Dual of linear map into a tensor product

Question

Suppose we have (finite dimensional) vector spaces $U$, $V$ and $W$ and a linear map

$f: U\otimes V \to W; u\otimes v\mapsto f(u\otimes v)$.

Is there a standard way to write the dual linear map? I mean, it is of course

$f^*: W^* \to (U\otimes V)^*; w^* \mapsto f^*(w^*)$

but since $(U\otimes V)^*\simeq U^* \otimes V^*$ in this case, can we somehow put emphasis on the fact that $f^*(w^*)$ is a linear sum of tensor components? Like

$f^*(w^*)=\sum_{something-I-dont-see} f^*_{?}(w^*)\otimes f^*_{?}(w^*)$

Answer 1

As usual, the dual of a linear map $f:V\to W$ is $f^*:W^* \to V^*, w^* \mapsto w^* \circ f$. So we have

$$f^*(w^*)\left(\sum_{i,j}a_i \otimes b_j\right)= \sum_{i,j} w^*(f(a_i\otimes b_j))$$

Answer 2

First, there is some major problem with your question. Notice that there is no chance to write

$$f^*(w^*)=\sum_{something-I-dont-see} f^*_{?}(w^*)\otimes f^*_{?}(w^*),$$

because the left hand side is linear in $w^*$ and the right hand side is quadratic!

But something can be done!

You said that $(U\otimes V)^*\simeq U^* \otimes V^*.$ However, it is only true, because $U$ and $V$ have finite dimensions. In general, for $U,V$ with arbitrary dimensions we only have an injective map $\phi:U^* \otimes V^*\to(U\otimes V)^*$ given by formula $$\phi\left(\sum_i\alpha_i\otimes\beta_i\right)\left(\sum_jx_j\otimes y_j\right)=\sum_{i,j}\alpha_i(x_j)\beta_i(y_j).$$ Here you have a related discussion about this problem.

So your question can be reformulated as

Problem. Let $f:U\otimes V\to W$ be given and let $f^*:W^*\to(U\otimes V)^*$ denote the dual map. Is there a map $\bar{f}:W^*\to U^*\otimes V^*$ such that $$f^*=\phi\circ \bar{f}?$$

Answer. If $U,V$ have finite dimension, then there is such $\bar{f}$ but not defined in canonical manner (it depends on basis of $U$ and $V$).

In fact, if $u_1,\dots,u_n$ and $v_1,\dots,v_m$ are basis in $U$ and $V$ respectively, then we have a dual basis $\omega_1,\dots,\omega_n$ and $\eta_1,\dots,\eta_m$ in $U^*$ and $V^*$ respectively. As a result $\lbrace \omega_i\otimes\eta_j\rbrace_{i,j}$ form a base in $U^*\otimes V^*$ and $\lbrace \phi(\omega_i\otimes\eta_j)\rbrace_{i,j}$ form a base in $(U\otimes V)^*$. Thus, for any $w^*\in W^*$ we have that

$$f^*(w^*)=\sum_{i,j}a_{i,j}(w^*)\phi(\omega_i\otimes\eta_j).$$

Putting $\bar{f}=\phi^{-1}\circ f^*$ we obtain that

$$\bar{f}(w^*)=\sum_{i,j}a_{i,j}(w^*)\omega_i\otimes\eta_j.$$

This is all I can tell for certain! Notice also that we do not need $W$ to have finite dimension.