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math/0007121 | We will construct a pseudotensor functor MATH from MATH to MATH. On objects MATH we let MATH. In order to define it on polylinear maps, we need to find out how MATH can be recovered from MATH and the action of MATH. Denote by MATH the embedding MATH, and let MATH be the composition MATH . Explicitly, MATH is given by the formula MATH . This is a homomorphism of both left MATH-modules and of right MATH-modules. Moreover, for MATH, MATH, we have: MATH . The crucial observation, which will allow us to invert MATH, is that for any MATH, MATH, we have: MATH . Here MATH acts diagonally on MATH from the right; the left-hand side of REF is invariant under MATH. Given a polylinear map MATH in MATH, we can extend it to a map MATH . (Note, however, that MATH is not MATH-linear.) Now we define MATH by the formula: MATH . It is easy to check that MATH is MATH-linear, so it is a polylinear map in MATH. Moreover, MATH. For a polylinear map MATH in MATH, it is immediate from REF and the MATH-linearity of MATH that MATH. Therefore, MATH is a pseudotensor functor inverse to MATH. |
math/0007121 | Can be found in CITE. |
math/0007121 | By the construction of the filtrations it is evident that we can assume MATH. Then MATH, and MATH acts on it by linear differential operators. The rest of the proof is clear. |
math/0007121 | For a proof of REF see CITE. In order to prove REF , notice that if MATH is a derivation of MATH, then its action on MATH is given by MATH for all MATH, where the MATH form a topological basis of MATH and the MATH are continuous derivations of MATH. Subtracting MATH from MATH, we get a derivation MATH acting trivially on MATH. We are going to show that if MATH is simple then MATH is of the form MATH where MATH. Let us fix MATH. Then MATH can be written as MATH, where MATH are continuous MATH-endomorphisms of MATH. From MATH we see that MATH for every MATH. This means that MATH commutes with MATH for all MATH. By a NAME 's lemma argument CITE and simplicity of MATH, we conclude that the MATH are multiples of the identity map, hence MATH for some MATH and all MATH. It is now immediate to check that the mapping MATH is indeed a derivation of MATH, proving REF . In order to prove the rest of the statements, denote by MATH the MATH-th component of the canonical MATH-gradation of MATH, MATH, MATH or MATH. This is a reductive subalgebra of MATH, hence MATH, where MATH. But MATH, hence MATH, that is, any element MATH defines an endomorphism of MATH viewed as a MATH-module. Since MATH and MATH, we conclude that MATH also preserves the canonical gradation of these NAME algebras and we may assume that MATH acts trivially on the MATH-st component. Using the transitivity of MATH and MATH, we conclude that MATH. By REF , all components of the canonical MATH-gradation of MATH and MATH are inequivalent MATH-modules, hence MATH preserves this gradation in this case as well. Subtracting from MATH a multiple of MATH, we may assume that MATH acts trivially on the MATH-st component and, using transitivity, we conclude that MATH is a multiple of MATH. Thus REF is proved. Since MATH acts completely reducibly on the space MATH of MATH-cocycles on MATH with values in MATH, and since MATH acts trivially on cohomology, we may choose a subspace MATH of MATH, complementary to the space of trivial MATH-cocycles, on which MATH acts trivially. Hence for any MATH-cocycle MATH we have: MATH if MATH, MATH and MATH are irreducible non-contragredient MATH-submodules of MATH. Let MATH for short, where MATH is the MATH-th component of the canonical gradation. It follows from REF that all pairs of MATH-submodules in MATH are non-contragredient, except for the adjoint MATH-submodules in MATH. Thus, we have: MATH . Taking now MATH, MATH and MATH, the cocycle condition MATH gives MATH. Since MATH, we conclude that MATH. Hence all central extensions of MATH are trivial. Likewise, MATH is zero on any pair of subspaces MATH, unless MATH, and on the pair MATH (see REF ). Choosing MATH, MATH, MATH, we obtain, as above, from the cocycle condition, that MATH is zero on the pair MATH. It follows from REF applied to the subalgebra MATH of MATH (see REF ) that MATH is zero on this subalgebra if MATH. Thus any cocycle on MATH is trivial. In the case of MATH the cocycle MATH is trivial. The case of MATH is similar. In the remaining case of MATH we show, as above, that the cocycle MATH is trivial on any pair MATH if MATH. Using the cocycle condition for MATH, MATH and MATH, and the fact that MATH, we conclude that MATH is trivial on any pair MATH as well, unless MATH. It is easy to see that this implies that MATH. |
math/0007121 | REF follows from REF. The skew-commutativity of the bracket REF follows immediately from that of MATH. The proof of the NAME identity is straightforward by using REF. Let us check for example that the associativity of MATH is equivalent to that of MATH; the case of the NAME identity is similar. We will use the notation from REF - REF, and we will write MATH for MATH, MATH. Then we want to compute the products MATH and MATH. By definition, if we have REF, then MATH . Similarly, if we have REF, then MATH . Now recalling REF, we see that the associativity of MATH is equivalent to that of MATH. |
math/0007121 | We have MATH for every MATH. Since the right action of MATH on MATH is surjective (see REF ), it follows that MATH for any MATH. |
math/0007121 | Let us choose bases of MATH and MATH, and let us fix expressions of elements from the first basis as MATH-linear combinations of elements from the second basis. Denote by MATH the highest degree of the coefficients of all these expressions. Using REF, we see that MATH for all MATH. Repeating the same reasoning after switching the roles of MATH and MATH, we get MATH for some MATH and all MATH. |
math/0007121 | CASE: The linear-compactness follows from REF , since REF a filtration by finite-codimensional subspaces with trivial intersection and MATH is complete with respect to this filtration. The continuity of the MATH-action follows from REF: MATH . CASE: It only remains to check that the NAME bracket of MATH is continuous. Let MATH be a finite-dimensional (over MATH) subspace of MATH which generates it over MATH. For MATH, we can write MATH for some MATH and MATH. Then the NAME bracket in MATH, for MATH, is given by: MATH . We can find a number MATH such that all coefficients MATH occurring in pseudobrackets of any elements MATH belong to MATH. Then REF imply: MATH where MATH. This shows that the NAME bracket is continuous. |
math/0007121 | Follows immediately from REF . |
math/0007121 | First of all, notice that we can assume MATH is torsion-free, since by REF , MATH where MATH is the torsion submodule of MATH. The proof of the proposition is then based on REF and the following two lemmas. The map MATH induced by the inclusion MATH constructed in REF is uniformly continuous, that is, for every MATH we have: MATH where MATH and MATH are independent of MATH. The same is true for MATH where MATH is the embedding MATH from REF . Let us choose finite-dimensional vector subspaces MATH of MATH generating MATH over MATH, and MATH of MATH generating MATH over MATH and containing MATH. Let us also choose a second finite-dimensional vector subspace MATH of MATH containing MATH and generating MATH over MATH. We will denote the filtrations induced by these subspaces by MATH, MATH, and MATH, respectively. Up to identifying MATH with MATH, we have constructed injective maps MATH such that the composition MATH is a multiplication by MATH. These maps induce maps MATH which are surjective, as one can see by tensoring by MATH and using that MATH if MATH is a torsion MATH-module (see REF ). The above maps are also continuous with respect to the common topology defined by any of the above constructed filtrations. In fact, by construction, one has MATH . The second inclusion proves that MATH for some MATH independent of MATH, because the filtrations MATH and MATH are uniformly equivalent by REF . Applying MATH to the first inclusion, we get MATH. On the other hand, MATH, and REF implies that MATH where MATH is such that MATH but MATH. Therefore MATH for all MATH. A similar argument works for MATH. If MATH is a surjective uniformly continuous map of filtered modules, then MATH. By REF for all MATH and some MATH independent of MATH. The induced map MATH is a surjective map of finite-dimensional vector spaces. Hence MATH. Using the above lemmas, now we can complete the proof of REF . We have constructed an embedding MATH of MATH into a free MATH-module MATH, and we have shown that the induced map MATH is surjective and uniformly continuous. This implies MATH. Similarly, the inclusion MATH gives us MATH. Therefore, MATH. It remains to note that, for any (nonzero) free MATH-module MATH of finite rank, one has MATH. This follows from the fact that MATH because MATH. |
math/0007121 | It is well known that MATH is trivial for MATH and MATH-dimensional for MATH; see for example, CITE. |
math/0007121 | The proof is by induction on the number of MATH not contained in MATH, the basis of induction being trivial. Consider MATH, with all MATH. We can write MATH so that MATH is a linear combination of NAME - NAME - NAME basis elements of MATH not containing MATH in their expression, and MATH. Then: MATH . Since the third summand in the right-hand side belongs to MATH, it follows that the first and second summands lie in MATH too. This implies: MATH. Hence MATH and we can apply the inductive assumption. |
math/0007121 | First of all, it is easy to check that the elements REF indeed belong to MATH. REF is easy, and the computation of the pseudobrackets is straightforward using REF, reformulated as MATH . Now let us consider an element MATH, MATH. We will prove that MATH can be expressed as MATH-linear combination of the above elements REF by induction on the maximal degree MATH of the MATH. Since MATH, then MATH, hence MATH. By REF , we can find elements MATH such that MATH . Therefore the difference MATH still lies in MATH and its first tensor factor terms have degree strictly less than MATH. By inductive assumption, we are done. |
math/0007121 | Straightforward computation, using REF . |
math/0007121 | Similar to that of REF. If MATH, there is nothing to prove. Let MATH be elements that complement MATH to a basis of MATH. If MATH is not in MATH we take it to be one of the MATH's. Write out REF as MATH . Now, if MATH involves some MATH's, there is no way to cancel out the terms MATH. This proves that MATH. Similarly, REF reads MATH . If, for example, MATH involves MATH's, then the terms MATH cannot be cancelled. Therefore MATH. If MATH involves MATH's, then the terms MATH can be cancelled only with terms coming from MATH. This shows that MATH. |
math/0007121 | Let us write MATH and MATH (summation over repeated indices). Then REF is equivalent to MATH where ``cyclic" means summing over cyclic permutations of the indices MATH. Multiply this equation by MATH and sum over MATH. Using that MATH, we get MATH where now the cyclic permutations are over MATH. This is exactly REF . Conversely, multiplying REF by MATH and summing over MATH, we get REF. Similarly, since MATH, we can write MATH. Then REF is equivalent to MATH which after multiplying by MATH and summing over MATH becomes MATH or MATH. Conversely, REF gives REF after multiplying by MATH and summing over MATH. Now start with a solution MATH of REF . Above we have deduced REF and MATH. On the other hand, since MATH, we have MATH, and REF implies MATH. Together with MATH this gives MATH, which is REF. If we start with a solution MATH of REF , the above arguments can be inverted to show that MATH, and we get REF . |
math/0007121 | It remains to show that, conversely, any vector field that preserves the form MATH is equal to MATH for some MATH. Indeed, let MATH be such that MATH. Since MATH and MATH, this is equivalent to MATH which implies MATH for some MATH. In other words, MATH for any MATH. Using MATH, we get MATH for MATH. This, together with REF, implies MATH since the MATH-form MATH is nondegenerate. |
math/0007121 | The proof is very similar to that of REF . There we showed that MATH is equivalent to REF, and the same argument applies here. Similarly, REF is equivalent to REF . Now if MATH, then MATH implies MATH, which together with REF leads to MATH. |
math/0007121 | Given MATH, above we have defined the MATH-form MATH such that MATH, MATH and MATH. Since MATH is nondegenerate, we have MATH. Conversely, starting with a contact MATH-form MATH, we can define MATH and MATH satisfying REF - REF. |
math/0007121 | It remains to show that, conversely, any vector field from MATH is equal to MATH for some MATH. Indeed, let MATH be such that MATH for some MATH. Let us write MATH for some MATH. Then MATH and MATH, while MATH. Therefore MATH, which implies MATH, MATH, and MATH. |
math/0007121 | CASE: We have seen in REF that MATH. Let MATH be one of the pseudoalgebras MATH, MATH or MATH and MATH be its natural embedding in MATH. We have shown in REF that in this case the image of MATH in MATH under MATH is MATH, MATH or MATH, respectively. REF below implies that MATH is a central extension of its image in MATH. Moreover, since MATH is simple, it is equal to its derived subalgebra, and therefore MATH is equal to its derived subalgebra (see REF). Hence, MATH is an irreducible central extension of its image in MATH. Now REF implies that MATH, MATH in the cases MATH, MATH respectively, and MATH is a quotient of MATH in the case MATH. However, since MATH is a free MATH-module of rank one, MATH is isomorphic to MATH as a topological MATH-module. Therefore, MATH. CASE: Note that MATH maps surjectively to MATH with kernel isomorphic to MATH. Moreover MATH, MATH, and hence MATH. We have: MATH and MATH. |
math/0007121 | We define MATH in terms of its NAME coefficients - the MATH-products MATH. We have already seen, when we discussed associativity, that REF is equivalent to the following equation (compare REF): MATH . This can be inverted to find (compare REF): MATH . The MATH-sesqui-linearity properties of MATH with respect to MATH and MATH are easy to check by a direct calculation. By REF of the filtration MATH, and locality of MATH, it follows that for each fixed MATH, MATH there is a MATH such that MATH for MATH. Therefore, for each MATH we have defined MATH. In order that MATH be well defined, we need to check that MATH satisfies locality, that is, that MATH when MATH with MATH. By the locality of MATH and MATH, for each MATH there is a MATH such that MATH for MATH and all MATH. Since MATH is finite, we can choose a MATH that works for all MATH belonging to a set of generators of MATH over MATH. Now the MATH-sesqui-linearity of MATH with respect to MATH (for fixed MATH) implies that MATH for all MATH and MATH. Hence MATH for MATH. |
math/0007121 | REF is an immediate consequence of REF . Indeed, the only thing that remains to be checked is the associativity of MATH, and it follows from REF: MATH . To prove REF , we associate with each MATH the MATH-pseudolinear map MATH given by MATH. Then MATH which shows that MATH. |
math/0007121 | Since MATH, we can identify MATH with MATH so that its action on MATH is given by REF. To prove REF, we use REF and the explicit definition of associativity from REF. Due to MATH-bilinearity, we can assume that MATH. Then: MATH . On the other hand, we have: MATH . Now REF follows from the uniqueness from REF . |
math/0007121 | Since MATH is a MATH-submodule of MATH, after replacing MATH with MATH, we can assume that MATH. By definition, MATH. Since, by REF , MATH, we have MATH. On the other hand, since MATH is finite over MATH and MATH satisfies REF, there exists a finite-dimensional subspace MATH of MATH such that MATH. It follows that MATH, which is finite dimensional. Since MATH is injective, MATH is finite dimensional. |
math/0007121 | REF follows from REF : if MATH with linearly independent MATH, then MATH can be zero only if all MATH, which implies MATH. Similarly, REF follows from REF, since we can write MATH uniquely in the form MATH. |
math/0007121 | REF are obvious. CASE: By NAME 's REF, MATH with MATH. If MATH is finite dimensional, any continuos homomorphism MATH must contain some MATH in its kernel. Let MATH but MATH. Then, by REF , MATH so that for each MATH, MATH for some MATH. This implies MATH, since MATH, proving REF . Similarly, REF follows from the fact that MATH for any nonzero MATH. |
math/0007121 | This can be verified by straightforward but rather tedious computations. To illustrate them, let us check REF. By definition, we have: MATH while MATH . Hence REF is a consequence of the following identity: MATH which can be checked by pairing both sides with MATH. Indeed, MATH . A more conceptual proof can be given by noticing that REF is the same as the formula for the commutator of MATH-pseudolinear maps. For MATH consider the family MATH indexed by MATH. It is easy to see that it satisfies REF. So, if it also satisfies REF, it would give a MATH-pseudolinear map from MATH to itself. Although this is not true in general, the argument still works because all infinite series that appear will be convergent. (In other words, we embed MATH in a certain completion of MATH.) |
math/0007121 | To construct the inverse of MATH, identify MATH with MATH and consider MATH . Here, as before, MATH, MATH are dual bases in MATH and MATH, and MATH for MATH. This is well defined, that is, the sum is finite, because MATH for all but a finite number of MATH and MATH. Using identity REF, it is easy to see that MATH is MATH-linear. Next, we have for MATH: MATH showing that MATH. In particular, MATH is surjective. Assume that MATH for some MATH. This means that MATH for any MATH. But then for any MATH, we have: MATH which implies MATH. Hence MATH is injective. |
math/0007121 | It remains to show that a non-torsion element MATH does not lie in the kernel of MATH. Consider the map MATH constructed in REF . Then MATH. The map MATH induced by MATH maps the NAME coefficient MATH of MATH to the corresponding NAME coefficient MATH of a nonzero element in the free MATH-module MATH. Now, it is clear from REF that MATH for some MATH, hence MATH must be nonzero too. |
math/0007121 | MATH means that MATH, hence MATH is torsion. |
math/0007121 | To show the equivalence of REF , it is enough to tensor the exact sequence MATH with MATH, getting the exact sequence MATH, and to apply REF . Assume that REF holds but MATH is not torsion. Then it contains a nonzero element MATH which generates a free MATH-module. Then MATH has growth MATH, which is a contradiction. |
math/0007121 | Since MATH is torsionless, MATH is injective by REF . Hence, for MATH, MATH, one has MATH iff MATH for all MATH. By REF, this is equivalent to MATH. Hence, for MATH, MATH for all MATH iff MATH lies in the center of MATH. Now MATH is in the kernel of MATH iff MATH for all MATH and MATH, which means that MATH is central for all MATH. |
math/0007121 | Let MATH be in the kernel of MATH; then by the previous lemma MATH is contained in the center of MATH. The latter is finite dimensional by assumption, so the kernel MATH of MATH is of finite codimension in MATH. This implies that MATH is open in MATH, and it contains MATH for some MATH. Let MATH but MATH; then by REF , MATH. Since MATH is MATH-linear, MATH is a MATH-submodule of MATH. Then MATH, therefore MATH and MATH. |
math/0007121 | The only thing that remains to be checked is the locality property for MATH. It follows from that of MATH, since in this case MATH is injective. |
math/0007121 | The kernel of MATH acts trivially on MATH and hence on MATH. |
math/0007121 | Assume that MATH belongs to MATH; then all MATH, MATH. This implies that all NAME coefficients MATH of MATH are zero, hence MATH. |
math/0007121 | Follows easily from REF. Say that MATH lies in the kernel of MATH. Since MATH is a homomorphism of NAME MATH-differential algebras, its kernel is a MATH-stable ideal of MATH. Then by REF, MATH for all MATH, MATH, because in the right-hand side all elements MATH lie in MATH. This means that MATH for all MATH, hence MATH for every MATH. Now, use this in REF to obtain that MATH is central. |
math/0007121 | By REF, MATH is a current NAME algebra over MATH, where MATH is either a finite-dimensional simple NAME algebra MATH (for MATH), or one of the NAME algebras of vector fields MATH, MATH, MATH or MATH (MATH). In particular, we know that MATH. By REF , a sufficiently high power of any nonzero element MATH maps any given open subspace of MATH surjectively onto MATH. This cannot hold if MATH belongs to MATH, therefore MATH is injective and the image of MATH intersects MATH trivially. Comparing the dimensions, we get that MATH is an isomorphism. |
math/0007121 | As was already remarked, MATH is a subalgebra of MATH. Since MATH is a free MATH-module, we have MATH. For MATH, write MATH with MATH and linearly independent MATH. Then for any MATH we have (compare REF): MATH . This is zero for any MATH iff MATH for all MATH, which means MATH. |
math/0007121 | For MATH, both sides of REF are trivial. For MATH, we will proceed by induction on MATH. First of all, note that the compatibility or solvability of the systems REF or REF does not change when we apply an automorphism of MATH. The same is true when we make an elementary transformation: multiply one equation by a function (an element of MATH) and add it to another equation. For example, we can replace all MATH REF by MATH, and correspondingly MATH by MATH and MATH by MATH, as long as we do not violate REF . Any vector field MATH satisfying MATH can be brought to MATH after an automorphism of MATH, so we will assume that MATH. Replacing MATH REF by MATH, we can assume in addition that MATH for MATH. Now it makes sense to put MATH in the equations with MATH in REF. Let us denote MATH, MATH, MATH, MATH. Consider the reduced system MATH for an unknown MATH. Note that, since MATH, we have: MATH for any MATH, hence MATH does not contain MATH. In particular, putting MATH we see that the operators MATH satisfy REF. The other assumptions of the proposition are also easy to check, so by induction the system REF has a solution MATH. The equation MATH has a unique solution MATH satisfying the initial condition MATH . We claim that this MATH is then a solution of the system REF. Indeed, it satisfies REF for MATH. Next, we compute for MATH (using REF, and the compatibility of REF): MATH . This shows that MATH. We can apply the same argument with MATH instead of MATH, and so on, to show that all derivatives with respect to MATH vanish at MATH. |
math/0007121 | Let us choose a basis MATH of MATH and write MATH REF where MATH and MATH, MATH. Note that MATH and MATH is a NAME algebra homomorphism. We can choose the basis MATH in such a way that MATH for MATH, and MATH for MATH. Then MATH is a basis of MATH. Moreover, note that MATH is given by MATH. Let MATH be another transitive embedding of MATH into MATH. Since, by REF , the homomorphism MATH is uniquely determined by the choice of MATH, we can assume that MATH. Then MATH for some MATH. By assumption, MATH, hence MATH for MATH. Now we want to find an automorphism MATH such that MATH and MATH. This equation is equivalent to REF, and it is easy to see that all conditions of REF are satisfied: for example, the system REF is compatible because MATH is a homomorphism. This completes the proof. |
math/0007121 | It is enough to show that the reconstruction functor MATH gives the same result on the topological MATH-modules MATH and MATH. In order to do so, we must show that every MATH-linear continuous homomorphism of MATH to MATH can be obtained from a unique MATH-linear continuous homomorphism of MATH to itself by composing with the canonical projection MATH. Since MATH is linearly compact, by REF there is a bijection between MATH and MATH. The space MATH is nothing but the augmentation ideal MATH. Therefore we are reduced to show that every MATH-linear map MATH is a restriction of a unique MATH-linear map MATH. A MATH-linear MATH is determined by its value on MATH. If MATH, then MATH, hence MATH. Let MATH be the maximal degree of MATH for MATH. Then MATH modulo MATH. This means that there exists some MATH such that for every MATH, MATH modulo MATH. Then the difference between MATH and right multiplication by MATH is still MATH-linear, and its maximal degree on elements from MATH is strictly less than MATH. The proof now follows by induction. |
math/0007121 | REF is standard. REF follows from REF and the fact that MATH. CASE: If MATH has an abelian ideal MATH, then the preimage of MATH under the natural projection MATH must be solvable and strictly bigger than MATH, which is a contradiction. |
math/0007121 | REF has already been proved, when MATH is finite, in REF . In the general case, it can be deduced from REF . Let MATH, then MATH for any MATH. Hence the action of MATH on MATH is trivial, and by REF, MATH for any MATH. In order to prove REF , notice that MATH. Then build a NAME MATH-pseudoalgebra structure on MATH by letting MATH act on the abelian ideal MATH via the adjoint action. Then by REF , MATH is central in MATH, hence MATH acts trivially on MATH. This means MATH. |
math/0007121 | REF is immediate from REF . Assume that REF holds but MATH is not simple. Then MATH has a nontrivial central regular ideal. If MATH is central in MATH for every MATH, then by REF MATH for every MATH, MATH. When MATH is either finite or free, MATH for all MATH if and only if MATH (compare REF ). Therefore MATH for all MATH, MATH, and by REF we get MATH for any MATH. Hence MATH. |
math/0007121 | It is easy to check that MATH satisfies the assumptions of REF (see REF). |
math/0007121 | Let MATH be non-central. Assume that MATH for all MATH. Note that by REF , we have: MATH for MATH. This implies: MATH, which gives MATH for any MATH, MATH. Then we can use REF to show that MATH is central in MATH, which is a contradiction. Therefore, there is some MATH such that MATH has a nonzero NAME coefficient, that is, MATH. Since MATH, and MATH is MATH-stable, we see that all NAME coefficients of MATH lie in MATH. Then, due to REF, all elements in the ideal MATH of MATH generated by MATH have all of their NAME coefficients in MATH, that is, MATH. |
math/0007121 | First of all, we observe that MATH is a linearly compact NAME algebra with respect to the topology defined in REF, see REF . Consider the extended annihilation algebra MATH, obtained by letting MATH act on MATH according to its MATH-module structure. MATH is a linearly compact NAME algebra possessing a fundamental subalgebra, that is, an open subalgebra containing no ideals of MATH. Indeed, if MATH is a finite-dimensional subspace of MATH generating it over MATH, then because of REF, MATH is a subalgebra of MATH for MATH. None of MATH contains ideals of MATH, since every such ideal is stable under the action of MATH and MATH, which implies MATH. The center MATH of MATH is a MATH-stable closed ideal. The quotient MATH is a linearly compact NAME algebra possessing a fundamental subalgebra MATH. REF will be deduced from REF applied for MATH. By REF , any nonzero MATH-stable ideal of MATH contains the image of a nonzero regular ideal of MATH. Since MATH is simple, this means that the only nonzero MATH-stable ideal of MATH is the whole MATH. Then every nonzero ideal of MATH contained in MATH must equal MATH. Hence MATH is a minimal closed ideal of a linearly compact NAME algebra satisfying the assumptions of REF , and is therefore of the form stated in REF of this proposition. Therefore, MATH is a central extension of a current NAME algebra over a simple linearly compact NAME algebra. Moreover, MATH equals its derived subalgebra (otherwise we would have a proper nonabelian subideal of MATH). Hence it is an irreducible central extension. Consider the canonical filtration MATH, where MATH is the canonical filtration of MATH defined in REF (if MATH we put MATH for MATH). Then the growth of MATH (with respect to this filtration) equals MATH. It is clear from REF that any irreducible central extension of MATH has the same growth. On the other hand, with respect to the filtration defined by REF, the growth of MATH is equal to MATH (see REF ). We have to show that the two different filtrations give the same growth. Recall that by REF , a sufficiently high power of any nonzero element MATH maps any given open subspace of MATH surjectively onto MATH. Then the same argument as in the proof of REF shows that MATH intersects MATH trivially, where MATH is induced by the canonical filtration on MATH. This implies MATH. To show the inverse inequality, note that since MATH is open in MATH, it contains some MATH. Now REF implies MATH, which together with REF leads to MATH for all MATH. This implies MATH. This completes the proof of REF. |
math/0007121 | The only place in the proof of REF where we used the simplicity of MATH was where we deduced that any nonzero regular ideal of MATH must equal the whole MATH. This argument is modified as follows. Let MATH be a nonabelian regular ideal of MATH contained in MATH. Then the minimality of MATH implies that MATH and MATH. The proof is concluded again by applying REF . |
math/0007121 | By REF, the annihilation algebra MATH of MATH is an irreducible central extension of a current NAME algebra MATH, where MATH is a simple linearly compact NAME algebra of growth MATH. We have surjective maps MATH where MATH is the universal central extension of MATH. By REF, MATH is either finite dimensional (when MATH) or one of the NAME algebras MATH, MATH, MATH or MATH. By REF , we have MATH in all cases, except MATH in which case the center of MATH is REF-dimensional. Note that MATH, and therefore, by REF , we have MATH. This implies MATH. Then the action of MATH on MATH induces actions on MATH and MATH. The argument from the proof of REF shows that these actions are transitive. Now, let us apply the reconstruction functor MATH to the maps in REF. By REF, MATH, and MATH is one of the NAME pseudoalgebras described in REF - REF above. Moreover, by REF , we have MATH, and hence MATH. We therefore obtain MATH-linear maps MATH whose composition is the identity. Hence MATH is isomorphic to MATH, which is a simple NAME pseudoalgebra REF . The homomorphism MATH given by REF is injective because MATH is centerless REF . The action of MATH on MATH built in REF shows that the image of MATH is an ideal of MATH. Since MATH is simple, it follows that MATH is an isomorphism. |
math/0007121 | By NAME 's Lemma, it is enough to show that MATH for any collection of finite ideals MATH of MATH such that MATH for MATH, where MATH is a totally ordered index set. Assume that MATH. Then there is a chain of ideals MATH of the same rank whose intersection is zero. Fix some MATH. Then for any MATH, MATH, the module MATH is torsion, so by REF , MATH acts trivially on it. This implies MATH for each such MATH, hence MATH is central. |
math/0007121 | By REF , the annihilation algebra MATH. By REF , the latter is isomorphic to MATH, since MATH. Then MATH (see REF). Now by REF , MATH. |
math/0007121 | Let MATH be a current pseudoalgebra over MATH, and MATH be one of the primitive pseudoalgebras of vector fields, where MATH is a NAME subalgebra of MATH. Then, by REF, the annihilation algebra MATH is a current NAME algebra over MATH: MATH, and MATH is isomorphic to MATH, MATH, MATH or MATH, where MATH, MATH. As in the proof of REF , we have MATH. By REF , we have: MATH, and MATH, MATH, MATH or MATH is a NAME subalgebra of MATH. In particular, we see that MATH, and hence MATH is a subalgebra of MATH. So, we have: MATH, MATH. Take any two nonzero elements MATH, MATH. Then we claim that MATH implies MATH. This follows easily from REF , using that MATH is a subalgebra of MATH (see the proof of REF below for a similar argument). Therefore MATH. For MATH, we can write MATH for some MATH and MATH such that the classes MATH are linearly independent in MATH. Then if MATH, it is easy to see that all MATH must belong to MATH. Hence MATH. But MATH, so MATH. |
math/0007121 | Consider the set MATH of all minimal nonzero ideals of MATH. This set is nonempty by REF , and finite because MATH is a NAME MATH-module. The adjoint action of MATH on MATH gives a homomorphism of NAME pseudoalgebras MATH, compare REF . We claim that the direct sum of these homomorphisms is an injective map. Indeed, let MATH be the set of all elements that act trivially on all MATH. This set is an ideal of MATH. If it is nonzero it must contain some minimal ideal MATH. But then this MATH is abelian, which contradicts the semisimplicity of MATH. Therefore we have embeddings MATH. By REF all MATH are simple NAME pseudoalgebras. If MATH is not a current pseudoalgebra over a finite-dimensional NAME algebra, then by REF , MATH. For MATH, we have MATH. Any subalgebra of MATH containing MATH is of the form MATH, where MATH is a subalgebra of MATH. |
math/0007121 | Let us write MATH where MATH and MATH is a basis of MATH. Denote by MATH (respectively MATH) the maximal degree of the MATH (respectively MATH). We have (compare REF): MATH . Assume that MATH. Notice that only the third summand contains coefficients from MATH of degree MATH, hence it must be zero modulo MATH. Since the MATH are linearly independent, the same is true for each term MATH. If we choose MATH such that MATH is of degree exactly MATH, we get a contradiction. |
math/0007121 | Assume that MATH is not a direct sum of pseudoalgebras of vector fields. If MATH is not semisimple, then MATH contains nonzero abelian elements. If MATH is semisimple, REF implies that MATH contains a subalgebra of the form MATH with MATH, and therefore contains nonzero abelian elements (for example MATH for any MATH). The converse statement follows from REF . |
math/0007121 | By REF , a pseudoalgebra MATH of vector fields does not contain nonzero abelian elements, and hence is semisimple. Then, by REF, MATH is a direct sum of finite simple NAME pseudoalgebras and of pseudoalgebras of the form MATH. In fact, there is only one summand, as MATH for any two nonzero elements MATH. Furthermore, MATH cannot be of the form MATH with MATH, because MATH contains nonzero abelian elements. |
math/0007121 | We have already noticed (see the paragraph before REF) that any nonzero ideal MATH of MATH contains MATH. Then MATH is an ideal of MATH, but MATH is simple by REF. |
math/0007121 | Any homomorphism MATH leads to abelian elements in MATH, and therefore is zero (see REF ). Let MATH be a homomorphism from MATH to MATH. Then MATH induces a homomorphism of NAME algebras MATH. By REF, MATH is simple, so MATH where MATH is a primitive MATH-pseudoalgebra of vector fields (MATH and MATH is a subalgebra of MATH). By REF, the annihilation algebra MATH is isomorphic to a current NAME algebra MATH over MATH. Moreover, the quotient of MATH by its center is infinite dimensional and simple. On the other hand, the annihilation algebra MATH is a current NAME algebra over MATH, which is a projective limit of finite-dimensional NAME algebras MATH. Hence the adjoint action of MATH on MATH maps trivially to each of them via MATH. But since MATH, this implies that each MATH is trivial. Therefore MATH, and by REF , we get MATH. |
math/0007121 | As a MATH-module, MATH is isomorphic to a direct sum of several copies of MATH. Any ideal MATH of MATH is in particular a MATH-module, hence it is spanned over MATH by elements of the form MATH where MATH and MATH is a root vector in MATH. If MATH is such that MATH for some nonzero MATH, then MATH, since MATH is an ideal of MATH and MATH is simple. Setting MATH, we see that MATH is an ideal of MATH and MATH. |
math/0007121 | Define a map MATH by the formula: MATH . Then for MATH, we have: MATH . It is easy to see that MATH is a surjective NAME algebra homomorphism. Any isomorphism MATH induces an isomorphism of NAME algebras MATH. By REF , MATH for some proper ideal MATH of MATH. Recall that MATH is isomorphic as a topological algebra to MATH (MATH), and MATH has a unique maximal ideal MATH. Noting that MATH corresponds to MATH via the isomorphism MATH, we deduce that MATH. If MATH, then MATH is a nontrivial ideal of MATH, which is impossible because MATH is simple. It follows that MATH. Now fix MATH and write MATH for some MATH and linearly independent MATH. Assume that, say, MATH. Then we can find MATH such that MATH. Then, by REF, the element MATH is mapped by MATH to MATH, which is a contradiction. This shows that MATH, completing the proof. |
math/0007121 | The statement of the proposition is equivalent to saying that all solutions MATH of REF with MATH are either trivial or of the form MATH. It is easy to check that the latter is indeed a solution (compare REF ). Let us choose a basis MATH of MATH, and write MATH and MATH for some MATH, MATH. We will assume throughout the proof that MATH, and denote by MATH the maximal degree of the MATH. Substituting the above expressions for MATH and MATH in REF, we get: MATH . If MATH, expressing all MATH in the NAME - NAME - NAME basis relative to the basis MATH, we see that MATH-coefficients of degree MATH in the second summation in the left-hand side cannot cancel with terms from other summations, which only contribute lower degree terms. Therefore MATH . This implies that MATH for every MATH, which gives a contradiction, since we can choose MATH to have degree exactly MATH. So MATH, and we can write MATH with MATH, MATH. Substituting this into REF and comparing degree four terms we get MATH for all MATH. Since MATH we conclude that MATH for all MATH. We are only left with showing that MATH. Substitute MATH and MATH into REF, and then use REF to obtain: MATH . Notice that MATH is the only term lying in MATH and everything else belongs to MATH. Hence MATH, which is only possible if MATH or MATH. In the latter case we are done. In the former, the left-hand side of REF becomes zero, and MATH since MATH. Thus MATH and MATH. |
math/0007121 | REF is an immediate consequence of REF. By the above remarks, it remains to prove REF in the cases when MATH is a current pseudoalgebra over either MATH or MATH, where MATH is a subalgebra of MATH. In the former case (MATH, MATH), note that MATH is spanned over MATH by elements MATH for MATH. Then MATH, and by REF we know that the only nonzero homomorphism of the NAME MATH-pseudoalgebra MATH to MATH maps MATH to MATH. Hence any embedding of MATH in MATH maps each MATH to the corresponding element MATH of MATH. Now let MATH be a current pseudoalgebra over MATH. We will give the proof in the case when MATH, the case of currents being completely analogous. We are going to make use of the following lemma. If MATH is a finite-dimensional NAME algebra of dimension MATH, then there exist MATH-dimensional subalgebras MATH such that MATH for every MATH. If MATH has a semisimple element MATH, we complement it to a basis of MATH eigenvectors MATH. The subalgebras MATH then satisfy the statement of the lemma. If MATH has no semisimple elements, then from NAME 's theorem we know that MATH must be solvable. In this case it has a MATH-dimensional ideal MATH. Complementing MATH to a basis MATH, we conclude as before. Now consider a MATH-dimensional subalgebra of MATH with basis MATH. Then the element MATH from REF depends on the choice of basis only up to multiplication by a nonzero element of MATH. Moreover, the MATH-span of this element is a (free) rank one subalgebra of MATH, as can be easily checked (compare REF ). Let MATH be the rank one subalgebras of MATH associated as above with REF-dimensional subalgebras MATH of MATH constructed in REF . Then by comparing second tensor factors, we see that MATH for each MATH. Therefore the sum of all MATH is a free MATH-submodule MATH of MATH of rank MATH. Since the rank of MATH is also MATH, we see that MATH is a torsion MATH-module. Denote by MATH the subalgebra of MATH generated by MATH. Since MATH is a torsion MATH-module, we conclude, by REF , that MATH is an ideal of MATH. Hence MATH by simplicity of MATH. Now by REF , each subalgebra MATH embeds uniquely in MATH. Hence MATH embeds uniquely in MATH. This completes the proof of REF. |
math/0007121 | CASE: If MATH then all its coefficients in front of MATH are zero for different MATH. Since MATH, it follows that all MATH. REF are clear by REF. REF follows from REF - REF and the fact that MATH is MATH-invariant. (MATH is MATH-invariant because MATH is MATH-equivariant, see REF.) |
math/0007121 | Follows from REF . |
math/0007121 | By REF , MATH is simple iff MATH is not abelian and has no nontrivial MATH-invariant ideals. In particular, MATH is semisimple. Using REF and the fact that MATH is an ideal of MATH if MATH is an ideal, we see that MATH is a direct sum of isomorphic finite simple NAME MATH-pseudoalgebras. |
math/0007121 | It only remains to check REF, which is straightforward. |
math/0007121 | The second statement is obvious by the definitions. Since MATH REF and MATH REF act irreducibly on MATH, we only have to check that the action of MATH on MATH is irreducible for MATH. Using diagonal matrices, we see that it suffices to check that MATH acts irreducibly on MATH. Recall that this action is given by (see REF): MATH . For MATH, let MATH. Then MATH. If MATH is a MATH-submodule, and MATH, MATH, then the previous formula implies MATH. Therefore MATH. |
math/0007121 | Let MATH be a left ideal, MATH, and MATH with linearly independent MATH. Then: MATH . Taking MATH, we see that all MATH. In particular, MATH, and hence each element from MATH is a MATH-linear combination of elements of the form MATH. For such MATH, we have MATH. For MATH, MATH, we get that MATH. This proves the first part of the proposition. REF is obvious, and REF follows easily from REF . |
math/0007121 | Assume that MATH for some MATH, MATH. Let REF be a relation of this form with MATH so that MATH is minimal. We call MATH the degree of the relation REF. Assume that MATH's are linearly independent, so that the degree of REF is positive. We can find MATH, MATH such that MATH. Applying MATH to REF and using REF, we obtain MATH . Subtracting this from REF, we get a relation of lower degree than REF, because MATH for MATH. |
math/0007121 | Fix nonzero elements MATH, MATH, and let MATH. Let MATH be the linear span of MATH; we set MATH for MATH. For MATH, MATH, we have: MATH, and by induction, MATH . In particular, all MATH are MATH-modules. Since MATH is a NAME MATH-module, there exists MATH such that MATH. In particular, MATH for some MATH. Writing REF for MATH and using REF, we get MATH . On the other hand, applying MATH to both sides of REF and using the MATH-sesqui-linearity gives MATH . Comparison of the last two equations gives MATH . If MATH, then the module MATH is torsion, hence MATH acts on it as zero by REF . This gives MATH for all MATH, MATH. Then REF implies MATH. Since MATH, it follows that MATH, which contradicts to the assumption MATH. Therefore MATH. This is possible only when MATH. Then for any MATH, one has: MATH . This implies that MATH, proving REF. |
math/0007121 | Using REF , we can assume that MATH. The proof will be by induction on the length of the derived series of MATH. First consider the case when MATH is abelian. By a NAME 's Lemma argument, it is enough to find an eigenvector when MATH is abelian generated by one element MATH. We may assume that MATH; then by REF all MATH are finite dimensional. Let MATH be such that MATH. Then the statement follows from the usual NAME Theorem applied to the MATH-module MATH. Now let MATH be nonabelian, MATH. By the inductive assumption, MATH has a space of eigenvectors MATH. If MATH, then MATH is a MATH-submodule of MATH on which MATH acts as zero. The abelian MATH-pseudoalgebra MATH has an eigenvector in MATH, which is also an eigenvector for MATH. Now assume that MATH. By REF , we have for MATH, MATH: MATH . On the other hand, we can compute: MATH . It follows that MATH . Assume that MATH for some MATH, MATH, and write MATH for short. For MATH, MATH, the above equation becomes: MATH (recall that MATH for MATH). Note that MATH if MATH for sufficiently large MATH. Take the minimal such MATH, and let MATH be such that MATH. By REF , there exists MATH such that MATH. Then MATH and MATH, which is a contradiction. It follows that all MATH, hence MATH preserves MATH. By REF , MATH, and therefore MATH has an eigenvector by the usual NAME Theorem for MATH. |
math/0007121 | Assume that MATH is not semisimple, that is, it has a nonzero abelian ideal MATH. Then, by REF, MATH has an eigenvector in MATH. If MATH is the corresponding eigenspace in MATH, then, by REF , MATH is a MATH-submodule of MATH. The irreducibility of MATH implies that MATH. Now, by REF , MATH is a free MATH-module, since MATH by the faithfulness of MATH. REF and the faithfulness of MATH also show that MATH embeds into MATH. In particular, MATH is finite. Then MATH exists, and we can assume that MATH is an eigenspace for MATH. For each element MATH and MATH we have MATH, which means that MATH is identified with MATH for MATH. Note that MATH is of rank one, because MATH and MATH has no zero divisors. Let MATH. Notice that MATH is a subalgebra of MATH and MATH is a semidirect sum, because MATH. Since MATH, this also implies that MATH is contained in MATH. Then MATH embeds in MATH, that is, MATH. Hence MATH is torsion, and by REF , MATH is an ideal of MATH. But MATH is semisimple, and by REF a finite semisimple NAME pseudoalgebra does not have proper ideals of the same rank. Therefore, MATH and MATH is a semidirect sum of pseudoalgebras. Finally, to show that MATH is of the form MATH, notice that MATH is an ideal of MATH, since MATH is an ideal of MATH. This ideal is generated over MATH by abelian elements, so by REF if it is nonzero it is of the form MATH for some semisimple subalgebra MATH of MATH. Hence MATH. But any subalgebra of MATH containing MATH is equal to MATH for some subalgebra MATH of MATH. This completes the proof. |
math/0007121 | First of all, note that the property that any element MATH of MATH acts nilpotently on MATH remains valid when we replace MATH by any quotient of MATH by an ideal. In particular, MATH will have that property. However, MATH is semisimple, and from the classification of finite semisimple NAME pseudoalgebras we see that this is impossible, unless MATH. Therefore MATH is solvable. The nilpotence of all MATH imply that all eigenvalues for MATH are zero. Now REF implies that MATH is a nilpotent NAME pseudoalgebra. |
math/0007121 | In order to prove the statement, it is enough to show that all MATH-module extensions between generalized weight modules relative to distinct weights are trivial. The strategy is to consider first the case when MATH is the NAME pseudoalgebra generated by one element MATH. Then in the general case, we show that the generalized weight spaces MATH relative to some element MATH are MATH-invariant. Let MATH be a finite MATH-module, MATH, and MATH be a nilpotent NAME pseudoalgebra. If MATH contains a MATH-generalized weight module MATH and MATH with MATH, then MATH as MATH-modules. Since MATH for any MATH, it suffices to prove the statement when MATH. Let us first consider the case when MATH is a cyclic MATH-module. In order to prove that the extension is trivial, we need to find a lifting MATH of MATH such that MATH and to show that MATH is a direct sum of MATH-modules (here and below, we write just MATH instead of MATH). We will prove this by induction on the depth of MATH, the basis of induction being trivial. Let thus the statement be true for all MATH-generalized weight modules of depth MATH and consider a module MATH of depth MATH. Fix an arbitrary lifting MATH of MATH; then: MATH . Set MATH . Then we claim that for MATH, MATH implies MATH. We are going to show this first in the case when MATH, the proof for MATH only requiring minor changes. So, let MATH. Then MATH is a free MATH-module, because it is generated by its MATH-eigenspace and we can apply REF . We pick some MATH-basis MATH for MATH modulo MATH. If MATH is some MATH-basis of MATH compatible with its filtration, we write MATH where MATH. Notice that for MATH, MATH belongs to MATH where MATH is the derived algebra of MATH, hence all weights are zero on it. This means that MATH . We have: MATH . We compute the right-hand side, using REF - REF, and obtain: MATH . Now the assumption MATH implies that coefficients of all MATH must be zero. Let us choose the highest degree MATH for which there is some MATH of degree MATH such that MATH for some MATH. Then we get MATH for all MATH and all MATH of degree MATH, hence MATH, giving a contradiction. This proves that all MATH, and therefore MATH. Now, because of nilpotence of MATH, MATH for MATH, and obviously MATH. Thus we can pull the statement back to MATH to obtain that MATH maps any lifting MATH of MATH inside MATH. Now we can choose the lifting MATH of MATH so that MATH. Indeed, performing the same computation as above, using instead of REF MATH for some MATH, we get MATH. This shows that MATH for some choice of MATH. Now choose MATH to be the lifting of MATH minimizing the top degree MATH of MATH such that some MATH is nonzero. Then if we replace MATH by MATH, all coefficients MATH in degree MATH vanish, against minimality of MATH. This contradiction shows that the lifting MATH can be chosen in such a way that MATH for all MATH, and MATH modulo MATH. This shows that MATH is indeed a submodule of MATH, and it satisfies the hypotheses of our claim. Moreover, MATH is of depth MATH and we can apply the inductive assumption to show that MATH decomposes as a direct sum of MATH-submodules. This means that we can find a lifting MATH of MATH for which MATH holds exactly. We have found a lifting MATH of MATH proving that MATH. We are left with showing that this is a direct sum of MATH-modules. This is clear if MATH since in this case MATH is free, hence projective. If instead MATH, assume the sum not to be free. This means that some multiple MATH of MATH lies in MATH. Since MATH is killed by MATH, so is MATH, showing MATH as no other vector in a generalized weight module of nonzero weight MATH is killed by MATH. This concludes the proof in case MATH. If MATH, then we choose a MATH-basis of MATH modulo MATH, and use in REF coefficients of the form MATH. The rest of the proof is the same. Finally, consider the general case of a non-cyclic MATH-module MATH. We distinguish two cases. If MATH, then MATH is free by REF , and it decomposes as a direct sum of cyclic modules to which we can apply the above argument independently. If MATH, then we choose generators MATH of MATH over MATH, lift them to elements MATH of MATH in such a way that each of them is mapped by MATH to zero, and then argue that if MATH then MATH is an element of MATH killed by MATH, hence is zero. Therefore the extension of MATH-modules splits, and so does that of MATH-modules, by the above computation. Now let MATH be any nilpotent NAME MATH-pseudoalgebra, MATH be a finite MATH-module, and MATH, MATH. Every MATH-generalized weight submodule of MATH is stabilized by the action of MATH. We set MATH and MATH . Then the MATH are MATH-submodules of MATH whose union is MATH (since MATH is nilpotent), and the MATH are vector subspaces of MATH whose MATH-span is the MATH-generalized weight space MATH (because MATH for all MATH). It is easy to show by induction on MATH that MATH . Indeed, the basis of induction (say MATH) is trivial, and the inductive step follows from REF and the identity MATH. REF implies that MATH, as desired. We are now able to complete the proof of REF. Let MATH be finest among all decompositions into direct sum of MATH-submodules of MATH such that all of the MATH-torsion of MATH is contained in one of the MATH. Note that such a finest decomposition always exists, because any decomposition defines a partition of MATH into non-negative integers, and finer decompositions define finer partitions. We claim that each MATH is a generalized weight module for MATH. Otherwise, there must be some element MATH for which some of the MATH is not a MATH-generalized weight module. But if so, then MATH decomposes into a direct sum of its MATH-generalized weight submodules, and all torsion elements lie in the MATH-eigenspace of eigenvalue MATH. Since all MATH-generalized weight submodules are MATH-invariant, we obtain a contradiction. Therefore MATH is a direct sum of its generalized weight submodules. |
math/0007121 | The proof is an application of REF , using the following fact. Let MATH be a vector space and MATH REF be subspaces of MATH such that all MATH are finite dimensional, then MATH is finite dimensional. It is enough to show this for MATH, in which case it follows from the isomorphism MATH . |
math/0007121 | Any MATH is contained in some MATH, which is finite dimensional by REF , and MATH-invariant because MATH. |
math/0007121 | CASE: Let MATH be an extension of MATH-modules, which is split over MATH. Choose a splitting MATH as MATH-modules. The fact that MATH and MATH are homomorphisms of MATH-modules implies (MATH, MATH, MATH): MATH . It is clear that MATH and MATH is MATH-linear; in other words, MATH. For MATH, MATH, we have (compare REF): MATH . Subtracting these two equations and using REF, we get: MATH (recall that the action of MATH on MATH was defined in REF ). The last equation means that MATH. If we choose another splitting of MATH-modules MATH, then it will differ by an element MATH of MATH: MATH. Then the corresponding MATH . Since MATH (see REF ), we get MATH, that is, MATH. Conversely, given an element of MATH, we can choose a representative MATH and define an action MATH of MATH on MATH by REF, which will depend only on the cohomology class of MATH. This proves REF . The proof of REF is similar. Write MATH as MATH-modules. Denoting the pseudobracket of MATH by MATH, we have for MATH, MATH: MATH . It is clear that MATH, and the NAME identity for MATH implies MATH. REF A first-order deformation of MATH is the structure of a NAME MATH-pseudoalgebra on MATH, where MATH acts trivially on MATH, such that the map MATH given by putting MATH is a homomorphism of NAME pseudoalgebras. This means that MATH is an abelian extension of NAME pseudoalgebras, so REF follows from REF . |
math/0007121 | Since the adjoint action of MATH on MATH is trivial, we obtain that MATH maps to zero in the exact sequence REF. Therefore we have exact sequences MATH which lead to isomorphisms REF follows from REF . |
math/0007121 | Let MATH be a central extension of MATH with a pseudobracket MATH where MATH for MATH. The skew-symmetry of MATH is equivalent to REF. The NAME identity is equivalent to NAME identity for MATH together with the following cocycle condition for MATH (compare REF ): MATH . With the usual notation MATH, MATH, etc., we have: MATH . From here it is easy to see that REF is equivalent to REF. Let now MATH be nonzero. Rewrite REF in the form MATH . If MATH, then the degree of the left-hand side equals MATH while that of the right-hand side is at most MATH, giving a contradiction. So MATH, and REF shows that MATH. |
math/0007121 | CASE: The NAME pseudoalgebra MATH is free of rank one, with MATH, MATH, hence we can use REF . In this case MATH, and REF becomes MATH for MATH. We choose a basis MATH of MATH such that MATH, and express MATH in a NAME - NAME - NAME basis as MATH, MATH (see REF). Then the above equation becomes: MATH . Comparing terms of the form MATH REF we find that MATH is zero unless MATH for some MATH. Hence MATH, MATH. Substituting and comparing coefficients, we obtain that MATH. This obviously satisfies REF. The trivial cocycles are multiples of MATH, hence MATH is the unique central extension up to scalar multiples. This is the well-known NAME central extension. CASE: Choose a basis of MATH and let MATH be a cocycle representing a central extension of MATH. Then for each basis element MATH, MATH restricts to a cocycle of MATH, which is a current NAME pseudoalgebra over MATH. By REF we can then add to MATH a trivial cocycle as to make MATH, MATH, for every such basis element MATH. Denoting MATH, the NAME identity for elements MATH then gives: MATH . Let MATH be distinct elements in the above basis, which we extend to a basis MATH of MATH with MATH, MATH. We substitute the NAME - NAME - NAME basis expression MATH in REF, to get: MATH . Comparing coefficients of the form MATH for MATH, we find that MATH can be nonzero only when MATH, in which case MATH, and when MATH for some MATH. This means that MATH for some polynomial MATH. We can repeat the same argument after switching the roles of MATH and MATH, to get: MATH. Then the skew-symmetry MATH implies: MATH. This is possible only when MATH, MATH. Therefore MATH is identically zero. |
math/0007121 | MATH is free of rank one and MATH, hence REF becomes MATH, MATH. This is satisfied only by multiples of MATH if MATH and by all elements of MATH otherwise. Since MATH is abelian, then MATH and trivial cocycles are multiples of MATH. |
math/0007121 | MATH is free of rank one, and MATH where MATH is a basis of MATH with the only nonzero commutation relations MATH, MATH (see REF). It is immediate to check that MATH for all MATH. Moreover, the element MATH from REF equals MATH. Then, if MATH with MATH, MATH, REF becomes: MATH . All solutions MATH of this equation are multiples of MATH. Trivial cocycles are multiples of MATH, hence all cocycles are trivial. |
math/0007121 | By REF , MATH is spanned over MATH by elements MATH satisfying the relations MATH and MATH . The pseudobrackets are (see REF): MATH . Trivial cocycles MATH are determined by the identity (see REF): MATH where MATH, which gives: MATH . Let MATH be a cocycle for MATH representing a central extension. Write MATH for short. REF give the identities: MATH . Using this, NAME identity for the elements MATH gives the following equation for MATH: MATH . This is a homogeneous equation and can be solved degree by degree. If MATH is homogeneous of degree one, then the left-hand side is zero, and we immediately see that MATH. Then, by REF, MATH. If MATH is homogeneous of degree other than one, then MATH, since MATH restricts to a cocycle of the free rank one NAME pseudoalgebra-MATH, which has been shown in REF to take values in MATH. Then REF give MATH. Hence if MATH and MATH are linearly independent, we can find some MATH such that MATH . We substitute this into REF and after simplification obtain MATH. Therefore, MATH, hence the only nonzero solutions to REF occur in degree two. Now using REF, we get: MATH . The skew-symmetry MATH gives the equations MATH and MATH . From this we obtain that MATH lies in the linear span of MATH. Comparing the coefficients in front of MATH in REF, we see that the coefficient of MATH in MATH is equal to the coefficient of MATH in MATH. Call this coefficient MATH; then MATH. Then comparison of other coefficients in REF shows that MATH for all MATH. Substitute this in REF to obtain that MATH is trivial (compare REF). |
math/0007121 | Let MATH be a cocycle for MATH. We will write MATH for MATH. Then NAME identity leads to the equation MATH . This immediately implies: MATH. Since MATH this shows that MATH for all MATH. We can now solve the homogeneous REF degree by degree. Solutions of degree zero are cocycles of the NAME algebra MATH, hence they are all trivial. Solutions of degree one satisfy MATH, and skew-symmetry implies MATH. Therefore every such MATH is an invariant symmetric bilinear form on MATH with values in MATH. Any such bilinear form can be written as MATH where MATH is the Killing form on MATH and MATH is some element of MATH. Such cocycles MATH are nontrivial, hence inequivalent central extensions are in one-to-one correspondence with elements of MATH. |
math/0007121 | REF follow from REF , and REF from a direct application of REF . For any other simple pseudoalgebra MATH, the strategy is to construct a continuous family of pseudoalgebras MATH, indexed by MATH endowed with the NAME topology, that are all isomorphic to MATH when MATH, and whose fiber at MATH is one of the pseudoalgebras already considered in REF . Then, since MATH for MATH, it will follow that MATH whenever MATH lies in a neighborhood of MATH, hence for all MATH. In the case of a current pseudoalgebra over a MATH or MATH type NAME pseudoalgebra, choose a basis MATH of MATH that contains a basis of MATH, and construct the family MATH of NAME algebras indexed by MATH generated by elements MATH with NAME bracket MATH. Then for MATH we have an isomorphism MATH, whereas MATH is an abelian NAME algebra. Then MATH, where MATH, MATH, is a family of pseudoalgebras all isomorphic to MATH for MATH. The fiber of this family at MATH has been shown in REF to have no nontrivial central extensions. In the same way, if we set MATH, then MATH is a family of pseudoalgebras all isomorphic to MATH for MATH, and the fiber at MATH is MATH where MATH are isomorphic to MATH as vector spaces but have trivial bracket. If MATH is a current pseudoalgebra over MATH, for finite-dimensional NAME algebras MATH, choose a basis MATH of MATH as in REF , and complete it with MATH to a basis of MATH. Then a continuous family MATH of NAME algebras can be constructed for MATH as MATH spanned by MATH, MATH, MATH, MATH, and by setting MATH to span a NAME algebra, and all brackets involving MATH to be zero. Define MATH by MATH, MATH. Then MATH is the limit of the NAME pseudoalgebras-MATH, which are all isomorphic to MATH, and the former NAME pseudoalgebra is of the type treated in REF . |
math/0007127 | Since MATH and MATH are discrete, and MATH is totally disconnected, there exists a compact, open subgroup MATH of MATH, such that MATH. Let MATH and MATH, so MATH and MATH are finite-index, open subgroups of MATH, and define MATH by MATH for MATH and MATH. |
math/0007127 | Let MATH and MATH be finite-index subgroups of MATH, such that MATH is an isomorphism from MATH to MATH. Then MATH induces isomorphisms MATH . By identifying each of MATH and MATH with MATH in the natural way (and noting that MATH), we may think of MATH and MATH as MATH-subspaces of MATH. By replacing MATH and MATH with finite-index subgroups, we may assume that these subspaces are contained in MATH. Then, because MATH is an isomorphism, we see that the conditions of Notation REF are satisfied, so REF below implies that there exist CASE: a standard automorphism MATH of MATH, and CASE: a finite-index subgroup MATH of MATH, such that MATH, for all MATH. Because MATH is an arithmetic lattice, it is commensurable with MATH. Thus, replacing MATH with a finite-index subgroup, we may assume that MATH. Then we may define MATH by MATH. |
math/0007127 | By assumption, there exists a lattice MATH such that MATH. Since MATH, and MATH is abelian, we see that MATH. Therefore MATH, so, by the choice of MATH, we have MATH. By assumption, MATH is a lattice in MATH, so MATH is closed CITE, hence discrete. Thus, there is an open compact subgroup MATH, such that MATH. Let MATH, and extend MATH to a homomorphism MATH by defining it to be trivial on MATH. |
math/0007127 | We may assume MATH. (Otherwise, we must have MATH, which means MATH is abelian, so REF applies.) From REF , we may assume there exist CASE: a standard automorphism MATH of MATH, and CASE: a homomorphism MATH, such that MATH, for all MATH. From REF , we may assume that there is a finite-index subgroup MATH of MATH, such that MATH contains MATH, and MATH extends to a homomorphism MATH that is trivial on MATH. Let MATH. Define MATH by MATH, for MATH, so MATH is a continuous homomorphism that extends MATH. Because MATH is trivial on MATH, we know that MATH. Also, because MATH, we know that MATH for all MATH. Thus, MATH induces an automorphism of MATH, and an isomorphism MATH, so MATH is an isomorphism. |
math/0007127 | Because MATH and MATH have finite codimension in MATH and MATH, respectively, we see that MATH has finite codimension in MATH. Thus, in proving REF, we may assume that MATH. CASE: This follows from our proof of REF below. CASE: There are some nonzero MATH, such that MATH. Let MATH. We claim that MATH. Otherwise, we have MATH so MATH. This implies MATH, which is a contradiction. This completes the proof of the claim. For any MATH, we have MATH so MATH contains MATH, which is of finite codimension in MATH. CASE: We may write MATH (uniquely) in the form MATH, with MATH, and such that we may write MATH with MATH whenever MATH. (Note that we do not assume MATH.) For MATH, we have MATH . Whenever either MATH or MATH is large, it is obvious that MATH is much smaller than MATH. Also, we may assume MATH (otherwise, we have MATH, as desired), and, from the definition of MATH, we know that MATH, so MATH . Therefore, we conclude that MATH must be congruent to either MATH or MATH, modulo MATH. Thus, because of our assumption that MATH, we see that MATH does not contain elements of all large degrees, so it does not have finite codimension in MATH. Then, from REF, we conclude that it does not have finite codimension in MATH. |
math/0007127 | CASE: Choose MATH. Then REF implies the desired conclusion. CASE: From REF, we have MATH and MATH, so MATH. |
math/0007127 | Assume, for the moment, that MATH. For MATH, define MATH iff MATH. For nonzero MATH, we see, from Notation REF, that MATH has finite codimension in MATH if and only if MATH has finite codimension in MATH. Therefore, REF implies that MATH iff MATH. The equivalence classes are precisely the sets of the form MATH, for some MATH-separable MATH, so the desired conclusion is immediate. We may now assume MATH. Let MATH. There is some MATH-separable MATH, such that MATH. By induction on MATH, we know that there is some MATH-separable MATH, such that MATH. From the definition of MATH, we know there is some MATH, such that MATH. Then, because MATH is MATH-separable, we know that MATH is MATH-separable. Define MATH, MATH, MATH, and MATH as in REF (with MATH, MATH, and MATH in the places of MATH, MATH, and MATH, respectively). Because MATH is MATH-separable, we know, from the case MATH in the first paragraph of this proof, that there is some MATH-separable MATH, such that MATH. Therefore MATH as desired. |
math/0007127 | Choose MATH, such that MATH and MATH; let MATH. For MATH, we have MATH so MATH contains MATH. Because MATH and MATH contain codimension-MATH subspaces of MATH and MATH, respectively, this implies that MATH contains a codimension-MATH subspace of MATH. Because both MATH and MATH belong to MATH, the desired conclusion follows. |
math/0007127 | Choose MATH as in REF , and let MATH and MATH. It suffices to show MATH and MATH . Let MATH be the irreducible factors of MATH. Then we may write MATH where MATH. From the NAME Remainder Theorem, we know that the natural ring homomorphism from MATH to MATH is an isomorphism. Thus, we may work in each factor MATH, and add up the resulting codimensions. Define MATH by MATH. Then, letting MATH, we have MATH so MATH . We have MATH, so MATH and MATH . From REF, we have MATH so MATH . This establishes REF. Because MATH, and from REF, we have MATH so MATH . This establishes REF. |
math/0007127 | For any MATH, we have MATH . Then the proof is completed by induction on MATH. |
math/0007127 | Let MATH be the codimension of MATH. Choose a power MATH of MATH so large that MATH is MATH-separable. Then REF implies MATH. Choose MATH, such that MATH and MATH. We have MATH . So MATH has small codimension in MATH. Therefore MATH must have small codimension in MATH, so MATH must be small, as desired. |
math/0007127 | Apply REF to the map MATH of REF . |
math/0007127 | Let MATH be the codimension of MATH in MATH, and choose MATH so large that, for every MATH, the subspace MATH contains elements of degree MATH whose leading coefficients span MATH. For any element of MATH of degree MATH, we show that there is a MATH-separable element of MATH of degree MATH with the same leading coefficient. Let MATH be the leading coefficient of some element of MATH of degree MATH. Then MATH contains exactly MATH elements of degree MATH with leading coefficient MATH. On the other hand, if MATH is an element of MATH that is of degree MATH and is not MATH-separable, then MATH must be of the form MATH, where MATH is an element of MATH of some degree MATH, and MATH is an element of MATH of degree MATH. Thus, the number of such elements MATH of degree MATH is no more than MATH . Therefore, not every element of MATH of degree MATH whose leading coefficient is MATH can be such an element MATH, so MATH has a MATH-separable element of degree MATH with leading term MATH, as desired. |
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