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I have a couple of questions about tensor products:

Why is $\text{Hom}(V,W)$ the same thing as $V^* \otimes W$?

Why is an element of $V^{*\otimes m}\otimes V^{\otimes n}$ the same thing as a multilinear map $V^m \to V^{\otimes n}$?

What is the general formulation of this principle?

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The result is generally wrong for infinite-dimensional spaces: see this question.

For finite dimensional space $V$, let's build an isomorphism $f : V^* \otimes W \to \hom(V,W)$ by defining

$$f(\phi \otimes w)(v) = \phi(v) w$$

This clearly defines a linear map $V^* \otimes W \to \hom(V,W)$ (it's bilinear in $V^* \times W$). Reciprocally, take a basis $(e_i)$ of $V$, then define $g : \hom(V,W) \to V^* \otimes W$ by:

$$g(u) = \sum_i e_i^* \otimes u(e_i)$$

Where $(e_i^*)$ is the dual basis to $(e_i)$ (I will use a few of its properties in $\color{red}{red}$ below). This is well-defined because $V$ is finite-dimensional (the sum is finite). Let's check that $f$ and $g$ are inverse to each other:

  • For $u : V \to W$, $$f(g(u))(v) = \sum e_i^*(v) u(e_i) = u \left( \sum e_i^*(v) e_i \right) \color{red}{=} u(v)$$ and so $f(g(u)) = u$.

  • For $\phi \otimes w \in V^* \otimes W$, $$g(f(\phi \otimes w)) = \sum e_i^* \otimes f(\phi \otimes w)(e_i) = \sum e_i^* \otimes \phi(e_i) w = \sum \phi(e_i) e_i^* \otimes w \color{red}{=} \phi \otimes w$$

And so $f$ and $g$ are isomorphisms, inverse to each other.

It is known that for finite dimensional $V$, then $(V^*)^{\otimes m} = (V^{\otimes m})^*$. Then an element of $V^{* \otimes m} \otimes V^{\otimes n}$ is an element of $(V^{\otimes m})^* \otimes V^{\otimes n} = \hom(V^{\otimes m}, V^{\otimes n})$. So by definition / universal property of the tensor product, it's a multilinear map $V^m \to V^{\otimes n}$.

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    $\begingroup$ That's great! I think another way to see that they are the same is as follows: given a basis $\{w_i\}$ for $W$, one can specify a linear transformation by picking any $n$ elements of $V^*$, and writing $$Tu = \varphi_1(u)w_1 + \dotsb + \varphi_n(u)w_n.$$ This is a bilinear map from $V^*\times W$ to $\text{Hom}(V,W)$, and it has "basis" $u_i^* \otimes w_j, \, 1\leq i\leq n, \, 1\leq j \leq m$. Note that $\varphi_i = \rho_i \circ T$, where $\rho_i$ is the projection onto the $i$th basis vector in $W$. $\endgroup$
    – Eric Auld
    Commented Feb 22, 2014 at 23:41
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    $\begingroup$ I will also add that the general principle is that there is an embedding $$\theta: \text{Hom}(A, A')\otimes \text{Hom}(B,B') \hookrightarrow \text{Hom}(A\otimes B , A'\otimes B'),$$ which is an isomorphism if the spaces are finite-dimensional. This particular correspondence comes from letting $U=A,\,U'=\mathbb{F},\,B=\mathbb{F},\,B'=W.$ $\endgroup$
    – Eric Auld
    Commented Apr 22, 2014 at 16:40
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    $\begingroup$ Above should read $A=U$, $A’ = \mathbb{F}$, $B = \mathbb{F}$, $B’=W$. $\endgroup$
    – Eric Auld
    Commented May 3, 2023 at 15:20
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One general form of that assertion (noting that it cannot be quite as simple as one might imagine) is the Cartan-Eilenberg adjunction $$ \mathrm{Hom}(X\otimes Y,Z)\;\approx\;\mathrm{Hom}(X,\mathrm{Hom}(Y,Z)) $$ in some reasonable additive (or whatever) category, where, significantly, the tensor product must be a genuine categorical tensor product, as opposed to a "projective" or "injective" tensor product, which have only half the properties of a genuine tensor product. So, for example, there is no genuine tensor product of (infinite-dimensional) Hilbert spaces in any reasonable category of topological vector spaces. (The thing often called the "Hilbert-space tensor product" has only half the requisite properties.

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  • $\begingroup$ Useful! Might help people to provide links to various concepts e.g. projective product $\endgroup$
    – Eric Auld
    Commented Jun 9, 2021 at 20:00
  • $\begingroup$ @EricAuld, I'm afraid I don't know a source I'd approve about those various fractional notions of "tensor product". If I were left to myself, I'd look on Wikipedia... Most of the discussions I can recall were definitely not categorically oriented, etc. If you find something good, please feel free to edit my remarks. $\endgroup$
    – paul garrett
    Commented Jun 9, 2021 at 20:04
  • $\begingroup$ Sorry, I don't understand how this adjunction (which holds in the category of vector spaces without any restriction — only saying that the internal Hom exists) leads to the result which holds only when $V$ is dualizable. $\endgroup$
    – Yai0Phah
    Commented Aug 18, 2022 at 21:01
  • $\begingroup$ @Yai0Phah, well, yes, it is more a variant than including the simplest case as a special case. For example, taking $Y=W^*$ and $Z=k$ scalars, it only gives $(X\otimes W^*)^*\approx \mathrm{Hom}(X,W^{**})$. If we knew from a different logical order of development that the dual on the left-hand-side is $X^*\otimes W^{**}$, and so on... but, yes, proof of the latter amounts to the same as the short proof of the simpler thing. :) $\endgroup$
    – paul garrett
    Commented Aug 18, 2022 at 21:34
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Well, the actual result is $V^*\otimes W\cong \{\varphi\in \mathcal{L}(V,W):\dim \text{range }\varphi<\infty\}$, via the canonical map $\Phi: \varphi\otimes w\mapsto (v\mapsto \varphi(v)w)$ as given in the answer above (note that this is always an embedding of $V^*\otimes W$ into $\mathcal{L}(V,W)$).

On one hand, every element of $V^*\otimes W$ can be written as $\sum^{n}_{i=1} \varphi_i\otimes w_i$, and its image under $\Phi$ is included in $\mathrm{span}(w_1,\cdots,w_n)$; on the other hand, if $\dim \text{range }\varphi<\infty$, pick a basis $w_1,\cdots,w_n$ of $\text{range }\varphi$, then $\varphi = \sum^{n}_{i=1} \varphi_iw_i$ for $\varphi_i\in V^*$, and $\sum^{n}_{i=1} \varphi_i\otimes w_i$ is a preimage of $\varphi$.

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