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math/0107109 | REF implies that the NAME class for the sum of two bundles is the smash product of NAME classes. In particular, MATH. |
math/0107109 | Since MATH is quasi-projective we can find an affine torsor MATH such that MATH is affine. Since motivic cohomology of MATH and MATH are the same it is sufficients to prove the lemma for an affine MATH. Over an affine MATH the sequence MATH splits and we have MATH. Let MATH be the obvious morphism. REF implies that MATH which in turn implies the statement of the lemma. |
math/0107109 | Let MATH be a generic point of MATH, MATH the fiber of MATH over MATH and MATH the fiber of MATH over MATH. Then, for a generic point MATH of MATH the fiber of MATH over MATH is MATH. Given another pair MATH consider the isomorphism MATH and let MATH be the corresponding isomorphism of NAME spaces. We claim that MATH on the level of actual cycles. Indeed, the fiber of MATH over a generic point MATH of MATH is MATH (in MATH) which coincides with the fiber of MATH over MATH. Using REF and applying REF to MATH we get the statement of our lemma. |
math/0107109 | Direct comparison. |
math/0107109 | By REF we have MATH where MATH . Therefore by REF : MATH . Since MATH the NAME isomorphism theorem implies REF . |
math/0107109 | Take MATH as a model of MATH such that MATH is given by the tautological section MATH of MATH on MATH. Applying the construction of MATH to the corresponding diagonal cycle MATH we get the restriction to MATH of the morphism MATH corresponding, by Construction REF, to the tautological cycle on MATH over MATH. In view of REF MATH represents the NAME class of the trivial bundle MATH. Applying NAME isomorphism to get MATH we conclude that MATH is the restriction of the NAME class of MATH with respect to the morphism MATH. |
math/0107109 | Follows from REF . |
math/0107109 | Let MATH be the fiber of MATH over our distinguished point MATH. Then, if we compute the analog of our composition using MATH instead of MATH, we get the map MATH which is of the form MATH where MATH is the tautological cycle on MATH and MATH is Construction REF with respect to the isomorphism MATH. Our result follows now from REF since MATH is the trivial bundle. |
math/0107109 | The morphism MATH can be represented by the composition MATH . The first of these maps is the zero section of a vector bundle MATH and gives an isomorphism on motivic cohomology by homotopy invariance. The second map is an open embedding of smooth schemes and the codimension of the complement is at least MATH. By REF it defines an isomorphism on MATH. |
math/0107109 | Given any sequence of maps of pointed simplicial sheaves MATH with the colimit MATH we have a long exact sequence of the form MATH . The limit MATH is the kernel of the fourth arrow and therefore to prove the corollary it is sufficient to show that the map MATH is an epimorphism. It follows from the proposition. |
math/0107109 | We have MATH. The projection MATH is invariant under the action of MATH and therefore gives a map MATH. One verifies that this map is isomorphic to the complement to the zero section of the line bundle MATH on MATH. |
math/0107109 | Let MATH with the standard action of MATH and the corresponding action of MATH. The square MATH is pull-back. The fact that the right vertical arrow is MATH is standard (for example, it is not MATH because it has a section other than the zero one). |
math/0107109 | Existence follows from the exact sequence MATH and the fact that MATH in MATH. The exact sequence REF around MATH shows that MATH. This implies the uniqueness. |
math/0107109 | The standard argument shows that it is sufficient to consider the case when MATH is of the form MATH for a smooth scheme MATH. The same reasoning as we used to establish REF applies to the motivic cohomology groups of MATH for any smooth scheme MATH and we get the following result. For any smooth scheme MATH there is a long exact sequence of MATH-modules of the form MATH . For MATH-coefficients we have MATH and REF becomes a short exact sequence of MATH-modules of the form MATH . Let MATH be an element in MATH such that the image of MATH in MATH is MATH and the restriction of MATH to MATH is zero. Since MATH is the image of MATH, the short exact sequence REF implies that the monomials MATH and MATH form a basis of MATH over MATH. On the other hand, the image of MATH in MATH is not zero and hence MATH where MATH. This implies that the monomials MATH, MATH also form a basis. |
math/0107109 | We need to compare two motivic cohomology classes in MATH. Let MATH be the embedding of the generic point. Since the base field MATH may be assumed to be perfect, the NAME long exact sequence in motivic cohomology implies that the kernel of the induced map in MATH is covered by a direct sum of groups of the form MATH. Since such groups are zero it is a monomorphism. Therefore, it is sufficient to show that MATH in MATH. By CITE, this group is isomorphic to MATH and we conclude by the well known relation MATH in the NAME 's NAME. |
math/0107109 | We have a natural transformation from the motivic cohomology to the etale cohomology with MATH-coefficients. For a class MATH in the etale MATH we have MATH where MATH is the NAME homomorphism. Since MATH is separably closed and in particular contains MATH, the NAME in the etale cohomology commutes with the NAME in the motivic cohomology and we conclude that the image of MATH in the etale cohomology coincides with the image of MATH in the etale cohomology. An etale analog of the long exact sequence REF shows that the image of MATH in the etale cohomology is non zero. |
math/0107109 | Let MATH be the line bundle on MATH corresponding to the tautological REF-dimensional representation MATH of MATH. Then MATH is the NAME class of MATH and MATH is the only element in MATH which is zero at MATH and which maps to MATH under the map MATH. The automorphism MATH takes the tautological REF-dimensional representation MATH to MATH and, therefore, it takes MATH to MATH. Our result follows now from REF . |
math/0107109 | We will use the notations established at the beginning of REF. Choose a linear representation MATH. The map REF is defined by the collection of maps MATH with respect to the identification of REF . The maps MATH are finite etale of degree MATH and the fundamental cycle on MATH over MATH defines a map of freely generated sheaves with transfers MATH . The composition of MATH is the multiplication by MATH. Replacing MATH by its the standard simplicial resolution by coproducts of representable sheaves we may assume that terms of MATH are coproducts of sheaves of the form MATH. Then, MATH where MATH is the normalized chain complex functor from simplicial sheaves with transfers to complexes of sheaves with transfers. In particular these groups are functorial in MATH which implies the first claim of the proposition. The second claim follows from the fact that the normalizer of MATH in MATH acts on MATH over MATH and the map MATH, being defined by the fundamental cycle, is invariant under this action. |
math/0107109 | The element MATH is the NAME class of MATH where MATH corresponds to the regular representation of MATH under our isomorphism MATH. Therefore, we have MATH where MATH is the line bundle corresponding to the tautological REF-dimensional representation of MATH. By REF we get MATH . |
math/0107109 | For MATH we have MATH and our result follows from REF . Assume that MATH. The transfer argument shows that to prove the theorem for MATH it is sufficient to prove it for a separable extension of MATH of degree prime to MATH. In particular, we may assume that MATH contains a primitive MATH-th root of unity MATH. To prove the existence of MATH we need to show that MATH is a MATH-torsion element in MATH. For any ring of coefficients where MATH is invertible, the map MATH defines, by REF , a split injection: MATH . In particular, since REF is an injection for coefficients in MATH localized at MATH, it is sufficient to prove that the image of MATH in MATH is an l-torsion element. This follows from REF . To show that MATH is unique, it is sufficient to check that the map MATH is injective. Injectivity of REF for MATH-coefficients implies that it is sufficient to show that the map MATH is injective. REF implies that for MATH-coefficients the right hand side of REF is of the form MATH where MATH and MATH. This descriptions shows that MATH generated by MATH. We have MATH. Therefore, REF is injective. |
math/0107109 | We may assume that MATH. Then, the description REF implies that MATH for an element MATH. Since MATH and MATH we conclude that MATH. |
math/0107109 | For MATH we have MATH and our result follows from REF . Assume that MATH. We need to show that the map MATH is an isomorphism. By the transfer argument we may assume that MATH contains a primitive l-th root of unity. By REF the homomorphism MATH defined by a choice of a primitive root MATH is a mono. Therefore, it is sufficient to show that the composition MATH is an isomorphism. The fact that REF is an isomorphism follows from the analog of REF for MATH and REF . |
math/0107109 | By the transfer argument we may assume that MATH contains a primitive l-th root of unity. In view of REF it is sufficient to prove our equality in the motivic cohomology of MATH. By REF we have MATH. The vector bundle MATH restricted to MATH splits into the sum of line bundles MATH where MATH is the line bundle corresponding to the tautological REF-dimensional representation of MATH. By REF we get: MATH . Since the restriction of MATH to MATH is MATH (by REF ) this finishes the proof. |
math/0107109 | Direct comparison. |
math/0107109 | By REF we have MATH or, equivalently, MATH . By REF we rewrite it as MATH . By REF we get MATH . By REF we get MATH . By NAME isomorphism REF, we get MATH . Multiplying both sides by MATH we get by REF MATH . |
math/0107109 | Replacing MATH by an affine torsor we may assume that it is affine. Let MATH be the linear representation of MATH corresponding to MATH and MATH the open subset of MATH where MATH acts freely. Then, for some MATH, there exists a MATH-equivariant map MATH. By REF it is sufficient to prove the corollary for MATH and MATH. REF together with REF shows that we have MATH on MATH. By REF , multiplication with MATH is injective and we conclude that MATH. |
math/0107109 | The action of MATH on MATH extends to a free action of the semidirect product MATH. The permutational representation MATH also extends to a permutational representation of MATH. Therefore, MATH factors through the map MATH which implies that it is symmetric. |
math/0107109 | Follows from REF . |
math/0107109 | Replacing MATH by an affine torsor we may assume that it is affine. Let MATH be a faithful liner representation of MATH and MATH the open subset of MATH where MATH acts freely. Then, for some MATH, there exists a MATH-equivariant map MATH. By REF it is sufficient to show that MATH is symmetric for MATH. This follows from REF since we may choose MATH such that the codimension of MATH is larger than MATH. |
math/0107109 | Let MATH be the tautological motivic cohomology class of MATH. We need to show that MATH is invariant under the permutation of two copies of MATH. REF imply that MATH where MATH is the power operation corresponding to MATH. We conclude by REF and, again, REF . |
math/0107109 | The pointed sheaf MATH is a quotient sheaf of the sheaf MATH and the power operation for integral coefficients defines a map MATH . It is sufficient to show that the composition of this map with the projection to MATH factors through MATH that is, that for two cycles MATH be cycles on MATH with integral coefficients such that MATH is divisible by MATH the cycle MATH is in MATH. Let MATH be the cycle on MATH whose pull-back to MATH is MATH. It is sufficient to show that MATH is in MATH where MATH is the projection or, equivalently, that MATH for some MATH. We have MATH and the left hand side can be rewritten as MATH. Since any MATH-invariant cycle with coefficients divisible by MATH is of the form MATH it is sufficient to consider the summands in this expression with coefficients not divisible by MATH. They are of the form MATH. The sum of all such cycles is, up to multiplication by MATH, of the form MATH. |
math/0107109 | Consider the standard simplicial resolution MATH (see REF). Since a cycle with MATH-coefficients on a smooth scheme lifts to an integral cycle, the map MATH which represents MATH lifts to a map with values in MATH. Let MATH be such a lifting. Then MATH lands in MATH and MATH gives a morphism of complexes MATH which defines MATH. REF implies immediately that MATH lifts to a morphism with values in the image of the transfer map which implies the statement of the lemma. |
math/0107109 | The composition of the transfer map with the restriction map is the multiplication with the degree of the covering, in our case MATH. Hence it is sufficient to show that the restriction map MATH is surjective. Since motivic cohomology of MATH are trivial and class on the right can be written as MATH where MATH is in MATH and MATH is the NAME class. Any such MATH is clearly in the image of the restriction map. |
math/0107109 | By REF we have MATH. By REF we get MATH and by the NAME isomorphism theorem we conclude that MATH. |
math/0107109 | Follows immediately from REF . |
math/0107109 | By REF we have MATH. In view of REF and the fact that MATH we get MATH that is, MATH for all MATH, MATH and MATH for MATH. Applying REF we get REF . |
math/0107109 | This follows from REF . |
math/0107109 | REF implies that MATH where MATH is a constant. REF applied to the canonical line bundle on MATH implies that MATH. |
math/0107109 | Follows immediately from REF , the fact that MATH and the product REF for the NAME homomorphism. |
math/0107109 | Follows immediately from REF and the vanishing result REF. |
math/0107109 | Follows from REF . |
math/0107109 | Let MATH and MATH. Then MATH is in MATH. By REF we get MATH . By our assumption MATH and the right hand side is zero because the diagonal map MATH is zero in MATH. |
math/0107109 | By transfer argument we may assume that MATH contains a primitive l-th root of unity MATH. Consider the map MATH defined by MATH. Then MATH is a mono (by REF ) and by REF one has MATH where MATH and MATH are the generators from REF . We get: MATH where the last equality holds by REF . The right hand side equals to MATH which implies REF . For MATH we get MATH . By REF MATH, by REF and since MATH, MATH. By REF , MATH and therefore our expression equals MATH . This implies REF . |
math/0107109 | Consider the class MATH for MATH. Denote by MATH the generators of the cohomology of MATH appearing when the first MATH is applied and by MATH the generators of the cohomology of MATH appearing when the second MATH is applied. According to the symmetry REF the resulting expression is symmetric with respect to the exchange of MATH and MATH. We have (to simplify the notations we sometimes omit MATH from our expressions): MATH . Consider the coefficients in this expression at MATH, MATH, MATH and MATH. At MATH we have: MATH . At MATH we have: MATH . At MATH we have: MATH . At MATH we have: MATH . Set MATH, MATH. Then coefficient at MATH is MATH . Coefficient at MATH is MATH . Coefficient at MATH is MATH . Coefficient at MATH is MATH . Consider the coefficient at MATH where MATH for sufficiently large MATH and MATH. For MATH of this form, the coefficient MATH is non-zero if and only if MATH (follows from CITE) and we conclude that our coefficient is MATH. By symmetry it equals to the coefficient at MATH. Setting MATH, MATH and MATH and using the fact that MATH we can write the later as MATH . From the standard relation MATH and the fact that MATH if MATH is even and MATH is odd we get the first of the equalities stated in the theorem. A very similar argument starting with the equality between the coefficients at MATH and MATH gives the third equality - the case of even MATH and odd MATH. To prove the case when both MATH and MATH are even consider the coefficient at MATH. Consider the second part of this coefficient that is, the sum MATH . This is the coefficient at MATH which is equal to the coefficient at MATH that is, to MATH . Equating our new expression for the coefficient at MATH with the old expression for the coefficient at MATH we get MATH . Setting again MATH, MATH and MATH and using the standard relations between the binomial coefficients one recovers the identity for MATH when both MATH and MATH are even. Finally, to get the identity in the case when MATH is odd and MATH is even one uses REF , the identity for MATH and MATH even and the fact that MATH. |
math/0107109 | These relations are exactly the same as the NAME relations in the topological NAME algebra for odd coefficients. The proof of these relations given in CITE works in exactly the same way in the motivic context as in the topological one if one replaces the reference to CITE with the reference to our symmetry REF . The apparent difference in signs between our situation and the situation of CITE is explained by the fact that the image of MATH in the cohomology of MATH is MATH. |
math/0107109 | This follows from the NAME relations and the NAME REF . |
math/0107109 | It is sufficient to show that there exists MATH such that MATH are linerly independent in MATH with respect to the MATH-module structure. One starts with REF and uses exactly the same reasoning as in the proof of CITE . |
math/0107109 | Exactly parallel to the proof of CITE where REF is used to prove uniqueness. |
math/0107109 | By uniqueness part of REF it is enough to check that for any MATH one has: MATH . This follows immediately from definitions. |
math/0107109 | Let MATH be two elements of MATH which become equal modulo MATH. To check that MATH it is sufficient, in view of REF , to check that for any MATH and any MATH one has MATH. Let MATH be the cup product. Then by definition of MATH we have MATH . Our assumption on MATH implies that MATH. To prove that MATH is a ring homomorphism we have to check that for MATH and MATH we have MATH . This follows immediately from definitions. |
math/0107109 | The associativity follows immediately from the definition. To verify commutativity it is sufficient, in view of REF , to verify it on generators of the algebra MATH that is, on operations MATH and MATH. For this operations commutativity follows directly from the NAME formulas REF . |
math/0107109 | REF follow from REF and the fact that MATH. The other statement follows from REF and multiplicativity of MATH. |
math/0107109 | The value of MATH is a homogeneous element of MATH of degree zero. Hence, it is sufficient to show that the image of MATH in MATH is MATH or MATH depending on whether MATH or MATH. This is done using the action of MATH on MATH described in REF in exactly the same way as in the proof of CITE. |
math/0107109 | Elements MATH are exactly the admissible monomials. They form a basis of MATH by REF . The fact that elements MATH form a basis of MATH follows now from REF . |
math/0107109 | We already know by REF that MATH has a basis which consist of monomials in MATH which are of degree MATH in each MATH. The relation MATH for odd MATH is a corollary of graded commutativity. Thus we may assume that MATH in which case we have only to show that MATH. The required relation follows immediately from REF , the multiplicativity of MATH and the relation MATH in the motivic cohomology ring of MATH. |
math/0107109 | Let MATH be any elements of MATH. If MATH we have, by definition of product in MATH: MATH . This implies by induction starting with REF that for any MATH one has MATH . For MATH all the terms except for the ones which show up in the right hand side of REF cancel out since we work with MATH-coefficients. |
math/0107109 | We have by REF MATH . On the other hand by REF we get: MATH . Comparing coefficients at powers of MATH we get the first of the required equalities. To get the second one we write MATH by REF . And by REF we get: MATH . Comparing coefficients we get the second equality. |
math/0107109 | One verifies easily that MATH is indeed well defined by REF . To prove that it is an isomorphism consider the basis MATH in MATH and let MATH be the dual basis in MATH. The elements MATH clearly generate MATH as a left MATH-module. The image of MATH with respect to MATH is the functional which equals one on MATH and zero on all other elements of the form MATH. This implies that MATH are linearly independent in MATH and hence form a basis of this left MATH-module. It also implies that MATH maps this basis to a basis of MATH and therefore MATH is an isomorphism. |
math/0107109 | It follows from a direct computation and REF . |
math/0107109 | Follows from REF . |
math/0107109 | Follows immediately from REF . |
math/0107109 | We have to compute the pairing of MATH with MATH and show that it is MATH for MATH, MATH and zero otherwise. By REF we have: MATH where MATH . We can choose our representation MATH such that MATH are of the form MATH and, in particular, MATH are in MATH. Then, REF depends only on the class of MATH in MATH where MATH is the ideal generated by MATH for MATH. In this quotient ring we have: MATH . This easily implies that REF is non-zero if and only if MATH and MATH. |
math/0107109 | Since operations MATH form a basis we can write MATH as a formal linear combination MATH. Since the weight of MATH is zero this implies that MATH for MATH. Since MATH, REF implies that MATH. |
math/0107109 | We have to show that MATH that is, that the only monomials which pair non-trivially with MATH are MATH and MATH and that for those monomials the pairing gives MATH. Using REF we see that it is sufficient to show that the only monomials MATH such that MATH appears in the decomposition of MATH are MATH and MATH and that for those monomials MATH appears with coefficient MATH. The later follows immediately from REF . To prove the former note that the question of whether or not MATH appears in the expression for MATH depend only on the class of MATH in MATH where MATH is generated by MATH for MATH. In this quotient MATH and MATH. This shows that the only way to get MATH is to consider MATH or MATH. |
math/0107109 | It follows in the standard way from REF . |
math/0107109 | Since MATH is quasi-projective the NAME trick (see CITE) shows that there exist an affine scheme MATH and a MATH-weak equivalence MATH. Therefore, we may assume that MATH is affine. By CITE[] and REF , any element of MATH is of the form MATH for a line bundle MATH. Since MATH is affine there is a map MATH for some MATH such that MATH. On the other hand the reduction of MATH modulo MATH is MATH where MATH is the standard morphism. This proves the lemma. |
math/0107109 | By REF it is sufficient to prove our statement for MATH and MATH. In this case our result follows from REF . |
math/0107109 | Using standard argument we can reduce the problem to the case MATH and MATH. The restriction of MATH to MATH is zero. Hence, MATH for some MATH. The restriction of MATH to MATH is MATH. Hence MATH. Since MATH is not a zero divisor, we conclude that MATH. |
math/0107109 | Consider the standard projection MATH. As shown in REF, it defines a monomorphism on motivic cohomology. Together with REF this immediately implies that MATH if MATH for some MATH. As shown in the proof of REF we have MATH. Hence, by REF we have MATH . Since MATH is a monomorphism we conclude that MATH. |
math/0107110 | In REF below we show that the statement holds for MATH and a vector bundle MATH on MATH of dimension MATH. Let MATH be a closed embedding and MATH the normal bundle to MATH. Consider the composition MATH . If MATH is the canonical map and MATH is a class in MATH then by REF one has MATH . Let MATH be the tautological class in MATH, then MATH where the second equality holds by REF and the third one by REF . Finally, in MATH we have MATH and therefore, MATH . |
math/0107110 | Let MATH where MATH is the tangent bundle on MATH and MATH is its dual. The dimension of MATH is MATH. MATH . We have to show that MATH. Consider the standard exact sequence MATH (see CITE) and its dual MATH . The first sequence implies that MATH, hence MATH . Form the second sequence MATH, hence MATH . Consider the incidence hyperplane MATH in MATH where the first projective space is thought of as the projective space of a vector space MATH and the second one as the projective space of MATH. The complement MATH considered as a scheme over MATH by means of a projection MATH is an affine bundle over MATH. On the other hand, the NAME embedding MATH gives MATH as the divisor at infinity for an appropriate choice of the intersecting hyperplane MATH. Therefore, MATH is an affine variety. This construction is known as the NAME trick (see CITE). Consider the fiber product: MATH . The open embedding MATH defines a section MATH of MATH. Let MATH be the normal bundle to the NAME embedding MATH and let MATH be the normal bundle to MATH. There is an integer MATH and an isomorphism of vector bundles MATH on MATH. Since MATH is affine, two vector bundles give the same class in MATH if and only if they become isomorphic after the addition of MATH for some MATH. Therefore, to prove the lemma it is sufficient to show that one has MATH in MATH. Let MATH be the second of the two projections. One can easily see from the definition of MATH that MATH. Hence MATH is the restriction to MATH of the tangent bundle on MATH. On the other hand the short exact sequence which defines MATH shows that over MATH we have MATH which finishes the proof of the lemma. There exists a map MATH such that the homomorphism MATH defined by MATH, MATH and the NAME isomorphisms coincides with the degree map. Consider the pointed sheaf MATH . There is an obvious map MATH from MATH to MATH. We claim that MATH is isomorphic in MATH to MATH and that the composition of MATH with this isomorphism satisfies the condition of the lemma. The same reasoning as in the proof of the homotopy purity theorem in CITE, shows that in MATH one has MATH . Consider again the pull-back square REF . The map MATH is clearly a MATH-weak equivalence. On the other hand one verifies easily that MATH is contained in MATH and that this embedding is a MATH-weak equivalence. Hence we have a weak equivalence from MATH to MATH . The later quotient is isomorphic to MATH and since the normal bundle to MATH in MATH is MATH we conclude by CITE that it is weakly equivalent to MATH. This finishes the construction of the map MATH. It remains to check that for MATH we have MATH. Consider the following diagram MATH where the arrows going up are weak equivalences. One verifies immediately that it commutes. Denote the upper horizontal map of the diagram by MATH. Since this map coincides with MATH where MATH is the embedding of a point (see CITE), REF shows that MATH where MATH is the tautological class in MATH. Therefore, it it sufficient to show that for a class MATH we have MATH . Let MATH denote the map MATH defined by MATH. Then MATH . The commutative diagram MATH implies further that MATH and therefore MATH . The equality REF follows now from REF . We are ready now to finish the proof of REF . We take MATH and MATH. The map MATH defines a map of NAME spaces MATH. The isomorphism REF defines an isomorphism MATH . Composing these maps with the m-fold suspension of MATH we get a map MATH . For MATH we have MATH which shows that MATH satisfies the condition of the proposition. |
math/0107110 | Follows easily from the long exact sequences in the motivic cohomology defined by the short exact sequence MATH. |
math/0107110 | For MATH our condition means that MATH has a rational point over a separable extension of MATH of degree MATH where MATH. By REF this implies that MATH and by REF we conclude that MATH. Assume that MATH. Set MATH. Let MATH be a vector bundle on MATH and MATH a map satisfying the conclusion of REF . Consider the cofibration sequence MATH corresponding to MATH. The long exact sequence of motivic cohomology corresponding to this cofibration sequence shows that ther exists a unique class MATH in MATH whose restriction to MATH is the NAME class MATH. Multiplication with this class gives us a map MATH . Consider the map MATH. We claim that this map is a weak equivalence. Indeed, it is a part of a cofibration sequence and to verify that it is a weak equivalence it is enough to check that MATH is contractible. This follows from the cofibration sequence MATH and REF . Since MATH is a weak equivalence, REF defines a map MATH . We claim that MATH is something like a contracting homotopy for the complex MATH. More precisely we have the following lemma which clearly imply the statement of the theorem. There exists MATH such that for any MATH one has MATH . Let MATH be the pull-back of the tautological motivic cohomology class on MATH with respect to MATH. Since the map MATH is a weak equivalence it is sufficient to verify that there exists MATH such that MATH . Let us show that for MATH we have MATH. Indeed, the restriction of MATH to MATH is zero by CITE for any MATH. On the other hand for MATH the restriction map is a mono since there are no motivic cohomology of negative weight. This fact together with CITE implies that we have MATH and therefore MATH . We now have two possibilities. If MATH has a point over an extension of degree prime to MATH then MATH and the statement of our lemma obviously holds. Assume that MATH has no points over extensions of degree prime to MATH. Let us show that under this assumption MATH for MATH. Note first that since MATH has no points over extensions of degree prime to MATH, MATH does not have points over such extensions either. Since MATH restricts to zero on MATH by CITE it is sufficient to show that MATH. By CITE we have MATH. Since MATH is a reduction of an integral class we have MATH and to show that it is non-zero we have to check that MATH can not be lifted to cohomology with MATH-coefficients. If it could there would be a lifting MATH of MATH to the cohomology with MATH-coefficients such that MATH. Our condition that MATH has no points over extensions of degree prime to MATH implies, by REF , that the value of MATH does not depend on the choice of MATH. On the other hand by CITE we know that MATH is the reduction of the integral class MATH which we assumed is non-zero mod MATH. |
math/0107110 | For the hypersurface given by a generic section of a vector bundle MATH the normal bundle is canonically isomorphic to MATH. In particular we have a short exact sequence of the form MATH . The tangent bundle on MATH fits into an exact sequence of the form MATH . Therefore, in MATH we have MATH . Since MATH is an additive characteristic class which on line bundles is given by MATH we get for MATH where MATH. By REF we conclude that MATH . |
math/0107110 | We are going to show that if MATH has a rational point over MATH then MATH is divisible by MATH in MATH. Since any variety has a pojnt over its function field this implies that MATH is divisible by REF in the generic point of MATH. Let MATH denote the quadric given by the equation MATH. For any MATH the following two conditions are equivalent CASE: MATH has a MATH-rational point CASE: MATH has a MATH-rational point The first condition implies the second one because the form MATH is a subform in MATH and therefore MATH is a subvariety in MATH. Assume that the second condition holds. By CITE it implies that the form MATH is hyperbolic. Hence, for any rational point MATH of MATH there exists a linear subspace MATH of dimension MATH which lies on MATH and passes through MATH. The quadric MATH is a section of MATH by a linear subspace MATH of codimension MATH in MATH. The intersection of MATH and MATH is a rational point on MATH. To prove REF we proceed by induction on MATH. Consider first the case MATH. Then MATH is given by the equation MATH. We may assume that it has a point of the form MATH (otherwise MATH is a square root in MATH and the statement is obvious). Then MATH is the norm of the element MATH from MATH and thus the symbol MATH is divisible by REF. Suppose that the proposition is proved for sequences MATH of length smaller than MATH. The quadric MATH is given by the equation MATH . The form MATH is of the form MATH. By induction we may assume that our point MATH belongs to the affine part MATH. Consider the plane MATH generated by points MATH and MATH. The restriction of the quadratic form MATH to MATH is of the form MATH for some MATH (the idea is that MATH is a ``subfield" in the vector space where MATH lives). Consider the field extension MATH. The form MATH and therefore the form MATH represents zero over MATH and thus by the inductive assumption MATH in MATH. On the other hand by the construction MATH represents MATH and therefore we have MATH which proves the proposition. |
math/0107110 | Denote by MATH the category of NAME motives over MATH and let MATH be the n-th NAME motive in this category. The proof of the theorem is based on the following important result. There exists a direct summand MATH of MATH in MATH together with two morphisms MATH such that CASE: the composition MATH is the morphism defined by the projection MATH CASE: for any field MATH over MATH where MATH has a point the pull-back of the sequence MATH to MATH is split-exact. See CITE, CITE. The NAME moving lemma for families of cycles shows that for any MATH there is a functor from the category of NAME motives over MATH to MATH (it is shown in CITE that this functor is a full embedding if MATH is perfect). Therefore, the NAME motive MATH can be also considered in MATH where it is a direct summand of the motive MATH of the norm quadric. To show that it fits into a distinguished triangle of the form REF we need the following lemmas. Let MATH be a perfect field, MATH an object of MATH and MATH a smooth variety over MATH such that for any generic point MATH of MATH the pull-back of MATH to the residue field MATH is zero. Then one has MATH . Since MATH is perfect, the group MATH is the hypercohomology group of MATH with coefficients in the complex of sheaves with transfers with homotopy invariant cohomology sheaves MATH which represents MATH. Our condition on MATH and MATH implies easily that the cohomology sheaves MATH of MATH vanish on the generic points of schemes etale over MATH. Since MATH are homotopy invariant sheaves with transfers we conclude by CITE that they vanish on all schemes etale over MATH. Therefore, MATH. To prove that MATH it is sufficient to show that the class of objects MATH such that MATH for all MATH, contains MATH. Since this class is a localizing subcategory, it is sufficient to show that for any smooth MATH over MATH and any MATH one has MATH . If MATH and MATH satisfy the conditions of the lemma then for any MATH and any MATH, MATH and MATH satify these conditions. Therefore, REF follows from the first assertion of the lemma. Let MATH be a perfect field. Then the sequence of MATH is split-exact. Observe first that MATH is a split epimorphism. Indeed, REF implies that the morphism MATH where the first arrow is defined by the diagonal and the second by the projection MATH is a section of MATH. It remains to show that REF extends to a distinguished triangle. Let MATH be a cone of the morphism MATH. Since there are no motivic cohomology of negative weight, the morphism MATH factors through a morphism MATH. Let MATH be the a cone of MATH. Standard properties of triangles in triangulated categories imply that the sequence REF extends to a distinguished triangle if and only if MATH. It follows from REF . Let MATH be a smooth scheme over MATH and MATH an object of the localizing subcategory generated by MATH. Then one has CASE: the morphism MATH is an isomorphism CASE: the homomorphism MATH is an isomorphism. It is clearly sufficient to prove the lemma for MATH. In this case the first statement follows immediately from REF and the fact that MATH takes simplicial weak equivalences to isomorphisms. Let MATH be the cone of the morphism MATH. To prove the second statement we have to show that any morphism MATH is zero. The morphism MATH can be written as the composition MATH which is zero because the first part of the lemma implies that MATH . (Proof of REF continues) The morphism MATH has a canonical lifting to a morphism MATH by REF . Together with the composition MATH this lifting gives us a sequence of morphisms MATH . The composition MATH is zero by REF and the fact that MATH . Let MATH be a cone of MATH. The morphism MATH factors through a morphism MATH and we have to show that MATH is an isomorphism. The category MATH of objects MATH such that MATH is an isomorphism is a localizing subcategory. By REF , the morphism MATH is an isomorphism and we conclude by REF that MATH contains MATH and therefore it contains MATH. On the other hand we have a commutative diagram MATH with both vertical arrows are isomorphisms by REF . This finishes the proof of REF . |
math/0107110 | Since MATH any purely inseparable extension of MATH is of odd degree and the transfer argument shows that it is sufficient to consider the case of a perfect MATH. Denote by MATH the sheaf on MATH such that for a connected smooth scheme MATH over MATH the group MATH is the subgroup in the n-th NAME of the function field of MATH which consists of elements MATH such that all residues of MATH in points of codimension MATH are zero. The proof is based on the following result. The natural homomorphism MATH is a monomorphism. See CITE. The following lemma shows that the cohomology group on the left hand side of REF can be replaced by a motivic cohomology group. Let MATH be a smooth scheme over a field MATH. Then for any MATH there is a canonical homomorphism MATH which is an isomorphism if MATH. Considering MATH as a limit of smooth schemes over the subfield of constants of MATH we may assume that MATH is perfect. Let MATH denote the cohomology sheaves of the complex MATH. Since MATH for MATH the standard spectral sequence which goes from cohomology with coefficients in MATH and converges to the hypercohomology with coefficients in MATH shows that there is a canonical homomorphism MATH . The same spectral sequence implies that the kernel and cokernel of this homomorphism are bulit out of groups MATH and MATH respectively, where MATH. Since MATH we get MATH and the cohomological dimension theorem for the NAME topology implies that these groups are zero. It remains to show that MATH. Since MATH is a homotopy invariant sheaf with transfers for any smooth connected MATH the restriction homomorphism MATH is a monomorphism CITE. It was shown in CITE that for any field MATH one has canonical isomorphisms MATH. In particular for any MATH the group MATH is a subgroup in MATH and one verifies easily that it coincides with the subgroup MATH of elements with zero residues. By REF and the suspension isomorphism (see CITE) we have an exact sequence MATH . Since the dimension of MATH equals MATH, the left hand side group in REF is isomorphic to the group MATH by REF and we obtain a natural monomorphism MATH . Let MATH be the lifting of MATH to the algebraic closure of MATH. Since MATH is a direct summand of MATH we conclude that the map MATH is injective. Since MATH we conclude that the second arrow in REF is zero. On the other hand since MATH we have MATH . |
math/0107110 | In order to prove the proposition we will need to do some preliminary computations. Fix an algebraic closure MATH of MATH. Since MATH has no extensions of degree prime to MATH there exists a primitive root of unit MATH of degree MATH. Let MATH be a cyclic extension of MATH of degree MATH. We have MATH where MATH for an element MATH in MATH. Denote by MATH the generator of the NAME group MATH which acts on MATH by multiplication by MATH and by MATH the class in MATH which corresponds to the homomorphism MATH which takes MATH to the canonical generator of MATH (one can easily see that this class is determined by MATH and MATH and does notdepend on MATH). Let MATH be the projection. Consider the etale sheaf MATH. The group MATH acts on MATH in the natural way. Denote by MATH the kernel of the homomorphism MATH. One can verify easily that MATH and that as a MATH-module MATH has dimension MATH. In particular we have MATH. Note that the extension MATH represents the element MATH in MATH. Let MATH be the element in MATH defined by the exact sequence MATH where MATH and MATH. One has MATH where MATH is an invertible element of MATH and MATH for MATH. The fact that MATH for MATH follows from the commutativity of the diagram MATH . To compute MATH note first that since the action of MATH on MATH factors through MATH it comes from a well defined element in MATH. This element is not zero for trivial reasons. On the other hand the group MATH is generated by the element MATH where MATH is the canonical generator of MATH and MATH is the NAME homomorphism. Thus we conclude that up to multiplication by an invertible element of MATH our class MATH equals MATH. It remains to show that MATH which follows by simple explicite computations from the fact that MATH has a lifting to an element of MATH. Assume that MATH holds. Then for all fields MATH of characteristic MATH, all MATH and all MATH one has: CASE: The sequence MATH where the first homomorphism is given on the second summand by MATH is exact. CASE: The homomorphisms MATH given by the canonical morphisms MATH, MATH are surjective. We proceed by induction on MATH. Consider first the case MATH. The first statement follows immediately the assumption that MATH holds. Let us prove the second one. The image of MATH in MATH coincides with the kernel of the norm homomorphism MATH . The first statement implies then that MATH maps surjectively to this image. It is sufficient therefore to show that an element MATH which goes to zero in MATH belongs to the image of MATH. Any such element is a composition of a cohomology class in MATH with the canonical extension MATH . Thus we may assume that MATH and MATH is the element which corresponds to this extension. Let MATH be the image of MATH (where MATH is as in REF ) under the homomorphism MATH. The composition MATH where the later homomorphism corresponds to the extension MATH equals to multiplication by MATH. We conclude now by REF that the image of MATH in MATH is zero. Then it lifts to MATH which proves our Lemma in the case MATH. Suppose that the lemma is proved for all MATH. Let us show that it holds for MATH. The first statement follows immediately from the inductive assumption and the commutativity of the diagram MATH . The proof of the second one is now similar to the case MATH with a simplification due to the fact that MATH for MATH REF . The statement of the proposition follows immediately from REF . |
math/0107110 | It is essentially a version of the proof given in CITE for MATH and in CITE for MATH. Let us define a homomorphism MATH as follows. Let MATH be an element in MATH of the form MATH and let MATH be an element in MATH such that MATH . We set MATH. Since MATH holds the element MATH does not depend on the choice of MATH and one can easily see that MATH is a homomorphism from MATH to MATH. To show that it is a homomorphism from MATH it is sufficient to verify that MATH takes an element of the form MATH such that say MATH to zero. Let MATH be a preimage of MATH in MATH. We have to show that MATH. Assume first that MATH is not in MATH and let MATH be an element in MATH such that MATH. Let further MATH. Then by MATH one has MATH since MATH. The proof for the case when MATH is similar. Clearly MATH is a section for the obvious morphism MATH . It remains to show that it is surjective. It follows immediately from the fact that under our assumption on MATH the group MATH is generated by symbols of the form MATH where MATH and MATH (see CITE). |
math/0107110 | The MATH part follows from the projection formula. Since MATH holds we conclude that there is a commutative square with surjective horizontal arrows of the form MATH and thus the cokernel of the left vertical arrow is the same as the cokernel of the right one. By REF it gives us an exact sequence MATH and since the last arrow clearly factors through MATH it is zero. Lemma is proved. |
math/0107110 | This proof is a variant of the proof given in CITE for MATH. Since MATH has no extensions of degree prime to MATH it is separable and its NAME group is an l-group. Therefore it is sufficient to prove the lemma in the case of a cyclic extension MATH of degree MATH. By REF we have an exact sequence MATH . Let MATH be an element in MATH and let MATH be an element such that MATH. Then MATH and we conclude that the endomorphism MATH is surjective. Since MATH this implies that MATH. |
math/0107110 | Let MATH be an element of MATH. We have to show that MATH. By REF and obvious induction we may assume that MATH vanishes on a cyclic extension of MATH. Then by REF it is of the form MATH where MATH is the element which represents our cyclic extension. Thus since MATH holds it belongs to the image of the homomorphism MATH and therefore is zero. |
math/0107110 | We will use the following two lemmas. Let MATH be a field of characteristic not equal to MATH. Then for any MATH and any MATH-module MATH, the complex MATH has homotopy invariant cohomology sheaves. Since the sheaf associated with a homotopy invariant presheaf with transfers is homotopy invariant (CITE and CITE) it is sufficient to show that the functors MATH are homotopy invariant that is, that for any smooth MATH the homomorphism MATH given by the restriction to MATH is an isomorphism. Consider the universal coefficients long exact sequence relating NAME motivic cohomology with coefficients in MATH to NAME motivic cohomology with coefficients in MATH and MATH. The cohomology with MATH-coefficients is homotopy invariant by REF . The cohomology with MATH-coefficients is isomorphic to the etale cohomology by REF and therefore homotopy invariant as well. Our claim follows now from the five lemma. The canonical homomorphisms MATH are isomorphisms. It is sufficient to show that for any MATH and any smooth scheme MATH one has MATH . Since the MATH are sheaves with transfers it is a particular case of CITE. REF implies that MATH is a homotopy invariant sheaf with transfers and by the assumption that MATH holds we know that it vanishes over fields. Therefore, by CITE we conclude that MATH. In order to show that MATH is quasi-isomorphic to zero it remains to verify that for any smooth scheme MATH over MATH and any MATH the homomorphism MATH is an isomorphism. REF and the universal coefficients long exact sequence imply that it is sufficient to verify that the homomorphisms MATH are isomorphisms for MATH. Since we assume MATH, we conclude that the group MATH is infinitely divisible and using the trick of CITE we conclude that the map REF is surjective for MATH and MATH where MATH is a field. By the analog of CITE for MATH-coefficients and the comparison between the higher NAME groups and the etale cohomology we conclude that REF is an isomorphism. |
math/0107110 | We have to show that the complex MATH is canonically quasi-isomorphic in the etale topology to the sheaf MATH. In the category of complexes of sheaves with transfers of MATH-modules in the etale topology with homotopy invariant cohomology sheaves MATH we have MATH . Since MATH in the etale topology we have MATH which proves our claim. |
math/0107110 | By REF the homomorphisms MATH are isomorphisms for MATH. By REF , for MATH and MATH the homomorphism REF is isomorphic to the homomorphism REF . |
math/0107110 | One verifies easily that this complex becomes exact after tensoring with MATH. It remains to show that it becomes eact after tensoring with MATH. Recall that for a smooth scheme MATH over MATH we let MATH denote the free sheaf with transfers generated by MATH. Consider the complex of presheaves with transfers of the form MATH where the second arrow is the transfer map. Denote this complex with the last MATH placed in degree zero by MATH. One can easily see that it is exact in the etale topology and therefore MATH where MATH is the category of etale sheaves with transfers. Since MATH the map MATH is injective and we conclude by REF that MATH . We have MATH and in particular for a separable extension MATH of MATH we have MATH . Our result follows now from REF and the standard spectral sequence which computes morphisms in a triangulated category from a complex in terms of morphisms from its terms. |
math/0107110 | For MATH the theorem is proved in CITE. Assume that MATH. By induction on MATH we may assume that MATH for all MATH. Let MATH be a field which has no extensions of degree prime to MATH and such that MATH is REF-divisible. By REF our inductive assumption implies that MATH holds. By REF we conclude that MATH . Together with REF this shows that the group MATH is torsion free. On the other hand MATH by REF and we conclude that MATH. For a finite extension MATH of degree prime to MATH the homomorphism MATH is a monomorphism by the transfer argument. Thus in order to prove MATH it is sufficient to show that for any element MATH there exists an extension MATH such that MATH is divisible by MATH in MATH and the homomorphism MATH is a monomorphism. Since any element in MATH is a sum of symbols it is sufficient to construct MATH for MATH of the form MATH. Let us show that the function field MATH of the norm quadric has the required properties. The fact that MATH becomes divisible by MATH in the NAME of MATH proved in REF . It remains to show that the map REF is injective. We do it in two steps - first we prove that the kernel of REF is covered by the group MATH and then that the later group is zero. For the first step we will need the following three lemmas. Let MATH be a non empty smooth scheme over MATH. Then the homomorphisms MATH defined by the morphism MATH are isomorphisms for all MATH. By definition, (see REF) we can rewrite the homomorphism REF as the homomorphism MATH where the morphisms are in the derived category of sheaves of abelian groups in the etale topology on MATH. We claim that the morphism of complexes MATH is a quasi-isomorphism. To check this statement we have to verify that for any strictly henselian local scheme MATH the map of complexes of abelian groups MATH is a quasi-isomorphism. Since MATH is non-empty there exists a morphism MATH and therefore, the simplicial set MATH is contractible (see the proof of REF ). Therefore we have MATH . Let MATH be the complex of sheaves on MATH defined in REF. Let MATH be a field of characteristic not equal to MATH and assume that MATH holds. Then for any smooth scheme MATH the map MATH defined by the projection MATH, is an isomorphism. Recall from CITE that for a functor MATH from schemes to abelian groups we denote by MATH the functor MATH where the map is defined by the projection. To prove the lemma we have to show that MATH. This is clearly equivalent to checking that the standard map MATH is an isomorphism. For a complex of sheaves with transfers MATH there is a complex MATH (defined up to a canonical quasi-isomorphism) such that MATH. By CITE, if MATH is a complex of sheaves with transfers with homotopy invariant cohomology sheaves MATH then MATH . Therefore, it is sufficient to check that the maps MATH are isomorphisms. Since both sides are zero for MATH and since MATH are the sheaves associated with the presheaves MATH it remains to check that the maps MATH are isomorphisms for MATH. In this range the map MATH is an isomorphism and therefore it remains to check that the map MATH is an isomorphism for MATH. Consider the commutative diagram MATH where the vertical arrows are defined by the multiplication with the canonical class MATH. The upper horizontal arrow is an isomorphism by our assumption that MATH holds and REF . The left hand side vertical arrow is an isomorphism by the supension theorem CITE. It remains to check that the right hand side vertical arrow is an isomorphism. This we can verify separately for rational coefficients and MATH-coefficients. In the former case the result follows from REF and again CITE. In the later case it follows from REF and the corresponding result for the etale cohomology. Assume that MATH holds and let MATH be a field of characteristic not equal to MATH, MATH a smooth scheme over MATH and MATH a dense open subscheme in MATH. Then the map MATH is an isomorphism. Considering MATH to be a limit of smooth schemes (possibly of greater dimension) over the subfield of constants in MATH we may assume that MATH is perfect. By obvious induction it is sufficient to show that the statement of the lemma holds for MATH where MATH is a smooth closed subscheme in MATH. Moreover, one can easily see that it is sufficient to prove that for any point MATH of MATH there exists a neighborhood MATH of MATH in MATH such that MATH is an isomorphism. This follows from the easiest case of the NAME distinguished triangle CITE or from the homotopy purity theorem CITE and REF . (Proof of REF continues) Let MATH be an element in the kernel of REF . By REF it is sufficient to show that the image MATH of MATH in MATH belongs to the image of the group MATH that is, that under our assumptions the image of MATH in the hypercohomology group MATH is zero. Since MATH becomes zero in the generic point of MATH there exists a nonempty open subscheme MATH of MATH such that the restriction of MATH to MATH is zero. By REF we conclude that the restriction of MATH to MATH is zero. The canonical morphism MATH factors through the morphism MATH which is a part of the distinguished triangle of REF . Since MATH has a section, our class MATH becomes zero on MATH and by REF we conclude that it belongs to the image of the group MATH. Since MATH this group is zero by REF . It remains to show that MATH. Consider the pointed simplicial scheme MATH defined by the cofibration sequence MATH . Since MATH for MATH the homomorphisms MATH defined by the third arrow of this sequence are isomorphisms for MATH. Thus it is sufficient to verify that MATH. Since there exists an extension of degree two MATH such that MATH . REF implies that all the motivic cohomology groups of MATH have exponent at most MATH. Thus it is sufficient to show that the image of MATH in MATH is zero. Let MATH be an element of this image. Consider the composition of cohomological operations MATH. It maps MATH to an element of MATH . Let MATH be a class in MATH which the image of an integral class. Then MATH is the image of an integral class for any MATH. By CITE we have MATH. Since MATH is the image of an integral class we have MATH and MATH. On the other hand, the NAME homomorphism MATH can be written as the composition MATH where the first map is the connecting homomorphism for the exact sequence MATH and the second map is the reduction modulo MATH. Therefore, any element of the form MATH is the image of an integral class. REF implies that this element belongs to the image of the corresponding integral motivic cohomology group. Therefore by REF it is zero. It remains to verify that the composition MATH is a monomorphism that is, that the operation MATH acts monomorphically on the group MATH for MATH. By REF , the motivic NAME homology of MATH are zero for MATH with MATH. Therefore, the kernel of MATH on this group is covered by the image of MATH. The later group is zero by the inductive assumption that MATH holds, REF . |
math/0107110 | Since MATH, MATH is the image of an element MATH in MATH with respect to the NAME homomorphism MATH. By REF there exists a dense open subset MATH of MATH such that MATH on MATH is in the image of the canonical map MATH. Since this map factors through the integral cohomology group we have MATH on MATH. |
math/0107110 | Let MATH be the subcomplex of MATH such that MATH . Since the complexes MATH are bounded the usual argument shows that MATH for MATH and all MATH. On the other hand one has an exact sequence of complexes MATH which expresses the fact that MATH. This short exact sequence defines a long exact sequence of groups of morphisms in the derived category which shows that MATH for MATH. |
math/0107110 | Let MATH be a smooth scheme over MATH. Then the simplicial set MATH is the simplex generated by the set MATH which is contractible if and only if MATH. This implies immediately that for any MATH the map of simplicial sets MATH is a weak equivalence. |
math/0107110 | Let MATH be an extension of MATH and MATH considered as a smooth scheme over MATH. Then there are homomorphisms MATH and MATH and the composition of the first one with the second is multiplication by MATH. Let MATH be an extension of degree MATH such that MATH. Since MATH the pointed sheaf MATH is contractible and therefore, MATH . We conclude that MATH. |
math/0107111 | We begin by making the identification MATH in the process make a choice of generators MATH, MATH for the MATH copies of MATH. Now recall that there are self-diffeomorphisms of MATH which act trivially on MATH, but for which MATH for any desired choice of signs. Thinking of MATH as the MATH-fold symplectic blow-up of MATH, and then moving the symplectic structure via these diffeomorphism, we thus obtain MATH distinct symplectic structures on MATH, with first NAME classes MATH for all possible choices of signs. Applying REF to MATH and MATH, we thus conclude that all the classes of the form MATH are monopole classes, where MATH . Now, given any particular metric MATH on MATH, let us make a new choice MATH of generators for our MATH copies of MATH in such a way that MATH . The resulting monopole class MATH then satisfies MATH REF therefore tells us that any metric MATH on MATH satisfies MATH . (The inequality is strict because MATH, and this guarantees that MATH is certainly not the first NAME class of a symplectic structure.) Now for any metric MATH on our compact orientable MATH-manifold MATH, we have the NAME type formula CITE MATH where MATH is the traceless NAME tensor. If MATH is NAME, MATH, and REF then becomes MATH . For MATH, we have MATH and MATH . REF therefore asserts that a necessary condition for the existence of an NAME metric on MATH is that MATH or, in other words, that MATH . By contraposition, this shows that there cannot be an NAME metric if MATH exactly as claimed. For MATH, we instead have MATH and MATH so that REF instead tells us that a necessary condition for the existence of an NAME metric on MATH is that MATH or in other words that MATH . We thus conclude that there cannot be an NAME metric if MATH and this finishes the proof. |
math/0107111 | Set MATH, and consider the simply connected MATH-manifolds MATH . Each of these smooth oriented MATH-manifolds is homeomorphic to MATH . However, since MATH and MATH all reduce to MATH, and since MATH . REF asserts that none of these smooth manifolds MATH can admit an NAME metric. Now, for any fixed MATH, observe that the sequence MATH must contain infinitely many different diffeotypes. Indeed, REF asserts that MATH and MATH are both monopole classes on MATH. However, the difference MATH is divisible by MATH, and the bandwidth of MATH, as defined in REF , therefore satisfies MATH . Thus, for any fixed MATH, MATH and it follows that no individual MATH has maximal bandwidth, for any given MATH. Hence MATH runs through infinitely many different diffeotypes for each MATH, and the claim follows. |
math/0107116 | We consider the reflection tiling of MATH corresponding to MATH and MATH. If MATH is a codimension-MATH cell in this tiling, then there are exactly MATH maximal cells containing MATH, and the stabilizer MATH is isomorphic to MATH. If MATH is torsion-free, then the natural map MATH must be injective; hence, the index of MATH is at least MATH. If the index is equal to MATH, then for any MATH-cell MATH, the map MATH is a bijection. We let MATH be the composition MATH . It is not a priori a homomorphism, but since the defining relations for MATH also hold for the images MATH, there is an induced epimorphism MATH defined by MATH, for all MATH. It is clear that if MATH and MATH agree on all but one element in the stabilizer of a cell, then they must also agree on the entire stabilizer. Using this fact and the fact that MATH and MATH agree on the generators of MATH, it follows that MATH. In particular, MATH is an epimorphism and MATH is its kernel. That MATH is a characteristic function follows from the fact that MATH must be a bijection for every MATH. |
math/0107116 | The stabilizer MATH fixes MATH pointwise and is mapped isomorphically by MATH onto a codimension-MATH subspace MATH (where MATH). Each facet of MATH can be expressed uniquely as the intersection MATH where MATH is a facet of MATH that is orthogonal to MATH, and we let MATH denote the subgroup of MATH generated by the corresponding reflections MATH. The characteristic function MATH induces a characteristic function MATH for the polytope MATH with kernel MATH. It follows that MATH. |
math/0107116 | It is clear that equivalent small covers are isomorphic as cell complexes. Conversely, suppose MATH is an isomorphism of cell complexes. Then the lift MATH is an automorphism of the tiling of MATH such that MATH. Since the automorphism group of the tiling is a semi-direct product of MATH and MATH, there exists a MATH such that MATH for some element MATH. It follows that MATH (they are both normal), hence MATH for some MATH. |
math/0107116 | Let MATH be a connecting path with endpoints lying on the (disjoint) facets MATH and MATH. Because MATH is regular, it has a ``cubical" decomposition obtained by cutting each edge with the orthogonal hyperplane through its midpoint (see REF ). We define two open neighborhoods MATH and MATH of the facet MATH as follows. MATH is the union of the interiors of all cubes that intersect MATH, and MATH is the set of all points of MATH whose distance to MATH is less than MATH. We claim that MATH. To see this, it suffices to note that if MATH is an edge meeting MATH othogonally at a vertex and MATH is the hyperplane bisecting MATH, then MATH lies entirely on one side of MATH (since the edge MATH, being a common perpendicular to MATH and MATH, realizes the shortest distance between the corresponding hyperplanes). Similarly, if MATH and MATH are the corresponding neighborhoods of MATH, we have MATH. Since MATH and MATH are not adjacent, MATH and, hence, MATH. This means MATH and since the edge MATH is the unique path of minimal length joining MATH to MATH, the equality MATH is only possible if MATH is an edge of MATH. |
math/0107116 | First we show that if MATH is any small cover of MATH, then any closed geodesic of minimal length must be an edge loop. Let MATH be a closed geodesic in MATH, and let MATH be a codimension-one cell in MATH that intersects MATH transversely. Lifting MATH to the universal cover MATH, we obtain a geodesic segment MATH that connects two lifts MATH and MATH of the cell MATH (see REF ). Let MATH and MATH be the hyperplanes spanned by MATH and MATH, respectively. Since MATH and MATH are both mapped to the face MATH under the projection MATH, they are hyperparallel, thus have a (unique) common perpendicular MATH. Moreover, MATH must pass through at least two copies of the tile MATH. It follows from REF that the length of MATH is MATH with equality holding only if MATH is the lift of an edge. Since the geodesic MATH is at least as long as MATH with equality holding only if MATH, the closed geodesic MATH will have minimal length only if it is an edge loop. Now suppose MATH is an isometry. Since MATH takes minimal length closed geodesics to minimal length closed goedesics, it must take edge loops to edge loops. The MATH-cells in a small cover can be characterized as the points where edge loops intersect, thus MATH must take MATH-cells to MATH-cells and, therefore, MATH-cells to MATH-cells. Since any cell of dimension MATH in a small cover can be characterized as the convex hull of its bounding MATH-cells, the isometry MATH must take cells to cells. Thus, by REF , MATH and MATH are equivalent. |
math/0107117 | MATH is homeomorphic to the topological union MATH of MATH discs, with a band glued between MATH and MATH for every MATH. |
math/0107117 | The `only if' part is trivial. NAME, it suffices to prove that each connected edge-ordered graph MATH with MATH vertices and MATH edges can be transformed, by using elementary moves and their inverses, into a canonical form dependent only on MATH. Let us denote by MATH the cardinalities of the non-trivial orbits generated by any permutation of MATH and let MATH for each MATH, then we can choice as a canonical representative of MATH the permutation MATH given by the product MATH . On the other hand, there exists a numbering MATH of the vertices of MATH, such that MATH, where MATH is the transposition associated to the edge MATH, for every MATH. We want to transform MATH by elementary moves, leaving the numbering of the vertices fixed, in such a way that the sequence MATH becomes MATH where the first two rows contain the transposition sequence defining MATH, with additional pairs of consecutive equal transpositions inserted between disjoint cycles, and the last two rows consist of pairs of equal consecutive transpositions. Moreover: REF if MATH is the identity then the first two rows are empty and MATH; REF if MATH then the third row is empty; REF the fourth row contains MATH pairs of transpositions. We proceed by induction on MATH. If MATH, then MATH itself has the required form, in fact the only possibility is MATH and MATH, since MATH is connected. In the rest of the proof, we deal with the inductive step, assuming MATH. To begin with, we show how to perform moves on MATH in order to obtain a sequence MATH of the type MATH, with MATH and MATH for each MATH. First of all, by using REF, it is easy to get a sequence having the form MATH, with MATH and MATH for each MATH. Then, since MATH change the pair MATH with MATH into the pair MATH, we can limit ourselves to consider only the case MATH. If MATH, we have done. Otherwise, if MATH, we have that MATH, by connectedness, and that MATH is even, since MATH can only assume the value MATH or MATH. At this point, we could get the desired form by the sequence of elementary moves MATH if MATH was MATH. So, it remains to show how to obtain MATH without changing the edges MATH with MATH. By connectedness, there exists a chain of edges MATH of minimum length MATH connecting MATH and MATH, with MATH. If MATH, then MATH and we can finish by using REF again. If MATH and MATH, then we can decrease by one the length of the chain by performing the move MATH with MATH. On the other hand, if MATH, then we can reduce by one the difference between MATH and MATH, by performing either MATH if MATH or MATH if MATH. So we can conclude this part of the proof by induction on MATH and MATH. From now on, we assume that the first MATH edges of MATH do not contain MATH and that all the last MATH edges of MATH join MATH and MATH. If MATH, then we have finished (MATH and either MATH or MATH depending on the parity of MATH). Let us consider the case MATH. We denote by MATH the subgraph having vertices MATH and edges MATH. Since MATH is connected and the permutation MATH is of the type requested for MATH, we can apply the induction hypothesis, in order to transform MATH into the canonical form, by a sequence of elementary moves and inverse of them. The same sequence of moves also transforms MATH in a canonical form, possibly except for the presence of more than two transpositions MATH immediately before the MATH transpositions MATH. In fact, the canonical form for MATH contains either one transposition MATH if MATH or two of them if MATH. Hence, to complete the proof, it suffices to change all the MATH exceeding transpositions MATH into MATH. Taking into account that such transpositions are preceded by at least one more MATH and followed by MATH and that their number MATH is even, we can realize the wanted change by the sequence of elementary moves MATH. |
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