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:

paragraph from Dospinescu, Le-Bras paper


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.
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    $\begingroup$ This is a combination of two facts which workfor the real and complex fields, presumably also in the general situation: a linear mapping on a finite dimensional tvs is automatically continuous and such a mapping on an inductive limit space is continuous if and only it is so on each component. $\endgroup$
    – terceira
    Commented Jul 10 at 12:12
  • $\begingroup$ Thank you! I hadn't used anywhere that the inductive limit topology is the final topology. I can see how these facts prove that $(Y')' = (Y')^*$ because there we are interested in maps from an inductive limit into $K$. However, I can't see how they establish that $Y' = Y^*$ because there we are looking at maps from a projective limit (which is equipped with an initial topology) and so its harder to use the first fact you mentioned about finite dimensional vector spaces. I'm probably missing something obvious. It wouldn't be the first time! I'll think through what you have written again :) $\endgroup$
    – Tom Adams
    Commented Jul 10 at 12:54

1 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 $c=|\mathbb R|$.

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    $\begingroup$ Thank you, Jochen! It looks like my mistake was misinterpreting the green line. All the authors assert there is (essentially) the statement in your first paragraph but I, mistakenly, thought they were claiming something stronger. Now I know to be a little bit more careful with limits! $\endgroup$
    – Tom Adams
    Commented Jul 10 at 14:16

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