5
$\begingroup$

What is an explanation for what the theory of symplectic rigidity is and what kind of questions it can answer? I was led to this after reading about the symplectic non-squeezing theorem of Gromov.

$\endgroup$
12
$\begingroup$

Rigidity, as used throughout mathematics and not just symplectic geometry, indicates that some structure attached to an object captures more data than one would "expect" from the underlying object itself. There are a number of helpful examples already in this Wikipedia entry. (I write "expect" in quotes because often times the "expectation" is far from correct, and might not even match your initial intuition.)

Gromov's non-squeezing theorem is typically the first example most people encounter in symplectic geometry. It states if there exists a symplectic embedding of the standard ball $B^{2n}(R)$ into $\mathbb{R}^{2n}$ such that its image lies completely in the cylinder $Z^{2n}(r) = B^2(r) \times \mathbb{R}^{2n-2}$, then $R \leq r$. However, there certainly exist volume-preserving embeddings $B^{2n}(R) \hookrightarrow Z^{2n}(r)$. Hence, this is an example of symplectic rigidity if you take the setting of volume-preserving geometry as your expectation. In other words, symplectic geometry is more rigid, or stronger, than volume-preserving geometry, since it sees more information.

Volume-preserving geometry is not the only setting from which you can build your "expectation." Depending on the situation, you may take:

  • Almost symplectic geometry: $(M^{2n},\omega)$ with $\omega \in \Omega^2(M)$ such that $\omega^n \neq 0$ (we drop the closedness condition).
  • The underlying smooth geometry

Obviously, if you have symplectic rigidity when compared with almost symplectic geometry, then you also have symplectic rigidity with respect to volume geometry and smooth geometry, for example. On the other hand, sometimes the problem at hand doesn't have an interpretation in volume-preserving geometry, and so you have to settle for rigidity with respect to smooth geometry.

Here's another example of rigidity: there is no closed exact Lagrangian submanifold embedded in standard $\mathbb{R}^{2n}$. However, in the smooth category, there are many $n$-dimensional closed submanifolds of $\mathbb{R}^{2n}$. So symplectic geometry is more rigid than smooth geometry. (Here's an example where it wouldn't make sense to compare this to volume-preserving or almost symplectic geometry.)

There's a whole industry of rigid invariants nowadays in symplectic geometry; the industrial revolution occurred in 1985 with Gromov's famous paper Pseudo holomorphic curves in symplectic manifolds. In fact, both examples mentioned so far first appeared in that paper. Often, these rigid invariants are tied up intimately with counts of pseudo-holomorphic curves (e.g. Gromov-Witten theory and Quantum Cohomology). Gromov non-squeezing can be seen as a zeroth order approximation to these subtle curve counts: it boils down to the existence of a particular pseudo-holomorphic curve. There's also the Fukaya Category, which is by now a big machine which can be thought of as encoding rigid invariants coming from Lagrangian submanifolds, and is itself intimately tied up with physics via mirror symmetry (this is all a long story in itself).

Finally, I should mention on the flip side that you can ask if symplectic geometry exhibits phenomena in which the weak "expectation" actually matches. This does happen, and is called flexibility. (This is particularly fruitful when combined with contact geometry, in which there are overtwisted discs and loose Legendrians.) As a purely symplectic example of flexibility, for any symplectic manifold $(M^{2n},\omega)$ and any manifold $L$ with $\dim L < n$ there is a forgetful map from the space of isotropic embeddings $\{f \colon L \hookrightarrow M \mid f^*\omega = 0\}$ to the space of formal isotropic embeddings, i.e. embeddings $f \colon L \hookrightarrow M$ and bundle monomorphisms $F \colon TL \rightarrow f^*TM$ (over $L$) with $F^*\omega = 0$. The forgetful map is just given by taking the same underlying embedding and $F = df$. This forgetful map induces a homotopy equivalence (and is hence referred to as an h-principle, where the "h" stands for "homotopy").

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the very nice answer! $\endgroup$ – David Mar 29 at 21:46

Your Answer

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

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