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math/9908094 | CASE: Observe that MATH is dense in MATH, as MATH raises MATH. By induction, it follows that MATH is dense in MATH. Because MATH, we must have MATH, whence MATH. REF follows from the fact that MATH (respectively, MATH) if MATH is the source of an edge with target MATH and type MATH (respectively, MATH, MATH). CASE: By REF , the product maps MATH are birational for MATH. It follows that the morphism MATH has image MATH; moreover, its degree is the product of the degrees of the MATH that is, MATH. Let MATH. We show that MATH. Otherwise, there exists MATH such that MATH. Thus, MATH, and MATH. Let MATH, then MATH raises MATH to MATH. This contradicts the assumption that MATH. A similar argument shows that MATH. CASE: If MATH, then there exists a point MATH, a simple root MATH and a surjective group homomorphism MATH where MATH is the normalizer of a torus in MATH. Since MATH consists of semisimple elements, it is a quotient of MATH. By assumption, the latter is isomorphic to a subgroup of MATH. Thus, MATH is abelian, a contradiction. |
math/9908094 | Recall that MATH (closure in MATH) is a product of minimal parabolic subgroups. Thus, MATH. The converse has just been proved. |
math/9908094 | Since MATH, we have MATH if and only if MATH. In that case, MATH raises MATH if and only if MATH, that is, MATH. Then MATH and the map MATH is the pull-back of MATH under the map MATH. This yields REF . But if MATH, then MATH has length MATH, so that MATH . Thus, MATH raises MATH if and only if MATH raises MATH. Then, as MATH, we can join MATH to MATH by two paths: one beginning with MATH edges of type MATH followed by an edge from MATH to MATH, and another one beginning with an edge from MATH to MATH followed by MATH edges of type MATH. Using REF , this yields REF . |
math/9908094 | By induction on MATH, the case where MATH being evident. If there exists MATH such that MATH, then MATH raises MATH, and MATH, MATH are in MATH. Now the induction assumption for MATH concludes the proof in this case. Otherwise, we can find distinct MATH such that MATH. Then MATH and MATH raise MATH to subvarieties of MATH. Let MATH be the common codimension of MATH and MATH in MATH, then we have MATH . Choose MATH, then MATH contains MATH and, similarly, MATH, as neighbors. Moreover, MATH contains MATH and MATH, whereas MATH contains MATH and MATH. Now we conclude by the induction assumption for MATH and MATH. |
math/9908094 | Write MATH and MATH. Let MATH . We claim that MATH equals MATH. Otherwise, MATH and MATH. By the strong exchange condition (REF applied to MATH), one of the following cases occurs: CASE: MATH where MATH. Comparing both expressions for MATH, we obtain MATH. Thus, there exists MATH such that MATH. But MATH, for MATH. It follows that MATH and MATH are in MATH. Thus, MATH. Since MATH, we have MATH, a contradiction. CASE: MATH where MATH is obtained from MATH by deleting a simple reflection. Then the equality MATH leads to a braid relation of length at most MATH, a contradiction. CASE: MATH where MATH. Then MATH. But MATH, for MATH; a contradiction. By the claim and REF , we have MATH and MATH. Write MATH where MATH and MATH; then MATH. Since MATH and MATH, it follows that MATH by REF . Thus, MATH and MATH. Since MATH and MATH, we must have MATH, so that MATH. |
math/9908094 | Write MATH and MATH as in the definition of neighbors. Then MATH and MATH are neighbors in MATH. Moreover, MATH, whence MATH. Thus, we may assume that MATH. Let MATH and MATH, then we obtain similarly: MATH and MATH. If MATH, write MATH where MATH and MATH. Then MATH and MATH have rank MATH and are raised to MATH by MATH. Thus, MATH and, by induction on MATH, we obtain MATH. This subvariety is stable under MATH. Applying REF , we may assume that MATH (that is, MATH), MATH and MATH where the center of MATH is trivial and MATH has finite index in its normalizer. Moreover, we have MATH, for MATH and MATH do not stabilize MATH. We claim that any MATH can be written as MATH where MATH satisfies MATH. For this, we argue by induction on the codimension of MATH in MATH. We may assume that MATH raises MATH. By the induction assumption, we have MATH . In the latter case, let MATH (MATH terms). Since MATH and MATH, it follows that MATH. In the former case, MATH is stable under MATH and hence equal to MATH; in particular, MATH has codimension MATH in MATH. Now MATH (MATH terms), so that we are in the previous case. By the claim, all MATH-orbit closures in MATH have the same rank, and MATH is the unique closed MATH-orbit. Let MATH; we may assume that MATH. Since the MATH-orbit in MATH corresponding to the MATH-orbit MATH in MATH is closed, the connected isotropy group MATH is a NAME subgroup of MATH. It follows that MATH. On the other hand, MATH by assumption. Thus, MATH. If MATH then MATH is a parabolic subgroup of MATH (in fact, a NAME subgroup as MATH.) Moreover, MATH is the MATH-fixed point in MATH. But then MATH consists of a unique element (of maximal length in MATH), a contradiction. If MATH then MATH as well. Using the classification of homogeneous spaces of rank MATH under semi-simple groups of rank MATH (see for example, REF), this forces MATH and MATH embedded diagonally in MATH. As a consequence, the simple roots MATH and MATH are orthogonal, and MATH is generated by MATH. If MATH then MATH, that is, MATH is unipotent. Since MATH is spherical, MATH is a maximal unipotent subgroup of MATH. This contradicts the assumption that MATH has finite index in its normalizer. |
math/9908094 | CASE: Let MATH (respectively, MATH) be the source (respectively, target) of MATH, and let MATH be another oriented path from MATH to MATH. By REF , it suffices to show that MATH. Joining MATH to MATH by an oriented path, we reduce to the case where MATH; then MATH and MATH are in MATH. By REF , we may assume moreover that MATH and MATH are neighbors. Using REF , we reduce to the case where the center of MATH is trivial, MATH, MATH where MATH has finite index in its normalizer, MATH and MATH for some MATH. Since MATH is simply-laced, we have either MATH and MATH, or MATH and MATH. In particular, MATH. If MATH then MATH by REF . If MATH then MATH. Using REF , we may assume moreover that MATH is not contained in any NAME subgroup. Then we see by inspection that MATH is conjugate to MATH or to MATH. In the latter case, here is MATH: Thus, MATH. In the former case, we have MATH, since MATH is as follows: CASE: Let MATH be an oriented path joining MATH to MATH. We may assume that MATH contains double edges. Consider the lowest maximal subpath MATH of MATH that consists of double edges only; we may assume that the endpoint of MATH is not MATH. Let MATH be the source of the top edge of MATH, and let MATH (respectively, MATH) be the label of that edge (respectively, of the next edge of MATH, a simple edge by assumption.) We claim that there exists an oriented path MATH joining MATH to MATH and beginning with a simple edge; then REF will follow by induction on MATH. To check the claim, it suffices to join MATH to MATH by an oriented path MATH' beginning with a simple edge. As above, we reduce to the case where MATH equals MATH or MATH, and MATH is not contained in a NAME subgroup of MATH; Moreover, MATH has finite index in its normalizer. Using the fact that MATH contains a double edge followed by a simple edge, one checks that MATH is a product of subgroups of MATH if MATH; and if MATH, then MATH is conjugate to the subgroup of REF , or to its transpose. The path MATH exists in all these cases. |
math/9908094 | CASE: Note that MATH is MATH-stable if and only if MATH is an eigenvector of MATH, that is, MATH extends to a character of that group. This amounts to: MATH. Let MATH be the MATH-stable divisor in MATH corresponding to the MATH-stable divisor MATH. Then MATH is the zero scheme of a section of the homogeneous line bundle on MATH associated with the character MATH of MATH. Let MATH be the natural map, then MATH equals the degree of the restriction MATH. The latter degree is the intersection number of MATH with a fiber of MATH, that is, MATH. CASE: For the first assertion, it suffices to show existence of MATH containing MATH and not containing MATH; but this follows from REF . For the second assertion, note that MATH stabilizes MATH if it stabilizes MATH. Thus, MATH if MATH. On the other hand, if MATH then MATH is birational. Restricting to MATH, it follows that MATH is birational if generically finite. |
math/9908094 | CASE: The product map MATH is an isomorphism; moreover, MATH. Therefore, the product map MATH is an isomorphism. The assertion follows by intersecting with MATH. REF The union of all MATH-orbits in MATH that contain MATH in their closure is a MATH-stable open subset of MATH. Thus, we may assume that MATH is the unique closed MATH-orbit in MATH. Let MATH be the boundary divisors, then MATH. Moreover, MATH is isomorphic to affine space MATH with coordinate functions MATH, equations of MATH. The compositions MATH are equations of MATH; they generate the ideal of MATH in MATH. The map MATH identifies to MATH. The intersections of MATH-orbit closures with MATH are the pull-backs of coordinate subspaces of MATH. By REF , MATH is irreducible. We check that MATH. For this, note that the product map MATH is an isomorphism. Moreover, since MATH meets MATH properly, with MATH as an irreducible component, it follows that MATH is equidimensional, with MATH as an irreducible component. The latter is isomorphic to MATH. Thus, the MATH-stable set MATH is finite, so that it consists of MATH-fixed points. Since MATH is the unique MATH-fixed point in MATH, our assertion follows. The map MATH identifies with MATH. We just saw that the set-theoretical fiber of MATH is MATH. Since MATH is the unique closed MATH-orbit in MATH, all fibers of MATH are finite. Thus, MATH contains a dense MATH-orbit. Since MATH is affine and contains a MATH-fixed point MATH, it follows that MATH is finite and that the pull-back of any MATH-orbit in MATH is a unique MATH-orbit. This implies REF . Finally, we check that the degree of MATH equals MATH, that is, the degree of the natural map MATH. For this, note that the map MATH is an open immersion. Thus, MATH is the degree of the product map MATH, or, equivalently, of its restriction MATH . The latter map fits into a commutative diagram MATH where the bottom horizontal map is MATH ; indeed, MATH by REF . Moreover, the fibers of the right (respectively, left) vertical map are isomorphic to MATH (respectively, to MATH.) Thus, the diagram is cartesian, and the degree of MATH equals the degree of MATH. |
math/9908094 | We have MATH . Moreover, since MATH, the product map MATH is an isomorphism. Now the equality MATH follows from the definitions. Together with REF , it implies that MATH is finite surjective of degree MATH. |
math/9908094 | By REF , MATH is the union of the MATH. Choose MATH and MATH, then MATH is an irreducible component of MATH. The latter is isomorphic to MATH, and MATH is a unique MATH-orbit, by REF . It follows that MATH is irreducible, so that MATH is uniquely determined by MATH. Equivalently, the MATH are pairwise disjoint. Let MATH be a closed MATH-orbit in MATH, then MATH where the former equality follows from CITE, and the latter from CITE Moreover, we have by REF : MATH and MATH. Thus, MATH . Using the fact that MATH is irreducible, together with associativity of intersection multiplicities (see CITEEF), we obtain MATH . |
math/9908094 | The first assertion follows from REF . By REF , MATH is finite surjective of degree MATH, hence an isomorphism because MATH is smooth. |
math/9908094 | We may assume that MATH. If MATH is a closed MATH-orbit, then the assertion follows from the description of MATH in terms of MATH, together with REF . Indeed, for any MATH such that MATH, we have MATH, where MATH is the base point of MATH. For arbitrary MATH, let MATH be a closed MATH-orbit in MATH. Let MATH, MATH be unions of irreducible components of MATH such that MATH. Then MATH and MATH are unions of irreducible components of MATH (for any irreducible component MATH of MATH meets MATH properly in MATH); Moreover, their intersection has codimension MATH in MATH and MATH, by the first step of the proof. It follows that MATH has codimension MATH in both MATH and MATH. |
math/9908094 | We begin with a reduction to the case where no simple normal subgroup of MATH fixes points of MATH. For this, we may assume that MATH is the direct product of a torus with a family of simple, simply connected subgroups; let MATH be one of them. If MATH is not of type MATH, MATH or MATH, then there exists a simple, simply connected group MATH together with a maximal proper parabolic subgroup MATH such that a NAME subgroup MATH has the same adjoint group as MATH (indeed, add an edge to the NAME diagram of MATH to obtain that of MATH.) Then MATH is the quotient of MATH by a finite central subgroup MATH. We may assume moreover that MATH maps injectively to MATH, that is, MATH is trivial. Then the first projection MATH is injective. We claim that the second projection MATH is injective as well. Indeed, as MATH is simply connected, its NAME group is trivial; as some open subset of MATH is the direct product of MATH with an affine space, the NAME group of MATH is trivial as well. But MATH is the total space of the line bundle over MATH associated with the character MATH of MATH, minus the zero section. Thus, MATH is the quotient of MATH by the class of that line bundle. Moreover, MATH is isomorphic to the character group of MATH, as MATH is simply connected. Therefore MATH generates the character group of MATH. Since MATH is abelian, the claim follows. By that claim, MATH is trivial; thus, MATH embeds into MATH as its derived subgroup. We shall treat MATH as an inclusion, which defines an action of MATH on MATH. On the other hand, MATH acts on MATH via MATH, and this action commutes with that of the remaining factors of MATH. Thus, MATH is a variety with an action of the product MATH with the remaining factors of MATH. This variety is spherical and fibers equivariantly over MATH, with fiber MATH. Thus, we may assume that the action of MATH on MATH extends to an action of MATH. Now the parabolically induced variety MATH contains MATH as a multiplicity-free subvariety REF but contains no fixed point of MATH. Iterating this argument removes the fixed points of all simple normal subgroups of MATH. We now reduce to the case where MATH is projective. For this, we use embedding theory of spherical homogeneous spaces, see CITE. We may assume that MATH contains a unique closed MATH-orbit MATH (for MATH is the union of MATH-stable open subsets, each of which contains a unique closed MATH-orbit.) Together with REF , the assumption that no simple factor of MATH fixes points of MATH amounts to: MATH contains no simple factor of MATH. Let MATH be the set of all colors MATH that contain MATH; then we can find an equivariant projective completion MATH of MATH such that MATH for any MATH-orbit closure MATH in MATH. By REF , it follows that MATH, and that no simple factor of MATH fixes points of MATH. We next reduce to an affine situation, in the following standard way. Choose an ample MATH-linearized line bundle MATH over MATH. Replacing MATH by a positive power, we may assume that MATH is very ample and that MATH is projectively normal in the corresponding projective embedding. Let MATH be the affine cone over MATH. This is a spherical variety under the group MATH, and the origin MATH is the unique fixed point of any simple normal subgroup of MATH, since MATH. Moreover, the affine cone MATH over MATH is stable under the NAME subgroup MATH of MATH, and is multiplicity-free. Thus, we may assume that MATH is affine with a fixed point MATH, and we have to show that MATH has rational singularities outside MATH. By CITE, the MATH-variety MATH is spherical, with rational singularities, so that we may assume that MATH. We argue then by induction on the codimension of MATH in MATH. Let MATH be the set of all MATH such that MATH. This is a proper standard parabolic subgroup of MATH, acting on MATH by automorphisms. Let MATH be a MATH-equivariant resolution of singularities. Denote by MATH (respectively, MATH) the algebra of regular functions on MATH (respectively, MATH). Then MATH is a finite MATH-module. Moreover, we have an exact sequence of MATH-modules MATH where the support of MATH is the non-normal locus MATH of MATH, by NAME 's main theorem. Note that MATH acts on MATH compatibly with its MATH-module structure. We first show that MATH is supported at MATH, that is, MATH is normal outside MATH. Let MATH be a simple root raising MATH and let MATH. Let MATH be the fiber bundle with fiber the MATH-variety MATH; let MATH be the natural morphism. Then the map MATH is injective, and makes MATH a finite MATH-module. Since MATH is multiplicity-free, MATH is birational and MATH is multiplicity-free as well. By the induction assumption, MATH is normal outside MATH. Therefore, the cokernel of MATH is supported at MATH, by NAME 's main theorem again. The MATH-equivariant resolution MATH induces a MATH-equivariant resolution MATH . Composing with MATH, we obtain a MATH-equivariant birational morphism MATH . As above, the map MATH is injective and its cokernel is supported at MATH. We shall treat MATH and MATH as inclusions. We have MATH . Moreover, MATH is the MATH-linearized sheaf on MATH associated with the (rational, infinite-dimensional) MATH-module MATH . We shall use the notation MATH . Then MATH and these MATH-modules coincide outside MATH. Consider the exact sequence of MATH-linearized sheaves on MATH: MATH . Since the restriction map MATH is surjective, the MATH-module MATH is the quotient of a rational MATH-module. Since MATH is a projective line, it follows that MATH. Thus, we have an exact sequence of MATH-modules MATH . It follows that MATH is supported at MATH. Now normality of MATH outside MATH is a consequence of the following Let MATH be a finite MATH-module with a compatible action of MATH, such that the MATH-module MATH is supported at MATH for any minimal parabolic subgroup MATH that raises MATH. Then MATH is supported at MATH. Otherwise, choose an irreducible component MATH of the support of MATH. Let MATH be the ideal of MATH in MATH. Define a submodule MATH of MATH by MATH . Observe that the support of MATH is MATH (indeed, the ideal MATH is a minimal prime of the support of MATH; thus, this ideal is an associated prime of MATH.) Note that MATH stabilizes MATH and acts on MATH. Moreover, MATH is a MATH-module supported at MATH (as a MATH-submodule of MATH.) We claim that MATH is MATH-stable. Otherwise, let MATH be a simple root raising MATH; then MATH raises MATH. Define as above the maps MATH . The MATH-module MATH with a compatible MATH-action induces a MATH-linearized sheaf MATH on MATH, and we have MATH as MATH-linearized sheaves on MATH. It follows that the MATH-module MATH is supported at MATH. On the other hand, we have MATH. Moreover, the map MATH is generically finite (as MATH raises MATH), and the support of MATH is MATH (as the support of MATH is MATH). Thus, the support of MATH is MATH, and the same holds for the support of MATH. This contradicts the assumption that MATH. The claim is proved. Let MATH be the NAME subgroup of MATH containing MATH, then MATH. Since MATH is MATH-stable, it is not fixed pointwise by MATH (here we use the assumption that no simple normal subgroup of MATH fixes points of MATH.) Since MATH is affine, MATH acts non trivially on MATH. Thus, we can find an eigenvector MATH of MATH in MATH of positive weight with respect to the coroot MATH. Then MATH, so that MATH acts nilpotently on MATH. But MATH does not act nilpotently on MATH, for the support of this module is MATH. Therefore we can choose a finite-dimensional MATH-submodule MATH of MATH such that MATH for any large integer MATH. For such MATH, all weights of MATH in MATH are positive. It follows that MATH. But MATH . Since MATH is supported at MATH, we have MATH for large MATH, a contradiction. Next we fix MATH and consider MATH, a MATH-linearized coherent sheaf on MATH. Since MATH is affine, this sheaf is associated with the MATH-module MATH endowed with a compatible action of MATH. We claim that the MATH-module MATH is supported at MATH. For this, note that the map MATH is a resolution of singularities. By the induction assumption, MATH has rational singularities outside MATH; thus, the MATH-modules MATH are supported at MATH, for all MATH. Moreover, MATH (recall that MATH denotes the MATH-equivariant extension of MATH.) And the fibers of MATH identify to closed subsets of projective line, as the map MATH is a closed immersion. Thus, MATH for any MATH and for any coherent sheaf MATH on MATH. It follows that the NAME spectral sequence MATH degenerates at MATH: then MATH is a quotient of MATH. In particular, the MATH-module MATH is supported at MATH. Moreover, MATH is the MATH-linearized sheaf on MATH associated with the MATH-linearized sheaf MATH. Thus, MATH . This proves the claim. By REF , it follows that the MATH-module MATH is supported at MATH. Thus, MATH has rational singularities outside MATH. |
math/9908094 | For rationality of singularities of MATH, it is enough to check that MATH satisfies the assumption of REF . We may assume that MATH acts effectively on MATH. If a simple normal subgroup MATH of MATH fixes points of MATH, let MATH be a component of the fixed point set. Then MATH is MATH-stable: it is the closure of some orbit MATH. Since MATH is regular, the normal space MATH is a direct sum of MATH-invariant lines. Since MATH is simple and fixes pointwise MATH, it fixes pointwise MATH as well. It follows that MATH fixes pointwise MATH, a contradiction. For the remaining assertions, observe that the local equations of MATH at any point MATH are a regular sequence in MATH. Moreover, as noted above, the scheme-theoretical intersection MATH is equidimensional of codimension MATH, and generically reduced. Since MATH is NAME, then MATH is reduced, and the local equations of MATH at any point MATH are a regular sequence in MATH. |
math/9908094 | Note that the singularities of MATH are rational by REF ; thus, the same holds for MATH. Let MATH be a resolution of singularities; consider the quotient map MATH, the preimage MATH in MATH, and the fiber product MATH. Then MATH is smooth, since MATH and MATH are; the projection MATH is proper and birational, since MATH is; and MATH for MATH, since cohomology commutes with flat base extension. Therefore the singularities of MATH are rational. Now MATH and MATH is a locally trivial fibration, so that the singularities of MATH are rational as well. For the second assertion, we identify MATH to its image MATH in MATH. Since MATH is MATH-equivariant, it is enough to check the statement at MATH. Let MATH be the boundary divisors containing MATH, with local equations MATH in MATH. It follows from REF that the pull-backs MATH are a regular sequence in MATH and generate the ideal of MATH. Moreover, the restriction of MATH to MATH is flat with reduced fibers, as MATH is MATH-equivariant. Now we conclude by a local flatness criterion, see REF . |
math/9908094 | Let MATH be the label of a double edge with source MATH. Denote by MATH the natural map and by MATH its restriction to MATH; then MATH is a projective line bundle, and MATH is generically finite of degree MATH. Choose an ample line bundle MATH on MATH; then MATH is an effective line bundle on MATH. Now our assertion is a direct consequence of the following claim: the restriction map MATH is not surjective for large MATH. To check this, note that MATH and that MATH, by the projection formula. Thus, MATH identifies with the map MATH defined by the inclusion of MATH into MATH. Since MATH has degree MATH, the quotient MATH has rank MATH as a sheaf of MATH-modules. Moreover, since MATH is ample, the cokernel of MATH is isomorphic to MATH for large MATH. This proves the claim. |
math/9908094 | CASE: We prove that MATH by induction over MATH, the case where MATH being evident. If MATH, we can write MATH for some simple root MATH and some MATH such that MATH; then MATH. Then MATH. Since MATH, it follows that MATH raises MATH and that MATH. Because MATH has maximal rank, MATH consists of two MATH-orbits, both of maximal rank. But MATH so that MATH is a unique MATH-orbit of maximal rank and of codimension MATH in MATH. By the induction assumption, we have MATH, that is, MATH. If moreover MATH for some MATH, then a similar induction shows that MATH. If MATH then there exists MATH such that MATH. Thus, MATH, so that MATH is contained in MATH. But MATH; therefore, MATH, and MATH. It follows that MATH, a contradiction. CASE: Let MATH be a representative of MATH in the normalizer of MATH. By assumption, the map MATH is surjective. Thus, it induces an injective homomorphism from the ring MATH of regular functions on MATH, to MATH. The group of invertible regular functions MATH is mapped into MATH. NAME by MATH and taking ranks, we obtain MATH by REF , whence MATH. If moreover MATH, we show that MATH by induction over MATH; we may assume that MATH. Then we can write MATH where MATH, MATH and MATH. It follows that MATH. We begin by checking that MATH. Otherwise, by REF , there exists MATH fixed by MATH. Thus, MATH is fixed by MATH. Since the unipotent radical of MATH acts freely on MATH by REF , it follows that MATH. Then MATH which contradicts the assumption that MATH. As above, it follows that MATH is a MATH-orbit of maximal rank and of dimension MATH; moreover, MATH. We can write MATH where MATH, MATH, and MATH. Thus, MATH, and MATH as MATH. By the induction assumption, MATH. Moreover, MATH, for MATH and MATH. It follows that MATH; in particular, MATH. But MATH is in MATH as well. Thus, MATH and MATH. Let MATH be a simple root of MATH. Then we see as above that MATH. We have MATH with MATH and MATH. Thus, MATH which forces MATH (as MATH) and MATH (as MATH.) We conclude that MATH is a simple root of MATH. Conversely, let MATH such that MATH is a simple root of MATH. Then MATH, whence MATH . Let MATH be a MATH-orbit in MATH. Then MATH. By REF , we have MATH, whence MATH and MATH. |
math/9908094 | With notation as in REF, recall that MATH where MATH is fixed pointwise by MATH. Now REF implies that MATH fixes pointwise MATH and normalizes MATH. Thus, MATH stabilizes MATH. Moreover, intersecting that space with those boundary divisors that contain a given closed MATH-orbit, we obtain MATH-stable hypersurfaces meeting transversally at a fixed point. Arguing as in the proof of REF , it follows that MATH fixes pointwise MATH. By REF , MATH meets each MATH-orbit along a unique MATH-orbit. As a consequence, the intersection of MATH with each MATH-orbit is contained in a unique MATH-orbit. We apply this to MATH, the open MATH-orbit in MATH. Since MATH equals MATH by REF , we see that the product map MATH is an isomorphism. Moreover, MATH is a unique MATH-orbit of dimension equal to the rank of MATH. It follows that each MATH-orbit in MATH is a unique orbit of MATH. Indeed, any MATH-orbit is isomorphic to some affine space, and its projection to MATH is a morphism to a torus, hence is constant. Choose MATH and let MATH. Since MATH is dense in MATH, we have MATH. The latter equals MATH by the previous step. Because MATH acts freely on MATH, it follows that MATH is open in MATH. But both are affine spaces, so that they are equal. Thus, MATH is MATH-stable. It is even MATH-stable, because MATH by REF . Since MATH meets each MATH-orbit along a unique MATH-orbit, MATH meets each MATH-orbit along a unique MATH-orbit. Let MATH, then MATH and, therefore, MATH. By REF again, we have MATH. The remaining assertions follow from REF together with REF . |
math/9908094 | Clearly, MATH is contained in MATH. And since MATH stabilizes MATH, the set MATH is stable under right multiplication by MATH. This implies the first assertion. Let MATH and observe that MATH with equality if and only if MATH (indeed, a reduced decomposition of MATH defines a non-oriented path in MATH with endpoints MATH and MATH). Let MATH. Since MATH, we have MATH. Moreover, MATH. Since we can change MATH in its right MATH-coset, it follows that MATH. Conversely, let MATH such that MATH. Then MATH, whence MATH and MATH. Since MATH, this forces MATH and then MATH. This proves the first assertion. Together with REF , this implies the second assertion. |
math/9908094 | Let MATH. We choose a reduced decomposition MATH and we argue by induction on MATH. If MATH then MATH is a reflection in MATH, so that MATH. Now we conclude by the induction assumption. If MATH then MATH has codimension REF in MATH. Let MATH be the largest integer such that MATH. Let MATH, then MATH and MATH. If MATH then MATH by definition of the MATH-action and maximality of MATH. Let MATH. Then MATH is a reflection of MATH, and MATH. If MATH, then MATH by REF . Otherwise, MATH is isomorphic to MATH or to MATH; it follows that MATH, and that MATH. Now we conclude by the induction assumption. If MATH then MATH raises MATH to (say) MATH. Choose MATH, then MATH and MATH. Moreover, MATH. We have MATH for some MATH such that MATH. Thus, MATH. Therefore, we may assume that there exist MATH and MATH such that MATH. By REF , we may assume moreover that MATH and MATH are neighbors. Then we conclude by REF . |
math/9908102 | Since this is a local statement, we may assume that MATH, where MATH is open, with MATH the projection onto the first factor, and with right action MATH given by MATH . Hence, MATH via the map MATH and the projection MATH is given explicitly by right translation MATH . We identify the map MATH with MATH by MATH and the map MATH with MATH by MATH. We have the following identifications: MATH so that MATH . Then, MATH where the bracket is the commutator of matrices as MATH is a matrix group. Hence, MATH, a MATH-valued (vertical-valued) MATH-form. Now, MATH, so using REF , MATH. (We make the identification MATH.) So for any vector field MATH on MATH, MATH . Let MATH be the local connection MATH-form associated to MATH. Then using REF , MATH . Now to obtain the formula for MATH, we use the injective correspondence between MATH and MATH. Hence, MATH . It is standard (see CITE) that MATH is given by MATH, so we need only use REF to compute the horizontal lift MATH. We have that MATH so that MATH, and this completes the proof. |
math/9908102 | We use the same notation as in the proof of REF . Let MATH be a basis of MATH and MATH its dual basis. Let MATH be any section of MATH, and let MATH be any section of MATH. Using REF , we have that MATH where MATH is the natural pairing between MATH and MATH. Hence, the operator MATH satisfying REF is MATH . This expression can be defined globally and it is straightforward to verify REF. |
math/9908102 | CASE: REF is a standard argument in the calculus of variations. For REF, we use that MATH with REF . For REF, we have that MATH . REF of REF gives that MATH so that MATH . As MATH has compact support, by NAME theorem, MATH, so we conclude that MATH for all sections MATH of MATH with compact support. Thus, we obtain the NAME equations. |
math/9908103 | It is clear that MATH is closed under right composition; hence, we must show that MATH is a submanifold of MATH. To do so, we shall use the transversal mapping theorem (see, for example, CITE) which states that if MATH is transversal to the zero section of MATH, then MATH is a submanifold of MATH. Since our manifolds are NAME, in order to establish the transversality of MATH with MATH, it suffices to prove that MATH is a surjection. The NAME derivative on MATH induces, by a pointwise lift, natural (weak) covariant derivatives MATH on MATH and MATH (see REF). We compute that for all MATH in MATH, MATH where MATH denotes the covariant derivative in the pull-back bundle MATH . Next, we compute the covariant derivative of MATH. For all MATH, and MATH, along the boundary MATH, MATH where MATH denotes the tangential component. Hence, MATH is symmetric with respect to the inner-product on MATH. Now, by definition, for MATH, MATH so setting MATH in REF shows that MATH . It follows that for all MATH, MATH where MATH. It remains to show that for every MATH, there exists MATH such that MATH. By right translation to the identity, it suffices to find MATH such that MATH for every MATH. To do so we obtain a solution to the following elliptic boundary value problem: For MATH, MATH and MATH, find MATH such that MATH where in Cartesian components MATH, that is, MATH is the component-wise NAME operator. Note that by definition of MATH, MATH on MATH and MATH. A weak solution to REF in the class of MATH divergence-free vector fields that are parallel to MATH is supplied by REF of the proof of REF. Noting that in coordinates MATH, REF provides a strong solution in the class of MATH divergence-free vector fields that are parallel to the boundary and satisfy the boundary condition MATH on MATH. REF then provides the elliptic regularity required to obtain MATH whenever MATH, and this completes the proof of the theorem. |
math/9908103 | Restricting MATH to the algebra MATH, we compute the first variation of the action function MATH for constrained variations of the form MATH. Integrating by parts, we obtain that MATH . The boundary term vanishes for MATH and MATH in MATH, so another integration by parts yields MATH . Since MATH is arbitrary, MATH is a fixed-point of the action if and only if MATH . Using the definition of the NAME projector MATH concludes the proof. |
math/9908103 | We compute the first variation of the action function MATH which we decompose as MATH and MATH . By definition of MATH, the boundary terms appearing from integration by parts vanish; hence, we restrict our computations to the interior. We have MATH . Let MATH be a smooth curve in MATH such that MATH and MATH. Then MATH . Integration by parts yields MATH . We use the product rule to get that MATH . Integrating by parts, noting that the boundary terms vanish by virtue of the subgroup MATH, we have that MATH . Computing the first variation of MATH, we obtain MATH . Setting MATH, and using the projector MATH given by REF gives MATH where MATH . Let us prove that the above expression is well-defined; namely, we shall show that it makes for the NAME projector to act on both MATH and MATH. To see this, notice that MATH, and that MATH, where MATH. The NAME operator acts on MATH by MATH whose domain is MATH, and this operation is well-defined as both MATH and MATH are in the domain of MATH, since MATH on MATH. We may reexpress the above equation as MATH thus the right-hand-side is also well-defined as the image of MATH is the domain of MATH. Denoting the right-hand-side of the above equation by MATH, we have that MATH . We rewrite this equation as the system MATH . We shall prove that MATH and that MATH is a MATH bundle map. Then the standard theorem for existence and uniqueness of ordinary differential equations on a NAME manifold provides the existence of a unique MATH curve MATH solving the above system on MATH, that depends smoothly on the initial data MATH; the time-reversal symmetry allows us to extend the interval to MATH. That MATH is MATH follows from the fact that MATH is of class MATH whenever MATH is in MATH (because MATH forms a multiplicative algebra when MATH), so that MATH is in MATH. That MATH is of class MATH follows from REF together with the smoothness of the NAME projector. The fact that MATH is of class MATH whenever MATH follows from similar arguments (see the proof of REF for details). |
math/9908103 | The proof of this theorem is essentially identical to the proof of REF so will not be repeated. |
math/9908109 | When MATH is thought of as a MATH-form field on MATH, REF may be reexpressed as MATH . Taking the exterior derivative of this equation and setting MATH yields the vorticity equation MATH . On two-dimensional manifolds, we may identify the MATH-form MATH with its scalar density, in which case the above equation takes the particularly simple form MATH with the corresponding weak form MATH for all MATH. REF is equivalent to the pointwise conservation of vorticity along the Lagrangian trajectory MATH this is, of course, just the coadjoint action of MATH acting on MATH by right composition. REF gives a time interval MATH of existence of solutions to REF for MATH. This, in turn, gives the existence of a weak solution MATH in MATH. The conservation law REF yields the NAME MATH . Thus, we have that MATH is a conserved quantity, and by standard elliptic estimates we have that MATH for all MATH. Thus, the MATH-norm of MATH does not blow up, so MATH. The usual bootstrap argument now yields the result for MATH, MATH. |
math/9908109 | This follows from REF , the trace theorem, and the fact that MATH is smooth, as MATH and MATH are MATH. |
math/9908109 | The result follows from REF . |
math/9908109 | Acting on divergence-free vector-fields, MATH. Thus, the proof that MATH is a MATH subgroup of MATH shows that the NAME REF has a unique solution MATH for any MATH, MATH. It is easy to verify that the summands in REF are MATH-orthogonal, so it only remains to show that MATH is smooth. For MATH, let MATH, and let MATH solve REF . By REF , it suffices to prove that MATH is smooth. Letting MATH, we have the equivalent expression for MATH given by MATH which is a MATH bundle map by REF together with REF . |
math/9908109 | That MATH is MATH on MATH follows from REF . That MATH is a positive-definite symmetric bilinear form is proven as follows: MATH so for any of the boundary conditions prescribed on elements of MATH, we have that MATH so that integrating by parts (and noting that the boundary terms vanish), we may express MATH in the equivalent form MATH . Since MATH is a self-adjoint positive operator (on MATH vector fields that are divergence-free), this shows that MATH is a well defined MATH weak invariant Riemannian metric on MATH. |
math/9908109 | From REF , the reduced Lagrangian is given by MATH, so that MATH is a geodesic of MATH on MATH if MATH is a fixed point of the reduced action function (on an arbitrary interval MATH) MATH given by MATH . Let MATH be a smooth curve in MATH such that MATH and MATH; the map MATH is the variation of the curve MATH on the interval MATH and MATH. The curve MATH induces a curve MATH in the single fiber MATH such that MATH and MATH. The NAME reduction theorem gives the relation MATH . Computing the first variation of the action MATH, we have that MATH . Since MATH and MATH satisfy the boundary conditions prescribed to elements of MATH, the boundary term in the above equation vanishes, leaving only MATH . Using the formula MATH and integrating by parts, we obtain MATH where again MATH. Since right translation is an isomorphism, MATH is arbitrary, so MATH is a fixed point of MATH iff MATH and this is precisely REF , as MATH since MATH. |
math/9908109 | First notice that for MATH, MATH is a MATH vector field on MATH whose trace vanishes on MATH; thus, it makes sense for the operator MATH to act on MATH. Recall that MATH, so we begin by computing the commutator of MATH. Let MATH be a local orthonormal frame, and write the NAME Laplacian MATH acting on MATH-forms (identified with vector fields) as MATH, so that MATH . Using the definition of the Riemannian curvature operator, we compute that MATH . Expressing MATH as MATH, we see that MATH; hence, one may easily verify that MATH so that MATH . Using the fact that div-MATH, and combining terms involving the NAME curvature gives the result. |
math/9908109 | We first set MATH. NAME differentiating MATH yields MATH . Using REF , we obtain that MATH . Now MATH, for some MATH; hence, MATH if and only if MATH for some MATH, and by REF , this is precisely REF with MATH. Adding the term MATH to REF produces REF . |
math/9908109 | For each MATH, the metric MATH induces a natural inner-product, say MATH, on elements of MATH, and hence a weak MATH metric on MATH given by MATH. There exists a unique NAME covariant derivative associated with this weak MATH metric which we denote by MATH. The covariant derivative MATH is induced by the connector MATH which is the functorial lift of the connector MATH uniquely associated with the metric MATH thru the fundamental theorem of Riemannian geometry (see REF). Let us denote the map MATH by MATH, that is, MATH. Continuity of MATH is immediate. Thus, we shall show that MATH is of class MATH. Let MATH be a smooth curve in MATH such that MATH and MATH; then, MATH is computed as MATH where MATH denotes the unique NAME covariant derivative in the pull-back bundle MATH associated to the metric MATH on MATH. Specifically, for MATH and MATH, MATH has the local expression MATH where MATH denotes the NAME symbol of the metric MATH evaluated at the point MATH. We compute the operator norm of MATH which we shall denote by MATH. We have that MATH . Computing the supremum of MATH in a neighborhood of MATH yields the MATH topology; as the supremum is finite, we have established that MATH is a MATH map. To see that MATH is of class MATH, we compute in a local chart MATH . Since MATH is in the multiplicative algebra MATH, and MATH, the same argument as above shows that MATH is MATH. In particular, we see that the MATH-th derivative of MATH is a rational combination of MATH and derivatives of MATH, which combined with our argument showing that MATH is MATH together with the fact that multiplication of MATH maps is smooth, shows that MATH is MATH for any integer MATH, and hence that MATH is MATH. |
math/9908109 | We identify MATH with MATH, respectively. It then suffices to prove that MATH is in MATH, and hence that MATH is in MATH (since MATH is MATH). But this follows since MATH is a multiplicative algebra, and MATH is a MATH bundle map. |
math/9908109 | By the MATH orthogonal NAME decomposition, MATH where MATH denotes the Harmonic fields. Hence, MATH . Let MATH denote the MATH orthogonal projection of MATH onto MATH, and let MATH denote the restriction of MATH to MATH, so MATH. Since MATH is a finite dimensional subspace of MATH consisting of MATH elements, REF asserts that MATH is a smooth bundle map, and that MATH and hence MATH is a subbundle. We may thus form the following exact sequence MATH . Since MATH is a MATH bundle map, this shows that MATH and MATH are subbundles. Now let MATH be the restricted orthogonal projector. Then by the same argument MATH is a smooth bundle map and MATH is a subbundle. Hence, we may form the exact sequence MATH and thus obtain that MATH and MATH are subbundles. Using REF , we may restrict the domain and range to ensure that the maps MATH and MATH are isomorphisms. To find the inverse of MATH between these vector spaces, first let MATH. Then MATH therefore, MATH so that MATH is the inverse of MATH. Similarly, we find that MATH is the inverse of MATH. Next, let MATH so MATH. Now MATH is a smooth bundle map by REF , and since MATH is a subbundle, we may form the exact sequence MATH . Thus, the MATH is a subbundle from which it follows that MATH is a subbundle, so that it makes sense to define MATH as a smooth bundle isomorphism. A similar argument allows us to define MATH as smooth bundle isomorphism. We have shown that the bundle map MATH covering the identity is the inverse of MATH which is smooth; hence, by the inverse function theorem, the bundle map MATH is also smooth. On the other hand, MATH is the inverse of MATH, and by the same argument is smooth. Since MATH and MATH are MATH, then MATH is MATH on MATH, and hence MATH is MATH on MATH again by the inverse function theorem. Thus far, we have been working with sections of differential MATH-forms over the boundaryless manifold MATH. We shall now restrict our attention to MATH class sections of MATH. Letting MATH denote the outward-pointing normal vector field on MATH, for MATH, we define the closed subspace of MATH by MATH and for MATH, set MATH . Note that the restriction operator to these subspaces is a continuous linear map. MATH is a self-adjoint linear unbounded nonnegative operator on MATH with MATH, and MATH is an isomorphism. It follows that MATH is an isomorphism. Since MATH and since we have proven that MATH, MATH, MATH, and MATH are MATH bundle maps, it follows that MATH is a MATH bundle isomorphism covering the identity, so that by the inverse function theorem, MATH is MATH as well. This proves the theorem in the case that MATH. In the case that MATH, simply set MATH, and for MATH, set MATH in the definition of MATH. |
math/9908109 | The proof follows trivially from REF . |
math/9908109 | From REF , it is clear that the proof is identical to the proof of REF once we show that MATH is a MATH bundle map. The result follows from the fact that for MATH, MATH is a multiplicative algebra, so that the terms MATH and MATH are of class MATH whenever MATH. This observation together with the results of REF complete the proof. |
math/9908109 | For MATH, MATH; thus to prove that MATH is a smooth curve in MATH, we need only copy the proof of REF , and show that MATH is a smooth map into the second tangent bundle MATH. We leave the trivial details to the reader. |
math/9908109 | Again, right invariance of MATH follows from the right invariance of MATH. Extend MATH to smooth right invariant vector fields MATH on MATH and let MATH, and MATH. Let MATH. Then MATH where MATH denotes the commutator of operators, and MATH. Since MATH is in MATH for MATH and MATH in MATH, the remainder of the proof follows exactly the proof of REF . |
math/9908110 | Since MATH is conjugate to some diagonal matrix MATH, its minimal polynomial is just MATH where the MATH are the distinct diagonal entries of MATH. By the previous proposition, MATH, hence MATH must have MATH distinct diagonal entries. Thus all of the diagonal entries of MATH are distinct. |
math/9908110 | For MATH, the proof is by direct computation using MATH and MATH. For example, for MATH: MATH . The first of the two quotients is easily seen to be real. For the second quotient, MATH . Multiply the numerator and the denominator by MATH to see that this is still MATH . For the case MATH, note that MATH, so MATH . Hence MATH, and MATH is real. |
math/9908110 | Suppose MATH . Multiply by MATH both on the left and on the right. The only term of the sum that survives is MATH . Let MATH be an eigenvector of MATH corresponding to MATH. Then MATH, so MATH. Hence MATH. For MATH, multiplying by MATH on the left and by MATH on the right shows MATH . But MATH so MATH. Hence MATH. So MATH is linearly independent. It has MATH elements, hence it is a basis of the MATH-dimensional space MATH. |
math/9908110 | It is sufficient to prove that MATH acts as an antihomomorphism on the elements of the basis MATH. MATH has two different types of elements, therefore we will have four different cases. Since each can verified directly by a simple computation, we will show the details for only one: CASE: MATH CASE: MATH CASE: For MATH, MATH CASE: MATH . Also MATH . That MATH follows immediately from the definition. |
math/9908110 | First note that MATH. Multiply MATH by MATH on both sides: MATH into MATH . |
math/9908110 | MATH . Similarly, MATH . |
math/9908110 | Note that MATH. That is MATH and MATH satisfy the braid relation. So the group homomorphism MATH defined by MATH and MATH is another representation of MATH on MATH. Once again, the braid relation implies that MATH and MATH are conjugates. Hence they have the same eigenvalues, namely MATH. But MATH only permutes the basis MATH of MATH. Hence MATH and MATH and MATH generate the algebra MATH. That is MATH is also a simple representation of MATH . Now, MATH (recall MATH). By REF , the eigenvalues MATH (if d=MATH) or the eigenvalues together with MATH (if MATH) uniquely determine a simple representation of MATH on MATH up to isomorphism. But we already know such a representation, namely MATH and MATH. Hence there exists MATH such that MATH and MATH. Then MATH is in the centralizer of MATH. MATH . Call the quantity in parentheses MATH. Note that MATH is the eigenprojection to the subspace spanned by the eigenvector MATH of MATH with eigenvalue MATH. But the eigenvectors MATH of MATH are also eigenvectors of MATH and span MATH (the eigenvalues are distinct). Hence MATH and MATH for MATH. That is MATH as their action on the basis MATH is identical. Then REF shows MATH. Hence conjugation by MATH is a MATH-algebra isomorphism that fixes MATH and MATH. But MATH and MATH generate the basis MATH of MATH, hence they generate the algebra MATH. So conjugation by MATH must fix every element of MATH. In particular, MATH. |
math/9908110 | Suppose MATH . Note that MATH. Since MATH is invertible MATH if and only if MATH. Multiply by MATH on the left. Then MATH . But MATH, so MATH. Now, multiply by MATH REF on the left. Then MATH. We know MATH by simplicity, so MATH and MATH. Hence MATH is a linearly independent set, and we can conclude that it is a basis of the MATH-dimensional vector space MATH. |
math/9908110 | Suppose MATH for all MATH. Consider the action of MATH on MATH. The sesquilinear form defined above is invariant under the action of MATH. So it is sufficient to show that it is an inner product. That is we need to prove is that it is positive definite. On the basis MATH: MATH . Hence MATH for MATH by assumption, and MATH. We claim that MATH is orthogonal with respect to MATH. Let MATH and MATH: MATH . We used MATH in the last computation just like in REF . Hence MATH is a positive definite form. Then MATH is a MATH-vector space with inner product MATH and the action of MATH on this space is unitary. Conversely, suppose MATH is unitarizable. So there exists MATH a MATH vector space with inner product MATH and MATH such that MATH and MATH act as unitary operators on MATH. Let MATH be the transpose induced by MATH. We know MATH and MATH. Let MATH be an eigenvector of MATH with eigenvalue MATH. Then MATH and MATH . Hence MATH. We know MATH for MATH by simplicity, so MATH in this case. |
math/9908111 | Consider the norm MATH on MATH. We will show that on MATH satisfies the triangle inequality. Since by REF on MATH the conclusion then is a simple consequence of REF : MATH . For the proof of the triangle inequality for MATH show - in a first step and in complete analogy to the proof of the usual NAME inequality - that for each NAME REF function space MATH and MATH with MATH (see also CITE). In particular, we obtain with MATH that MATH . Now simulate for MATH the proof of the NAME inequality: MATH . This obviously completes the proof. |
math/9908111 | MATH by REF is MATH-convex, hence MATH by REF is a NAME function space. Moreover, MATH and MATH by REF are isomorphic NAME function spaces, and MATH by REF is REF-convex (since the space MATH is REF-convex). Assume now that MATH is not MATH-order continuous - then it contains MATH as a topological subspace (note first that NAME function spaces are MATH-complete and see for example, CITE or CITE). But this contradicts the fact that MATH by REF is REF-concave. |
math/9908111 | The proof is based on a standard separation argument. Define the weak-MATH-compact and convex set MATH and for MATH and MATH the affine and continuous function MATH (put MATH). Note first that the set MATH of all these functions is convex: the sum of two such functions belongs to MATH, and for MATH and MATH . We will now show that for each MATH there is MATH with MATH indeed, by the NAME theorem and REF there are MATH such that MATH hence (recall that MATH for MATH and MATH) MATH . By NAME 's lemma (see for example, CITE) there is MATH such that MATH . This easily gives the conclusion: define for MATH . Then MATH if MATH, then MATH for all MATH, hence MATH. |
math/9908111 | Take MATH, MATH and MATH the identities, and recall from REF that MATH is MATH-convex and MATH-convex with constants REF. |
math/9908111 | We may assume without loss of generality that MATH and MATH is the identity (otherwise replace MATH by MATH). By assumption REF MATH is MATH-convex with constant REF for some MATH, and by REF we may assume that MATH. Consider for MATH a new scalar multiplication on MATH: for MATH and MATH define MATH then it is easy to check that the mappings MATH with respect to this multiplication are homogeneous, and for MATH . From now on MATH will always be endowed with its natural quasi norm MATH which by REF is even a norm since MATH by REF . Hence, the homogeneous form MATH is defined, and by NAME 's inequality, REF on MATH, and the definition of the norm on MATH it satisfies MATH for all MATH and MATH. By REF MATH hence it follows from REF that there are two functionals MATH such that for all MATH, MATH . Clearly, for MATH . On the other hand REF MATH but since MATH and hence also its MATH-th power by REF are MATH-order continuous, we obtain from REF MATH . Moreover, by REF for all MATH, MATH . It remains to give this inequality the form of REF: note first that for MATH the multiplication operator MATH by REF is defined, and for all MATH, MATH (obviously, MATH-a.e. for all MATH). Since MATH has dense range (use for example, the NAME theorem), for all MATH . Finally, it remains to prove the norm estimates REF for the weights MATH . The estimate in REF and the fact that MATH is norming in MATH (see REF ) give MATH and if MATH is MATH-order continuous, then REF gives MATH . This completes the proof. |
math/9908111 | Clearly, the proof of the first statement is an immediate consequence of REF (represent MATH as in REF and MATH as in REF ), and for the factorization define MATH . |
math/9908111 | If one replaces the MATH's in REF by the class of all MATH's MATH, then the equivalence MATH is a well-known result of CITE (see also CITE). Since MATH follows from REF , it remains to check the implication MATH: note first that by NAME 's result (just mentioned) there is a MATH such that MATH for all operators MATH. The use of conditional expectation operators shows that this inequality also holds for all MATH, MATH an arbitrary measure. But then NAME 's localization technique from CITE (see also CITE) assures that every MATH, MATH an abitrary NAME lattice, is MATH-convex. |
math/9908111 | Again, represent MATH as in REF , MATH as in REF and use REF . It remains to prove that for MATH-order continuous MATH linear MATH factorize as indicated: with the function MATH representing MATH define MATH, and MATH . Then MATH, and since MATH and MATH, the operator MATH has an extension MATH to all of MATH with MATH. Clearly, MATH. |
math/9908111 | As mentioned in REF by trace duality is equivalent to the fact that each operator MATH is MATH-summing CITE, and by the definitions an operator MATH is MATH-summing if and only if it is MATH-concave. On the other hand, by NAME 's localization technique from CITE (see again the proof of CITE) each MATH is MATH-concave if and only if REF holds. This shows that MATH . Clearly, REF implies REF by REF . In order to prove the implication MATH we only have to check that each MATH is MATH-concave. But since MATH is MATH-convex, this follows by an easy calculation from the inequality in REF . |
math/9908120 | We construct a functor MATH and show that it is an isomorphism. Definition of MATH. Objects. First we define the homotopy class of a connected NAME cube manifold, denoted by MATH, associated to the MATH-homotopy type MATH of a connected MATH-dimensional NAME manifold MATH. Choose an at most MATH-dimensional connected locally finite polyhedron MATH and a (proper) MATH-homotopy equivalence MATH (see CITE, CITE). Let MATH denote a countable connected MATH-complex obtained from MATH by killing MATH-dimensional homotopy groups of MATH for each MATH. This means that MATH is a countable connected MATH-complex such that MATH and MATH for each MATH. By CITE, there exists a homotopy equivalence MATH, where MATH is a countable connected and locally finite polyhedron. By NAME 's theorem CITE, the product MATH is a MATH-manifold. Finally let MATH be the homotopy type of the NAME cube manifold MATH. Clearly, MATH for each MATH. Let us show that the homotopy type MATH is well defined, that is, it does not depend on the choices of a NAME manifold MATH (within the MATH-homotopy type MATH of MATH), of a polyhedron MATH, of a MATH-homotopy equivalence MATH, of MATH-complexes MATH, MATH and of a homotopy equivalence MATH. Indeed let MATH be a MATH-dimensional NAME manifold and MATH be a MATH-homotopy equivalence. Let also MATH be a connected at most MATH-dimensional locally finite polyhedron and MATH be a MATH-homotopy equivalence. Let MATH denote a countable connected MATH-complex obtained from MATH by killing all MATH-homotopy groups of MATH for each MATH and let MATH be a homotopy equivalence where MATH is a connected locally finite polyhedron. Our goal is to show that the MATH-manifolds MATH and MATH are homotopy equivalent. Let MATH, MATH and MATH be MATH-homotopy inverses of the maps MATH, MATH and MATH respectively. Observe that MATH and MATH . This shows that the composition MATH is a MATH-homotopy equivalence. Consequently, MATH is a homotopy equivalence. Next denote by MATH and MATH the homotopy inverses of MATH and MATH respectively and observe that MATH . Similarly, MATH . This shows that the composition MATH is a homotopy equivalence. Then the product MATH is also a homotopy equivalent as needed. This shows that the homotopy type MATH is well defined. Morphisms. Let now MATH and MATH be two connected MATH-dimensional NAME manifolds and MATH be a map. We need to define the homotopy class MATH. As before let MATH be a MATH-homotopy equivalence where MATH is at most MATH-dimensional locally compact polyhedron, MATH. Let MATH denote the MATH-homotopy inverse of MATH, MATH. The composition MATH admits an extension MATH. Next let MATH denote a homotopy equivalence where MATH stands for a connected locally compact polyhedron, MATH. Let also MATH denote the homotopy inverse of MATH, MATH. Next consider the composition MATH and the product MATH . We let MATH . Finally let MATH be the homotopy class of the product map MATH . Let us show that this definition is correct. Indeed let MATH is a map such that MATH. Then MATH. As was noted earlier extensions MATH are homotopic. Consequently MATH and MATH as required. This completes the definition of the functor MATH. Next we show that MATH is an isomorphism. CASE: MATH is surjective. Indeed, let MATH be a connected MATH-manifold such that MATH for each MATH. By the triangulation theorem for MATH-manifolds CITE, there exists a locally compact polyhedron MATH such that MATH is homeomorphic to the product MATH. Since the projection MATH is a homotopy equivalence, it follows that MATH for each MATH. Let MATH. According to the resolution theorem for NAME manifolds CITE there exists a MATH-homotopy equivalence MATH, where MATH is a connected MATH-dimensional NAME manifold. It easily follows from the construction of the functor MATH that MATH. CASE: MATH is injective. Let MATH and MATH be two MATH-dimensional NAME manifolds such that MATH. According to the definition of the functor MATH this means that there exist: CASE: A MATH-homotopy equivalences MATH, where MATH is at most MATH-dimensional locally compact polyhedron, MATH. CASE: A homotopy equivalence MATH, where MATH is a connected locally finite polyhedron, MATH. CASE: A homotopy equivalence MATH . Since the projection MATH is a homotopy equivalence (MATH), it follows that there exists a homotopy equivalence MATH. Then the composition MATH, where MATH is the homotopy inverse of MATH, is a homotopy equivalence. Without loss of generality we may assume that MATH for each MATH. In particular, MATH . This defines a map MATH. It is easy to see that MATH and therefore, MATH is a homotopy equivalence. This obviously implies that MATH is a MATH-homotopy equivalence. Finally the composition MATH, where MATH is the MATH-homotopy inverse of MATH, is also a MATH-homotopy equivalence. This proves that MATH. CASE: MATH is surjective. Let MATH be a map between connected MATH-manifolds such that MATH for each MATH and MATH. By the triangulation theorem for MATH-manifolds, MATH, where MATH is a connected locally compact polyhedron, MATH. Let MATH denote a section of the projection MATH. Observe that MATH is a homotopy equivalence. Let MATH, where MATH is the projection onto the first coordinate. Note also that MATH is homotopic to the product map MATH. Without loss of generality we may assume that MATH for each MATH. Next consider MATH-dimensional NAME manifolds MATH and MATH and MATH-homotopy equivalences MATH, MATH. Let MATH denote the MATH-homotopy inverse of MATH and MATH. A straightforward verification shows that MATH. CASE: MATH is injective. Let MATH be two maps between MATH-dimensional NAME manifolds such that MATH. Choose a MATH-homotopy equivalence MATH, where MATH is at most MATH-dimensional locally compact polyhedron, MATH. Let MATH denote the MATH-homotopy inverse of MATH. Also let MATH be a homotopy equivalence, MATH. Finally let MATH stands for the homotopy inverse of MATH. According to our assumption the product maps MATH and MATH are homotopic. This implies that MATH. By MATH, MATH. Since MATH and MATH, MATH, are MATH-homotopy equivalences it follows that MATH as required. |
math/9908120 | In the light of REF it suffices to show that the categories MATH and MATH as well as the categories MATH and MATH are isomorphic. In order to define the functor MATH recall CITE that for any MATH-manifold MATH the product MATH can be embedded into the NAME cube as an open subspace. Based on this observation we let MATH and MATH . A straightforward verification shows that the functor MATH is indeed an isomorphism. In order to define the functor MATH we use concept of the MATH-homotopy kernel MATH of a MATH-dimensional NAME manifold MATH (see, CITE, CITE). It is important to note that MATH plays the role of the product MATH in the category of MATH-dimensional Manger manifolds. Let MATH be a connected MATH-dimensional NAME manifold. According to CITE, CITE the MATH-homotopy kernel MATH of MATH can be topologically identified with an open subspace of the MATH-dimensional universal NAME compactum MATH. The complement MATH is a MATH-set in MATH and consequently MATH and MATH are MATH-homotopy equivalent. This allows us to let MATH . Let now MATH be a map between MATH-manifolds. Since MATH is a MATH-set in MATH, it follows that there is a map MATH as close to MATH as we wish (considered as maps into MATH). In particular, we may assume CITE that MATH and MATH are MATH-homotopic (in MATH). Considerations similar to the argument used in the proof of CITE, CITE guarantee that the MATH-homotopy class of the restriction MATH of MATH is uniquely defined by the MATH-homotopy class of MATH. This observation suffices to complete the definition of the functor MATH by letting MATH . The fact that the functor MATH is an isomorphisms can be extracted directly from the above definition. |
math/9908120 | By REF , it suffices to show that the categories MATH and MATH as well as the categories MATH and MATH are isomorphic. In order to define the functor MATH we recall that the separable NAME space MATH can be embedded into the NAME cube MATH in such a way that the complement MATH forms a MATH-skeletoid in MATH CITE. Now for a connected open subspace MATH of MATH consider an open subspace MATH of MATH such that MATH. It follows from the elementary properties of MATH-skeletoids that the inclusion MATH is a homotopy equivalence. Consequently, if MATH is another open subspace of MATH such that MATH, then MATH and MATH have the same homotopy type. This observation allows us to let MATH . For a continuous map MATH of open subspaces of MATH consider the composition MATH, where MATH stands for the homotopy inverse of the inclusion MATH and MATH denotes the inclusion Obviously the homotopy class MATH is completely determined by the homotopy class MATH and we let MATH . Note that if MATH and MATH are homotopy equivalent open subspaces of the NAME cube MATH, then the open subspaces MATH and MATH of MATH are also homotopy equivalent and according to CITE, CITE are even topologically equivalent. This is the only non-trivial observation involved in the verification of the fact that the functor MATH is an isomorphism. We define functor MATH similarly. First of all we note that the MATH-dimensional universal NAME compactum MATH also admits CITE a MATH-skeletoid MATH such that its complement MATH is homeomorphic CITE, CITE to the MATH-dimensional universal NAME space MATH. Now we proceed as above. We let MATH where MATH stands for an open subspace of MATH such that MATH. Also we let MATH where MATH has the same meaning as above. The proof of the fact that the functor MATH is an isomorphisms is based on CITE which state that MATH-homotopy equivalent connected open subspaces of MATH are homeomorphic. |
math/9908125 | The claim that MATH follows immediately from MATH and MATH. Because MATH is a homeomorphism and MATH is continuous, the continuity claim only needs to be confirmed at points of MATH. The restriction of MATH to MATH is the projectivization MATH of a linear isomorphism, so this restriction is a MATH diffeomorphism on MATH. Because MATH, where MATH as MATH, MATH is continuous on the normal line MATH through MATH: MATH which tends to MATH as MATH, with convergence uniform in MATH. Therefore, by the triangle inequality, MATH is continuous at every point of MATH. |
math/9908125 | REF shows that origin - preserving coordinate changes with MATH - linear derivatives act as automorphisms of MATH. If MATH are local coordinate systems REF on a neighborhood MATH of MATH in MATH then MATH gives a change of coordinates on the model blowup. The homomorphism properties established in REF show that blownup coordinate change maps satisfy the cocycle condition, yielding a consistent pasting construction for MATH from the data defining MATH. |
math/9908125 | Away from MATH we know that MATH and MATH may be identified, so the regularity issue only arises along MATH. Along MATH we have defined MATH to be the projectivization of the linear map MATH, so MATH is infinitely differentiable in those directions. If for MATH near MATH we have MATH, where MATH as MATH, then MATH . The division by MATH inside the homogeneous coordinates gives a zero - th order term of MATH, similarly reduces the degree of the other homogeneous terms in the NAME expansion, and reduces by one the order of vanishing for the remainder term. The resulting expansion of MATH near MATH shows that we lose one partial derivative of MATH along the fibers of the normal bundle MATH, compared to the degree of smoothness of MATH at MATH. (Recall from REF that MATH is MATH along MATH.) The partial derivatives of MATH along the singular locus and normal to it are continuous, through order MATH, so MATH is MATH at points of MATH by the familiar theorem deducing REF differentiability from continuous partial derivatives. REF and surrounding discussion show that MATH is a homomorphism. Once we know that MATH is continuous, it follows that MATH is injective, since MATH determines MATH on the dense subset MATH. MATH is continuous because the MATH distance between diffeomorphisms on MATH majorizes the MATH distance between their blowups on MATH and the MATH distance between those blowups on MATH. |
math/9908125 | REF defines MATH in the model case, so the restriction of MATH to MATH is the projectivization of the derivative MATH. Therefore, fixed points of MATH in MATH are solutions of MATH, that is, projective equivalence classes of tangent vectors MATH for which there exist scalars MATH satisfying MATH, that is, projective equivalence classes of eigenvectors of MATH. REF describes the set of all such projective classes as a disjoint union of projectivized subspaces of MATH. |
math/9908125 | In each case a MATH conjugacy as described in REF between MATH and another diffeomorphism MATH with a hyperbolic fixed point at MATH yields the topological blowup. Our job here is to allocate eigenvalues for MATH and apply REF to MATH. If every eigenvalue MATH then MATH has no real eigenvectors and the classical blowup's MATH is empty. If MATH has MATH distinct real eigenvalues and MATH complex eigenvalues appearing in conjugate pairs, then MATH must be even but may otherwise assume any value between MATH and MATH. In this case we see MATH isolated fixed points for MATH on MATH. Positive - dimensional fixed point sets arise from repeated real eigenvalues for MATH. These may appear in combinations so that the multiplicities of real eigenvalues form partitions of some even numbers MATH, MATH: MATH where MATH, MATH are the multiplicities of unstable, respectively stable, real eigenvalues of MATH. Each of these real eigenvalues produces a component of the fixed point set MATH which is diffeomorphic to a projective space: If MATH is a real eigenvalue of multiplicity MATH then MATH. The largest dimension arising in this way MATH. |
math/9908125 | This is a consequence of the following version of the homogeneity of manifolds: for any MATH there exists an isotopy from MATH to a homeomorphism which carries MATH to MATH. In more detail, MATH can be built in segments which are pasted together. We may apply a homeomorphism of the form MATH, where MATH is a homeomorphism, to arrange that MATH for all MATH. Continue the argument with the sequence MATH replacing MATH. Let MATH and let MATH be an isotopy over MATH such that MATH and MATH. Define the first two pieces of MATH by MATH and MATH. Subsequent segments are defined by taking an isotopy MATH over MATH which starts with the MATH already selected and ends at a homeomorphism which carries MATH to MATH. |
math/9908125 | Suppose that MATH are distinct points and that MATH, MATH are sequences in MATH such that MATH and MATH. We may assume that both sequences lie in a neighborhood MATH of MATH which admits a homeomorphism MATH and that the sequences MATH and MATH are strictly increasing and converge to MATH. Form a third sequence MATH such that each MATH is one of the MATH, each MATH is one of the MATH, and the real sequence MATH is strictly increasing and converges to MATH. MATH is not convergent in MATH, but all three of the sequences MATH, MATH, and MATH converge to MATH in MATH. Since MATH, REF implies that there is a homeomorphism MATH such that for every MATH, MATH. If MATH is covered by MATH then MATH is divergent although MATH converges to MATH, so MATH is not continuous. |
math/9908127 | Both claims are proved by induction on MATH. The existence of MATH in the base case follows from the regularity of MATH and the induction step is trivial. |
math/9908127 | The Lemma is proved by induction on MATH. This is trivial if MATH. Otherwise, suppose that MATH. If MATH, then MATH by induction, because each MATH. Furthermore, if MATH are elements of MATH, then MATH, with the inequality holding because MATH is hereditarily monotone and MATH, so MATH for all MATH. |
math/9908127 | The Proposition is proved by induction on MATH for all MATH. If MATH, then MATH is just the functional that is constantly the identity on MATH, which is easily seen to be hereditarily monotone. Suppose that MATH. First we must verify that MATH maps MATH to itself. Suppose that MATH and MATH. Then since MATH by the induction hypothesis and MATH is hereditarily monotone by assumption, MATH is hereditarily monotone, and so MATH. We must also verify that if MATH and MATH are hereditarily monotone, MATH, then MATH, which is just as easy to do. Second, we must verify the monotonicity of MATH: if MATH, MATH are such that MATH, then MATH. Fix any MATH. Then MATH; the first inequality follows from the fact that MATH (induction) and the second from the fact that MATH. This takes care of the successor case. If MATH is a limit, then MATH by definition; but this MATH is hereditarily monotone by induction and REF . |
math/9908127 | CASE: In general, the MATH of a subsequence is always greater than or equal to the MATH of the sequence, and vice-versa for MATH, so if MATH, then MATH. But for any sequence MATH, MATH, so this implies that MATH. CASE: Since MATH is non-decreasing, MATH and MATH . REF is similar to REF . |
math/9908127 | Both claims have similar proofs, so we just do the first. If MATH, then MATH, so by monotonicity of MATH, MATH. Since MATH was chosen arbitrarily, MATH. |
math/9908127 | Each clause is proved by a similar induction on MATH; we do just the first. Throughout the proof, we make silent use of REF to identify the limit of a sequence with the limit of a tail of that sequence, provided the former exists. If MATH, the claim is vacuous. Suppose that MATH; by induction, it suffices to show that MATH, and we do this by induction on MATH. If MATH, then this is just the hypothesis that MATH. The successor case is straightforward. Suppose that MATH is a limit. By the main induction hypothesis, MATH is a non-decreasing sequence, so MATH by REF . Now applying REF , MATH . This last sequence is a subsequence of MATH, so it is non-decreasing, and therefore by REF its supremum is a limit, and by REF the limit is the same as that of the original sequence: MATH. So MATH. This completes the induction step for successor MATH. Finally, suppose that MATH is a limit. Then by induction MATH is non-decreasing, so for any MATH, MATH. |
math/9908127 | This follows from REF (taking MATH), because the order on MATH is total. |
math/9908127 | All three clauses are proved by induction on MATH; we do REF as an example. Fix any ordinal MATH. If MATH, then MATH. If MATH, then MATH, where the second equality follows from the induction hypothesis and the third from REF . Suppose that MATH is a limit. By REF , MATH. Since MATH is a subsequence of MATH and the limit of the latter sequence exists, MATH completing the proof. |
math/9908127 | REF are immediate, and REF are proved by induction on type. We provide details for REF . Note that this is not a trivial claim, as it is an assertion about the h.p. order, not the pointwise order. The claim is true for the base cases because the two orders are the same. Suppose MATH for MATH. If MATH and MATH is h.p., then since MATH, we have MATH, and hence MATH, with the second equality following from the induction hypothesis. |
math/9908127 | The lemma follows from the special case MATH, since the MATH of a sequence is the same as the MATH of any tail of that sequence. The proof is by induction on MATH, using REF . The claim is trivially true in the base cases. Suppose that MATH. CASE: If MATH is h.p., then for all MATH, MATH is h.p., so MATH is h.p. by the induction hypothesis. CASE: If MATH is h.p., then for all MATH we have MATH, so MATH. CASE: If MATH are h.p., then for all MATH we have MATH, so MATH. MATH . |
math/9908127 | The proof is by induction on MATH. If MATH, then MATH is the identity function, which is easily seen to be h.p. Suppose that MATH. First we must show that if MATH is h.p., then so is MATH, using the fact that MATH is h.p. by the induction hypothesis. CASE: If MATH is h.p., then MATH is h.p.because MATH is h.p. by the induction hypothesis and MATH maps h.p.functionals to h.p. functionals by assumption. CASE: If MATH is h.p., then MATH, so MATH. The first inequality follows from the fact that MATH is hereditarily inflationary, the second from the fact that MATH is hereditarily monotone. CASE: If MATH are h.p., then MATH. The second inequality follows from the facts that MATH is h.p. and MATH is hereditarily monotone. To show that MATH is hereditarily inflationary, it suffices to show that if MATH and MATH are h.p., then MATH, which we did above. To show that MATH is hereditarily monotone, fix MATH and MATH and note that MATH, repeatedly using the induction hypothesis and hereditary monotonicity of h.p. functionals. If MATH is a limit, then MATH is h.p. by REF because MATH is h.p. for all MATH by the inductive hypothesis. |
math/9908127 | The proposition is proved by induction on MATH for all MATH. Note that it is true when MATH, even though MATH is not itself hereditarily positive. If MATH, then the claim is vacuously true. Suppose that MATH and fix any MATH. By the induction hypothesis MATH, so it suffices to show that MATH. To do so, fix h.p. functionals MATH and MATH. Since MATH and MATH are h.p. (notice that this is true even when MATH, since then MATH), MATH. Since MATH and MATH were chosen arbitrarily, MATH. Suppose that MATH is a limit and fix any MATH. By the induction hypothesis the sequence MATH is non-decreasing with respect to MATH. Thus, by REF , MATH. Since MATH, there is some MATH such that MATH, which, by the induction hypothesis applied to MATH, implies that MATH. |
math/9908127 | By definition, MATH. Since any set of ordinals attains its infimum, there is some MATH such that MATH. |
math/9908127 | By the choice of MATH, we have MATH. For the reverse inequality, fix any MATH such that MATH; then MATH by the choice of MATH. So for any MATH, MATH, and therefore there is some MATH such that MATH, which by REF implies that MATH. Keeping in mind that MATH is fixed while MATH was chosen arbitrarily, MATH; the final inequality follows from REF . Since MATH was chosen arbitrarily between MATH and MATH, this implies that MATH. |
math/9908127 | Fix MATH as in REF and set MATH. First, suppose that for all MATH there is MATH such that MATH. Then since MATH for all MATH, MATH. On the other hand, MATH and MATH, so MATH, and therefore MATH. If there is some MATH such that MATH for all MATH, then we can still conclude that MATH. Note that in this situation, MATH must be a limit. To show that the reverse inequality holds, fix any MATH; then there is MATH such that MATH, so by REF MATH. But since MATH was chosen arbitrarily, this implies that MATH. The inequality follows from REF by considering MATH as a single h.p. functional bounding all of the MATH. |
math/9908127 | Each part is proved by induction on MATH; we do REF as an example. If MATH and MATH is h.p., then MATH. If MATH, then MATH. The second equality is the induction hypothesis and the third is an application of REF . If MATH is a limit, then MATH, with the middle equality following from the induction hypothesis and the last one by REF . |
math/9908131 | The if direction is given by calculation REF. (only if): Replacing MATH with MATH in the defining REF shows that in an NAME sequence, each polynomial MATH can be recovered from the sequence of values MATH. Choosing an umbra MATH that represents this sequence guarantees MATH. |
math/9908131 | This is equivalent to the observation that if MATH is in MATH, the ring of formal power series in MATH, and if MATH, then MATH and the coefficient of MATH in the last is a polynomial in MATH. |
math/9908131 | By definition, and the substitution lemma REF , it suffices to prove that MATH for any distinct umbrae MATH, that is, that the MATH-th powers of each side of the displayed equation are umbrally equivalent for all MATH. Letting MATH be the polynomials from REF , it suffices to show, for all MATH, that the identity MATH holds purely on the level of polynomials in variables MATH. But this follows since REF says this identity holds with MATH replaced by any pair of positive integers. |
math/9908131 | The result holds when MATH is any integer. Repeating the argument in the proof of REF shows the identity holds when interpreted in terms of polynomials in MATH. |
math/9908131 | It suffices to check that MATH. If MATH are given by MATH, MATH, and MATH, then this amounts to observing that each side is umbrally equivalent to the composition MATH. |
math/9908131 | By the remarks preceding the theorem, it suffices to show that any sequence of binomial type can be so represented. By standard results, which are briefly sketched below, it suffices to show that choosing MATH appropriately allows us to choose the sequence MATH arbitrarily. It is enough to observe that REF tells us that the coefficient of MATH in MATH is MATH where MATH is depends only on MATH and MATH. |
math/9908131 | To start with, assume that MATH is always an integer and that MATH are distinct umbrae all exchangeable with MATH. We start by interpreting MATH as a generating function for sequences of length MATH on MATH symbols. By the NAME correspondence (see for example CITE) this is a generating function for the number of labeled free trees on MATH vertices where each tree is counted with weight MATH where vertex MATH has degree MATH. So MATH is the generating function for labeled rooted trees on MATH vertices where the same weight indicates that vertex MATH has outdegree MATH. This says that MATH is the generating function where the coefficient of MATH counts the number of labeled trees on MATH vertices where the root has degree MATH and each non-root vertex with outdegree MATH can be colored in any of MATH ways. Equivalently, MATH counts the number of planted forests on MATH vertices where each vertex with outdegree MATH can be colored in any of MATH ways and where each tree in the forest can itself be colored in any of MATH ways. Let's call this structure a MATH degree-colored forest on MATH vertices. So counting the number of ways to form a MATH degree-colored forest on MATH vertex by the number of vertices, MATH, in the trees which were colored in one of the first MATH ways gives MATH . This fact for all positive integers MATH implies REF as a polynomial identity. To see that indeed all normalized sequences of binomial type arise in this fashion, it suffices, by the remarks after REF , to observe that the sequence MATH can be chosen arbitrarily. Indeed MATH where MATH is a linear combination of MATH. |
math/9908131 | If MATH is monic or MATH, the result follows immediately from the preceding proposition and the remarks after REF . If the sequence is of binomial type and MATH, then so is the sequence whose MATH-th term is MATH; if MATH is the series associated to this new sequence, then MATH works for the original. The converse follows similarly. |
math/9908131 | We start by showing that all such presentations are of binomial type. By induction we have that MATH . As before, all normalized sequences of binomial type arise in this fashion, since the sequence MATH can be chosen arbitrarily. Observe that MATH where each term of MATH has degree less than MATH. |
math/9908131 | Evaluating MATH on MATH, we find MATH . Here the last line follows from the substitution lemma by replacing MATH with MATH. |
math/9908131 | It suffices to observe that we can construct such a tree by specifying a function mapping each vertex to its parent and then choosing colors. Choosing MATH from the MATH-th multiplicand corresponds to requiring the function to map vertex MATH to vertex MATH. Choosing MATH of course indicates that vertex MATH is a root. |
math/9908135 | Clearly stably isomorphic bundle gerbes have the same NAME class because trivial bundles have the zero NAME class and the NAME class is additive over tensor products. So REF implies REF . If MATH then MATH. Hence MATH is trivial CITE. So REF implies REF . Finally if MATH is trivial then MATH is also trivial as it has zero NAME class and then MATH so MATH and MATH are stably isomorphic. So REF implies REF . |
math/9908135 | This is clear because they have the same NAME class but it is instructive to also show that MATH is trivial. To see this note that the fibre of MATH at a point MATH in MATH is MATH . Whereas if we define MATH over MATH by MATH then MATH at MATH is MATH . The gerbe multiplication can be used to define an isomorphism MATH . |
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