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Let $X$ be a complex-analytic manifold. Consider the sheaf of holomorphic functions $\mathcal{O}_X$ as a sheaf with values in the category of locally convex vector spaces. For $U\subseteq X$ open, we give $\mathcal{O}_X(U)$ the topology of uniform convergence on compact subsets of $U$. The precosheaf $\mathcal{O}'_X$ of analytic functionals is defined by taking the continuous dual of $\mathcal{O}_X$, that is, $\mathcal{O}'_X(U) := \mathcal{O}_X(U)'$ where the latter denotes the vector space of continuous functionals.

Is $\mathcal{O}'_X$ a cosheaf of vector spaces?

Edit: More specifically, let $Y$ be an open subset of $X$ and let $\mathfrak{U}$ be an open cover of $Y$. The cosheaf condition for $\mathcal{O}'_X$ says that $$ \mathcal{O}'_X(Y) \longleftarrow \bigoplus_U \mathcal{O}'_X(U) \overset{r'}{\longleftarrow} \bigoplus_{U,V} \mathcal{O}'_X(U\cap V) $$ is a cokernel sequence in the category of vector spaces. The maps are the duals of the maps from kernel sequence $$ \mathcal{O}_X(Y) \longrightarrow \prod_U \mathcal{O}_X(U) \overset{r}{\longrightarrow} \prod_{U,V} \mathcal{O}_X(U\cap V) $$ in locally convex vector spaces from the sheaf condition for $\mathcal{O}_X$ where $r$ sends $(f_U)_U$ to $(f_U|_{U\cap V} - f_V|_{U\cap V})_{U,V}$. Let $A$ and $B$ denote the source resp. the target of $r$. Let $R: A/\ker r \rightarrow \mathrm{im}\ r$ be the map induced by $r$ where $\mathrm{im}\ r$ has the subspace topology in $B$. The answer to the above question would be yes if the dual map $R'$ were surjective. This condition appears when trying to prove that the natural map $\mathrm{coker} (r') \rightarrow (\ker r)' = \mathcal{O}'_X(Y)$ is injective by using the Hahn-Banach theorem. Surjectivity follows from the Hahn-Banach theorem. In the case of $Y$ an open subset of $\mathbf{C}$ and a cover consisting of two open subsets $R'$ is an isomorphism because $R$ is a bijective continuous map between Fréchet spaces and the open mapping theorem applies.

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    $\begingroup$ I have added the functional analysis tag because the question concerns duality in locally convex spaces. $\endgroup$ – Jochen Wengenroth Mar 20 at 17:50
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    $\begingroup$ Which cosheaf axiom makes problems? $\endgroup$ – Jochen Wengenroth Mar 20 at 17:51
  • $\begingroup$ @JochenWengenroth I have added more details. $\endgroup$ – Daniel Bruegmann Mar 21 at 9:29
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    $\begingroup$ I think that $R'$ is surjective if and only if $R$ is an isomorphism if and only if $r$ is open onto its range. This smells like the Cousin problem for $U$, and if this is so one should consider Stein manifolds. $\endgroup$ – Jochen Wengenroth Mar 21 at 12:32
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The answer to your question is yes. I found it instructive to consider firstly the one-dimensional case. This can be seen more concretely using the Silva-Köthe duality theorem which states that the dual of the space of holomorphic functions on an open subset $U$ of the complex plane is (naturally and functorially) representable as the space of germs of functions which are analytic on the complement of $U$ in the Riemann sphere and which vanish at $\infty$. It is then a simple exercise to show that the cosheaf conditions are fulfilled. By the way, I think that it is natural to consider your question, not in the category of vector spaces but in that of locally spaces. In the above special case, the spaces of holomophic functions is a nuclear Fréchet space, its dual a nuclear Silva space or $DFN$-space. In the general case, the same result holds. The positive answer to your question follows from the completely symmetric duality between nuclear Fréchet spaces and nuclear Silva spaces——in fact, as has been often observed, the nearest infinite dimensional analogy to the finite dimensional situation. Perhaps OT but relevant is the fact that you can embed $U$ into the above dual space in a natural way and it satisfies a corresponding universal property.

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  • $\begingroup$ I should add the proviso that I was tacitly assuming that we are dealing with open subsets of $n$-dimensional complex spaces. As pointed out above, the case of a general complex manifold could involve some subtle problems. $\endgroup$ – user131781 Mar 21 at 15:43
  • $\begingroup$ What do you mean by "completely symmetric duality" and how do you intend to apply this here? In particular, it is not true in general that (ker f)' is isomorphic to coker (f') if f is a continuous homomorphism which is not strict. $\endgroup$ – Daniel Bruegmann Apr 23 at 13:50

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