Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
    
For 1: you only get the finite rank linear maps. –  Pierre-Yves Gaillard Aug 13 '11 at 10:25
2  
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
    
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

1 Answer 1

up vote 6 down vote accepted
  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.

share|improve this answer
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.