Relation between $\text{Bil}(V^*,W^*)$ and $\text{Bil}(V,W)^*$ I was wondering if there is a relation between $\text{Bil}(V^*,W^*)$ and $\text{Bil}(V,W)^*$ (i.e. between the space of bilinear forms on $V^*\times W^*$ and the dual of the space of bilinear forms on $V\times W$). For example, is there is a isomorphism between them?
I can't find much on this subject, but there are probably a lot of books where this is explained, so I would also be happy if you have a suggestion for a good book about the topic.
Edit: An argument for why there is no isomorphism when $V$ and $W$ are infinite-dimensional can be found at Why is the inclusion of the tensor product of the duals into the dual of the tensor product not an isomorphism?
 A: There is a natural linear map $F:\mathrm{Bil}(V,W)^*\to\mathrm{Bil}(V^*,W^*)$ defined as follows.  Given $\alpha\in V^*$ and $\beta\in W^*$, we define an element $f_{\alpha,\beta}\in\mathrm{Bil}(V,W)$ by $f_{\alpha,\beta}(v,w)=\alpha(v)\beta(w)$.  Given $\varphi\in \mathrm{Bil}(V,W)^*$, then, we define $F(\varphi)(\alpha,\beta)=\varphi(f_{\alpha,\beta})$.
When $V$ and $W$ are finite dimensional, this $F$ is an isomorphism.  Indeed, you can see this by just explicitly picking bases.  If $B$ is a basis for $V$ and $C$ is a basis for $W$, then there are canonical bases for $\mathrm{Bil}(V,W)^*$ and $\mathrm{Bil}(V^*,W^*)$ which are in bijection with $B\times C$, and you can explicitly compute that $F$ sends each basis element to the corresponding basis element.
This story is greatly clarified using the language of tensor products.  We have $\mathrm{Bil}(V,W)=(V\otimes W)^*$, so our $F$ is a map $(V\otimes W)^{**}\to (V^*\otimes W^*)^*$.  In fact, $F$ is just the dual of the map $G:V^*\otimes W^*\to (V\otimes W)^*$ which corresponds to the pairing between $V^*\otimes W^*$ and $V\otimes W$ using the evaluation maps $V^*\otimes V\to k$ and $W^*\otimes W\to k$.  When $V$ and $W$ are finite dimensional, this pairing is perfect, so $G$ is an isomorphism, and hence $F$ is an isomorphism as well.
