes.
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math/0004109
|
We know MATH is the number of negative entries in the coordinate expression for MATH, in the coordinate system dictated by MATH. Let us suppose MATH is generated by the second through MATH standard basis elements plus one additional vector. Based on REF there are two possibilities. First, the additional generator can be of the form MATH; the number of MATH's is MATH and in this case MATH is not exceptional. The change of coordinates to the coordinate system of MATH preserves the last MATH entries of MATH. Hence MATH. In the remaining case, the additional generator of MATH is MATH where the number of MATH's is MATH. In this case, MATH is exceptional. If MATH, in the coordinates of MATH, is MATH then in the coordinates of MATH the coordinate expression is MATH . So MATH, with equality only if MATH, with additionally MATH for MATH and either MATH or MATH.
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math/0004109
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Let MATH be a torus-invariant genus MATH stable MATH-pointed map, which stabilizes (upon forgetting the map to MATH) to MATH distinct points on a single irreducible component MATH, such that the MATH marked point maps into MATH for MATH and such that the image of the MATH point is MATH. By REF , MATH, and in fact (exercise) there exist MATH for some MATH such that MATH, MATH, and MATH for each MATH (for the last assertion, use REF ). So it suffices to prove MATH. We induct on degree of MATH. The base case is the inequality MATH for every MATH such that MATH. This is immediate from the characterization of MATH as the number of negative entries in the corresponding coordinate expression for MATH. Equality holds only when the coordinate expression for MATH has exactly MATH positive entries, each close to MATH, and MATH negative entries, each small in magnitude. We divide the inductive step into two cases. Suppose MATH. For the first case, assume the MATH marked point MATH does not lie on the distinguished component MATH. Let MATH denote the connected component of MATH containing MATH, with the MATH terminating in MATH deleted. Assume that this MATH maps to the toric curve MATH with MATH . Let MATH denote the homology class of MATH. Then, by induction, MATH. By REF , MATH. So the inequality is established. If equality holds, then MATH is exceptional, and MATH is equal to the union of MATH and a MATH mapping with degree one to MATH. By induction, MATH is equivalent in homology to a chain MATH of toric curves, each exceptional, joining a point on MATH to the point MATH. Also, equality implies that there are precisely MATH divisors MATH having positive intersection with MATH, and for any of these, the corresponding MATH is a generator of MATH whose corresponding entry in the coordinate expression of MATH is positive. It follows that each of these MATH has nonpositive intersection with every component of MATH. The second case is when MATH. As before, let MATH denote the image of the MATH marked point. Choose coordinates on MATH so that the generators of MATH are the standard basis elements, and order these so that MATH has negative first coordinate, MATH, with MATH. Let MATH be the cone generated by the second through MATH basis elements; we have MATH for some (unique) MATH. Let MATH. Then MATH, so in particular MATH. Let MATH, where MATH is the tree of MATH's joining MATH to MATH, as given in REF . The degree of MATH is MATH. So, the union of MATH and MATH is (more precisely, determines) a torus-invariant genus zero MATH-pointed stable map, whose homology class MATH satisfies MATH, by REF . Moreover, the MATH marked point now does not lies on the distinguished component. By the previous case, we have MATH, and the desired equality holds. In case of equality we must have MATH and MATH equal to the sum of the homology classes of the curves joining MATH to MATH, MATH, MATH of REF , and then we find MATH. So, we are reduced to the pervious case.
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math/0004113
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Let us start with a combinatorial description for the objects involved in REF: By the NAME - NAME interpretation of NAME functions as generating functions of nonintersecting lattice paths, we may view the left - hand side of the equation as the weight of all pairs MATH, where MATH and MATH are MATH-tuples of nonintersecting lattice paths. The paths of MATH are coloured green, the paths of MATH are coloured blue. The MATH-th green path MATH starts at MATH and ends in MATH. The MATH-th blue path MATH starts at MATH and ends in MATH. For an illustration, see the upper left pictures in REF , where green paths are drawn with full lines and blue paths are drawn with dotted lines. For the right - hand side of REF, we use the same interpretation. We may view the first term as the weight of all pairs MATH, where MATH is a MATH-tuple of nonintersecting lattice paths and MATH is a MATH-tuple of nonintersecting lattice paths. The paths of MATH are coloured green, the paths of MATH are coloured blue. The MATH-th green path MATH starts at MATH and ends in MATH. The MATH-th blue path MATH starts at MATH and ends in MATH. For an illustration, see the upper right picture in REF . In the same way, we may view the second term as the weight of all pairs MATH, where MATH and MATH are MATH-tuples of nonintersecting lattice paths. The paths of MATH are coloured green, the paths of MATH are coloured blue. The MATH-th green path MATH starts at MATH and ends in MATH. The MATH-th blue path MATH starts at MATH and ends in MATH. For an illustration, see the upper right picture in REF . In any case, the weight of some pair of paths MATH is defined as follows: MATH . What we want to do is to give a weight - preserving bijection between the objects on the left side and on the right side: MATH . Clearly, such a bijection would establish REF. The basic idea is very simple and was already used in CITE and in CITE: Since it will be reused later, we state it here quite generally: Let MATH be two arbitrary families of nonintersecting lattice paths. The paths MATH of the first family are coloured with colour blue, the paths MATH of the second familiy are coloured with colour green. Let MATH be the ``two - coloured" graph made up by MATH and MATH in the obvious sense. Observe that there are the two possible orientations for any edge in that graph: When traversing some path, we may either move ``right - upwards" (this is the ``original" orientation of the paths) or ``left - downwards". A changing trail is a trail in MATH with the following properties: CASE: Subsequent edges of the same colour are traversed in the same orientation, subsequent edges of the opposite colour are traversed in the opposite orientation. CASE: At every intersection of green and blue paths, colour and orientation are changed if this is possible (that is, if there is an adjacent edge of opposite colour and opposite orientation); otherwise the trail must stop there. CASE: The trail is maximal in the sense that it cannot be extended by adjoining edges (in a way which is consistent with the above conditions) at its start or end. Note that for every edge MATH, there is a unique changing trail which contains MATH: For example, consider some blue edge which is right - or upwards - directed and enters vertex MATH. If there is an intersection at MATH, and if there is a green edge leaving MATH (in opposite direction left or downwards), then the trail must continue with this edge; otherwise it must stop at MATH. If there is no intersection at MATH, and if there is a blue edge leaving MATH (in the same direction right or upwards), then the trail must continue with this edge; otherwise it must stop at MATH. Note that a changing trail is either ``path - like", that is, has obvious starting point and end point (clearly, these must be the end points or starting points of some path from either MATH or MATH), or it is ``cycle - like", that is, is a closed trail. Let us return from general definitions to our concrete case: Starting with an object MATH from the left - hand side of REF, we interpret this pair of lattice paths as a graph MATH with green and blue edges. (See the upper left pictures in REF .) Next, we determine the changing trail which starts at the rightmost endpoint MATH: Follow the green edges downward or to the left; at every intersection, change colour and orientation, if this is possible; otherwise stop there. Clearly, this changing trail is ``path - like". (See REF for an illustration: There, the orientation of edges is indicated by small arrows in the upper pictures; the lower pictures show the corresponding changing trails.) Now we change colours green to blue and vice versa along this changing trail: It is easy to see that this recolouring yields nonintersecting tuples of green and blue lattice paths. Note that there are exactly two possible cases: CASE: The changing trail stops at the rightmost starting point, MATH, of the lattice paths. In this case, from the recolouring procedure we obtain an object MATH; see the upper right picture in REF . CASE: The changing trail stops at the the leftmost endpoint, MATH, of the lattice paths. In this case, from the recolouring procedure we obtain an object MATH; see the upper right picture in REF . It is clear that altogether this gives a mapping of the set of all objects MATH into the union of the two sets of all objects MATH and MATH, respectively. Of course, this mapping is weight - preserving. It is also injective since the above construction is reversed by simply repeating it, that is, determine the changing trail starting at the rightmost endpoint MATH (this trail is exactly the same as before, only the colours are exchanged) and change colours. For an illustration, read REF from right to left. So what is left to prove is surjectivity: To this end, it suffices to prove that if we apply our (injective) recolouring construction to an arbitrary object MATH or MATH, we do always get an object MATH; that is, two r - tuples of nonintersecting lattice paths, coloured green and blue, and with the appropriate starting points and endpoints. We do have something to prove: Note that in both cases, A (see REF ) and B (see REF ), there is prima vista a second possible endpoint for the changing trail, namely the leftmost starting point, MATH, of the lattice paths, where the leftmost blue path starts. If this endpoint could actually be reached, then the resulting object would clearly not be of type MATH. So we have to show that this is impossible. (NAME left out this indispensable step in CITE, but we shall close this small gap immediately.) The following properties of changing trails are immediate: CASE: If some edge of a changing trail is used by paths of both colours green and blue, then it is necessarily traversed in both orientations and thus forms a changing trail (which is ``cycle - like") by itself. CASE: Two changing trails may well touch each other (that is, have some vertex in common), but can never cross. Now observe that in Case A, there is also a second possible starting point of a ``path - like" changing trail, namely the left - most endpoint MATH of the lattice paths (see the left picture in REF ). Likewise, in Case B, there is a second possible starting point of a ``path - like" changing trail, namely the rightmost starting point MATH of the lattice paths (see the right picture in REF ). In both cases, if the changing trail starting in MATH would reach the leftmost starting point MATH of the lattice paths, it clearly would cross this other ``path - like" changing trail; a contradiction to REF . (The pictures in REF shows these other changing trails for the examples in REF , respectively.) This finishes the proof.
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math/0004113
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Without loss of generality we may assume that all even points are white and all odd points are black in MATH. By recolouring changing trails, all the points MATH are matched with points of opposite colour and parity. So if MATH is odd and black, then the recolouring trail starting at MATH connects it which some other point MATH which is even and white: After recolouring, MATH is odd and white, and the recolouring operation altogether yields some two - coloured graph object belonging to some MATH. Now if we apply the recolouring operation to an arbitrary object from MATH, the only possible partners for ``wrongly - coloured" MATH (odd, but white) is another ``wrongly - coloured" MATH (even, but black). Hence this operation takes objects from MATH back to MATH.
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math/0004113
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In the notation of REF, let MATH and MATH; and choose horizontal offset MATH. That is, interpret MATH as the generating function of two - coloured graph objects consisting of two MATH-tuples of nonintersecting lattice paths, coloured green and blue, respectively, where green path MATH starts at MATH and ends at MATH, and where blue path MATH starts at MATH and ends at MATH. Observe that this setting obeys the assumptions of REF . Now consider the set of green endpoints MATH, where MATH. (Here, MATH is the fixed list of integers from REF .) NAME changing trails which start at these points amounts to determining the set MATH of respective endpoints of the changing trails, and changing colours. Assume that MATH, then in terms of the associated NAME functions REF directly leads to the identity: MATH where the notation of the summands means that REF were exchanged with parts MATH, respectively. By the NAME - NAME identity REF are in fact equivalent.
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math/0004114
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Let's prove the statement by induction on MATH. When MATH, by CITE MATH is a group algebra and if MATH then MATH and MATH. Now assume the statement is true for MATH. Consider MATH of dimension MATH. MATH and thus, by CITE, there exists a central grouplike of order MATH in MATH and therefore MATH contains a normal subHopfalgebra MATH. Thus we get a short exact sequence of NAME algebras MATH where MATH. Dualizing MATH we get another short exact sequence of NAME algebras: MATH where MATH and MATH. Thus we get MATH and MATH is not cyclic unless MATH. In the first case we are done since it implies that MATH is not cyclic. In the second case, since MATH is normal in MATH, MATH is isomorphic as an algebra to a twisted group ring MATH where MATH. It is easy to show that, since-MATH is a group algebra of a cyclic group, MATH is commutative. Thus MATH is cocommutative and the only possible MATH with a cyclic group of grouplikes is MATH.
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math/0004114
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In REF we have described all possible NAME ring structures of non-commutative semisimple NAME algebras of dimension MATH and there are exactly MATH of them. Only one of these MATH-rings is not commutative, namely MATH, which corresponds to nonabelian MATH. Therefore, by CITE all NAME algebras with non-commutative MATH-ring are twistings of each other with a MATH-pseudo-cocycle. Moreover, by CITE, NAME algebras with non-commutative MATH-rings are not twistings of group algebras. If MATH is abelian then there are MATH possibilities for the MATH-ring structure, all of which are commutative, namely MATH, MATH, MATH, MATH, MATH and MATH. Thus by CITE MATH is a twisting of one of these group algebras with a MATH-pseudo-cocycle. Since MATH is semisimple, MATH is also semisimple by CITE. Therefore, if MATH is commutative, as algebras MATH when MATH and MATH when MATH. If MATH is not commutative, that is MATH, it is easy to see that MATH and thus MATH.
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math/0004117
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First of all, choose an open covering MATH of MATH such that MATH and MATH have transition functions MATH and MATH respectively, both relative to the same open cover MATH. Then, as we have already seen, choices for MATH and MATH are given by MATH and MATH where the MATH are defined as above. If one performs the calculation, using the group law in the abelian group MATH given in REF, then one finds that MATH . But this is precisely the classifying map we would have obtained for MATH using the recipe of REF . The other statement of the proposition is proved similarly.
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math/0004117
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Consider first the case of a map MATH which is onto. Here MATH and MATH only have to be sets and MATH is a function. Denote as usual by MATH the MATH-fold fibre product over MATH. Then define MATH by MATH where MATH. In the case that MATH is a point there is a standard proof that this complex has no cohomology, see for example CITE. This can be adapted completely to the present case by choosing a section MATH of MATH. Notice that because we are dealing with sets the question of existence of a section depends only on the fact that MATH is onto. Indeed if MATH and MATH then define MATH by MATH and it is straightforward to check that MATH. Notice that if MATH then MATH means that MATH is constant on the fibres of MATH so MATH descends to a map MATH. Clearly MATH. Now return to the case at hand and define MATH to be the space of all singular simplices in MATH and MATH to be the space of all singular simplices in MATH. MATH induces a surjective map MATH. Notice that the space of all singular simplices in MATH is just MATH with respect to MATH. To prove the proposition we just have to notice that the space of all singular cochains in MATH is just MATH.
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math/0004117
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Choose an open cover MATH of MATH such that there exist local sections MATH of MATH. Let MATH. Define a MATH bundle MATH on MATH by MATH, where MATH sends MATH to MATH, where MATH. The section MATH of MATH induces a section MATH of MATH and hence an isomorphism MATH. The coherency condition on MATH ensures that the MATH satisfy the glueing condition MATH. Hence we can use the standard clutching construction to form a MATH bundle MATH from the MATH and one can check that there is an isomorphism MATH.
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math/0004117
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We first define a map MATH. If MATH then choose MATH and define MATH . Here MATH is the bundle gerbe multiplication in MATH. Clearly MATH is independent of the choice of MATH. We need to show that MATH is a descent isomorphism, that is, that the following diagram commutes MATH . So let MATH, MATH, and MATH. Then MATH . Therefore MATH descends to a bundle MATH.
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math/0004117
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The bundle gerbe product in MATH defines an isomorphism MATH. We need to check that this isomorphism is compatible with the descent isomorphism defined in REF . That is, we need to check that the following diagram over MATH commutes: MATH where MATH is the descent isomorphism constructed in REF . Let MATH, MATH, and MATH. Then MATH .
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math/0004117
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First of all define a map MATH by mapping a point MATH of MATH to the point MATH of MATH. We then have MATH. Note that MATH. We therefore have the following series of isomorphisms of MATH bundles MATH where MATH denotes the projection MATH and so on. The MATH bundle map MATH pulls back to define an isomorphism MATH . We need to show that MATH is compatible with the descent isomorphisms: MATH . This is clearly true, since MATH is induced by the bundle gerbe morphism MATH. Therefore there is an isomorphism MATH which is compatible with the descent isomorphisms. Therefore we must have MATH.
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math/0004117
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To show this we need to use the group multiplication in MATH using the non-homogenous coordinates on MATH as reviewed in CITE and see also REF. We get MATH where we have used REF-cocycle REF to write MATH equal to MATH and so on.
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math/0004117
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We will calculate the NAME class of MATH relative to the same open cover MATH of MATH used above. We choose sections MATH of MATH above each MATH and form the pullback MATH bundles MATH over MATH. Since the sections MATH and MATH are related by the transition cocycles MATH, we see that a classifying map for MATH is given by MATH. Thus choosing sections MATH of MATH corresponds to choosing lifts MATH of the MATH. We have a canonical choice for the MATH given in the non-homogenous coordinates by MATH where again MATH and the MATH are defined as above. The MATH need not be cocycles and indeed their failure to be is measured exactly by REF-cocycle MATH for a calculation similar to that in the proof of REF above shows that MATH where MATH is the homomorphism induced by the inclusion of the subgroup MATH. In other words, the lifting bundle gerbe MATH has NAME class equal to MATH.
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math/0004117
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First of all note that a singular REF-chain in MATH is a linear combination of REF MATH with integer coefficients. Since MATH is just a point, it follows that we may identify a singular REF-chain of MATH with a formal linear combination MATH. A singular REF-cochain is a linear map MATH. There is a canonical choice of a singular REF-cochain in MATH, namely the linear map which sends a formal linear combination MATH to the integer MATH. Let us call this singular REF-cochain MATH. It has the property that MATH, where MATH and MATH are the projections on the first and second factors of MATH respectively, and where MATH is addition. This is because a singular REF-chain of MATH may be identified in the same way as above with a formal linear combination MATH and it is easy to see that the value of MATH on such a REF-chain is equal to the value of MATH on such a REF-chain. We also have MATH, where MATH is the singular coboundary, since a singular REF-chain of MATH must be a linear combination of constant maps MATH with integer coefficients from which it follows that MATH evaluated at such a REF-chain must be zero. Next we have the short exact sequence of groups MATH, where MATH is the inclusion and MATH is the map sending MATH to MATH. We will think of MATH as a `principal MATH bundle' on MATH. Form the fibre product MATH. We then have a canonical map MATH. Form a REF-cochain MATH of MATH by pullback with MATH. The property MATH of MATH shows that MATH in MATH, where MATH is the coboundary defined in REF by adding the pullback maps MATH with alternating signs. It follows from REF that there exists MATH such that MATH. Since MATH, we have MATH so by REF again we get MATH for some MATH, where MATH is the projection. Denote multiplication in MATH by MATH and denote multiplication (addition) in MATH by MATH. We have a commutative diagram MATH where now MATH and MATH denote the projections on the first and second factors in MATH respectively and MATH is addition in MATH. From this diagram we get MATH and since MATH, this becomes MATH. Using REF again, we see that there exists MATH such that MATH, where MATH is the projection MATH. From this last equation, it is easy to see that MATH satisfies REF above. Also from the last equation we see that MATH, in other words REF is satisfied.
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math/0004117
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The problem is to show that MATH is a representative in singular cohomology of the NAME class of MATH. A singular representative for the NAME class of MATH can be constructed by observing as above that MATH gives rise to a MATH bundle gerbe MATH on MATH - the lifting MATH bundle gerbe associated to the short exact sequence of groups MATH. As we saw above, the NAME REF-cocycle MATH relative to some open cover MATH of MATH associated to the MATH bundle gerbe MATH maps to the NAME class of MATH under the isomorphism MATH. To construct the singular cocycle representing the NAME class of MATH we use the fact (see CITE and the note following REF ) that the singular NAME sequence for the open covering MATH is exact. Thus we view MATH as living in MATH and use the exactness of the singular NAME sequence to solve MATH for some MATH. Since MATH is locally constant we have MATH and so MATH. Applying the exactness of the singular NAME sequence again implies that there is a MATH so that MATH. Applying MATH to this equation gives us MATH and so there exists MATH in MATH with MATH and MATH. We want to show that MATH and MATH represent the same class in cohomology. We use the same open cover MATH as above and form the principal MATH bundles MATH over MATH. We have the following commutative diagram: MATH which yields by composition the commutative diagram MATH . We let MATH denote the pullback MATH and we let MATH denote the pullback MATH. By pulling back REF to MATH we get MATH where MATH. Since MATH, we may rewrite MATH as MATH where MATH. Now the lifts MATH of the MATH furnish us with sections MATH of MATH. Also it follows that we have MATH. Hence from the equation above we get MATH . It follows from the definition of MATH that MATH. This is because MATH so if we can show that MATH for a REF-simplex MATH and MATH then we will be done. By REF. Hence we have MATH . Therefore MATH plays the role of MATH. Note that MATH equals MATH. It follows from this that MATH and MATH represent the same class in cohomology.
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