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A tensor product of $V$ and $W$ is a couple $(T,h)$ where $T$ is a vector space and $h:V\oplus W\to T$ is a linear surjective map satisfying the following universal property :

Let $f:V\oplus W\to U$ be a bilinear map. There exists a linear map $g:T\to U$ such that $f=g\circ h$.

In the following, $L_k(V_1,\ldots V_k;R)$ is the set of $k$-multilinear maps $V_1\times \ldots V_k$ to R.

Finite dimensional

I assume every $V_i$ is finite dimensional vector space (over the real numbers if that matters).

My guess

I guess that $L_k(V_1,\ldots V_k;R)$ is a tensor product for $V=L_{k-1}(V_1,\ldots V_{k-1};R)$ and $W=V_k^*$.

This is what I want to prove.

I guess that the maps $h$ is the map $h:L_{k-1}(V_1,\ldots,V_{k-1};R)\oplus V_k^*\to L_k(V_1,\ldots V_k;R)$ is given by

$$ h(\xi,\xi_k)(v_1,\ldots v_k)=\xi(v_1,\ldots v_{k-1})\xi_k(v_k). $$

Question

I can prove that $h$ is bilinear.

QUESTION: how to prove the universal property of $h$ ?

EDIT: answer in the case $V^*\times W^*$ here: tensor product and space of multilinear forms

EDIT: $h$ is probably not surjective, but does not need to.

Laurent Claessens
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  • Are your vector spaces finite-dimensional? – Paul Frost Mar 26 '23 at 10:07
  • Yes: finite dimensional. I edit the question to state it explicitely. – Laurent Claessens Mar 26 '23 at 10:26
  • How do you prove that $h$ is surjective? – Gribouillis Mar 26 '23 at 10:44
  • Also note that $h:V\oplus W\to T$ must be required to be a bilinear map (which is in general not surjective). Moreover you must require "There exists a unique linear map $g$". – Paul Frost Mar 26 '23 at 10:48
  • It's true for infinite dimensions as well https://planetmath.org/tensorproductofdualspacesisadualspaceoftensorproduct – Tychus Findlay Mar 26 '23 at 10:58
  • @Gribouillis To prove surjectivity, I use theorem 1.3 here. But I have to massively change the notations because --it's the point of my question-- I am not ok with his tensor product notation. – Laurent Claessens Mar 26 '23 at 12:13
  • @LaurentClaessens Theorem 1.3 proves that the domain and the codomain of $h$ have the same dimensions, but $h$ is not linear, so it doesn't suffice to prove surjectivity (also note that this paper takes $k$ copies of the same space $V$). – Gribouillis Mar 26 '23 at 12:58
  • @Gribouillis It's surjective on a basis. I think that being multilinear is even better than linear to span the whole vector space from a basis. I'll double-check. In fact, in the given paper, it is said that the space of multilinear maps IS the tensor product. So I'm quite sure that my "guess" is correct. But well ... without a full proof ... – Laurent Claessens Mar 26 '23 at 15:20
  • @Gribouillis You are right. The map $h$ is probably not surjective. But surjectivity is not required. My double-bad. – Laurent Claessens Apr 02 '23 at 05:04

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