From: palmieri@math.washington.edu (John H. Palmieri) Subject: Re: duals and tensor products Date: 05 Jul 2000 22:25:05 -0700 Newsgroups: sci.math Summary: [missing] "Norm Dresner" writes: > Where can I find a counterexample for infinite dimensional spaces? > Norm > John H. Palmieri wrote in message > news:s5tpuosfpll.fsf@goedel1.math.washington.edu... > > Luca ugaglia writes: > > > > > Let V and W be two vector spaces and V^*, W^* be their duals. > > > Is it true that the dual of the tensor product of V and W is equal to > the > > > tensor product of V^* and W^*? > > > > If V and W are finite-dimensional, yes. (First, maybe I should have said "or": if V or W is finite-dimensional, things should be okay.) Now, I've interpreted the question in a certain way; in particular, the word "equal" can be troublesome. If all you care about is whether (V^* tensor W^*) and (V tensor W)^* have the same dimension (so that they're "isomorphic") maybe that's always true. If you want these two vector spaces to be isomorphic in a "natural" way, then you really want either V or W to be finite-dimensional. Assume that k is the ground field, and all Homs and tensors are over k. Then: (V tensor W)^* = Hom (V tensor W, k) = Hom (V, Hom (W,k)) = Hom (V, W^*). (All of these = signs mean actual equality or natural isomorphism.) Now, there is a vector space map from (V^* tensor W^*) to this last thing: it sends (f tensor g) to the map (v --> f(v) g). This is always one-to-one, but it isn't onto if V and W are infinite-dimensional. Any example will do. This is important if V and W have some extra structure, and you want to know if they are isomorphic with that extra structure preserved; for example, V and W may be modules over a group algebra kG, and then you might want to know if (V^* tensor W^*) and (V tensor W)^* are isomorphic as kG-modules. They won't be in general if V and W are infinite-dimensional. -- John H. Palmieri Dept of Mathematics, Box 354350 palmieri@math.washington.edu University of Washington http://www.math.washington.edu/~palmieri Seattle, WA 98195-4350 ============================================================================== From: wcw@math.psu.edu (William C Waterhouse) Subject: Re: duals and tensor products Date: 7 Jul 2000 21:36:08 GMT Newsgroups: sci.math In article , "Norm Dresner" writes: > Where can I find a counterexample for infinite dimensional spaces? > Norm > John H. Palmieri wrote in message > news:s5tpuosfpll.fsf@goedel1.math.washington.edu... > > Luca ugaglia writes: > > > > > Let V and W be two vector spaces and V^*, W^* be their duals. > > > Is it true that the dual of the tensor product of V and W is equal to > the > > > tensor product of V^* and W^*? > > > > If V and W are finite-dimensional, yes. Here's an explicit example that could be extended to work in general (when both V and W are infinite-dimensional). Suppose V and W have countably infinite bases e_1, e_2, ... and f_1, f_2, .... Then the elements (e_i tensor f_j) = g_{ij} are a basis for the tensor product. Define an element of the dual (a linear function on the tensor product) by setting B(g_{ij}) to be 1 when i = j and 0 otherwise. I claim this does not come from anything in the tensor product of the dual spaces. The point is that it would have to be a FINITE sum of tensor product elements. That is, for some positive integer p, there would be elements s_k in V^* and t_k in W^* with Sum _1 ^p [ s_k(e_i) t_k(f_j)] = B(g_{ij}) for all i, j. To see that this is impossible, let v_i = , w_j = . The formula for B then tells us that the dot product of v_i and w_j is 1 when i = j and 0 when i is different from j. But now it's trivial to see that this forces the v_i to be independent, which they can't be in a p-dimensional space. William C. Waterhouse Penn State ============================================================================== From: palmieri@math.washington.edu (John H. Palmieri) Subject: Re: duals and tensor products Date: 07 Jul 2000 16:16:29 -0700 Newsgroups: sci.math wcw@math.psu.edu (William C Waterhouse) writes: > In article , > "Norm Dresner" writes: > > Where can I find a counterexample for infinite dimensional spaces? > > Norm > > John H. Palmieri wrote in message > > news:s5tpuosfpll.fsf@goedel1.math.washington.edu... > > > Luca ugaglia writes: > > > > > > > Let V and W be two vector spaces and V^*, W^* be their duals. > > > > Is it true that the dual of the tensor product of V and W is equal to > > the > > > > tensor product of V^* and W^*? > > > > > > If V and W are finite-dimensional, yes. > > Here's an explicit example that could be extended to work in general > (when both V and W are infinite-dimensional). > > Suppose V and W have countably infinite bases e_1, e_2, ... and > f_1, f_2, .... I agree with your example, but to clarify: I still think it depends on what you mean by "equal". In your example, (V^* tensor W^*) and (V tensor W)^* have the same dimension, so they are isomorphic. The isomorphism is not very pleasant, though, and in particular, it is not given by the obvious map from V^* tensor W^* to (V tensor W)^*. So I wouldn't call them equal. -- John H. Palmieri Dept of Mathematics, Box 354350 palmieri@math.washington.edu University of Washington http://www.math.washington.edu/~palmieri Seattle, WA 98195-4350 ============================================================================== From: wcw@math.psu.edu (William C Waterhouse) Subject: Re: duals and tensor products Date: 11 Jul 2000 21:26:32 GMT Newsgroups: sci.math In article , palmieri@math.washington.edu (John H. Palmieri) writes: > wcw@math.psu.edu (William C Waterhouse) writes: > ... > > > > > > > > > Let V and W be two vector spaces and V^*, W^* be their duals. > > > > > Is it true that the dual of the tensor product of V and W is equal to > > > the > > > > > tensor product of V^* and W^*? >... > I agree with your example, but to clarify: I still think it depends on > what you mean by "equal". In your example, (V^* tensor W^*) and (V > tensor W)^* have the same dimension, so they are isomorphic. The > isomorphism is not very pleasant, though, and in particular, it is not > given by the obvious map from V^* tensor W^* to (V tensor W)^*. So I > wouldn't call them equal. Yes, of course I agree. To be more formal, there is a natural linear mapping from the tensor product of the duals to the dual of the tensor product. It is always one-to-one, and the example shows that it is not onto unless one of the spaces has finite dimension. I tend to think of the natural mapping here as an inclusion, but sometimes we do have to notice that this is not formally true (and of course I would have to be much more careful if these were modules over more general rings). William C. Waterhouse Penn State ============================================================================== From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik) Subject: Re: In the most abstract sense... Date: 1 Aug 2000 18:03:32 -0400 Newsgroups: sci.math In article <8m78al$ief$1@news.panix.com>, Edward Green wrote: :What is a tensor? : :I don't mean in terms of transformation properties. Is it a map? Sticking to the request for utmost abstraction (still in the context of vector spaces): it is a construction of a universal bilinear mapping. Suppose V and W are vector spaces, we will call a vector space Z a tensor product of V and W (denoted V ox W as in another reply) and a bilinear mapping VxW ->Z , denoted (v , w) -> v ox w the corresponding tensor product, if for every vector space L, and every bilinear mapping f: VxW -> L, there exists a unique linear mapping g: Z -> L such that f(v,w) = g(v ox w) Two concrete constructions of tensor products were described elsewhere; a proof is still needed (and performed in courses of advanced algebra) to show that (1) these constructions are indeed universal (at least in finite dimensions), and (2) all tensor products (satisfying the universal description) of V and W are isomorphic, so that your favorite representation will do the job. Let me contribute a third construction, suitable for finite-dimensional V and W: Suppose is a basis of V, and is a basis of W. Then we define Z as the vector space with m*n elements of its basis, denoted formally (just as a mnemonic) h_(j,k) = e_j ox f_k , j=1,...,m and k=1,...,n. and the tensor multiplication is defined by imposing distributive law, that is (sum a_j*e_j) ox (sum b_k*f_k) = sum((a_j*b_k)*h_(j,k)). Every bilinear mapping f from VxW to L is well-defined by naming the results of f(e_j, f_k) = l_(j,k) (we have bases, that's why). So, we define the linear mapping g from the universality requirement by g(h_(j,k)) = l_(j,k). Linearity and uniqueness can be proved. So, in this twisted way, a tensor product is a map. For tensor products of three and more vector spaces: another "abstract nonsense" fact can be proved, namely that (V ox W) ox Z is (explicitly, "canonically") isomorphic to V ox (W ox Z) so that we can drop the parentheses if we are not finicky about isomorphisms of this kind. Whether it helps or not, you asked for it :-) Cheers, ZVK(Slavek). ============================================================================== From: grubb@math.niu.edu (Daniel Grubb) Subject: Re: In the most abstract sense... Date: 2 Aug 2000 13:59:55 GMT Newsgroups: sci.math >In article <8m78al$ief$1@news.panix.com>, Edward Green wrote: >:What is a tensor? >: >:I don't mean in terms of transformation properties. Is it a map? >Sticking to the request for utmost abstraction (still in the context of >vector spaces): it is a construction of a universal bilinear mapping. >Suppose V and W are vector spaces, we will call a vector space Z a tensor >product of V and W (denoted V ox W as in another reply) >and a bilinear mapping VxW ->Z , denoted (v , w) -> v ox w the >corresponding tensor product, if > for every vector space L, and every bilinear mapping f: VxW -> L, > there exists a unique linear mapping g: Z -> L >such that > f(v,w) = g(v ox w) >Two concrete constructions of tensor products were described elsewhere; a >proof is still needed (and performed in courses of advanced algebra) to >show that (1) these constructions are indeed universal (at least in finite >dimensions), and (2) all tensor products (satisfying the universal >description) of V and W are isomorphic, so that your favorite >representation will do the job. You have to be a little bit careful here. In diff. geo., a tensor is a multilinear map on a product of vector spaces. Such a tensor can be regarded as an element of the tensor product of the duals of the corresponding vector spaces as above. A tensor field is an object that consists of a tensor at every point in the base space (i.e. the manifold) such that the components vary differentiably and where the vector spaces used are either the tangent space or it's dual. A tensor field may *also* be regarded as a multilinear map on the vector space of differential functions and it's dual. A good differential geometry book should cover this. --Dan Grubb