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math/0011123 | We have already produced the right hand square. The map MATH is just the obvious inclusion. The map MATH sends MATH to MATH; it was observed in the proof of REF that this makes MATH into a free module of rank MATH over MATH, so MATH is faithfully flat of degree MATH. The points of MATH are the divisors of degree MATH contained in MATH, so MATH is a closed subscheme of MATH; we write MATH for the inclusion, and note that MATH. As MATH is faithfully flat and MATH factors through MATH we see that MATH factors through MATH, so there is a unique map MATH such that MATH. As MATH is a closed inclusion, the same is true of MATH. A point of the pullback of MATH and MATH is a list MATH of points of MATH such that the divisor MATH lies in MATH, and thus satisfies MATH. It follows from the definitions that this pullback is just MATH as claimed. As MATH is faithfully flat we have MATH. By construction, MATH is just the inclusion of the MATH-invariants in MATH, so MATH as claimed. |
math/0011123 | We can certainly regard MATH as a module over the subring MATH, and the map MATH respects this structure. This makes MATH into a module over MATH. If MATH and MATH then MATH but MATH so MATH. This shows that MATH is annihilated by MATH, so it is a module over MATH, as claimed. |
math/0011123 | Put MATH for brevity. Note that MATH . In particular, we see that MATH is free of rank MATH over MATH, and MATH is free of rank one over MATH. We also know from REF that MATH is free of rank MATH over MATH, so MATH is also free of rank MATH over MATH. Suppose for the moment that MATH is a homomorphism of MATH-modules. It is clear that MATH and this element generates MATH, so MATH is surjective. As the source and target are free of the same finite rank over MATH, we deduce that MATH is an isomorphism as claimed. We still need to prove that MATH is linear over MATH. By the argument of REF we reduce to the case where MATH is the divisor with equation MATH defined over the ring MATH and we can invert the discriminant MATH. We reuse the notation in the proof of that lemma, so MATH and MATH and MATH. We see from REF that MATH is the image of MATH in MATH, which is the ring MATH of symmetric functions from MATH to MATH. If we write MATH then MATH as MATH-sets so MATH. On the other hand, MATH is also a quotient of MATH, which is the ring of symmetric power series in MATH variables over MATH; a symmetric power series MATH corresponds to the function MATH. If MATH we put MATH, so these elements form a basis for MATH over MATH. Similarly, the set MATH is a basis for MATH. Using the previous paragraph we see that MATH, which tells us the MATH-module structure on MATH. We next analyse MATH. This is a quotient of the ring MATH . It is not hard to check that the relevant ideal is a product of terms MATH, where MATH is spanned by the elements MATH that do not lie in the list MATH. Thus MATH . Let MATH be the element of this module whose MATH'th component is MATH, and whose other components are zero. Clearly MATH is a basis for MATH over MATH. As a symmetric power series MATH corresponds to the function MATH and MATH is concentrated in the MATH'th factor we have MATH. It is also easy to see that MATH, and it follows that MATH is MATH-linear as claimed. |
math/0011123 | For any increasing sequence MATH we write MATH. We also write MATH and MATH, and we put MATH. We certainly have MATH for some MATH. To analyse these elements, put MATH, which is freely generated over MATH by MATH. Consider the element MATH . This clearly annihilates MATH, so multiplication by MATH induces a map MATH. As MATH is monic of degree MATH, we see that MATH. It follows that MATH. On the other hand, we can expand MATH in the form MATH, where MATH runs over sequences MATH. We have MATH if MATH and MATH are related as in the statement of the proposition, and MATH otherwise. It follows that MATH, and thus that MATH. Let MATH be the matrix whose MATH'th entry is the coefficient of MATH in MATH; it is then clear that MATH. On the other hand, we have MATH, so MATH, and the proposition follows. |
math/0011123 | Put MATH and let MATH be the set of lists MATH with MATH. We have MATH and the set MATH is a basis for this ring over MATH. Put MATH and MATH, and let MATH be the obvious projection. This clearly induces an isomorphism MATH. We take as our basic input (proved by CITE) the fact that when MATH is a point, the map MATH induces an isomorphism MATH. In particular, this means that MATH is a finitely generated free Abelian group, concentrated in even degrees. By duality we see that MATH, and thus that the map MATH is an isomorphism. Using an NAME spectral sequence we see that MATH is an isomorphism for any MATH. Now let MATH be arbitrary. If MATH is trivialisable with fibre MATH then MATH and it follows from the above that MATH is an isomorphism. If MATH is not trivialisable, we can still give MATH a cell structure such that the restriction to any closed cell is trivialisable, and then use NAME sequences, the five lemma, and the NAME sequence to see that MATH is an isomorphism. We next claim that the maps MATH give rise to a cocommutative coproduct. To see this, let MATH denote the following diagram: The claim is that the diagram MATH commutes. Let MATH be the two inclusions. The map MATH induces a map MATH, and it follows easily from our previous discussion that this is surjective. It will thus be enough to show that the two ways round MATH become the same when composed with the map MATH . It is standard that MATH is homotopic to MATH through linear isometries, so MATH is fibre-homotopic to MATH. Similarly, the identity map of MATH is homotopic to MATH. It is thus enough to check that the two composites MATH in the following diagram are the same: This is easy to see directly. We now see that the map MATH factors through MATH. As the map MATH and its composite with MATH are both isomorphisms, we deduce that MATH is an isomorphism as claimed. |
math/0011123 | We refer to CITE for background on free commutative formal groups; the results there mostly state that the obvious methods for constructing such objects work as expected under some mild hypotheses. Given a formal scheme MATH over a formal scheme MATH, we use the following notation: CASE: MATH is the free commutative monoid over MATH generated by MATH. This is characterised by the fact that monoid homomorphisms from MATH to any monoid MATH over MATH biject with maps MATH of schemes over MATH. It is clear that if there exists a MATH with this property, then it is unique up to canonical isomorphism. Similar remarks apply to our other definitions. In reasonable cases we can construct the colimit MATH and this works as MATH; see CITE for technicalities. CASE: MATH is the free commutative group over MATH generated by MATH. CASE: If MATH has a specified section MATH, then MATH is the free commutative monoid scheme generated by the based scheme MATH, so homomorphisms from MATH to MATH biject with maps MATH such that the composite MATH is zero. In reasonable cases MATH can be constructed as MATH. CASE: If MATH has a specified section we also write MATH for the free commutative group over MATH generated by the based scheme MATH. The one surprise in the theory is that often MATH; this is analogous to the fact that a graded connected NAME algebra automatically has an antipode. It is easy to check that MATH, where MATH is regarded as a discrete group scheme in an obvious way. We first suppose that MATH has a one-dimensional summand, so MATH for some bundles MATH and MATH with MATH. Note that for each MATH there is a canonical isomorphism MATH giving MATH. This gives an evident inclusion MATH with MATH. We define MATH by MATH, and note that MATH for all MATH. We also have an evident map MATH splitting the projection MATH. Left multiplication by MATH gives a map MATH. We also define MATH to be the restriction of MATH to MATH. Using the fact that MATH we see that MATH. Thus, if we define MATH to be the homotopy colimit of the spaces MATH, we get a map MATH of spaces over MATH. Using the usual bases for MATH we find that the maps MATH are surjective. It follows using the NAME sequence that MATH and thus that MATH. We claim that this is the same as MATH; this is clear modulo some categorical technicalities, which are covered in CITE. In the case where MATH is a point, it is well-known and easy to check (by calculation in ordinary homology) that the map MATH is a weak equivalence. In the general case we have a map between fibre bundles that is a weak equivalence on each fibre; it follows easily that the map is itself a weak equivalence, and thus that MATH. On the other hand, as MATH is actually a group bundle, we see that MATH is a formal group scheme, so MATH. We now turn to the groups MATH. We define MATH, viewed as a bundle of groups over MATH in the obvious way. This can be identified with MATH so the determinant map gives rise to a homomorphism MATH. Given MATH we have a homomorphism MATH given by MATH. This construction gives us a map MATH with MATH and thus a splitting MATH and thus an isomorphism MATH. One can check that the various uses of the map MATH cancel out and that the standard inclusion MATH is implicitly identified with the map coming from the inclusion MATH. This proves the corollary in the case where MATH has a one-dimensional summand. Now suppose that MATH does not have such a summand. We have an evident coequaliser diagram MATH, giving rise to a coequaliser diagram MATH of schemes over MATH, in which the map MATH is faithfully flat. The pullback of MATH to MATH has a tautological one-dimensional summand, which implies that MATH has the required universal property in the category of formal group schemes over MATH. Similar remarks apply to MATH. It follows by a descent argument that MATH itself has the required universal property, as one sees easily from CITE. |
math/0011123 | Let MATH be the set of functions MATH for which the function MATH is continuous. Using the above algebraic properties, we see that MATH is a subalgebra of MATH containing the functions MATH and MATH. By the NAME theorem, it is dense in MATH. Now suppose we have MATH, MATH and MATH. Put MATH, which is compact. As MATH is dense we can choose MATH with MATH on MATH. As MATH can choose MATH such that MATH whenever MATH. We may also assume that MATH, which means that when MATH we have MATH. Now if MATH on MATH and MATH then MATH as required. |
math/0011123 | Let MATH be a simple closed curve in MATH and let MATH be an integer. Let MATH be the set of endomorphisms of MATH that have precisely MATH eigenvalues (counted according to multiplicity) inside MATH, and no eigenvalues on MATH. A standard argument with NAME 's theorem shows that MATH is open in MATH. Given real numbers MATH, consider the rectangular contour MATH with corners at MATH and MATH. Clearly MATH iff MATH has at least MATH eigenvalues inside MATH for some MATH. It follows that MATH is open, as is MATH by a similar argument. This implies that MATH is continuous. |
math/0011125 | Suppose that MATH. Then the distributive property shows MATH . Conversely, suppose MATH. Then MATH . |
math/0011125 | Using distributivity we obtain Thus MATH is equivalent to MATH for all MATH. |
math/0011125 | Let MATH be an infinite loop space. For each MATH, the NAME operations MATH are defined. For our purposes, we need the following properties (from CITE and CITE): CASE: MATH CASE: MATH if dim-MATH. CASE: The homology suspension homomorphism MATH yields MATH. In our case, MATH, so MATH . By REF , MATH . |
math/0011125 | For any elements MATH with MATH, we may use the distributive property to obtain MATH, using the coproduct for MATH. Thus MATH . We use this fact and consequences of relation REF with MATH to compute MATH . |
math/0011125 | CASE: By virtue of the fact that MATH, we obtain MATH. CASE: The facts that MATH and MATH may be used to prove this equation. |
math/0011125 | There is only one nontrivial element in the bidegree of MATH so either this is zero or it is as claimed above. If it is zero then if we take circle multiplication by MATH it should still be zero. However, Which we know to be non-zero. Thus the relation has to hold. |
math/0011125 | We first prove MATH . By definition of MATH, the homotopy element MATH classifies the reduced canonical line bundle. Thus MATH is the image under MATH of the fundamental class MATH, which is nonzero by the NAME theorem. The only possible element in this dimension (in MATH) is MATH, proving relation REF . We return to the proof of relation REF . Using relation REF and the fact that MATH gives MATH . By REF , MATH . The distributive property completes the proof: MATH . |
math/0011125 | We refer to MATH, the unitary NAME spectrum. As we will prove in REF, MATH . We start by proving the relation MATH where MATH is the fundamental class in MATH. Since MATH with MATH, relation REF is proved using the same argument as relation REF above. Relation REF may now be proven using the forgetful map, MATH. We build on relation REF to yield MATH. Applying the forgetful map gives MATH . In much the same way as the proof of relation REF , we may use NAME ring relations to obtain the remaining equalities MATH . |
math/0011125 | We use the forgetful map MATH, again building on relation REF to yield MATH . |
math/0011125 | By relation REF and by the fact that MATH we have MATH . The argument to show MATH is much the same as in relation REF . |
math/0011125 | We first find the coproduct for MATH; Thus MATH, as MATH. For any MATH we may use distributivity to simplify MATH . Thus, for all MATH, MATH . In particular, MATH. Next, Since MATH, and since MATH, we now have MATH . To finish, we need only prove MATH. We have already established that MATH for all MATH, and so MATH. Using distributivity and the MATH- product for the MATH shows To find MATH, we calculate MATH. Next, we use the NAME ring property that MATH. Since MATH, we have MATH . Thus MATH . We may now solve for MATH . By virtue of these facts, we have obtained MATH, completing our proof. |
math/0011127 | Let MATH contain MATH exactly once. There are two possibilities: either the only occurrence of MATH in MATH does not contain MATH, or it contains MATH. In the first case we get immediately that any entry of MATH to the right of MATH is less than any entry of MATH to the left of MATH, since otherwise one gets an occurrence of MATH involving MATH. In the second case, let MATH be the occurrence of MATH in MATH. First, we have MATH, since otherwise either MATH or MATH would be a second occurrence of MATH in MATH. Next, MATH immediately precedes MATH in MATH, since if MATH lies between MATH and MATH, then either MATH or MATH would be a second occurrence of MATH in MATH. Finally, any entry to the right of MATH is less than any entry between MATH and MATH, which in turn, is less than any entry to the left of MATH (the proof is similar to the analysis of the first case). |
math/0011127 | By REF , we have exactly three possibilities for the block decomposition of an arbitrary MATH. Let us write an equation for MATH with MATH. The contribution of the first decomposition above is MATH. Here the first term corresponds to the case MATH avoids MATH and MATH avoids MATH, while the second term corresponds to the case MATH avoids MATH but contains MATH, and MATH. The contribution of the second possible decomposition is MATH; here MATH contains MATH, and hence is always distinct from MATH. Finally, the contribution of the third possible decomposition is MATH . Here the first term corresponds to the case MATH avoid MATH, MATH avoids MATH, while the second term corresponds to the case MATH avoids MATH, MATH avoids MATH but contains MATH, and MATH. Solving the obtained linear equation and using REF , and well known identities involving NAME polynomials (see for example, CITE), we get the desired expression for MATH, MATH. In the case MATH, the contributions of the first and the second decompositions degenerate to MATH each, while the contribution of the third decomposition degenerates to MATH (which means that the only permutation having this decomposition is MATH itself). The result follows immediately. Let us consider now the case of MATH with MATH. The contribution of the first decomposition is MATH. Here the first term corresponds to the case MATH avoids MATH and MATH avoids MATH, while the second term corresponds to the case MATH avoids MATH but contains MATH, and MATH avoids MATH. The contribution of the second decomposition is MATH. Finally, the contribution of the third decomposition is MATH . Here the first term corresponds to the case MATH avoid MATH, MATH avoids MATH, the second term corresponds to the case MATH avoids MATH, MATH avoids MATH but contains MATH, and MATH avoids MATH, while the third term corresponds to the case MATH avoids MATH but contains MATH, MATH, and MATH avoids MATH. The expression for MATH follows easily from this equation and REF . The case MATH is treated similarly. For general MATH, MATH, the contribution of the first decomposition equals MATH, the contribution of the second decomposition equals MATH, and the contribution of the third structure decomposition MATH . The final result again follows from REF . |
math/0011127 | The three possible block decompositions of permutations containing MATH exactly once are described in REF . Let us find the recursion for MATH. It is easy to see that the contribution of the first decomposition equals MATH the contribution of the second decomposition equals MATH while the contribution of the third decomposition equals MATH . Solving the obtained recursion with the initial condition MATH and using REF , and REF, we get the desired result. |
math/0011127 | Let MATH, then either MATH (which means that MATH), or MATH, where both MATH and MATH avoid MATH. According to this dichotomy, we get the following recursion: MATH . Solving this recursion with the initial condition MATH and using REF , we get the desired result. |
math/0011127 | The main ingredient of the proof is the following description of the block decompositions of permutations in MATH. Let MATH, then there exist MATH such that either MATH, or MATH, where MATH is a permutation of the numbers MATH, MATH is a permutation of the numbers MATH, and MATH is a permutation of the numbers MATH. |
math/0011128 | By REF , there exist MATH a complete fundamental set of invariants under MATH defined near MATH. Define MATH for MATH . We can view MATH as a polynomial in MATH whose coefficients are functions of MATH. In fact, these coefficients are invariant under MATH. Moreover, since all MATH's are invariant under MATH, their coefficients are also invariant under MATH. Observe that MATH. In other words, there exists a non-trivial functional relationship between MATH and the coefficients of MATH. This means that, locally, MATH can be written as a function of the coefficients of MATH. Since MATH are functionally independent, there must be MATH functionally independent functions among the coefficients of the MATH's. This shows the first part of the statement. To prove the second part, write MATH, with MATH for MATH. By freeness of the action of MATH in a neighborhood of MATH, we can choose MATH so that MATH acts on MATH regularly. We have MATH, for some MATH if and only if MATH, or equivalently MATH. By REF , this happens if and only if MATH, for all MATH. which is equivalent to saying that MATH, for all MATH and the conclusion follows. |
math/0011128 | By REF . |
math/0011128 | The necessity of the first and third statements follow from the invariance of the signature. For the second statement, we also use REF . To prove sufficiency of the first statement, assume that MATH. Then by REF , there exists MATH such that MATH. So MATH. To prove sufficiency of the second statement, assume that MATH for some MATH. By REF , this means that there exists MATH and MATH such that MATH and therefore MATH. The proof for sufficiency of the third statement is similar. |
math/0011128 | Follows from REF . |
math/0011128 | Since the points of the signatures are functions of the basic MATH-invariants, they are MATH-invariant themselves. Moreover the order of the points is chosen in an invariant way, except for the starting point. Therefore if two polygons are equivalent under the action of MATH, then their signature will be identical, up to cyclic permutation. Now suppose that MATH and MATH are two polygons with the same MATH. Assume that MATH corresponds to MATH and MATH, that MATH corresponds to MATH and MATH, and so on. Since the signature of the two polygons is the same, we have MATH (by (MATH)) . So we can find MATH which maps MATH to MATH and MATH to MATH. Moreover since MATH is uniquely prescribed by MATH and the value of MATH and MATH (by (MATH)), we have that MATH also maps MATH to MATH, and MATH to MATH, and so on. Therefore MATH. |
math/0011128 | The polygon MATH has a MATH-fold symmetry if and only if there exists MATH such that MATH. By invariance of the signature, that means MATH which proves the necessity of the statement. Now if MATH, then by property REF and (MATH), there exists MATH such that MATH. This proves sufficiency. |
math/0011128 | By invariance of the functions chosen to parameterize it, the MATH of two equivalent polygons must be the same, modulo the choice of starting point and direction. This proves the necessity of the first statement. To prove necessity of the second statement, we use REF and the fact that MATH commutes with rotations. If MATH, let MATH. Property REF allows us to conclude that MATH such that MATH. Property REF implies that MATH. and therefore MATH. This proves the sufficiency of the first statement. The proof of the sufficiency of the second statement is similar. |
math/0011128 | For simple polygons, rotations are the only MATH transformations which preserve the traveling direction on the vertices, since they are the only transformations which preserve orientation. So the proof is the same as for MATH symmetries. |
math/0011128 | For simple polygons, rotations are the only MATH-symmetries which reverse the traveling direction on the vertices, since they are the only transformations which reverse orientation. By invariance of the MATH and since the MATH commutes with rotations, if MATH for some MATH, then MATH. Now if MATH, then REF imply that there exists MATH such that MATH. In particular, if MATH, then MATH is fixed so we have an axe of reflection passing through MATH. Also if MATH, then MATH is mapped to MATH and MATH is mapped to MATH, so we have an axe of reflection passing through the middle of the edge joining MATH to MATH. (Of course other cases can be obtained by relabeling the vertices.) |
math/0011128 | By REF . |
math/0011128 | By REF . |
math/0011128 | Let MATH and MATH. Consider the signatures MATH and MATH parameterized by MATH. It is enough to show that REF on MATH imply that if MATH, then there exists MATH such that MATH. Assume MATH. If MATH, let MATH. Then by REF implies that there exists MATH such that MATH . Therefore MATH. If MATH, then by REF implies that there exists MATH such that MATH . By (MATH), we also have MATH . Therefore MATH. |
math/0011128 | By invariance of the MATH's and MATH's, ` REF ' is true. To show ` REF ', assume MATH, for MATH. Then there exists MATH such that MATH. By (MATH), REF implies that MATH. |
math/0011128 | REF tells us that there are exactly MATH fundamental MATH-point joint invariants and exactly MATH fundamental (MATH)-point joint invariants. The difference is MATH . Let MATH. Observe that MATH, unless MATH. It is shown in CITE that MATH if and only if MATH. So if MATH, then MATH, as defined in REF , is strictly smaller than MATH, otherwise MATH. Let MATH. Assuming that MATH, then we can build a local moving frame MATH in a neighborhood of MATH. Consider the group action equation MATH. According to CITE, setting MATH into this equation gives MATH a vector made of MATH-point invariants. Among those MATH invariants, there are exactly MATH, say MATH, such that MATH are functionally independent on an open subset of MATH. If MATH, then a local moving frame doesn't exists in a neighborhood of MATH. However, we can obtain a map MATH such that setting MATH into the equation MATH will give MATH a vector made of MATH-point invariants containing the MATH invariants we are looking for. |
math/0011128 | Obtain MATH-point functionally independent invariants MATH as described in the proof of REF . We claim that, on an open subset of MATH, we can express MATH as a function MATH . This is because if that were not the case, then the Jacobian matrix MATH would contain a sub-matrix MATH with rank strictly smaller than MATH, which would contradict the fact that, since MATH, the invariants MATH are functionally independent of invariants defined on MATH. |
math/0011128 | Follows from the fact that MATH where MATH is a MATH matrix, and that, since there are no MATH-point joint invariants, the sub-matrix MATH has maximal rank MATH. |
math/0011128 | Follows from the fact that the rank of the sub-matrix MATH is equal to MATH. |
math/0011128 | By normalizing the equations MATH, for MATH and MATH as described in CITE, we obtain functionally independent invariants MATH defined on an open subset of MATH. We set MATH . We then normalize the equation MATH to obtain MATH invariants MATH defined on an open subset of MATH. By REF , among those MATH invariants, there are exactly MATH, say MATH, such that MATH are functionally independent. By REF , there exists exactly MATH invariants among MATH, say MATH, such that MATH contains a complete fundamental set of invariants on an open subset of MATH. We set MATH . So we have now defined a total of MATH of the MATH's. Similarly, if we normalize the equation MATH . We obtain MATH functionally independent invariants out of which MATH, say MATH, are such that MATH are functionally independent. By REF , there exists exactly MATH invariants among MATH, say MATH, such that MATH contains a complete fundamental set of invariants on an open subset of MATH . In other words, MATH contains a complete fundamental set of invariants on an open subset of MATH. We set MATH . So we have now defined MATH of the MATH's. Following this procedure MATH times, we obtain MATH functionally independent invariants MATH defined on some open subset of MATH (although in fact they are defined on the smaller set MATH.) We claim that MATH explicitly contains a complete fundamental set of invariants on MATH. This is because, by construction, the set MATH which is contained in MATH, contains exactly MATH functionally independent invariants on MATH. Observe that MATH . |
math/0011128 | We use the notation of the previous theorem. Start with constructing MATH as in the previous theorem. Then we normalize the equation MATH, for MATH and MATH , and repeat the same procedure as in the previous theorem to obtain a total of MATH invariants MATH. More precisely, we choose some MATH invariants among MATH, say MATH, such that if MATH are a complete fundamental set of invariants defined on some open subset of MATH, then MATH are functionally independent. We set MATH this obtaining MATH invariants MATH. Since MATH, the set MATH explicitly contains a complete fundamental set of invariants on MATH. Moreover, since MATH, we have MATH. By the same argument as in the proof of REF , we can show that MATH are MATH point reductive. |
math/0011131 | If MATH, then MATH . Since any eigenfunction of MATH associated with an eigenvalue MATH changes sign, MATH somewhere. Let MATH. Then MATH and MATH by REF. Since MATH in MATH, MATH must also be the first NAME eigenvalue of MATH on MATH, so it follows from REF that MATH. Thus MATH. We complete the proof by showing that if MATH with MATH, then MATH and MATH. We have MATH . If MATH, then MATH satisfies MATH so MATH, contradicting REF. Thus MATH and satisfies MATH . |
math/0011131 | Since MATH is the only critical point of MATH, MATH . As MATH is positive homogeneous, MATH is contractible and MATH is homotopic to MATH, so REF follows from the reduced homology sequence of the pair MATH and REF. The second statement is proved similarly. |
math/0011131 | Taking MATH gives MATH . |
math/0011131 | Take MATH. Then MATH is isolated in MATH and MATH has no critical points in MATH, so MATH by REF , that is, MATH. Thus we have the commutative diagram MATH induced by inclusions, where MATH is an isomorphism by REF as MATH has no critical points in MATH. Since MATH, the conclusion follows. |
math/0011131 | If MATH is nontrivial, set MATH where MATH denotes the support of the singular MATH-chain MATH. Clearly, MATH. We will show that MATH has a critical point in MATH. We follow the proof of REF. Consider the subspace MATH of MATH consisting of all continuous deformations MATH such that CASE: MATH, CASE: MATH where MATH is the NAME metric on MATH, CASE: MATH. MATH equipped with the metric MATH is a complete metric space. For any MATH, the assumption that there is no path in MATH joining MATH and MATH together with REF above imply that, for each MATH, the restriction of MATH to MATH is a map of the pair MATH (that is, MATH and MATH). Thus, REF implies that MATH is homotopic to the identity on MATH, so we have that MATH for any MATH. Fix MATH, take MATH such that MATH and define a continuous function MATH by MATH . Let MATH be the identity in MATH (that is, MATH for all MATH), and note that MATH . Applying the NAME 's principle, we get a MATH such that MATH . Let MATH. Since MATH satisfies REF and MATH for all MATH, it is enough to show that there is a MATH such that MATH. Indeed, in view of REF, any such point necessarily satisfies MATH. Suppose now that MATH for all MATH. Applying REF, we get MATH, MATH, and MATH such that CASE: MATH, CASE: MATH, CASE: MATH. For MATH, let MATH. Clearly, MATH. Since MATH by REF above, REF gives MATH. Since MATH is compact, there is a MATH such that MATH, so we have MATH . In particular, if MATH is any cluster point of MATH as MATH, then MATH, and hence MATH by REF. On the other hand, MATH by REF, and combining this with REF gives MATH which contradicts REF. |
math/0011131 | We will show that every point MATH can be connected to MATH by a path in MATH. Denote by MATH the component of MATH containing MATH. By REF, MATH is achieved at a critical point MATH. Since MATH is path-connected by REF, it is enough to show that MATH can be connected to MATH by a path in MATH. If MATH, then MATH since every critical point of MATH other than MATH changes sign, and MATH can be connected to MATH by a path in MATH since MATH. So suppose that MATH, and let MATH . Taking MATH in MATH gives MATH so MATH . Thus MATH achieves the infimum and hence is a critical point of MATH. Since MATH, MATH, so MATH is a path in MATH connecting MATH to MATH. |
math/0011131 | CASE: Since MATH REF gives MATH so MATH and the conclusion follows from REF . By REF , MATH . We apply REF with MATH and MATH. The only critical point of MATH in MATH is the global minimizer MATH, and MATH by REF . Since MATH, there is a MATH such that MATH by the continuity of MATH, and an argument similar to that in the proof of REF shows that MATH . Thus MATH . CASE: By REF, there are MATH and MATH such that MATH . Since MATH has no critical values in MATH, MATH by REF . As REF implies that MATH and MATH are isolated in MATH, MATH and MATH by REF . By REF , MATH and the conclusion follows since MATH is path-connected by REF . |
math/0011131 | If both MATH and MATH are below MATH, then the conclusion follows from REF . So, by REF , it suffices to show that for any MATH that lies above MATH, MATH for all MATH sufficiently close to MATH. Choose MATH so small that MATH, which is possible since MATH is closed, and suppose that MATH. Then MATH so we have the inclusions MATH which induce homomorphisms MATH . Since the points MATH and MATH (respectively, MATH and MATH) are in MATH, MATH (respectively, MATH) has no critical values in MATH (respectively, MATH), so MATH (respectively, MATH) is an isomorphism by REF . Thus MATH is an isomorphism. Finally, MATH as MATH has no critical values between MATH and MATH. |
math/0011131 | Since MATH a.e., MATH, so by uniform convexity it suffices to prove norm to weak continuity, that is, MATH implies MATH. We have MATH in MATH, so by weak compactness MATH in X for a subsequence. Hence MATH and MATH for the whole sequence. |
math/0011131 | We only consider a local minimizer MATH of MATH as the argument for MATH is similar. We have to show that for every sequence MATH in MATH, MATH for sufficiently large MATH. By REF, MATH for some constant MATH, so MATH . Since MATH by REF , MATH for sufficiently large MATH. We will show that MATH . First we note that the measure of the set MATH goes to zero. To see this, given MATH, take a compact subset MATH of MATH such that MATH and let MATH. Then MATH where MATH, so MATH. Since MATH and MATH is arbitrary, the claim follows. If REF does not hold, setting MATH, MATH is bounded for a subsequence, so MATH in MATH and a.e. for a further subsequence, where MATH and MATH. But then MATH has positive measure for all sufficiently small MATH and MATH a contradiction. |
math/0011131 | Let MATH, and MATH so that MATH . Since MATH, MATH and MATH satisfy REF and have no critical points on MATH, so MATH . By homogeneity, MATH while it follows from REF that MATH as MATH and MATH, so MATH for all sufficiently small MATH and sufficiently large MATH. Take smooth functions MATH such that MATH and set MATH . Since MATH REF holds with MATH replaced by MATH also. The conclusion follows. |
math/0011133 | We give a proof of REF below. From this proof it follows that if MATH, where MATH is an isomorphism and MATH has finite rank, then MATH is NAME. To prove the converse, choose some orthonormal bases MATH and MATH in MATH and MATH respectively, using REF . Define MATH . Clearly MATH has finite rank, and MATH. Let us prove that MATH is an isomorphism. If this is done, then REF is proved. We need to prove that MATH and MATH. It is known (NAME 's theorem), that if MATH is a linear injection and MATH, then MATH is a bounded operator, so MATH is an isomorphism in the sense defined above. Suppose MATH. Then MATH (so that MATH), and MATH (because, according to REF , MATH is orthogonal to MATH). Since MATH, is a linearly independent system, REF implies MATH for all MATH, that is, MATH is orthogonal to MATH. If MATH and at the same time it is orthogonal to MATH, then MATH. So, MATH. Let us now prove that MATH: Take an arbitrary MATH and, using REF , represent it as MATH where MATH and MATH are orthogonal. Thus there is a MATH and some constants MATH such that MATH. We choose MATH orthogonal to MATH. This is clearly possible. We claim that MATH where MATH. Indeed, using the orthonormality of the system MATH one gets MATH. Thus we have proved that MATH. REF is proved. |
math/0011133 | If MATH is NAME, then REF are equivalent to REF , since REF follows from REF . Let us prove that if MATH, where MATH is an isomorphism and MATH has finite-rank, then MATH is NAME. Both REF are known for operators in finite-dimensional spaces. Therefore we will prove that MATH is NAME if we prove that REF are equivalent to linear algebraic systems in a finite-dimensional space. Let us prove this equivalence. We start with REF , denote MATH and get an equivalent to REF equation MATH where MATH is a finite rank operator which is of the same rank MATH as MATH because MATH is an isomorphism. REF is equivalent to REF : each solution to REF is in one-to-one correspondence with a solution of REF since MATH is an isomorphism. In particular the dimensions of the null-spaces MATH and MATH are equal, MATH and MATH is closed . The last claim is a consequence of the NAME alternative for finite-dimensional linear equations, but we give an independent proof of the closedness of MATH at the end of the paper. Since MATH is a finite rank operator, the dimension of MATH is finite and is not greater than the rank of MATH. Indeed, if MATH and MATH has finite rank MATH, then MATH, where MATH is an orthonormal basis of MATH, and MATH, so that MATH belongs to a subspace of dimension MATH. Since MATH and MATH enter symmetrically in the statement of REF , it is sufficient to prove REF for MATH and check that the dimensions of MATH and MATH are equal. To prove REF , let us reduce REF to an equivalent equation of the form: MATH where MATH is the adjoint to MATH and MATH . Since MATH is an isomorphism, MATH. Applying MATH to REF , one gets an equivalent REF and MATH is a finite-rank operator of the same rank MATH as MATH. The last claim is easy to prove: if MATH is a basis in MATH then MATH, and MATH, so MATH. By symmetry one has MATH and the claim is proved. Writing explicitly the linear algebraic systems, equivalent to REF , one sees that the matrices of these systems are adjoint. The system equivalent to REF is: MATH where MATH and the one equivalent to REF is: MATH where MATH and MATH is the matrix adjoint to MATH. For linear algebraic systems REF the NAME alternative is a well-known elementary result. These systems are equivalent to REF , respectively. Therefore the NAME alternative holds for REF , so that REF are proved. REF is proved. |
math/0011137 | To prove uniqueness, we begin by observing that if MATH and MATH are period mappings with the same associated nilpotent orbit and MATH then, for any MATH, we may consider the sections MATH and MATH of the canonical extension MATH. Clearly, MATH and MATH. On the other hand, since MATH, it follows that MATH and, consequently, MATH is a MATH-flat section which extends to the origin and takes the value zero there. Hence, MATH is identically zero and MATH for all values of MATH. To complete the proof of the Theorem it remains to show the existence of a period mapping with given nilpotent orbit and MATH. This amounts to finding a solution to the differential REF with MATH assuming that the integrability REF is satisfied. Set MATH and MATH. Then REF may be rewritten as MATH where MATH denotes the identity, while REF takes the form: MATH . By considering the MATH-graded components of REF we obtain a sequence of equations: MATH . Assume inductively that, for MATH, we have constructed MATH satisfying REF and such that MATH . Then, the initial value problem MATH has a solution which verifies MATH . Thus we may, inductively, construct a solution of REF . REF now implies that the map MATH is the desired period map. |
math/0011137 | Let MATH denote the filtration MATH, where MATH is an arbitrary element in the cone MATH positively spanned by MATH. Note that MATH leaves MATH invariant. Moreover, since the monodromy logarithms are MATH-morphisms of the mixed NAME structure MATH it follows easily that MATH where MATH denotes the canonical bigrading of the MHS. This implies that the associated bigrading of the NAME algebra MATH and, in particular, that the subalgebra MATH defined in REF are independent of the choice of coordinates. According to REF , MATH, therefore MATH . This identity, in turn, implies REF since the group elements in both sides of REF lie in MATH. |
math/0011137 | This follows considering the MATH-component in REF , given the observation that this subspace is invariant under coordinate changes. |
math/0011137 | Let MATH be an arbitrary system of coordinates around MATH. We can write MATH where MATH are holomorphic in a neighborhood of MATH and MATH. The transformation REF now implies that the coordinate system MATH is canonical. Moreover, that same formula shows that it is unique up to scaling. |
math/0011137 | We note, first of all, that the quasi-homogeneity of MATH follows from the assumption that MATH is a graded product. Moreover, REF is a consequence of the fact that MATH is the unit for MATH, while REF is the associativity condition for MATH. On the other hand, REF and the fact that MATH is NAME imply that MATH and REF follows. The converse is immediate since REF defines a commutative structure whose associativity follows from REF ; the quasi-homogeneity assumption implies that the product is graded; MATH is the unit because of REF . The compatibility between MATH and MATH comes from MATH. |
math/0011137 | It is clear that REF define a graded product whose unit is MATH. Commutativity follows immediately from the symmetry of MATH and the commutativity of the operators MATH. There are two non-trivial cases to check in order to prove the associativity of the product. When all three factors lie in MATH this follows, again, from the commutativity of the operators MATH. On the other hand, given MATH: MATH . Thus, MATH is a graded, commutative, associative algebra with unit MATH. It is straightforward to check that MATH is compatible with the product. |
math/0011137 | Let MATH be a graded, polarized, real NAME algebra of weight four. Let MATH be an adapted basis such that MATH is a framing of MATH. Let MATH be the tube domain MATH . We view MATH as the universal covering of MATH via the map MATH, MATH. Let MATH denote the trivial bundle over MATH with fiber MATH and MATH the trivial subbundle with fiber MATH. Given a potential MATH and elements MATH, MATH, the quantum product MATH may be thought of as a MATH-valued function on MATH. Let MATH denote its restriction to MATH. It follows easily from REF that MATH depends only on MATH and, therefore, it descends to a MATH-valued function on MATH, that is, a section of MATH. This allows us to define a connection MATH on MATH by MATH where in the left-hand side MATH represents the constant section defined by MATH. As shown in CITE the WDVV equations for the potential MATH imply that MATH is flat. We can compute explicitly the connection forms relative to the constant frame MATH. We have, for MATH and MATH: MATH where the coefficients MATH are defined as in REF . Note that these equations imply that the bundles MATH satisfy the horizontality REF . Moreover, suppose we define a bilinear form MATH on MATH as in REF and extend it trivially to a form MATH on MATH, then it is straightforward to check that MATH for all MATH and all MATH. Hence the form MATH is MATH-flat. We may also deduce from REF that MATH has a simple pole at the origin with residues MATH where MATH denotes multiplication by MATH relative to the constant product MATH. It then follows from CITE that, written in terms of the (multivalued) MATH-flat basis MATH the matrix of the monodromy logarithms MATH coincides with the matrix of MATH in the constant basis MATH, that is, with MATH. Because the operators MATH are real, so is the monodromy MATH and therefore we can define a flat real structure MATH on MATH. Since MATH are a framing of the polarized NAME algebra MATH, it follows from REF that the map MATH is the period map of a VHS (a nilpotent orbit) in the bundle MATH. Since the bundles MATH are already known to satisfy REF , we can apply REF to conclude that they define a polarized VHS on MATH. In order to complete the asymptotic description of the PVHS defined by MATH on MATH, we need to compute the holomorphic function MATH. Because of REF , it suffices to determine the component MATH. Moreover, it follows from REF that we may choose canonical coordinates MATH on MATH, so that, in terms of the basis MATH, MATH has the form: MATH . Thus, MATH is completely determined by the MATH-matrix MATH. On the other hand, as noted in REF, MATH is the expression of the NAME bundles MATH in terms of the canonical REF : MATH . The matrix MATH, in the basis MATH, is the matrix expressing the canonical sections MATH in terms of the constant frame. Therefore MATH and it suffices to compute MATH to determine MATH (and MATH). It is straightforward to show, using the formulae REF , that MATH . Hence, to obtain MATH it suffices to apply REF , together with the fact that the matrix of MATH in the basis MATH coincides with that of MATH-multiplication by MATH relative to MATH. Thus, MATH and MATH . This, together with the fact that MATH is a homogeneous polynomial of degree MATH, implies that MATH and MATH . Hence MATH . Thus, MATH . Conversely, suppose now that MATH is a polarized VHS of weight four over MATH, that the origin is a maximally unipotent boundary point, and that the limiting MHS is of NAME type. Let MATH denote the associated nilpotent orbit and set MATH. It follows from REF that we can define a product MATH, and a bilinear form MATH - as in REF - turning MATH into a polarized, real, graded NAME algebra with unit MATH. This structure is determined by the choice of unit and the fact that, relative to an adapted basis MATH as in REF , MATH, MATH, and MATH are the coefficients REF of the associated classical potential MATH. We have already noted that in canonical coordinates, the holomorphic function MATH associated with the PVHS takes on the special form REF . Moreover, since MATH satisfies the differential REF , we have from REF that MATH for MATH. Consequently, MATH . We now define a potential on MATH by MATH . Since we already know that the classical potential MATH satisfies REF , the WDVV equations for MATH reduce to REF . But this is a consequence of the integrability REF ; indeed, note that given REF , if we let MATH be the MATH-matrix of one forms MATH REF reduces to MATH which, in view of REF , is easily seen to be equivalent to REF . |
math/0011139 | Since MATH is compact, the image MATH is compact. Therefore the set of vectors MATH, such that the function MATH is strictly positive on MATH, is open. Hence we may assume that MATH lies in the integral lattice MATH of the torus MATH. Let MATH be the corresponding circle subgroup of MATH. Let MATH and let MATH. The form MATH is another MATH-invariant contact form with MATH. The moment map MATH defined by MATH satisfies MATH. Therefore MATH for all MATH. Since the function MATH is nowhere zero, the action of MATH on MATH is locally free. Consequently the induced action of MATH on the symplectization MATH is locally free as well. Hence any MATH is a regular value of the MATH-component MATH of the moment map MATH for the action of MATH on the symplectization MATH. Note that MATH. Now MATH is the MATH level set of MATH. Therefore MATH is a (compact connected) symplectic orbifold with an effective Hamiltonian action of MATH. The orbit map MATH makes MATH into a NAME bundle over MATH. A dimension count shows that the action of MATH on MATH is completely integrable. |
math/0011139 | By REF the manifold MATH is a NAME bundle over a compact connected symplectic toric orbifold MATH. A generic component of the moment map on MATH is a NAME function with all indices even. The NAME inequalities hold rationally for NAME functions on orbifolds (see CITE). Therefore the first cohomology MATH is zero. Next we apply the NAME sequence to the map MATH. Since the NAME sequence comes from the collapse of the NAME spectral sequence for MATH and since rationally the "fibration" MATH is a circle bundle, the NAME sequence does exist. We have MATH. Since MATH and since MATH, the result follows. |
math/0011139 | By REF the fibers of the contact moment map MATH are connected. Let MATH denote the symplectic moment map for the Hamiltonian action of MATH on the symplectization MATH of MATH. It is given by MATH. Thus MATH and the fibers of MATH are connected. The triple MATH is a symplectic toric manifold. Since the image of MATH is contractible and since the fibers of MATH are connected, it follows from a result of NAME, NAME and NAME REF , that the image of MATH determines the symplectic toric manifold MATH uniquely. In particular the image determines the first cohomology group of MATH. A standard argument for Hamiltonian MATH-spaces implies that the subspace MATH is the annihilator of the NAME algebra MATH of a subtorus MATH of MATH. Since MATH is a subtorus, the exact sequence MATH splits. Let MATH be a subtorus in MATH complementary to MATH and let MATH be the corresponding splitting of the duals of the NAME algebras. Then MATH and MATH where MATH is a proper cone. It follows from REF that the exists a basis MATH of weight lattice of the torus MATH so that the edges of the cone MATH are of the form MATH. In particular the representation of MATH on MATH defined by the infinitesimal characters MATH has the property that the image of the corresponding moment map is MATH. Consequently we can realize MATH as the image of the moment map for the product action of MATH on MATH. Therefore MATH is MATH-equivariantly symplectomorphic to MATH, which is homotopy equivalent to MATH. Since MATH and since MATH, MATH. |
math/0011139 | As we remarked earlier the symplectization MATH is a symplectic toric manifold. CITE showed that for symplectic toric manifolds all the isotropy groups are connected and all fixed points are isolated. If a point MATH is fixed by the action of MATH then the line MATH is fixed by the action of MATH on MATH. Therefore the action of MATH on a contact toric MATH-manifold has no fixed points (one can also give a direct proof of this fact). Next we use the fact that MATH and MATH. By the above observations the isotropy groups for the action of MATH on MATH are either trivial or circles. If the isotropy group of MATH is trivial, then a neighborhood of the orbit MATH in MATH is MATH-equivariantly diffeomorphic to MATH for some small epsilon. If the isotropy group of a point MATH is a circle MATH, then a neighborhood of the orbit MATH in MATH is MATH-equivariantly diffeomorphic to MATH for a small disk MATH. Moreover the action of MATH on MATH must be effective; hence we may identify MATH with MATH in such a way that the action of MATH on MATH is given by MATH. We conclude that locally MATH is homeomorphic to either MATH or to MATH. Thus if the action of MATH on MATH is not free, then MATH is a one-dimensional manifold with boundary. Since MATH is compact and connected, MATH has to be an interval. Therefore, by REF and NAME REF , MATH as a MATH-space is uniquely determined by the isotropy representations at the points in MATH above the endpoints of the interval MATH. It is easy to see that in this case MATH is diffeomorphic to two solid tori glued along their boundaries. We conclude that if MATH is a three dimensional compact connected contact toric manifold and if the action of a REF-torus MATH is not free, then MATH is a lens space. In particular MATH is not diffeomorphic to MATH. |
math/0011142 | It follows from the hypothesis that there is a sequence MATH of partial isometries in MATH such that the algebra generated by MATH is equal to MATH. Let MATH be the algebra generated by MATH and let MATH be the MATH-algebra generated by MATH. Then MATH is a finite dimensional MATH-algebra. Moreover, it contains a masa which is spanned by the operators MATH where MATH is any finite product of operators from the set MATH. Let MATH be the algebra generated by MATH and this masa. Then MATH is a digraph algebra and the system MATH associated with the inclusions MATH is a regular system determining MATH, and the masa MATH is a standard regular masa in MATH. |
math/0011142 | We have MATH where MATH is the AF MATH-algebra MATH, MATH is a regular AF masa in MATH and MATH is the NAME algebra. It follows from REF that MATH is not standard regular. |
math/0011142 | Since MATH and MATH are standard regular embeddings we can decompose each into a maximal direct sum of multiplicity one maps, MATH . We first note that since MATH and MATH are unitarily equivalent, there is indeed the same number of summands in each decomposition, as written. (This number is simply the number of edges in the NAME diagram for the C*-algebra extension map.) Let MATH, MATH and MATH. Clearly, each MATH and MATH is a rank one projection in MATH and MATH, MATH. Now, for each MATH, the map MATH is a multiplicity one summand of MATH and by the uniqueness of the decomposition, is unitarily equivalent to MATH for some MATH. We note that since MATH is block diagonal, MATH for each block projection MATH in MATH. Thus by REF , we can find a unitary MATH such that MATH for all MATH. Thus the maps MATH and MATH agree on the diagonal matrix units. Since these maps havemultiplicity one and are star extendible it follows that they are conjugate by a unitary in MATH. Thus, MATH and MATH are conjugate by a unitary in MATH and the desired conclusion follows on combining these equivalences. |
math/0011142 | Since MATH is regular, there exists a unitary MATH such that MATH is standard regular. Thus the maps MATH and MATH are unitarily equivalent standard regular maps. By REF we can find a unitary MATH such that MATH. Set MATH and the map MATH is standard regular with MATH, completing the proof. |
math/0011142 | Let MATH, MATH. By functoriality, we can find subsystems MATH and maps MATH such that the following diagram commutes: MATH where the maps MATH and MATH are compositions of the given embeddings and the MATH are regular, but not necessarily standard regular, for each MATH. We construct sequences of unitaries, MATH with the following properties: CASE: MATH. CASE: the maps MATH are standard regular, CASE: MATH for all MATH. Firstly, since MATH is regular, there exists a unitary MATH such that MATH is standard regular. Let MATH. Now suppose we have constructed unitaries MATH with the above properties. We now construct MATH and MATH. Set MATH. Since MATH is unitary, MATH is unitary (or may be extended to one if MATH is non unital) and for each MATH, MATH . Since MATH is regular and the maps MATH and MATH are standard regular, by REF , we can find a unitary MATH such that MATH is standard regular and MATH. Now set MATH, and we obtain a unitary with the required properties. Similarly, set MATH and repeat the above argument to find a unitary MATH such that MATH is standard regular and MATH . Now let MATH. The required map MATH is determined by the new commuting diagram that is MATH with inverse, MATH. It is clear from the construction of MATH. That it is a star extendible isomorhism which is approximately inner equivalent to MATH and that MATH. |
math/0011142 | As before let MATH be algebras in Alglim-MATH with standard regular masas arising from direct systems MATH, MATH respectively. Let MATH be a star extendible isomorphism. Then MATH is induced by a commuting diagram: MATH (after relabelling subsystems for notational simplicity) in which the horizontal maps belong to MATH and the crossover maps are merely star extendible. We wish to construct subsequences MATH, MATH such that the crossover maps in the induced diagram: MATH all belong to MATH. Let MATH be the isomorphism corresponding to the first diagram above. We have that MATH is a standard regular masa in MATH and by the uniqueness hypothesis there exists a sequence of unitaries, MATH, with MATH for all MATH which determines an automorphism MATH with MATH. Since MATH it follows that MATH and we can find sequences MATH and corresponding crossover maps (as in the diagram of REF ) such that MATH for each MATH, where the maps MATH and MATH are compositions of the given embeddings. Since MATH and MATH is a masa in MATH it follows that MATH and MATH are regular for all MATH. Since MATH is saturated these maps belong to MATH, as required. |
math/0011142 | REF and the discussion above shows that the spectrum is a well defined invariant. The completeness of the invariant follows from REF simplified to the algebraic case. |
math/0011142 | That MATH is functorial if MATH is functorial is immediate. Suppose then that MATH is functorial and let MATH be a star extendible isomorphism. In particular MATH is induced by a commuting diagram of maps MATH as in the statement of REF . Let MATH be a MATH-module isomorphism with lifting equal to MATH. First (irrespective of this lifting) we construct a natural map MATH so that the following diagram commutes. MATH . Let MATH be the identity map, with class MATH in MATH. Let MATH denote the corresponding class in MATH with image MATH in MATH. Then MATH for some element MATH in MATH which in turn is the image of a class MATH in MATH, for some MATH, where MATH belongs to MATH. Since the natural maps MATH are MATH-module maps and MATH is a generator for MATH as a MATH-module, the desired diagram follows. Since MATH is induced by MATH, by hypothesis, we may increase MATH, if necessary, so that MATH agrees with the induced map MATH given by MATH where MATH is the restriction of MATH in MATH and MATH is suitably large. Thus in particular, MATH is a class in MATH which is to say, since MATH is closed under unitary equivalence, that MATH belongs to MATH. Note that MATH has the form MATH . Since MATH is induced by MATH we deduce that MATH is functorial. |
math/0011142 | The lemma follows readily from the special (triangular) case in which MATH is the digraph algebra MATH with MATH a connected reduced digraph. Let MATH be the (rank one) atomic projections of MATH and let MATH denote the block projections in MATH. Note that if MATH is a multiplicity one star extendible embedding, then there is an associated index map MATH such that MATH, for MATH. Moreover it is elementary to prove that MATH is a complete inner conjugacy invariant for MATH. We next note the following simple algebraic criterion that a regular map MATH should possess a multiplicity one summand inner conjugate to MATH in its decomposition. Let MATH be a sequence of matrix units taken from MATH or MATH with the following two properties. CASE: the product MATH is a non zero projection. CASE: each MATH is the initial or final projection of at least one MATH. Then, writing MATH for the block projection MATH and denoting the star extension of MATH by MATH also, we have MATH . On the other hand, if MATH is a multiplicity one embedding which is not conjugate to MATH then MATH . Moreover, if MATH is a direct sum of multiplicity one embeddings then the product MATH splits as a corresponding direct sum of products, one for each summand, and so we conclude that this ``test product" is non zero precisely when MATH has a summand conjugate to MATH. Now, if MATH then, for each MATH, the test product for MATH is at most distance MATH from the test product for MATH. Since test products have norm zero or one, the statement will follow by a simple induction argument on the number of summands in the decomposition for MATH. |
math/0011143 | Assume MATH . REF then provides a projection MATH with MATH . Let MATH be the polar decomposition for MATH and note that each of MATH and MATH is an element in the abelian MATH-algebra MATH . Now let MATH . We show that MATH is the required partial isometry. Firstly, MATH is indeed a partial isometry: MATH, which, since MATH and MATH commute, is itself a projection. Secondly, we estimate MATH . The proof is concluded by noting that MATH and the square root map is continuous on the positive cone of a MATH-algebra. |
math/0011143 | Firstly, write MATH where MATH and MATH. By hypothesis, MATH is a block diagonal projection: MATH from which we see that MATH and MATH are projections. It follows that MATH and MATH are partial isometries and since MATH itself is a partial isometry, MATH and MATH must be orthogonal. Now we claim that MATH is an approximate partial isometry. To show this we estimate MATH . Since MATH is a projection, MATH and thus; MATH . Provided MATH, REF ensures the existence of a partial isometry MATH with MATH . Now set MATH. We obtain a partial isometry with range orthogonal to MATH and with MATH. Set MATH and let MATH . We aim to show that MATH is close to MATH . Firstly we claim that MATH and MATH are MATH-approximately orthogonal. Since they have orthogonal ranges, MATH . Now, MATH . We now estimate the distance between MATH and MATH: MATH . Now MATH need not be a partial isometry but we claim it is close to one. This is immediate from the following estimate: MATH since MATH is a partial isometry. Provided MATH, MATH is close to a partial isometry MATH dominated by MATH . Thus MATH is orthogonal to MATH . Now let MATH which by construction is a partial isometry of the required form. To complete the proof we estimate: MATH . |
math/0011143 | MATH from which we infer that MATH is an approximately block diagonal projection, that is there exists a block diagonal projection MATH with MATH say. REF now provides a unitary MATH such that MATH and MATH . Now set MATH . Then MATH a block diagonal projection. Thus MATH-is a partial isometry with the required properties and MATH . |
math/0011143 | REF gives the result for MATH, so we assume the result to hold for all MATH satisfying the conditions with MATH block decomposition for MATH and consider MATH as in the statement. Write MATH with the obvious identifications. We shall also refer to the decomposition MATH where, MATH and MATH . As in the proof of REF , MATH is block diagonal, implying that MATH-and MATH are orthogonal partial isometries, a fact we make use of later. Now, MATH and the block diagonality assumption yields MATH . Thus MATH is a block diagonal projection. Since MATH is small, we infer that MATH is close to a partial isometry. Also, clearly MATH is approximately block diagonal, from which REF guarantees the existence of a partial isometry MATH with block diagonal final projection and with MATH . Now MATH with the operators below the block diagonal having norm MATH. We now apply the induction hypothesis to MATH to discover a block upper triangular partial isometry MATH close to MATH, within MATH say. Let MATH . Now MATH need not be a partial isometry, since MATH and MATH need not be orthogonal. However they are approximately orthogonal: MATH since MATH, MATH and MATH . Let MATH where MATH . Then MATH-and MATH are orthogonal. Now MATH need not be a partial isometry, but it is close to one which it dominates, MATH, with MATH say. This follows since MATH is clearly an approximate partial isometry. Now MATH is a partial isometry by construction with the correct form and MATH . The result follows by induction. |
math/0011143 | Let MATH and let MATH-ran-MATH-ran-MATH for MATH . Set MATH. Then MATH and MATH is a unitary operator with reduced matrix MATH where MATH . Now MATH and MATH with MATH-for all-MATH . REF now supplies a unitary operator MATH with matrix MATH . This operator is upper triangular with respect to the same decomposition and with MATH . Since MATH is unitary, MATH-and MATH . Now in the original matrix for MATH replace each MATH by MATH for MATH and MATH for MATH to find the required partial isometry. |
math/0011143 | Let MATH and MATH be nest algebras each of which is determined by MATH non-zero nest projections and suppose MATH. Denote the matrix unit system for MATH by MATH. We wish to create a matrix unit system, MATH for a copy of MATH inside MATH satisfying MATH . We then set the star extendible map MATH to be the linear extension of the correspondences MATH for each MATH . By hypothesis and REF , MATH and it is a standard self adjoint result that this ensures the existence of a matrix unit system for a copy of MATH in MATH, satisfying MATH for all-MATH . Now since MATH is a nest algebra, its reduced digraph is generated by a bilateral tree and every non-selfadjoint matrix unit may be factorized uniquely into a product of matrix units associated with the tree. Label the matrix units corresponding to the edges of the tree by MATH . We need to find partial isometries MATH close to each MATH respectively and having the correct initial and final projections, that is, if MATH and MATH, then we demand that MATH and MATH . Having found suitable partial isometries as described, we may define the remaining matrix units without ambiguity by multiplication. More precisely suppose that the matrix unit MATH satisfies MATH (so that MATH lives in the MATH block) and suppose that we have found a partial isometry MATH with the correct initial and final projections as described above and with MATH . Then for MATH, and MATH (that is MATH corresponding to the MATH block) we define MATH . Then for such MATH we have MATH . Having generated all the matrix units we can using the matrix units MATH and the partial isometries MATH, the remaining matrix units are determined in a well defined way. Thus a full matrix unit system for a copy of MATH in MATH has been created. Note that the quantity MATH depends only on MATH and MATH; the error does not increase within a single block. The remainder of the proof shall be concerned with finding suitable candidates for MATH . We demonstrate the technique only for MATH as the other cases are similar. Consider MATH. We keep the same notation as above for its projections-MATH . By hypothesis there exists MATH with MATH . Let MATH , MATH . We use these projections to cut down MATH . Set MATH so that MATH and we have MATH (where MATH). By the polar decomposition, write MATH . Then MATH since, MATH and MATH . Also we have MATH . Provided MATH was originally small enough so that MATH REF, we have that MATH is invertible in MATH, from which MATH is invertible in MATH. Thus MATH . Similarly, MATH . We now show that MATH and MATH are close and hence that MATH and MATH are close: MATH . We see that MATH is a partial isometry with MATH and MATH (which are block diagonal) and so would be a candidate for our required partial isometry MATH, but we cannot guarantee that MATH . However MATH is a partial isometry in MATH and since it is within MATH of MATH which is in MATH, it must be the case that the operators in the zero blocks for MATH in the representing matrix of MATH have norm MATH . REF now provides a partial isometry MATH with the same initial and final projections as MATH and with MATH . Setting MATH completes the proof. |
math/0011143 | Firstly we show that for any MATH is an approximate projection. By hypothesis, there exists MATH with MATH . Then MATH . Thus MATH approximately commutes with any MATH . Next, since MATH is a MATH-approximate partial isometry, we show MATH . MATH . Then we have, MATH . Thus, MATH . We are now ready to estimate MATH . Thus for each MATH is an approximate projection and similarly for MATH . Now, MATH is MATH-close to an element MATH of MATH which clearly is an approximate projection in a MATH-algebra and so is close to a projection MATH with MATH . Then clearly we have MATH and the proof is complete. |
math/0011143 | The approximately normalising hypothesis easily yields the following facts. Each entry of MATH is close in modulus to REF or REF. If MATH is close to REF, then every other entry in row MATH and column MATH is close to REF. If row MATH is such that MATH then there exists a unique index MATH for which MATH is close to REF. We now define MATH to be the index set: MATH and define MATH . The summary above implies that MATH is a partial isometry. By REF , for each projection MATH there exist MATH with MATH and MATH both less than MATH . We assume that MATH is sufficiently small so that MATH . If MATH is a minimal projection, then since MATH has rank REF or REF and since MATH is a projection in MATH, MATH must either be minimal or zero, else MATH . Also, we note that if MATH then MATH which implies MATH is the minimal projection MATH satisfying MATH . Thus, if MATH we have, by choice of MATH . We now make use of the estimates MATH and MATH to estimate MATH . Consider MATH . We have MATH and similarly for MATH . Now MATH . REF now provides MATH with MATH . Next we need to estimate MATH . Now MATH is MATH with the rows MATH for which MATH removed. Since MATH is MATH close to a REF projection, we have MATH . We use this fact firstly to show that MATH, and therefore MATH, is an approximate partial isometry. Since MATH is a MATH-approximate projection. Thus MATH is an approximate partial isometry in the abelian MATH-algebra MATH . Replace MATH by a nearby partial isometry MATH with MATH, so that MATH . Then we have: MATH . Noting that MATH concludes the proof. |
math/0011143 | Suppose to the contrary that MATH . Since each MATH is an abelian MATH-algebra, and by the functional calculus, we can find a projection MATH with (necessarily), dist-MATH . Since MATH there exists MATH with MATH . Now consider the expectation map MATH where the MATH 's are the minimal projections in MATH and MATH is the identity in MATH . Then we have MATH and so MATH and since MATH, we have a contradiction. |
math/0011143 | Let MATH be a matrix unit system for MATH compatible with the masa MATH . Take any MATH and MATH . Then we can select a sufficiently large index, MATH and an element MATH of MATH with MATH . Then MATH and MATH thus MATH and since MATH we are done. |
math/0011143 | Take MATH . Then MATH and thus for all MATH with MATH there exists MATH with MATH and similarly for MATH . Since MATH we can find MATH with MATH . Our aim is to show that MATH is an approximate partial isometry which approximately normalises MATH. The former property is clear from the fact that MATH is close to MATH, an exact partial isometry. Also MATH approximately normalises MATH, again because it is close to the exactly normalising element MATH, so for all MATH there exist MATH with MATH . Now take MATH . Since MATH . We first need to estimate MATH . Now MATH since MATH and where MATH . Then MATH . Now MATH and we estimate MATH . Thus MATH approximately normalises MATH . REF now provides a normalising partial isometry MATH with MATH . Thus MATH . |
math/0011143 | Firstly, we observe that if MATH are permutation type partial isometries with MATH then MATH and MATH have the same support, for if not we could find minimal projections MATH-and MATH with MATH . Secondly if MATH then MATH is another permutation type partial isomerty. With these preliminary observations in mind, we proceed with the proof. Suppose MATH is a digraph algebra on MATH vertices, then any cycle within the digraph will have length no greater than MATH . Let MATH be any spanning tree for the digraph of MATH fixed throughout, and let MATH be the matrix units corresponding to the edges of MATH . Since MATH has no cycles, any matrix unit in MATH can be written uniquely as a word of minimal length, no greater than MATH using the alphabet MATH . To fix notation, let MATH be matrix unit systems for MATH and MATH respectively, compatible with the given masas. Since MATH, each diagonal matrix unit MATH can be written as a sum of the MATH's. Denote this sum for each MATH by MATH for MATH . Now take the matrix unit MATH . Since MATH there exists MATH with MATH . If MATH and MATH then, MATH and MATH . Since MATH and MATH are both standard projections and MATH is a unimodular sum of the MATH's , provided MATH and MATH and so MATH has the correct initial and final projections. Now set MATH . Similarly, we create MATH having the right initial and final projections (in the above sense) and with MATH for each MATH . We now form the corresponding alphabet MATH . Now take any other matrix unit MATH in MATH and let MATH denote its unique word of minimal length in MATH . Define MATH to be the element with corresponding word in MATH . Note that MATH . We now need to show that MATH for each MATH . Firstly by construction, each MATH is a permutation type partial isometry in the containing MATH-algebra. Since MATH we can find MATH with MATH . Then MATH . Provided we choose MATH sufficiently small so that MATH, MATH must have the same support as MATH and thus MATH . In this way we create a matrix unit system MATH for a copy of MATH in MATH with MATH-for each MATH . Set MATH to be the linear extension of the correspondences MATH . Then MATH is regular, since MATH and it is clear that given MATH we can choose MATH sufficiently small so that MATH . |
math/0011143 | By REF , we choose MATH so that MATH . Now REF implies the existence of MATH such that if MATH and MATH there exists a regular star extendible injection MATH with MATH . By REF there exists MATH for which given MATH we have MATH . |
math/0011143 | Necessity of the local condition is clear, so suppose the local condition holds. Choose a dense sequence in the unit ball of MATH . Let MATH be a summable sequence, with MATH for each MATH . By hypothesis, there exists a pair MATH with MATH and MATH . Now, given MATH and MATH, REF implies that we can find MATH such that if MATH is another digraph algebra such that MATH and MATH then there exists a regular star extendible algebra injection MATH with MATH . We now demonstrate how the local condition provides MATH . Since MATH is finite dimensional, we can select a finite MATH net for the unit ball of MATH-we assume MATH. Consider the finite subset MATH . By the local condition there exists a digraph algebra and a regular star extendible injection MATH with MATH and MATH for MATH from which MATH and MATH . Continuing in this way we construct a sequence of finite dimensional operator algebras MATH and regular star extendible injections MATH with MATH and MATH for each MATH . Now consider the diagram MATH where each MATH is the restriction to MATH of the identity map. We estimate MATH for each MATH . Since MATH is summable, the diagram commutes asymptotically, thus MATH and MATH and the proof is complete. |
math/0011148 | We need to prove that if MATH and MATH are isotopic for MATH then MATH . We can assume that MATH and MATH differ by one crossing change only (and repeat the argument as many times as necessary). By REF and by REF MATH and MATH . Therefore MATH . |
math/0011148 | Assume first that MATH is a framed oriented knot and MATH is one of its two possible splittings. We need to prove that MATH . This holds in REF since then MATH is symmetric. In REF we have MATH because MATH in MATH and MATH for any MATH . Now, equality REF follows from the fact that MATH is symmetric. Assume now that MATH has MATH connected components, MATH . In order to establish REF it is enough to prove that MATH will not change if we change the splitting MATH by one connected component. Let MATH for some MATH and let the new splitting be MATH . We need to show that MATH . By REF , the left side equals MATH and the right side is equal to MATH . Note that MATH is isotopic to MATH in MATH and hence MATH . Similarly, MATH . Therefore, by REF , the right and the left side of REF are equal. |
math/0011148 | If MATH is the knot MATH with the opposite orientation, then have MATH by the bilinearity of MATH and of MATH . |
math/0011148 | CASE: Two framed oriented knots represent the same MATH-class in MATH if one can be obtained from the other by isotopy, crossing changes, or by the reversal of orientation. By REF , the reversal of orientation of a knot MATH does not change MATH . Suppose that framed knots MATH differ by a crossing change. Then the MATH-component links, MATH and MATH differ by two crossing changes of the same sign. Hence, MATH . Hence, by REF , MATH . CASE: By definition of MATH the left side of REF is MATH . By REF , MATH . Since MATH in MATH the above expression is equal to MATH . Hence MATH . MATH is the linking number between the components labeled by MATH and MATH in . Hence, it is equal to MATH where MATH for MATH denotes the components labeled by MATH in . Smoothing crossings in MATH and in MATH does not change MATH . Therefore, the above expression is equal to MATH where MATH denotes the sublink composed of components labeled by MATH in . Hence, we have proved that MATH . By applying this equation to REF we get REF : MATH . We prove identities REF , and REF in the same way. |
math/0011148 | Let MATH be an an algebra homomorphism given by MATH where MATH is an arbitrary framed unoriented knot which represents the conjugacy class of MATH . By REF , MATH is well defied for each MATH and MATH . We also have MATH since MATH . Therefore, in order to show that MATH can be factored to a homomorphism MATH we need to prove that MATH . Let MATH and MATH be two disjoint oriented framed knots representing MATH . We may assume (by isotoping MATH if necessary) that MATH passes very close to MATH . Let MATH represent knots, which as in REF , are obtained by replacing in MATH by and respectively. Now REF takes the form MATH . By REF this equation reduces to the skein equation: MATH . Therefore, we proved that MATH induces a MATH-algebra homomorphism MATH . |
math/0011148 | CASE: By substituting MATH to REF , we get MATH and, therefore, MATH . CASE: By REF , MATH . CASE: Observe that MATH and MATH . Since MATH is skew-symmetric, we get MATH . We use here the fact that MATH and, therefore, MATH . |
math/0011148 | We need to show that if MATH is a permutation on MATH symbols then MATH . Since each permutation is a product of transpositions, it is enough to assume that MATH is itself a transposition, MATH . Notice that MATH . Observe also, that the sums MATH differ only by one term, that is, the first sum contains MATH and the second sum contains MATH . Therefore the above equality reduces to MATH which follows from REF . |
math/0011148 | We need to show that REF belong to the kernel of MATH . Let us look at REF first: if a link MATH is unlinked with then MATH since MATH and MATH by REF . Hence, skein REF lies in MATH . Now, we will show that REF also belongs to MATH by analyzing two cases, depending on how the ends of the tangle are connected in the ambient space: CASE: Suppose that in REF represents a link of MATH components, MATH whose the first two components are depicted in the diagram below: . Let MATH and MATH be as in REF , , . Then we need to show that MATH . The above equation, can be rewritten as MATH . By REF , MATH for MATH . Therefore the above equation simplifies to MATH . By REF , the above equation reduces to REF . This completes the proof of REF . CASE: The second possibility is that in REF represents a link of MATH components, MATH whose the first component is depicted in the diagram below: . Denote and by MATH and MATH respectively. By the method used for proving REF , we prove that MATH . We use identities REF in the proof. |
math/0011148 | Since MATH is spanned by knots, it is enough to show that MATH is equal to MATH for any knots MATH . We have MATH . |
math/0011152 | As we have noted before, there exists a reflexive NAME space MATH such that MATH is a subsemigroup of MATH. We can therefore suppose that MATH. We have MATH CITE. Since MATH is a topological group, so is MATH, that is, MATH with the topology it inherits as a subspace of MATH. To see that the action of MATH on MATH is (jointly) continuous, it suffices to prove that the action of MATH on the dual NAME space MATH is continuous. This fact has been proved in CITE, and easily follows also by observing that the topological groups MATH and MATH are naturally isomorphic. |
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