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math/9910022 | Let MATH denote the Right-hand side of REF . If MATH, then REF reduces to the standard evolution equation for MATH. It is easily checked that both sides of REF vanish if MATH, provided that MATH and MATH are constant. If MATH but MATH, then MATH . Since MATH, it follows that REF holds for MATH and MATH if and only if MATH hence if and only if MATH . If MATH but MATH, we recall that MATH and compute MATH . Since MATH, it follows that REF holds for MATH and MATH if and only if MATH . When MATH, this equation is the same as MATH because MATH. To complete the proof, it suffices to note that MATH and MATH because the exterior derivative is independent of the metric and satisfies MATH. |
math/9910022 | Because MATH we observe that when REF holds, we have MATH . |
math/9910022 | By REF , we have MATH . Next we observe that MATH and MATH . Hence by REF , MATH . If MATH, this becomes MATH . |
math/9910022 | Because MATH is symmetric, one computes directly from the definition that MATH . Hence by REF and the NAME identities, MATH . On the other hand, the second NAME identity implies that MATH while straightforward calculations reveal that MATH and MATH . Because REF is readily verified for MATH, we may assume without loss of generality that MATH. Then we can combine the identities above to obtain MATH because MATH and MATH. Now if we regard MATH as a (globally-defined) space-time MATH-tensor MATH, we may write MATH yielding MATH where MATH here denotes the MATH-tensor MATH. We claim REF is equivalent to REF . Indeed, it follows from REF - REF that for MATH one has MATH . So one need only check that if MATH denotes the number of space-like components of MATH, the Right-hand side of REF contains MATH terms of the form MATH, while the Right-hand side of REF contains MATH such terms. |
math/9910022 | The evolution equations are straightforward calculations. Together with REF , they imply that MATH and MATH remain closed if they are initially. To prove the assertions about exactness, suppose that MATH and MATH. Let MATH and MATH be solutions of MATH and MATH respectively. Then MATH and similarly MATH . By uniqueness of solutions to parabolic equations, we have MATH and MATH for as long as MATH exists. |
math/9910022 | By REF , there are closed solutions MATH and MATH of REF , respectively, existing for MATH. Set MATH. Then by REF , MATH . Since MATH at MATH, it will suffice to apply the space-time maximum principle REF to MATH for MATH. Notice that for MATH, MATH and define a generalized symmetric connection MATH on MATH for MATH by MATH where the MATH are determined by the NAME connection of MATH, and MATH . Let MATH be the space-time metric on MATH induced by MATH: namely, MATH and MATH for MATH. By REF , MATH is parallel with respect to MATH. One verifies by straightforward calculation that MATH and MATH and MATH where MATH. Thus REF can be written in the form MATH whence the theorem follows by applying REF to MATH for MATH . |
math/9910022 | Suppose MATH exists for MATH, and define MATH where MATH is the curvature of the generalized connection on MATH with MATH and MATH. By hypothesis, MATH at MATH for all MATH and MATH. So MATH for all MATH and all MATH by REF . Now since MATH, it follows from REF that MATH for all MATH, all MATH, and all MATH. In particular, at each fixed MATH, we have MATH for all MATH and MATH, and hence MATH . |
math/9910022 | For an orthonormal frame MATH, take MATH and trace over MATH. |
math/9910022 | If MATH, then direct computation gives MATH . The NAME - NAME Laplacian of MATH is MATH . The identity MATH implies in particular that MATH, and hence MATH . Thus MATH . But since the NAME tensor is MATH-invariant, we have MATH so that MATH . Hence MATH. The converse is proved similarly. |
math/9910022 | The choice MATH satisfies MATH by the preceding lemma, and MATH . If MATH for some smooth function MATH, then MATH will satisfy MATH if and only if MATH . The choice MATH satisfies MATH . So to apply REF , we need only check that MATH at MATH for any MATH-form MATH and MATH-form MATH. Noting that MATH, we compute MATH . This is clearly non-negative at MATH whenever the curvature operator is. Hence REF implies in particular that MATH for all MATH and MATH at any MATH such that the solution exists. |
math/9910022 | Write the area element of MATH as MATH, and suppose that MATH for some smooth function MATH. Then using the standard fact that MATH, we get MATH . Because MATH, it follows that MATH satisfies the NAME heat equation MATH if and only if MATH evolve by REF . Now suppose MATH for some smooth function MATH. Then since MATH it follows that MATH satisfies MATH if and only if MATH . Hence we can apply REF with any solution MATH of REF . To apply REF , we first calculate MATH . Then we need only note that MATH and set MATH in the resulting expression: MATH . |
math/9910022 | The choice MATH satisfies MATH . Then if MATH, we get MATH . Hence the trace LYH quadratic takes the form MATH which is certainly positive at MATH. |
math/9910024 | We show that the NAME class MATH is invertible. The isomorphism between MATH and MATH is then given by multiplication by this NAME class. The class in MATH represented by the map MATH induced by the classifying map MATH is the multiplicative inverse of MATH. The product of this class with MATH is homotopic to the unit map MATH. |
math/9910024 | The left-hand side MATH is a ring because MATH is a MATH-space via the equivalence MATH . To compute the left-hand side, apply MATH to the identification MATH. After applying the suspension isomorphisms MATH, the maps in the resulting directed system are multiplication by the MATH. |
math/9910024 | The restriction map is a fibration whose fiber at a given point is the space of MATH-maps which are specified on MATH. Using the skeletal filtration of MATH, we can then filter this mapping space by spaces MATH such that the maps are specified on the boundary of MATH, and where MATH is a proper subgroup of MATH. A standard change-of-groups argument yields that this mapping space is homeomorphic to MATH, again with the map specified on the boundary. But MATH is contractible, and thus so are these mapping spaces. Thus, the fiber of the restriction map is contractible. |
math/9910024 | From the definition of MATH, consider MATH . Applying REF , the restriction from this mapping space to MATH is a homotopy equivalence. Choosing MATH, we see that MATH is an entry of MATH. The bonding maps clearly commute with these restriction to fixed sets maps, so we have an equivalence of spectra. |
math/9910024 | The universality of MATH implies that MATH is a classifying space for MATH-dimensional complex MATH-vector bundles over trivial MATH-spaces. Using REF we see that this classifying space is weakly equivalent to MATH . Over each component of this union, the universal bundle decomposes as MATH, where MATH is the universal vector bundle over the factor of MATH corresponding to the trivial representation. The fixed set MATH is all of MATH while the fixed set MATH is the zero section. The result now follows by passing to NAME spaces. |
math/9910024 | This theorem is simple computation after REF . We use the computation MATH as rings, where MATH is represented by MATH mapping to MATH via its inclusion into MATH, which is standard as in CITE. Because MATH is a free MATH-module, it follows from the NAME theorem that MATH is a polynomial algebra as well. To finish the computation, we note that the NAME class MATH maps to the class in of MATH which is the generator on the MATH-th summand. |
math/9910024 | Let MATH be a complex MATH-bundle over MATH whose underlying real bundle is MATH, as given by the tangential unitary MATH-structure of MATH. Then by REF , MATH decomposes as a complex MATH-bundle MATH where MATH has trivial MATH-action. But we can identify MATH as having underlying real bundle equal to MATH. So the normal bundle MATH underlies MATH, which gives MATH the desired complex structure. |
math/9910024 | We use homogeneous coordinates. There are two possible components of the fixed sets. The points whose coordinates ``in MATH" are zero, constitute a fixed MATH, whose normal bundle is the tautological line bundle over MATH tensored with MATH. As a class in MATH, this manifold with reference map to MATH represents MATH. Alternately, when all other coordinates are zero the resulting submanifold is the space of lines in MATH, which is an isolated fixed point when MATH is one-dimensional and is a projective space with no fixed points, as MATH has no non-trivial invariant subspaces, when MATH has higher dimension. As classes in MATH, isolated fixed points represented negative powers of NAME classes. |
math/9910024 | Recall that from REF we have that for nice groups MATH . It suffices to consider the image of MATH in this ring. The NAME classes and their inverses are in this image by definition. And by REF , the classes MATH are in this image modulo negative powers of NAME classes. |
math/9910024 | We construct the appropriate NAME sequence and show that it breaks up into short exact sequences. Apply MATH to the cofiber sequence MATH to get the long exact sequence MATH . As MATH has suspension isomorphisms for any representation, MATH. By REF , MATH. The map MATH is by definition multiplication by MATH. To compute MATH, we note that for a non-trivial irreducible representation of a torus MATH is homeomorphic to the orbit space MATH. But maps from this orbit space to MATH are in one-to-one correspondence with maps from a single point in the orbit to the MATH-fixed set of MATH, which is homeomorphic to the MATH-fixed set of MATH. We deduce that MATH and that MATH is the restriction map. By NAME 's theorem REF , both MATH and MATH are concentrated in even degrees. Hence the long exact sequence above yields the short exact of the theorem. |
math/9910024 | By REF , any class in MATH can be multiplied by an NAME class to give a class in MATH modulo the kernel of the canonical map from MATH to MATH, where MATH is the multiplicative set of non-trivial NAME classes. By REF , the kernel of the map from MATH to the ring one obtains by inverting a single NAME class is injective, so it follows that the map to MATH is injective. Hence, any class in MATH is the quotient of some class in MATH by an NAME class. Thus, we may filter MATH exhaustively as MATH where MATH is obtained from by adjoining to MATH all MATH such that MATH for some MATH. By REF the set of all such MATH for a given MATH is MATH. The kernel of the restriction map is clearly generated by all classes MATH. So we may obtain MATH from MATH by applying MATH to every class in MATH, which proves that MATH is generated over the operations MATH by MATH. Next, we note that the relations are readily verifyable. Relation REF holds by definition. And we may use the fact that multiplication by non-trivial NAME classes is a monomorphism to verify relations REF by multiplying them by MATH, and REF by multiplying it by MATH. We are left to show that the relations are complete. For convenience, if MATH is a MATH-tuple of representations let MATH. We fix an ordering the representations of MATH. We claim that a multiplicative generating set for MATH is given by classes MATH where MATH is a generator of MATH and MATH is a (possibly empty) MATH-tuple of representations which respects the ordering we have imposed, as well as classes MATH, where MATH is the minimimal representation in MATH. By the relation REF, to construct a generating set it suffices to consider classes MATH, where MATH is either primitive itself or a product of classes in the image of MATH. And by relation REF, it suffices to consider within those classes only the ones MATH where MATH is greater than any of the representations in MATH. Next we give an additive basis for MATH. Fix an ordering on the generators of MATH in which MATH if MATH is in the image of some MATH where MATH is less than any MATH such that MATH is in the image of MATH. We define an additive basis for MATH in two families. Basis elements in the first family are the monomials MATH, where MATH is a monomial in MATH, MATH respects our ordering on the representations of MATH, MATH is a generator of MATH which is not in the image of MATH where MATH is the minimal element of MATH, under our ordering MATH is greater than any of the MATH, and for each representation MATH and each MATH we have that MATH. Basis elements in the second family are MATH, where MATH respects our ordering on the representations of MATH and each MATH is a generator of MATH in the image of MATH where MATH is the minimal element of MATH. We check that this basis is linearly independent by mapping to MATH. Define a multiplicative basis of MATH using the images of elements of MATH along with the multiplicative inverses of NAME classes. Extend our ordering of generators of MATH to an ordering of generators of MATH in which the inverses of NAME classes are less than any generator of MATH. Now order the monomials in MATH by a dictionary ordering. Then the image of an additive basis element as defined above MATH . These images are clearly linearly independent. Finally, we show that the relations suffice to reduce any product of multiplicative generators to a linear combination of additive basis elements. Let MATH be a monomial in the multiplicative generators defined above. If MATH with MATH, we may use relation REF to express MATH as a sum of terms MATH and then use relation REF to simplify. We may thus reduce so that for each representation MATH we do not have both MATH and MATH appearing in MATH. Next, note that REF gives rise to the following relation MATH . We may use this relation repeatedly so that all of the operations which appear in MATH are applied to a single generator of our choosing. So we reduce to terms of the form MATH, where MATH is greater than any MATH in our ordering of generators of MATH. Finally, we use the relation REF to reorder the representations which appear in MATH. Note that when we do so we may get terms MATH which violate one of our conventions in that MATH could be less than some generators which appear in MATH. We may then use REF so that the operations are being applied to a maximal generator, followed by relation REF to reorder. This process terminates. At each stage we may associate a monomial in MATH to a product of our generators of MATH by forgetting all operations MATH. After an application of the relation REF this associated monomial will be strictly smaller for each term than the associated monomial for the original product. Once the associated monomials are of the minimal form MATH where MATH is minimal among representations appearing in the indexing set for operations, we may use REF to equate the term with an additive basis element in the second family. |
math/9910024 | As the localization map is injective it suffices to check the equality in MATH. By REF we can compute the image of MATH, MATH and MATH in the localization at a full set of NAME classes by computing fixed sets with normal bundle data. The result follows easily as the fixed sets of MATH are those of MATH crossed with MATH (when in the notation of REF above, MATH is fixed and MATH) along with a MATH crossed with MATH (when MATH). In the localization, crossing with MATH coincides with multiplying by MATH. |
math/9910024 | By REF the image of MATH under completion is MATH . For any MATH we have that MATH because MATH is strongly mulitplicative and by REF is the difference between a twisted product and a trivial product of MATH and MATH. |
math/9910024 | As the map from MATH to its completion is a map of complex-oriented equivariant cohomology theories, the NAME class of the bundle MATH over a point gets mapped to the NAME class of MATH over MATH. For MATH, MATH the resulting bundle is the MATH-th-tensor power of the tautological bundle over MATH, whose NAME class is by definition the MATH-series. |
math/9910024 | The first condition on MATH says that the image contains all images of classes in MATH. Indeed, MATH is the image of the NAME classes. And we check that the image of MATH in MATH is MATH, which follows from the fact that the MATH action on MATH is pulled back from the MATH action on MATH by the degree MATH homomorphism from MATH to itself. By REF we may build any class in MATH by dividing classes in MATH by NAME classes. The second condition on MATH accounts for all possible quotients by NAME classes in the image. |
math/9910024 | For convenience, let refer to the NAME class MATH by MATH. A complex MATH manifold MATH with three isolated fixed points defines a class in MATH whose image under MATH is MATH for some integers MATH. We let MATH denote MATH . Without loss of generality, assume MATH is greatest of the integers MATH in absolute value. As MATH is divisible by MATH in MATH, REF implies that MATH restricted to MATH must be zero. The NAME class MATH restrict non-trivially to MATH unless MATH. Therefore one of MATH, say MATH must be equal to MATH. We first claim that this number must be MATH. Let MATH denote the multiplicative set generated by all the NAME classes associated to irreducible representations except for MATH. By localizing the modules in REF , we find that MATH is generated over the operation MATH by MATH. Suppose that MATH and that MATH. Then MATH is in the image of the canonical map from MATH to MATH, as it is actually in the image of MATH. Then we must have that MATH is divisible by MATH and thus is zero in MATH where MATH here is the multiplicative set of all NAME classes of MATH. This localization of MATH is the the target of the restriction map from MATH. And by abuse we are using the same names for NAME classes for different groups. But because MATH, MATH, MATH and their sum are non-zero in MATH. It is straightforward to rule out cases where some of MATH are equal to MATH. Next, consider the class MATH where MATH is the canonical map to this localization. Its image under the map to the full localization is MATH which implies that MATH is divisible by MATH in MATH or that MATH. But because MATH we have that MATH depending on whether MATH is positive or negative. Finally, as MATH and MATH consider MATH, which will be equal MATH. Case analysis of necessary divisibilities as we have been doing implies that this difference must be zero, so that the fixed-set data of MATH is isomorphic to that of MATH. Finally, by because the localization map MATH is injective, this fixed-set data determines MATH as in MATH-equivariant homotopical bordism uniquely, so that MATH must equal MATH in MATH. But REF says that the NAME map from MATH to MATH is injective for abelian groups MATH. Hence MATH is cobordant to MATH. |
math/9910024 | We use an analog of the simple fact that if MATH is a MATH-manfold and MATH has a free MATH-action then MATH in MATH, where MATH is the boundary of a tubular neighborhood around the fixed set MATH. The null-bordism is defined by MATH. If the fixed points of MATH are isolated, this will give rise to a relation among spheres with free MATH-actions. Let MATH where MATH is a quotient of MATH by MATH in MATH. Inductively, let MATH be a quotient of MATH by MATH note that this quotient is not unique as we are working in MATH equivariant bordism. Then the ``fixed sets" of MATH are given by MATH . As MATH corresponds to a tubular neighborhood of an isolated fixed point in geometric bordism, we can deduce via transversality arguments for free MATH-actions that MATH in MATH. But the image of MATH in MATH is MATH from which we can read off that MATH. Note that our expressions in MATH are independent of the indeterminacy in choosing MATH and the MATH. |
math/9910025 | Any singular manifold with reference to MATH must have free boundary. Conversely, given a MATH-manifold with free boundary, there is no obstruction to constructing a reference map to MATH. Applying these observations to manifolds which play the role of bordisms, we see that these correspondences are well-defined up to bordism and inverse to each other. |
math/9910025 | Let MATH, with ``straightened angles". Then MATH is free outside of MATH and MATH, so MATH is the required bordism. |
math/9910025 | Use the identifications we have made so far to equate MATH with the bordism module of MATH-vector bundles over trivial MATH-spaces where the action is free away from zero. The fiber of a MATH-vector bundle over a trivial MATH-space is a representation. Because MATH has only one non-trivial representation, the action on any fiber and thus the total space is completely determined. Hence the forgetful map from this bordism module of MATH-vector bundles which are free away from the zero section to the the bordism module of vector bundles is an isomorphism. The result follows from the fact that MATH is the classifying space for vector bundles. Note that we must grade according to the dimension of the total space of the bundle in question. |
math/9910025 | Consider the following diagram: MATH . Given a representative MATH with reference map MATH to MATH, pull back the principal MATH-bundle MATH to get MATH, which is in fact a free MATH-manifold. Conversely, starting with a free MATH-manifold MATH, there is no obstruction to constructing a map MATH to MATH. Pass to quotients to obtain MATH. These maps are well-defined, as we apply the previous argument to the manifolds which act as bordisms. The composites of these maps are clearly identity maps. |
math/9910025 | The MATH homology of MATH is well known to be MATH in every positive dimension. The class in dimension MATH is the image of the fundamental class of MATH under inclusion. Under the identifications of REF these classes correspond to spheres with antipodal action. |
math/9910025 | Apply MATH to the identification MATH. After applying the suspension isomorphisms MATH, the maps in the resulting directed system are multiplication by the MATH. |
math/9910025 | Recall the definition of MATH and consider the space of maps from MATH to MATH, for any representation MATH. First we show that for any MATH-spaces MATH and MATH, the restriction map MATH is an equivalence. First note that this restriction is a fibration. Over a given component of MATH a fiber is going to be the space of maps from MATH to MATH which are specified on MATH. We filter this mapping space by filtering MATH. Because the maps are already specified on MATH, we need only adjoin cells of the form MATH, where MATH denotes MATH acting on itself by left multiplication. Hence the subquotients in this filtration will be spaces of equivariant maps from MATH to MATH whose restriction to the boundary of MATH is specified. Because MATH is a free MATH-space, it suffices to consider the restriction of such a map to one copy of MATH. But MATH is contractible, hence so is this mapping space. Therefore the fibers of our restriction map are contractible. Applying this argument for MATH, MATH we see that our computation follows from knowledge of MATH. We claim that MATH. We show this by analysis of the fixed set of MATH. Any fixed point of MATH must lie over a fixed point of MATH. But the fixed set of MATH is the classifying space for MATH-vector bundles over trivial MATH-spaces. A vector bundle over a trivial MATH-space decomposes as a direct sum according to decomposition of fibers according to representation type. As there are only two representations types for MATH, we deduce that MATH. Restricted to a component of this fixed set MATH will be MATH where MATH fixes all points in the first factor and acts by multiplication by MATH on fibers in the second factor. Hence one component of the fixed set of MATH will be MATH. Passing to NAME spaces we find MATH. Using this result along with our first reduction we see that MATH . The theorem follows by passing to direct limits. |
math/9910025 | The multiplication on MATH is defined by the maps MATH which are the passage to NAME spaces of the map classifying the product of universal bundles. These maps restrict to MATH as the standard multiplication on MATH factors smashed with the classifying map for NAME sum on MATH factors. Passing to the direct limit and neglecting grading, we are computing MATH, where MATH has a MATH-space structure which is the product of the group structure on MATH and the MATH-space structure on MATH arising from NAME sum. The computation follows from the NAME theorem, as MATH is a NAME polynomial ring, which by our grading conventions is generated by a class we call MATH in degree MATH, and MATH is a polynomial ring in classes MATH where MATH is the image of the generator of MATH under the inclusion from MATH to MATH. That MATH is the image of MATH follows directly from their definitions, chasing through the identifications of REF . |
math/9910025 | This theorem is immediate as an application of NAME 's transfer. But in the spirit of giving elementary proofs, we argue geometrically as follows. From the definition of MATH consider a MATH-map from MATH to MATH. Because the latter space has a free MATH-action away from the basepoint, MATH must map to the basepoint. If we pass to the map from the quotient MATH, we have a MATH-map between MATH-spaces which are free MATH-manifolds away from their basepoints. Because transversality is a local condition, it is easy to verify that transversality arguments hold in the presence of free MATH-actions. Given a MATH-map from MATH to MATH we may homotop it locally to a map which is transverse regular to the zero section of MATH and pull back a sub-manifold of MATH which must necessarily be free. So following classic techniques we identify MATH with the bordism module of free MATH-manifolds. The theorem follows from REF . |
math/9910025 | The proof of this proposition is almost immediate, as NAME map MATH is a natural transformation of equivariant homology theories and our exact sequence of a pair from which we defined the NAME sequence coincides with the cofiber sequence from which we defined the tom NAME sequence. That the NAME map is an isomorphism when smashed with MATH follows from the fact that transversality arguments carry through in the presence of a free MATH-action. That the third vertical map is the standard inclusion follows from close analysis of the NAME map in this setting. |
math/9910025 | We show that the images of MATH and MATH along with MATH generate MATH. We do so using the explicit description of MATH from REF . Consider the following diagram: MATH which combines results of REF . To compute the image of MATH under localization it suffices to look at fixed-set data, because it is a geometric class. There are two components of the fixed set MATH. Using homogeneous coordinates MATH, these components are defined by the conditions MATH and MATH. The condition MATH defines a MATH dimensional projective space. Its normal bundle is the tautological line bundle. The condition MATH defines an isolated fixed point which has a MATH-dimensional normal bundle. Using the generators named in REF we have MATH . In REF we also noted that the image of MATH under localization was MATH. It thus follows that the images of MATH and MATH under localization, along with MATH, generate MATH, which is what was to be shown. |
math/9910025 | The exact sequence in question is a NAME sequence. Apply MATH to the cofiber sequence MATH where the first map is projection of MATH onto the non-basepoint of MATH. The resulting long exact sequence is MATH . By the periodicity of MATH, MATH. By REF is multiplication by MATH. From the fact that MATH is homeomorphic to MATH for any MATH-space MATH, we see that MATH and MATH is the augmentation map MATH. As we remarked after the statement of REF , the augmentation map MATH is split. Hence our long exact sequence breaks up into short exact sequences, and the result follows. |
math/9910025 | First we verify relations. Then we show that the classes listed generate MATH. Finally we show that the relations are a complete set of relations. It is convenient to view MATH as a subring of MATH, which we can do as the previous theorem implies that MATH is not a zero divisor. In this way, we may verify the second family of relations by direct computation. The first family of relations holds by definition. For convenience, rename MATH as MATH. By REF , any class in MATH when multiplied by some power of MATH is equal to a class in MATH modulo the annihilator ideal of MATH, which is zero. Hence we may filter MATH exhaustively as MATH where MATH is obtained from by adjoining to MATH all MATH such that MATH. By REF the set of all such MATH is MATH. The kernel of the augmentation map is clearly generated by all classes MATH. So we may obtain MATH from MATH by applying MATH to every class in MATH. Since MATH it suffices to apply MATH only to primitive elements. It follows that MATH and MATH constitute multiplicative generators. Finally, to show these relations are complete we identify an additive basis of MATH. There are two types of monomials in the additive basis, those of the form MATH, MATH and those of the form MATH, MATH. We may check that these classes are additively independent by mapping to MATH. Define the complication of a monomial in our basis elements to be the sum of the number of times both MATH and MATH appear in the monomial and the sum of all MATH where MATH appears for some MATH where MATH is not minimal among the MATH which appear. And define the complication of a sum of monomials to be the greatest of their individual complications. We may use our two families of relations to decrease complication, which inductively allows us to reduce to our additive basis whose members have zero complication. |
math/9910025 | Once again we use the fact that the map from MATH to MATH is a faithful representaion, along with direct computation. By REF , we may compute the image of MATH under localization by analyzing fixed-set data. Recalling the definition of MATH we see two types of fixed points MATH under the MATH-action, those with MATH and those with MATH and MATH. The first fixed set is MATH, with a trivial normal bundle. The second fixed set is the fixed set of MATH, whose normal bundle is the normal bundle of this fixed set in MATH crossed with a trivial bundle. By the fact that multiplication by MATH in MATH corresponds geometrically to crossing with a trivial bundle, this fixed set is the fixed set of MATH. |
math/9910025 | By analysis identical to that in the proof of REF , the map MATH is injective. We deduce from the comparison of exact sequences in REF that the NAME map from MATH to MATH is injective. So the image of MATH is the image of MATH in the subring MATH of MATH. The images of MATH generate this image, so these classes generate MATH. |
math/9910025 | This theorem follows almost immediately from REF . Any monomial of the form MATH, MATH is in fact in the image of the NAME map. Monomials of the form MATH, MATH are generated over MATH by MATH which we denote by MATH. The module relations for the quotient follow from the ring relations for MATH. |
math/9910026 | Suppose that MATH, then since MATH is simply connected MATH, that is we have a homotopy MATH agreeing with MATH and MATH on the boundary. Thus taking MATH with the obvious identifications with MATH at on the boundary we see we have a morphism from MATH to MATH. It is clear this morphism is invertible and hence MATH. Conversely, suppose that MATH via a morphism MATH with inverse MATH . Then the composition MATH must be identified with the identity morphism on MATH. In particular the glued surface MATH must be diffeomorphic to MATH in which case MATH and MATH must be made up of cylinders and so MATH. |
math/9910026 | First assume MATH is closed, connected of genus MATH. There is a cofibration MATH where MATH is the ``commutator" map. Applying MATH we get a cofibration MATH . The long exact homotopy sequence gives MATH is trivial since MATH is null by homotopy commutativity. Since MATH is trivial it follows that the middle map is an equivalence. Moreover for simply connected MATH we have MATH so MATH. Next assume that MATH is connected, genus MATH and MATH with MATH boundary components. Let MATH be the surface obtained from MATH by sewing in MATH discs. Let MATH be a point in the MATH'th disc. There is a fibration MATH whence an exact sequence MATH . The extremities are trivial so MATH. Thus MATH . Finally for a general surface each of its MATH connected components (closed or not) corresponds to an element of MATH by the above. |
math/9910026 | Gluing of two connected components corresponds to multiplication in MATH, and for the MATH'th component of MATH we multiply all components of MATH and MATH forming that component. |
math/9910026 | Let MATH be the subcategory of MATH with objects those of MATH and morphisms MATH . Note that the morphisms of MATH are closed under composition by REF . The correspondence MATH then induces an equivalence of categories MATH. The composition MATH gives a NAME algebra structure on MATH by REF. |
math/9910027 | For MATH, we have MATH . |
math/9910027 | Since MATH commutes with MATH, MATH is a DGA. Consider the sequence MATH where MATH is the inclusion, and MATH is the quotient map. Both MATH and MATH are homomorphisms of DGA's. It suffices to show that MATH induces the zero homomorphism on MATH. This is easy to see since every element of MATH is represented by an element in MATH. Furthermore, they are quasi-isomorphisms since all the cohomology groups can be identified with MATH. It is trivial for the first and third DGA's. For the second DGA, it follows from the decomposition MATH . By REF , MATH. This completes the the proof. |
math/9910028 | From REF , it is easy to see that REF follows easily from REF . MATH . |
math/9910028 | From REF we get MATH . The logarithm of the last term is MATH . When MATH, MATH only if MATH and MATH, hence MATH . |
math/9910028 | We take MATH in REF . For MATH even, we have MATH . For MATH odd, we have MATH . |
math/9910028 | We use REF for MATH, where MATH for some MATH: MATH . |
math/9910029 | By splitting principle, we may assume that MATH, where MATH's are line bundles. Then MATH . Consider the index set MATH. The action of MATH or equivalently the cyclic group MATH on MATH corresponds to the cycling of the indices: MATH . For each MATH, let MATH, and MATH be the subbundle spanned by line bundles MATH, MATH. Then MATH is invariant under the action of MATH. Since MATH is a direct sum of such MATH's, it suffices to find MATH for all MATH. When MATH, MATH is fixed by MATH. In other words, MATH has eigenvalue MATH on MATH. When MATH are not all identical, there are two cases to consider. If the orbit of MATH has length MATH, then MATH is the direct sum of MATH-copies of MATH, and each fiber is a regular representation of MATH. Therefore, MATH has eigenvalues MATH, MATH. Hence MATH . Another case is that the orbit of MATH has length MATH. Then it is easy to see that MATH, MATH, that is, MATH is of the form MATH . MATH is isomorphic to the direct sum of MATH copies of MATH. Furthermore, MATH acts on MATH via MATH-action which correpsonds to MATH. Therefore, one sees that MATH has eigenvalues MATH, MATH. Hence MATH . This completes the proof. |
math/9910029 | Denote by MATH and MATH the NAME roots of MATH and MATH respectively, then we have MATH for some homogeneous polynomial MATH of degree MATH. Now from REF we have MATH . |
math/9910029 | Combining REF with REF , we get MATH . |
math/9910029 | Combining REF with REF for MATH, we get MATH . |
math/9910029 | By splitting principle, we may assume that MATH, where each MATH's is a graded line bundle of degree MATH. Then MATH . For each multiple index MATH, agian let MATH, MATH the subbundle spanned by line bundles MATH, MATH. Then MATH is invariant under the action of MATH. Since MATH is a direct sum of such MATH,s, it suffices to find MATH for all MATH. When MATH, MATH and MATH acts by the multiplication of MATH . In other words, MATH . When MATH are not all identical, we modify the proof of REF by computing the characters as explained in REF to show MATH. |
math/9910029 | As in REF, we have MATH . From REF , we get MATH . Without loss of generality, we may assume that MATH, where each MATH's is a graded line bundle of degree MATH. Let MATH. Also let MATH be the NAME roots of MATH. Since MATH we have MATH . Here for the second equality we have used the same argument as in the proof of REF . This completes the proof. |
math/9910033 | If MATH, MATH, one can choose MATH such that MATH and MATH by making the support of MATH sufficiently small. Then we can take MATH to be the zero operator, showing that MATH. Thus, REF proves the corollary. |
math/9910033 | Let MATH be a metric defining the topology on MATH. Suppose MATH is not equicontinuous. That is, suppose that there exists MATH, sequences MATH, MATH in MATH, MATH, such that MATH but MATH. Since MATH is compact, one can pass to subsequences (which we do not show in the notation) such that MATH and MATH converge to points MATH and MATH respectively. Note that MATH since MATH. For any MATH-invariant function MATH, MATH is uniformly NAME (that is, the NAME constant is independent of MATH), so MATH shows that MATH; MATH independent of MATH. But MATH is continuous, so MATH, MATH, so we conclude MATH. But let MATH be such that MATH, MATH. All functions on MATH, pulled back by the bundle projection to MATH, are MATH-invariant, so we see that MATH as well. But MATH is MATH-invariant near MATH for all MATH, so this shows MATH, a contradiction. |
math/9910033 | Let MATH be a MATH-invariant function, MATH, MATH, and let MATH . We need to show that MATH. We only consider MATH for the sake of definiteness. So let MATH; we need to prove that there exists MATH such that for all MATH, MATH. But by the continuity of MATH on MATH, there exists a neighborhood MATH of MATH in MATH such that MATH for all MATH, MATH, MATH. Next, by the uniform convergence of the MATH, there exist MATH and MATH such that for MATH, MATH, MATH. Let MATH . Thus, for MATH, MATH, MATH satisfies MATH, hence MATH is non-decreasing, so for MATH, MATH, MATH . Now given MATH, simply choose MATH such that MATH, MATH, which is possible since MATH is continuous, so MATH for all MATH. Thus, using this particular value of MATH and REF, we conclude that MATH . This holds for every MATH, completing the proof. |
math/9910033 | Since MATH is equicontinuous and MATH is compact, any sequence of broken bicharacteristics in MATH has a convergent subsequence, converging uniformly over MATH, by REF. But the uniform limit of broken bicharacteristics is a broken bicharacteristic, proving the proposition. |
math/9910033 | Let MATH be given by MATH, where MATH is chosen sufficiently small. By the previous proposition, there exists a subsequence of MATH that converges to a broken bicharacteristic MATH. But then MATH, so MATH gives the desired extension of MATH to MATH. The other endpoint can be dealt with similarly. |
math/9910033 | For the sake of definiteness we take MATH, all other cases are very similar. Since MATH is monotone decreasing, and MATH is a bounded function on MATH, MATH exists (we wrote MATH). Now, let MATH be any sequence such that MATH for all MATH. Let MATH given by MATH. By REF , MATH has a subsequence, which we write as MATH, which converges uniformly to a generalized broken bicharacteristic MATH. Thus, for MATH, MATH, so MATH is constant on MATH. Since MATH is NAME, it is equal to the integral of its a.e. defined derivative, MATH, so we conclude that MATH a.e. along MATH. But then the arclength of the projection of MATH to MATH, which is the integral of MATH, is also zero, so MATH is a constant curve, and hence, as shown above, there exists MATH such that MATH for all MATH. Thus, MATH . Also, MATH, and MATH, so MATH. We still need to show that if MATH is a sequence in MATH, and MATH converges in MATH, say MATH, then either MATH, or MATH. Now, if MATH has a subsequence converging to some MATH, then MATH converges to MATH, so we are done. We may thus assume that MATH as MATH. By the argument of the previous paragraph, there is a subsequence MATH such that MATH converges uniformly (over MATH) to a generalized broken bicharacteristic MATH, which is a constant curve, MATH for MATH, in MATH. In particular, MATH converges to MATH. But on the other hand, MATH, so MATH. This proves that MATH . |
math/9910033 | The main step in the proof is the construction of an operator which has a microlocally positive commutator with MATH near MATH. In fact, we construct the symbol of this operator. This symbol will not be a scattering symbol, that is, it will not be in MATH, only due to its behavior as MATH corresponding to its MATH-invariance. This will be accommodated by composing its quantization with a cutoff in the spectrum of MATH, MATH, MATH supported near MATH, as discussed in REF . This approach simply extends the one taken in CITE CITE, though the actual construction is different due to the more complicated geometry. We introduce some notation and then fix MATH. We define MATH by MATH . Note that at MATH, where MATH, MATH and if MATH, then MATH when MATH. Thus, if MATH, then MATH . Now fix MATH and MATH to a neighborhood of MATH in MATH such that MATH, MATH (this is possible since MATH is closed), MATH is inside a fixed system of local coordinates, and MATH . If MATH, we make an additional definition. Namely, we let MATH be given by MATH . We will use MATH as a small parameter that microlocalizes in a neighborhood of MATH if MATH, and in a neighborhood of MATH if MATH; if MATH we will always take MATH. Employing an iterative argument as usual, we may assume that MATH and we need to show that MATH. (We can start the induction with a MATH such that MATH.) Our positive commutator estimates at a point MATH will arise by considering functions MATH where MATH localizes in the tangential variables MATH, MATH, MATH. In fact, if MATH, then MATH, so we construct MATH to be a `quadratic distance' (in MATH) from the MATH integral curve through MATH constructed so that MATH. That is, we define MATH on a hypersurface through MATH that is transversal to MATH, for example, MATH, to be given by a positive definite quadratic form in some local coordinates centered at MATH, for example, MATH, and extend it to MATH to be constant along the MATH integral curves; compare CITE. We can then extend MATH to a function on MATH defined near MATH as discussed before REF, and then MATH . On the other hand, if MATH, we can take MATH now MATH so MATH. Thus, for MATH sufficiently large, we see that for MATH since MATH on MATH, hence near MATH. Since MATH depends on the tangential variables only, we conclude in either case that for all MATH, MATH . Moreover, MATH under the decomposition MATH, so MATH. Now suppose that MATH, MATH . Since MATH and MATH, the first of these inequalities implies that MATH, and the combination of these two gives MATH as MATH. Then we conclude that MATH . Note that for all MATH with MATH, MATH at MATH. Thus, if MATH we see that MATH . Thus, using REF as well, we deduce that there exists MATH (independent of MATH) such that MATH . Note that here MATH is arbitrary. Similarly, suppose that MATH, MATH . Then we conclude that MATH . Thus, from REF, MATH . The positive commutator estimate then arises by considering the following MATH-invariant symbol MATH and quantizing it as in REF . Let MATH be equal to MATH on MATH and MATH for MATH. Thus, MATH, MATH, and MATH, MATH. Let MATH be MATH on MATH, MATH on MATH, with MATH and MATH on some small interval MATH, MATH. Furthermore, for MATH large, to be determined, let MATH if MATH, and let MATH if MATH. Thus, MATH, and on MATH we have MATH and MATH which are exactly REF, so MATH is a subset of REF respectively. We also see that as MATH decreases, so does MATH, in fact, if MATH then MATH on MATH. Note that in REF , by reducing MATH, we can make MATH supported in an arbitrary small neighborhood of a compact backward bicharacteristic segment through MATH, and in REF , by reducing MATH, we can make MATH supported in an arbitrary small neighborhood of MATH. We at once obtain positivity estimates for MATH. The following argument works similarly for both MATH and MATH; we consider the slightly more complicated case MATH. Thus, if MATH, then MATH . We break up the first term by using a cutoff that ensures that the hypothesis in REF is satisfied. Thus, let MATH so MATH . Then MATH with MATH . Hence, with MATH using REF, we deduce that MATH . Moreover, MATH since MATH on MATH, so MATH . On the other hand, MATH is supported where either MATH so near the backward direction along MATH bicharacteristic through MATH, or MATH . But by our assumption MATH, MATH, so the same holds for a sufficiently small neighborhood of MATH as MATH is closed. By choosing MATH sufficiently small, we can thus make sure that the region defined by REF is disjoint from MATH. Moreover, by further reducing MATH if necessary and using our second assumption, we can also make sure that the region REF is also disjoint from MATH, so that MATH is disjoint from MATH for all MATH. Moreover, by REF, for MATH sufficiently small, we deduce from the inductive hypothesis that MATH (hence MATH) is disjoint from MATH. Moreover, with MATH denoting a partial derivative with respect to one of MATH, MATH . At any MATH with MATH, defined by MATH, MATH, as above, MATH is independent of MATH at MATH so outside MATH . In fact, outside MATH, but in the set where MATH is positive, MATH so the uniform bounds of REF also follow. In addition, at any cluster MATH, MATH, MATH, and MATH, MATH are MATH-tangential variables, so MATH, hence MATH has the form REF around each MATH. Let MATH be identically MATH near MATH and supported close to MATH. We also define MATH . Thus, MATH is a MATH-invariant function satisfying REF. Let MATH be the operator given by REF with MATH in place of MATH, so in particular its indicial operators are MATH. Note that REF holds with MATH. So suppose that MATH and MATH. Choose MATH so large that MATH, and let MATH be the complement of MATH in MATH, and let MATH; so MATH by our choice of MATH. By REF we deduce that there exists MATH, MATH is supported in MATH, MATH near MATH, MATH, MATH with MATH, MATH such that MATH . Let MATH so MATH for MATH and it is uniformly bounded in MATH. The last statement follows from MATH being uniformly bounded as a REFth order symbol, that is, from MATH uniformly (MATH independent of MATH). We also define MATH . Then, with MATH identically MATH near MATH, MATH where MATH is uniformly bounded in MATH. Note that MATH arises by commuting MATH, powers of MATH and MATH through other operators, but as the indicial operators of MATH and MATH are a multiple of the identity, MATH, MATH and MATH commute with these operators to top order, and in case of MATH, the commutator is uniformly bounded as an operator of one lower order. Then, multiplying REF by MATH from the left and right and rearranging the terms we obtain the following estimate of bounded self-adjoint operators on MATH: MATH where MATH and MATH is uniformly bounded in MATH as MATH. Now, MATH is uniformly bounded in MATH, hence as a bounded operator on MATH. Thus, if MATH is chosen sufficiently large, then MATH for all MATH, so MATH . Adding this to REF shows that MATH . The point of the commutator calculation is that in MATH the pairing makes sense for MATH since MATH. Now apply REF to MATH and pair it with MATH in MATH. Then for MATH . Letting MATH now keeps the right hand side of REF bounded. In fact, MATH remains bounded in MATH as MATH. Similarly, by REF, MATH remains bounded in MATH as MATH since MATH. Also, MATH is bounded in MATH, so MATH stays bounded by REF as well since MATH. These estimates show that MATH is uniformly bounded in MATH. Since MATH strongly on MATH, we conclude that MATH. By REF this implies that for every MATH, MATH in fact that MATH for all MATH for which MATH. This is exactly the iterative step we wanted to prove. In the next step we decrease MATH slightly to ensure that MATH is disjoint from MATH, just as NAME decreases MATH in the proof of CITE. |
math/9910033 | As usual, broken bicharacteristic means generalized broken bicharacteristic in this proof. We only need to prove that for every cluster MATH, if MATH then MATH . In fact, if REF holds for all MATH with MATH, let MATH and put the natural partial order on MATH, so MATH if the domains satisfy MATH and MATH. Then MATH is not empty (due to REF) and every non-empty totally ordered subset of MATH has an upper bound, so an application of NAME 's lemma gives a maximal broken bicharacteristic of MATH in the intersection of MATH with MATH which passes through MATH. A similar maximal statement holds if we replace MATH by MATH. Indeed, it suffices to show that for any MATH, if MATH then MATH for the existence of a broken bicharacteristic on MATH can be demonstrated similarly by replacing the forward propagation estimates by backward ones, and, directly from REF , piecing together the two broken bicharacteristics gives one defined on MATH. We proceed to prove that REF implies REF by induction on MATH. This is certainly true for MATH by REF : there are no normal variables MATH, MATH, so MATH in the notation of that Proposition, showing that a segment of the backward bicharacteristic through MATH must be in MATH. Of course, this is simply NAME 's propagation theorem CITE. In addition, if MATH is arbitrary and MATH, then the constant curve MATH through MATH is a broken bicharacteristic, so REF holds with this MATH. So suppose that REF has been proved for all MATH with MATH and that MATH satisfies REF. As noted in the first paragraph, we thus know that the intersection of MATH with MATH is a union of maximally extended broken bicharacteristics of MATH. We use the notation of the proof of REF below. Let MATH be a neighborhood of MATH in MATH as in REF ; we may assume that MATH. By REF , either every point on the non-constant backward MATH-bicharacteristic segment through MATH is in MATH (in the neighborhood MATH of MATH), in which case we have proved REF, so we are done, or there is a sequence of points MATH such that MATH, MATH as MATH, and MATH for all MATH. Since MATH, MATH. Thus, by the inductive hypothesis there exist broken bicharacteristics MATH, MATH, with MATH, MATH for all MATH, and MATH is maximal with this property in MATH. That is, if MATH is an extension of MATH, then MATH, so MATH. By REF we can fix some MATH and extend (or restrict) each MATH to a broken bicharacteristic defined on MATH, which we keep denoting by MATH; by the previous remark MATH. But MATH is non-decreasing (for sufficiently large MATH) along broken bicharacteristics by the argument that preceeds REF , so we conclude that MATH for MATH, hence MATH, so MATH, MATH for all MATH. By REF , there is a subsequence of MATH converging uniformly to a broken bicharacteristic MATH. Since MATH is closed, MATH. In particular, MATH and MATH for all MATH, providing the inductive step. |
math/9910033 | The existence of MATH in MATH, MATH follows from the NAME estimate, as presented in CITE. That MATH implies MATH (in fact, MATH where MATH), follows from the work of CITE; see CITE to see how the proof would proceed with our notation. By REF , MATH is a union of maximally extended generalized broken bicharacteristics. So suppose that MATH, and let MATH be a generalized broken bicharacteristic in MATH with MATH. Then, by REF , MATH exists. If MATH then MATH for large negative MATH, contradicting that MATH on MATH. Thus, MATH, and hence by REF , MATH, so MATH. We thus conclude that REF holds. |
math/9910033 | As mentioned above, the first part follows from the self-adjointness of MATH, so that for MATH, MATH, MATH, we have MATH; recall that the distributional pairing is the real pairing, not the complex (that is, MATH) one. The wave front statement of REF and the assumption on MATH show the existence of the limit MATH in MATH and that in addition MATH for every MATH. The statement MATH follows from a microlocalized version of the NAME estimate due to CITE; see CITE or CITE for a detailed argument. The final part of the conclusion follows from REF , much as in the previous proof. |
math/9910033 | For each point MATH there exists a unique MATH such that MATH. Also, there exists MATH such that MATH lies in MATH, where MATH denotes the open ball of radius MATH in MATH with respect to the standard metric. Now, MATH is an open cover of the compact set MATH, so it has a finite subcover corresponding to some points, say, MATH, MATH. Let MATH. If MATH is as above, let MATH be such that MATH. For any MATH, the distance of MATH and MATH is bounded by the length of MATH, hence by MATH, so the image of MATH lies in MATH which in turn lies in MATH for some MATH. |
math/9910033 | We recall from CITE the explicit arclength parametrization, MATH, of the integral curves of MATH with kinetic energy MATH. In terms of this parameterization for MATH, MATH where MATH varies in a subinterval of MATH, and MATH is such that MATH is an integral curve of MATH with kinetic energy MATH. Since MATH (which we usually just write as MATH) is monotone decreasing, this shows that the total length of the segments of MATH which are integral curves of MATH with any given kinetic energy MATH, is at most MATH. If MATH, the bicharacteristic segment is constant. Since MATH must be such that MATH, there are MATH possible values of MATH, which proves our estimate. |
math/9910033 | In three-body scattering kinetic energy is constant along generalized broken bicharacteristics (essentially because there are no positive energy bound states), so the proof of the previous lemma applies and gives the desired conclusion. |
math/9910033 | Let MATH be a generalized broken bicharacteristic of length MATH, so the image of MATH lies in a region MATH for some MATH. Thus, we can use local coordinates around MATH for describing MATH. By REF and the argument preceeding it (showing that MATH cannot change sign more than once), there exist some points MATH such that on MATH and on MATH the image of MATH is disjoint from MATH, and on MATH, MATH is an integral curve of MATH, where some intervals may be empty or reduce to a point. Consider the interval MATH for the sake of definiteness, and assume that it is non-empty. Let MATH be given by MATH so MATH is the solution of the ODE MATH where we wrote MATH. Thus, MATH is MATH and its derivative is positive, so the same holds for its inverse function, MATH, defined on an interval MATH. We denote by MATH the characteristic variety of the (proper) subsystem Hamiltonian MATH at energy MATH. Now let MATH, given by MATH so in terms of Euclidean coordinates, MATH . It is straightforward to check that MATH is a generalized broken bicharacteristic of MATH; the change of parameters accounts for the change in normalization of the rescaled NAME vector fields. In fact, MATH, while its analog in the subsystem is MATH, and the quotient MATH is MATH. Thus, by the hypothesis, MATH is a broken bicharacteristic. Since MATH is constant along generalized broken bicharacteristics, we conclude that on MATH, MATH is a broken bicharacteristic if and only if MATH is, so by the hypothesis it is a broken bicharacteristic, and it has as many breaks as MATH, so if we assume uniform bounds in the proper subsystems, at most MATH. A similar estimate holds for MATH, so on MATH, MATH is a broken bicharacteristic, and if we also assume uniform bounds in the proper subsystems, it can have at most MATH breaks, proving the lemma. |
math/9910033 | If MATH is a generalized broken bicharacteristic, MATH compact, then its total arclength is finite, so dividing it up into segments of length MATH and applying the previous lemma proves the proposition. |
math/9910033 | Due to the previous corollary, every generalized broken bicharacteristic of MATH is a broken bicharacteristic. By REF , its length MATH satisfies the inequality MATH. Dividing it up into pieces of length MATH, of which (we can arrange that) there are at most MATH, shows that the total number of breaks is at most MATH which is independent of MATH, proving the proposition. |
math/9910033 | We again use the explicit arclength parametrization, MATH, of the integral curves of MATH with MATH. Thus, MATH for MATH where MATH varies in a subinterval of MATH. This gives MATH . If MATH then both MATH and MATH vanish. Summing over MATH and applying the NAME inequality to the left hand side proves the lemma. |
math/9910033 | We only need to show that there exists MATH as above such that for every compact interval MATH, MATH has at most MATH breaks. Let MATH denote the number of breaks in MATH. Note that MATH is monotone decreasing and it is bounded, with a bound given by MATH (since MATH in MATH). By REF , the total length of bicharacteristic segments with kinetic energy at least MATH is at most MATH, while by the proof of REF , the total length of bicharacteristic segments with kinetic energy less than MATH is at most MATH. Thus, MATH . On the other hand, dividing MATH into segments of length MATH, of which we can arrange that there are at most MATH, and applying REF , shows that MATH . Dividing through by MATH gives the desired estimate for MATH, uniform in the energy MATH as long as MATH stays in a bounded set. |
math/9910033 | Note first that MATH is closed, so for MATH, MATH. The kinetic energy MATH on any bicharacteristic segment satisfies MATH, so MATH. Applying REF with MATH, MATH, completes the proof. |
math/9910033 | The proof proceeds by induction. Thus, let MATH we need to show that for each MATH, given that MATH is smooth Lagrangian, so is MATH. By the usual argument, this certainly follows if the intersection of MATH with the partial diagonal MATH in MATH is transversal, that is, for MATH, MATH . Here `partial' means that we take the diagonal in the central factor MATH, and for the sake of simpler notation we wrote MATH . In this case the tangent space of the composite relation, MATH is given by the projection of MATH to the first and last factors, that is, to MATH. The transversality is equivalent to MATH over MATH, and then MATH is given by the projection of MATH to the first and third factors. Transversality certainly follows if we can find a Lagrangian subspace MATH of MATH such that MATH is in the range of the differential of the projection MATH, and MATH and in this case the Lagrangian subspace MATH given by the projection of MATH to MATH, is in the range of the differential of the projection MATH. Since the previous statements involving MATH referred to tangent vectors, they are essentially equivalent to mapping properties of MATH on Lagrangian submanifolds MATH of MATH. Thus, for the sake of convenience in notation, we will consider these mapping properties, that is, we will take MATH; it is simple to reinterpret results in the desired form. To be concrete, for MATH we take MATH . Note that MATH is certainly in the range of the differential of the projection MATH by REF (the projection is to the unprimed variables in the notation of this equation!). If MATH, an equally good choice is MATH, which simply corresponds to the Lagrangian given by plane waves MATH. Before continuing, we make some general remarks about Lagrangian submanifolds MATH of MATH. Suppose that MATH is (locally) a smooth graph over MATH; that is, that for every MATH, the bundle projection MATH has a surjective differential at MATH (which is thus an isomorphism). Then MATH is (locally) the image of a smooth bundle map MATH, that is, the composite of MATH with the projection to the base is the identity map. Since MATH naturally, MATH has the form MATH where MATH. Moreover, MATH can be naturally identified with the vector space MATH itself; so for each MATH, MATH is a linear map from MATH to MATH. Let MATH be the induced endomorphism of MATH via the metric identification of MATH and MATH. That MATH is Lagrangian means that for any two tangent vectors MATH, MATH, where MATH is the standard symplectic form given by MATH, so writing MATH as MATH, MATH, MATH, so in our case, with MATH, MATH, the Lagrangian condition becomes MATH. Since the metric identification means that MATH, the Lagrangian condition amounts to the statement that MATH is self-adjoint. A similar discussion also applies to MATH with the twisted symplectic form MATH, and shows that for MATH, which is a graph by REF, with the notation of REF, MATH and MATH are self-adjoint on MATH and MATH respectively (this of course follows from REF as well), while MATH. Now we return to transversality. In fact, we show the following stronger statement. Let MATH be arbitrary, MATH, MATH, MATH, MATH, and suppose that MATH is a Lagrangian submanifold of MATH which is (locally) a smooth graph over MATH of the form MATH where for each MATH, MATH is given by a positive operator MATH on MATH (that is, MATH; MATH is automatically self-adjoint as discussed above), and suppose in addition that MATH is positive definite if MATH. Then we show that the relation MATH intersects MATH transversally, and maps MATH to a Lagrangian MATH with similar properties, that is, MATH is (locally) a graph of the form MATH, MATH is positive for every MATH, and MATH is positive definite if MATH. Note that the transversal intersection property automatically shows that MATH is smooth and Lagrangian. Once this is shown, the desired result follows by induction starting from MATH. Indeed, to keep the induction going we need to check that MATH is positive definite if MATH, but since this implies MATH, that is, MATH, as discussed before, we see that our claim, that MATH is positive definite if MATH, ensures that this holds. We remark that for MATH, the Lagrangian subspace MATH of MATH defined in REF certainly has the desired positivity; the operator MATH is given by MATH, hence it is positive definite unless MATH. If instead we take MATH (corresponding to plane waves), then MATH, so it is still positive (though not positive definite), which suffices if MATH. We proceed to show the claimed mapping property of MATH. We need to show that for MATH . But, due to the first summand, MATH, MATH are in the left hand side, hence so is MATH, MATH, due to REF. Since the vectors MATH, MATH, are also in the left hand side due to the first summand, REF will follow if the block matrix MATH is invertible, that is, if MATH is invertible. But MATH, in fact negative definite if MATH, while MATH, in fact positive definite if MATH, we conclude that MATH is positive definite, hence invertible. This proves REF. Next, we need to find MATH . Dimension counting shows that we will have found the whole intersection if for every MATH we find an element of the intersection which is of the form MATH . To do so, fix MATH, and consider elements of MATH which are of the form MATH . For these to be in the intersection REF, we need that there exist MATH such that the projection of REF to MATH is equal to MATH, that is, such that MATH . The existence of such MATH and MATH thus again follows from the invertibility of MATH, which holds since MATH is positive definite, and we see that in particular MATH. The projection of the corresponding tangent vector in REF to MATH is given by MATH where we used MATH. Thus, the projection of the intersection MATH to MATH is a smooth Lagrangian MATH with tangent vectors of the form MATH . Since MATH is positive definite, the same holds for its inverse, so MATH. Moreover, MATH, so we conclude that MATH. If in addition MATH, then MATH is positive definite, hence the same holds for MATH. This provides the necessary inductive step and proves REF . |
math/9910033 | (Proof of REF .) With the notation of REF, if MATH, then MATH is an unbroken bicharacteristic, so the endpoints of MATH are related by the antipodal relation, which is certainly Lagrangian. So we may assume MATH. In addition, due to REF , the MATH bicharacteristics will only break upon hitting MATH with MATH (that is, MATH-tangential bicharacteristics cannot become normal to MATH without hitting MATH). Similarly, MATH. The only change from the proof of REF is that we need to replace the first and last NAME in the composition REF by the NAME MATH and MATH given by the (twisted) graphs of MATH and MATH respectively; here we wrote MATH and MATH for the coordinates on MATH and MATH respectively. Thus, on MATH, MATH, so the differential of the projection of MATH to MATH is surjective since MATH. In particular, we can choose a Lagrangian subspace MATH of MATH which is in the range of the differential of this projection, and which is a graph over MATH of the form MATH with MATH positive definite (for example, the identity). Then the proof of REF shows that the composite relation MATH is a smooth Lagrangian. Since the differential of the projection of MATH to MATH is also surjective as MATH, the composition of MATH with REF is transversal, so the result is a smooth Lagrangian submanifold of MATH. In view of REF , this and REF complete the proof. |
math/9910036 | See CITE. |
math/9910036 | The space MATH corresponds to the moduli space MATH of semi-stable parabolic MATH-bundles on MATH. Moreover MATH corresponds to the stable subspace and is open and dense in MATH CITE. |
math/9910036 | A direct calculation from REF shows that MATH is a real algebraic subvariety with positive co-dimension. The result then follows from the fact that MATH is smooth and has dimension MATH. |
math/9910036 | If MATH is a MATH representation, then up to conjugation: MATH . Note that: MATH . The other direction follows from the uniqueness of characters CITE. This proves REF . Suppose MATH is MATH, but not MATH then at least two of the following MATH and MATH are in MATH. Since MATH implies MATH, at least two of the three global coordinates of MATH must be zero. See CITE for a similar proof in the case of the one-holed torus. The other direction follows similarly from the uniqueness of characters CITE. This proves REF . |
math/9910036 | See CITE. |
math/9910036 | Since the ellipses MATH are (possibly degenerate) of uniformly bounded circumferences, there exists MATH such that for any MATH, the MATH-orbit is MATH-dense in MATH. |
math/9910036 | This result follows directly from REF which gives MATH, the center of the ellipse MATH . |
math/9910036 | The result holds by the continuous dependence of MATH on MATH and MATH by the continuous dependence of MATH on MATH, and by the geometry of MATH for MATH as described in REF . |
math/9910036 | Suppose that MATH, MATH, MATH and MATH all have zero trace. If MATH preserves MATH, then MATH (if not, note that the twist in MATH preserves MATH and MATH so MATH remains generic). By REF , MATH can be assumed non-zero. Hence MATH . Then the equation for the four-holed sphere MATH implies that: MATH . By REF and the fact that the left-hand side of REF is invariant under MATH it must be that MATH . By the formula MATH (see CITE) and the fact that MATH we have that MATH; this implies that MATH as matrices. Suppose MATH. Conjugating by an element in MATH, we may assume that MATH and MATH . The equations: MATH together with the matrix equation MATH and MATH have solutions that must include one of the following: MATH or MATH. By REF , one may assume that MATH and MATH. This leaves us with the case MATH, which implies that MATH contradicting the fact that MATH is generic. |
math/9910036 | Let MATH. Let MATH be a generic representation. By REF , MATH has a generic handle MATH (we adopt the notation presented in REF ). By REF , without loss of generality we may take MATH and show that there exists MATH such that MATH . Let MATH be a representation class with MATH and MATH . Let MATH etc. The map MATH is a submersion on MATH. By the implicit function theorem, there exists MATH such that the MATH-tubular neighborhood of MATH looks like MATH . Cutting along MATH and MATH gives a pants decomposition of MATH. Hence, by REF , we only need to show that there exists MATH such that MATH satisfies MATH and MATH. In other words, the goal is to move the MATH- and MATH-coordinates near MATH and MATH. The strategy is to use MATH to first move the MATH-coordinate near zero so that the moduli space of the four-holed sphere (obtained by cutting along MATH) contains MATH-coordinates near MATH. At the same time one needs to ensure that MATH has enough points on its orbit so that the MATH-coordinate can actually be moved near MATH (this may require a sequence of twists in MATH followed by a sequence of twists in MATH followed again by a sequence of twists in MATH). This can be accomplished by using the generic MATH to move the MATH-coordinate to a sufficiently general position (that is, outside of MATH for some very large MATH). By REF , this will produce a MATH-orbit in MATH with a sufficiently large number of points. Under such conditions, this MATH-orbit will potentially contain points that have a large number of points on their MATH-orbit in MATH. The exceptional case being when the MATH-action fixes MATH. In this case, MATH is a point, which implies that MATH . If this occurs, one may try to use the MATH-action. However in order to use the MATH-action, one must first ensure that MATH itself is not MATH . Hence under the assumption that MATH fixes MATH we have the following two exceptional cases: CASE: MATH is MATH. This leads to either MATH or MATH. This implies that either MATH or MATH . CASE: MATH is not MATH but MATH is fixed by MATH . Then one must have MATH . This leads to MATH . Since MATH, we have MATH . Moreover, for non-zero MATH, we have MATH. This implies that MATH . If MATH and MATH then either MATH or MATH . This contradicts our assumptions on MATH and MATH. So, MATH . Now we choose a generic representation MATH that allows us to avoid the special cases presented above. The following lemma follows immediately from REF and the fact that the MATH actions fix MATH. For any integer MATH, there is MATH such that the MATH-orbit of MATH has at least MATH points satisfying the following conditions: CASE: The MATH-coordinates of these MATH points have MATH small enough so that MATH and MATH where MATH is provided in REF for MATH . CASE: These MATH points do not belong to the subvarieties defined by REF , MATH, and MATH. Note that REF ensures that if MATH fixes MATH then MATH is not MATH and MATH does not fix MATH. Now choose MATH with MATH sufficiently large in REF so that one of the MATH points on the MATH-orbit of MATH denoted MATH has a MATH-orbit (or MATH-orbit) with at least MATH points with distinct MATH-coordinates inside MATH . Since MATH and MATH the sphere MATH (with MATH), has points with MATH-coordinates inside MATH . Since the MATH-orbit (respectively, MATH) of MATH has at least MATH points with distinct MATH-coordinates, one such point MATH has MATH-coordinate not in MATH with non-degenerate ellipse MATH . Thus, the MATH orbit of MATH is MATH-dense. Thus, the MATH-orbit of MATH has at least MATH points with MATH-coordinates inside MATH . One such point, MATH has MATH-coordinate not in MATH with non-degenerate ellipse MATH . Thus, the MATH-orbit of MATH is MATH-dense in MATH. This fact, together with the properties of MATH provided in REF imply that at least one point, MATH, in the MATH-orbit of MATH has MATH-coordinate MATH that comes within MATH of MATH . The one-holed torus MATH is generic so long as MATH and MATH (see REF ). This can be accomplished by replacing MATH by MATH at the start of the argument. Since MATH is generic, we may apply MATH and MATH to obtain MATH so that the MATH-coordinate MATH of MATH is within MATH of MATH . Note that both MATH and MATH fix MATH, thus MATH remains generic. The result now follows by REF . To summarize, one first uses MATH to obtain points with MATH-coordinates near zero having MATH (respectively, MATH) actions that generate points having MATH-actions with a sufficiently large number of points. Next, one uses MATH-action to get the MATH-coordinate to be near zero on the moduli space of the four holed sphere obtained by cutting at MATH . This ensures that one can move the MATH-coordinate near MATH (see REF ). After these twists, the handle MATH is shown to remain generic. Finally move the MATH-coordinate near MATH by using REF . |
math/9910036 | We first treat the case of MATH. Let MATH and MATH be two boundary loops separated from MATH by MATH . Let MATH be the remaining boundary loop (see REF ). Since MATH is generic, we have that MATH . The goal, therefore, is to find MATH such that MATH. CASE: NAME torus. Suppose MATH . Then MATH and MATH . If MATH, then we may apply REF to the two-holed torus bounded by MATH and MATH. Hence, if MATH there exists an element in MATH that fixes MATH with MATH such that MATH generic. This implies MATH. The same conclusion can be drawn if MATH. Suppose that MATH and MATH. Since MATH we have that MATH . Hence, MATH. This is only possible if MATH or MATH . The latter case is ruled out by the generic assumption on MATH. Hence MATH . We will show that in this special case, we can obtain MATH. Suppose that MATH commutes with MATH. If MATH also commutes with MATH, then MATH all belong to the same one parameter subgroup of MATH. Hence MATH and MATH commute. Since MATH is generic, we have that MATH does not commute with MATH . Suppose that MATH (MATH and MATH anti-commute). Then, MATH which implies that MATH . Again, since MATH is generic, we may assume that MATH . To summarize, since MATH is generic, one may arrange that MATH neither commutes nor anti-commutes with MATH (that is, MATH). Now consider the curve MATH . The NAME twist in MATH preserves MATH and MATH. Hence it fixes MATH (consider the pants bounded by MATH and MATH). On the four-holed sphere bounded by MATH and MATH, the action of MATH on the curve MATH is MATH so MATH . Hence MATH . Observe that by REF we may assume that MATH is initially not in MATH. Since MATH preserves MATH and MATH, MATH is also generic, note that MATH is not MATH since MATH and MATH . This proves the case MATH. CASE: Getting rid of MATH on MATH for MATH. The above argument may be repeated iteratively, starting with loops in MATH that bound two boundary loops and working inward towards MATH . We demonstrate using the case MATH with notation provided in REF . First consider the three-holed torus bounded by MATH, MATH and MATH . Since MATH is generic, if MATH then MATH (otherwise MATH). Use the previous argument for MATH to arrange MATH. Next, the three-holed torus bounded by MATH, MATH and MATH is used to make MATH, etc. |
math/9910036 | Let MATH. Suppose MATH with MATH generic. We adopt the usual practice of omitting the symbol MATH and using the subscript MATH to denote the values of MATH. We will also adopt the notation of REF . By REF , without loss of generality we may take MATH and show that there exists MATH such that MATH . The map MATH is a submersion on MATH. By the implicit function theorem, there exists MATH such that the MATH-tubular neighborhood of MATH looks like MATH . By REF , assume that MATH is generic. By REF , we may assume MATH and MATH are not MATH . Cutting along MATH, MATH and MATH yields a pants decomposition. Hence, by REF , we only need to show that there exists MATH such that MATH satisfies MATH, with MATH generic. The goal is to move MATH near MATH. The strategy is to first move MATH and MATH near zero so that the moduli space of the four-holed sphere bounded by MATH and MATH contains MATH-coordinates near MATH. At the same time, one needs to ensure that either MATH or MATH has enough points on its orbit in MATH, so that the MATH-coordinate can actually be moved near MATH. This can be accomplished by using the generic MATH to move the MATH-coordinate to a general position (that is, outside of MATH for some large MATH). This will potentially produce large MATH-orbits in MATH and MATH. Under such conditions, the MATH-orbit in MATH and MATH will potentially contain points such that the MATH- and MATH-orbits at these points have a sufficiently large number of points in MATH. For this strategy to work, however, we must first deal with the special case where both MATH and MATH are MATH . These are the representations that are fixed by MATH (see REF ). In addition, one must also deal with the analogous situation with the four-holed sphere (bounded by MATH and MATH) to ensure that either MATH or MATH does not fix the MATH-coordinate. We begin with a detailed analysis of these exceptional cases where both of the following hold: CASE: MATH is MATH or MATH fixes MATH CASE: MATH is MATH or MATH fixes MATH . CASE: Both MATH and MATH are MATH and MATH. This implies that MATH and MATH . Note that MATH . Note that the application of MATH fixes the moduli space of the two-holed torus bounded by MATH and MATH that is, MATH fixes: MATH, MATH, MATH, MATH, and MATH . For the NAME twist MATH to fix MATH with MATH amounts to MATH and MATH . In the special case where both MATH the defining equation for MATH yields MATH . CASE: MATH is MATH while MATH is not. If MATH fixes MATH then MATH . CASE: MATH is MATH while MATH is not, with MATH fixing MATH then MATH. The argument for this case is symmetric to that of REF . CASE: Neither MATH nor MATH is MATH. If both MATH and MATH fix MATH, then MATH . These lead to MATH . A special case of the above equation is when MATH . Then REF becomes MATH . Note this quadratic equation is degenerate in MATH only if MATH or MATH . The case MATH leads to the equations MATH and MATH . Therefore, MATH that is, MATH . Similarly, MATH leads to MATH . Note that the non-degenerate solutions of REF together with MATH are: MATH and MATH . Now we carefully choose a generic representation of the two-holed torus bounded by MATH and MATH that allows us to avoid the special cases presented above. For any integer MATH, there is MATH such that MATH has MATH-orbit containing at least MATH points satisfying all of the following: CASE: Each of the MATH points have distinct MATH and MATH-coordinates inside MATH, where MATH is given as in REF applied to the four-holed sphere that bounds MATH, MATH, MATH, and MATH. CASE: MATH are small so that MATH . CASE: If MATH or MATH, then these MATH points do not belong to the subvariety defined by either REF or REF . CASE: If MATH, then these MATH points do not belong to the subvariety defined by REF CASE: The MATH points do not belong to the subvariety defined by REF . CASE: The MATH points do not belong to the subvariety defined by REF if MATH and do not belong to the subvarieties defined by REF if MATH. Note that the MATH points provided in REF are obtained by using the mapping class action on the two-holed torus bounded by MATH and MATH. Hence the value MATH remains fixed. We begin by noting that the two-holed torus bounded by MATH and MATH is generic. REF follows from REF . Note that both MATH and MATH can be made simultaneously near zero, since MATH can always be made arbitrarily close to zero for any value of MATH (see REF ) and since the MATH values of MATH for MATH take on values inside MATH which, for MATH near zero, contains the value MATH . By REF , for MATH having a MATH-dense orbit for MATH sufficiently small, we may find a point with a MATH-orbit with at least MATH points not belonging to any of the proper subvarieties described in REF. Moreover, after a possible application of MATH we also have that one of MATH or MATH is not MATH. REF ensure that for every point on the MATH-orbit of MATH either MATH is not MATH and MATH does not fix MATH or MATH is not MATH and MATH does not fix MATH . We assume without loss of generality assume the former. Finally, we start with a point MATH with MATH sufficiently large so that one of the points on the MATH-orbit of MATH has a MATH-action containing at least MATH points. The argument now follows similarly to that presented in REF to produce MATH by obtaining MATH . This completes the proof. To summarize, one first makes MATH. Then uses the fact that the two-holed torus bounded by MATH is generic to move the MATH-coordinates near zero. This allows the MATH-coordinate to be moved near MATH. Finally the generiticity of the two-holed torus allows one to move the MATH-coordinates near MATH. |
math/9910036 | The cases MATH and MATH have previously been established. For MATH the argument follows by an induction process similar to that in the proof of REF . Let MATH. Let MATH be a generic representation. By REF , MATH has a generic handle MATH . We demonstrate how to proceed in the case MATH (see REF below). By REF , without loss of generality we may take MATH and show that there exists MATH such that MATH . The map MATH is a submersion on MATH. By the implicit function theorem, there exists MATH such that the MATH-tubular neighborhood of MATH looks like MATH where MATH denotes the MATH-open box neighborhood of the set MATH. CASE: The induction on REF-holed torus. By REF , we have that MATH . By induction, REF is true for the four-holed torus bounded by MATH and MATH. We use this four-holed torus to arrange for MATH and MATH (see REF ) to have near zero traces. This ensures that the MATH-coordinate of MATH is accessible. We then arrange MATH . We now cut MATH at MATH and use REF to get the MATH-coordinate of MATH within MATH of MATH . Next, we get the MATH-coordinate of MATH within MATH of MATH by using the four-holed torus obtained by cutting at MATH. Finally, REF yields the result. The situation is identical for any MATH . |
math/9910036 | We first show that there is MATH so that MATH is generic, for all MATH and MATH that bound a three-holed sphere that forms a one-holed torus in MATH . Consider MATH and MATH as in REF , where MATH bound a pants that forms a one-holed torus in MATH . CASE: Suppose that MATH . In this case, at least one of MATH, MATH or MATH is not in MATH . Without loss of generality, assume that MATH (otherwise, NAME twist in MATH). Thus, we may cut MATH at MATH and MATH and apply REF to the resulting three-holed torus to obtain MATH with MATH . Note that in the special case MATH we apply REF along with REF . Thus the handle MATH is generic. CASE: Suppose that MATH . Then both MATH and MATH are not MATH . Moreover, MATH since MATH . Thus, we may cut MATH at MATH and MATH and then apply REF to reduce the situation to REF . CASE: Suppose that MATH . Then, necessarily, MATH since MATH is generic. If either MATH or MATH we may cut at MATH (or MATH) and apply REF as in REF . Since MATH, If both MATH and MATH then MATH . Note that the action of MATH does not affect the generic handle MATH . We now cut at MATH and apply REF as in REF . This argument can be applied independently to each of the MATH handles of MATH. Having obtained that all handles are generic, we now apply REF to the remaining curves in MATH that are interior to the MATH-holed torus obtained by cutting off each of the MATH handles. In the special case MATH and MATH . REF applies to those curves of MATH that are separated from MATH by the curve MATH . Moreover, MATH since MATH is assumed generic. |
math/9910036 | Suppose that MATH for all MATH . Since MATH is generic, there exists MATH for MATH . The handle MATH has the desired property. |
math/9910036 | Since MATH is generic, by the previous Lemma, there exists MATH for MATH . Then MATH defines a one-parameter subgroup MATH of MATH. Move MATH to the intersection point of MATH and MATH. Since MATH is generic, there exists MATH. This implies that MATH is non-Abelian, hence, not MATH. |
math/9910036 | Suppose MATH is MATH, but not MATH. The group MATH acts on MATH via its quotient MATH and preserves a circle MATH. There are two cases. CASE: MATH preserves a unique circle MATH. Since MATH is generic, there exists MATH not preserving MATH. This implies that MATH does not preserve any circle in MATH, hence, not MATH. CASE: MATH preserves three circles: MATH. Then MATH and MATH . This implies that MATH, MATH and MATH . Suppose further that MATH and MATH are MATH or MATH preserving three circles. Then up to conjugacy, MATH . Let MATH with MATH be such that MATH is not contained in the isomorphic copy of MATH. Since MATH is either MATH or MATH preserving three circles, we have that MATH . However, this implies that MATH so MATH . For MATH we see that MATH . For MATH we see that MATH . Likewise, since MATH is either MATH or MATH preserving three circles, we have that MATH . The case MATH leads to MATH while the case MATH leads to MATH . Taken together, we see that in all cases three of MATH are zero. This implies that MATH is equal to one of: MATH or MATH which is a contradiction. |
math/9910036 | By REF , there exists MATH such that MATH is not MATH. Suppose MATH is MATH. By applying NAME twists, and in light of the analysis carried out in CITE, one may assume that MATH has character MATH on MATH with signs taken together. We will only address representations with character MATH as the arguments for MATH are practically identical. Up to conjugation, we may choose: MATH . The group MATH consists of: MATH REF matrices of the form MATH (signs in the second row determined by those in the first), and REF matrices of the form MATH . Let MATH with MATH be such that MATH is not contained in the isomorphic copy of MATH. First, consider the case where each of MATH and MATH are MATH. This means that MATH which implies that MATH . Similarly, MATH . This means that MATH which is a contradiction. Next, suppose that MATH is MATH and that MATH is MATH. Then MATH and MATH implies that MATH and MATH . The equation MATH implies that MATH . Finally, MATH implies that either MATH or MATH is not real, a contradiction. A similar argument holds in the event that MATH is MATH and MATH is MATH . We indicate how to proceed in the case where MATH is MATH and MATH is MATH. Since MATH is MATH both MATH and MATH must take on the values MATH or MATH . The case where both are zero corresponds to a MATH representation which was already handled. Suppose that MATH and MATH . Then MATH and MATH so MATH and MATH . The equation MATH yields MATH or MATH. If MATH then MATH is not real. This implies that MATH and MATH . However, this implies that MATH which is a contradiction. Similar arguments hold in the other REF cases, as well as REF cases where MATH is MATH and MATH is MATH. We now address the remaining cases where MATH and MATH are either MATH or MATH. In these cases, MATH, MATH, and MATH must take on the values MATH or MATH . We explicitly work out a MATH case in which MATH and MATH are all zero. These lead to the equations MATH . Therefore, MATH and MATH which implies that MATH and MATH . However, the trace of the matrix MATH is MATH . Thus the handle MATH is neither MATH nor MATH. (This handle is obtained by first twisting in MATH on MATH, then using the loop MATH . ) The remaining REF cases are obtained by varying the values of MATH and MATH in MATH . The arguments proceed similarly to those exemplified above and lead to: CASE: matrices MATH inside MATH (a contradiction), CASE: a handle MATH that is neither MATH nor MATH CASE: one or more of MATH is not real (a contradiction). |
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