paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9907007 | By REF , MATH is conjugate to MATH as claimed. To prove REF , it is clearly sufficient to show the following: suppose that MATH and MATH are two MATH-triples which are conjugate in MATH. Then MATH in MATH. By REF , there is a MATH such that MATH where MATH. Moreover we can assume that MATH. Clearly MATH, and we must show that MATH. First note that, since MATH, MATH commutes with MATH, and thus MATH acts on MATH and hence on MATH. In particular MATH acts on MATH. By REF , the center MATH surjects onto MATH, and so the induced action of MATH on MATH is trivial. Since MATH, the inner automorphism MATH defines an automorphism of MATH. Moreover, MATH is the image of the corresponding element of MATH under the natural homomorphism MATH. We must show that MATH. We may write MATH, where the MATH are simple groups of MATH type and MATH is either trivial or a simple group not of type MATH. Clearly MATH preserves the factors MATH and MATH. There is a corresponding direct sum decomposition MATH where the MATH are the projections of MATH, MATH to the factor MATH. We analyze the action on each of these factors separately. The group MATH is a subgroup of MATH. Thus the map MATH factors through MATH. On the other hand, MATH. Since MATH, the map MATH factors through MATH. But as we have seen, MATH acts trivially on MATH. Thus, for any element MATH, we have MATH. Now consider the action of MATH on MATH. Since MATH defines an automorphism of MATH which sends MATH to an element of the form MATH, where the MATH, it follows from REF that the action of MATH on MATH is trivial. Thus, for MATH, MATH. It follows that MATH, and finally that MATH as claimed. To see REF , if the conjugacy class of MATH is in the same component of MATH as that of MATH, then after conjugation we can assume that MATH for MATH. Then by the definition of MATH and the fact that it is well-defined, we see that MATH. To see REF , set MATH. Then MATH. On the other hand, MATH is a MATH-triple in the same component of MATH as MATH, so by REF , the order of MATH is MATH. By definition, this is the order of MATH in MATH. Finally, we prove REF . Suppose that MATH. Then the conjugacy class of MATH is in the same connected component of MATH as that of MATH, say, and the conjugacy class of MATH is in the same connected component of MATH as that of MATH, where MATH and MATH lie in MATH and differ by an element of MATH. Hence the classes of MATH and MATH are in the same connected component of MATH, as are those MATH and MATH. Thus MATH is injective on the set of components, and its image is contained in the set of elements in MATH of order MATH. To see that it is surjective, choose any element of order MATH in MATH, represented by MATH, say. Then MATH is a MATH-triple of order MATH, and by construction MATH. |
math/9907007 | Let MATH. There is an action of MATH on the set of all conjugacy classes of rank zero MATH-triples in MATH. By REF , the action of MATH on the first two factors is trivial. We analyze the action of MATH on the last factor. Write MATH with MATH and MATH. There is an inclusion MATH. Since MATH decomposes as a product of groups of MATH-type. It is easy to see that MATH acts freely on the moduli space of MATH-triples in MATH. According to REF the stabilizer of a conjugacy class of MATH-triples in MATH is the intersection MATH, which is a subgroup MATH of index given by the gcd of the quotient coroot integers for MATH. The two possibilities for MATH when MATH are MATH of type MATH or MATH. In both these cases, MATH is of order MATH and may be described explicitly. We let MATH be the intersection MATH. Clearly, MATH has order at most MATH, and is trivial if MATH is trivial. Thus, the stabilizer in MATH of any MATH-triple of rank zero in MATH is equal to MATH where MATH is trivial if MATH is trivial, and MATH has order at most MATH if MATH is non-trivial. Finally, it follows from REF that the NAME group of MATH acts trivially on the set of conjugacy classes of rank zero MATH-triples in MATH. The theorem now follows from REF . |
math/9907007 | Each MATH vanishes on MATH and hence on MATH. Similarly, MATH. This proves that MATH is the vertex of MATH opposite the face MATH. |
math/9907007 | Suppose first that MATH is reduced. Since MATH normalizes MATH, it normalizes MATH and hence it also normalizes the root system MATH on MATH. Thus, MATH is the vertex of an alcove MATH for MATH containing the origin. There is an element MATH such that MATH. Then MATH is an automorphism of MATH. Since MATH is the unique element of MATH for which the coroot integer MATH is divisible by MATH, the root MATH is fixed by any automorphism of MATH. Thus, MATH is fixed by MATH. Since MATH fixes MATH and normalizes MATH it normalizes MATH. If MATH is not reduced, then MATH is of type MATH for some MATH. The fundamental relation among the coroots in this case is of the form MATH where MATH. Thus, MATH. (The difference here is that the coefficient of MATH is not one, as it is for reduced root systems.) This implies that MATH is the face of MATH given by the equation MATH, and so MATH is a linear subspace. Hence MATH, and the statement of the lemma is obvious. |
math/9907007 | We may assume that MATH for some simple group MATH with maximal torus MATH. We set MATH and let MATH. Let MATH and let MATH be a commuting triple of order MATH and rank zero in MATH. Then MATH has MATH as a maximal torus. Let MATH and let MATH. Then MATH is a commuting triple of order MATH in MATH. Then according to REF , the point MATH is conjugate in MATH and hence in MATH to a point which is the exponential of a point in MATH. Hence, MATH is conjugate under MATH to a point of MATH. |
math/9907007 | First let us consider the case when MATH is reduced. Clearly, MATH acts on MATH normalizing MATH and MATH is one of the elements of this decomposition. Suppose that MATH is an element of this decomposition. Let MATH be an interior point of MATH. According to the previous lemma, there is an element MATH which conjugates MATH to a point of MATH. Since there are only finitely many elements of MATH which conjugate MATH so as to meet MATH, it follows that there is some element MATH which conjugates an open subset of MATH into MATH. This element normalizes MATH and hence sends MATH onto MATH. This proves the transitivity statement in the reduced case. If MATH is not reduced, then it is of type MATH for some MATH, MATH and MATH is the linear subspace spanned by the coroots of the subsystem MATH. The induced alcove decomposition of MATH is exactly the alcove decomposition for MATH. Since MATH contains the affine NAME group of MATH, the lemma is clear in this case. |
math/9907007 | Let MATH be an element of MATH such that MATH. Let MATH be an alcove for MATH containing MATH in its closure, and let MATH. Then MATH also contains MATH, and hence MATH contains MATH. But the unique element of MATH taking MATH to MATH fixes pointwise the intersection MATH. Thus, MATH is the identity, and consequently MATH is the identity. |
math/9907007 | Consider an element MATH for the decomposition MATH which shares a codimension-one wall with MATH. Let MATH be an element of MATH which sends MATH to MATH. This affine isometry of MATH fixes the intersection of MATH and MATH, which is a codimension-one affine subspace. Thus, it is a reflection in this subspace. This shows that the reflections in the walls of MATH are elements of MATH. Since MATH acts simply transitively on the elements in the decomposition MATH, it follows that these reflections generate MATH. |
math/9907007 | The follows immediately from the fact, established in the previous proposition, that MATH is a NAME group. |
math/9907007 | Since the walls of the alcove MATH are the subspaces of MATH orthogonal to the MATH for MATH, the first statement is clear. Clearly, the image under the differential of MATH is contained in the NAME group of MATH. By REF this map is onto. By REF , MATH is generated by the reflections in the walls of MATH. The image under the differential of these reflections is the set of reflections of MATH in the MATH for MATH. |
math/9907007 | Since MATH spans MATH and since MATH if MATH, this is clear. |
math/9907007 | The coroot lattice of MATH is identified with the group of translations of MATH which occur as restrictions of elements of MATH normalizing MATH. Let MATH. Translation by MATH carries MATH to an affine subspace MATH of MATH which meets MATH in a point of the form MATH where MATH is an element in the dual to the root lattice of MATH. In particular, MATH is an element of MATH. But since MATH is the vertex of MATH opposite the wall MATH, and MATH is the unique element in MATH whose coroot integer is divisible by MATH, it follows that the action of MATH on MATH fixes MATH. Thus there is an element MATH such that MATH. The composition of translation by MATH followed by MATH is an element of MATH normalizing MATH and acting on it by translation by MATH. This shows that MATH is contained in the coroot lattice of MATH. Conversely, suppose that MATH normalizes MATH and acts on it by a pure translation. We write MATH where MATH and MATH is in the NAME group of MATH. The element MATH normalizes MATH and hence MATH so that MATH for some MATH. Thus, MATH. Since this element restricts to MATH to give a translation, its restriction is translation by MATH. Since MATH, we have MATH. |
math/9907007 | Everything except the last statement is contained in REF . Since MATH and since these form, up to positive multiples, a set of extended coroots, to complete the proof we need only see that the MATH are indivisible elements of MATH. This will follow from the next lemma. Writing MATH, one of the following holds: CASE: The cardinality of MATH is one and the unique MATH is one; CASE: There are at least two MATH for which MATH. We have MATH . Thus, for MATH, MATH, and the remaining MATH are not divisble by MATH. The result then follows from REF . Now let us return to the proof of the theorem. Since MATH is a set of simple roots for MATH, we have an exact sequence MATH . We can rewrite this sequence as MATH . According to REF , the image of MATH in MATH is MATH where either the cardinality of MATH is one or there are at least two MATH for which MATH. In the first case MATH is a point. In the second case, it is easy to see that the image of each MATH is indivisible in MATH. |
math/9907007 | Set MATH and set MATH. According to REF , there is a MATH-triple MATH of rank zero in MATH, and MATH is a maximal torus of MATH. By REF we can assume that MATH is a MATH-pair in MATH in weak normal form with respect to MATH. Let MATH normalize MATH. Then MATH also normalizes MATH and MATH is another MATH-triple of rank zero and order MATH in MATH. Thus, by REF this triple is conjugate by an element MATH to a triple of the form MATH for some MATH. In particular, MATH. Since MATH is a triple of rank zero in MATH, the centralizer MATH is a torus. Suppose that a root of MATH with respect to MATH annihilates MATH and MATH. This root then annihilates MATH, and hence is a root of MATH with respect to MATH annihilating MATH and MATH. But we have just seen that there are no such roots. Thus, MATH is also a torus. Since MATH is a MATH-pair in weak normal form, REF implies that MATH is a maximal torus. Because MATH fixes MATH and normalizes MATH, it normalizes MATH. Clearly, the image in MATH of MATH is equal to that of MATH. |
math/9907007 | The subspace MATH is the span of the coroots inverse to the roots represented by nodes of MATH. As such it decomposes as an orthogonal sum of the subspaces MATH indexed by the connected components of MATH. The factor MATH corresponding to a component is the subspace of MATH spanned by the coroots inverse to the roots represented by the nodes of that component. The coroot MATH is contained in MATH if and only if it is orthogonal to all the coroots inverse to the roots corresponding to nodes of MATH. This is equivalent to the node of MATH corresponding to MATH not being connected in MATH to any node of MATH. The second item is clear. If MATH, then MATH and MATH. If MATH and MATH are adjacent nodes of MATH, then since MATH is contractible, MATH and MATH are not connected to a common component of MATH. This means that the projections MATH and MATH are orthogonal, where MATH denotes orthogonal projection to MATH. The third item follows. Suppose that MATH and MATH are not adjacent and are not connected in MATH to a common component of MATH. Then there is no factor MATH of MATH with the property that the orthogonal projections of both MATH and MATH into MATH are both non-trivial. Thus, MATH. Lastly, if MATH and MATH are not adjacent in MATH then MATH and hence MATH. Thus, we complete the proof by showing that, under the hypothesis of REF , the inner product MATH. Let MATH be the subspace spanned by the coroots inverse to the roots corresponding to the nodes of the component of MATH connected to both MATH and MATH. By REF , the component of MATH connected to both MATH and MATH is of type MATH for some MATH. Furthermore, if MATH is the set of simple coroots for this component given by the nodes of MATH, then MATH is connected to a unique MATH and MATH is connected to a unique MATH. Thus, MATH is a negative multiple of the fundamental coweight MATH for the root system corresponding to MATH, and likewise MATH is a negative multiple of MATH. The following computation in MATH shows then that these vectors have positive inner product. Let MATH be the simple roots in MATH, ordered so that MATH for MATH, where MATH is the standard NAME invariant inner product. Let MATH be the fundamental weight corresponding to MATH. Then for MATH, MATH . Proof. This is a straightforward computation. |
math/9907007 | The NAME diagram MATH of a simple, non-simply laced group is a chain with a single multiple bond. Therefore, MATH is either a chain with at most two multiple bonds or has one multiple bond and one trivalent vertex. The proof is then an elementary argument involving the possible diagram automorphisms of MATH. |
math/9907007 | The inner product MATH is invariant under the NAME group of MATH, and thus is a multiple of MATH. Clearly, this multiple is MATH. |
math/9907007 | We begin by showing that MATH only depends on MATH and not any of the choices made above. First fix the maximal torus MATH. Then the choice of MATH is determined up to an element MATH in the coroot lattice of MATH. Since MATH projects to an element of the center of MATH, we see that MATH. Thus MATH is independent of the choice of MATH. Now fix MATH and vary MATH. If MATH is another lift of MATH, then MATH is an element of the coroot lattice of MATH. Since MATH projects under the exponential mapping to a central element of MATH, we see that MATH. Finally consider another maximal torus MATH in MATH. There is an element MATH conjugating MATH to MATH. Since MATH and MATH are central in MATH, conjugation by MATH fixes MATH and MATH. This establishes that MATH is well-defined. Clearly, then, it is a conjugacy class invariant. |
math/9907007 | Choose a maximal torus MATH of MATH. Choose a maximal torus MATH for MATH with MATH. Let MATH. Then MATH and there is a rank zero MATH-triple MATH in MATH and elements MATH such that MATH. It suffices by REF to show that MATH. We fix a maximal torus MATH for MATH such that MATH. Clearly MATH. Let MATH be lifts of MATH, let MATH, and let MATH project under the exponential mapping to MATH. If MATH exponentiates to MATH, then MATH. We can lift MATH to MATH, where MATH is the identity component of the inverse image of MATH in MATH, and similarly for MATH. Thus MATH and hence MATH. Lastly, replace MATH by MATH. Clearly, a lift of MATH to MATH is given by MATH where MATH exponentiates to MATH. Thus, MATH. But MATH under the pairing MATH, so that MATH. Hence MATH. This completes the proof. |
math/9907007 | The fact that MATH descends to a function MATH is immediate from REF . To see that it is a homomorphism, since MATH is cyclic by REF , it suffices to show that MATH for all MATH. Let MATH be lifts of MATH, let MATH, and let MATH project to MATH. Finally, let MATH be a lift of MATH to MATH. Then MATH. Since MATH, the element MATH is a lift of MATH. Clearly, then, MATH. |
math/9907007 | This follows easily from REF . |
math/9907007 | By REF , we know that MATH is isomorphic to a product of simply connected groups of type MATH and that MATH is a product of elements MATH generating the center of MATH. It follows immediately that MATH. If MATH is of type MATH for some MATH, then MATH is a product of simple groups of type MATH where MATH, and so REF hold in this case. Assume that MATH is not of type MATH for any MATH. Then MATH is contractible and has at most two vertices of order MATH. Furthermore, if it has a vertex of order MATH, then it is MATH. Any diagram automorphism of such a diagram has order MATH, or MATH. Since the center acts faithfully on MATH, MATH is divisible by at most one prime. It follows that MATH for some MATH. This proves REF . We number the MATH so that MATH, and we let MATH. Suppose MATH is simply laced. Let MATH be an element of order MATH and let MATH be a MATH-pair in MATH. Clearly, MATH and hence MATH. Let MATH lift MATH and let MATH lift MATH. Since MATH is simply laced, it follows from REF that MATH. Since MATH and MATH each generate MATH, REF shows that MATH is of order MATH. By REF , MATH divides the order of MATH in MATH. Since MATH, the order of MATH in MATH divides MATH by REF . If MATH is the orbit containing the extended root, then MATH divides MATH. It follows that all of the above divisibilities are in fact equalities. In particular MATH and the order of MATH is MATH. Thus we have proved REF , as well as REF in the simply laced case. To treat the non-simply laced case, we need the following. Suppose that MATH is non-simply laced and that MATH. Let MATH be defined as follows: Choose MATH with MATH, and set MATH. Then MATH is well-defined. Finally, MATH for every MATH-triple MATH in MATH if and only if MATH if and only if MATH if and only if MATH is connected. Since MATH, it follows immediately that varying MATH and MATH by elements in MATH changes MATH by an integer, showing that MATH is well-defined. By REF MATH. Hence, by REF , all the MATH are of type MATH. Let MATH be the simple roots of MATH. Then a representative for MATH is MATH. Another representative for MATH is MATH for some simple root MATH of MATH. Since MATH represents an element of MATH, MATH, and thus MATH and MATH is a long root of MATH. Let MATH be a MATH-triple in MATH. By definition, for any lifts MATH and MATH of MATH to MATH, we have MATH. Thus MATH. Hence, if MATH for every MATH-triple MATH in MATH, then MATH. By REF , MATH . Thus MATH is zero if MATH is distinct from all the MATH. Using the fact that MATH is a long root and hence MATH if MATH, it follows that MATH is equal to MATH if MATH is equal to one of the MATH. In particular, if MATH, then MATH is orthogonal to all the MATH and thus is orthogonal to MATH. In this case, MATH and hence MATH. Thus, MATH implies that MATH. By REF , MATH. On the other hand, by REF , the map MATH is surjective. If MATH, then MATH and hence MATH is connected. Finally suppose that MATH is connected. Then, by REF , MATH. Let MATH be a MATH-triple in MATH. Then by REF , the order of MATH divides MATH and hence is MATH. Since the order of MATH divides the order of MATH, by REF , it follows that MATH. Returning to the proof of the proposition, let us suppose that MATH is non-simply laced and that MATH, so that each MATH is of type MATH. Let MATH be the simple root of MATH. Let MATH be the unique simple root such that MATH. Since MATH, the root MATH is long. Let MATH be any lift of MATH and let MATH. Then MATH . Since MATH projects to a generator of the center of MATH, we see that MATH . On the other hand, using the lift MATH for MATH shows that MATH . Since MATH is long, MATH. It follows that MATH. In other words, every MATH distinct from MATH is short. Hence either no root of MATH is a long root of MATH, in which case MATH, or exactly one simple root of MATH is a long root of MATH and MATH. In the first case, MATH is connected, and hence MATH, and in the second case MATH has two components and so MATH. The argument also shows that, if MATH, then the order of MATH is exactly MATH. Of course, if MATH, then the order of MATH is also MATH. This proves REF , as well as REF in the non-simply laced case. The proofs of REF are very similar, and we shall just prove REF . Suppose that MATH is not simply laced and that one of the MATH, say MATH, is a long root of MATH. Since the NAME diagram for MATH has a unique multiple bond, the long roots form a connected chain. Thus, if there is another long simple root MATH of MATH, there is a long simple root MATH of MATH with MATH. Since the MATH are short roots for MATH, and hence MATH, we have that MATH for all MATH. It follows that MATH. This is impossible since MATH. Thus MATH is the unique long simple root in MATH. |
math/9907007 | Fix MATH dividing exactly one of the integers MATH. By REF it suffices to exhibit a single MATH-triple MATH such that MATH has order MATH. Let MATH be a MATH-triple of rank zero and order MATH. By REF , MATH is conjugate to the exponential of a vertex of the alcove MATH. Moreover, MATH is semi-simple and contains the rank zero MATH-pair MATH. Thus by REF , the universal cover MATH is a product of groups MATH of type MATH and there is a lift MATH of MATH to MATH such that the image of MATH generates the center of every simple factor. Let MATH be the orthogonal direct sum decomposition induced by the decomposition of MATH. We write MATH where MATH exponentiates to a generator of MATH. Let MATH be a lift of MATH to MATH. Then MATH lies in the center of MATH. Hence, if MATH is the least common multiple of the integers MATH, then MATH. Since the image of MATH in the group MATH has order MATH, it follows that MATH. For every MATH the triple MATH is a MATH-triple. We shall find a MATH such that MATH has order MATH. Supposing this, by REF , it follows that MATH divides the order of MATH. Thus, MATH divides the order of MATH, which in turn divides at least one of the MATH. It follows that MATH and that the order of MATH is MATH. Thus, the order of MATH is equal to MATH. It remains to construct the required element MATH. For any MATH in MATH, let MATH be a lift of MATH. We write MATH with MATH. Clearly, MATH . Since MATH projects to a generator of the center of MATH and MATH is isomorphic to MATH, it follows from REF that, for every MATH of order dividing MATH, there is an element MATH, exponentiating to an element contained in the center of MATH, such that MATH. For appropriate choices of elements MATH of order dividing MATH, the element MATH has order MATH. Consequently, there is an element MATH such that MATH is of order MATH modulo MATH. If MATH is simply laced, then all the roots of MATH are long roots of MATH, and hence the factors MATH are all one. In this case, the element MATH constructed in the last paragraph exponentiates to an element MATH in the center of MATH and MATH is of order MATH modulo MATH. Now suppose that MATH is non-simply laced and MATH, so that MATH. We write MATH where MATH is the factor containing the highest root of MATH. Then, by REF , the factor MATH is of type MATH and the image of MATH in this factor is the center of MATH. Since one of the roots of this factor is the highest root of MATH, the roots of this factor are long roots of MATH. Choose a generator MATH of MATH, and let MATH be the image of MATH under the projection MATH. Since MATH is of type MATH and MATH generates the center of MATH, there is a MATH-triple MATH in MATH of order MATH. By REF , MATH is also of order MATH. For each MATH there is a MATH-triple MATH in MATH of order one. Clearly, the product MATH is a MATH-triple in MATH, automatically of rank zero. It projects to a commuting triple MATH of rank zero in MATH. Clearly, MATH. Since the order of MATH is one for all MATH, and since the roots of MATH are long roots of MATH we see that MATH and hence has order MATH. This proves the proposition in case MATH or MATH is simply laced. There remains the possibility that MATH is not simply laced and MATH. However, as the next lemma shows, there is just one possible MATH in this case: Suppose that MATH is not simply laced and MATH. If there is a rank zero MATH-triple MATH in MATH, then MATH is of type MATH. By REF , MATH. Since MATH is a product of groups of type MATH, MATH, and hence MATH is the exponential of a vertex of the alcove MATH contained in a wall of MATH corresponding to the highest root. Let MATH be the simple root of MATH so that the face MATH of MATH is opposite to the vertex exponentiating to MATH. By REF , MATH is either a chain with two multiple bonds at the ends or has one multiple bond meeting one leaf and one trivalent vertex, two of whose ears are the remaining two leaves. Moreover, the complement of the node corresponding to MATH is a diagram which is a union of diagrams of type MATH. It is easy to see that the only possibilities for such extended diagrams MATH are MATH, MATH, or MATH. The three possibilities for the quotient coroot integers are MATH in the case of MATH, MATH in the case of MATH, and MATH in the case of MATH. Thus, by REF , only in case MATH does MATH contain a rank zero MATH-triple. Returning to the proof of REF in the case where MATH is of type MATH, let MATH be any rank zero MATH-triple. Up to conjugation, it follows that MATH is the exponential of the vertex opposite the wall defined by MATH, where MATH is the unique short simple root of MATH. Hence MATH, where each MATH is of type MATH and the extended root MATH is a simple root for one of the MATH, say MATH. Moreover MATH, where the MATH is the nontrivial central element of MATH, and MATH is a product of rank zero MATH-pairs MATH in MATH. Furthermore MATH is the exponential of MATH. By REF , MATH. Since MATH, MATH. |
math/9907007 | We begin with the following lemma on the structure of MATH: Suppose that MATH. With MATH as above, MATH is properly contained in MATH and MATH has a unique component MATH which is not of MATH-type for some MATH. Write MATH. If MATH is not simply laced, then MATH is of type MATH and all simple factors of MATH are of type MATH whose roots are short roots of MATH. By REF , since MATH does not divide MATH, MATH is properly contained in MATH and MATH has a unique simple factor MATH not of MATH-type. Clearly if MATH is not simply laced, then MATH is also not simply laced. In particular the two nodes of the double bond for MATH are simple roots for MATH. According to REF , one of these nodes lies in MATH. It follows that the projection MATH of MATH to MATH is nontrivial. Since MATH contains a rank zero MATH-pair, it follows from REF that MATH is of type MATH. By REF , MATH is a product of groups of type MATH and has at most one simple root which is a long root of MATH. If there is a long simple root in MATH, the corresponding node is a node of the double bond of MATH. Hence it is a root of MATH. It follows that all of the simple factors of MATH are of type MATH whose roots are short roots of MATH. Returning to the proof of REF , first assume that MATH is simply laced. Let MATH be the product decomposition of MATH in simple factors, where MATH is the factor which is not of type MATH. Let MATH be the corresponding decomposition of MATH. For each MATH, suppose that we are given a MATH-triple of rank zero MATH in MATH of order MATH. Then MATH is a MATH-triple in MATH. Moreover, since MATH is simply laced, MATH. Let MATH be a MATH-triple in MATH whose order in MATH is MATH. Let MATH be the image of MATH in MATH and let MATH be its order as a MATH-triple in MATH. Note that MATH has rank zero in MATH. Thus by REF , the order of MATH is the order of MATH as a MATH-triple in MATH, and hence MATH for some integer MATH relatively prime to MATH. The order of MATH, as a MATH-triple in MATH, is the least common multiple MATH of the MATH. By REF , MATH. It is an elementary number-theoretic argument that there exist MATH such that MATH has order MATH. For MATH, MATH is of type MATH, and thus, given the integer MATH, there exists a rank zero MATH-triple MATH in MATH such that MATH. Note that, by REF , the order of MATH in MATH divides MATH. Replace MATH by the MATH-triple MATH. Then MATH is of order MATH. On the other hand, the order of MATH in MATH, which is the least common multiple of the orders of the MATH, divides MATH. By REF , the order of MATH divides the order of MATH as a MATH-triple in MATH, which in turn by REF divides the order of MATH as a MATH-triple in MATH which divides MATH which is the order of MATH. Therefore, the order of MATH in MATH is the order of MATH, namely MATH. We write MATH. Then MATH is a MATH-triple of order MATH in MATH such that the order of MATH is also MATH. Next suppose that MATH. In this case MATH is simple. By REF , there is a simple root for MATH which is a long root of MATH. Choose a rank zero commuting triple MATH in MATH of order MATH in MATH. By REF , the order of MATH is MATH. We may thus assume that MATH is not simply laced and that MATH. By REF , MATH is of type MATH and all simple factors of MATH are of type MATH whose roots are short roots of MATH. Thus MATH. Let MATH be a MATH-triple in MATH. By REF , MATH. Choose a rank zero MATH-triple MATH in MATH. Then its order is MATH and by REF , the order of MATH is also MATH. Choose any rank zero MATH-triple MATH in MATH. Its order divides MATH. Thus the order of MATH in MATH is MATH. Since the order of MATH is MATH, it follows that the order of MATH in MATH is also MATH. This concludes the proof. |
math/9907007 | Fix a MATH dividing at least one of the MATH. There is the corresponding group MATH containing a rank zero MATH-triple of order MATH in MATH. It follows from REF that there exists a rank zero MATH-triple MATH in MATH, such that every component MATH of MATH of order MATH in MATH contains the conjugacy class of MATH for exactly one MATH where MATH and MATH is relatively prime to MATH. By REF , MATH. Since MATH is constant on connected components, it clearly suffices to find, for every MATH, a MATH-triple MATH of order MATH in MATH such that the order of MATH is also MATH. This follows from REF . |
math/9907007 | Since MATH is connected, every principal MATH-bundle over MATH is isomorphic to the trivial bundle. REF follows. A lifting of the MATH-bundle isomorphism MATH to an automorphism of the trivial MATH-bundle MATH is the same as a lifting of MATH to MATH. This proves REF . |
math/9907007 | Let MATH be the MATH-bundle map obtained from MATH by dividing out by MATH. The connection MATH on MATH lifts the connection MATH on MATH. Since MATH and since MATH, it follows that MATH. Thus, MATH lifts MATH. But MATH is the unique lifting of MATH. It follows that MATH, and hence the holonomies of MATH and MATH are conjugate in MATH. |
math/9907007 | Given a flat MATH-connection on the two-torus with holonomy MATH around the coordinate circles, let MATH be lifts of these elements. Then the commutator MATH lies in MATH and is equal to the characteristic class MATH. Applying this to the various coordinate two-tori inside the MATH-torus, shows that the MATH-tuple associated to MATH proves the first statement. The rest of the proposition is a straightforward exercise. |
math/9907007 | Since the question only involves the homotopy type of the pair MATH, we may assume that MATH. Over MATH, we can take MATH. Let MATH be the corresponding map. Since MATH is trivial, so is MATH, and hence the fundamental group of MATH acts trivially on the fundamental group of the fiber of MATH, which is isomorphic to MATH. Thus the obstructions to extending MATH to MATH lie in MATH. The only nonzero such group is MATH. The value of the obstruction on a relative MATH-cell MATH is MATH. For the standard relative cell decomposition of MATH for which the MATH-skeleton is the union of the MATH and the corresponding cells are MATH, the elements MATH are commutators in MATH. Since MATH is abelian, MATH is trivial. Hence the automorphism extends. |
math/9907007 | By a result of REF , MATH . On the other hand, the action of MATH on MATH has eigenvalue MATH on the root space MATH. The result is then a direct computation. |
math/9907007 | Let MATH be the MATH-bundle obtained from MATH by gluing the ends together by MATH. The difference MATH is equal to MATH. We choose a trivialization MATH of MATH such that there is an isomorphism MATH to MATH whose underlying MATH-bundle isomorphism is the identity. Then in this trivialization MATH is given by a continuous map MATH. The fact that MATH comes from an automorphism of the enhanced bundle means that MATH lifts to a map MATH. This means that MATH lifts to a MATH-bundle, and hence, since MATH is connected and simply connected, MATH is trivial. A standard obstruction theory argument then shows that MATH is isomorphic as a MATH-bundle to the connected sum of the product bundle MATH and a MATH-bundle MATH. Thus, MATH. But since MATH is simply connected, MATH lifts to a MATH-bundle MATH and thus MATH takes an integral value on MATH. |
math/9907007 | Since the critical points of the functional MATH are the flat connections, MATH. In particular, MATH. The corollary now follows from the additivity formula of the previous lemma. |
math/9907007 | Clearly, there exists a flat connection MATH with the required holonomy on the bundle MATH. The connections MATH and MATH have the same holonomy. It is then easy to see that there is an automorphism MATH of the trivial bundle, supported near MATH, such that MATH satisfies the conclusions of the lemma. |
math/9907007 | The composition MATH pulls back the trivial connection to MATH. The composition is given by left multiplication by MATH, and moreover MATH. By changing the trivialization MATH by a constant change of gauge, we can arrange that MATH. It follows that MATH. |
math/9907007 | Let MATH be the connection on MATH which is trivial over MATH in the trivialization MATH and which is given by the one-form MATH over MATH in the trivialization MATH. As before, one sees easily that these two descriptions match over MATH. Let us compute MATH. First notice that MATH over both MATH and MATH. Clearly, then MATH. Over MATH the connection MATH is trivial. Thus, on this patch MATH, and consequently the NAME integrand vanishes on this patch. Over MATH we have MATH so that MATH. Lastly, on this patch MATH so that MATH. This shows that MATH and hence by REF MATH . Now we compute MATH . First let us show that the NAME integrand for this invariant is zero over MATH. Over MATH, and with the trivialization MATH, MATH is the pullback of a connection MATH on MATH and MATH is trivial. Thus MATH, MATH and MATH are all pulled back from forms on MATH. Thus, we see that over MATH the NAME integrand vanishes identically. Now we compute the NAME integral over MATH using the trivialization MATH. The connection MATH is trivial on this patch and the one-form MATH is MATH. It follows that MATH. Thus, MATH . Since MATH, we can rewrite this as MATH . Doing the MATH-integration and using the fact that MATH yields MATH . Since MATH, and using REF , we have MATH . |
math/9907007 | By REF it suffices to show that if MATH and MATH are connections on MATH pulled back from connections on MATH then MATH. But this is clear - under this hypothesis the NAME integrand (which is a three-form) is pulled back from a form on the two-torus MATH and hence vanishes identically. |
math/9907007 | The above construction shows that, given a MATH-triple MATH, there is a MATH-bundle MATH, a trivialization MATH of MATH and a flat connection MATH on MATH such that the MATH-holonomy of MATH measured using the trivialization MATH, which is the union of the trivialization MATH on MATH with the product trivialization around the third coordinate circle, is MATH. Since MATH is a MATH-pair, there is an isomorphism MATH. By REF , there is a bundle isomorphism from MATH to MATH which carries the trivialization MATH to MATH. Let MATH be the pullback of the connection MATH. Then MATH is a flat connection on MATH whose MATH-holonomy is the conjugacy class of MATH, and hence the MATH-holonomy of MATH is the conjugacy class of MATH. By REF MATH. More generally, suppose that MATH is a flat connection on MATH such that the MATH-holonomy of MATH is the conjugacy class of MATH. Then by REF there is an enhanced automorphism of MATH carrying MATH to MATH. By REF this implies that MATH and the result follows. |
math/9907007 | Since MATH, the MATH-bundle MATH given in the statement of REF is trivial. Furthermore, an enhancement of a MATH-bundle is no extra information. The result is now immediate from this corollary and REF . |
math/9907007 | Suppose that MATH is a rank zero MATH-triple in MATH. After conjugation, we may assume that MATH is a MATH-pair in normal form. Thus MATH is a maximal torus of MATH. The element MATH commutes with MATH and MATH up to an element of the center. Since the fixed subgroup of conjugation by MATH on MATH has rank zero, it follows from CITE II REF that MATH is a torus and hence MATH. By inspection, all of the integers MATH, for the action of MATH on the coroot diagram of MATH are MATH. Hence by REF , MATH. The triple MATH is a MATH-triple in MATH. By REF , MATH is a maximal torus of MATH. This implies that MATH acts trivially on MATH. Since MATH acts on this torus without fixed points, it follows that MATH acts by MATH on MATH, and hence MATH for all MATH. By REF , the torus MATH is conjugate to MATH. It will be convenient to conjugate MATH so that MATH. Of course, in this case MATH for all MATH. Write MATH and MATH for elements MATH and MATH. Since MATH, it follows that MATH for some MATH. This implies that MATH is the exponential of an element of the form MATH mod MATH. The same computation for MATH shows that MATH is also the exponential of such an element. Thus, both MATH and MATH are the images under the exponential mapping of elements of the form MATH mod MATH. It follows easily that, for MATH, there is a root of MATH annihilating MATH and annihilating MATH and MATH. This root then annnihilates both MATH and MATH, contradicting the fact that MATH is a torus. Hence MATH. Now let us describe all conjugacy classes of rank zero MATH-triples in MATH. Let MATH be a rank zero MATH-triple in MATH. As above, we arrange that MATH with MATH and MATH. Then MATH. According to REF , the conjugacy class of MATH depends only on the the pair MATH. It is easy to see that MATH is the quotient of MATH by MATH. The image in the NAME group of MATH of elements in MATH which act trivially on MATH is MATH, where one of the factors acts by MATH on MATH and the other acts by switching the coordinates. The argument above shows that MATH is the exponential of an element of the form MATH mod MATH. Let MATH be a lift of MATH to MATH. The element MATH normalizes MATH and acts by MATH on MATH MATH sends MATH to MATH. Since MATH contains a regular element of MATH, the element MATH normalizes MATH and hence normalizes the unique maximal torus of MATH containing MATH. Since MATH is the exponential of MATH, it follows that MATH is the exponential of one of MATH and hence of one of MATH or MATH. The fact that MATH implies that MATH is the exponential of MATH. By conjugation by an element commuting with MATH we can arrange that MATH is the exponential of MATH. A computation shows that MATH must be congruent to MATH mod MATH. After conjugating by an element of the normalizer of MATH, we may assume that MATH is congruent to MATH mod MATH. Since there is no root of MATH annihilating both MATH and MATH, this means that MATH is either MATH or MATH mod MATH. After conjugation by an element of the normalizer of MATH which interchanges the two coordinates, and hence fixes MATH, we may assume that MATH is congruent to MATH mod MATH. This proves that, given a rank zero MATH-triple MATH in MATH, the MATH-pair MATH is determined up to conjugation in MATH. The representatives that we have chosen are: MATH and MATH where MATH is an element of MATH normalizing the unique maximal torus of MATH containing MATH, and the image of MATH in the NAME group of MATH is the product of the non-trivial elements in each factor of MATH. Next we show that there is MATH such that MATH is a rank zero MATH-triple. Write MATH as a product, where MATH. Similarly write MATH. Let MATH where MATH commutes with MATH and represents the NAME element which is multiplication by MATH on MATH. Direct computation shows that MATH is a rank zero MATH-triple. Suppose that MATH is a rank zero MATH-triple in MATH. Any other rank zero MATH-triple is conjugate to MATH, where MATH for some MATH. Since MATH is cyclic of order two and since MATH on MATH, there are exactly two conjugacy classes of rank zero MATH-triples in MATH, one of which is represented by MATH and the other by MATH, where MATH but MATH. Direct computation shows that MATH is of order MATH and that MATH is in the non-trivial component of MATH. Thus, MATH represent the two conjugacy classes of rank zero MATH-triples. |
math/9907007 | This is immediate from the explicit description given above. |
math/9907007 | Let MATH. Then by REF , since MATH, MATH. We may write MATH, where each MATH is of type MATH and MATH is either trivial or of type MATH for some MATH. For MATH, let MATH denote projection to the MATH factor. The projection of MATH to MATH is a rank zero MATH-triple. Thus, if MATH, then MATH is of type MATH. If MATH is not trivial, then MATH is the full center of MATH. By REF , MATH. Since MATH, it is clear that the only possibilities for MATH which satisfy the above conditions are either MATH or MATH. |
math/9907007 | The pair MATH is a MATH-pair in MATH. Also, MATH and hence lies in MATH. The result follows from REF . |
math/9907007 | By REF , the number of components of MATH with maximal torus conjugate to MATH is given by the number of conjugacy classes of rank zero MATH-triples in MATH modulo the action of MATH and of MATH, where MATH. Let us first consider the action of MATH. It is easy to see that MATH . Straightforward computation shows that the action of MATH on the space of conjugacy classes of rank zero MATH-triples in MATH preserves the invariant MATH and acts transitively on the set of conjugacy classes with a given MATH. The image of MATH in the outer automorphism group of MATH is easily checked to be the subgroup of all permutations of the MATH factors. Thus the action of MATH fixes MATH. This completes the proof of the lemma. |
math/9907007 | This is immediate from REF . |
math/9907007 | By REF , the number of components of MATH with maximal torus conjugate to MATH is given by the number of conjugacy classes of rank zero MATH-triples in MATH modulo the action of MATH and of MATH, where MATH. Let us first consider the action of MATH. Let MATH be the emdedding which sends MATH to MATH. Then it is easy to check that MATH . Note that, by REF , if MATH is an element of MATH and MATH is a rank zero MATH-triple in MATH, then MATH. Conversely, it is clear that, if MATH, then MATH and MATH are in the same MATH-orbit. Finally, the image of MATH in the outer automorphism group of MATH is the permutation group of the factors MATH for MATH. Hence MATH for all MATH. The lemma follows. |
math/9907007 | This is immediate from REF . |
math/9907007 | Since MATH is constant on components of MATH, the first statement is clear. To prove the second it suffices to compute MATH for one flat connection MATH whose MATH-holonomy is of order MATH. By REF , a MATH-triple MATH of order MATH is given by replacing MATH in MATH by MATH where MATH is the non-trivial element in MATH. Since the inclusion of each of the MATH-factors of MATH into MATH induces an isomorphism on MATH, to compute the NAME invariant MATH, it suffices to compute the relative NAME of the MATH-connections obtained taking the images of MATH and MATH under projection the first MATH-factor of MATH. REF applied to MATH show that this relative NAME invariant is MATH. This shows that MATH. |
math/9907007 | Let MATH be a rank zero MATH-triple. Write MATH, where MATH is a rank zero MATH-triple in MATH. Let MATH be a flat connection on MATH whose MATH-holonomy is the conjugacy class of MATH. The MATH-triple MATH obtained by replacing MATH by MATH is also of rank zero. Clearly, for MATH, MATH. By REF , MATH, and hence, by REF , the conjugacy classes of MATH and MATH lie in different components of order MATH of MATH. On the other hand, given the MATH-triple MATH defined above, let MATH, where MATH, and similarly for MATH. Clearly MATH . (Here MATH are the nontrivial elements of MATH.) Thus by REF , MATH and MATH represent points in the component MATH. Let MATH be the diffeomorphism of the three-torus which is the identity on the first two factors and is inversion on the third. Note that MATH so that MATH is an enhanced bundle over MATH. Then MATH is the MATH-holonomy of MATH on MATH, and hence by the above computations MATH represents a point in MATH. Likewise MATH represents a point in a different component from MATH. Since MATH is an orientation-reversing diffeomorphism, we have MATH . Thus, MATH takes opposite values modulo MATH on the two components of order MATH of MATH. With MATH as above, let MATH be the triple MATH. Then MATH is a MATH-triple. Likewise, let MATH be the MATH-triple MATH. The MATH-triple MATH is in the non-trivial component of the modulo space of MATH-triples in MATH. The MATH-triple MATH is in the trivial component of the modulo space of MATH-triples in MATH. It follows from the proof of REF that the square of MATH in MATH is MATH which is in the non-trivial component of MATH. Direct computation shows that MATH is connected, and hence is equal to MATH. Thus MATH lies in the non-trivial component of MATH. Since MATH, by REF , the MATH-triple MATH represents a point in the non-trivial component of the moduli space of MATH-triples in MATH. By REF , MATH. Since the MATH lie in MATH, it follows that MATH is congruent modulo MATH to MATH which lies in MATH. Hence MATH, and so the MATH-triple MATH lies in the trivial component of the moduli space of MATH-triples in MATH. Let MATH be the map on the three-torus which double covers the last coordinate. Note that MATH induces a map from the pair MATH to itself. Thus MATH is an enhanced MATH-bundle. The MATH-triple MATH is the MATH-holonomy of MATH while MATH is the MATH-holonomy of MATH. Since the NAME invariant is given by integrating a local expression involving connections and curvature, MATH . Since the conjugacy classes of MATH and MATH lie in opposite components of the moduli space of MATH-triples, by REF applied to the case of MATH-triples, MATH . It follows that MATH . From the fact that the value of this invariant on one component of order MATH is the negative of its value on the other such component, it now follows that on one of these components the value is MATH modulo MATH and on the other it is MATH modulo MATH. |
math/9907008 | In MATH, the twosided NAME component of MATH is MATH, that of MATH is MATH and that of MATH is MATH, whence REF . We now check what it means for MATH to map (each twosided NAME component of) MATH to itself: As MATH, we get MATH . Applying MATH to MATH yields MATH which is in MATH if and only if MATH. Similarly applying MATH to MATH yields the condition MATH. Now MATH if and only if MATH. Finally, for MATH with MATH, one has MATH and REF follows. If MATH with MATH and the remaining coefficients in MATH, then the inner derivation MATH vanishes on the primitive idempotents if and only if MATH in MATH. Calculating its value on the arrows yields MATH and REF follows. |
math/9907008 | It suffices to recall that the NAME bracket of two first order differential operators on MATH is given by MATH . Now the ideal MATH is mapped to itself by the vectorfield MATH if and only if MATH. The final assertion follows from the inclusions MATH . |
math/9907008 | If MATH, then the right hand side of REF evaluates to MATH, if MATH it evaluates to MATH, if MATH it evaluates to MATH. The second case simply rewrites MATH. |
math/9907008 | NAME 's determinant. |
math/9907008 | The condition on (the characteristic of) MATH ensures that MATH is well defined. If MATH is a differential operator on MATH with MATH, then MATH . Now assume MATH for some matrix MATH of elements from MATH. Multiplying by MATH and subtracting yields an equation MATH for each pair of elements MATH. If MATH is a third element from MATH, there result equations MATH in MATH. The (classes of) exponential functions involved are MATH and MATH for MATH with MATH. By the assumption on the set of differences, if MATH are pairwise distinct, the set MATH has cardinality equal to MATH, and as MATH, the corresponding classes MATH are linearly independent in MATH by REF . It follows in particular that MATH for MATH. As the choice of MATH was arbitrary and as there are at least MATH distinct elements in MATH, it follows that MATH whenever MATH. In turn, the system REF evaluates now at MATH to MATH and so MATH implies that the diagonal elements MATH vanish as well. Finally, REF shows that MATH and any equation in REF yields MATH. |
math/9907008 | The conditions guarantee that MATH satisfy the hypotheses of REF and that MATH is a unit in MATH, whence MATH. As MATH is upper semicontinuous on MATH it remains to show that the minimal value is taken on generically, for example over the purely transcendental field extension MATH of MATH. There REF applies. |
math/9907008 | Consider first the integers MATH with MATH. As MATH, each MATH is a nonzero polynomial. As MATH the order satisfies MATH whence MATH and the classes MATH are MATH-linearly independent in MATH. For the differential operator MATH one has MATH by construction and for MATH one finds MATH in view of MATH and MATH. It follows that MATH for each MATH, whence the subset MATH is MATH-linearly independent of cardinality MATH. Now consider the integers MATH with MATH. The polynomials MATH satisfy then MATH and MATH, whence the set of (scalar) differential operators MATH defines a MATH-linearly independent subset of MATH of cardinality MATH. As the union MATH is clearly still linearly independent over MATH, the lower bound on MATH follows. For the final assertion observe that MATH whence MATH, with equality on the right if and only if MATH. Similarly MATH and so MATH with equality on the right if and only if MATH or MATH. The claim follows. |
math/9907011 | Introduce MATH, MATH, and MATH as in REF; note that MATH are independent and distributed like MATH; we have MATH as well as MATH . However, MATH. |
math/9907011 | First, the following three properties of MATH are equivalent: CASE: MATH; CASE: MATH; CASE: MATH. Indeed, REF since MATH; REF since MATH; REF since MATH. The set MATH is a closed linear subspace of MATH. By REF , if MATH then MATH. Also, MATH contains constants. It is well-known that such a space is the whole MATH where MATH is the MATH-field generated by MATH. |
math/9907011 | Each element of MATH satisfies REF for the partition into atoms of MATH. Therefore each element of MATH satisfies REF for every partition, and we get REF . For proving REF assume that MATH satisfies REF and prove that MATH for all MATH. From now on MATH and MATH run over MATH. By REF, MATH, MATH; we have to prove that MATH unless MATH contains exactly one atom. By REF again, MATH . We see that MATH appears twice in MATH, but only once in MATH, therefore MATH. If MATH contains at least two atoms, we can choose MATH such that MATH intersects both MATH and MATH; then MATH does not appear in MATH, but appears in MATH, therefore MATH. |
math/9907011 | Denoting MATH, MATH, we have MATH and the same for MATH. Therefore MATH . The space MATH contains some of the terms, and is orthogonal to others. Only the term MATH is contained in MATH for all MATH. |
math/9907011 | MATH by REF; MATH by REF; note that the space decreases when MATH increases. Similarly, MATH and MATH. By REF it suffices to prove that MATH for MATH . However, MATH is (by definition) the direct sum of MATH over atoms MATH of MATH, MATH, and MATH is such a sum over MATH, MATH. It remains to note that MATH. |
math/9907011 | Denote by MATH the MATH-field generated by MATH. It suffices to prove that MATH for all MATH, since MATH (see REF). I give a proof for MATH; it has a straightforward generalization for higher MATH. We have to prove that MATH. For each MATH, MATH (since MATH is invariant under all MATH). However, MATH decreases to MATH, and MATH is orthogonal to MATH. Therefore the union of MATH is dense in MATH; it remains to prove that MATH for all MATH. Note that MATH is the direct sum of MATH over atoms MATH of MATH, MATH (since, again, MATH is invariant under all MATH). REF reduces the needed inclusion to an evident fact, MATH. |
math/9907011 | CASE: Let MATH, then MATH by REF . CASE: It suffices to prove that MATH for all MATH orthogonal to MATH. By REF we may assume that MATH is orthogonal to MATH for some MATH (since such vectors are dense in MATH). For such MATH, MATH. |
math/9907011 | If MATH is not contained in MATH then there exists MATH, MATH, orthogonal to MATH, that is, sensitive. The case is impossible, if MATH for all sensitive MATH or, equivalently, for a dense set of such MATH; the more so, if MATH for all these MATH. By REF, MATH is the average of MATH over MATH distributed MATH; here MATH. It suffices to choose MATH such that MATH for a dense set of sensitive MATH. For each sensitive MATH and each MATH, by REF, MATH for MATH. Therefore MATH for MATH, if MATH grow fast enough. Diagonal argument gives a single sequence MATH that serves a given sequence of vectors MATH. It remains to choose a sequence dense among all sensitive vectors. |
math/9907011 | Similarly to REF, for every MATH, MATH where MATH runs over atoms of MATH, and MATH, MATH are subspaces of stable and sensitive, respectively, elements of MATH. For every MATH, MATH, the operator MATH (recall REF) on MATH is the unit (identity) if MATH, otherwise it vanishes. Assuming MATH we get MATH on MATH, where MATH. Therefore MATH on MATH, MATH (note that MATH), while on MATH we have MATH; so, MATH provided that MATH for some MATH. The general case, MATH, will not be used, and I leave it to the reader. |
math/9907011 | Choose MATH by REF , then MATH for MATH-almost all MATH; here MATH. On the other hand, for every MATH and every MATH, MATH by REF , MATH therefore MATH for all sensitive MATH. Applying REF to the increasing sequence MATH we get MATH for all MATH, therefore MATH for MATH-almost all MATH. So, each sensitive MATH satisfies MATH, that is, MATH, which is impossible unless MATH. |
math/9907012 | Since the inclusion MATH is cofinal, we have MATH . Since MATH is complete, MATH is a NAME space and MATH . It follows that MATH where the second isomorphism follows from REF and the last isomorphism from REF . |
math/9907012 | Assuming MATH, we have MATH where the second isomorphism follows from REF and the third from REF . |
math/9907012 | For any object MATH of MATH, we have MATH where the first and last isomorphisms follow from REF . The conclusion follows from the theory of representable functors. |
math/9907012 | Applying MATH to the canonical morphisms MATH and using the characterization of inductive limits, we get the canonical morphism MATH . Let MATH be a closed absolutely convex bounded subset of MATH. It follows from for example, CITE that, for some MATH, MATH is the image of a closed absolutely convex bounded subset MATH of MATH by the canonical morphism MATH. Since MATH is injective, it induces the isomorphism of semi-normed spaces MATH . Hence, we get the isomorphism of NAME spaces MATH . Composing with the morphism MATH we get a canonical morphism MATH . Finally, using the characterization of inductive limits, we obtain a canonical morphism MATH . A direct computation shows that this morphism is a left and right inverse of REF . |
math/9907012 | Keeping in mind the properties of bornological spaces, it is clear from the definition of MATH that MATH . Since any ball of MATH is included in the closure of a semi-ball of MATH, any bounded subset of MATH is included in the closure of a bounded subset of MATH. This property and the completeness of MATH shows that MATH . Hence the conclusion. |
math/9907012 | For any MATH, set MATH . Clearly, MATH belongs to MATH. Moreover, if MATH is closed in MATH, then MATH is closed in MATH and one checks easily that MATH as NAME spaces. Hence, one has successively MATH where the second isomorphism follows from the fact that the inclusion MATH is cofinal. |
math/9907012 | We have successively MATH where the isomorphism REF follows from REF from REF from REF . |
math/9907012 | For MATH and MATH, denote MATH the absolutely convex hull of MATH . This is clearly a bounded absolutely convex subset of MATH. As a matter of fact, MATH . Moreover, we have a canonical linear map MATH . This map is clearly continuous since MATH when MATH, MATH. Applying the completion functor, we get a morphism MATH and hence a morphism MATH . Using the definition of inductive limits, we get a morphism MATH . |
math/9907012 | Since MATH is nuclear, there is a bounded sequence MATH of MATH, a bounded sequence MATH of MATH and a summable sequence MATH of complex numbers such that MATH . Since MATH is summable, one can find a sequence MATH of non-zero complex numbers converging to zero such that MATH is still summable. One checks easily that the maps MATH and MATH defined by MATH have the requested properties. |
math/9907012 | Consider the set MATH . The relation MATH defined by setting MATH if MATH or MATH turns MATH into a filtering ordered set. By REF , for any MATH, we may choose a continuous linear map MATH and a nuclear map MATH making the diagram MATH commutative. For any MATH, we set MATH and MATH. If MATH, we set MATH . The map MATH being nuclear, MATH is also nuclear. An easy computation shows that if MATH, then MATH. Consider the functors MATH defined by MATH and MATH. They are clearly cofinal and if MATH in MATH, the diagrams MATH are commutative. Hence, we get the two morphisms MATH . Since these morphisms are easily checked to be inverse one of each other, the proof is complete. |
math/9907012 | Since MATH is nuclear, we can find a bounded sequence MATH of MATH, a bounded sequence MATH of MATH and a summable sequence MATH of complex numbers such that MATH for any MATH. Using the isomorphism MATH, we see that MATH . Therefore, MATH and the sequence MATH is summable. Moreover, MATH where the permutation of the sums is justified using REF . |
math/9907012 | By REF , we may assume that MATH for any MATH and that the transition morphisms MATH are nuclear. One checks easily that MATH is a well-defined continuous map. By REF , we know that the sequence MATH is summable and that MATH . Therefore, we may define a continuous linear map MATH by setting MATH . One sees easily that the morphisms MATH and MATH induce morphisms of pro-objects MATH and MATH . A direct computation shows that these morphisms are inverse one of each other. |
math/9907012 | Let MATH and let MATH be a semi-norm of MATH. It follows from our assumptions, that there is MATH and MATH such that MATH and MATH . Fix MATH. Since the maps MATH, MATH and MATH are continuous, we can find a semi-norm MATH of MATH, a semi-norm MATH of MATH and a semi-norm MATH of MATH such that MATH . Consider MATH and let MATH be an element of MATH of the type MATH where MATH, MATH. Using REF above, we obtain MATH and MATH such that MATH and MATH . For MATH, we get MATH . Since any element of MATH is a finite sum of elements of the type considered above, we see that for any MATH, MATH . The conclusion follows directly since MATH is dense in MATH. |
math/9907012 | Since MATH is a FN space, there is a cofinal increasing sequence MATH of continuous semi-norms of MATH such that the canonical map MATH is nuclear. For such a sequence, the canonical map MATH is also nuclear and has a dense range. Moreover, it is well-known (see for example, CITE) that MATH . Clearly, MATH where MATH is the inductive dual of MATH. Recall that an absolutely convex subset MATH is a neighborhood of MATH in MATH if it absorbs any equicontinuous subset of MATH. Hence, it is clear that a neighborhood of MATH in MATH is a neighborhood of MATH in MATH. We know that MATH is reflexive (see for example, CITE). Hence, MATH is bornological (see for example, CITE). The space MATH being itself bornological, the bounded subsets of MATH are equicontinuous. So, any neighborhood of MATH in MATH is a neighborhood of MATH in MATH and MATH. Since REF follows directly from REF , the proof is complete. |
math/9907012 | Since MATH is a DFN space, there is a FN space MATH such that MATH . Let MATH be a projective system of the kind considered in REF . We have MATH . Since the transition morphisms MATH are injective and MATH is complete, REF show that MATH . Using REF , we find a nuclear projective system MATH with MATH such that MATH . It follows that MATH . Hence, we have successively MATH where the isomorphism REF follows from the fact that filtering inductive limits are exact in MATH, REF follows from REF follows from the fact that MATH is projective in MATH, REF follows from REF follows from REF . By REF , we have the isomorphism MATH . Forgetting the topologies and applying the derived projective limit functor for pro-objects (see CITE), we obtain the isomorphism MATH . Since MATH satisfies Condition ML, it is MATH-acyclic in MATH (see REF ). It follows that MATH is also MATH-acyclic in MATH and, hence, satisfies Condition ML. Using REF , we see that MATH is concentrated in degree MATH. It follows that the projective system MATH is MATH-acyclic and the conclusion follows. |
math/9907012 | It is sufficient to show that MATH for MATH. This will be the case if MATH is concentrated in degree MATH for any set MATH. Let MATH be an arbitrary set. Since MATH is a projective object of MATH and since MATH is complete, we have MATH where MATH is the NAME space formed by the bounded families MATH of MATH (a fundamental system of semi-norms being given by MATH where MATH). Therefore, we have the chain of isomorphisms MATH and the conclusion follows from REF . |
math/9907012 | Work as for REF . |
math/9907012 | Work as for REF using REF . |
math/9907012 | If MATH is a projective resolution of MATH, we have successively MATH . |
math/9907012 | Using REF , we may assume that MATH where MATH is a filtering inductive system with MATH, the transition morphisms MATH being nuclear. We may also assume that MATH where MATH is a filtering inductive system. Then, we have MATH where the isomorphism REF follows from REF from the fact that MATH is projective in MATH. |
math/9907012 | Let MATH be an open subset of MATH and let MATH be an open covering of MATH. Consider the sequence MATH where MATH and MATH are the continuous applications defined by MATH where MATH and MATH are the canonical projections and MATH is the restriction map. Since MATH is a sheaf of vector spaces, this sequence is algebraically exact. Let us show that it is strictly exact. CASE: If MATH is countable, MATH, MATH and MATH are NAME spaces. Then, by the homomorphism theorem, the sequence REF is strictly exact. CASE: Assume that MATH is not countable. Since MATH has a countable basis, there is a countable set MATH of open subsets of MATH such that for any open MATH of MATH, MATH . Then, consider the countable set MATH . For any MATH, we may assume that MATH, with MATH. It follows that MATH covers any MATH in MATH and therefore is a covering of MATH. Hence, by REF , the sequence MATH is strictly exact. Now, consider a map MATH such that MATH for any MATH. Then, consider the commutative diagram MATH where MATH and MATH are respectively defined by MATH and MATH . To prove that the sequence REF is strictly exact, it is sufficient to establish that MATH is a kernel of MATH. Let MATH be a morphism of MATH such that MATH. Since MATH and since MATH is a kernel of MATH, there is a unique morphism MATH such that MATH. Set MATH . We clearly have MATH and MATH. Fix MATH. For any MATH such that MATH, we have MATH . It follows that MATH . Since MATH is a covering of MATH and since MATH is a sheaf of vector spaces, we get MATH . It follows that MATH and that MATH. Since MATH is injective, MATH is the unique morphism of MATH such that MATH. Therefore, MATH is a kernel of MATH and the sequence REF is strictly exact. Finally, since the functor MATH preserves projective limits of complete objects of MATH (see REF ), the sequence MATH is strictly exact in MATH . Hence, the conclusion. |
math/9907012 | We know that MATH has a fundamental system MATH of relatively compact open neighborhoods such that MATH for any MATH. Replacing, if necessary, MATH by the union of those of its connected components which meet MATH, we may even assume that any connected component of MATH meets MATH. In this case, it follows from the principle of unique continuation that the restriction MATH is injective. Moreover, by cofinality, MATH . Hence, by REF , it follows that MATH . Since MATH is a taut subspace of MATH, a cofinality argument shows that MATH and the conclusion follows. |
math/9907012 | For any object MATH of MATH, denote MATH the functor defined by setting MATH. Using the techniques developed in CITE, one shows easily that MATH for any projective object MATH of MATH. Therefore, the result will be true if the sheaf of abelian groups MATH is soft. This follows from the fact that it has clearly a canonical structure of MATH-module. |
math/9907012 | As is well-known, since MATH is a soft sheaf, the NAME complex MATH is a MATH-acyclic resolution of MATH. Therefore, MATH is given by the complex MATH . Moreover, since MATH for MATH, the sequence MATH is algebraically exact. Since MATH and MATH are FN spaces, the last sequence is strictly exact in MATH. Using CITE, one sees easily that the sequence MATH is strictly exact in MATH. For any open ball MATH of MATH, NAME 's Theorem B shows that MATH . Hence, the sequence MATH is strictly exact in MATH. Filtering inductive limits being exact in MATH, we see that MATH is a strictly exact sequence of sheaves with values in MATH. Moreover, since, by REF , MATH is MATH-acyclic, MATH is given by MATH . The sequence REF being strictly exact, we get MATH . |
math/9907012 | It is well-known that MATH has a fundamental system MATH of NAME open neighborhoods. By tautness, it follows that for MATH, we have MATH where the second isomorphism follows from REF . Hence, using REF , we get MATH . |
math/9907012 | This follows directly from the fact that the application MATH defined by MATH is continuous and bijective. |
math/9907012 | This follows directly from the fact that the canonical restriction morphism MATH may be factored through MATH for MATH. |
math/9907012 | Let MATH, MATH be open subsets of MATH and MATH. The map MATH defined by setting MATH is clearly bilinear and continuous. Hence, it induces a morphism MATH and by REF , we get a morphism MATH which is clearly well-behaved with respect to the restriction of MATH or MATH. Therefore, we get a canonical morphism MATH . To show that it is an isomorphism, it is sufficient to work at the level of germs and to prove that MATH is an isomorphism. The problem being local, we may assume MATH, MATH, MATH, MATH. In this case, REF shows that MATH and MATH . A direct computation shows that through these isomorphisms MATH corresponds to the inductive limit of the maps MATH associated to the continuous bilinear maps MATH defined by MATH . Since the diagram MATH is clearly commutative, to prove that MATH is an isomorphism, it is sufficient to prove that MATH is an isomorphism. Thanks to REF , this fact is an easy consequence of the well-known isomorphism MATH . By REF , MATH . Since MATH is a DFN space, REF , shows that MATH . Therefore, MATH as requested. |
math/9907012 | The first part is a direct consequence of REF and the NAME theorem for sheaves with values in MATH. The second part follows from the first using REF and the fact that MATH is a DFN space. |
math/9907012 | Recall that integration along the fibers of MATH (that is, on MATH) defines morphisms MATH which are compatible with MATH and MATH. Fix MATH, MATH. Let MATH be a compact subset of MATH and let MATH be an open subset of MATH. One checks easily that the morphism MATH is continuous for the canonical topologies. Applying MATH, we get a morphism MATH . Taking the inductive limit on MATH, we get a morphism MATH and hence a morphism MATH of sheaves with values in MATH. Thanks to the compatibility of REF with MATH and MATH, we also get a morphism of complexes MATH . Using REF resolutions, we get the requested integration morphism MATH . |
math/9907012 | The problem being local, it is sufficient to treat the case MATH and to show that the morphism MATH obtained by adjunction from MATH is an isomorphism for any open interval MATH of MATH. This follows directly from REF below with MATH reduced to a point. |
math/9907012 | Let MATH be an open interval of MATH and assume that MATH and MATH are isomorphisms. Then, we have successively MATH where REF follow from our assumptions, REF is obtained by adjunction and REF comes from REF . Using REF , we check easily that the composition of the preceding isomorphisms is equal to MATH. Hence, an induction on MATH reduces the problem to the case where MATH. This will be dealt with in REF below. |
math/9907012 | For MATH, sheaf theory gives us the two distinguished triangles MATH and MATH where MATH denotes the boundary of MATH. If we apply the functor MATH to the last triangle, we obtain the morphism of distinguished triangles MATH where MATH and MATH are isomorphisms of the type considered in REF below (MATH is a finite union of closed intervals of MATH). It follows that MATH is an isomorphism. |
math/9907012 | Assume first that MATH is a closed interval of MATH. Since MATH is closed in MATH, we have the distinguished triangle MATH . By NAME 's Theorem B and REF , we have the isomorphisms MATH and MATH . Hence, the long exact sequence associated to the preceding distinguished triangle ensures that MATH and that the sequence MATH is strictly exact. Applying the functor MATH to the sequence of REF below, we get the split exact sequence MATH in MATH. Therefore, MATH and MATH . Combining these results with REF , we obtain successively MATH . Thanks to REF below, it follows easily that the canonical morphism MATH is an isomorphism. Assume now that the result has been established when MATH is a union of MATH closed intervals of MATH and let us prove it when MATH where MATH REF is a closed interval of MATH. Set MATH and MATH. By the NAME theorem associated to the decomposition MATH, we have the distinguished triangle MATH . Applying the functor MATH, we obtain the distinguished triangle MATH . Now, consider the NAME distinguished triangle MATH . Since MATH is a union of MATH closed intervals of MATH, the canonical morphisms MATH are isomorphisms. The canonical diagram MATH being commutative, the canonical morphism MATH is also an isomorphism and the conclusion follows. |
math/9907012 | Let MATH be an injective resolution of MATH. Denote MATH a morphism extending MATH. The class MATH of MATH in MATH is represented by MATH where MATH extends MATH. Let MATH be a function of class MATH on MATH equals to MATH on MATH and to MATH on MATH, MATH and MATH being compact intervals such that MATH, MATH. Then, it is clear that MATH and that MATH . Therefore, MATH and MATH give the same class in MATH. It follows that MATH corresponds to the class of MATH in MATH. Since MATH represents the image of MATH by the canonical map MATH we see that MATH . Hence the conclusion. |
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