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math/0107157 | Suppose MATH is a nest satisfying the above conditions. Then we can choose a sequence of clopen subsets MATH and MATH such that MATH and MATH. For each MATH, applying the condition to MATH and MATH, we have clopen sets= MATH and MATH with MATH and MATH defined accordingly. Let MATH be the groupoid defined by the nest MATH. Since MATH and MATH, it suffices to prove that each MATH is AF. Since the nest MATH satisfies the conditions in REF , the result follows. |
math/0107157 | Clearly, REF follows from REF. Suppose REF holds. Let MATH and MATH such that MATH. Then MATH exists and the limit is independent of the choice of MATH. Therefore, we can extend MATH to a continuous map MATH on MATH. It follows from b that MATH is a homeomorphism on MATH. |
math/0107160 | We use the local description explained after REF . We write MATH REF and MATH (MATH). Then the map MATH is described by the following formulas: MATH for MATH and MATH for MATH. We note that the actions of MATH and MATH are rivial on the MATH and MATH while those on the MATH and MATH are translations. Since MATH for any MATH, the fibers of MATH are homeomorphic to MATH . The restriction of MATH on MATH for sufficiently small MATH is homeomorphic to the first projection of MATH to MATH. By gluing together, we obtain our result. |
math/0107160 | Since the boundary MATH has no self-intersection, MATH is normal, hence MATH is toroidal and quasi-smooth. MATH is smooth and the normal bundle MATH is the direct product of line bundles. Since MATH is homeomorphic to the complement of a MATH neighborhood of MATH in MATH, we obtain our assertion. |
math/0107160 | We assume first that MATH. Then we have MATH. By REF , MATH induces a MATH-bundle MATH, and we have a map MATH . Since the fiber of MATH is reduced, we may take the homeomorphism to the MATH neighborhood as in REF in a suitable way to conclude that MATH is bijective and a homeomorphism. In the general case, if we restrict MATH to MATH, we obtain a homeomorphism MATH . By applying REF again to MATH, we conclude the proof. |
math/0107160 | We consider the restriction of the locally constant sheaf MATH on MATH. Since MATH, the weight filtration induces a spectral sequence MATH . By REF , MATH is a constant sheaf for any MATH. Hence we have our result. |
math/0107160 | CASE: This is a special case of REF. For example, we check the case where MATH. Let MATH for MATH. If MATH then we have MATH for all MATH. Since MATH for MATH, we conclude that MATH by the descending induction on MATH. On the other hand, we can solve the equations MATH by the descending induction on MATH to find MATH such that MATH. CASE: We check the case where MATH and MATH, that is, MATH and MATH. The case MATH follows from the projection formula, and the general case is similarly proved. The first equality follows from the fact that MATH is multi-valued on MATH near MATH. We have an exact sequence MATH where MATH is a locally constant sheaf on MATH such that the stalk MATH at MATH has a MATH-basis MATH corresponding to the MATH with the monodromy action given by MATH . Then MATH and MATH. Hence MATH and MATH. REF is a consequence of REF . This follows also from REF, because we have MATH for the inclusion MATH by REF . |
math/0107160 | CASE: We check the case where MATH, MATH, and MATH. The general case is similar. Let MATH for MATH. If MATH in MATH, then we have MATH for any MATH. Since MATH for MATH or MATH, we prove by the descending induction on MATH that there exist MATH for MATH such that MATH, and if we expand MATH in a power series MATH, then we have MATH unless MATH. Therefore, MATH. On the other hand, we can solve the equations MATH by the descending induction on MATH to find MATH such that MATH in MATH. CASE: We check the case where MATH, MATH, MATH and MATH, that is MATH and MATH. The general case is similar. We have MATH and MATH. The map MATH is given by MATH. Let MATH be a local section of MATH, where MATH. The monodromies of the multi-valued functions MATH and MATH along the fibers of MATH which are homeomorphic to MATH's are given by MATH . The function MATH is single valued along these MATH's if and only if there exist MATH for MATH such that MATH. Therefore we have the first equality. We have an exact sequence MATH where MATH and MATH are the locally constant sheaves on MATH and MATH such that the stalks MATH and MATH at MATH and MATH have MATH-bases MATH and MATH whose the monodromies MATH around the first and second factors and MATH are given by MATH and the homomorphism MATH is given by MATH . In other words, we have MATH . We have MATH hence MATH . Since MATH we have MATH . Therefore, we have the desired result. REF is obtained by combining REF . |
math/0107160 | Since the truncation is a canonical functor, this is a consequence of REF and CITEEF. |
math/0107160 | The first formula is CITEEF. REF. The second is similar. |
math/0107160 | We have to prove that MATH is a NAME structure of weight MATH: MATH . By REF , we can write MATH for some set of the strata MATH. Thus the problem is reduced to prove MATH . But this is the usual NAME theorem on MATH. |
math/0107161 | It follows from the equality MATH where MATH is the inclusion map. |
math/0107161 | Given a subsheaf MATH of MATH such that MATH, let us consider the complementary subcurve MATH of MATH in MATH, that is, the closure of MATH. Since MATH is torsion free, we have MATH with MATH so that MATH and the result follows. |
math/0107161 | See CITE, NAME Partie, REF . |
math/0107161 | NAME proves this lemma in REF when MATH is a tree-like curve, that is, its irreducible components are smooth, but her proof is valid for our curve. |
math/0107161 | Let us consider the subsheaves MATH of MATH, MATH, MATH being the subcurves of MATH given by REF . By the stability of MATH, we get MATH . Considering the subsheaves MATH of MATH, MATH, yields MATH . Since MATH and MATH, MATH meet only at MATH, we have MATH, MATH and MATH. Then, REF give MATH . We have MATH so that MATH is an integer. Then, if MATH for some MATH, REF becomes a contradiction. Thus MATH for all MATH and there is only one possibility for MATH, namely MATH . From MATH and the exact sequence MATH we deduce that MATH for all MATH and the proof is complete. |
math/0107161 | CASE: If MATH is a connected subcurve of MATH, let us denote by MATH, MATH, (respectively, MATH) the component of MATH (respectively, MATH) which contains the point MATH (it is possible to have MATH for MATH). Suppose first that MATH for MATH. Then, the subcurve MATH of REF is the connected component of MATH which contains MATH. Let us see that MATH. Since the inclusion MATH is clear we have only to prove that MATH contains no component of MATH and that MATH if MATH . If MATH for some MATH, then MATH and MATH. Hence, MATH where we denote by MATH (respectively, MATH) the connected component of MATH that contains MATH (respectively, MATH). Analogously, MATH, MATH are contained in MATH and MATH where MATH (respectively, MATH) is the connected component of MATH containing MATH (respectively, MATH). We deduce that MATH, MATH and MATH. Thus, MATH and MATH, which is absurd. Therefore, none of the components MATH of MATH is in MATH and, since MATH are connected, no other component of MATH is either. Moreover, a similar argument shows that MATH and MATH have no common components for MATH . Let now MATH be the number of irreducible components in MATH. Then, MATH is equal to MATH. On the other hand, the number of intersection points in MATH is MATH, where MATH is the number of intersection points of MATH, MATH. Since MATH is the number MATH of intersection points in MATH, we get MATH and MATH. Then, MATH, MATH and MATH . By definition of MATH, for every MATH, we have MATH . Taking into account that MATH, MATH and MATH, we obtain the result in this case. Assume now that MATH. Then, MATH is the connected component of MATH which contains MATH. Since MATH is connected, MATH contains MATH and then MATH contains MATH and MATH for MATH. This implies that MATH for MATH. If we had MATH for some MATH, then MATH would be contained in MATH as before and MATH, which is absurd. Thus, MATH for MATH. In this situation, one has that MATH, where MATH and MATH intersects MATH in MATH points. Thus, we have relations REF for MATH and, arguing as in the former case, we obtain MATH . By definition of MATH, one has MATH which together with REF proves the statement. If the subcurve MATH is not connected, the inequality holds for every connected component and then it is easy to deduce it for MATH. CASE: By REF , it is enough to show that MATH for every proper subcurve MATH of MATH. Since MATH the result follows from REF. |
math/0107161 | By REF , MATH is not empty only if MATH is not integer for every MATH, and in this case it is equal to MATH. We prove now the remaining statements of the theorem. If MATH is a strictly semistable line bundle on MATH of degree MATH then, MATH is equal to one of the two extremal values of REF . In particular, MATH is an integer for some MATH. Let MATH be the first index such that MATH is integer. Then, there are two possibilities for MATH: CASE: MATH REF MATH . Let us construct a NAME filtration for MATH in both cases. Since REF are the same but with the roles of MATH and MATH intertwined, we give the construction in REF . We have that MATH. Then, MATH and MATH are semistable with respect to MATH and, by REF , they are semistable with respect to MATH and MATH respectively. For simplicity, we shall write MATH with MATH and MATH, which are again generalized tree-like curves. Let us see when the sheaves MATH and MATH are stable. We can fix an ordering for MATH as in REF , so that MATH and we obtain subcurves MATH of MATH for MATH. CASE: The sheaf MATH is stable if and only if MATH is not an integer for MATH. Proof. Since the residue class of MATH modulo MATH is MATH, the numbers MATH are not integers for MATH. Then, from REF , we have only to prove that MATH is in MATH, where MATH are the integer numbers defined as MATH but with the new ordering of MATH and MATH runs through the irreducible components of MATH. This is equivalent to proving that MATH . Actually, since MATH is semistable and MATH is a proper subcurve of MATH, arguing as in REF , we obtain MATH where MATH is the number of intersection points of MATH and its complement in MATH. We have that MATH and MATH is not equal to the extremal values of REF because MATH. Moreover, if it were MATH since MATH, MATH and MATH, MATH being the complement of MATH in MATH, then MATH which contradicts the semistability of MATH. Thus, MATH and the proof of REF is complete. On the other hand, the irreducible components of MATH are ordered according the instructions in REF and the subcurves MATH, where MATH runs through the irreducible components of MATH and MATH, are equal to either MATH or MATH. CASE: The sheaf MATH is stable if and only if MATH is not an integer for every MATH. Proof. Since the residue class of MATH modulo MATH is MATH and MATH , by the hypothesis, the numbers MATH are not integers for MATH and, by the choice of MATH, they aren't for MATH either. Then, by REF , it is enough to prove that MATH . Since MATH is semistable, we have that MATH . Moreover, if MATH then, MATH and if MATH then, MATH. We obtain the desired result in both cases and the proof of REF is complete. We return now to the proof of the theorem. If MATH and MATH are not integers for MATH and MATH, then MATH is a NAME filtration for MATH and the MATH-equivalence class of MATH belongs to MATH. On the other hand, if MATH is integer for some MATH, the sheaf MATH is strictly semistable and we have to repeat the above procedure with MATH in the place of MATH and the curve MATH in the place of MATH. Similarly, if MATH is integer for some MATH, the sheaf MATH is strictly semistable. Then, we have to repeat the above procedure for MATH. By iterating this procedure, we get a NAME filtration for MATH: MATH and another for MATH: MATH . Therefore, a filtration for MATH is given by MATH . Thus, the MATH-equivalence class of MATH belongs to MATH, where MATH are the integer numbers constructed with the algorithm. Finally, let us consider a torsion free sheaf MATH on MATH of rank REF and degree MATH which is not locally free. When MATH is locally free at the intersection points MATH for all MATH, calculations and results are analogous to the former ones. If MATH is not locally free at MATH for some MATH, then there is a natural morphism MATH where MATH, MATH are the connected components of MATH, that is clearly an isomorphism outside MATH. But this is an isomorphism at MATH as well because by REF , MATH and this is precisely the stalk of MATH at MATH. We conclude that if MATH is strictly semistable, then MATH y MATH are integers, MATH and MATH are given by MATH and MATH and MATH are semistable as well. Then, the construction of a NAME filtration for MATH can be done as above and thus the MATH-equivalence class of MATH belongs to MATH. |
math/0107166 | We verify only some of the important identities. The other identities are similar to check. CASE: MATH . CASE: MATH . CASE: MATH . CASE: MATH . CASE: MATH . MATH . CASE: MATH . |
math/0107166 | We show that MATH commutes with the cosimplicial and cyclic operators: MATH CASE: MATH . CASE: MATH . CASE: MATH . MATH CASE: MATH . MATH . |
math/0107166 | We define a map MATH, by MATH . One can check that MATH is a cocyclic map and MATH. For example MATH . |
math/0107166 | We prove this for MATH the general case being similar. The operators are well defined on the coinvariant subspaces. For example, for MATH, since MATH and MATH if MATH then MATH . To show for example MATH we see that in the coinvariant subspace we have, MATH . So we have, MATH . Therefore, MATH . Therefore, in the coinvariant subspace we have MATH and then we conclude that MATH . Also when MATH, then MATH . So MATH is the coinvariant subspace. |
math/0107166 | Since MATH is cosemisimple, we have MATH for MATH and the spectral sequence collapses. The first column of MATH is exactly MATH. |
math/0107167 | We have MATH by REF . We need to check that MATH for all MATH, where MATH. To this end, we compute, using REF : MATH . Since MATH is non-degenerate, the identity in question is equivalent to MATH which reduces to REF by REF . The proof of the statement regarding MATH is completely similar. |
math/0107167 | Clearly, MATH is unimodular, since twisting preserves integrals in MATH. NAME of MATH is equivalent to MATH by NAME 's theorem. By REF , we have MATH where we used REF . |
math/0107167 | Let MATH and MATH. We will show that MATH is a coseparability pairing for MATH. To this end we need to check that MATH satisfies the conditions of REF . First, we check that MATH for all MATH : MATH by REF . Next, we compute, for all MATH : MATH and also MATH where we used REF . Comparing the results of the above computations and using the non-degeneracy of MATH, we conclude that the following equations MATH imply the identity MATH . From REF we see that REF are equivalent to MATH . These, in turn, are equivalent to REF , therefore the proof is complete. |
math/0107167 | The proof is analogous to CITE, where we refer the reader for details. Note that MATH and MATH are full squares, because of the minimality of the twists MATH and MATH, so the right hand side of REF is an integer. |
math/0107167 | This follows from REF . |
math/0107167 | Write MATH. The map MATH is left MATH-linear, where MATH acts on MATH by the NAME action, since we have MATH . It follows from CITE that MATH is cleft right MATH-Galois with coinvariants MATH (by counting dimension), and MATH is an isomorphism of right MATH-comodule algebras. From REF we see that the bijective map between MATH and MATH mentioned above coincides with MATH. It also follows from the construction of this map that MATH. |
math/0107167 | Form the biproduct MATH where the generator MATH of MATH acts on MATH by parity; it is an ordinary NAME algebra (see CITE). Consider the algebra MATH . On one hand, MATH as an algebra, and on the other hand, MATH as an algebra. Since MATH is cosemisimple, the result follows. |
math/0107167 | It remains to show that the two assignments are inverse to each other. Indeed, this follows from the results of CITE. |
math/0107167 | Follows from REF since for any twist MATH for MATH is cosemisimple triangular by REF . |
math/0107168 | The proof is a straightforward application of the NAME spectral sequence of the map MATH, noting that the fibers are spaces of the form MATH, which have trivial reduced cohomology with MATH - coefficients. |
math/0107168 | We know that MATH is a finite, almost free MATH - CW complex. It follows from CITE that there is a spectral sequence converging to MATH, with MATH - term equal to zero if MATH is odd and equal to MATH where MATH denotes the collection of MATH - dimensional cells in MATH and MATH denotes the complex representation ring of the stabilizer of MATH in MATH. In fact the MATH term is simply the homology of a chain complex assembled from these terms. By our hypotheses, each MATH is finite, and there are finitely many such cells, hence each term is finitely generated as an abelian group and there are only finitely many of them. We conclude that MATH satisfies the required finiteness conditions, and so must its subquotient MATH, whence the same holds for MATH. |
math/0107168 | As has been remarked, we can assume that MATH is a finite, almost free MATH - CW complex. Now, as in CITE and CITE, the main idea of the proof is to construct a natural NAME character for any MATH - space as above, and then prove that it induces an isomorphism for orbits of the form MATH, where MATH is finite. Using induction on the number of orbit types and a NAME - NAME sequence will complete the proof. To begin we recall the existence (see REF ) of a ring homomorphism MATH in this much more elementary setting it can be defined by its value on vector bundles, namely MATH for any MATH - vector bundle MATH. We make use of the natural maps MATH as well as the NAME map MATH . Note that the isomorphism above is due to the crucial fact that all the fibers of the projection map MATH are rationally acyclic, as they are classifying spaces of finite groups. Finally we make use of the ring map MATH, with kernel the ideal of elements whose characters vanish on all generators of MATH. Putting all of this together, and using the restriction map, we obtain a natural ring homomorphism MATH . Here we have taken invariants on the right hand side, as the image naturally lands there. Verification of the isomorphism on MATH is an elementary consequence of the isomorphism MATH and details are left to the reader. |
math/0107168 | We begin by considering the situation locally. Suppose that we have a chart in MATH of the form MATH, mapping onto MATH in MATH, where we assume MATH is a finite group. Then MATH . Let us now define a MATH action on MATH by MATH. We define a map MATH by MATH. We check that this is well - defined; indeed if MATH then there exists a MATH with MATH, so MATH. This means that MATH and so MATH as MATH. Now suppose that MATH; then MATH; hence we have a well-defined map on the orbit space MATH . This map turns out to be injective, indeed if MATH for some MATH, then MATH and MATH, hence MATH and MATH. The image of MATH consists of the MATH - equivalence classes of pairs MATH where MATH and MATH is conjugate to MATH in MATH. Putting this together and noting that MATH unless MATH is conjugate to an element in MATH, we observe that we obtain a homeomorphism MATH . To complete the proof of the theorem it suffices to observe that by the compatibility of charts, the local homeomorphisms on fixed-point sets can be assembled to yield the desired global homeomorphism on MATH. |
math/0107168 | Indeed we recall that in CITE, the orbifold cohomology MATH is defined to be additively isomorphic to MATH. Hence via the NAME character map we get the isomorphism above. |
math/0107168 | This result is the analogue of the untwisted version (see CITE, page REF). The natural map MATH can be combined with the map MATH (which gives any vector bundle the trivial MATH - action) to yield a natural isomorphism MATH which covers the restriction to the MATH - action; the result follows from looking at inverse images of the subgroup generated by the scalar representation. |
math/0107168 | Fix the class MATH; for any subgroup MATH, we can associate MATH. Note the special case when MATH, a cyclic subgroup. As MATH, MATH is isomorphic to MATH. Now consider MATH, a MATH - twisted bundle over MATH; it restricts to a MATH - twisted bundle over MATH. Recall that we have an isomorphism MATH. Let MATH denote the MATH - map MATH described previously, where the centralizer acts on the projective representations as described above. Then the composition MATH has image lying in the invariants under the MATH - action. Hence we can put these together to yield a map MATH . One checks that this induces an isomorphism on orbits MATH; the desired isomorphism follows from using induction on the number of MATH - cells in MATH and a NAME argument (as in CITE). |
math/0107172 | We may choose MATH sufficiently small so that MATH action is conjugate to a linear action. For example, we can choose the normal neighborhood. This is REF. |
math/0107172 | We will assume that MATH is compact. If MATH is only locally compact, the proof is similar. First cover MATH by a finite collection of normal neighborhoods. For a point MATH of a model pair MATH of a normal neighborhood, there exists a radius MATH, such that the ball MATH of radius MATH in MATH has the property that for each pair of points MATH and MATH of MATH, there exists a unique geodesic segment connecting MATH and MATH. Thus, using the NAME number, we can find a real number MATH so that for each point MATH of MATH, the ball MATH of radius MATH, MATH in MATH has the generic convexity property; that is, for each generic pair of points MATH and MATH of MATH, there exists a unique geodesic connecting MATH and MATH. Given two balls MATH and MATH, their intersection MATH, which is open, can have only one component. By an induction, we can show that any finite intersection of balls MATH is connected. The collection consisting of MATH, MATH, covering MATH is a nice covering. |
math/0107172 | For the model pairs, the groups have to be isomorphic. The rest is straightforward. |
math/0107172 | Again a local consideration proves this in a straightforward manner. |
math/0107172 | The first one has the same proof as the ordinary topological theory since liftings are determined up to the action of local groups and we only need to match them. We use the open and closeness for the second one. |
math/0107172 | For each model pair MATH of MATH, we define MATH to be MATH. |
math/0107172 | First, we consider the case when MATH has no mirror points. Then we claim that MATH is connected: In the model pair of a singular point of MATH, if there are no element fixing a hypersurface, then the set of regular points in the model open set MATH is connected since the actions are conjugate to linear actions on sufficiently small open sets. Thus, each point of MATH can be a boundary point of only one component of MATH. Therefore, MATH can have only one component. (By same reason, MATH is connected since the identity map is an orbifold-covering map.) Now, MATH is a topological covering automorphism. Thus, MATH is one-to-one and onto. By REF , MATH is a covering isomorphism. If MATH has mirror-points, then we form the two-fold orbifold-covering map MATH. Then an orbifold-map MATH covering MATH is also a covering morphism of MATH from MATH to itself. Since MATH has no mirror-points, MATH is a covering isomorphism. Therefore, so is MATH obviously. |
math/0107172 | Let MATH be a model neighborhood of MATH, and MATH the model pair, and MATH an orbifold-covering map. For a path MATH in MATH with the base point MATH, we can lift MATH to a path in MATH easily by using the elementary neighborhoods. Two homotopic paths MATH and MATH lift to homotopic path-classes again using elementary neighborhoods. By taking a base point in MATH and path-classes from the base point to all points of MATH, we can lift MATH to a map MATH so that MATH. Since any path in MATH can be lifted, MATH is obviously a surjective orbifold-map. (This works in the same manner as in the covering space theory.) From here, it is straightforward to verify that MATH is of form MATH quotient out by a finite subgroup MATH of MATH. That is, we show that the inverse image of every point of MATH is an orbit of MATH by an open and closedness argument. Thus, MATH is elementary. |
math/0107172 | CASE: The first part follows from REF . The second part is a consequence of REF . CASE: The morphism lifts to an orbit-preserving map MATH, which sends each MATH-orbit to a MATH-orbit. Again MATH is an element MATH of MATH covering the identity map of MATH. Thus MATH for each regular point MATH of MATH. Thus, MATH. The elements MATH and MATH induce a same map MATH if and only if MATH for MATH and MATH. REF follows from REF . |
math/0107172 | Let MATH be obtained by a list of coverings MATH with base point MATH over a base point MATH which lists one-element from each isomorphism class of covering maps of MATH preserving base-points. Let MATH be the base point from the above construction. Given any covering MATH with a base point MATH, since we obviously have our MATH isomorphic to say MATH, MATH, there exists a covering morphism MATH where MATH. Let MATH be a point of MATH different from MATH. We show that there exists a deck transformation sending MATH to MATH: Clearly, MATH with MATH as a base point is in the list of all covering maps of MATH. Thus, there exists a morphism MATH sending MATH to MATH. By REF , MATH is a deck transformation. Now, let MATH be a cover of MATH with a base point MATH mapping to a base point MATH of MATH, perhaps different from above MATH. Let MATH be a point of MATH of MATH (deemed to be our new base point). Find a path MATH from MATH to MATH, which maps to a path MATH from MATH to MATH on MATH, and find a path MATH on MATH from MATH ending at a point MATH lifting MATH. Then MATH. By above construction of fiber-products, there exists a morphism MATH so that MATH for some MATH since MATH and MATH are connected. By precomposing with MATH a deck transformation, we may assume that MATH. Since MATH goes to MATH, we see that MATH maps to MATH. Therefore MATH satisfies the definition of a universal cover. |
math/0107172 | Since MATH is a covering map, MATH is the universal covering of MATH. Since MATH is also a universal covering of MATH, there is a morphism MATH so that MATH by the uniqueness of the isomorphism class of universal covering orbifolds. The uniqueness follows since MATH restricts to MATH a lift of MATH restricted MATH and the ordinary covering space theory. The freedom of choice follows by the transitivity of automorphism group in the fiber of MATH. The final statement follows trivially. |
math/0107172 | CASE: This follows from definition, that is, the condition on base-points. The covering map MATH induces an orbifold-map MATH which is one-to-one over the regular part. MATH restricts to a homeomorphism over the regular part since it is proper and is a local homeomorphism. Therefore, MATH is an orbifold-diffeomorphism by REF . CASE: There is a morphism MATH. Since MATH covers any cover of MATH also, MATH is again a universal cover of MATH. Since MATH is a quotient of its universal cover, REF follows. CASE: If two covering spaces MATH and MATH are isomorphic, then there exists a morphism MATH. MATH lifts to a morphism MATH by REF . Since MATH is a morphism, MATH satisfies MATH, and so MATH covers identity on MATH. Thus, the lift MATH of MATH is a deck transformation of MATH. In order that MATH descends to a map MATH, we need that for each MATH, there exists MATH so that MATH. Thus, MATH. A converse argument shows that a conjugate of MATH is in MATH. Hence, MATH and MATH are conjugate. CASE: This follows from REF . CASE: Given a base point MATH of MATH mapping to a base point MATH of MATH, each point of MATH is of form MATH for MATH. Given MATH for a subgroup MATH of MATH, and two points in the fiber of MATH, we see that there exists a deck transformation MATH sending MATH to MATH for some MATH if and only if MATH normalizes MATH. It is easy to see that all coset representative of MATH has to occur. Thus MATH is normal. CASE: Let MATH and MATH be orbifold-covering maps. Then we have MATH. The map MATH is a universal covering map of MATH since MATH can cover any orbifold-covering MATH using MATH and deck transformation group of MATH. Thus, there exists an isomorphism MATH. If we identify MATH and MATH by MATH and MATH and MATH by MATH, then MATH becomes a morphism. By REF , MATH is an isomorphism. |
math/0107172 | Clearly, the lift MATH of MATH, which is a homotopy between MATH and a map MATH, exists. We give a local description. Let MATH be a point of MATH. There is a model open subset MATH of MATH where MATH acts so that MATH for MATH where MATH is a small open interval containing MATH. MATH lifts to MATH where MATH is a model open subset where a finite group MATH also acts on, and there is a homomorphism MATH so that MATH. We also have MATH for MATH and MATH. Let MATH be the neighborhood of MATH in MATH corresponding to MATH. Take a component MATH of the inverse image of MATH in MATH. Then MATH maps it into a component MATH of the inverse image of MATH in MATH. Moreover, we may assume without loss of generality that MATH has the same model as above MATH and MATH and an associated homomorphism MATH where MATH is a subgroup of MATH and MATH one of MATH. The map MATH is defined on MATH mapping to MATH and it extends MATH. Clearly, MATH lifts to MATH on MATH with an associated homomorphism MATH restricted to MATH. This means that MATH induces an orbifold-map MATH. Since MATH extends MATH in the neighborhood, by considering every neighborhood, we can show that MATH extends to a homotopy MATH. Obviously, a lift MATH is obtained by restricting MATH. |
math/0107172 | Let MATH be a deck transformation of MATH. We have that MATH is homotopic to MATH, and let MATH be the homotopy between them given by MATH for MATH. Then MATH is a deck transformation for each MATH. (To see this simply post-compose MATH with the covering map of MATH.) Since the group of deck transformations is discrete in MATH-topology, MATH and MATH are equal. |
math/0107172 | This is found in REF (see also CITE and CITE). We rewrite it here for the reader's convenience: Let MATH be a neighborhood of MATH, and MATH the model pair for MATH an open set in MATH and MATH the associated finite group acting on MATH. We form MATH and give an action of MATH by MATH. Then MATH is a manifold and has a projection MATH induced by the projection to the second factor. (Here, MATH is an orbifold-map.) We find a nice locally finite cover of MATH by such neighborhoods MATH. If MATH and MATH meet, then MATH has an inclusion map MATH. Then there is a connected open subset MATH of MATH and a subgroup MATH acting on it being a model for MATH. We form MATH where MATH denotes the fact we used MATH as a model and find a map MATH induced by MATH defined by MATH. The map MATH is a well-defined imbedding since a different choice MATH gives us MATH, MATH, and so MATH is replaced by a map MATH. We find an open subset MATH of MATH corresponding to MATH, and form MATH similarly, and find an imbedding MATH. Since MATH and MATH can be identified by the identification of the model pairs, we see that MATH and MATH can be pasted by this relation. We can easily show that such identifications of MATH are possible, and obtain a manifold MATH from the identifications. The foliation of MATH with leaves that are images of MATH for MATH gives rise to a foliation on MATH whose leaves meet the fibers of MATH at unique points. Take a leaf MATH in MATH. Then MATH is an orbifold-covering map and MATH is a manifold. Take a universal cover MATH of MATH with covering map MATH. Then MATH is a universal covering map of MATH. MATH has a MATH-structure since it covers MATH: one can induce charts. Then MATH has a MATH-structure. Since by REF MATH is a universal cover of MATH, REF implies that MATH for the fundamental group MATH is MATH-diffeomorphic to MATH by a map induced by MATH. As MATH is a MATH-manifold, MATH has a developing map MATH (which follows from the geometric structure theory for manifolds). For a deck transformation MATH, MATH is also a MATH-map, and this means that MATH for some MATH. We can clearly verify that MATH is a homomorphism. The rest of the conclusion follows in the same way as the geometric structure theory for manifolds. |
math/0107172 | We take a product MATH and let MATH be a vector field in the positive MATH-direction in the product space. Then MATH acts smoothly on MATH by sending MATH to MATH for MATH. We average MATH for MATH to obtain a smooth MATH-invariant vector field MATH. The integral curve MATH of MATH starting from MATH is mapped to an integral curve MATH of MATH starting from MATH by the MATH-action. Thus, the endpoint MATH is sent to MATH, and so MATH. Hence, let MATH equal the point of the path from MATH at time MATH. Then MATH for the projection MATH is a desired diffeomorphism and MATH is the desired open ball. |
math/0107172 | Parameterize MATH by MATH for a semi-algebraic set MATH with MATH corresponding to MATH and, for each MATH, there is a map MATH from the above REF . Again, we obtain a smooth MATH-invariant vector field MATH on MATH as above, and MATH depends continuously on MATH. From this, we see that MATH corresponding to a representation corresponding to MATH depends continuously on MATH. |
math/0107172 | Let MATH denote the space of all MATH-equivariant isotopies of MATH with the above parametrization MATH. The maps MATH are in MATH. There exists a section MATH where MATH is a neighborhood of MATH by REF . Hence, there exists MATH, MATH, so that MATH for MATH for all MATH. We define MATH for MATH to be MATH. This defines a function MATH so that MATH is an isotopy for each MATH. We modify MATH to be MATH-equivariant. We define MATH to be given by MATH. Then MATH for a given MATH is an integral curve of a vector field MATH on MATH with the component in the MATH-direction equal to MATH. Since MATH is a MATH-equivariant isotopy, we see that MATH restricted to the image of MATH is MATH-invariant. We may now average the image vector fields of MATH under MATH, and call MATH the resulting MATH-invariant vector field on MATH. Again MATH restricted to the image equals MATH on the image and the second component equals MATH. The integral curves of MATH give us a MATH-equivariant isotopy MATH extending MATH. Since the section MATH is continuous, and we do averaging and integration, it follows that MATH is continuous. Now restrict MATH to MATH. |
math/0107173 | This is very well known, especially when translated into the language of flags via the connection mentioned in REF. NAME can be proved by an explicit construction, and injectivity is easy by induction. |
math/0107173 | Clearly the map in REF is MATH-stable. So we need only note that since MATH, MATH, and all MATH are connected, MATH. |
math/0107173 | Since MATH is connected, MATH, so MATH . Thus it suffices to show that MATH . This follows easily from the fact that MATH. |
math/0107173 | As with REF , this is well known when translated in terms of MATH-orbits on the flag variety, and easy to prove (see for instance CITE). |
math/0107173 | Again, this follows from REF , since MATH, MATH, and MATH are all connected. |
math/0107173 | Since MATH is connected, MATH, so MATH . Thus it suffices to show that MATH is generated by MATH . This follows easily from our description of MATH. |
math/0107173 | The proof is exactly analogous to that of REF . |
math/0107173 | Clearly MATH so MATH which proves the result. |
math/0107173 | As with REF , this is better known as a statement about MATH-orbits on the flag variety (see CITE). It is easy to prove. |
math/0107173 | By REF , we may identify MATH with MATH. Under this identification, an involution is in the image of MATH precisely when the corresponding MATH-stable MATH-orbit on MATH contains a MATH-fixed point (by connectedness of MATH). Since MATH is odd, this MATH-orbit is a single MATH-orbit, so this is automatic. |
math/0107173 | Since the image of the NAME map on MATH is MATH, the number of MATH - MATH double cosets in MATH is the same as the number of orbits of MATH on MATH for the action MATH. Using NAME 's theorem, one sees that only the ``disconnected part" of MATH contributes. So this is the same as the number of orbits of MATH for the action MATH. In other words, it is the index in the latter group of the subgroup where MATH, for all MATH. Now since MATH is odd, MATH is non-empty. The result follows. |
math/0107173 | We first note that MATH, as may be seen directly (using formulas for MATH-rank such as those below) or deduced by the method of CITE. So MATH . Thus it suffices to observe that MATH is generated by MATH . |
math/0107173 | As in the proof of REF , an involution is in the image of MATH precisely when the corresponding MATH-stable MATH-orbit on MATH contains a MATH-fixed point. If the involution has a fixed point, this MATH-orbit is a single MATH-orbit, so this is automatic. Suppose MATH. It is in the image of MATH precisely when there exists a decomposition of MATH into lines MATH such that CASE: MATH, and CASE: MATH (here we have identified MATH with MATH in some way). We must prove that this happens if and only if MATH. Since MATH is multiplicative with respect to MATH-stable orthogonal direct sums, we may assume that MATH acts transitively on MATH. So if MATH, then MATH has two orbits on MATH which MATH interchanges; in this case the existence of MATH as above is clearly equivalent to the existence of a decomposition MATH into MATH-stable Lagrangian subspaces, which indeed happens if and only if MATH. On the other hand, if MATH, then MATH is a MATH-cycle and MATH; in this case the existence of MATH as above is easily seen to be equivalent to the existence of a Lagrangian subspace MATH of MATH such that MATH, which indeed happens if and only if MATH. |
math/0107173 | The method of proof of REF applies again here. |
math/0107173 | The proof is exactly the same as that of REF . |
math/0107173 | This follows by combining REF . |
math/0107173 | This is deduced in the same way as REF . |
math/0107173 | The proof is the same as that of REF . |
math/0107173 | We first prove that MATH. The MATH-rank of MATH is MATH . If MATH has eigenvalue MATH on MATH, then MATH has MATH-rank which differs from that of MATH by MATH . Since MATH is odd, MATH is odd, so this difference is zero. Thus MATH. The rest of the proof follows that of REF . |
math/0107173 | The proof is mostly the same as that of REF . Note that if MATH, MATH . So in REF of the proof of REF , MATH should be replaced by MATH. Since MATH if MATH, we get the result. |
math/0107173 | The proof of REF applies again here. |
math/0107173 | By the same argument as the proof of REF , we see that MATH . Hence MATH if and only if MATH and for all MATH, MATH . Clearly this is equivalent to MATH. |
math/0107174 | For the first three categories, and the NAME, the statement is precisely CITE. Let us check that the inclusion MATH induces an isomorphism in NAME. By CITE, which shows that REF is satisfied, we can apply CITE, in the situation where MATH is the category of MATH-equivariant quasicoherent MATH-Modules, MATH the category of cohomologically bounded complexes in MATH, MATH the category of MATH-equivariant quasicoherent flat MATH-Modules, MATH is the natural inclusion. In particular, any complex in MATH receives a quasi-isomorphism from a complex in MATH. That is, CITE, applied to the inclusion MATH, is satisfied; since the other REF is obviously satisfied, we conclude by CITE. |
math/0107174 | The proof follows closely NAME 's proof of CITE, therefore we will only indicate the changes we need for that proof to adapt to our situation. Let us denote by MATH the NAME category consisting of pairs MATH where MATH is a bounded above complex of MATH-equivariant quasi-coherent flat MATH-Modules with bounded coherent cohomology, MATH is a bicomplex of MATH-equivariant quasi-coherent flat MATH-Modules with bounded coherent total cohomology such that MATH for MATH, any MATH and also MATH for MATH, for some integer MATH, any MATH; finally MATH is an exact augmentation of the bicomplex MATH. In particular, for any MATH, the horizontal complex MATH is a flat resolution of MATH. The morphisms, cofibrations and weak equivalences in MATH are as in CITE. CITE shows that the forgetful functor MATH from MATH to the category MATH of bounded above complexes of MATH-equivariant quasi-coherent flat MATH-Modules with bounded coherent cohomology induces a homotopy equivalence between the associated NAME MATH-theory spectra. In other words, by REF , we can (and will) use MATH as a ``model" for MATH. With these choices, the morphism of spectra MATH can be represented by the exact functor MATH which sends MATH to the total complex of the bicomplex MATH. The rest of the proof is exactly the same as in CITE. One first consider functors MATH sending an object MATH to the total complex of the bicomplex MATH which results from truncating all the horizontal complexes of MATH at the MATH-th level. The functors MATH are zero for MATH and come naturally equipped with functorial epimorphisms MATH whose kernel MATH has the property that MATH is quasi isomorphic to MATH CITE, essentially because each horizontal complex in MATH is a flat resolution of MATH. Therefore, by induction on MATH, starting from MATH, each MATH has values in MATH and preserves quasi-isomorphisms. Moreover, the arguments in CITE, show that MATH actually preserves cofibrations and pushouts along cofibrations; hence each MATH is an exact functor of NAME categories. As in CITE, the quasi-isomorphism MATH shows that the canonical truncation morphism MATH, MATH being the codimension of MATH in MATH, is a quasi-isomorphism, that is, the morphism of spectra MATH can also be represented by the exact functor MATH. Now, the Additivity Theorem CITE shows that the canonical exact sequences of functors MATH yield up-to-homotopy equalities MATH between the induced map of spectra. And finally, recalling that a shift MATH induces multiplication by MATH at the level of spectra, by induction on MATH we get equalities up to homotopy MATH of morphisms of spectra MATH CITE. |
math/0107174 | To prove REF we may restrict the action of MATH to its toral component. By NAME 's generic slice theorem CITE there are only finitely many possible diagonalizable subgroup schemes of MATH that appear as stabilizers of a geometric point of MATH. Then we can take the MATH to be the toral components of the MATH-dimensional stabilizers. REF follow from REF and CITE. |
math/0107174 | Let us denote by MATH the NAME spectrum associated with the category of coherent equivariant MATH-sheaves on a noetherian separated algebraic space MATH. There is a homotopy equivalence MATH . The commutative diagram MATH induces a morphism of spectra MATH . By the self-intersection formula REF there is a homotopy MATH on the other hand MATH is homotopic to zero, because MATH is trivial. So we have that MATH where MATH is the suspension of MATH. We define the specialization morphism of spectra MATH by composing the morphism REF with the canonical projection MATH . Finally, MATH is defined to be the homomorphism induced by MATH on homotopy groups. Let us check compatibility; it suffices to show that the diagram of spectra MATH commutes up to homotopy. The essential point is that the diagram of algebraic spaces MATH is NAME, that is, MATH for all MATH. Write MATH and MATH for the NAME categories of MATH-equivariant complexes of quasicoherent MATH-modules and MATH-modules with bounded coherent cohomology, while MATH and MATH will denote the NAME categories of complexes of MATH-equivariant quasicoherent MATH-modules and MATH bounded coherent cohomology, that are respectively degreewise MATH-acyclic and degreewise MATH-acyclic. By the NAME of the diagram above, we have that MATH gives a functor MATH, and the diagram MATH commutes. By CITE this concludes the proof of the theorem. |
math/0107174 | REF imply REF . To prove REF , notice that, by CITE, MATH is regular, and so is MATH. Let MATH be the pullback to MATH of a local equation for the section at infinity of MATH, and let MATH be a point of MATH. Since the conormal space to MATH in MATH has no nontrivial MATH-invariants, clearly the differential of MATH at MATH can not lie in this conormal space, hence MATH is not zero in any neighborhood of MATH in MATH. This implies that MATH is a regular NAME divisor on MATH, as claimed. |
math/0107174 | This follows immediately from the compatibility of specializations REF . |
math/0107174 | Choose a splitting MATH; by CITE, we have MATH . If MATH is the eigenspace decomposition of MATH, we have that MATH corresponds to the element MATH of MATH, so it enough to show that MATH is not a zero-divisor in MATH. But we can write MATH where MATH is a unit in MATH. Now we can apply the following elementary fact: suppose that MATH is a ring, MATH, , MATH central elements of MATH such that MATH is a unit, MATH a monomial different from REF. Then the element MATH is not a zero-divisor in MATH. |
math/0107174 | From the self-intersection formula REF we see that the composition MATH is multiplication by MATH, so MATH is injective. We get REF from this and from the localization sequence. REF follows easily from REF , together with the following elementary fact. Let MATH, MATH and MATH be rings, MATH and MATH ring homomorphisms. Suppose that there exist a homomorphism of abelian groups MATH such that: CASE: The sequence MATH is exact; CASE: the composition MATH is the multiplication by a central element MATH which is not a zero divisor. Then MATH and MATH induce an isomorphism of rings MATH where the homomorphism MATH is the projection, and the one MATH is induced by the isomorphism MATH and the projection MATH. |
math/0107174 | This follows from REF , and REF . |
math/0107174 | Let us start with a lemma. The restriction homomorphism MATH is surjective, and its kernel is the ideal MATH. Since the complement of the zero section MATH coincides with MATH REF , we can apply REF to the normal bundle MATH, and conclude that there is an exact sequence MATH . Now, MATH is an isomorphism, and the composition MATH is multiplication by MATH, because of the self-intersection REF , and this implies the thesis. Therefore the restriction homomorphism MATH induces an isomorphism of MATH with MATH; the Proposition follows from this, and from REF . |
math/0107174 | By multiplying each MATH by MATH we may assume that MATH has the form MATH with MATH and MATH a central unit in MATH. We will show that for any MATH the relation MATH, MATH implies MATH; from this the thesis follows with a straightforward induction. We may assume that MATH, MATH and MATH. Since MATH, MATH are MATH-linearly independent elements of MATH, we may complete them to a maximal MATH-independent sequence MATH of MATH; this sequence generates a subgroup MATH of finite index. Suppose at first that MATH, so that MATH. Replacing MATH by MATH, we may assume that MATH. If MATH, we can multiply this equality by a sufficiently high power of MATH and assume that MATH and MATH are polynomials in MATH. Since MATH is a polynomial in MATH with central coefficients and invertible leading coefficient, the usual division algorithm allows us to write MATH, where MATH is a polynomial whose degree in MATH is less than MATH. By comparing the degrees in MATH in the equality MATH we see that MATH must be zero, and this proves the result. In the general case, choose representatives MATH, , MATH for the cosets of MATH in MATH; then any element MATH of MATH can be written uniquely as MATH with MATH. Then from the equality MATH we get MATH for all MATH, because MATH and MATH are in MATH; hence the thesis follows from the previous case. |
math/0107174 | REF is obvious. REF follows from the fact that MATH is noetherian, so formation of MATH commutes with direct sums and direct limits. Let us prove REF . From REF we see that we may assume that MATH is cyclic. If MATH, then MATH, and the statement follows from the facts that MATH is NAME - NAME, and that the height of a prime ideal is at least equal to its fiber codimension. Assume that MATH, so that MATH. The associated primes of MATH are the generic points of the fibers of MATH over the primes dividing MATH, so REF is satisfied. Take a prime MATH of MATH of fiber codimension at least REF, and consider the exact sequence MATH . If the characteristic of MATH is positive, then the height of MATH is at least REF, so MATH, because MATH is NAME - NAME, and we are done. Otherwise, we have an exact sequence MATH but the height of MATH is at least REF, so MATH. From this we deduce that multiplication by MATH is injective on MATH, and this concludes the proof of REF . For REF , notice first of all that if MATH is a finitely generated MATH-module of fiber codimension at least REF then we can filter MATH with successive quotients of type MATH, where MATH is a prime of fiber codimension at least REF, so MATH. If MATH is not finitely generated and MATH is an exact sequence of MATH-modules, MATH is a finitely generated submodule of MATH, and MATH is the pullback of MATH to MATH, then the sequence MATH splits; but because of REF we have MATH, hence there is a unique copy of MATH inside MATH. Hence there is a unique copy of MATH inside MATH, and the sequence splits. This completes the proof of the Proposition. |
math/0107174 | We need some preliminaries. Suppose that MATH acts on MATH with stabilizers of constant dimension MATH. Then the support of MATH as a MATH-module has pure fiber dimension MATH, and any associated prime of MATH has pure fiber dimension MATH. Suppose first of all that MATH is REF. Then it follows easily from NAME 's localization theorem CITE that the support of MATH has fiber dimension REF, and from this we see that every associated prime must have fiber dimension REF. In the general case, we may assume that the action is connected (that is, MATH is not a nontrivial disjoint union of open invariant subspaces); then there will be a splitting MATH, where MATH is a diagonalizable group scheme of finite type acting on MATH with finite stabilizers, and MATH is a totally split torus that acts trivially on MATH. In this case MATH . The proof is concluded by applying the following lemma. Let MATH be a flat NAME - NAME MATH-algebra of finite type, MATH a smooth homomorphism of finite type with fibers of pure dimension MATH. Suppose that MATH is a MATH-module whose support has fiber dimension REF. Then MATH has support of pure dimension MATH, and each of its associated primes has fiber dimension MATH. Since tensor product commutes with taking direct limits and MATH is flat over MATH, we may assume that MATH is of finite type over MATH. By an obvious filtration argument, we may assume that MATH is of the form MATH, where MATH is a prime ideal of fiber dimension REF. In this case the only associated primes of MATH are the generic components of the fiber of MATH over MATH, and this proves the result. Suppose that MATH and MATH are algebraic spaces on which MATH acts with stabilizers of constant dimension respectively MATH and MATH. If MATH is a MATH-submodule of MATH and MATH, then there is no nontrivial homomorphism of MATH-modules from MATH to MATH. Given such a nontrivial homomorphism MATH, call MATH its image. The support of MATH has fiber dimension at most MATH, so there is an associated prime of fiber dimension at most MATH in MATH, contradicting REF . Now we prove REF . Let MATH be the dimension of MATH, so that MATH. First of all, let us show that the natural projection MATH is an isomorphism. This will be achieved by showing that for all MATH with MATH the natural projection MATH is an isomorphism, where we have set MATH . For MATH this is our main REF , so we proceed by induction. If MATH and the projection above is an isomorphism, we have an exact sequence MATH where the last arrow is the difference of the composition of the projection MATH with the pullback MATH, and of the specialization homomorphism MATH. If we call MATH the image of this difference, we have an exact sequence of MATH-modules MATH and the support of MATH is of fiber dimension at most MATH by REF , hence it is of fiber codimension at least REF. It follows from the fact that MATH is sufficiently deep and from REF that this sequence splits. From the fact that MATH has only associated primes of fiber dimension REF we see that the pullback map MATH must be injective, and from REF that a copy of MATH living inside MATH must in fact be contained in MATH; this implies that the projection MATH is an isomorphism. So the projection MATH is an isomorphism. Then the kernel of the specialization homomorphism from MATH to MATH maps injectively in MATH, so it must be REF, again because MATH has only associated primes of fiber dimension REF. Furthermore MATH is the disjoint union of the MATH when MATH ranges over all finite subtori of MATH of codimension REF, and similarly for MATH. On the other hand, because of our main theorem applied to the action of MATH on MATH, we have the natural isomorphism MATH, and this completes the proof of REF . |
math/0107174 | Let MATH, and call MATH the NAME group of MATH over MATH. Choose a one parameter subgroup MATH as before. Let MATH, , MATH be the connected components of MATH, MATH as in NAME 's theorem. The MATH correspond to the orbits of the action of MATH on MATH; obviously MATH also permutes the MATH, so we let MATH, , MATH be the smooth subvarieties of MATH corresponding to the orbits of MATH on MATH. The MATH descend to morphisms MATH. REF are obviously satisfied, because they are satisfied after passing to MATH. We have to prove REF . Let MATH be the inverse image of MATH in MATH, MATH the ideal of MATH in the algebra MATH. Because MATH is affine, MATH is a projective MATH-module, and MATH is diagonalizable, the projection MATH has a MATH-linear and MATH-equivariant section MATH. This induces a MATH-equivariant morphism of MATH-schemes MATH, sending MATH to REF-section, whose differential at the zero section is the identity (notice that MATH is also the restriction to MATH of the relative tangent bundle MATH). We want to show that this is an isomorphism; it is enough to check that this is true on the fibers, so, let MATH be one of the fibers of MATH on some point MATH. According to REF ynicki-Birula, the fiber of MATH on MATH is MATH-equivariantly isomorphic to MATH; hence an application of the following elementary lemma concludes the proof of REF . Suppose that MATH acts linearly with positive weights on a finite dimensional vector space MATH over a field MATH. If MATH is an equivariant polynomial map whose differential at the origin is an isomorphism, then MATH is also an isomorphism. First all, notice that because of the positivity of the weights, we have MATH. By composing MATH with the inverse of the differential of MATH at the origin, we may assume that this differential of MATH is the identity. Consider the eigenspace decomposition MATH, where MATH acts on MATH with a character MATH, and MATH. Choose a basis of eigenvectors of MATH; we will use groups of coordinates MATH, , MATH, where MATH represents the group of elements of the dual basis corresponding to basis elements in MATH, so that the action of MATH is described by MATH. Then it is a simple matter to verify that MATH is given by a formula of the type MATH and that every polynomial map of this form is an isomorphism. |
math/0107174 | We us the same notation as in CITE; in particular, for each cone MATH of the fan of MATH, we denote by MATH the corresponding affine open subscheme of MATH. First of all, assume that the fan MATH consists of all the faces of a MATH-dimensional cone MATH. Call MATH a part of a basis of MATH that spans MATH: we have an action of MATH on the MATH-vector space MATH generated by MATH, by letting each REF-parameter subgroup MATH in MATH act by multiplication on the corresponding line in MATH, and an equivariant embedding MATH. Then MATH is MATH-equivariantly isomorphic to the MATH-equivariant vector bundle MATH in such a way that the zero section corresponds to MATH. Since MATH and MATH is a vector bundle over MATH, we get a canonical isomorphism MATH, and from this a canonical isomorphism MATH . It follows that the deformation to the normal bundle of MATH in MATH is also isomorphic to the product MATH, and from this we get the second part of the statement. In the general case we have MATH, and if MATH is a cone of dimension MATH in MATH, the intersection of MATH with MATH is precisely MATH. From this we get the first part of the statement in general. The second part follows by applying the compatibility of specializations to the morphism of deformation to the normal bundle induced by the equivariant morphism MATH. |
math/0107174 | First of all, let us show that the MATH and their inverses generate MATH . Set MATH, and let MATH be an element of MATH; we want to show that MATH can be expressed as a NAME polynomial in the MATH evaluated in the MATH. The ring MATH is a ring of NAME polynomials in the images of the MATH with MATH, so we can find a NAME polynomial MATH, in which only the MATH with MATH appear, such that the image of MATH in MATH equals the image of MATH in the same ring. By subtracting MATH, we see that we may assume that the projection of MATH into MATH is zero. Now, let us repeat the procedure for the maximal cone MATH: find a polynomial MATH, in which only the MATH with MATH appear, such that the image of MATH in MATH equals the image of MATH in the same ring. The key point is that the restriction of MATH to MATH is zero, so in fact MATH can only contain the variables MATH with MATH not in MATH. Hence the restriction of MATH to MATH is also zero, and after having subtracted MATH from MATH we may assume that the restriction of MATH to both MATH and MATH is zero. We can continue this process for the remaining cones MATH, , MATH; at the end all the projections will be zero, and therefore MATH will be zero too. Now we have to show that the kernel of the homomorphism MATH sending each MATH to MATH equals the ideal MATH, where MATH varies over all subsets of MATH not contained in any maximal cone. The kernel of the projection MATH equals the ideal MATH generated by the MATH, where MATH varies over the set of REF cones in MATH not contained in MATH, hence the kernel of the homomorphism REF is the intersection of the MATH. The result is then a consequence of the following lemma. Let MATH be a (not necessarily commutative) ring, MATH a finite set of central indeterminates; consider the ring of NAME polynomials MATH. Let MATH, , MATH be subsets of MATH; for each MATH, , MATH call MATH the ideal of MATH generated by the elements MATH with MATH. Then the intersection MATH is the ideal of MATH generated by the elements MATH, where MATH varies over all subsets of MATH that meet each MATH. When each MATH contains a single element this is a particular case of REF . The obvious common generalization should also hold. We proceed by induction on MATH; the case MATH is clear. In general, take MATH; by induction hypothesis, we can write MATH where MATH varies over all subsets of MATH whose intersection with each of the MATH, , MATH is not empty. We can split the sum as MATH the first summand is in MATH and is of the desired form, so we may subtract it from MATH and suppose that MATH is of the type MATH . Now, consider the ring MATH of NAME polynomials not involving the variables in MATH; it is a subring of MATH, and there is also a retraction MATH sending each MATH with MATH to MATH, whose kernel is precisely MATH. The elements MATH are in MATH, and MATH so we can write MATH . Then we write each MATH as a linear combination of the polynomials MATH with MATH; this concludes the proofs of the lemma and of the theorem. |
math/0107174 | If the base scheme MATH is the spectrum of a field, this is a particular case of the main theorem of CITE. If MATH is a torus, the proof of this statement given in CITE goes through without changes, because it only relies on NAME 's generic slice theorem for torus actions CITE. In the general case, choose an embedding MATH into some totally split torus MATH over MATH, and consider the quotient space MATH by the customary diagonal action of MATH; this exists as an algebraic space thanks to a result of NAME CITE. The same argument as in the beginning of REF shows that MATH is separated. Now observe that if MATH is an essential subgroup relative to the action of MATH on MATH, we have MATH unless MATH is an essential subgroup; hence the least common multiples of the orders of all essential subgroups are the same for the action of MATH on MATH and the action of MATH on MATH. Also, if MATH is an essential subgroup we have MATH, and therefore, by NAME equivalence CITE, we get an isomorphism MATH which is an isomorphism of MATH-algebras, if we view MATH as a MATH-algebra via the restriction homomorphism MATH. Moreover, MATH is exactly the preimage of MATH under the natural surjection MATH; therefore we have compatible MATH-localized NAME isomorphisms MATH and (for MATH equal to the trivial subgroup) MATH hence the theorem for the action of MATH on MATH follows from the theorem for the action of MATH on MATH. |
math/0107174 | Consider the commutative diagram of group schemes MATH by taking representation rings we get a commutative diagram of rings MATH (without the dotted arrow). But it is easy to see that in fact the composition MATH factors through MATH, so in fact the dotted arrows exists; and this proves the thesis. |
math/0107174 | To simplify the notation, we will implicitly assume that everything has been tensored with MATH. We apply REF together with REF . According to REF we have an injection MATH whose image is the subring of sequences MATH with the property that for each MATH, , MATH the pullback of MATH to MATH coincides with MATH. Moreover, by REF , we can decompose further each MATH as MATH, where MATH varies in the (finite) set of essential subgroups of MATH of dimension MATH. By compatibility of specializations, for any MATH, the following diagram is commutative MATH where MATH is induced by the obvious pullbacks. On the other hand, the following diagram commutes by definition of MATH . Then the Theorem will immediately follow if we show that MATH is an isomorphism. This is true because of the following Lemma. Fix an essential subgroup MATH of dimension MATH. Then for any MATH with MATH, the scheme MATH is open in MATH; furthermore, MATH is the disjoint union of the MATH for all essential MATH with MATH, MATH. We will show that MATH is the disjoint union of the MATH; since each MATH is closed in MATH, and there are only finitely many possible MATH, it follows that each MATH is also open in MATH. Let us first observe that if MATH and MATH are essential subgroups in MATH of dimension MATH to which MATH is subordinate, and MATH, then MATH, therefore MATH is equal to MATH. But this implies that MATH, since MATH and MATH are both equal to the inverse image in MATH of the image of MATH. According to REF , if MATH, , MATH are the essential MATH-dimensional subtori of MATH, then MATH is the disjoint union of the MATH. Clearly, if MATH is not contained in MATH, then MATH is empty, so MATH is the disjoint union of the MATH, where MATH ranges over the essential MATH-dimensional subtori of MATH with MATH. But there is a bijective correspondence between MATH-dimensional dual semi-cyclic subgroups MATH with MATH and MATH-dimensional subtori of MATH with MATH: in one direction we associate with each MATH its toral part MATH, in the other we associate with each MATH the subgroup scheme MATH. The proof is concluded by noticing that if MATH and MATH are as above, with MATH, then MATH. |
math/0107174 | By the definition of MATH, we need to check that MATH commutes. By the projection formula (see CITE) we see that the group homomorphisms MATH and MATH are multiplications by MATH respectively: so we have to prove that the diagram MATH commutes. Since MATH is a ring homomorphism, this is equivalent to saying that MATH . But MATH is the restriction of MATH, so we have to show that the restriction of MATH to MATH is MATH; and this follows immediately from the fact that the square MATH is cartesian and NAME, by REF . |
math/0107174 | Consider the natural embedding of deformations to the normal bundle MATH generically, that is, over MATH, they coincide. On the other hand it follows from REF in the paper that the inverse image of MATH in MATH is scheme-theoretically dense in MATH, and this shows that this embedding is an isomorphism. Since the fibers over MATH of MATH and MATH are MATH and MATH respectively, this concludes the proof. |
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