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Problem:

I'm wondering what could be a realization for algebraic tensor product of infinite dimensional vector spaces?
Any ideas are welcome, of course!

Attempts:

My first guess would be the space of bilinear forms:
$$\phi:U\times V\to\mathcal{B}(U^*\times V^*; \mathbb{R}):\phi(u,v)(\mu,\nu):=\mu(u)\nu(v)$$ Within this realization the tensor product becomes the linear span of the image: $$U\otimes V\cong\langle\mathrm{im}\phi\rangle$$ This description is rather abstract. So how to prescribe it concretely?

Disclaimer:

I'm neither concerned with the topological tensor product nor with monoids, groups, rings or modules but only with the algebraic tensor product of vector spaces. My interest in here is not about a categorical description (though it should of course satisfy the universal property).

I'm asking since I'd like to have a realization applicable in general since many authors give a variety of realizations - most of them not working for infinite dimensional vector spaces.

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  • $\begingroup$ A linear functional has rank $1$ or $0$. $\endgroup$
    – Berci
    Feb 14, 2014 at 0:29
  • $\begingroup$ Hmm, I see ...so how can one save the day so a similar realization holds? $\endgroup$
    – C-star-W-star
    Feb 14, 2014 at 0:34
  • $\begingroup$ Your claim is false for finite-dimensional vector spaces. The LHS has dimension $\dim V \dim W$, the RHS has dimension $\dim V + \dim W$. $\endgroup$
    – Zhen Lin
    Feb 14, 2014 at 8:05
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    $\begingroup$ Oh I read $V^* \otimes W^*$ on the right hand side, because a product just doesn't fit at all :). $\endgroup$
    – Martin Brandenburg
    Feb 14, 2014 at 13:54
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    $\begingroup$ I'd be interested in a refinement of what is desired... since the usual constructions certainly succeed. For free modules over a ring a basis of the tensor product is given by tensor products of basis pairs, etc. Maybe there is some agenda, some desiderata, insufficiently described by saying that "the usual construction" isn't sufficient. $\endgroup$
    – paul garrett
    Mar 31, 2014 at 2:05

3 Answers 3

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If $M,N$ are modules over a commutative ring $R$, there is a canonical linear map $$M \otimes N \to M^{**} \otimes N^{**} \to (M^* \otimes N^*)^*.$$ It maps $m \otimes n$ to $(\phi \otimes \psi \mapsto \phi(m) \psi(n))$. It is an isomorphism when $M$ and $N$ are finitely generated projective (the usual proof: test it for $M=R$ and $N=R$, then use direct sums and direct summands). If $M,N$ are infinitely generated free, we cannot expect there to be an isomorphism. This is also true if we replace $M^*$ by $\{f \in M^* : \mathrm{im}(f) \text{ finite}\}$, the right hand side is just too big.

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  • $\begingroup$ But that is not my question; I don't want to find isomorphisms between V and V^* resp for tensor products but V(x)W and Hom(V^*xW^*;R) $\endgroup$
    – C-star-W-star
    Feb 14, 2014 at 1:11
  • $\begingroup$ btw for reflexive spaces theres a canonical isomorphism, isn't there? $\endgroup$
    – C-star-W-star
    Feb 14, 2014 at 1:14
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You could always pick bases $B$ and $B'$ for $V$ and $W$, and define $V \otimes W$ to be the vector space with basis $B \times B'$.

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  • $\begingroup$ That is a way of saying what we're looking for. However the problem is the existence of such a space with desired properties; and that boils down to find a realization... Besides when one tries to define such in object in an abstract way the problem arises what addition means in there:<br>$(a_i,b_i)_{i\in I}+(\tilde{a}_i,\tilde{b}_i)_{i\in I}=?$<br> Though one might save the day by some tedious construction - quotiening out desired properties - however then working with those equivalence classes becomes seemingly unpractical. $\endgroup$
    – C-star-W-star
    Mar 26, 2014 at 20:44
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    $\begingroup$ @Freeze_S: I'm not sure what you mean by the notation $(a_i, b_i)_{i \in I}$. There is a simple description of a vector space with basis $B$: recall that the coordinates of a vector are simply a function that maps basis elements to the corresponding coefficient, and only finitely many coefficients are nonzero. Thus, such a vector space is isomorphic to the space of all functions $B \mapsto \mathbf{R}$ that are nonzero for only finitely many $B$. The vector space operations are ordinary function operations. $\endgroup$
    – user14972
    Mar 27, 2014 at 5:35
  • $\begingroup$ Ah so u mean: $U\cong\mathcal{l}_0(I), V\cong\mathcal{l}_0(J)\Rightarrow U\otimes V\cong\mathcal{l}_0(I\times J)$ $\endgroup$
    – C-star-W-star
    May 20, 2014 at 2:36
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Ok the guess seems to be wrong, though it might be saved be regarding rather bilinear functionals of finite cokernel...

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