6
$\begingroup$

lately I have been interested in functional analysis, especially with a view towards its applications in the world of (complex) analytic geometry. I have been using R. Taylor's book Several complex variables with connections to algebraic geometry and Lie groups (as it is suggested in this MO question) and J.P. Demailly excellent book [D]: http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf.

However, my poor background in functional analysis doesn't allow to me to answer to a very natural question I have: in [D], definition 5.32 the notion of transverse nuclear $A$-modules over a nuclear Fréchet algebra $A$ is introduced. In the subsequent Proposition 5.33, it is shown (in particular) that if $X$ is a (finite dimensional) Stein space and $\mathcal F$ is a coherent sheaf over $X$, then for every pair of open Stein subspaces $U' \subset U \Subset X$, $\mathcal F(U)$ and $\mathcal O_X(U')$ are transverse over $\mathcal O_X(U)$.

So far so good, but the proof leaves me astonished: I would expect that $\mathcal O_X(U')$ is "analytically flat" over $\mathcal O_X(U)$ (in the sense that the functor $\mathrm{Tôr}^{\mathcal O_X(U)}_q(\mathcal O(U'), -)$ is identically zero. The reason I would expect that is simply that $U'\to U$ is an open immersion in analytic geometry, and I would expect it to be flat "in the correct sense" (of course it is flat stalkwise, but I would expect more).

Can someone explain to me if the intuition coming to me from Algebraic Geometry leads me to some mistake or if it is possible to prove the above statement?

$\endgroup$
2
  • $\begingroup$ when you say flat in the locally ringed sense do you mean stalkwise? $\endgroup$ Commented Jul 30, 2014 at 23:39
  • $\begingroup$ Yes, precisely. I edited! $\endgroup$ Commented Jul 31, 2014 at 6:34

1 Answer 1

3
$\begingroup$

Hope this answer still will be useful: For an open embedding of smaller polydisk into a bigger polydisk $U \rightarrow V$, the $\mathcal{O}(V)$-module $\mathcal{O}(U)$ is not Frechet flat. The problem is that there are much more Frechet modules over $\mathcal{O}(V)$ than one would expect from comparison with the algebraic situation. For example, there are modules of ($L^2$, smooth, ...) functions on a boundary of a smaller polydisk.

One reference to this is Pirkovskii, On certain homological properties of Stein algebras. He shows that there's a nontrivial first Frechet Tor over the algebra of entire functions on a line $\mathcal{O}(\mathbb{C})$ between functions on a unit disk $\mathcal{O}(D)$ and the module $L^2/H^2$, where $L^2$ is the space of $L^2$-functions on the unit circle, and $H^2$ is the Hardy space ($L^2$-functions holomorphically extendable into the inside of a circle).

$\endgroup$

You must log in to answer this question.