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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 MATH is finite, it is surjective. If MATH, by surjectivity MATH cannot be the product of two non trivial elements of MATH, so has to be in MATH by REF . As MATH is a bijection of MATH, the above proof shows that MATH is a left multiple of all elements of MATH. So it is the left lcm of MATH. |
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, and we have arranged for elements of MATH to be atoms, so REF is satisfied. The length inherited from MATH is still compatible with the product so we have REF (but note that the elements of MATH may have length greater than REF even if all elements of MATH had length REF). The lcm of two elements of MATH is MATH-fixed by its uniqueness and is in MATH by REF, whence REF . If MATH and MATH are in MATH and have a lcm MATH and if MATH is such that MATH and MATH are in MATH then MATH is the lcm of MATH and MATH so is in MATH by REF, and is clearly MATH-fixed, whence REF . The monoid MATH is by definition a submonoid of MATH, so REF holds. Let MATH; the uniqueness of its normal form implies that each term is in MATH. On the other hand, if MATH, as MATH is the unique maximal element in MATH dividing MATH it is also the unique maximal element in MATH dividing MATH, so, by the definition of normal forms, the normal form of an element of MATH is a normal form in MATH. This shows that MATH identifies with MATH. The same argument shows that MATH identifies with MATH when MATH exists, as MATH is in MATH. |
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 discuss by cases: CASE: Either MATH. Then let MATH. Clearly, MATH, and any partition coarser than MATH and MATH must be coarser than MATH. CASE: Either MATH. Let MATH (the non-trivial part may have three or four elements). Clearly MATH. Now let MATH. Consider MATH the part of MATH in which MATH lies, MATH the part of MATH in which MATH lies. Because MATH and MATH, we have MATH and MATH. As MATH is non obstructing, we must have MATH (otherwise we would have MATH, which contradicts MATH). Thus MATH is finer than MATH. REF is proved similarly. To prove REF , consider the natural morphism MATH and its composition with the epimorphism MATH. Let MATH, MATH such that either MATH or MATH. Denote by MATH the image of MATH in MATH. We have MATH or MATH. As MATH is a group, this implies MATH. By REF , the restriction of MATH to MATH is injective, thus we have as required MATH. |
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 presentation of MATH where the generators corresponds to the elements of MATH. A set of defining relations is for example given in REF .: CASE: MATH when MATH is non obstructing, CASE: MATH for MATH coming in clockwise order. These relations are valid in MATH, thus in MATH, and the morphism MATH is an isomorphism. We conclude using the natural commutative diagram: MATH and the injectivity of the map MATH REF . |
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 consequence of REF. |
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 have proved that the image of MATH is MATH. |
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: MATH . According to REF ., the first line consists of homeomorphisms. Consider the following commutative diagram of group morphisms: MATH where MATH and MATH are defined by functoriality, as in the introduction. According to our REF , MATH is an isomorphism. What we want to prove is that MATH is also an isomorphism. This will result from the fact that both MATH and MATH are isomorphisms. CASE: The map MATH is an isomorphism: by an easy (and almost classical) argument, one can see that MATH is injective and identifies MATH with the subgroup of MATH consisting of elements whose associated permutations of MATH fix the point MATH. As clearly MATH is included in this subgroup, MATH is an isomorphism. CASE: The map MATH is an isomorphism: as noticed in CITE (fact MATH, used in the proof of REF , page REF), there is an exact sequence MATH where MATH is the free group on MATH generators and such that the action of MATH permutes without fixed points the images of the generators of MATH. As a consequence, the intersection of MATH with MATH is trivial, and MATH is injective. The surjectivity of MATH results from the surjectivity of MATH and from the MATH-equivariance of the diagram. |
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 following two statements. CASE: There exists a unique MATH-sided ideal MATH of MATH such that MATH and MATH. CASE: For such MATH, there exists a unique MATH-equivariant quantization map MATH. Notice that in REF , MATH follows automatically by uniqueness. To prove REF we need only elementary facts about MATH (see for example, CITE). The natural MATH-representation on MATH is multiplicity free. One can get a very quick abstract proof of REF just from this, but we will be more concrete. MATH is irreducible and carries the MATH-th NAME power MATH of the adjoint representation. Since the representation MATH occurs just once in MATH, MATH has a unique graded MATH-stable complement MATH in MATH; then MATH identifies with MATH. We define a vector space isomorphism MATH where MATH is the usual symmetrization map; here we only assume that MATH. Let MATH be the corresponding map from MATH to MATH. Then clearly MATH is a MATH-equivariant quantization map. If MATH is another such map, then the composition MATH satisfies: MATH implies MATH where MATH. But also MATH is MATH-linear and so the MATH-decomposition of MATH forces MATH. Thus MATH. The proof of REF breaks into two cases. If MATH is different from MATH, then as in CITE we take MATH to be the NAME ideal constructed in CITE. We may characterize MATH as the unique MATH-sided ideal in MATH whose associated graded is MATH. This is not the most familiar characterization, but it is immediate from the fact CITE that MATH is the only completely prime MATH-sided ideal such that MATH, and the equality CITE MATH . Then uniqueness of MATH implies that MATH. Now suppose that MATH, MATH. Let MATH be the algebra of global sections of the sheaf of twisted differential operators acting on local sections of the MATH-th power of the canonical bundle on complex projective space; this makes sense for any complex number MATH. We have a natural algebra homomorphism MATH. It is easy to write nice formulas for the twisted vector fields MATH, MATH, in local coordinates on the big cell MATH; see for example, CITE. Let MATH be the kernel of MATH. Then MATH is surjective and MATH (see CITE). All MATH-sided ideals MATH in MATH with MATH arise in this way. The principal anti-involution MATH carries MATH to MATH. So MATH satisfies the two conditions in REF . Assume MATH. Then we claim that MATH is MATH-stable iff MATH. To show this, we consider copies of the adjoint representation MATH. Since MATH appears (exactly) once in MATH, we see that MATH occurs twice in MATH and once in MATH. The copy of MATH in MATH corresponds, uniquely up to scaling, to a MATH-linear map MATH, MATH. Put MATH. Then the copy of MATH in MATH consists of elements MATH, where MATH is some function of MATH. A simple computation using the formulas for MATH mentioned above gives (for an appropriate choice of MATH) MATH. We have MATH while MATH. So MATH. Thus, if MATH then the unique copy of MATH in MATH is not MATH-stable and consequently MATH is not MATH-stable. This proves the claim and finishes the proof of REF . |
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 coefficients of MATH, we find MATH. Computing the coefficients of MATH, we get the relation MATH where MATH; so the MATH transform in the adjoint representation of MATH. |
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 on MATH which annihilate MATH. Also MATH annihilates MATH. It follows that for any MATH we have MATH and MATH. Now we can compute MATH. We have MATH where MATH and MATH. Then we find MATH. Let MATH. Then MATH by REF . Also, since MATH and MATH is a vector field we find (as in CITE) MATH . Using this we find MATH where MATH. Now we obtain MATH where MATH and MATH. This proves REF . We can prove REF by applying a certain automorphism. Let MATH be the NAME group homomorphism corresponding to the NAME algebra inclusion MATH where MATH is the span of MATH, MATH, and MATH. The adjoint action of MATH defines a NAME algebra automorphism MATH of MATH. Then MATH, MATH and MATH. Clearly MATH preserves MATH and hence induces algebra automorphisms of MATH and of MATH which we again call MATH. Then MATH and MATH. Now applying MATH to REF we get REF . |
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 REF ; choosing MATH and MATH we find MATH. |
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 may assume now, by rechoosing MATH if needed, that MATH; see CITE. Then by REF we have MATH . This number is positive, since MATH is positive by REF . For MATH we have, by REF , the operator MATH on MATH, where MATH and MATH. Then plainly MATH . This formula easily implies that the adjoint of (ordinary) left multiplication by MATH is MATH. Hence the operators MATH are skew-adjoint. But also the operators MATH are skew-adjoint since they correspond to the action of MATH. Thus, using REF , we see the operators MATH are all skew-adjoint. |
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. It follows, as in CITE, that MATH . So REF gives REF . |
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.) We will show that MATH. Certainly MATH lies in MATH since MATH is locally finite on MATH. Now MATH identifies with the space MATH of holomorphic functions on MATH which are homogeneous degree MATH polynomials on the cotangent fibers. This follows from compactness of MATH. We will show that any MATH in MATH extends to a holomorphic function MATH in MATH. Now MATH is the (finite-dimensional) space of holomorphic sections of a finite rank vector bundle MATH over MATH and MATH is the space of smooth sections of MATH. Both MATH and MATH are homogeneous vector bundles where MATH is the base fiber of MATH. Consequently we can identify MATH and MATH with certain spaces of functions MATH and MATH in the familiar way. If MATH is simply connected, every smooth MATH-finite function on MATH extends uniquely to a holomorphic (MATH-finite) function on MATH. Using this, it follows easily that MATH extends to MATH as desired. CASE: We have MATH by CITE - their result goes through to the twisted case with the same proof. So MATH. Now MATH lies in the MATH-finite part MATH of MATH, since MATH is locally finite on MATH. Using principal symbols, we find MATH, since MATH lies in MATH. (In fact this proves MATH.) Clearly MATH, and so MATH. |
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 and then MATH. Now MATH is a valid choice for MATH. If MATH is multiplicity free, then MATH is unique for each MATH, and so MATH is the unique choice for MATH. Notice that uniqueness of MATH does not require REF . |
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. Using this and the commutativity of holomorphic and anti-holomorphic operators we find MATH and so we get REF . For positive definiteness, we just need to show that MATH is MATH-MATH on MATH. We expect there is a geometric proof of this, but we have not worked that out. Instead, we will use results from representation theory. This argument will be clear for experts in these matters and too technical for everyone else; so we just sketch it briefly. MATH is MATH-equivariant and so MATH maps MATH into MATH. We have MATH where MATH. We can show that MATH is the same (up to scaling) as the map from MATH to MATH defined by NAME and NAME in CITE. (This is the ``MATH" case in their notation.) This follows easily because both maps are MATH-equivariant; here MATH acts on MATH by the twisted vector fields MATH. Next we need a suitable criterion for injectivity of the NAME - NAME map. We find it in NAME 's result CITE on injectivity of certain maps of induced modules into produced modules; see also CITE. |
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. Iteration gives MATH. This proves MATH . |
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 that the adjoint of MATH is MATH. Suppose MATH. Then the highest degree term in MATH, namely MATH, occurs in degree MATH. If MATH and MATH occurs in MATH, then MATH occurs in MATH and so MATH. Similarly MATH. So MATH. |
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 we know MATH is non-degenerate on MATH; this follows because MATH is MATH-stable and MATH is positive if MATH. So there is only one MATH-orthogonal splitting. This proves our claim. |
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 . Thus for MATH we have MATH where MATH denotes the graded star product defined by MATH and MATH. We find, directly from REF or using CITE, that MATH is given by REF . We know MATH where MATH. Now, for MATH and MATH, REF gives 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. Similarly, REF gives MATH where MATH. We put MATH. We put MATH. Let MATH and MATH. Now REF gives MATH where MATH . Each MATH, and so also their sum MATH, is a bidifferential operator on MATH with polynomial coefficients. That is, MATH lies in MATH where MATH and MATH. Now MATH is invariant under MATH; this is clear since MATH, MATH and MATH are all invariant. It follows by projective geometry (as in CITE) that MATH extends uniquely to a global MATH-invariant bidifferential operator on MATH. We have MATH where the arrows indicate that one set of values completely determines the next set. The middle arrow follows because any bidifferential operator on MATH is completely determined by its values on MATH CITE. Clearly MATH extends naturally (and uniquely) to an algebraic differential operator MATH on MATH; this amounts to replacing our NAME coordinates MATH by their holomorphic counterparts MATH. Then MATH is MATH-invariant and (by projective geometry again) extends to MATH. |
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 in REF are still equal. This proves REF , for all MATH, where MATH. Then MATH extends to an algebraic differential operator on MATH; this follows since both MATH and MATH so extend. The MATH, for MATH, all have the same order. This follows because the MATH, like the MATH, transform in the adjoint representation of MATH. We can choose MATH (the choice of MATH is arbitrary). Let MATH be the corresponding operator MATH. Using REF we find after some calculation MATH . So MATH. Clearly MATH has order MATH. Using principal symbols, we see that MATH has no left factors of the form MATH if MATH. For MATH, REF gives MATH, and so the only such factor is MATH. |
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 equal. |
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, the exact graded complex in the proof of CITE is precisely MATH where the matrices are MATH . And the last complex is the result of applying the same graduation to the resolution of MATH too, because MATH . |
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 the left MATH-module associated to the right MATH-module MATH, that is to say, MATH . Using REF , we deduce that MATH. The regularity of MATH follows from the regularity of MATH (compare CITE). |
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 than zero, it is enough to show that a pair of functions MATH such that MATH does not exist, that is to say, that MATH. Let us take MATH. As MATH, REF we will prove that, or MATH and MATH have no lineal parts, or that after derivation those lineal parts become REF. Of course MATH has no quadratic part: in that case, because of the classification of the singularities in two variables, MATH would be equivalent to a quasi homogeneous curve MATH, for some MATH. Then we can suppose that MATH where MATH and MATH. We will write MATH where the linear part MATH is MATH, and MATH is a matrix MATH with complex coefficients. If MATH, we have finished. Otherwise, the possibilities of the NAME form of MATH are MATH . As MATH is not an NAME vector (because MATH is not quasi homogeneous), we deduce: CASE: If we take the first NAME form, then (see the cited demonstration of CITE) MATH y MATH. After a sequence of changes of coordinates we have that MATH with MATH, that contradicts that MATH is reduced. CASE: For the second NAME form with MATH, it has to be MATH, that contradicts that MATH has its initial part of grade MATH. CASE: For the second option with MATH we have MATH and, in this situation, the linear of MATH is MATH. If we precisely apply MATH, we obtain REF. In a similar way, you prove the same for MATH. |
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, neither their NAME complexes are isomorphic, that is MATH using the NAME correspondence of NAME. |
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 sufficient to show that REF , which is weaker than REF , implies that for every infinite cardinal number MATH we have MATH. First we show that for two infinite cardinal numbers MATH and MATH we have MATH. For this, let MATH and MATH be such that MATH and MATH. Now we get MATH . It is easy to compute, that MATH and MATH . Now we apply REF to the cardinal MATH. If MATH, we get MATH which implies (by the above, according to the NAME Theorem) MATH. By the result of NAME mentioned above we get MATH. The case when MATH is similar. So, if REF holds, then we get MATH for every cardinal number MATH and therefore, each set MATH can be well-ordered, which is equivalent to the axiom of choice. |
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 MATH corresponds to a set MATH defined by MATH. |
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 MATH because otherwise we have MATH and we find again a MATH such that for some MATH, MATH. The case when MATH is co-finite is similar. |
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, which implies that the function MATH does not belong to MATH. |
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 MATH is not a function or that MATH is not a support of MATH. In both cases we get a contradiction to our assumption. The fact that MATH we get by MATH (see REF ) and by MATH (which will be shown in REF ). CASE: Because MATH, by REF it remains to show that MATH. Assume there exists a function MATH from MATH into MATH and let MATH be a support of it. - If for each MATH we have MATH, then we find MATH and MATH in MATH with MATH, and a permutation MATH such that MATH and MATH (for MATH). Now we get MATH and MATH, which contradicts our assumption. - Otherwise, there exists a set MATH with MATH, hence we find in the sequence MATH an element MATH which does not belong to MATH. Now let MATH be such that MATH (for MATH) and MATH, then MATH and MATH, which contradicts again our assumption. CASE: Because MATH is infinite we have (by REF ) MATH, which implies (by REF ) that MATH and it remains to show that MATH. Assume there exists a function MATH from MATH into MATH and let MATH be a support of MATH with MATH. Consider MATH, then, because MATH, it is easy to compute that MATH, which implies (by REF ) that there exists a MATH such that MATH is not a support of MATH. Let MATH, then MATH is a support of MATH, too. Choose a MATH and a permutation MATH such that MATH. Now, because MATH and MATH we have MATH, and by construction we get MATH. This implies either that MATH is not a function or that MATH is not a support of MATH and in both cases we get a contradiction to our assumption. CASE: By MATH and MATH we get MATH, and the inequality MATH follows from MATH and MATH. |
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 sequence of MATH, which implies that MATH, but this is a contradiction to REF . MATH: Because MATH is infinite, by REF we have MATH. MATH: For a set MATH, let MATH is a support of MATH, then MATH is a support of MATH, too; in fact, it is the smallest support of MATH. Using the order-relation MATH on the set of atoms MATH, we can define an ordering on the set of finite subsets of MATH as follows. For two finite sets MATH and MATH of MATH, where MATH and MATH (for MATH and MATH), let MATH iff either MATH or for MATH we have MATH. The ordering MATH on the finite subsets of MATH induces an ordering on the power-set of MATH (because every subset of MATH has a well-defined smallest finite support). Further, the order-relation MATH induces in a natural way an ordering on the set of all permutations of a given finite subset of MATH and we identify a permutation MATH of a finite subset MATH with MATH. Now we choose MATH distinct atoms MATH of MATH and define a function MATH from MATH into MATH as follows. For MATH with MATH, let MATH be the MATH-th permutation of MATH, where MATH is the MATH-th subset of MATH with smallest support MATH (this we can do because for MATH we have MATH). If MATH for MATH (where MATH), then we choose the first REF elements (with respect to MATH) of MATH which are not in MATH, say MATH and put MATH, where MATH is the MATH-th permutation of MATH and MATH is the MATH-th subset of MATH with smallest support MATH. By REF , the function MATH is a well-defined one-to-one function from MATH into MATH. If there exists a one-to-one function from MATH into MATH, then, because MATH for MATH, we can build an one-to-one MATH-sequence of MATH, which is a contradiction to REF . MATH: Because each one-to-one sequence of MATH is a sequence of MATH, we have MATH. Now assume that there exists also a one-to-one function MATH from MATH into MATH. Choose an arbitrary atom MATH and let MATH, where MATH denotes the sequence of MATH of length MATH. Because for every MATH, the sequence MATH is a one-to-one sequence of MATH, for every MATH there exists a MATH and a MATH such that MATH occurs in MATH but for MATH, MATH does not occur in MATH. Because a sequence is an ordered set, with the function MATH we can build an one-to-one MATH-sequence of MATH, which contradicts REF . |
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 one-to-one function MATH from MATH into MATH and let MATH be a finite support of MATH. Without loss of generality we may assume that if MATH, then also MATH. Let MATH and for MATH let MATH. Let MATH and let MATH, where MATH is the least natural number such that for every coloring MATH we find a MATH-element subset MATH such that MATH is constant. (If MATH are natural numbers such that MATH and MATH, then by the NAME Theorem ( compare CITE), MATH is well-defined.) Choose MATH distinct elements MATH, let MATH and let MATH REF be an enumeration of MATH. We define a coloring MATH as follows. For MATH such that MATH let MATH where for MATH we define MATH . By the definition of MATH we find MATH elements MATH with MATH such that for MATH, MATH is constant on MATH. So, for MATH with MATH and for MATH, we are at least in one of the following cases: MATH . If we are in REF or REF , then MATH, and therefore MATH is not a one-to-one function. If we are in REF , then MATH is also not a one-to-one function because MATH. If we are in REF , let MATH be such that MATH and MATH. Assume MATH (the case when MATH is symmetric). Then we have MATH, but MATH, and therefore MATH is not a function in MATH. If we are in REF , let MATH be such that MATH and let MATH. Take an arbitrary MATH and let MATH be such that MATH and MATH. Then we get MATH but MATH, and therefore MATH is not a function in MATH. If we are in REF , let MATH be such that MATH, thus MATH for some MATH. Let MATH be such that MATH. Then we have MATH but MATH, and therefore MATH is not a function in MATH. So, in all the cases, MATH is either not a function or it is not one-to-one, which contradicts our assumption and completes the proof. |
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 function from MATH into MATH. Such a one-to-one function would generate a function MATH from MATH into MATH, but because MATH - as an element of MATH - has a finite support, this is impossible. |
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 holds in MATH. Take an arbitrary MATH in MATH and let MATH be a support of MATH. If MATH are such that MATH, then we find (by construction of MATH and MATH) a permutation MATH such that MATH and therefore we have MATH if and only if MATH. Hence, the set MATH is determined by MATH, which is a finite set of finite sets of atomic formulas. Now we show that if MATH and MATH are two distinct supports of a set MATH, then MATH is also a support of MATH. If MATH or MATH, then it is obvious that MATH is a support of MATH. So, assume that MATH and MATH are both non-empty and let MATH. Take an arbitrary MATH and let MATH. Let MATH be any atom such that MATH. We have to show that also MATH. If MATH belongs to MATH (and thus MATH), then MATH and we have MATH. So, assume that MATH does not belong to MATH (for any MATH). If MATH does not belong to MATH, let MATH. Otherwise, if MATH belongs to MATH, because MATH and MATH we have MATH. By construction of MATH we find a MATH such that MATH and MATH, hence, MATH. Now let MATH . Because MATH we find (again by construction of MATH) a MATH such that MATH and MATH. Now, by MATH we have MATH, by MATH we have MATH, and because MATH we finally get MATH. Hence, MATH is a support of MATH and by construction it is unique. |
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 it is enough to find a one-to-one function from MATH into MATH which lies in MATH. For each finite set MATH, let MATH be an enumeration of the set MATH. (The function MATH exists as MATH is a linear order on the finite set MATH.) Then by REF and its proof, for each finite set MATH, MATH induces a mapping from MATH into MATH, for some MATH. Now fix two distinct atoms MATH and let MATH be defined as follows: MATH such that MATH and the length of the sequence MATH is equal to MATH, where MATH maps MATH to MATH. The function MATH is as desired, because it is a one-to-one function from MATH into MATH which lies in MATH. |
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. Now we define an ordering on the set MATH as follows: MATH . This is a total order on the set MATH. Let MATH, then for each MATH the set MATH is a set of MATH-sequences of MATH. The order relation MATH defines an ordering on each MATH and we must have one of the following two cases: Case: For each MATH, MATH is well-ordered by MATH. Case: There exists a least MATH such that MATH is not well-ordered by the relation MATH. If we are in casethen we find a well-ordering on MATH. Let the ordinal MATH denote its order-type, then MATH (otherwise the MATH-sequence MATH would not be one-to-one) and therefore we can build a one-to-one MATH-sequence MATH of MATH. If we define MATH, then the set MATH is a set of pairwise disjoint subsets of MATH of cardinality MATH. Therefore, the cardinality of MATH is MATH and because for MATH the function MATH is a one-to-one function, we get MATH. If we are in caselet MATH be the least natural number such that MATH is not well-ordered by MATH. Let MATH. Then MATH has no smallest element, too. For MATH we define MATH as follows. If MATH, then MATH; otherwise, MATH. By construction, for every MATH, the set MATH is not the empty set and it is not well-ordered by MATH. Thus, for every MATH there exists a MATH such that MATH is a proper subset of MATH. Now let MATH be such that for all MATH we have MATH and let MATH. Then the set MATH is again a set of pairwise disjoint subsets of MATH of cardinality MATH and we can proceed as above. |
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 MATH and let MATH. We can order the set MATH as follows: MATH iff either MATH or MATH. Because MATH and MATH, we have MATH and hence there exists a first MATH (with respect toMATH), such that the second component of MATH does not belong to MATH. Now we define MATH and the MATH-sequence MATH is a one-to-one sequence of MATH. Repeating this construction, we finally get an one-to-one MATH-sequence of MATH. But this is a contradiction to MATH. So, our assumption was wrong and we must have MATH. |
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 (for MATH) and MATH, then MATH is a set of pairwise disjoint, non-empty subsets of MATH such that MATH. Let MATH be a one-to-one sequence of MATH of length MATH. With respect to the sequence MATH we define an equivalence relation on MATH as follows. MATH if and only if for all MATH: MATH. For MATH let MATH and let MATH be such that MATH if and only if MATH. Notice that we have MATH if and only if MATH. We define an ordering on the set of equivalence classes by stipulating MATH if there exists a MATH such that MATH and for all MATH we have MATH. Further let MATH, then MATH is a set of pairwise disjoint, non-empty subsets of MATH such that MATH and MATH. Let us assume that we already have constructed a set MATH (for some MATH) where MATH and MATH for some MATH. Every partition of MATH induces a partition of MATH (this is because of the properties of MATH) and hence we get a one-to-one mapping MATH from MATH into MATH. Notice that the ordering MATH on MATH induces an ordering on MATH. Because MATH we have MATH and therefore we find a first partition MATH of MATH (first in the sense of the ordering on MATH) such that the set MATH is not the union of elements of MATH. We define MATH, MATH and MATH. Repeating this construction, we finally get an one-to-one MATH-sequence of MATH. But this is a contradiction to MATH and therefore we have MATH and by MATH we get MATH. |
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 line into MATH disjoint non-empty sets. If MATH, then MATH. Hence, in the model of NAME and NAME (compare CITE) - in which the real numbers are the countable union of countable sets - we find a decomposition of the real line into more than MATH disjoint non-empty sets (see also CITE). |
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 from the formula MATH and all the terms but MATH belong to the two-sided ideal generated by the MATH's and the MATH's. |
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 space MATH. It is clear that in this way we indeed obtain an action of MATH on MATH, and hence on MATH-comod, by endo-functors. |
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 every MATH, compatible with the tensor structure on MATH in the same sense as in the definition of MATH. The sought-for maps are defined as follows: MATH where the second arrow comes from the braiding on the category MATH. The MATH-module MATH is defined as the fiber at MATH of MATH viewed as a quasi-coherent sheaf on MATH. It comes equipped with a data of MATH by construction. It is easy to see that the functor MATH admits a natural associativity constraint, which makes MATH into a monoidal category. Moreover, if both MATH and MATH are finitely generated as MATH-modules, then so is MATH. Hence, this monoidal structure preserves the sub-category of finite-dimensional MATH-comodules, which is the same as MATH. |
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 the MATH-coaction and the MATH-action by construction. Thus, we obtain a map MATH . |
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 vector spaces can be factored as MATH where MATH is the corresponding functor for MATH. Therefore, it suffices to show that MATH is exact and faithful. However, the triple MATH satisfies assumption (iii b), and we already know that MATH induces an equivalence between MATH and the category of vector spaces. In particular, MATH is exact and faithful. |
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 MATH-action and MATH-coaction, and MATH surjects onto MATH. Hence, MATH. |
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 above the co-action map defines a map MATH, and the composition MATH is the identity map. Hence, MATH is a surjection. |
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 applying MATH to MATH we then obtain that MATH projects to the MATH element in MATH, that is, belongs to MATH. Using the fact that MATH i surjective for any MATH-comodule, we can find a MATH-comodule MATH with a surjection MATH. However, from REF , we obtain that in MATH the composition is an isomorphism. Hence MATH, which implies that MATH for any MATH. In particular, if MATH and MATH are two MATH-comodules, any map MATH respecting the MATH-coaction, respects also the MATH-coaction. Applying this to the composition MATH, we obtain that MATH is a MATH-subcomodule. |
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 . It is easy to see that this map defines the sought-for isomorphism MATH. Now let us show that all MATH are ``good". Suppose not. Since MATH is an embedding , there exists a collection MATH of proper sub-schemes of MATH defined over MATH, such that all points in MATH are ``good". By what we proved above, the translation by a ``good" MATH maps the collection MATH to itself. Let us make a field extension MATH. Over this field, MATH has the canonical generic point, which is clearly ``good". However, this generic point cannot map a proper sub-scheme defined over MATH to another proper sub-scheme defined over MATH, which is a contradiction. |
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 equivalence of REF, the functor MATH goes over to MATH. Therefore, MATH . |
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 an equivalence. |
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 is a non-zero direct summand of MATH, which means that MATH. Let MATH be an object of MATH. Let MATH be a block decomposition in MATH. Let us show that MATH and MATH are in fact MATH-sub-comodules. Without restricting the generality, we can assume that MATH is a sub-comodule of MATH for some MATH. However, as we have just seen, the block decomposition of MATH coincides with the block decomposition of MATH. Therefore, the block decomposition of MATH is ``coarser" than that of MATH. However, by our assumption on MATH, its block decomposition is ``coarser" than the block decomposition of MATH. This implies the assertion of the proposition in view of REF. |
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 is the natural forgetful functor. Note that for MATH we have: MATH . (Recall that MATH is a MATH-dimensional MATH-module, and, hence, we are allowed to tensor any MATH-module by it on the right.) In particular, MATH. Therefore, two irreducible objects MATH and MATH of MATH belong to the same block if and only if there exists MATH, such that MATH and MATH belong to the same block of MATH, that is, MATH and MATH belong to the same orbit of the extended affine NAME group MATH. Thus, we obtain that the functor MATH maps MATH to MATH. We claim now that this functor has a left quasi-inverse. Namely, it is given by MATH where MATH denotes the functor of projection onto the regular block in MATH. Indeed, for MATH we have: MATH because for MATH, the object MATH is non-zero only if MATH, which follows from the description of blocks of MATH in terms of MATH. To finish the proof of the proposition it remains to show that if MATH is a non-zero object in MATH, then MATH is non-zero either. For that, it is enough to suppose that MATH is irreducible, that is, of the form MATH, and our assertion follows from the explicit description of MATH given above. |
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 and MATH. As this is a contradiction, we get MATH, as desired. Conversely, if MATH is a spectral pair such that MATH and MATH, then MATH is a singleton and so this face is an exposed point. Finally, by REF and the fact that MATH is faithful, we have that MATH exactly when the face MATH has positive dimension. |
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 on the boundary of MATH. Further, if MATH, then we have MATH and so MATH. Hence, MATH is a hyperplane of support for MATH, and MATH is an exposed face of MATH. |
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 is the unique minimal exposed face containing MATH. |
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 such that MATH. As MATH and MATH are in MATH, we may find operators MATH and MATH in MATH such that MATH and we may write MATH where MATH. With this we have MATH by MATH. As MATH we get MATH and so MATH is injective. Hence, MATH implements an affine isomorphism of MATH onto MATH and therefore MATH . |
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 hyperplanes of the form MATH that contain MATH and such that their normal vectors MATH are linearly independent. In fact we must have that the MATH's are linearly independent. To see this, suppose that there are real numbers MATH such that MATH . In this case we get MATH . But, then MATH would be a linear combination of the remaining normal vectors contradicting their linear independence. For the converse, if real scalars MATH and linearly independent vectors MATH exist such that MATH for all MATH, then it immediately follows that every MATH in MATH is orthogonal to each vector MATH and so the subspace spanned by MATH has dimension at most MATH. |
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 one, then by REF MATH has dimension one. Thus by REF there are MATH linearly independent vectors MATH and real numbers MATH such that MATH. Since the MATH's are linearly independent, we get that each MATH is a multiple of MATH. Hence assertion MATH holds. Now suppose MATH has dimension two so that MATH has dimension two by REF . In this case, by REF , there are MATH linearly independent vectors MATH and real numbers MATH such that MATH for MATH. This means that there is a vector MATH such that MATH and so MATH is abelian. |
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-MATH such that if MATH and MATH denote the associated cut down spectral interval projections, then we have MATH . Hence, by REF we have MATH . If MATH, then we are done. Otherwise, suppose that spectral pairs MATH and sequences MATH have been determined as above such that if MATH and MATH then we have MATH . With this, we may write MATH, MATH and let MATH denote the associated spectral interval projections. As MATH is exposed in MATH, we may apply REF to get the affine map MATH such that MATH and MATH is an exposed face in MATH. Hence, applying REF we get a spectral pair MATH such that if we write MATH and let MATH denote the associated spectral interval projections, then we have MATH . Applying MATH to MATH, yields the desired result. |
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 and MATH and so MATH has the desired form. Thus, MATH. Now suppose MATH holds. As MATH, we get MATH and so 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 MATH. Thus, MATH is a basis for MATH. Next, since each vector in MATH is orthogonal to the difference of any two vectors in MATH, REF is clear. Finally, REF follows immediately from the definitions of a sharp face and the degree of MATH. |
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