paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0002064 | REF has already been remarked. We show REF . In light of REF (applied with MATH) we can assume that MATH. Let MATH be a MATH-module and MATH an almost MATH-module; we have natural bijections MATH which proves REF . Now REF follows by inspecting the proof of REF , or by CITE (ch. III REF ). |
math/0002064 | The functor MATH is right adjoint to an exact functor, hence it preserves injectives. Now, let MATH be an injective envelope of MATH; to show that MATH is an injective envelope of MATH, it suffices to show that MATH is an essential extension of MATH. However, if MATH and MATH, then MATH, hence MATH, but MATH does not c... |
math/0002064 | We recall that the categories MATH and MATH are both complete and cocomplete. Now let MATH be any small indexing category and MATH be any functor. Denote by MATH the composed functor MATH. We claim that MATH. The proof is an easy application of REF . A similar argument also works for limits and for the category MATH. |
math/0002064 | REF follows easily from REF follows easily from REF . |
math/0002064 | By REF it follows that MATH is right exact. To show that it is also left exact when MATH is a flat MATH-module, it suffices to remark that MATH is left exact. Now, by REF , the functor MATH is right adjoint to an exact functor, so REF is clear. |
math/0002064 | Let MATH be a MATH-algebra, MATH a MATH-algebra and MATH a morphism of MATH-algebras. By REF we obtain a natural MATH-linear morphism MATH. Together with the structure morphism MATH this yields a map MATH which is easily seen to be a ring homomorphism. It is equally clear that the ideal MATH defined above is mapped to ... |
math/0002064 | REF : we remark only that MATH for all MATH and leave the details to the reader. We prove REF . From CITE REF it follows easily that MATH. It then suffices to show that MATH is almost zero. We have MATH and the product MATH is again a NAME product. Next let MATH. If MATH then MATH and MATH. In view of REF , MATH is alm... |
math/0002064 | REF is easy and we leave it to the reader. To prove REF , take a finitely generated ideal MATH such that MATH, pick a morphism MATH whose cokernel is annihilated by MATH, and apply the following REF . |
math/0002064 | We need the following Let MATH be a finitely generated MATH-module and suppose that we are given MATH and a (not necessarily commutative) diagram MATH such that MATH, MATH. Let MATH be an ideal such that MATH has a finitely generated submodule containing MATH. Then MATH has a finitely generated submodule containing MAT... |
math/0002064 | The ``only if" part in REF (respectively, REF ) is first checked when MATH is finitely generated (respectively, finitely presented) and then extended to the general case. We leave the details to the reader and we proceed to verify the ``if" part. For REF , choose a set MATH and an epimorphism MATH. Let MATH be the dire... |
math/0002064 | These facts can be deduced from REF , or proved directly. |
math/0002064 | Left to the reader. |
math/0002064 | REF : for given MATH, we consider any MATH-module MATH and we apply the functor MATH to REF : MATH which implies MATH for all MATH. Since MATH is arbitrary, REF follows from REF . REF : by hypothesis, for arbitrary MATH we can find MATH and a morphism MATH such that MATH. Let MATH be the image of MATH, so that MATH fac... |
math/0002064 | We have a spectral sequence : MATH . On the other hand we have also natural isomorphisms MATH . Hence : MATH which is the claim. |
math/0002064 | REF : let MATH be such a MATH-module. Let MATH and pick a three term complex MATH such that MATH. Set MATH; this is a finitely presented MATH-module and MATH factors through a morphism MATH. Let MATH; from REF we see that MATH is the image of some element MATH. If we define MATH and MATH, MATH by MATH and MATH, then cl... |
math/0002064 | Pick an indexing set MATH large enough, and an epimorphism MATH. For every MATH we have the standard morphisms MATH such that MATH and MATH. For every MATH choose MATH such that MATH. It is easy to check that MATH is generated by the almost elements MATH (MATH, MATH). REF follows already. For REF , the ``only if" is cl... |
math/0002064 | If MATH for some finite set MATH, then MATH and the claims are obvious. More generally, if MATH is almost projective and almost finitely generated, for any MATH there exists a finite set MATH and morphisms MATH such that MATH. We apply the natural transformation MATH to REF : an easy diagram chase allows then to conclu... |
math/0002064 | REF is an easy consequence of REF . To prove REF , we we apply the natural transformation REF to REF : by diagram chase one sees that the kernel and cokernel of the morphism MATH are killed by MATH. |
math/0002064 | Let MATH. For any MATH let MATH. Then MATH. For MATH set MATH and let MATH be the natural morphism. An easy induction shows that MATH for all MATH. Since MATH we obtain MATH for all MATH. Therefore MATH for all MATH. Since MATH, the claim follows. |
math/0002064 | Let MATH. By the above REF it suffices to show that MATH vanishes for all MATH and all MATH-modules MATH. The maps MATH define a map MATH such that we have a short exact sequence MATH. Applying the long exact MATH sequence one obtains a short exact sequence (cp. REF ) MATH . Then REF implies that MATH for all MATH and ... |
math/0002064 | Let MATH be a flat MATH-module and MATH an injective map of MATH-modules. Denote by MATH the kernel of the induced map MATH; we have MATH. We obtain an exact sequence MATH. But one sees easily that MATH and MATH, which shows that MATH is a flat MATH-module. Similarly, let MATH be two MATH-modules. Then the natural map ... |
math/0002064 | Using the idempotent MATH we get a MATH-linear decomposition MATH where the bimodule structure on MATH is given by the zero morphisms and the bimodule structure on MATH is given by MATH and MATH. We have to prove that MATH is equivalent to a one-point category. By REF we can assume that MATH or MATH. By REF we have MAT... |
math/0002064 | REF : the hypothesis MATH implies that MATH is a morphism of (non-unitary) MATH-monoids. We can then replace MATH by MATH and thereby assume that MATH, MATH and MATH is a (non-unitary) MATH-bimodule and the right and left actions on MATH coincide. The assertion to prove is that MATH lifts to a unique idempotent MATH. H... |
math/0002064 | Of course REF is an immediate consequence of REF . To show REF , let MATH be any object of MATH. Using REF one sees easily that the sequence MATH is right exact; MATH won't be exact in general, unless MATH (and therefore MATH) is an exact algebra. In any case, the kernel of MATH is almost zero, so we get an extension o... |
math/0002064 | Using the adjunction REF we are reduced to showing that the natural map MATH is a bijection for all MATH-modules MATH. Given MATH we construct MATH. We extend MATH to MATH by setting it equal to zero on MATH. Then it is easy to check that the resulting map descends to MATH, hence giving a MATH-derivation MATH. This pro... |
math/0002064 | To ease notation, set MATH and MATH. We have natural isomorphisms : MATH . But it is easy to see that the natural map MATH is an isomorphism. |
math/0002064 | With the notation of the proof of REF we have natural isomorphisms MATH where the last isomorphism follows directly from REF and the subsequent REF . Finally, REF shows that MATH, as required. |
math/0002064 | It follows directly from CITE |
math/0002064 | REF gives an isomorphism : MATH (and likewise for MATH). Then the claim follows from REF . |
math/0002064 | For any almost MATH-algebra MATH we let MATH denote the complex of MATH-modules MATH placed in degrees MATH; we have a distiguished triangle MATH . By the assumption, the natural map MATH is a quasi-isomorphism and MATH. On the other hand, for all MATH we have MATH . In particular MATH for all MATH. As MATH is flat ove... |
math/0002064 | Let us remark that the functor MATH : MATH commutes with tensor products; hence the same holds for the functor MATH (see REF ). Then, in view of REF , the theorem is reduced immediately to REF . |
math/0002064 | Since MATH for all MATH, REF applies (with MATH and MATH), giving the natural isomorphisms MATH . Since MATH, the same theorem also applies with MATH, MATH, MATH, and we notice that in this case MATH; hence we have MATH . Next we apply transitivity to the sequence MATH, to obtain (thanks to REF) MATH . Applying MATH to... |
math/0002064 | We replace MATH by MATH and apply the functor MATH (which commutes with tensor products by REF ) thereby reducing the assertion to the above mentioned REF . |
math/0002064 | For REF use the Tor sequences. In view of REF , to show REF it suffices to know that MATH is an almost finitely presented MATH-module; but this follows from the existence of an epimorphism of MATH-modules MATH defined by MATH. Of the remaining assertions, only REF are not obvious, but the proof is just the ``almost ver... |
math/0002064 | Suppose that MATH is unramified. We start by showing that for every MATH there exist almost elements MATH of MATH such that MATH . Since MATH is an almost projective MATH-module, for every MATH there exists an ``approximate splitting" for the epimorphism MATH, that is, a MATH-linear morphism MATH such that MATH. Set MA... |
math/0002064 | Left to the reader as an exercise. |
math/0002064 | Let MATH and MATH be given. Obviously we have MATH and MATH. Pick morphisms MATH and MATH as in REF. Using the foregoing notation, we can write : MATH from which the claim follows directly. |
math/0002064 | REF is left to the reader as an exercise. For REF we compute using REF : MATH. |
math/0002064 | Suppose first that there exists a splitting MATH for MATH, so that we can view MATH as a matrix MATH, where MATH. By additivity of the trace, we are then reduced to show that MATH. By REF , this is the same as MATH, which obviously vanishes. In general, for any MATH we consider the morphism MATH and the pull back morph... |
math/0002064 | REF are left as exercises for the reader. We verify REF . For given MATH pick morphisms MATH and MATH such that MATH and MATH. If we set MATH, MATH, MATH and MATH then we have MATH. Define MATH . Using REF we can write MATH which implies immediately the claim. |
math/0002064 | Under the stated hypotheses, MATH is an almost projective MATH-module (by REF ). Let MATH and MATH the trace morphism for the morphism of almost MATH-algebras MATH. By faithful flatness, the natural morphism MATH is a monomorphism, hence it suffices to show that MATH is an epimorphism (here MATH is considered as a MATH... |
math/0002064 | Suppose that MATH is étale. Let MATH be the idempotent almost element of MATH provided by REF . We define a morphism MATH by MATH. To start with, we remark that both MATH and MATH are MATH-linear morphisms (for the natural MATH-module structure of MATH defined in REF ). Indeed, let us pick any MATH, MATH and compute di... |
math/0002064 | Under the stated hypothesis, MATH is an almost projective MATH-module (by virtue of REF ). Hence, for given MATH, pick a sequence of morphisms MATH such that MATH; with the notation of REF, define MATH by MATH, so that MATH. One verifies easily that MATH for all integers MATH. Now, suppose that MATH. It follows that MA... |
math/0002064 | Let MATH be any MATH-extension of MATH by MATH. The composition MATH of the structural morphism for MATH followed by MATH coincides with the projection MATH. Therefore MATH and MATH. Hence MATH factors through MATH; the restriction of MATH to MATH defines a morphism MATH and a morphism of MATH-extensions MATH. In this ... |
math/0002064 | It follows directly from the (almost version of the) local flatness criterion (see REF ). |
math/0002064 | By induction we can assume MATH. Then REF follows directly from REF . We show REF : by REF (and again REF ) a given weakly étale morphism MATH can be lifted to a unique flat morphism MATH. We need to prove that MATH is weakly étale, that is, that MATH is MATH-flat. However, it is clear that MATH is weakly étale, hence ... |
math/0002064 | Under the assumptions, we can find a finitely generated MATH-module MATH such that MATH. By REF , there exists a finite filtration MATH such that each MATH is a quotient of a direct sum of copies of MATH. This implies that, for every MATH-module MATH, we have MATH . REF follows easily. Notice that if MATH and MATH, the... |
math/0002064 | REF : we have to show that MATH is almost zero for every MATH-module MATH. Let MATH; by REF is nilpotent, so by the usual devissage we may assume that MATH. If MATH is represented by an extension MATH then after tensoring by MATH and using the flatness of MATH we get an exact sequence of MATH-modules MATH. Thus MATH co... |
math/0002064 | As usual we reduce to MATH. Then REF applies with MATH, MATH, MATH, MATH, MATH and MATH. We obtain a class MATH which gives the obstruction to the existence of a flat MATH-module MATH lifting MATH. Since MATH is almost projective, we know that MATH, which says that MATH for all MATH. In other words, for every MATH we c... |
math/0002064 | Since both MATH and MATH are almost projective and MATH are epimorphisms, there exist morphisms MATH and MATH such that MATH and MATH. Then we have MATH and MATH, that is, the morphism MATH (respectively, MATH) has image contained in the almost submodule MATH (respectively, MATH). Since MATH this implies MATH and MATH.... |
math/0002064 | Let MATH be such a morphism. Under the assumption, we can find a finite MATH-module MATH such that MATH. One sees easily that MATH is a faithful MATH-module, so by CITE REF , MATH satisfies the following condition : MATH . Now let MATH be a MATH-module such that MATH is flat. Pick an epimorphism MATH with MATH free ove... |
math/0002064 | In the weakly étale case, we have to show that the multiplication morphism MATH is flat. As MATH is nilpotent, the local flatness criterion reduces the question to the situation over MATH. So we may assume that MATH is a monomorphism. Then MATH is a monomorphism, but MATH is the multiplication morphism of MATH, which i... |
math/0002064 | General nonsense. |
math/0002064 | Indeed, MATH is the natural morphism. So, the assertions follow by applying MATH to the short exact sequence of MATH-modules MATH where MATH and MATH. |
math/0002064 | Indeed, in this case, REF is split exact as a sequence of MATH-modules, and it remains such after tensoring by MATH. |
math/0002064 | To fix ideas, suppose that MATH is an epimorphism. Consider any object MATH of MATH. Let MATH; we deduce a natural morphism MATH such that MATH. It follows that MATH is injective, hence it is an isomorphism, by REF . We derive a commutative diagram with exact rows : MATH . From the snake lemma we deduce MATH . Since MA... |
math/0002064 | Suppose that MATH is an epimorphism and let MATH be its kernel. Let MATH; it suffices to show that MATH is a flat MATH-module. However, in view of REF , the assumption implies that MATH is a flat MATH-module. MATH is the kernel of the epimorphism MATH. Moreover, MATH identifies naturally with an ideal of MATH and MATH.... |
math/0002064 | The assertion for flat almost modules follows directly from REF . Set MATH. To establish the second equivalence, it suffices to show that, if MATH is a MATH-module such that MATH is almost projective over MATH, then MATH is almost projective over MATH, or which is the same, that MATH for all MATH and any MATH-module MA... |
math/0002064 | For any MATH, denote by MATH (respectively, MATH) the MATH-fold tensor product of MATH (respectively, MATH) with itself over MATH (respectively, MATH), and by MATH the natural morphism. First of all we claim that, for every MATH, the natural diagram of almost algebras MATH is cartesian (where MATH and MATH are MATH-fol... |
math/0002064 | Let us say that MATH is an epimorphism with kernel MATH. Then MATH is also an ideal of MATH and we have MATH and MATH. We intend to apply REF to the morphism MATH. However, the induced morphism MATH in MATH has a section, and hence it is of universal effective descent for every fibred category. Thus, we can replace in ... |
math/0002064 | REF : suppose that MATH for a finite MATH-algebra MATH; then MATH is a finite MATH-algebra and MATH. It remains to show that MATH is nilpotent. Suppose that MATH is generated by MATH elements as a MATH-module and let MATH (respectively, MATH) be the NAME ideal of MATH (respectively,of MATH); we have MATH (see CITE (Cha... |
math/0002064 | REF : if MATH is an inverse of MATH up to MATH and MATH is an inverse of MATH up to MATH, then MATH is an inverse of MATH up to MATH. REF : given an inverse MATH of MATH up to MATH and an inverse MATH of MATH up to MATH, let MATH. We compute : MATH . REF is similar and REF is an easy diagram chasing left to the reader.... |
math/0002064 | REF : given MATH, construct a morphism MATH using the morphism MATH coming from MATH and MATH. REF : the assertion for MATH is clear, and the assertion for MATH follows from REF . |
math/0002064 | REF : we first show that MATH is faithfully flat. Since MATH is flat, it remains to show that if MATH is a MATH-module such that MATH, then MATH. It suffice to do this for MATH, for an arbitrary ideal MATH of MATH. After base change by MATH, we reduce to show that MATH implies MATH. However, MATH is invertible up to MA... |
math/0002064 | We first consider REF for the special case where MATH. The functor MATH induces a functor MATH, and by restriction (see REF ) we obtain a functor MATH; by REF , the latter is isomorphic to the functor MATH of the lemma. Furthermore, from REF we derive a natural ring isomorphism MATH, hence an essentially commutative di... |
math/0002064 | In light of REF , it suffices to show that MATH is invertible up to a power of MATH. For this, factor the identity morphism of MATH as MATH and argue as in the proof of REF . |
math/0002064 | Suppose first that MATH is an isomorphism; in this case we claim that MATH is an epimorphism and MATH for any MATH-algebra MATH. Indeed, since by REF, MATH acts on MATH, hence MATH annihilates MATH, hence annihilates its image MATH, whence the claim. If, moreover, MATH is a MATH-algebra, REF implies that MATH is invert... |
math/0002064 | REF : let MATH be a quasi-isomorphism of MATH-algebras. Clearly the induced morphism MATH is still a quasi-isomorphism of MATH-algebras. But by REF , MATH and MATH are exact simplicial almost MATH-algebras; moreover, it follows from REF that MATH is a quasi-isomorphism of MATH-modules. Then the claim follows easily fro... |
math/0002064 | We show by induction on MATH that VAN REF MATH . For MATH the claim follows immediately from REF . Therefore suppose that MATH and that VAN REF is known for all almost isomorphisms of MATH-algebras MATH and all MATH. Let MATH. Then by transitivity REF we have a distinguished triangle in MATH . We deduce that VAN REF an... |
math/0002064 | Apply the base change theorem (CITE II. REFEF) to the (flat) projections of MATH onto MATH and respectively MATH to deduce that the natural map MATH is a quasi-isomorphism in MATH. By REF the induced morphism MATH is still a quasi-isomorphism. There are spectral sequences MATH . On the other hand, by REF we have MATH f... |
math/0002064 | By transitivity we may assume MATH. Let MATH be the standard resolution of MATH (see CITE II. REFEF). Each MATH contains MATH as a direct summand, hence it is exact, so that we have an exact sequence of simplicial MATH-modules MATH. The augmentation MATH is a quasi-isomorphism and we deduce that MATH is a quasi-isomorp... |
math/0002064 | Let MATH be the map corresponding to MATH under the bijection REF. By inspection, the compositions MATH and MATH are induced by scalar multiplication. Pick any MATH and lift it to an element MATH; define MATH by MATH for all MATH. Then MATH and MATH. This easily implies that MATH annihilates MATH and MATH. In light of ... |
math/0002064 | Fix an element MATH. For each MATH-algebra MATH and each element MATH we get a map MATH by MATH, hence a map MATH. For varying MATH we obtain a map of sets MATH : MATH. According to the terminology of CITE, the system of maps MATH for MATH ranging over all MATH-algebras forms a homogeneous polynomial law of degree MATH... |
math/0002064 | It is deduced directly from REF by applying MATH. |
math/0002064 | This is deduced from REF applied to MATH which is a quasi-isomorphism of chain complexes of flat MATH-modules. We note that CITE deals with a more general mixed simplicial construction of MATH which applies to bounded above complexes, but one can check that it reduces to the simplicial definition for complexes in MATH. |
math/0002072 | This is an immediate consequence of REF , since the second term on the right-hand side is positive semi-definite, and zero iff MATH. |
math/0002072 | Suppose MATH is a NAME critical point; that is, a point at which MATH is stationary. (There is no loss of generality in omitting a possible gauge transformation on one side of this equation.) By considering just MATH of the form MATH, for MATH, it is clear from REF that MATH is a necessary condition for MATH to be a cr... |
math/0002072 | A simple calculus of variations computation shows that MATH extremizes MATH iff MATH . On the other hand, this condition also ensures that the covariantly constant curvatures MATH, related by the non-Abelian analog of NAME Theorem mentioned previously, are continuous at MATH. This was the only place MATH might have fai... |
math/0002075 | Write MATH, where the prime denotes the imaginary part. Then MATH . All these products, except for the cross-product, are commutative, and REF follows. From the same formula with MATH we obtain MATH. This vanishes if and only if MATH is real or purely imaginary. Together with REF we obtain REF. |
math/0002075 | CASE: If MATH and if MATH is a unit vector orthogonal to REF, then MATH. Hence MATH works for MATH, and the uniqueness, up to sign, follows easily from MATH and MATH. If MATH is arbitrary, and MATH then put MATH. Clearly, MATH. Moreover, MATH works for MATH if and only if MATH works for MATH. CASE: If MATH, and MATH is... |
math/0002075 | The tangent vector MATH is identified with the homomorphism MATH, that maps MATH into MATH. |
math/0002075 | We consider MATH as a (right) complex vector space with imaginary unit MATH. Then MATH is MATH-linear and has a (complex) eigenvalue MATH. If MATH, then MATH . Hence MATH. We choose a basis MATH of MATH such that MATH, that is, MATH for some MATH, and MATH. Then MATH implies MATH . For the affine parametrization MATH w... |
math/0002075 | Choose non-zero basis vectors MATH. Then elements in MATH and endomorphisms of MATH or of MATH are represented by quaternionic MATH-matrices, and therefore the assertion reduces to that of REF . |
math/0002075 | The vector bundle MATH with fibre MATH has a total space of real dimension MATH. Therefore there exists MATH such that MATH has no zero. REF yields a unique holomorphic structure MATH such that MATH. Now any MATH is of the form MATH for some MATH. Then, by REF, for any section MATH we have MATH . Note that MATH implies... |
math/0002075 | Recall from REF MATH . Therefore, using MATH, MATH . But MATH by the following type argument: Using that MATH we have MATH . Similarly MATH, because MATH is left MATH and MATH is right MATH. |
math/0002075 | We first need a formula for the derivative of MATH-forms MATH which stabilize MATH, that is, MATH. If MATH, then for MATH where we wedge over composition. Note that the composition MATH makes sense, because MATH, and MATH is annihilated by MATH. We apply this to MATH and MATH. Since MATH we have on MATH, by REF , MATH ... |
math/0002075 | We have MATH, and therefore MATH . Then, from REF, MATH . Similarly, MATH . But, by a property of the real trace, MATH . |
math/0002075 | Let MATH be a variation of MATH in MATH with variational vector field MATH. Then MATH and MATH . Using the wedge REF and MATH, we get MATH . Thus MATH . Therefore MATH is harmonic if and only if MATH is normal. For the other equivalences, first note MATH . Now, together with MATH and MATH, this implies MATH . |
math/0002075 | Recall from REF MATH . Further MATH by type. Therefore MATH and similarly for MATH. |
math/0002075 | We may assume MATH. Then MATH, and MATH implies MATH. Let MATH, and MATH . Then MATH, and MATH implies MATH . Therefore MATH is imaginary, too. It follows MATH, and MATH . |
math/0002075 | MATH . Because MATH we similarly have MATH see REF. Because MATH, we have MATH . This proves REF. The positivity follows from REF . |
math/0002075 | CASE: We consider REF-form on MATH defined by MATH . Then MATH is a linear combination of terms of the form MATH . But if MATH, we get MATH hence MATH. Therefore, if MATH is the inclusion, MATH . CASE: We have MATH and REF yields the formula. The topological invariance under deformations of MATH follows from NAME theor... |
math/0002075 | Let MATH be a variation, and MATH its mean curvature sphere. Note that for MATH to stay conformal the complex structure, that is, the operator MATH, varies, too. The variation has a variational vector field MATH given by MATH . As usual, we abbreviate MATH by a dot. Note that for MATH . We now compute the variation of ... |
math/0002075 | For MATH because MATH. But MATH is right MATH, and MATH is left MATH. Hence, by type, MATH . This shows the right hand inclusion. Also, MATH . |
math/0002075 | We know from REF that MATH is an involution with the tangent space as its fixed point set: MATH . Its MATH-eigenspace is the normal space, so we need to compute MATH . But differentiation of REF yields MATH or MATH . |
math/0002075 | By definition of the trace, MATH but MATH . If follows that MATH and MATH . Similarly for MATH. |
math/0002075 | MATH . Therefore MATH . This proves the formula for MATH. Using REF and the NAME equation MATH we find, after a similar computation, MATH . On this we use REF to obtain REF. |
math/0002075 | REF give MATH . |
math/0002075 | We only have to consider the reformulation of MATH. But MATH . |
math/0002075 | MATH . Now see REF and, for the second equality, REF. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.