2
$\begingroup$

I found the following paragraph in the paper " Intro to symplectic field theory " which I don't understand what does it mean precisely?

Suppose W is a symplectic (or Kahler) manifold. D, smooth divisor in it.

then $\tilde{W}= W-D $ is a Weinstein manifold which contains an isotropic deformation retract $\Delta$.

Does any body know what kind of object is $\Delta$ here? and what does isotropic deformation mean?

$\endgroup$
5
  • $\begingroup$ "isotropic" probably means that the tangent planes of $\Delta$ are all isotropic with respect to the symplectic structure of W (if $\omega$ is the symplectic form for W, then $\omega$ restricted to the tangent planes of $\Delta$ is zero) (this is based on the notion of isotropic subspaces in symplectic vector fields). For "Deformation retract" see en.wikipedia.org/wiki/Deformation_retract $\endgroup$
    – j.c.
    Apr 11, 2010 at 19:26
  • $\begingroup$ For a good reference for basic definitions like the first, see Arnol'd and Givental's survey on "symplectic geometry": www.maths.ed.ac.uk/~aar/papers/arnogive.pdf $\endgroup$
    – j.c.
    Apr 11, 2010 at 19:34
  • $\begingroup$ thanks, got it. Do you know by any chance that why such a statement is true? $\endgroup$ Apr 11, 2010 at 22:19
  • $\begingroup$ Unfortunately I don't, I'm still just learning this field as well. If you edit your question to include more background and definitions, perhaps a more knowledgeable person will come by and answer. $\endgroup$
    – j.c.
    Apr 12, 2010 at 1:17
  • $\begingroup$ Here's a link to the paper so that it will be picked up by the arxiv trackback scraper: arxiv.org/abs/math/0010059 $\endgroup$
    – j.c.
    Apr 14, 2010 at 15:42

1 Answer 1

4
$\begingroup$

This story begins with the Lefschetz hyperplane theorem - the fact that if $D$ is a smooth, very ample divisor in a closed Kaehler manifold $X$ then $D$ carries all the homology and homotopy of $X$ below the middle dimension of $X$.

One way to understand this theorem is via Morse theory (the Andreotti-Frankel approach). There are explanations in Milnor's "Morse Theory", Griffiths-Harris and elsewhere. The line bundle $\mathcal{O}(D)$ has a canonical section $s$ whose zeroes cut out $D$. One looks at the function $f= - \log \|s\|^2$ on $M=X-D$ (defined via a hermitian metric on the line bundle). This is a bounded-below, proper, strictly plurisubharmonic function; it makes $M$ a Stein manifold. Its critical points (if non-degenerate) all have index $\leq \dim_{\mathbb{C}}X$.

Assuming $f$ Morse, the isotropic skeleton of $M$ with respect to $f$ is the union of the unstable manifolds of the critical points of $f$. Each of these unstable manifolds is isotropic with respect to the Kaehler form (their union may be singular). The downward gradient flow of $f$ will (after a bit of tweaking) define a deformation retraction from $M$ to this skeleton. E.g. for $\mathbb{CP}^{n-1}\subset \mathbb{CP}^n$, the skeleton will be a point.

Very ample divisors are very special; none of this applies to more general divisors. Nonetheless, the story generalizes to symplectic manifolds whose symplectic class in $H^2$ is integral. In this case, Donaldson ("Symplectic submanifolds and almost complex geometry") showed that one can find distinguished symplectic divisors $D$ representing high multiples of the symplectic class. The discussion of the hyperplane complement $M$ is valid also for the complement of a Donaldson hypersurface. The "Stein" condition has to be replaced by something more suitable for almost complex manifolds ("Weinstein") but by a really startling theorem of Eliashberg, Weinstein structures can always be deformed to Stein structures.

For further reading, try Paul Biran's paper Lagrangian Non-Intersections.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.