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math/0004117
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Suppose in the above construction of the singular REF-cocycle MATH and the singular REF-cochain MATH that we had been unable to define a set map MATH which was a section of MATH and which satisfied REF . We rewrite REF as MATH where MATH denotes the projection from MATH to MATH (compare with REF ) and where MATH is defined by MATH . If we can show that MATH in MATH then we will have MATH for some MATH and hence MATH. Also it is easy to see that we must have MATH in MATH. We see that MATH as follows. Recall that MATH. Hence we get MATH where MATH is defined by MATH . It is a tedious but straightforward calculation to verify, using REF , that MATH and we will omit it. Notice that we could have considered a simplicial line bundle REF on an arbitrary simplicial manifold, the constructions above will still work but we will replace the MATH by the face operators MATH for whatever simplicial manifold we are working with. Of course we will not be able to use REF , as that only applies for the special simplicial manifold MATH with MATH. So it follows that we could consider the central extension MATH as defining a simplicial line bundle (or simplicial MATH bundle rather) on the simplicial manifold MATH as per REF and constructed MATH representing the NAME class of MATH in MATH, MATH such that MATH and MATH such that MATH and MATH. (Thus the triple MATH represents a class in MATH - the cohomology of the total complex of the double complex MATH of REF - and hence a class in MATH). We have the natural map MATH defined by MATH for MATH and MATH in the same fibre of MATH. In fact MATH gives rise to a simplicial map MATH where MATH is the simplicial manifold with MATH, a portion of which is pictured in the following diagram: MATH . By pulling back MATH, MATH and MATH to MATH, MATH and MATH respectively by this simplicial map and using the conditions MATH, MATH and MATH it follows from REF that we can adjust MATH by a coboundary MATH so that MATH satisfies MATH in MATH. It follows that there exists MATH such that MATH and hence there exists MATH with MATH and MATH. We need to show that MATH is the transgression of MATH. Recall from CITE that the transgression MATH of MATH is characterised by the property that there exists MATH such that MATH, where MATH is a representative of MATH, and MATH, where MATH is the injection of the fibre MATH into MATH and where MATH denotes the restriction of MATH to the fibre MATH. So we need to show that MATH. The injection MATH is defined by choosing a point MATH of MATH and mapping MATH to MATH (this needs MATH connected). If we consider MATH then MATH gives us MATH. Thus MATH is the transgression of MATH. Hence from REF we have that the transgression and the NAME class are equal.
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math/0004117
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To show that this does indeed define a gerbe we first of all need to show that the assignment MATH defines a stack in groupoids on MATH - that is the gluing laws for objects and arrows are satisfied - and that MATH is locally non-empty and locally connected. Note that by REF is a groupoid. Also MATH defines a presheaf of categories by definition of a morphism of gerbes. Explicitly if MATH are open subsets of MATH then there are natural restriction functors MATH obtained by restricting a morphism MATH to MATH and similarly for natural transformations. We have to show that in fact MATH defines a sheaf of categories. Suppose then that we are given an open set MATH together with an open cover MATH of MATH by open subsets MATH of MATH such that there exist objects (that is, morphisms of gerbes) MATH of MATH for every MATH and arrows (that is, natural transformations) MATH in MATH which satisfy the gluing condition MATH in MATH. We need to construct a morphism of gerbes MATH which is isomorphic to MATH when restricted to MATH. Let MATH and suppose MATH is an object of MATH. Consider the objects MATH of MATH. The natural transformations MATH provide arrows MATH which are descent data in the gerbe MATH because of the condition MATH. Let the descended object of MATH be MATH say. We put MATH. In a similar way one can define the action of MATH on arrows of MATH and it follows from the uniqueness of gluing in sheaves that this defines a functor. It is also not hard to see that that this functor is compatible with the restriction functors in MATH and MATH and induces the identity morphism on the bands of the gerbes MATH and MATH. Finally, note that since MATH is the descended object for the descent data MATH and MATH there are isomorphisms MATH compatible with the MATH. These isomorphisms induce natural transformations, also denoted MATH from MATH to MATH. We now have to show that arrows glue properly. Suppose we are given morphisms of gerbes MATH, that is, objects of MATH, and suppose also that we are given an open cover MATH of MATH by open subsets MATH of MATH such that there exist natural transformations MATH for each MATH such that MATH. Let MATH be an object of MATH and let MATH. The MATH provide arrows MATH in MATH. Since MATH and MATH are morphisms of gerbes we can think of the MATH as arrows MATH. We have MATH. Since MATH is a gerbe and hence a sheaf of groupoids, there is a unique arrow MATH such that MATH. The assignment MATH defines a natural transformation. Hence MATH is a sheaf of groupoids. We will omit the proof that MATH is locally non-empty and locally connected and refer instead to CITE where it is also shown that MATH has band equal to MATH. To show that there is an equivalence of gerbes between MATH and MATH, we calculate the NAME class of MATH. Choose an open cover MATH of MATH such that there exist morphisms MATH where MATH and MATH denote the restriction of MATH and MATH to MATH respectively. We can choose the open cover MATH so that there exist natural isomorphisms MATH in MATH. Then the NAME class of MATH will be given by choosing an object MATH and forming the cocycle MATH. We want to relate this cocycle to the cocycles representing the NAME classes of MATH and MATH. To do this we first need to choose objects MATH for each MATH. Again we can assume that there exist arrows MATH in MATH. Since MATH is a natural transformation we get the commutative diagram MATH . This induces the following diagram of arrows in MATH (We have omitted some restriction functors for convenience of notation). MATH . The diagonal arrow is a cocycle representing the NAME class of the gerbe MATH, while the bottom horizontal arrow is a cocycle representing the class of MATH. Since the left vertical arrow represents the NAME class of MATH, we must have MATH.
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math/0004117
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Recall that the stable morphism MATH corresponds to the trivialisation MATH of MATH which is obtained by descending the MATH bundle MATH on MATH to MATH. In particular, this implies that the following isomorphism is true on MATH: MATH . The two trivialisations MATH and MATH of MATH differ by the pullback to MATH of a MATH bundle MATH on MATH. So we have MATH . MATH is then a section of the bundle MATH on MATH. From REF we have MATH where we have denoted the various projections to MATH by MATH. Using REF again we get MATH where again we have denoted the projections to MATH by MATH. The transformation MATH provides an isomorphism MATH which is compatible with the trivialisations MATH and MATH. It follows from the series of equations above and the property of MATH just mentioned that there is an isomorphism MATH over MATH. The cocycle condition on MATH is that the section MATH of MATH is mapped to the section MATH of MATH under the this isomorphism. We now define the bundle gerbe MATH as follows. We put MATH equal to the disjoint union MATH with the obvious projection MATH. The fibre product MATH is equal to the disjoint union MATH. Define a MATH bundle MATH on MATH by putting MATH. We now need to define a bundle gerbe product on MATH. This will be a MATH bundle isomorphism MATH covering the identity on MATH which satisfies the associativity condition. Note that from REF we have the isomorphism MATH of MATH bundles on MATH. If we embed MATH inside MATH by sending MATH to MATH, then using the section MATH of MATH and the identity section of MATH over the diagonal, we can construct an isomorphism MATH. The coherency condition on MATH shows that this product is associative and hence MATH is a bundle gerbe. We now need to show that there is a stable morphism MATH, that is, a trivialisation MATH of MATH over MATH. The MATH bundle MATH lives over MATH. Let MATH be the disjoint union of the MATH bundles MATH. Since MATH the isomorphism MATH becomes MATH which shows that MATH trivialises MATH. Finally we have to define transformations MATH, compatible with MATH in the appropriate sense. We have two trivialisations MATH and MATH of MATH and hence there is a MATH bundle MATH on MATH such that MATH where MATH denotes the projection MATH. Therefore we have MATH, an isomorphism of MATH bundles on MATH. On each constituent MATH of the above disjoint union this isomorphism takes the form MATH . From REF we have MATH . This shows that there is an isomorphism MATH on MATH. Therefore MATH induces a section MATH of MATH which descends to a section MATH of MATH. Because of the coherency condition satisfied by MATH, it follows that MATH will be compatible with MATH.
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math/0004117
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Choose an open cover MATH of MATH such that there exist local sections MATH of MATH. Form the pullback bundle gerbes MATH. The stable morphism MATH induces a stable morphism MATH by pullback: MATH, and the transformation MATH induces a transformation MATH by pullback as well: MATH. The transformation MATH satisfies the condition MATH . Therefore we can apply REF to find a bundle gerbe MATH on MATH and a stable morphism MATH and transformations MATH which are compatible with the transformations MATH (the fact that the stable morphisms MATH go from MATH to MATH rather than from MATH to MATH is of no consequence). We need to construct a stable morphism MATH. Let MATH denote the map MATH. MATH induces a map MATH by MATH. Therefore we can define a stable morphism MATH by composition: MATH . Next we define transformations MATH by composition as indicated in the following diagram: MATH that is, MATH. Here MATH is the map MATH. Because of the compatibility of MATH with MATH we have MATH. Hence we can glue the stable morphisms MATH together to form a stable morphism MATH. We now need to define the transformations MATH and prove they are compatible with MATH. It is sufficient to define transformations MATH which are compatible with the gluing transformations MATH. MATH is induced from MATH by pullback with the map MATH which sends MATH to MATH where MATH. We try to indicate this in the following diagram: MATH . One can check that the MATH are compatible with the MATH and also that the glued together transformation MATH is compatible with MATH.
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math/0004117
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We first need to define the bundle gerbe morphism MATH which maps MATH . Define MATH covering the identity on MATH by sending MATH to the path MATH given by MATH . Next, we need to define a MATH equivariant map MATH covering MATH and check that it commutes with the bundle gerbe product. So take pairs MATH and MATH where MATH and MATH and MATH are homotopies with endpoints fixed between paths MATH and MATH respectively. Then we put MATH where MATH is the homotopy with endpoints fixed between MATH and MATH given by MATH . We need to check firstly that this map is well defined - that is it respects the equivalence relation MATH - and secondly that MATH commutes with the bundle gerbe products. So suppose MATH and MATH, where MATH and MATH are homotopies with endpoints fixed between paths MATH and MATH and where MATH and MATH are homotopies with endpoints fixed between paths MATH and MATH. We want to show that MATH . Therefore we want to show that for all homotopies MATH with endpoints fixed between MATH and MATH we have MATH . Note that if MATH is a homotopy with endpoints fixed between MATH and MATH and MATH is a homotopy with endpoints fixed between MATH and MATH, then by integrality of MATH we have MATH . Therefore we are reduced to showing that MATH . We have MATH . By the bi-invariantness of MATH, we get MATH, hence MATH which implies the result. Hence MATH is well defined. It is a straightforward matter to verify that MATH respects the bundle gerbe products. It remains to show that there is a transformation of the bundle gerbe morphisms MATH and MATH over MATH which satisfies the compatibility criterion over MATH. This has already been done above for the tautological bundle REF-gerbe and the proof given there carries over to this case.
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math/0004117
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Because MATH commutes with the bundle gerbe products on MATH and MATH, MATH commutes with the bundle gerbe products on MATH and MATH. Since MATH is a bundle gerbe connection, so is MATH and therefore MATH and MATH are bundle gerbe connections on MATH. Since any two connections differ by the pullback of a complex valued one form on the base, we have MATH, for some MATH. Because MATH and MATH are both bundle gerbe connections, we must have MATH and so MATH for some MATH.
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math/0004117
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Define a map MATH by sending the point MATH of MATH to the point MATH of MATH. The bundle gerbe product on MATH gives the following isomorphisms of line bundles with connection over MATH. MATH . Here the line bundle MATH is equipped with the connection MATH, while the line bundles MATH and MATH are endowed with the connections MATH and MATH respectively. Hence we can combine the isomorphisms REF to get an isomorphism of line bundles MATH . Note that this is an isomorphism of line bundles with connection. We have the following sequence of isomorphisms of line bundles MATH . The descent isomorphism MATH of REF is the composition of these line bundle isomorphisms. Here the isomorphism REF is induced by the canonical trivialisation of MATH, the isomorphism REF is induced by the isomorphisms MATH and MATH and the isomorphism REF is induced by the bundle gerbe product MATH (in other words by the isomorphisms REF). From this, and REF we see that the difference MATH is equal to MATH.
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math/0004117
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The isomorphism MATH gives rise to an isomorphism over MATH: MATH . Also by definition we have an isomorphism of line bundles with connection MATH where each bundle is equipped with the obvious pullback connection. Next, the diagram MATH commutes. This is really the definition of MATH. Therefore we will have MATH .
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math/0004117
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Same as for REF .
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math/0004117
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Suppose we are given a bundle REF-gerbe MATH. Let MATH be a bundle gerbe connection on MATH with curving MATH such that MATH, where MATH is the curvature of MATH with respect to MATH. The three curvature MATH then satisfies MATH. Let us look at what happens over MATH. We have induced connections MATH on the bundle gerbes MATH. Let MATH denote the curving for MATH induced by MATH. Let MATH denote the tensor product bundle gerbe connection on MATH. The bundle gerbe morphism MATH gives rise to a MATH-bundle isomorphism MATH clearly commutes with the respective bundle gerbe products. By REF there is a one form MATH such that MATH . Hence we get MATH and hence MATH . Therefore, there exists MATH such that MATH where MATH denotes the projection MATH. It is easily checked that MATH . Now let us look at the situation over MATH. We will use the notation introduced in REF so that we will denote MATH by MATH, MATH by MATH, MATH by MATH and so on. As mentioned in REF, the bundle gerbe morphisms MATH and MATH are defined as in the following diagram: MATH . So MATH and MATH. Recall from REF that MATH is the line bundle on MATH which descends to the associator line bundle MATH on MATH. Furthermore MATH has a section MATH which descends to a section MATH of MATH - the associator section. MATH inherits the pullback connection MATH (here MATH denotes the induced connection on MATH) relative to which MATH has curvature equal to MATH . Now MATH . Similarly we have MATH . Hence we have MATH . On the other hand we can write MATH for some one form MATH on MATH. Thus we must have MATH where MATH is the one form MATH . Let MATH denote the descent isomorphism of REF . By Lemma's REF we have MATH . Therefore the connection MATH on MATH descends to a connection MATH on MATH. Let MATH for some MATH. Then we have MATH. Hence we get MATH . If we can show MATH then we are done. To do this we need to examine the situation over MATH. MATH induces five bundle gerbe morphisms MATH for MATH. We can form five MATH bundles MATH, MATH,, MATH on MATH, all of which descend to MATH bundles MATH,MATH,, MATH on MATH. Since each pair of maps MATH has MATH factoring through MATH and MATH factoring through MATH or vice versa, we can identify MATH with MATH, MATH with MATH, MATH with MATH, MATH with MATH and MATH with MATH as explained in REF. Also note that the associator section MATH induces sections MATH of MATH. Note that the bundle gerbe product in MATH supplies an isomorphism MATH and the coherency condition on MATH is that MATH is mapped to the identity section of MATH. Let MATH denote the bundle gerbe connection on MATH induced by MATH. Then MATH induces a connection MATH on the (line) bundle MATH by pullback: MATH. Since MATH is a bundle gerbe connection, under the isomorphism REF above, the connection MATH on MATH is mapped to the connection MATH on MATH. Hence we have that MATH . We now need to calculate MATH. Using REF - REF one finds MATH . Here we have abused notation and denoted by MATH the projection MATH. If one does the calculation then one finds that MATH and so we conclude that MATH, completing the proof.
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math/0004117
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MATH induces an isomorphism MATH as pictured in the following diagram: MATH . This diagram lives over MATH. Clearly over MATH we get the following diagram, induced from the one above by tensor product and pullback, MATH . Let MATH and MATH denote the sections of MATH and MATH induced by the bundle gerbe products respectively. In the second diagram above, the isomorphism MATH maps MATH to MATH, since this isomorphism is induced by a bundle gerbe morphism. Similarly, the vertical isomorphism MATH maps MATH to MATH and the vertical isomorphism MATH maps MATH to MATH. Here MATH and MATH are the sections induced by the canonical trivialisations of MATH and MATH respectively. It follows now that if we define a section MATH of MATH using the isomorphism MATH from the first diagram above, then MATH satisfies MATH and hence MATH descends to MATH.
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math/0004117
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All that one really needs to do is to check that MATH does indeed define a cocycle. This follows from the coherency condition on MATH. One can also check that another choice of the sections MATH, or the line bundles MATH, or the sections MATH changes MATH by a coboundary.
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math/0004117
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We have MATH. But MATH and so MATH. Hence we get MATH. Also we have MATH . Therefore we can construct a section MATH of MATH by taking the section MATH of MATH and choosing sections MATH of MATH and forming the tensor product section MATH. From here it is not too hard to show that the cocycle MATH is cohomologous to MATH.
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math/0004117
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We calculate the NAME four class of the bundle REF-gerbe MATH and get it exactly equal to the NAME cocycle defined by NAME and NAME in CITE and CITE. We can then apply REF to conclude that this NAME four class is MATH. We set out to calculate the NAME cocycle MATH. First choose an open cover MATH of MATH relative to which MATH has local sections MATH. Since MATH is a fibration, we can calculate the NAME cocycle representing the four class of MATH using the second construction of a NAME cocycle as detailed in REF. Recall that we first choose sections MATH. This is equivalent to choosing maps MATH such that MATH and MATH. Next we have to choose sections MATH. This amounts to choosing maps MATH such that MATH, MATH, MATH and MATH (This uses the fact that MATH and the cocycle condition MATH satisfied by the transition functions MATH). Define the section MATH of the bundle MATH by MATH. If we denote a homotopy with end points fixed MATH from a path MATH to a path MATH by a REF-arrow MATH then we see that MATH amounts to composing horizontally REF-arrows in the following diagram MATH . The bundle gerbe product MATH is obtained by composing REF-arrows MATH and MATH vertically. In a similar manner, one constructs the section MATH of the bundle MATH. We make MATH into a section of MATH by using the associator section: MATH. Finally we define the cocycle MATH by MATH. We can get an explicit formula for MATH as follows: we choose a homotopy with endpoints fixed MATH such that MATH, MATH, MATH, MATH, MATH and MATH and we set MATH. This is just the integral of MATH over the tetrahedron described in CITE and CITE. Thus our cocycle must agree with that defined by NAME and NAME.
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math/0004117
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We need to show that MATH is associative, that is MATH. We have MATH where we have denoted composition by bundle gerbe product by juxtaposition and composition via the bundle REF-gerbe product with MATH. We also have MATH . We have to show these two expressions are the same. For this we will need the following lemma. Under the hypotheses of the above proposition, we have MATH . To prove this one needs to use the associativity isomorphism to regroup brackets and use the coherency condition on MATH. The first of the above two expressions is equal to MATH . This proves that MATH is associative and hence the triple MATH is a bundle gerbe.
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math/0004117
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Let MATH be a `nice' open cover of MATH. So each non-empty finite intersection MATH is contractible and there exist local sections MATH of MATH. Define maps MATH by MATH. Pullback the bundle gerbe MATH with this map to obtain a new bundle gerbe MATH. So we get a manifold MATH and an induced projection map MATH. If MATH then MATH. Since MATH is contractible the bundle gerbe MATH is trivial. So let MATH be a MATH bundle trivialising MATH. Let MATH be a bundle gerbe trivialising the bundle REF-gerbe MATH. Pullback the bundle gerbe MATH with the sections MATH to get a new bundle gerbe MATH on the contractible open set MATH. Let MATH be a MATH bundle trivialising the bundle gerbe MATH. The bundle gerbe morphism MATH induces a bundle gerbe morphism MATH over MATH. Consider the MATH bundle MATH on MATH. Since the bundles MATH, MATH and MATH trivialise the bundle gerbes MATH, MATH and MATH respectively, and because MATH is a bundle gerbe morphism and hence commutes with the bundle gerbe products on MATH and MATH, it follows that there is a section MATH of MATH over MATH satisfying the coherency condition MATH. Hence the MATH bundle MATH descends to a bundle MATH on MATH. From the construction of the NAME cocycle MATH associated to the bundle REF-gerbe MATH given in REF we have that the MATH bundle MATH descends to a MATH bundle MATH on MATH and that furthermore there is an isomorphism MATH over MATH, where MATH. Also the NAME REF-cocycle MATH was defined by choosing sections MATH of MATH and then comparing the section MATH with the image of the section MATH under the above isomorphism. It is not difficult to show that there is an isomorphism MATH of MATH bundles over MATH, where MATH denotes the pullback of the bundle MATH on MATH via the map MATH defined by sending a point MATH of MATH to MATH. From here we see that if we choose sections MATH of MATH then we can define a section MATH of MATH by MATH, where MATH. Comparing MATH with MATH and using the coherency condition MATH, we see that the cocycle MATH must be trivial.
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math/0004117
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We first prove that the sheaf of bicategories MATH is locally non-empty. Choose an open cover MATH of MATH such that there exist local sections MATH of MATH. Form the maps MATH sending MATH to MATH. Let MATH. So MATH is a bundle gerbe on MATH. Now pullback the stable morphism MATH using the map MATH, MATH where MATH. Let MATH denote this pullback stable morphism MATH. To construct a transformation MATH we pullback the transformation MATH using the map MATH, MATH where again MATH. The coherency condition MATH satisfied by MATH ensures that MATH. Therefore for each MATH we have constructed objects MATH of MATH. Therefore MATH is locally non-empty. We now show that MATH is locally connected. Let MATH and MATH be objects of MATH. So MATH and MATH are bundle gerbes MATH and MATH. We need to show that at least locally on MATH these two objects are connected by a REF-arrow. We have stable morphisms MATH and MATH corresponding to trivialisations MATH and MATH of the bundle gerbes MATH and MATH respectively. Form the bundle gerbe MATH. This is trivialised by the MATH bundle MATH which lives over MATH, where MATH . Consider the MATH bundle MATH which lives over MATH. One can show that there is a section MATH of MATH satisfying the coherency condition MATH. MATH corresponds to the isomorphism given fibrewise by composition MATH . One can show this isomorphism satisfies a coherency condition given a third point MATH of MATH lying in the same fibre as the points MATH and MATH. Therefore by REF MATH descends to a MATH bundle MATH on MATH and it is not difficult to show that MATH trivialises the bundle gerbe MATH over MATH. Therefore the trivialisation MATH corresponds to a stable morphism MATH. One can also show that there is a transformation MATH satisfying the coherency condition MATH. We will omit the proof of this as the calculation gets rather messy. The calculation exploits the fact that the transformations MATH and MATH satisfy MATH and MATH and so MATH in a sense one has to make precise. Therefore we can apply REF to conclude there is a bundle gerbe MATH on MATH and a stable morphism MATH over MATH corresponding to a trivialisation MATH. There is also a transformation MATH compatible with MATH. By shrinking MATH if necessary, we may assume that the bundle gerbe MATH is trivial - say MATH. Therefore MATH is trivial as well. In fact, if one considers the MATH bundle MATH on MATH then one can show that there is a coherent section MATH of MATH and hence MATH descends to a MATH bundle MATH on MATH. One can also show that there is a trivialisation MATH. So we have a stable morphism MATH. As well, one can show that the transformations MATH give rise to transformations MATH and which satisfy the compatibility condition with MATH and MATH. Therefore MATH is a REF-arrow. So MATH is locally connected. Since the bicategory of bundle gerbes and stable morphisms is a bigroupoid REF , all REF-arrows of MATH are coherently invertible and all REF-arrows of MATH are invertible. We now have to show that REF-gerbe MATH is bound by the sheaf of abelian groups MATH. For this, we first need to show that any two REF-arrows can be locally connected by a REF-arrow. So let MATH and MATH be objects of MATH joined by REF MATH. So MATH and MATH correspond to trivialisations MATH and MATH respectively of MATH. Therefore there is a MATH bundle MATH on MATH such that MATH. It is easy to check that MATH and MATH provide a section MATH of the bundle MATH over MATH. The coherency conditions satisfied by MATH and MATH ensure that MATH and hence by REF the bundle MATH descends to a MATH bundle MATH on MATH. By shrinking MATH if necessary we may assume that MATH is trivial. Hence there is a section of MATH which lifts to a section MATH of MATH which is compatible with MATH. In other words, MATH is a REF-arrow MATH. In a similar way one can show that for a given REF-arrow MATH the sheaf of REF MATH is isomorphic to MATH and that this isomorphism is natural with respect to composition of REF. Therefore MATH is a REF-gerbe on MATH bound by MATH. If one calculates the NAME cocycle MATH representing the class of MATH in MATH then it is not hard to show that MATH represents the same class as the four class of the stable bundle REF-gerbe MATH.
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math/0004117
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We have already seen REF that a trivial stable bundle REF-gerbe has zero class in MATH. Conversely, given a stable bundle REF-gerbe with zero class in MATH, the associated REF-gerbe also has zero class in MATH and hence has a global object. This global object trivialises the stable bundle REF-gerbe MATH.
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math/0004119
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REF is obvious (use induction). To prove REF , suppose that MATH are finite metric spaces, MATH, and let MATH be an isometric embedding. Pick an isometric embedding MATH, and use MATH-homogeneity of MATH to find an isometry MATH of MATH such that MATH extends the isometry MATH. Then MATH is an isometric embedding that extends MATH. For REF , enumerate dense countable subsets in MATH and MATH and use the ``back-and-forth" (or ``shuttle") method to extend the given isometry between MATH and MATH to an isometry between dense subsets of MATH and MATH. Then use completeness to obtain an isometry between MATH and MATH. Applying REF in the case when MATH, we see that every complete separable NAME space is MATH-homogeneous. Thus REF and uniqueness in REF follow from REF - REF . The existence of MATH is a special case of REF that we shall prove later; the idea of the proof is due to NAME. The existence of a non-complete NAME space easily follows from NAME 's methods presented in this paper; we refer the reader to CITE for details.
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math/0004119
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There exists a family MATH of continuous left-invariant pseudometrics on MATH which determines the topology of MATH and has the cardinality MATH. Replacing, if necessary, each MATH by MATH, we may assume that all pseudometrics in MATH are bounded by REF. For every MATH let MATH be the metric space associated with the pseudometric space MATH, and let MATH be the disjoint sum of the spaces MATH. There is an obvious metric on MATH which extends the metric of each MATH: if two points of MATH are in distinct pieces MATH and MATH, define the distance between them to be REF. The left action of MATH on itself yields for every MATH a natural continuous homomorphism MATH. The homomorphism MATH thus obtained is a homeomorphic embedding. It remains to note that the group MATH can be identified with a topological subgroup of MATH.
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math/0004119
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Let MATH be the NAME space of all bounded real functions on MATH which are uniformly continuous with respect to the right uniformity. The natural left action of MATH on MATH, defined by the formula MATH, yields an isomorphic embedding of MATH into MATH. The weight of MATH may exceed the weight of MATH, but it is easy to find a MATH-invariant subspace MATH of MATH such that MATH determines the topology of MATH and MATH. Then the natural homomorphism MATH still is a homeomorphic embedding.
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math/0004119
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It suffices to show that for every MATH the map MATH from MATH to MATH is continuous at the unity. If MATH, then MATH, and we have MATH. Let MATH be given. If MATH is close enough to the unity, then MATH, MATH, and therefore MATH.
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math/0004119
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Since MATH is a NAME function, for every MATH we have MATH and MATH. Hence MATH, and at MATH the equality is attained.
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math/0004119
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We may assume that MATH is a subspace of MATH and that MATH for every MATH. Let MATH be the NAME function defined by MATH for every MATH, where MATH is the metric on MATH. Let MATH. We claim that the extension of MATH over MATH which maps MATH to MATH is an isometric embedding. It suffices to check that MATH for every MATH. Fix MATH. Let MATH be the restriction of MATH to MATH. According to REF we have MATH. Since MATH, MATH and the map MATH is distance-preserving, it follows that MATH, as claimed.
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math/0004119
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It is clear that every isometry MATH between any two metric spaces can be extended to an isometry MATH. Such an extension is unique, since every point in MATH (or, more generally, in MATH) is uniquely determined by its distances from the points of MATH REF , and similarly for MATH. In particular, every isometry MATH uniquely extends to an isometry MATH. The map MATH is a homomorphism of groups. We show that this homomorphism is continuous. Fix MATH and MATH. Pick a finite subset MATH of MATH and MATH so that MATH. Let MATH be the set of all MATH such that MATH for every MATH. Then MATH is a neighbourhood of unity in MATH. It suffices to show that MATH for every MATH. Fix MATH. Let MATH. Then MATH. Thus for every MATH we have MATH . Since MATH it follows that MATH whence MATH.
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math/0004119
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For every MATH pick a continuous homomorphism MATH such that MATH extends MATH for every MATH. By transfinite recursion on MATH define a homomorphism MATH such that MATH extends MATH for every MATH and MATH. Let MATH be the identity map of MATH. If MATH, put MATH. If MATH is a limit ordinal, let MATH be the isometry of MATH such that for every MATH its restriction to MATH is equal to MATH. We prove by induction on MATH that each homomorphism MATH is continuous. This is obvious for non-limit ordinals. Assume that MATH is limit. To prove that MATH is continuous, it suffices to show that for every MATH the map MATH from MATH to MATH is continuous. Fix MATH. Pick MATH so that MATH. Then MATH for every MATH. The map MATH is continuous by the assumption of induction, hence the map MATH also is continuous. Thus MATH is a continuous homomorphism such that MATH extends MATH for every MATH. This means that MATH is MATH-embedded in MATH.
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math/0004119
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Let MATH be a finite metric space of diameter MATH, and let MATH be an isometric embedding. Pick MATH so that MATH. In virtue of REF , there exists an isometric embedding of MATH into MATH which extends MATH. This means that MATH is NAME. The second assertion of the proposition follows from REF .
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math/0004119
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It suffices to show that for every metric space MATH the weight of MATH is equal to the weight of MATH. Let MATH, and let MATH be a dense subset of MATH of cardinality MATH. Let MATH. Then MATH is a family of separable subspaces of MATH, MATH and MATH is dense in MATH (see the proof of REF). Hence MATH has a dense subspace of cardinality MATH.
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math/0004119
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Let MATH be a metric space, MATH be its completion. Every isometry MATH uniquely extends to an isometry MATH. We show that the homomorphism MATH is continuous. Let MATH be the metric on MATH. Fix MATH and MATH. Pick MATH so that MATH. Let MATH. Then MATH is a neighbourhood of unity in MATH. If MATH, then MATH. This implies the continuity of the homomorphism MATH.
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math/0004119
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Let MATH be a complete metric space containing a dense NAME subspace MATH. We must prove that MATH is NAME. Let MATH be a finite subset of MATH, and let MATH be a NAME function. It suffices to prove that there exists MATH such that MATH for every MATH. Pick a sequence MATH such that: CASE: if MATH and MATH, MATH, then the series MATH converges; CASE: every MATH is a cluster point of the sequence MATH. To construct such a sequence, enumerate MATH as MATH, and for every MATH and MATH (MATH, MATH) pick a point MATH such that MATH. Then MATH for every MATH and MATH, and the sequence MATH has the required properties. Let MATH. We construct by induction a sequence MATH of points of MATH such that: CASE: if MATH, then MATH; CASE: MATH for every MATH. Pick MATH so that MATH. This is possible since MATH is NAME. Suppose that the points MATH have been constructed so that REF are satisfied. Consider two NAME functions MATH and MATH on the set MATH: let MATH for every MATH, and let MATH. By REF , the functions MATH and MATH coincide on MATH, hence the distance between them in the space MATH is equal to MATH . Let MATH be the metric space MATH, considered as a subspace of MATH. In virtue of REF , the map of MATH to MATH which leaves each point of MATH fixed and sends MATH to MATH is an isometric embedding. Since MATH is NAME, this map can be extended to an isometric embedding of MATH to MATH. Let MATH be the image of MATH. Then MATH. In virtue of REF , for every MATH we have MATH. Thus REF are satisfied, and the construction is complete. Since the series MATH converges, it follows from REF that the sequence MATH is NAME and hence has a limit in the complete space MATH. Let MATH. By REF , we have MATH for every MATH. Since MATH is contained in the closure of the set MATH, it follows that MATH for every MATH.
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math/0004119
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Let MATH be a metric space of diameter MATH, and let MATH be the NAME extension of MATH constructed above. Consider the completion MATH of MATH. REF implies that MATH is NAME. REF shows that MATH. Finally, MATH is MATH-embedded in MATH REF and MATH is MATH-embedded in MATH REF , so MATH is MATH-embedded in MATH. Thus MATH has the properties required by REF .
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math/0004119
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Let MATH be a metric space. We first construct a MATH-embedding of MATH into a metric space MATH such that MATH and every isometry between finite subsets of MATH extends to an isometry of MATH. For every isometry MATH between finite non-empty subsets of MATH consider the graph MATH of MATH, and let MATH be the set of all such graphs. Thus a non-empty finite subset MATH is an element of MATH iff for any two pairs MATH we have MATH. Equip MATH with the metric MATH defined by MATH, and let MATH be the corresponding NAME metric on the set of finite subsets of MATH. If MATH and MATH are two non-empty finite subsets of MATH and MATH, then MATH iff for every MATH there exists MATH such that MATH, and for every MATH there exists MATH such that MATH. Let MATH be the non-expanding function on MATH defined by MATH. Let MATH be the free group on MATH, equipped with the NAME pseudometric MATH. To avoid confusion of multiplication in MATH with composition of relations, we assign to each MATH a symbol MATH, and consider elements of MATH as irreducible words of the form MATH, where MATH. Similarly, we consider elements of the semigroup MATH as words of the same form. Let MATH be the diagonal of MATH. The set MATH is a symmetrical subsemigroup of the semigroup of all relations on MATH. Let MATH be the map defined by the following rule: if MATH is a non-empty irreducible word, then MATH. If MATH, then MATH iff there exists a chain MATH of points of MATH such that for every MATH we have either MATH and MATH or MATH and MATH. For the empty word MATH we put MATH. Note that the definition of MATH makes sense also without the assumption that the word MATH be irreducible, so we can assume that MATH is defined on the set MATH of all words of the form MATH. Recall that MATH denotes the word obtained by writing MATH after MATH (without cancelations). We have MATH. If MATH and MATH is the irreducible word equivalent to MATH, then MATH. It suffices to prove that MATH if MATH is obtained from MATH by canceling one pair of letters. Let MATH and MATH. Since MATH is a functional relation, we have MATH and hence MATH. For every MATH we have MATH. We claim that MATH for every MATH. Indeed, the product MATH is the irreducible word equivalent to MATH, therefore MATH by REF . For every MATH let MATH be the set of all MATH such that MATH. We claim that MATH and MATH for every MATH. This follows from the properties of MATH established in the preceding paragraph. Indeed, pick MATH and MATH. Then MATH and MATH, hence MATH and MATH. This proves the inclusion MATH. The equality MATH is proved similarly. Note that MATH if and only if MATH, since MATH. Note also that MATH if and only if MATH, since MATH. Consider the following equivalence relation MATH on MATH: a pair MATH is equivalent to a pair MATH iff MATH. Since MATH, MATH and MATH for all MATH, the relation MATH is reflexive, symmetric and transitive and thus is indeed an equivalence relation. Let MATH be the quotient set MATH. The group MATH acts on MATH by the rule MATH. The relation MATH is invariant under this action, so there is a uniquely defined left action of MATH on MATH which makes the canonical map MATH into a morphism of MATH-sets. Let MATH be the map which sends each point MATH to the class of the pair MATH. If MATH, then the pairs MATH and MATH are not equivalent, since MATH. The map MATH is therefore injective, and we can consider MATH as a subspace of MATH, identifying MATH with MATH. Every MATH can be written in the form MATH (or simply MATH), where MATH and MATH. Let MATH. The set of all MATH such that MATH is equal to MATH. If MATH is a relation containing the pair MATH, then MATH and hence MATH. It follows that the action of MATH on MATH is transitive. Moreover, for every isometry MATH between finite subsets of MATH there exists MATH such that the self-map MATH of MATH extends MATH. Indeed, if MATH is the graph of MATH, then MATH and hence MATH for every MATH. Thus MATH has the required property. We now define a MATH-invariant pseudometric MATH on MATH which extends the metric MATH on MATH. Let MATH be the NAME seminorm on MATH corresponding to the pseudometric MATH. We have MATH for every MATH. For every MATH let MATH . Then MATH is a pseudometric on MATH. Since the seminorm MATH is invariant under inner automorphisms, the pseudometric MATH is MATH-invariant. Indeed, for MATH and MATH we have MATH. We claim that MATH extends the metric MATH on MATH: MATH for every MATH. Since for MATH the condition MATH is equivalent to MATH, we have MATH. If MATH, then MATH and MATH. It follows that MATH. It remains to prove the opposite inequality, which is equivalent to the following assertion: If MATH and MATH, then MATH. Let MATH. We argue by induction on MATH, the length of MATH. If MATH, then MATH, and we noted that MATH implies MATH. If MATH, then MATH and MATH. Since MATH, the relation MATH contains either MATH or MATH and hence MATH. Assume that MATH. It suffices to show that there exists MATH of length MATH such that MATH. We use the construction of the NAME seminorm MATH described in REF. Let MATH be a MATH-pairing for which MATH is attained. In other words, MATH is a disjoint system of two-element subsets of the set MATH such that for the NAME sum MATH we have MATH. Considering the pair MATH with the least possible value of MATH (``the shortest arc"), we see that at least one of the following three cases must occur: CASE: there exists a MATH such that MATH; CASE: there exists a MATH such that MATH and MATH; CASE: there exists a MATH such that MATH. In REF or REF we replace the subword MATH of MATH by the letter MATH, where MATH. In REF we replace the subword MATH of MATH by the letter MATH, where MATH. In all cases we get a word MATH of length MATH. To justify the usage of the symbol MATH, we must show that MATH, which reduces to the fact that MATH. Had MATH been empty, the same would have been true for MATH. On the other hand, since MATH, we have MATH. Let MATH be the irreducible word equivalent to MATH. The length of MATH is less than MATH. We show that MATH and MATH. By REF we have MATH. Plainly MATH. Since MATH, we have MATH. Thus MATH and MATH, as required. We prove that MATH. As in REF, we define MATH even if the word MATH is reducible, and we have MATH, where MATH runs over the set of all MATH-pairings. The MATH-pairing MATH in an obvious way yields a MATH-pairing MATH, which coincides with MATH outside the changed part of MATH and leaves the new letter MATH unpaired. The NAME sums MATH and MATH differ only by the term MATH in the sum MATH and the terms MATH REF or MATH REF or MATH REF in the sum MATH. According to REF below, we have MATH. Thus MATH. Let MATH. CASE: If MATH and MATH is non-empty, then MATH; CASE: if MATH and MATH is non-empty, then MATH; CASE: if MATH and MATH is non-empty, then MATH. Since MATH and MATH for every MATH, we may assume that MATH. Pick MATH so that MATH. REF follows from REF (take for MATH in REF a sufficiently large finite part of MATH), so let us consider REF . There exist MATH such that MATH, MATH and MATH. Since MATH, there exists a pair MATH such that MATH. The relation MATH, being an element of MATH, is the graph of a partial isometry, so from MATH and MATH it follows that MATH. Note that MATH. Thus we have MATH, as required. REF is easy: there exists a point MATH such that MATH and MATH, hence MATH. We have thus proved that the pseudometric MATH on MATH extends the metric MATH on MATH. Let MATH be the metric space associated with the pseudometric space MATH. The metric space MATH can be naturally identified with a subspace of MATH. We show that MATH is MATH-embedded in MATH. In virtue of the functorial nature of the construction of MATH, every isometry MATH of MATH naturally extends to an isometry MATH of MATH. The map MATH from MATH to MATH is a homomorphism of groups. We claim that this homomorphism is continuous. This follows from the fact that at each step of our construction new spaces are obtained from the old ones via functors ``with finite support": every element of MATH is a finite relation on MATH, and every word MATH involves only finitely many elements of MATH. Given an isometry MATH, the isometry MATH can be obtained step by step in the following way. First we consider the isometry MATH of the metric space MATH corresponding to MATH; the isometry MATH preserves the function MATH on MATH and gives rise to the automorphism MATH of the group MATH which preserves the NAME pseudometric MATH; then we get the isometry MATH of MATH which maps the class of each pair MATH (MATH, MATH) to the class of the pair MATH; finally we get the isometry MATH of MATH. We show step by step that MATH depends continuously on MATH. For MATH this is straightforward: use the fact that MATH consists of finite subsets of MATH. For MATH apply REF with MATH. Let us consider the case MATH. Pick a point MATH (MATH, MATH). It suffices to check that MATH is small if MATH is close to the identity. We have MATH. By the definition of MATH, the first term of the last sum does not exceed MATH and hence is arbitrarily small if MATH is close enough to the identity. The same is true for second term, and we are done. Finally, MATH is the image of MATH under the natural morphism MATH, and the case MATH follows. We have thus proved that MATH is MATH-embedded in MATH. We saw that each isometry between finite subsets of MATH extends to an isometry of MATH and hence also to an isometry of MATH. It is easy to see that MATH. If the diameter MATH of MATH is finite, replace the metric MATH of MATH by MATH. This operation can make the group MATH only larger, and the diameter of MATH becomes equal to that of MATH. To finish the proof of REF , iterate the construction of MATH. We get an increasing chain MATH of metric spaces such that each MATH is MATH-embedded in MATH, every isometry between finite subsets of MATH extends to an isometry of MATH, MATH and MATH, MATH. Consider the space MATH. We have MATH and MATH. In virtue of REF , each MATH is MATH-embedded in MATH. Since every finite subset of MATH is contained in some MATH, it is clear that MATH is MATH-homogeneous.
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math/0004119
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Consider two cases. CASE: MATH is separable. According to REF , there exists a complete separable NAME space MATH such that MATH is a MATH-embedded subspace of MATH. According to REF , MATH is MATH-homogeneous. CASE: MATH is not separable. Let MATH. Applying in turn REF , construct an increasing continuous chain MATH of metric spaces of weight MATH and diameter MATH such that MATH, each MATH is MATH-embedded in MATH (MATH), and MATH is complete NAME for MATH even and MATH-homogeneous for MATH odd. Let MATH. REF implies that each MATH is MATH-embedded in MATH. The space MATH is NAME, being the union of the increasing chain MATH of NAME spaces. For similar reasons the space MATH is MATH-homogeneous. Finally, since every countable subset of MATH is contained in some MATH, MATH, and all spaces MATH are complete, every NAME sequence in MATH converges, which means that MATH is complete. Thus MATH has the properties required by REF .
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math/0004119
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Let MATH be a topological group. According to REF , there exists a metric space MATH such that MATH and MATH is isomorphic to a subgroup of MATH. We may assume that MATH has diameter MATH: otherwise replace the metric MATH by MATH. REF implies that there exists a complete MATH-homogeneous NAME metric space MATH such that MATH and MATH is isomorphic to a subgroup of MATH. Then MATH and MATH is isomorphic to a subgroup of MATH, as required.
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math/0004119
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The condition MATH (respectively, MATH) holds if and only if the function MATH (respectively, MATH) is non-expanding for every MATH. The condition MATH (respectively, MATH) holds if and only if MATH (respectively, MATH) for all MATH.
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math/0004119
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We claim that every non-empty closed subset of MATH has a maximal element. Indeed, if MATH is a non-empty linearly ordered subset of MATH, then MATH has a least upper bound MATH in MATH, and MATH belongs to the closure of MATH. Thus our claim follows from NAME 's lemma. The set MATH is a closed subsemigroup of MATH. Let MATH be a maximal element of MATH. For every MATH we have MATH, whence MATH. It follows that MATH is idempotent and that MATH. Thus MATH is the greatest element of MATH.
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math/0004119
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Let MATH be a closed subset of MATH. It is clear that MATH. If MATH, then MATH for every MATH. Thus MATH is an idempotent. The same is obviously true if MATH. Conversely, let MATH be an idempotent in MATH such that MATH. Let MATH. The function MATH, being non-expanding in each argument, is continuous, hence MATH is closed in MATH. We claim that MATH. We first show that MATH. This is evident if MATH, so assume that MATH. For every MATH we have MATH, since the functions MATH and MATH are non-expanding. It follows that MATH. We prove that MATH. Fix MATH. We must show that MATH. This is evident if MATH, so assume that MATH. Fix MATH so that MATH. Since MATH, for every MATH we have MATH. Hence we can construct by induction a sequence of points MATH in MATH such that MATH . Adding the first MATH inequalities, we get MATH . It follows that the series MATH converges. Since MATH, the series MATH also converges. This implies that the sequence MATH is NAME and hence has a limit in MATH. Let MATH. Since the series MATH converges, we have MATH and hence MATH. Thus MATH. Since MATH is an idempotent, it satisfies the triangle inequality: MATH for all MATH. REF therefore implies that MATH for every MATH. Passing to the limit, we get MATH. Thus MATH. Since MATH was arbitrary, it follows that MATH.
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math/0004119
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We have MATH. Taking MATH, we see that the right side is MATH. On the other hand, for every MATH we have MATH, whence the opposite inequality. The continuity of the left action easily follows from the explicit formula that we have just proved. The argument for the right action is similar.
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math/0004119
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Let MATH be invertible. Let MATH be a MATH-triple corresponding to MATH. This means that MATH is a metric space, MATH and MATH are distance-preserving maps from MATH to MATH, MATH and MATH for all MATH. We saw that elements of MATH correspond to triples MATH satisfying the condition MATH. Thus we must verify this condition. Let MATH be the inverse of MATH. Then MATH. For every MATH we have MATH and hence MATH. This means that MATH belongs to the closure of MATH in MATH. Since MATH is complete and MATH is an isometric embedding, MATH is closed in MATH. It follows that MATH. Since MATH was arbitrary, we have MATH. Similarly, MATH and therefore MATH.
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math/0004119
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The inequality MATH is obvious: if MATH is a pseudometric on MATH extending MATH and MATH and MATH, then at least one of the numbers MATH and MATH must be MATH. To prove the reverse inequality, we construct a pseudometric MATH on MATH extending MATH and MATH such that MATH . The function MATH is defined by these requirements on MATH, MATH, and the set MATH. To see that MATH can be extended to a pseudometric on MATH, it suffice to verify that for any sequence MATH of points of MATH such that all the expressions MATH REF and MATH are defined the inequality MATH holds. Then the required extension is given by the formula MATH where the infimum is taken over all chains MATH such that all the terms MATH are defined. An easy argument using induction shows that REF follows from its special case: for any ``quadrangle" in MATH of the form MATH each of the four numbers MATH, MATH, MATH, and MATH does not exceed the sum of the three others. This case is obvious: for example, since MATH, we have MATH .
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math/0004119
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If MATH is a finite subset of MATH and MATH, let MATH. Let MATH be the set of all pairs MATH such that MATH for all MATH. The sets of the form MATH constitute a base of entourages of the uniformity on MATH. If MATH, we say that MATH and MATH are MATH-close. Our proof proceeds in three parts. CASE: We prove that MATH is dense in MATH. Let MATH, and let MATH be a neighbourhood of MATH in MATH. We must prove that MATH for some MATH. We may assume that MATH is the set of all MATH such that MATH is MATH-close to MATH: MATH where MATH is a finite subset of MATH and MATH. Let MATH. We claim that there exist points MATH such that MATH and MATH, MATH. Indeed, since MATH is bi-Katětov, the formulas above define a pseudometric on the set MATH, where MATH are new points. Since MATH is NAME, the embedding of MATH into MATH extends to a distance-preserving map from MATH to MATH. Since MATH is MATH-homogeneous, there exists an isometry MATH of MATH such that MATH, MATH. Let MATH. For every MATH we have MATH. Thus MATH. This proves that MATH is dense in MATH. CASE: We prove that the uniformity MATH is coarser than MATH. Whenever a topological group MATH acts continuously on a compact space MATH (on the left), for every MATH the orbit map MATH from MATH to MATH is right-uniformly continuous. We saw that MATH acts continuously on MATH REF . The embedding MATH can be viewed as the orbit map corresponding to MATH, the neutral element of MATH. It follows that MATH is MATH-uniformly continuous. Similarly, MATH is MATH-uniformly continuous (use the right action of MATH on MATH, or, alternatively, use the involution on MATH to deduce MATH-uniform continuity of MATH from its MATH-uniform continuity). Therefore, the uniformity MATH is coarser than both MATH and MATH and hence coarser than MATH. CASE: We prove that MATH is finer than MATH. It suffices to show that for every MATH there exists an entourage MATH of the uniformity on MATH (in other words, a neighbourhood of the diagonal of MATH) with the following property: if MATH are such that MATH and MATH are MATH-close, then MATH. Assume that MATH. We claim that MATH has the required property. Let MATH be such that MATH and MATH are MATH-close. This means that MATH . Let MATH, MATH and MATH, MATH. We have MATH and MATH for all MATH and MATH. In virtue of REF , there exist points MATH such that the correspondence MATH, MATH is distance-preserving and MATH, MATH. Since MATH is MATH-homogeneous, there exists an isometry MATH of MATH such that MATH and MATH, MATH. We have MATH, since each MATH is moved by less than MATH. Put MATH. For every MATH we have MATH, hence MATH. Thus MATH, as required.
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math/0004119
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Let MATH, MATH, MATH be neighbourhoods of MATH, MATH and MATH, respectively. We must show that MATH meets the set MATH. We may assume that for some finite set MATH and MATH we have MATH . We saw in the last paragraph of the preceding section that there exist a metric space MATH and isometric embeddings MATH REF such that MATH, MATH and MATH for all MATH. Let MATH. Since MATH is NAME, there exists an isometric embedding of MATH into MATH which extends the isometry MATH. It follows that there exist points MATH such that MATH, MATH, MATH and MATH for all MATH. Since MATH is MATH-homogeneous, there exists an isometry MATH such that MATH, MATH. Let MATH. Using again the MATH-homogeneity of MATH, we find an isometry MATH such that MATH, MATH. Note that MATH and MATH for all MATH. We claim that MATH, MATH and MATH. Indeed, we have MATH for all MATH and hence MATH. The other two cases are considered similarly. Thus MATH.
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math/0004119
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Let MATH and MATH be the left and the right uniformity on MATH, respectively. In each of REF - REF the map MATH under consideration is an automorphism of the uniform space MATH. This is obvious for REF . For REF , observe that the uniformities MATH and MATH are invariant under left and right shifts, hence the same is true for their greatest lower bound MATH. It follows that in all cases MATH extends to an automorphism of the completion MATH of the uniform space MATH.
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math/0004119
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According to REF , every idempotent MATH is of the form MATH for some closed MATH. If MATH is invariant under inner automophisms, then MATH and hence MATH for every MATH. Since the action of MATH on MATH is transitive, no proper non-empty subset of MATH is MATH-invariant. Thus either MATH or MATH. Accordingly, either MATH or MATH.
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math/0004119
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Let MATH be a normal subgroup of MATH. We show that MATH is not compact. Fix MATH and MATH such that MATH. Let MATH, and let MATH be the sphere of radius MATH centered at MATH. We claim that the orbit MATH contains MATH. Fix MATH. Since MATH is MATH-homogeneous, there exists an isometry MATH which leaves the point MATH fixed and maps MATH to MATH. Let MATH. Since MATH is normal, we have MATH and hence MATH. Thus MATH, as claimed. Since MATH is NAME, we can construct by induction an infinite sequence MATH of points in MATH such that all the pairwise distances between distinct members of this sequence are equal to MATH. Since MATH, it follows that MATH is not compact. Hence MATH is not compact.
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math/0004119
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It suffices to show that for every MATH the map MATH from MATH to MATH is continuous. Fix MATH, MATH and MATH. Let MATH, MATH and MATH. If MATH and MATH, then MATH.
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math/0004119
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Let MATH be the set of all self-maps of MATH, equipped with the product uniformity. The group MATH can be considered as a subset of MATH. The uniformity MATH on MATH induced by the product uniformity on MATH coincides with the left uniformity MATH. Indeed, a basic entourage for MATH has the form MATH, where MATH is the metric on MATH, MATH is a finite subset of MATH and MATH. Let MATH. Then MATH is a basic neighbourhood of unity in MATH, and MATH is a basic entourage for MATH. Thus MATH. It follows that the map MATH from MATH to MATH induces the right uniformity on MATH, and the map MATH defined by MATH induces the upper uniformity MATH. Since MATH is complete, to prove that MATH is complete it suffices to show that MATH is closed in MATH. Let MATH be the set of all non-expanding self-maps of MATH. Then MATH is closed in MATH. The map MATH from MATH to MATH is continuous REF . Since MATH, it follows that MATH is closed in MATH and hence in MATH.
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math/0004119
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For every metric space MATH we have MATH. If MATH is homogeneous, then for every MATH the map MATH from MATH to MATH is onto, whence MATH.
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math/0004119
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We saw that MATH is NAME REF . REF shows that MATH is complete, and REF shows that MATH. Let MATH be a continuous onto homomorphism. According to REF , to prove that MATH is minimal and topologically simple, it suffices to prove that either MATH is a homeomorphism or MATH. Since MATH is NAME, so is MATH. Let MATH be the NAME compactification of MATH. The homomorphism MATH extends to a continuous map MATH. Let MATH be the unity of MATH, and let MATH. CASE: MATH is a subsemigroup of MATH. Let MATH. In virtue of REF , there exist filter bases MATH and MATH on MATH such that MATH converges to MATH (in MATH), MATH converges to MATH and MATH is a cluster point of the filter base MATH. The filter bases MATH and MATH on MATH converge to MATH, hence the same is true for the filter base MATH. Since MATH is a cluster point of MATH, MATH is a cluster point of the convergent filter base MATH. A convergent filter on a NAME space has only one cluster point, namely the limit. Thus MATH and hence MATH. CASE: The semigroup MATH is closed under involution. In virtue of REF , the inversion on MATH extends to an involution MATH of MATH. Since MATH for every MATH, the same holds for every MATH. Let MATH. Then MATH and hence MATH. CASE: If MATH and MATH, then MATH. We saw that the left shift MATH of MATH extends to a continuous self-map MATH of MATH defined by MATH REF . According to REF , the self-map MATH of MATH extends to a self-homeomorphism MATH of MATH. The maps MATH and MATH from MATH to MATH coincide on MATH and hence everywhere. Replacing MATH by MATH, we see that MATH. Thus MATH. Using right shifts, we similarly conclude that MATH. CASE: MATH is invariant under inner automorphisms of MATH. We have just seen that MATH for every MATH, hence MATH. Let MATH. Note that MATH. According to REF , there is a greatest element MATH in MATH, and MATH is idempotent. Since inner automorphisms of MATH preserve the order on MATH and the unity MATH, REF implies that MATH is invariant under inner automorphisms. In virtue of REF , either MATH or MATH. We shall show that either MATH is a homeomorphism or MATH, according to which of the cases MATH or MATH holds. Consider first the case MATH. CASE: If MATH, then all elements of MATH are invertible in MATH. Let MATH. Then MATH and MATH, since MATH is a symmetrical semigroup. According to REF , we have MATH and MATH. Since MATH, there are no elements MATH in MATH. Thus the inequalities MATH and MATH are actually equalities. It follows that MATH is the inverse of MATH. CASE: If MATH, then MATH. REF imply that MATH is a subgroup of MATH. This subgroup is normal REF and compact, since MATH is closed in MATH. REF implies that MATH. CASE: If MATH, then MATH is a homeomorphism. REF imply that MATH and that the map MATH is bijective. Since MATH is a map between compact spaces, it is perfect, and hence so is the map MATH. Thus MATH, being a perfect bijection, is a homeomorphism. Now consider the case MATH. CASE: If MATH, then MATH. Let MATH and MATH. We have MATH. On the other hand, REF implies that MATH. Thus MATH.
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math/0004120
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In the context of NAME operators, REF is proved in CITE; in the corresponding NAME situation REF is proved in CITE. REF then follows by combining REF , taking into account the asymptotic expansions (along any ray with MATH) MATH in the case of NAME operators (compare CITE) and MATH in the case of NAME operators (compare CITE).
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math/0004120
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In order to prove REF, let MATH be the fundamental system of solutions of REF as defined in REF and observe that MATH in the NAME operator case. Thus, any nonnormalized solutions MATH, MATH of MATH for MATH are of the type MATH for some nonsingular MATH matrices MATH. Thus, MATH by REF. In particular, MATH is independent of the normalization chosen for MATH. Varying the reference point MATH then yields REF. Similarly, in order to derive REF, let MATH be the fundamental system of solutions of REF as defined in REF. Any other fundamental system of REF of the form MATH with MATH a basis of MATH for MATH, necessarily must be of the form MATH . Here MATH is of the type MATH and MATH are invertible MATH matrices. In particular, MATH and thus, MATH by REF. Consequently, MATH is independent of the normalization chosen for MATH. Varying MATH one verifies REF.
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math/0004120
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In order to prove REF one only needs to observe MATH for a.e. MATH, using REF. By continuity of the right-hand side of REF with respect to MATH, REF extends to all MATH. REF then proves REF, and using REF again, REF proves REF. Similarly, to prove REF, one computes MATH for a.e. MATH, using REF.
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math/0004120
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Suppose there is a MATH such that MATH. Then there is a sequence MATH, MATH, MATH, such that MATH . Any NAME function MATH with values in MATH admits the representation MATH where MATH, MATH, MATH is a NAME space, MATH, MATH is a self-adjoint operator in MATH, and MATH with MATH. By REF, MATH where MATH is invertible in MATH, MATH, MATH. Thus, REF together with REF imply MATH and hence MATH. Applying REF again one infers MATH, contradicting the hypothesis MATH. Since MATH, MATH is invertible, the numerical range MATH of the operator MATH is a subset of the half-plane MATH. By the spectral inclusion theorem (see, for example, CITE), the spectrum of MATH is contained in the closure of the numerical range MATH and hence REF holds.
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math/0004120
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Introducing the map MATH the estimate MATH combined with REF shows that MATH maps MATH into itself. Moreover, using MATH REF implies that MATH is a strict contraction. An application of NAME 's fixed point theorem then proves the existence of a unique fixed point MATH. Moreover, the fixed point MATH can be obtained as the limit of the sequence MATH, where MATH and any fixed MATH, MATH . Taking MATH proves representations REF, which in turn completes the proof since any fixed point of MATH is a solution of the matrix-valued NAME REF .
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math/0004120
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For MATH the estimate MATH shows that MATH . Analogously, one proves that MATH . Combining REF yields REF.
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math/0004120
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Since MATH uniquely determines MATH (that is, MATH or MATH) by REF , it suffices to show that MATH and MATH uniquely determine MATH in REF and similarly, MATH, MATH, and MATH, MATH uniquely determine MATH in REF . We start with the NAME REF . By REF, and REF, MATH and hence MATH by REF. Applying REF (with MATH and MATH) to REF (with MATH) then shows that MATH are uniquely determined by MATH and MATH. In the corresponding NAME REF one rewrites REF (with MATH) in the form MATH . Thus, for MATH sufficiently large, we can again apply REF (with MATH, MATH, and MATH the r.-h.s. of REF) since clearly MATH for MATH sufficiently large. Hence MATH is uniquely determined for MATH sufficiently large. Since MATH is analytic in MATH, one concludes that MATH is uniquely determined for all MATH. Since MATH is known by hypothesis, this also uniquely determines MATH for all MATH. In either case an application of REF completes the proof.
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math/0004120
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In the NAME context one observes from REF , and REF (for MATH) that MATH . Here MATH is a closed and counterclockwise oriented NAME contour which encircles MATH with winding number MATH and MATH with winding number MATH, and whose existence is guaranteed by REF (compare REF). Using the asymptotic expansion (compare CITE) MATH REF and a standard MATH-argument yield MATH along a ray with MATH. Hence one obtains REF by REF . In the case of NAME operators one infers from REF that MATH and then proceeds in exactly the same manner as in REF - REF replacing REF by (compare CITE) MATH .
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math/0004120
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Combining REF, MATH yields REF after straightforward matrix computations.
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math/0004120
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Using REF, and the asymptotic representations MATH one infers MATH . Here REF is explicitly needed in the case of the MATH-sign in REF. By REF implies MATH and hence MATH . Introducing the orthogonal polynomials MATH associated with the normalized spectral measures MATH, MATH, REF implies that MATH and MATH which proves REF using REF.
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math/0004120
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REF . A direct computation utilizing REF shows that MATH . In addition, the matrix MATH can be read off from the asymptotic expansion of MATH as MATH by a simple NAME series-type argument. Indeed, by REF MATH and hence by REF MATH along any ray with MATH, MATH. Therefore, combining REF, and REF one infers that MATH along any ray with MATH, MATH, in complete analogy with the corresponding scalar case (see, for example, CITE, CITE). Given MATH and using REF one can easily solve REF for MATH, MATH which in turn determines MATH, MATH for all MATH by REF . CASE: The data REF are perhaps the closest discrete analog of the continuous version in REF . As before (compare REF), one can determine MATH from the asymptotic expansion of MATH, MATH . Next, using REF, one observes that MATH is the unique solution of the following NAME matrix equation MATH (compare REF) by REF . Determining MATH from REF completes the proof of REF by REF as in REF . CASE: Since MATH is known, the data REF, and REF imply MATH . Together with REF for MATH this leads to the following equation for MATH, MATH . Using the asymptotics REF and the representation MATH one infers MATH for MATH sufficiently large. Thus, by REF , MATH is uniquely determined by the data REF for MATH large enough and hence for all MATH by analytic continuation. The NAME matrix MATH is then determined by (compare REF) MATH completing the proof.
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math/0004120
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CASE: Since MATH in REF implies that MATH and we denote the latter by MATH. By REF the MATH . NAME matrices MATH associated with MATH, MATH can be represented as MATH and they admit the estimate MATH . Using the asymptotics MATH the estimate REF combined with REF proves MATH . Since MATH REF yield MATH . REF yield MATH . The asymptotic expansions REF together with REF then prove REF by REF . CASE: Since MATH in REF again implies that MATH and we denote the latter by MATH. Given MATH, let MATH be a closed clockwise oriented NAME contour MATH such that MATH has winding number MATH and MATH . Next, given MATH, MATH, introduce the contour MATH in the upper half-plane by MATH . Using REF one concludes that MATH has winding number MATH with respect to the contour MATH if MATH lies on the ray MATH for MATH sufficiently large. For such MATH the matrix-valued functions MATH admit the representation (compare REF) MATH where MATH . Introducing MATH one infers by REF that MATH . The asymptotics MATH along the ray MATH, and the representation MATH then yield MATH along the ray MATH, uniformly with respect to MATH. As a consequence of REF we have MATH along the ray MATH, uniformly in MATH. Using REF, and REF one concludes that MATH and hence by REF that MATH along the ray MATH. This proves REF by REF . CASE: One observes that MATH satisfy (compare REF) MATH . Taking into account the asymptotic expansions REF implies REF by REF . This proves REF applying REF again.
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math/0004127
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Without loss of generality we can assume that MATH. Let MATH be the NAME isomorphism witnessing that MATH. Find NAME sets MATH and MATH and the NAME isomorphism MATH extending MATH (exercise in CITE). By removing a countable set from MATH we can arrange that MATH is MATH-dimensional. Let MATH be a meager set such that CASE: MATH is not meager in MATH, CASE: MATH is continuous on MATH, CASE: MATH is homeomorphic to MATH. Apply NAME 's theorem to get REF . Without loss of generality we can assume that MATH is a MATH set which is dense. By REF, a MATH subset of a MATH-dimensional Polish space which has empty interior is homeomorphic to MATH. Call this homeomorphism MATH. Now the required mapping is MATH.
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math/0004127
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Suppose that MATH. By REF there exists MATH, a NAME set MATH and NAME isomorphism MATH such that CASE: MATH is continuous, CASE: MATH is not meager in MATH. Apply Axiom MATH to to find a compact set MATH such that MATH is not meager in MATH. Set MATH and note that MATH is a homeomorphism between MATH and MATH. Under this homeomorphism MATH is the image of MATH thus it is not meager in MATH. Since MATH it follows that MATH. As the other inclusion is obvious, the theorem follows.
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math/0004127
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Let MATH be a continuous increasing sequence such that MATH . By properness, MATH, has the required property.
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math/0004127
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For the sake of clarity we will break the proof into three lemmas. The main idea of the proof is already present in CITE. Let MATH be the canonical name for a MATH-generic real and let MATH be the NAME forcing represented as MATH with MATH being the canonical name for the NAME real. For MATH let MATH. Note that MATH is a compact set in MATH. For two sequences MATH we say that MATH is good if MATH and MATH for MATH. Suppose that MATH, MATH, and MATH is good. Then there is a MATH-name MATH for an element of MATH such that CASE: MATH, CASE: MATH . Let MATH be a NAME real over MATH. Working in MATH define MATH . It is enough to check that MATH. In fact, we will show that if MATH and MATH then MATH. Suppose that MATH and MATH. In particular, MATH is good. Let MATH. For MATH let MATH . We show that MATH is dense in MATH below MATH. Take any MATH and let MATH. If MATH then put MATH. Otherwise, let MATH be such that MATH and MATH . It is clear that MATH. By genericity, we conclude that the set MATH is infinite in MATH. In particular, MATH. Since MATH was arbitrary, it finishes the proof. Suppose that MATH, MATH, and MATH is good. Let MATH be a MATH-name for a closed nowhere dense subset of MATH. There exists a MATH-name MATH for an element of MATH such that CASE: MATH, CASE: MATH CASE: MATH. Let MATH be a MATH-generic real over MATH, and let MATH be the interpretation of MATH using MATH. In MATH define sequences MATH such that CASE: MATH. CASE: MATH, Since MATH is nowhere dense this definition is correct. Going back to MATH we conclude that there is a MATH-name MATH such that CASE: MATH, CASE: MATH. For each MATH find a condition MATH such that CASE: there is a sequence MATH such that MATH. CASE: MATH, CASE: MATH. Observe that by the choice of MATH it follows that MATH . Let MATH be a NAME real over MATH. Working in MATH define MATH . We will show that MATH is infinite in MATH. For MATH let MATH . We show that MATH is dense in MATH below MATH. Suppose that MATH and let MATH. Pick MATH and define MATH such that MATH and MATH . Note that by the properties of MATH it follows that MATH. By genericity, for every MATH there is MATH such that MATH, which implies that MATH is infinite. Let MATH. Define in MATH, MATH and let MATH. Since the nodes corresponding to the elements of MATH were good, it follows that MATH and MATH. In addition, MATH (by the choice of MATH) and MATH (by the choice of MATH). Since MATH was arbitrary, the proof is finished. Finally we show: Suppose that MATH, MATH, and MATH is good. Let MATH be a MATH-name for a sequence of closed nowhere dense subsets of MATH. There exists a MATH-name MATH for an element of MATH such that CASE: MATH, CASE: MATH CASE: MATH. The proof is a refinement of the proof of the previous lemma. Suppose that MATH is a MATH-name for a sequence of closed nowhere dense subsets of MATH. Without loss of generality we can assume that MATH. Find a MATH-name MATH such that CASE: MATH, CASE: MATH. Build by induction a sequence of REF such that CASE: MATH, CASE: MATH, CASE: if MATH and MATH then there exists a sequence MATH such that CASE: MATH, CASE: MATH. As in the previous lemma, it follows that for MATH and MATH, MATH is good. The construction is straightforward; the first step is essentially described in the previous lemma. Let MATH and let MATH be a NAME real over MATH. Working in MATH define for each MATH and MATH: MATH . As before, it follows that MATH is infinite in MATH for every MATH. Finally, let MATH be defined so that for every MATH, MATH. It follows from the definition of MATH that MATH. On the other hand, for every MATH, MATH. Thus MATH . Since the sets MATH are increasing, we conclude that MATH . Now we are ready to prove REF . Suppose that MATH, MATH is not meager. We will show that MATH. Suppose otherwise and let MATH be a MATH-name for a meager set in MATH such that for some MATH, MATH . Let MATH be a countable elementary submodel containing MATH, MATH, MATH, etc. Since MATH is not meager there exists a real MATH which is NAME over MATH. By REF there exists a condition MATH, MATH such that MATH . This contradicts the choice of MATH and finishes the proof.
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math/0004130
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For the sake of simplicity we will write MATH, instead of MATH, to denote a canonical divisor of MATH. By contradiction, assume that MATH does not impose independent conditions to MATH. Let MATH be a minimal REF-dimensional subscheme of MATH for which this property holds and let MATH. This means that MATH and that MATH satisfies the NAME condition (see, for example, CITE). Therefore, a non-zero element of MATH corresponds to a non-trivial rank REF vector bundle MATH; so, one can consider the following exact sequence MATH . This implies that MATH-that is, MATH . We now want to compute MATH in cases MATH and MATH. In the first one, MATH by MATH. In the other case, using MATH and the Index Theorem, MATH since we supposed MATH. In both cases, the vector bundle MATH is MATH (see CITE or CITE), that is, there exist MATH and a REF-dimensional scheme MATH (possibly empty) such that MATH holds and MATH. We recall that MATH denotes the ample divisor cone of MATH. This means that MATH . The exact sequence MATH ensures us that MATH. If we consider the tensor product of the exact sequence MATH by MATH, we get MATH . We state that MATH; otherwise, MATH would be an effective divisor, therefore MATH for each ample divisor MATH. From MATH, it follows that MATH, so, by MATH and MATH, MATH . Thus, for every ample divisor MATH, MATH . Furthermore, from REF . and MATH . , it immediately follows that MATH and MATH. Indeed, MATH and MATH. Since MATH is irreducible, this implies that MATH is a nef divisor; from MATH and from NAME 's criterion, we get MATH . It follows that MATH so, since MATH is nef, MATH can not be effective. If we consider the cohomological exact sequence associated to MATH, we deduce that there exists a divisor MATH such that MATH. If the irreducible nodal curve MATH, whose sets of nodes is MATH, were component of MATH, then MATH would be an effective divisor. By applying MATH and by using the fact that MATH and hypothesis MATH . , one determines MATH which contradicts the effectiveness of MATH, since MATH is nef. NAME 's theorem implies that MATH . On the other hand, taking MATH maximal, we may further assume that the general section of MATH vanishes in a MATH-codimensional locus MATH of MATH. Thus, MATH. By standard computations on NAME classes, we obtain MATH-which implies MATH . By applying the Index theorem to the divisor pair MATH, we get MATH . From MATH and from the fact that MATH is positive, it follows that MATH . We observe that the left side member of REF is non-negative, since MATH, where MATH is effective and, by (MATH), MATH. Thus, REF still holds when we square both its members and, together with (MATH), this gives MATH . On the other hand, using MATH, we get MATH-that is, MATH . Putting together MATH and MATH, we get MATH . Summarizing, the assumption on MATH, stated at the beginning, implies MATH. We want to show that our numerical hypotheses hold if and only if the opposite inequality is satisfied. To this aim, observe that the discriminant of the equation MATH is MATH, so, by REF . and MATH . , it is positive. The inequality MATH is verified iff MATH, where MATH we have to show that, with our numerical hypotheses, MATH. From MATH . , it immediately follows that MATH, since, as we shall see in the sequel, the bound in MATH is smaller than MATH. Note also that MATH. Indeed, if MATH, then MATH, which contradicts the Index theorem, since MATH. Observe that MATH if and only if MATH . To simplify the notation, we put MATH so that MATH becomes MATH . Two cases can occur. If MATH, there is nothing to prove since the right side member of REF is always positive. Note, before proceeding to consider the other case, that in this situation we want that MATH in order to have at least an effective positive integral value for the number of nodes; but MATH if and only if MATH. By squaring both members of the previous inequality, we get MATH, which is our REF .; so the upper-bound for MATH is surely greater than REF. Moreover, the expression for such bound is the one in MATH and it can not be written in a better non-trivial form. On the other hand, if MATH, by squaring both members of MATH, we get MATH, which is our REF . Therefore, MATH; moreover, the condition MATH is trivially satisfied, since it is equivalent to MATH. From MATH, we can write MATH so we can replace the bound MATH with the more "accessible" one MATH, which is the bound in MATH. Observe that MATH-so, it is not correct to directly write MATH. Therefore, MATH is the right approximation. In conclusion, our numerical hypotheses contradict (MATH), therefore the assumption MATH leads to a contradiction.
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math/0004132
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We prove the result by induction on MATH. The cases MATH are nothing but REF respectively. Suppose the statement is true for MATH. Then replacing MATH by MATH and noting MATH, we see that MATH . Then by REF , we conclude that MATH .
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math/0004132
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Recall REF and the fact that MATH for positive integers MATH (compare REF ). We have MATH . Now the result follows by induction, noting that MATH .
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math/0004132
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Since the proofs of REF are almost identical we only prove REF . The special case MATH follows from REF . So we can proceed by induction on MATH and assume that MATH . Then by REF we have MATH as required.
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math/0004132
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If MATH this is REF , so we may assume that MATH . First we deal with the case MATH . Following REF we have MATH . Continuing in this fashion, the result follows.
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math/0004132
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Recall from CITE the vertex operator MATH . It is easy to see that MATH for MATH for all MATH . So it is immediate that MATH .
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math/0004132
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We prove this by induction on MATH . If MATH the assertion is clear. If MATH the assertion follows from the fact that MATH on MATH . Now we assume that MATH . Using REF gives MATH . By induction, both MATH and MATH converge to holomorphic functions in MATH whence so does MATH . Indeed, MATH . It is easy to express MATH in terms of the MATH and MATH, and so obtain a recursive formula. We leave this to the reader. Since MATH and MATH are independent of MATH is also independent of MATH .
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math/0004132
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First we take MATH for MATH with MATH . Then from REF we have MATH for some scalars MATH and quasi-modular forms MATH . Now the result follows from REF . For arbitrary MATH we can assume that MATH is a monomial in MATH for MATH and MATH . Use the result for MATH and REF again to see that MATH is a linear combination of functions of the form MATH where MATH belong to MATH are non-negative integers and MATH . It is easy to see that MATH is a linear combination of MATH for MATH . This completes the proof of the theorem.
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math/0004132
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After REF it suffices to show that if MATH in MATH satisfies MATH then the function MATH lies in MATH . This is essentially established in the paper CITE. In the notation of CITE, the present function MATH is denoted MATH where MATH is the integral quadratic form which corresponds to MATH . If MATH is odd then the function is identically zero. If MATH is even then REF of CITE says that the function MATH is a holomorphic modular form in MATH . In REF , MATH is a certain non-zero constant equal to MATH when MATH . Moreover MATH is just the theta function of MATH . Using this information, it follows from REF and induction on MATH that MATH indeed lies in MATH as required.
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math/0004132
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It is enough to prove sufficiency. Let MATH of weight MATH be expressed in the form MATH with each MATH a form in MATH of weight MATH . As MATH is also a generalized modular form of weight MATH each of its MATH-transforms REF has a MATH-expansion. In particular, the MATH-transform of MATH yields an equality of the shape MATH . The MATH-transform of MATH is well-known. In particular, we have MATH . Comparing REF - REF we obtain an equality MATH . Since each MATH has a MATH-expansion, an equality like REF can only hold if all MATH are zero for MATH . Thus MATH is indeed modular.
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math/0004132
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Recall that MATH is an orthonormal basis of MATH . Note that MATH and both MATH for MATH and MATH annihilate MATH. Since all summands of MATH contain one of these as a factor, we immediately see that MATH.
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math/0004132
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It suffices to consider the condition MATH. Since MATH and MATH annihilates MATH if MATH we see that MATH is equivalent to MATH. Now since MATH, MATH acts as MATH on MATH.
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math/0004137
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It is enough to show that MATH is the sum of the signed monomials MATH for tableaux MATH with no integers larger than MATH. Now number the diagonals of MATH from south-west to north-east as described in the previous section and let MATH be the free Abelian group with one basis element for each partition containing MATH. Imitating REF and CITE, we define a linear action of MATH on MATH as follows. If a partition MATH has an outer corner in the MATH'th diagonal, then we set MATH where MATH is obtained by adding a box to MATH in this corner. If MATH has an inner corner in the MATH'th diagonal, and if the box in this corner is not contained in MATH, then we set MATH. In all other cases we set MATH. We claim that if MATH is any permutation such that MATH then MATH for some partition MATH and MATH. This claim is clear if MATH. If MATH, write MATH where MATH. Since MATH we can assume by induction that MATH for some partition MATH, and MATH. It is enough to show that MATH has no inner corner in the MATH'th diagonal. If it had, then the box in this corner would be outside MATH since MATH. But then we could write MATH such that MATH and no reduced expression for MATH contains MATH, MATH, or MATH. This would mean that MATH, contradicting that MATH. Using REF it follows from the claim that the coefficient of the basis element MATH in MATH is MATH. Finally, it is easy to identify the terms of MATH which take MATH to MATH with set-valued tableaux on MATH. In fact, this product expands as a sum of terms of the form MATH . From any such term which contributes to the coefficient of MATH we obtain a tableau on MATH by joining the integer MATH to the set in the inner corner in diagonal number MATH of the partition MATH for each MATH.
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math/0004137
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Since MATH is super-symmetric, it is enough to prove that MATH for any number of variables MATH. Now MATH has a MATH-linear automorphism which sends MATH to MATH for each MATH. Since this automorphism takes MATH to MATH and MATH to MATH, the lemma follows from REF.
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math/0004137
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Suppose MATH was bumped out of box number MATH in MATH, counted from the top. Then since MATH is a tableau, MATH is less than or equal to the set in box number MATH in MATH. This implies that the MATH'th box of MATH is equal to the MATH'th box of MATH for all MATH. For MATH, the MATH'th box of MATH can contain elements from the MATH'th box of MATH and from MATH. Since both of these sets are larger than or equal to the MATH'th box in MATH, this shows that MATH is a tableau.
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math/0004137
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Notice at first that MATH, which follows directly from REF . Since all of the integers in MATH come from MATH, it suffices to show that all integers from MATH which are MATH are also strictly greater than MATH. Let MATH be the number of the box in MATH (counted from the top) which contains MATH. There are then two possibilities. Either MATH is the largest element in this box of MATH, in which case all integers in MATH must come from the boxes MATH through MATH of MATH. Furthermore, the boxes in MATH strictly below the MATH'th box contain the same sets as the corresponding boxes of MATH, so they are strictly greater than MATH. Since these are also the only boxes that can be greater than or equal to MATH, the statement follows in this case. Otherwise MATH is not the largest element in the MATH'th box of MATH, which implies that all integers in MATH come from the boxes MATH through MATH in MATH. Since the integers in MATH which are strictly greater than MATH all come from the MATH'th box in MATH or from boxes below this box, the statement is also true in this case.
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math/0004137
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Suppose the southernmost box of MATH occurs in column MATH. Then let MATH be the tableau consisting of the leftmost MATH columns of MATH and let MATH be the rest of MATH. We will write MATH to indicate this. Similarly we let MATH and MATH, that is, MATH and MATH are the leftmost MATH columns of these products. Finally set MATH and MATH. We then have MATH and MATH. Since MATH has one more box in the first column than MATH, this box of MATH must contain a subset of MATH. Since MATH by REF , this means that MATH consists of MATH with MATH attached below the first column (or MATH if MATH is empty). The lemma follows from this.
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math/0004137
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Suppose at first MATH is a single integer, and let MATH. Then by the definition of MATH, the minimal element of MATH will bump out MATH from MATH, after which the remaining elements of MATH will be added to the same box as the minimal element went into. This recovers the tableau MATH. If MATH has more than one element, let MATH and MATH, and write MATH and MATH. Then MATH where MATH. By induction we can assume MATH. There are two cases to consider. Either MATH. In this case MATH is obtained from MATH by joining MATH to the box containing MATH. This means that if MATH then we have MATH. This proves that MATH in this case. Otherwise we have MATH, in which case MATH. We then know that MATH. Since MATH, it is enough to show that MATH. Here it is enough to check that none of REF or REF are used to define MATH. But since MATH is bumped out when MATH is formed, this would imply that MATH and MATH are in the same box of MATH. This contradicts the fact that MATH is bumping MATH but not MATH out when multiplied to MATH. The proof of the second statement is similar and left to the reader.
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math/0004137
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Notice at first that if MATH has shape MATH then MATH must be defined by one of REF or REF . Let MATH be the largest subset of the form MATH such that none of REF - REF are used in the definition of MATH, and let MATH. The lemma is easy to prove if MATH is empty, so we will assume MATH. Let MATH and MATH. Then MATH where MATH. Furthermore we have MATH by REF . There are two cases to consider. First suppose MATH is defined by REF . Then MATH is equal to MATH with MATH attached in a new box at the bottom, and MATH is empty. This means that MATH, so MATH. By REF we therefore have MATH as claimed. Otherwise MATH is defined by REF . If the bottom box of MATH contains MATH with MATH, then MATH and MATH is obtained by removing MATH from the bottom box of MATH and attaching MATH in a new box below it. It follows that MATH where MATH is obtained from MATH by moving the elements of MATH from the bottom box to a new box below it. We conclude that MATH, so MATH by REF . The proof of the second statement is similar and left to the reader.
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math/0004137
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It follows from REF that the maps MATH define an inverse to the map of REF when MATH. If MATH and MATH is any element, there are unique rook strips MATH which split the vertical strip MATH up into disjoint intervals from north to south, such that MATH contains the columns of the extra boxes in MATH. Then set MATH and MATH for MATH, and let MATH be the column whose MATH'th box contains MATH. An argument similar to the proof of REF shows that MATH which implies that MATH is a tableau. Finally REF show that the map MATH gives an inverse to REF .
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math/0004137
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Let MATH be any positive integer. It is enough to show that for any partition MATH there exists a polynomial MATH such that MATH is a linear combination of NAME polynomials MATH for partitions MATH of length MATH. Define a partial order on partitions by writing MATH if MATH, or MATH and MATH. Notice that given a partition MATH there are only finitely many partitions MATH such that MATH and MATH. We shall prove that the polynomials MATH exist by induction on this order. Since the smallest partitions MATH have only one column, the existence of MATH is clear for these partitions. Let MATH be a partition of length MATH and assume that MATH exists for all MATH. Let MATH be the partition obtained by removing the first column of MATH. Then MATH. By REF we furthermore have MATH where the sum is over all partitions MATH properly containing MATH such that MATH is a vertical strip. Notice that any such partition MATH for which MATH must satisfy MATH. We can therefore define MATH . This finishes the proof.
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math/0004137
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If MATH is a reverse lattice word, then the rectification of MATH in the plactic monoid is the semistandard NAME tableau MATH of shape MATH in which all boxes in row MATH contain the integer MATH CITE. This implies that the identity MATH holds even with the weaker relations of the plactic algebra. On the other hand, if MATH then one can obtain the sequence MATH from MATH by replacing subsequences in the following ways: MATH . Since all of these moves preserve reverse lattice words, MATH must be a reverse lattice word with content MATH.
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math/0004137
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Start by noticing that the only tableau of shape MATH whose word is a reverse lattice word is the tableau MATH. It follows from this that the coefficient of MATH in the right hand side of REF is MATH. On the other hand, the left hand side is equal to MATH. If MATH is a tableau on MATH, then MATH is equal to MATH exactly when MATH is a reverse lattice word with content MATH by the lemma. The theorem follows from this.
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math/0004137
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By replacing each non-commutative variable MATH with MATH in REF , we obtain the identity MATH for single stable NAME polynomials. Since MATH is super symmetric, the same equation must hold for the double polynomials as well.
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math/0004137
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By REF the left hand side is the signed sum of monomials MATH for all tableaux MATH of shape MATH such that all contained integers are between MATH and MATH. If MATH is such a tableaux, let MATH be the subtableau obtained by removing all integers strictly larger than MATH (as well as all boxes that become empty), and let MATH be obtained by removing integers less than or equal to MATH. Then the shape of MATH is MATH for some partition MATH, while the shape of MATH is of the form MATH, and we have MATH. Since MATH is the shape where MATH and MATH overlap, this skew shape must be a rook strip. Finally, since MATH this gives a bijection between the terms of the two sides of the claimed identity.
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math/0004137
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Number the rows of MATH such that the top box in the leftmost column of MATH is in row number one, and so that the numbers increase from top to bottom. Then suppose MATH and MATH are tableaux of shapes MATH and MATH for which all contained integers are less than or equal to MATH. Then all integers in row MATH of MATH will be greater than or equal to MATH because they have at least MATH boxes above them. Similarly the integers in row MATH of MATH will be smaller than or equal to MATH because they have MATH boxes below them. Therefore MATH and MATH fit together to form a tableau MATH of shape MATH. This shows that the terms of MATH and MATH are in bijective correspondence.
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math/0004137
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In this proof we will write MATH for the variables MATH and MATH for MATH. Then by REF we have MATH where the sum is over all partitions MATH such that MATH is a rook strip. Notice that when MATH has more than MATH columns, then MATH by REF . Therefore we only need to include terms for which MATH in the sum. For such terms REF implies that MATH. The lemma follows from this by applying REF to MATH.
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math/0004137
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Let MATH be the first MATH columns of MATH and let MATH be rest like in REF. Then since MATH we get MATH. The statement therefore follows from the proposition.
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math/0004137
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Let MATH denote MATH and let MATH denote MATH. Then REF implies that MATH. Now using REF we obtain MATH.
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math/0004137
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It is enough to show this for finitely many variables MATH and MATH, as long as MATH and MATH can be arbitrarily large. Let MATH be a rectangle with MATH rows and MATH columns. If MATH is a partition such that MATH occurs with non-zero coefficient in MATH then first of all MATH. Furthermore, REF shows that MATH is non-zero only if MATH has the form MATH for partitions MATH and MATH. By these observations we get MATH . Since REF shows that MATH, this proves the theorem.
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math/0004137
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These statements are clear from REF , and REF .
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math/0004137
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We will start by comparing the coefficients MATH and MATH. Suppose MATH is a tableau of shape MATH such that MATH is a partial reverse lattice word with respect to the intervals MATH and MATH and with content MATH. Then all integers in MATH which come from the interval MATH must be located in the upper-left corner in MATH of shape MATH. Furthermore, any such integer MATH can occur only in row MATH. Now let the skew shape MATH be the region in which the integers larger than MATH are located in MATH. Since this region can only overlap MATH in a rook strip, MATH must be a rook strip. If you remove all integers smaller than or equal to MATH from MATH and subtract MATH from the rest, then the result is a tableau of shape MATH whose word is a reverse lattice word with content MATH. Since MATH is uniquely determined by MATH and this skew tableau, we obtain MATH where this sum is over all partitions MATH such that MATH is a rook strip. To finish the proof we set MATH. Then we obtain MATH . By comparing with REF and noting that the transition matrix between the MATH and the MATH for fixed MATH is invertible, we conclude that MATH.
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math/0004137
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It follows from REF that the single polynomial MATH is a possibly infinite linear combination of polynomials MATH for partitions MATH with at most MATH columns: MATH . Using that MATH is super symmetric, this implies that REF is true with the same coefficients. Let MATH be the longest permutation of MATH. Using REF we then deduce that MATH is zero unless MATH. We conclude that MATH is a finite linear combination of the polynomials MATH.
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math/0004137
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It is enough to show that if MATH is not a rectangle and MATH then the absolute value of some coefficient MATH is at least two. The assumptions on MATH and MATH imply that we can find a partition MATH containing MATH such that MATH is a vertical strip with MATH boxes and at least three columns. Then it follows from NAME 's NAME formula that MATH, which means that there exist two different tableaux MATH and MATH of shape MATH such that the concatenations MATH and MATH are reverse lattice words with the same content MATH. But then MATH and MATH must also be reverse lattice words with the same content, so MATH if MATH is the content of MATH.
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math/0004137
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Start by locating the leftmost shared box MATH of MATH which contains MATH. To construct MATH we start by removing MATH from this box. Then look for the nearest box MATH below or to the left of MATH which contains MATH, such that the box above MATH does not contain MATH and the box to the left of MATH does not contain MATH. If no such box exists, then MATH stays a reverse lattice word even if MATH is removed from MATH. If we can locate a box MATH satisfying these requirements, we replace MATH with MATH in this box. Notice that MATH can't be a shared box by the assumptions. We then continue in the same way, with MATH in the role of MATH and MATH in the role of MATH. MATH is the tableau resulting when no new box MATH can be obtained.
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