paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0010020 | We begin with noting that a line MATH in MATH has REF if and only if it is the graph MATH of a nonzero homomorphism MATH. It is clear that there are MATH such lines. They come in MATH pairs: we have MATH if and only if MATH. In that case MATH is isotropic and so MATH correspond to MATH-orbits of MATH-vectors that are p... |
math/0010020 | We only prove the statements involving a MATH-vector, the proof of the one about a MATH-vector is similar and easier. We begin with the last clause. Let MATH be as in the proposition. By NAME 's result, a unitary transformation will map this into a pair with second component MATH and so we may assume that MATH. Then MA... |
math/0010020 | We apply this to the case at hand: MATH, MATH and MATH spanned by the MATH-vector MATH. Then MATH, where in the latter case the skew form takes the value MATH on a generator. One easily verifies that the group of unitary transformations of MATH is MATH. As this is also the group of unitary transformations of MATH, it f... |
math/0010020 | Suppose first MATH spanned by the MATH-vector MATH and a MATH-vector MATH. We have MATH for some MATH. Since MATH is positive definite, we must have MATH. Since MATH, this implies that up to a unit MATH equals MATH, MATH, MATH or MATH. By multiplying MATH with a unit we may assume that MATH acually equals one of these ... |
math/0010020 | By REF we may assume that MATH is spanned by MATH and that MATH in case MATH and MATH in case MATH. This identifies MATH with MATH. A MATH-vector in MATH maps to a MATH-vector in MATH of the same MATH-invariant and the MATH-vector of MATH lies in MATH or in MATH. In case MATH it is clear that any MATH-vector in MATH ha... |
math/0010020 | Let us first assume that MATH is isomorphic to an orthogonal product MATH with MATH. Since MATH and MATH, we see that it is enough to investigate the corresponding issue in MATH. The MATH-vectors in MATH are all unitary equivalent, and so we can assume that MATH. The assertions regarding the classification now follow f... |
math/0010020 | For MATH or MATH this is part of the statement of the previous proposition. The cases MATH and MATH are handled in a similar way. |
math/0010025 | To prove bijectivity, we show that the inverse assignment is given by taking derived toric manifolds; to each MATH-translation MATH we associate the MATH-equivariant diffeomorphism MATH, where MATH if and only if MATH. It follows directly from the definitions that MATH descends to the original MATH-translation MATH of ... |
math/0010025 | Given MATH and MATH, we consider tangents to the orbit MATH at MATH. One such has direction vector MATH, whose inner product with MATH is given by MATH. Since this quantity is strictly positive the orbit meets MATH only at MATH, (and the intersection is transverse). On the other hand, given any point MATH in the open c... |
math/0010025 | Since exponentiation is a diffeomorphism, MATH is a tubular neighbourhood of the embedding; each choice of basis for MATH therefore trivialises the normal bundle. |
math/0010025 | Following REF, we must identify MATH with MATH. Any choice of basis for the NAME algebra of MATH will have this effect; since the space of bases has two connected components, it actually suffices to give an orientation. But MATH is the kernel of an epimorphism MATH and MATH is canonically oriented, so it remains to ori... |
math/0010025 | The existence of MATH is assured by the fact that MATH and MATH agree on MATH; we write their common restriction as MATH. Our data implies that MATH maps diffeomorphically onto MATH under the projection of MATH onto MATH, so that MATH is isomorphic to MATH in MATH. Since MATH is given by MATH in MATH, it is equivariant... |
math/0010025 | The data yields a diffeomorphism MATH, with MATH; we then apply REF to MATH with MATH, and to MATH with MATH. |
math/0010025 | Considering MATH as a face of MATH, we note that MATH in MATH, that MATH in MATH, and that MATH in MATH by REF. Thus MATH, and MATH acts as MATH. It follows that the normal bundle MATH of MATH in MATH has total space MATH, and therefore that restriction and induction commute for MATH. Considering MATH as a face of MATH... |
math/0010025 | The first statement follows from the proof of REF by choosing MATH, and letting MATH be a tubular neighbourhood of MATH; then MATH lies in MATH, and MATH is the corresponding coordinate line bundle in the trivial summand MATH. For the second statement, we note that the proof of REF identifies the total space of MATH wi... |
math/0010025 | Since MATH restricts to a preferred section for MATH, so MATH pulls MATH back to MATH. From REF , we deduce that MATH is isomorphic to MATH over MATH, and is trivial over the complement. But these properties characterise MATH. |
math/0010025 | Pulling REF back along MATH yields the complex structure on MATH, which REF converts to an isomorphism MATH. Different choices of preferred section MATH yield homotopic isomorphisms, and therefore homotopic stably complex structures, because the corresponding diffeomorphisms MATH are isotopic. By REF , the complex stru... |
math/0010025 | By definition, there is a canonical diffeomorphism between MATH and MATH, and a canonical isomorphism between MATH and MATH which preserves their respective actions on MATH and MATH. We obtain a diffeomorphism MATH which repects the induced stably complex structures, and the result therefore follows by pullback along i... |
math/0010025 | The first statement follows from our introductory remarks, in view of the cell decomposition defined above. Since there are MATH submanifolds MATH, but only MATH two-cells in the decomposition, NAME duality shows that there are MATH linear relations amongst the cobordism classes of the inclusions. |
math/0010025 | We first extend REF to MATH-fold products by induction. For MATH, the inductive hypothesis provides a combinatorial equivalence MATH where the MATH are products of simplices; iterating REF replaces the right-hand expression by MATH and applying REF yields MATH, where the MATH are products of simplices as required. As i... |
math/0010025 | Since the subtori MATH generate MATH, we may define an automorphism MATH of MATH by mapping MATH onto MATH, preserving orientation, for each MATH. We conclude by replacing MATH with MATH, and appealing to REF . |
math/0010025 | We consider the halfspace MATH of MATH given by MATH for some small MATH, and note that its inverse image under the projective transformation MATH is a halfspace MATH, which intersects MATH in a closed neighbourhood MATH of the vertex MATH. We then apply REF by choosing MATH, MATH, MATH, and MATH; we take MATH to be MA... |
math/0010025 | This is a direct consequence of REF , since the connected sum of any two stably complex structures is cobordant to their disjoint union. |
math/0010029 | Let MATH be a cycle representing the class MATH. There is a smooth (see CITE) curve MATH on MATH that contains the support of MATH. Hence it is enough to show that there is a homologically trivial cycle MATH on MATH so that MATH for every cycle MATH on MATH such that the image under MATH lies in MATH. Let MATH denote t... |
math/0010029 | By the functoriality of the NAME map we have a commutative diagram MATH . By projection we can replace the bottom right corner with MATH. Now we tensor this with the NAME map for MATH to obtain, MATH . The required commutative diagram now follows from the definition of MATH. |
math/0010029 | The following composite map is multiplication by REF MATH . By the divisibility of MATH for a surface MATH we see that the lemma follows from the following result. |
math/0010029 | Let MATH, MATH be points on MATH; we get points MATH and MATH of MATH. The image of MATH in MATH is MATH. Thus, we have an expression MATH . Now, any cycle MATH in MATH can be written as MATH. The NAME variety of MATH is MATH and the image of MATH under the NAME map is MATH. Thus, if the cycle is in MATH, then MATH. No... |
math/0010029 | By NAME 's result there are non-trivial classes in MATH. The Jacobian variety MATH is spanned by decomposable elements. Moreover MATH. Thus there is a pair of elements of MATH of the form MATH, MATH such that the image of their tensor product in MATH is non-zero. By a result of NAME, for any fixed class MATH in MATH, t... |
math/0010030 | Consider the first identity. By definition both sides are degree MATH super-derivations on MATH so it suffices to check that they agree on generators. Clearly, both sides give MATH when evaluated on MATH and for MATH we have MATH A similar argument proves the second identity. |
math/0010030 | Clearly MATH when MATH or MATH would be a vertex-idempotent whence in MATH. Let MATH be the starting point of MATH and MATH the end point of MATH and assume that MATH, then MATH from which the assertion follows. |
math/0010030 | Define the NAME derivation MATH on MATH by the rules that MATH . By induction on the length MATH of an oriented path MATH in the quiver MATH one easily verifies that MATH. By induction one can also proof that MATH. This implies that MATH is a bijection on each MATH, where MATH and on MATH, MATH has V as its kernel. By ... |
math/0010030 | Let MATH be the MATH-space spanned by all necklace words MATH in MATH and define a linear map MATH for all oriented paths MATH in the quiver MATH, where MATH is the necklace word in MATH determined by the oriented cycle MATH. Because MATH it follows that the commutator subspace MATH belongs to the kernel of this map. C... |
math/0010030 | If MATH is not a cycle, then MATH and so vanishes in MATH so we only have to consider terms MATH with MATH an oriented cycle in MATH. For any three paths MATH and MATH in MATH we have the equality MATH whence in MATH we have relations allowing to reduce the length of the differential part MATH so MATH is spanned by ter... |
math/0010030 | We know that MATH. By a result of CITE one knows that the geometric points of the quotient scheme MATH are the isomorphism classes of MATH-dimensional semi-simple representations of MATH. Moreover, the MATH-stabilizer of a point in MATH is trivial if and only if it determines a simple MATH-dimensional representation of... |
math/0010030 | Assume that MATH is a point of semi-simple representation type MATH, that is, MATH and MATH a simple MATH-representation. We have MATH . But then, the dimension of MATH is equal to MATH from which it follows immediately that MATH . On the other hand, as MATH we know that MATH . But then, equality occurs if and only if ... |
math/0010030 | As MATH we know that MATH has dimension MATH which is equal to MATH. By the hyper-Kähler correspondence so is the dimension of MATH, whence the open subset of MATH consisting of MATH-semistable representations has dimension MATH as there are MATH-stable representations in it (again via the hyper-Kähler correspondence).... |
math/0010030 | As MATH we know that the local quiver MATH in a simple representation MATH corresponding to MATH is a one vertex quiver having MATH loops. That is, MATH . But then, for MATH a point corresponding to MATH, the local quiver is still MATH but this time the local dimension vector MATH. If MATH lies in the smooth locus, the... |
math/0010031 | For a fixed REF-PS of MATH, there are two possibilities in the previous proposition: either all the MATH's vanish for MATH and we are done, or it is not so. Assume that all MATH for MATH. Because the sum MATH, it follows that there must exist a MATH. We know that MATH and therefore the restriction MATH. Because MATH is... |
math/0010031 | The line bundle MATH is again very ample and its associated linear system gives an embedding MATH . The MATH-action on MATH induces one on MATH. For a stable map MATH which is MATH-semi-stable and MATH a REF-PS of MATH, REF ensures the existence of sections MATH whose restrictions to MATH give a basis of MATH; in parti... |
math/0010031 | By the NAME criterion, MATH . Since MATH is assumed irreducible, REF implies that for any REF-PS MATH of MATH, the image of MATH intersects the MATH-unstable locus of MATH in finitely many points; denote by MATH the NAME open subset of MATH consisting of points which are mapped by MATH into the MATH-semi-stable locus o... |
math/0010031 | The geometric invariant quotient MATH is a projective subvariety of MATH. This last geometric quotient can be described as MATH . Let's denote by MATH the quotient map. There is an invertible sheaf MATH such that MATH for some MATH. It has the additional property that for large enough values of MATH, MATH . The assumpt... |
math/0010031 | There are two things to prove in this statement: the first one is that the stabilizer of MATH in MATH is finite, and the second one is that this point is indeed stable. When MATH, for any MATH we have MATH, and therefore the stabilizer of the map is indeed finite. Let us prove that it is so in general. Consider a repre... |
math/0010031 | We may assume as usual that MATH. Recall that a map MATH is stable if and only if MATH is ample. Also, there is an integer MATH with the property that MATH is very ample. In this way, any stable map MATH gave rise to an embedding MATH into a space isomorphic to MATH, where MATH. The ambiguity in the choice of this isom... |
math/0010031 | Because MATH are all MATH-invariant, it follows that MATH is also. For proving that MATH is a moment map, we need to show that its differential is the same as the contraction of the NAME form MATH on MATH with the vector field MATH. It what follows, the symbol MATH will always denote the contraction of a differential f... |
math/0010031 | We claim that for MATH there is a point MATH with the property that MATH that is, MATH. Suppose that it is not the case, so either Image-MATH or Image-MATH. Let assume that we are in the first case. At a point MATH, MATH . Recall that the smooth real-valued function MATH defined on MATH is the ``quotient" MATH, where M... |
math/0010031 | According to the NAME criterion, is sufficient to prove the statement for every REF-PS MATH. For a fixed REF-PS MATH of MATH, the point MATH is MATH-stable, if its MATH-orbit meets the zero-level set of the moment map MATH on MATH. Assume, for instance, that MATH. By hypothesis, Image-MATH, so that under the MATH-actio... |
math/0010031 | REF follow immediately from the irreducibility of MATH and from the initial assumption that the class MATH is can be represented by a morphism MATH. CASE: If both MATH and MATH are convex, MATH and MATH are open and dense (see REF , page REF); convexity implies also that we are working in the expected dimension. The on... |
math/0010031 | I start noticing that MATH is generically injective on its image. Indeed, by REF , we may restrict our attention to maps whose image is contained in MATH: if MATH are morphisms such that MATH, it follows from REF that there is a morphism MATH such that MATH for all MATH. Since MATH is affine, the morphism MATH must be ... |
math/0010031 | We have seen in REF that MATH and MATH . Moreover, MATH are birational for all MATH. Let MATH and MATH be respectively the images of the MATH-point evaluation maps on MATH and MATH. Then MATH is the closure of the image of the evaluation map on MATH and there is a natural map MATH compatible with the other arrows in RE... |
math/0010031 | It is clear that REF just extends the action of the real gauges by composition on the right. We shall prove the formula for the action of MATH on MATH searching a connection MATH on MATH which makes MATH-holomorphic the map MATH. For doing computations we use a local trivialization of MATH, so that MATH itself may be a... |
math/0010031 | Clearly, we may assume that MATH. The uniqueness part is immediate. For the existence part, notice that if MATH is exact, that is, MATH for MATH, then MATH does the job. Homotopy classes of maps MATH are parameterized by MATH, so for MATH there exists MATH such MATH and MATH. By the discussion above, there exists MATH ... |
math/0010031 | The curvature of MATH is MATH. For MATH and MATH, MATH, and MATH evaluates zero on other pairs of vectors. It is easy to see that for any MATH-form MATH on MATH and any tangent vector MATH to MATH, MATH. This implies that the connection MATH defines indeed a holomorphic structure in MATH because its curvature is a MATH... |
math/0010031 | REF implies that the formula above is indeed an action. It is also holomorphic because MATH preserves the connection MATH; indeed, MATH . CASE: The statement is a direct consequence of REF. |
math/0010031 | CASE: MATH has a holomorphic structure because MATH is a holomorphic bundle, according to REF . That it is also a projective variety follows from the fact that the NAME torus is projective. REF implies that given MATH there is a unique MATH (depending on MATH) such that MATH . The composed map MATH is MATH-equivariant ... |
math/0010031 | Consider a MATH-holomorphic, MATH-equivariant map MATH together with MATH marked points MATH and let MATH. REF says that this data induces a morphism MATH and moreover, it does not depend on the MATH-orbit of MATH. So we get a map MATH . This map is clearly MATH-invariant and therefore descends to the quotient MATH . B... |
math/0010031 | Notice that in the diagram MATH. Indeed, for MATH there is a unique MATH such that MATH, so we may identify MATH. Consequently, MATH and we obtain a MATH-equivariant map MATH which covers the identity of MATH. This one must be an isomorphism. |
math/0010031 | Let me start explaining the guiding idea: we have learned in REF that we should transfer the integrals on MATH used for defining NAME invariants of MATH to integrals on MATH because MATH. All the evaluation maps involved for making these computations live on the right-hand-side of REF. On the other hand, the invariant ... |
math/0010033 | The first statement is well known. To prove the second part of the lemma we have to show that there exists a ray with infinite diameter in any graph MATH that does not contain a star ball. If there is no star ball in MATH then the complement of any vertex MATH in MATH must contain a connected component MATH with infini... |
math/0010033 | For two sets MATH and MATH in MATH we have MATH . MATH is a vertex-cut and MATH the set of ends whose rays are completely contained in MATH and MATH from some index onwards. Thus MATH, and finally MATH . |
math/0010033 | As MATH is connected and a superior set of MATH any vertex MATH in MATH can be connected by a path MATH of vertices in MATH with a vertex MATH in MATH. By induction we get a path MATH such that the initial vertex of MATH is adjacent to the last vertex of MATH and MATH is contained in MATH for every natural MATH greater... |
math/0010033 | Under the given preliminaries there exists a vertex MATH in MATH with infinite degree, which is adjacent to infinitely many connected components of MATH containing elements of MATH. Every element MATH of the base MATH which contains MATH must contain almost all of these components, since the vertex-boundary of MATH is ... |
math/0010033 | Let MATH be a sequence of pairwise different elements of MATH. If MATH is the only vertex-cut, then every element of MATH is an accumulation point of the sequence. Otherwise let MATH be a vertex in the complement of a vertex-cut MATH, in which lie infinitely many elements of MATH. If the sequence has no accumulation po... |
math/0010033 | We may assume that an open cover MATH consists of base elements. Given a set MATH of the cover, we choose a vertex MATH in MATH. Since MATH is finite there exists a finite subcover MATH of MATH in MATH. The set MATH of vertices in MATH is a vertex-cut. Every ray in MATH must lie in one of those vertex-cuts MATH for whi... |
math/0010033 | A sequentially compact NAME space is compact. Thus we can conclude from REF , that MATH is compact. It remains to show that at least one of two elements MATH and MATH of MATH has an open environment, in which the other element is not contained. We distinguish: CASE: One of the two elements is a vertex. If MATH is a ver... |
math/0010033 | We define the MATH-image of an end in MATH as the end in MATH that contains it. The preimage of every base element of the edge-topology is open in the vertex-topology. |
math/0010033 | The statement is a consequence of REF . |
math/0010033 | Let MATH and MATH be open and closed in their corresponding topologies. They can be represented as a finite union of elements of the base, because they are open and compact. Hence MATH is an edge-cut and MATH is a vertex-cut. The same argument holds for MATH and therefore MATH is a vertex-cut, too. Thus MATH is an edge... |
math/0010033 | For any closed set MATH not containing some point MATH we can find a neighbourhood MATH of MATH, which is element of the base and contained in the complement of MATH. As the base elements in MATH are open and closed the indicator function on MATH is continuous. Thus the proper metric end topology is MATH or completely ... |
math/0010033 | Let MATH be an open cover of base elements. It must contain sets MATH and MATH such that MATH is a locally complete set and MATH is a global star-set. After removing these base elements from MATH we obtain a set MATH such that MATH and MATH. MATH is a set of vertices, that contains only finitely many elements of any ba... |
math/0010033 | To show that every sequence in MATH has an accumulation point we can assume that it contains an infinite partial sequence MATH consisting of pairwise different elements. Otherwise the existence of an accumulation point is immediate. If MATH contains a bounded subsequence of vertices, the local end is an accumulation po... |
math/0010033 | MATH implies MATH . Let MATH be a quasi-inverse to MATH. If MATH is infinite, then, by REF , this also holds for MATH. Now MATH implies that MATH has no upper bound. |
math/0010033 | By connecting the MATH-images of adjacent vertices of a metric ray MATH in MATH with geodesic paths with lengths, that are smaller or equal MATH, we obtain a path MATH in MATH. Its diameter is infinite by REF . If it had infinite subset MATH with finite diameter, we also could find infinitely many elements in MATH that... |
math/0010033 | Let MATH be an element of MATH. As MATH the vertex MATH must not be contained in MATH, so MATH and MATH. Hence MATH . Now let MATH be an element of MATH which is adjacent to MATH and choose an element MATH of MATH such that MATH . For a MATH in MATH with MATH we now have MATH . By the boundedness of MATH we have MATH a... |
math/0010033 | As the support of MATH generates MATH we can choose four distinct elements MATH, MATH, MATH and MATH in MATH such that MATH and MATH are positive. We define a random walk MATH on the free group MATH with generating system MATH by choosing MATH, MATH, MATH and MATH as the probabilities for the right multiplication with ... |
math/0010036 | The construction above gives sets of linear evolution data MATH and MATH from MATH and MATH, where MATH lie in vector spaces MATH. It is easy to see that MATH is a vector subspace of MATH, since the NAME algebra MATH of MATH is a vector subspace of MATH. Let MATH be the annihilator MATH of MATH in MATH. Then MATH. Now ... |
math/0010036 | Since MATH by REF , one can show from MATH that MATH, MATH and MATH are linearly independent if and only if MATH, or equivalently if and only if MATH and MATH are linearly independent. But MATH is an immersion near MATH if and only if MATH, MATH and MATH are linearly independent. |
math/0010036 | As MATH span a vector space of dimension REF, they satisfy MATH for some MATH not all zero. We may think of the MATH as giving a linear map MATH with kernel MATH, and so MATH defines a point in the kernel in MATH. Under the action of MATH on MATH defined in REF - REF , the vector MATH transforms under the natural actio... |
math/0010036 | Suppose for simplicity that MATH and MATH are linearly independent in MATH. Consider the action of MATH in MATH upon MATH and MATH given in REF - REF , where MATH. We can think of this as an action of MATH on MATH, where MATH in MATH is mapped to MATH. This action preserves the quadratic form MATH upon MATH. Now there ... |
math/0010036 | REF follows immediately. One can verify that REF - REF are solutions to REF by subtituting them in, and using MATH. The six solutions of REF this gives, with coefficients MATH, are easily seen to be linearly independent. But as REF is a well-behaved first-order o.d.e., its solutions are determined by the initial data M... |
math/0010036 | When MATH, we have MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH. Thus REF holds at MATH if and only if REF holds, so by REF - REF hold when MATH if and only if REF holds. But REF - REF hold at MATH if and only if they hold for all MATH. |
math/0010036 | Substitute REF into the first equation of REF , and use the values for MATH in REF . We get MATH . But MATH as MATH. So dividing by MATH gives MATH which is equivalent to the first line of the equation we have to prove. The other lines follow in the same way. |
math/0010036 | From REF we see that MATH, where MATH . Thus the eigenvalues of MATH are the same as those of MATH, which are the roots of the polynomial MATH. Direct calculation shows that MATH . Now MATH and MATH with MATH. Thus we may substitute MATH, MATH, for MATH in MATH. Then MATH becomes MATH which factorizes as MATH . Thus th... |
math/0010036 | The MATH matrix MATH studied above is not symmetric, because the signs of the MATH terms are not right. However, if we define MATH then MATH is a symmetric complex matrix, and REF shows that MATH are eigenvectors of MATH, with eigenvalues MATH and MATH respectively. Now if MATH are eigenvectors of a symmetric matrix wi... |
math/0010037 | CASE: Use the surjectivity of the map MATH. CASE: It follows from the fact that, by MATH-invariance, MATH contains the vertical part of the tangent space to the orbit of MATH under the action of MATH. |
math/0010037 | CASE: We start observing that the sheaf MATH is globally generated. Indeed, by NAME 's vanishing REF we have MATH . Therefore, by NAME 's m-regularity theorem ( REF , page REF) the maps MATH are surjective for MATH. The result now follows immediately recalling the isomorphism MATH and the surjection MATH . CASE: Again ... |
math/0010037 | By REF , we have MATH . Let MATH be a smooth point of MATH. Since the bundle MATH is generated by its global sections, there exists a section MATH such that MATH . Since MATH is generically an immersion, we obtain from the above a non zero element in MATH coming from MATH. |
math/0010037 | Recall that MATH can be naturally interpreted as the kernel of the NAME map MATH. Hence one easily verifies that MATH contains MATH, for all MATH and MATH. Then MATH contains elements of the form MATH for all MATH and MATH, coming from the wedge product of elements in MATH. Since we are supposing that MATH is in the ba... |
math/0010037 | Since the distribution MATH is integrable, the brackets induce a map MATH which is given at the point MATH by MATH . Since we are supposing that MATH contains MATH but not the whole ideal MATH, there is a canonical isomorphism MATH and hence MATH identifies with a map MATH which we also denote by MATH. To prove the int... |
math/0010037 | CASE: Assume MATH . Then the assertion follows from the fact that the contact condition MATH for a line MATH through MATH with respect to MATH becomes MATH . REF A dimension count shows that all hypersurfaces MATH in MATH of degree MATH admit a point through which passes a line having contact with MATH of maximal order... |
math/0010041 | First, we will show that MATH is generated as a MATH-module by MATH . Let MATH be a homogeneous endomorphism in the MATH-centre of MATH. Then, MATH for some MATH and all MATH in MATH. Since MATH is homogeneous, MATH is some homogeneous MATH of the same degree as MATH. Then MATH. Since MATH has the same value on MATH as... |
math/0010041 | It is clear that MATH is contained in MATH so we only need to show that every MATH is in the MATH span of MATH. To this end, suppose for some MATH and all MATH that MATH. Then we have for any MATH, MATH . Since MATH is an automorphism of MATH, MATH. Hence MATH. Since MATH is a group, we may put MATH. Then MATH for any ... |
math/0010041 | Since MATH is a bracket, MATH is a derivation. Hence, for any MATH, we have MATH . Inductively, we have that if MATH then MATH. Hence MATH is in the MATH-span of MATH, and so it is in MATH. It follows that for any MATH, MATH. Hence MATH. |
math/0010041 | Indeed, if MATH, then MATH for homogeneous MATH. Corresponding to each MATH, there exists MATH such that MATH. Hence, MATH . Now induction completes the lemma. |
math/0010041 | Let MATH. By REF , it is enough to show MATH. For any MATH, MATH . Thus MATH for any MATH. Hence MATH, as required. |
math/0010041 | This follows from the fact that MATH. |
math/0010041 | It is enough to prove the lemma when MATH is a monomial, for if the lemma holds for monomials MATH, MATH, and MATH, then we can find differential operators MATH of the prescribed form such that MATH. Putting MATH yields the desired result. Now, suppose MATH where each MATH. We will show by induction on MATH that there ... |
math/0010041 | Let MATH be the MATH-module generated by MATH . Since MATH commutes with MATH up to multiplication by a scalar in MATH, we have MATH for any MATH. We will show by induction on MATH that MATH. Since MATH, we have MATH. This proves the base case. Now suppose that MATH. By REF above, we know that MATH. Thus, we must show ... |
math/0010041 | Let MATH be an ideal in MATH. Let MATH. Then MATH can be written as MATH where MATH are polynomials in MATH, MATH and the multi indices MATH have entries in MATH. We induct on MATH, to claim that the ideal containing MATH should contain REF. Assume that MATH. Then MATH where MATH and MATH. Now MATH have fewer monomials... |
math/0010041 | It is clear that MATH, MATH, and MATH are homogeneous of degree REF. We must show that they generate all of MATH, that they commute, and that they have no other relations among them. Suppose that MATH. Then by REF , MATH can be written as MATH where each MATH is a monomial in MATH, MATH, MATH, and MATH. Moreover, each ... |
math/0010041 | First we show that MATH cannot be a zero divisor. It is obvious that MATH if MATH. Suppose MATH, then MATH for all positive integers MATH. Also note that MATH. Hence MATH does not satisfy REF . Suppose MATH. Let MATH and MATH be the highest degree parts of MATH and MATH of degrees MATH and MATH respectively. Then MATH.... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.