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math/9908073 | The implication MATH follows from CITE. Let us prove the implication MATH . According to REF it suffices to construct a MATH-invertible map MATH where MATH is a metrizable compactum with MATH. Since MATH is finitely dominated, there exist a finite complex MATH and two maps MATH and MATH such that MATH. By CITE there exists a compactum MATH and a map MATH with the following properties: CASE: For each map MATH, defined on a separable metrizable space with MATH, there exists a map MATH such that MATH. CASE: For each map MATH, where MATH is a closed subset of MATH, there exists a map MATH such that MATH. Observe that, by REF , MATH. Consequently, by the Homotopy Extension Theorem, the map MATH has a continuous extension over MATH. This in turn means that MATH. By REF and CITE, the map MATH is MATH-invertible. |
math/9908073 | The implication MATH follows from CITE. In order to prove the implication MATH it suffices to show that MATH for each normal space MATH with MATH (see REF ). Let MATH be a map defined on a closed subset MATH of MATH. Since MATH is a MATH, there exists an extension MATH of MATH, where MATH is an open neighborhood of MATH in MATH. Clearly MATH and MATH. Since MATH there exists a map MATH such that MATH. An argument similar to CITE shows that MATH is homotopic to a map MATH such that MATH is compact. Consequently MATH has a continuous extension MATH onto the whole MATH. Now consider the two maps MATH and MATH. Their restrictions MATH and MATH are homotopic. Since MATH is normal it follows that MATH. By CITE, the restriction operator provides a bijection of homotopy classes MATH and MATH. Consequently, MATH. By the homotopy extension theorem (recall that MATH has an extension MATH onto MATH and that MATH is a MATH), MATH also has an extension onto MATH which serves as an extension of the originally given map MATH. |
math/9908074 | Let MATH be a map of a closed subset MATH. Since MATH is metrizable, MATH is compact. Hence MATH is contained in some finite subcomplex MATH. But every finite complex is a MATH for any normal space. Thus, there is an extension MATH of MATH defined on an open neighborhood MATH of MATH in MATH. |
math/9908074 | MATH. By REF , we need to check that for every closed set MATH any map MATH has an extension MATH. By REF , there is an extension MATH, where MATH is a neighborhood of MATH in MATH. Let MATH be a smaller neighborhood of MATH in MATH such that MATH. Set MATH and let MATH. By REF , there is an extension MATH. As in the proof of REF , MATH is contained in some finite complex MATH. But as was noted above MATH is compact. Hence the map MATH can be extended to a map MATH. It remains to show that MATH. But MATH. Hence, since MATH is dense in MATH, we have MATH. On the other hand, MATH. MATH. Let MATH be a closed subset of MATH and let MATH be a map. Set MATH. Since MATH is closed in a normal space MATH, MATH. Then MATH can be extended to a map MATH because MATH lies in some finite complex MATH. Now, by REF, the map MATH can be extended to a map MATH. It only remains to note that the map MATH extends MATH. REF is proved. |
math/9908074 | By REF , we need to verify that for every closed set MATH any map MATH has an extension MATH. Let MATH. Since MATH, the map MATH can be extended to a map MATH. Define the map MATH by letting MATH and MATH. Clearly, MATH is continuous. By REF , the map MATH has an extension MATH, where MATH is a neighborhood of MATH in MATH. Take a neighborhood MATH of MATH such that MATH and let MATH, MATH, MATH. Then MATH is closed in MATH and MATH is closed in MATH. By REF. Hence, the map MATH has an extension MATH. Finally, define a map MATH by letting MATH . Evidently, MATH is well defined and continuous. It is also clear that MATH. REF is proved. |
math/9908074 | We use the scheme of the proof of CITE, where a similar result was obtained for countable complexes. The only difference is that in our situation we can not apply auxiliary results for countable complexes which were used in CITE. Consider a complex MATH. By definition of MATH, there is a metrizable compactum MATH such that MATH . Set MATH, where MATH is a manifold from REF . We claim that this is a required manifold. First of all, MATH is countably compact. Hence, in view of REF , MATH . Further, by REF , we have MATH . Now we apply REF to the pair MATH. Since MATH is open in MATH, REF yields MATH . Finally, let us apply REF to the pair MATH. Since MATH for any compactum MATH, by REF , we obtain MATH . REF yield MATH . Thus, equality REF finishes the proof of REF . |
math/9908074 | Because each manifold is a MATH-space (being first countable) it follows from CITE that MATH is countably compact. Hence, by NAME 's theorem CITE, MATH is pseudocompact and MATH. By REF , MATH. So we have to find out the exact value of MATH. Let MATH, MATH and MATH. By NAME 's theorem CITE, MATH where MATH . It is clear, that each MATH from REF is contained in some MATH, where MATH is a finite sum of MATH-dimensional cubes, MATH. By NAME 's theorem CITE, MATH whenever MATH is a paracompact space and MATH is a compact polyhedron. By the finite sum theorem for MATH, REF yields MATH . Consequently, MATH . Equality REF finishes the proof of REF . |
math/9908074 | According to REF , we only need to check that MATH. But for any compact spaces MATH and MATH we have (see CITE) MATH . Hence, MATH . REF is proved. |
math/9908074 | Let MATH be a two-dimensional metrizable compactum such that MATH. Such a compactum was constructed by CITE. Let MATH. Then in accordance with REF , MATH . Let MATH be a manifold from REF . We claim that MATH is a required manifold. Indeed, by the properties of MATH, the set MATH is homeomorphic to an open subset of MATH. Consequently, REF yield MATH. In this situation REF finishes the proof of REF . |
math/9908074 | We follow the proof of REF . First let us recall the following result CITE. CASE: For all natural numbers MATH and MATH such that MATH and MATH, there exist metrizable compacta MATH and MATH such that MATH and MATH. Set MATH, where MATH and MATH are the above mentioned compacta with MATH and MATH satisfying inequalities MATH. From REF we get MATH. On the other hand, for MATH from REF we have MATH. In view of REF we have MATH. REF is proved. |
math/9908074 | Let MATH, where MATH is a manifold from REF and MATH is well known NAME 's infinite-dimensional compactum with no positive-dimensional compact subsets CITE. By REF , MATH. Now let MATH be a closed subset of MATH such that MATH. Then in view of REF , MATH . But, since MATH is normal, MATH. Set MATH . By REF for the pair MATH, we have MATH. But in accordance with REF , MATH is a disjoint sum of MATH and some open subset of MATH. Hence, MATH . Therefore, by REF 's compactum, MATH. Consequently, MATH . REF is proved. |
math/9908074 | By REF , there is a metrizable compactum MATH of extension dimension MATH. Let MATH be the NAME compactification of the discrete sum MATH of these compacta. Now we define MATH as a manifold MATH from REF . Since MATH is a discrete family in a compact space MATH, there is a disjoint family of neighborhoods MATH of MATH in MATH. We may assume also, that MATH . Now, in REF , we set MATH . To realize REF we take an open metrizable subset MATH and set MATH . Claim. MATH. Proof. By REF , it suffices to verify that MATH . But MATH . On the other hand, according to REF , MATH is dense in MATH. Hence, MATH . Let MATH and let MATH. Since MATH is dense in MATH, we have MATH . Hence, MATH . For every compactum MATH we have MATH. On the other hand, MATH and MATH whenever MATH. Hence, MATH. Now we apply REF to the pair MATH. We have MATH . Finally, REF give us the required equality REF . This finishes proof of our Claim. In order to prove the equality MATH we need more general version of REF . Its proof is based on the fact that a countable simplicial complex is a MATH for the class of all normal spaces. CASE: Suppose that a normal space MATH is the union of its closed countably compact subsets MATH, MATH. If MATH for each MATH, then MATH. In order to finish the proof of REF represent MATH as the union of an increasing sequence MATH and apply our claim. REF is proved. |
math/9908075 | Since the Abelian group MATH is a metrizable MATH-compactum, it is a torus group (the equivalence MATH of Theorem A). Since torus groups are injective objects in the category of compact Abelian groups (the equivalence MATH of Theorem A), there exists a continuous retraction MATH which is a homomorphism. Obviously, there is an isomorphism MATH (here is the explicit formula: MATH, MATH). Now observe that MATH is an isomorphism. Consequently, the following diagram MATH commutes. This proves the lemma. |
math/9908075 | MATH. Note that MATH is a MATH-compactum. MATH. Apply a straightforward transfinite induction to conclude that MATH is topologically and algebraically equivalent to the product MATH. REF show that such a product is isomorphic to the torus group MATH. MATH. We may assume that MATH is a closed subgroup of a product MATH, MATH, of compact metrizable Abelian groups MATH, MATH, each of which is connected and locally connected (clearly each of this MATH's can be chosen to be the circle group MATH but in order to avoid a confusion we deliberately use a different notation). By CITE, there exists an open MATH-invertible map MATH, where MATH is a MATH-dimensional compactum of weight MATH. Consider the inverse image MATH of MATH and the map MATH. Since MATH and since MATH, according to REF , is a MATH-compactum, there exists a map MATH such that MATH. Next let us denote by MATH and MATH the natural projections onto the corresponding subproducts (MATH). We call a subset MATH admissible (compare with the proof of CITE) if the following equality MATH is true for each point MATH. We need the following properties of admissible sets. CASE: The union of arbitrary collection of admissible sets is admissible. Indeed let MATH be a collection of admissible sets and MATH. Let MATH. Clearly MATH for each MATH and consequently MATH . Obviously, MATH and it therefore suffices to show that the set MATH contains only one point. Assuming that there is a point MATH such that MATH we conclude (having in mind that MATH) that there must be an index MATH such that MATH. But this is impossible MATH . CASE: If MATH is admissible, then the restriction MATH is MATH-soft. Let MATH and MATH be two maps defined on a MATH-dimensional compactum MATH and its closed subset MATH respectively. Assume that MATH. We wish to construct a map MATH such that MATH and MATH, that is, MATH makes the diagram commutative. Since, according to our choice, all MATH's are MATH-compacta, so is the product MATH. This implies the MATH-softness of the projection MATH and hence of its restriction MATH . Then there exists a map MATH such that MATH and MATH. Since MATH is MATH-invertible (and MATH), there exists a map MATH such that MATH. Now let MATH. Since MATH, we have MATH. Finally observe that the admissibility of MATH implies MATH as required. CASE: For each countable subset MATH there exists a countable admissible subset MATH such that MATH. Since MATH, it follows (consult CITE) that MATH can be represented as the limit space of a MATH-spectrum MATH consisting of metrizable compacta MATH, MATH, and continuous surjections MATH, MATH, MATH. Consider also the standard MATH-spectrum MATH consisting of countable subproducts of the product MATH and corresponding natural projections. Obviously the full product coincides with the limit of MATH. One more MATH-spectrum arises naturally. This is the spectrum MATH the limit of which coincides with MATH. Consider the map MATH. Applying the Spectral Theorem CITE there is a cofinal and MATH-closed subset MATH of MATH such that for each MATH there is a map MATH such that MATH. Moreover, these maps form a morphism MATH limit of which coincides with MATH. Since MATH is open (and closed), we may assume without loss of generality (considering a smaller cofinal and MATH-subset of MATH if necessary) that the above indicated morphism is bicommutative. This simply means that MATH for any MATH and any closed subset MATH of the product MATH. Similarly applying the Spectral Theorem to the map MATH we obtain a cofinal and MATH-closed subset MATH of MATH and the associated to it morphism MATH limit of which coincides with the map MATH. By CITE, the intersection MATH is still a cofinal and MATH-closed subset of MATH. It therefore suffices to show that each MATH is an admissible subset of MATH. Consider a point MATH. First observe that bicommutativity of the morphism associated with MATH implies that MATH. Since the maps MATH and MATH coincide on MATH we have MATH as required. CASE: If MATH and MATH are admissible subsets of MATH and MATH, then the map MATH is MATH-soft. This property follows from REF and CITE. After having all the needed properties of admissible subsets established we proceed as follows. Since MATH we can write MATH. By REF , each MATH is contained in a countable admissible subset MATH. Let MATH. We use these sets to define a transfinite inverse spectrum MATH as follows. Let MATH and MATH for each MATH. REF of the spectrum MATH are satisfied by construction. Since, MATH is a metrizable MATH-compactum, it follows from the equivalence MATH of Theorem A, that MATH is a metrizable torus. Finally, REF is a consequence of REF . Proof is completed. |
math/9908075 | By REF , we may without loss of generality assume that MATH is the product MATH and the homomorphism MATH coincides with the projection MATH. Now consider the following commutative diagram (compare it with REF ): in which MATH is the characteristic map of the previous diagram CITE. We therefore need to show the MATH-softness of MATH. A straightforward verification shows that in this situation MATH is isomorphic to the product MATH and the homomorphism MATH coincides with the product MATH. Observe also that with these identifications MATH becomes isomorphic to the projection MATH. Since MATH is a metrizable MATH-compactum, the map MATH is an open surjection between completely metrizable separable spaces (equivalence MATH). By CITE, MATH is MATH-soft. This implies MATH-softness of MATH. |
math/9908075 | Since the statement is known to be true in the metrizable case we assume that MATH. MATH. Represent MATH as the limit of a well-ordered inverse spectrum MATH with the properties listed in REF . The space MATH is then the limit of the associated spectrum MATH and the map MATH is the limit of the morphism MATH . According to REF this morphism is MATH-soft. MATH-softness of the limit map MATH follows now from CITE (applied with MATH). MATH. The proof of this part follows the proof of CITE. Only two things have to be taken into account. The first has already been mentioned: the space MATH is isomorphic to MATH and hence is a MATH-space in the sense of CITE. The second: MATH being a compact group is also a MATH-compactum (Theorem C). Following the indicated proof and making slight adjustments (as in the proof of REF ) we arrive to the following situation. There exists a well-ordered spectrum MATH, satisfying REF and a part of REF which guarantees that the fiber MATH is metrizable. Also there exists a morphism (we keep notation of the first part of the proof) MATH between this spectra which is MATH-soft in the sense that characteristic maps of all the square diagrams MATH are MATH-soft. In addition, it follows from the above remarks that typical projection MATH is isomorphic to the projection MATH. Since the characteristic map of the above diagram is MATH-soft, it is surjective CITE. This means that MATH for any point MATH. Apply the latter to the point MATH. We have MATH. Since, as was noted, the map MATH is isomorphic to the projection MATH, it follows that MATH is isomorphic to MATH. Consequently a metrizable compact Abelian group MATH being a surjective image of MATH (through the map MATH) is arcwise connected. Equivalence MATH applied to MATH shows that the latter is a metrizable torus. This finishes verification of REF . Since MATH is obviously a metrizable torus, it follows REF that MATH is a MATH-compactum. Proof is completed. |
math/9908075 | Apply the Spectral Theorem for MATH-spectra of compact Abelian groups (see CITE where the non Abelian and non compact version is proved, consult also CITE), note that MATH-smooth decompositions are nothing else but MATH-spectra and apply duality. |
math/9908075 | Let MATH be a subgroup of an Abelian group MATH. If MATH is free it has a smooth decomposition MATH with the properties indicated in REF . Then MATH is a smooth decomposition of MATH. Since MATH and MATH, MATH, we conclude (by REF ) that MATH satisfies conditions of REF and hence MATH is free. |
math/9908076 | Implications REF are trivial. CASE: It follows from the proof of Theorem B CITE that there exists a continuous homomorphism MATH where MATH is the commutator subgroup of MATH, MATH is the connected component of the center of MATH, and MATH is a zero-dimensional compact group isomorphic to the (closed and central) subgroup MATH of MATH. Further, there exists a continuous homomorphism MATH where each MATH, MATH, is a simple, connected and simply connected compact NAME group, and MATH is zero-dimensional. Obviously, the kernel of the homomorphism MATH, defined as the composition MATH is zero-dimensional. Let us show that MATH. Indeed, assuming that MATH and remembering that MATH, we can find a compact group MATH and two continuous homomorphisms MATH such that MATH, MATH is finite and MATH. Clearly MATH is a bundle (since MATH is a finite group) and MATH, as a MATH-compactum, is arcwise connected and locally arcwise connected. Also, by REF , MATH. Consequently, by CITE, MATH is a trivial bundle, that is, MATH. Now observe that MATH, as a continuous image of MATH, is connected. This implies that MATH is a singleton which is impossible. This contradiction shows that MATH and hence MATH. From this we conclude that MATH, as a retract of MATH, is a MATH-compactum with MATH. An abelian group which is a MATH-compactum is a torus (Theorem B). But the only simply connected torus is the singleton. Thus MATH and MATH. CASE: Suppose that MATH is a product MATH of simple, connected and simply connected compact NAME groups. By Theorem of CITE, MATH, MATH. By the local contractibility of MATH, this means that MATH is a MATH-compactum, MATH. Then MATH, as a product of MATH-compacta, is a MATH-compactum. The proof is complete. |
math/9908076 | CASE: By Theorem C, MATH is isomorphic to the product MATH of simple, connected and simply connected compact NAME groups. Let MATH. By REF and a result of CITE, we then have MATH . This obviously is possible precisely when MATH which implies that MATH is isomorphic to some MATH and hence is a simple, connected and simply connected compact NAME group. CASE: Every connected and simply connected NAME group is a MATH-space. Since MATH is simply connected, it is non abelian. Since MATH is simple, by the above cited result of NAME, MATH. |
math/9908076 | Let MATH be a compact MATH-group. Since every MATH-compactum is a MATH-compactum, it follows from Theorem C, that MATH is a product MATH of simple, connected and simply connected compact NAME groups. Since each retract of a MATH-compactum is a MATH-compactum, it follows that MATH is a MATH-compactum, MATH. Then MATH for each MATH. This contradicts REF and completes the proof. |
math/9908076 | Let MATH be a locally compact MATH-group. By CITE, MATH is homeomorphic to the product MATH for some integer MATH, a cardinal number MATH and a compact group MATH. Since each retract of a MATH-space is a MATH-space, we conclude that MATH is finite. Therefore it suffices to show that MATH is metrizable. By CITE, MATH is isomorphic to the limit of a MATH-spectrum MATH consisting of metrizable compact groups MATH and continuous homomorphisms MATH. Observe that MATH, as a retract of MATH, is a MATH-compactum. Consequently, by CITE, there exists a cofinal (in particular, non-empty) subset MATH of the indexing set MATH such that the limit projection MATH is MATH-soft for each MATH. Then, by CITE, the fiber MATH is a compact MATH-group. By REF , MATH. Therefore MATH is an isomorphism. Since MATH is metrizable, so is MATH. |
math/9908077 | Indeed, we only have to note that MATH, together with its identity map MATH, forms the projective system MATH. So the collection MATH is in fact a morphism MATH. The rest follows from the definitions given above. |
math/9908077 | The continuity of MATH and the definition of the topology on MATH guarantee that there exists an index MATH and an open subset MATH such that MATH where MATH denotes the norm of the NAME space MATH. Since MATH is an NAME algebra, MATH can be identified with a closed subalgebra of the product MATH of NAME algebras MATH, MATH REF . Let MATH denote the norm of the NAME space MATH, MATH. For MATH, let MATH denote the natural projection onto the corresponding subproduct. If MATH is a finite subset of MATH, then MATH for each MATH. Since MATH is open in MATH, the definition of the product topology guarantees the existence of a finite subset MATH and of a number MATH such that MATH . Combining REF , we have MATH . It then follows that if MATH and MATH, then MATH and consequently MATH . Since MATH we must have (by REF) MATH and hence MATH . Let us now show that the map MATH is well defined. Assuming the contrary, suppose that for some MATH there exist two points MATH such that MATH. Consequently, MATH. On the other hand, MATH. Then REF implies that MATH . This contradiction shows that the map MATH is indeed well defined. Note that MATH which implies that the map MATH is linear. Next consider points MATH and MATH such that MATH. By REF , MATH . This shows that MATH is bounded and, consequently, continuous. Next let us show that MATH is multiplicative. Let MATH and consider a point MATH such that MATH and MATH. Then, by REF , MATH . Since MATH is complete, MATH admits the linear continuous extension MATH . Since the multiplication on MATH is jointly continuous, we conclude that MATH is also multplicative. Finally, define the map MATH as the composition MATH . Obviously MATH is a continuous homomorphism satisfying the required equality MATH. |
math/9908077 | By REF , MATH can be identified with a closed subalgebra of the product MATH of some collection of NAME algebras. If MATH, then MATH itself is a NAME algebra and therefore our statement is trivially true. If MATH, then consider the set MATH of all countable subsets of MATH. Clearly, MATH is MATH-complete set (see Subsection REF). For each MATH, let MATH (closure is taken in MATH), where MATH denotes the natural projection onto the corresponding subproduct. Also let MATH where MATH is the natural projection, MATH, MATH. The straightforward verification shows that MATH is indeed a projective NAME system such that MATH. Conversely, let MATH be a projective NAME system. Clearly, MATH can be identified with a closed subalgebra of the product MATH (see Subsection REF). Each MATH, MATH, can obviously be identified with a closed subalgebra of the product MATH of a countable collection of NAME algebras MATH. Then MATH, as a closed subalgebra of the product MATH is, according to REF , an NAME algebra. |
math/9908077 | We perform the spectral search by means of the following relation MATH . Let us verify the conditions of REF . Existence Condition. By assumption, MATH is a NAME algebra. Therefore MATH can be identified with a closed subspace of a countable product MATH of NAME algebras. Let MATH denote the MATH-th natural projection. For each MATH, by REF , there exist an index MATH and a continuous homomorphism MATH such that MATH. By REF , there exists an index MATH such that MATH for each MATH. Without loss of generality we may assume that MATH. Let MATH, MATH. Next consider the diagonal product MATH . Obviously, MATH. It only remains to show that MATH. First observe that MATH. Indeed, let MATH. Then MATH . Finally, MATH . NAME Condition. The verification of this condition is trivial. Indeed, it suffiecs to consider the composition MATH. MATH-closeness Condition. Suppose that for some countable chain MATH in MATH with MATH and for some MATH with MATH, the maps MATH have already been defined is such a way that MATH for each MATH (in other words, MATH for each MATH). Next consider the composition MATH where MATH is the topological isomorphism indicated in REF . Observe that for each MATH . This shows that MATH and finishes the verification of the MATH-closeness condition. Now denote by MATH the set of all MATH-reflexive elements in MATH. By REF , MATH is a cofinal and MATH-closed subset of MATH. One can easily see that the MATH-reflexivity of an element MATH is equivalent to the existence of a continuous homomorphism MATH satisfying the equality MATH. Consequently, the collection MATH is a morphism of the cofinal and MATH-closed subspectrum MATH of the spectrum MATH into the cofinal and MATH-closed subspectrum MATH of the spectrum MATH. It only remains to remark that the original map MATH is induced by the constructed morphism. This finishes the proof of the first part of our Theorem. The second part of this theorem can be obtained from the first as follows. Let MATH be a topological isomorphism. Denote by MATH the inverse of MATH. By the first part proved above, there exist a cofinal and MATH-closed subset MATH of MATH and a morphism MATH such that MATH. Similarly, there exist a cofinal and MATH-closed subset MATH of MATH and a morphism MATH such that MATH. By REF , the set MATH is still cofinal and MATH-closed in MATH. Therefore, in order to complete the proof, it suffices to show that for each MATH the map MATH is a topological isomorphism. Indeed, take a point MATH. Also choose a point MATH such that MATH. Then MATH . This proves that MATH. Similar considerations show that MATH for each MATH. Since MATH is dense in MATH and MATH is dense in MATH REF , it follows that MATH and MATH. It is now clear that MATH, MATH, is a topological isomorphism (whose inverse is MATH). |
math/9908077 | Consider a trivial projective NAME system MATH, where MATH and MATH for each MATH. By REF (applied to the identity homomorphism MATH), there exist an index MATH and a continuous homomorphism MATH such that MATH. Clearly, in this situation, MATH is an embedding with a closed image. But this image MATH is dense in MATH REF . Therefore MATH (and, consequently MATH for any MATH) is a topological isomorphism. |
math/9908077 | By REF , there exists a cofinal and MATH-closed subset MATH of MATH and a morphism MATH consisting of continuous homomorphisms MATH, MATH, such that MATH. Let us show that MATH, MATH is actually a MATH-homomorphism. Indeed, let MATH and MATH such that MATH. Then MATH . |
math/9908077 | If MATH, then MATH . This shows that the formula MATH defines a continuous linear map MATH. Moreover, MATH is a topological isomorphism between MATH and MATH considered as topological vector spaces (to see this observe that the continuous and linear map MATH defined by letting MATH for each MATH, has the following properties: MATH and MATH). We now show that MATH is an isomorophism of the category of topological algebras as well. Let MATH. We need to show that MATH. Since MATH we can write MATH, where MATH and MATH, MATH. Observe that since MATH it follows from REF that MATH. Consequently, by REF , MATH. Then MATH . Consequently, MATH . This shows that MATH is a homomorphism and, consequently, a topological isomorphism as required. |
math/9908077 | Let MATH be a complemented subalgebra of the uncountable product MATH of NAME algebras MATH, MATH, where MATH is an indexing set with MATH. There exists a continuous homomorphism MATH such that MATH for each MATH. A subset MATH is called MATH-admissible if MATH for each point MATH. CASE: The union of an arbitrary family of MATH-admissible sets is MATH-admissible. Let MATH be a collection of MATH-admissible sets and MATH. Let MATH. Clearly MATH for each MATH and consequently MATH for each MATH. Obviously, MATH and it therefore suffices to show that the set MATH contains only one point. Assuming that there is a point MATH such that MATH we conclude (remembering that MATH) that there must exist an index MATH such that MATH. But this is impossible MATH . CASE: If MATH is MATH-admissible, then MATH is a closed subalgebra of MATH. Indeed, let MATH be the canonical section of MATH (this means that MATH). Consider a continuous linear map MATH. Obviously, MATH for any point MATH. Since MATH is MATH-admissible the latter implies that MATH . This shows that MATH is closed in MATH. CASE: Let MATH and MATH be MATH-admissible subsets of MATH and MATH. Then the map MATH is topologically isomorphic to the natural projection MATH. Obviously MATH. Consider the map MATH. Also let MATH. Observe that MATH. Indeed, if MATH, then MATH. Since MATH is MATH-admissible, we have MATH. Consequently, MATH . Next observe that MATH for any point MATH. Indeed, since MATH is MATH-admissible and since MATH we have MATH . Application of REF (with MATH, MATH, MATH, MATH and MATH) finishes the proof of REF . Every countable subset of MATH is contained in a countable MATH-admissible subset of MATH. Let MATH be a countable subset of MATH. Our goal is to find a countable MATH-admissible subset MATH such that MATH. By REF , there exist a countable subset MATH of MATH and a continuous homomorphism MATH such that MATH and MATH. Consider a point MATH. Also pick a point MATH such that MATH. Then MATH . This shows that MATH. It also follows that MATH is closed in MATH. In order to show that MATH is MATH-admissible let us consider a point MATH. By the observation made above, MATH. Finally MATH which implies that MATH is MATH-admissible. Since MATH, we can write MATH. Since the collection of countable MATH-admissible subsets of MATH is cofinal in MATH (see REF ), each element MATH is contained in a countable MATH-admissible subset MATH. According to REF , the set MATH is MATH-admissible for each MATH. Consider the projective system MATH where MATH . Since MATH, it follows that MATH. Obvious transfinite induction based on REF shows that MATH . Since, by the construction, MATH is a countable MATH-admissible subset of MATH, it follows from REF that MATH and MATH, MATH, are NAME algebras. This finishes the proof of REF . |
math/9908077 | The implication MATH is trivial. MATH. The NAME space MATH can be identified with a closed linear subspace of the space MATH for some set MATH. By REF , there exists a linear continuous map MATH such that MATH for each MATH. This obviously implies that MATH is a complemented subspace of MATH. MATH. First let us show that the NAME space MATH (for any set MATH) is an injective object of the category MATH. Let MATH where MATH, MATH, stands for a copy of MATH, be a continuous linear map defined on a closed linear subspace MATH of a NAME space MATH. Since MATH is bounded, for each MATH we have MATH . This shows that MATH, where MATH denotes the canonical projection onto the MATH-th coordinate. By the NAME Theorem, the linear map MATH, MATH, admits a continuous linear extension MATH such that MATH. Consequently, MATH . This shows that the map MATH, given by letting MATH is well defined. Obviously MATH is continuous, linear and extends the map MATH. Therefore MATH is indeed an injective object of the category MATH. Next consider a complete locally convex topological vector space MATH, its closed linear subspace MATH and a continuous linear map MATH, where MATH is a complemented subspace of the NAME space MATH for some set MATH. We need to show that MATH admits a continuous linear extension MATH. Identify MATH with a closed linear subspace of the product MATH of NAME spaces MATH, MATH. Clearly MATH is a closed linear subspace of MATH. By REF , there exist a finite subset MATH and a continuous linear map MATH such that MATH. Clearly MATH is a closed linear subspace of the NAME space MATH. The first part of the proof of this implication, coupled with condition MATH, implies that MATH is an injective object of the category MATH. Therefore the map MATH admits a continuous linear extension MATH. It only remains to note that the map MATH is a continuous linear extension of MATH. |
math/9908077 | MATH. By REF , MATH is an injective object of the category MATH for any set MATH. Obviously the product of an arbitrary collection of injective objects of the category MATH is also an injective object of this category. Consequently, the NAME space MATH, MATH, as a complemented subspace of MATH is injective. Finally, the space MATH, as a product of injectives, is an injective object of the category MATH. MATH. The space MATH can be identified with a closed linear subspace of the product MATH of NAME spaces MATH, MATH. Each of the spaces MATH can in turn be identified with a closed linear subspace of the space MATH for some set MATH, MATH. REF implies in this situation that MATH is a complemented subspace of the product MATH. The required conclusion now follows from REF . |
math/9908077 | Consider the following relation MATH on the set MATH: MATH . Next we perform the spectral search with respect to MATH. Existence Condition. We have to show that for each MATH there exists MATH such that MATH. We begin with the following observation. Claim. For each MATH there exists a finite subset MATH such that MATH. The unit ball MATH (here MATH denotes a norm of the NAME space MATH) being bounded in MATH is, by CITE, bounded in MATH. Continuity of MATH guarantees that MATH is also bounded in MATH. Applying CITE once again, we conclude that there exists a finite subset MATH such that MATH. Finally the linearity of MATH implies that MATH. Let now MATH. For each MATH, according to Claim, there exists a finite subset MATH such that MATH. Let MATH. Observe that MATH. Clearly MATH for each MATH. The linearity of MATH guarantees in this situation that MATH . Finally let MATH. Since, by our assumption, MATH, it follows that MATH. Therefore MATH. NAME Condition. Let MATH, MATH and MATH. REF implies that MATH. The inclusion MATH implies that MATH. Consequently MATH which means that MATH. MATH-closeness Condition. Let MATH, MATH, where MATH is a countable chain in MATH. We need to show that MATH, where MATH. REF implies that MATH, MATH. Also observe that MATH . Consequently, by the linearity of MATH, MATH . By REF , the set of MATH-reflexive elements of MATH is cofinal and MATH-closed in MATH. It only remains to note that an element MATH is MATH-reflexive if and only if MATH. |
math/9908077 | Let MATH be a complemented subspace of an uncountable sum MATH of NAME spaces MATH, MATH. Clearly there exists a continuous linear map MATH such that MATH for each MATH. A subset MATH is called MATH-admissible if MATH. For a subset MATH, let MATH. We state properties of MATH-admissible subsets needed below. CASE: If MATH is a MATH-admissible, then MATH. Indeed, if MATH, then there exists a point MATH such that MATH. Since MATH is MATH-admissible, it follows that MATH . Clearly, MATH. This shows that MATH. Conversely. If MATH, then MATH and hence, by the property of MATH, MATH. Since MATH, it follows that MATH. CASE: The union of an arbitrary collection of MATH-admissible subsets of MATH is MATH-admissible. This is matter of a straightforward verification of the definition of the MATH-admissibility. CASE: Every countable subset of MATH is contained in a countable MATH-admissible subset of MATH. This follows from REF applied to the map MATH. CASE: If MATH is a MATH-admissible subset of MATH, then MATH for each point MATH, where MATH. This follows from the corresponding property of the map MATH. CASE: If MATH and MATH are MATH-admissible subset of MATH and MATH, then MATH is isomorphic to the sum MATH, where the map MATH is defined by letting MATH for each MATH. Observe that MATH for each MATH. CASE: If MATH and MATH are MATH-admissible subsets of MATH and MATH, then there exists a continuous linear map MATH such that MATH. Let MATH. Then MATH and consequently, MATH . This shows that by letting MATH for each MATH, we indeed define a map MATH. Finally observe that MATH . In other words, MATH as required. Let MATH. Then we can write MATH. Since the collection of countable MATH-admissible subsets of MATH is cofinal in MATH (see REF ), each element MATH is contained in a countable MATH-admissible subset MATH. According to REF , the set MATH is MATH-admissible for each MATH. Consider the inductive system MATH where MATH and MATH denotes the natural inclusion for each MATH. For a limit ordinal number MATH the sum MATH is topologically isomorphic to the limit space of the direct system MATH, where MATH is the natural inclusion. This observation, coupled with REF , implies that MATH for each limit ordinal number MATH. In particular MATH is topologically isomorphic to the limit space of the inductive system MATH. For each MATH, according to REF , the inclusion MATH is topologically isomorphic to the inclusion MATH. In this situation the straightforward transfinite induction shows that MATH is topologically isomorphic to the direct sum MATH . Since, by construction, the set MATH is countable, REF guarantees that MATH is a MATH-space. Similarly, since the set MATH is countable, REF guarantees that the space MATH is a MATH-space for each MATH. This completes the proof of REF . |
math/9908081 | Let MATH be a Polish MATH homotopically equivalent to MATH and MATH and MATH be two maps such that MATH is homotopic to MATH and MATH is homotopic to MATH. For extension dimension with respect to MATH this lemma was proved in CITE. So, for a given (complete) separable metric space MATH there is a (complete) separable metric space MATH with MATH and a MATH-soft map MATH. According to next claim, MATH is MATH-soft. Claim. If MATH is normal, then MATH . Suppose MATH is a normal space. Since every Polish MATH is a MATH for normal spaces, MATH is equivalent to MATH. Take a map MATH, where MATH is closed and consider the map MATH. Because MATH can be extended to a map from a neighborhood MATH of MATH into MATH, MATH can be extended to a map from MATH to MATH. Since MATH, there is an extension MATH of MATH. Then the restriction MATH is homotopic to MATH. Finally, using that the Homotopy Extension Theorem holds for normal spaces and Polish MATH's, we conclude that MATH is extendable to a map from MATH into MATH. Hence MATH. It remains only to show that MATH. And this follows from MATH and the fact that the Homotopy Extension Theorem holds for metric spaces and MATH-complexes CITE. |
math/9908081 | It suffices to prove this corollary when MATH is the space MATH. Fix a compatible metric MATH on MATH and an uniformly REF-dimensional surjection (with respect to MATH) MATH with MATH a separable metric space. By REF , there exists a separable metric space MATH with MATH and a MATH-soft map MATH. Let MATH be the fibered product of MATH and MATH with respect to MATH and MATH, and let MATH and MATH denote the corresponding projections of this fibered product. If MATH is any metric on MATH, then MATH is uniformly REF-dimensional with respect to the metric MATH (see CITE). Hence, by REF , MATH. The MATH-softness of MATH implies that MATH is also MATH-soft. It remains only to show that MATH is completely metrizable. To this end let MATH be the space obtained from MATH by making the points of MATH isolated. According to CITE, MATH is paracompact, and obviously, MATH is first countable. Claim. MATH . Let MATH be an arbitrary map, where MATH is closed. Since MATH, there exists an extension MATH of MATH. Now we need the following result CITE: any contractible MATH-complex is an absolute extensor for all spaces admitting a perfect map onto a first countable paracompact space. This statement implies that MATH is an absolute extensor for MATH. Therefore, there exists an extension MATH of MATH. Let MATH, where MATH. Fix a retraction MATH. Since MATH is clopen in MATH, it follows that MATH can be extended to a map MATH. Then MATH is an extension of MATH and, consequently, MATH. Now let us go back to the proof of the completeness of MATH. Considering MATH as a closed subset of MATH and using the fact that MATH is an absolute extensor for paracompact spaces, we can find a map MATH such that MATH. Then, since MATH is MATH-soft and MATH, there exists a retraction from MATH onto MATH. Finally, applying the argument from CITE, we conclude that MATH is complete. |
math/9908081 | If MATH is countable, the proof follows directly from REF with MATH. If MATH is torsion, by CITE, there exists a countable family MATH of countable groups such that MATH for any metrizable space MATH. Then, according to CITE, for each MATH there is a countable complex MATH with MATH if and only if MATH, MATH is any metrizable space. Finally, apply REF to MATH. |
math/9908083 | We have MATH . Also, calculate MATH which was required as well. |
math/9908083 | Calculate, by using the defining relations of a pre-operad: MATH which is the required formula. |
math/9908083 | Use REF . Note that MATH and calculate, MATH which is the required formula. |
math/9908083 | First note that MATH . By transposing the arguments, we have MATH . Now introduce the new summation indices MATH and MATH by MATH . Then we have MATH which proves the required formula. |
math/9908083 | Use the NAME identity REF. |
math/9908083 | Evidently, MATH which is the required formula. |
math/9908083 | Indeed, in MATH the following identity holds: MATH so the NAME identity implies the graded NAME identity: MATH. |
math/9908083 | For MATH calculate MATH . Now note that NAME 's identity REF implies MATH and the NAME identity REF implies MATH. So, superfluous terms cancel out and we obtain MATH which proves the required formula. |
math/9908083 | First use the definitions of MATH and MATH: MATH . Now, note that MATH and use NAME 's identity REF with REF . So it follows that MATH which is the required formula. |
math/9908083 | See REF. |
math/9908083 | See REF. |
math/9908083 | Combine the previous REF with REF . |
math/9908085 | The universal deformation of a spin curve with underlying stable curve in MATH is dependent only upon MATH and MATH; in particular, if MATH and MATH then the forgetful map from the universal deformation of the spin curve to the universal deformation of the underlying curve is of the form MATH (see REF ). Moreover, for MATH the morphism MATH is ramified along MATH to order MATH. Similarly, over MATH, the map MATH is ramified along MATH to order MATH. In either case the proposition follows. |
math/9908085 | The proof is straightforward and very similar to the proof of REF . Only note that MATH lies over MATH if MATH or if MATH. |
math/9908085 | The method of proof is simply to recall from CITE that there are families of stable curves MATH, with MATH a smooth and complete curve, such that the vectors MATH are independent. After suitable base change, one can install a spin structure on the families in question. (This follows, for example, from the fact that, after base change, a spin structure can be installed on the generic fibre of MATH, and since the stack is proper over MATH, this structure can be extended - again after possible base change - to the entire family). Since the effect of base change is to multiply the vectors' entries by a constant, the vectors are still independent, and thus the elements MATH and MATH are all independent in MATH. |
math/9908085 | Again the idea is to install a spin structure on a family of curves MATH over a smooth, complete curve, where the degrees of MATH, and the MATH are known. In particular, given two curves MATH and MATH of genera MATH and MATH, respectively, fix MATH and let MATH vary in MATH. Consider the family MATH constructed by joining the two curves at the points MATH and MATH CITE. Then the degrees of MATH and MATH are all zero on MATH, except when MATH, and then MATH. We can construct a MATH-th root of MATH on MATH and a MATH-th root of MATH on MATH, and thus a MATH-spin structure on MATH. Moreover, the two MATH-th roots can be chosen to be of any index in MATH or MATH, respectively; and thus MATH can be endowed with a spin structure of any index MATH in MATH along every fibre. Because the MATH are disjoint, the degree of MATH for any other index MATH must be zero on MATH, but MATH implies that MATH is non-zero. Consequently, in any relation of the form MATH, the coefficients MATH must all vanish. And thus a relation must be of the form MATH. But a similar method shows that the coefficients MATH must also be zero. In particular, consider the family MATH constructed by taking a general curve MATH of genus MATH and identifying one fixed point MATH with another, variable point MATH CITE. Again one may produce a MATH-spin structure on MATH of any type. And MATH whenever MATH, but MATH is equal to MATH, which is non-zero (since MATH). Thus MATH, and so also MATH. |
math/9908085 | For MATH smooth, MATH is canonically isomorphic to MATH, and thus in the smooth case, MATH and MATH are trivial. In the general case these products are all integral linear combinations of boundary divisors. To compute the coefficients we evaluate degrees on families of curves MATH parameterized by a smooth curve MATH and having smooth generic fibre. In this case, since MATH is supported on the exceptional locus of MATH, whereas the canonical bundle has degree zero on the exceptional locus, the product MATH must be trivial. Computing the coefficients of MATH requires that we consider the local structure of MATH near an exceptional curve. Recall from CITE that for any singularity where MATH has order MATH with MATH, and such that if MATH, MATH, MATH, and MATH, the underlying singularity of MATH is analytically isomorphic to MATH, where MATH is an element of MATH. MATH is generated by two elements, say MATH and MATH, with the relations MATH and MATH. Moreover, over such a singularity, MATH is locally given as MATH where MATH is the local ring of MATH at the singularity. The exceptional curve, call it MATH, is defined by the vanishing of MATH and MATH, and we have a situation like that depicted in REF . We need to express MATH in terms of a divisor, but this is easy since it is supported completely on the exceptional locus. MATH is locally of the form MATH, for some integer MATH. And any NAME divisor of the form MATH is NAME if and only if MATH and MATH both divide MATH. Moreover, it is easy to see that if MATH is NAME, then when restricted to the exceptional curve MATH, the degree of MATH is MATH. Finally, since MATH has degree MATH on MATH, we have MATH so that MATH . Now, to compute the coefficients of MATH just note that for families MATH over a smooth base curve MATH with smooth generic fibre, the degree of MATH is just the intersection number MATH. Thus if MATH indicates the exceptional curve over a singularity MATH, and if MATH indicates the order of MATH near MATH, then MATH . The second line follows because MATH and because over MATH we have MATH and MATH. This proves the theorem for MATH, and the result for MATH is just the pullback of the relation for MATH from MATH. |
math/9908085 | Since MATH, we have MATH NAME and NAME now give MATH . Here the last equality follows from the well-known NAME isomorphism: MATH (see CITE). |
math/9908085 | If the proposition were false and the element in question were zero, then in MATH this element would be a sum of boundary divisors. In particular, the element in question would be of the form MATH. In the first case, multiplication by two, and in the second case, multiplication by three, allows us to replace the element in question with a sum (from REF ) consisting exclusively of boundary divisors. Thus for the first case we have a relation between boundary divisors where the sum on the right has all coefficients divisible by two: MATH . And for the second case, the sum on the right has all coefficients divisible by three: MATH . Thus by REF the coefficients on the left must also be divisible by MATH or MATH, respectively. However, in both cases the coefficient of MATH for any MATH and MATH is MATH, which has no divisors in common with MATH; a contradiction. The third and fourth cases are similar, but first we must subtract MATH times the second equation of REF .bis from the first. Now reduction mod MATH and mod MATH give the necessary contradictions in the third and fourth cases, respectively. |
math/9908085 | First recall that if MATH is the locus MATH and MATH is the locus MATH, then the stacks MATH and MATH have a representation as the space of cubic forms MATH, that is MATH and MATH respectively, modulo the ``standard" MATH action MATH CITE. We denote this action by MATH and the action of the proposition by MATH. We have commutative diagrams of stacks MATH and MATH where the top morphism is given by the fact that there is a MATH-equivariant choice of a line bundle MATH on the curve MATH and a MATH-equivariant isomorphism MATH. The bundle MATH is generated by a MATH-th root of MATH, the invariant differential. This makes sense because MATH has no zeros or poles. Alternately, we may simply take the trivial line bundle MATH on MATH generated by an element MATH, with the action MATH defined as MATH. If MATH is the projection to MATH, then we define MATH to be MATH, and the homomorphism MATH to be MATH. This homomorphism is MATH-equivariant since MATH; and it is well-known that for this family MATH generates the relative dualizing sheaf MATH. The proposition now follows since both the left and right-hand vertical morphisms are étale of degree MATH, and the bottom morphism is an isomorphism. Thus the top morphism is étale of degree MATH. It is clearly an isomorphism on geometric points, thus also an isomorphism of stacks. |
math/9908085 | By CITE for any smooth quotient stack MATH the NAME ring MATH is the equivariant NAME ring MATH, and MATH. Thus it suffices to compute MATH, where MATH is the MATH or MATH, and MATH in MATH with the action MATH of REF . Choosing a MATH-dimensional representation MATH of MATH with all weights MATH, and letting MATH be the open set where MATH acts freely, then the diagonal action of MATH on MATH is free, and MATH is defined CITE to be the usual NAME group MATH of the quotient scheme MATH, which is isomorphic to the complement of the zero section of the vector bundle MATH over MATH. Thus MATH. Moreover, since the form MATH has weighted degree MATH with respect to the action, the MATH equivariant fundamental class MATH of MATH is MATH, and MATH . Similarly, the class MATH is the intersection of MATH and MATH. The theorem follows. |
math/9908086 | Define MATH by MATH . Then MATH is approximately convex and MATH for MATH. So MATH for all MATH, which gives REF. |
math/9908086 | The left-hand inequality just says that MATH is concave (to see this note that MATH). To prove the right-hand inequality, first consider the case MATH. For fixed MATH and MATH, let MATH . Then MATH . Thus MATH is decreasing on MATH and attains its maximum at MATH. But MATH . Thus, if MATH, then MATH . Similarly, if MATH, then MATH . |
math/9908086 | MATH is a sum of concave functions (by REF ) and so MATH is concave. For MATH and MATH in MATH and MATH, we can use REF for the first inequality to get MATH . The function MATH is concave and symmetric about MATH. Thus, MATH with equality in the last inequality only if MATH. |
math/9908086 | The upper bound is due to CITE. For the lower bound, since MATH, we have MATH . |
math/9908086 | We may assume that the right-hand side of REF is finite, otherwise there is nothing to prove. Observe that the effect of replacing MATH by MATH is to multiply both sides of REF by MATH. So, by choosing MATH appropriately, we may assume that MATH . The right-hand estimate for MATH is due to CITE. For completeness we recall the proof. Let MATH (MATH). First note that MATH is MATH-Lipschitz and non-negative. To see that MATH is approximately convex, note that for MATH, MATH, and MATH, we have MATH . Taking the infimum of this expression over all choices of MATH and MATH yields MATH . Now suppose that MATH. By NAME 's Theorem (see for example, CITE), MATH, a convex combination of MATH elements MATH. Then REF yields MATH since MATH for all MATH. The left-hand inequality uses the entropy functions MATH. Let MATH be an NAME basis for MATH (see for example, CITE). Recall that this means that MATH for all scalars MATH. Set MATH so that MATH is a MATH-simplex. For each MATH, we define a set MATH thus: MATH . First let us verify that MATH is approximately convex. Suppose that MATH and that MATH and MATH belong to MATH, where MATH and MATH belong to MATH. Then MATH also belongs to MATH, where MATH. Since MATH is a unit vector and MATH is approximately affine REF, we have MATH and so MATH is approximately convex. Note that MATH . We shall show that MATH as MATH. To see this, fix MATH. By continuity of MATH there exists MATH such that if MATH then MATH, whence by REF MATH . Now suppose, on the other hand, that MATH. By REF MATH as MATH. Thus, for all sufficiently large MATH, we have MATH. Since MATH is arbitrary, this gives the lower bound MATH. |
math/9908086 | For MATH, we have MATH. The argument used to prove REF (too lengthy to recall here) shows that the result will follow provided MATH is large enough to ensure that MATH . This holds for MATH. |
math/9908086 | We shall use the following consequence of REF . Let MATH and MATH. Then every normed space of sufficiently large dimension contains a compact MATH-convex set MATH such that MATH. Using this fact repeatedly, a routine argument (compare CITE) shows that MATH contains a subspace MATH with a finite-dimensional decomposition MATH and sets MATH REF such that MATH is a MATH-convex set containing zero and MATH. Let MATH be the collection of all vectors of the form MATH, where MATH and only finitely many of the MATH's are nonzero. First let us verify that MATH is approximately convex. Suppose that MATH and MATH are in MATH and that MATH. Since MATH is MATH-convex and compact, there exists MATH with MATH. Moreover, we may choose the MATH's so that only finitely many are nonzero, ensuring that MATH belongs to MATH. By the triangle inequality MATH . Let us verify that MATH. Since MATH is a finite-dimensional decomposition, the natural projection maps from MATH onto MATH are uniformly bounded in operator norm by MATH, say. Since MATH, there exists MATH such that MATH, and since MATH, we have MATH . Thus, MATH. |
math/9908086 | Using the notation of REF , we prove that MATH has the required property. It was shown in REF that MATH is approximately convex and MATH-Lipschitz. Choose MATH so that MATH. Suppose that MATH is a convex function satisfying MATH. Since MATH for all MATH, it follows that MATH for all MATH, and hence MATH for all MATH. But MATH, and so MATH. |
math/9908086 | In order to simplify notation we shall prove the result for all normed spaces MATH of dimension MATH (with MATH replacing MATH in REF). Note that MATH contains a subspace MATH of codimension one such that MATH (since MATH contains subspaces isometric to MATH). Let MATH be a linear functional in MATH of unit norm such that MATH. Let MATH be an unit vector in MATH which is normed by MATH, that is, such that MATH. Note that by the triangle inequality MATH for all MATH REF and MATH. Since MATH, MATH has a basis MATH satisfying MATH for all choices of scalars MATH. For each MATH, define MATH by MATH . It was proved in REF that MATH is approximately convex for all choices of MATH. Observe also that MATH . In order to verify REF, it suffices to show that MATH for a suitable choice of MATH. To that end, fix MATH and fix MATH. Let MATH and set MATH (MATH). Then, by REF for the first inequality and the left-hand side of REF for the second, we have MATH where at the last step we use the fact that MATH for MATH. Now set MATH. There are two cases to consider. First, if MATH, then MATH . Secondly, if MATH, then MATH . Hence MATH. Setting MATH we see that REF is satisfied (with MATH replaced by MATH) by MATH whenever MATH is large enough to ensure that MATH. Finally, the right-hand side of REF yields MATH for all sufficiently large MATH, and so MATH satisfies REF . |
math/9908086 | Setting MATH, we follow REF taking advantage of some simplifications in the proof which we now indicate. First, we may choose MATH to be the standard unit vector basis of MATH, so that REF becomes simply MATH, for all MATH. The estimate for MATH then becomes MATH . Setting MATH, we obtain MATH . Setting MATH we see that REF is satisfied (with MATH replaced by MATH) by MATH whenever MATH is large enough to ensure that MATH. Finally, MATH which yields REF. |
math/9908086 | Fix MATH. CITE (compare also CITE) proved that every MATH-dimensional normed space MATH contains a subspace MATH, with MATH, satisfying MATH where MATH is a constant. Set MATH. Then, for MATH, MATH . Applying REF to MATH and to MATH yields an approximately convex set MATH satisfying REF (from REF) and REF (from REF). |
math/9908086 | First we prove the result for MATH. Let MATH be one of the vertices for which the corresponding barycentric coordinate of the origin is at most MATH. Let the line segment through the origin joining MATH to the opposite MATH-face MATH intersect MATH in a point MATH, say. Then the origin divides the line joining MATH to MATH into two segments bearing a ratio of not less than MATH to MATH. Since MATH lies on the unit sphere, it follows that MATH. Thus, MATH, which completes the proof for the case MATH. The proof for MATH is by induction on MATH. Suppose that the result holds for MATH and for MATH. Let MATH be a MATH-face of MATH nearest to the origin and let MATH be the point in MATH nearest to the origin. Then MATH. The largest Euclidean ball inscribed in MATH with center the origin touches MATH at MATH. Hence MATH is in the interior of the MATH-simplex MATH whose vertices lie on the MATH-sphere with center MATH and radius MATH. Fix MATH. By the inductive hypothesis applied to MATH and MATH there exists a MATH-face MATH of MATH such that MATH . So MATH where MATH. The right-hand side is greatest when MATH, which gives MATH . |
math/9908086 | By slightly perturbing the elements, if necessary, we may assume that the set MATH is affinely independent and that the origin lies in the interior of the simplex MATH. Let MATH. Clearly, MATH for all MATH. Now REF applied to the simplex MATH with vertices MATH yields the desired result. |
math/9908086 | Assuming (as we may) that MATH is compact, there exists MATH with MATH. By translating MATH, we may assume that MATH. Thus, MATH and MATH. The fact that MATH now implies that MATH for all MATH. By NAME 's Theorem, there exist MATH in MATH such that MATH. Let MATH, then by REF there exists MATH such that MATH and MATH . Let MATH be the point in MATH nearest the origin. Because MATH is approximately convex, the function MATH is an approximately convex function which vanishes at each MATH. So, by REF , MATH for MATH and if MATH then MATH and MATH. Thus MATH for MATH. Therefore MATH which yields MATH where this defines MATH. If MATH is a non-negative integer with MATH then MATH. Therefore MATH where MATH and MATH as MATH. For each MATH and MATH there is an integer MATH so that MATH, and for MATH . Therefore if MATH is sufficiently large and MATH is chosen so that MATH then MATH and thus MATH. For any MATH and MATH where MATH. If MATH, then MATH implies MATH. Now assume that MATH, and that MATH. Then MATH, so MATH. Therefore the argument above implies that for MATH the bound MATH holds. For this lower bound to be nontrivial we also require MATH. However this holds for all MATH and so the lower bound on MATH holds and is nontrivial for all MATH. |
math/9908086 | Clearly, MATH . To establish the reverse inequality, we show that MATH for MATH. Observe that MATH and also that MATH where MATH . Hence MATH . But MATH, where MATH . (Note that MATH satisfies the constraints for the optimization problem.) Since MATH is minimized by MATH (see REF ), we get MATH . Hence MATH . Thus, MATH. The estimate for MATH is straightforward. |
math/9908086 | Let MATH be the infimum of MATH taken over all MATH which satisfy the constraints. There exist MATH REF satisfying the constraints such that MATH as MATH. By replacing each MATH by its non-increasing rearrangement, we may assume that each MATH is right-continuous and non-increasing. By NAME 's selection theorem (see for example, CITE), we may also assume (by passing to a subsequence) that MATH pointwise. Since MATH, it follows from the Bounded Convergence Theorem that MATH satisfies the constraints and that MATH. Finally, let MATH be the right-continuous modification of MATH. |
math/9908086 | In the notation of REF , we may assume that MATH is a step function minimizing MATH over all step functions of the form MATH satisfying the constraints. A value MATH taken by MATH must satisfy the following NAME multiplier equation for a local minimum: MATH where MATH is a constant. It is easily seen that this equation has at most two roots in MATH. By the pointwise convergence of MATH to MATH, it follows that MATH takes at most two values in MATH. Therefore we may apply REF multipliers again to deduce that these values must also satisfy REF (with MATH replaced by MATH). Equivalently, setting MATH, MATH . Suppose that there are two distinct roots, MATH and MATH, with MATH, and suppose that MATH takes one of these values, MATH say, on an interval MATH. (The argument is similar if MATH takes the value MATH.) Let MATH take the value MATH on the complement of MATH, and the values MATH and MATH on the left-hand and right-hand halves of MATH, respectively. Since MATH, it follows that MATH satisfies the constraints, provided MATH is sufficiently small. Moreover, MATH . Since MATH minimizes MATH, MATH . To derive a contradiction, suppose that MATH also takes the value MATH on an interval. Then, by the same argument, MATH . Since REF is satisfied by MATH and MATH, the Mean Value Theorem implies the existence of MATH such that MATH . Thus, MATH . But this contradicts REF. Thus, MATH cannot take the value MATH, which completes the proof. |
math/9908086 | For MATH, we have MATH . Suppose that MATH takes the value MATH on MATH and the nonzero value MATH on an interval of length MATH. If MATH then MATH, which contradicts REF. So we may assume that MATH. Now MATH . So MATH . Since MATH minimizes MATH, it follows that MATH, as desired. |
math/9908086 | By REF , MATH takes only one nonzero value MATH on an interval of length MATH. So MATH . Thus, MATH with equality if and only if MATH. Since MATH minimizes MATH, it follows that MATH, which gives the result. |
math/9908086 | We may assume (compare REF ) that MATH, that MATH, and that MATH for all MATH. Since MATH there exist MATH, MATH and MATH (MATH), with MATH and MATH. Let MATH be a sequence of independent identically distributed MATH-valued random variables defined by MATH . Then MATH REF and MATH. Thus, CITE yields (for each MATH) MATH . So there exist MATH REF with MATH . Since MATH is approximately convex, MATH . So MATH . Put MATH (noting that MATH since MATH by assumption) so that MATH. Then MATH which yields REF. |
math/9908086 | It is known that MATH is NAME if and only if MATH has type MATH for some MATH CITE. Thus, MATH follows from REF . Now suppose that MATH is not NAME. By definition (see REF), MATH contains `almost isometric' copies of MATH for all MATH. So by REF MATH contains approximately convex sets MATH such that MATH and MATH, where MATH is an absolute constant. Clearly, REF cannot hold in MATH, and so REF . |
math/9908086 | It is known that MATH has type MATH. Setting MATH in REF gives the result. |
math/9908086 | Let MATH and let MATH and MATH satisfy the conclusion of REF . Then MATH has the required properties. |
math/9908086 | Let MATH (MATH). Then MATH is a continuous approximately NAME function, that is, MATH . By CITE MATH is a MATH-convex function, which implies that MATH is a MATH-convex set. |
math/9908086 | From the definition of MATH we see that MATH if and only if MATH . Indeed, if MATH satisfies REF, then for every MATH, we have MATH and so MATH. Conversely, if MATH, then MATH and REF is satisfied. The condition MATH is clearly equivalent to REF . Since MATH is a weighted MATH norm, the condition MATH is equivalent to the condition MATH . Suppose that MATH. Then MATH and MATH, and so REF becomes MATH. Now suppose that MATH. Then MATH and MATH, and so REF becomes MATH and so REF is satisfied. Conversely, if MATH satisfies REF , then the mapping MATH will extend linearly to an element of MATH. |
math/9908086 | We define MATH recursively. First define MATH from MATH into MATH to be an arbitrary extension of the restriction of MATH to MATH. Suppose that MATH and that MATH has been defined on MATH to extend the restriction of MATH to MATH. Let MATH. Then MATH, and so MATH and MATH have already been defined. If MATH, then MATH, and so MATH and MATH. It follows from REF that REF will be satisfied with MATH. If MATH, define MATH, so that REF is trivially satisfied. This completes the definition of MATH on MATH. |
math/9908086 | First we define a mapping MATH. If MATH, let MATH, and if MATH, let MATH so that REF are satisfied. Now extend to the rest of MATH recursively as follows. Suppose that MATH and that MATH has been defined on MATH to satisfy REF. Let MATH. Then MATH for some MATH. Define MATH . If MATH, then, as MATH and MATH, we have MATH so that MATH that is, REF is satisfied by MATH. Also, if MATH, then REF is trivially satisfied. On the other hand, if MATH, then REF is trivially satisfied by MATH. In order to verify REF, suppose that MATH. Then both MATH and MATH. Moreover, both MATH and MATH satisfy REF by the recursive hypothesis. Thus, MATH . Thus, REF is satisfied by MATH, which completes the recursive definition of MATH. Now MATH and MATH (replacing MATH) satisfy the hypotheses of REF . Let MATH be the extension of MATH given by REF . |
math/9908086 | Suppose that MATH. By REF , MATH for all MATH, and so REF follows from the NAME Theorem. Suppose that MATH. Then MATH, and by REF MATH which gives REF . To prove REF , note that REF implies that MATH (since MATH is convex), and hence MATH . So it suffices to prove that MATH. Fix MATH and choose distinct elements MATH. We shall prove that MATH where MATH as MATH. Let MATH. If MATH and MATH, then MATH. Since MATH, it follows that MATH has cardinality at most MATH. Thus MATH . Let MATH be the function associated to MATH defined in REF , and let MATH be the linear functional corresponding to MATH. If MATH, then MATH by REF. If MATH and MATH, then MATH by REF. Hence if MATH, then MATH. Moreover, MATH by REF, since MATH. So MATH and so MATH where MATH as MATH as desired. |
math/9908086 | Let MATH denote the space constructed above for MATH. Then the MATH-sum MATH has the required property. |
math/9908086 | By REF there exists MATH such that MATH is MATH-convex and MATH. Then MATH has the required properties. |
math/9908087 | We already know the existence of the right-coprime factorization over MATH of MATH. On the other hand, the existence of the left-coprime factorization over MATH of MATH is derived from REF of this paper and REF. |
math/9908087 | Since for every MATH, MATH is a nonzerodivisor of MATH, every MATH is nonsingular over MATH. It is known that the following equality holds: MATH . Directly from this, we have REF . The remaining relations REF are obtained directly from REF . |
math/9908087 | Observe that the following matrix equation holds: MATH . The determinants of the matrices in the right hand side of the equation above are MATH, MATH, MATH, respectively. Hence the left hand side is nonsingular over MATH if and only if MATH is nonsingular over MATH. By REF , this completes the proof. |
math/9908087 | We prove only REF by showing that REF for any matrix MATH, there exists a matrix MATH such that MATH, and that REF for any matrix MATH, there exists a matrix MATH such that MATH. Once REF is obtained, REF is obvious by REF . We first prove REF . We rewrite REF as follows: MATH . Since both of the inverse matrices in REF are unimodular (compare REF), we let MATH . Then REF can be rewritten further as follows: MATH . Partition MATH as MATH where MATH, MATH, MATH, MATH. Then REF can be rewritten again as follows: MATH which is equal to MATH. Therefore the matrix MATH is the matrix MATH satisfying MATH. Next we prove REF . From the proof of REF , letting MATH as MATH where MATH's denote arbitrary matrices, we obtain directly MATH. |
math/9908087 | In order to prove this theorem, it is sufficient to show the following: CASE: For any matrix MATH, if MATH is nonsingular, there exists a stabilizing controller MATH such that MATH. CASE: Conversely, for any stabilizing controller MATH, there exists a matrix MATH such that MATH. We first prove REF . Suppose that MATH is nonsingular. Assume without loss of generality that MATH is a nonzerodivisor with MATH. Then by REF , there exists a matrix MATH over MATH such that MATH. By REF , there exists a MATH-stabilizing controller MATH. Observe now that MATH, so that MATH is over MATH. Since MATH, MATH is over MATH, which implies that MATH is a stabilizing controller of MATH. Next we prove REF . Suppose that MATH is stabilizable. As in REF, let MATH's be in MATH for MATH such that MATH with a sufficiently large integer MATH. Let MATH be an arbitrary but fixed stabilizing controller of MATH. Since MATH is also a MATH-stabilizing controller, by REF there exists MATH over MATH such that MATH for each MATH. By REF there exists a matrix MATH over MATH such that MATH for each MATH, so that MATH over MATH. Hence we have MATH over MATH. From REF , the equation above can be rewritten as follows: MATH . Letting MATH, we have proved REF . |
math/9908088 | We employ the notations in REF. Thus, MATH, say MATH and MATH. Then the generalized elementary factors of the plant are given as follows: MATH . It is obvious that MATH. On the other hand, MATH is a member of MATH since the ratio MATH in REF can be rewritten as MATH. Since MATH is coprime, we have MATH. Hence by REF the plant is stabilizable. |
math/9908088 | From REF , and REF , the plant MATH is stabilizable and does not have right-/left-coprime factorization by REF . |
math/9908088 | In the case MATH, the plant MATH is obviously stabilizable. Hence in the following we assume without loss of generality that MATH. Let us consider a plant MATH is expressed as MATH, where MATH. Let MATH denote MATH over MATH. Let MATH be an integer such that MATH. We note that MATH does not have MATH as a factor. To show the stabilizability, we here apply REF. Since the plant is of the single-input single-output, the set MATH defined in REF is equal to MATH, say MATH and MATH. Then the generalized elementary factors MATH and MATH are given as follows: MATH . Then one can check that MATH and MATH. Since MATH over MATH, we have MATH. Hence any plant MATH is stabilizable by REF. |
math/9908088 | Suppose that MATH is a causal plant in MATH expressed as MATH with MATH and MATH. Let MATH be MATH over MATH rather than over MATH. We assume without loss of generality that MATH is expressed as MATH with MATH. Let MATH and MATH be polynomials MATH and MATH, respectively, in MATH and further MATH and MATH be polynomials in MATH defined as follows: MATH (Note here that MATH above can be other values). Then one can check that MATH, MATH and the greatest common divisor of MATH and MATH over MATH is a unit. Let MATH, MATH be in MATH such that the following equation holds over MATH: MATH where MATH is an arbitrary element in MATH. Let MATH, MATH, MATH, and MATH denote the constant terms of MATH, MATH, MATH, and MATH, respectively. Similarly let MATH, MATH, MATH, and MATH denote the coefficients of MATH, MATH, MATH, and MATH, respectively, with the degree MATH with respect to the variable MATH. In order to show that REF holds over MATH, we want to find MATH such that MATH. To do so, we let MATH. Then it is easy to check that MATH hold from the relations that MATH and MATH. Now that MATH, MATH, we have MATH, so that the plant is stabilizable. |
math/9908094 | CASE: Let MATH be an eigenvector of MATH in MATH with weight MATH. Then MATH restricts to an invertible regular function on MATH, and is uniquely determined by MATH up to a constant. Conversely, let MATH be an invertible regular function on the MATH-orbit MATH. Then MATH pulls back to an invertible regular function on MATH, that is, to a scalar multiple of a character of MATH. Thus, MATH is an eigenvector of MATH in MATH. CASE: Choose MATH. Let MATH (respectively, MATH) be the isotropy group of MATH in MATH (respectively, MATH). Since MATH, we have MATH. Thus, MATH acts transitively on MATH, the flag variety of MATH. Using for example, CITE, it follows that MATH contains a maximal connected semisimple subgroup of MATH, that is, a conjugate of MATH. REF follows from REF ; it can be deduced from REF as well. |
math/9908094 | Let MATH be the complement in MATH of the union of all irreducible MATH-stable divisors that do not contain MATH. Then MATH is an open affine MATH-stable subset of MATH, and MATH equals MATH; see REF . Let MATH be the stabilizer of MATH in MATH, then MATH consists of all MATH such that MATH for all MATH. Clearly, MATH is a standard parabolic subgroup, contained in MATH. It follows that MATH. Let MATH be the standard NAME subgroup of MATH. By CITE and REF, there exists a closed MATH-stable subvariety MATH of MATH such that the product map MATH is an isomorphism; moreover, MATH acts trivially on MATH. In particular, for any MATH, the product map MATH is an isomorphism. Since MATH and since MATH fixes points of MATH, it follows that MATH, whence MATH. Moreover, the character group of MATH is isomorphic to that of the torus MATH, whence MATH. |
math/9908094 | We begin with two lemmas that reduce the ``local" study of MATH to simpler situations. Let MATH and let MATH be a standard parabolic subgroup of MATH, with radical MATH. Let MATH be the set of all closures in MATH of MATH-orbits in MATH; in other words, MATH is the set of all MATH such that MATH. Let MATH be the oriented graph with set of vertices MATH, and with edges those edges of MATH that have both endpoints in MATH and labels in MATH. The quotient MATH is a MATH-homogeneous spherical variety with graph MATH. Since MATH is a unique MATH-orbit and MATH is a normal subgroup of MATH contained in MATH, the quotient MATH exists and is homogeneous under MATH; moreover, any MATH-orbit in MATH pulls back to a unique MATH-orbit in MATH. Let MATH be a MATH-orbit in MATH and let MATH. Then MATH contains MATH, the square MATH is cartesian, and the map MATH is an isomorphism. Thus, the type is preserved under pull back. Assume now that MATH is homogeneous under MATH; write then MATH. Let MATH be a closed subgroup of the normalizer MATH such that MATH contains MATH, and that the quotient MATH is connected. Let MATH be the center of MATH. Let MATH, a homogeneous spherical variety under the adjoint group MATH. The natural MATH-equivariant map MATH is the quotient by the right action of MATH on MATH. The pull-back under MATH of any MATH-orbit in MATH is a unique MATH-orbit in MATH. This defines an isomorphism of MATH onto MATH. The first assertion follows from REF . The second assertion is checked as in the proof of REF . Let MATH, MATH, and let MATH. If MATH then MATH is orthogonal to MATH, and the derived subgroup of MATH fixes pointwise MATH. Let MATH be the isotropy group in MATH of a point of MATH. Since MATH, we have MATH. Equivalently, the map MATH is an isomorphism. But since MATH, we have MATH, so that the image of MATH in MATH is a proper subgroup. It follows that MATH is solvable. Thus, MATH is the flag variety of MATH. Now the connected automorphism group of this flag variety is the quotient of MATH by its center. On the other hand, the connected automorphism group of MATH is MATH if MATH is not orthogonal to MATH (this follows for example, from CITE.) In this case, we have MATH so that MATH is a unique MATH-orbit, a contradiction. Thus, MATH is the product of MATH with MATH, and the map MATH is an isomorphism. It follows that the derived subgroup of MATH is contained in MATH. We now prove REF . Applying REF to MATH and MATH, we may assume that MATH for some subgroup MATH of MATH and that MATH. If MATH has type MATH, then MATH whence MATH has type MATH as well. We claim that MATH consists of MATH and MATH. Indeed, if MATH and MATH, then MATH is connected to MATH by an oriented path in MATH. Let MATH be the source of the top edge of this path. That edge cannot have MATH as its target (otherwise MATH would be stable under MATH or MATH); thus, it raises MATH to MATH. Since MATH and MATH have type MATH, it follows that MATH, whence MATH. Thus, MATH; then MATH and MATH are orthogonal by REF . If MATH has type MATH, then MATH, whence MATH has type MATH or MATH. In the former case, we see as above that MATH. Thus, MATH and MATH are orthogonal by REF . Using REF , we may assume that MATH and that MATH contains a copy of MATH. Then MATH is conjugate to MATH embedded diagonally in MATH. But then both MATH and MATH have type MATH, a contradiction. If MATH has type MATH and MATH has type MATH, then there exists MATH such that MATH is contained in MATH. Since the homogeneous spaces MATH and MATH are affine, the same holds for MATH. It follows that MATH is pure of codimension MATH in MATH. But MATH meets both MATH-orbits of codimension MATH in MATH, so that MATH. This case is excluded as above. Thus, type MATH does not occur. |
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