paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9907012 | Note that the definition of MATH is meaningful since the right hand side clearly does not depend on the choices of MATH, MATH and MATH. It is also clear that the function MATH is holomorphic on MATH and that the operator MATH is linear. Let us show that MATH is continuous. Let MATH be a continuous semi-norm of MATH. We may assume that there is a bounded subset MATH of MATH and a compact subset MATH of MATH such that MATH . For MATH, set MATH. By cofinality, we have MATH . Moreover, for any MATH, MATH is a NAME space and the restriction MATH is injective. Hence, by CITE, there is MATH and a bounded subset MATH of MATH such that MATH. Choosing a compact interval MATH of MATH such that MATH and MATH, we see that MATH and we can find MATH such that MATH . Let us consider the linear map MATH defined by setting MATH . Let us check that MATH is continuous. Consider a compact subset MATH of MATH and a compact subset MATH of MATH. The set MATH being bounded in MATH, MATH is a bounded subset of MATH and we have MATH . For any MATH and MATH, there is an open MATH of MATH, containing MATH and MATH such that MATH. Let MATH be a closed interval included in MATH and such that MATH and let MATH be the oriented boundary of MATH. For any MATH, we have using the continuity of MATH and NAME representation formula MATH . It follows that MATH or, in other words, that MATH is a section of MATH. Let us consider the continuous linear map MATH defined as follows. Let MATH and MATH. Consider MATH such that MATH . Then, for any MATH, we set MATH where MATH is the oriented boundary of MATH. Since, for any MATH and any MATH, we have MATH we see that MATH. The map MATH is thus a retraction of MATH. Thanks to a well-known result of homological algebra, the proof will be complete if we show that MATH . To this end, consider MATH and MATH. Fix MATH such that MATH and denote MATH the oriented boundary of MATH. Let MATH be the oriented boundary of a closed interval MATH such that MATH and MATH. Denoting MATH the oriented boundary of MATH and using NAME integral formula, we get MATH . |
math/9907012 | We have successively MATH where MATH denotes the dualizing complex on MATH for sheaves with values in MATH. Note that REF follow from REF and that REF comes from REF . |
math/9907012 | Transposing to sheaves with values in MATH a classical result of the theory of abelian sheaves, we see that MATH if MATH and MATH are objects of MATH and MATH. This formula combined with REF gives the first part of the result. The second part follows from REF using REF . |
math/9907012 | This follows directly from the general isomorphism MATH which holds for any objects MATH and MATH of MATH and MATH. |
math/9907012 | Since the problem is local, it is sufficient to show that MATH for MATH if MATH and MATH are NAME open neighborhoods of MATH in MATH and MATH. In this situation, MATH is a holomorphically convex compact subset of MATH and we get from REF that MATH . The conclusion follows directly. |
math/9907012 | Using REF , we see that MATH . Applying MATH, we get successively MATH . This gives the first part of the result. To get the second one, it is sufficient to use REF , if we remember that, following CITE, we have MATH . |
math/9907012 | For any MATH, it is clear that MATH since MATH. The canonical morphism MATH induces by adjunction a morphism MATH . If MATH, MATH and MATH and the preceding morphism is an isomorphism. It follows that it is also an isomorphism for MATH and, hence, if MATH is a direct summand of a free MATH-module of finite type. Thanks to the local structure of perfect complexes, the conclusion follows easily. |
math/9907012 | Applying MATH to the morphism MATH we get a canonical morphism MATH since MATH. The conclusion follows by working as in the proof of REF . |
math/9907012 | Thanks to REF , this is an easy consequence of REF . |
math/9907015 | A simple induction argument shows that REF reduces to MATH so the case MATH is realized using the golden mean shift itself. For the general case, let MATH where MATH is a set with MATH elements, and define MATH by MATH. Then MATH so we are done. |
math/9907015 | From NAME and NAME, CITE, we have that MATH mod MATH if MATH. The identities MATH mod MATH and REF imply that MATH mod MATH. |
math/9907018 | For convergence, it suffices to show that each factor in the square brackets above is a MATH-unit in MATH, which follows from REF . A straightforward manipulation of the factors permits us to prove that MATH satisfies the identity in (REF c) and thus represents the sigma function. See CITE for further details. |
math/9907018 | Because the value of MATH is independent of a chosen NAME equation, we are free to fix a NAME equation which is minimal at all places of MATH extending MATH. Furthermore, it suffices that the theorem holds on MATH. Given the ways we have normalized our pairings, we need to show that MATH . By the definition of MATH along with REF , we have MATH and thus the terms from REF corresponding to the finite places exactly match those from REF . Finally we need to show for each MATH that, before taking norms, we have MATH, which follows from REF and the ultrametric inequality. |
math/9907018 | Recalling the ramification hypotheses on MATH and MATH set earlier in the section, the equation for MATH will be minimal at all places of MATH extending them, explaining REF . REF is then a restatement of REF . |
math/9907018 | Using REF , we find MATH and then the result follows by induction on REF . |
math/9907018 | Taking MATH, we note that MATH for all MATH and MATH. By REF we see that MATH where MATH is given by the product in REF . By the construction of MATH we know that MATH. The fact that MATH does not divide MATH combined with REF then implies that MATH . Letting MATH and MATH denote the positive parts with respect to MATH and MATH, we then have MATH where MATH is the product of all the local factors at the finite places. Furthermore, from the construction of MATH we have that MATH where MATH is the same in both equations. As we will only be interested in these quantities up to MATH-th powers, it is important to note that, for all MATH, MATH modulo MATH, and likewise for MATH. Now let MATH be a set of representatives of the embeddings of MATH. Up to MATH, we know from REF that MATH where for some fixed place MATH the point MATH is defined in REF . Likewise, we can show MATH for some fixed place MATH and corresponding MATH. By REF we then have MATH . Likewise, we have MATH . For both of these two congruences, the right-hand sides are actually elements of MATH. Now by REF the points MATH and MATH are simultaneously MATH-conjugates, and since MATH is not a MATH-th power, we conclude that MATH is also not a MATH-th power by REF . In particular, MATH and the theorem follows. |
math/9907018 | It suffices to prove the theorem for points MATH, MATH in MATH. For REF , choosing a MATH so that MATH we need to show that MATH satisfies REF . Fixing places MATH and MATH, we need to show that for all embeddings MATH, the points MATH and MATH are simultaneously MATH-conjugate. Let MATH, which is torsion-free by the hypotheses on MATH. By the elementary divisors theorem there is a MATH-basis MATH for MATH so that MATH . Note that as MATH, our hypothesis that the MATH-torsion MATH for all MATH implies that MATH. Thus via the NAME pairing with chosen basis MATH, (REF c) becomes MATH where MATH is the kernel of the NAME MATH. By expressing each MATH as a linear combination of the MATH, we need to show that there is a MATH so that MATH . First, if MATH, we apply (REF b). Then there is an element MATH accomodating REF , which we lift back to MATH. Second, if MATH, then for each MATH we set MATH to be the unique element of MATH. Indeed, because the MATH-torsion MATH by assumption, these two sets meet at only one point. Using the assumptions on the MATH-torsion on the reductions, we have that MATH for each MATH. Finally, in the general case we proceed by combining the above two arguments. For each MATH, MATH is a largest integer MATH for which MATH. As in the preceding paragraph we have MATH. Thus from REF we find that there is then an automorphism MATH which satisfies REF . For REF , the proof is much the same as in REF . By taking MATH so that MATH, an argument similar to the proof of REF shows that we need to have MATH and MATH satisfy REF simultaneously. That is, we require that there exist MATH so that MATH for all MATH. This is achieved by applying the arguments of REF to MATH. |
math/9907018 | Because the determinant of the NAME height is non-zero, we know MATH is non-degenerate on MATH. Thus we suppose MATH for some MATH, MATH. By taking suitable multiples of MATH and MATH we can assume that both are in MATH, and then MATH for all MATH large enough. We write MATH where MATH, MATH. Then MATH, and by REF , we have that MATH is not a MATH-th power in MATH. Moreover, MATH and we are done. |
math/9907019 | Let MATH be the algebraic closure. Let MATH be an automorphism of MATH which fixes MATH. Then MATH is also a zero of MATH of the same absolute value. Thus Conjecture REF implies that MATH actually equals MATH. This is equivalent to the total inseparability of MATH. |
math/9907019 | We begin by assuming that MATH is a real valued character. Let MATH be a zero of MATH. Then complex conjugation, and the functional equation, imply that MATH, MATH, MATH and MATH are also zeroes of MATH with of course the same absolute value. If MATH is non-real and non-purely imaginary, then these elements are distinct which is a violation of the conjecture. The purely imaginary zeros of MATH are the real zeroes of MATH of which there are only finitely many by the functional equation. We postpone the proof when MATH is non-real to the next section. |
math/9907019 | By REF , Conjecture REF implies that almost all zeroes of MATH are totally inseparable over MATH. This is then obviously true for MATH. |
math/9907019 | As above we deduce that Conjecture REF implies that almost all zeroes of MATH are real. We therefore immediately deduce the same for MATH. |
math/9907019 | Let MATH be the automorphism taking MATH to MATH. Extend MATH to the full algebraic closure. Let MATH be a zero of MATH where MATH is sufficiently large so that Conjecture REF applies for both MATH and MATH. Then MATH by REF . On the other hand, MATH is a zero of MATH. Thus by Conjecture REF it must be the unique zero of its absolute value. |
math/9907019 | Assume GRH. Then, using complex conjugation and the functional equation as before, we see that the MATH and MATH fill out all zeroes of a fixed absolute value. Thus the result follows immediately from GSC. |
math/9907019 | The first part of Conjecture REF (applied to MATH) implies that almost all the zeroes are simple. Combining this with Conjecture REF, and a simple argument using NAME Polygons, we find that almost all zeroes of MATH are actually in MATH. On the other hand these zeroes are also totally inseparable over the subfield MATH of MATH by REF . Note that MATH is separable over MATH. Thus, the only way that this can happen is that the zeroes belong to MATH. |
math/9907019 | As MATH, it is clear that MATH and MATH are isomorphic. Moreover, without loss of generality, we can set MATH. As the NAME polygon does not depend on the choice of uniformizer, we choose our positive uniformizer to be MATH and we begin by letting MATH be a positive integer divisible by MATH. Now the coefficient of MATH in MATH is precisely the sum of MATH where MATH and MATH is monic. On the other hand, the coefficient of MATH in MATH is the sum of MATH such that MATH is monic of degree MATH and MATH. This last condition is the same as saying that MATH has non-vanishing constant term. The set MATH, where MATH is monic, ranges over all polynomials MATH in MATH with constant term MATH and degree (in MATH) MATH. Moreover, as MATH is divisible by MATH, the set MATH is the same as the set MATH where MATH is a monic polynomial of degree MATH and has non-vanishing constant term. Let us denote by MATH the function obtained by replacing MATH by MATH in MATH and applying the isomorphism MATH given by MATH. The above now implies that MATH . The result for positive MATH divisible by MATH follows immediately. The general result then follows by passing to the limit. |
math/9907020 | Suppose the contrary that MATH. Since MATH, MATH is either MATH or MATH. According CITE and its notation, MATH with MATH (respectively, MATH) when MATH (respectively, MATH). Thus MATH and MATH. Hence MATH. Write MATH . Then MATH. Since MATH, we have MATH. We may assume that MATH and MATH. By the topological NAME fixed point formula, we have the diagonalization MATH, relative to some basis. By considering minimal polynomial of MATH over MATH, we have either MATH (identity matrix), or MATH . If the second case for MATH occurs, then simultaneously diagonalize MATH and MATH on MATH, we would get a diagonalization of MATH whose diagonal entries consist of a few MATH and between MATH and MATH entries of MATH primitive roots of the unity, which is impossible because MATH is of order MATH and the NAME number MATH. If the first case for MATH occurs, then we get the following diagonalizations, relative to two possibly different bases (up to re-ordering): MATH . Thus MATH by the topological NAME fixed point formula. In particular, MATH contains a curve. On the other hand, since MATH, we have MATH, so that MATH consists of finitely many points, a contradiction. |
math/9907020 | Since MATH is a primitive sublattice of the unimodular lattice MATH, we have a natural isomorphism MATH. Noting that MATH, we can apply the same argument as in CITE and CITE for the pair MATH (instead of MATH there) to get MATH and MATH for some integer MATH with MATH. Since MATH, we have MATH . By the classification of indefinite unimodular even lattices, the case MATH, MATH is impossible and in the case MATH we have MATH. By CITE, a MATH-elementary REF even hyperbolic lattice of rank MATH, is determined uniquely by its rank and discriminant. So, MATH when MATH. Suppose that rank MATH. Write MATH, where MATH for integers MATH. Then det-MATH (mod REF) and hence the case MATH is impossible. We consider the case where MATH, MATH. Note that MATH is generated by a MATH-basis MATH where MATH . Here MATH's form the basis of MATH with MATH as the intersection matrix. Since MATH, each MATH (and hence each MATH) is divisible by MATH. So MATH with an indefinite even unimodular lattice MATH. Thus MATH under a suitable basis. |
math/9907020 | If MATH, the result follows from CITE. Let us consider the case MATH. Since MATH, MATH admits a MATH-stable Jacobian fibration MATH by CITE. Let MATH and MATH be a general fiber of MATH and the unique MATH-stable section of MATH. Here the uniqueness of the MATH-stable section follows from the fact that if MATH is also a MATH-stable section then MATH and MATH . We see then these equalities imply MATH and MATH. Let MATH be the automorphism of the base space MATH induced by MATH. Since there are no elliptic curves admitting NAME automorphism of order REF, MATH is also of order REF. We may then adjust an inhomogeneous coordinate MATH of MATH so that MATH. We note that MATH and MATH are both irreducible, because the irreducible component MATH of MATH meeting MATH is MATH-stable so that MATH unless MATH. Since MATH, an easy local coordinate calculation shows that neither of MATH is of NAME 's type MATH. Moreover, noting that MATH permutes the other singular fibers, we have MATH for some positive integer MATH. Thus after suitable change of inhomogeneous coordinate MATH if necessary, MATH is of type MATH and the set of the other singular fibers is either REF MATH, all of NAME 's type MATH, REF MATH (MATH), all of NAME 's type MATH, or REF MATH, all of NAME 's type MATH. |
math/9907020 | Assuming the contrary that REF occurs, we denote by MATH the irreducible component of MATH meeting MATH. Since MATH is MATH-stable, we have MATH. Now MATH implies that MATH and hence MATH. This leads to MATH which is a contradiction. |
math/9907020 | Assuming the contrary that REF happens, we will determine the NAME equation MATH of MATH. Since the singular fibers of MATH are all of type MATH, the MATH-function MATH as a rational function. Thus, MATH and the equation is MATH. Let us consider the discriminant divisor MATH . Since the singular fibers of MATH over MATH are MATH and these are all of type MATH, we have MATH for some nonzero constant MATH. Then MATH for some nonzero constant MATH. Changing MATH by suitable multiples, we finally find that MATH is given by the equation MATH which is isomorphic to MATH in REF . In particular, MATH by CITE. Thus MATH, a contradiction. The referee pointed out that the argument above is similar to CITE; we keep this argument for readers' convenience. |
math/9907020 | Again we will determine the NAME equation MATH of MATH, where MATH are polynomials in MATH. First note that MATH and MATH by the canonical bundle formula. Since MATH has singular fibers MATH of type MATH, the discriminant divisor MATH is equal to MATH for some non-zero constant MATH. Since the MATH-function MATH is MATH-invariant, MATH (and hence MATH) are also MATH-semi invariant. Thus MATH where MATH are constants, MATH, and MATH when MATH. Comparing coefficients of the equality MATH we see that MATH . Noting that MATH because of the existence of singular fibers of type MATH. We have also MATH, otherwise, MATH is birational to a product of a fibre and the parameter space MATH and hence is not a MATH surface, absurd! We can, by a suitable coordinate change, normalize the NAME equation of MATH as MATH . Here MATH is a constant, and MATH for otherwise MATH by CITE. Conversely, by the standard algorithm to finding out the singular fibers CITE, we see that this elliptic surface MATH has REF singular fibers of type MATH and a singular fiber of type MATH if and only if MATH. Moreover, MATH admits an automorphism MATH of order MATH given by MATH . Since MATH and MATH make the fibration MATH and the section MATH stable and satisfy MATH we have MATH. Now the condition that MATH implies that MATH and MATH, by the maximality of MATH and by considering MATH where MATH acts on MATH as the involution around MATH. Write MATH with an involution MATH. Since MATH, we see that MATH and MATH are both MATH-stable, and MATH and MATH are two MATH-fixed points. In other words, MATH does not switch MATH and MATH, because the fibres MATH and MATH are of different types: MATH, MATH. If MATH acts on the base space MATH as an involution, MATH permutes REF singular fibers of type MATH as well as REF roots of the discriminant divisor MATH whence MATH, a contradiction. Thus, MATH is the involution of MATH around MATH, that is, MATH. This means MATH and we are done. |
math/9907020 | The proof is almost identical to the situation where MATH, except that MATH does not admit MATH-stable sections and we use the assumption that MATH and the fact that MATH is smooth. The type of the action is determined by an elementary local coordinate calculation of the normalization of MATH and the fact that MATH. Actually, we have one more possible case in which MATH is smooth with MATH and MATH is of type MATH with MATH. But then the relatively minimal model of MATH is a rational elliptic surface with no multiple fibers and hence has a section MATH. Now the pullback on MATH of MATH is a MATH-stable section, which contradicts MATH. |
math/9907020 | The first assertion is clear. We use the notation, like MATH, MATH in the Introduction for MATH. If MATH, then MATH is either MATH or MATH by REF . However, then MATH contains a curve by the main REF (proved already when MATH) and REF , which contradicts the fact that the canonical covering map is étale in codimension one. Thus MATH. Next, we show that MATH is maximal. Assume that MATH and MATH also satisfies the condition in the Introduction. By REF , MATH. By CITE, it is enough to eliminate the case where MATH. Assume the contrary that this case happens. We may assume that the order MATH element MATH is as in the Introduction. Since rank MATH, MATH and MATH is either MATH or MATH, we have either MATH and MATH equals one of: MATH in the case where MATH, or MATH in the case where MATH. Since MATH and MATH, we have MATH, noting that the actions of MATH around two points MATH are different. Thus the topological NAME formula shows that the only possible case is: MATH . Let MATH. Then, MATH . In particular, MATH. This, together with the fact that MATH, implies that MATH consists of smooth curves and at least one of them is a smooth rational curve, say MATH. Write the (disjoint) irreducible decomposition of MATH as MATH . Since MATH, the MATH acts on the set MATH. First assume that MATH. Then MATH would be mutually disjoint REF rational curves with MATH where both sides of the inclusion are of rank REF, whence they are equal. However, MATH then contains no ample classes, a contradiction. Thus MATH and MATH. But, this can not happen, because the action of MATH around MATH are of types MATH if MATH and MATH if MATH, and there are no MATH with MATH. Therefore, MATH is maximal and REF is proved. |
math/9907021 | Consider MATH for all MATH. If the character of MATH is the same for all MATH, one can identify the MATH as vector spaces. Their inner product matrix is smooth (in fact analytic) in MATH, and positive definite at MATH since we consider the compact case. This implies that all eigenvalues are positive on MATH: assume to the contrary that the matrix were not positive definite for some MATH. Then it would have a zero eigenvalue for some MATH, which implies that its null space is a submodule of MATH. But this is impossible, since the MATH are irreducible by definition. For MATH, some eigenvalues may vanish; but then MATH is the quotient of MATH modulo its null space, which again has a positive definite inner product. In particular, assume that MATH. Let MATH be the smallest integer MATH, which is MATH. Then MATH is associated to MATH as defined in REF. Therefore by REF , MATH at MATH has the same character as for MATH, since MATH. For all other roots of unity MATH, the character is again the same since the associated MATH is larger than MATH. Thus the above argument applies. |
math/9907021 | By using an element MATH of the NAME group if necessary, we can assume that MATH is an anti - dominant weight. Then MATH with MATH, and MATH. This implies that MATH where MATH is a hyperplane as defined in REF , and equality holds precisely if MATH and MATH for some MATH. The desired set of simple roots is obtained by applying MATH to the original simple roots. Then all (new) simple roots MATH for MATH are compact according the definition in REF . |
math/9907021 | Assume that MATH is as required. Then there are MATH such that MATH where MATH, thus MATH is dominant integral. For MATH, let MATH, MATH, and MATH. We claim that for sufficiently large MATH, the character of MATH is given by NAME formula for all MATH between REF and MATH. Then the first part of the proposition follows from REF . Let MATH with MATH, with associated MATH as in REF. By the strong linkage principle CITE, the character of MATH can differ from MATH only by the sum of classical characters REF with dominant MATH, which are ``strongly linked" to MATH by a series of reflections by hyperplanes MATH defined as in REF using MATH, but shifted by MATH. They again divide weight space into (shifted) alcoves, with corresponding special points for MATH, also shifted by MATH. Now MATH implies that MATH and MATH are in the same shifted alcove for any MATH between REF and MATH, because MATH for all positive MATH. Moreover, the union of the alcoves which have MATH as a corner is a convex set of weights, and invariant under the NAME group action with center MATH. Therefore all weights in that set which are strongly linked to MATH are obtained by the action of the classical NAME group with center MATH; in particular, MATH is not. Thus if MATH is large enough, all dominant MATH strongly linked to MATH can be obtained by reflections of MATH by those MATH which contain the special point MATH. However using the assumption, the character of MATH is not affected by these MATH: indeed, by a shift by MATH as in REF, MATH can be related to MATH for MATH. The special point MATH is then moved to MATH, and is only relevant for large MATH if MATH, when it becomes MATH. However by the assumption on the first reduction point, the character of the classical MATH is not affected by the hyperplanes through MATH for MATH. Using the fact that MATH is the same as MATH as algebra over MATH CITE, the character of MATH is not affected by the hyperplanes through MATH either. Combining all this, it follows that the character of MATH is given by NAME formula for all MATH between REF and MATH. In particular, this holds if MATH for all positive noncompact roots MATH, by REF . This is certainly satisfied for compact MATH if MATH is sufficiently large. Using MATH and the fact that MATH for the non - simply laced cases, this bound follows from the given condition. |
math/9907030 | When MATH we have by the previous lemma MATH . So we can extend the inclusion of MATH in MATH to a contraction MATH from MATH to MATH. Now we have to prove the injectivity of MATH. Suppose MATH for MATH. Then we have MATH for all MATH. Because the inclusion of MATH in MATH is completely isometric (see for example, CITE), we may conclude that MATH. |
math/9907030 | It is sufficient to do this for elements in MATH. If MATH then MATH and so MATH and this last expression will tend to MATH. |
math/9907030 | We will first show that the maps MATH and MATH are closed. Take MATH and assume that MATH in MATH and MATH in MATH. Then for any MATH we have MATH . Then MATH. By the closed graph theorem we have that MATH is bounded. Similarly MATH is bounded. Now, because of REF we have MATH . As for MATH we have MATH we get that the imbedding of MATH in MATH is isometric. Finally we show that MATH is complete. Take a NAME sequence MATH in MATH. Write MATH. Then MATH and MATH are NAME sequences in MATH. They have limits and we get maps MATH defined by the pointwise limit. Obviously they still satisfy MATH for all MATH. By a standard technique we get that MATH and MATH also in norm. This completes the proof. |
math/9907030 | Take MATH. As MATH we can clearly define a left multiplier from MATH to MATH by MATH. We will show that this is bounded and therefore extends to a left multiplier from MATH to MATH. To see this, take a bounded approximate unit MATH of MATH and observe that MATH . Now MATH and so MATH . Similarly, MATH defines a bounded right multiplier of MATH. And obviously MATH so that indeed MATH. |
math/9907030 | Now, if MATH we can define two maps from MATH to MATH by MATH where we use now MATH for the multiplication from MATH to MATH. Because MATH this pair of maps gives an element in MATH. |
math/9907030 | Take MATH and observe that MATH in MATH REF . On the other hand also MATH as MATH in norm in MATH. |
math/9907030 | This is quite easy to see. We look at the case of MATH. Then the formulas in the proposition define an element in MATH. Now write MATH with MATH and MATH and observe that MATH. This gives the result. |
math/9907030 | First we need to verify that MATH leaves the NAME norm invariant. This is more or less straightforward. Then MATH is extended to MATH. Finally the formulas in the proposition will yield the extension of MATH to MATH and also show the strict continuity. |
math/9907030 | Let MATH. Then the net MATH is norm convergent, and the net MATH is bounded. So, just as in the remark after REF you get that the net MATH is norm convergent in MATH. We denote the limit by MATH. Analogously we define MATH . Then it is clear that MATH belongs to MATH, and by definition the net MATH converges strictly to MATH. |
math/9907030 | First observe that MATH is naturally identified with MATH and that MATH is hence identified with MATH, the MATH-algebra of bounded continuous complex functions on MATH. Therefore, MATH as defined in the formulation of the proposition is a map from MATH to MATH. It is clearly a MATH-homomorphism and it is not hard to show that MATH is non-degenerate. The coassociativity comes from the associativity of the product in MATH. |
math/9907030 | Take MATH and write MATH. By using REF we get MATH . So we can conclude that MATH. Analogously MATH. |
math/9907030 | It is clear by the NAME theorem that there exists a locally compact space MATH and a continuous associative binary operation on MATH such that MATH and such that MATH is given by the formula above under this isomorphism. We will prove now that MATH is a group and that the inverse is continuous. First choose a non-empty open subset MATH of MATH. We claim that the set MATH is dense in MATH. Indeed, suppose that it is not dense. Then there is an element MATH such that MATH but MATH for all MATH and MATH. Choose an element MATH such that MATH and such that MATH has support in MATH. Then MATH for all MATH. Hence MATH and by assumption we may conclude that MATH or MATH. So we get a contradiction and this gives the claim. Now let MATH. For any pair MATH of open neighborhoods of MATH and MATH respectively, we have, by the property above, points MATH, MATH and MATH such that MATH. These pairs of neighborhoods give an index set MATH when ordered by opposite inclusions. By definition the net MATH converges to MATH and MATH to MATH. For every MATH we have MATH and this converges to MATH. Now choose MATH and MATH such that MATH. Because MATH by assumption, we can take compact sets MATH and MATH such that MATH is smaller then MATH outside MATH. Then, for MATH large enough, we will have MATH and so MATH. Then we can find a subnet of MATH that converges to a point MATH. Hence we find MATH such that MATH. Because MATH were chosen arbitrarily we get MATH for all MATH. By symmetry we also get MATH for all MATH. Then it is an easy exercise to prove that MATH is a group. In particular we have the cancellation property. Therefore the linear span of MATH is a MATH-subalgebra of MATH that separates points of MATH and hence is dense in MATH. Now for any MATH the map MATH is continuous. By density we get that the map MATH is continuous for all MATH. From this it follows that MATH is continuous and that concludes the proof. |
math/9907030 | The injectivity of the extensions is an immediate consequence of these defining formulas. If for example, MATH for MATH, then MATH for all MATH by the injectivity of MATH on MATH and hence MATH. So it essentially remains to show that the above formulas yield maps into MATH. So, let MATH. We define a left multiplier from MATH to MATH by MATH. When MATH is an approximate identity in MATH, we have MATH . This implies that MATH . Taking the limit we get MATH . This shows that we have a left multiplier of MATH with norm less than MATH. Similarly, we define MATH as a right multiplier by MATH . A similar argument as above, by looking at MATH, together with the non-degeneracy of MATH, will give that MATH is also a bounded right multiplier of MATH. Hence MATH and MATH. From the formulas above it follows immediately that MATH extended to MATH will be strictly continuous on bounded sets. We can give a similar argument for MATH. |
math/9907030 | We first make some introductory remarks. Let MATH and MATH be elements of MATH. Define MATH . Then we have MATH . From this we can conclude that MATH, and we extend MATH to a contraction from MATH to MATH by continuity. If now MATH and MATH, we define, consistently with the previous notation if MATH . Then we have, with MATH an approximate identity for MATH . So again we can extend MATH to a contraction from MATH to MATH. If finally MATH we define MATH by MATH for all MATH. Suppose now MATH. Take MATH satisfying MATH and MATH. Put MATH. Then we have MATH and we compute MATH. Let MATH. Then MATH . If MATH we have MATH . By continuity we may conclude from this and the previous formula that MATH . So we get MATH and hence MATH. |
math/9907030 | This is analogous to the proof of REF . Just remark that by REF MATH . So we get MATH for all MATH and that proves the result. |
math/9907030 | Choose MATH. Let MATH be the universal representation of MATH on the NAME space MATH. Take MATH such that MATH for all MATH. Let MATH be an orthonormal basis for MATH. Let MATH be a finite subset of MATH. Then we have MATH . So the increasing net of positive numbers MATH converges to MATH . Now we can apply NAME 's theorem to the compact NAME space MATH, and the increasing pointwise converging net of positive continuous functions on it given by REF . So we get uniform convergence of this net and hence the lemma follows. |
math/9907030 | By continuity it is enough to verify this for MATH. So let MATH and choose MATH. Let MATH act on its universal NAME space MATH and take vectors MATH and MATH in MATH such that MATH and MATH. Let MATH be an orthonormal base for MATH. By the previous lemma we get that for every MATH the net MATH is norm convergent in MATH with limit MATH. Using the remark after REF we may conclude from this and an analogous statement for the second leg that MATH with norm convergence. The norm of this element is majorized by MATH . Because the net in REF is also norm convergent in MATH with limit MATH the result follows. |
math/9907030 | The extension of MATH to MATH is completely analogous to the extension of the maps MATH and MATH. If MATH we can obtain MATH as a right multiplier by putting MATH equal to MATH. One shows just as in the proof of REF that we obtain a bounded right multiplier of MATH this way. In a similar way we define a bounded left multiplier. This gives us the extension of MATH. Now write for every MATH, MATH and MATH . At the moment we only have MATH. We claim that MATH and MATH . When MATH, when MATH acts on a NAME space MATH with MATH, MATH and when MATH is an orthonormal basis for MATH, then we have as in the proof of the previous proposition MATH with norm convergence in MATH. Further we get MATH . This proves our claim that MATH and also proves REF . Now let MATH, MATH, MATH and MATH. Then MATH and MATH . Because MATH is non-degenerate we may conclude that for MATH, MATH and MATH we have MATH. We still denote this element with MATH and we have MATH . From the formulas above we can also conclude that for every MATH and MATH the map MATH is strict-norm continuous on bounded sets from MATH to MATH. Finally let MATH and MATH. Write MATH with MATH and MATH. Then we get for every MATH and MATH . Analogously one can prove that MATH with norm majorized by MATH. So we get indeed MATH with norm majorized by MATH. With the notation as above we can write MATH and this depends strict-norm continuous on bounded MATH by one of the remarks above. One proves an analogous statement on the other side and gets the strict continuity on bounded sets of MATH from MATH to MATH. |
math/9907030 | We will prove this formula by applying MATH and by showing that MATH. Then, the injectivity of MATH will give us the result. Now observe that MATH . This can easily be shown by using the concrete formulas for MATH obtained in the proof of REF . Then, using coassociativity, we get MATH . Then, using the strict continuity on bounded sets of all the maps involved, we get MATH . As MATH, the righthand side is MATH . Then the proof is complete. |
math/9907030 | Take MATH. Then MATH . We now use the injectivity of MATH (on MATH) to find MATH for all MATH. This is only possible when MATH. |
math/9907030 | Take MATH and MATH such that MATH. Take MATH and consider the formula MATH. Apply multiplication on both sides. For any MATH we have MATH . Again this can be proved by writing out the formula for MATH that we obtained in the proof of REF . We get by continuity that MATH . As this is true for all MATH we must have MATH . By the lemma MATH. So we can define MATH by MATH when MATH and MATH. Now we show that MATH is a homomorphism. To do this, take MATH and MATH such that MATH and MATH. Use the notation of the proof of REF and define MATH. We have seen that MATH and MATH. Let MATH. Then we have MATH . But for MATH we have MATH . So we can conclude that MATH . This gives the required result. |
math/9907030 | Let MATH and MATH satisfy MATH. Apply the slice map MATH and then MATH to the formula MATH as obtained in REF . This yields MATH . Now, when MATH we have with the same technique as in the proof of the previous proposition MATH . By continuity this is still true when we replace MATH by MATH. We obtain MATH . This gives MATH. Now apply the slice map MATH to get MATH . By REF we get MATH. As this is true for all MATH, we must have MATH. As in the remark following REF we see that actually MATH. Then we can define MATH by MATH when MATH and MATH satisfy MATH. Next we show that MATH is an anti-homomorphism. As before take MATH and MATH such that MATH and MATH. Use again the notation of the proof of REF . Put MATH. Then MATH and MATH. Let MATH. Then we get MATH . Now for MATH one has MATH . This gives MATH and so MATH. Finally MATH where MATH is the map from REF . As MATH, we have MATH. Also MATH and this proves the result. |
math/9907030 | Before we can really start the proof we have to make some introductory remarks. Let MATH and MATH. Then we have MATH . Now we would like to have the same formula for arbitrary MATH instead of MATH. So we have to extend the map MATH. Because MATH is completely bounded as a map from MATH to MATH with norm MATH we get that MATH can be uniquely extended to a continuous map from MATH to itself with norm majorized by MATH. See the discussion in the beginning of REF. For every MATH and MATH we can, by the same argument, extend MATH to a continuous map from MATH to itself with norm majorized by MATH. We write MATH for this extension applied to MATH. If now MATH, MATH, MATH and MATH we get MATH . So by the non-degenerateness of MATH it is possible to extend MATH uniquely to a map MATH such that MATH for MATH, MATH, MATH and MATH. It is clear from the formulas above that for every MATH and MATH the map MATH from MATH to MATH is strict-norm continuous on bounded sets. Let now MATH. Write MATH with MATH and MATH. Then define MATH for every MATH and MATH. Then for every MATH the map MATH is strict-norm continuous on bounded sets and MATH . Analogously the map MATH can be extended uniquely to a map from MATH to MATH which is strict-norm continuous on bounded sets. So we have proved that MATH can be extended to a bounded map from MATH to itself which is strictly continuous on bounded sets. We denote this (unique) extension by the same symbols. Now we start the real proof. Let MATH, MATH and MATH. Then we have MATH so that for all MATH we get MATH. Now the computation in the beginning of this proof, together with the strict continuity of the maps involved gives MATH . It follows from this that MATH for MATH satisfying MATH . Now MATH . So if we replace MATH by MATH we get MATH and so MATH. |
math/9907030 | Take MATH such that MATH. It follows from the proof of REF that MATH. Hence we have MATH. Now for MATH we get MATH so that MATH. This gives the result. |
math/9907030 | Start again from the basic formula MATH and now apply the slice map MATH. For MATH we have MATH . Exactly as in the proof of the previous proposition we can extend MATH to a map from MATH to itself which is strictly continuous on bounded sets. So applying the formula above to MATH in the place of MATH we get MATH . Hence MATH and MATH . Again, with MATH replaced by MATH we get MATH . So MATH . This proves the result. |
math/9907030 | This follows immediately when one takes MATH and MATH. |
math/9907030 | First observe that MATH as for example, MATH and MATH (for all MATH). Let MATH, MATH and MATH be as before. Then by REF we have MATH . So MATH is a contraction for the NAME norm. Now suppose MATH and MATH. Let MATH and apply MATH. This formally gives MATH. To give a meaning to this we have to verify that the map MATH from MATH to MATH is completely bounded. But this is true because MATH is completely bounded from MATH to MATH and MATH from MATH to MATH. Then the formula MATH is easily checked for MATH and then extended by continuity. So we get indeed MATH for all MATH because MATH is injective. But then MATH for all MATH. Because MATH is injective we get MATH for all MATH and hence MATH. |
math/9907030 | Before this proposition we already remarked that MATH . Because MATH we get from this that MATH . Now for MATH we have MATH and so by continuity we have MATH. This proves the formula for the counit. Now starting again with REF and using the definition of the antipode (see REF) we get MATH . But for all MATH and so we get MATH . This gives the first formula for the antipode. Now choose MATH and apply MATH to this equation. This gives MATH . But for MATH we have MATH . In REF we have seen that MATH extends to a bounded map from MATH to itself. A slight modification of that proof gives that this map is even completely bounded. But this allows us to conclude that the map MATH can be extended to a bounded map from MATH to MATH. We will denote this extension by MATH. So we get in MATH the equality MATH . Then we may conclude that MATH . Observe now that MATH . By continuity we conclude from REF that MATH . This is the second formula for the antipode we had to prove. Next we observe that MATH . Now we have to remark that MATH is a contraction from MATH to MATH and this allows us to write MATH . Combined with REF we get MATH and so MATH. We already observed that MATH and so we get MATH. Clearly MATH and this concludes the proof of the proposition. |
math/9907030 | In a completely similar fashion we get for elements MATH formulas for MATH and MATH as in the previous proposition. From these formulas, it immediately follows that MATH and MATH. |
math/9907030 | That MATH induces a comultiplicaton on MATH is proved in for example, REF . The only thing that remains to be shown is that the maps MATH and MATH are injective on MATH. We will prove it for MATH. Then the injectivity of MATH can be obtained easily as MATH (where MATH is the flip) so that MATH and we have seen that MATH is an isometry of the NAME tensor product to itself. Now remark that for MATH, MATH and any right convolution operator MATH on MATH we have MATH where we used the obvious fact that left and right convolution operators commute. It is clear that for any MATH we can define a bounded map MATH from MATH to MATH, with norm majorized by MATH and such that MATH. If now MATH and MATH we conclude by continuity from the computation above that MATH for all MATH of the form MATH. But the space MATH is norm dense in the space of multiplication operators by functions of MATH. So the linear span of the elements MATH with MATH and MATH a right convolution operator, is norm dense in the compact operators on MATH. So we can conclude that MATH for all compact operators MATH on MATH. In particular we get, for all MATH where we used the notation MATH for the rank one operator from MATH to MATH given by MATH. From this it follows that MATH and that completes the proof. |
math/9907030 | First remark that it is sufficient to look at the case MATH with MATH. Because MATH we can take a representation MATH of MATH on another NAME space MATH such that MATH with MATH. Then we replace MATH by MATH. So we may suppose MATH with MATH. Let MATH be an orthonormal basis for MATH and define MATH . We have MATH for all MATH and by a slight modification of REF (or by REF) we get that MATH converges strictly to MATH. Also MATH is a bounded strictly converging net. So by REF we can define an element MATH by MATH the sum being strictly convergent. Then we have MATH . So indeed MATH. Next we have MATH and MATH so that MATH. This completes the proof. |
math/9907030 | Let MATH where MATH denotes the flip from MATH to MATH. Then MATH. If MATH and MATH one verifies, as in the proof of REF, that MATH and that MATH . Hence MATH. Also MATH . Together with the previous proposition this gives the result. |
math/9907030 | Take MATH and MATH. Just as in the proof of REF we may suppose that MATH and MATH. Write MATH and MATH. Use the flip map MATH and denote MATH. Then MATH and MATH . From the previous proposition it follows that MATH, and because both are subalgebras of MATH we get MATH. To show that MATH we consider MATH . We claim that for all MATH we have MATH and MATH . Let MATH be an orthonormal basis for MATH and MATH for MATH. Define MATH . As in the proof of REF we get that the nets MATH and MATH are both bounded and strictly converging. So also MATH will be bounded and strictly converging. Analogously MATH is bounded and strictly converging. It is clear that MATH is a bounded (infinite) MATH by MATH matrix over MATH. Then it is not hard to see that for every MATH the net MATH is norm convergent in MATH (see also the estimate in the proof of REF ). The limit in MATH is given by MATH. Applying MATH to this element gives, after some computations, MATH where we used MATH . But now observe that the element in REF equals MATH . So we get that MATH belongs to the range of MATH on MATH for all MATH. This gives MATH. |
math/9907030 | Let MATH and assume that MATH. Now observe that for all MATH, MATH and MATH . By continuity we may conclude that MATH . By the faithfulness of MATH we get MATH for all MATH. So MATH. |
math/9907030 | Let MATH be a NAME for MATH. Let MATH be an orthonormal basis for MATH and define for MATH the operator MATH for MATH. Now we will use the map MATH introduced in REF. This map takes values in MATH, the space of adjointable maps between the NAME MATH-modules MATH and MATH. Of course, for MATH and MATH we have MATH for all MATH, but it is possible to extend MATH to a much larger domain, see CITE. Then one has MATH for all MATH and MATH in the domain of MATH. With the help of this we define MATH . Then MATH for all MATH. But if we fix MATH we can approximate MATH in norm by elements of the form MATH with MATH. This equals MATH and hence belongs to MATH by REF. So MATH for all MATH. Now consider the net MATH . This net is strictly convergent to MATH . Analogously MATH is bounded and strictly convergent. So by REF we can define, with strict convergence MATH . Then we get MATH by left invariance. All these manipulations are very rigorous when interpreted in the sense of CITE. So we may conclude that MATH. Then we can make analogous computations as above to get the stated formulas for MATH and MATH. Assume now the extra conditions stated in the proposition. Let MATH such that MATH for all MATH. By the NAME theorem it is enough to prove MATH. By the result above we have MATH for all MATH. Because MATH is faithful we conclude that MATH for all MATH and hence for all MATH by density and continuity. Then we get MATH for all MATH and MATH. By the assumed density we get MATH. |
math/9907030 | Again, as in the proof of the previous proposition, we will use the results of REF. Implicitly a lot of weight theory techniques that appeared in that paper will be used. Also remark that a more elementary but very long proof for this result is given in CITE. So we can consider MATH . If now MATH it is possible to define MATH such that MATH where MATH . This should not be too surprising. If MATH we have on the one hand MATH and on the other hand MATH and this agrees with the usual identification of MATH and MATH. So we can consider the element MATH given by MATH . Now let MATH and MATH be orthonormal bases for MATH and MATH respectively. Define for MATH and MATH: MATH . Then we have that MATH. But for MATH and MATH we have MATH . By the results of CITE one has MATH and so the expression above belongs to MATH. Then we get MATH for all MATH and MATH. From its definition it follows now immediately that MATH is a bounded matrix over MATH. Define also MATH . As in the proof of the previous proposition we get MATH and MATH strictly converging and bounded. But then we have for any MATH that MATH converges in norm in MATH, because for MATH and MATH finite we have MATH . In order to apply MATH to this we have to realize that (use the abbreviation MATH for MATH) MATH . Then it is not too hard to see that with strict convergence MATH . So we get indeed that MATH and for all MATH we have MATH . Because we can write MATH and because MATH we get MATH by the previous proposition. Analogously MATH. The formulas to be proven for MATH can be obtained from REF with a similar calculation as above. |
math/9907030 | We use of course the notations and conventions from CITE and CITE. So let MATH be a reduced locally compact quantum group. We let MATH act on the NAME space MATH, which is the NAME of the left NAME weight MATH. We first proof that MATH is a NAME MATH-algebra. We will work along the same lines as in the proof of REF . Let MATH be the multiplicative unitary associated with MATH and observe that for any MATH and any MATH we have MATH . We can define for any MATH a bounded map MATH from MATH to MATH such that MATH for all MATH. Then it is clear that for any MATH the map MATH is strongly continuous on bounded subsets of MATH. Let now MATH and suppose MATH. Then it follows from the computation above that MATH for all MATH of the form MATH, with MATH and MATH. Now we claim two things. First, the closed linear span of MATH is a MATH-algebra that acts non-degenerately on MATH. Secondly this MATH-algebra is strongly dense in MATH. Here MATH denotes the modular conjugation of the weight MATH, and so MATH. After these two claims have been proven we may conclude that MATH for all MATH and exactly as in the proof of REF this is enough to conclude MATH. To prove the first claim denote the stated closed linear space by MATH. It is then enough to prove that for all MATH . But this expression equals MATH where MATH and MATH is the modular conjugation of the dual weight MATH. Here we used the equality MATH. Now observe that MATH . Hence we can conclude that the expression in REF belongs to the closed linear span of MATH because MATH. So we have proved the first claim. To prove the second claim first observe that MATH. If MATH then MATH because MATH, and with MATH the extension of MATH to the NAME algebra MATH. But this implies MATH. By taking the commutant we get MATH and from this follows the second claim. So we have proven the injectivity of MATH. The injectivity of MATH follows easily by using the unitary antipode MATH. Just observe that for MATH where MATH denotes the flip map extended to MATH. Because MATH is an anti-automorphism of MATH we can extend MATH to an isomorphism of MATH (compare to REF ). Then the injectivity of MATH follows from the already proven injectivity of MATH. From REF it follows that MATH is a core for the antipode of the locally compact quantum group MATH, which we shall denote by MATH. Because MATH is an anti-homomorphism also the linear span of the products of two such elements will give a core. It is not hard to finetune the proofs of REF to obtain that MATH is a subspace of the domain of MATH. All these remarks, together with REF give that MATH is a core for MATH. |
math/9907032 | First, let MATH be an interior vertex. Suppose that there are MATH triangles MATH incident to MATH. The sum of all of their angles is MATH . The cone angle at MATH is the sum of the angles of the triangles MATH at MATH . On the other hand, the sum of the dihedral angles of the edges MATH incident to MATH is the sum of all of the angles of MATH not incident to MATH. Thus, MATH . The result follows by rearranging the terms. If MATH is a boundary vertex, and there are MATH faces incident to MATH, then there are MATH edges, and REF becomes: MATH and the result follows by rearranging terms, as above. |
math/9907032 | Every angle of every face of MATH is opposite to exactly one edge of MATH. |
math/9907032 | The first inequality of REF is a restatement of REF , applied to the subcomplex MATH. To show the second inequality, let MATH . Evidently, MATH while MATH . Applying REF to MATH, we see that MATH while applying them to MATH, we see that MATH . Note now that MATH while MATH . Adding REF , and applying REF , we obtain MATH where the last equality is obtained by applying the equality case of REF . |
math/9907032 | CASE: Note that if there is a MATH as required in the statement of the Theorem, the objective function is, indeed, equal to MATH. This is nothing other than the equality case of REF . Now, let MATH. Let MATH, and let MATH, for all values of the indices. The new variables are still feasible for MATH, and by REF above, this transformation does not change the value of the objective. Furthermore, if MATH for all MATH, then the objective is non-positive, and is equal to zero only if all of MATH are equal to zero as well - this is so, since all of the MATH must be non-positive, and all of their coefficients are positive, by REF . Assume, then, that MATH for MATH; MATH is a proper subset of MATH by construction. CASE: Suppose now that MATH, for all MATH. Then MATH by REF. Now, let MATH, and let MATH if MATH, and otherwise MATH. Likewise, MATH if MATH, and otherwise MATH. This still leaves us in the feasible region of MATH and strictly increases the value of the objective (by REF ). The new nonzero set MATH is a proper subset of MATH, and we can repeat this process. In the end, we will wind up with a feasible point MATH, with MATH, where the value of the objective is non-positive (by REF ), and strictly greater than the value of the objective at MATH (by REF ), thus completing the proof. |
math/9907032 | Suppose the contrary. Then MATH, for some MATH, by REF . However, in that case the last inequality of MATH is not satisfied, since the left hand side of the last constraint of MATH vanishes. Indeed, it is equal to MATH . Observe, however, that MATH counts each non-boundary edge with multiplicity MATH, and each boundary edge with multiplicity MATH . |
math/9907032 | We will show that REF ; the converse is immediate. MATH . The last sum of REF can be rewritten thus: MATH . Finally, using the definition of MATH, and combining REF it follows that MATH . Now, assume that the dihedral angles satisfy REF . This means that the middle term on the right hand side of REF is non-negative (and strictly positive unless MATH). |
math/9907032 | By REF , it is enough to check the connected subcomplexes of MATH. If such a subcomplex MATH is simple, then we are done. Otherwise, its complement is not connected. Let MATH be a connected component of the complement of MATH. lemma The complex MATH is simple. proof By construction, MATH is connected. Also, every point of MATH can be connected by a path to a point of MATH. Since MATH is assumed connected, the lemma follows. Consider MATH. By assumption, MATH . Now MATH . By REF (more precisely, a version for simple complexes), MATH thus, MATH . Thus, in order to check that REF holds for MATH, it is enough to check that it holds for MATH. The proof of REF is finished by the obvious induction argument on the number of connected components of the complement of MATH. |
math/9907032 | Consider a cut MATH of MATH. This will have some edges at level MATH, removing which which will disconnect a subset MATH of the faces of MATH from the source. It is then not necessary to remove any edges of level MATH emanating from MATH. Let MATH. Let MATH be those faces MATH for which the cutset contains all three edges of level MATH emanating from MATH. Finally, MATH. All of the edges of level MATH (indirectly) emanating from MATH must be in the cutset MATH. These are precisely the edges corresponding to the edges of the subcomplex of MATH whose faces are in MATH. What is the capacity of MATH? Evidently, it is equal to MATH . If we want the flow through MATH to be MATH, we must have MATH . Or, noting that MATH, MATH . In the special case where MATH, it follows that MATH . Since MATH could have been any subset of MATH, it follows that REF is necessary (this is, in any event, self evident). On the other hand, if REF holds for any subcomplex of MATH, substituting REF into REF, we see that MATH . |
math/9907032 | The argument parallels very closely that of REF. The existence of the desired structure is, as before, equivalent to a negative objective of a linear program, and as before, we set up a slightly simpler linear program first. To wit, the program MATH is: CASE: All angles are strictly positive. CASE: For every face MATH, MATH . CASE: For every boundary edge MATH incident to the triangle MATH, MATH where the slack variables (see the comments following the statement of REF ) MATH are also non-negative. CASE: For every interior edge MATH incident to triangles MATH and MATH, MATH . We relax the program MATH to a program MATH by dropping the requirement that the angles be strictly positive, make the objective REF, as before, and we see that the dual to the new weakened linear program MATH is the following: CASE: Maximize MATH, subject to the conditions CASE: MATH whenever MATH is an edge of MATH. CASE: Whenever MATH is a boundary edge of MATH, MATH . Since the constraints of the above program MATH are a superset of the constraints of MATH, REF still tells us that the objective function is maximized if there exists a MATH, such that MATH, for all MATH, MATH. Now, this is not enough to guarantee that the objective is zero, since the equality case of REF (when MATH) no longer exists. Since the new inequalities require MATH to be non-negative, it follows that for the objective function to equal MATH, MATH must be MATH. Now, we follow REF again, to define a program MATH in the same way as before that is, since we want the angles to be strictly positive, we set MATH, etc, and to also define its dual MATH. By the same reasoning as before, it follows that the objective of MATH, and hence of MATH, is negative. In fact, we can do more: we can also require all of the slack variables MATH to be strictly positive. The (yet another) new dual program MATH will have the form: Third dual. Maximize MATH, subject to the conditions MATH whenever MATH is an edge of MATH. MATH, when MATH is a boundary edge of MATH. MATH. If we omit the last constraint, we remain with the dual program MATH, and as the discussion above showed, the objective of that can only be MATH if MATH, which is at odds with the last constraint of MATH . |
math/9907032 | This is an immediate consequence of the theorem of Cosines, or the theorem of Sines. |
math/9907032 | This is immediate from the definition of excess. |
math/9907032 | Suppose that the conclusion of the theorem does not hold. Assume, without loss of generality, that the edge between the vertices MATH and MATH is the shortest one (and thus of length MATH). We construct a family of disks MATH, all centered on MATH, and such that the radius of MATH is equal to MATH . Let MATH, and let MATH to be the maximal closed subcomplex of MATH contained in MATH. The hypothesis of the theorem ensures that at least one of the annuli MATH contains no vertices of MATH, let this annulus be MATH. Then we claim that the excess of MATH is smaller than MATH. Indeed, consider a triangle MATH of MATH adjacent to MATH along an edge MATH. The vertex MATH of MATH lies outside MATH, and thus the lengths of MATH and MATH are at least MATH, while the length of MATH is at most MATH. Thus, MATH by REF . Thus, by REF , it follows that MATH contradicting the hypothesis of the theorem. |
math/9907032 | We only need to show that the positive excess conditions are sufficient, since they are obviously necessary. In addition, we may assume that the MATH-skeleton of the NAME dual MATH is connected (if not, we prove the theorem for each connected component separately). Now, we pick a pair of adjacent base vertices MATH and fix MATH. Now, for MATH we define MATH to be equal to the combinatorial distance between MATH and MATH in the MATH-skeleton of MATH. Now, define MATH to be the span of all vertices MATH, such that MATH. The complex MATH is finite by local finiteness of MATH. In addition, MATH, so every finite subcomplex of MATH belongs to some MATH. For each MATH we consider a geometric realization MATH, whose existence is guaranteed by REF (there are many such realizations, we pick any one of them). Now, enumerate the faces of MATH, in such a way that MATH contains the edge MATH, and, for any MATH, the faces MATH and MATH are adjacent (that is, share an edge) in MATH. For each triangle, we have the space of shapes (similarity classes), given (for example) by the complex parameter MATH, obtained by placing the first two vertices of the triangle at the points MATH and MATH in the complex plane, and reading off the position of the third point (in the upper half-plane, assuming the triangle is positively oriented). REF tells us that for any face MATH of MATH, the set of shape parameters of realizations of MATH is contained in a compact set MATH (since the ratio of lengths of any two sides is bounded by some constant, depending on the function MATH). We can think of each MATH as being an element of MATH, which is a compact set by NAME 's theorem, and hence we can extract a convergent subsequence from MATH. Call the limit of that subsequence MATH. Since the dihedral angles are obviously continuous functions of the triangle parameters, the dihedral angles of MATH will be given by MATH, and so MATH is the sought-after realization. |
math/9907032 | Let an old vertex of MATH be one that was already a vertex of MATH, while a new vertex be one that was added at stellation. The set MATH of new vertices of MATH corresponds to the set of faces of MATH. For any vertex MATH, recall that MATH denotes the cone angle at MATH. The NAME theorem tells us that MATH . Or, recombining the terms: MATH . Note that every edge of MATH is incident to an old vertex, and to at most one new vertex. Combining this observation with REF , we see that for any NAME triangulation combinatorially equivalent to MATH, it must be true that MATH . Combining REF with REF , it follows that MATH . Note now that MATH. By a standard computation using NAME 's formula for triangulations of the sphere, we know that MATH thus MATH . Thus, MATH . REF together imply that the average cone angle at an old vertex of MATH is at least MATH . If MATH it follows that the average cone angle at an old vertex is greater than MATH, which contradicts the assumption that the cone angles were positively curved. |
math/9907032 | We will use the method of REF. First, let us write the dual program of the linear program (referred to as MATH in the sequel) for weak hyperbolic structure: Our variables are MATH, where MATH ranges over all the simplices of MATH, and MATH where MATH ranges over all the edges of MATH. The dual program MATH is: CASE: Maximize MATH. CASE: Subject to MATH for all faces MATH and pairs of opposite edges MATH. In order for the primal program to have a non-empty feasible region, the objective of the dual must be non-positive. Consider now a normal surface MATH (see, for example, CITE for rudiments of normal surface theory). The surface MATH intersects each simplex MATH in a collection of disks, which are combinatorially either triangles (cutting off one vertex of MATH from the other three), or quadrilaterals (separating one pair of vertices from another). For each simplex MATH, define MATH to be the number of triangular components of MATH, and MATH to be the number of quadrilateral components. For each edge MATH of MATH, define MATH to be the number of intersections of MATH with MATH. The intersections of MATH with the simplices of MATH induces a triangulation MATH of MATH, where each triangular disk contributes one triangle, and each quadrangle contributes two. Define MATH to be the number of triangles of MATH sitting inside a simplex MATH. Evidently, MATH. Note that, by NAME 's formula, MATH. This is seen to be very similar in form to the objective function of MATH, so let us set MATH, and MATH. lemma The assignment of the variables as above satisfies the inequality constraints of MATH. We need to check that for MATH a simplex and MATH a pair of disjoint edges of MATH. We need to check that MATH . By linearity, we need just check REF for connected components of MATH. If that component is a triangle MATH, then MATH contributes MATH to MATH (since a ``normal triangle" intersects exactly one of each pair of opposite edge). Also, MATH contributes MATH to MATH, so for a triangular face, the right and left hand sides of REF are equal. Suppose now we have a quadrilateral component MATH. It contributes MATH to the right hand side of REF . As for the left hand sides, MATH hits two pairs of opposite sides of MATH, so if MATH and MATH is one of those pairs, then we have a contribution of MATH to the left hand side, and otherwise we have a contribution of MATH. remark Notice that if MATH is such that all of the components of MATH, for all MATH are triangles, then all of the constraints of MATH are equalities with the assignment of variables as above. Any such MATH is easily seen to be a union of boundary tori. REF concludes the proof of the ``weak" part of REF , since if any MATH had positive NAME characteristic, the program MATH would have a positive objective, and thus the program MATH would have no solution. For the proof of the ``strong part" we use the same trick as in REF. Define new variables MATH, etc. Our primal linear hyperbolicity program MATH is now: CASE: Minimize: MATH CASE: subject to face constraints MATH . CASE: and to edge constraints MATH, where MATH is the valence of MATH. The dual program MATH is then CASE: Maximize MATH. CASE: Subject to the old constraints MATH for all faces MATH and pairs of opposite edges MATH. CASE: and the new constraint MATH . In order for MATH to be linearly hyperbolic, the objective must be strictly negative. Observe that the sum of the left hand sides of the old constraints is equal to the left hand side of the new constraint. Indeed, each MATH occurs three times (once for each pair of opposite edges), and each MATH occurs the number of times equal to the valence of MATH. Hence, the new constraints simply says that in at least one of the old constraints the inequality must be strict. Keeping in mind REF , this implies that every non-boundary-parallel normal surface must have strictly negative NAME characteristic, thus proving the second part of REF . |
math/9907033 | Let MATH be an index pair of MATH. We first notice that MATH. Moreover, the pair MATH is an index pair for MATH (with respect to the flow MATH, of course). Denote MATH and MATH. There is a map of pairs MATH which induces a map of quotients: MATH. Notice that this map induces one of connected simple systems in the following way. First, let MATH be another index pair of MATH and let, as above, MATH be the index pair of MATH with MATH, MATH. There are standard, flow defined CITE, comparison maps, MATH and MATH. By construction, these maps commute with the projections MATH and MATH. Now assume that MATH is another index pair of MATH (and not necessarily one obtained as a preimage of one of MATH) and consider also an index pair of MATH, MATH. We define a map MATH as the composite MATH where the first and last maps in this composition are the standard comparison maps. The homotopy class of this map does not depend on the choice of the pair MATH. To prove naturality with respect to attractor-repellor sequences consider an attractor-reppellor decomposition MATH for an invariant set MATH and a triple MATH such that the pair MATH is an index pair for MATH, the pair MATH is an index pair for MATH and MATH is an index pair for MATH. Following the notations above we denote by MATH the respective preimages by MATH of these sets. It is clear that MATH and MATH form an attractor-repellor pair. The result now follows from the commutativity of the diagram: MATH . The vertical arrows are given, as above, by projection. The rows are the cofibration sequences that represent the attractor-repellor cofibration sequences of connected-simple systems. Moreover, note that the right vertical map up is induced on cofibres by the left and middle projections and this implies that, by extending the cofibrations to the right, we will continue to get commutative squares. |
math/9907033 | The naturality given by the previous lemma implies the existence of the maps in the statement. These maps give a cofibration sequence because of the general fact that if in the diagram: MATH the top two horizontal rows as well as the columns are cofibration sequences, then the third horizontal row is also a cofibration sequence. In our case, the top row contains the spaces representing the NAME index of MATH . The spaces in the middle row are obtained from the spaces giving the NAME indexes with respect to MATH by pasting the reduced cylinders of the projections. The vertical maps are inclusions in the free ends of these cylinders. |
math/9907033 | Suppose MATH is an index pair for MATH with respect to the flow MATH. Let MATH be the total space of the restriction of the unit disk bundle associated to MATH restricted to MATH, MATH. Let MATH be the total space of the restriction of MATH to MATH. It is easy to see that the pair MATH is an index pair for MATH with respect to the flow MATH. Notice that if MATH is the unit disk in MATH and MATH its boundary, then the region where MATH is negative or null in MATH has the structure of a cone over the boundary. This implies that the region MATH where MATH is negative or null is homeomorphic to MATH the identification used making MATH correspond to the point MATH for each MATH. On the other hand we have a map of pairs MATH. This induces a map between the respective quotients: MATH. Because of the description of MATH above, the domain of this map is a space in MATH, the target is a space in MATH. This map is a homotopy equivalence as both inclusions MATH and MATH are homotopy equivalences (the first because the domain and the target are homotopy equivalent to MATH and the second for a similar reason). It is easy to verify, as in the lemma above, that the map MATH induces an isomorphism of connected simple systems and that this morphism is natural with respect to attractor-repellor pairs. |
math/9907033 | Let MATH and let MATH. Notice that MATH and MATH is contractible. With the notations in the proof of the proposition, we have MATH, MATH, MATH. The statement is an immediate consequence of the construction of the NAME index and of the fact that the cofibre of the projection MATH is, in a natural way, homotopy equivalent to MATH. With this identification we have the commutative diagram below: MATH the columns being (homotopy) cofibration sequences. By definition the NAME index of MATH is represented by the cofibre of the bottom horizontal map which is MATH. |
math/9907033 | We will construct a duality between two cofibration sequences representing those of the statement. Afterwards we show that the induced duality of cofibrations of connected simple systems does not depend on the choice of the representing sequences. The proof has four steps. CASE: Let MATH be a tubular neighborhood of MATH. In the following we identify it to MATH. Let MATH be a smooth function which locally has the form MATH where MATH is the coordinate along the fibre of MATH and MATH the coordinate along the base. Consider on MATH the flow MATH induced by the sum of the canonical horizontal lift of MATH and of the flow induced by MATH. Notice that MATH and MATH. Applying REF to MATH we obtain isomorphisms MATH, MATH. This reduces the proof to showing that MATH and MATH are NAME duals. CASE: Extend MATH to a flow on MATH which is constant outside a neighborhood of MATH. We continue to denote this flow by MATH. Let MATH be the maximal attractor of MATH that does not contain MATH , let MATH be the complementary repellor. Let MATH be the attractor of MATH obtained as the union of MATH with the set of the points of MATH or situated on flow lines originating in MATH and let MATH be the complementary repellor. Similarly, let MATH be given by the union of MATH with the set of the points of MATH or situated on flow lines originating at MATH, let MATH be the complementary repellor. For each pair MATH for MATH we may construct by classical techniques (see, for example, CITE) smooth NAME functions, MATH such that MATH and MATH. By NAME 's theorem we may find regular values MATH and MATH of respectively MATH and MATH such that MATH and MATH intersect transversely. Notice that MATH is the maximal invariant set inside MATH. Moreover let MATH and MATH. Then MATH and the pairs MATH and MATH are index pairs for MATH for, respectively, the flows MATH and MATH. We now choose a regular value MATH of MATH such that MATH cuts transversely MATH and such that the inequality MATH implies MATH (this is possible because MATH). Let MATH, MATH, MATH, MATH and MATH, MATH. With these notations the pairs MATH, MATH and MATH are index pairs of MATH for the flow MATH; MATH and MATH are index pairs for MATH with respect to the flow MATH; MATH is an index pair for MATH with respect to MATH; MATH, MATH are index pairs for MATH with respect to MATH. CASE: The basic argument that will be used is very classical in nature and is a variant of one that appears in CITE. For completeness we will formulate it as a lemma and we will indicate the idea of the proof. Assume that MATH is a MATH manifold with boundary emdedded in MATH such that MATH admits a decomposition MATH with MATH and MATH being MATH-dimensional manifolds such that MATH. Then MATH and MATH are MATH-Spanier-Whitehead duals. Proof of the lemma. NAME MATH in MATH as the boundary of a disk MATH. Denote by MATH the complementary disk. Consider a copy of MATH, MATH, in the exterior of MATH and paste it, in MATH, to MATH following MATH. (We use here and below the existence of collared neighborhoods of MATH, MATH, MATH and MATH .) The result of this pasting, MATH, is homotopy equivalent to MATH. As MATH is the cone over the boundary of MATH it is easy to see that MATH deforms to the union MATH with MATH a slightly smaller disk than MATH and MATH a copy of MATH that does not touch MATH. Hence, MATH. Notice that the argument of the lemma can be also used to prove the duality of some maps. Indeed assume that MATH is a MATH-dimensional submanifold of MATH such that MATH decomposes, similarly to MATH, as MATH and such that MATH, MATH separates and is an NDR in MATH, MATH and MATH are NDRs in MATH. Then the maps MATH and MATH are NAME duals. The reason is that, as MATH and MATH, by the construction of the lemma the two maps are identified to a pair of dual inclusions. We return now to the proof of the theorem. We intend to show that the cofibration sequences: MATH and MATH are NAME duals. Consider first the pair of maps MATH. We have MATH and the duality is immediate by the above. Similarly, for the pair MATH we have MATH and MATH. CASE: The proof is completed by showing that the duality induced by the two cofibration sequences described above does not depend on the choice of the NAME functions or on the choice of the constants MATH, MATH and MATH. It is easely seen that this statement reduces to proving that if MATH, MATH and MATH are constructed in the same way as, respectively, MATH, MATH and MATH and we also have MATH, then the map MATH is homotopic to the composition MATH . Here MATH and MATH are duality maps (here and above all duality maps given by the lemma are constructed with respect to the same embedding of MATH in MATH); the maps MATH and MATH are the standard comparison maps given, respectively, by the connected simple systems MATH and MATH. This can be seen as follows. Let MATH, MATH and MATH. Notice that MATH and MATH are homeomorhpic as well as the pairs MATH, MATH and MATH, MATH. The comparison map MATH factors using these homeomorphisms MATH with MATH being induced by the inclusions. We apply the general duality argument in the lemma to obtain that MATH is dual to the map MATH. This map is clearly homotopic to the comparison map MATH. Hence, if we denote by MATH the comparison map which is the homotopy inverse of MATH we obtain that MATH and MATH are duals and this implies the homotopy we look for. |
math/9907034 | Consider the evaluation map MATH defined by MATH. Let MATH be the projection from the product to the first factor, then the definition of MATH above is equivalent to MATH where MATH denotes integrating over the fibres MATH. Since MATH and MATH, and MATH is compact without boundary, MATH . |
math/9907034 | We first have to find the map MATH arising from the action of MATH. For a map MATH this action is composition with the diffeomorphism MATH: MATH . Under the evaluation map, we have MATH . Thus MATH acts on MATH by the vector field MATH where MATH . Now write MATH in a neighbourhood of MATH. We have MATH and MATH is still MATH invariant. From REF we obtain MATH . Now by the definition of the symplectic form MATH, we have, near MATH, MATH and now MATH since MATH is MATH invariant. But MATH so from REF MATH . But MATH since MATH is only REF-dimensional, so MATH . Thus from REF , MATH . Thus for MATH, MATH is exact and the class of MATH (which restricted to MATH is the form MATH for which MATH) is the moment map. |
math/9907034 | Differentiating the two conditions MATH and MATH, we see that MATH is tangent to MATH if and only if MATH when pulled back to MATH. But MATH is a REF-form on MATH of type MATH so that MATH and hence MATH . From REF we see that if MATH is tangent to MATH then so is MATH, so that MATH is complex. |
math/9907034 | On an oriented Riemannian manifold with volume form MATH, let MATH be the vector field dual to REF-form MATH (that is, MATH for all vector fields MATH). Then MATH . Now take MATH. The vector field dual to this in the NAME metric on MATH is MATH, which is tangent to MATH, so this is its dual on MATH in the induced metric. Now since MATH restricts to the volume form MATH on MATH, MATH from REF , which completes the proof. |
math/9907034 | To address the theorem, let us for simplicity take just two points MATH and MATH. Then these are linearly equivalent if and only if REF-form MATH has integral periods in MATH, where MATH and MATH . If MATH is a closed REF-form with compact support in MATH and integral cohomology class in the compactly supported cohomology of MATH, then this is equivalent to the condition MATH for all such MATH. Now let MATH be a path from MATH to MATH. As a current we define MATH but as a current MATH also has a NAME decomposition CITE MATH where MATH is the harmonic part, which by elliptic regularity is a smooth globally defined harmonic REF-form. Now by NAME 's theorem MATH so that MATH . Hence from REF MATH and so MATH. Now since the boundary of MATH is MATH, MATH is an integral relative cycle in MATH, and so if MATH is a compactly supported REF-form with integral cohomology class, MATH . Now since MATH, MATH so from REF MATH and hence since MATH, MATH if and only if MATH. Now let MATH be a harmonic REF-form on MATH, then since MATH, MATH . But MATH is cohomologous to a closed form MATH with compact support in MATH: all we have to do is solve MATH in the open sets MATH and MATH, extend to a global function using a partition of unity and write MATH. It follows, since MATH is closed, that MATH . Thus finally, MATH and MATH are linearly equivalent if and only if MATH is an integer for all harmonic REF-forms with integral cohomology class. |
math/9907034 | The map MATH is clearly smooth and commutes with the projections onto MATH, the moduli space of special Lagrangian tori, so it remains to prove that the map on the fibres is a diffeomorphism. Consider a fibre MATH. The corresponding fibre on MATH is the moduli space of degree one gerbes on MATH, also a REF-torus. If we choose a basepoint MATH, and make MATH an origin in the moduli space of gerbes, then the map MATH can be interpreted as the map MATH where MATH is the holonomy of the flat gerbe MATH. We can easily see from the proof of REF that this is the NAME map MATH (in particular it is clear that MATH if and only if MATH and MATH are linearly equivalent). Let us consider the kernel of the derivative of MATH at MATH. By the definition of MATH, this consists of the tangent vectors MATH such that MATH for all harmonic REF-forms MATH on MATH. But as we noted, for a fibration the harmonic REF-forms are linearly independent at each point so the kernel is always zero and MATH is a local diffeomorphism. Since MATH is compact it is a covering map. But now the NAME map is defined for all metrics, and for the flat metric on the torus, MATH is a diffeomorphism. In particular the induced map on MATH is an isomorphism. Any metric can be continuously connected to the flat metric hence the induced map: MATH is an isomorphism for all metrics. It follows that the covering map is a diffeomorphism and the theorem is proved. |
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