# Universal property of the tensor product

Assume $\Phi:V_1^*\times ...\times V_k^* \rightarrow L(V_1,...,V_k;\mathbb{R})$ is a multilinear map. $$\Phi (w^1,...,w^k)(v_1,...,v_k)=w^1(v_1)...w^k(v_k)$$ By the universal property of the tensor product this descends to a linear map $\tilde{\Phi}: V_1^*\otimes ... \otimes V_k^* \rightarrow L(V_1,...,V_k;\mathbb{R})$ $$\tilde{\Phi}(w^1 \otimes ... \otimes w^k)(v_1,...,v_k)=w^1(v_1)...w^k(v_k)$$

And now I hate myself for asking, but how is it we conclude that $\tilde{\Phi}$ is an isomorphism, s.t. $V_1^*\otimes ... \otimes V_k^* \overset{\sim}{=} L(V_1,...,V_k;\mathbb{R})$

• A thought: $\tilde{\Phi}$ is linear, injective as well as surjective ? – mort May 31 '15 at 20:37

## 1 Answer

It's not surjective in general. Let's take $k=2$ to simplify the notation. By the definition of $\tilde\Phi$, $$\tilde\Phi\Big(\sum_{i=1}^n w_i^1\otimes w_i^2\Big)(v_1,\cdot) = \sum_{i=1}^n w_i^1(v_1)w_i^2 \in V_2^\ast$$ As $v_1$ varies in $V_1$ (but the $w_i^j$ are fixed), you just get different linear combinations of $w_1^2,\dotsc,w_n^2$. Thus \begin{align*} V_1 &\to V_2^\ast \\ v_1 &\mapsto \tilde\Phi\Big(\sum_{i=1}^n w_i^1\otimes w_i^2\Big)(v_1,\cdot) \end{align*} has finite rank. Since every element of $V_1^\ast\otimes V_2^\ast$ can be written as a sum of finitely many elementary tensors, this shows that the range of $$\tilde\Phi\colon V_1^\ast\otimes V_2^\ast \to L(V_1,L(V_2,\mathbb R)) \cong L(V_1,V_2;\mathbb R)$$ is (a subspace of) $F(V_1,L(V_2,\mathbb R))$, where $F(X,Y)$ denotes the space of finite-rank linear maps from $X$ to $Y$. If $V_1$ and $V_2$ are both infinite-dimensional, there are more maps in $L(V_1,L(V_2,\mathbb R))$ than that.

It is true that $\tilde\Phi$ is injective. I wrote up some short notes on tensor products of vector spaces (pdf) a few years back that include this: see §40 for the statement about dual spaces; the relevant proof, for a more general statement, is in §30. It uses the idea of hoisting linear independence from the vector spaces up to the tensor product.

• Thanks for the thorough answer, as well as the notes :) – mort May 31 '15 at 22:07