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math/0102169 | Denote MATH . Clearly, MATH is a totally antisymmetric tensor. It follows from REF that MATH . First we show that MATH . One has MATH . Using REF one obtains MATH. Similarly, MATH . Finally, MATH . Therefore MATH from whence REF follows. REF takes the form MATH due to the fact that REF is now satisfied. Using REF and the condition that MATH respects the metric MATH we rewrite REF as follows: MATH . Using REF once more, we get MATH and rewrite REF in the form MATH . It follows from REF that MATH. Therefore one gets from REF that MATH . Cyclicly permuting the indices MATH in REF and adding the resulting equation to REF one obtains MATH . The last equality in REF follows from REF . Thus MATH . Summing up REF over all the cyclic permutations of indices MATH, one gets from REF that MATH. Therefore MATH. Now the statement of the proposition follows from REF . |
math/0102173 | It suffices to consider two cases: MATH is separable over MATH, or MATH. In case MATH is normal and separable over MATH, let MATH be a separable polynomial such that MATH (which exists by the primitive element theorem). Choose a lift MATH of MATH to MATH, and define MATH to be MATH. Then MATH is a complete discrete valuation ring with residue field MATH, so it is henselian. In particular, if MATH is any other separable polynomial such that MATH and MATH is any lift of MATH to MATH, then MATH has a root in MATH, so MATH is well-defined up to isomorphism. To extend MATH to MATH, set MATH to be the unique root of the polynomial MATH congruent to MATH modulo MATH (which again exists by NAME 's Lemma). It is easily verified that this definition is also independent of the choice of MATH and its lift MATH, which is precisely to say that the desired canonicality holds. In case MATH, take MATH to be a ring isomorphic to MATH and let MATH denote the isomorphism. Now map MATH into MATH by sending MATH to MATH. NAME is immediate in this case: if MATH is analogously defined to be isomorphic to MATH via MATH, we map MATH to MATH by sending MATH to MATH. |
math/0102173 | We will only give the arguments for MATH, as the arguments for MATH are the same. Also, we may assume MATH does not map MATH into MATH. Pick any residual uniformizer MATH in MATH, and let MATH. Then for all MATH, MATH. (First check this for powers of MATH, then extend by linearity.) We claim that MATH, which is to say MATH. Suppose that the contrary holds. Recalling that MATH for some MATH, we have MATH. By hypothesis, MATH, which is to say MATH. If we write MATH, then MATH. Since MATH is not divisible by MATH by hypothesis, we must have MATH for MATH and MATH, which is an absurdity. Thus MATH as claimed. Now for MATH and MATH, define MATH as follows. Choose MATH such that MATH, and set MATH . For MATH, we have MATH, so the sequence MATH converges in MATH to a limit MATH such that MATH. Moreover, MATH so that MATH. It is easily verified that the assignment MATH yields a derivation, that this derivation extends MATH, and that any derivation extending MATH must agree with this one on MATH for each MATH; from this final assertion follows the uniqueness. |
math/0102173 | It suffices to consider the cases in which MATH is separable but tamely ramified, purely inseparable, or an NAME extension, as any MATH can be expressed as a tower of these. In the first case, we can choose MATH, where MATH, and MATH is clearly overconvergent with respect to MATH. In the second case, we can choose MATH such that MATH. In the third case, choose MATH such that MATH for some MATH not divisible by MATH; then MATH exists and is a residual uniformizer, with respect to which MATH is overconvergent. Now MATH is a root of the polynomial MATH, which has coefficients which are overconvergent with respect to MATH. However, Crew CITE has shown that the ring of overconvergent series with respect to a given residual uniformizer is henselian. Thus MATH is also overconvergent with respect to MATH. |
math/0102173 | We first establish that MATH for all MATH. From the definition of the NAME map, MATH is congruent modulo MATH to MATH for any MATH lifting MATH. In particular, we can pick MATH to be a series MATH such that MATH for MATH, and then MATH will have the same property, so MATH will not be less than MATH (and not greater, since MATH). To establish the desired formula for MATH arbitrary, let MATH, and let MATH be the smallest nonnegative integer such that MATH. On one hand, we have MATH by the previous paragraph together with REF (or rather, by its analogue for MATH). On the other hand, modulo MATH, the coefficient of MATH receives zero contribution from MATH, a contribution divisible by MATH but not by MATH from MATH, and contributions divisible by MATH from MATH. Therefore the coefficient of MATH in MATH is nonzero modulo MATH, and MATH. We conclude MATH, proving the desired result. |
math/0102173 | Let MATH be such that MATH for MATH. (This is possible because MATH is overconvergent with respect to MATH and because MATH.) We will construct MATH for MATH such that MATH for MATH, and MATH; the existence of such MATH suffices to prove the desired assertion with MATH. We may start with MATH. Now suppose MATH have been constructed. Write MATH, and set MATH . Then MATH so that MATH. On the other hand, MATH, and MATH (the estimate MATH following from the bound MATH). Thus MATH for all MATH, and the construction of MATH is complete. |
math/0102173 | The first, second, and third assertions follow from the fourth one, since in the diagram MATH (with MATH) the left vertical arrow is injective by flatness (MATH is an unramified extension of discrete valuation rings, hence flat). Thus we concentrate our attention on proving that MATH is injective. Suppose MATH is a nonzero element of MATH such that MATH in MATH, and such that MATH is minimal for the existence of such an element. Then the MATH are linearly independent over MATH, otherwise we could replace one of them by a combination of the others and decrease MATH. Now each MATH can be uniquely written in the form MATH, and MATH implies MATH for all MATH. Writing MATH, we find (by the uniqueness of the decomposition of MATH) that MATH for each MATH. However, since the MATH are linearly independent over MATH, we have MATH for all MATH and MATH, contradicting the fact that MATH is nonzero. To prove the final assertion, suppose MATH is a nonzero element of MATH such that MATH in MATH, and such that MATH is minimal for the existence of such an element. By the first assertion, we have that MATH maps to zero in MATH, which means that the MATH are linearly dependent over MATH. Choose MATH in MATH, not all zero, such that MATH. Without loss of generality, suppose that MATH and MATH for all MATH. Then MATH maps to zero in MATH, and by the minimality of MATH, we must have MATH for all MATH, which is to say MATH. Since the MATH lie in MATH, we can apply the second assertion to deduce that MATH in MATH and hence also in MATH. Thus MATH as well. |
math/0102173 | Suppose MATH is a basis of MATH. After reordering the MATH suitably, one can find a basis MATH of MATH such that if one writes MATH with MATH, then MATH if MATH and MATH (by Gaussian elimination). Now simply replace MATH by MATH and one gets a set of vectors in MATH, and we can take MATH to be their span. |
math/0102173 | For short, we write MATH for MATH (with MATH). Choose MATH such that MATH, and choose linearly independent elements of MATH on which MATH acts via a matrix MATH with MATH. This step is accomplished using MATH: given MATH on which MATH acts via a matrix MATH, MATH acts on MATH via the matrix MATH. Thus repeated application of MATH will eventually produce the desired elements. Now for any MATH and any MATH, we shall show that MATH . (It is this point that is unclear in CITE.) Of course it suffices to work with MATH equal to one of our chosen elements. In that case, modulo MATH, MATH acts simply as the derivation MATH, so it suffices to show that MATH . But this is evident: applying MATH maps MATH to MATH, and the product of MATH consecutive integers is always divisible by MATH. With REF now proved, we conclude that for MATH with MATH, MATH as desired. |
math/0102173 | The argument is motivated by the fact that for all MATH, MATH . To check this, verify that the right side is a ring endomorphism and notice that the two sides agree when MATH. Given MATH a MATH-crystal over MATH equipped with MATH, we wish to define a linear map from MATH to MATH, with which we can then compose MATH to get a linear map from MATH to MATH. We will show that MATH is such a map, except that in general it maps MATH into MATH. But that will suffice, because then a suitable isogeny will produce an actual MATH-crystal structure on MATH equipped with MATH. First of all, we must show that the series in REF converges in MATH. This follows from REF : if MATH, then MATH, and applying REF with MATH gives the desired convergence. This completes the proof in the case MATH. In case MATH, we must also show that the series in REF converges to a limit MATH defined over MATH and not just over MATH. Let MATH be the matrix through which MATH acts on our chosen system of elements, and MATH be such that MATH and MATH, where MATH. As in the previous paragraph, we may deduce from the lemma that there exist constants MATH such that all but (at most) the first MATH terms of the series have absolute value less than MATH. Putting MATH as in the lemma, we have MATH . Thus MATH is overconvergent. |
math/0102173 | Choose an isobasis MATH of MATH such that MATH for some MATH. Now define MATH as an element of MATH. Then MATH . Thus the sequence MATH converges MATH-adically in MATH to a limit which we call MATH. We claim that MATH is actually defined over MATH. To show this, let MATH be the matrix by which MATH acts on the basis MATH, and let MATH. Now we have MATH, which we may rewrite as MATH . Putting MATH, we have MATH. Now write MATH, MATH, and MATH. Because MATH, for any MATH there exists MATH such that MATH and MATH for all MATH. Moreover, there exists MATH such that MATH for MATH. REF becomes MATH, giving the estimate MATH . A straightforward induction and the fact that MATH for MATH gives the conclusion MATH for all MATH. This bound is not strong enough to give the desired conclusion, but for large enough MATH it can be substantially improved. To be precise, we show that if we put MATH, then MATH for MATH. Again, this is by induction on MATH. For MATH, in case MATH we have MATH . Otherwise, using the earlier estimate MATH for MATH, we have MATH . From this and the estimate REF , we deduce the desired inequality by induction. All that remains is to show that the change of basis matrix from the MATH to the MATH is invertible. In fact, the inverse matrix can be constructed as an analogous change of basis matrix for the dual crystal MATH. |
math/0102173 | We define convergent sequences MATH such that MATH. We start with MATH. To define MATH and MATH, write MATH with MATH and MATH, and put MATH and MATH. Then the sequences MATH and MATH both converge MATH-adically to the desired MATH and MATH. To establish the final assertion, write MATH, and note that since MATH, MATH and so if MATH, then MATH; the proof that MATH for MATH is similar. |
math/0102173 | We first prove the statement with a weaker form of REF , namely that MATH has integral entries and determinant not divisible by MATH. For this, we may induct on the valuation of MATH. The case where this valuation is zero is trivial, so we assume MATH. By REF , there exists an invertible matrix MATH over MATH satisfying REF , and such that the entries of the first column of MATH are divisible by MATH. (Write MATH in the formulation of the lemma as a product of elementary matrices over MATH and lift each to an elementary matrix over MATH. Then MATH is also a product of such matrices.) We now can multiply MATH by a diagonal matrix on the right so as to divide the entries in the first column of MATH by MATH. This reduces the valuation of MATH while maintaining the integrality of the entries, so application of the induction hypothesis completes the proof of the weaker assertion. To prove the original assertion, it suffices to note that by REF again, the reduction of MATH is the product of elementary matrices, so again it can be lifted to a product MATH of elementary matrices so that MATH and MATH have entries which are finite sums of powers of MATH. Replacing MATH with MATH gives the desired result. |
math/0102173 | Put MATH. Let MATH be a positive rational number strictly less than MATH. For the moment, we enlarge the ring of scalars by replacing MATH with MATH. The operation of tilting consists of replacing a matrix MATH with the new matrix MATH. The tilted matrix MATH has coefficients in MATH and, for each MATH, is congruent modulo MATH to a finite sum MATH. There is no loss of generality in assuming that MATH has integral entries (by multiplying MATH by an appropriate scalar matrix, which in the end can be divided from MATH and MATH). Apply REF with MATH, and let MATH be the matrix MATH in the conclusion of the lemma. Then REF gives a decomposition MATH, where MATH has only positive powers of MATH (and constant term REF), and MATH has only negative powers of MATH (and invertible constant term). We now wish to untilt MATH and conclude that we still have the decomposition MATH. For this, it suffices to show that MATH have entries in MATH. This is obvious for MATH because it has entries in MATH, and for MATH because untilting a matrix with only negative powers of MATH only decreases the absolute values of its coefficients. As for MATH, note that for each MATH, there exists MATH such that MATH for MATH large enough, and so MATH for MATH large enough. Applying the final assertion of REF , we get that MATH for MATH large enough, so MATH for MATH large enough. We conclude that MATH has entries in MATH. Set MATH and MATH. We now have the desired factorization MATH, except that we have enlarged MATH and the assertion of the lemma does not permit such an enlargement. On the other hand, if MATH is a basis for MATH over MATH, with MATH, and MATH, then MATH. Since MATH has entries in MATH, MATH has entries in MATH for MATH. In particular, MATH has entries in MATH, and MATH is a decomposition of the desired form. |
math/0102173 | Clearly if MATH is bounded above, then MATH is invertible. Conversely, suppose MATH but MATH is not bounded above. Choose MATH such that MATH and MATH are bounded above for MATH. Let MATH be the sequence of indices MATH such that MATH for MATH, which by assumption is infinite. Put MATH then MATH as MATH. So we may choose MATH such that MATH. Now tilt MATH and MATH to obtain MATH and MATH. On one hand, we have MATH. On the other hand, we may assume without loss of generality that MATH and MATH are elements of MATH with nonzero reduction; then MATH and MATH are congruent modulo MATH to polynomials which are not both constant. (By construction, MATH and MATH are nonzero modulo MATH.) Thus their product is congruent to a nonconstant polynomial modulo MATH, contradiction. |
math/0102173 | By REF , we have MATH. The desired result then follows immediately from the NAME preparation theorem CITE. |
math/0102173 | The proof will resemble that of REF . By the previous corollary, MATH has finitely many zeroes in the formal unit disc, and we induct on the number of these zeroes, counted with multiplicities. If MATH has no zeroes in the disc, its inverse is rigid analytic (again by the previous corollary) and so we may use the trivial factorization MATH. Otherwise, let MATH be a zero of MATH, let MATH be the minimal polynomial of MATH over MATH, and let MATH be the reduction of MATH modulo MATH. By assumption, MATH, so the rank of MATH is less than MATH. Therefore there exists an invertible matrix MATH over MATH such that MATH, and likewise MATH, has its first column identically zero modulo MATH. Now let MATH be the diagonal matrix with MATH as its first entry and REF in its other diagonal positions. The matrix MATH now satisfies the same hypotheses as MATH, but MATH has fewer zeroes than does MATH. By the induction hypothesis, we have MATH, with MATH invertible over MATH and having constant term REF, and MATH defined over MATH. We now factor MATH as MATH, and the two terms again have the desired properties. |
math/0102173 | By REF , we have a decomposition MATH, with MATH having entries in MATH (and constant term REF) and MATH having entries in MATH. In particular, MATH is invertible in MATH, so REF allows us to write MATH as MATH, with MATH invertible over MATH (and having constant term REF), and MATH having entries in MATH. The decomposition MATH now has the desired form. |
math/0102174 | We proceed by induction on MATH. For MATH REF is true by REF . So, suppose that the transformation is already shown for MATH. Then, by using some elementary identities from CITE, MATH . Now we expand the last factors (those involving MATH) by applying the MATH, MATH, MATH, MATH, MATH, and MATH case of the MATH-Pfaff - NAME summation in REF . We obtain MATH . By the MATH, MATH, case of the inductive hypothesis we obtain MATH . Now, we evaluate the inner multiple sum by the MATH, MATH, MATH, MATH, MATH, and MATH case of the MATH-Pfaff - NAME summation in REF , and obtain MATH which is the right side of REF. |
math/0102174 | In REF, we let MATH and MATH, for MATH. In this case the MATH series on the right side terminates from below, and from above, and evalutes to one. In particular, the appearance of the factor MATH makes the terms in the series vanish unless MATH, MATH. Similarly, the appearance of the factor MATH ensures that if MATH, the terms of the series are zero. In total, only the term where MATH survives, and that term is just one. |
math/0102174 | We proceed by induction on MATH. For MATH REF is true by REF . So, suppose that the transformation is already shown for MATH. Then (again using some elementary identities from CITE), MATH . Now we expand the last factors (those involving MATH) by applying the MATH, MATH, MATH, and MATH case of the MATH summation in REF . We obtain MATH . By the MATH case of the inductive hypothesis we obtain MATH . Now, we evaluate the inner multiple sum by the MATH, MATH, MATH, and MATH case of the MATH summation in REF . We obtain MATH which, after an elementary manipulation of MATH-shifted factorials, gives us the right side of REF, as desired. |
math/0102174 | We have, for MATH, MATH by the MATH-IPD type transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by MATH and simplify to obtain REF. |
math/0102174 | We have, for MATH, MATH by the MATH-IPD type transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by MATH and simplify to obtain REF. |
math/0102174 | We have, for MATH, MATH by the MATH-IPD type transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by REF and simplify to obtain REF. |
math/0102174 | We have, for MATH, REF by the MATH-IPD type transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by REF and simplify to obtain REF. |
math/0102175 | With the notation as above, we prove by induction that MATH for all MATH . To establish the inequality for the case MATH let MATH . Define MATH . Then MATH . Suppose the proposition holds for some MATH . Let MATH be MATH-admissible. Without loss of generality, we may assume that MATH are successive integer intervals, that for all MATH . MATH-supp-MATH for some MATH and that MATH . Also let MATH be the integer interval MATH(MATH). Let MATH and MATH . For MATH set MATH and MATH . Then define MATH . Note that MATH for all MATH . If MATH set MATH . It follows from REF that MATH is MATH-admissible. Finally, MATH-let MATH . Now MATH . If MATH then either MATH for some MATH or MATH for some MATH . Therefore MATH . Hence MATH . Thus MATH whenever MATH is MATH-admissible. It follows that MATH . This completes the induction. |
math/0102175 | The proof is by induction on MATH . The case MATH is clearly true by definition. Suppose the result holds for some MATH . Let MATH . Then MATH where MATH and MATH is MATH-admissible. For any MATH . Since MATH is arbitrary, the result follows. |
math/0102175 | The proof is by induction on MATH . The case MATH holds by REF . Assume the result holds for some MATH . Let MATH . Suppose MATH is MATH-admissible. Let MATH and MATH . For each MATH choose MATH such that MATH . Now MATH . Hence MATH . On the other hand, for each MATH there exist MATH-admissible sets MATH such that MATH . Now MATH . Hence MATH is MATH admissible. Thus MATH . From the inductive hypothesis and REF we get MATH where both suprema are taken over all MATH-admissible sets MATH . This completes the induction. |
math/0102175 | Let MATH then MATH can be written as MATH where MATH, and MATH . In particular, there exists MATH such that MATH for all MATH . Therefore, MATH . Note that as MATH is spreading, MATH . From REF , MATH . But from REF , MATH . Hence MATH . Conversely, we prove by induction that MATH for all MATH . The cases where MATH is clear. Suppose the claim is true for some MATH . Let MATH with standard representation MATH as an element of MATH . Choose a sequence MATH in MATH that converges nontrivially to MATH . We may assume that MATH for all MATH . Now we may write MATH where MATH, MATH and MATH . Let MATH. Note that MATH . We consider two cases. CASE: There exists MATH such that MATH . In this case, MATH . It is clear that MATH is the standard representation of MATH as an element of MATH . By REF , the standard representation of MATH as an element of MATH has the form MATH . In particular, MATH . Thus MATH as required. CASE: Suppose MATH for all MATH . In this case, MATH for all MATH . Hence MATH for all MATH . Furthermore, MATH converges to MATH nontrivially. Thus MATH . Therefore MATH as required. Suppose MATH is a limit ordinal and the result holds for all MATH . Let MATH have standard representation MATH as an element of MATH . By the inductive hypothesis, for each MATH where MATH and MATH . By the argument in REF above, if there exists MATH such that MATH then MATH . Otherwise, MATH for all MATH . Hence MATH . This completes the induction. |
math/0102175 | The proof is by induction on MATH . The case MATH is clear. Suppose the result is true for some MATH . Then MATH . Suppose the proposition is true for all MATH where MATH is some limit ordinal. Clearly, MATH . On the other hand, let MATH have standard representation MATH as an element of MATH . It is clear that MATH is also the standard representation of MATH as an element of MATH for any MATH . In particular, MATH for all MATH . Hence MATH . It follows that MATH . This completes the proof. |
math/0102175 | Let MATH . The proof is by induction on MATH . If MATH then MATH for some MATH . Therefore MATH and hence MATH . Suppose the proposition holds for some MATH . Let MATH be a well founded block tree with MATH . For each MATH let MATH . According to CITE, MATH . Therefore, there exists MATH such that MATH . By the inductive hypothesis, MATH . Let MATH supp MATH . Then MATH implies MATH . Thus MATH . Since MATH is spreading, MATH for all MATH . It follows that MATH . Hence MATH . Suppose MATH where MATH is a countable limit ordinal and the proposition holds for all MATH. Since MATH for all MATH . By the inductive hypothesis, MATH for all MATH . Hence MATH . This completes the induction. |
math/0102175 | If MATH then MATH for some MATH . Hence there exists a MATH-block tree MATH on MATH such that MATH . Given MATH there exists MATH such that MATH-supp-MATH . According to REF , MATH where MATH is the unit vector basis of MATH . Since MATH-basis-MATH . Therefore, MATH-basis for all MATH . Since it is clear that MATH for all MATH whenever MATH is strictly increasing, it follows that MATH-basis for all MATH . By REF , MATH . Thus by CITE, there exists MATH such that MATH . Hence, for all MATH and all MATH . Choose MATH such that MATH . According to REF , MATH for all MATH . Applying CITE, we obtain MATH such that MATH for all MATH . By CITE, there exists MATH and MATH such that MATH and if MATH with MATH then MATH . Consider MATH . If MATH and MATH then MATH . Hence MATH . It follows that MATH for MATH . By REF , MATH contrary to REF . This proves that MATH . On the other hand, according to REF , for any MATH, MATH for any MATH . By REF , MATH . Therefore, there exists a MATH-block basis tree MATH on MATH with MATH . Hence MATH . Thus MATH . We conclude that MATH . As MATH it follows from CITE that MATH . |
math/0102175 | If MATH then MATH is a successor ordinal and MATH CITE. If MATH is a successor ordinal, let MATH and MATH . By REF , MATH . |
math/0102176 | The number of MATH in MATH with descent set MATH and MATH-cycles is the coefficient of MATH on the left hand side of the first equation. Let MATH be the partition with MATH parts of size MATH and let Lie-MATH be the symmetric function associated with the corresponding NAME character (for background on NAME characters and relevant symmetric function theory see CITE). By CITE, the number of MATH in MATH with descent set MATH and MATH-cycles is equal to the inner product MATH . From CITE it follows that Lie-MATH is the coefficient of MATH in MATH . This proves the first assertion. For the second assertion, note that MATH. This follows from the fact that if a permutation MATH has RSK shape MATH and descent set MATH, then its reversal has RSK shape MATH and ascent set MATH. Thus MATH as desired. |
math/0102176 | Given a length MATH word MATH on the symbols MATH, let MATH be the number of occurrences of symbol MATH in MATH respectively. Define a permutation MATH in two line form by putting MATH in the positions occupied by the MATH's of MATH from left to right, then putting the next MATH numbers in the positions occupied by the MATH's of MATH from left to right, and so on. For instance the word MATH corresponds to the permutation MATH . It is easy to see that in general the recording tableau of MATH under the RSK algorithm is equal to the recording tableau of MATH under the RSK algorithm. Arguing as in CITE, if the entries of the random word MATH are chosen independently with probability MATH of symbol MATH, then the resulting distribution on permutations MATH is the same as performing a MATH biased riffle shuffle. As in CITE, the combinatorial definition of the NAME function immediately implies that the chance that MATH has recording tableau MATH is MATH. |
math/0102176 | Fix any permutation MATH such that Des-MATH and such that MATH has RSK shape MATH (this is possible if MATH). Let MATH be the probability of obtaining MATH after a biased MATH shuffle. Since all MATH with Des-MATH are equally likely, MATH where MATH is the probability that a biased MATH shuffle leads to a permutation with inverse descent set MATH and RSK shape MATH, and MATH is the number of permutations with inverse descent set MATH and RSK shape MATH. Now MATH is the probability that after a biased MATH shuffle one obtains a permutation MATH with Des-MATH, shape-MATH. Note that MATH is simply MATH, since the insertion tableau can be any standard NAME tableau of shape MATH and descent set MATH, and the recording tableau can be any standard NAME tableau of shape MATH. |
math/0102176 | Let MATH be a fixed permutation such that Des-MATH; then Prob-MATH will denote the probability of obtaining MATH after a biased riffle shuffle with parameters MATH. Using REF one concludes that the sought cycle index is MATH REF implies that MATH is MATH multiplied by the probability that the recording tableau of a permutation obtained after a biased MATH shuffle has shape MATH. By REF , this latter probability is MATH. Hence the sought cycle index is simply the inner product MATH . Applying the NAME identity yields MATH . Since MATH this simplifies to MATH . |
math/0102176 | Let MATH be a fixed permutation such that Asc-MATH; then Prob-MATH will denote the probability of obtaining MATH after a MATH biased riffle shuffle followed by reversing the order of the cards. Using REF one concludes that the sought cycle index is MATH . From the proof of REF , MATH . Consequently the sought cycle index is simply the inner product MATH . Applying the dual NAME identity yields MATH . Since MATH this simplifies to MATH . |
math/0102176 | The generating function for fixed points is given by setting MATH for all MATH in the cycle index. This yields MATH . Multiplying and dividing by MATH gives MATH . Observe that MATH since this is what one obtains by setting all MATH in the cycle index. Hence the generating function for fixed points is MATH . Then one differentiates with respect to MATH, sets MATH, and takes the coefficient of MATH. |
math/0102176 | This follows because the coefficient of MATH in MATH is MATH. |
math/0102176 | As noted after REF , the cycle index of a MATH-shuffle followed by reversing the order of the cards is MATH . The proof of REF gives that MATH . Dividing these equations implies that MATH . This proves the first assertion of the theorem. The second assertion follows from dividing both sides of this equation by MATH and applying REF . (Note that if all but finitely many MATH, only finitely many terms in the generating function remain. Since MATH the NAME series converges at MATH provided that the remaining MATH's aren't too much larger than REF). |
math/0102176 | Given the cycle index for k-shuffles followed by a reversal, this follows from minor modifications of either the arguments in CITE or CITE. |
math/0102176 | Given a length MATH word MATH on the symbols MATH, let MATH be the number of occurrences of the symbol MATH in MATH respectively. Define a permutation MATH in two line form by putting MATH in the positions occupied by the MATH's of MATH from left to right, then putting the next MATH numbers (arranged in decreasing order) in the positions occupied by the MATH's of MATH from left to right, then the next MATH numbers (arranged in increasing order) in the positions occupied by the MATH's of MATH from left to right, etc. For instance the word MATH corresponds to the permutation MATH . If the word entries are chosen independently with MATH having probability MATH, the resulting distribution on permutations is the same as performing a MATH shuffle of this section and forgetting about signs. It is easy to see that the recording tableau of MATH under the RSK algorithm is equal to the recording tableau of MATH under our variant of the RSK algorithm. Let MATH be the number of occurrences of symbol MATH in a tableau MATH. By REF , the probability that MATH has recording tableau MATH under our variant of RSK is equal to MATH where MATH has shape MATH and satisfies REF shows that this sum is equal to MATH. |
math/0102176 | Let MATH be a fixed permutation such that Des-MATH and let Prob-MATH be the probability of obtaining MATH after a MATH unsigned type MATH shuffle. Using REF and the fact that the probability of MATH depends only on MATH through Des-MATH, it follows that the sought cycle index is MATH . Arguing as in REF shows that MATH . Thus the sought cycle index is simply the inner product MATH . Applying the third identity in the introduction (due to CITE) yields MATH . Since MATH, this simplifies as desired to MATH . For the second assertion, REF shows that the cycle index after reversing the card order at the end is given by MATH . It is easy to see that the MATH signs all drop out. |
math/0102176 | A permutation on MATH symbols is unimodal with maximum at position MATH if and only if it has descent set MATH. Hence REF implies that MATH . This can be further simplified using NAME 's identity (page REF) MATH with MATH replaced by MATH to yield MATH . Note that we have used the identity MATH . |
math/0102176 | First suppose that MATH. As indicated earlier in this section, each length MATH word MATH on the symbols MATH defines a permutation MATH. From this construction, it is easy to see that the recording tableau of MATH under the BRKV variation of the RSK algorithm is equal to the recording tableau of MATH under the RSK algorithm. Thus it is enough to prove that the probability that the word MATH has BRKV recording tableau MATH is MATH. This is immediate from REF . Now the case MATH can be handled by introducing MATH extra symbols between MATH and MATH - call them MATH and choosing each with probability MATH. Thus the random word is on MATH and these extra symbols. Each word defines exactly one permutation - the symbols MATH are treated as positive. By the previous paragraph, the probability of obtaining recording tableau MATH is equal to MATH where the associated MATH are defined by MATH . As MATH, this distribution on permutations converges to that of a MATH shuffle, and the generating function of the MATH converges to MATH . |
math/0102176 | For each MATH, let MATH be a random MATH matrix formed by letting each entry equal MATH with probability MATH, MATH with probability MATH, and MATH with probability MATH. Let MATH be the first time that all rows of MATH containing no zeros are distinct; from the inverse description of MATH shuffles this is a strong uniform time in the sense of REFB-REFD of NAME MATH, since if all cards are cut in piles of size one the permutation resulting after riffling them together is random. The separation distance after MATH applications of a MATH shuffle is upper bounded by the probability that MATH CITE. Let MATH be the event that rows MATH and MATH of MATH are the same and contain no zeros. The probability that MATH occurs is MATH. The result follows because MATH . |
math/0102176 | Given the results of REF, the proof of the first part runs along exactly the same lines as in the proof of REF . The second assertion follows from the observation that a MATH shuffle followed by reversing the order of the cards is conjugate (by the longest length element in the symmetric group) to a MATH shuffle. Alternatively, arguing as in the proof of REF , one sees that the effect of reversing the cards on the cycle index of a MATH shuffle is to get MATH . |
math/0102176 | A MATH shuffle corresponds to choosing at random a word where MATH appears MATH times, and each word has probability MATH. It is easy to see that the RSK recording tableau of the word and the corresponding permutation obtained after the shuffle are identical. Now the number of words of length MATH where MATH appears MATH times and with recording tableau MATH is equal to MATH, since such words biject with the possible insertion tableau which have shape MATH and weight MATH. |
math/0102176 | From REF and the description of iterations of top to random shuffles as mixtures of MATH shuffles, it follows that the sought probability is MATH . Finally observe the MATH, since the MATH ones must appear in the first row and what remains is a standard NAME tableau. |
math/0102183 | The results are invariant under orientation-preserving diffeomorphisms of MATH. Thus it is sufficient to prove the theorem assuming a conformal parametrization. Because REF implies that tangent vectors to the two surfaces have the same lengths, and lie within MATH or MATH as appropriate, it is clear that solutions are isometric surfaces in MATH and MATH. Equivalently, it is straightforward to check algebraically that any quaternionic solutions to REF have the following two properties: first, MATH is constant if and only if MATH is constant; second, the metrics induced by MATH and MATH are conformal, with lengths scaling by the factor MATH. Now assume MATH exists and write REF as MATH. This is integrable for MATH exactly when MATH vanishes. In conformal coordinates, the first term becomes MATH . Now MATH and MATH are pure imaginary, being tangent to MATH at MATH. It follows that MATH and so MATH . Comparing with REF, we see that our system is integrable for MATH exactly when MATH is minimal. Conversely, let MATH be conformal, and set MATH, so that REF becomes MATH, or MATH. Using the machinery of NAME - NAME, the integrability condition for such an equation is MATH (See CITE for a discussion of this fundamental theorem of calculus for NAME derivatives.) But we calculate MATH, and MATH, using REF. Thus the integrability condition is precisely REF, so a solution MATH exists exactly when MATH is CMC, and it is determined up to left translation. In either case, MATH and MATH are conjugate cousins; either could be derived from the other via REF, and both integrability conditions are satisfied. This verifies that MATH is minimal and MATH is CMC, no matter which we started with. |
math/0102183 | By definition of the shape operator, MATH, using MATH. Substituting MATH and MATH, we get MATH . The first term on the right can be simplified: MATH takes values in the tangent plane to MATH; since multiplication with the normal MATH simply rotates by MATH, we get MATH . Therefore, MATH as desired. |
math/0102183 | The cousin of MATH is a CMC surface of revolution with meridians contained in mirror planes with normal MATH. Since MATH is a maximal interval for which these planes are nonhorizontal, the upper half MATH is indeed conjugate to MATH on the claimed domain. The axes of the helicoid give circles of planar reflection on the CMC surface, contained in planes with normal MATH. In particular, the first axis is parametrized with speed MATH and so gives a neck circle with circumference MATH, while the polar axis is parametrized with speed MATH and so gives a circle with larger circumference MATH. Since MATH, this latter bulge circumference is less than MATH, showing that the CMC surface is an unduloid, not a nodoid. Note that this description shows easily that the meridians of any unduloid have length MATH between neck and bulge. Since unduloids are embedded, MATH is transverse to the vertical lines, and so MATH immerses MATH by REF . The bounding rulings MATH are MATH-Hopf circles through MATH, and hence MATH-Hopf project to single points MATH, which are at spherical distance MATH. |
math/0102183 | Let MATH be the MATH end of MATH, and MATH. REF states that MATH is exponentially asymptotic to a NAME unduloid MATH of necksize MATH. By REF, conjugation preserves exponential convergence, and so the bounding rays of MATH are asymptotic to the great circles bounding the helicoid MATH. However, by REF the two rays in the boundary of MATH are themselves great-circle rays. Thus they must cover the great circles bounding the cousin of the asymptotic unduloid. REF shows that these NAME to points at spherical distance MATH. |
math/0102183 | By REF the immersed open disk MATH is transverse to the vertical lines in MATH. It follows from REF that MATH immerses MATH into MATH. Finally, MATH is well defined up to left translation, so as in REF, MATH is well defined up to rotation. |
math/0102183 | By joining two copies of the metric, we obtain a spherical metric on the disk MATH whose completion boundary develops to the two endpoints MATH of the original arc. Any path in MATH then lifts to MATH, so MATH is the universal isometric covering space of MATH, that is, a line of slit spheres. The arc across which we joined the two copies of the original metric develops to MATH, and thus decomposes MATH into two rays of slit spheres. |
math/0102183 | The unduloid MATH is the cousin of the spherical helicoid MATH from REF . The half-bubble MATH is the image under MATH of the rectangle MATH, MATH, and we want to consider its NAME image. The two MATH-coordinate lines in MATH with MATH are MATH-Hopf and project, as before, to the points MATH . The NAME projection of MATH is the minimizing great-circle arc MATH of length MATH (with MATH as midpoint) connecting these points. Indeed the MATH-coordinate lines with MATH and MATH, which correspond to the necks of the unduloid, form (antipodal) arcs of length MATH along the axis of the helicoid. Since this axis is MATH-Hopf and MATH, these arcs project to the great-circle arc MATH with doubled length MATH. The MATH-coordinate line with MATH, on the other hand, corresponds to the bulge of the unduloid, and lies along the polar axis of the helicoid. This NAME to the complementary arc MATH of the same great circle (with MATH as midpoint). The MATH-coordinate lines foliate MATH and are great semicircles along the rulings of the helicoid, corresponding to meridian arcs from neck to neck on the unduloid. Since the NAME field of each is MATH, its MATH-Hopf projection is a round circle MATH in MATH. The points MATH, MATH and MATH are equally spaced along the ruling semicircle. The points MATH project to the same point MATH, while MATH projects to the point MATH. Because of the equal spacing, MATH and MATH must be diametrically opposite in MATH. As MATH increases from MATH to MATH, the points MATH and MATH each move at constant speed from MATH to MATH, along MATH and MATH, respectively. It follows that the circles MATH foliate MATH, and thus that MATH embeds MATH onto the slit sphere MATH. |
math/0102183 | Let MATH denote the disk MATH with the spherical metric induced from NAME projection of its cousin. Consider the horizontal boundary MATH of MATH, which has three connected components. The simply connected surface MATH has a cousin with three boundary components consisting of NAME fibers. Consider its quotient space MATH obtained by identifying each boundary component to one point of some three-point set MATH. Observe that the NAME projection descends to, and thus defines a spherical metric on, MATH. Moreover, MATH . NAME projects to the triple MATH. We claim MATH is the completion boundary of MATH; that is, that MATH. We first show that MATH is complete, by decomposing it into a (nondisjoint) union of four complete sets. Choose open sets MATH representing the ends of MATH; removing them from MATH leaves us with a compact set MATH. Its quotient is a compact, and hence complete, subset of MATH. On the other hand, according to REF each end MATH is exponentially close to some unduloid MATH of necksize MATH. Thus MATH is also exponentially close to MATH. Moreover, as shown in the proof of REF , these two surfaces are bounded by rays on the same MATH-Hopf circles. Choosing a smaller representative MATH if necessary, it follows that, when each is endowed with the spherical pullback metric from the cousin NAME projection, MATH is isometric to a subset of MATH. Similarly, the closure of MATH, taken within MATH, becomes isometric to a closed subset of the quotient of MATH. By REF , the latter quotient is the metric completion of a ray of spheres (with slit length MATH); in particular it is complete. Therefore, the closures of the three ends MATH are complete subsets of MATH. It remains to show that in the completion MATH the three points of MATH are indeed distinct. But the endpoints of the completion boundary arc of a ray of slit spheres have positive distance, and so each pair of points in MATH also has positive distance. |
math/0102183 | Let MATH be the developing map. Any path in MATH lifts, by completeness, to some path in MATH. Thus if we remove the preimage MATH from MATH, then MATH restricts to a covering map. Let us first deal with the case that the three arcs of MATH are disjoint from each other. The complement MATH then consists of two embeddable triangles. The completion MATH of the preimage MATH is a graph in MATH. Its vertices are the three completion points and the points of MATH, which we call the finite vertices. The graph MATH has valence two at any finite vertex, since MATH is an immersion there. The complement of MATH consists of embeddable triangles isometric to those in MATH. We claim there is a unique triangle MATH with no finite vertices. First, note that if some triangle had two (respectively, three) finite vertices, then it would be adjacent to the same triangle across all three of its edges. The join of these two triangles would be a once-punctured sphere (respectively, an entire sphere). But neither of these is a three-point metric, and in neither case is there a way to join further triangles. Second, if every triangle had a single finite vertex, then these vertices would all develop to the same point of MATH, since this is true for adjacent triangles. But then our metric would be a line of slit spheres. So there must be a triangle MATH with no finite vertices. Its complement in MATH consists of three rays of slit spheres, by REF ; in particular MATH is unique. This proves the claim and establishes the desired decomposition. Finally, we must deal with the case where one arc MATH of MATH includes the third point MATH of the triple in its interior. In this case, we replace MATH by its complementary arc in the same great circle. The combinatorial argument above applies to this new MATH, with the embeddable triangles replaced by domains isometric to hemispheres, whose completion boundaries develop to MATH. So there is a unique hemisphere whose three vertices form the completion boundary of MATH. We join this chosen hemisphere, across its nonminimizing edge, to the adjacent hemisphere. This gives an embeddable triangle MATH of area MATH, whose completion boundary develops to the original MATH. |
math/0102183 | We use REF to decompose the spherical metric on the disk MATH into an embeddable triangle MATH with three rays of spheres, and similarly, for MATH. The three vertices MATH uniquely determine the three minimizing arcs bounding the triangle MATH, except in the case when MATH contains an antipodal pair; then we have agreed to choose MATH to have area MATH. In any case, MATH is isometric to the triangle in MATH which sits to the left of MATH when traveling in the order of increasing label. Consequently, given MATH, MATH with MATH, the two embedded triangles MATH and MATH are isometric. This isometry extends to the spherical metrics on the entire disks MATH and MATH by mapping corresponding rays of slit spheres onto one another. |
math/0102183 | Except at isolated points, two nonidentical minimal surfaces intersect along a curve transversely, that is, with different conormals. From our assumption that MATH is empty to one side of MATH, and the fact that the surfaces are both transverse to the MATH-Hopf field, we conclude that MATH has the same sign on a dense subset of MATH. The claim then follows from REF. |
math/0102183 | Consider MATH with MATH; let MATH, MATH, MATH denote the common necksizes of MATH and MATH. We regard the cousin disks MATH and MATH as sections of MATH. Because MATH consists of three curvature lines of infinite length in the horizontal plane, the completion boundary of MATH must include the three MATH-lines added in MATH; the same is true for MATH. Therefore, the lemma will apply. The base disk of the bundle MATH consists, by REF , of a central embeddable triangle and three rays of spheres. The bundle over the MATH ray MATH is foliated by spherical helicoids, namely all vertical translates of the cousin of a half-unduloid with necksize MATH. By the asymptotics result of CITE and REF , each cousin is asymptotic, over MATH, to one of these helicoid leaves. Thus, there is a well defined asymptotic height difference between MATH and MATH at each end. If these three height differences are equal, then a vertical translation can be used to make MATH asymptotic to MATH over all three ends. We then claim MATH. These sections have the same smooth MATH-lines as boundary, and MATH is foliated by minimal surfaces: the vertical translates of MATH. We can find a leaf of the foliation which has a one-sided contact with MATH; the interior and boundary maximum principles then prove the claim. If the height differences are unequal, then after a vertical translation, MATH lies asymptotically above MATH on one end, but below it on some other end. Let us then consider the open set MATH consisting of those points over which MATH lies strictly above MATH. By assumption, MATH has a component MATH which includes one end. Then MATH is the projection of a piecewise smooth intersection curve MATH, connecting two of the completion MATH-lines. This curve has finite length. Moreover, MATH and MATH do not intersect over MATH. Thus we can apply REF to find that the periods of MATH and MATH cannot simultaneously vanish. This contradicts the fact that both MATH and MATH are half triunduloids. |
math/0102183 | We choose the points MATH such that on MATH we have MATH on balls MATH, whose radii satisfy MATH (see CITE). Due to this uniform curvature bound and the area bound, the MATH subconverge to a minimal surface MATH, together with their normals. On MATH, we have MATH, so MATH is not a plane. To bound the total curvature of MATH, we follow CITE in choosing connected subsets MATH with uniformly bounded area such that MATH and MATH consists of geodesic loops (around necks of the surface). Since the MATH are triunduloids, the number of loops needed is at most three; the NAME theorem then uniformly bounds the total curvature of MATH. Because total curvature is invariant under rescaling, and the rescaled MATH converge to MATH, we find that MATH has finite total curvature MATH. (See CITE and CITE for details.) The MATH are NAME with respect to the horizontal plane MATH. Suppose the symmetry planes of the MATH are not at bounded distance from the origin. NAME symmetry implies that on MATH the NAME map takes values in the upper hemisphere. Thus by convergence of normals, the minimal surface MATH also has NAME image in this hemisphere, contradicting the fact that MATH is nonplanar. So instead, the MATH have symmetry planes at bounded distance from the origin; a subsequence will have convergent symmetry planes. Convergence of normals guarantees that MATH has NAME symmetry in the limit plane. |
math/0102183 | An NAME surface meets the horizontal symmetry plane MATH perpendicularly. An end not meeting MATH must be horizontal; an end meeting MATH is vertical and is itself NAME. At any planar end of a minimal surface, the NAME map has a branch point CITE, so a vertical planar end contradicts NAME symmetry. Similarly, a vertical end with higher winding number cannot be decomposed into symmetric halves on the two sides of MATH, so it cannot be NAME. Thus a vertical end is catenoidal. If MATH is not a union of compact curves, there must be an end MATH meeting MATH. By the above, MATH is vertical and catenoidal, so it has nonzero force. Consider a simple closed loop around MATH, which is nontrivial in MATH. Because MATH has negative NAME curvature except at isolated points, there is a unique simple closed geodesic MATH homotopic to this loop. The reflection of MATH in MATH is also a geodesic, and so coincides with MATH. Thus MATH is symmetric; its force equals that of MATH. |
math/0102183 | Suppose we have MATH for some sequence of triunduloids MATH with necksizes greater than MATH. Then the rescaled surfaces MATH, as in REF , converge to an NAME minimal surface MATH. The MATH have genus MATH, so there are no closed loops in MATH. Since the MATH converge to MATH with finite multiplicity, there are no closed loops in MATH either. But MATH meets MATH, and hence does so in an unbounded curve. Thus, we can apply REF to get a closed geodesic MATH in MATH with nonzero force. This is the limit of closed curves MATH on MATH, whose lengths and forces converge to those of MATH. So on the unrescaled MATH, the corresponding curves MATH have nonzero forces, which converge to zero. Because MATH has nonzero force, it is nontrivial; a simple nontrivial curve on a triunduloid is homologous to the boundary of one of the ends. By the necksize bound, the force of REF is bounded away from zero, a contradiction. |
math/0102183 | Consider a compact subset MATH. By REF there is a MATH such that each surface MATH has necksizes bounded below by MATH, and thus uniformly bounded curvature by REF . Suppose two triunduloids in MATH are sufficiently close in NAME distance on an open domain in MATH with compact closure. Because of the uniform curvature bounds, either one can then be written as a normal graph over the other in any closed subset of the domain. By elliptic regularity, these subsets of the triunduloids are then also MATH (or MATH) close. Thus, up to left translation in MATH, the corresponding subsets of the cousins are again close, and so are the NAME projections of their boundaries. We conclude that MATH is continuous. To show that MATH is compact, let us now prove that any sequence MATH subconverges with respect to the topology on MATH. The MATH are given only up to horizontal translation; to get convergence, we choose representatives such that the enclosing balls MATH from REF are centered at the origin. Due to the area and curvature estimates we find a subsequence of the MATH that converges on each compact subset to some CMC surface MATH. By smooth convergence, it is clear that MATH is NAME. Because a sequence of contractible loops has a contractible limit, MATH has genus zero and at most three ends. On the other hand, by the enclosure theorem, on each MATH there are three closed curves in MATH which bound the three ends. These three sequences of curves subconverge to three closed curves which are disjoint in the limit surface MATH, and thus bound three different ends. Thus MATH is a triunduloid, as desired. |
math/0102183 | Note that a trinoid MATH has total curvature MATH. Hence the NAME map has degree two and so by the NAME formula at most two branch points, which are the umbilic points of MATH. Therefore the theorem applies. If MATH were degenerate, there would be a MATH . NAME field, which would be bounded, and thus induced by a translation. But a translation of a catenoidal end (even if orthogonal to the axis) has normal component which fails to be MATH. |
math/0102183 | Because MATH is a continuous injection, it is an embedding if and only if it is a closed (or open) map onto its image. Given a closed set MATH, we claim MATH is closed in MATH, and hence in MATH; applying the claim to MATH will confirm the last statement of the lemma. We check the claim by showing MATH is closed for all compact MATH. Since MATH is proper, MATH is compact, and so the closed subset MATH is also compact. Thus its image MATH is compact, and (since MATH is NAME) closed. |
math/0102183 | Consider the union of the closed MATH-simplices in MATH and remove its MATH-skeleton. Let MATH be the closure (in MATH) of some connected component of this set. Its MATH-simplices have valence at least two because MATH is borderless. Since MATH is closed, the restriction MATH is still proper. We will show this restriction is surjective. (This implies that MATH and that every valence is exactly two.) By REF , MATH is an embedding and MATH is closed in MATH. In particular, the image of the MATH-skeleton MATH is a properly embedded MATH-complex in the connected MATH-manifold MATH; therefore its complement MATH is connected. Letting MATH be the complement of MATH in MATH, we have that MATH is a closed subset of MATH. We claim that MATH is also open in MATH. It then follows by the connectedness of MATH that MATH maps MATH onto MATH. Since MATH and MATH is closed, MATH, as desired. To prove the claim, suppose first that MATH is interior to a MATH-simplex. Since the dimension of MATH is also MATH, and MATH is an embedding, MATH contains an open neighborhood of MATH. Otherwise MATH is interior to a MATH-simplex MATH. Because MATH is borderless, MATH is in the interior of the union of two MATH-simplices with common face MATH, so again MATH contains a neighborhood of MATH. |
math/0102183 | To apply REF to MATH, we need to verify its assumptions. The space MATH of triples is clearly a connected three-manifold. The classifying map MATH is proper and injective by REF . From the first part of REF , the moduli space MATH of triunduloids is locally a real analytic variety; injectivity of MATH shows it has dimension at most three. So REF implies MATH is a borderless MATH-complex for some MATH. It remains to check that MATH has dimension MATH. By REF there is a nondegenerate triunduloid, so by the second part of REF , MATH is a three-manifold in some neighborhood of this triunduloid. REF now applies to conclude MATH is a homeomorphism. |
math/0102185 | By REF, MATH for any point MATH. Then, by REF, there exists a MATH immersion MATH whose hyperbolic NAME map, secondary NAME map and NAME differential are MATH, MATH and MATH, respectively. Moreover, by REF, the induced metric is complete. |
math/0102185 | By REF, we have MATH which implies the first item of the theorem. In particular, if MATH, then REF imply MATH, and since MATH by REF and MATH by REF, the second item of the theorem follows. NAME more particularly, if MATH, then MATH and MATH. But the proof in CITE of REF (in this paper) shows that MATH cannot be MATH or MATH as well, implying the third item of the theorem. Suppose MATH. Then REF implies MATH and we consider two cases: CASE: If MATH, then REF implies that MATH for all MATH, so MATH for all MATH, and hence REF implies that MATH for all MATH. But by REF, we have MATH, and so MATH for all MATH. CASE: If MATH, then MATH holds because of REF. Hence by REF, MATH, but REF implies MATH, so MATH. This implies MATH and MATH, by REF. Suppose MATH. Then MATH, and then by REF, the only end MATH is regular and embedded. Then REF implies that the end has nonzero flux, contradicting REF, so MATH. |
math/0102185 | We only have to show that a MATH surface with MATH of genus MATH and with MATH regular end satisfying MATH cannot exist. By REF, such a surface would satisfy MATH and hence the hyperbolic NAME map MATH is a meromorphic function on a compact NAME surface MATH of genus MATH with MATH. Therefore MATH can be only MATH or MATH, and so MATH, a contradiction. |
math/0102185 | Assume MATH is of class MATH. In this case, MATH by REF, and then MATH is MATH-reducible (the case of REF). Then by REF, the dual immersion MATH is also well-defined on MATH, with dual absolute total curvature MATH. Such a surface cannot exist because of the results in CITE (see REF ). Now suppose MATH is of class MATH. If MATH, then for the same reason as in the MATH case, such a surface does not exist. Now assume MATH. Then the surface is in class REF: MATH. By the same argument as in the case MATH of MATH in the proof in CITE of REF , such a surface cannot exist. |
math/0102185 | Assume such an immersion MATH exists. By REF, there is the only umbilic point MATH. We set the ends MATH and the umbilic point MATH. Then the NAME differential MATH has a zero only at MATH with order MATH, and two poles at MATH and MATH with orders MATH and MATH, respectively. Thus, MATH is written as MATH . On the other hand, by REF, we have MATH, and MATH . The secondary NAME map branches at MATH and MATH with branch orders MATH, MATH, MATH and MATH, respectively. Then by REF , we can take the secondary NAME map MATH such that MATH where MATH and MATH REF are mutually distinct points. Without loss of generality, we may assume MATH (if not, we can take MATH instead of MATH). Then by REF , we have MATH so MATH or MATH is an odd integer. Then by REF, we have MATH and MATH or MATH. First, we assume MATH, and we set MATH. Then MATH . Such a MATH exists if and only if the residues of the right-hand side of REF vanish: MATH (compare REF). Since MATH, these equations are equivalent to MATH and MATH: MATH . On the other hand, let MATH, and consider the equation MATH which is named REF. The roots of the indicial equation of REF at MATH are MATH and MATH. By REF, the log-term coefficient of REF at MATH must vanish if the surface exists: MATH . (See REF or REF). Here, the solution of REF is MATH a contradiction. Hence the case MATH is impossible. Next, we consider the case MATH. Then the secondary NAME map MATH satisfies MATH . The residue at MATH vanishes if and only if MATH . On the other hand, in the same way as the first case, the log-term coefficient of REF at MATH vanishes if and only if MATH . However, there is no pair MATH satisfying REF simultaneously. Hence this case is also impossible. |
math/0102185 | Assume such an immersion MATH exists. Then we have MATH because of REF, and by REF, it holds that MATH . We set MATH. Since MATH, there are no umbilic points, by REF. Then the NAME differential MATH has a pole of order MATH at MATH, a pole of order MATH at MATH, and a zero of order MATH at MATH. Hence MATH is written as MATH . The secondary NAME map MATH branches at MATH, MATH and MATH with orders MATH, MATH and MATH, respectively. Then by REF, MATH can be put in the following form: MATH where MATH are mutually distinct numbers, MATH is a positive real number, and MATH or MATH, and MATH . The second case of REF is impossible because of REF. Hence MATH, and then MATH and MATH. Thus we have the form MATH . Such a MATH exists if the residues at MATH and MATH vanish. This is equivalent to MATH . By direct calculation, we have MATH where MATH. Consider REF with MATH . Then, the indicial equation at MATH has the two roots MATH and MATH with difference MATH. By direct calculation again, the log-term at MATH vanishes if and only if MATH which is impossible because MATH. |
math/0102185 | For MATH and MATH as in REF, set MATH . Since MATH, the right-hand side is not identically zero. Hence there exists a MATH immersion MATH with hyperbolic NAME map MATH, secondary NAME map MATH and NAME differential MATH, by REF . Moreover, one can easily check that MATH is complete and of type MATH with MATH. Conversely, suppose such a surface exists. Then without loss of generality, we set MATH and MATH. By REF, there are no umbilic points. Then the NAME differential MATH has poles of order MATH at MATH and MATH, and has no zeros. Hence we have MATH . Hence the secondary and hyperbolic NAME maps branch only at the ends. Then by a similar argument as in CITE, we have MATH where MATH . Here, MATH because of REF. Hence we have MATH . Thus we have MATH . On the other hand, REF imply that MATH or MATH. If MATH, REF implies that MATH. Then by REF, MATH for MATH. Hence using REF, we have MATH . Here MATH and MATH are integers, hence MATH. The hyperbolic NAME map MATH is a meromorphic function on MATH. Then the NAME relation implies that MATH . This is impossible, and hence we have MATH. When MATH, by similar arguments, we have MATH and MATH. This implies that MATH. Thus by REF, we have MATH . We may assume that MATH (if not, exchange the ends MATH and MATH). Assume MATH. In this case, MATH which is impossible. Hence, using also REF, we have MATH. Moreover, MATH, so MATH. Putting all this together, we have MATH . If MATH, the secondary NAME map MATH is a meromorphic function on MATH with only one branch point, which is impossible. Hence MATH. In this case, the pseudometric MATH branches on the divisor MATH because there are no umbilic points (see REF). Thus the secondary NAME map MATH satisfies REF. One possibility of such a MATH is in the form REF with MATH. On the other hand, since the surface is MATH-reducible, MATH can be normalized as in REF because of REF. Since MATH, the Schwarzian derivative of the hyperbolic NAME map MATH is uniquely determined, and MATH is determined up to NAME transformations. Then such a surface is unique, with given MATH and MATH. |
math/0102185 | The residue of MATH in REF at MATH and the residue of MATH in REF at MATH vanish. Thus there exist MATH and MATH such that REF hold. Moreover, by direct calculation, we have MATH . Then by REF , there exists a MATH immersion MATH. One can easily check that MATH is complete and MATH. Conversely, assume such an immersion MATH exists. Then by REF, there is only one umbilic point of order one. By REF, we have MATH or MATH. When MATH, by REF, MATH holds, and then MATH. Assume MATH and MATH. If one of MATH is an integer, by REF, the other is also an integer, and hence the surface is MATH-reducible. Thus both MATH and MATH are non-integers. By REF, we have MATH . Set MATH and the umbilic point MATH. Then the secondary NAME map MATH branches at MATH, MATH, MATH and MATH with orders MATH, MATH, MATH and MATH, respectively. Then by REF, MATH can be chosen in the form MATH where MATH and MATH are mutually distinct points. We may assume MATH (if not, take MATH instead of MATH). Then by REF, MATH holds. This implies that MATH or MATH is an even integer. Then by REF, we have MATH . Hence we have MATH . Such a map MATH exists on the universal cover of MATH if and only if the residue at MATH of the right-hand side of REF vanishes, that is, if MATH . The NAME differential of such a surface is written in the form MATH because it has poles of order MATH at MATH and MATH, a pole of order MATH at MATH and a zero of order MATH at MATH. Let MATH, MATH and MATH be the branch orders of the hyperbolic NAME map at MATH, MATH and MATH, respectively. Then by REF, we have MATH where MATH . Then MATH and MATH are positive real numbers. Without loss of generality, we may assume MATH. Then we have MATH . Hence MATH, that is, MATH. Since MATH by REF, the hyperbolic NAME map branches at MATH, MATH, MATH and MATH with branching order MATH, MATH, MATH and MATH, respectively. Then the NAME relation implies that MATH . On the other hand, we have MATH. Hence we have MATH . We set MATH. By a suitable NAME transformation, we may set MATH. Since MATH is a point of multiplicity MATH, MATH has no pole except MATH. Then MATH is written in the form MATH and we can choose MATH by a suitable NAME transformation. Moreover, the NAME differential MATH is as in REF if and only if MATH . Thus we have the conclusion. |
math/0102185 | Since the residue of MATH in REF at MATH vanishes, there exists a meromorphic function MATH satisfying REF. Since the metric MATH is non-degenerate and complete on MATH. Hence there exists a MATH immersion MATH with hyperbolic NAME map MATH and NAME differential MATH as in REF. (In fact, there exists a MATH immersion MATH with NAME data MATH. Then taking the dual yields the desired immersion.) Let MATH be the lift of MATH. Then MATH is a solution of REF, and there exists a representation MATH as in REF. The components MATH and MATH of MATH satisfy the equation MATH in CITE: MATH . By a direct calculation, the roots of the indicial equation of REF at MATH are MATH and MATH, and the log-term coefficient at MATH vanishes (see REF). Hence MATH and MATH are meromorphic on a neighborhood of MATH, and then, the secondary NAME map MATH is meromorphic at MATH. Hence, by REF, MATH, where MATH is a representation corresponding to the secondary NAME map MATH, and MATH is a deck transformation corresponding to a loop surrounding MATH. Moreover, the difference of the roots of the indicial equation at MATH is MATH. This implies that one can choose the secondary NAME map MATH such that MATH, where MATH is a deck transformation of MATH corresponding to a loop surrounding MATH. Then MATH. Hence the representation MATH lies in MATH, since MATH and MATH generate the fundamental group of MATH. Then by REF , the immersion MATH is well-defined on MATH, and by REF , MATH is a complete immersion. Using REF, we have MATH . Then by REF, we have MATH. In the second case, we can prove the existence of MATH in a similar way. Conversely, we assume a complete MATH-reducible immersion MATH of type MATH with MATH exists. Without loss of generality, we may set MATH and the only umbilic point MATH. As shown in the proof of REF , we have MATH. Thus, by REF and the assumption MATH, we have MATH. Hence by REF, we can set MATH . Without loss of generality, we may assume MATH. Then by the NAME relation, we have MATH . On the other hand, MATH. Thus we have MATH or MATH. We set MATH. Assume MATH. Then by REF, MATH. Hence, the hyperbolic NAME map MATH branches at MATH, MATH, MATH and MATH with branch orders MATH, MATH, MATH and MATH, respectively. By a suitable NAME transformation, we assume MATH. The multiplicity of MATH at MATH is MATH. Then MATH has no other poles on MATH. Thus, MATH can be written in the form MATH where MATH. By a suitable NAME transformation, we may set MATH. On the other hand, the NAME differential MATH is written in the form MATH because MATH is type MATH and MATH is the umbilic point of order MATH. Thus, by REF, we have MATH where MATH . Thus we have REF. Next, we assume MATH. Then by REF, we have MATH. If we set MATH, then MATH has only one simple pole other than the pole MATH, since the multiplicity of MATH at MATH is MATH. So MATH is written in the form MATH where MATH, which can be set to MATH by a suitable NAME transformation. The residue of MATH at MATH vanishes if and only if MATH . On the other hand, the relation REF also holds in this case. Thus we have REF. |
math/0102185 | Assume such an immersion MATH exists. Then there are no umbilic points, and by REF, MATH holds. Let MATH be the hyperbolic NAME map. Then MATH because MATH. Hence by the NAME relation, MATH holds. This implies that MATH is an even number not less than MATH: MATH . Since MATH, we have MATH . Hence by REF, MATH . We set MATH . We may assume MATH. (In fact, if MATH, the NAME transformation MATH maps the ends MATH to MATH.) The NAME differential can be written as MATH . By the relation REF, we have MATH where MATH for MATH. Since MATH and MATH is a non-negative integer, we have MATH and consequently, MATH (MATH). Let MATH . Then MATH which implies that the MATH REF are real numbers. And then MATH are real numbers. Here, without loss of generality, we may set MATH. (In fact, if MATH, applying the coordinate change MATH, we may set MATH. Moreover, if MATH, by the transformation REF, we may set MATH.) We choose the sign of MATH as MATH. Then, we have MATH . Moreover, by REF, we have MATH . Using this, we have MATH . To prove REF, by REF we have MATH for MATH. Then MATH. Hence by REF, MATH. By definition, this implies that MATH . Thus we have MATH . As the MATH are non-negative integers, MATH, which implies REF. By REF and the definition of MATH, we have MATH . Since MATH, this implies that MATH . Now, defining MATH (MATH), REF imply MATH . Moreover, MATH holds. To prove this, if MATH, then MATH. This implies that MATH is an odd number, contradicting REF. Using REF, the equality REF is written as MATH . We shall prove that REF cannot hold, making a contradiction. Let MATH, MATH, MATH be positive integers which satisfy REF. Define MATH on the closure MATH of the open domain MATH in the MATH-plane. Then it holds that MATH . To prove this, note that since MATH is a continuous function on a compact set MATH, it takes a minimum on MATH. By a direct calculation, we have MATH . So MATH does not take its minimum in the interior MATH of MATH, but rather on MATH. Similarly, MATH on MATH, so the minimum occurs at MATH, where MATH if the line MATH does not intersect MATH. If the line MATH does intersect MATH, then MATH holds, and the minimum occurs somewhere on this line with MATH in the interval MATH. We have MATH which minimizes at the endpoints of the interval MATH, where its values are MATH . Hence MATH on MATH, contradicting REF and proving the theorem. |
math/0102185 | Assume MATH is reducible. Then one can choose the developing map MATH such that MATH is diagonal. Then for each deck transformation MATH, MATH holds. Hence we have MATH. Differentiating this, MATH holds. Hence MATH is single-valued on MATH. Conversely, we assume MATH is well-defined on MATH for a developing map MATH. Then MATH is a constant. Hence we have MATH for some constant MATH. Then MATH is diagonal. |
math/0102185 | If MATH is MATH-reducible, then the developing map MATH is a meromorphic function on MATH. So the branch orders must all be integers. Conversely, assume all conical singularies have integral orders. Then by REF, MATH for each MATH, where MATH is the deck transformation on MATH corresponding to the loop surrounding MATH. Since MATH is generated by MATH, MATH is the trivial representation. |
math/0102185 | Let MATH be a developing map such that MATH is diagonal. Here, as in the proof of the previous lemma, we have MATH (MATH). Then we have MATH because MATH. Hence MATH is also a diagonal matrix. |
math/0102185 | If MATH is MATH-reducible, MATH is a meromorphic function on MATH which branches at MATH with branch orders MATH. Hence MATH is a zero of order MATH or a pole of order MATH of MATH for each MATH. Let MATH be the simple poles of MATH on MATH, then each MATH is a pole of order MATH of MATH. (The MATH are not branch points of MATH.) The zeros and poles of MATH are the branch points and the simple poles of MATH. Hence we have REF for MATH. By a suitable change MATH (which is a special form of the change REF), we can choose MATH such that MATH. Since MATH is a zero of order MATH or a pole of order MATH of MATH, we have REF. Next we assume MATH is MATH-reducible. Without loss of generality, we may assume MATH and MATH. Then by REF , we can choose the developing map MATH as MATH where MATH is a rational function. By REF, we have MATH . Differentiating this, we have MATH where MATH is a rational function and MATH . Since each MATH REF is a branch point of MATH of order MATH, we have REF by an argument similar to the MATH-reducible case. Moreover, since MATH, we have REF. |
math/0102185 | Let MATH be a lift of MATH. Then MATH is well-defined on MATH if and only if MATH. This is equivalent to MATH being the trivial representation, by REF. |
math/0102186 | We claim equality instead of the mere existence of an isomorphism because the direct product can be interpreted as an inner direct product as follows. The maps MATH and MATH both are non-degenerate and injective, which yields monomorphisms MATH and MATH, respectively, by REF . We have to prove that each projectivity in the join can be written as a product of a projectivity in MATH with a projectivity in MATH. The simplices of MATH are written as MATH, implying that MATH and MATH. Note that the distinct facets MATH and MATH are adjacent if and only if MATH and MATH adjacent to MATH in MATH, or MATH adjacent to MATH in MATH and MATH. Moreover, any two facets MATH and MATH with MATH adjacent to MATH and MATH adjacent to MATH are contained in the star of the codimension-MATH-face MATH. There are precisely two more facets contained in this star, namely MATH and MATH. Applying REF to MATH yields MATH and thus MATH . Invoking the identity REF several times, allows to ``sort" a projectivity: Each projectivity MATH from MATH onto itself can be written as the product MATH . |
math/0102186 | Fix an arbitrary facet MATH of MATH and an arbitrary coloring of the vertices of MATH. For each facet path MATH from MATH to some other facet MATH the projectivity MATH induces a coloring of the vertices of MATH. Two such colorings induced by facet paths MATH and MATH, respectively, coincide if and only if the projectivity MATH, which is induced by the facet loop MATH based at MATH, is the identity. Observe that, in general, the color of a vertex MATH does depend on the choice of the facet MATH. Since, however, the star of MATH in MATH is also strongly connected, this color is the same for all facets containing MATH. |
math/0102186 | Let MATH be an arbitrary facet loop based at MATH which is null-homotopic. Without loss of generality let MATH be the vertex of MATH corresponding to the barycenter of MATH. It is known that MATH can be contracted to the constant map MATH at MATH within the MATH-skeleton of MATH. NAME a suitable homotopy from MATH to MATH yields a sequence MATH of closed paths in the MATH-skeleton from MATH to MATH in the MATH-skeleton of MATH such that MATH, MATH, and MATH coincides with MATH outside some MATH-face MATH of MATH; see REF . The dual of MATH in MATH is a codimension-MATH-face MATH. Because the facet paths MATH and MATH are the same outside MATH we have that the projectivity MATH coincides with some projectivity MATH, where MATH is the common initial segment of MATH and MATH up to some facet MATH and MATH is a facet loop in MATH based at MATH. In particular, by REF , MATH is either a transposition or trivial, depending on the parity of MATH. An induction on MATH establishes the theorem. |
math/0102186 | The boundary complex of a polytope is homeomorphic to a sphere, and thus the fundamental group is trivial, provided that the dimension of the polytope is at least MATH. The group of projectivities coincides with the reduced group of projectivities. For MATH-dimensional polytopes the only MATH-face is the polytope itself and the result follows from REF . A MATH-dimensional polytope does not have any MATH-face, its dual graph consists of two isolated points, and hence the group of projectivities is trivial. |
math/0102186 | Each MATH-face of a simplex is a triangle. Each MATH-face of the dodecahedron and the MATH-cell is a pentagon. Each MATH-face of the MATH-cube is a quadrangle. |
math/0102186 | The product MATH is again a simple polytope. Its boundary complex is dual to the join of the duals of the boundary complexes of MATH and MATH. |
math/0102186 | Let MATH be an even simple MATH-polytope. Due to REF we know that this is characterized by the property that the group of projectivities vanishes. A proper coloring of the facets of MATH clearly corresponds to a proper coloring of the vertices of the dual MATH. The existence of such a coloring now follows from REF . This proves the equivalence of the first, the third and the fourth statement. The equivalence of the first and the second statement is known. We indicate a short proof in the Appendix. |
math/0102186 | The dimension MATH of a freely acting subgroup is bounded from above by MATH according to CITE. The same number is bounded from below by MATH by a result of CITE, see CITE. But REF enforces MATH. |
math/0102186 | Let MATH be an edge of MATH. If MATH is properly MATH-colored, then the two facets MATH and MATH have the same color. Assign this color to the edge MATH. Evidently, this procedure requires exactly MATH colors. Assume that this edge coloring is not proper, that is, there are vertices MATH, MATH, MATH such that MATH and MATH are edges of the same color. Then we have MATH, but the facets MATH and MATH both contain the vertex MATH. This contradicts the assumption that MATH is a proper coloring of the facets. |
math/0102190 | The implication ` REF ' is trivial. Conversely, suppose MATH , and choose a polynomial MATH such that MATH has polynomial coefficients. Then MATH ; since MATH is a bounded operator on MATH , it follows that MATH . Hence MATH , as required. |
math/0102190 | The corresponding result for the closures MATH is trivial, so the Corollary follows at once from REF and the fact that the operators in MATH have rational coefficients. |
math/0102190 | Everything is trivial, except possibly to check that if MATH and MATH are anti-automorphisms, and MATH , then MATH . For this, we need to prove that MATH . But because of REF , it is equivalent to prove that MATH which is easy. |
math/0102190 | It is obvious that MATH . The converse is a consequence of REF . Let MATH : then MATH for some MATH , so if MATH , then MATH is a polynomial, since MATH . Hence MATH . |
math/0102190 | By REF , the functions REF all belong to MATH . They span MATH because the projection MATH is an isomorphism, and the MATH-th function REF has the form MATH (lower terms) for large MATH . |
math/0102190 | Note first that MATH (we used the rule in REF at the last step). So REF is equivalent to MATH . That is true if and only if the operator acting on MATH on the left has negative order; equivalently, if the operator MATH has negative order. If MATH is differential, this means exactly that MATH . Since MATH and MATH (and hence also MATH) have rational coefficients, and MATH has polynomial coefficients, it is clear that MATH . |
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