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math/0010254 | The lcm exists by the preceding proposition. Let MATH be this lcm. Any divisor of MATH divides MATH, whence the result. |
math/0010254 | Implication ``only if" is clear. The converse is an immediate application of REF, taking MATH. |
math/0010254 | It is sufficient to define the automorphism on MATH. As MATH is a right multiple of all elements of MATH, for any MATH there exists a unique MATH such that MATH and there exists a unique MATH such that MATH, so that MATH. The map MATH is injective by cancellability in MATH and is thus compatible with the product. As MA... |
math/0010254 | The statement follows CITE and the same proof applies. |
math/0010254 | REF holds as it holds in MATH and as the product of two MATH-fixed elements is MATH-fixed. Let MATH, and let MATH such that MATH; then MATH is divisible by all elements in the orbit of MATH, so is divisible by their lcm (which exists). So MATH is generated by the lcms of MATH-orbits in MATH which exist, thus by MATH, a... |
math/0010254 | REF is a consequence of the associativity in the monoid MATH. REF are easy consequences of the existence of the length function MATH introduced in REF. The atoms are clearly, by construction, the elements of MATH. |
math/0010254 | Since MATH is convex, MATH is an lcm of MATH, as explained in REF. This eliminates the need to check REF . Let us now prove REF . Using the isomorphism in REF , and noticing that any pair of elements of MATH defines a non obstructing edge, we have to prove: for all MATH, the set MATH has a minimum element for MATH. We ... |
math/0010254 | Consider the group MATH given by the group presentation corresponding to the monoid presentation of MATH. As the partial product in MATH is a restriction of the one in MATH, there is a natural morphism MATH . By REF , this morphism is surjective. Note that it is easy to adapt NAME 's presentations to find a presentatio... |
math/0010254 | It is an immediate consequence of REF that REF holds for a generator of the form MATH where MATH and MATH are two consecutive points in a numbering of MATH. It follows that it holds for any generator MATH by using that MATH (which follows also from REF), and it follows thus for any element of MATH. REF is then a conseq... |
math/0010254 | The first three statements are obvious consequences of the previous definition. As MATH, we have, using REF , MATH . By REF , we have MATH . It is clear that MATH is equivalent to MATH. We have proved REF . |
math/0010254 | The injectivity part of the proposition has been proved in CITE. Let us prove the surjectivity. Let MATH. Multiplication by MATH coincides with conjugating by MATH, so, by REF , we know that MATH is generated by the MATH such that MATH. Using REF , this means that MATH. Thus any such MATH is in the image of MATH. So we... |
math/0010254 | For all integer MATH, let us denote by MATH the space of configurations of MATH points in MATH, with its natural topology. There is a natural inclusion MATH and a natural injection MATH defined by MATH. The action of MATH on MATH restricts to an action on MATH. Thus we have a commutative diagram of continuous maps: MAT... |
math/0010257 | The discussion in REF reverses easily to give a converse to REF . Precisely, if MATH is a MATH-sided MATH-stable ideal of MATH such that MATH and MATH is a MATH-equivariant quantization map such that MATH, then the formula MATH defines a MATH-equivariant graded star product on MATH. Thus it suffices to prove the follow... |
math/0010257 | MATH is generated by a maximal ideal in the center of MATH, it follows directly that MATH is MATH-stable and MATH iff MATH. |
math/0010257 | The NAME ideal is maximal by CITE, and this means MATH is simple. If MATH, MATH, then MATH is isomorphic to MATH, which is simple by CITE. |
math/0010257 | We have MATH where MATH is graded of degree MATH. Then MATH defines a MATH-linear map MATH. We know MATH - see the proof of REF . An easy fact about representations (from highest weight theory) is that if MATH appears MATH then MATH lies in MATH. So MATH if MATH. Thus we get REF where MATH. We have MATH. Computing the ... |
math/0010257 | CASE: Once we know REF , it is easy to compute MATH by induction on MATH. CASE: This is easy, in fact MATH is the coefficient of MATH in MATH. Notice that MATH is graded in negative degrees, so that MATH corresponds to MATH. |
math/0010257 | MATH is equivalent to the natural representation MATH of MATH on MATH; indeed MATH is an intertwining map. REF implies that MATH is simple (and vice versa). |
math/0010257 | This is a summary of the following results in CITE: REF , and REF . |
math/0010257 | This occupies REF. |
math/0010257 | Let MATH. We have two MATH-linear maps MATH defined by MATH and MATH. These must be proportional because MATH is MATH-dimensional. We know that MATH is non-zero by REF . So there is a unique scalar MATH such that MATH. |
math/0010257 | We have to go back into our explicit construction of MATH in CITE. We worked over the NAME open dense set MATH in MATH. We constructed MATH as the quotient MATH where MATH is a certain differential operator on MATH. More precisely, MATH where MATH and MATH is an explicit noncommutative polynomial in some vector fields ... |
math/0010257 | REF is immediate from REF . This gives REF if MATH (since MATH if MATH). Since the MATH transform in the adjoint representation, we get REF for all MATH. Finally REF follows because the map MATH, MATH, is MATH-invariant and so must be a multiple MATH of our normalized Killing form (see REF). Then MATH is non-zero by RE... |
math/0010257 | Suppose one of MATH or MATH left divides MATH so that the quotient is a differential operator MATH on MATH. Since MATH has order MATH REF , MATH has order MATH. But then the MATH span a copy of the adjoint representation in MATH which is different from the copy spanned by the MATH. This contradicts uniqueness in REF . |
math/0010257 | The function MATH is MATH-invariant and so vanishes on MATH. Now REF give MATH. This means (see REF) that MATH lies in MATH, and so MATH acts by MATH. |
math/0010257 | The pairing MATH is clearly sesquilinear and MATH-invariant with MATH. It follows by MATH-invariance that MATH is orthogonal to MATH if MATH. Now to show MATH is hermitian positive definite, it suffices to show that each number MATH, MATH, is positive. We will use REF so that we can treat all cases simultaneously. We m... |
math/0010257 | This follows by REF since MATH is an admissible MATH-module where MATH acts by MATH and MATH acts corresponding to the operators MATH. |
math/0010257 | Going back to REF , we find MATH where we are using the classical notation MATH. By definition, MATH where MATH is an orthonormal basis of MATH with respect to MATH. On the other hand, MATH where MATH is an orthonormal basis of MATH with respect to the hermitian inner product MATH. This is positive definite since MATH.... |
math/0010258 | CASE: We just need to show that MATH restricts to MATH, that is, if MATH and MATH belong to MATH then MATH belongs to MATH. MATH-invariance of MATH implies that MATH restricts to the MATH-finite part MATH of MATH. (MATH is the subalgebra consisting of functions which lie in a finite-dimensional MATH-stable subspace of ... |
math/0010258 | This follows since the momentum functions MATH, MATH, form a complete set of functions (that is, their differentials span the cotangent spaces) over some open dense set MATH in MATH. Indeed, the image of the moment map MATH is the closure of a single nilpotent orbit MATH, and we can choose MATH. |
math/0010258 | The proof occupies REF. |
math/0010258 | By complete reducibility, we can find a MATH-stable complement MATH to MATH inside MATH. This gives a MATH-stable splitting of the order filtration; let MATH be the corresponding quantization map. The spaces MATH may fail to be stable under MATH and/or complex conjugation. To remedy this, we ``correct" MATH by putting ... |
math/0010258 | MATH-invariance means that the operators MATH are skew-hermitian, or equivalently, the adjoint of MATH is MATH. So we want to show MATH . We have MATH; the last equality holds because MATH for any half-densities MATH. MATH-invariance of MATH means that MATH kills MATH if MATH, or equivalently MATH kills MATH if MATH. U... |
math/0010258 | The NAME module is MATH. We just established injectivity of MATH. NAME follows by CITE or by directly checking that the source and target contain the same irreducible MATH-representations with the same multiplicities. |
math/0010258 | Since the MATH generate MATH, we may assume MATH. Then REF gives MATH where MATH. This map MATH gives a well-defined anti-linear algebra involution of MATH; indeed we have MATH. |
math/0010258 | Clearly MATH is MATH-invariant. Then MATH is the unique invariant linear projection because the MATH-action on MATH is completely reducible and the constants are the only MATH-invariants in MATH (since the constants are the only MATH-invariants in MATH). MATH is MATH-invariant, that is, MATH. We write this as MATH. Ite... |
math/0010258 | MATH-invariance of MATH implies that MATH is MATH-stable. The other two follow easily using MATH and the properties MATH, MATH. |
math/0010258 | Since MATH is graded, it suffices to consider MATH. Let MATH and MATH denote respectively left and right MATH-multiplication by MATH. It is easy to check that the map MATH extends to a graded anti-linear algebra involution MATH of MATH; this follows because the nilpotent orbit MATH is MATH-stable. It follows by REF tha... |
math/0010258 | Certainly REF implies REF where MATH. Now suppose MATH and MATH. Because of orthogonality of the spaces MATH we find MATH. |
math/0010258 | Suppose MATH satisfies REF . Then MATH satisfies REF and so MATH for MATH. Equivalently, MATH if MATH. We claim that this uniquely determines MATH among all MATH-stable splittings of the order filtration of MATH. For it implies that the spaces MATH are orthogonal with respect to the symmetric bilinear pairing MATH. But... |
math/0010259 | Let MATH be the canonical algebra anti-isomorphism and let MATH be the NAME algebra anti-involution defined by MATH if MATH. Then MATH by CITE. This implies MATH. So we have parity if MATH. Otherwise parity is violated, already for MATH. Indeed, if MATH and MATH, then MATH, and so MATH while MATH. |
math/0010259 | It suffices to check that MATH maps MATH onto MATH (which is stated for MATH in CITE). This follows easily in any number of ways. For instance, the formula for MATH in CITE implies MATH. But MATH and MATH are complete sets of lowest weight vectors in MATH and MATH. |
math/0010259 | Let MATH be the normal ordering quantization map. The construction of MATH in CITE gives MATH where MATH and MATH are certain operators. Here MATH identifies with MATH in the usual way. In CITE they give a very nice formula for the MATH when MATH. Going back to CITE, we get a similar formula for all MATH. We find MATH ... |
math/0010259 | We just showed that MATH if MATH and MATH. This implies, if MATH is bidifferential, that MATH is a differential operator on MATH. Looking at our expression for MATH, we deduce MATH for MATH. But this forces MATH and MATH. By parity, MATH. |
math/0010259 | We return to the proof of REF . Let MATH and MATH, with MATH. Writing out REF termwise, we get, for MATH, MATH . More succinctly, MATH. For MATH, REF simplifies in that MATH factors cancel out. Then MATH where MATH and MATH. Then MATH. This is a formal relation, valid for MATH odd since then MATH is invertible. Similar... |
math/0010259 | Suppose MATH is odd. For MATH, REF gives MATH . The last equality follows because the operator MATH is graded of degree MATH. For MATH even, REF is still true on account of REF , except in the case where MATH and MATH. But if MATH then both MATH and MATH vanish for degree reasons and so the first and third expressions ... |
math/0010259 | This follows because MATH. |
math/0010266 | First we find an expression of the MATH from the MATH: MATH . Besides, MATH . To prove that MATH, we will establish that MATH . We have MATH . Therefore MATH where MATH . Multiplying by the transposed adjoint matrix and by MATH we obtain MATH and hence the equality follows. In a similar way MATH. Then both ideals are e... |
math/0010266 | To check the exactness of the resolution above, it is enough to consider a discrete filtration on that complex and to verify the exactness of the resulting resolution (see REF ). The same argument is used in CITE, REF to prove that the complex MATH is a free resolution of MATH (as a left MATH-module). But, for MATH, th... |
math/0010266 | We take the free resolution of MATH see CITE, CITE MATH where MATH is a basis of the MATH-module NAME(log-MATH), where MATH and, on the other hand, MATH is the syzygy matrix MATH . Applying the MATH functor to calculate the dual module, we obtain the sequence MATH where MATH is the right product by MATH. Hence, MATH is... |
math/0010266 | The proof of this result contains, as an essential ingredient, a re-reading of the demonstration of REF. As a matter of fact, we include some tricks of this demonstration. By REF , a free resolution of MATH is MATH where MATH is the matrix MATH . Hence, MATH. To guarantee that this vector space has dimension greater th... |
math/0010266 | As we pointed, if MATH is quasi homogeneous then MATH and therefore MATH is an isomorphism. Reciprocally, if MATH is an isomorphism, then MATH. Because of a result of CITE, we have MATH and, if we take into account REF , we obtain that MATH has to be quasi homogeneous. |
math/0010266 | If MATH is quasi homogeneous we have pointed yet that MATH is isomorphic to MATH. By REF MATH and then we have MATH where the first isomorphism is obtained in CITE (see also CITE) and the second one could be found in CITE. Reciprocally, if MATH is not quasi homogeneous then MATH and, as both are regular holonomic, neit... |
math/0010268 | The proof is essentially the same as NAME 's proof that the axiom of choice follows if MATH for all infinite cardinals MATH (compare CITE). NAME proved in CITE (compare also CITE) the following relation for infinite cardinals MATH: MATH . Notice that MATH implies that every set MATH can be well-ordered. Therefore it is... |
math/0010268 | First take an arbitrary set MATH. For REF note that a set MATH corresponds to the set MATH. For REF note that a finite one-to-one sequence MATH of MATH can always be written as MATH, which is an element of MATH. The relation REF is trivial. For REF let MATH, where MATH. Because MATH is assumed to be infinite, every MAT... |
math/0010268 | The only non-trivial part is MATH, which follows by the NAME Theorem. |
math/0010268 | Let MATH with support MATH. Because MATH is a support of MATH, for all MATH and every MATH we have MATH if and only if MATH. If MATH is neither finite nor co-finite, the sets MATH and MATH are both infinite and hence we find a MATH such that for some MATH, MATH. Now, if MATH is finite, then MATH must be a subset of MAT... |
math/0010268 | Assume there exists a one-to-one function MATH which belongs to MATH. Then, because MATH is symmetric, there exists a finite set MATH (a support of MATH) such that MATH. Now let MATH be such that MATH and let MATH be such that MATH. With REF we get that MATH and therefore MATH. So, MATH cannot be a support of MATH, whi... |
math/0010268 | CASE: Assume first that there exists a function MATH from MATH into MATH and let MATH be a support of MATH. Choose two arbitrary distinct elements MATH and MATH of MATH such that MATH and put MATH. Choose a MATH and a permutation MATH such that MATH and MATH (for MATH). Now, MATH but MATH, which implies either that MAT... |
math/0010268 | Let MATH be the set of atoms MATH of the ordered NAME model. MATH: It is obvious that the function MATH, defined by MATH, is a one-to-one function from MATH into MATH. Now assume that there exists also a one-to-one function MATH from MATH into MATH. Let MATH and MATH (for MATH). The MATH-sequence MATH is a one-to-one s... |
math/0010268 | First we show that MATH. For this it is sufficient to find a one-to-one function MATH from MATH into MATH. We define such a function as follows. For MATH where MATH and MATH let MATH . For any MATH and MATH we have MATH and therefore, the function MATH is as desired and belongs to MATH. Now assume that there exists a o... |
math/0010268 | Let MATH denote the cardinality of the set of atoms of the permutation model MATH constructed in the proof of CITE. Then in MATH we have MATH: The inequality MATH is CITE and because MATH is infinite, by REF , we also get MATH. To see that also MATH holds in MATH, assume that there exists (in MATH) a one-to-one functio... |
math/0010268 | If MATH, MATH and MATH, then we can define a MATH such that MATH, MATH, MATH, MATH and for any MATH we have MATH. |
math/0010268 | For a proof see for example, Theorem. REF. |
math/0010268 | For MATH let MATH be a finite set of atoms. Further, let MATH the set of all atomic MATH-formulas MATH such that we have MATH is either the formula MATH (for some MATH) or MATH (for some MATH) or MATH (for MATH and MATH). For an atom MATH let MATH thus, MATH is the set of all atomic formulas in MATH such that MATH hold... |
math/0010268 | First we show MATH. For MATH let MATH . By the construction of MATH, the function MATH belongs to MATH and is a one-to-one mapping from MATH to MATH. Hence, MATH and because (by REF ) MATH is provable in ZF, we get MATH. To see that MATH, notice first that by REF , the inequality MATH is provable in ZF, and therefore i... |
math/0010268 | Take an arbitrary MATH. Because MATH, we find an one-to-one MATH-sequence MATH of MATH. Define an equivalence relation on MATH by MATH and let MATH. For MATH let MATH, then, for every MATH, we have MATH and MATH if and only if MATH. We can consider MATH as a MATH-sequence of MATH by stipulating MATH if MATH and MATH if... |
math/0010268 | We will give the proof only for the former case, since the proof of the latter case is similar. Let MATH be an arbitrary natural number. It is obvious that we have MATH. So, for a MATH, let us assume that we also have a one-to-one function MATH from MATH into MATH. For MATH let MATH be a one-to-one MATH-sequence of MAT... |
math/0010268 | For a finite MATH, it is easy to compute that MATH. So, let us assume that MATH is infinite and take MATH. Because MATH has more than MATH elements we have MATH. Now let us further assume that there exists a one-to-one function MATH from MATH into MATH. First we choose MATH distinct elements MATH from MATH. Let MATH (f... |
math/0010268 | This follows from REF and the fact that MATH. |
math/0010268 | Note that MATH and therefore, by MATH, we must have MATH, and by MATH we get MATH. |
math/0010268 | NAME showed in CITE that if the real numbers are the countable union of countable sets, then MATH and MATH are incomparable. Furthermore, NAME gave in CITE a proof that MATH (see also CITE). Therefore we can decompose effectively the interval MATH into MATH disjoint non-empty sets and obtain a decomposition of the real... |
math/0010270 | From the short exact sequence MATH we obtain the MATH-action on MATH comes from a MATH-action. In particular, the element MATH acts on MATH as the NAME algebra element MATH. We define the action of MATH and MATH as MATH and MATH, respectively, and we need just to check that the relation MATH holds. But this follows fro... |
math/0010270 | Let MATH be the kernel of the map MATH and let MATH be the dual of MATH. From the axioms on the MATH's, we obtain that the composition MATH is on the one hand zero, and on the other hand equals the natural map MATH, which is a contradiction. The surjectivity of MATH is proved in the same way. |
math/0010270 | Let us view MATH-modules as objects of MATH via REF. Given an object MATH and an element MATH we define a new object MATH as follows: The underlying MATH-module is the same, that is, MATH. However, the corresponding morphism MATH is the old MATH composed with MATH, where MATH is viewed as an automorphism of the vector ... |
math/0010270 | Given two objects MATH and MATH in MATH we have to define their tensor product MATH as a new object of MATH. Consider first their naive tensor product MATH as a MATH-module. We claim that the algebra MATH acts on it by endomorphisms. Indeed, to define such an action, it is enough to define MATH-module maps MATH for eve... |
math/0010270 | We need to construct adjunction maps MATH for MATH and MATH in MATH and MATH, respectively. Let MATH be as above. Consider the composition MATH. By construction, this is a map of MATH-comodules and it obviously factors through MATH . Therefore, we obtain a map MATH. For MATH, consider the map MATH . This map respects t... |
math/0010270 | For an object MATH, consider the tensor product MATH. This is a left MATH-module and a left MATH-comodule via the diagonal co-action. Thus, we obtain a functor MATH, which is exact and faithful, since MATH was assumed faithfully flat over MATH. Now, the functor MATH considered as a functor from MATH to the category of ... |
math/0010270 | One direction is clear: if MATH is finite as a MATH-module, then MATH is finite-dimensional. Conversely, assume that MATH is finite dimensional, and let MATH be a finite-dimensional MATH-subcomodule, which surjects onto MATH under MATH . Then the MATH-submodule MATH in MATH generated by MATH is stable under both the MA... |
math/0010270 | First, it is known (compare REF or REF ) that the functor MATH is exact. Therefore, it is sufficient to show that the map MATH is surjective when when MATH is an irreducible MATH-module. However, it is known (compare REF ) that every irreducible MATH-module is of the form MATH for an irreducible MATH. Now, for MATH as ... |
math/0010270 | Set MATH, MATH. Set MATH. Then MATH is a co-algebra and a right MATH-module, and we have a sequence of epimorphisms MATH respecting both structures. We must show that the inclusion MATH is an equality. For this, it is enough to show that MATH is MATH-stable, that is, that the composition MATH maps to MATH. Indeed, by a... |
math/0010270 | Let MATH be a point over which the embedding MATH induces an injection on fibers MATH. We will call such MATH's ``good". First, we claim that for a ``good" MATH we do obtain an isomorphism MATH . Indeed, let MATH be any linear functional MATH which extends the evaluation map MATH corresponding to MATH. Consider the map... |
math/0010270 | Using REF, we will think of MATH in terms of MATH and we will construct a functor MATH. Let MATH be the action map. By assumption, we obtain the map MATH . The axioms on the MATH's imply that MATH acts on MATH as an associative algebra. We set MATH. Let us show now that MATH is canonically isomorphic to MATH. Under the... |
math/0010270 | Let MATH be a MATH-equivariant object in MATH. By definition, the underlying MATH-comodule has an additional commuting structure of a MATH-equivariant quasi-coherent sheaf on MATH. By taking its fiber at the point MATH, we obtain a MATH-comodule. Thus, we have constructed a functor MATH and it is easy to see that it is... |
math/0010270 | First, let us observe that if we put MATH, MATH, MATH, the corresponding triple would satisfy REF. Hence, the general REF is applicable as well as REF. Therefore, the category MATH is equivalent to the category of MATH-equivariant objects in MATH. However, the latter is by definition the same as MATH. |
math/0010270 | First, observe that MATH preserves each MATH, by assumption, since MATH. Secondly, let us show that MATH maps each MATH to itself. Indeed, let MATH and let MATH be a MATH-stable direct summand of MATH, which belongs to some MATH. Then MATH is preserved by the MATH-action, and thus defines a sub-object of MATH. But then... |
math/0010270 | For MATH, let us denote by MATH the restriction of MATH to MATH. By construction, it depends only on the class of MATH in MATH. Let us consider the forgetful functor MATH. Note that in terms of MATH and MATH, it acts as follows: MATH . This functor has a right adjoint, which we will denote by MATH. On the level of MATH... |
math/0010271 | The first conclusion follows immediately from REF and MATH of REF . Suppose MATH is an exposed point so that by REF of this corollary we have MATH where MATH is a spectral pair. Write MATH and MATH . If MATH, then there is a nontrivial projection MATH in MATH such that MATH. But in this case, we have MATH so that MATH ... |
math/0010271 | Fix MATH. We have then that MATH for each MATH and so MATH . Hence MATH and therefore, MATH . Now fix MATH so that MATH . If we had MATH for some MATH, then since MATH we would have MATH and therefore MATH contradicting MATH. Hence, MATH and so MATH . Finally, since MATH by hypothesis, we get that MATH contains a point... |
math/0010271 | Let MATH denote the family of all exposed faces that contain MATH. We have that MATH by REF above. Write MATH . We have that MATH is the intersection of a finite number of faces in MATH by REF . Since MATH this intersection is not empty and so MATH is an exposed face by REF . It is clear from its definition that MATH i... |
math/0010271 | The assertions are easily established using REF and a standard induction argument. |
math/0010271 | Observe that we have the relation MATH . Now fix MATH so that we have MATH for some MATH. If we write MATH then we have MATH and therefore MATH. Further, since every vector in MATH has the form MATH for some MATH, this same calculation shows that MATH is surjective. Finally, suppose MATH and MATH are vectors in MATH su... |
math/0010271 | Suppose MATH is a hyperplane in MATH that contains MATH and observe that since MATH, we may take MATH. In this case, we have MATH for each MATH. Now select MATH such that MATH. Setting MATH we get MATH and since MATH is faithful MATH. Similarly, we get MATH and so MATH. If MATH has dimension MATH, then there are MATH h... |
math/0010271 | We have MATH. Suppose that MATH has dimension one, that is, that there are scalars MATH such that MATH for each MATH. In this case, if MATH so that MATH, where MATH, then we have MATH . Since MATH, we get that MATH is the line segment with end points MATH and MATH and so MATH has dimension one. If MATH has dimension on... |
math/0010271 | Let us proceed by induction. As MATH is an exposed face in MATH, there is a spectral pair MATH such that MATH . If MATH, we are done. Otherwise, write MATH and let MATH denote the affine map defined in REF for the face MATH of MATH. We have that MATH is an exposed face in MATH and so by REF , there is a spectral pair-M... |
math/0010271 | Suppose MATH holds and MATH is the hyperplane of support that contains MATH. Since MATH we get MATH by REF and MATH of REF . Hence MATH. It is also clear from the above that MATH. Now suppose MATH holds. In this case since, MATH are the spectral interval projections for MATH and MATH we get MATH . Further we have MATH ... |
math/0010271 | The assertions are easily established using the definition of a sharp face and REF . |
math/0010271 | All of the assertions are simple consequences of REF and MATH of REF . |
math/0010271 | We have that MATH is a singleton if and only if MATH. In this case, in the notation of REF , we get MATH and so MATH has the desired matrix decomposition by REF . We have that MATH is an exposed point if and only if MATH, where MATH. Since MATH by REF , the final assertion follows. |
math/0010271 | For MATH, it is clear that there can be no more than MATH such vectors. Now suppose MATH is a maximal set of linearly independent vectors in MATH and fix any vector MATH in MATH . We must have MATH, where MATH. This means that both MATH and MATH are linear combinations of MATH and so MATH is a linear combination of MAT... |
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