# isotropic deformation retract of Weinstein manifolds?

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?

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"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 –  j.c. Apr 11 '10 at 19:26
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 –  j.c. Apr 11 '10 at 19:34
thanks, got it. Do you know by any chance that why such a statement is true? –  Mohammad F. Tehrani Apr 11 '10 at 22:19
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. –  j.c. Apr 12 '10 at 1:17
Here's a link to the paper so that it will be picked up by the arxiv trackback scraper: arxiv.org/abs/math/0010059 –  j.c. Apr 14 '10 at 15:42

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.

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thanks.It was good –  Mohammad F. Tehrani Apr 12 '10 at 16:45