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math/0106063 | Suppose MATH is transcendental over MATH. (This will lead to a contradiction.) Let MATH be a transcendence basis for MATH over MATH. (That is, MATH is algebraic over MATH, and MATH is transcendental over MATH, for each MATH.) By replacing MATH with MATH, we may assume MATH. Therefore, MATH, where each MATH is algebraic over MATH. From the proof of REF, we see that there exists MATH, such that MATH are integral over MATH. Choose some irreducible MATH, such that MATH . We have MATH so MATH, for some MATH. Therefore, MATH is integral over MATH. From the proof of REF, we conclude that MATH. (The ring MATH is integrally closed.) This contradicts the choice of MATH. |
math/0106063 | CASE: Let MATH. Then MATH is a field (because MATH is maximal), so REF implies MATH is algebraic over MATH. Since MATH is algebraically closed, we conclude that MATH, as desired. CASE: From REF, we know there exists MATH, such that MATH; let MATH . From the choice of MATH, we have MATH. On the other hand, it is easy to see that MATH is a field, so MATH must be a maximal ideal. Hence, it is equal to MATH. |
math/0106063 | Let MATH be a maximal ideal that contains MATH, and choose MATH as in REF. Then MATH for all MATH, so, since MATH, the desired conclusion follows. |
math/0106063 | We have MATH, for some MATH. There is a homomorphism MATH defined by MATH. Let MATH be the kernel of MATH, and choose a maximal ideal MATH that contains MATH. Then MATH, so there is a natural homomorphism MATH (see REF ). |
math/0106063 | We prove the contrapositive: suppose MATH is reducible over MATH. Clearing denominators, we may write MATH, for some nonzero MATH, with MATH and MATH. Dividing MATH by an integer constant, we may assume MATH, where MATH denotes the content of MATH, that is, the greatest common divisor of the coefficients of MATH. Then, letting MATH, we have MATH (see REF). Thus, letting MATH, we have MATH so MATH is reducible over MATH, as desired. |
math/0106063 | Suppose MATH is reducible over MATH. (This will lead to a contradiction.) Then MATH is also reducible over MATH (see REF), so we may write MATH, with MATH and MATH. Then MATH . From the unique factorization of polynomials in MATH (recall that MATH is a Euclidean domain, because MATH is a field), we conclude that there exist MATH and MATH, such that MATH. Since MATH and MATH, we conclude that MATH. Therefore MATH, so MATH. This is a contradiction. |
math/0106063 | There is (up to isogeny) only one complex group of each type (see REF ). One can verify, under the given restrictions on MATH, that none of these complex groups are isogenous to any of the others. For the classical cases MATH, it is not too difficult to establish this directly. In any case, one can use the theory of roots: the conclusion follows from the observation that none of the NAME diagrams are isomorphic. Note that if MATH is complex, then MATH is isomorphic to MATH. If MATH is a real (non-complex) simple group, then MATH is a complex simple group, so it is isogenous to one of the groups in REF. It is easy to verify, by inspection of REF, that MATH and MATH have the same type. Indeed, this is precisely the criterion that led to the groupings in REF Thus, when MATH and MATH are either both real or both complex, we see that MATH and MATH have the same type if and only if MATH is isogenous to MATH. On the other hand, when MATH is real and MATH is complex, we see that MATH and MATH have the same type if and only if MATH is isogenous to MATH. |
math/0106063 | Suppose MATH and MATH are two of the groups listed in REF, with the restrictions on MATH given in REF. Assume that MATH is isogenous to MATH. From REF, we know that MATH and MATH have the same type. We may assume that neither MATH nor MATH is complex. (Otherwise, they must both be complex. Since REF lists only one complex group of each type, we conclude that MATH, as desired.) The rest of the proof is based heavily on the observation that MATH (see REF for a discussion of real rank, and see REF for a record of the real rank of each classical group). For MATH, MATH, and MATH, we always assume that MATH (see REF, and REF). Assume MATH and MATH are of type MATH, for some MATH. Let us assume MATH (see REF). The groups MATH and MATH all have different real ranks, so no one is isogenous to any of the others. Thus, we may assume that MATH (which implies that MATH is odd). Then MATH (recall that MATH), so we must have MATH, for some MATH and MATH with MATH. Now the maximal compact subgroup of MATH is MATH, which is simple, but the maximal compact subgroup of MATH is MATH which is not simple. This is a contradiction. Assume MATH and MATH are of type MATH, for some MATH. The groups MATH all have different real ranks, so none are isogenous. Assume MATH and MATH are of type MATH, for some MATH. The groups MATH and MATH all have different real ranks, so none are isogenous. Assume MATH and MATH are of type MATH, for some MATH. The groups MATH all have different real ranks, so none are isogenous. Thus, we may assume that MATH (and that MATH). The maximal compact subgroup of MATH is MATH, which is isogenous to MATH. Since the maximal compact subgroup of MATH is MATH, we conclude that MATH, and that MATH is isogenous to MATH. Now MATH is of type MATH, and MATH is of type MATH. Therefore, REF implies that MATH, so MATH. (Recall that, by assumption, we have MATH.) Hence, Conclusion REF applies. |
math/0106063 | The general case is a rather difficult theorem of Algebraic Number Theory, so let us assume that MATH is prime. This suffices to construct many interesting examples, and we will see that it follows easily from REF below. CASE: Suppose MATH, with MATH. Because MATH (see REF, we may assume MATH (by replacing MATH with MATH). From our simplifying assumption that MATH is prime, we conclude that MATH; thus, MATH. So MATH.) Therefore, MATH is not a division algebra. Alternatively, one can easily give a very elementary proof. We have MATH, for some MATH. Then MATH so MATH so MATH is a zero divisor. Therefore, MATH is not a division algebra. CASE: Because MATH is central simple see REF, NAME 's REF implies that MATH, for some central division algebra MATH, and some MATH. Because MATH, and MATH is prime, there are only two possibilities to consider. If MATH, then MATH is a division algebra, as desired. If MATH, then MATH, so MATH. From REF , we conclude that MATH. |
math/0106063 | CASE: Note that MATH via the map MATH. Then the desired conclusion follows from the observation that MATH see REF. CASE: We have MATH, so there is some subfield MATH of MATH, and some MATH, such that CASE: MATH is isomorphic to MATH; CASE: the map MATH given by MATH generates MATH; and CASE: MATH. The NAME REF implies that we may assume MATH (after conjugating by an automorphism of MATH). Then, replacing MATH by a power, we may assume that MATH for all MATH. So MATH centralizes MATH. Since MATH is a maximal subfield see REF, we conclude that MATH, for some MATH. Then MATH so MATH. |
math/0106063 | From REF , we know that MATH, for some cyclic extension MATH of MATH, and some MATH. From the cohomological approach to division algebras (see REF), it is not difficult to see that MATH in MATH. Therefore, REF implies that MATH if and only if MATH. So the desired conclusion follows from REF . |
math/0106063 | CASE: We have the antiinvolutions MATH and MATH. CASE: Any antiinvolution is (by definition) an isomorphism from MATH to MATH. CASE: Since MATH in MATH (see REF), and MATH, we have MATH in MATH. CASE: From REF , we see that MATH has degree MATH. |
math/0106063 | Let us assume MATH. If MATH is any central division algebra over MATH, then MATH is a simple algebra, so, by NAME 's REF , there is a corresponding central division algebra MATH over MATH. Thus, we have a natural homomorphism MATH. It is well-known that there are only two central division algebras over MATH, namely MATH and MATH, so MATH. For each prime MATH, we have a MATH-adic field MATH. Since MATH is a central simple algebra over MATH, there is a natural homomorphism MATH. Classifying the central division algebras over MATH yields the conclusion that MATH. By combining the preceding paragraphs, we see that there is a natural homomorphism MATH . To complete the proof, one shows that the map is injective, and that its image is MATH where we realize MATH as the subgroup MATH of MATH. These last steps are the hard part. |
math/0106063 | From REF , it is clear that there are infinitely many elements of order MATH in MATH. Each of these elements corresponds to a central division algebra of degree MATH over MATH (see REF). |
math/0106063 | From the pf. of REF , we see that the central division algebras that split over MATH correspond to the elements of the Right-hand side of REF, such that MATH. Any element with MATH has even order, so all central division algebras of odd degree split over MATH. (This conclusion can also be proved easily by elementary means.) On the other hand, if MATH is even, then it is easy to construct infinitely many elements of order MATH with MATH, and infinitely many elements of order MATH with MATH. So infinitely many central division algebras split over MATH, and infinitely many do not split. |
math/0106063 | There is some MATH, such that MATH, for all MATH (see REF). Let MATH. We have MATH for all MATH, so MATH. Furthermore, MATH so MATH. Let MATH so MATH is a subring of MATH. Because MATH, we see, by arguing as in the proof of REF , that MATH, where MATH. Then MATH so MATH. Therefore MATH. Hence, MATH spans MATH over MATH. The set MATH is linearly independent over MATH (as a subset of MATH). (For example, this follows from REF .) Therefore, MATH is linearly independent over MATH. Now, it is clear that MATH. |
math/0106063 | CASE: This is known as NAME 's Lemma. Let MATH. It is straightforward to verify that MATH is a MATH-submodule of MATH (because MATH). Hence, MATH is either MATH or all of MATH (because MATH is irreducible). Thus, MATH is either invertible or MATH. Because MATH is an arbitrary element of MATH, this means that MATH is a division algebra. REF Given a MATH-basis MATH of MATH, and any MATH, we wish to show that there exists MATH with MATH for MATH. That is, we wish to show that the MATH-submodule MATH of MATH generated by MATH is all of MATH. Let us consider only MATH, rather than MATH. The proof can be completed by induction (see REF). Given MATH, such that MATH we wish to show that MATH . Note that MATH is (obviously) a MATH-submodule of MATH. The projection of MATH to MATH is a MATH-submodule of MATH. It is nontrivial (because it contains MATH), so the irreducibility of MATH implies MATH . Also, the intersection MATH is a MATH-submodule of MATH. Because MATH is irreducible, there are only two possibilities to consider. Assume MATH. This means that MATH, so MATH . On the other hand, REF can be restated as MATH. Therefore MATH, as desired. Assume MATH. For each MATH, we know, from REF, that there exists MATH with MATH. Furthermore, the assumption of this case implies that MATH is unique. Thus, MATH is the graph of a well-defined function MATH. Because the graph of MATH is a MATH-submodule of MATH, it is straightforward to verify that MATH. Furthermore, we have MATH (because MATH). Therefore MATH. This contradicts the fact that MATH and MATH are linearly independent over MATH. |
math/0106063 | Let MATH be a maximal left ideal of MATH. (Since MATH is finite dimensional, it is obvious that maximal ideals exist. However, if MATH is a division algebra, then MATH.) Then MATH is an irreducible left MATH-module. The identity element MATH acts nontrivially on MATH, so the annihilator of this module is a proper ideal of MATH. Since MATH is simple, we conclude that the annihilator is trivial, so the module is faithful. Thus, the NAME Density REF implies that MATH. |
math/0106063 | Given MATH, define a MATH-algebra MATH with multiplication given by CASE: MATH for MATH and MATH, and CASE: MATH. (The cocycle identity REF implies that MATH is associative (see REF).) Now MATH may not be a division algebra, but it turns out that it is always a central simple algebra over MATH. So the NAME REF tells us that it is isomorphic to MATH, for some MATH, and some central division algebra MATH. Define MATH. It is easy to see that MATH if MATH and MATH are in the same coset of MATH (see REF), so MATH is well defined as a map MATH. A reasonable effort shows that MATH is injective. The proof of REF shows that MATH, so MATH splits over MATH. This means that the image of MATH is contained in MATH. An argument similar to the proof of REF shows that every element of MATH is in the image of MATH. |
math/0106064 | We shall prove the statement MATH. The proof of MATH is similar. Find an open locally finite covering MATH of MATH such that closures of elements of MATH form strong star-refinement of MATH and order of MATH is MATH. Put MATH. Let us construct a sequence of mappings MATH such that MATH extends MATH and MATH . Then we can let MATH since MATH. Suppose MATH has been already constructed. Since MATH holds, it suffices to define MATH on the "interior" MATH of each "simplex" MATH. Since MATH is locally finite and the "interiors" of "closed MATH-dimensional simplices" are mutually disjoint we can consider each simplex independently. Since MATH is a star refinement of MATH, there exists MATH such that MATH . Since the triple MATH is MATH-connected refinement of the triple MATH, there exists MATH such that the pair MATH is MATH-connected. Let MATH. For any MATH we have MATH and the property MATH implies MATH. Hence MATH and therefore MATH can be extended over MATH to a map MATH. We let MATH. Let us check property MATH. Since MATH refines MATH, for all MATH we have MATH. By MATH, any point MATH is contained in some "interior" MATH. Since MATH, we have MATH. |
math/0106064 | We shall construct MATH and MATH by reverse induction on MATH starting from MATH. Since all inductive steps are similar we shall show the constructions only for MATH. Since MATH is stable, for each MATH there exist open neighbourhoods MATH of MATH in MATH and MATH of MATH in MATH such that MATH. Since MATH is MATH-filtration there exist open neighbourhoods MATH of MATH and MATH of MATH such that MATH and the pair MATH is MATH-connected. We may assume that the covering MATH refines MATH. Let MATH be a locally finite strong star-refinement of MATH. For each MATH find MATH such that MATH. We shall also use notations MATH and MATH. For each MATH we put MATH. Let MATH be a strong star-refinement of MATH. Let us check that MATH is a stable singular neighbourhood of MATH. Consider any MATH. Find open neighbourhood MATH of MATH which intersects only finitely many elements of MATH. We may assume that MATH for some MATH. Put MATH. Since MATH it follows by the choice of MATH that for all MATH such that MATH we have MATH. Hence, using the fact MATH we obtain MATH. Finally, we have MATH by the definition of MATH. Let us show that MATH is MATH - connected refinement of the triple MATH. Consider any MATH. Find MATH such that MATH. There exists MATH with MATH. Take MATH. Then MATH and the pair MATH is MATH - connected. Finally, observe that by the choice of MATH and MATH we have MATH. |
math/0106064 | We shall prove statement MATH. The proof of MATH is similar. Let MATH be an arbitrary stable neighbourhood of the graph of MATH. Since MATH is stable, for each MATH there exist open neighbourhoods MATH of MATH and MATH of MATH such that MATH. Let MATH be a strong star refinement of MATH. For each MATH we let MATH. Let us check that MATH is a stable singular neighbourhood of MATH. Fix MATH and consider MATH which contains MATH. Then MATH where MATH is chosen so that MATH. Using REF , construct a MATH-connected sequence MATH. Observe that since MATH is stable singular neighbourhood of MATH, the graph MATH contains an open stable neighbourhood MATH of MATH. Suppose that MATH is a singlevalued continuous mapping such that graph of MATH is contained in MATH. Then MATH for all MATH. Hence we can apply REF and obtain singlevalued continuous mapping MATH extending MATH such that MATH for each MATH. This fact and the definition of MATH imply that graph of MATH is contained in MATH. |
math/0106064 | We shall construct MATH, MATH and MATH by reverse induction on MATH starting from MATH. Since all inductive steps are similar we shall show the constructions only for MATH. Since MATH is stable, for each MATH there exist open neighbourhoods MATH of MATH in MATH and MATH of MATH in MATH such that MATH. Since MATH is MATH-filtration there exist open in MATH neighbourhood MATH of MATH and open neighbourhood MATH of MATH such that MATH and the pair MATH is MATH-connected. We may assume that the collection MATH refines MATH. Put MATH. Let MATH be a locally finite covering of MATH which is a strong star-refinement of MATH. For each MATH find MATH such that MATH. For any MATH we put MATH. Let MATH be a strong star-refinement of MATH. Then similarly to the proof of REF we obtain that MATH is a stable singular neighbourhood of MATH and the triple MATH is MATH-connected refinement of the triple MATH . |
math/0106064 | Consider an embedding of MATH into the projection MATH of metric spaces where MATH and fix an arbitrary neighbourhood MATH of MATH in MATH. Let MATH be a compact MATH - connected MATH-filtration of MATH. Then the mapping MATH admits a compact MATH - connected MATH-filtration MATH. Since the mapping MATH is compact, MATH is a stable neighbourhood of the graph MATH. For each MATH find open neighbourhood MATH of MATH in MATH and open subset MATH of MATH such that MATH. Let MATH and MATH be a strong star refinement of MATH. We can define a stable singular neighbourhood MATH of MATH letting, as before, MATH for all MATH. By REF we can find MATH-connected sequence of triples MATH where MATH is a stable singular neighbourhood of MATH. Put MATH and show that the pair MATH satisfies lifting property. Consider an arbitrary mapping MATH where MATH is a paracompact space with MATH. We may assume that MATH is embedded into a projection MATH for some NAME space MATH such that MATH. For each MATH we let MATH and define open in MATH covering MATH of MATH and multivalued mapping MATH letting MATH for all MATH. It is easily seen that the sequence MATH is also MATH-connected. Hence we can apply REF to obtain a map MATH such that MATH for all MATH. Now we can define lifting map MATH on MATH letting MATH. Clearly MATH. It is easel seen from the construction and definition of MATH that MATH maps MATH into MATH. |
math/0106064 | Since the mapping MATH is lower MATH-continuous, there are positive MATH and a neighbourhood MATH of the point MATH such that MATH for every point MATH. Put MATH. Then for every MATH and every MATH we have inclusions MATH and MATH. Therefore, MATH. |
math/0106064 | For every point MATH take a number MATH and neighbourhoods MATH of the point MATH and MATH of the point MATH by REF . Choose a finite subcovering MATH of the cover MATH of compactum MATH and consider the corresponding numbers MATH and neighbourhoods MATH of the point MATH. Clearly, we can put MATH . The lemma is proved. |
math/0106064 | Using REF , we can find for every point MATH a number MATH and open neighbourhoods MATH of the point MATH and MATH of the compactum MATH such that MATH for any points MATH and MATH. Moreover, we may take a neighbourhood MATH to be so small that MATH is contained in MATH and MATH. Let us refine a locally finite cover MATH into the cover MATH and for every MATH take a point MATH such that MATH is contained in MATH. Let MATH be a continuous positive function such that for every point MATH we have MATH. Put MATH. Since MATH is contained in MATH and the sets MATH cover MATH, then MATH is a neighbourhood of the graph MATH. Consider an arbitrary point MATH. By the construction of MATH, there is a set MATH containing MATH such that MATH. Then MATH. Therefore, since MATH and MATH, we have MATH. |
math/0106064 | Consider a point MATH. For every point MATH we fix its open neighbourhood MATH refining MATH. Take a finite subcover MATH of the cover MATH of the compactum MATH and let MATH be its NAME number. We put MATH . Then for any points MATH and MATH the set MATH refines MATH. Consider an open locally finite cover MATH refining the cover MATH. For every MATH we fix an element MATH of the cover MATH such that MATH. Since the cover MATH is locally finite, the function MATH is locally positive. Let MATH be any positive continuous function which is less than MATH. Then we define MATH. |
math/0106064 | For any point MATH we have MATH. Then the distance between points MATH and MATH is less than MATH. Clearly, every element of the cover MATH containing the point MATH lies in the set MATH. Therefore, the star MATH is contained in the product MATH. The lemma is proved. |
math/0106064 | By REF there exist a neighbourhood MATH of the graph MATH and continuous positive function MATH such that MATH and for any points MATH and MATH we have MATH. By REF there is a covering MATH of the graph MATH such that the star MATH is contained in MATH. Define a continuous positive function MATH by the equality MATH. Consider a covering MATH which is starlike refined into MATH and MATH and such that the function MATH vary within any element of the covering MATH less than by half. Then for every point MATH the star MATH is contained in MATH. Indeed, the star MATH is contained in some element MATH of the cover MATH. Take a point MATH. By the construction of the cover MATH the set MATH is contained in MATH. By REF the star MATH is contained in MATH. Consider an arbitrary point MATH and suppose that the intersection of the set MATH with the fiber MATH is not empty and contains a point MATH. Then this intersection is contained in MATH. Since the point MATH lies in MATH, then MATH. Fix an element MATH of the cover MATH containing the star MATH. Clearly, the element MATH of the cover MATH contains the star MATH (we apply REF ) and the set MATH. |
math/0106064 | The multivalued mapping MATH which assigns the small star MATH to a point MATH has the open graph in the space MATH. Indeed, for a point MATH there is an element MATH containing the image MATH. Then by the upper semicontinuity of MATH, for some neighbourhood MATH of the point MATH, the image MATH is contained in MATH. Then the set MATH is an open neighbourhood of the point MATH in the graph MATH. Now the completeness and the lower MATH-continuity of mapping MATH imply these properties for the mapping MATH by the openness of the graph MATH. |
math/0106064 | By REF , it suffices to check property MATH for Polish spaces. Since any Polish space MATH with MATH admits closed embedding into Polish MATH-space of extension dimension MATH CITE, we may assume that MATH. Let MATH be a closed subspace of MATH and MATH be a continuous mapping. There is an open covering MATH of MATH with the following property: MATH for any point MATH and any its neighbourhood MATH in MATH there exists a neighbourhood MATH of MATH in MATH such that for all MATH if MATH then MATH CITE. Since MATH there exists an open refinement MATH of MATH where MATH is a countable discrete system of open disjoint sets CITE. For each MATH choose MATH such that MATH and define a mapping MATH on MATH as follows: MATH and MATH. It is easily seen that MATH is continuous. By induction on MATH we shall find neighbourhoods MATH of MATH in MATH and using MATH we shall extend MATH to MATH. Since MATH covers MATH the mapping MATH extends MATH to the neighbourhood MATH of MATH in MATH. Suppose that MATH has been already constructed. Since MATH, for each MATH there exists a neighbourhood MATH of MATH in MATH such that MATH is MATH-homotopic to a constant map in MATH. Applying to MATH property MATH of MATH find neighbourhood MATH. Put MATH and MATH. Observe that for all MATH we have: MATH is MATH-homotopic to a constant map in MATH provided MATH. We shall define MATH as an extension of MATH from the set MATH. Since the system MATH is disjoint, we can define MATH independently on every MATH. Consider an arbitrary MATH such that MATH. If MATH is open in MATH, choose a point MATH such that MATH and define MATH. Otherwise let MATH be an open neighbourhood of MATH in MATH such that MATH. Let MATH. Observe that MATH is MATH as an open subspace of MATH-space MATH. Hence MATH is MATH and therefore inclusion of MATH into the base of the cone can be extended to a map of MATH into this cone. By MATH there exists an extension of MATH to the set MATH. Let MATH be an extension of MATH such that MATH. Since MATH is discrete system it suffices to check continuity of MATH at every point MATH. Fix MATH. Since MATH and MATH is continuous mapping there exists neighbourhood MATH of MATH in MATH such that MATH is MATH-homotopic to a constant map in MATH-neighbourhood of MATH. Applying property MATH of MATH to MATH find neighbourhood MATH of MATH. Additionally, we may assume that MATH for some MATH such that MATH. For all MATH such that MATH and MATH we have MATH by the choice of MATH. Therefore construction of MATH and choice of MATH imply MATH. If MATH is open in MATH then by the construction we have MATH where MATH. Hence MATH in this case. Otherwise MATH was obtained as an extension of MATH from nonempty set MATH and it follows that MATH. Therefore MATH as required. |
math/0106064 | Suppose that the mapping MATH is not lower MATH-continuous at the point MATH of its graph. Then there exist a positive MATH and a sequence of mappings MATH, where MATH is a closed subset of paracompact space MATH of extension dimension MATH, such that MATH, the images MATH converges to the point MATH, the images MATH converges to the point MATH, and the mapping MATH can not be extended to a mapping of MATH into MATH. We consider a topological space MATH formed by the discrete union of all spaces MATH and a point MATH with the following topology: an open base at the point MATH consists of unions of this point and all but finite number of spaces MATH. Clearly, the space MATH is paracompact and MATH, while the set MATH is closed in MATH. Let MATH be a mapping such that MATH and MATH. Also, let MATH be a mapping such that MATH and MATH. These mappings are continuous and MATH. It is easy to see that we can not extend the mapping MATH over neighbourhood of MATH in MATH to a lifting of MATH with respect to MATH. Therefore, MATH is not locally MATH-soft. The first part of our lemma is proved. Let the mapping MATH be MATH-soft. We consider a point MATH and a mapping MATH of a closed subset MATH of some paracompact space MATH with MATH. Since MATH is MATH-soft, the constant mapping MATH admits a lifting MATH extending MATH. Thus MATH. |
math/0106064 | Consider MATH. Using REF choose sequence MATH of positive numbers and neighbourhoods MATH of MATH such that for all MATH and for any points MATH and MATH the pair MATH is MATH-connected. Let MATH be a finite covering of compactum MATH such that MATH for all MATH and choose MATH such that MATH. Let MATH. Fix MATH and consider MATH such that MATH where MATH has extension dimension MATH. Let MATH be an open covering MATH of MATH. Find an open locally finite covering MATH of MATH such that closures of elements of MATH form strong star-refinement of MATH and order of MATH is MATH. For each MATH find MATH such that MATH and pick MATH. Note that MATH. Letting MATH we shall inductively construct a sequence of mappings MATH, where MATH was defined in the beginning of REF, such that MATH extends MATH and MATH . Since MATH and MATH, MATH implies MATH. Moreover, MATH is MATH-close to MATH, since for any MATH we have MATH. Therefore, letting MATH we shall obtain desired mapping. Suppose that MATH has been already constructed. It suffices to define MATH on the "interior" MATH of each "simplex" MATH. Let MATH. By REF of MATH we have MATH. Further, since MATH for any MATH and MATH, we have MATH. Since MATH, we therefore obtain MATH. Therefore MATH . By the choice of MATH and MATH the pair MATH is MATH-connected. Hence the map MATH can be extended to a map MATH such that MATH. Let us check the property MATH. For any point MATH by the construction of MATH we have: MATH, as required. |
math/0106064 | Embed MATH into NAME space MATH and consider MATH as a mapping into MATH. Fix MATH and take a neighbourhood MATH of MATH in MATH. By REF there exist MATH and a neighbourhood MATH of the point MATH such that for any point MATH, for any space MATH of extension dimension MATH and its closed subset MATH, and for any mapping MATH there exists a mapping MATH such that MATH and MATH. Applying Homotopy Extension Theorem (see for example CITE) to MATH, we find a number MATH such that for any space MATH, any closed subspace MATH of MATH, and any two MATH-close maps MATH such that MATH has an extension MATH, it follows that MATH also has an extension MATH which is MATH-close to MATH. Using the MATH-property of the pair MATH in MATH, we take a number MATH such that the pair MATH is MATH-connected. By REF there exists MATH such that for any space MATH of extension dimension MATH and for any mapping MATH there is a mapping MATH with MATH. Put MATH. Consider a point MATH, a space MATH of extension dimension MATH and its closed subspace MATH. Now any mapping MATH is MATH-close to some mapping MATH which can be extended to a mapping MATH. Since MATH and MATH are MATH-close maps into MATH, MATH can also be extended to a mapping MATH which is MATH-close to MATH. Finally, there is another extension MATH of the mapping MATH. Thus, the pair MATH is MATH-connected. |
math/0106064 | If MATH is MATH-pair in MATH, consider a multivalued mapping MATH of the unit interval MATH defined as follows: MATH and MATH for any positive MATH. Clearly, MATH is lower MATH-continuous. Now REF implies the MATH-property of the pair MATH in MATH. Assume that MATH is MATH-pair in MATH. Take an open neighbourhood MATH of MATH in MATH and consider an open neighbourhood MATH in MATH such that MATH. By REF there exists MATH such that for any space MATH of extension dimension MATH and its closed subset MATH, and for any mapping MATH there exists a mapping MATH such that MATH and MATH. Using the MATH-property of the pair MATH in MATH, we can find a neighbourhood MATH of MATH in MATH. Put MATH. Now any mapping MATH of closed subset MATH of space MATH of extension dimension MATH can be extended to a mapping MATH. And by the choice of MATH there is an extension MATH of the mapping MATH. |
math/0106064 | There exists an embedding MATH which can be extended to an embedding of any Polish space containing MATH (see REF). If the pair MATH is MATH-connected in a Polish space MATH, then we can extend MATH to an embedding of MATH in MATH and the pair MATH is MATH-connected in MATH by REF . Consider any Polish MATH-space MATH, containing MATH. Extending MATH to an embedding of MATH into MATH, we obtain MATH-connectedness of the pair MATH in MATH by REF . |
math/0106064 | Consider MATH as a submapping of the projection MATH. Let MATH be some neighbourhood of a compact space MATH in MATH. We must find a neighbourhood MATH for MATH such that the pair MATH is MATH-connected. By the MATH-connectedness of the pair MATH, we fix an open neighbourhood MATH of MATH such that the pair MATH is MATH-connected. By approximate MATH-invertibility of the mapping MATH there exists a neighbourhood MATH of MATH such that any mapping MATH of the space MATH of extension dimension MATH admits a lifting map MATH. Now if MATH is a mapping of closed subset MATH where MATH, we take a lifting map MATH and extend it to a mapping MATH. Define an extension of MATH as MATH. |
math/0106064 | Let a sequence of compact sets MATH from the MATH-exponential of the pair MATH be convergent with respect to the NAME metric to a compact set MATH. Consider a neighbourhood MATH of MATH. There exists MATH such that MATH. Now MATH-connectedness of the pair MATH allows us to find a neighbourhood MATH of the compact set MATH such that the pair MATH is MATH-connected. |
math/0106064 | Since the exponential of an open set is open and the exponential of an intersection coincides with the intersection of exponentials, the exponential of a MATH-set is a MATH-set. Since the exponential of a closed set is closed, the exponential of a fiber closed in a MATH-set is closed in the exponential of a MATH-set. |
math/0106064 | By REF, there is a compactum MATH of extension dimension MATH and a continuous mapping MATH of MATH onto MATH such that every fiber MATH is MATH - compactum. By REF , the mapping MATH is approximately MATH - invertible. There exists MATH-compactum MATH containing MATH such that MATH CITE. It is easy to see from REF that the pair MATH is MATH-connected. Since the pair MATH is MATH-connected, we can extend the mapping MATH to a mapping MATH. Put MATH. Then the pair MATH is MATH-connected by REF . |
math/0106064 | Denote MATH, and for a point MATH fix a compact set MATH. Fix a positive number MATH. By REF there are number MATH and neighbourhood MATH of the point MATH such that the pair MATH is MATH-connected for any point MATH. Since MATH is lower semicontinuous and MATH is compact, there exists a neighbourhood MATH of the point MATH such that MATH for any points MATH and MATH (apply REF ). Let MATH be a neighbourhood of MATH such that MATH and MATH for every point MATH. Take any point MATH. By REF there exists a compactum MATH such that the pair MATH is MATH-connected, and therefore MATH. It remains to enlarge (if necessary) the compactum MATH to obtain a compactum MATH with MATH. By the choice of the neighbourhood MATH there is a finite set of points MATH in MATH such that MATH. We put MATH. |
math/0106064 | Consider MATH. According to REF , the mapping MATH has nonempty fibers. By REF , MATH is lower semicontinuous. By REF , MATH is fiberwise closed in MATH, and therefore, the completeness of this mapping follows from the completeness of the latter, which was established in REF . Then by the compact-valued selection theorem from CITE, the mapping MATH admits a compact selection MATH. Define a compact mapping MATH by the equality MATH. Since for any MATH, the pair MATH is MATH-connected, then the pair MATH is also MATH-connected. |
math/0106064 | The construction of filtration MATH is performed by induction with the use of REF . The initial step of induction consists in the construction of a compact submapping MATH. This can be done by the use of the compact-valued selection theorem from CITE since the initial term of the filtration MATH is lower semicontinuous. If compact MATH-connected filtration MATH has been constructed such that MATH for MATH, then the pair MATH satisfies the conditions of REF , and according to this lemma, we complete the construction of the filtration. |
math/0106064 | By REF we take for every point MATH an open neighbourhood MATH of the point MATH and an open neighbourhood MATH of the set MATH such that the set MATH is contained in MATH and the pair MATH is MATH-connected for every point MATH. Fix a closed neighbourhood MATH of the set MATH such that MATH. Let MATH be a locally finite open (in MATH) cover of MATH refining the cover MATH. For every MATH we take a set MATH such that MATH. Let MATH be a locally finite open cover starlike refining MATH. For MATH we define MATH . Since the cover MATH is locally finite, the set MATH is an intersection of finitely many open sets, and, therefore, MATH is open. Since for every MATH the pair MATH is MATH-connected, then the pair MATH is MATH-connected. Since the cover MATH is locally finite, then for every point MATH there is a neighbourhood MATH such that for any point MATH we have MATH. Therefore, for every MATH we have MATH. Thus, the set MATH is open. |
math/0106064 | Given an arbitrary number MATH and an open neighbourhood MATH of the graph MATH in the product MATH, consider a covering MATH of the graph MATH such that the star MATH is contained in MATH (REF is applied), while the function MATH does not exceed MATH. For a MATH-continuous mapping MATH and for its compact submapping MATH, applying successively REF , we construct the coverings MATH such that MATH is starlike MATH-connectedly refined into MATH for any MATH. By REF there is a neighbourhood MATH of the graph MATH in the product MATH such that for any point MATH, the star of this point relative to the covering MATH intersects the fiber MATH. By REF , there is a continuous singlevalued mapping MATH whose graph is contained in MATH. We fix a MATH-connected MATH-filtration MATH given fiberwise by the equality MATH. Since the projection of the star MATH onto MATH has the diameter less than MATH, then MATH. By REF the filtration MATH is complete and lower MATH-continuous. Finally, REF allows us to find a compact MATH-connected MATH-subfiltration MATH. |
math/0106064 | Let MATH be MATH-filtration of MATH. Denote MATH by MATH and take an arbitrary neighbourhood MATH of the graph MATH. Consider a MATH-subset MATH such that all fibers of MATH are closed in MATH and fix open sets MATH such that MATH. By induction with the use of REF , we construct a sequence of MATH-filtered mappings MATH and of open neighbourhoods of graphs of these mappings MATH such that for any MATH, the gauge MATH does not exceed MATH, while the graph MATH together with its neighbourhood MATH is in MATH. It is not difficult to choose the neighbourhood MATH of the graph MATH in such a way that the fibers MATH have the diameter not more than MATH. Then for any MATH and for any point MATH, MATH; this implies that MATH is a NAME sequence. Since MATH is complete, there exists a limit MATH of this sequence. The mapping MATH is singlevalued by the condition MATH and is upper semicontinuous (and, therefore, is continuous) by the upper semicontinuity of all the mappings MATH. Clearly, for any MATH the point MATH lies in MATH and is a limit point of the set MATH. Since MATH is closed in MATH, then MATH, that is, MATH is a selection of the mapping MATH. |
math/0106064 | Apply REF to the mapping MATH defined as follows: MATH . |
math/0106064 | For every MATH define a multivalued mapping MATH as follows: MATH . Then MATH is lower MATH-continuous, complete, and fiberwise MATH-connected MATH-filtration. By REF the mapping MATH contains a compact MATH-connected MATH-subfiltration. And application of REF completes the proof. |
math/0106064 | Put MATH. Using REF we find open neighbourhoods MATH of the graph MATH in MATH such that for any MATH the pair MATH is MATH-connected for every MATH. Let MATH be a closed neighbourhood of MATH contained in MATH. For every MATH define a multivalued mapping MATH by equality MATH. Then MATH is fiberwise MATH-connected MATH-filtration. As a closed subset of MATH, the space MATH is paracompact of extension dimension MATH. It is easy to see that every mapping MATH is lower MATH-continuous and complete. Applying REF we extend MATH to a selection of MATH over MATH. |
math/0106064 | The part "only if" is proved in REF . For the "if" part, consider a paracompact space MATH with MATH, its closed subset MATH, and continuous mappings MATH and MATH such that MATH. Then the multivalued mapping MATH defined as MATH is lower MATH-continuous and complete. By REF a selection MATH of MATH admits an extension MATH on some open neighbourhood MATH of the set MATH. If every set MATH is MATH, then filtration MATH is fiberwise MATH-connected and by REF we can assume that MATH is defined on MATH. Clearly, MATH is a lifting of MATH and theorem is proved. |
math/0106064 | Using REF , we can construct a sequence MATH of open in MATH neighbourhoods of the graph MATH such that MATH and for every MATH the pair MATH is MATH-connected for all MATH from some open neighbourhood MATH of the set MATH. We may assume that the set MATH is contained in MATH-neighbourhood of the graph MATH (for metric spaces MATH and MATH we equip the product MATH with a metric MATH). Take a sequence MATH of closed neighbourhoods of the set MATH such that MATH and MATH for every MATH. Put MATH. Define the maps MATH by the rule MATH for all MATH. Using REF , we obtain a continuous singlevalued selection MATH of the map MATH. Let the map MATH be given by MATH on MATH and by MATH on MATH. Since the graph MATH over the set MATH is contained in MATH (and, therefore, in MATH-neighbourhood of the graph MATH), we see that MATH is upper semicontinuous. |
math/0106064 | Consider proper continuous mapping MATH of separable metric spaces such that every fiber MATH is MATH-compactum and MATH (see REF). Denote by MATH the set MATH. Using REF we can find u.s.c. MATH-valued extension MATH of the mapping MATH which is singlevalued and continuous on MATH. Let MATH be positive continuous function on MATH such that MATH. Using REF from CITE, we can change the mapping MATH on MATH in such a way that new mapping MATH has the following properties: CASE: the restriction of MATH to the fiber MATH is an embedding for all MATH; CASE: MATH for all MATH. Upper semicontinuity of MATH easily follows from REF . Let the map MATH be given by MATH for all MATH. From REF it follows that MATH is homeomorphic to MATH-compactum MATH for all MATH. Clearly, MATH is upper semicontinuous. |
math/0106064 | Let MATH and MATH be given maps. Consider a mapping MATH of Polish space MATH with MATH and a mapping MATH of a closed subset MATH such that MATH. A multivalued mapping MATH is complete and lower MATH-continuous by REF . We have u.s.c. MATH-valued submapping MATH of the map MATH. By REF there is u.s.c. MATH-valued submapping MATH of MATH defined on some neighbourhood MATH of MATH such that MATH and MATH is continuous and singlevalued. Clearly, if the map MATH is MATH-soft, we may assume MATH. Then the mapping MATH extending MATH is singlevalued and continuous, and MATH. |
math/0106064 | Apply REF to the composition MATH, where MATH is a constant map. |
math/0106064 | Suppose MATH is closed and MATH is a map. We are going to find a continuous extension MATH of MATH. Let MATH be the cone over MATH with a vertex MATH. Denote MATH - a closed subspace of MATH. We define a multivalued map MATH as follows: MATH . Claim. MATH admits continuous singlevalued selection. If MATH is a continuous selection for MATH, then the mapping MATH defined by MATH is continuous. Since MATH for every MATH, we have MATH. Now if MATH denotes the natural retraction, then MATH is the desired continuous extension of MATH. Proof of the claim. Since MATH is Polish space, the space MATH is also Polish as well as its closed subspace MATH. Clearly, the graph of MATH is open in MATH, therefore MATH is complete. Lower MATH-continuity of MATH easily follows from the facts that the space MATH is locally contractible and MATH has open graph. Let us prove that the inclusion MATH is fiberwise MATH-connected. Fix a point MATH and consider a mapping MATH of closed subspace MATH of a space MATH with MATH. Since MATH is Polish space, by REF we may assume that MATH is a Polish space. It defines a mapping MATH by the formula MATH. Extend MATH to a set MATH letting MATH. Clearly, MATH takes the set MATH into MATH. Since MATH, we can extend MATH over the set MATH to take it into MATH. Finally extend MATH over MATH as a mapping into MATH-space MATH. Now define an extension MATH of the mapping MATH by the formula MATH. To find a continuous selection of MATH we apply REF to a MATH-filtration MATH. |
math/0106064 | Consider the NAME realcompactification MATH of the space MATH. Note that MATH (see CITE, CITE). By CITE, the realcompact space MATH can be represented as the limit space of a Polish spectrum MATH such that MATH for each MATH. Since MATH is MATH-embedded in MATH it follows that MATH coincides with the NAME realcompactification MATH of MATH. Next consider the inverse spectrum MATH, where MATH for each MATH with MATH. Since MATH is closed in MATH it follows that MATH. It is clear that MATH is MATH-embedded in MATH. This observation, combined with the fact that the spectrum MATH is factorizing, guarantees that the spectrum MATH is also factorizing. Now consider a continuous mapping MATH. Since MATH is Polish there exists a continuous extension MATH. NAME of the spectrum MATH implies that we can find an index MATH and a continuous mapping MATH such that MATH. Now recall that the pair MATH is MATH-connected and that MATH is a Polish space such that MATH. Consequently there exists a continuous extension MATH of MATH. Finally consider the composition MATH and let MATH. Straightforward verification shows that MATH. |
math/0106065 | Let MATH give an irreducible decomposition of MATH. Then, for the rigidity copairing MATH, we have MATH where MATH is the transposed map of MATH. By the associativity and the naturality of MATH-actions, we see that the composite morphism MATH is equal to MATH where MATH is given by the composition MATH . If we replace this with MATH and then compute MATH, we obtain the formula MATH . On the other hand, the definition of multiplication in MATH gives MATH for MATH. By using the obvious identity MATH the above expression takes the form MATH or equivalently we have another formula MATH proving the assertion. |
math/0106065 | We just check the compatibility of left and right isomorphisms: Given MATH-modules MATH and MATH, we shall prove the commutativity of the diagram MATH . By the associativity of the monoidal functor MATH the problem is reduced to the equality of compositions MATH . By an easy manipulation of transposed morphisms (no spherical normalization is needed here for rigidity), we see that these are the ones associated to the following composite NAME transforms MATH . Now the coincidence of these is further reduced to the equality of left and right transposed morphisms, which is a consequence of the involutiveness of antipodes for finite-dimensional semisimple NAME algebras REF . Given a vector MATH in the middle vector space, we need to identify the map MATH where MATH are NAME transforms of MATH and MATH respectively. Now REF shows that the morphism MATH is obatined by applying NAME transforms to MATH repeatedly. |
math/0106065 | Let MATH be an element associated to a simple MATH-module MATH. Then the composition MATH is given by MATH which is, by the naturality of MATH, equal to the composition MATH . We now compute how the operation works on vector spaces: MATH . Here the families MATH, MATH are chosen so that MATH and set MATH. Note that, if we denote by MATH the dual basis of MATH, then the family MATH is the dual basis of the basis MATH of MATH. By the relation MATH the above operation on vector spaces ends up with MATH . Since the morphism MATH is associated to the pairing MATH the above formula gives the result. To see the MATH-linearity, we again use the functoriality of trivializing morphisms and the problem is reduced to check the commutativity MATH that is, MATH, which is an immediate consequence of hook identities. |
math/0106065 | The first relation is obvious from definitions. On the tensor product MATH, the morphism MATH is given by MATH . According to this sequence of morphisms, we compute MATH as follows: MATH . Now, letting MATH and MATH be NAME transforms of MATH and MATH respectively, we have MATH because of MATH . Thus we have MATH which gives rise to the morphism MATH. |
math/0106065 | Let MATH belong to MATH, that is, MATH for any MATH. The commutativity is then equivalent to MATH . Removing the MATH factor, we have MATH for any MATH, MATH and MATH, which means the equality MATH for any MATH. If we take MATH and MATH with MATH, then the condition is reduced to MATH which is equivalent to MATH for any MATH, that is, MATH for any MATH. Thus, it is proportional to the identity morphism MATH. |
math/0106065 | The composite morphism MATH is given by MATH where the hook identity is used to get the expression MATH . Now we apply the associativity of MATH, MATH, to obtain MATH . |
math/0106065 | Consider the commutativity of the diagram MATH . The composite morphism MATH is given by MATH . By the naturality of the trivialization MATH, this composition can be described by MATH whence the problem is reduced to showing MATH . The commutativity of this diagram is then a routine work of NAME transforms: The longer circuit is given by MATH . By replacing the summation indices MATH and MATH by their NAME transforms MATH and MATH, we have MATH which is used to get MATH . A bit of care is needed for the right action: MATH . By using the previous lemma, the composite morphism MATH is given by MATH . By the naturality of trivialization, this is equal to MATH . If we compare this with the other composite morphism MATH then the problem is reduced to the commutativity of MATH which is now easily checked as before. A similar computation works for the MATH linearity. For example, the commutativity of MATH is reduced to that of MATH which holds if we define the morphism MATH without weights. |
math/0106065 | We shall check MATH. By the commutativity of left and right actions on MATH, we see that the composition MATH is given by MATH . From the definition of MATH, the morphism MATH is equal to MATH. Since MATH, we obtain the relation MATH and hence MATH by taking the summation over the set MATH. |
math/0106065 | We have just checked the former relation. By NAME reciprocity, this implies MATH and hence MATH by semisimplicity. |
math/0106067 | CASE: For MATH we have MATH hence MATH is left MATH-colinear. CASE: Let MATH be a MATH-linear retraction (respectively, section) of MATH. Then MATH is a left MATH-colinear retraction (respectively, section) of MATH. Assume first that MATH is a retraction of MATH. Then, for MATH hence MATH is a left MATH-colinear retraction of MATH. On the other hand, if MATH is a section of MATH, then for MATH that is, MATH is a left MATH-colinear section of MATH. |
math/0106067 | MATH . Let MATH be a total integral. We have to construct a natural transformation MATH that splits MATH. Let MATH and MATH, be the MATH-linear retraction of MATH given by MATH, for all MATH and MATH. We define MATH, that is, MATH for all MATH, MATH. It follows from REF that the map MATH is a left MATH-colinear retraction of MATH. It remains to prove that MATH is a natural transformation. Let MATH be a morphism in MATH. We have to prove that the diagram MATH is commutative. For MATH and MATH, using that MATH is right MATH-linear, we have MATH and using that MATH is left MATH-colinear MATH that is, MATH is a natural transformation that splits MATH. MATH . Assume that for any MATH the MATH-coaction MATH splits in the category MATH of left MATH-comodules and the character of the splitting is functorial. In particular, MATH splits in MATH, and let MATH be a left MATH-colinear retraction of it. Using the naturality of MATH, we will prove that MATH is also right MATH-colinear, where MATH and MATH are right MATH-comodules via REF. First, let MATH be a MATH-module and MATH. Then MATH via the structures arising from the ones of MATH, that is, MATH for all MATH, MATH and MATH. Using the naturality of MATH, we shall prove now that MATH . Let MATH and MATH . Then MATH is a morphism in MATH. From the naturality of MATH we obtain that MATH . Hence MATH that is, REF holds. In particular, let us take MATH and MATH viewed only as a MATH-module. Then MATH via the structures arising from the ones of MATH, that is, MATH for all MATH, MATH, MATH, MATH. With these structures the map MATH is a morphism in MATH. From the naturality of MATH the following diagram MATH is commutative, that is, MATH is also right MATH-colinear. MATH . The left MATH-coaction MATH is a MATH-bicomodule map. Let MATH be a split of MATH in MATH. In particular, MATH for all MATH, MATH. We define, MATH for all MATH, MATH. We will prove that MATH is a total integral. First, MATH that is, REF holds. We will prove that REF also holds. For MATH, MATH, the left hand side of REF is MATH . In order to compute the right hand side of REF we adopt the temporary notation MATH . Now, MATH . Hence, REF is equivalent to MATH for all MATH, MATH. Now, it is time to use the fact that MATH is also right MATH-colinear. Denoting MATH and evaluating REF at MATH, we obtain MATH hence, MATH . Now we apply MATH to the second factor of both sides. Using the fact that MATH, we obtain MATH . REF follows after we let the second factor act on the first one. |
math/0106067 | MATH is viewed as an object in MATH with the structures arising from the ones of MATH, that is, MATH for all MATH, MATH, MATH and MATH. First we shall prove that MATH is a MATH-split (even a MATH-colinear split) surjection. Let MATH, MATH, for all MATH. Then MATH is left MATH-colinear (but is not right MATH-linear) and for MATH we have MATH that is, MATH is a left MATH-colinear section of MATH. For MATH, MATH, MATH we have MATH that is, MATH is right MATH-linear. It remains to prove that MATH is also left MATH-colinear. First we compute MATH and MATH that is, MATH is left MATH-colinear. Hence, we proved that MATH is an epimorphism in MATH and has a MATH-colinear section. Now, taking a MATH-free presentation of MATH in the category of MATH-modules MATH we obtain an epimorphism in MATH where MATH. Hence MATH is a generator in MATH. |
math/0106067 | Using REF for MATH, the map MATH for all MATH, MATH is a left MATH-colinear retraction of MATH. In particular, MATH and MATH for all MATH and MATH. We define now MATH for all MATH, MATH. Then, for MATH we have MATH that is, MATH is still a retraction of MATH. Now, for MATH, MATH, MATH we have MATH hence MATH is right MATH-linear. It remains to prove that MATH is also left MATH-colinear: MATH that is, we proved that MATH is a retraction of MATH in MATH. |
math/0106067 | CASE: We shall prove that there exists a well defined left trace given by the formula MATH for all MATH. Taking MATH in REF we obtain MATH, that is, MATH, for all MATH. Now, for MATH and MATH hence MATH is a left MATH-module map and finally MATH hence MATH is a left MATH-module retraction of the inclusion MATH. CASE: Similarly, we can prove that the map given by the formula MATH for all MATH, is a well defined right trace of the inclusion MATH. |
math/0106067 | Straightforward: the unit and the counit of the adjointness are given by MATH for all MATH, MATH and MATH for all MATH, MATH and MATH. |
math/0106067 | MATH . Trivial. MATH is standard from the categorical point of view: a pair of adjoint functors (as MATH and MATH are) gives an equivalence of categories iff one of them is faithfully exact (or both of them are exact) and the adjunction maps in the key objects of categories (MATH in MATH and MATH in MATH) are bijective. We point out that MATH for all MATH, and MATH for all MATH, can be constructed from MATH and MATH using the naturality condition: for details, in a more general frame, we refer to REF or CITE (for NAME modules) or REF (for entwining modules). |
math/0106067 | Using the left quantum trace MATH we shall construct an inverse of MATH. We define MATH for all MATH. As MATH, MATH. Let now, MATH. Then MATH . It follows that (after we apply first MATH and then the flip map MATH) MATH . Now, if we multiply the last factors we get MATH . Hence we obtain MATH that is, MATH is an inverse of MATH. |
math/0106067 | As MATH is cosemisimple, there exists a left integral MATH on MATH with MATH CITE and the antipode of MATH is bijective. Then, using REF , MATH, MATH for all MATH, MATH is a total quantum integral and REF applies. |
math/0106067 | In REF we have shown that, under REF , the adjunction map MATH is an isomorphism for all MATH. It remains to prove that the other adjunction map, namely MATH, MATH is an isomorphism for all MATH. For this we shall try to adapt the proof from REF. Let MATH be a MATH-module. Then MATH via the structures induced by MATH that is, MATH for all MATH, MATH and MATH. In particular, for MATH, MATH via MATH for all MATH, MATH, MATH. We will prove first that the adjunction map MATH is an isomorphism for any MATH-module MATH. First, MATH and MATH via the usual MATH-actions MATH, and MATH for all MATH, MATH, MATH. The flip map MATH, MATH is an isomorphism in MATH. On the other hand MATH via MATH for all MATH, MATH and MATH. The flip map MATH, MATH, is an isomorphism in MATH. Applying REF for MATH, we obtain the following isomorphisms in MATH . The adjunction map MATH for MATH is an isomorphism, as it is the composition of the canonical isomorphisms MATH . Let MATH for all MATH, MATH. As MATH is surjective, MATH is surjective, because the diagram is commutative, where MATH is the canonical surjection. Let us define now MATH for all MATH, MATH. The map MATH is surjective, as MATH and MATH are. We will prove that MATH is a morphism in MATH, where MATH and MATH are quantum NAME modules via REF. Indeed, MATH and MATH for all MATH, MATH, MATH. Hence MATH is a surjective morphism in MATH. MATH is projective over MATH; hence MATH is projective as a right MATH-module, where MATH is a right MATH-module in the usual way, MATH, for all MATH, MATH, MATH. On the other hand, the map MATH is an isomorphism of right MATH-modules: here the first MATH has the usual right MATH-module structure and the second MATH has the right MATH-module structure given by REF. The MATH-linear inverse of MATH is given by MATH . In fact, MATH is the isomorphism given by REF, associated to MATH. We obtain that MATH, with the MATH-module structure given by REF, is still projective as a right MATH-module. It follows that the surjective morphism MATH splits in the category of right MATH-modules. In particular, MATH is a MATH-split epimorphism in MATH. Let now MATH. Then MATH via the structures arising from MATH, that is MATH for all MATH, MATH, MATH, MATH. On the other hand, MATH via the structures arising from the ones of MATH, that is, MATH for all MATH, MATH, MATH, MATH. We obtain that MATH is a MATH-split epimorphism in MATH. Applying REF for MATH we obtain that the map MATH is a MATH-split epimorphism in MATH. Hence, the composition MATH is a MATH-split epimorphism in MATH. We note that the structure of MATH as an object in MATH is of the form MATH, for the MATH-module MATH. To conclude, we have constructed a MATH-split epimorphism in MATH such that the adjunction map MATH for MATH is bijective. As MATH is MATH-split and there exists a total quantum integral MATH, we obtain that MATH also splits in MATH. In particular, the sequence MATH is exact. Continuing the resolution with Ker REF instead of MATH, we obtain an exact sequence in MATH which splits in MATH and the adjunction maps for MATH and MATH are bijective. Using the Five lemma we obtain that the adjunction map for MATH is bijective. |
math/0106076 | Let MATH be a two-rowed array in MATH, represented as in REF. Suppose that there are MATH entries in the first row of MATH that are larger than MATH, and that there are MATH entries in the second row of MATH that are larger than or equal to MATH, MATH. Equivalently, we have MATH and MATH . In particular, we have MATH. From REF it is immediate that we must have MATH. Conversely, given integer vectors MATH and MATH with MATH and MATH, by REF there are MATH possible choices for the entries MATH and MATH, MATH, in the first and second row of a two-rowed array which satisfies REF, and thus REF. This establishes REF. |
math/0106076 | We show this recurrence relation by decomposing an array MATH in MATH - the generating function of which is the left-hand side of REF - into two parts. Let MATH be the smallest integer with MATH, or, if all MATH are smaller than MATH, let MATH. Now we have to distinguish between two cases. If MATH, we decompose such an array into the array MATH in MATH, and the array MATH in MATH. Clearly, this is a pair of two-rowed arrays enumerated by the first sum in the right hand side of REF, with the summation index MATH equal to MATH. If MATH, we decompose REF into the array MATH in MATH, and a single row MATH . Note that, if MATH, this row is empty. These pairs are enumerated by the second sum on the right hand side of REF, with the summation index MATH equal to MATH. |
math/0106078 | By the NAME - NAME inequality we have that MATH . Thus MATH . On the other hand, equality holds in REF if and only if MATH is constant on the boundary and thus, by NAME 's result, MATH must be a ball. |
math/0106078 | Let MATH be the maximum of MATH in MATH, and assume that this is attained at an interior point. Define MATH . From the continuity of MATH we have that MATH is closed in MATH and that MATH is a continuous function of the radius. Let MATH be a point in MATH, and take MATH such that MATH is in MATH. We have that either there exists MATH in MATH such that MATH or not. In the second case, MATH for all MATH in MATH, which immediately yields that MATH is an interior point of MATH. In the first case, let MATH . Again by continuity, we have that MATH is positive and also that MATH. Hence MATH and thus MATH, contradicting the hypothesis. |
math/0106078 | Let MATH be such a continuous function. Then, in some point MATH in the interior of some ball MATH, MATH is bigger then the solution MATH of NAME 's problem for the the NAME equation in MATH having MATH as boundary data. Denote MATH. Let MATH and MATH and consider the usual mollifiers MATH supported in MATH, MATH a continuous cut-off function which is MATH in MATH and MATH in MATH and extend MATH by MATH in the complement of MATH. Then NAME 's inequality holds for MATH in MATH. To check this statement let MATH and notice that MATH . As MATH uniformly on MATH as MATH, we have, for sufficiently small MATH, MATH on MATH and MATH. Hence MATH has all the desired properties in MATH. |
math/0106078 | Assume MATH. NAME 's inequality can be rewritten as MATH . Let now MATH . As MATH we have that MATH which in turn equals MATH provided the latter limit exists. Now MATH where the last step follows by applying the divergence theorem. Thus, MATH . Since MATH we finally obtain that MATH and thus if MATH, it follows that MATH. |
math/0106078 | Again assume MATH. Assume NAME 's inequality holds for some smooth function MATH with MATH. Let MATH. Then MATH and MATH . Hence NAME 's property holds for MATH with MATH, hence it holds for MATH for any MATH and with MATH and consequently MATH for any MATH. But then MATH. |
math/0106082 | Note first that if we see MATH as a MATH-vector space it is clear that MATH. Pick a non-null polynomial MATH and let MATH. So MATH if MATH or MATH if MATH is a dependent set of MATH. If MATH then MATH. Suppose now that MATH is a dependent set of MATH. Then there is a circuit MATH. From REF we know that MATH. It is clear that MATH and so we have also MATH. |
math/0106082 | From REF it is enough to prove that MATH is a reduced NAME of MATH. Let MATH be any element of MATH, we have from REF that MATH . Let now remark that MATH and that these terms are all different. We have then clearly that MATH . Given an arbitrary MATH it is clear that MATH. So, MATH . Let MATH be a circuit of MATH such that MATH. So we have that MATH divides MATH and MATH is a NAME basis. Suppose for a contradiction that MATH is not a reduced NAME basis: that is, there exists two circuits MATH and MATH in MATH and an element MATH such that MATH divides MATH. First we can say that MATH because the sets MATH and MATH are incomparable. This in particular implies that MATH and MATH. On the other hand we have MATH so MATH a contradiction. |
math/0106082 | From REF , the reduced NAME bases constructed for the different orders MATH are all contained in MATH. We prove the minimality by contradiction. Let MATH be a circuit of MATH and let MATH be a permutation such that MATH. Then MATH it is not a NAME basis because MATH is not in MATH. |
math/0106082 | MATH . Let MATH be a term order of MATH. Since MATH is a universal NAME basis of MATH (see REF ) it is trivially a NAME basis relatively to MATH. We have already remarked that the leading term of MATH is MATH where MATH. From REF we conclude that MATH. MATH . Suppose that MATH. Let MATH be the degree lexicographic order of MATH determined by the permutation MATH. Note that MATH. From REF we know that MATH is the reduced NAME basis of MATH with respect to the term order MATH. Then MATH is the canonical basis for the reduced NAME basis MATH. MATH . It is a consequence of REF . |
math/0106082 | By a reordering of the elements of the matroid MATH we can suppose that MATH. It is clear that MATH so the proposition is a consequence of REF . |
math/0106082 | From the definitions we know that the composite map MATH, is the null map so MATH. We will prove the equality MATH. By a reordering of the elements of MATH we can suppose that MATH. The minimal broken circuits of MATH are the minimal sets MATH such that either MATH or MATH is a broken circuit of MATH (see REF). Then MATH . So MATH. There is a morphism of modules MATH . It is clear that the composite map MATH is the identity map. From REF we conclude that the exact sequence REF splits. |
math/0106082 | The first equivalence is easy to prove in both direction. To obtain the expression of MATH we just need to iterate MATH times the residue. This gives: MATH . After simplification we obtain the announced formula. The last result is clear. |
math/0106082 | Pick two elements MATH. Note that MATH (the NAME delta), from REF . The elements of MATH are linearly independent: suppose that MATH then MATH a contradiction. It is clear also that MATH is the dual basis of MATH. |
math/0106082 | By REF and MATH are true. We claim that MATH verifies MATH. Suppose for a contradiction that MATH and there is MATH such that MATH. Set MATH and MATH and suppose that MATH and MATH. Then there is a circuit MATH of MATH such that MATH . If MATH [respectively, MATH] we conclude that MATH a contradiction. So MATH is a diagonal basis of MATH. From REF we conclude that MATH is the dual basis of MATH. Suppose now that MATH where MATH and MATH. Then MATH and the remaining follows from REF . |
math/0106082 | We have for any MATH-subset of MATH. (This is the development of a determinant with two lines of REF.) For any rank MATH unidependent MATH of the matroid MATH we have MATH . Since the sum of the coefficients in these relations is REF and that these relations are generating, see REF , we can deduce that the sum of the coefficients in any relation in MATH is also equal to REF which concludes the proof. |
math/0106085 | MATH: Assume that MATH, MATH, are open covers of MATH. For each MATH, replacing each open member of MATH with all of its clopen subsets we may assume that all elements of MATH are clopen, and thus we may assume further that they are disjoint. For each MATH enumerate MATH. As we assume that the elements MATH, MATH, are disjoint, we can define a function MATH from MATH to MATH by MATH . Then MATH is continuous. Therefore, MATH is not dominating. Let MATH be a witness for that. Then for each finite MATH, MATH (that is, MATH) for infinitely many MATH. Thus, MATH is a MATH-cover of MATH. MATH: This was proved in CITE. For completeness, we give a minor variation of the original proof. Since MATH is continuous, MATH also satisfies MATH CITE. Consider the basic open covers MATH defined by MATH. Then there exist finite MATH, MATH, such that either MATH for some MATH, or else MATH is a MATH-cover of MATH. The first case can be split into two sub-cases: If MATH for infinitely many MATH, then for these infinitely many MATH, the set MATH is finite. Thus MATH cannot be dominating. Otherwise MATH for only finitely many MATH, therefore we may replace each MATH satisfying MATH with MATH, so we are in the second case. In the second case, since MATH and MATH is a MATH-cover of MATH, we have that each finite subset of MATH is contained in infinitely many elements of MATH. Define MATH by MATH. For each finite MATH, we have that MATH and thus MATH for infinitely many MATH. Then MATH witnesses that MATH is not dominating. |
math/0106085 | This follows from REF , together with the equivalences MATH, and the fact that MATH is closed under taking NAME images, see CITE. |
math/0106085 | We will prove the less trivial implication. Assume that for each MATH, MATH is not MATH-dominating. For each MATH, let MATH witness that MATH is not MATH-dominating. Since the collection MATH is countable, there exists MATH bounding it with respect to eventual dominance MATH. Then MATH witnesses that MATH is not dominating. |
math/0106085 | This follows from REF , see CITE for the proof of the corresponding NAME version. |
math/0106091 | By translation invariance we may set MATH. Also by a time translation and some redefinition of MATH if necessary we may set MATH. Fix MATH, MATH, MATH, MATH. In the region away from the light cone, that is, MATH or MATH, we can control MATH using REF and MATH using energy estimates to easily obtain REF. Now suppose we are near the light cone, so that MATH. We shall write the null form MATH in terms of the null frame MATH, MATH, MATH developed in the previous section. More precisely, we claim the pointwise estimate MATH . To see this, we substitute in the identities MATH and MATH, and observe that in each of the three null forms MATH, MATH, MATH the MATH term (which is the only potentially bad term) always vanishes. From the previous, REF and NAME we have MATH . The claim now follows from estimating MATH using decay estimates REF, and MATH using the standard energy estimate and REF. |
math/0106091 | Since MATH has frequency MATH, it will suffice to show that MATH for some suitable MATH which we can choose at will. Fix MATH. From REF we have the regularity estimate MATH for any MATH. We now divide the region MATH into several sub-regions and prove the localization estimate REF separately for each sub-region. CASE: MATH . In this case we are away from the light cone and one can use the fact that the vector fields MATH span. By REF we have MATH and hence by the regularity estimate REF MATH . Since MATH, we thus obtain REF for this region by setting MATH sufficiently large. CASE: MATH, and MATH. Without loss of generality we may take MATH, thus MATH. In particular MATH. The idea is now to use the fact that MATH together with MATH spans. We now use the identity MATH since MATH and MATH form a null frame, we thus informally have MATH . Since MATH, we thus obtain after repeated applications of the NAME rule MATH for any MATH. Applying the regularity estimate REF and substituting MATH, we thus obtain MATH . The claim follows by setting MATH sufficiently large. CASE: MATH, and MATH. In this case we use the fact that MATH together with MATH span. Specifically, we use the identity MATH . Since MATH by hypothesis, we thus informally have MATH . One then argues as in REF . |
math/0106092 | As before, we induct on the complexity of MATH. We may assume that the proposition is true for all graphs MATH of lower complexity and all heirarchies MATH. The inductive hypothesis is allowed to be vacuous when the complexity of MATH is zero. By the arguments in the preceding section, we may freely reverse edges in MATH (but not in MATH!), and assume that MATH contains no circuits as an undirected graph. We may assume that MATH does not contain any edge which has the same initial and final vertices as one in MATH, since the edge from MATH is either redundant or can be used to replace the edge in MATH, and in either case we can reduce the complexity of MATH. We divide into two cases, depending on whether MATH is non-empty or not. CASE: Suppose that MATH is non-empty. Then MATH contains an edge MATH for some MATH. By replacing MATH with its dual if necessary we may assume that MATH . If MATH (that is, the path MATH crosses the vertex MATH,) then REF and the fact that MATH is a heirarchy forces MATH to contain a directed circuit of positive weight, so that MATH vanishes. Hence we may assume that MATH. In particular we have MATH. Let MATH be the link of MATH. From REF we have MATH . The graph MATH can be written as the union of MATH and MATH. Since MATH is a heirarchy, one can verify using REF that MATH is also a heirarchy (in fact, the depth of any vertex either stays constant or increases). Thus this summand is acceptable by the induction hypothesis. Now consider the contribution of MATH. If the superior MATH of MATH is equal to MATH then one can see using REF that this graph contains a directed circuit of positive weight, so the contribution of this graph vanishes. Now suppose MATH. Let MATH be the link of MATH. Observe from REF that the edge MATH is redundant in MATH: MATH . The graph MATH can be written as the union of MATH and MATH. The latter graph can be verified to be a heirarchy using REF by similar arguments as before, and so this case is also treatable by the induction hypothesis. This concludes the treatment of REF . CASE: Suppose that MATH is empty. Define the total ranking to be the sum of rankings of all the vertices in MATH which are adjacent to MATH. We induct on the total ranking. If the total ranking is zero then MATH is semi-direct with dominant frequency MATH. Now suppose the total ranking is positive. Then MATH is adjacent to at least one vertex MATH in MATH. By reversing edges if necessary, we may assume that MATH contains an edge MATH of the form MATH. Let MATH be the link of MATH, and let MATH denote the edge MATH. By repeating the derivation of REF we have the triangle inequality MATH . The first paraproduct on the right-hand side is of the same form as MATH, but has the total ranking reduced by one, and so can be handled by the induction hypothesis. Since MATH is a tree, the second paraproduct is separable. This concludes the treatment of REF . |
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