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math/9907006 | Define a collection MATH of subsets of MATH by MATH . The collection MATH has the inclusion property, because say that we have a set MATH and a set MATH with MATH such that MATH. Then MATH, so MATH, which is impossible as both MATH and MATH are idempotents. But MATH is not necessarily an ultrafilter (which holds only when MATH is irreducible). However, there exists a finite maximal partition MATH of MATH with the property that MATH for all MATH, and furthermore, such that the collections MATH are all MATH-complete ultrafilters over MATH. Assume for the moment that we have such a partition of MATH. Observe first that MATH if and only if there exists MATH such that MATH. As MATH is a set of less than measurable cardinality, so are the sets MATH, and so by REF all filters MATH are based on (unique) points MATH. Thus MATH if and only if there exists MATH such that MATH. Now set MATH. Certainly MATH is a finite subset of MATH. As MATH and MATH, it follows by the intersection property of MATH that MATH, so MATH. For MATH we thus get MATH so MATH is equal (and so a fortiori equivalent) to MATH and MATH. Therefore we are done if we can prove that a finite partition MATH of MATH as described above does indeed exist. We first prove that if MATH then the collections MATH are all MATH-complete ultrafilters over MATH. To this end we first note that if MATH with MATH, then MATH. This follows because MATH as otherwise MATH, MATH and MATH would be a partition of MATH satisfying REF and which is strictly larger than the maximal one. Hence for any MATH, we either have MATH or MATH. Now proceed as in the proof of REF with MATH replaced by MATH, to prove that the collections MATH are all MATH-complete ultrafilters over MATH. Hence we are left with proving the existence of a finite partition MATH with REF . As MATH, this is clearly equivalent to saying that MATH . Using the inclusion property of MATH, one sees that the negation of this statement is equivalent to saying that MATH . Going for ad absurdum we may complete the proof by showing that REF leads to a contradiction. We will obtain this contradiction by constructing a countable partition MATH of MATH such that MATH. This is sufficient, because say we have such a partition MATH of MATH. By REF we conclude that MATH for all MATH. Define MATH by MATH and note that MATH . Thus MATH, which is a contradiction as the spectrum of an operator in a NAME space MATH is bounded. Thus assume that REF holds for MATH. Now to construct the desired countable partition of MATH, we rely on the following property: Namely, if MATH and MATH form a partition of a set MATH having REF , then MATH or MATH have REF . Assume by absurdum that neither MATH nor MATH have REF . Then there exist numbers MATH such that MATH are maximal partitions of MATH such that MATH for all MATH. Let MATH be MATH. Thus for MATH there are no partitions MATH of MATH such that MATH for all MATH. Since MATH has REF , we may choose a partition MATH of MATH with MATH. Then define the sets MATH by MATH. They give partitions of MATH, respectively, and, for any MATH there exists MATH, depending on MATH, such that MATH, namely there exists MATH such that at least MATH elements of the partition MATH have the property that MATH. By adding the complement to one of the elements of the partition and using the inclusion property for MATH, we therefore obtain a partition MATH of one of the sets MATH such that MATH for all MATH, a contradiction with the above. Thus either MATH or MATH have REF . Consider now again the set MATH. It has REF , so split it up in two parts, say MATH and MATH with MATH. Then by the above, (at least) one of them have REF , say MATH, so we may split this one up in two parts MATH and MATH with MATH. Again one of these must have REF , say MATH, so we may split it up in two parts MATH and MATH with MATH, and one of them must have REF again, say MATH. This way we construct an infinite sequence of disjoint subsets MATH of MATH such that MATH. Now set MATH . Clearly MATH, so, by the inclusion property of MATH, we have that MATH. Define for MATH the sets MATH. Then clearly the collection MATH is a countable partition of MATH such that MATH for all MATH. This concludes the proof. |
math/9907006 | The theorem is an immediate consequence of REF . Note that the MATH's in the statement may be assumed to be nonzero. |
math/9907006 | Let MATH be the MATH-complete ultrafilter over MATH given by the representation MATH, see REF . According to REF , since MATH has less than measurable cardinality, MATH is principal. Thus for MATH having less than measurable cardinality there exists MATH such that MATH . For this MATH, we have by definition of MATH that MATH, where we consider the unit MATH of MATH as an element of MATH. As MATH, we have MATH, and thus for all MATH where MATH is considered an element of MATH and MATH denotes the element MATH. Thus MATH is a finite dimensional irreducible representation of MATH on MATH. But the finite dimensional algebra MATH is isomorphic to a matrix algebra, and we know, compare for instance CITE, that every finite dimensional irreducible representation of a matrix algebra is equivalent to the identity representation. Therefore there exists an invertible linear operator MATH such that MATH for all MATH. Hence MATH for all MATH, saying that MATH is equivalent to MATH, as desired. |
math/9907007 | All of these elementary facts are proved in CITE. |
math/9907007 | Let MATH be the component of the identity of the center of MATH. Clearly, since MATH is contained in the center of MATH, MATH. Since MATH, we have MATH. Since MATH is a maximal torus of MATH, this implies that MATH. Since MATH is the identity component of its centralizer, MATH is conjugate to MATH for some MATH. Uniqueness of MATH is clear. |
math/9907007 | If the MATH are all contained in the torus MATH, then they are mutually commuting. Moreover, if MATH is a maximal torus containing MATH, then MATH. Thus, since MATH is a maximal torus of MATH, MATH. Conversely, suppose that MATH is a maximal torus of MATH. Since each MATH commutes with MATH, it lies in MATH. |
math/9907007 | The proof is by induction on MATH. The case MATH is deduced from the following lemma. Suppose that MATH is finite. There are only finitely many conjugacy classes of elements MATH for which the center of MATH is finite. In case MATH is connected and simply connected, and thus semi-simple, the center of MATH is finite if and only if MATH is conjugate to the exponential of a vertex of an alcove MATH. Fix a maximal torus MATH. Let MATH be the normalizer of MATH in MATH modulo its centralizer. It is a finite group, whose action on MATH is covered by a linear action on MATH. Fix MATH. According to CITE, II REF Proposition MATH (see also CITE) there is a regular element in MATH fixed under MATH. Thus, after conjugation we can assume that MATH normalizes MATH and a positive NAME chamber MATH. This implies that MATH contains a regular point of MATH. The torus MATH is a maximal torus of MATH and the normalizer of MATH in MATH is contained in the normalizer of MATH. Let us suppose that the center of MATH is finite. This implies that MATH acts on MATH and that there are only finitely many points fixed by the action of MATH. Equivalently, MATH acts on the quotient torus MATH with only finitely many points fixed by MATH. It suffices to show that there are only finitely many conjugacy classes of elements MATH which REF are congruent to MATH modulo MATH, REF such that MATH and REF have the property that MATH is equal to MATH. Any element MATH satisfying REF is an element of the form MATH for some MATH. The conjugacy class of MATH in MATH depends only on MATH. For each MATH there is an element MATH such that MATH and MATH. Suppose that MATH satisfies REF for some MATH. Then, for all MATH there exists MATH, MATH, commuting with MATH. Then MATH or MATH . Thus MATH is fixed by MATH for every MATH. This means that there are only finitely many possibilities for MATH and hence only finitely many possibilities for the conjugacy class of MATH in MATH. In the case when MATH is connected and simply connected, MATH is connected. Thus, it has a finite center if and only if it is semi-simple. This occurs exactly when MATH is conjugate to the exponential of a vertex of an alcove MATH. There is the following which we shall need later: Let MATH be connected and simply connected, let MATH be a rank zero subset of almost commuting elements of MATH, and suppose that MATH for all MATH. Then MATH is semi-simple and MATH is conjugate in MATH to the exponential of a vertex of an alcove. The MATH-tuple MATH is a rank zero almost commuting MATH-tuple in MATH. Hence the center of MATH is finite, so that MATH is semi-simple. It follows from REF that MATH is conjugate to the exponential of a vertex of an alcove. Proof of REF . CASE: An almost commuting MATH-tuple is a single element MATH. Rank zero means that MATH is finite and a fortiori that its center is finite. Thus, the result in this case is immediate from REF . General Case. We prove the general case by induction on MATH. Suppose that the result is known for all groups with finite center and for all almost commuting MATH-tuples for MATH. Consider an ordered almost commuting MATH-tuple MATH of rank zero. Let MATH be the subgroup of elements of MATH whose commutator with MATH lies in MATH. Since MATH and MATH is of rank zero, we see that the center of MATH must be finite. Let MATH be the quotient of MATH by its center and let MATH be the image of MATH in MATH. Then we have an exact sequence MATH and it follows that the center of MATH is finite. Applying the previous lemma to MATH, we see that there are only finitely many possibilities for MATH up to conjugation in MATH. Hence, there are only finitely many possibilities for MATH up to conjugation. Let MATH. This is an ordered almost commuting MATH-tuple in MATH. Consider the center of the centralizer MATH of MATH in MATH. Clearly, MATH is a subgroup of finite index of MATH. Since the center of MATH is finite, it follows that the center of MATH is finite. This means that MATH is of rank zero. Thus, by the inductive hypothesis, there are only finitely many possibilities for MATH up to conjugation in MATH, and hence only finitely many possibilities for MATH in MATH up to conjugation by MATH. This completes the inductive step. |
math/9907007 | Consider the composition of MATH and the affine linear map which is the action of MATH on MATH. The translation part of this map is given by an element of MATH which exponentiates to the identity in MATH and hence is contained in the coroot lattice for MATH. The linear part is a composition of elements of the NAME group and hence is an element of the NAME group. That is to say this composition is an element of the affine NAME group of MATH. Of course, it sends MATH to MATH. This means that it is the trivial element of the affine NAME group. |
math/9907007 | Suppose MATH. If MATH stabilizes MATH, then MATH for some MATH projecting to MATH, and so MATH. Thus MATH defines a homomorphism from MATH to MATH. If in the above notation MATH, then by REF , MATH is in the group generated by reflections in the roots which are integral on MATH. Thus MATH, where MATH. From MATH it follows that MATH and hence that MATH. Thus, MATH is trivial, so that MATH is trivial as well. This shows that MATH is injective, and, if MATH is its image, then MATH. Now suppose that MATH fixes MATH. Thus MATH for some MATH. Hence MATH and MATH are two alcoves, both of which contain MATH. After transforming MATH by an element in the group generated by reflections in the walls of MATH containing MATH, we can then assume that MATH. By REF , it follows that MATH. This says that every element of MATH can be written as a product of an element in MATH times an element of the group generated by reflections in the walls of MATH containing MATH. But by REF , this second group is exactly MATH. This proves that MATH and that MATH. Since MATH is a normal subgroup of MATH and the product decomposition is unique, by the first paragraph of the proof, we see that MATH. |
math/9907007 | We have MATH . The coefficients MATH and MATH, are positive integers. Since MATH is the highest root, respectively, MATH is the highest short root, the MATH, respectively,MATH are also nonnegative. REF follows. Clearly, REF implies REF . Moreover, since the passage to an inverse system is conformal and permutes the roles of MATH and MATH, REF is equivalent to REF . It remains to see that REF implies REF . Assume REF holds. Let MATH be the simple roots not orthogonal to MATH. There exists a root MATH which is a sum of distinct simple roots including MATH. (Namely, MATH is the sum of the simple roots corresponding to the nodes of the interval connecting MATH and MATH in the NAME diagram for MATH.) Then MATH. But since MATH is a root at least as long as all other roots, the NAME integer MATH implies that MATH, and REF follows. This proves REF . By REF , there is a unique simple root MATH for which MATH meaning that MATH is a multiple of MATH. Of course, MATH. Since MATH is short, MATH, with MATH if and only if MATH. Since MATH is not of type MATH, this can never happen, proving REF . |
math/9907007 | CASE: If MATH is simple and not of type MATH, then by REF there is a fundamental weight MATH which is a root and hence kills MATH. Conversely, if MATH, the fundamental representations are the exterior powers MATH, MATH, of the standard representation, and their highest weights are nontrivial on MATH. If MATH, then MATH. Thus, REF follows from REF and the fact, again easily verified by direct inspection, that, for a proper subgroup MATH of the center of MATH, there is a fundamental weight which kills MATH. Finally, REF follows by examining a generator for the center of MATH. |
math/9907007 | Recall that MATH and that MATH if and only if MATH. From these facts REF are clear. If MATH, then MATH for some element MATH in the real linear span of the coroots MATH for MATH. Since the element MATH differs by an element of the coroot lattice from MATH, we see that MATH for all MATH. That is to say MATH. |
math/9907007 | We have MATH with MATH, MATH, and MATH where MATH is an alcove for MATH. The condition that MATH be finite implies that for each MATH the action of MATH on the alcove MATH permutes transitively the vertices of MATH. Hence, every vertex of the alcove MATH is an element of the center of MATH. This means that the highest root is the sum of the simple roots. The only groups with this property are groups of type MATH. This shows that REF implies REF . It is clear that REF implies REF . The fact that REF implies REF follows from REF . |
math/9907007 | First suppose that MATH. By Subsection REF, MATH is a set of simple roots for MATH and MATH. Of course, in the expression MATH all the coefficients MATH are non-integral. Hence by REF MATH is a product of groups MATH where MATH is isomorphic to MATH for some integer MATH and MATH where MATH generates the center of MATH. Conversely, suppose that MATH for some MATH, that MATH and that MATH is a product of groups MATH where MATH is isomorphic to MATH for some integer MATH and MATH where MATH generates the center of MATH. By REF , MATH. On the other hand, by REF , no coefficient of MATH, expressed as a linear combination of the MATH, is integral, and hence MATH. Thus MATH and so MATH. |
math/9907007 | Let MATH. It is the NAME algebra of a maximal torus for MATH. Let MATH be the alcove in MATH associated with the set of simple roots MATH for MATH with respect to MATH. Let MATH be the element of MATH whose restriction to MATH induces the action of MATH on MATH. By REF MATH fixes a unique point, say MATH, of MATH. Let MATH be the extension of MATH to MATH by the identity on MATH. Thus MATH is the image of MATH in MATH. A root of MATH which is integral on the affine space MATH must vanish on MATH and hence be a root of MATH. But since MATH is a regular element for MATH, it follows that there are no roots of MATH taking integral values on MATH. Thus there is an open dense subset of MATH consisting of regular elements for MATH. In particular, there is MATH with the following three properties: CASE: MATH is a regular element for MATH; CASE: MATH is fixed by MATH; CASE: the unique alcove MATH for the affine NAME group of MATH containing MATH also contains MATH. The point MATH lies in the alcove MATH, and MATH contains the origin. REF above implies that MATH contains the origin. It follows from REF that MATH sends MATH to itself. By REF we see that MATH is the action of MATH on MATH. Thus, the fixed point set of the NAME part MATH of MATH is exactly MATH and exponentiates onto MATH. The proposition now follows since the NAME part MATH of the action of MATH on MATH is conjugate to MATH, and hence MATH is conjugate to MATH. |
math/9907007 | The group MATH is a semi-simple subgroup of MATH for which MATH is a set of simple roots. Let MATH be the subspace spanned by the coroots MATH for MATH. Then MATH is the NAME algebra of a maximal torus of MATH. The element MATH is contained in MATH and is also in MATH since it is equal to MATH and by REF for every MATH. This means that every root of MATH takes integral values on MATH, and hence that MATH exponentiates to an element MATH contained in the center of the simply connected form MATH of MATH. By REF , MATH. The definition of MATH implies that, when MATH is expressed as a linear combination of the basis MATH, all the coefficients of the simple coroots for MATH are non-integral. Hence, by REF , MATH is a product MATH, where for each MATH the group MATH is isomorphic to MATH for some MATH, and MATH is of the form MATH where for each MATH the element MATH generates the center of MATH. Since MATH, each of the MATH has order dividing MATH. Since MATH generates the center of MATH, its order is MATH. We conclude that MATH for each MATH. Consider now the component of MATH that contains MATH. (Since MATH, we have MATH.) We index the MATH so that this component corresponds to MATH. Since the expression for MATH as a linear combination of coroots has MATH as the coefficient of MATH, it follows that the order of MATH in MATH is divisible by MATH, that is, that MATH. Since we have already shown the opposite divisibility, it must be the case that MATH, showing that MATH is isomorphic to MATH. Moreover, the order of MATH is divisible by MATH and hence is equal to MATH, and MATH is the least common multiple of the MATH. The fundamental group of MATH is cyclic, by REF , and contains the element MATH, which is of order MATH. Thus MATH divides the order of MATH. On the other hand, MATH is identified with a cyclic subgroup of MATH, and hence its order divides the least common multiple of the MATH, namely MATH. Since MATH has order MATH, it generates the fundamental group of MATH. |
math/9907007 | Let MATH . Then MATH lies in the MATH-span MATH of MATH in MATH as well as in MATH. Since MATH is a proper subset of MATH, it is a set of simple roots for a root system on MATH with the property that the given inner product on MATH restricts to a NAME invariant inner product on MATH. Let MATH be the inverse root system on MATH, so that MATH is a set of simple coroots for MATH. Since MATH, MATH has integral inner product with the lattice spanned by MATH and hence MATH. Moreover, all of the coefficients of MATH are non-integral with respect to the set of simple coroots given by MATH. Thus, by Addendum REF, MATH is a product of irreducible root systems MATH of type MATH for some integers MATH and MATH projects into each factor MATH as a generator. Since MATH, it follows that MATH. Let MATH be the cyclic subgroup of MATH of order MATH. The result now follows immediately from REF . |
math/9907007 | It suffices to consider the case where MATH is reduced and to show that the MATH such that MATH are exactly the integers MATH such that MATH. Let MATH be the smallest positive integer such that MATH. Thus MATH for MATH, and hence, by REF , MATH for MATH. If MATH, since MATH is reduced, there exists a MATH with MATH and with MATH not divisible by MATH. Choose MATH to be the smallest such positive integer. In particular, MATH, and, by REF , MATH. Since MATH it follows that MATH. For MATH, MATH and so if MATH and MATH, then MATH. This says that MATH. But MATH and MATH, contradicting the choice of MATH. Hence MATH. Applying REF with MATH and MATH, we see that MATH for all MATH with MATH. |
math/9907007 | Suppose that MATH and that MATH is relatively prime to MATH. Then by REF , MATH. Assume that MATH. Then MATH a contradiction. Thus MATH or MATH, and so MATH. |
math/9907007 | We may assume that MATH is reduced and that MATH. Since MATH, MATH. First suppose that MATH. By REF , MATH. Since MATH, by REF , it follows that MATH, which as we have just seen is at least MATH. Now suppose that MATH. Then MATH or MATH. If MATH, then as before MATH and so MATH. If MATH, then MATH. But MATH and so again MATH. |
math/9907007 | Using the identity MATH, we have MATH . |
math/9907007 | Let us first show that it suffices to consider the the case where MATH is reduced. For a general pair MATH, let MATH be the associated reduced pair. Thus MATH. For each MATH, write MATH, where MATH. Then MATH, where MATH. Moreover, it is easy to see that MATH for all other MATH. An elementary argument shows that the set of rational numbers of the form MATH, with MATH and MATH relatively prime to MATH, and MATH, is exactly the set of rational numbers of the form MATH, with MATH and MATH relatively prime to MATH and with MATH, where MATH. Thus, MATH for MATH is invariant under translation by MATH and the image of MATH in MATH is the corresponding subset for the pair MATH. Hence it suffices to consider the reduced case. Let MATH be the NAME sequence of rational numbers between MATH and MATH whose denominator is at most MATH, written in increasing order. We call integers MATH and MATH adjacent with respect to MATH if there exist MATH with MATH such that MATH and MATH are consecutive terms in MATH. If MATH and MATH are consecutive terms in MATH, then it is a standard fact that MATH. The conclusions of REF are easily seen to be equivalent to the following statement: For all consecutive terms MATH and MATH in MATH, MATH . Using the fact that MATH, this condition is equivalent to: For all integers MATH and MATH which are adjacent with respect to MATH, MATH . Another way to write the conclusions of REF is as follows: for all MATH and MATH with MATH, MATH . By the symmetry MATH it is sufficient to check these conditions for MATH. The case MATH follows from MATH. Thus, for MATH, it is enough to check the first two conditions: MATH . Let us consider the first condition. By REF , MATH . Since MATH, we see that it suffices to show that MATH . On the other hand, by REF , MATH . The condition then follows from the elementary lemma: For every positive integer MATH, and every positive divisor MATH of MATH, MATH . Fix MATH. Then that MATH where the second equality follows from MATH . This proves that the first condition holds under under REF . A very similar argument handles the second condition. The result follows for MATH. The case MATH can be checked directly. |
math/9907007 | Let MATH be a MATH-pair in MATH. Conjugation by the element MATH normalizes the connected group MATH. Thus, by CITE II REF, since MATH is finite, MATH is a torus. Hence MATH is a regular element of MATH. By symmetry, MATH is also regular. Recall that MATH is the alcove containing the origin associated to the set of simple roots MATH for MATH. By conjugation we can assume that MATH, so that MATH. Conjugation by MATH normalizes MATH, and hence MATH. Finally, conjugation by an element of MATH makes MATH the image under the exponential mapping of a point MATH. Since MATH is regular, MATH is an interior point of MATH. Let MATH be the NAME element defined by conjugation by MATH. The relation MATH yields MATH for some MATH such that MATH. We denote by MATH the affine linear map MATH. The map MATH normalizes the alcove structure for MATH and MATH, where MATH is an interior point of MATH. Thus, MATH. It follows that MATH and, by REF , that MATH is the action of MATH on MATH. This means that MATH. Since MATH, the fact that MATH is of rank zero implies that MATH is a single point. Thus, REF implies that MATH is a product of groups MATH, where, for each MATH, the group MATH is isomorphic to MATH for some MATH, and MATH projects to a generator of MATH. The alcove MATH is a product of alcoves MATH for the simple factors MATH of MATH. The unique fixed point of the MATH-action on MATH is the product of the barycenters of the MATH. Thus, MATH is the image under the exponential mapping of the product of the barycenters of the MATH. Let MATH be the subgroup of MATH generated by MATH. In fact, it is a subgroup of MATH and MATH is the cyclic group generated by MATH. This element is of order MATH, the order of MATH. Also, MATH is a product of barycenters in the MATH, and so by inspection MATH has order MATH modulo MATH. From this it is clear that MATH is isomorphic to MATH. Lastly, reversing the roles of MATH and MATH and replacing MATH by MATH we see that MATH is also conjugate to the product of the barycenters of the MATH. Hence, MATH and MATH are conjugate in MATH. Let MATH be any MATH-pair in MATH and let MATH be a maximal torus of MATH. By conjugation we can assume that MATH for some subset MATH. Then MATH, and hence, by REF , MATH. Since MATH is a product of simple factors of type MATH and MATH projects to a generator of every factor, it follows from REF that MATH. Thus, MATH and consequently, MATH is trivial, which means that MATH is of rank zero. By what we proved above, we see that MATH is conjugate to a pair MATH where MATH is the image under the exponential mapping of the product of the barycenters of the MATH and MATH. But MATH operating by inner automorphism on MATH is transitive since the component of the identity of MATH is trivial. [Proof: Consider the map MATH defined by MATH. The isotropy group of MATH is MATH, which is finite.] Thus, all such pairs are conjugate by elements of MATH. Suppose that MATH. Since MATH is a regular element of MATH, MATH and MATH. Since MATH is a product of groups of type MATH and the image of MATH in each factor generates the center of that factor, it follows by inspection that MATH. Conversely, suppose that MATH is a product of simple groups isomorphic to MATH and MATH projects to a generator of the center of each factor. Set MATH equal to the image under the exponential mapping of the product of the barycenters of the alcoves for the simple factors of MATH and take MATH. Then MATH is a MATH-pair of rank zero. |
math/9907007 | All of these statements were established in the course of the proof of the previous proposition, except the statements about the orders of MATH and MATH. This statement follows by inspection of the order of the image under the exponential mapping of the barycenter in a group of type MATH. |
math/9907007 | Set MATH. According to REF , MATH and there is a rank zero MATH-pair in MATH. Since MATH is simply connected, it follows from REF that there are integers MATH such that MATH is isomorphic to MATH and under this isomorphism MATH where MATH generates the center of MATH. According to REF , writing MATH no MATH is an integer. It then follows from REF that MATH, and hence that any maximal torus of MATH is conjugate to MATH. It also follows from REF that the MATH-pair in MATH is unique up to conjugation. |
math/9907007 | According to REF , the moduli space MATH of MATH-pairs of rank zero in MATH is empty unless MATH and the moduli space MATH is a single point. Thus, the actions of MATH and of MATH on MATH are trivial. The result is now immediate from REF . |
math/9907007 | We know by REF that MATH is isomorphic to a product of groups of the form MATH and that, if MATH is the component of MATH in the MATH factor, then MATH generates the center of MATH. The vector space MATH is a quotient of the vector space MATH by the one-dimensional space spanned by MATH. The element MATH acts on MATH by permuting the MATH and is the identity on MATH. For each orbit MATH, the subspace MATH is MATH-invariant, and the eigenvalues of MATH on this subspace are MATH, where MATH is a primitive MATH root of unity. On the other hand, MATH is conjugate to the product of the MATH, and the eigenvalues of MATH on the subspace of MATH corresponding to the simple factor MATH are MATH. The proposition follows by comparing the two forms for the set of eigenvalues. |
math/9907007 | By REF , after conjugation, we can assume that the maximal torus of MATH is MATH. There is a rank zero MATH-pair MATH in MATH and elements MATH such that MATH . The intersection MATH is the maximal torus of MATH. According to REF we can assume that MATH is the barycenter of an alcove MATH of MATH and that MATH normalizes this torus and has image in MATH equal to the NAME part MATH of the action of MATH on the alcove MATH for MATH. Thus, MATH and MATH normalizes MATH and projects to the image of MATH in MATH. Since this image is the NAME part of the action of MATH with respect to any alcove MATH for MATH containing MATH, the result now follows by conjugating MATH and MATH by an element of MATH which sends MATH to MATH. |
math/9907007 | Let MATH have rank zero. By REF , MATH is semi-simple and MATH is conjugate to the image under the exponential mapping of a vertex MATH. Letting MATH be the simple root such that MATH defines the wall of MATH opposite MATH, we see that the order of MATH modulo MATH is MATH. In particular, MATH divides the order of MATH. The pair MATH is a commuting pair in MATH of rank zero. Lift MATH to elements MATH in the universal covering MATH of MATH. Then MATH is a rank zero MATH-pair for some MATH. According to REF , the group MATH is cyclic of order MATH and is generated by MATH, where MATH . By REF , the existence of a rank zero MATH-pair implies that MATH generates MATH and that, if MATH, then, in the expression of MATH as a linear combination of the simple coroots, all coefficients are non-integral. Since MATH is a power of MATH, the same is true for MATH. This implies that, for each MATH, the integer MATH is not divisible by MATH. By REF , the elements MATH and MATH are conjugate in MATH, and hence MATH and MATH are conjugate in MATH. It follows from the same proposition that the subgroup of MATH generated by MATH is isomorphic to MATH. In particular, MATH and MATH have the same order MATH in MATH. Interchanging the roles of MATH and MATH in this construction, we see that MATH and MATH are conjugate. Thus, MATH are all conjugate in MATH and hence all have the same order, MATH. Since we have already shown that the order of MATH is divisible by MATH and since MATH, it follows that MATH. To see the converse, suppose that MATH for exactly one MATH. Then, by REF , MATH. Let MATH be the image under the exponential map of the vertex of the alcove opposite the face MATH. By REF , the universal cover MATH is a product of groups of type MATH, and MATH, where MATH has order MATH and projects to a generator of every factor. By REF , there is a rank zero MATH-pair MATH in MATH. It suffices to take MATH to be the image in MATH of MATH. |
math/9907007 | The first statement follows from REF . Suppose that MATH is rank zero commuting triple of order MATH. According to REF , MATH is conjugate to the image under the exponential mapping of the vertex of the alcove opposite the face of MATH defined by MATH where MATH is the unique element of MATH such that MATH. Then MATH is a rank zero commuting pair in MATH. Let MATH be a lift of MATH to the universal covering MATH of MATH. This is a rank zero MATH-pair for some MATH generating MATH. It follows that MATH is MATH with MATH isomorphic to MATH for an integer MATH and MATH where MATH generates the center of MATH. The element MATH dependes only on the conjugacy class of MATH in MATH. By REF , MATH determines the conjugacy class of MATH in MATH and hence the conjugacy class of MATH in MATH. On the other hand, again by REF , for each MATH there is a MATH-pair MATH in MATH. Moreover, the MATH-pair MATH is of rank zero if and only if MATH generates MATH, and in this case MATH is unique up to conjugation in MATH. The image MATH is a rank zero commuting pair in MATH. The group MATH is cyclic of order MATH and hence has MATH generators. This shows that there are exactly MATH conjugacy classes of commuting pairs of rank zero in MATH, and hence MATH conjugacy classes of rank zero commuting triples in MATH. Clearly, if MATH then MATH. This proves the last statement. |
math/9907007 | Let MATH be a rank zero commuting triple in MATH. Let MATH be the order of MATH. There is a unique MATH such that MATH. After conjugation we can assume that MATH is the exponential of the vertex MATH of the alcove MATH opposite the face MATH. After composing MATH with a suitable inner automorphism, we can assume that MATH normalizes MATH. The action of MATH on the set of simple roots preserves the integers MATH. Hence MATH and therefore MATH and MATH. Thus MATH acts on MATH, and by REF , since MATH preserves the coroot integers MATH, the action is trivial. Hence MATH acts trivially on the conjugacy class of MATH in MATH, by REF , and therefore on the conjugacy class of MATH in MATH. |
math/9907007 | Let MATH be a rank zero commuting triple of order MATH. First consider the action of MATH on MATH. We can assume that MATH is the image under the exponential mapping of the vertex of the alcove MATH opposite the face MATH where MATH is the unique element with MATH. For any MATH let MATH be the NAME element which is the linear part of the action of MATH on MATH. The NAME element MATH normalizes MATH and, according to Subsection REF, preserves the MATH in the sense that MATH. This implies that MATH. Thus, the affine automorphism MATH of MATH fixes MATH. By REF , this means that if MATH projects to MATH, then MATH. Thus the triples MATH and MATH are conjugate. Conjugation by MATH normalizes MATH and induces the identity automorphism of its fundamental group. Since MATH the first statement is clear. By REF , the exponential map identifies the subgroup of MATH generated by MATH with the fundamental group of MATH. The element MATH normalizes the set MATH and the subset MATH. Since MATH, for MATH, it is clear that MATH fixes MATH, and hence acts trivially on MATH. It follows immediately from the claim that MATH lifts to a MATH-pair in MATH, where MATH for any two lifts of MATH to MATH. By REF , MATH is conjugate in MATH to MATH. Hence MATH is conjugate to MATH, and so the action of MATH on the first factor of commuting triples induces the trivial action on the space of conjugacy classes of commuting triples of rank zero. The situation is completely symmetric in MATH and it then follows that the action of MATH on the space of conjugacy classes of commuting triples of rank zero is trivial. |
math/9907007 | Let MATH be a commuting triple, let MATH be a maximal torus of MATH and let MATH. By REF , there is a commuting triple MATH of rank zero in MATH such that MATH. Let MATH be the order of MATH. If MATH, then MATH is trivial. In this case, MATH and hence they are all contained in a maximal torus. Thus MATH itself is a maximal torus for MATH. Conversely, if MATH are contained in a maximal torus MATH, then MATH is a maximal torus for MATH and hence MATH is trivial. For the rest of the proof, we assume that MATH is not trivial, or equivalently that MATH is not a maximal torus. MATH is a simply connected, simple group, not of type MATH for any MATH. If MATH is not simply laced, then MATH is not simply laced. Let MATH be the decomposition of MATH into its simple factors. Since MATH, MATH is simply connected and the NAME diagram of MATH is identified with a subdiagram of the NAME diagram of MATH. Thus, at most one of the components MATH is not of type MATH. Since MATH has a commuting triple of rank zero, it follows that each of the MATH has such a triple. By REF this implies that no MATH is of type MATH. Thus, MATH is simple. The last statement is clear since, if MATH is not simply laced, then MATH is a chain, and hence every simply laced subdiagram is of type MATH for some MATH. For a generic MATH, MATH. In particular, MATH is semi-simple. Suppose that MATH is generic in MATH. The roots in MATH are the roots that annihilate MATH. Since MATH is generic, a root annihilates MATH if and only if it annihilates MATH and annihilates MATH. This shows that the roots of MATH annihilating MATH are exactly the roots of MATH annihilating MATH and hence that MATH. Since MATH is a rank zero commuting triple in MATH, the center of MATH is finite. Since MATH is connected, it is semi-simple and equal to its own derived group. Thus MATH. Let MATH be generic. After conjugation, we can assume that MATH is the the image under the exponential mapping of a point in MATH. We let MATH be the subset consisting of all the roots vanishing on MATH. The subset MATH forms a set of simple roots for the root system of MATH with respect to the maximal torus MATH. It now follows from REF that MATH is a cyclic group of order MATH, and hence MATH is cyclic of order MATH. By REF , this means that MATH divides MATH for every MATH. That is to say MATH. Hence MATH is the image under the exponential of a point of MATH. Since MATH is generic, the same conclusion holds up to conjugation for MATH. Since MATH is a rank zero triple in MATH, it follows from REF that if MATH is any set of simple roots for the root system MATH and if MATH has the property that its image under the exponential mapping is a central element in MATH generating MATH, then no MATH is an integer. But according to REF , MATH is a set of simple roots for the root system MATH and the exponential of MATH generates the fundamental group of MATH. It follows that MATH for every MATH. This implies that MATH, and hence that MATH. Hence MATH is contained in the interior of the face MATH of MATH and MATH has MATH as a set of simple roots. Consequently, MATH has MATH as a set of simple roots. Since MATH is semi-simple, MATH. Thus, MATH is the perpendicular space to MATH. It follows that MATH and MATH. |
math/9907007 | We begin with the following result about root systems: Let MATH be a reduced and irreducible root system on a vector space MATH, and suppose that MATH is an integer such that MATH for some MATH. Define MATH and MATH as before, and let MATH be the set of all roots which annihilate MATH. Then MATH is an irreducible root system. Moreover, if the coroot integers for MATH are of the form MATH, then MATH for exactly one MATH, and in this case MATH. We may assume that MATH is the root system of a simple and simply connected group MATH. Thus there is a torus MATH corresponding to MATH. Let MATH. Then MATH is the set of roots for MATH, and in particular it is a root system. Let MATH be the sublattice of MATH generated by the coroots of MATH. Then MATH is a primitive sublattice of MATH, by REF . Let MATH be the lattice spanned by MATH. Then MATH is a sublattice of MATH. By REF , MATH is a cyclic group of order equal to the gcd of the MATH such that MATH, and by REF , this gcd is MATH. Since MATH is a primitive sublattice of MATH, it follows that MATH is also cyclic of order MATH. Next, we have the following description of the root system MATH: Let MATH be the set of positive roots for MATH corresponding to MATH and let MATH. Then MATH is a set of positive roots for MATH. Let MATH be the corresponding set of simple roots. Then MATH for some root MATH. The root system MATH is irreducible and the highest root MATH for MATH is also a highest root for MATH. Choose any MATH contained in the interior of MATH. Then the roots of MATH which are positive on MATH are exactly those in MATH. Thus, MATH is a set of positive roots with respect to some set of simple roots of MATH. Clearly, the elements of MATH are positive roots. Since none of these can be written as a non-trivial linear combination of positive roots of MATH, a fortiori none of these can be written as a non-trivial linear combination of positive roots of MATH. Thus MATH is a subset of the set of simple roots MATH determined by MATH. Since MATH, MATH implying that the cardinality of MATH is one less than the dimension of the span of MATH. Thus, there is a root MATH with the property that MATH. Let MATH be the highest root of MATH with respect to the positive roots MATH. Since MATH, MATH. Write MATH. Since MATH is a basis for MATH and MATH is a basis for MATH, it follows that MATH is cyclic of order MATH. Thus MATH. Since MATH for all MATH, it follows that MATH for all MATH. This proves that all the coefficients is this expression are non-trivial, and hence MATH is irreducible. Since the sum of MATH and any positive root in MATH is not a root of MATH, it follows that MATH is the highest root of MATH with respect to the set of simple roots MATH. Returning to the proof of REF , we see that we have proved that MATH is irreducible and that MATH and MATH for MATH. This completes the proof of REF . Finally, let us finish the proof of REF . Since MATH is a simple group and MATH is equal to exactly one of the coroot integers of MATH, it follows by REF that MATH contains a commuting rank zero triple of order MATH. Of course, such a triple will also be a commuting triple of order MATH in MATH. |
math/9907007 | The first statement follows from REF . Clearly, the order is a conjugacy class invariant, and it is locally constant on MATH by REF . Now suppose that MATH divides at least one of the MATH. By REF , there is a commuting triple MATH of order MATH in MATH. By REF , we can assume after conjugation that MATH is a maximal torus of MATH. By REF , there is a rank zero commuting triple of order MATH in MATH. It then follows from REF there are exactly MATH conjugacy classes of commuting triples of rank zero in MATH. By REF , the center of MATH acts trivially on the space of conjugacy classes of commuting triples in MATH or, in the notation of REF , the group MATH acts trivially on the space of conjugacy classes of commuting triples in MATH. By REF , the NAME group of MATH acts trivially on the set of conjugacy classes of rank zero commuting triples in MATH. REF now implies that there are exactly MATH components of MATH of triples of order MATH, and each of these components is homeomorphic to MATH proving the first sentence in REF. Let MATH be a component of MATH of order MATH. By REF , MATH . It follows directly from the definition of MATH that MATH is equal to the number of MATH such that MATH. The second statement of REF then follows from the first statement of REF . |
math/9907007 | Since MATH contains an interior point of an alcove, no wall of MATH can contain MATH. The second and third statements are now clear. As to the last, if MATH is an alcove of MATH normalized by MATH, then MATH contains an interior point of MATH, and hence MATH is the closure of its interior in MATH. Clearly, MATH. Conversely, let MATH. Since MATH contains a non-empty open subset of MATH, it contains a element of MATH. This shows that MATH is contained in a unique alcove MATH of MATH. Clearly, MATH and MATH. |
math/9907007 | Clearly, it suffices to establish this result in the case that MATH is irreducible, so that MATH is a simplex. Suppose that MATH are in the same MATH-orbit. Then their restrictions to MATH are equal. Furthermore, the walls MATH and MATH of MATH determined by MATH and MATH are in the same MATH-orbit. This means that MATH. Since the restrictions of the walls of MATH to MATH cut out a compact convex body, there must be at least MATH distinct and non-parallel walls. But this is exactly the cardinality of MATH. Hence, it follows that distinct MATH-orbits in MATH cut out distinct and non-parallel walls in MATH, and hence have distinct, nonempty restrictions to MATH. |
math/9907007 | Since MATH normalizes MATH, it normalizes the group generated by reflections in the walls of MATH, that is, the affine NAME group MATH. To establish REF , clearly MATH. Conversely, suppose that MATH. Then MATH is an element of MATH meeting MATH in an interior point. Thus for all MATH, MATH. By REF , MATH and hence MATH. Since this is true for all MATH, MATH. The final statement is now clear since MATH acts simply transitively on the set of all alcoves. |
math/9907007 | The wall MATH is a common wall between two alcoves MATH of MATH. Let MATH be the MATH-invariant alcoves of MATH containing MATH. Let MATH be the unique element carrying MATH to MATH. Then MATH is a product of reflections about walls separating MATH and MATH, and hence it is the identity on MATH and a fortiori on MATH. Since MATH contains a nonempty open subset of MATH, MATH is an isometry fixing MATH and sending MATH to MATH. It is then the reflection in MATH. |
math/9907007 | We may assume that MATH is irreducible. By REF , there is a NAME group with MATH as fundamental domain. By the general classification result for such NAME groups CITE, it follows that this NAME group is isomorphic to the affine NAME group of a (reduced) root system MATH. That is to say there is a linear structure on MATH compatible with its given affine structure, such that under this identification the NAME group becomes the affine NAME group of a root system. Of course this linear structure is determined by choosing a point MATH to identify MATH with MATH. The point MATH must be a vertex of an alcove. In fact, we can choose it to be a vertex of the alcove MATH. When we do this, the restricted roots defining the walls of MATH become the set of extended simple roots. By REF these roots are exactly the walls of the restrictions of the orbits MATH. The proof of REF shows that MATH has MATH walls. Hence MATH is a simplex, so that MATH is irreducible. The last statement follows since a root is determined up to a multiple by the wall it defines, and it is easy to see in this case that the multiple must be positive. |
math/9907007 | By REF , we may assume that MATH is the image under the exponential mapping of a point MATH and that MATH projects to MATH in MATH. By REF applied to the affine automorphism MATH of REF, there is a MATH, commuting with MATH, such that MATH. Let MATH project to the element of MATH which is the linear part of MATH. Then MATH, MATH satisfy the conclusions of the lemma. |
math/9907007 | By the remarks before the statement of the lemma, it suffices to consider the case where MATH is irreducible. If MATH is of type MATH, then the corresponding group of diagram automorphisms is a cyclic group or a dihedral group and the lemma follows by inspection. Otherwise, MATH is a connected, contractible diagram which is not a single orbit. Let MATH be the proper subdiagram of MATH defined by MATH and suppose that MATH is a union of MATH connected subdiagrams MATH. Then the MATH are all isomorphic and the stabilizer of MATH in MATH acts transitively on the nodes of MATH. It follows that each MATH is of type MATH or to MATH. The first case corresponds to an ordinary orbit, the second to an exceptional orbit. The final statement of REF follows easily from the NAME fixed point formula. |
math/9907007 | Choose a point MATH which is contained in no other wall. By REF , the set of roots in MATH which take integral values on MATH is exactly MATH. By REF , since MATH, the set of roots MATH in MATH such that MATH is integral on MATH is a root system with simple roots equal to MATH. If MATH is ordinary, then this root system exactly is a product of root systems of type MATH and one of MATH is contained in MATH. Otherwise MATH is exceptional and this root system is a product of root systems of type MATH. Every root of this system, up to sign, is either in MATH or is the sum of an exceptional pair in MATH. |
math/9907007 | Since MATH, the wall MATH corresponding to MATH meets MATH in a hyperplane. Thus there is MATH such that MATH defines a wall of MATH. Suppose that MATH corresponds to the orbit MATH of simple roots. If MATH is the linear part of MATH, then MATH commutes with MATH and hence sends MATH-orbits to MATH-orbits. Then result now follows from REF . |
math/9907007 | Elements MATH restrict to give the same root in MATH if and only if MATH. Clearly, MATH is a positive real multiple of MATH and by REF MATH. The proves the first statement. The second follows from the fact that MATH. |
math/9907007 | Clearly the MATH span the dual space to MATH and hence the roots of MATH span this space. As we saw in the above claim, for any MATH the inner product MATH is MATH. It is clear from the definitions that for MATH we have MATH. Thus it suffices to show that, for all MATH, we have MATH . We fix MATH, respectively, MATH, such that restriction of MATH, respectively, MATH to MATH is MATH, respectively, MATH. Then MATH . First assume that MATH is an ordinary orbit. Since MATH for all MATH and since MATH, we have MATH . Suppose that MATH where the MATH are pairwise distinct. Since the MATH are mutually orthogonal, this last equation can be rewritten as MATH . (Notice that the MATH commute, so that the composition is independent of the ordering.) Clearly, then, MATH. In case MATH is an exceptional orbit MATH with the MATH being exceptional pairs we have MATH since MATH and MATH have the same restriction to MATH. In this case, MATH is a root and MATH since MATH, MATH, and MATH have the same length. Thus MATH . (Here, the MATH are orthogonal, so that the order of the reflections is again irrevelant.) This proves that MATH in this case also. The walls in MATH defined by the elements of MATH are the same as the walls of MATH, viewed as linear hyperplanes. Thus, the NAME groups are the same. Finally, if MATH is irreducible, we cannot divide the set of walls for MATH into two nonempty, mutually orthogonal subsets. Thus the same is true for MATH, and hence MATH is irreducible. |
math/9907007 | In case MATH is simply laced, the inner product identifies MATH with MATH and MATH with MATH. Thus the roots of MATH are identified with the coroots in MATH. |
math/9907007 | First we claim that there is fixed point for the action of MATH on the topological space MATH associated to MATH. The proof is by induction on the number of vertices. Clearly MATH has a fixed point if there are MATH or MATH vertices. Otherwise, let MATH be the subgraph obtained by deleting the leaves. Then MATH is a nonempty contractible proper subgraph on which MATH acts, so by induction there is a fixed point of the action of MATH on MATH and hence on MATH. Choose a point MATH fixed by MATH. If MATH is an interior point of an edge MATH whose boundary is MATH, then it is easy to see that MATH is the set of MATH such that MATH. Assume that we are not in this case. There is a unique path MATH in MATH joining MATH to MATH. Possibly after switching MATH and MATH, we can assume that MATH does not lie on this path. Hence the unique path MATH from MATH to MATH is the union of MATH with the edge connecting MATH and MATH. If MATH fixes MATH, then MATH. Since MATH, MATH. Thus MATH. It follows that MATH. |
math/9907007 | Let MATH have the property that MATH for all MATH and all MATH. The two subspaces of MATH spanned by the MATH such that MATH, respectively, the MATH such that MATH, are orthogonal, and hence MATH and MATH in MATH are also orthogonal. Thus MATH and there is no bond between MATH and MATH in the affine diagram associated with the generalized NAME matrix of these elements. Suppose that there exist MATH and MATH such that MATH. By the previous lemma, either MATH or MATH. Since the root inverse to MATH is MATH we have MATH . If MATH, we see that MATH and MATH in this case. On the other hand if MATH, then MATH and thus MATH . |
math/9907007 | Since MATH is dihedral, the stabilizer of an element has either one or two elements. The only case not covered by REF is the case where there exist two non-orthogonal coroots MATH and MATH such that MATH and MATH are both nontrivial. In this case, the product of the two nontrivial elements is a rotation which either has order MATH, if MATH is even, or MATH, if MATH is odd. In the first case, MATH, and in the second case either MATH or MATH is of type MATH, the rotation subgroup MATH of MATH has order exactly MATH, and there is an involution in MATH fixing two vertices. In this case, MATH, and the quotient coroot diagram is of type MATH. |
math/9907007 | It suffices to consider the case when MATH is irreducible. Clearly, the result holds if MATH is the trivial group. Thus, we assume that MATH. We have seen that we can realize the elements of the NAME group of MATH or of MATH as elements of the NAME group MATH normalizing MATH. Thus there is a homomorphism from the NAME group of MATH to MATH, and since the NAME group of MATH acts faithfully on MATH, this homomorphism is injective. We must show that its image is all of MATH. Given MATH, represent MATH by an element of MATH which normalizes MATH. Since MATH permutes the set MATH of restricted roots and preserves the inner product on MATH, it defines an automorphism of the root system MATH. We claim that MATH is either non-simply laced or MATH. Assuming this, since every automorphism of a root system which is either non-simply laced or MATH is given by a NAME element, it follows that MATH is given by an element of the NAME group of MATH, or equivalently of MATH. Note that that MATH is non-simply laced if there is an exceptional orbit. Thus we may assume that there are no exceptional orbits. First assume that MATH is not of MATH type. In particular, the numbers MATH cannot all be equal since the diagram MATH is contractible. In particular there must exist two coroots MATH which are not orthogonal and such that MATH. If MATH is simply laced, it follows from REF that MATH, and hence that MATH is not simply laced. If MATH is not simply laced, MATH is a short coroot, and MATH is a long coroot, it is easy to see that MATH, and thus that MATH. Thus MATH, and MATH . Thus MATH is non-simply laced in this case also. Direct inspection then handles the case where MATH is of MATH type and MATH is not cyclic. |
math/9907007 | REF follows since MATH contains a regular element. To see REF , note that, if MATH and MATH are such that MATH, then MATH. By REF , this can only happen if MATH and MATH lie in the same orbit or MATH lies in an exceptional orbit and MATH. In this case MATH. Thus, if MATH and MATH and MATH do not lie in the same orbit, then MATH for some MATH. This contradicts REF . Moreover, it follows that MATH is reduced if and only if there are no exceptional orbits. It suffices to prove REF , and REF under the assumption that MATH is irreducible. To see REF , and REF , note that every positive root MATH can be written as a positive integral linear combination MATH. Thus MATH, and similarly for negative roots. Since the cardinality of MATH is the dimension of MATH, it follows that the MATH are linearly independent and hence a set of simple roots for MATH. Taking MATH, we see that the coefficients of MATH, in terms of the simple roots, are at most those of MATH, and hence MATH is a highest root for MATH. Hence the root integers are as claimed. To see REF , since MATH is the highest root for MATH, it follows that the alcove for MATH is the alcove for MATH. Thus if MATH is reduced we must have MATH. If MATH is not reduced, it is easy to see that the root system associated to the alcove is exactly the set of roots MATH such that MATH is not a root. The remaining statement in REF follows by a direct inspection. |
math/9907007 | Clearly MATH normalizes MATH, and since MATH contains a regular element, the action of MATH on MATH is faithful. Thus MATH. Conversely, if MATH, choose a regular element MATH. Then, for all MATH, MATH. Thus since MATH is regular MATH for all MATH, so that MATH. To see REF , the coroot lattice for MATH is spanned by elements of the form MATH. If MATH is ordinary, this is just MATH, and if MATH is exceptional, corresponding to the pair MATH, then MATH is again in the coroot lattice and the corresponding orbit sum is MATH. Thus the coroot lattice is generated by orbit sums. Since MATH permutes an integral basis for MATH, the coroot orbits generate MATH. Finally, it follows from the definition that the coroot lattice for MATH is MATH. |
math/9907007 | Let MATH. We claim that no root of MATH is integral on MATH and hence MATH is in the interior of an alcove MATH of the root system MATH. In fact, if MATH is integral on MATH, then MATH is a root of MATH which vanishes on MATH and hence is integral on MATH. But this contradicts our assumption that MATH contains an interior point of an alcove for MATH. The group MATH normalizes MATH and the alcove structure for the root system MATH. The point MATH is the unique fixed point for MATH and is in the interior of MATH. Thus MATH is the barycenter of MATH. |
math/9907007 | Let MATH. Then its differential is an element of the NAME group of MATH which normalizes MATH. The elements of the NAME group of MATH are exactly the restrictions to MATH of the differentials of elements MATH. This proves that the NAME group of MATH is identified with a subgroup of MATH. Let MATH. Then MATH normalizes MATH and the alcove decomposition of MATH for MATH. As such, setting MATH, the point MATH is the barycenter of some alcove for this alcove decomposition. Thus, there is an element MATH of the affine NAME group MATH with MATH. Then MATH. Let MATH be the NAME part of MATH. Since MATH, the element MATH normalizes MATH. Hence, MATH normalizes MATH, and hence by REF is an element of MATH. Thus, the restriction to MATH of its differential is an element of the NAME group of MATH. The element MATH, and consequently also its differential, centralize MATH. Thus, the restriction to MATH of the differential of MATH agrees with that of MATH. This proves that the NAME group of MATH is all of MATH. |
math/9907007 | Clearly, we may assume that MATH is irreducible. The only irreducible case not covered by REF is when MATH is of type MATH and MATH is a group of rotations of the extended NAME diagram. We choose a representation of MATH on the subspace MATH in MATH such that the set of simple roots is MATH. Suppose that MATH is a group of order MATH and let MATH. Then MATH is the intersection of the kernels of the roots MATH. Hence, this case is covered by REF . |
math/9907007 | We decompose MATH and let MATH. Suppose that MATH and that MATH is translation by an element MATH. We write MATH for MATH in the NAME group of MATH. Since MATH, MATH, and hence MATH. Conversely, if MATH, write MATH for some MATH, with MATH. Clearly, MATH is contained in the center of the root system MATH, and hence MATH is the barycenter of an alcove for MATH. This means that there is an element MATH such that MATH. The composition of translation by MATH followed by MATH is an element of MATH whose restriction to MATH is translation by MATH. |
math/9907007 | By REF , for each MATH, the fixed point subspace MATH is conjugate under the NAME group to MATH. Since MATH satisfies the hypothesis of REF , so does MATH. Since MATH is the linearization of MATH, and MATH, the lemma follows. |
math/9907007 | By REF , MATH and MATH have the same coroot lattices. By REF , their NAME groups are the same. Now there is the following general lemma on two root systems with the same coroot lattice: Let MATH and MATH be two root systems on a vector space MATH, and suppose that MATH is reduced. Suppose that the coroot lattice MATH is equal to the coroot lattice MATH and that MATH. Then MATH is the set of all non-multipliable roots in MATH. In particular, if MATH is reduced, then MATH. Let MATH be the subroot system of MATH consisting of the non-multipliable roots. Then MATH is the sub-coroot system of MATH consisting of all indivisible coroots in MATH. In particular, the coroot lattice of MATH is equal to that of MATH. Of course, MATH and MATH have the same NAME group. Thus, it suffices to assume that MATH is reduced. Let MATH. For each MATH, MATH is an indivisible element of MATH. Since reflection in the wall defined by MATH is an element of MATH, there exists a MATH such that MATH for some real number MATH. But MATH is also indivisible, so that MATH. Thus MATH. By symmetry MATH, and hence MATH. Applying the lemma to MATH and MATH completes the proof of REF . |
math/9907007 | We have an identification of MATH with the non-multipliable roots of MATH. We also know by REF that, up to positive multiples, the image under orthogonal projection of MATH forms the set of coroots inverse to an extended set of simple coroots for MATH. If MATH is reduced, then it is equal to MATH and its coroots are indivisible elements in the coroot lattice. Thus, as before, the multiples are all MATH and the image of MATH under orthogonal projection forms the set of coroots inverse to an extended set of simple roots for MATH. If MATH is not reduced, then it is of type MATH for some MATH and MATH is the subsystem of type MATH. In the extended set of simple coroots for MATH all but two are neither divisible nor multipliable in MATH and the last two are multipliable by MATH in MATH. Thus, the image of MATH contains all the former coroots and, for each of the latter two, contains either the coroot in MATH or twice it. A priori there are three possibilities: CASE: MATH is equal to the extended set of coroots for MATH (that is, consists of indivisible coroots); CASE: MATH is equal to the extended set of coroots for MATH (that is, contains one indivisible but multipliable coroot and one non-multipliable but divisible coroot); or REF contains two non-multipliable but divisible coroots. In REF the lattice spanned by MATH is of index two in the coroot lattice of MATH and hence this case is ruled out by REF . Since by REF MATH is the subroot system of MATH consisting of the non-multipliable roots, REF are exactly the two cases listed in the statement of the proposition. |
math/9907007 | By REF, MATH. On the other hand, by REF, and the MATH are relatively prime integers. Thus the corollary is clear. |
math/9907007 | Consider the exact sequence MATH . Taking the associated long exact cohomology sequence gives MATH . By the remarks above, MATH and under this identification the image of MATH is identified with MATH. Thus the kernel of the map MATH is exactly the torsion subgroup. Since MATH is connected, it follows that MATH. If MATH is simply connected, then MATH and MATH acts as a permutation of the set of simple roots MATH. Thus MATH. The remaining statements are clear. |
math/9907007 | First we claim that, if MATH is fixed by the action of MATH, then MATH. Since MATH, it clearly suffices to show that, if MATH is fixed by MATH, then MATH. The action of MATH on MATH normalizes the root spaces of MATH. The action of MATH on a root space MATH is given by the character MATH. By REF , MATH contains a regular element of MATH. Thus, none of these characters is trivial. Hence, if MATH is fixed by MATH, then MATH. It follows that the fixed subspace of the action of MATH on MATH is MATH, and hence that the fixed subspace of the action of MATH on MATH is MATH. Thus MATH is a maximal torus of MATH. Also, the roots for the MATH-action on MATH are given by restrictions of roots of MATH to MATH and so are elements of MATH. Hence the roots for MATH with respect to MATH form a subset of MATH. To see that they are a sub-root system, in other words that the corresponding coroots are the same, it is enough to show that the inner product on MATH induced by the NAME invariant inner product on MATH is invariant under the NAME group of MATH. Since MATH contains a regular element of MATH, every element in the NAME group of MATH is the restriction of a NAME element of MATH, and thus the last statement is clear. |
math/9907007 | Let MATH. Then MATH is a maximal torus of MATH and hence there is MATH such that MATH. Clearly, MATH and MATH and MATH lie in the same component of MATH. If MATH and MATH are in the same component of MATH and both normalize MATH, then MATH and normalizes MATH. Thus MATH for some MATH. |
math/9907007 | The induced map MATH to MATH is surjective by REF . Its kernel is MATH. Any element in this intersection normalizes MATH and its action on MATH commutes with MATH. Thus, it also normalizes MATH. Conversely, since MATH contains a regular element of MATH, any element of MATH normalizing MATH normalizes MATH. |
math/9907007 | By definition, there is an exact sequence MATH . Now MATH is a normal subgroup of MATH, and the image of MATH in MATH is MATH. Thus MATH contains MATH as a normal subgroup, and there is a surjection MATH . By a diagram chase, the kernel of this map is then MATH . This establishes the exact sequence of the statement. |
math/9907007 | Since MATH is an linearly independent subset of MATH invariant under MATH, the quotient set MATH is a linearly independent subset of MATH. For each MATH, order MATH so that MATH (by convention MATH). Then the action of MATH on MATH sends the root space MATH to MATH. Hence, MATH sends MATH to itself for every MATH. This means that there is MATH such that MATH is multiplication by MATH on each root space MATH. By equivariance under MATH, we see that MATH for all MATH. We denote this common value by MATH. Choose an element MATH such that MATH. Since the elements of MATH are linearly independent in MATH, it follows that there is an element MATH such that MATH for every MATH. We set MATH. By construction, if MATH, then MATH is the identity. Choose a non-zero element MATH. Set MATH . Then MATH is a non-zero element of MATH invariant under MATH, and hence MATH. Moreover, for all MATH, MATH. Thus, each MATH is a root of MATH with respect to MATH. Since every root of MATH is the restriction of a root of MATH to MATH and since MATH is a set of simple roots for MATH, it follows that every root of MATH can be written uniquely as a linear combination of the elements in MATH and the coefficients of this linear combination are either all positive integers or all negative integers. Consequently, MATH is a set of simple roots of MATH with respect to MATH. |
math/9907007 | According to REF , the indivisible roots of MATH form a reduced root system with MATH a set of simple roots. By REF , MATH is also a set of simple roots for MATH. Since the latter is also a reduced root system, the two root systems agree. By REF , MATH is then MATH. |
math/9907007 | By REF , MATH. By the previous lemma, MATH, and MATH. Thus the quotient MATH is trivial, so that MATH. |
math/9907007 | According to REF , these root systems have the same NAME group, which implies that they have the same set of walls. Since each of these root systems is reduced, the result follows. |
math/9907007 | Since MATH, it follows that MATH and that MATH and MATH have the same action on MATH so that MATH. Since MATH has no exceptional orbits, it follows from REF that MATH is a reduced root system. Thus, by REF each MATH in MATH is a root of of MATH with respect to MATH. The roots of MATH with respect to MATH are a subset of MATH. Let MATH be an orbit of order MATH. Let MATH be this orbit, ordered so that MATH (by convention MATH). For each MATH let MATH be the root space for MATH. Since MATH is a root of MATH, it follows that the action of MATH identifies the root space MATH with MATH in such a way that the action of MATH is the identity on each of these root spaces. Since MATH, we have MATH for all MATH. Thus, MATH if and only if the action of MATH is trivial on all these root spaces if and only if MATH is a root of MATH. Of course MATH if and only if MATH. |
math/9907007 | Since MATH has no exceptional orbits, MATH is a reduced root system and the root of this system corresponding to MATH is MATH. Given this the result is immediate from the previous lemma. |
math/9907007 | From the long exact cohomology sequence we see that MATH is an injective homomorphism from MATH to MATH. Since MATH is a compact group, the image of MATH is finite, and hence is contained in MATH. |
math/9907007 | The only thing that needs to be established is that if MATH is a root of MATH and MATH then MATH. The root lattice MATH has as a basis the roots MATH. It suffices to show that these take integral values on MATH. Let MATH be the orbit MATH and lift MATH to MATH. Then MATH. |
math/9907007 | Since MATH, we see that MATH if and only if MATH or equivalently MATH . Equivalently, in additive notation, MATH . Since every lift of MATH to an element of MATH is of the form MATH for some MATH, we see that MATH lifts to an element of MATH if and only if MATH, or equivalently MATH. |
math/9907007 | By REF , MATH is the stabilizer in the NAME group of MATH of the point MATH. By REF , the NAME group MATH is the group MATH generated by reflections in walls defined by the roots in MATH which are integral on MATH. By REF (and the remarks immediately following it in case MATH is not reduced), the quotient MATH is isomorphic to the stabilizer of MATH under the action of MATH on MATH, and this is the statement of the proposition. |
math/9907007 | Since MATH is simply connected, MATH is trivial, and hence MATH is trivial. By REF , MATH is also trivial. Thus MATH is trivial, so that MATH is connected. |
math/9907007 | Fix a component of MATH, and find, by REF , an element MATH lying in this component. Let MATH be the image of MATH. Then by REF and MATH. By REF , there is a MATH whose image in MATH is also MATH. Thus MATH for some MATH. Let MATH be a lift of MATH to the covering MATH of MATH and let MATH be a lift of MATH to MATH. Then the element MATH is a lift of MATH to MATH. The calculation of REF shows that MATH and identifies MATH (additively) with MATH, up to an element in MATH. It then follows from the definition of MATH that MATH is the image of MATH in MATH. Thus, letting MATH denote the orthogonal projection onto MATH, by REF , the image MATH of MATH in MATH is represented by MATH modulo MATH. Of course, MATH. Thus MATH lifts MATH. In other words, MATH. Let MATH; it is an alcove in MATH for MATH. Now there is a unique MATH such that MATH and MATH. Let MATH be the NAME part of MATH. It is a product of reflections in walls defined by roots in MATH which are integral on MATH. By REF , such walls are walls of roots of MATH. Hence MATH. Let MATH be the affine transformation MATH, where MATH mod the coroot lattice of MATH. Then by construction MATH normalizes the alcove MATH, and so by REF. Since MATH, we see that MATH and that MATH. This completes the proof. |
math/9907007 | Let MATH be the homomorphism MATH. By REF , the image of MATH is exactly the stabilizer of MATH in MATH. Since the image of MATH contains the kernel of the map from MATH to MATH, it follows that the image of MATH is the stabilizer of MATH in MATH. The second statement follows immediately from the fact that every affine automorphism of a simplex has a fixed point. |
math/9907007 | We have an exact sequence MATH . Since MATH, we see that there is an exact sequence MATH . Conjugation by MATH induces an action on this sequence. On the second term it is the action induced by the permutation action of MATH with quotient MATH. Hence, taking coinvariants yields an exact sequence MATH . Clearly, then, the torsion subgroup of MATH is as claimed. Since MATH for all MATH, we see that MATH. The two expressions for MATH are equal since, in MATH,we have the relation MATH. |
math/9907007 | Fix a MATH-pair MATH in normal form. Suppose that MATH exponentiates to a regular element MATH. Then MATH, MATH, and MATH. According to REF , since MATH for every MATH, MATH is a cyclic group of order MATH. Applying this to MATH instead of MATH, we see that MATH, which is the group of components of MATH, is a cyclic group of order MATH. Furthermore, the natural map MATH induces an isomorphism on the torsion subgroups of the MATH-coinvariants. It follows from REF that the inclusion MATH induces a bijection on components, and in particular that MATH has MATH components. |
math/9907007 | The groups MATH and MATH are connected and both have MATH as a maximal torus. Since MATH, MATH, and hence MATH. The natural surjection MATH is the map on fundamental groups induced by the inclusion MATH. It follows immediately from the description of the torsion subgroup and its generator given in REF that, if MATH is cyclic of order MATH and MATH is cyclic of order MATH, then MATH and that the map MATH induced by MATH on coinvariants sends a generator of MATH to MATH times a generator of MATH. Thus it is injective on the torsion subgroup. |
math/9907007 | This is immediate from REF . |
math/9907007 | The fact that MATH is of rank zero means that MATH is finite. If MATH is not a torus, then, by CITE II REF, MATH is of positive dimension. Thus MATH is a torus, and since MATH is a maximal torus of MATH they are equal. If MATH, then MATH which by REF is a torus. Thus, the action of MATH on MATH is trivial. By REF this implies that MATH is a point, which means that MATH is a MATH-pair of rank zero. The statements about MATH and MATH now follow from REF . By REF , MATH is an extension of a cyclic group by a torus. Thus MATH is conjugate to MATH only if MATH is in the same component as MATH. The converse follows easily from the fact that MATH acts on MATH with isolated fixed points. |
math/9907007 | Suppose that MATH acts on MATH interchanging two roots MATH which are not orthogonal. The extended NAME diagram MATH is thus symmetric about the bond connecting the nodes corresponding to MATH and MATH. This symmetry implies that there are three possible types of diagrams to consider: CASE: MATH is simply laced and has no trivalent vertices; CASE: MATH is simply laced and has two trivalent vertices; and REF MATH has two double bonds. In the first case MATH is isomorphic to MATH for some MATH. This case is ruled out since the center of MATH acts freely on the extended NAME diagram. In the second case, the subdiagram of MATH which contains the chain connecting the trivalent vertices together with all nodes adjacent to the trivalent vertices is the extended NAME diagram for MATH, and hence there is a non-trivial linear relation between the elements of MATH corresponding to the nodes of this subdiagram. Since any proper subset of MATH is linearly independent, it follows that this is the entire extended diagram for MATH. Thus, MATH is of type MATH for some MATH. Direct examination of the action of the center in this case shows that MATH is a generator of the center and that the integers MATH are all equal to MATH. In the third case, MATH has no trivalent vertices and hence is a chain. Since the highest root is a long root and since according to REF it must be at one end of the chain, it follows that the MATH and MATH are short roots. The subchain that contains MATH and MATH and contains one long root on each side of MATH is the extended diagram for MATH for some MATH. As before, since there is a nontrivial linear relation between these roots, it follows that MATH is of type MATH. Direct inspection shows that the MATH are all equal to MATH in this case. What we have seen is that if MATH acts on MATH with an exceptional orbit, then the integers MATH are all equal and the number of orbits is at least MATH. Now suppose that MATH is a MATH-triple of rank zero in MATH. Then we know that MATH. Since all the MATH are equal, by REF MATH. Consequently, the action of any MATH on MATH is trivial. That is to say the rank of MATH is equal to that of MATH. But the rank of MATH is one less than the number of orbits and hence is positive. This contradicts the fact that MATH is of rank zero. |
math/9907007 | Since MATH is of rank zero, it follows from REF that MATH and that conjugation by MATH normalizes MATH and its action on this torus has isolated fixed points. Since MATH is abelian, its NAME group is trivial. Thus, by REF there is a well-defined action of the element MATH on MATH, and, by REF , it is given as the NAME element MATH. The element MATH preserves MATH and fixes only the origin there. By REF , MATH is an element of the NAME group of MATH. It follows that MATH has finite center and thus is semisimple. This proves REF . Since MATH, the affine action of MATH on the alcove MATH has isolated fixed points, and hence fixes only the barycenter of MATH. Since MATH fixes MATH, by REF , it follows that MATH is the barycenter of MATH, proving REF . Moreover, by REF , no coefficient MATH of MATH is integral, proving REF . Any two sets of simple roots of MATH are related by an element of the NAME group of MATH. Since the NAME group acts trivially on the center, and since MATH projects to an element of the center, REF follows. Since MATH is semisimple, MATH has cardinality equal to the dimension of MATH. Thus MATH is the image under the exponential map of a vertex MATH of MATH, opposite the face MATH, say. By REF , MATH is cyclic of order MATH. Since the class of MATH is an element of order MATH in MATH, and since MATH is injective by REF , MATH is an element of order MATH in MATH. Consequently, MATH. |
math/9907007 | By REF , a generator for MATH is MATH, where MATH. Thus MATH and we can write MATH for some integer MATH of the form MATH, where MATH and MATH are relatively prime. A representative for MATH is then given by MATH . Clearly, then, no coefficient of MATH is integral if and only if, for all MATH, MATH. |
math/9907007 | This is immediate from the previous lemma and REF of the previous proposition. |
math/9907007 | The only if part of the first statement is REF . Conversely, suppose that exactly one of the MATH is divisible by MATH. Choose MATH so that, for MATH, we have MATH. By REF , the order of MATH is divisible by MATH and hence is divisible by MATH. Let MATH be such that MATH is a special MATH-pair in normal form. Let MATH be the barycenter of the alcove MATH for the root system MATH and let MATH. We set MATH. Fix an element MATH of order MATH. Since MATH is the barycenter of MATH, MATH fixes MATH. Hence by REF , there is a MATH whose image in MATH under the map MATH is MATH. Since MATH is a regular element for MATH, it follows that MATH. Lifting the image of MATH in MATH to an element MATH, MATH. It follows by construction that no coefficient MATH of a MATH is integral. Hence, by REF , MATH acts with isolated fixed points on MATH. It follows that the conjugation action of MATH on MATH also has isolated fixed points, and hence MATH is of rank zero. Clearly, it is of order MATH. Now suppose that MATH does not divide MATH for MATH. Let MATH be a MATH-triple of rank zero and order MATH and let MATH be a rank zero MATH-triple whose order MATH divides MATH. After conjugation we can assume that MATH and MATH are MATH-pairs in normal form. Then MATH. Each of MATH and MATH is the image under the exponential mapping of the vertex of MATH opposite the face MATH for the unique MATH for which MATH. Thus, MATH. By REF there is MATH such that MATH is a special MATH-pair in normal form. We write MATH and MATH for elements MATH which are the images of MATH and MATH in MATH. It follows from REF that MATH and MATH are barycenters for the alcove decomposition of MATH associated with the root system MATH. Thus, there is an element in the NAME group of MATH on MATH which conjugates MATH to MATH. By REF , the NAME group of MATH is the NAME group of MATH. Thus, there exists a MATH conjugating MATH to MATH, and hence MATH and MATH are conjugate MATH-pairs in MATH. This allows us to assume further that MATH. By REF , MATH. By REF , the group MATH is cyclic of order dividing MATH, and MATH is an element of this group of order exactly MATH. Hence MATH generates MATH. Thus there is a unique integer MATH with MATH such that MATH and MATH are in the same component of MATH. Since MATH acts on MATH with isolated fixed points, this implies that MATH and MATH are conjugate in MATH and hence that MATH and MATH are conjugate in MATH. |
math/9907007 | Since MATH is a maximal torus for MATH, there is an inner automorphism MATH, MATH, such that MATH normalizes MATH. Since MATH is connected, the inner automorphism MATH of MATH induces the identity on its fundamental group. Thus, without loss of generality, we can assume that MATH normalizes MATH. Let MATH be the lift of MATH lying in the alcove MATH. Then MATH for some element MATH lifting MATH. Since MATH, the affine automorphism MATH of MATH normalizes the alcove decomposition associated with the root system of MATH. Since MATH, MATH is an alcove containing MATH. Hence there is an element MATH of the affine NAME group of MATH which fixes MATH and sends MATH to MATH. The composition MATH then normalizes MATH. Since the translational part of this affine transformation is congruent to MATH modulo the coroot lattice MATH, we see by REF that this composition is MATH, which we denote by MATH. On the other hand, since MATH fixes MATH, the NAME part MATH of MATH is an element of the NAME group of MATH. Since MATH, multiplying MATH by an element of MATH allows us to assume, without loss of generality that MATH. As noted in REF, we have MATH for all MATH. Since MATH and MATH commute, we also have MATH for all MATH. Now we see directly from the expression for a generator of MATH given in REF , that MATH acts trivially on MATH. This completes the proof of the first part of the lemma. Since MATH contains a generic element of MATH, MATH normalizes MATH. The equation MATH says that MATH commutes with MATH. Let MATH be the lift of MATH. There is a NAME element MATH commuting with MATH such that MATH and such that MATH. Since MATH, it follows that MATH for all MATH. Since MATH commutes with MATH, it is also the case that MATH for all MATH. Thus, MATH acts trivially on MATH. The first part of the lemma implies that MATH acts trivially on this group. Thus, it follows that MATH acts trivially on MATH. The final statement follows since the inclusion MATH of REF is equivariant with respect to the automorphism MATH. |
math/9907007 | We can assume that MATH is a MATH-pair in normal form. Since MATH, the triple MATH is also a MATH-triple, clearly of rank zero. By REF , composing MATH with an inner automorphism allows us to assume that MATH for some integer MATH. Since MATH is of rank zero, by REF , MATH. Hence, MATH. Applying REF we see that MATH acts trivially on MATH. Hence, by REF , it follows that MATH and MATH are in the same component of MATH, and hence that MATH is conjugate to MATH. |
math/9907007 | Note that MATH. It follows that the order of MATH is equal to that of MATH, and it clearly has rank zero. We can assume that MATH is in normal form with respect to MATH. By REF there is MATH which conjugates MATH to MATH. Since MATH and MATH, MATH induces an automorphism of MATH. By REF , the induced action of MATH on MATH is trivial. In particular, MATH and MATH lie in the same connected component of MATH. Hence they are conjugate in MATH. It follows that MATH and MATH are conjugate in MATH. |
math/9907007 | By symmetry between MATH and MATH (at the expense of replacing MATH by MATH), the result in this case follows from the previous one. |
math/9907007 | Let MATH be a MATH-triple of rank zero such that MATH is a MATH-pair in normal form. Then MATH. For any MATH, MATH and MATH are in the same component of MATH if and only if MATH. This proves that MATH. Lastly, we show that the order of MATH is equal to MATH. By REF , MATH is the number of components of MATH. Since MATH and since MATH is the component of the identity of MATH, this shows that the order of MATH divides MATH. To complete the proof, we need to show that the inclusion MATH is onto on the level of components. We state this as a separate lemma: Let MATH be a simple group and let MATH. The inclusion MATH induces a surjection MATH. Proof. Applying cohomology to the MATH-actions on the exact sequence MATH and considering the torsion subgroups, we have an exact sequence MATH . Thus, it suffices to show that MATH is injective. Set MATH. There is the exact sequence MATH where MATH is the root lattice of MATH. Dualizing gives MATH where the second map is obtained by sending MATH to MATH. (Here the MATH are the dual basis to the basis MATH of the free MATH-module MATH.) Hence, the group MATH is a cyclic group of order MATH. Clearly, if MATH contains the extended root, MATH is the order of MATH and MATH so that MATH is the order of MATH. Thus the order of MATH divides the order of MATH. On the other hand, a generator for MATH is represented by MATH for any MATH for which MATH. Choose MATH to be the root mapped by MATH to MATH. Then the corresponding element of MATH is represented by MATH, and its image in MATH is therefore equal to MATH. Since the order of MATH is at most that of MATH, the map MATH is injective, and in fact an isomorphism. This proves the lemma. |
math/9907007 | Since MATH is generic, it follows that MATH. On the other hand, for every MATH in the interior of MATH, the triple MATH is a MATH-triple with the same centralizer as the triple MATH. This means that MATH is contained in MATH. Comparing this with the dimension estimate shows that MATH is the linear subspace parallel to MATH. |
math/9907007 | After conjugation, we may assume that MATH is a MATH-pair in normal form. Since MATH is of order MATH, the order of MATH is divisible by MATH, and hence MATH has order divisible by MATH. By REF , this means MATH is divisible by MATH, and in particular MATH for some MATH. Moreover MATH. Thus, MATH is the image under the exponential mapping of a point MATH contained in MATH. Take MATH to be a generic element with the property that MATH for some MATH. Let MATH be the face of MATH containing MATH in its interior. By the first part of this lemma, MATH is a face of MATH. According to the previous lemma MATH is parallel to MATH. |
math/9907007 | The ``only if" direction follows from REF . Conversely, suppose that MATH divides at least one of the MATH. Choose MATH to be any element in the face MATH. Then MATH divides the order of MATH. It then follows by REF that there is a choice of MATH such that MATH is a MATH-pair and such that MATH divides the order of MATH. Choose a MATH mapping to an element of order MATH in MATH. Then MATH is a MATH-triple in MATH of order MATH. |
math/9907007 | Write MATH. Let MATH be a maximal torus for MATH. By REF , possibly after conjugating MATH, we can assume that MATH is the torus whose NAME algebra is parallel to MATH and that MATH is the exponential of an element of MATH. Thus MATH for some MATH. There are MATH such that MATH and MATH, where MATH. After a further conjugation in MATH, we can assume that MATH is a MATH-pair in normal form for MATH. Note that MATH has rank zero in MATH. It then follows from REF , applied to the simple factors of MATH, that MATH is as described in the statement of the proposition. |
math/9907007 | As in the case of commuting triples, since the NAME diagram of MATH is a proper subdiagram of the NAME diagram of MATH, MATH can have at most one simple factor which is not of MATH-type. This proves the last statement. To prove the equivalences of the proposition, first note that, by REF , MATH if and only if MATH divides MATH for every MATH. Thus REF is equivalent to REF . If MATH, then by REF. Thus REF implies REF . Conversely, if MATH, then MATH is the torus associated to the intersections of the kernels of the roots of MATH, and thus MATH. Since, by REF , all simple factors of MATH are of type MATH, REF implies REF . Finally, we show that REF implies REF . Let MATH be a MATH-triple of rank zero in MATH such that MATH, where the MATH. Write MATH, where the MATH lie in the simple factors MATH of MATH, and similarly for MATH. Since MATH is of type MATH for some MATH, MATH is a rank zero MATH-pair in MATH. It follows that MATH contains the rank zero MATH-pair MATH. Since MATH, it follows by REF that MATH. Since MATH contains a rank zero MATH-pair, it follows that MATH. Thus MATH. |
math/9907007 | Since MATH, we have MATH. Moreover, MATH is a product of groups of MATH-type, and MATH projects to a generator of the center of each factor. The result now follows from REF . |
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