paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0011008 | The construction of MATH is complete by the algorithm of REF, and the fixing of all parameters in the present section. We have to prove that every structure MATH is a mazery. The proof is by induction. We already know that the statement is true for MATH: it was handled in REF . Assuming that it is true for all MATH, we prove it for MATH. REF is satisfied by the argument before REF . REF is satisfied by the form of the definition of the new types. REF is satisfied as shown in REF . REF has been proved in REF . In REF has been proved in REF has been proved in REF has been proved in REF is proved for emerging walls in REF , and for compound walls in REF . REF is satisfied trivially for the emerging type, (as pointed out in REF ), and proved for the compound type in REF . REF is satisfied via REF (the Grate Lemma), as discussed after REF . There are some conditions on MATH required for this lemma. Of these, REF follows from REF, while REF follows from REF . Let us show that the conditions preceding the Main REF hold. REF is implied by REF . REF is implied by REF . REF follows immediately from the definition of cleanness. Finally, REF follows from REF. |
math/0011010 | We argue along the lines of the proof of CITE. Let MATH, MATH and observe that MATH, where MATH it follows that MATH . Now compute MATH . In the above calculation equality REF holds because the permutation MATH is dominant CITE, and the sum REF is over all MATH-tuples MATH of nonnegative integers. Observe that MATH where MATH. The action of MATH relevant to our computation is explained by NAME and NAME; if MATH for MATH then CITE: MATH for any function MATH symmetric in the MATH variables. Moreover CITE states that if MATH and MATH then MATH vanishes unless MATH for some MATH, in which case MATH . Using the above facts and REF we deduce that MATH where the ranges of summation are as in the statement of the theorem and MATH. Finally, note that MATH and the signs fit to complete the proof. |
math/0011010 | Let MATH; then the analysis in CITE gives MATH for every MATH. We now use REF to compute MATH where the above sums are over all MATH with MATH and MATH. It follows that for any MATH, MATH . Applying CITE gives MATH . We now use REF and apply REF to obtain MATH which is the desired result. |
math/0011010 | It is shown in CITE that MATH is a free MATH-module with basis MATH, for MATH, MATH. Furthermore, this basis has the orthogonality property CITE MATH . REF follows immediately by applying REF . To prove REF , factor MATH (where MATH) and observe that MATH . Now use REF and the corresponding properties of the inner products defined by MATH and MATH (stated in CITE and CITE, respectively). |
math/0011010 | We use REF and the definitions in REF to get MATH . The result now follows from REF . |
math/0011010 | REF above shows that MATH is stable when MATH. It follows from the definition of the polynomials MATH by divided difference operators that MATH . This implies that if MATH is stable, then so is MATH for any MATH. |
math/0011010 | Recall that if a polynomial MATH is homogeneous of degree MATH, then MATH is homogeneous of degree MATH for each MATH. Hence, the definition of the polynomials MATH and their normalization in REF imply that MATH is homogeneous of degree MATH. In particular MATH unless MATH, when MATH. REF (with MATH) and REF thus give MATH . We deduce that MATH for all MATH. The result now follows from REF . |
math/0011010 | The vanishing property for MATH from CITE gives MATH . On the other hand, MATH for all MATH. |
math/0011010 | We will show that the above statement is equivalent to the corresponding result in CITE REF . Note that, as in CITE, the proof is obtained by pulling back the corresponding equality on the flag bundle MATH. In order to avoid notational confusion define new alphabets MATH and MATH with MATH in agreement with the conventions in CITE. Identify the NAME group MATH with its image MATH and let MATH denote the divided difference operators with respect to the MATH variables, defined as in CITE For example MATH, where MATH sends MATH to MATH. REF gives MATH where MATH . Using the criterion in CITE, we see that one can replace the kernel MATH in REF with MATH where the permutation MATH acts on the MATH variables (compare with the analysis in CITE). Next we apply CITE to replace MATH in REF by MATH, as these two polynomials are congruent modulo the relations in the NAME ring of the flag bundle of MATH. Note also that we have the identities MATH where we define MATH. The map MATH interchanges MATH with MATH for all MATH; it follows from REF that MATH as required. |
math/0011010 | For each MATH there is an equality of MATH matrices MATH . Taking determinants in REF shows that MATH . Now apply REF and use REF to obtain MATH . The proof is completed by noting that MATH which follows from REF and the NAME expansion of a determinant. |
math/0011010 | Given the NAME polynomial MATH one can recover MATH by using REF . It follows easily from this that if MATH is MATH-stable then MATH is MATH-stable. For the converse, apply CITE to obtain MATH for every MATH. Assuming that MATH is MATH-stable, we deduce that MATH is MATH-stable from REF and a straightforward induction. |
math/0011010 | We apply the permutation MATH to the polynomials in REF , then use REF to obtain MATH and compare this with REF . |
math/0011010 | Given MATH define MATH and MATH as above and associate to MATH a partition MATH of length MATH as follows: MATH . Note that MATH, where MATH hence MATH . We compute as in the proof of REF : MATH where the sum REF is over all MATH-tuples MATH of nonnegative integers. Observe that MATH since MATH mod MATH. The operator MATH is a NAME symmetrizer; hence for any partition MATH we have MATH unless MATH, when MATH is a NAME MATH-polynomial. In the latter case we can apply CITE and deduce that for any partition MATH, MATH unless each part of MATH occurs in MATH. In this case, the image is MATH where MATH (note that the operator MATH in CITE differs from ours by a sign). Moreover, if MATH is a MATH-tuple of distinct nonnegative integers and MATH is such that MATH then MATH hence the previous analysis applies, up to MATH. Noting that MATH for MATH and MATH-mod MATH, use REF to compute MATH where the ranges of summation are as in the statement of the theorem and MATH. Finally, observe that MATH so the signs fit to complete the proof. |
math/0011010 | We argue along the lines of the proof of REF , using the operator MATH in place of MATH. We claim that the image of MATH is a free MATH-module with basis MATH. This statement is a type MATH analogue of CITE, and the proof is similar. The only difference is that here we use the relation MATH, and thus MATH, for each MATH. Moreover, CITE implies that MATH (note that MATH). Now the vanishing property for MATH from CITE and the same argument as in CITE show that MATH for every MATH. Observe, since MATH, that MATH. We deduce as in the proof of REF that MATH . For any permutation MATH, apply the operator MATH to both sides of REF . We have seen that MATH so it remains to show that MATH for all MATH. To prove this, note that for each MATH, the operator MATH either vanishes or equals MATH, for a unique MATH. Moreover, different elements MATH lead to different elements MATH. |
math/0011010 | If MATH, then the claim follows directly from REF . Assume that MATH, fix a partition MATH and equate the coefficients of MATH in the sums that occur in REF : MATH . To prove this equality, we expand each determinant MATH in REF along the last (MATH-th) column, and compare with REF . The result then follows by using the identity in the next Proposition (for varying MATH and MATH). For each partition MATH of length MATH, we have MATH . The proof of REF , while elementary, uses additional algebraic formalism, and is given in CITE. |
math/0011010 | Let MATH be the universal subbundle on MATH, and let MATH, where MATH refers to the standard skew-symmetric form MATH on MATH, and hence also on MATH. The first claim is that MATH fits into an exact sequence REF . This is clear, since the (dualized, twisted) NAME sequence on MATH identifies MATH with MATH. On MATH there is the natural map of vector bundles MATH . Let MATH be the projectivization of the indicated vector bundle. We follow the convention of CITE: points of MATH are points of the base together with one-dimensional subspaces of the fiber. Let MATH be the universal subbundle of MATH on MATH. Then we have a morphism of bundles on MATH . Over any affine open subset of MATH, this lifts to a morphism MATH such that, fiberwise, the map on global sections has image MATH. To phrase this mathematically, let MATH denote the projection MATH, and equally, the restriction over an affine open of MATH. Then MATH applied to REF should be an isomorphism onto MATH. This requirement determines the lift, uniquely up to automorphisms of the fibers of MATH. Since the Quot functor is a sheaf, these maps, defined locally, patch to give a vector bundle MATH and injective map of sheaves MATH on MATH, such that the quotient sheaf is flat over MATH and on every fiber of MATH has rank REF and degree REF. The resulting morphism MATH is an isomorphism since it is a birational morphism of smooth varieties, which is one-to-one on geometric points. |
math/0011011 | One can check that the fibre components of the MATH are independent of MATH and equal to MATH . However, the limit of the MATH as MATH is not defined, that is, in coordinates, all the terms with components along the base blow up in the limit. Using the bundle isomorphisms MATH defined by each of the nondegenerate forms MATH, we can construct the dual bivectors MATH . (We include the negative sign because if we associate to MATH a nondegenerate skew-symmetric matrix, then to the dual bivector at MATH we associate the negative inverse of this matrix.) These bivectors are nondegenerate NAME structures and we may rewrite REF as MATH . In contrast to the MATH the NAME structures MATH do have a well defined limit, MATH. This is a degenerate NAME structure whose symplectic leaves are the fibres of MATH. As well, for MATH we have MATH. Hence, MATH where MATH denotes the exterior derivative with respect to just the fibre variables. This can be rewritten as MATH . Indeed, this equation defines the limiting vector field MATH globally and we note that the convergence of MATH to MATH is MATH for any MATH. By REF the flow of MATH is fibrewise quasiperiodic and we have at least MATH periodic orbits in each fibre. |
math/0011011 | First we show that MATH implies that MATH for some MATH. If MATH is a constant solution then MATH since MATH. Hence, we only have to discount those critical points with MATH . In fact, because MATH depends only on the fibre variable past MATH, we get MATH for all MATH in MATH . When MATH, we also have MATH . Hence, MATH . The last two inequalities follow from the fact that MATH for MATH, and MATH. Next we show that MATH. A critical point of MATH is a one-periodic solution of the Hamiltonian dynamical system defined on MATH by MATH and MATH. Thus, MATH has period one and because of our splitting of MATH it satisfies MATH . Now, since MATH has zero mean, MATH . Hence, MATH . It is easily checked for our choice of MATH that MATH is of order MATH for all MATH. Hence, MATH remains sufficiently close to the origin and the proof of the claim is complete. |
math/0011011 | First we label the parts of MATH as follows MATH . For MATH recall that MATH . On MATH we have points of the form MATH so that MATH . For the other parts of the boundary we need to employ half of REF , namely MATH . This yields MATH which when restricted to MATH becomes MATH . Overall, on MATH we now have MATH . For our nonvanishing section MATH it is clear that there exists a real constant MATH such that MATH for all MATH. We now choose MATH to be greater than MATH. Then for MATH large enough to satisfy both MATH and MATH, we have MATH . |
math/0011011 | Since the functions MATH are equal to zero on a neighborhood of MATH, the functions MATH satisfy MATH, MATH, and MATH for all MATH in MATH. (Here `` MATH " denotes a fibrewise derivative.) Restricting to the spaces MATH we see then that MATH . |
math/0011011 | Let MATH, MATH and MATH be the projection maps corresponding to the splitting MATH. Consider the maps MATH where MATH. Letting MATH be the zero section in MATH it follows easily that MATH . The maps MATH are smooth and we claim that they are also nonlinear NAME maps with index equal to zero. This is easily confirmed for MATH and so must also hold for MATH with MATH sufficiently small. Then, since MATH for any MATH, it also holds for arbitrary MATH. One may also easily verify that MATH is transversal to MATH and that MATH. We now consider a fixed MATH. Note that if MATH is not transversal to MATH, then the proof is complete since transversality must fail at some MATH with MATH. Hence, we may assume that MATH is transversal to MATH. By REF (see CITE) we can then perturb the homotopy MATH, MATH, from MATH to MATH to a transversal NAME homotopy MATH such that MATH and MATH . The sign of MATH on MATH and MATH ensures that MATH for all MATH. Consequently, MATH. With this, REF implies that MATH is a MATH dimensional submanifold of MATH with boundary equal to MATH . Upon projecting to MATH we see that these boundary components must generate the same homology class in MATH. It is easy to see that the first component generates the fundamental class and hence the second component must be nonempty. |
math/0011011 | We will make use of two distance functions on the bundle MATH, which we now define. Consider the two fibrewise norms, MATH and MATH, which are given by MATH for MATH and MATH, respectively, where MATH. Each of these norms yields a fibrewise metric which, when coupled with the base metric MATH, define a MATH and a MATH metric on MATH. We are interested in the distance functions corresponding to these metrics which we will denote by MATH and MATH. Given a sequence MATH such that for the MATH-gradient we have MATH with respect to the MATH metric, we need to show that there exists a convergent subsequence with respect to MATH. (Since MATH is compact we already know that there is a convergent subsequence MATH.) In what follows we focus entirely on the fibre component of the gradient (equal to the gradient of the restriction to a fibre), which we again denote by MATH. The fibre component, MATH, is a function MATH, where we identify a tangent space to the fiber MATH with the fibre itself. Clearly, the norm of the fibre component does not exceed the norm of the gradient and hence REF still holds for the fibre components. Consider the special form that REF takes on MATH. Recall that the functional MATH is given as MATH, where MATH . With respect to the fibrewise orthogonal splitting MATH (see REF), it is straightforward to check that MATH and, since MATH denotes the fibre component of the gradient, MATH . As in CITE, one can also show that MATH where MATH is the formal adjoint of the inclusion MATH and is a compact map. Indeed, our second REF implies that MATH takes bounded sets in MATH to bounded sets so that the map MATH defined as MATH, is also compact. We may now rewrite REF, in a slightly weakened form, as MATH . Assume first that the MATH are bounded for an infinite subsequence of points. Without loss of generality we may assume that MATH is this subsequence. By the compactness of the map MATH we know that MATH is relatively compact. Thus, the sequence MATH has a convergent subsequence. Note that MATH and MATH are orthogonal to each other in MATH. After passing if necessary to subsequences, this implies that each of the sequences MATH and MATH converges for the same subsequence of points MATH. As before, we may restrict our attention to this subsequence. Finally, since MATH is a bounded sequence in a finite dimensional space it too has a convergent subsequence and the proof in this case is finished. Looking for a contradiction, we assume that MATH are unbounded for some infinite sequence on which we now focus. Set MATH . REF now takes the form MATH . By REF , the sequence MATH is bounded. Indeed, MATH . The compactness of the operators MATH then implies that MATH is relatively compact in MATH with respect to MATH. Hence, the sequence MATH is also relatively compact in MATH with respect to MATH. Just as above we then get a convergent subsequence MATH . Note also that MATH . Now MATH where MATH is the quadratic term in the definition of MATH. The second term on the right hand side goes to zero as MATH because MATH is continuous. As for the first term, we have MATH . However, by the construction of MATH, the difference MATH has compact support, and hence, as is easy to see, MATH is bounded on MATH. Thus the first term also goes to zero and MATH with respect to MATH. This means that MATH with respect to MATH . Accordingly, MATH satisfies MATH . This is equivalent to MATH . However, MATH is a self adjoint map on MATH with spectrum MATH. Since MATH is not a positive even integer, this forces MATH and we get a contradiction to MATH. |
math/0011014 | The statement is an easy consequence of the commutative diagram MATH where the two vertical morphisms on the left are injective by assumption. |
math/0011014 | Let MATH. We have : MATH as expected. |
math/0011014 | By direct computation. |
math/0011014 | This is a direct computation again. |
math/0011014 | Keeping in mind that MATH-invariants are precisely homogeneous sections of null weight, the result is a direct consequence of REF . |
math/0011014 | We recall MATH is obviously stable by exterior differentiation. Since MATH is reductive, on can find one-dimensional subtori MATH of MATH such that MATH. Then, by REF , we have : MATH . And therefore MATH . By REF , all the terms in the intersection above are stable by MATH, so we can conclude that MATH is stable by MATH too. |
math/0011014 | By a direct calculation, using the explicit definitions of MATH and MATH. |
math/0011014 | Let MATH be a homogeneous section of MATH. Then by REF we have MATH. Therefore if MATH the class of MATH in MATH vanishes. Since MATH is a direct factor of MATH, the equality is proved. |
math/0011014 | Easy, by arguments on weights. |
math/0011014 | By REF the hypotheses of REF are satisfied for the map MATH. Taking MATH-invariants in the diagram of REF together with the isomorphism MATH gives the result. |
math/0011014 | By REF we have MATH. Therefore MATH vanishes for all the complexes involved in the first isomorphism and this proves the first statement. For the second one, take MATH-invariants in the first diagram and use REF . |
math/0011014 | Since MATH, by REF we have inclusions MATH of torsion-free modules. Moreover, by REF CITE, these are isomorphisms outside the closed subset MATH, therefore outside a closed subset of codimension MATH. Thus the modules involved have isomorphic biduals and we obtain : MATH . |
math/0011014 | It is then a direct consequence of the fact that MATH has rational singularities CITE. Indeed, this implies that MATH. |
math/0011014 | After deleting a closed subset of codimension MATH we may assume that the morphism MATH is surjective in degree MATH, that is, that we have a surjection MATH and we want to prove that under this hypothesis the singular locus of MATH has codimension MATH. The proof, now divides in five steps. CASE: Quite generally, let MATH be a map of reductive algebraic groups and MATH an affine MATH-scheme together with a MATH-equivariant map MATH. We let MATH act on MATH in the following way : MATH and denote by MATH the quotient by MATH. Observe that since MATH acts freely on MATH, the map MATH is a principal fibration and therefore is smooth. We obtain commutative diagram of MATH-schemes : MATH where the vertical maps are quotients by MATH, the horizontal maps in the left-square are quotients by MATH and MATH is the factorization of the MATH-invariant map MATH acts trivially on MATH. For MATH a closed point, we denote by MATH the unique closed orbit over MATH. Let MATH be a closed point with (necessarily) reductive stabilizer MATH. The NAME Slice theorem of Luna CITE, asserts the following : There exists a smooth locally closed, MATH-stable subvariety MATH of MATH such that MATH, MATH is an open set and such that in the natural commutative REF the right-square is cartesian with etale horizontal maps (that is, an etale base change diagram). Moreover, letting MATH be the normal space at MATH of the orbit MATH, understood geometrically as a rational representation of MATH, there is a natural map of MATH-schemes MATH, etale at MATH, which induces a commutative diagram : MATH where the two squares are cartesian and the horizontal maps are etale neighbourhoods. CASE: We again refer to CITE. Let MATH be a reductive subgroup and MATH a MATH-module. We have a commutative diagram : MATH which realizes MATH as the total space of a MATH-equivariant vector bundle over the affine homogeneous space MATH with fiber at MATH equals to MATH. Conversely let MATH be a MATH-equivariant vector bundle over an affine MATH-homogeneous base MATH. Let MATH be a closed point then MATH is a MATH-module and MATH is reductive. Thus we have an equivalence between the set MATH up to conjugacy and the isomorphism classes of MATH-equivariant vector bundles over affine homogeneous bases. We denote by MATH any of those sets and classes by brackets MATH. By the preceding, we thus have a map MATH which sends MATH to the isomorphism class MATH or equivalently to the ``conjugacy class" MATH with the notations of the preceding section. Let MATH, then the set MATH is a locally closed subset of MATH, smooth with its reduced scheme structure. We will denote by MATH this smooth locally closed subscheme of MATH. Moreover the collection MATH is a finite stratification of MATH (in particular MATH has finite image). Therefore, the map MATH can be extended to all the points of MATH : Let MATH be an irreducible closed subset, then there exists a unique MATH such that MATH is dense in MATH and one can set MATH. Observe that MATH is the slice type of a general point of MATH. Another important fact about MATH is that it is compatible with strongly etale (also called excellent) morphisms : Given such a map MATH between smooth affine MATH-schemes, we have MATH. We now look closer to MATH-schemes of the kind MATH and their quotients by MATH. Write MATH for the canonical complementary submodule to MATH in MATH : MATH. Then in the construction of MATH, MATH is a trivial MATH-module and therefore the diagram obtained when MATH is replaced by MATH in the left square of REF reads : MATH . Let MATH be the class of MATH, then MATH. One can convince oneself of this fact through the description of MATH as an equivariant vector bundle over MATH. CASE: First, it is harmless to assume that the singular locus of MATH, MATH is irreducible. Let MATH and let MATH be a general closed point. By standard etale base change arguments in REF , our hypothesis and our conclusion hold for MATH at MATH if and only if they respectively hold for MATH at MATH. We can therefore assume that MATH, MATH and MATH. Now, with the notations of REF , it is clear that MATH. On the other hand MATH and, since MATH, the closed subset MATH should cut a dense open set on MATH. Consequently, we must have MATH and thus MATH. Let MATH be the quotient map by MATH, then clearly MATH. Let MATH be an integer, then the map MATH is diagonal with respect to the decompositions : MATH . Therefore MATH is surjective in degree MATH if and only if MATH is surjective in all degrees MATH. To conclude, we can therefore make the extra assumption that MATH has only an isolated singularity at MATH. And one should notice that the theorem remains in fact only to be proved when MATH or MATH, since, otherwise REF the statement is obviously true. CASE: We keep in mind all the identifications and assumptions made previously. Recalling REF and applying REF to the fibration MATH, we have an exact sequence MATH . Taking MATH-invariants together with REF for MATH leads to the exact sequence : MATH . Therefore, we have proved that MATH. Taking MATH-invariants, we obtain MATH . One can then conclude, that the hypothesis and the conclusion of the theorem hold for MATH if and only if they respectively hold for MATH. Thus we are reduced to prove the theorem in the case where MATH is a rational representation of MATH with MATH having only an isolated singularity at the origin. CASE: Carrying on, MATH is now a rational MATH-module with quotient MATH, such that MATH has only an isolated singularity at the origin. We recall the hypothesis in the theorem : The morphism MATH is surjective in degree MATH. We must prove that MATH is smooth in codimension MATH. Thus we have to prove that if MATH or MATH then MATH is smooth. The one dimensional torus MATH acts on MATH by homothety and this action commutes with the action of MATH. Thus MATH is a MATH scheme and MATH is a MATH-scheme. Both MATH and MATH are quasi-conical and MATH. Let MATH. Applying REF to MATH and MATH we obtain an injective morphism of exact complexes (the kernel of MATH is exactly the torsion of MATH, compare REF ) : MATH . From this diagram, we deduce that if MATH is surjective in degree MATH, then it is also surjective in degree MATH. Therefore we have an isomorphism MATH. Moreover, by REF we know that MATH. Thus MATH is a reflexive module. Recall that by REF CITE, MATH has rational singularities and in particular is normal and NAME and that MATH is then the dualizing module of MATH. The fundamental class map MATH CITE, CITE and REF, in degree MATH, factors through : MATH . But MATH is reflexive and, since MATH is normal, MATH is an isomorphism in codimension MATH. Therefore MATH is necessarily surjective. We now invoke REF and NAME CITE to conclude that MATH is smooth. The proof of REF is complete. |
math/0011014 | Using REF , we can a give a straightforward proof of the result : By REF the hypotheses of REF are satisfied for the same integer MATH. |
math/0011014 | By the preceding remarks, we are reduced to the case where MATH is a rational representation of MATH as a small group of automorphism. So that the map MATH is unramified in codimension one. We recall that, MATH being finite, we have MATH. Moreover by REF we deduce that MATH is smooth in codimension REF and we can assume that MATH. Thus we can assume that MATH and purity of the branch locus implies that MATH is unramified in codimension REF. From now on we proceed by induction on MATH. Since MATH is abelian, MATH decomposes as a product of representation : MATH with MATH. We have a diagram MATH where the vertical maps are quotient by MATH and the horizontal ones are embeddings. This induces a commutative diagram : MATH where all the morphisms are surjective. Thus, by the induction hypothesis, MATH is smooth. Now, if MATH were not trivial, the origin being a fixed point, the map MATH should have to be ramified and, by purity of the branch locus again, its ramification locus should have codimension one. But then MATH should be ramified in codimension REF. It is a contradiction. Thus, MATH is trivial and therefore MATH is smooth. |
math/0011018 | The proof we present below is well-known. The argument appeared in REF , p. REF, where it was attributed to NAME. Let MATH be a vector field on MATH that leaves MATH invariant, say induced by MATH. Then REF holds for a certain MATH. Since MATH, NAME 's formula, MATH implies that REF is equivalent to MATH where MATH for MATH. Note that MATH is also induced by MATH. Let MATH. Let MATH be the ideal generated by the partial derivatives of MATH: MATH . As MATH, by NAME 's formula, MATH. Since MATH is smooth, MATH. Hence, the partial derivatives MATH form a system of parameters for the local ring MATH, where MATH is the localization of MATH at MATH. Now, since MATH is regular, it is NAME by REF , p. CASE: So, the sequence MATH is regular on MATH by REF , p. CASE: It follows that MATH is regular on MATH as well, whence its NAME complex is an exact sequence by REF , p. CASE: In particular, the module of relations, MATH is generated by the trivial ones, MATH . As MATH by REF, and MATH is induced by MATH, the proof is finished. |
math/0011018 | By REF , the vector field MATH is induced by REF for certain homogeneous MATH of the appropriate degrees. Since MATH, we have MATH for certain MATH. Since MATH has degree MATH, we have MATH. As MATH, we get MATH. |
math/0011018 | Let MATH be the normalization of MATH and MATH the adjoint to the induced pullback map. Since MATH has at most ordinary nodes for singularities, we can check locally that MATH is surjective. As MATH is birational, the kernel of MATH is the torsion sheaf of MATH. So MATH . By NAME 's duality REF on MATH, the projection formula, and NAME 's duality on MATH, MATH . Now, as MATH is ample on MATH, the degree of MATH on each component of MATH is negative. Thus MATH, and so MATH. |
math/0011018 | The normal bundle of MATH in MATH is MATH. By REF , p. REF, the dualizing sheaf MATH of MATH satisfies MATH . Since MATH, the sheaf MATH is ample, and hence MATH by REF . |
math/0011018 | The theorem was proved by NAME and NAME for MATH; see REF and the remark thereafter on p. CASE: We will prove the more general statement above in a forthcoming work. In the next remark, we will show what happens when MATH. |
math/0011018 | Let MATH and MATH be linear subspaces of MATH of dimensions MATH and MATH satisfying MATH. Assume MATH and MATH are in general position. We claim that the following three conditions hold. CASE: The intersection MATH is empty. CASE: The projection of MATH with center MATH has degree MATH. CASE: For each irreducible component MATH, the linear subspace of MATH spanned by MATH and a general MATH does not contain the direction given by MATH at MATH. In fact, since MATH has pure codimension MATH, a general linear subspace MATH of dimension MATH does not meet MATH. Thus, REF holds. As for REF , let MATH be a general linear subspace of dimension MATH containing MATH. As MATH is of pure codimension MATH, the intersection MATH is finite, and contains at least one general point MATH on each irreducible component MATH. Since MATH is a general subspace of MATH of dimension MATH, we have MATH and, for each irreducible component MATH, the linear subspace of MATH spanned by MATH and a general MATH intersects MATH only at MATH. Now, as MATH has codimension REF in MATH, and MATH is of pure dimension, for each irreducible component MATH, the general MATH satisfies MATH. Thus, since MATH is general of dimension MATH, the linear subspace of MATH spanned by MATH and MATH intersects MATH only at MATH. So, for each irreducible component MATH and each general MATH, the linear subspace of MATH spanned by MATH and MATH intersects MATH only at MATH, and with multiplicity REF. Projecting MATH with center MATH, we obtain a finite map which is an isomorphism over a dense open subscheme of the image. Hence, the image has degree equal to that of MATH, that is, degree MATH. Finally, by hypothesis, for each irreducible component MATH, the general MATH is non-singular for MATH. Since MATH, and MATH is general of dimension MATH, the direction given by MATH at MATH does not meet MATH. Our claim is proved. Let MATH. After changing coordinates, we may assume MATH is cut out by MATH, and MATH is cut out by MATH in MATH. As MATH, there is a linear form in MATH which is not identically zero on any irreducible component of MATH. Up to subtracting from MATH a certain multiple of this form, we may assume MATH is not identically zero on any irreducible component of MATH. Let MATH. Let MATH denote the saturated, homogeneous ideal cutting out MATH, and put MATH. Let MATH. As MATH cut out the empty set in MATH, we have MATH. Then the quotient MATH is a vector space of finite dimension over MATH. By hypothesis, MATH is NAME of pure dimension MATH. So, MATH is a local, NAME ring of dimension MATH. As MATH generate in MATH an ideal of dimension zero, they form a regular sequence by REF , p. CASE: Now, the multiplicity of MATH is the degree of MATH; see REF , p. CASE: Since MATH form a regular sequence in MATH, by REF , p. REF, the multiplicity of MATH is equal to the dimension of MATH over MATH, which is the same as the dimension of MATH. So MATH. Let MATH be monomials in MATH generating MATH over MATH. We may assume MATH. Let MATH for MATH. By REF , p. REF, the regularity of MATH is equal to that of MATH, which is MATH. Since MATH has finite length, using REF , p. REF, we get MATH . We may assume MATH. As MATH generate MATH, for each MATH there are homogeneous MATH for MATH such that MATH . Let MATH and MATH. As MATH, and REF holds for each MATH, it follows that MATH. Clearly, MATH is monic in MATH, hence non-zero. In addition, MATH is homogeneous of degree MATH, because the degree of MATH is MATH for each MATH. As MATH, the projection MATH of MATH with center MATH is contained in the hypersurface cut out by MATH in MATH. By REF above, also MATH has degree MATH. So, MATH cuts out MATH. If MATH, also MATH. In this case, let MATH. Then MATH is non-zero, homogeneous of degree MATH, and the following property holds. MATH . Suppose now that MATH. We claim there is MATH non-zero, homogeneous of degree MATH, satisfying REF . Indeed, for each MATH, let MATH be the determinant of the matrix obtained from MATH by removing the MATH-th column and the last row. Put MATH. For each MATH, let MATH be the matrix obtained from MATH by exchanging the last row with the row MATH, where MATH for MATH and MATH . By REF, modulo MATH, the matrix MATH has eigenvalue MATH, with MATH being one eigenvector if MATH and MATH one if MATH. Since MATH, it follows that MATH. Expanding MATH by the cofactors of the last row, we get MATH and MATH if MATH. Now, as MATH is not identically zero on any irreducible component of MATH, and MATH is arithmetically NAME, MATH is regular in MATH. So MATH. As MATH, and MATH generate MATH, REF holds. We show now that MATH, whence MATH. Indeed, if MATH, as MATH, we have MATH. Then, as REF holds, MATH for each MATH as well. Since MATH, there is MATH such that MATH. Fix such MATH. As MATH is generated by MATH, and MATH, the degree of MATH would be at least MATH. However, as MATH, the degree of MATH is at most MATH. We have a contradiction. Hence MATH. As MATH, we have MATH, whence MATH. So, MATH is as claimed. Let MATH be the projection of MATH with center MATH. If MATH cuts out a subscheme containing MATH, then MATH is left invariant by the vector field on MATH induced by MATH, which is non-zero because MATH, and has degree MATH. The theorem is proved in this case. Suppose now that MATH is not contained in the subscheme cut out by MATH. As MATH is reduced, there is an irreducible component MATH such that MATH for the general MATH. Let MATH homogeneous of degree MATH such that MATH is induced by MATH. By REF , for each MATH, there is MATH such that MATH. Let MATH be the vector field on MATH induced by MATH. If MATH, then MATH has degree MATH. Moreover, since MATH leaves MATH invariant, MATH leaves MATH invariant; see REF. It remains to show that MATH. Suppose MATH. Then there is MATH such that MATH, whence MATH, for each MATH. For the general MATH we have MATH; hence MATH for each MATH. So, the linear subspace of MATH spanned by MATH and MATH contains the direction given by MATH at MATH. We get a contradiction with REF . So MATH. |
math/0011018 | (Theorem MATH) Let MATH be the degree of MATH. Let MATH be the normal-crossings hypersurface obtained from MATH by projection with general center. The degree of MATH is MATH as well. By REF , the hypersurface MATH is left invariant by a non-zero vector field on MATH of degree MATH. Then MATH by CITE. So MATH. |
math/0011018 | Suppose first that MATH. Then MATH. So, the desired inequality follows from REF if MATH is reducible or MATH, and REF otherwise. We may thus assume MATH. Let MATH be the projection of MATH with center a general linear subspace MATH of dimension MATH. Since MATH has at most ordinary nodes for singularities, and MATH is general, MATH has at most ordinary nodes as well. (The proofs given to REF , p. REF , p. REF apply.) By REF with MATH and MATH, there is a non-zero vector field on MATH of degree MATH leaving MATH invariant. Now, apply REF or REF to MATH. |
math/0011018 | CASE: Let MATH. Since MATH is arithmetically NAME, and MATH does not contain any irreducible component of MATH, also MATH is arithmetically NAME by REF , p. CASE: It follows from REF , p. REF that the degree of MATH is MATH. In addition, the regularity of MATH is MATH. (The proof given to REF , p. REF applies.) Since MATH is smooth at all but finitely many points of MATH, the tangent space of MATH at all but finitely many points has dimension at most REF. From REF , we get MATH with equality only if MATH is reducible or MATH. |
math/0011020 | We first define classes which generate MATH as a NAME algebra. Pick a basepoint in MATH, say with MATH for definiteness. There are MATH generators of MATH, corresponding to distinct pairs MATH, which we now realize geometrically. We define MATH as the class represented by the composite of two maps. First, we collapse MATH onto MATH by sending the ``southern hemisphere" of MATH to MATH through the height function. Next, choose a path MATH from MATH to the point MATH in the complement of the other configuration points, and let MATH denote the map which sends MATH to the unit sphere about the point MATH. To define MATH we compose the collapse map above with the map MATH to MATH which sends MATH to MATH as MATH and MATH to MATH as MATH. To see inductively that these classes are generators of MATH, we simply note that MATH is equal to the image under MATH of the generator of MATH defined by the inclusion of the MATH-th wedge factor. It is simple to check that these MATH satisfy the relations for MATH in the definition of MATH. Note that from the usual graded commutativity of the NAME product, brackets in MATH anti-commute when MATH is odd and commute when MATH is even. Note that MATH and MATH so that relations REF are satisfied. We next verify that the MATH satisfy relation REF . Recall that if MATH and MATH are elements of MATH then MATH if and only if MATH extends to MATH. If MATH, the map MATH may be so extended by sending MATH where MATH is the composite of the collapse map of MATH onto MATH with MATH. Informally we say that MATH can travel around MATH and MATH can travel around MATH without having their paths (the images of MATH under the projection onto the MATH-th and MATH-th coordinates) intersect. Next, we verify that the MATH satisfy relation REF . Equivalently, we claim that MATH. Informally, we say that MATH is represented by a map in which MATH travels around MATH and MATH but no other points in the configuration, and this may happen simultaneously as MATH travels around MATH, giving an extension of MATH to MATH similar to the one given for MATH. We claim that relations REF through REF are a complete set of relations for MATH. This follows from the fact that these relations may be used to reduce to an additive basis of NAME algebra monomials of the form MATH, where MATH. We exhibit this claim algorithmically when discussing the computations in REF; see in particular REF . |
math/0011020 | The NAME characteristic of the complex MATH is by REF, which after applying REF , reversing the order of summation, and ignoring zero terms, is equal to MATH . We claim that MATH, which can be verified by computing the coefficient of MATH of MATH. Hence the NAME characteristic is equal to MATH, which is zero if MATH. |
math/0011021 | The obvious projection MATH restricts to a distance-increasing local diffeomorphism MATH since MATH is everywhere spacelike. As MATH is closed, the induced Riemannian metric is complete, and thus MATH must be proper. This makes MATH a covering map, hence a global diffeomorphism. Finally MATH is a graph because MATH is connected. If MATH and MATH are two points of the image which are null or timelike separated, consider the path between them given by the intersection of MATH and an indefinite two-dimensional plane containing MATH and MATH. The ``secant line" joining MATH and MATH in this plane has slope greater than MATH, so the mean value theorem implies that some tangent vector to this path is null or timelike, a contradiction. |
math/0011021 | The first part is straightforward by REF - since MATH, MATH has finite cohomological dimension and is therefore torsion-free. Discreteness follows because MATH acts properly discontinuously on MATH. Because all pairs of points in MATH are spacelike-separated, it is clear that MATH consists only of spacelike vectors. It is isomorphic to MATH since it is a discrete subgroup of MATH and MATH cannot be MATH or else MATH would have to contain a non-spacelike vector. For the last part, if MATH is given by MATH and MATH is translation by MATH, then MATH is easily computed to be the element MATH of MATH. |
math/0011021 | REF (see CITE) says that MATH is always solvable and is non-trivial since MATH is indiscrete. The closed, connected, solvable, non-trivial NAME subgroups of MATH are easy to write down as in CITE; in particular, the set of points MATH on the sphere at infinity MATH fixed by MATH must consist of one or two points. The stabilizer in MATH of a point at infinity is isomorphic to the group MATH of orientation-preserving similarities of MATH, which is a solvable group. Since MATH normalizes MATH, it leaves MATH invariant and therefore has a subgroup of index at most two fixing MATH pointwise, and therefore conjugate into MATH. The lemma follows. |
math/0011021 | While a much more general result is proved in CITE, we give a simpler proof sufficient for our needs. Consider the cover MATH of MATH corresponding to the MATH subgroup. Since MATH, MATH is a MATH hence is homotopy equivalent to MATH. This implies that MATH, thus MATH is closed and the covering MATH is finite. Finally, appealing to CITE, we have MATH. |
math/0011021 | The ``if" half is easy as each of the spaces MATH, MATH, and MATH embeds in MATH (though there can be geometrically distinct embeddings as we will see for MATH). We have CASE: MATH CASE: MATH CASE: MATH. If MATH is modelled on a model space MATH, then it can be realized as MATH, where MATH is a discrete, cocompact subgroup of the isometry group MATH. It is easy to see that if MATH is one of the three examples above embedded in MATH, then there is a corresponding embedding of its isometry group in MATH, and that any discrete subgroup of MATH acts discontinuously on a regular neighborhood of MATH in MATH. The quotient of a small regular neighborhood is therefore a flat spacetime containing MATH as a spacelike slice, as desired. For the ``only if" half, let MATH be a spacelike slice of a flat spacetime with holonomy group MATH. The proof is broken down into four cases, depending on the rank of the translational subgroup MATH REF . Note that the cases get easier as we go along, because the presence of a large normal abelian subgroup of MATH typically puts strong topological constraints on MATH. Suppose MATH. This means that MATH injects MATH into MATH, that is, MATH. Our assumption that spacelike slices have trivial normal bundles means that MATH actually lies in the orthochronous subgroup MATH which coincides with the full isometry group of MATH. Now MATH is either discrete or indiscrete. If it is discrete, then it is also cocompact for cohomological reasons. Since MATH is aspherical, it is homotopy equivalent to the closed hyperbolic MATH-manifold MATH and a result of CITE implies that MATH is itself hyperbolic. In fact, it will turn out that this is the only possibility for the holonomy when MATH is hyperbolic. Thus in all remaining cases we will be proving that MATH is modelled on MATH or MATH. In light of REF , it suffices to show that MATH has a finite cover homeomorphic to MATH, where MATH is a closed, orientable surface, MATH. We will exploit this fact in all that follows by freely passing to finite covers of MATH without changing notation. Also note that MATH is reducible in all remaining cases, yielding REF . If MATH is indiscrete, REF shows that MATH is virtually solvable. By the main result of CITE, we may pass to a finite cover and assume MATH is a torus bundle over MATH with monodromy MATH represented by a matrix in MATH, by abuse of notation also denoted MATH. Clearly MATH has finite order if and only if MATH is finitely covered by a MATH-torus, we will assume MATH has infinite order. Let MATH denote an element inducing the monodromy, that is, MATH for all MATH. The image of the fiber subgroup MATH of MATH under MATH must consist either of elements leaving invariant a geodesic in MATH (generated by loxodromics or irrational elliptics) or of parabolics with a common fixed point at infinity. In the first case, since MATH normalizes MATH, MATH must leave the geodesic invariant (indeed it must fix it pointwise since MATH has infinite order). But this implies that MATH commutes with MATH which means MATH is the identity and MATH. This contradicts the hypothesis that MATH has infinite order, or alternatively shows directly that MATH in this case by the proof of REF . In the second case, identify MATH and conjugate so that the parabolics fix MATH. If MATH has a MATH-eigenvalue, let MATH denote an element of MATH such that MATH and write MATH as MATH for some MATH. Since MATH normalizes MATH, MATH must also fix MATH; write it as MATH for some MATH. But then with this notation, the relation MATH becomes MATH. Thus MATH, MATH is parabolic, MATH commutes with MATH, and we are done as above. The final possibility is that MATH is of infinite order and has no MATH-eigenvalue, in which case it has two real eigenvalues of absolute value not equal to one. In this case, there is, up to conjugacy, only one possibility for MATH, and we use this in the Appendix to show that MATH cannot be discrete, contradicting REF . This completes Case MATH. Suppose MATH. We may assume MATH is orientable by passing to a double cover. Though we can get away with less, we might as well use the NAME Fiber Space Theorem of Mess CITE, CITE, and CITE which states that a closed, orientable, irreducible MATH-manifold whose fundamental group contains a normal MATH is a NAME fiber space. The proof of this result amounts to showing that the quotient group MATH (namely MATH in our setup) is the fundamental group of a closed MATH-orbifold MATH, which by passing to a finite cover, we can assume is a closed, orientable surface of genus MATH. It follows that MATH is modelled on MATH, MATH, MATH, or MATH. Excluding the final two possibilities amounts to showing that the NAME number MATH of MATH is zero. Our original proof that MATH amounted to finding the place where the NAME number appears in the NAME sequence for MATH; we have chosen to give a more geometric argument here. The group MATH has a presentation of the form CITE MATH the notation is meant to be obvious: MATH generates the normal subgroup MATH and the other generators project to MATH, the fundamental group of the base MATH. Write MATH for the translation MATH, and MATH, MATH for the other generators. If we let MATH be the subspace left invariant by MATH, then the commutation relations imply that MATH and MATH, and so MATH. Now focus on the final relation: the left-hand side is the translation MATH while the right-hand side's translational part is a combination of the MATH and MATH: MATH where MATH. The linear map being applied to MATH (respectively, MATH) in the expression above is called the NAME derivative MATH (respectively, MATH) of the usual surface group relator MATH (see CITE, CITE). These partial derivatives can be combined neatly into a single NAME differential MATH, in which case MATH . In CITE, NAME has a nice argument showing that the image of MATH is exactly MATH. Thus MATH and MATH, so MATH . Since MATH is spacelike (this is essential - see CITE), MATH, and so MATH. This completes Case MATH. Suppose MATH. A theorem in CITE shows that MATH has two ends and therefore has a finite index subgroup isomorphic to MATH. As usual, we will pass to a finite cover without changing notation and assume MATH. NAME 's theorem CITE implies that MATH fibers overs the circle with torus fibers. Let MATH be a generator of MATH. It leaves invariant the spacelike MATH spanned by MATH, and in fact, acts by linear isometries. Since MATH normalizes the lattice MATH, the action of MATH on MATH must have finite order. It follows that MATH is finitely covered by the MATH-torus, and we conclude that MATH is modelled on MATH. This completes Case MATH. Suppose MATH. REF implies that MATH is finitely covered by the MATH-torus and is therefore modelled on MATH. This completes the proof of the main theorem. |
math/0011021 | In fact, we will show that the subgroup MATH must be indiscrete. The idea of the proof is to view MATH as obtained from the indiscrete group MATH by adding translations, and to show that there is no way of doing so which makes MATH discrete. Such a choice of translations is a cocycle in MATH; that is, a function MATH satisfying the cocycle relation MATH for all MATH. The coboundaries (change of basepoint) are cocycles of the form MATH for some fixed MATH. There is, of course, a restriction map to the fiber subgroup MATH where the notation is meant to indicate that restricted classes are invariant under the action of MATH: MATH . The lemma will follow from the fact that the group MATH vanishes. The cocycle relation applied to MATH means that for any cocycle MATH we must have MATH . REF (really two equations for MATH and MATH) and REF are linear in MATH and MATH and can be solved explicitly by choosing a basis for MATH and elements of MATH representing the generators of MATH. For instance, following CITE, we can choose the first two basis elements to be null vectors fixed by MATH (the second fixed by MATH) and the last two basis elements to be spacelike (the final one also fixed by MATH). This yields: MATH . Some linear algebra shows that there is a one-dimensional solution set to REF ; namely by taking MATH and MATH. This solution is a coboundary however, since MATH for MATH as a simple calculation shows. Thus MATH, proving the lemma. |
math/0011028 | We assert that MATH is a rank-MATH partial isometry if and only if so is MATH. By the order preserving property of MATH we readily have MATH which verifies our claim in the case MATH. Suppose that we have the assertion for MATH. Let MATH be a partial isometry on MATH. It is clear that MATH has rank MATH if and only if for every MATH with MATH, MATH it follows that the rank of MATH is less than or equal to MATH and there is such a MATH whose rank is MATH. The order preserving property of MATH now implies that the rank of MATH is MATH. Our next aim is to show that MATH is completely orthoadditive. The meaning of this property will be clear in a moment. Let MATH be a set of pairwise orthogonal partial isometries. It is well-known that the series MATH is convergent in the strong operator topology and its sum MATH is a partial isometry. The sum MATH is also a partial isometry, so there exists a MATH such that MATH . Since MATH for every MATH, it follows that MATH. On the other hand, since MATH, we have MATH for every MATH and we obtain that MATH. This yields MATH. Therefore, we have MATH and this means the complete orthoadditivity of MATH. Take MATH. Let MATH be MATH-dimensional subspaces of MATH. Denote by MATH the set of all partial isometries on MATH whose initial space is a subset of MATH and whose final space is a subset of MATH. Let MATH be a partial isometry on MATH whose initial space is MATH and whose final space is MATH. Then MATH is a rank-MATH partial isometry. Let MATH and MATH. It is easy to see that an operator MATH belongs to MATH if and only if for every MATH which is orthogonal to MATH it follows that MATH is orthogonal to MATH. By the orthogonality preserving property of MATH we deduce that MATH maps MATH onto MATH. Clearly, MATH and MATH are both isomorphic to the space MATH of all partial isometries on a MATH-dimensional NAME space MATH. Therefore, our map MATH induces a transformation MATH on MATH which has the same preserver properties as MATH. Obviously, MATH is unitary. Multiplying MATH by a fixed unitary element of MATH, we can assume that MATH. This latter property implies that MATH sends projections to projections. In fact, for any partial isometry MATH we have MATH if and only if MATH is a projection. So, MATH is a bijective transformation of the set of all projections in MATH which preserves the order and the orthogonality in both directions. The form of such transformations is well-known. It follows, for example, from CITE that there exists an either unitary or antiunitary operator MATH on MATH such that MATH for every projection MATH in MATH. Let MATH. We state that MATH is a scalar multiple of the identity. To see this, first observe that if MATH is a rank-REF partial isometry, then MATH is a scalar multiple of MATH (which scalar obviously has modulus REF). This follows from the fact that any partial isometry in MATH is orthogonal to MATH if and only if it is orthogonal to MATH and that MATH, MATH are both of rank REF. Now, one can characterize the scalar operators in MATH in the following way. The partial isometry MATH is equal to the identity multiplied by a scalar from MATH if and only if for every rank-one projection MATH we have that MATH for some MATH. Therefore, we obtain that MATH preserves the scalar partial isometries. Consequently, there is a function MATH such that MATH . We know that if MATH is any rank-one projection and MATH, then there is a MATH such that MATH. As MATH, we obtain that MATH. This implies that MATH, so we have MATH for every rank-REF projection MATH and MATH. By the orthoadditivity of MATH (this follows form the orthoadditivity of MATH) we infer that the previous equality holds true for every projection MATH without any restriction on its rank. If MATH is any partial isometry, then MATH can be written in the form MATH, where MATH is unitary and MATH is a projection. Considering the transformation MATH on MATH and applying what we have proved above, it follows that there is a function MATH such that MATH . The function MATH might depend on MATH since the function MATH appearing above depends on MATH. However, we prove that MATH for every partial isometry MATH. This will be done below. Let MATH be a rank-REF partial isometry. Pick a rank-REF projection MATH which is orthogonal to MATH. By the orthoadditivity of MATH, for any MATH we compute MATH . On the other hand, we have MATH . These imply that MATH. So we have MATH for every rank-REF partial isometry MATH. Since every partial isometry is the sum of mutually orthogonal rank-REF partial isometries, by the orthoadditivity of MATH we obtain that the above equality holds also for every partial isometry MATH on MATH. Since for every MATH we have MATH it follows that MATH is a multiplicative bijection. We supposed that our original transformation is norm-continuous at a point MATH. We can find pairwise orthogonal rank-REF partial isometries MATH for which MATH. By the complete orthoadditivity of MATH we obtain that MATH . As we have seen above, the operator MATH is a scalar multiple of MATH for every MATH. Let MATH be a sequence in MATH converging to REF. This implies that MATH and then we have MATH for every MATH. Pick a MATH and let us suppose that MATH. We state that MATH is continuous. In fact, by REF we can deduce that MATH is continuous at REF. Since MATH is multiplicative, we obtain that MATH is a continuous character of MATH. It is well-known that the continuous characters of MATH are exactly the functions MATH. Since MATH is bijective, we obtain that MATH is either the identity or the conjugation on MATH. Suppose that the operator MATH in REF is unitary and suppose that MATH is the conjugation on MATH. Using the spectral theorem and the orthoadditivity of MATH, we obtain from REF that for every unitary MATH in MATH we have MATH . If MATH is arbitrary, then there is a partial isometry MATH on MATH which is orthogonal to MATH such that MATH is unitary for every MATH. It follows that MATH for every MATH. This obviously implies that MATH. Since MATH, we obtain MATH. Examining the remaining cases concerning MATH and MATH, we find that MATH is of one of the following forms: CASE: there exists a unitary MATH on MATH such that MATH CASE: there exists a unitary MATH on MATH such that MATH CASE: there exists an antiunitary MATH on MATH such that MATH CASE: there exists an antiunitary MATH on MATH such that MATH . Going back to our original map MATH, we see that on MATH the function MATH is of one of the following forms: CASE: there exist unitaries MATH on MATH such that MATH CASE: there exist unitaries MATH on MATH such that MATH CASE: there exist antiunitaries MATH on MATH such that MATH CASE: there exist antiunitaries MATH on MATH such that MATH . Pick two mutually orthogonal unit vectors MATH and consider the partial isometries MATH, MATH, MATH. Denote by MATH the subspace generated by MATH. On MATH is of one of the above forms. Suppose that we have REF . It follows that MATH, MATH and MATH. Consider a finite dimensional subspace MATH of MATH which contains all the initial and final spaces of MATH. It is easy to see that on MATH the transformation MATH must be of the form REF . In fact, the other possibilities can be easily excluded considering the relations among MATH and MATH. Let us take any finite rank operator MATH. By spectral theorem and polar decomposition, MATH can be written as a finite sum MATH where MATH's are finite rank partial isometries and MATH's are scalar. We define MATH . We claim that MATH is a conjugate-linear transformation on MATH extending the restriction of MATH onto the set of all finite rank partial isometries on MATH. First, we have to show that MATH is well-defined. Let MATH be another resolution of MATH similar to what was given in REF. Let MATH be a finite dimensional subspace of MATH which contains the initial and finial spaces of all MATH appearing above. It follows that on MATH the transformation MATH is of the form REF . Therefore, we can write MATH which means that MATH is well-defined. Now, the additivity and conjugate-linearity of MATH is trivial to verify. So, we have a conjugate-linear map MATH which extends the restriction of MATH onto the set of all finite rank partial isometries. Using the 'local' form REF of MATH, it is obvious that MATH satisfies MATH, that is MATH is a conjugate-linear NAME homomorphism. Since every finite rank partial isometry is in the range of MATH, it follows that MATH is surjective. If MATH, then we can write MATH in the form REF where MATH's are pairwise orthogonal finite rank partial isometries. By the orthogonality preserving property of MATH we readily have MATH. Therefore, the transformation MATH is a linear NAME automorphism of MATH. Now, we can apply the result CITE on the form of linear NAME isomorphisms between standard operator algebras (in that paper we used the term 'triple isomorphism'). This gives us that, on the set of all finite rank partial isometries, MATH is one of the forms REF . Using the complete orthoadditivity of MATH we find that the same holds true on the whole set MATH. The last part of the proof has begun with supposing that MATH is of the form REF on a certain subset MATH of the set of all partial isometries. In any of the other cases one can apply very similar arguments. |
math/0011028 | As we have noted in the proof of our first theorem, if MATH, then MATH can be written as a finite sum MATH where MATH's are rank-one partial isometries and MATH's are scalars. We define MATH . We claim that MATH is well-defined. Indeed, if MATH is another resolution of MATH of the above kind, then we compute MATH and, similarly, MATH . Therefore, it follows that MATH for every MATH. Since MATH is surjective, we obtain that MATH. So, MATH is well-defined. It is now obvious that MATH is a linear transformation on MATH. Since the range of MATH contains every rank-REF partial isometry, it follows that MATH is surjective. An operator MATH has rank REF if and only if it is a nonzero scalar multiple of a partial isometry. It follows that MATH preserves the rank-REF operators in both directions. The form of all surjective linear (even additive) transformations on MATH which preserve the rank-REF operators is known. Namely, it follows from CITE that either there are bijective linear operators MATH on MATH such that MATH or there are bijective conjugate-linear operators MATH on MATH such that MATH . Depending on the actual case, we easily obtain from REF that MATH are scalar multiples of unitaries or antiunitaries. It is now trivial to complete the proof. |
math/0011029 | For any finite sets MATH and MATH we define MATH . We have to show that MATH is well-defined. If MATH, where MATH and MATH are finite subsets, then for any MATH we compute MATH . Therefore, we have MATH for every MATH. By the linearity of the trace functional it follows that we have similar equality if we replace MATH by any finite linear combination of MATH's. This gives us that MATH . The operator MATH, being the square of a self-adjoint operator, is positive. Since its trace is zero, we obtain that MATH which plainly implies that MATH . This shows that MATH is well-defined. The real-linearity of MATH now follows from the definition. The uniqueness of MATH is also trivial to see. From REF we immediately obtain REF. One can introduce an inner product on MATH by the formula MATH (the norm induced by this inner product is called the NAME norm). The equality REF shows that MATH is an isometry with respect to this norm. Thus, MATH is injective. It follows from REF that MATH which clearly implies that MATH . This completes the proof of the lemma. |
math/0011029 | Since every finite-rank projection is the finite sum of pairwise orthogonal rank-REF projections, it is obvious that MATH preserves the finite-rank projections. It follows from CITE and the spectral theorem that MATH is a NAME homomorphism (we note that CITE is about self-adjoint operators on finite dimensional complex NAME spaces, but the same argument applies for MATH even if it is infinite dimensional and/or real). We next prove that MATH can be extended to a NAME homomorphism of MATH. To see this, first suppose that MATH is complex and consider the transformation MATH defined by MATH . It is easy to see that MATH is a linear transformation which satisfies MATH. This shows that MATH is a NAME homomorphism. If MATH is real, then the situation is not so simple, but we can apply a deep algebraic result of NAME as follows (compare the proof of CITE). Consider the unitalized algebra MATH (of course, we have to add the identity only when MATH is infinite dimensional). Defining MATH, we can extend MATH to the set of all symmetric elements of the enlarged algebra in an obvious way. Now we are in a position to apply the results in CITE on the extendability of NAME homomorphisms defined on the set of symmetric elements of a ring with involution. To be precise, in CITE NAME homomorphism means an additive map MATH which, besides MATH, also satisfies MATH. But if the ring in question is REF-torsion free (in particular, if it is an algebra), this second equality follows from the first one (see, for example, the proof of CITE). The statements CITE in the case when MATH and CITE if MATH imply that MATH can be uniquely extended to an associative homomorphism of MATH into itself. To be honest, since the results of NAME concern rings and hence linearity does not appear, we could guarantee only the additivity of the extension of MATH. However, the construction in CITE shows that in the case of algebras, linear NAME homomorphisms have linear extensions. To sum up, in every case we have a NAME homomorphism of MATH extending MATH. In order to simplify the notation, we use the same symbol MATH for the extension as well. As MATH is a locally matrix ring (every finite subset of MATH can be included in a subalgebra of MATH which is isomorphic to a full matrix algebra), it follows from a classical result of CITE that MATH can be written as MATH, where MATH is a homomorphism and MATH is an antihomomorphism. Let MATH be a rank-REF projection on MATH. Since MATH is also rank-REF, we obtain that one of the idempotents MATH is zero. Since MATH is a simple ring, it is easy to see that this implies that either MATH or MATH is identically zero, that is, MATH is either a homomorphism or an antihomomorphism of MATH. In what follows we can assume without loss of generality that MATH is a homomorphism. Since the kernel of MATH is an ideal in MATH and MATH is simple, we obtain that MATH is injective. We show that MATH preserves the rank-REF operators. Let MATH be of rank REF. Then there is a rank-MATH projection MATH such that MATH. We have MATH which proves that MATH is of rank at most MATH. Since MATH is injective, we obtain that the rank of MATH is exactly REF. From the conditions of the lemma it follows that MATH sends rank-REF projections to rank-REF projections. Therefore, the range of MATH contains an operator with rank greater than REF. We now refer to NAME 's work CITE on the form of linear rank preservers on operator algebras. It follows from the argument leading to CITE that either there are linear operators MATH on MATH such that MATH is of the form MATH or there are conjugate-linear operators MATH on MATH such that MATH is of the form MATH . Suppose that we have the first possibility. By the multiplicativity of MATH we obtain that MATH . This gives us that MATH for every MATH. On the other hand, since MATH sends rank-REF projections to rank-REF projections, we obtain that for every unit vector MATH we have MATH. These imply that MATH is an isometry and with the notation MATH we have MATH for every MATH. We show that the possibility REF cannot occur. In fact, similarly to REF we have MATH . Fixing unit vectors MATH in MATH and considering the operators above at MATH, we find that MATH giving us that MATH is of rank REF. Since MATH sends rank-REF projections to rank-REF projections, we arrive at a contradiction. This completes the proof of the lemma. |
math/0011029 | By the spectral theorem it is obvious that the real linear span of MATH is MATH. Then, by REF we see that there is a unique real-linear extension MATH of MATH onto MATH which preserves the rank-REF projections and, by REF, MATH also preserves the orthogonality between the elements of MATH. REF applies to complete the proof. |
math/0011029 | Since the real-linear span of MATH is MATH, it is sufficient to show that every rank-REF projection is a real-linear combination of rank-MATH projections. To see this, choose orthonormal vectors MATH in MATH. Let MATH and define MATH . Clearly, every MATH can be represented by a MATH diagonal matrix whose diagonal entries are all REF's with the exception of the MATH one which is REF. The equation MATH gives rise to a system of linear equations with unknown scalars MATH. The matrix of this system of equations is a MATH matrix whose diagonal consists of MATH's and its off-diagonal entries are all REF's. It is easy to see that this matrix is nonsingular, and hence MATH (and, similarly, every other MATH) is a real-linear combination of MATH. This completes the proof. |
math/0011029 | Let MATH. Since MATH is a projection whose range is contained in the range of MATH, it follows that MATH is a projection which is orthogonal to MATH. If MATH is a unit vector in the range of MATH, then we have MATH. Since MATH is a vector whose norm is at most REF and its image under the projection MATH has norm REF, we obtain that MATH is a unit vector in the range of MATH. Similarly, we obtain that MATH is a unit vector in the range of MATH and, finally, that MATH is a unit vector in the range of MATH. Therefore, MATH belongs to the range of MATH and MATH. Since MATH was arbitrary, we can infer that the range of MATH is included in the range of MATH. Thus, we obtain that MATH is a projection which is orthogonal to MATH. Next, using the obvious relations MATH we deduce MATH . Since MATH implies MATH for any MATH, we obtain from REF that MATH. The second part of the assertion is now easy to check. |
math/0011029 | By REF , MATH can be uniquely extended to an injective real-linear transformation MATH on MATH. The main point of the proof is to show that MATH preserves the rank-REF projections. In order to verify this, just as in the proof of REF , we consider orthonormal vectors MATH in MATH, define MATH and set MATH . We show that the ranges of all MATH's can be jointly included in a MATH-dimensional subspace of MATH. To see this, we first recall that MATH has the property that MATH (see REF ). Next we have the following property of MATH: if MATH are orthogonal rank-REF projections, then MATH. Indeed, if MATH are orthogonal, then we can include them into two orthogonal rank-MATH projections. Now, referring to the construction given in REF and having in mind that MATH preserves the orthogonality between rank-MATH projections, we obtain that MATH. (Clearly, the same argument works if MATH.) Since the rank-MATH projections MATH are commuting, by the preserving property of MATH and REF , it follows that the projections MATH are also commuting. It is well-known that any finite commuting family of operators in MATH can be diagonalized by the same unitary transformation (or, in the real case, by the same orthogonal transformation). Therefore, if we resctrict MATH onto the real-linear subspace in MATH generated by MATH, then it can be identified with a real-linear operator from MATH to MATH for some MATH. Clearly, this restriction of MATH can be represented by a MATH real matrix MATH. Let us examine how the properties of MATH are reflected in those of the matrix MATH. First, MATH is trace preserving. This gives us that for every MATH the sums of the coordinates of the vectors MATH and MATH are the same. This easily implies that the sum of the entries of MATH lying in a fixed column is always REF. As we have already noted, MATH holds for every MATH. For the matrix MATH this means that the coordinatewise product of any two columns of MATH is zero. Consequently in every row of MATH there is at most one nonzero entry. Since MATH sends rank-MATH projections to rank-MATH projections, we see that this possibly nonzero entry is necessarily REF. So, every row contains at most one REF and all the other entries in that row are REF's. Since the sum of the elements in every column is REF, we have that in every column there is exactly one REF and all the other entries are REF's in that column. These now easily imply that if MATH is such that its coordinates are all REF's with the exception of one which is REF, then MATH is of the same kind. What concerns MATH, this means that MATH sends every MATH to a rank-REF projection. So, we obtain that MATH preserves the rank-REF projections and the orthogonality between them. Now, by REF we conclude the proof. |
math/0011029 | First suppose that MATH, MATH. If MATH, then we can apply the method followed in the proof of REF concerning the infinite dimensional case. If MATH, then consider the transformation MATH on MATH. We learn from CITE that if MATH, then there exists a unitary operator MATH such that MATH and MATH. It follows from the preserving property of MATH that for any MATH we have MATH for some unitary operator MATH on MATH. This gives us that MATH . In that way we can reduce the problem to the previous case. So, there is an either unitary or antiunitary operator MATH on MATH such that MATH . It follows that MATH, and hence we have the result for the considered case. Next suppose that MATH, MATH. If MATH, then once again we can apply the method followed in the proof of REF . If MATH, then using the 'dual method' that we have applied right above we can reduce the problem to the previous case. If MATH, consider a fixed rank-MATH projection MATH. Clearly, if MATH is any rank-MATH projection orthogonal to MATH, then the rank-MATH projection MATH is orthogonal to MATH. Therefore, MATH induces a transformation MATH between MATH-dimensional spaces (namely, between the orthogonal complement of the range of MATH and that of the range of MATH) which preserves the principal angles between the rank-MATH projections. Our 'dual method' and the result concerning REF-dimensional subspaces lead us to the conslusion that the linear extension of MATH maps rank-REF projections to rank-REF projections and preserves the orthogonality between them. This implies that the same holds true for our original transformation MATH. Just as before, using REF we can conclude the proof. In the remaining case MATH we apply the 'dual method' once again. |
math/0011029 | Let MATH be any projection on MATH commuting with MATH. By REF , it is easy to see that we can choose a monotone decreasing net MATH of projections with finite corank such that MATH converges weakly to MATH and MATH commutes with MATH for every MATH. Since MATH commutes with MATH for every MATH, we obtain that MATH commutes with MATH. Interchanging the role of MATH and MATH, we obtain that any projection commutes with MATH if and only if it commutes with MATH. Let MATH be any unit vector from the range of MATH. Consider MATH. Since MATH commutes with MATH, it must commute with MATH as well. By REF we obtain that MATH belongs either to the range of MATH or to its orthogonal complement. It follows that either MATH, or MATH. Since the set of all unit vectors in the range of MATH is connected and the distance function is continuous, we get that either every unit vector in MATH belongs to MATH or every unit vector in MATH belongs to MATH. Interchanging the role of MATH and MATH, we find that either MATH or MATH. This gives us that either MATH or MATH. |
math/0011029 | We first prove that MATH is injective. If MATH and MATH, then by the preserving property of MATH we have MATH . Putting MATH, we see that MATH is unitarily equivalent to MATH. We distiguish two cases. First, let MATH be of infinite corank. By REF, we deduce that for every MATH we have MATH if and only if MATH. This gives us that MATH. As the second possibility, let MATH be of finite corank. Then MATH can be written in the form MATH and MATH, where, by the equivalence of MATH, the projections MATH and MATH have finite and equal rank. Let MATH be any finite rank projection on MATH. It follows from MATH that there is a unitary operator MATH on MATH such that MATH . This implies that MATH . Taking traces, by the equality of the rank of MATH and MATH, we obtain that MATH . Since this holds for every finite rank projection MATH on MATH, it follows that MATH and hence we have MATH. This proves the injectivity of MATH. Let MATH be of infinite corank. Then there is a projection MATH such that MATH. By the preserving property of MATH, this implies that MATH which means that MATH is of infinite corank. One can similarly prove that if MATH is of infinite corank, then the same must hold for MATH. This yields that MATH is of finite corank if and only if so is MATH. Denote by MATH the set of all finite rank projections on MATH. It follows that the transformation MATH defined by MATH is well-defined and bijective. Since MATH is unitarily equivalent to MATH for every MATH (this is because MATH), it follows that MATH is rank preserving. We next show that MATH . This can be done following the argument leading to REF. In fact, by the preserving property of MATH there is a unitary operator MATH on MATH such that MATH . This gives us that MATH . Taking traces on both sides and referring to the rank preserving property of MATH, we obtain REF. According to REF , let MATH denote the unique real-linear extension of MATH onto MATH. We know that MATH is injective. Since MATH is in the range of MATH, we obtain that MATH is surjective as well. It is easy to see that REF can be applied and we infer that there exists an either unitary or antiunitary operator MATH on MATH such that MATH . Therefore, we have MATH for every projection MATH with finite corank. It remains to prove that the same holds true for every MATH with infinite corank as well. This could be quite easy to show if we know that MATH preserves the order between the elements of MATH. But this property is far away from being easy to verify. So we choose a different approach to attack the problem. Let MATH be a projection of infinite corank. By the preserving property of MATH we see that for every MATH the operator MATH is a projection if and if MATH is a projection. By REF , this means that MATH commutes with MATH if and only if MATH commutes with MATH. Therefore, for any MATH of finite corank, we obtain that MATH commutes with MATH (this is equivalent to that MATH commutes with MATH) if and only if MATH commutes with MATH. By REF we have two possibilities, namely, either MATH or MATH. Suppose that MATH. Consider a complete orthonormal basis MATH in the range of MATH and, similarly, choose a complete orthonormal basis MATH in the range of MATH. Pick nonzero scalars MATH, MATH with the property that MATH and MATH. Define MATH . Clearly, MATH is of finite corank (in fact, its corank is REF). Since MATH is unitarily equivalent to MATH, it follows that the spectrum of MATH is equal to the spectrum of MATH. This gives us that the spectrum of MATH is equal to the spectrum of MATH. By the construction of MATH this means that MATH which is an obvious contradiction. Consequently, we have MATH, that is, MATH. Thus, we have proved that this latter equality holds for every MATH and the proof is complete. |
math/0011030 | We first show that MATH is injective. Indeed, if MATH are such that MATH, then from REF we obtain that MATH for every MATH. This implies that MATH . It is an easy fact that if MATH, then MATH . Since in the one-dimensional case our statement is trivial, in what follows we assume that MATH. From REF we infer that MATH for every MATH. It follows that MATH which proves the injectivity of MATH. Observe that MATH preserves the rank-one operators. In fact, this follows from the following characterization of rank-one elements of MATH. The operator MATH has rank one if and only if MATH, MATH and MATH for every MATH (MATH denotes cardinality). Observe that if MATH, then MATH. Our next step is to show that MATH is linear. The easiest way to verify this is the use of the trace functional as follows. Since the trace of a rank-one operator MATH is MATH, we obtain from REF that MATH for every MATH and rank-one operator MATH. If MATH are arbitrary and MATH is any rank-one operator, then we compute MATH . By the arbitrariness of MATH we obtain that MATH is additive. One can check that MATH is homogeneous in a similar way. So, MATH is a linear bijection of MATH preserving the rank-one operators. The form of such transformations is well-known. It follows from the argument in CITE leading to CITE that we have two possibilities: CASE: there exist bijective linear operators MATH and MATH such that MATH CASE: there exist bijective linear operators MATH and MATH such that MATH . Suppose first that we have REF . According to REF we obtain MATH . Consequently, there is a scalar MATH such that MATH . By the closed graph theorem we readily obtain that the bijective linear operators MATH are bounded and hence we infer that MATH. Thus, MATH and this implies that MATH for every finite rank operator MATH. Using REF of MATH we have MATH, that is, either MATH or MATH. Suppose that MATH. Let MATH be arbitrary. Pick any rank-one operator MATH. From REF it follows that MATH . By the arbitrariness of MATH we obtain that MATH for every MATH. Assume now that we have REF . Similarly to REF one can prove that MATH is a bounded invertible linear operator and MATH where MATH denotes the natural embedding of MATH into MATH. Since MATH, we obtain that in this case MATH is of the form MATH for every finite rank operator MATH. Just as above, we infer that MATH and then obtain the form of MATH on the whole MATH. To see that in the infinite dimensional case this second possibility REF cannot occur, we refer to CITE stating that on an infinite dimensional NAME space MATH every point spectrum preserving surjective linear map is an automorphism (not an antiautomorphism). Since, as it can be seen, MATH or MATH satisfies these conditions, that result applies. To verify that the finite dimensional case is different, that is, REF can really occur, we remark that in that case the injectivity, surjectivity, bijectivity of an operator are all equivalent and that it is true for any elements MATH of any NAME algebra that MATH. Consequently, for every MATH we have MATH. The proof is complete. |
math/0011030 | One can argue in a very similar way as in our first result. This can be done since, by the NAME alternative, for any finite rank operator (in fact, even for any compact operator) MATH we have MATH . To exculde the appearence of the second possibility REF in the proof of REF choose a nonsurjective isometry MATH on MATH. Let MATH (the NAME space adjoint of MATH) and set MATH. Then we see that MATH is invertible while MATH is not surjective. So, MATH. |
math/0011030 | The linearity of MATH can be proved in the very similar way as above. Since the norm and the spectral radius of a selfadjoint operator in MATH are equal, it follows from MATH that MATH. Consequently, MATH is a surjective linear isometry of MATH. The form of such transformations is well-known. Namely, to every surjective linear isometry MATH there exist unitaries MATH such that MATH is either of the form MATH or of the form MATH . If MATH is of infinite dimension, then the appearence of this second possibility can be excluded just as in the last part of the proof of REF . |
math/0011030 | We have MATH. The spectrum of an element of MATH equals its range. Therefore, MATH and considering the transformation MATH, we can and do assume that our function MATH satisfies MATH. We obtain from REF that MATH for every MATH. So, MATH preserves the range of functions. This implies that MATH maps real functions to real functions and it sends nonnegative functions to nonnegative functions. We prove that MATH is injective. This will follow from the following characterization of the equality between functions. Let MATH. Then MATH if and only if MATH for every nonnegative function MATH. Indeed, suppose that MATH for some MATH. We can assume that MATH. Let MATH be an open disk centered at MATH which does not contain MATH and let MATH be a neighbourhood of MATH such that MATH for every MATH. Let MATH be a continuous function such that MATH and MATH. Such a function exists by NAME 's lemma. Then we obtain MATH but MATH. Therefore, MATH. So, we have the injectivity of MATH. Therefore, MATH and MATH are bijective functions having the same properties concerning the spectrum. Our next claim is that MATH preserves the usual ordering between real functions. This will follow from the following characterization of that ordering. If MATH are real functions, then MATH if and only if CASE: MATH for every MATH and MATH and CASE: MATH for every MATH and MATH. To see this, suppose that MATH for some MATH. Clearly, there exists a positive number MATH such that either MATH does not belong to the MATH-neighbourhood of MATH or MATH does not belong to the MATH-neighbourhood of MATH. Suppose that we have the first possibility. Choose a continuous function MATH for which MATH, and the support of MATH is a subset of a neighbourhood of MATH in which MATH takes its values in the MATH-neighbourhood of MATH. Then we find that MATH is a subset of the MATH-neighbourhood of MATH but MATH is not a subset of that set. It is easy to see that there is a constant MATH such that MATH but MATH. So, the above characterization really holds and then we get that MATH if and only if MATH. Observe that by REF we have MATH if and only if MATH. Now, if MATH and MATH, then we find that MATH . To any point MATH there exists a continuous function MATH such that MATH and MATH if MATH. In fact, by the first countability of MATH there is a sequence MATH of neighbourhoods of MATH which forms a base of neighbourhoods of that point. For every MATH there is a continuous function MATH whose support is in MATH and MATH. Now, set MATH. This function fulfils our requirements. Denote by MATH the set of all such function MATH. We assert that if MATH, then MATH belongs to MATH for some MATH. In fact, since MATH preserves the range of functions, it follows that MATH maps into MATH and it takes the value REF. Suppose that MATH equals REF at two different points, say MATH. It follows that there are functions MATH and MATH such that MATH and MATH. Let MATH and MATH. Then we have MATH and by the previous sections of the proof we infer that MATH. Since MATH takes the value REF at at least two points, the same must be true for MATH which is a contradiction. This means that MATH for some MATH. We next show that the point MATH does not depend on the particular choice of MATH. Indeed, let MATH. Then MATH and this implies that MATH for some MATH. This proves that MATH and MATH take their maximum at the same point. So, we have a function MATH such that MATH implies MATH. Since MATH and MATH share the same properties, we obtain that MATH is a bijection. We assert that MATH is homogeneous. Let MATH and MATH. For any MATH we have MATH which implies that MATH. Let MATH, MATH and let MATH. There exists MATH such that MATH. Then we have MATH. This gives us that MATH . Since MATH has the same properties as MATH, it follows that MATH that is, we also have MATH. Therefore, we obtain MATH for every MATH and MATH. We show that MATH is a homeomorphism. We need only to show that MATH is continuous. Let MATH be a sequence in MATH converging to the point MATH. Suppose that MATH does not converge to MATH. Then there is a neighbourhood MATH of MATH such that MATH for infinitely many indices. Let MATH be a continuous function with support in MATH such that MATH. Let MATH be such that MATH. Then we have MATH for infinitely many MATH's and this contradicts MATH. So, MATH is a homeomorphism of MATH and we have MATH for every nonnegative MATH. Finally, for any MATH and MATH we compute MATH which gives us that MATH. This completes the proof. |
math/0011030 | Similarly as in the proof of REF one can verify that MATH is injective. Indeed, if MATH are such that MATH, then we have MATH for every MATH which implies that MATH. Observe that we have MATH which implies that MATH is a function of modulus REF. Considering the transformation MATH, we can and do assume that our function MATH satisfies MATH. We have MATH for every MATH. Therefore, MATH is self-bijection of set MATH of all real valued continuous functions on MATH which satisfies MATH . Since, as it turns out from the proof of the previous result, REF remains valid for the function algebra MATH as well, we obtain that there is a homeomorphism MATH such that MATH . If MATH, then we have MATH for every MATH which yields MATH. The proof is complete. |
math/0011031 | The idea of the proof is very simple. We first extend MATH from MATH to a linear transformation on the operator algebra MATH and then apply a result of NAME and NAME on linear preservers. Notice that our approach is completely different from the usual proofs of NAME 's theorem and, in particular, this is the case also with the proof presented in CITE. First suppose that MATH is a complex NAME space. We define MATH whenever MATH are rank-one idempotents and MATH are scalars. We assert that MATH is well-defined. To see this, we prove that MATH implies MATH . Indeed, from REF we obtain that MATH . By the linearity of the trace and the equality REF, it follows that MATH . Since MATH maps onto MATH, we can infer that MATH . So MATH is well-defined. Clearly, MATH is a linear transformation on MATH. Since every matrix MATH is a linear combination of rank-one idempotents, it follows that every finite-rank operator belongs to the linear span of MATH. This yields that MATH is defined on the whole MATH and maps onto MATH (in fact, one can prove that it is injective as well). So we have a surjective linear transformation MATH on MATH which preserves the rank-one idempotents. Now, we can apply a result of NAME and NAME describing the form of all such maps. In view of CITE, we distinguish two cases. First suppose that MATH is infinite-dimensional. Then for the form of MATH, and hence for the form of MATH, we have the following two possibilities: CASE: There exists a bijective bounded linear operator MATH such that MATH . CASE: There exists a bijective bounded linear operator MATH such that MATH . Recall that our map MATH is linear. This is the reason that the two remaining possibilities in CITE do not appear here. As for the finite dimensional case, the form of all surjective linear maps on MATH preserving rank-one idempotents is given in CITE. Since every linear transformation on MATH is continuous and the only continuous ring automorphisms of MATH are the identity and the conjugation, we obtain from that result that in this case our map MATH is either of the form MATH or of the form MATH (MATH stands for the transpose) with some nonsingular matrix MATH. But under the identification of matrices and operators, we obtain the forms appearing in REF once again. If MATH is a real NAME space, then one can argue in the same way but since in that case CITE holds without any assumption on the dimension of MATH, there is no need to refer to CITE. |
math/0011031 | We first show that the range of MATH linearly generates MATH. As usual, denote by MATH the matrix whose MATH entry is REF and its all other entries are REF. Define MATH . Suppose that MATH for some scalars MATH. Fix indices MATH. We have MATH . Taking trace, we obtain MATH . By the preserving property of MATH, it follows that MATH . Since MATH, from this equality we easily deduce that MATH. As MATH were arbitrary, it follows that the matrices MATH, MATH form a linearly independent set in MATH. This implies that the range of MATH linearly generates MATH. Similarly to the proof of REF we define a transformation MATH by MATH where the MATH's are rank-one idempotents and the MATH's are scalars. First, MATH is well-defined. Indeed, let MATH be rank-one idempotents and MATH be such that MATH. Then we have MATH . By the preserving property of MATH we infer that MATH . This implies that MATH and, as the matrices MATH linearly generate MATH, we can conclude that MATH. Therefore, MATH is well-defined. Since the rank-one idempotents linearly generate MATH, we obtain that MATH is a linear transformation from MATH into itself which preseves the rank-one idempotents. The form of such transformations is described in CITE and this gives us the form of MATH. |
math/0011031 | The proof is quite similar to the proof of our previous result. Define MATH . Just as in the proof of REF , one can show that the matrices MATH, MATH are linearly independent in MATH. This gives us that the range of MATH linearly generates MATH. Next we define MATH for every finite system MATH of scalars and symmetric rank-one idempotents MATH. One can prove that MATH is well-defined in a way very similar to the corresponding part of the proof of REF . Since the symmetric rank-one idempotents linearly generate MATH, MATH is a linear transformation from MATH into itself which preserves the rank-one idempotents in MATH. The form of such transformations on MATH is described in CITE and this gives us the form of MATH. |
math/0011036 | The action property of MATH follows from REF . Let MATH. Then it is also a MATH bimodule endomorphism of MATH, hence MATH, MATH, for some MATH. Hence MATH for all MATH, therefore MATH. Similarly, one can prove that MATH. Since MATH and MATH are sections of MATH, they are injective and this proves the Lemma. |
math/0011036 | The idea of the proof is to show that MATH factors through a strongly monoidal forgetful functor MATH. Here the ring MATH is the additive group MATH together with the multiplication MATH and unit MATH. Every object MATH in MATH carries the MATH-bimodule structure defined by the left and right actions MATH NAME monoidality of the functor MATH follows from exactness of REF . |
math/0011036 | Since MATH is a finite progenerator, MATH. NAME of MATH determines the comonoid MATH where MATH . Using also the split monoidality structure we can define MATH which form a comonoid MATH in MATH and can be verified to obey the properties REF . The maximality REF holds automatically by the very definition of MATH as MATH. Thus MATH is a bialgebroid with separable base MATH endowed with multiplication as in REF . |
math/0011036 | Need to show that MATH is an isomorphism for objects MATH, MATH of MATH. Choosing a direct sum decomposition MATH we can explicitely write down the inverse as MATH for MATH. This proves strong monoidality of MATH. From the construction of MATH it is clear that MATH . , as categories. Use this equivalence to define a monoidal structure on MATH . Now apply REF to conclude that MATH is a bialgebroid over MATH and the equivalence is that of monoidal categories. The monoidal factorization through MATH holds by the very definition of the monoidal structure of MATH . |
math/0011038 | Consider MATH with coordinates MATH and let MATH be given by MATH; consider in MATH the metric MATH, with MATH, and MATH given by: MATH . The choice of the sign MATH in the above expressions is made according to the desired causal character of MATH. It is easily checked that the NAME symbols of the NAME - NAME connection of MATH in the canonical basis vanish along MATH; this implies that MATH is a geodesic and that MATH gives a parallel trivialization of the normal bundle MATH. |
math/0011038 | It follows easily from REF that every nondegenerate symplectic differential system is isomorphic to one whose component MATH is constant. We may thus assume without loss of generality that MATH is constant (and nondegenerate). To conclude the proof we must exhibit a smooth curve MATH in the NAME group MATH and a smooth curve MATH of symmetric MATH matrices such that the righthand side of REF vanishes. It suffices to take MATH and MATH to be the solution of MATH with MATH. In order to see that MATH takes values in MATH simply observe that MATH is in the NAME algebra MATH of MATH given by: MATH . |
math/0011038 | REF is obtained by straightforward verification. For REF , we describe how to construct the symplectic differential system MATH from the abstract symplectic system MATH. Choose a smooth curve MATH where each MATH is a symplectomorphism from MATH to MATH (endowed with the canonical symplectic form) such that MATH for all MATH. Define MATH to be the unique symplectic differential system whose fundamental matrix MATH is given by MATH; more explicitly, take MATH. It is easy to check that MATH is an isomorphism from MATH to MATH. |
math/0011038 | Let MATH be such that the open ball MATH of center MATH and radius MATH is contained in MATH and choose a smooth curve MATH such that MATH and MATH. Set MATH and choose MATH small enough such that MATH and MATH . Now, let MATH be a smooth non decreasing reparameterization of MATH such that MATH and MATH. Choose smooth functions MATH with MATH and such that the support of MATH is contained in MATH and the support of MATH is contained in MATH. Finally set: MATH and define MATH. To check that such MATH works observe that MATH is less than or equal to the left hand side of REF. |
math/0011038 | Simply apply REF to the following objects: CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH is chosen small enough so that MATH, where MATH is such that the open ball MATH is contained in MATH. Finally, define MATH for MATH. |
math/0011038 | Let MATH be a Lagrangian complementary to both MATH and MATH; it's easy to see that MATH can be chosen such that MATH equals any prescribed nondegenerate bilinear form on MATH. In particular, we may assume that MATH and MATH have the same index. Let MATH be such that MATH is contained in the domain of the chart MATH and define MATH by MATH. The conclusion follows by an application of REF to MATH, keeping in mind that if MATH then: CASE: MATH; CASE: MATH; CASE: MATH is just a push-forward of MATH by an isomorphism between MATH and MATH. |
math/0011038 | By REF , it suffices to find an abstract symplectic system MATH of index MATH with MATH whose set of conjugate instants is MATH. Consider the space MATH endowed with the canonical symplectic form and set MATH; given MATH, we'll construct a smooth curve MATH such that MATH is nondegenerate for all MATH and MATH iff MATH. The desired curve MATH will then be obtained by applying REF . The curve MATH will take values in the domain of the chart MATH where MATH; we define MATH, where MATH is defined by: MATH and MATH is a smooth map such that MATH. The condition MATH implies that MATH is always nondegenerate and therefore also MATH is nondegenerate; moreover, MATH iff MATH. The existence of the required function MATH follows by taking MATH in REF below. |
math/0011038 | Write MATH as a disjoint union of open intervals MATH. For each MATH let MATH be a smooth map such that: CASE: MATH is zero outside MATH; CASE: MATH is positive on MATH; CASE: MATH for MATH, where MATH denotes the MATH-th derivative of MATH. To conclude the proof set MATH. |
math/0011048 | The original deformation REF satisfies the NAME REF . In each order MATH REF for the deformation REF has a solution MATH which is a homogeneous polynomial in MATH of degree MATH. Replacing in MATH each monomial MATH by MATH, one, obviously, gets a solution MATH. |
math/0011048 | This is an immediate consequence of the fact that all the NAME products of the two first non-trivial cohomology classes vanish: MATH and so do the obstructions. |
math/0011048 | This formula readily follows from REF . |
math/0011048 | By definition, MATH, a straightforward computation then yields REF . |
math/0011048 | Let us use the existence of modules MATH. Each term MATH in REF polynomially depends on MATH. The operator MATH defines a MATH-action and, so, satisfies the homomorphism REF . A term of degree of schift MATH in REF is again a polynomial in MATH, more precisely, a sum of the terms MATH with operator coefficients. But, all the monomials REF with MATH given by REF are, obviously, linearly independent and, so, REF has to be satisfied independently for the operator coefficients of all monomials REF . These conditions are therefore independent on MATH. |
math/0011048 | REF coincide with REF from REF that are sufficient for integrability. |
math/0011048 | The cocycles REF are precisely of this form. The cup product REF of two such linear maps is a bilinear map which is also homogeneous in MATH and MATH. Finally, the coboundary operator MATH preserves the homogeneity in the same way. |
math/0011050 | NAME that a l.c.K. structure MATH is equivalent to another MATH. By the definition, MATH. As MATH, MATH, we have MATH on the intersection MATH. Let MATH on MATH. Similarly for MATH as above. Thus there is a global function MATH on MATH such that MATH. By the definition, the l.c.K. metrics MATH, MATH satisy MATH. Hence MATH for each MATH. Thus the equivalence class of MATH determines the conformal class MATH. Conversely, if MATH for a l.c.K. metric MATH where MATH is some positive function, note that the fundamental two-form MATH satisfies MATH where MATH. As MATH, MATH is also a l.c.K. manifold. Let MATH and MATH as above. Since MATH, there is some constant MATH such that MATH on MATH. In particular, MATH. By the defintion, MATH. So the induced l.c.K. structures MATH, MATH are equivalent. |
math/0011050 | A l.c.K. MATH action lifts to a MATH or a MATH-action by homotheties on MATH. Let us show that, owing to the purely conformal character of MATH in MATH, it is impossible to lift to a MATH-action. Let MATH be such that, for a given MATH, MATH (such a MATH must be small enough, close to MATH.) Then MATH with MATH on MATH. This MATH lifts to a MATH acting on MATH and we have MATH on MATH. The NAME metric MATH is related to the lifted globally conformally NAME metric MATH by MATH, hence REF implies (recall that MATH is constant on MATH): MATH . Now, if MATH is an isometry, then MATH, contradiction. Thus, if MATH is the homomorphism defined on the set of all MATH-homotheties of MATH with values in MATH by the formula MATH, we derive MATH. Hence, the lift of MATH cannot be MATH, otherwise its image by MATH would be a compact non-trivial subgroup of MATH, contradiction. |
math/0011050 | Suppose, ad absurdum, that MATH admits a purely conformal MATH action. According to REF (and to its proof), this action lifts to a MATH-action by MATH-homotheties and MATH is injective. But MATH can be understood both as being defined on MATH and on MATH, because MATH and MATH are both contained in the group of MATH-homotheties of MATH. Further, for a fixed MATH, consider the evaluation map MATH, MATH. We denote also by MATH the restriction of this map to the considered MATH, so obtain the commutative diagram: where the vertical arrow is injective. Taking into account the previous observation, at the homotopy level we then have the commutative diagram: Hence, MATH is infinite in MATH. But, MATH being semi-simple, it has no torus as a direct summand and has finite MATH. Thus, chasing on the upper side of the diagram, the image of MATH in MATH through MATH is finite. This contradiction completes the proof. |
math/0011050 | As the deck group of the covering MATH preserves MATH and commutes with MATH, MATH projects to a one-parameter subgroup on MATH. This one is global, by compactness of MATH and lifts to a global one-parameter subgroup on MATH which clearly coincides with MATH. |
math/0011050 | Note that MATH . As MATH, MATH. On the other hand, from MATH we obtain for MATH: MATH . This proves that MATH is onto and MATH is a codimension-MATH, smooth submanifold of MATH. |
math/0011050 | Let MATH be a component of MATH and let MATH be the set MATH. As MATH acts freely and MATH, we have MATH for MATH. Thus MATH is an open subset of MATH. We now prove that it is also closed. Let MATH be the closure of MATH in MATH. We choose a limit point MATH. Then MATH. We denote MATH, MATH, so MATH. Since MATH is regular (that is, closed with respect to the relative topology induced from MATH), so is its component MATH. Hence MATH. Therefore MATH, proving that MATH is closed in MATH. In conclusion, MATH. Now, if MATH is another component of MATH, the same argument shows MATH. Thus, MATH. This implies MATH, in other words MATH is connected. |
math/0011050 | Let MATH be the inclusion and MATH be the canonical projection. For MATH we put MATH where MATH such that MATH. As MATH is a fiber bundle with MATH, MATH is well-defined. To prove that MATH is a contact form, we first note that by the definition MATH . Then MATH . Now recall that MATH, thus MATH and, moreover, MATH . Hence, MATH on MATH showing that MATH is a contact form. Let us now show that MATH is the characteristic field of MATH. For any distribution MATH on MATH, denote MATH the orthogonal distribution with respect to the metric MATH. Then MATH is an isomorphism and induces an isomorphism MATH. So, MATH, MATH for MATH. We now show that MATH for any MATH. We have MATH. Let MATH for some MATH. Then, using REF on MATH (that is, for MATH): MATH . But MATH and MATH because MATH. So: MATH . Since MATH we have MATH . If MATH is the exponential map with respect to MATH, then the subset MATH constitutes a coordinate neighborhood of MATH at MATH. In particular, the above MATH is a linear combination of coordinate vector fields around MATH without containing MATH. On the other hand, noting MATH on MATH from REF, it follows MATH. We finally deduce MATH on MATH and the proof of the lemma is complete. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.