5
$\begingroup$

Consider the categories $\mathbf{Vect}$ of vector spaces $X$ with linear maps and $\mathbf{TopVect}$ of topological vector spaces $(X, \tau)$ with continuous linear maps both over $\mathbb{R}$.

Taking the algebraic dual is a contravariant functor $* : \mathbf{Vect} \to \mathbf{Vect}$ with $X \mapsto X^*$ and $(f : X \to Y) \mapsto (f^* : Y^* \to X^*)$ (the transpose of $f$) with $(f^*(y^*))(x) := y^*(f(x))$.

Similarly, it is also clear that taking the topological dual leads to a contravariant functor $' : \mathbf{TopVect} \to \mathbf{Vect}$ with $X \mapsto X'$ and $(f : X \to Y) \mapsto (f' : Y' \to X')$ with $(f'(y'))(x) := y'(f(x))$. In order to see that this is indeed a functor it is enough to observe that $f'$ is just $f^*$ restricted to $Y' \subseteq Y^*$ and is well-defined, i.e. $f^*|_{Y'} \subseteq X'$ due to continuity of $f$.

We can also consider the "taking the weak* topological dual" as the contravariant functor $'_\sigma : \mathbf{TopVect} \to \mathbf{TopVect}$ with $X \mapsto X'_\sigma$ and $(f : X \to Y) \mapsto (f' : Y' \to X')$ with $(f'(y'))(x) := y'(f(x))$. Clearly, $'_\sigma$ is well-defined, i.e. $X'_\sigma$ is an object in $\mathbf{TopVect}$ and $f'$ is a morphism in $\mathbf{TopVect}$ since $f'$ is continuous (and linear) by the choice of the weak-* topology on both $X'$ and $Y'$. Similarly, one can also consider more general functors equipping $X'$ with some vector space topology type $\tau_{\text{source}}$ and $Y'$ with a topology type $\tau_{\text{target}}$ such that all the $f'$ are continuous.

Questions:

  1. Is it possible to decompose the $'_\sigma$-functor into $'_\sigma = E_\sigma \circ \, '$ where $E_\sigma$ is some kind of "equip with the weak* topology" functor? The problem is, that we loose information about the space $X$ when performing $'$, but we need $X$ in order to define $E_\sigma$. One plausible solution is to redefine $'$ to $\bar{'} := (', id) : \mathbf{TopVect} \to \mathbf{TopVect} \times \mathbf{Vect}$, $X \mapsto (X', X)$ and set $E_\sigma : im(\bar{'}) \to \mathbf{TopVect}$ that sends the pair $(X', X)$ to the topological vector space $(X', \sigma(X', X))$ where $im(\bar{'})$ is the particular subcategory of $\mathbf{TopVect} \times \mathbf{Vect}$ that we can reach by $\bar{'}$. This construction seems to be rather ugly and artificial. Maybe there is a better description for such a decomposition.

  2. All these "take some top. dual functors" seem to be special instances of the "take the algebraic dual" functor. Is there a better more abstract point of view for such a relation?

$\endgroup$
3
  • $\begingroup$ I'm more comfortable with locally convex spaces and am a bit biased, but to me (for LCSs at any rate) the answer to #2 comes through thinking of the weak (weak-$*$) topology as an example of a dual topology for $(X,X')$. If you're interested in LCSs, are you familiar with polar and dual topologies? It's not really categorical in nature, but the weak, weak-$*$, Mackey, strong, and more-or-less all the other locally convex topologies defined on a vector space $Y$ that relate it through a pairing map to some vector space $X$ are examples of polar topologies. $\endgroup$
    – Dan
    Feb 29, 2016 at 16:19
  • $\begingroup$ @Dan Yes, I'm familiar with polar topologies and dual topologies (involving a bilinear form). The moving between LCS topologies on an LCS seems a little bit categorical, therefore the question, whether there is a nice deeper and more abstract point of view on these movings / functors. $\endgroup$
    – yada
    Feb 29, 2016 at 18:12
  • $\begingroup$ Ok great. That was my impression from your post but wanted to double check. Hopefully someone can enlighten us about a more categorical approach/perspective. :) $\endgroup$
    – Dan
    Feb 29, 2016 at 20:23

0

Your Answer

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

Browse other questions tagged or ask your own question.