The correspondence between algebraic and analytic geometry is thought-provoking. The GAGA makes this precise to some extent. But there is more of this analogy. As I checked an article on analytic space, most of the notions there seems to have an algebraic analogue. I had been wondering why people studied analytic geometry separately to the extent it has been. It would be nice to know about some theorems in analytic geometry that have no analogue in algebraic geometry, and this would justify the study of this subject separately. Here I mean theorems that do need to use the full power of analytic geometry; not things like Hodge decomposition for which Kähler is sufficient.

  • $\begingroup$ I don't understand your comment re: Hodge decomposition. What is the "full power" of analytic geometry? $\endgroup$ – Kevin H. Lin Aug 10 '10 at 3:47
  • $\begingroup$ Hodge decomposition is true over Kahler manifolds. You do not need a complex analytic structure. $\endgroup$ – user977 Aug 10 '10 at 10:57
  • $\begingroup$ Hodge decomposition, as I know it, is true for compact Kähler manifolds. Unless I am mistaken, Kähler manifolds by definition have an integrable complex structure. Are you saying that Hodge decomposition also holds for almost Kähler manifolds? (By almost Kähler I mean Kähler minus the condition that the almost complex structure be integrable.) $\endgroup$ – Kevin H. Lin Aug 10 '10 at 20:23
  • $\begingroup$ Oops, I screwed up badly. Instead of saying "complex analytic structure", I should have said, "structure of an analytic set", ie that it is a zero set of some power series, instead of polynomials in the case of algebraic sets. $\endgroup$ – user977 Aug 10 '10 at 20:28
  • 2
    $\begingroup$ Sorry, I still don't get it: every complex manifold is an analytic space in the sense you linked to above. $\endgroup$ – Pete L. Clark Aug 11 '10 at 15:47

I'm not sure that the main motivation for studying analytic geometry is that there are theorems true there that are not true in algebraic geometry; if anything, the reverse is true, since any complex variety is also a complex analytic space.

But there are analytic varieties that are not algebraic, so analytic geometry is a broader subject than complex algebraic geometry, which has its own appeal. Thinking analytically also gives a source of techniques that are not available by purely algebraic methods (and maybe this is what you mean by theorems that are true in analytic geometry but not algebraic geometry).

Finally, let me also mention that very interesting phenomena can occur when you study the interaction between the algebraic and analytic worlds, for example of the following kind: the moduli space of analytic K3 surfaces is a connected 20 dimensional space. The moduli space of algebraic K3 surfaces is a closed subspace, which is the union of countably many 19 dimensional connected components. This shows that some problems, like the study of moduli spaces, can become simpler in the analytic regime than they are if you restrict to the algebro-geometric setting.

  • $\begingroup$ I meant that there are people who still study analytic spaces, even after algebraic geometry developed so much. What could be their motivation? $\endgroup$ – user977 Aug 10 '10 at 10:58
  • $\begingroup$ @LLN BBK: It could just be a matter of taste -- the analytic theory has a completely different flavor from the algebraic theory. It is beautiful on its own. Many people might also find it easier and more "concrete" than the algebraic theory. $\endgroup$ – Kevin H. Lin Aug 10 '10 at 19:00
  • $\begingroup$ Yes, but still it would be hear of some theorem which has no algebraic analogue. $\endgroup$ – user977 Aug 10 '10 at 19:15
  • $\begingroup$ There is a purely algebraic proof of the Hodge-de Rham degeneration, but in fact, Hodge decomposition itself has yet no purely algebraic proof. See the discussion here: mathoverflow.net/questions/28265/… $\endgroup$ – Kevin H. Lin Aug 10 '10 at 20:25
  • $\begingroup$ "The moduli space of K3 surfaces is a connected 20 dimensional space" is a statement that is true in the analytic category but not in the algebraic category. Does this count, or is it "cheating"? $\endgroup$ – Kevin H. Lin Aug 10 '10 at 20:50

I would argue that much of analytic geometry relates to non compact manifolds and is concerned with notions and results that have absolutely no analogue in algebraic geometry, like:

Reinhardt and Runge domains, pseudoconvexity, plurisubharmonic functions, analytic capacity, Levi problem, regularization of currents, Bochner-Martinelli kernels, $L^2$-estimates and regularity for $\bar \partial$, etc. etc.

One of the very first results in analytic geometry was Poincaré's proof that the open bidisk and the open ball in $\mathbb C^2$ , both simply connected, were not analytically isomorphic, shattering any hope that Riemann's uniformization theorem might generalize in dimension greater than one.
His proof was obtained by showing that the automorphism groups of these domains are non-isomorphic.
This result and its proof beautifully illustrate why algebraic geometry (which is of no help in that problem) doesn't render analytic geometry superfluous.

And the work of Fields medalists Charles Fefferman and Lars Hörmander (not to mention luminaries like Yum-Tong Siu) should convince the most hardened algebraic geometer that there is much, much room left for analytic geometry ...


Your Answer

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