1
$\begingroup$

Question regarding tensor products. Is this argument correct?

Let $b(V,\mathbb{R})$ be the vector space (in the natural way) of real valued bi-linear forms on the finite dimensional vector space $V$. By the universal property of the tensor product, each $b\in b(V,\mathbb{R})$ corresponds uniquely to a $\bar b\in (V\otimes V)^*$. I.e there exists $g$ such that $\bar b\circ g=b$, in fact $g(v,w)=v\otimes w$.

The map $b\rightarrow \bar b$ is an isomorphism since it is injective by the universal property and the dimension of the two spaces are the same.

Does this holds in general, i.e if $V$ is an arbitrary $R$-module and $b(V,R)$ denotes the $R-$module of $R$-bilinear maps from $V$ to $R$, and $(V\otimes V)^*$ is the $R$-module of $R$-linear maps from $V\otimes V$ to $R$ are these $R$-modules isomorphic? I.e how do one shows surjectivity?

$\endgroup$
1
$\begingroup$

It is safest here to assume the ring is commutative so that bilinear maps are guaranteed to be interesting.

One way to prove something is an isomorphism is to construct an inverse. By definition of the tensor product, for every $R$-bilinear map $b:V\times V\to R$ there is a unique homomorphism $\overline{b}:V\otimes_R V\to R$ such that composing $\overline{b}$ with the universal bilinear map $t:V\times V\to V\otimes_R V$ yields $b$, i. e. $b=\overline{b}\circ t$. Mapping $b\to \overline{b}$ gives a homomorphism from the module of bilinear maps $V\times V\to R$ into $\hom(V\otimes V,R)$. To construct an inverse, choose $c\in \hom(V\otimes V,R)$. Then we simply map $c$ to $c\circ t$, where again $t:V\times V\to V\otimes_R V$ is the universal bilinear map characterizing the tensor product. The fact that these constructions are inverses of each other follows from the universal property.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.