# The Double Dual and Morita Duality

Here is a fact from linear algebra: if $V$ is a finite dimensional vector space over a field $k$ then $V$ is naturally isomorphic to its double dual. Of course, it is also isomorphic to its dual as well, but it is the natural isomorphism that is more interesting.

Write $V^* = \mathrm{Hom}_k(V,k)$ for the dual space of $V$.

This is not true in general for infinite dimensional vector spaces. Before I actually talk about the subject matter, let us describe a counterexample:if we use the field $k = \mathbb{Q}$, and $I$ is a countable set, then $\mathbb{Q}^{(I)}$ is also countable (here, the notation $\mathbb{Q}^{(I)}$ refers to the direct sum of $I$ copies of $\mathbb{Q}$). The dual space of any vector space has cardinality at least as great as the original space, but $\mathrm{Hom}_k(\mathbb{Q}^{(I)},\mathbb{Q})\cong\prod_{i\in I}\mathbb{Q}$, which is uncountable.

Anyways, the point is $V\cong V^{**}$. Why is this? Of course, one can provide a short and direct proof: define $\Phi:V\to V^{**}$ by $\Phi(v)(f) = f(v)$. This map is injective and since $V$ is finite dimensional, it is surjective.

Is this really satisfying though? Sometimes in mathematics, it is not completely satisfying just to prove something. Instead, one must also find the proper context for results. The proper context should be enlightening in a variety of situations. For instance, that we know $V\cong V^{**}$ where $V$ is a finite-dimensional vector space isn't all that enlightening. We have a natural isomorphism, so we ought to ask: is there an equivalence of categories hiding somewhere? Indeed there is, and it's called Morita theory!
More »