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math/0008085 | Notice that if MATH for all MATH, the quantity REF equals MATH by CITE. The following argument is a simple generalization of that fact. Suppose MATH is a generic, REF-parameter family of perturbations. Let MATH be the parameterized moduli space. Recall that MATH is a smooth REF-manifold. If all the perturbations MATH in the path are sufficiently small, then MATH, hence for each MATH, MATH gives a REF-dimensional cobordism from MATH to MATH with orientations given by the spectral flow. Thus each sum MATH is independent of MATH, which proves the proposition. |
math/0008085 | Since invariance of MATH is proved in CITE, we only need to prove that MATH is independent of all choices made. Now MATH where MATH is a reducible flat MATH connection close to MATH. Of course MATH for some MATH, and making compatible choices for MATH and MATH, we see that MATH . The third step follows by additivity of the spectral flow together with the fact that MATH, since the NAME function is constant along connected components of flat connections. Now letting MATH and applying REF completes the proof. |
math/0008085 | We show that MATH for sufficiently small MATH whenever REF holds, that is, whenever MATH for all MATH. This cohomology assumption implies MATH also vanishes for every MATH for any small MATH . (Note that the assumption of smallness of MATH here is stronger than the assumption needed to define MATH . ) Thus MATH and MATH for all MATH. This shows that each summand in the definition of MATH vanishes for MATH sufficiently small. |
math/0008085 | In CITE, it is proved that MATH. So, the lemma follows once we show that MATH satisfies the same formula. Reversing the orientation of MATH changes the sign of the NAME function but has no effect on the perturbations. Therefore, there is a natural correspondence between the flat moduli spaces MATH and MATH. Obviously, if MATH is regular, then so is MATH. The odd signature operator MATH acts on MATH by MATH . Changing the orientation of MATH changes the sign of the NAME star operator. Replacing MATH by MATH as well, we see that MATH . Hence, if MATH, then MATH . Thus switching orientations and replacing MATH by MATH reflects the spectrum through zero. The following formula is a consequence of the MATH convention: MATH . Now suppose MATH is a small perturbation and MATH is regular. If MATH then MATH and MATH and so MATH . In this formula, on the left the spectral flow is taken from MATH to MATH, and on the right it is from MATH to MATH. Further, if MATH is flat and reducible, then it is a simple exercise to prove MATH. REF then implies that MATH . Similarly MATH. Thus MATH . |
math/0008085 | Since MATH was shown to satisfy a similar formula in CITE, it suffices to show additivity of MATH under connected sum. We first claim that the numbers MATH and MATH are additive under connected sum. To make this precise, we need to set up the notation. Set MATH . Then every connection MATH on MATH is of the form MATH, where MATH is a connection on MATH and MATH is the gluing parameter. Furthermore, if MATH is reducible and flat, then so are MATH and MATH. For MATH, let MATH be the components of MATH, where MATH is the component containing the trivial connection. The components of MATH are then given by the sets MATH for MATH and MATH. Note that MATH is now the component containing the trivial connection. For each MATH, we also choose an open set MATH containing MATH so the collection MATH is disjoint. For MATH and MATH let MATH and MATH be the quantities defined by REF . Writing MATH and MATH for same numbers defined with respect to the components MATH for MATH, we claim that MATH . To see this, suppose MATH is a reducible flat connection. Adding a REF-handle to MATH gives a flat cobordism from MATH to MATH. Since this cobordism has no REF-handles, its signature and twisted signature vanish, and so the NAME index theorem implies that MATH . In addition, the NAME principle implies MATH . REF now follow by applying REF above to REF . Suppose MATH is a small admissible perturbation on MATH for MATH, so that MATH is regular. Viewing MATH and MATH as perturbations on MATH set MATH and assume the perturbations are chosen small enough so that MATH . Given MATH, we can write MATH where MATH is a MATH-perturbed flat reducible connection on MATH and MATH is a MATH-perturbed flat reducible connection on MATH . Then MATH consists of two types of components CITE: CASE: MATH components of the form MATH where MATH for MATH and MATH is a gluing parameter with MATH reducible. CASE: Point components of the form MATH or MATH, where MATH is the trivial MATH connection over MATH and MATH for MATH. Note that the intersection MATH consists entirely of components of type REF unless MATH or MATH, in which case it consists of point components. We first argue that components of type REF do not contribute to MATH. To see this, suppose MATH is a component of type REF . Then MATH for fixed MATH. Let MATH be a perturbation so that the restriction of MATH to MATH is NAME. (The existence of such functions is shown in CITE.) Then, for small MATH, the contribution of MATH to MATH is given by MATH which vanishes since the sum evaluates to the NAME characteristic MATH and MATH. (This is similar to the proof of REF.) Thus, dropping all the MATH terms with MATH from the following sum and applying REF to the remaining terms, we conclude that MATH . |
math/0008088 | We begin by considering the vector function MATH defined for MATH by MATH where MATH represents the angle between MATH and the integration variable MATH and MATH where MATH represent some (initial) set of Cartesian coordinates. The vector MATH simply gives the center of mass in MATH of the hypersurface distribution on MATH with mass density given by MATH. Note that its dependence on the point MATH is entirely through the function MATH; indeed, if not for this function we would have only a single center of mass vector MATH. First we argue that what we should look for are points MATH such that MATH that is, such that MATH and MATH are linearly dependent. To see that finding such a point will suffice to prove our theorem, suppose MATH is a point where REF holds and let MATH be a MATH by MATH rotation matrix with MATH as its last row. Defining new Cartesian coordinates MATH via MATH we have (with MATH measured from MATH) MATH since MATH is an orthogonal matrix and hence its first MATH rows are orthogonal to its last row. Thus, the conclusion to the theorem will follow from finding a solution to REF and from here on we concentrate on finding such a solution. Now if MATH ever vanished for some MATH the conclusion REF would follow immediately with no need to rotate coordinates. So we may as well assume that MATH never vanishes on MATH. Under this assumption we can pass to consideration of the mapping MATH defined by MATH . The dependence REF then reduces to MATH so that we are seeking a fixed point or an ``anti - fixed point" of MATH. Under the assumptions of the theorem it suffices to seek only fixed points, that is, solutions to MATH . This follows from the symmetry of MATH about MATH which implies that MATH for all MATH and thus that MATH for all MATH. Hence if MATH is a solution to REF , then either MATH or MATH must be a solution to REF and we can therefore concentrate solely on finding solutions to REF . (Geometrically, too, it is more natural to view a point MATH satisfying REF as a center of mass of MATH than the point MATH satisfying MATH, even though there is no difference between the two as far as fulfilling the conditions of the theorem goes.) Finally we are at a point where we can use degree theory to conclude that MATH must have a fixed point, that is, a solution MATH to REF . The definition of MATH shows that it is continuous and our assumption that MATH does not vanish on MATH guarantees the continuity of MATH as defined by REF . Therefore, by standard theory see, for example, CITE, p. REF; REF; CITE, p. REF; CITE, p. REF; REF; CITE, p. CASE: MATH has a degree as a map from MATH to MATH. The fact that MATH for all MATH implies that MATH is even (simply observe that for an image point MATH the points in the preimage come in pairs MATH). Now if MATH has no fixed points, that is, if MATH for all MATH, then MATH would be homotopic to the antipodal map MATH defined by MATH via the homotopy MATH . But the degree of a mapping is a homotopy invariant and MATH, a contradiction. Hence MATH must have some fixed point MATH and the proof is complete. |
math/0008088 | Defining MATH as above REF either MATH vanishes somewhere and we are done or we can pass to MATH. Continuing with the latter case, if MATH has neither a fixed point nor an anti - fixed point (that is, there are no solutions MATH to MATH) then as above we can show that MATH is homotopic to the antipodal map and also to the identity map. But the antipodal map has degree MATH (see, for example, CITE, p. REF ; CITE, p. REF ; CITE, p. REF ; CITE, p. REF ; CITE, p. REF) and for MATH even MATH degree of the identity map. This is a contradiction since degree is a homotopy invariant (see CITE, p. REF; CITE, p. REF; CITE, p. REF; CITE, p. REF; CITE, p. REF; or CITE, p. REF), hence MATH must have either a fixed point or an anti - fixed point, and the conclusion of the theorem follows. |
math/0008088 | In the notation of the preceding paragraph we must show that MATH. The argument proceeds via a simple use of NAME 's theorem as applied to REF rewritten in the forms MATH and MATH . In particular, with MATH in the first of these we have MATH and with MATH in the second we have MATH . By NAME 's theorem, between any two zeros of MATH there is a zero of MATH and hence of MATH, since REF holds. Similarly, between two zeros of MATH there is a zero of its derivative and hence, by REF , of MATH. Thus for fixed MATH the zeros of MATH and MATH on MATH interlace. Now consider MATH and MATH for MATH. Since this makes MATH the first positive zero of MATH it is clear by what we have just proved that MATH has exactly one zero in MATH and that MATH is not a zero of MATH. This then implies, by the fact that the positive zeros of any MATH are decreasing functions of the parameter MATH (see, for example, CITE, p. REF, or CITE, p. REF), that MATH. That the multiplicity of MATH as an eigenvalue of MATH on the polar cap (= geodesic ball) with NAME boundary conditions is MATH follows from the details of separation of variables. It can be shown that the corresponding eigenfunctions can be taken as MATH (restricted to MATH) where MATH, MATH, and MATH is the eigenfunction of REF for the eigenvalue MATH. These functions form an orthogonal basis for the eigenspace of MATH corresponding to the eigenvalue MATH, showing that its multiplicity is MATH. This completes our proof. |
math/0008088 | MATH satisfies MATH in MATH (since MATH follows from the variational characterization of the eigenvalues of MATH via the NAME quotient MATH), which implies that MATH is decreasing in MATH. Hence MATH, which proves the lemma. |
math/0008088 | Using the product representation of MATH, that is, MATH, one has MATH (convergence of the series for MATH here is understood in the sense of symmetric partial sums). From REF we obtain the following representation for MATH . It follows from REF that MATH for MATH. Also from REF we have MATH which is positive for MATH. Thus, MATH is positive and MATH is decreasing for MATH. Taking derivatives again we find, MATH . The right-hand side of REF is positive for MATH. Hence, MATH is increasing and MATH is concave in MATH. It also follows from REF that MATH is increasing, and therefore MATH is convex in MATH. Lastly, the values of MATH at MATH, MATH, and MATH are found by explicit evaluation. |
math/0008088 | Since MATH (see REF above), we just need to prove that MATH . This inequality follows from REF . Note that it holds for all MATH. |
math/0008088 | Since MATH and MATH, to prove REF and therefore MATH reduces to showing (since MATH can be grouped as MATH) MATH where MATH. From REF this is equivalent to proving MATH which is obviously true for MATH since MATH is positive. For MATH it suffices to show MATH . The fact that MATH is an increasing function of MATH on MATH for MATH (see REF below) implies that MATH and MATH must have exactly one crossing in MATH. Since MATH is positive and increasing in MATH the desired REF follows by applying REF, p. REF; this inequality is also known as Bank's inequality CITE (see also CITE or CITE, p. REF). |
math/0008088 | The proof is similar to that of REF. We assume that MATH through most of the proof, returning to the case MATH (which can be treated explicitly in terms of elementary functions) only at the end. We shall use a suitable trial function (based on MATH) in the NAME quotient for MATH: MATH . It suffices that the trial function MATH be real and continuous on MATH, have MATH finite and MATH, and be such that all the integrals occurring above exist as finite real numbers. This includes, in particular, the case MATH, which we now adopt. With this choice we find MATH and, upon using the differential equation satisfied by MATH (that is, REF with MATH), we have MATH . That the expression on the right here has a finite limit as MATH follows from the fact that MATH can be taken as MATH and MATH as MATH (see REF ). It now follows from REF (since for MATH, MATH does not satisfy the equation that MATH does, we can write a strict inequality here) that MATH . Since we would like to show that MATH, it will be enough to show that the right-hand side of REF is less than or equal to MATH, or, equivalently, MATH or MATH . We now rewrite this integral using MATH (where MATH) and proceed to simplify the integrand using the relations REF developed for the MATH's. First, using REF with MATH the expression in square brackets becomes MATH or, since MATH, MATH . Finally, we use the recursion relation REF with MATH to see that MATH and hence, since we are taking MATH, REF may be rewritten simply as MATH which clearly holds if MATH for MATH since MATH and MATH on MATH. To see that MATH for MATH we use the relation REF , first for MATH and then for MATH. With MATH we have MATH showing that MATH for MATH since MATH on MATH and MATH for MATH by domain monotonicity of NAME eigenvalues (and the fact that MATH). This in turn yields MATH for MATH if MATH since by REF with MATH we have MATH and we know by domain monotonicity that MATH for MATH. Thus we have proved REF of our theorem for MATH (with strict inequality in REF ). It only remains to show that equality holds when MATH and this is elementary since in this case we can solve explicitly for all eigenvalues and eigenfunctions without even the need of special functions. One finds that MATH with eigenfunction MATH and that MATH with exactly MATH linearly independent eigenfunctions MATH for MATH. Clearly MATH and our proof is complete. |
math/0008088 | The proof is similar to that of REF. We use MATH in the NAME quotient for MATH (that is, in REF above). Since MATH we get from REF MATH (again, MATH does not satisfy the equation that MATH does, so REF is strict). Now the lemma follows by the monotonicity of MATH in MATH for MATH. |
math/0008088 | The analog of this result in the Euclidean case was proved in REF. That MATH and MATH as MATH follow from the boundary behavior of MATH. Using the raising and lowering identities REF with MATH and MATH respectively and with MATH fixed at MATH, that is, MATH one obtains MATH and in similar fashion MATH . The function MATH is finite at MATH, and thus MATH at MATH. Since MATH is positive and decreasing in MATH, MATH there. Hence, MATH in MATH and therefore by REF MATH is increasing there. Clearly MATH in MATH, since MATH and MATH is increasing there. Moreover, from REF we find MATH . From REF , and REF we obtain MATH in MATH for MATH. Since MATH at MATH, this implies that MATH in MATH and therefore, by REF , MATH there and the proof is complete. |
math/0008088 | Consider the function MATH. Using REF , and REF one can show that MATH satisfies the equation MATH . Since the function MATH is odd and analytic in MATH for MATH near MATH, we can expand it in odd powers of MATH. Inserting a series expansion in the NAME REF we find MATH for MATH near MATH. Substituting this into the definition of MATH we find MATH as MATH. Since MATH for MATH (which follows from the fact that MATH and MATH is decreasing in MATH), MATH is positive in a neighborhood of MATH. Now, MATH is a continuously differentiable function in MATH, it is positive in a neighborhood of MATH, and it goes to infinity as MATH. This implies that either MATH is an increasing function in MATH or that it has a positive local maximum. However, this latter possibility is ruled out by REF (any possible positive critical point of MATH must be a local minimum). Thus, MATH is increasing on MATH if MATH. If MATH, MATH, hence MATH, MATH, and finally MATH since MATH. |
math/0008088 | Let MATH, respectively MATH, be the lowest eigenfunction, respectively eigenvalue, of the NAME problem on MATH, that is, MATH . Define MATH and MATH. Let MATH, and MATH, where MATH denotes MATH - dimensional measure on MATH. Then we have (see, for example, CITE, p. REF ) MATH for almost every MATH. Applying NAME 's theorem to REF , we have MATH since the outward normal to MATH is given by MATH. Using the NAME - NAME inequality and REF we obtain MATH . As discussed above, if MATH is a domain in MATH, the classical isoperimetric inequality is given by MATH where MATH is a geodesic ball having the same volume as MATH. The MATH - dimensional measure of MATH, MATH, is given in terms of MATH, the geodesic radius of MATH, by MATH where MATH is the volume of the unit ball in MATH (and MATH is the volume of MATH). Therefore, from REF it follows that MATH . Hence, from REF we have MATH . Finally one uses the fact that MATH which follows directly from the definition of MATH, the decreasing rearrangement of MATH on the interval MATH (here MATH is a variable denoting volume and is related to the geodesic radial variable MATH via MATH, that is, MATH). Since MATH is the inverse function to MATH we have MATH which, combined with REF , yields MATH . Now, if we view MATH as a function of the volume MATH (here we will abuse notation and continue to call it MATH) rather than as a function of MATH (or MATH), with MATH, it satisfies REF with equality, that is, MATH . In fact, REF is an integrated version of REF with MATH and MATH in the variable MATH. Having obtained the relations REF for MATH and MATH respectively, we will prove that under the normalization condition imposed on them, they are either identical or they cross only once in the interval MATH (in the sense specified by REF ). All the arguments we give below depend on the continuity of MATH and MATH. The function MATH is in fact real analytic in MATH and, furthermore, it is decreasing there as can be seen from REF above (or from REF and the fact that MATH). The absolute continuity of MATH on MATH follows from arguments in CITE. Since MATH and MATH are normalized to have the same MATH - norm, they either are identical or they cross. If they coincide, then MATH, and REF is proved since any MATH will serve as MATH. Next, following CITE we conclude that MATH cannot exceed MATH. In fact, if MATH it follows by mimicking the proof of the main theorem in CITE that MATH for all MATH which, in turn, implies MATH and we are back in the previous case. Thus, if MATH, MATH in a neighborhood of MATH, and both functions being of the same norm they must cross at least once. Let MATH be the largest MATH such that MATH for all MATH. By the definition of MATH, there is an interval immediately to the right of MATH on which MATH. Indeed, by continuity and the definition of MATH for some MATH. It now follows that MATH at least on the interval from MATH to MATH since by the absolute continuity of MATH by virtue of REF . We will now show that MATH for all MATH, which will prove the theorem. If not, then the point MATH defined to be the largest MATH for which MATH for all MATH, would be less than MATH, and we would have MATH with MATH and MATH. In this case, we can define the function MATH . It follows from REF that MATH satisfies MATH with strict inequality for all MATH. From MATH define the function MATH for MATH where MATH is the polar angle (angle from the north pole) corresponding to MATH. Thus MATH is a radial function on MATH (assumed centered at the north pole). Because of REF (or REF or REF ), MATH cannot be the groundstate of the Laplacian with NAME boundary conditions on MATH (but it is certainly an admissible trial function for MATH). Therefore, MATH . By standard change of variables, MATH and MATH . Using REF (substitute for one of the MATH's in REF , using the fact that MATH) and integration by parts we get MATH . From REF , and REF we get a contradiction, and the theorem follows. |
math/0008089 | See CITE REF and also CITE REF . |
math/0008089 | CASE: is classical, REF. has been proved by CITE, and REF. was given by CITE. |
math/0008089 | Define MATH, MATH, and MATH, then the above functional equation is reduced to the equation from REF, for which it is well-known (compare for example, CITE) that there is only the differentiable function MATH (up to a multiplicative constant) which satisfies the latter equation. |
math/0008089 | It is a straightforward matter to check that the above elements lie in the kernel of MATH. Nevertheless, we give some interrelationships between the various equations. CASE: The symmetry of REF is equivalent to REF : We have to write the following relation MATH as a sum of MATH-term relations. On the left hand side of the equation we add the MATH-term relation in the following form MATH and we do the same on the right hand side with MATH replaced by MATH. This leaves us with another form of REF-term relation MATH (to see this we should replace, in REF , MATH by MATH and MATH by MATH and use REF ), thereby proving the first claim. The equivalence of REF is easily shown using the inversion and REF-term relation. CASE: The second family of five term relations is almost direct to deduce: the combination given is the sum of MATH times REF-term relation REF and its following equivalent formulation MATH . (replace in REF MATH and MATH by their inverses, respectively, then multiply the result by MATH and finally use the inversion relation on three of the ensuing terms). From this, we get a very simple proof of the five term relations in cocycle form, that is, REF : in each of the two versions REF of REF-term relation we put MATH and MATH. Introducing for the moment the notation MATH, we can rewrite the two equations in a concise way: MATH . Given MATH, there is a linear combination of the two equations such that the coefficient of MATH (which only occurs in the first equation) and of MATH (only occurring in the second equation) is MATH and MATH, respectively. If, for MATH and MATH, we compute the coefficients of the other three arguments, we obtain exactly the expressions given in the claim. For example, let us compute the coefficient of the first argument in REF the first equation is multiplied by MATH the second by MATH so the coefficient becomes MATH which is equal to MATH . |
math/0008089 | REF-term equation and REF will follow directly from the equation in the next proposition. The equivalence of the two NAME analogues becomes evident after applying the change of variables MATH, MATH, and multiplying the result by MATH. |
math/0008089 | Add REF-term equation to its variant where MATH is replaced by MATH. Four of the terms cancel and the remaining two give the inversion relation. |
math/0008089 | According to REF , the function MATH is a well-defined function on MATH, and as it is not identically zero on MATH, the dimension of MATH is non-zero. By CITE(NAME REF, p. REF), we know that MATH. But as the relations in MATH are given by REF-term equation (that is, the Fundamental Equation of Information Theory) and the relation MATH, (see REF and also Sah's Lemma in REF ), and as we further know, again by Sah (see the remark on p. REF in op. cit), that these two relations are independent, we can conclude that the kernel of the map MATH is generated by the element MATH. Evaluating MATH on this element shows that it is non-zero, which ends the proof. |
math/0008089 | This is a consequence of the following theorem, together with REF . |
math/0008089 | Set MATH, and suppose that MATH verifies MATH and the following identity in MATH . Differentiating the previous equation with respect to MATH gives, MATH with MATH and thus MATH. Setting MATH in the previous identity gives MATH . But as MATH, the previous equality implies MATH . In other words, since MATH we have MATH, which proves the claim. |
math/0008089 | It is a consequence of the following lemma Suppose that MATH is a sequence of integers with MATH (MATH an odd prime fixed), which fulfills the following rules MATH and MATH for all MATH. Then MATH for all MATH. The proof goes by descending induction starting from MATH. First we notice that by the third rule, we have MATH modulo MATH. Suppose that MATH modulo MATH for all MATH. Now compute MATH modulo MATH. Observe that we can assume MATH, since we can compute from the rules MATH and MATH. If MATH is odd then by the first rule we deduce directly MATH, but we still have to show that MATH modulo MATH. This is done via the If MATH is odd and MATH modulo MATH for all MATH. Then MATH modulo MATH. We have to show that, modulo MATH, MATH or equivalently, assuming MATH, that MATH . But MATH . Using the usual rule MATH if MATH, we have MATH . But MATH, and as, modulo MATH, we have MATH, we finally get, using the fact that MATH is odd, the desired identity. Now return to the proof of the lemma and suppose that MATH is even. If MATH then the process ends, so we can suppose that MATH. The idea is to show that we can compute directly MATH and to deduce MATH from the first rule (we will still need to show the desired property). As MATH is even, MATH is odd and thus MATH is even. Thus by the third rule we have MATH and by the second rule we have MATH . But there exists MATH such that MATH with MATH (because MATH). Hence, applying once again the third rule gives MATH . But MATH . And as MATH, we have MATH, which means, applying the induction, that MATH is already known. We then get the value of MATH and by applying the first rule to it we deduce the value of MATH. Now to finish the proof we need to show that, in this case MATH modulo MATH. Notice that we can also assume by the induction that MATH modulo MATH for all MATH. First we show that in the previous process, we get MATH modulo MATH. Indeed, by the induction we have MATH . Thus MATH . And finally MATH . To conclude we need to prove a variant of NAME REF. Suppose that MATH is even, MATH modulo MATH for all MATH and MATH modulo MATH. Then MATH modulo MATH. We have the equality MATH . Using the same arguments than in the proof of NAME REF, we get the following identities, MATH . And we finally have MATH from which we deduce MATH. Hence the proof of REF is complete. NAME to the proof of REF . Suppose that MATH verifies the conditions of the proposition. Then applying the three equations to MATH gives MATH, and the other coefficients MATH fulfill the rules described in REF . |
math/0008089 | The fact that MATH fulfills REF is a direct computation. For MATH, the family is free for degree reasons, since MATH. Furthermore MATH does not belong to this family for valuation reasons. |
math/0008089 | Let MATH be a polynomial of degree MATH, and suppose that MATH fulfills the following two equations MATH . Then observing that we can deduce the inversion formula as a consequence of REF-term, and taking the derivative with respect to these equations shows that MATH fulfills the inversion formula and the duplication formula. As, by hypothesis, MATH fulfills also REF-term equation, we conclude from REF that MATH is MATH up to a constant, which implies that MATH is MATH up to a constant. |
math/0008089 | We only need to show that if MATH, assumed to be of degree less than or equal to MATH, setting MATH, MATH fulfills REF-term equation. In order to do that let MATH denote the formal NAME analogue, then taking the derivative with respect to MATH, and rewriting the equation with MATH and finally specializing to MATH, we can see that, modulo the inversion formula for MATH (which we can get directly by deriving the inversion formula for MATH), we have the identity MATH. |
math/0008089 | First we observe that the map MATH is well defined. Indeed this is a consequence of the MATH property of the map MATH defined on the units and of the fact that MATH which implies that MATH. Then, the commutativity of the diagram is a direct check. |
math/0008089 | It is a direct consequence of REF and of the construction of MATH. |
math/0008089 | MATH is defined on MATH by assumption. For MATH, we have seen that the function is essentially unique: MATH and the resulting infinitesimal dilogarithm MATH vanishes on MATH (due to REF , it is enough to check that it vanishes on the four term relation, which is straightforward). Now suppose the claim is true for MATH. The maps MATH, MATH respectively, MATH, MATH, are well-defined by the inductive assumption respectively, by assumption (MATH is clean). Furthermore, an element MATH lies in the kernel of each of the ``components" of MATH, say MATH and MATH, and therefore MATH which shows that the function defined by MATH can be linearly extended to a well-defined function on MATH. |
math/0008089 | Again, the case MATH gives the unique choice for MATH (up to multiplicative constant). Inductively, starting from MATH and MATH, one can form an arbitrary linear combination of them using MATH and MATH which gives a candidate for MATH, with coefficients MATH, say; a subsequent ``integration" (putting MATH and successively MATH, MATH) provides a candidate MATH whose coefficients MATH have to satisfy the further REF - this gives a linear restriction on the possible MATH at each step. We thus obtain inductively an extra degree of freedom at each level. For example, normalizing MATH such that MATH, we obtain successively MATH . It remains to check that NAME 's choice REF does satisfy MATH which is straightforward. Also, the MATH satisfy REF since MATH . |
math/0008089 | The infinitesimal polylogarithm MATH vanishes on MATH by REF , and reducing mod MATH obviously conserves this vanishing property. NAME 's result now says that the reduction of MATH is equal to MATH. |
math/0008089 | CASE: Follows from CITE and CITE, respectively. CASE: This follows via the ``derivation map" (see ยง REF). CASE: MATH. |
math/0008089 | CASE: The inversion relation can be checked via a straightforward algebraic manipulation. CASE: In order to prove the distribution relation, let us fix a primitive MATH-th root of unity MATH. Dividing both sides by MATH and developing the fraction into a (finite) series leaves us to prove: MATH and this is true due to the character relations MATH . CASE: (Proof of the special values) MATH if MATH follows from the well-known fact that MATH for any polynomial MATH of degree MATH (here we apply it to the monomials MATH), the statement for MATH being obvious. The assertion for MATH is a direct consequence of the inversion relation. To prove the last formula of REF we only need to take MATH (the odd values correspond to the above identities). For this, one can use the special case MATH, in CITE REF , MATH: MATH and the fact that MATH is equal to the sum in REF : rewriting MATH one sees that the first sum is equal to MATH times the sum in REF , while the second one equals MATH and therefore is zero (for MATH) by the above special value. |
math/0008089 | CASE: We will apply the recipe above. We have MATH and as the degree of either polynomials is less than MATH, we conclude that MATH where MATH is a constant. This, in turn, implies that MATH (specialize MATH and MATH, respectively), and therefore we get as a by-product MATH (in characteristic MATH). CASE: The following proof is a slight variation of the recipe, in that it uses two iterated derivatives. Denote by MATH and MATH the derivatives to respect to MATH and MATH. We can check, using the differential equation for MATH and the rational REF for MATH, that MATH which is an expression that is symmetric in MATH and MATH. Thus MATH . But the maximum degree for each indeterminate in the polynomial MATH is never greater than MATH, and as a consequence the above identity implies that MATH where MATH. But setting MATH implies both MATH and MATH, and the construction of MATH and MATH shows directly that they are both zero (the coefficients of MATH and MATH in MATH are both equal to MATH). |
math/0008089 | CASE: Set MATH . We want to prove that MATH is REF in MATH. Computing MATH we get MATH . But by REF , MATH and MATH. Moreover by the inversion formula (see REF ) we have MATH. Hence, MATH . As MATH and MATH, we know that MATH and therefore MATH, but using the inversion relation one sees that MATH, which implies MATH. |
math/0008089 | The strategy of proof could be summarized as follows: CASE: Prove that MATH in MATH. CASE: Prove that MATH in MATH. CASE: Prove that the coefficient of MATH in MATH is REF. For the proof of this functional equation we will need several preliminary formulas. First we will use these two relations, in MATH, coming from REF-term equation for MATH . We use implicitly the following formal derivation rules, where MATH is an indeterminate and MATH a constant independent of MATH: MATH . We also point out that the following simple formula will be often used: MATH . For the convenience of the reader we will give detailled computations in order to make checking almost straightforward. Let's first split MATH into six pieces to facilitate the identification of the cancellation in the forthcoming computations: MATH . Set MATH. First step: prove that MATH. It is immediate that MATH. Using the rules described above, we get the following equalities: MATH . Then, applying consecutively REF to MATH, with MATH, MATH, with MATH, MATH, and with MATH, MATH, and to MATH with MATH, MATH, and using REF for simplification as well as the basic relations for MATH, we get MATH then MATH . It remains to transform MATH, but using REF , we have for example, MATH then MATH . Now by invoking the inversion formula we see that MATH . Second step: Prove that the relation is true for MATH. Putting MATH in MATH gives MATH and applying the inversion formula for MATH we get REF. Third step: Prove that the coefficient of MATH is REF. Notice first that if MATH is an expression independent of MATH, then the coefficient of MATH in the sum MATH is MATH. Using this fact, we can see that the coefficient of MATH in the expression MATH is given by MATH . But each of the previous lines are REF by using REF-term functional equation of MATH (see REF. ) and this completes the proof of REF-term functional equation for MATH. |
math/0008089 | we can compute explicitly the coefficients of the polynomial MATH. First of all, its zeroth coefficient is MATH. For MATH between MATH and MATH the MATH-th coefficient of MATH is equal to MATH . We use here the standard fact that MATH for any polynomial MATH of degree at most MATH. |
math/0008093 | By the definition of the MATH-th supersymmetric algebra we have MATH where MATH is the diagonal subgroup of MATH. By REF we have therefore MATH where MATH in the previous line is summed over all NAME diagrams satisfying the conditions MATH and MATH. The second to last equality follows from the well-known fact that MATH is a self-contragredient module. |
math/0008093 | By the definition of the MATH-th skew-symmetric algebra we have MATH where MATH is the diagonal subgroup of MATH and MATH is the subspace of MATH that transforms according to the sign character of MATH. By REF we have therefore MATH where MATH in the previous line is summed over all NAME diagrams satisfying the conditions MATH and MATH. The second to last equality follows from the following well known facts: MATH is a self-contragredient module and tensoring the module MATH with the sign character yields the module MATH. |
math/0008093 | Since MATH is a highest weight vector for MATH with weight MATH, it belongs to the subspace MATH of MATH which consists of vectors with weight MATH annihilated by the standard NAME in MATH. By REF , MATH is isomorphic to MATH as a MATH-module. There exists a unique vector (up to scalar multiple) in MATH which has weight MATH, which is the highest weight vector. By REF has weight MATH as a MATH-module, so it is a highest weight vector for MATH. |
math/0008093 | Our argument follows closely the one given in the proof of REF with REF replacing the classical NAME duality. Let MATH denote the the subgroup of MATH, which preserves the partition MATH of MATH. Note that MATH is isomorphic to a semidirect product of MATH and MATH, where MATH acts by interchanging MATH with MATH and MATH acts by permuting the pairs. Let MATH denote the character on MATH which is trivial on MATH, but transforms by the sign character on MATH. We observe that MATH . Thus using REF we obtain MATH . Now by REF the space MATH is non-zero if and only if MATH is constructed from nesting hooks of types MATH, in which case it is one-dimensional. |
math/0008093 | Observe that MATH has the same MATH-weight as the first NAME diagram of REF . Clearly MATH is non-zero. It is straightforward to verify that MATH is annihilated by the operators in REF . It follows from REF that MATH is also a highest weight vector for MATH. |
math/0008093 | Consider MATH and MATH where MATH. By REF the product MATH is a highest weight vector and thus is annihilated by all operators in REF . In particular applying the operator MATH to MATH and dividing by MATH we obtain the desired identity. |
math/0008093 | We first show that MATH indeed has the correct weight. First note that REF corresponds to the MATH-weight MATH. Let MATH, MATH, denote the MATH disjoint subsets of MATH defined by the condition that MATH if and only if MATH contains a marked box at its MATH-th column and MATH-th row. Put MATH and MATH. The weight of MATH is MATH. Now MATH has weight MATH. Hence each MATH has weight MATH, as required. Hence by REF it is sufficient to show that REF annihilates it, namely MATH . We will first establish REF . Note that the simple root vector MATH maps the vectors MATH to MATH and MATH to MATH. For a diagram MATH, let us denote by MATH a diagram obtained from MATH by moving each marked box in its MATH-th row to the box above it in the MATH-st row. Analogously we define MATH a diagram obtained from MATH by moving each marked box in the MATH-st row to the box below it in the MATH-th row. It is easy to check MATH . Thus we have MATH . But evidently MATH thanks to the equality MATH. Hence the right-hand side of REF is zero, proving REF . Our next step is to prove REF . In this case MATH maps the vectors MATH to MATH and MATH to MATH. For a diagram MATH such that MATH and MATH we denote by MATH the diagram obtained from MATH by removing MATH from MATH and adding MATH to MATH. For a fixed MATH we write MATH . First observe that MATH . This is because if MATH, for some MATH, then the term MATH where in general MATH is the matrix obtained from MATH by replacing the MATH-th row with the vector MATH. Now if MATH and MATH, then MATH . Let MATH be the same diagram as MATH, except MATH and MATH. Then we have MATH . Of course MATH and MATH anticommutes with MATH, so MATH . Next we observe that if MATH is a diagram such that MATH then MATH so that our task of proving REF reduces to proving that MATH where the sum is over all diagrams MATH with MATH. But the left-hand side of REF is equal to MATH . To complete the proof we now need to verify REF . The odd simple root vector MATH has the effect of changing the vectors MATH to MATH and MATH to MATH. If MATH is a diagram such that MATH with MATH, then MATH . Thus MATH . For a diagram MATH with MATH (respectively, with MATH) we denote by MATH (respectively, MATH) the diagram obtained from MATH by adding MATH to MATH (respectively, by removing MATH from MATH). Then REF becomes MATH . Setting MATH in the first sum of REF we may rewrite REF as MATH . |
math/0008093 | Denoting by MATH and MATH the right-hand side and the left-hand side of REF , respectively, we may regard MATH and MATH as functions of MATH. Since the group MATH acts on MATH, the space of MATH matrices, by left multiplication, it acts on functions of MATH. To be more precise if MATH, then MATH and MATH. We want to study the effect of this action on MATH and MATH. In order to do so, consider first the action of the three kinds of elementary matrices on them. Namely, those that interchanges any two rows, that multiplies a row by a scalar, and those that add a scalar multiple of a row to another. It is subject to a direct verification that if MATH is any of the three types of elementary matrices, we have MATH . Since every element in MATH is a product of elementary matrices, we conclude that REF holds for every MATH as well. Putting MATH, the identity MATH matrix, we see that MATH so that MATH by REF for all MATH. As MATH is a NAME open set in the space of MATH matrices we have MATH for any MATH matrix MATH. |
math/0008093 | REF is an obvious consequence of REF. For REF let MATH denote the complementary subset of MATH in MATH put in increasing order. We apply successively the differential operators MATH, MATH, MATH, MATH to REF and find that MATH . Dividing by MATH we obtain REF . By REF we have MATH . Thus if MATH is the permutation arranging MATH in increasing order, then MATH . Now REF follows from dividing the above equation by MATH and multiplying by MATH. |
math/0008093 | Given diagram MATH with MATH rows and MATH columns, for MATH, we want to know how to simplify the expression MATH . We move all MATH to the left and get MATH . Now each MATH is a product of MATH. We arrange all MATH together so that MATH appears to the left of MATH if and only if MATH and call the resulting expression MATH and move it to the left. We do the same thing to MATH and move MATH to the right of MATH etc. Then REF becomes MATH . We apply now REF to REF and obtain MATH . Since MATH, the proposition follows. |
math/0008093 | Since for a subset MATH of MATH, MATH is a scalar multiple of MATH by REF , it follows that the expression is divisible by MATH and independent of MATH, for MATH, after division. It is clear that it is annihilated by REF . |
math/0008093 | Let MATH, MATH denote the number of marked boxes in MATH that appear in the MATH-th row of some diagram MATH. Then MATH has weight MATH while the expression MATH has weight MATH . So the combined weight is MATH which of course is the weight of the NAME REF . |
math/0008093 | The only thing that remains to prove is that the vector in REF is indeed killed by REF . But this is because of the presence of MATH in the formula of MATH and so is an immediate consequence of REF . |
math/0008093 | It is enough to restrict ourselves to real symmetric MATH matrices MATH. Let MATH be an orthogonal MATH matrix such that MATH, where MATH is a diagonal matrix. We compute MATH where MATH and MATH. But the determinant of the matrix on the right-hand side of REF is zero. |
math/0008093 | First consider the action of the operator MATH, for MATH, on MATH given as in REF . Certainly MATH annihilates the first summand of REF , and furthermore it takes the summand MATH for MATH to MATH which is zero. MATH takes MATH to MATH . The first summand is zero, while the second summand remains. Now we verify similarly that MATH takes MATH to the identical expression as the second summand above with the difference that the MATH-st and MATH-th rows are interchanged. Thus MATH. Consider now the action of MATH on MATH. Note that MATH kills every term in REF except for the first and the last. The contribution from the first summand is MATH, while that from the last summand is MATH, and hence MATH. Finally we consider the action of MATH on MATH as in REF . MATH kills the first term in REF and the resulting vector is MATH which can be in a consistent form with our earlier notation written as MATH. Expanding along the first row we see that REF is the same as MATH which is zero by REF. |
math/0008093 | The fact that MATH is annihilated by MATH for MATH is a consequence of REF . Now the proof of REF shows that MATH, for every MATH. Thus MATH annihilates MATH as well. So it remains to show that MATH kills MATH. Given a summand in MATH of the form MATH there exists a summand of the form MATH, which is identical to it except at these two places. Now MATH takes the first of the two summands above to MATH and the second summand to MATH . But MATH and MATH differ by a transposition MATH and hence MATH and so these two terms cancel. Consider a summand in MATH of the form MATH. But in MATH we also have a summand of the form MATH. Applying MATH to these two terms, we again see that they cancel by the same reasoning as before. Now we look at a term of the form MATH. We also have a term of the form MATH. Again they will cancel each other after applying MATH. Finally a term of the form MATH is killed by MATH by REF . This completes the proof. |
math/0008094 | It suffices to check that MATH is invariant under a transposition of two adjacent MATH's. By the decomposition REF , this reduces to showing that MATH is symmetric in MATH and MATH, identically in MATH for any MATH. Let us single out one of the MATH's and write MATH where MATH, MATH, and the rest of the MATH's are considered as parameters. Our goal is to show that MATH identically. The function MATH is a rational function of MATH with at most first order poles at the following points MATH . First we show that, in fact, this function is regular everywhere. By symmetry, it suffices to consider the points MATH. Observe that on the divisor MATH the function MATH itself is regular. Indeed, a function which is symmetric in MATH and MATH cannot have a pole of exact order one on the diagonal MATH. Next, consider the divisor MATH. Only those functions MATH which satisfy MATH contribute to the pole of MATH at MATH. For such function MATH and any MATH we have either MATH or MATH. The condition MATH and REF yield that MATH . Therefore MATH where MATH denotes the restriction of MATH to MATH. It follows that the residue of MATH at MATH is proportional to MATH which by the condition MATH and induction on the number of MATH's is symmetric in MATH and MATH. In other words, MATH is regular at MATH. By the symmetry between MATH and MATH, it is also regular at MATH. Now consider the point MATH. Since for any function MATH we either have MATH or MATH, it follows from the definition of MATH that MATH where the summation is over all possible function MATH and where we use the following abbreviation MATH . From REF we have MATH . Similarly, from REF we obtain MATH . Therefore, MATH as MATH, which means that both the MATH and MATH terms are symmetric in MATH and MATH. In other words, not only is MATH regular at MATH but it also vanishes at MATH. Similarly, MATH is regular at MATH and thus, as a function regular everywhere and vanishing at one point, MATH is zero identically. |
math/0008094 | Suppose MATH is not row decreasing. Then there exist two squares MATH and MATH in MATH such that MATH is just left of MATH and MATH . In this case, MATH has the factor of MATH and hence MATH. |
math/0008094 | Recall that MATH has two poles MATH and two zeros MATH. The pole MATH is not an issue in limit REF and the zero MATH has been already accounted for in REF . The pole MATH means that MATH where MATH is the square right on top of the square MATH. This factor is present in MATH if MATH . Similarly, the zero MATH means that MATH where MATH is the square to the left of MATH. This factor enters MATH if MATH . Let MATH be the order of the pole of MATH at MATH. We will show that MATH and that this maximum is reached precisely for reverse tableaux. We have MATH . The contribution of any square MATH to REF is MATH or MATH. If the square MATH is not in the first column of MATH then its contribution to REF is MATH. Indeed, if MATH then since MATH is row decreasing we also have MATH . That is, only the squares in the first column can make positive contribution to REF which proves REF. It is also clear that if MATH is a reverse tableaux then MATH. Now suppose that MATH is not a reverse tableaux. In order to prove that MATH it suffices to find one square whose contribution to MATH is negative. By assumption, there exists a square MATH such that MATH. Additionally, we can assume that such a MATH is minimal in the sense that MATH. Then since MATH is row decreasing we conclude that MATH which means that the contribution of this square MATH to REF equals MATH. It follows that MATH. This concludes the proof. |
math/0008094 | Let us split the sum in REF in two MATH where MATH and MATH contains summands with nonnegative and negative number MATH, respectively. Using the identity MATH to perform the summation along the rows, we compute: MATH . Similarly, we compute MATH using REF instead of REF and we find that MATH . For any diagram MATH define MATH by MATH . Here the constant MATH compensates for the appearances of MATH in the expansion of MATH which occur precisely when MATH. Also set by REF . It follows trivially from this definition that MATH . It is clear that MATH . Note that by our construction we have MATH. Therefore the above sum is either zero or else the only possible value of MATH in it is MATH. In the latter case we have MATH which means that MATH . In other words, MATH . It now clear that MATH depends only on the first and last element in the chain REF which precisely means that REF does not depend on the particular choice of the tableau MATH on MATH. |
math/0008096 | For MATH the statement is obvious. Suppose it is established for some MATH and let MATH be a diagram on MATH chords. Recall that we think of knots and diagrams as being ``long" so the ends of chords of a chord diagram are naturally ordered. Consequently, the chords themselves can also be ordered according to the order of their larger ends. Let MATH be the last (that is, the largest) chord of MATH and MATH be the last chord of the diagram MATH obtained from MATH by deleting MATH. The diagram MATH can be presented on MATH strands, so it can also be presented on MATH strands in such a way that no chords have ends on any of the last two strands. Now it is possible to add the chord MATH to the presentation of MATH in such a way that REF it is horizontal; REF one of its ends is on one of the first MATH strands and the second end is on the MATH-nd strand. Similarly, one can add the chord MATH to the obtained diagram in such a way that its last end is situated on the last strand; this gives a presentation of MATH on MATH strands. |
math/0008096 | Say that a diagram with MATH chords on MATH strands is arranged if it has exactly one end of a chord on each of the first MATH strands. We describe an algorithm of connecting any diagram with MATH chords on MATH strands to an arranged one by relations of strand exchange; the lemma follows from the existence of such an algorithm. Let MATH be the number of ends of chords on the MATH-th strand of a given NAME diagram. Suppose the diagram is not arranged. Find such MATH that MATH and that MATH for all MATH. If MATH for all MATH and MATH is the smallest number such that MATH and MATH we move one chord end from the MATH-th strand to MATH-st strand by a strand exchange relation. (We can do it as MATH.) Otherwise, move one chord end from MATH-st strand to the MATH-th strand. The above manipulations can be repeated until we get an arranged chord diagram. It is a straightforward check that the algorithm always terminates on such a diagram. |
math/0008096 | A REFT relation can be thought of as a resolution of a ``singular" chord diagram whose two chords have a common end. So in order to show that every REFT relation for diagrams of knots up to isotopy comes from a REFT relation for NAME diagrams one can just check that every singular chord diagram on MATH chords can be presented on MATH strands. This can be easily seen using REF . The statement about the framing independence relations is similarly straightforward. |
math/0008099 | Using a NAME solid link, we see that for the left side of REF , MATH and for the right, MATH. (This is also seen by REF .) Thus REF are equivalent. Suppose REF . Attach MATH-handles to MATH and MATH. Then MATH and MATH change to MATH-spheres which split by isotopy, and hence MATH is null-bordant. Thus REF . It is obvious that REF . |
math/0008099 | Consider a tri-punctured sphere MATH embedded in MATH whose boundary is MATH with MATH and MATH. There exists an identification of a regular neighborhood MATH in MATH with MATH such that the NAME MATH-links MATH and MATH in MATH correspond to MATH, where MATH is an oriented NAME disk pair. The desired MATH is obtained as MATH. |
math/0008099 | It is a consequence of REF . |
math/0008099 | CASE: Let MATH be a NAME hypersurface for MATH which intersects MATH transversely. For each MATH (MATH), let MATH be the intersection MATH, which is the union of oriented simple loops in MATH (or empty). Let MATH be a regular neighborhood of MATH in MATH. For a component MATH of MATH, let MATH be a solid torus MATH in MATH and MATH the boundary of MATH. They are oriented by use of the orientation of a meridian disk MATH of MATH and the orientation of MATH. Then MATH corresponds to MATH by the isomorphism MATH obtained by the NAME and NAME dualities, where MATH is the exterior of MATH. Put MATH and MATH. The surface MATH is bordant to MATH in MATH, for they cobound a REF-manifold MATH, where MATH denotes the closure. Thus, without loss of generality, we may assume that MATH is MATH. Consider a NAME hypersurface MATH for MATH. For a component MATH, let MATH be a regular neighborhood of MATH in MATH such that the union MATH forms a NAME solid link in MATH with core MATH, see REF (the figure shows a section transverse to MATH). Let MATH be a regular neighborhood of MATH in MATH with MATH and let MATH be a REF-manifold MATH. By MATH, the MATH-component surface - link MATH is bordant to MATH. This MATH-component surface - link is ambient isotopic to MATH, where MATH is a NAME solid link obtained from the NAME solid link MATH by pushing out along MATH using MATH, see REF . We denote by MATH a loop obtained from MATH by pushing off along MATH so that MATH is disjoint from MATH and it is a core of the NAME solid link MATH. Put MATH, MATH, and MATH. The MATH-component surface - link MATH is bordant to MATH such that MATH . CASE: Let MATH be the NAME hypersurface for MATH used in REF . By the construction of MATH, MATH. For each MATH with MATH, we may assume that MATH intersects MATH and MATH transversely. Using MATH, we apply a similar argument to REF to modify MATH up to bordism. For a component MATH of MATH, let MATH be a solid torus MATH in MATH and MATH a solid torus MATH. The intersection MATH is MATH and the intersection MATH is a union of some meridian REF-disks of MATH. Let MATH be a MATH-component surface - link MATH such that MATH . By use of a REF-manifold which is MATH removed the above intersections, we see that MATH is bordant to a MATH-component surface - link MATH which is the union of MATH and some meridian MATH-spheres of MATH with labels MATH for MATH. The meridian MATH-spheres are the boundary of meridian REF-disks that are the intersection MATH. The surface - link MATH is bordant to the union of MATH and some standard NAME MATH-links MATH or MATH whose core loops are small trivial circles in MATH. REF is a schematic picture of this process (in projection in MATH), where an isotopic deformation and fission of a NAME MATH-link are applied. Since MATH belongs to MATH, we see that MATH is bordant to MATH modulo MATH. CASE: Inductively, we see that MATH is bordant to MATH modulo MATH such that MATH . Specifically, we see that MATH is bordant to a MATH-component surface - link which is a union of MATH and some meridian REF-spheres of MATH such that MATH and MATH. These REF-spheres are the boundary of the meridian REF-disks that are the intersection MATH. Since MATH belongs to MATH, MATH is bordant to MATH modulo MATH. Thus MATH is bordant to MATH modulo MATH such that MATH . It is a union of NAME MATH-links and the link bordism class is in MATH. |
math/0008100 | The proof that the NAME relations are preserved under the correspondence MATH and that MATH is sent to MATH is a simple modification of the proof of the quantum analogue of NAME 's theorem presented in REF . The classical analogue of MATH, obtained by specializing MATH to MATH, is easily seen to be injective. This taken together with REF and the fact that the monomials consisting of products of lexicographically ordered generators MATH form a basis for MATH over MATH proves injectivity of MATH. |
math/0008100 | In CITE it is shown that MATH and MATH are weakly separated precisely when, after interchanging MATH and MATH if neccessary, either: CASE: MATH and there do not exist three indices MATH such that MATH and MATH or REF MATH and there do not exist four indices MATH such that MATH and MATH REF above indicates that two MATH-subsets MATH and MATH are weakly separated precisely, when viewed as subpolygons, no diagonal of the subpolygon MATH disjoint from MATH crosses a diagonal of MATH disjoint from MATH. This property is clearly preserved under any dihedral symmetry of the MATH-gon. |
math/0008100 | Given MATH, the number of chambers in the first MATH strips of the corresponding double wiring arrangement is equal to the number of red and black crossings in the first MATH strips plus MATH - corresponding to the MATH far right chambers . The number of black (respectively red) crossings in the first MATH strips in turn is given by the number of simple reflections MATH (respectively MATH) occurring in the reduced word MATH with MATH. The number of MATH in MATH with MATH is MATH. The number of of MATH in MATH with MATH is MATH; if MATH is optimal this will be MATH. Consequently the number of chambers occurring in the first MATH strips of the double wiring arrangement for MATH optimal ( or equivalently the size of MATH ) is: MATH . |
math/0008100 | Take any quantum minors MATH and MATH in MATH. REF proves that if MATH and MATH are chamber sets of a single wiring arrangement then either MATH or MATH. This, taken together with the fact that the single wiring arrangements for the MATH and MATH components of MATH are oppositely labeled, proves the second part of the proposition. To prove that MATH and MATH quasi-commute we must show that MATH and MATH are weakly separated. We may assume, after exchanging MATH with MATH and MATH with MATH if neccessary, that MATH and MATH. This in turn is equivalent to MATH which demonstrates that MATH and MATH are weakly separated. The fact that MATH is maximal follows from REF . |
math/0008100 | This theorem follows from the well known fact that the any two triangulations are connected by a series of chord exchanges where the diagonal chord of an inscribed quadralateral is "flipped" to its crossing pair. The diagonal "flips" correspond to MATH-moves. |
math/0008100 | Immediate corollary of REF . |
math/0008100 | Since the MATH-action preserves MATH-moves it is enough to verify this assertion in the case where MATH. Proceed by induction on MATH. For MATH the statement is evident. Assume MATH. It is enough to verify the claim for the group elements MATH and MATH, which generate MATH, given by MATH . This follows from the observation that if MATH can be reduced by a sequence of MATH-moves to MATH then MATH can be reduced to MATH. The collection MATH contains the MATH-sets MATH, MATH, MATH, MATH, and MATH. Applying the MATH-move which replaces MATH with MATH we obtain MATH. By induction MATH can be reduced by a sequence of MATH-moves to MATH. Thus MATH can be reduced to MATH. To deal with MATH, notice that MATH contains REF-sets MATH, MATH, MATH, MATH, and MATH. We apply the MATH-move which replaces MATH with MATH. This new collection contains REF-sets MATH , MATH, MATH, MATH, and MATH. We may apply the MATH-move which replaces MATH with MATH. The resulting collection is exactly MATH. By the induction hypothesis MATH can be reduced by a sequence of MATH-moves to MATH. NAME MATH can be reduced to MATH. |
math/0008100 | For a MATH-subset MATH of MATH define the diameter of MATH to be the minimal cardinality of a boundary MATH-subset of MATH that contains MATH. Thus the boundary MATH-subsets are precisely those of diameter MATH. Let us call MATH-subsets of diameter MATH almost boundary subsets. It suffices to prove that every maximal collection MATH contains an almost boundary subset. Assume by contradiction that MATH does not contain an almost boundary MATH-subset. We make the following easy observation: Let MATH, MATH, MATH, and MATH be four consecutive vertices in MATH; then the MATH-subsets that are not weakly separated with an almost boundary subset MATH are precisely the non-boundary MATH-subsets containing MATH but not MATH. Therefore our assumption and maximality of MATH imply that for every two consecutive vertices MATH and MATH in MATH, there is a non-boundary MATH-subset in MATH which contains MATH but not MATH. Choose a non-boundary MATH-subset MATH in MATH of minimal possible diameter. Without loss of generality, we can assume that a boundary subset of minimal cardinality that contains MATH has MATH and MATH as its endpoints; let us denote this boundary subset by MATH. We can also assume that MATH is not a neighbor of MATH. Let MATH be the neighbor of MATH in MATH. Consider a MATH-subset MATH in MATH such that MATH contains MATH but not MATH. Since MATH is weakly separated from MATH it must be contained in MATH. But then MATH has smaller diameter than MATH which contradicts our choice of MATH. This proves the claim and hence the lemma as well. |
math/0008100 | We proceed by induction on the height. If MATH then we are already done. Assume inductively that the assertion is true for collections of height MATH and let MATH be a collection of height MATH. We need the following: Let MATH and suppose that MATH. Then there exists a unique index MATH such that both MATH and MATH are in MATH. We call MATH the pinch point over MATH and MATH. Let MATH be the maximal index with the property that MATH. Suppose, by contradiction, that MATH. By maximality of MATH this means there exists a non-boundary set MATH which is not weakly separated with MATH. Therefore there exist indices MATH such that one of the following holds: CASE: MATH REF. MATH and MATH REF. MATH and MATH REF Since MATH and MATH are weakly separated it follows that MATH. But then MATH will be weakly separated with MATH. CASE: Since MATH and since MATH is a non-boundary set containing MATH it follows that MATH. But then MATH will be weakly separated with MATH. CASE: Once again it must be the case that MATH. So MATH where MATH violating the maximalitly of MATH. Hence MATH. Suppose there was another pinch point MATH. Either MATH or MATH. If MATH then MATH will not be weakly separated from MATH. If MATH then MATH will not be weakly separated from MATH. Both possibilities violate that fact that MATH consists of only pairwise weakly separated MATH-sets. Uniqueness follows. Let MATH and assume MATH. Let MATH be the pinch point over MATH and MATH. Assume in addition that MATH. Then there exists MATH with MATH such that both MATH and MATH are in MATH. Consider the set of all MATH with the property that MATH and MATH. This set is clearly non-empty since MATH and MATH. Let MATH be the maximal index with this property. Suppose MATH. Then there exists MATH with MATH such that one of the following holds: CASE: MATH REF. MATH REF. MATH REF Since MATH it follows that MATH and MATH must be weakly separated. The only way this can happen is that MATH. But then MATH and MATH will be weakly separated. CASE: Since MATH and MATH are weakly separated it must be the case that MATH. Since MATH it follows that MATH and MATH are weakly separated. The only way this can be resolved is that MATH. But then MATH and MATH are weakly separated. CASE: Either MATH or not. Suppose MATH. Since MATH, and hence weakly separated from MATH, it follows that MATH in which case MATH and MATH will be weakly separated. Thus MATH. Since MATH we know that MATH and MATH are weakly separated. The only way this can happen is that MATH and hence MATH. But this violates the maximality of MATH since MATH. Thus MATH and MATH are in MATH as required. Returning to REF , let MATH be the pinch point of MATH - that is, the unique index MATH such that both MATH and MATH are in MATH. If MATH it follows that MATH. This, taken together with the fact that MATH, violates the hypothesis that MATH. Therefore MATH. Since MATH . REF implies that there exists MATH with MATH such that both MATH and MATH are in MATH. Thus MATH contains MATH, MATH, MATH, MATH, and MATH. The associated MATH-move for this quintuple replaces MATH with MATH. Let MATH be the resulting collection. Notice that MATH contains MATH and that MATH. By induction MATH can be further reduced by a sequence of MATH-moves into a collection of height MATH. NAME this MATH-reduction with the MATH-move transforming MATH to MATH we obtain the desired reduction for MATH. |
math/0008100 | Momentary consideration reveals that MATH consists of pairwise weakly separated MATH-subsets of MATH. In virtue of REF we know that MATH will be maximal if and only if MATH. Since MATH it follows that if MATH and MATH then either MATH or MATH. Consequently MATH. For MATH if MATH then either MATH or else there exists MATH such that, after interchanging MATH and MATH if neccessary, MATH and MATH. By REF , MATH is unique. Hence MATH as required. The inclusion MATH is also clear from the definitions. |
math/0008100 | Let MATH and suppose that all MATH are real and positive for MATH. We need to show that all other NAME coordinates MATH are real and positive. Take any MATH. Take any maximal collection MATH containing MATH. Since MATH is either MATH or MATH we know that Conjecture REF holds and thus MATH and MATH are connected by a sequence of MATH-moves. Claim: Suppose MATH is in MATH and is a positivity test. Let MATH be in MATH and assume that MATH is obtained from MATH by a single MATH-move. Then MATH is a positivity test. Indeed, since MATH and MATH differ by a single MATH-move there exist MATH and MATH, where MATH is empty if MATH and MATH if MATH, such that MATH, MATH, MATH, and MATH are in both MATH and MATH and such that, without loss of generality, MATH is obtained from MATH by replacing MATH with MATH. The fact that MATH is a positivity test is an immediate consequence of the short NAME relation MATH . Let MATH be the minimal number of MATH-moves required to join MATH and MATH. To prove the theorem proceed by induction on MATH and use the claim. |
math/0008101 | For MATH and MATH, we denote by MATH the vector in MATH whose MATH-coordinate is MATH if MATH and zero otherwise. Let MATH, MATH, and MATH be as above. Set MATH and define the sets: MATH . First, we concentrate our attention on the set MATH. We define the sets: MATH . Finally, set MATH . Note that MATH, MATH, and MATH are mutually disjoint. We also have from the triangle inequality that MATH . But, since MATH, we obtain from REF that MATH . Concentrating our attention now on the set MATH, we let MATH . Set MATH . Then MATH, MATH, and MATH are mutually disjoint and, as above, we have MATH . In general, at the MATH step of the induction, we set MATH . Set MATH . Then the sets MATH (MATH), MATH, and MATH are mutually disjoint, and we have MATH . This process must end at some stage MATH with MATH. Summing the inequalities MATH so obtained and simplifying, we have: MATH . But, for each MATH, we have MATH for MATH. Moreover, the mutually disjoint sets MATH REF are contained in MATH. Recalling that MATH, from these two observations REF reduces to MATH . Since the sets MATH are mutually disjoint, we have MATH and so, using REF: MATH . Using REF we have MATH, and thus from REF we have MATH . Also, by REF, we have MATH; thus, MATH . Since MATH, we may now obtain REF by adding REF . |
math/0008102 | Most of the details are contained in the paper CITE, and others will be in a later more detailed paper. But we sketch here the argument for orthogonality in the NAME module. Let the functions MATH be as stated in the lemma. Orthogonality refers to MATH whenever MATH. Hence we must calculate, for MATH: MATH using REF , and the sum under the integral vanishes by REF if MATH. It is important that the selection result mentioned in REF is generally not possible in the category of continuous functions. (See CITE.) Conversely, if MATH is given such that REF defines a representation of MATH on MATH, then the closed subspaces MATH REF are the submodules in a corresponding orthogonal MATH-module decomposition, MATH, that is, the MATH-closure of MATH is MATH for each MATH. Specifically, from REF , we get the identity MATH, where MATH. |
math/0008102 | Let MATH be given as in REF , and let MATH. Set MATH . Then MATH satisfies the same orthogonality relations. For MATH proving the assertion. We used the fact that MATH, see REF . Conversely, if MATH and MATH are given representations as described in REF , then MATH . For it follows from the NAME relations REF that the matrix in REF is unitary, and a computation shows that its matrix entries are multiplication operators. Indeed, MATH . Hence MATH and so MATH defines an element of MATH, and it satisfies REF by its very construction. This proves transitivity. |
math/0008107 | Assume first that MATH is a computable function as described in the proposition. Suppose that MATH is a triangulation of MATH with MATH REF-simplices. Let MATH be a triangulation of some closed MATH-manifold MATH containing MATH REF-simplices. Do all possible sequences of NAME moves on the triangulation MATH of length at most MATH, and check each time if the result is isomorphic to MATH. This gives an algorithm to determine whether MATH and MATH are PL homeomorphic. Conversely suppose that we have an algorithm to recognize MATH among all REF-manifolds. Now we need a complete (finite) list of all triangulations of all MATH-manifolds with a fixed number of MATH-simplices. In dimension three, such a list can be built algorithmically because there is an easy way of recognizing REF-sphere (the NAME characteristic suffices) as a link of a vertex. We can now create all triangulations of MATH with the specific number of REF by running the recognition algorithm for MATH (which exists by assumption) on the list of all REF-manifold triangulations with the specified number of REF. An algorithm, making all possible NAME moves on a triangulation of our MATH-manifold MATH with MATH REF-simplices will after a finite number of steps (by REF ) necessarily produce a given triangulation of MATH containing MATH REF-simplices. Since we can list all triangulations of MATH with MATH (respectively MATH) REF-simplices, this gives an algorithm to calculate the value of the function MATH as required. |
math/0008107 | Normal triangles and squares chop up any tetrahedron in MATH into several pieces. But at most six of these regions are not of the form MATH or MATH (see REF ). Let MATH be the maximal number of disjoint non-parallel normal REF-spheres in MATH. Then the complement of this family has precisely MATH components. Each of those components must contain at least one of the non-product regions. This is because any component, consisting only of product pieces, is bounded by two parallel normal REF-spheres. Since the total number of non-product regions is bounded by MATH, our lemma is proved. |
math/0008107 | First note that after the natural isotopy, all the non-normal arcs we get will give rise to elementary isotopies in the direction of MATH. After each isotopy both MATH and MATH will change, but we'll still denote both resulting spaces by MATH and MATH respectively. After the natural isotopy, MATH and MATH satisfy the following conditions: CASE: In each tetrahedron of MATH the component MATH consists of a family of REF, each one bounded by pieces of MATH and a (possibly disconnected) planar surface, contained in the boundary of REF-simplex. CASE: Each REF-ball from REF intersects any face of the tetrahedron it lies in, in at most one disc. An elementary isotopy moves a disc in MATH over a REF-ball in MATH, which intersects a single edge MATH in MATH. So the new MATH is just the old MATH without REF-ball we isotoped over. This REF-ball is a union of a family of REF, one in each tetrahedron of the star of MATH. In fact there can be more than one REF-ball from the same family in a single REF-simplex if this REF-simplex occurs more than once in the star of the edge MATH. This is perfectly feasible in a non-combinatorial triangulation, but it does not have any effect on the process we are studying. The elements of the above family are the ones that are going to determine the topology of the pieces of MATH in tetrahedra of MATH. In fact, each REF-ball from REF will after an elementary isotopy still satisfy both conditions if we substitute the old MATH with the new one. So after performing all possible elementary isotopies towards the interior of MATH, the surface MATH we end up with will intersect each triangle of MATH in normal arcs and simple closed curves which miss the boundary of the triangle. The region MATH will, after the isotopy, consist of REF in each REF-simplex. There is going to be a bijective correspondence between REF-balls in the end, and the ones we started with. By REF , every REF-ball will still intersect any face of REF-simplex it lies in, in at most one disc. It is also true that the number of these discs will not increase when we pass from REF-ball pieces of MATH at the beginning to REF-balls at the end. Let's look at the pieces of MATH in each tetrahedron. It is obvious that all the possibilities of the lemma can actually arise. We have to see that they are the only ones. Claim. A single piece of MATH can intersect a triangle of MATH in either a unique normal arc or in a single simple closed curve. Every piece of MATH is contained in the boundary of a REF-ball piece of MATH. This REF-ball intersects each triangle of REF-skeleton MATH in at most one disc. So no triangle can contain two simple closed curves or a simple closed curve and a normal arc, both belonging to the same piece of MATH. The same argument tells us that a triangle in MATH can either contain two normal arcs of intersection with a single piece of MATH or at most three of them, each one cutting off a vertex of the triangle in REF-skeleton. Now we need to prove that our piece of MATH can have at most one normal simple closed curve boundary component. So assume the opposite. Since the piece is a subset of the boundary of a REF-ball, no arc contained in it, running between two distinct boundary components of our piece, can be extended to a simple closed curve in REF-sphere bounding that REF-ball, without increasing the number of intersection points with the boundary of our piece. On the other hand, assuming we have at least two normal simple closed curves in the boundary, there surely exist two normal arcs, belonging to the distinct boundary components of our piece, that are contained in a single REF-simplex. Connecting them by an arc in the piece of MATH contradicts what was said before (because these two normal arcs are both contained in the boundary of a disc in REF-simplex they lie in). So now it follows that the piece of MATH we are looking at, can contain at most one normal boundary component which is of length at most eight. This is because the only normal curve of length REF, intersecting each REF-simplex (in the boundary of a tetrahedron) in REF normal arcs as above, consists of REF simple closed curves (one for each vertex of the tetrahedron). It is also well known that normal simple closed curves of lengths REF or REF do not exist. There are precisely three normal simple closed curves of length eight in the boundary of a tetrahedron. So if our piece of MATH is bounded by one such curve, then at least one of the faces of the tetrahedron intersects REF-ball piece of MATH (containing in its boundary the piece of MATH we are considering) in two discs. This is a contradiction that proves the claim. The claim implies the following seven possible boundaries for any piece of MATH: normal simple closed curve of length three, normal simple closed curve of length four, single simple closed curve, normal simple closed curve of length three and a simple closed curve, two simple closed curves, three simple closed curves, four simple closed curves. Since every piece of MATH in any tetrahedron is planar, it is up to homeomorphism determined by its boundary. This implies that all possible pieces of MATH are the ones listed in the lemma. The fact that all these planar surfaces are embedded as in REF (up to an isomorphism of the tetrahedron) follows from the observation that all the elementary discs are parallel to edges of REF-skeleton. |
math/0008107 | REF-ball from the lemma can be view as a union of the following two PL REF-balls: the star of the edge MATH and the cone on the disc in the bounding MATH, which is the complement of the star of the vertex MATH on REF-sphere. The triangulation we are aiming for is equal to the triangulation of the latter REF-ball. We therefore want to flatten the star of the edge MATH down to the cone on the link of MATH. This can be achieved by ``moving" the cone points of REF-simplices in the second of the two REF-balls described above, from our initial cone point to the vertex MATH. Such a REF-simplex, having a face in MATH, which is adjacent to the star of MATH, can be moved by a MATH move or its inverse, depending on the number of edges it has in common with the star of MATH. Repeating this for all (but one) REF-simplices in the cone on the disc MATH almost does the job. All we have to do at this stage, is to use a single MATH move on what's left of the two REF-balls described above. We should also note that the sequence of MATH moves and their inverses, we used to alter the initial triangulation, can always be found. This follows from the well-known fact that every combinatorial triangulation of a PL REF-disc is shellable. |
math/0008107 | Since the triangulation of our disc is shellable, we can index all the simplices in it by numbers from MATH to MATH, so that the increasing integers specify a way of reducing our triangulation down to a single triangle. REF-simplex that's left has index MATH. Let's make a MATH move on it. REF-simplex corresponding to MATH has to share a unique edge with it. Making a MATH move over this edge, changes our original triangulation in the last two REF-simplices to a cone on the boundary of the disc that they compose. The rest of the triangulation is unchanged at this stage. Noticing that the union of the last MATH REF-simplices in our sequence always gives a disc, makes the following induction possible. Say that we already have a cone on the boundary of the disc which is the union of the last MATH REF-simplices and that the rest of the triangulation we started with is unchanged. If the triangle corresponding to MATH has a single edge in common with our cone, we act as before (a single MATH move suffices). If it has two faces in common, a single MATH move finishes the proof. |
math/0008107 | We will divide the process into three steps. First, we glue a cone on the cone on the boundary of MATH onto the bottom part of the boundary of MATH REF . This is a reversed process to destroying an edge which connects the two cone points of the bit that we glued on. It can therefore, by REF , be accomplished by less than MATH . NAME moves. In the second step we perform the same move again, that is, we glue the cone MATH onto the space we've got so far REF . This again requires not more than MATH . NAME moves. The space we've created can be described as a suspension of MATH glued onto the cone on the disc MATH. We know that we can transform the cone triangulation of the disc MATH into the triangulation of MATH by using not more than MATH two dimensional NAME moves REF . It is also clear that in the suspension setting, each MATH move (or its inverse) can be realized by one MATH and one MATH move. A MATH . NAME move can be realized by a MATH and a MATH move. Putting all this together implies our bound. |
math/0008108 | As in CITE. See CITE as well. |
math/0008108 | As in CITE. |
math/0008108 | Let MATH, say MATH where MATH. Let MATH be the singularity type of MATH at MATH. Let MATH be the chain of rational, smooth curves on MATH that are contracted to MATH in MATH. Assume that MATH intersects MATH and MATH intersects MATH. Put MATH and MATH. For each MATH let MATH be the point of intersection of MATH and MATH (see REF ). For each MATH let MATH denote the canonical sheaf of MATH with focus on MATH. For each MATH let MATH and MATH be the correction numbers for MATH at MATH and MATH, respectively. Since MATH is smooth and rational, and its self - intersection is MATH, it follows that MATH has degree MATH. Hence, by NAME 's formula, the ramification divisor MATH of the limit canonical aspect of MATH with focus on MATH satisfies MATH . Put MATH and MATH. Then MATH for MATH. By CITE the limit NAME scheme MATH of MATH satisfies MATH . Now, MATH is the push - forward of MATH to MATH. Combining REF we get the expression for MATH claimed. |
math/0008108 | The existence of MATH and MATH satisfying REF is equivalent to the existence of MATH and MATH satisfying the following four conditions. MATH . In fact, suppose that MATH and MATH satisfy REF . Define MATH and MATH by letting for each MATH, MATH . Then MATH and MATH satisfy REF . Conversely, suppose that MATH and MATH satisfy REF . By REF , there is a proper subset MATH of MATH with MATH. Define MATH and MATH by letting for each MATH, MATH . Then MATH and MATH satisfy REF . Let's prove now the existence of MATH and MATH satisfying REF . Let MATH be the set of points MATH such that MATH where MATH and MATH satisfy REF . Let MATH . If MATH meets MATH, then there are MATH and MATH satisfying REF . In addition, MATH by REF . Thus, it is enough to show that MATH meets MATH. If MATH then MATH by REF . Since MATH, where MATH there is a point MATH closest to MATH. Fix this point MATH, and let MATH and MATH satisfying REF . NAME show that MATH. Suppose MATH. Let MATH be the distance function to MATH. Since the coordinate axes are not parallel to MATH and MATH, we have MATH for every MATH. If MATH for a certain MATH, then we would be able to produce a point of MATH closer to MATH by making MATH vary. So, either MATH or MATH for each MATH. In addition, MATH if MATH and MATH if MATH. Suppose that MATH and MATH for certain MATH. Define MATH and MATH by letting for each MATH, MATH . Then MATH for every MATH. However, since MATH and MATH, we would get a point of MATH closer to MATH by increasing MATH, reaching a contradiction. Therefore, either MATH or MATH. By REF , MATH, and thus MATH for every MATH. Now, let MATH be the closest point to MATH, and MATH. Then MATH. But MATH for every MATH because MATH. Hence MATH and MATH, reaching a contradiction. Thus MATH, and hence MATH and MATH satisfy REF . Let's prove now uniqueness of MATH and MATH meeting REF . Let MATH and MATH satisfying the same conditions as MATH and MATH. Suppose MATH for some MATH. So, MATH and hence MATH for every MATH, with equality only if MATH. Thus MATH for every MATH. Then MATH reaching a contradiction with REF . So MATH. Interchanging the roles of MATH and MATH, we get MATH. Suppose now that MATH for some MATH. So REF holds, and hence MATH for every MATH. Since MATH and MATH for every MATH, we reach a contradiction. So, MATH for every MATH. Interchanging the roles of MATH and MATH, we get MATH. Finally, if MATH for some MATH, then MATH for every MATH by REF . If so, MATH by REF . Now, if MATH but MATH for some MATH, then MATH and MATH by REF . |
math/0008108 | Note first that MATH where the second isomorphism holds because of REF . Now, for each MATH and each MATH, let MATH denote the dualizing sheaf of MATH. Then MATH . Since MATH is smooth and rational, MATH. So, MATH REF follow immediately. Since MATH by REF , it follows from REF that MATH . In addition, since MATH by REF follows from REF . By REF , if there is MATH such that MATH, then MATH, and hence MATH. Conversely, if MATH then MATH. So, since MATH by REF , by REF , MATH . In any case, we get REF . Since MATH, with equality only if MATH, by REF and NAME - NAME, MATH . Putting together REF we get REF . |
math/0008108 | Let's prove REF . Since REF holds, REF applies. Let MATH. Let's check first that MATH is injective. In fact, let MATH such that MATH. Then MATH for every MATH. So MATH for every MATH by REF , and hence MATH. By REF , MATH as well. So MATH, and thus MATH for every MATH by REF . Hence MATH. So MATH is injective. Let's prove now that MATH is the canonical sheaf of MATH with focus on MATH. We must check that REF are verified. REF holds because MATH is injective. Let's check REF . The limit canonical system of MATH associated to MATH has (affine) rank MATH. Hence, to show that REF holds it is enough to prove that the kernel MATH of the restriction map MATH has dimension strictly less than MATH for every irreducible component MATH of MATH other than MATH. We divide the proof in three steps. CASE: We show that MATH. In fact, if MATH then MATH for each MATH. So MATH for each MATH by REF , and hence MATH. If MATH then MATH for every MATH by REF , and hence MATH. So, MATH . Thus MATH by REF . CASE: Let MATH and MATH with MATH such that either MATH or MATH. We show that MATH. In fact, if MATH then MATH by REF . By REF , if MATH then MATH for every MATH, and thus MATH. If MATH as well, then MATH by REF . So, MATH . Since MATH, it follows from REF that MATH. By REF , MATH. CASE: Let MATH and MATH such that MATH. We show that MATH. In fact, if MATH then MATH by REF . If MATH, then MATH for every MATH by REF , and thus MATH. If MATH as well, then MATH by REF . So, MATH . By REF , MATH, with equality only if MATH. So MATH because MATH. Since MATH, we have MATH . So MATH by REF . REF is checked. So MATH is the canonical sheaf of MATH with focus on MATH. Let's prove now that MATH is the limit canonical aspect of MATH with focus on MATH. Since the aspect has rank MATH, it is enough to show that MATH. By definition of MATH, we have MATH. Hence, MATH by REF . So MATH by REF . So MATH is the limit canonical aspect of MATH with focus on MATH. Consider the remaining assertions in REF . Let's check first that MATH . In fact, if MATH then MATH for every MATH by REF . Therefore, MATH . Now, since MATH, from the definition of MATH we get MATH, and hence MATH by REF . So equality holds in REF , and thus REF follows. Let's check now that MATH is not a base point of MATH for any MATH. In fact, if there were MATH such that MATH, then the restriction MATH would be zero by REF , thus contradicting REF . In addition, if MATH for a certain MATH, then MATH by REF , thus contradicting REF . So, no point of MATH is a base point of MATH. Finally, MATH because MATH, and because of REF . The proof of REF is complete. REF follows by analogy. Let's prove now REF . By REF , the sheaves MATH and MATH are the canonical sheaves and MATH and MATH are the limit canonical aspects of MATH with foci on MATH and MATH, respectively. By REF , for each MATH the correction number for MATH at MATH is MATH. By REF , for each MATH the correction number for MATH at MATH is MATH. Then REF follows from REF . |
math/0008108 | Let MATH denote the reducible nodes of MATH. Let MATH be the versal formal deformation of MATH. Then MATH, where MATH for a certain integer MATH. In addition, we may assume that there is an isomorphism of MATH - algebras, MATH for each MATH, where MATH and MATH are the local equations of the branches of MATH meeting at MATH; see REF For each MATH choose MATH. Consider the surjection MATH, sending each MATH to MATH and each MATH to MATH. Let MATH and set MATH, where MATH is viewed as a closed subscheme of MATH by means of MATH. Since all MATH are non - zero, the surface MATH is regular. So MATH is a regular smoothing of MATH. In addition, the isomorphism REF restricts to an isomorphism of MATH - algebras, MATH for each MATH. Recall the notation in REF We need to show that we may choose MATH such that MATH for MATH. Fix MATH and MATH. For convenience, put MATH . Let MATH such that MATH. Suppose that MATH is the local equation of MATH at MATH and MATH that of MATH. The completion MATH of MATH at MATH is generated by MATH. In addition, MATH where MATH. For each MATH let MATH be the map defined by sending MATH to MATH. Then MATH. In addition, since MATH is an invertible MATH - submodule of MATH such that MATH, there is MATH such that MATH. We claim that we may choose MATH such that MATH, and hence MATH. In fact, there's clearly MATH such that MATH if MATH. On the other hand, if MATH then MATH by REF , and hence MATH. So any choice of MATH yields MATH. Now, let MATH vary in MATH and pick MATH for each MATH as in the above paragraph. By REF , MATH for every MATH such that MATH. It follows that there is MATH such that MATH for every MATH. Then MATH for each MATH. Finally, let MATH vary in MATH and pick MATH for each MATH as in the above paragraph. Now, for each MATH the sheaves of fractional ideals MATH and MATH coincide away from the reducible nodes of MATH because MATH . As we proved that MATH for each reducible node MATH of MATH and each MATH, we get MATH for MATH. |
math/0008108 | The ``only if" assertion is clear; see REF Let's prove now the ``if" assertion. Since MATH for every MATH, we may assume that MATH is an invertible MATH - submodule of MATH satisfying MATH. To finish the proof we apply REF with MATH, with MATH and MATH. So we check that the conditions for REF hold. First, since MATH, we have MATH as well. Second, since MATH, we need only verify REF for MATH. Now, if MATH are such that MATH, then there is MATH such that MATH. Since MATH, we have MATH as well. So REF is verified. By REF , there is a regular smoothing MATH of MATH such that MATH. |
math/0008108 | Let MATH be the set of irreducible components of MATH. Let MATH such that MATH . Let's prove REF . Consider first the ``only if" assertion. Let MATH be a regular smoothing of MATH and MATH, where MATH is the dualizing sheaf of MATH. By REF , MATH is the canonical sheaf with focus on MATH. If MATH, then REF holds by REF . Consider now the ``if" assertion. Preserve REF Let MATH denote the partition of MATH in maximal MATH - balanced subsets. Since MATH, by REF , MATH for every MATH. Hence MATH as well. So MATH, and hence MATH is treelike for every MATH. Let MATH. Since REF holds and MATH is treelike, MATH for every MATH. By REF , there is a regular smoothing MATH of MATH such that MATH, where MATH. Let MATH be the dualizing sheaf of MATH. Since REF holds, by REF the canonical sheaf MATH of MATH with focus on MATH satisfies MATH. So MATH. The proof of REF is complete. REF follows by analogy. Let's prove REF . For each pair MATH let MATH. Then MATH if and only if MATH. In addition, using that MATH it follows from REF that MATH where MATH and where MATH is a formal sum with integer coefficients of the rational curves MATH with MATH and MATH. Consider first the ``only if" assertion of REF . Let MATH be a regular smoothing of MATH, and MATH its dualizing sheaf. By REF , the canonical sheaves MATH and MATH of MATH with foci on MATH and MATH, respectively, are given by REF . By REF , MATH . So, it follows from REF that MATH. Consider now the ``if" assertion. Let MATH and MATH. We shall use the set - up of REF. Let MATH and MATH. By REF , we may view MATH and MATH as invertible MATH - submodules of MATH and MATH such that MATH and MATH. Multiplication by MATH gives automorphisms of MATH and MATH; let MATH and MATH denote the images of MATH and MATH under these automorphisms. It's clear that MATH and MATH. For each MATH, each MATH and each MATH let MATH. NAME choose MATH such that MATH and MATH satisfy REF . In other words, we'll pick MATH such that MATH for all MATH and all distinct MATH such that MATH. In fact, we need only obtain REF for MATH and MATH that intersect, that is, for MATH and MATH such that MATH for some MATH and some MATH. In addition, we need only obtain REF for MATH in a MATH - basis of MATH . Let MATH for each MATH and each MATH. Choose a MATH - basis MATH of MATH for each MATH and each MATH. Note that MATH if and only if MATH and MATH; in this case let MATH. In all other cases, MATH has rank REF. If MATH then MATH is a basis of MATH for every MATH; in this case let MATH. Consider distinct MATH with non - empty intersection. Then either MATH is treelike or MATH and there are distinct MATH such that MATH. In either case there is an isomorphism between MATH and MATH for each MATH. In fact, such an isomorphism exists in the former case because both sheaves have the same restriction to MATH and to MATH; and in the latter case because MATH and MATH. So, for each MATH there is a unique MATH such that MATH as subsheaves of MATH. Let MATH for each MATH and each MATH. Since MATH, it follows that MATH . Therefore, finding MATH such that REF holds for all MATH and all distinct MATH with MATH is equivalent to finding MATH such that, for each MATH, MATH . Set MATH and MATH. For MATH choose MATH inductively for MATH such that MATH . By REF , we may choose MATH such that MATH for every MATH. If MATH, set MATH and MATH. Then REF is achieved for every MATH. If MATH is such that MATH and MATH, then choose MATH inductively for MATH such that MATH and choose MATH inductively for MATH such that MATH . In addition, set MATH for MATH and MATH for MATH. Then REF is achieved. Let MATH such that either MATH or MATH or MATH. Then MATH has rank REF for every MATH. By REF , MATH. Now, since MATH and MATH are positive and MATH, it follows from REF as well that either MATH or MATH. So, there is MATH such that MATH . For MATH choose MATH inductively such that MATH . For MATH choose MATH inductively such that MATH . Now, let MATH and MATH. By REF , MATH and MATH are MATH - independent. So, we may choose MATH such that MATH . Then REF is achieved. So, we obtained MATH such that REF holds for all MATH and all distinct MATH such that MATH. By REF there is a regular smoothing MATH of MATH such that MATH . Since MATH and MATH, it follows from REF that MATH and MATH are the restrictions of the canonical sheaves of MATH with foci on MATH and MATH, respectively. |
math/0008108 | Let MATH be the map contracting the subcurve MATH for every MATH. For each MATH let MATH be the blow - up map, MATH and let MATH be the map contracting the subcurve MATH for every MATH. Let's prove REF . First, let MATH be a regular smoothing of MATH such that MATH is the limit canonical aspect of MATH with focus on MATH. Let MATH be the restriction to MATH of the canonical sheaf of MATH with focus on MATH. By REF , MATH . By REF , the sheaf MATH is torsion - free, rank - REF, and fails to be invertible precisely along the set MATH. Hence there are an invertible sheaf MATH on MATH and an isomorphism MATH. Restricting MATH to MATH and MATH, and removing torsion, we obtain isomorphisms MATH and MATH. So there is a commutative diagram, MATH where the vertical maps are restriction maps, and the horizontal maps are the isomorphisms induced by MATH. In addition, MATH and MATH by REF . Now, since REF is commutative, the restriction maps MATH and MATH have the same image. Hence MATH by REF , because MATH is the limit canonical aspect of MATH with focus on MATH. So MATH. Conversely, let MATH be an invertible sheaf on MATH as in REF . Then MATH is torsion - free, rank - REF, and fails to be invertible precisely along MATH. As MATH for each MATH, there are an invertible sheaf MATH on MATH such that REF holds and an isomorphism MATH. As before, MATH induces isomorphisms MATH and MATH. So REF holds. By REF , there is a regular smoothing of MATH whose canonical sheaf with focus on MATH restricts to MATH on MATH. As before, MATH is the image of the restriction map MATH. So MATH is MATH - smoothable. The proof of REF is complete. REF follows by analogy. Let's prove REF . First, let MATH be a regular smoothing of MATH such that MATH and MATH are the limit canonical aspects of MATH with foci on MATH and MATH, respectively. Let MATH and MATH be the restrictions to MATH of the canonical sheaves of MATH with foci on MATH and MATH, respectively. As observed in the proof of REF , there are invertible sheaves MATH and MATH on MATH and MATH, respectively, such that MATH and MATH. In addition, MATH and MATH satisfy REF , respectively. Since MATH, there is a map MATH. This map is injective because it is injective generically on MATH and MATH and because MATH is invertible. So MATH and MATH are isomorphic because they restrict to isomorphic sheaves on the irreducible components of MATH. Analogously, MATH. Let MATH and MATH be the pull - backs of MATH and MATH to MATH. Let MATH be given by REF . Since MATH by REF , we have MATH . Now, since MATH, it follows that MATH. So REF holds. Conversely, let MATH be an invertible sheaf on MATH and MATH an invertible sheaf on MATH as in REF . As observed in the proof of REF , there are invertible sheaves MATH and MATH on MATH such that MATH and MATH, and such that REF below hold. MATH . Now, since MATH, it follows from REF that REF holds. As shown above, MATH and MATH. Hence MATH. By REF , there is a regular smoothing of MATH whose canonical sheaves with foci on MATH and MATH restrict to MATH and MATH, respectively. As shown in the proof of REF , the restriction maps MATH and MATH have images MATH and MATH, respectively. So MATH is MATH - smoothable. |
math/0008108 | Let's prove REF . Since MATH, also MATH, and MATH for every MATH by REF . By NAME - NAME, MATH and MATH is surjective. Now, MATH is injective by REF . So MATH by REF . Thus, taking inverse images by MATH gives us a closed embedding MATH. Since MATH and REF holds, the description of MATH given in REF applies. Comparing this description with that of MATH given in REF, we get MATH. The proof of REF is complete. REF is proved analogously. Let's prove now REF . Since MATH and MATH are non - zero, also MATH and MATH are non - zero. Since REF hold as well, the description of MATH given in REF applies. Comparing this description with that of MATH given in REF, we get MATH. REF is proved. |
math/0008108 | We shall use the set - up of REF. Let's prove REF . Assume first that MATH. Then REF says that MATH is the limit canonical aspect with focus on MATH of any regular smoothing of MATH. Therefore MATH. Assume now that MATH. Then MATH and MATH. By REF , all the NAME coordinates of MATH in MATH are non - zero. In addition, MATH because MATH. Since the NAME coordinates of MATH are non - zero, MATH if and only if MATH for all subsets MATH and MATH of MATH satisfying MATH. Since MATH, given MATH distinct, there is a subset MATH with MATH such that MATH but MATH. Letting MATH, we have MATH if and only if MATH. It follows that the orbit map MATH factors through an isomorphism MATH, where MATH is viewed inside MATH under the diagonal embedding. So MATH is isomorphic to a torus of dimension MATH. Since MATH, REF says that MATH is the image of MATH under a closed embedding MATH. Thus MATH is locally closed in MATH and isomorphic to a torus of dimension MATH. The proof of REF is complete. REF follows by analogy. Let's prove REF . If MATH then MATH. If MATH then MATH. So REF follows from REF in the first case, and from REF in the second case. Assume now that MATH and MATH. Then MATH, MATH, MATH and MATH are non - zero. View MATH inside MATH and MATH under the diagonal embeddings. Let MATH, where MATH is given by REF . Since MATH and MATH are coprime, MATH is a subtorus of dimension MATH of MATH. In addition, MATH is a one - dimensional subtorus of MATH unless MATH, in which case MATH. Now, MATH and the orbit map MATH is the restriction to MATH of MATH. Since MATH and MATH factor through isomorphisms MATH and MATH, then MATH factors through an isomorphism MATH. So MATH is closed in MATH and isomorphic to a torus. In addition, MATH unless MATH, in which case MATH. Since MATH and MATH are non - zero, REF says that MATH is the image of MATH under a closed embedding MATH that sends MATH to MATH. Thus MATH is closed in MATH and isomorphic to a torus of the dimension prescribed in REF . |
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