Let $K$ be a field, and put $K \text{-vect}$ for the category of $K$-vector spaces. There is an adjunction $[-, K]_{K \text{-vect}} : K \text{-vect}^{op} \leftrightarrow K \text{-vect} : [-, K]_{K \text{-vect}}$ where $[-, K]_{K \text{-vect}}$ sends a vector space $V$ to $[V, K ]_{K \text{-vect}}$, the $K$ vector space of linear maps from $V$ to $K$.

Evidently, the vector space structure on $V^*$ alone is not enough to recover $V$. I wonder, though, about putting a topology on it. For each $a \in V$, there is a map $\hat{a} : V^* \rightarrow K$ sending $\phi$ to $\phi(a)$. Give $V^*$ the weakest topology such that $\hat{a}$ is continuous for each $a \in V$.

Question: form $[V^*, K]_{K \text{-topvect}}$, the vector space of continuous linear maps of topological vector spaces from $V^*$ to $K$. Is this vector space isomorphic to $V$?

If this turns out to be true, then we have a functor $F : K \text{-vect}^{op} \rightarrow K \text{-topvect}$ and a functor $G : K \text{-topvect} \rightarrow K \text{-vect}^{op}$, such that $G \circ F \cong 1_{K \text{-vect}^{op}}$


Yes, $V$ and $[V^*,K]_{\operatorname{K-topvect}}$ are indeed canonically isomorphic as vector spaces.

To see this we should first note, that the map $V\to [V^*,K]_{\operatorname{K-topvect}}$ is obviously injective. So it remains to prove that it is surjective. Suppose that a functional $f: V^*\to K$ is continuous, then it is bounded on a neighbourhood $U$ of zero in $V^*$. This neighbourhood can be chosen as a basic neighbourhood of zero in your topology, i. e. $$ U=\{g\in V^*: \forall i=1,...,n\quad |g(a_i)|\le 1\} $$ for some finite set of vectors $a_1,...,a_n\in V$. This implies that $f$ is bounded on the set of common zeroes of $\widehat{a_i}$, $$ \bigcap_{i=1}^n\operatorname{Ker}\widehat{a_i}=\{g\in V^*: \forall i=1,...,n\quad |g(a_i)|=0\} $$ and since this set is a vector subspace in $V^*$, $f$ vanishes on it: $$ f\Big|_{\bigcap_{i=1}^n\operatorname{Ker}\widehat{a_i}}=0 $$ This in its turn implies that $f$ is a linear combination of $\widehat{a_i}$: $$ f=\sum_{i=1}^n\lambda_i\cdot \widehat{a_i}, \qquad \lambda_i\in K. $$ And therefore $f$ is the image (under the mapping $a\mapsto\hat{a}$) of the vector $$ x=\sum_{i=1}^n\lambda_i\cdot a_i\in V. $$ (Another way to prove this is the reference to the Mackey-Arens theorem.)


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.