Algebraic and continuous duals of an inverse limit of finite dimensional vector spaces

Question

I have been trying to understand the following section of a paper "Revêtements du demi-plan de Drinfeld et correspondance de Langlands p-adique" by Gabriel Dospinescu and Arthur-César Le Bras:

Transcription:

La dualité de Serre pour les variétés Stein [13] montre que ce complexe est dual du complexe des sections globales du complexe de de Rham de $X$, tordu par $\Omega^d(X)$. Cela permet de montrer [47, th. 4.11] que si $X$ est pure de dimension $d$, alors pour tout $k$ on a des isomorphismes canoniques $$H^k_\text{dR}(X) \simeq H^{2d - k}_\text{dR,c}(X)^*\quad\text{et}\quad H^k_\text{dR,c}(X) = H^{2d - k}_\text{dR}(X)^*,$$ les duaux étant topologiques (comme toujours dans cet article). La preuve de [46, cor.3.2] montre que pour tout $k$ l'espace vectoril topologique $H^k_\text{dR}(X)$ est isomorphe à la limite inverse d'une suite $(V_n)_n$ d'espaces de dimension finie sur $K$. En particulier $H^k_\text{dR}(X)$ est un Fréchet réflexif et son dual topologique $H^{2d - k}_\text{dR,c}(X)$ est la limite inductive des $V_n^*$. On en déduit que $H^k_\text{dR}(X)$ est aussi le dual algébrique de $H^{2d - k}_\text{dR,c}(X)$. Puisque $\Omega$ est un espace Stein^{(23)}, il en est de même de $\Sigma_0$ et puisue $\Sigma_n$ est un revêtement étale fini de $\Sigma_0$, on obtient la …

(the full version of a preprint of their paper can be found here: https://arxiv.org/abs/1509.00606). Above, $K$ is a field of characteristic zero, complete with respect to a discrete valuation and $X$ is a smooth, rigid analytic $K$-space. I cannot prove the sentence underlined in green. My suspicion is that this is a general result: if $Y$ is the inverse limit¹ of finite dimensional $K$-vector spaces then its continuous dual $Y'$ and its algebraic dual $Y^*$ coincide. So far, I believe that I can prove that $(Y')' \cong (Y')^*$ as vector spaces, and have tried proceeding using reflexivity. However, I can't get to the final result and I'm now concerned that I am missing something silly. Any help would be hugely appreciated!

¹over the non-negative integers. We may also need to impose a density condition, i.e. that if $Y= \varprojlim_{n \geq 0} Y_n$ then the projection maps $Y \rightarrow Y_n$ have dense image for all $n \geq 0$. This allows us to apply results like [1, Theorem 11.6.1].

1. Perez-Garcia C, Schikhof WH. Locally Convex Spaces over Non-Archimedean Valued Fields. Cambridge University Press; 2010.

Answer 1

It seems that you confuse limits and colimits. If $X$ is a colimit (or inductive limit -- in the category of locally convex vector spaces) of finite dimensional Hausdorff spaces then the algebraic dual of $X$ equals the topological dual because a linear functional $f:X \to \mathbb R$ is continuous if (and only if) all compositions $f\circ i$ for the inclusions $i$ (colimit morphisms) of the spaces constituting $X$ are continuous. But this is automatic because finite dimensional spaces have a unique Hausdorff vector space topology.

On the other hand, the corresponding result is not true for limits (in the categorical sense, often called projective limits): The topological dual of $X=\mathbb K^{\mathbb N}$ is countably dimensional but the algebraic dual has at least the dimensionality of $X$ (in the sense of the cardinality of each Hamel basis) which is $c=|\mathbb R|$.