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?


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 $V^m \to V^{\otimes n}$.

  • 1
    $\begingroup$ 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 Feb 22 '14 at 23:41
  • 8
    $\begingroup$ 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 Apr 22 '14 at 16:40

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