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It has just occurred to me that most of my intuition for tensor products is derived from the special case of finite-dimensional vector spaces, so I'm wondering which properties I've taken for granted are true in general, and which are not.

  1. Let $U$ and $V$ be $k$-vector spaces, possibly infinite-dimensional. Does it remain true that $U^* \otimes V \cong \textrm{Hom}(U, V)$ naturally in $U$ and $V$?

  2. Let $A, B, C$ be objects in an abelian category, or better, a monoidal closed category. Is it true that $\textrm{Hom}(A, B \otimes C) \cong \textrm{Hom}(A, B) \otimes \textrm{Hom}(A, C)$ naturally in $A, B, C$? (Motivation: $\textrm{Hom}(A, -)$ preserves (cartesian) products.)

  3. In the same context as above, is there a bifunctor $\mathscr{F}(-, -)$ such that $\textrm{Hom}(A, C) \otimes \textrm{Hom}(B, C) \cong \textrm{Hom}(\mathscr{F}(A, B), C)$ naturally in $A, B, C$? (Motivation: $\textrm{Hom}(-, C)$ maps coproducts to products.)

Zhen Lin
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  • For 1: you only get the finite rank linear maps. – Pierre-Yves Gaillard Aug 13 '11 at 10:25
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    Well an element of $U^{\ast} \otimes V$ has finite rank as a linear map, no? For the other two: did you try to see what happens if you fix two variables and let the third vary, for instance in 2. take $A = \varinjlim A_i$ the functor on the left will then be $\varprojlim \operatorname{Hom}{(A_i, B \otimes C)}$ and on the right hand side you'd have to commute the tensor product with limits, so no chance without finiteness conditions. Similarly for 3. – t.b. Aug 13 '11 at 10:28
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    Of course, you still have a canonical isomorphism of $U^ *\otimes V$ onto the subspace of finite rank maps from $U$ to $V$. – Pierre-Yves Gaillard Aug 13 '11 at 10:37
  • @Theo, Pierre-Yves: Thanks. Seems like I have a lot to (un)learn. – Zhen Lin Aug 13 '11 at 11:18
  • I noticed that the most modest persons are the ones who have the least reasons for being so. – Pierre-Yves Gaillard Aug 13 '11 at 11:54
  • Is there a positive answer to 2 even for finite-dimensional vector spaces over a field? I get the isomorphism part, but where does the canonicity come from? – darij grinberg Aug 13 '11 at 13:22
  • @darij: Actually, I really didn't think through 2 and 3 very much at all. In the finite-dimensional case, for dimension reasons, 2 only works when $A$ is the base field, and even then it says something really trivial. – Zhen Lin Aug 13 '11 at 13:28

1 Answers1

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  1. No. As Theo says in the comments, the elements of $U^{\ast} \otimes V$ are precisely the finite-rank maps in $\text{Hom}(U, V)$.

  2. This isn't even true in finite dimensions. The dimension of the LHS grows linearly in $\dim A$ but the dimension of the RHS grows quadratically in $\dim A$.

  3. This also isn't even true in finite dimensions. The dimension of the LHS grows quadratically in $\dim C$ but the dimension of the RHS grows linearly in $\dim C$.

You seem to be under the mistaken impression that the tensor product is supposed to behave like a product. It isn't. Abstractly it comes from the tensor-hom adjunction

$$\text{Hom}(A \otimes B, C) \cong \text{Hom}(A, \text{Hom}(B, C))$$

and concretely it comes from wanting the free vector space functor $\text{Set} \to \text{Vect}$ to be (lax?) monoidal.

Working with naked infinite-dimensional vector spaces is asking for trouble. See topological tensor product for appropriate substitutes for topological vector spaces.

Qiaochu Yuan
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  • Your last point is interesting: I think the properties I've been ascribing to the hom functor are in fact properties of the product functor! (Because, after all, $\mathbf{Set}$ is also a monoidal closed category.) – Zhen Lin Aug 13 '11 at 13:53
  • @Zhen: all the properties you've ascribed to the Hom functor are correct, but the product in $\text{Vect}$ and related categories is the direct product, not the tensor product. – Qiaochu Yuan Aug 13 '11 at 13:56
  • And once again you show me that I should have thought a bit more before commenting... – t.b. Aug 13 '11 at 13:59
  • @Qiaochu: Yes, I'm aware. My comment was informal: I was observing that, in some sense, $\textrm{Hom}(A, -)$ preserves products because of the product is defined to make it so. But it is a little distressing that the tensor product is neither a pure limit nor a pure colimit in most categories. – Zhen Lin Aug 13 '11 at 14:07
  • @Zhen: oh. That was a confusing way of saying that. Okay, but the tensor product is something better than a limit or colimit: it's an adjoint of an internal hom, and those are very natural functors. – Qiaochu Yuan Aug 13 '11 at 14:13
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    @Zhen: another way to think about tensor products is that they are really a way to describe multicategories. In this case the relevant multicategory has morphisms given by multilinear maps. – Qiaochu Yuan Aug 13 '11 at 14:25