## Compact Kaehler manifolds that are isomorphic as symplectic manifolds but not as complex manifolds (and vice-versa)

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

1. What are some examples of compact Kaehler manifolds (or smooth complex projective varieties) that are not isomorphic as complex manifolds (or as varieties), but are isomorphic as symplectic manifolds (with the symplectic structure induced from the Kaehler structure)? Elliptic curves should be an example, but I can't think of any others. I'm sure there should be lots...

2. In the other direction, if I have two compact Kaehler manifolds (or smooth complex projective varieties) that are isomorphic as complex manifolds (or as varieties), then are they necessarily isomorphic as symplectic manifolds?

3. And one last question that just came to mind: If two smooth complex (projective, if need be) varieties are isomorphic as complex manifolds, then they are isomorphic as varieties?

-

## 5 Answers

1. Well, there are stupid examples like the fact that $\mathbb{P}^n$ has Kähler structures where any rational multiple of the hyperplane class is the Kähler class which are compatible with the standard complex structure (you just rescale the symplectic structure and metric). I think you should get similar examples with multi-parameter families on things like toric varieties with higher dimensional $H^2$.

2. I know some non-compact examples where you can deform the complex structure without changing the symplectic one. I don't know any compact examples, but they probably exist. The thing is, the only thing you can deform about a symplectic structure on a compact thing is its cohomology class (by the Moser trick), so anything with an big enough family of Kähler metrics will work.

3. This probably follows from GAGA, but you'd have to ask someone more expert than me to be sure. Edit: David's answer made me realize I forgot to say projective here. That's important.

-

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

Re 3: If you say projective, then yes. GAGA tells you that an analytic isomorphism is also an algebraic one.

If you don't say projective, then no. See the appendix to Hartshorne for a family of nonisomorphic algebraic structures on C^2/Z^2.

-

So here are some examples: When X has no continuous families of automorphisms (H^0(X, TX)=0), complex deformations of X to first order are given by H^1(X, TX). For compact Calabi-Yaus this is H^{(n-1, 1)} and moreover by Bogomolov-Tian-Todorov the deformations are unobstructed.

Symplectic deformations as Ben noted are controlled by H^2(X, R) by Moser's trick. If we want to deform while staying Kahler, then in H^{(1,1)}(X, R). In mirror symmetry (where this discussion is stolen from) one allows a B-field and correspondingly a complexified space of deformations H^{(1,1)}. Then for mirror manifolds these two spaces of deformations are switched.

This is discussed in Denis Auroux's notes on mirror symmetry (http://math.mit.edu/~auroux/18.969/, any misinterpretation is my fault).

Mirror symmetry is cool and all, but if we just stay on the same Calabi-Yau the deformation spaces for symplectic and complex structures can have different dimensions - with either one bigger, giving examples for both 1 and 2.

-

That's a very nice observation! – Kevin Lin Oct 16 2009 at 15:24

In case anybody is curious, there are still examples of (1) even if one replaces the requirement that the complex manifolds be nonisomorphic with the requirement that they be not even deformation equivalent. In fact in arXiv:0608110 Catanese showed that Manetti's examples of general type surfaces which are diffeomorphic but not deformation equivalent are symplectomorphic (with respect to their canonical Kahler forms).

-

If $M \to X$ is smooth and proper, and $M$ is K\"ahler, then the fibers are all symplectomorphic. (Proof: the Levi-Civita connection generates symplectomorphisms.) The family of elliptic curves was already mentioned, but another interesting one has every general fiber being $F_0$ and the special fiber $F_2$ (Hirzebruch surfaces).

A curious example is the family `$\{ xy = t \}$` of hypersurfaces in ${\mathbb C}^2$ as $t$ varies (away from $0$). There, the fibers are all holomorphic, and symplectomorphic, but not by the same diffeomorphism (their unique closed geodesics are of varying length).

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9459375739097595, "perplexity_flag": "middle"}