paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0101089 | It is enough to apply REF . |
math/0101089 | It is enough to show that MATH in the NAME metric for every monotone sequence MATH. Let us consider an increasing sequence MATH, and let MATH be the closure of the union of the sets MATH. Then MATH is a continuum contained in MATH and MATH in the NAME metric. By the continuity of the function MATH and by REF we have MATH, which implies MATH (see REF ). Let us consider now a decreasing sequence MATH, and let MATH be the intersection of the sets MATH. Then MATH is a continuum containing MATH and MATH in the NAME metric. The conclusion can be obtained as in the previous case. |
math/0101089 | For every MATH we have MATH . The conclusion follows dividing by MATH and taking the limit as MATH. |
math/0101089 | Let MATH be defined by MATH. By REF MATH is continuous in MATH for a.e. MATH and every MATH. By REF MATH . The equivalence between REF is now obvious. |
math/0101089 | Consider a minimizing sequence MATH. By the Compactness REF , passing to a subsequence, we may assume that MATH converges in the NAME metric to some continuum MATH containing MATH. For every MATH let MATH be a solution of the minimum REF which defines MATH. By REF we conclude that MATH converges strongly in MATH to MATH where MATH is a solution of the minimum REF which defines MATH. By NAME 's REF we have MATH. As MATH, we conclude that MATH. Since MATH is a minimizing sequence, this proves that MATH is a solution of the minimum REF . |
math/0101089 | Let us fix an integer MATH with MATH. From the absolute continuity of MATH we have MATH where the integral is a NAME integral for functions with values in MATH. This implies that MATH where the integral is a NAME integral for functions with values in MATH. As MATH and MATH q.e. on MATH, from the minimality of MATH we obtain, using REF , MATH where MATH . Iterating now this inequality for MATH we get REF with MATH. |
math/0101089 | As MATH is admissible for REF which defines MATH, by the minimality of MATH we have MATH, hence MATH for every MATH. As MATH is absolutely continuous with values in MATH the function MATH is integrable on MATH and there exists a constant MATH such that MATH for every MATH. This implies the former inequality in REF . The latter inequality follows now from REF . |
math/0101089 | By construction MATH is compact. To prove that it is connected, let MATH. Then MATH and, since MATH is the limit of MATH in the NAME metric as MATH along a suitable sequence, there exist sequences MATH, MATH converging to MATH and MATH, respectively. By the monotonicity of MATH, we may assume that MATH, so that MATH and MATH belong to the same continuum MATH. Passing to a subsequence, we may assume that MATH converges in the NAME metric to some continuum MATH. Then MATH converges to MATH, which implies that MATH, so that MATH. As MATH and MATH are connected in MATH by the set MATH, it follows that MATH is connected. The monotonicity of MATH implies that MATH for MATH. To prove the reverse inclusion, let now MATH and MATH. Then MATH and, as MATH is closed, MATH, which implies that MATH. Let us fix MATH. To prove that MATH converges to MATH, by the Compactness REF we may assume that MATH converges in the NAME metric to some compact set MATH as MATH along a suitable subsequence of the sequence considered for MATH, and we have only to prove that MATH. Let us prove the first inclusion. As MATH is closed, it is enough to show that MATH. Let MATH. Then there exists MATH such that MATH. As MATH in the NAME metric, there exists a sequence MATH such that MATH, and hence MATH. As MATH for MATH small enough, by the monotonicity of MATH, we have MATH. Since MATH we conclude that MATH. Let us prove that MATH. Let MATH. By the definition of MATH there exist MATH such that MATH. Then MATH and MATH. This implies MATH, hence MATH. It is easy to see that for MATH where the last equality follows from REF . This implies that MATH if and only if MATH. The equality MATH follows now from REF . The statement about MATH is a consequence of this equality and of the continuity properties of monotone functions. |
math/0101089 | As MATH is a solution of the minimum REF which defines MATH, and MATH strongly in MATH, the conclusion follows from REF . |
math/0101089 | Let us fix MATH and MATH with MATH. Since MATH converges to MATH in the NAME metric as MATH, by REF there exists a sequence of continua MATH, converging to MATH in the NAME metric, such that MATH and MATH as MATH. Let MATH and MATH be solutions of the minimum REF which define MATH and MATH, respectively. By REF MATH strongly in MATH. The minimality of MATH expressed by REF gives MATH, which implies MATH. Passing to the limit as MATH and using REF we get MATH. Adding MATH to both sides we obtain REF . A similar proof holds for REF . The only difference is that now we take MATH and use the fact that MATH, which implies MATH. Passing to the limit as MATH and using REF and NAME 's REF we obtain REF . |
math/0101089 | Let us fix MATH with MATH. Given MATH let MATH and MATH be the integers such that MATH and MATH. Let us define MATH and MATH. Applying REF we obtain MATH with MATH converging to zero as MATH. Since MATH converges to MATH in the NAME metric, by REF for every MATH we have MATH strongly in MATH as MATH. By REF we get MATH . Passing now to the limit in REF as MATH we obtain REF . |
math/0101089 | Let MATH. From the previous lemma we get MATH . On the other hand, by REF we have MATH, and by REF MATH where MATH is a solution of the minimum REF which defines MATH. Therefore MATH . Since there exists a constant MATH such that MATH and MATH for MATH, from REF we obtain MATH which proves that the function MATH is absolutely continuous. As MATH strongly in MATH when MATH, if we divide REF by MATH, and take the limit as MATH we obtain REF . Equality REF follows from REF . |
math/0101089 | Let us fix MATH, with MATH, and a continuum MATH. For MATH we have MATH, and from REF we obtain MATH. As the functions MATH and MATH are continuous, passing to the limit as MATH we get REF . |
math/0101089 | Let MATH. Since MATH in the NAME metric as MATH, and MATH by REF , it follows that MATH as MATH by REF . As the function MATH is continuous, we obtain REF . The proof of REF is analogous. |
math/0101089 | Let us fix MATH. For every MATH let MATH be a solution of the minimum REF which defines MATH. As MATH and MATH, from REF we obtain, adding and subtracting MATH, MATH . As MATH is a solution of REF with MATH, and MATH q.e. on MATH, we have MATH. By the monotonicity of MATH, for MATH we have MATH and MATH q.e. on MATH. By the minimum property of MATH we obtain MATH for MATH. Therefore MATH which concludes the proof. |
math/0101089 | Let MATH and MATH be the connected components of MATH. Since MATH and MATH have a NAME boundary, by REF MATH belongs to MATH and MATH. This implies that MATH, and hence MATH. The conclusion follows now from CITE, as shown in CITE. |
math/0101089 | It is enough to adapt the proof of CITE. |
math/0101089 | Let MATH with MATH, let MATH be a solution of the minimum REF which defines MATH, and let MATH be the function defined by MATH on MATH and by MATH on MATH. As MATH q.e. on MATH the function MATH belongs to MATH; using also the fact that MATH q.e. on MATH, we obtain that MATH q.e. on MATH. Therefore MATH . On the other hand, by the minimality of MATH, MATH where the last inequality follows from REF , since MATH is connected. From REF we obtain MATH and the minimality of MATH yields REF . |
math/0101089 | Let MATH be an arbitrary point in MATH and let MATH, MATH, be a family of open balls centred at the points MATH. If the radii are sufficiently small, we have MATH and MATH for MATH. Moreover we may assume that MATH, for suitable constants MATH with MATH, and that the arcs MATH intersect MATH only at the points MATH and MATH, with a transversal intersection. All these properties, together with REF , imply that MATH . In particular this happens for MATH, and for MATH close to MATH if MATH is continuous at MATH. By REF and by REF for every MATH we have that MATH . By REF this implies, taking MATH, MATH which yields MATH . REF follows now from REF applied with MATH. By REF for a.e. in MATH we have MATH. Moreover, for a.e. in MATH the derivative MATH exists for MATH. Let us fix MATH which satisfies all these properties. By REF for MATH close to MATH we have MATH where MATH and MATH. Note that the equality holds in REF for MATH. As the functions MATH and MATH are differentiable at MATH (by REF and by the existence of MATH), we conclude that MATH . By REF we have MATH for MATH, so that the previous equalities yield REF . |
math/0101092 | CASE: By the definition of the relations, REF is satisfied. CASE: The pair MATH is in the same relation as MATH because MATH. Note that the multiplication by MATH is an element of MATH. It follows that MATH for all MATH. Then we say that the association scheme is symmetric. CASE: For all MATH, MATH and MATH are in the same relation. Thus MATH. If MATH is in the same relation than MATH, then MATH such that: MATH . Then MATH . We conclude that MATH. CASE: Let MATH and let MATH for MATH. Then there is a MATH such that for any section MATH we have that MATH . Let MATH be defined as MATH. Then MATH. We will prove that MATH. For every MATH we can check that MATH and also that MATH. This means that MATH. Then MATH is a function from MATH to MATH. We still need to show that this function is injective. Let MATH and let us assume that MATH . CASE: By the symmetry of the association scheme we deduce that MATH. |
math/0101092 | Suppose that MATH is not a prime in MATH, that is, MATH. Then it is clear that the equivalence in MATH given by the quotient MATH is MATH-invariant, and therefore the action is imprimitive, hence by REF the association scheme is imprimitive. Conversely, if MATH is prime then is a well known fact in number theory (see CITE) that either it is MATH multiplied by an unit or: CASE: MATH, MATH an odd prime, and in this case the lattice MATH has MATH points, and clearly MATH. CASE: MATH and MATH, in this case MATH can be represented as MATH, and therefore has MATH points and MATH. In the case MATH it is obvious that the scheme is primitive, in the other two cases we have that the estabiliser of any point MATH in the lattice is given by the group of rotations around the point (that is, MATH, where MATH is the translation of vector MATH ). It is clear that in both cases above there is no group between MATH since the group generated by a single element in MATH and the group MATH is MATH (See remark below for a further explanation). Therefore the estabiliser is maximal, and by REF, the action is primitive, and so is the association scheme. |
math/0101092 | It follows directly from the reasoning above, that is, the decomposition of the group of translations in a product of cyclic groups. |
math/0101092 | It is obvious by the first part of the proof of theorem in REF. |
math/0101097 | We may as well assume that MATH with the standard filtration, where each MATH is finite dimensional. If we take a basis MATH of MATH, we can obtain a strictly increasing basis of MATH, and by throwing out unnecessary elements, we can choose a subsequence which spans a subspace mapping injectively to MATH. Let MATH be the subset of MATH so obtained, ordered in a strictly increasing manner, because by continuity, given any MATH, there can only be a finite number of the MATH whose image has order smaller than MATH. Next we claim that an element MATH lies in MATH precisely if it can be written as an infinite sum of the form MATH. To see this fact, first note that because the order of the MATH's are increasing, and MATH is the image of some MATH, we can form the element MATH which is well defined in MATH since it is complete, and the image of this element must be MATH. On the other hand, by our construction, if MATH, then by using the fact that for any finite combination of the MATH can be expressed in terms of the MATH, we can subtract off a linear combination MATH from MATH such that MATH is expressed as an image of terms of high order in MATH. This therefore must have high order in MATH, and thus we can express MATH as the limit of a NAME sequence in the MATH. |
math/0101097 | If MATH, then MATH, where MATH is the cocycle determining the extension to MATH. Similarly, if MATH is the obstruction to extending the deformation to MATH, we can express MATH, where MATH determines the extension MATH. Since MATH, the result follows immediately. |
math/0101099 | CASE: It is simply the definition of REF-norm in MATH as a NAME C*-bimodule. CASE: It suffices to observe that MATH is positive as an element of MATH. CASE: The isometry of the involution is obtained by embedding MATH via the composition law into the space of the bounded MATH-linear maps of right NAME C*-modules from MATH into MATH: if MATH then MATH is the adjoint of MATH as NAME C*-module map. |
math/0101099 | The first assertion is obvious. For the second one, let MATH; then MATH, MATH, MATH for some object MATH in MATH and we can define the isometries MATH, MATH. Thus the projection MATH is a direct sum of MATH and MATH. |
math/0101099 | Let MATH be a C*-category, MATH a C*-algebra morphism. We construct a pullback C*-functor MATH. Now there exists, for each object MATH in MATH, the C*-algebra morphism MATH we define MATH. Now, as MATH for each MATH in MATH, we define the bounded linear maps MATH by setting MATH . By REF, so that MATH. Let now MATH be a C*-functor; we want to define a C*-functor MATH. We observe first that for each object MATH in MATH the C*-algebra morphism MATH is induced, so we define MATH . Again, as MATH is bounded, we find that the maps MATH where MATH, can be extended to MATH. It is trivial to check that composition and involution are preserved. The proof for contravariant C*-functors follows the same lines by using C*-algebra antihomomorphisms. |
math/0101099 | Let MATH be objects in MATH. With the notation used in the proof of REF , it suffices to verify that MATH is a direct sum of MATH. |
math/0101099 | The computation above shows that the map REF preserves the composition of arrows in MATH. Using commutativity of MATH it is also verified that MATH preserves the involution. Now, thanks to the C*-identity, in order to verify that MATH is bounded, it suffices to show this just in the case where the domain object coincides with the codomain. We note that the map MATH defines a morphism MATH, that we extend to a morphism MATH tensoring by the identity on MATH. Thus, for MATH, the inequality MATH holds. Now for MATH we have MATH so that MATH . Thus we can extend MATH to MATH and define, for each pair MATH in MATH, the object MATH in MATH that can be expressed conveniently by the following formal expression MATH . We pass now to prove the second point. We define the tensor product on MATH as the C*-functor MATH induced by the tensor product on MATH; with an abuse of notation, we write MATH to indicate the tensor product of arrows in MATH. By REF , the only properties we have to prove are strict associativity and the existence of an identity object. Strict associativity follows by using the strict associativity of the tensor product of arrows in MATH. For the symmetry, we define MATH (in the previous expression we regarded MATH as the identity arrows of MATH); the properties of symmetry are verified with direct computations. For the existence of conjugates, we observe that MATH as defined at the beginning of this section obviously has conjugates. Then we apply REF , where it is shown that each subobject of an object having conjugates itself has conjugates. |
math/0101099 | The first and second points are obvious. For the third point, note that MATH can be regarded as continuous functions MATH. Then we consider in MATH the subobjects MATH defined by MATH, and prove that REF is a continuous field of NAME spaces. If MATH then the continuous function MATH is defined; we put MATH and observe that MATH is isometrically isomorphic as a NAME space to MATH via the map MATH, where MATH. Thus MATH and the function MATH is continuous. The equality REF implies also that MATH. We already know that MATH is a NAME MATH-module, so have proved that REF is a continuous field of NAME spaces. We postpone the proof of local triviality until the next lemma. Finally, we have to prove that the map MATH, MATH, MATH preserves the fibers, that is, MATH, but this is obvious regarding MATH as continuous map and recalling the definition of the fiber functor. |
math/0101099 | We define MATH . By hypothesis there exist isometries MATH intertwining MATH with the units arrows MATH. Thus the map MATH, defines the isomorphism between MATH and MATH. For the second point, since MATH, for each MATH there exists a closed neighborhood MATH such that MATH, MATH. This implies that MATH and the constant maps MATH are NAME equivalent as elements of the C*-algebras MATH. Applying REF , since MATH are constant maps, there must be an isomorphism MATH. |
math/0101099 | By compactness of MATH and continuity of MATH there exists, using the argument of the local triviality lemma, a finite open cover MATH such that MATH and MATH are NAME equivalent as elements of MATH if MATH and MATH belong to the same MATH. Now there must exist some MATH belonging to MATH, so that the NAME equivalence class of MATH is constant as MATH varies in MATH. MATH being connected, we obtain that MATH and MATH are equivalent for each MATH, so there exists an isometry MATH such that MATH. The claim of the lemma now follows observing that, by definition of MATH , MATH, so that MATH is a unitary arrow in MATH. |
math/0101099 | The existence of the trivialization follows from local triviality and the previous lemma. Now given the trivialization MATH, MATH then satisfies the conditions required in the second point. Regarding the last point, we fix a fiber MATH of MATH and consider two trivializations MATH, MATH, defining MATH, MATH. Then MATH, where MATH. |
math/0101099 | Define MATH to be the class in MATH of the cocycles coming from trivializations of MATH. If MATH and MATH are unitarily equivalent then we can choose the same fiber MATH for MATH and MATH, and if MATH is a cocycle for MATH, then MATH is a trivialization of MATH defining the cocycle MATH so that MATH. Let now MATH and MATH be such that MATH. Then there are two equivalent cocycles MATH, MATH associated respectively to MATH and MATH, and there exist continuous maps MATH from MATH into MATH such that MATH. We define MATH; then MATH and MATH so that we can clutch together the MATH's and obtain an isometry MATH intertwining MATH and MATH. |
math/0101099 | By REF follows that for each MATH there exist isometries MATH providing an (isometric) C*-algebra isomorphism MATH so that MATH . This proves the lemma. |
math/0101099 | By the first point of the local triviality lemma there exists an isometry MATH and a family of NAME space isomorphisms MATH, defined by MATH where MATH. Hence MATH defines an isomorphism of MATH-graded C*-algebras MATH . |
math/0101099 | NAME on the right with MATH gives the isometric embedding of continuous fields of NAME spaces MATH and we have the commutative diagram MATH so that the evaluation over MATH is well defined for MATH: MATH . The properties of continuous fields of NAME spaces are naturally inherited by the continuous fields MATH. For local triviality, it suffices to note the compatibility of the local charts defined above with the inductive limit structure. |
math/0101099 | The existence of MATH follows by the previous lemmas and the isomorphism is constructed with an isometry MATH such that MATH, MATH. We have also, for each MATH, isometries MATH, MATH implementing the following NAME space isomorphisms MATH the map defined by the upper horizontal arrow providing the isomorphism MATH that, by construction, satisfies the condition required. |
math/0101099 | Let MATH; then for MATH in a suitable neighborhood of MATH and the last term is a continuous function as MATH varies in MATH. To prove the result about the norm of MATH it suffices to check that MATH has the required property with respect to the circle action. But it is obvious that MATH. |
math/0101099 | Let MATH be a trivialization of MATH defining the cocycle MATH; then the trivialization MATH, with MATH, is induced on MATH. Thus MATH is a cocycle for MATH and for MATH we have MATH . This completes the proof of the first statement. The second follows by REF . |
math/0101099 | CASE: The identity REF implies that MATH where MATH. Hence for MATH we get MATH, where MATH. CASE: By REF it follows that MATH is generated as a C*-algebra by MATH. CASE: By CITE MATH is exact, so that (see CITE) for every C*-algebra MATH the maximal and spatial tensor products MATH, MATH each have a natural structure of a continuous field with fiber respectively MATH, MATH, where MATH is the rank of MATH in MATH. Since MATH is nuclear the maximal and spatial norms coincide over the algebraic tensor product of MATH by MATH. CASE: As MATH we find that MATH for each MATH. Then we apply REF . |
math/0101099 | CASE: Let MATH be a family of transition functions for MATH. Then the sphere bundle MATH can be conveniently described by the transition functions MATH, with MATH. Now, note that if MATH then we have a local description of MATH in terms of continuous functions MATH satisfying the relation MATH . Let now MATH be a set of local charts defining the transition functions MATH. We define continuous functions MATH by setting MATH. Now, as by REF the identity MATH holds, we find MATH. We denote by MATH the continuous function on MATH obtained by clutching the maps MATH. By extending the map MATH in the obvious way we obtain an isometric MATH-algebra monomorphism MATH. By applying for every local chart the NAME theorem on the factor MATH in MATH we find that the extended map MATH is also surjective. CASE: By REF it follows that MATH. An alternative way to verify the isomorphism is to observe that complex conjugation on the fibers induces a homeomorphism MATH: on each local chart we define MATH and the identity MATH holds, so that a global homeomorphism is defined by clutching the maps MATH. |
math/0101099 | CASE: Let MATH be a projection. As MATH is in the commutant of MATH the amplimorphism MATH is defined, so that MATH has subobjects. CASE: As MATH is a subcategory of MATH it suffices to prove the statement only in this first case. Let now MATH, MATH be amplimorphism. With the same notations used in the proof of REF we consider a direct sum MATH of MATH in MATH. Then we define MATH, MATH and the map MATH from MATH into MATH. We leave to the reader the verification that MATH is an amplimorphism and a direct sum for MATH,MATH. |
math/0101099 | By property of the spatial tensor product we obtain, tensoring the identity on MATH with the right tensor product with MATH, a morphism MATH defined by MATH. By composing with MATH we get a morphism MATH the image of which commutes with that of MATH defined by MATH. Thus, by using the universal property of the maximal tensor product, we define MATH. |
math/0101099 | We verify that MATH; by REF we obtain, with MATH, MATH . The same computation for MATH shows that MATH if MATH, MATH. The other properties of tensor product can be verified by routine computations. |
math/0101099 | The proof follows the same line as in CITE. For reader's convenience we expose some elementary facts about the first assertion. The embedding of the C*-algebras follows by funtoriality. Now, MATH is embedded in MATH as the space of arrows of the identity object and if MATH the multiplication by MATH is given by MATH and this proves that arrows between tensor powers of MATH commute with elements of MATH. |
math/0101099 | Let MATH. By definition of MATH it follows that MATH (we write MATH for MATH) belongs to MATH if and only if commutes with MATH; but MATH . Let now MATH; as MATH is finitely generated there exists a finite set MATH such that MATH and this implies that each MATH is finitely generated by elements of the type MATH. Now we have seen that MATH so that MATH and MATH belongs to MATH. For the second point, we define the map from MATH into MATH by MATH. As MATH the map is injective, and by the argument of REF if MATH we get a decomposition MATH, MATH, so we have the surjectivity. |
math/0101099 | As MATH is finitely generated it is isomorphic to a projective NAME MATH-module, hence isomorphic to the module of sections of a vector bundle MATH. Now MATH is inner if and only if the injective MATH-module map MATH from MATH to MATH is surjective, so that the definition of the left action MATH is forced by MATH. The equivalence with the third statement is consequence of REF applied to the projection MATH defined by MATH: we obtain in this way the isomorphism MATH, and a family of compatible NAME MATH-bimodule isomorphisms MATH. This implies the isomorphism between the algebraic tensor product MATH and MATH. Now observe that by nuclearity of MATH there is a unique C*-norm defining the tensor product MATH (in the sense of CITE, see also CITE), that coincides with the norm on MATH for which the circle action is isometric. |
math/0101100 | All the statements are obvious except the one concerning the irreducibility of MATH which will be proved in REF . |
math/0101100 | Since MATH is projective, there are characters MATH of MATH such that the associated line bundles MATH are very ample. Sections of MATH can be naturally identified with MATH-equivariant holomorphic functions on MATH, and a fortiori on MATH because MATH has codimension at least two. We say that MATH is MATH-equivariant if MATH and denote MATH the (finite dimensional) vector space of such functions. There is a natural map MATH having the properties: MATH it covers the projective embedding of MATH defined by the linear system of MATH; MATH for all MATH and MATH for all MATH; MATH for all MATH and MATH. In a certain basis of MATH, the MATH-action can be diagonalized, and we denote MATH the corresponding characters; they have the property that MATH for MATH. With respect to this basis, we are defining the MATH-action on MATH, together with a linearization in the standard ample line bundle over this projective space as follows: MATH acts on MATH by the characters MATH and on the MATH-term by MATH. Similarly, we let MATH act on MATH, in the standard fashion on MATH and trivially on the last component. The ample line bundle MATH we are looking for is obtained by restricting via MATH the natural (tensor product) ample line bundle on the product of these projective spaces. After replacing MATH with a sufficiently large power of it, one may check that the points MATH, MATH, are MATH-unstable, while the points MATH, MATH, are MATH-semi-stable. Consequently the MATH-semi-stable set of MATH is precisely MATH. |
math/0101100 | As expected, we put on MATH the final topology for the projection MATH; we must prove that the quotient is NAME when we consider on MATH the analytic topology given by small balls. For two sections MATH lying over the same point in MATH, which are not in the same MATH-orbit, we want to prove that there are neighborhoods MATH and MATH such that MATH (when MATH and MATH lie above two different points of MATH everything is clear). From the very definition of MATH we deduce that MATH with MATH a finite set. We are distinguishing two cases: CASE: When there is a point MATH such that MATH and MATH are not in the same MATH-orbit in MATH, the existence of the two disjoint MATH-invariant neighborhoods of MATH and MATH is immediate; CASE: It might happen that for any MATH, the evaluations MATH and MATH are in the same MATH-orbit (such a situation does appear in the simple case when we projectivize the space of sections of a line bundle). In this case, since MATH acts freely on MATH, there is a morphism MATH such that MATH. If there are no neighborhoods MATH and MATH as wanted, we deduce the existence of sequences MATH and MATH such that MATH and MATH, where the norm MATH on MATH is defined by MATH . Let us fix a closed disk MATH. For positive MATH, there is a rank MATH such that for MATH, MATH . It follows that MATH, which in turn implies that MATH for a possibly different choice of MATH (this is because we have restricted ourselves to the compact MATH). Since for MATH the MATH's take values in MATH, we deduce that MATH for all MATH; but this means that MATH is constant when restricted to MATH and consequently is constant everywhere. We conclude that MATH and MATH are in the same MATH-orbit in MATH, which contradicts our assumption. |
math/0101100 | According to REF there is an ample line bundle on MATH which linearizes the MATH-action on MATH, such that the corresponding MATH-semi-stable set is MATH, and consequently MATH. This one induces the relatively ample line bundle MATH and the action of MATH is still linearized in MATH. NAME MATH with a sufficiently ample line bundle on MATH we obtain an ample line bundle on MATH, together with a linearization of the MATH-action in it, which has the additional property that the MATH-semi-stable locus is precisely MATH. Identifying MATH with the space of morphisms from MATH into MATH which represent the class of a section, we find ourselves in the situation studied in REF, where is shown that in this case is possible to linearize (rather canonically) the MATH-action on MATH in an ample line bundle such that the corresponding semi-stable points have the property that their image is not completely contained in the unstable locus of MATH (according to REF loc.cit.). Denoting MATH the set of MATH-semi-stable points of MATH, we have found that MATH. We want to prove now that MATH is projective; quasi-projectivity comes for free from the very construction, so that remains to prove the compactness (completeness). Since the MATH-action on MATH covers the identity of MATH, MATH comes with the projection MATH . The first claim is that this map is surjective: indeed, according to REF, points MATH whose image is contained in MATH are MATH-semi-stable. Since the `bad' set MATH has codimension at least two and because the line bundles MATH are globally generated, we deduce that for any MATH there is a section MATH having the property that its image is disjoint from MATH. This proves the claim. Consequently is enough to prove the compactness of the fibres of the projection MATH. The following lemma describes the fibrewise situation. Suppose we are given a torus action on MATH as described at the beginning of REF, so that the quotient is a smooth and projective toric variety. Let us consider now the action MATH, where MATH and the torus MATH acts on the direct summands of MATH respectively by the same characters MATH as on MATH. Then the quotient is a smooth, compact toric variety. The compactness of the quotient can be easily seen using the moment map description of toric varieties. In coordinates, the moment map corresponding to the MATH-action on MATH is (see REF): MATH . Then MATH can be described as MATH for (a well chosen) MATH, where MATH denotes the real torus. Since MATH is compact, MATH is still compact and MATH acts freely on it. Let us move now to the new situation and denote MATH the points of MATH. The moment map in this case has the form MATH and we want to prove that the quotient MATH is smooth and compact. It is easy to see that the maps MATH and MATH have the same image, which is compact since MATH is so. The compactness of MATH is implied now by the compactness of standard spheres. The action of MATH on MATH is free because for any point MATH which solves the equation MATH, is possible to find a MATH-equivariant embedding MATH which pass through MATH. The conclusion follows now from the fact that MATH acts freely on MATH. Coming back to our proposition, we deduce from the lemma that the fibres of MATH are smooth and compact toric varieties, all isomorphic to MATH. Since MATH is quasi-projective, it follows that MATH is actually projective and so is the invariant quotient MATH. Remains to prove that MATH: the natural inclusion MATH being an open map, its image is both open and closed in MATH, so that MATH and therefore MATH. |
math/0101100 | The statement follows from the fact that when restricted to the fibres of MATH, the classes MATH generate the integral cohomology ring of MATH (recall that MATH is a toric variety). |
math/0101100 | The first part of the proposition being already proved, we are left with the second claim. Again, we are going to check it only for MATH, the general statement coming from the symmetry of the problem. We observe that MATH, because MATH is the locus where MATH acts freely. Therefore MATH and the statement follows because MATH acts componentwise. |
math/0101100 | This equality is just a rewriting of the main result in CITE, but it can be proved in a more elementary way as follows: MATH being a toric variety, its NAME characteristic coincides with the number fixed points under the MATH-action. Since this big torus contains MATH, the fixed point set is contained in the union of the MATH-fixed subvarieties. But fibrewise these are just products of projective spaces on which MATH acts in standard fashion. |
math/0101100 | Since MATH is a principal MATH-bundle, we have the following MATH-equivariant exact sequence on MATH where MATH denotes the MATH-invariant relative tangent bundle to the total space of MATH. But MATH is an open subset in a vector bundle over MATH, so that the relative tangent bundle is canonically isomorphic to MATH, for MATH the projection. As MATH preserves the decomposition of MATH, MATH . We observe now that MATH, the isomorphism being given by MATH . This proves the exactness of the middle row in REF . A similar argument proves the exactness of the first horizontal sequence, and the last row is now a simple consequence. The very important thing which must be clarified yet is the way how MATH acts on MATH. The sequence REF being MATH-equivariant, we have to describe the induced action on MATH under the isomorphism REF. For MATH, MATH so that we can see that indeed the MATH-action on MATH is trivial while the action on MATH is as in REF. |
math/0101100 | It is clear that MATH, for some MATH, and this element is precisely the weight of the action of MATH on the stalk MATH at some point MATH. Again, we shall make the computations for MATH only: in this case the assignment MATH gives an isomorphism MATH and is easy to see that MATH . A short computation shows that the above isomorphism is induced precisely by the choice of the dual basis to MATH, and the conclusion follows. |
math/0101100 | Indeed, for such a set MATH and consequently MATH for any MATH (otherwise MATH, a contradiction). The conclusion follows now from the relations REF. |
math/0101102 | For any MATH define MATH to be the colimit of the diagram MATH . Clearly MATH is the domain and MATH the codomain of MATH. Moreover it is easy to see that the pushout of MATH by MATH with the obvious attaching map is canonically isomorphic to MATH. Since MATH preserves colimits the assignment MATH is a transfinite composition. |
math/0101102 | We apply CITECITE with generating cofibrations MATH, generating trivial cofibrations MATH and weak equivalences the maps which are weak equivalences in MATH. By CITE[VI. REF ], MATH is complete and by CITE[REF ] cocomplete. REFCITE is clear, REF follow by adjunction from our smallness assumptions on the domains of MATH and MATH. Since each element of MATH is in MATH, hence a retract of a map in MATH, each element of MATH is in MATH, hence together with our assumption we see that REF is fulfilled. By adjunction MATH (respectively, MATH) is the class of maps in MATH which are trivial fibrations (respectively, fibrations) in MATH. Hence REF and the second alternative of REF are fulfilled. |
math/0101102 | As in CITECITE one shows that cofibrations have the left lifting property with respect to trivial fibrations. Then it follows that if MATH is a map in MATH with MATH cofibrant such that the maps MATH are (trivial) cofibrations then the map MATH is a (trivial) cofibration in MATH. So a good trivial cofibration (definition as in the proof of CITECITE) with cofibrant domain is a trivial cofibration and trivial cofibrations with cofibrant domain have the left lifting property with respect to fibrations. We then can construct functorial factorizations into a good trivial cofibration followed by a fibration for maps with cofibrant domain as in the proof of CITECITE and also the factorization into a cofibration followed by a trivial fibration (for the alternative definition for maps with cofibrant domain). It follows that a trivial cofibration with cofibrant domain is a good trivial cofibration. All other properties are immediate. |
math/0101102 | We define the weak equivalences (respectively, fibrations) as the maps in MATH which are weak equivalences (respectively, fibrations) as maps in MATH. By adjointness the fibrations are MATH and the trivial fibrations are MATH. We define the class of cofibrations to be MATH. Since the adjoint of MATH is the forgetful functor REF is clear. The bicompleteness of MATH follows as in the proof of REF . REF-out-of-REF and retract axioms for the weak equivalences and the fibrations hold in MATH since they hold in MATH, the retract axiom for the cofibrations holds because MATH is closed under retracts. So REF is fulfilled. The first half of REF is true by the definition of the cofibrations. By our smallness assumptions we have functorial factorizations of maps into a cofibration followed by a trivial fibration and into a map from MATH followed by a fibration. We claim that a map MATH in MATH whose domain is cofibrant in MATH is a trivial cofibration. MATH is a weak equivalence by assumption. Factor MATH as MATH into a cofibration followed by a trivial fibration. Since MATH has the left lifting property with respect to MATH, MATH is a retract of MATH by the retract argument, hence also a cofibration. Hence we have shown REF . Now let MATH be a trivial cofibration whose domain is cofibrant in MATH. We can factor MATH as MATH with MATH and MATH a fibration. MATH is a trivial fibration by REF-out-of-REF property, hence MATH has the left lifitng property with respect to MATH, so MATH is a retract of MATH and has therefore the left lifitng property with respect to fibrations. This is the second half of REF immediately follows from the assumptions, and REF is true since limits in MATH are computed in MATH. |
math/0101102 | This follows either by CITECITE applied to the adjunction of two variables MATH, MATH, or by REF . |
math/0101102 | Easy from CITECITE. |
math/0101102 | We claim that in all four cases the functors MATH define a monad the algebras of which are the operads in MATH. So we have to define in all four cases maps MATH and MATH satisfying the axioms for a monad. We will restrict ourselves to REF and leave the other cases to the interested reader. The domain of the map MATH is a coproduct over all MATH, MATH for all MATH over the MATH and the map MATH sends such an entry via the identity to the entry associated to the tree in MATH obtained by replacing every vertex MATH of MATH by the tree MATH in such a way that the numbering of the arrows starting at MATH and the numbering of the tails of MATH correspond. The map MATH sends MATH to the summand MATH in MATH which belongs to the tree with one vertex and MATH tails such that the labelling of the arrows coincides with the labelling of the tails (which are of course all arrows in this case) (that is, the labelling of the tails is the canonical one). It is clear that MATH is associative and MATH is a two-sided unit. To see that a MATH-algebra is the same as an operad one proceeds as follows: Let MATH be a MATH-algebra. Let MATH. The structure maps of the operad structure we will define on MATH are obtained from the algebra map by restricting it to the summands belonging to trees where every arrow starting at the root goes to a vertex which has only tails as outgoing vertices and where the labelling of the tails is the canonical one. The unit in MATH corresponds to the empty tree. The right action of a MATH on MATH is given by the algebra map restricted to the tree with one vertex and MATH tails such that the MATH-th arrow simultaneously is the MATH-th tail. That MATH acts as the identity is the unit property of MATH, and the associativity of the action follows from the associativity of MATH. It is easy to see that the associativity and symmetry properties of MATH also follow from the associativity of MATH. The unit properties follow from the behaviour of the empty tree. On the other hand let MATH be an operad. We define a MATH-algebra structure on MATH: Let MATH be a tree with canonical labelling of the tails. Then it is clear how to define a map from the summand in MATH corresponding to MATH to MATH by iterated application of the structure maps of MATH (the unit of MATH is needed to get the map for the empty tree). The map on the summand corresponding to MATH for MATH is the map for MATH followed by the action of MATH on MATH. One then can check that the associativity, symmetry and unit properties of the structure maps of MATH imply that we get indeed a MATH-algebra with structure map MATH just described. |
math/0101102 | Let MATH be the colimits described in the Proposition. First of all we check that this is well defined, that is, that firstly the MATH maps we have described glue together. This is the case because the processes of removing old vertices of valency MATH and/or changing a new vertex into an old one and concatenating commute with each other. Secondly this map factors through the quotient described in the Proposition because of the symmetry properties of MATH and because of the fact that in previous steps quotients with respect to analoguous equivalence relations have been taken. Next we have to equip MATH with an operad structure. The unit is the one coming from MATH. We define the structure map MATH REF in the following way: For MATH let MATH . First one defines for trees MATH, MATH, MATH, a map MATH . Therefore one glues the tree MATH to the tail of MATH with label MATH and concatenates such that one gets a tree MATH. Then by applying structure maps of MATH one gets a map MATH and composes this with the canonical map MATH. Let MATH. Suppose we have already defined for a MATH and for all trees MATH, MATH, a map MATH. From this data one then obtains the same data for MATH instead of MATH as follows: Let MATH, MATH. One defines the map MATH by transfinite induction on the terms of the MATH-sequence defining MATH: So let MATH be already defined and let MATH be the map by which MATH is a pushout of MATH. We define MATH on MATH by using the data described above for MATH to get the map after taking the appropriate quotient on the codomain of MATH. One has to check the compatibility of this map with the given map via the attaching map. To do this for one of the MATH summands of the domain of MATH one uses the fact that the same kind of compatibility is valid in MATH. Finally when arriving at MATH we get the desired structure map. The associativity of the structure maps follows by proving the corresponding statement for the MATH. This one gets by first glueing trees without concatenating and then observing that the concatenation processes at different places commute. The symmetry properties follow in the same way as for free operads, the unit properties are forced by the fact that in the MATH's the pushout product over the unit maps is taken. Hence MATH is an operad. It receives canonical compatible maps in MATH from MATH and MATH. In the end we have to show that our operad MATH indeed satisfies the universal property of the pushout by MATH. We need to show that a map MATH in MATH together with a map MATH compatible with the attaching map is the same as a map MATH. To get MATH from MATH and MATH one first defines for any MATH a map MATH using the structure maps of MATH. Then one checks that these maps indeed glue together to a MATH. To get MATH and MATH from MATH one composes MATH with MATH and MATH. These processes are invers to each other. |
math/0101102 | We restrict to the case MATH, the general case is done in the same way. Set MATH, MATH and MATH. We can assume that MATH and MATH (or MATH or MATH). Let MATH and MATH such that MATH is a pushout by MATH and MATH is a pushout by MATH. Then by REF MATH is a MATH-sequence, and the transition maps are pushouts by the MATH, MATH; MATH. We can modify this sequence to make it invariant under the MATH-action: Let MATH be the set of unordered sequences of length MATH with entries in MATH, and for MATH let MATH be the set of ordered sequences of length MATH with entries in MATH which map to MATH. Let MATH. In the following let us view MATH and MATH as monotonly increasing sequences of length MATH. We say that MATH if there is a MATH such that MATH for MATH and MATH. With this order MATH is well-ordered. Now MATH is a MATH-sequence with MATH-th transition map MATH, so MATH is the corresponding MATH-sequence with transition maps the MATH, MATH, MATH. Note that on these maps there is a MATH-action. Now to prove our claim it suffices to show that every MATH is a (trivial) cofibration in MATH, which can easily be seen by noting that every MATH and MATH is of the form MATH for MATH (or MATH). |
math/0101102 | Let MATH be the equivalence relation on MATH which identifies MATH and MATH in MATH if there is an isomorphism of directed graphs MATH which respects the labeling of the arrows starting at new vertices. Let MATH be an equivalence class of MATH in MATH. The MATH-action on MATH restricts to a MATH-action on MATH. We have to show that the part of the map in REF given as the appropriate quotient of MATH is a (trivial) cofibration in MATH. Let MATH be a doubly colored directed graph, where the arrows starting at new vertices are labelled, isomorphic to the objects of the same type underlying the objects from MATH. Set MATH . On MATH there is an action of MATH. Let MATH be the set of tails of MATH. There is an action of MATH on MATH. It is easily seen that the quotient of the map REF we are considering is isomorphic to MATH. Hence we are finished if we show that MATH is a (trivial) cofibration in MATH. This is done by induction on the depth of MATH. Let MATH be the different isomorphism types of doubly colored directed graphs, such that the arrows starting at new vertices are labelled, sitting at the initial vertex of MATH with multiplicities MATH and set MATH, MATH. Then, if the initial vertex of MATH is old, MATH, otherwise MATH, and the map MATH is given like the map MATH in REF . Now the claim follows from REF and the induction hypothesis. |
math/0101102 | We apply REF to the monad MATH which maps MATH to MATH. It is known that MATH is cocomplete. Since filtered colimits in MATH are computed in MATH, it follows from REF that those maps from MATH respectively, MATH whose domain is cofibrant in MATH are cofibrations (respectively, trivial cofibrations) in MATH. It is clear that MATH is right proper if MATH is. If MATH is left proper, then MATH is left proper, and the pushout in MATH by a cofibration whose domain is cofibrant in MATH is a retract of a transfinite composition of pushouts by cofibrations in MATH, hence weak equivalences are preserved by these pushouts. |
math/0101102 | The first part follows from the Remark above. Let MATH be compatible with all MATH-algebra structures. Then it can be checked directly that f is also compatible with the algebra structure on MATH. But since the maps MATH are coequalizers in MATH the claim follows. |
math/0101102 | We have to do the same steps as in the proof of REF . Let MATH be the colimit described in the Proposition. The attaching maps are again well-defined because the various concatenation processes commute with each other and because of the symmetry properties of MATH and the equivalence relations appearing in previous steps. We equip MATH with a MATH-algebra structure: Let us define the structure map MATH. For MATH let MATH be as in the proof of REF . For MATH let MATH . Let MATH and MATH, MATH. We obtain a tree MATH by glueing MATH to the tail of MATH labelled by MATH and then concatenating. By applying operad and algebra structure maps we get a map MATH. It is then possible by similar considerations as in the proof of REF to get from these maps the desired structure map of MATH. It is easy to see that these structure maps are associative and symmetric. Hence MATH is a MATH-algebra which receives a MATH-algebra map from MATH and MATH-algebra maps from MATH and MATH which are compatible with each other in the obvious way. We have to check that for a MATH-algebra MATH a map MATH is the same as a map of MATH-algebras MATH and a map MATH which are compatible with each other. We get the maps MATH and MATH from MATH by the obvious compositions. Given MATH and MATH we first obtain a map of MATH-algebras MATH. Moreover for any MATH there is a map MATH by applying the MATH-algebra structure maps of MATH. It is then easy to check that these maps glue together to give the map MATH. These processes are invers to each other. |
math/0101102 | Let MATH be the equivalence relation on MATH which identifies MATH and MATH in MATH if there is an isomorphism of directed graphs MATH which respects the labeling of the tails and of the arrows starting at new vertices. Let MATH be an equivalence class of MATH in MATH. We have to show that the appropriate quotient of the map MATH is a (trivial) cofibration in MATH (or lies in MATH under the assumptions of the last statement). This is done as in the proof of REF by induction on the depth of the trees in MATH. This time instead of using REF it is sufficient to use REF applied to rings of the form MATH. |
math/0101102 | We apply REF to the monad MATH which maps MATH to MATH. It is known that MATH is cocomplete. Since filtered colimits in MATH are computed in MATH we are reduced to show that the pushout of a MATH-algebra MATH which is cofibrant as an object in MATH by a map in MATH (respectively, in MATH) is a cofibration (respectively, trivial cofibration) in MATH. Since MATH is a retract of a cell operad (that is, a cell complex in MATH) such a pushout is a retract of a pushout of the same kind with the additional hypothesis that MATH is a cell operad. So let MATH be a cell operad. Then the pushout in question is a transfinite composition of maps MATH as in REF , hence by REF it is a (trivial) cofibration. It is clear that MATH is right proper if MATH is. The pushout in MATH by a cofibration whose domain is cofibrant in MATH is a retract of a transfinite composition of pushouts by cofibrations in MATH, hence if MATH is left proper weak equivalences are preserved by these pushouts, so MATH is also left proper. The last statement follows again from REF . |
math/0101102 | The whole Proposition is shown by induction on MATH, so suppose that it is true for ordinals less than MATH. We construct the map in REF, prove its properties and define the attaching map in REF by transfinite induction: Suppose MATH is a successor, that MATH is defined for MATH and that the map in REF is defined for all limit elements MATH with MATH. Let MATH with MATH and let MATH coincide with MATH except that MATH. The attaching map on the summand MATH of the domain of MATH is given as follows: Let MATH be defined by MATH, MATH, MATH for MATH and MATH for MATH. Let MATH. There is a canonical map MATH whose codomain maps naturally to the domain of MATH. So we get maps MATH the composition of which is the attaching map on the summand MATH. These maps glue together for various summands MATH: There are two cases to distinguish. In the first one the intersection of two summands contains MATH twice. Then the two maps on this intersection coincide because of the symmetric group invariance. In the second case the intersection MATH contains MATH and MATH with MATH. Let MATH be as above and MATH be similarly defined for MATH. Now the two REF of the maps MATH state that both maps MATH are equal the map induced by first mapping both MATH and MATH to MATH and then applying a suitable map MATH. Now suppose MATH is a limit element with MATH for some MATH. Define MATH as the colimit of the preceeding MATH, MATH. Let MATH, MATH and MATH be as in REF. We define MATH by induction on MATH and on MATH using the fact that MATH by induction hypothesis for the induction on MATH. For abbreviation set MATH. Let MATH be a successor and let a map MATH be already defined. MATH is a pushout of MATH by MATH . Let MATH. Then the codomain of MATH is MATH. Moreover by induction hypothesis for the MATH-induction there is a map MATH hence by plugging in MATH into the MATH-th place of MATH we get a map MATH . This map and MATH glue together to a map MATH: We have to show that they coincide after composition on domains of the form MATH for MATH a summand of the domain of MATH containing MATH for some MATH (the definition of MATH is similar to the one of MATH). To do this we can restrict for every MATH to objects MATH, MATH, for MATH of the same shape as MATH and MATH successors. Then the two possible ways to get from MATH to MATH can be compared by mapping MATH to MATH, unwrapping the definitions of MATH and MATH and using associativity of MATH. We arrive at a map MATH. That it factors through the MATH-quotient follows after replacing MATH by MATH (the MATH and MATH as above) in the domain of this map since then the MATH-relation is obviously also valid in MATH. Both REF follow easily by the technique of restricting any appearing MATH by a factor MATH. Now using the maps MATH and REF we can equip MATH with a MATH-algebra structure (to do this accurately we have to enlarge MATH a bit and the corresponding sequence by trivial pushouts). We are left to prove the universal property for MATH by transfinite induction on MATH. So let it be true for ordinals less than MATH. If MATH is a limit ordinal or the successor of a limit ordinal there is nothing to show. Let MATH, let MATH be a MATH-algebra and MATH a map in MATH and MATH a map in MATH such that these two maps are compatible via the attaching map. We define maps MATH by transfinite induction on MATH, starting with the given map on MATH. So let MATH be a successor. Since for any MATH there is a map MATH we have a natural map MATH using the algebra structure maps of MATH. We have to show that this is compatible via the attaching map from the domain MATH of MATH to MATH with the map MATH coming from the induction hypothesis. We check this again on a summand MATH of MATH containing some MATH. The attaching map on MATH is induced from MATH as above. The canonical map from the domain of MATH to MATH is compatible with MATH (as one checks again by replacing any MATH by essentially products of MATH's as above), which together with the fact that MATH and MATH coincide on MATH implies the compatibility. By construction and the definition of the algebra structure on MATH the map MATH just defined is a MATH-algebra map. If we have on the other hand a map of MATH-algebras MATH we can restrict it to get compatible maps MATH and MATH. These two assignments are inverse to each other. |
math/0101102 | Let MATH be cofibrant in MATH. We have to show that the pushout of a MATH-algebra such that the map from the initial MATH-algebra to MATH is in MATH by a map from MATH (respectively, MATH) is a cofibration (respectively, trivial cofibration) in MATH. We can assume that MATH is a MATH-algebra, since in the general situation all maps we look at are retracts of corresponding maps in this situation. But if MATH is a cell MATH-algebra our claim immediately follows from REF . |
math/0101102 | Let MATH and MATH. We denote by MATH the homomorphism object MATH. There is a map of operads MATH which is described on as follows: We give the maps MATH on MATH-valued points CASE : A map MATH is sent to the composition MATH where the second map is induced by the diagonal MATH. Hence for objects MATH and MATH the object MATH has a natural structure of MATH-algebra given by the composition MATH. We denote this MATH-algebra by MATH. For a fixed MATH the functor MATH, MATH, has a left adjoint MATH, which is given for a free MATH-algebra MATH, MATH, by MATH and which is defined in general by be requirement that MATH respects coequalizers (note that every MATH-algebra is a coequalizer of a diagram where only free MATH-algebras appear). So we have a functor MATH. Let now MATH be fixed. By a similar argument as above the functor MATH, MATH, has a left adjoint MATH, which sends a free MATH-algebra MATH, MATH, to the image of MATH in MATH. One checks that the functor MATH we constructed defines an action of MATH on MATH. It remains to show that this functor is a NAME bifunctor and that the unit property is fulfilled. So let MATH be a cofibration in MATH and MATH a fibration in MATH. We have to show that MATH is a fibration in MATH, that is, lies in MATH. By adjointness this means that MATH has the right lifting property with respect to the maps MATH for all MATH, which is by adjointness the case because MATH is a trivial cofibration. When MATH or MATH is trivial we want to show that MATH lies in MATH, so MATH should have the right lifting property with respect to the maps MATH for all MATH, which is again the case by adjointness. If MATH is cofibrant in MATH we are ready. In the other case the unit property follows by transfinite induction from the explicit description of algebra pushouts, and hence the structure of cell algebras, given in REF and the structure of cell algebras given in REF . |
math/0101102 | (Compare to CITE[REFEF. Lemma.]) A MATH-module structure on a MATH-module MATH is given by maps MATH for MATH such that the compositions with MATH equals the composition MATH. The same statement is true for a module structure under the described pushout algebra on a MATH-module. |
math/0101102 | This Proposition is proven in essentially the same way as REF except that this time we have to define associative algebra structures on the MATH and to verify the universal property stating the equivalence of module categories. For the associative algebra structure one uses the same formulas as for the tensor algebra and checks that they are compatible with the attaching maps. For the universal property one uses the fact that a MATH-module MATH is given by maps MATH which are compatible in various ways the explicit formulation of which we leave to the reader. |
math/0101102 | We prove the first part of the Proposition, the second one is similar. Let MATH, MATH and MATH a MATH-morphism from MATH to MATH in MATH. First of all it is clear that the pushforward functor MATH is a left NAME functor between MATH-semi model categories by the definition of the MATH-semi model structures. We have to show that MATH induces a natural isomorphism between MATH and MATH on the level of homotopy categories. So let MATH be a cosimplicial frame on MATH. MATH can be represented by a chain of MATH-simplices in MATH, and a homotopy between two representing chains by a chain of MATH-simplices. So we can assume that MATH is a MATH-simplex, that is, MATH. We have maps MATH, and MATH is an equivalence. Hence for MATH there is a unique isomorphism MATH with MATH. Then the MATH define a natural isomorphism between MATH and MATH. Now if we have a homotopy MATH, the three natural transformations which are defined by the three MATH-simplices of MATH are compatible, since on a given object they are the images in MATH of three compatible isomorphisms between the three possible images of MATH in MATH. |
math/0101102 | By REF a MATH-algebra MATH is the same as a MATH-algebra MATH together with a map MATH such that the structure map MATH is the composition MATH. Hence a unital MATH-algebra comes from a MATH-algebra. On the other hand if MATH is a MATH-algebra we have to show that the induced pointing MATH is a map of algebras. This follows easily from the fact that the map MATH has a right inverse (a pointing of MATH). For the first part of the Lemma it remains to prove that a MATH-algebra morphism between MATH-algebras is in fact a MATH-algebra morphism, which follows from REF . Consider the commutative square MATH of MATH-algebras and let MATH be the pushout of the left upper triangle of the square. We want to show that the canonical map MATH is an isomorphism. By the first part of the Lemma MATH is a MATH-algebra. Now again by the first part of the Lemma it is easily seen that MATH has the same universal property as MATH-algebra as MATH. So the above square is a pushout square in MATH, and hence by left properness of MATH over MATH REF the right vertical arrow is a weak equivalence. This implies that MATH is a weak equivalence. It remains to prove that MATH is cofibrant as object in MATH, which follows from REF . |
math/0101102 | The proof is along the same lines as the proof of REF . |
math/0101102 | In any MATH-semi model category MATH over MATH the category of objects under an object from MATH which becomes cofibrant in MATH is again a MATH-semi model category over MATH. |
math/0101102 | We can assume that MATH is a cell MATH-algebra. It is easy to see that the map MATH is compatible with the descriptions of MATH and MATH in REF as transfinite compositions, and the map MATH from the map of REF to the map of REF is induced by the map MATH, which itself is induced by the unit, the pointing and a structure map of MATH. Since MATH is a MATH-operad this is a weak equivalence, hence since the domains of the maps in MATH are cofibrant MATH is a weak equivalence. Now the claim follows by transfinite induction and left properness of MATH. |
math/0101102 | Let MATH be a cofibrant replacement. Then in the commutative square MATH the horizonrtal maps are weak equivalences (the upper one by REF ) and the left vertical arrow is a weak equivalence by the Lemma above, hence the right vertical map is also a weak equivalence. |
math/0101102 | This follows immediately from REF . |
math/0101102 | That MATH is a cofibrantly generated model category together with a closed action of MATH on it is true for any associative unital ring MATH in MATH which is cofibrant as an object in MATH. Let MATH and MATH be cofibrations in MATH. The MATH-pushout product of MATH and MATH is isomorphic to MATH. As a left MATH-module MATH is (non canonically) isomorphic to MATH, hence MATH is a cofibration MATH, and it is trivial if one of MATH or MATH is trivial. To show that for a cofibrant MATH-module MATH the map MATH is a weak equivalence we can assume that MATH is a cell MATH-module and we can take MATH. Then MATH is a transfinite composition where the transition maps are pushouts of maps MATH, where MATH is a cofibration in MATH with MATH cofibrant. But the composition MATH is a weak equivalence between cofibrant objects in MATH, hence the composition MATH is a weak equivalence between cofibrations in MATH. So by transfinite induction the composition MATH is a weak equivalence between cofibrant objects in MATH. |
math/0101102 | Follows from REF . |
math/0101102 | Similar to the proof of CITE. |
math/0101102 | MATH is a cofibrantly generated model category by REF . Let MATH and MATH be cofibrations in MATH. By REF the MATH-pushout product of the maps MATH and MATH is given by MATH, hence since MATH as MATH-modules this is a cofibration in MATH, and it is trivial if one of MATH or MATH is trivial. Note that MATH is cofibrant in MATH and that the map MATH is a weak equivalence by REF . So we have to show that for cofibrant MATH the map MATH is a weak equivalence, which follows from the fact that the maps of the form MATH for cofibrations MATH with cofibrant domain are weak equivalences between cofibrations in MATH. The first part of the last statement follows from REF , and the second part by REF . |
math/0101102 | Let MATH. One checks that both compositions are equal to the composition MATH, where the first arrow is a tensor product of obvious objectwise morphisms. |
math/0101102 | It suffices to show this for relative cell complexes MATH and MATH and cell complexes MATH and MATH, for which it follows for the first case by writing the pushout product of a MATH-sequence and a MATH-sequence as a MATH-sequence. Let MATH. Then if MATH is a MATH-sequence, MATH itself is a MATH-sequence in MATH. One concludes now by writing the products in REF again as appropriate sequences. The cases with group actions work in the same way. |
math/0101102 | This follows from the fact that for a cofibrant MATH-module MATH the map MATH is a weak equivalence. |
math/0101102 | This follows from the fact that the map MATH is a weak equivalence, which follows itself from the description of these algebras in terms of transfinite compositions as in REF . |
math/0101102 | Follows by the description of MATH as in REF . |
math/0101102 | Follows from REF and transfinite induction. |
math/0101102 | There is a natural inclusion MATH, and MATH maps its image to MATH in the natural way. Now let MATH with MATH for some MATH. There is a segment MATH starting at MATH which is isomorphic to MATH as a well-ordered set. Via this identification MATH corresponds to all MATH with MATH. Then MATH maps MATH to MATH if MATH and to MATH if MATH. It is easy to see that this way MATH is well-defined, bijective and order-preserving. |
math/0101102 | By REF we can work in MATH. So let MATH be cofibrant. Let us denote the coproduct in MATH by MATH. We have to prove the base change isomorphism for the diagram MATH . Let MATH be cofibrant. Then MATH is cofibrant in MATH by REF . Hence the base change morphism is represented by the morphism of MATH-modules MATH which is adjoint to the map MATH. We can assume that MATH is a cell module. Then by transfinite induction we are reduced to the following statement: Let MATH be cofibrant. Then the map MATH is a weak equivalence. By REF this follows if we show that the map of MATH-modules MATH (where we exchanged the roles of MATH and MATH) is a weak equivalence. It suffices to prove this for cell algebras MATH and MATH. So let MATH, where the transition maps are given by pushouts by cofibrations MATH in MATH with cofibrant domain as in REF . Similarly let MATH, where the transition maps are given by pushouts by cofibrations MATH in MATH with cofibrant domain. Then the map MATH is described as in REF by a MATH-sequence REF . Since the maps MATH respectively, MATH are MATH- respectively, MATH-sequences, the map MATH is a MATH-sequence REF by REF (this also holds in the case of a symmetric monoidal category with pseudo-unit). Let MATH be the isomorphism of well-ordered sets of REF . Let MATH and MATH be successors. Then MATH identifies MATH and MATH, and the relevant pushouts in the sequences REF are by maps MATH . It is easy to see by transfinite induction that the map MATH is compatible with sequences REF via the identification MATH on the indexing sets and with the above pushouts by the map induced by the tensor multiplication map MATH which inserts the second object into the last slot of the first object. This map is a weak equivalence because MATH is a MATH-operad, hence the claim follows by transfinite induction. |
math/0101102 | By REF we can assume that we have a cofibrant MATH, a cofibrant MATH and a cofibrant MATH and prove the projection isomorphism for MATH and the image MATH of MATH in MATH, where MATH is the image of MATH in MATH. Since MATH is cofibrant as MATH-module by REF the projection morphism is represented by the composition MATH where the isomorphism at the second place is from REF . So we have to show that the first map is a weak equivalence. We can assume that MATH is a cell module. Then by transfinite induction one is left to show that for a cofibrant MATH-module MATH the map MATH is a weak equivalence. But this map is the map from the free MATH-module on MATH to the free MATH-module on MATH, which is a weak equivalence by REF . Hence we are finished. |
math/0101102 | This proof is very similar to a part of the proof of REF . By REF we are reduced to the case where MATH is unital. It suffices to prove the claim for cell algebras MATH and MATH. So let MATH, where the transition maps are given by pushouts by maps MATH as in REF . Similarly let MATH, where the transition maps are given by pushouts by maps MATH. Then the map MATH is described by REF by a MATH-sequence REF . Since the maps MATH respectively, MATH are MATH- respectively, MATH-sequences, the map MATH is a MATH-sequence REF . Let MATH be the isomorphism of well-ordered sets of REF . Let MATH and MATH be successors. Then MATH identifies MATH and MATH. The relevant pushouts in the sequences REF are by maps MATH and again one shows by transfinite induction that the map MATH is compatible with sequences REF via the identification MATH on the indexing sets and with the above pushouts by the map induced by MATH where MATH inserts the pointing MATH into the last MATH slots of MATH and MATH inserts the pointing into the first MATH slots. This map is a weak equivalence since MATH is a MATH-operad, so our claim follows by transfinite induction and the assumptions. |
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