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

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?

• Have you thought about why it is $V^*\otimes W$ and not $V\otimes W$? Feb 17, 2014 at 15:31
• More general case: math.stackexchange.com/questions/1988633 Dec 27, 2018 at 20:33

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

• 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 $W$. Feb 22, 2014 at 23:41
• 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 $U=A,\,U'=\mathbb{F},\,B=\mathbb{F},\,B'=W.$ Apr 22, 2014 at 16:40

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.

• Useful! Might help people to provide links to various concepts e.g. projective product Jun 9, 2021 at 20:00
• @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. Jun 9, 2021 at 20:04

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 $$\varphi$$.