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This Wikipedia article states that for a vector space V:

Using the universal property, it follows that the space of (m,n)-tensors admits a natural isomorphism $T^m_n(V) \cong L(\underbrace{V^* \otimes \cdots \otimes V^*}_m \otimes \underbrace{V \otimes \cdots \otimes V}_n; F) \cong L^{m+n}(\underbrace{V^*, \ldots,V^*}_m,\underbrace{V,\ldots,V}_n; F).$

(With $T^m_n(V) = \underbrace{ V\otimes \dots \otimes V}_{m} \otimes \underbrace{ V^*\otimes \dots \otimes V^*}_{n}$).

It seems to me like there are two things wrong with this:

  • This is only true if V is finite-dimensional, which is not stated as of this writing (see e.g. this question).
  • This does not follow at all from the universal property, but rather from constructing the isomorphism similar to point 2 here - and using the finite dimensionality to prove that this is indeed an isomorphism. (Edit: When I say "this", I mean the asserted isomorphisms between the space of (m,n) tensors and the two spaces on the right hand side, since this is what the article is talking about.)

Am I wrong here, or does the article need fixing?

Christophe AGUETTAZ
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  • Why is it only true for finite dimensional $V$? I think it should be true in general – Yanko Dec 01 '22 at 00:00
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    The universal property is precisely what gives you the second natural isomorphism listed there, but you're correct the first requires finite-dimensionality. – blargoner Dec 01 '22 at 03:57
  • @blargoner Why not an official answer? – Paul Frost Dec 01 '22 at 08:47
  • @blargoner The way I understand it, the universal property gives you the existence and unicity of a linear function, but has nothing to say about whether it's a natural isomorphism - indeed the universal property applies in the infinite-dimensional case for which there is no natural isomorphism. I am missing something here? – Christophe AGUETTAZ Dec 01 '22 at 17:13
  • @ChristopheAGUETTAZ In general the isomorphism $L(E,F;H)\cong L(E\otimes F;H)$ induced by the UMP of the tensor product is natural. Why do you think that fails in the infinite-dimensional case? – blargoner Dec 01 '22 at 21:20
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    @blargoner Sorry I misunderstood, I thought that by "second natural endomorphism" you meant $T^m_n(V) \cong L^{m+n}(\underbrace{V^, \ldots,V^}_m,\underbrace{V,\ldots,V}_n; F)$ rather than the isomorphism on the right hand side. – Christophe AGUETTAZ Dec 01 '22 at 22:01

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