User sam lisi - MathOverflow

Orientations for pseudoholomorphic curves with totally real boundary condition

2013-01-09T20:59:02Z

I am trying to understand what the obstructions are to orienting moduli spaces of pseudoholomorphic curves with totally real boundary condition.

I believe that Fukaya-Oh-Ohta-Ono have shown that if a Lagrangian is relatively spin, the moduli spaces of disks with boundary in it can be oriented.

My question has 3 related parts:

is there any sense in which the FOOO relative spin condition is also necessary?
if I consider curves of higher genus and/or more boundary components, do I need to impose additional conditions on the Lagrangian to guarantee orientability of the moduli spaces?
There has been a fair bit of recent work in the case in which the Lagrangian is the fixed point set of an anti-symplectic involution (Crétois, Georgieva-Zinger). Are the orientation difficulties in this case the same as in the general case, or do some special features appear here?

EDIT: Penka Georgieva pointed out that I was mistaken. Her paper (arxiv/1207.5471) deals with the general case of curves with boundary on a Lagrangian.

Answer by Sam Lisi for Almost Complex Structures: 'Tame' versus 'Compatible'

2012-09-09T12:31:33Z

I think YangMills's answer is fantastic. I will go in a slightly different direction, and address the "usefulness" of each definition part of the question.

A very useful feature of tame acs is that tameness is an open condition. This means that it is more straightforward to talk about generic perturbations, e.g. so that somewhere injective curves are transverse. The whole discussion can be carried through with compatible, but it becomes more involved (most recently, I've seen this discussed in some detail in Massot-Niederkruger-Wendl on filling questions for higher dimensional contact manifolds.) The compatible case works, if I am not mistaken, because the space of compatible $J$ is a Banach manifold.

A useful feature of a compatible almost complex structure $J$ is that for a $J$-holomorphic curve, $u^* \omega = |du|^2_J d \operatorname{vol}$. If you have a tamed almost complex structure, you obtain an inequality that is good enough for compactness, but the proofs become a little more involved.

To the best of my knowledge, there is nothing that anyone has proved for $J$-holomorphic curves for compatible $J$ that is believed to be false for tame $J$. Many results, however, are only proved for compatible $J$ because it makes life easier. I personally would love to see an example of a result that wasn't overly technical for which the difference mattered.

Answer by Sam Lisi for When does a hypersurface have contact-type?

2012-08-24T09:22:05Z

There are also examples that are characterized dynamically. By Viterbo, every contact-type hypersurface of $\mathbb{R}^{2n}$ carries a closed Reeb orbit, i.e. a closed leaf of the characteristic line bundle. (Actually, Hofer and Zehnder show this is true for any stable hypersurface.) There are examples of autonomous Hamiltonians on $\mathbb{R}^{2n}$ that fail to have any closed orbits on a specific level set, thus providing examples of hypersurfaces that fail to be stable. These are due to Ginzburg and Gurel.

Answer by Sam Lisi for Given a vector field all of whose integral curves are closed, is the period a smooth function?

2012-08-06T18:40:54Z

You are confusing minimal period and period. The function $\tau(p)$ you computed on $M$ is the minimal period, which is a well-defined function, but is only lower semi-continuous. The period as discussed by Sebastian is only locally defined, and is actually multivalued if you think of it globally.

This multivalued period is what per is about. In your Moebius band example, every point has period 2 (also 4, 6, 8...), though the 0-section has minimal period 1 (I denote the 0 section by $\mathbf{0}$.

Then, per consists of $\bigcup_{k \ge 1} \{ 2k-1 \} \times \mathbf{0} \cup \bigcup_{k \ge 1} \{ 2k \} \times M \subset \mathbb{R} \times M $

Answer by Sam Lisi for Problem:Gromov-Witten;Moduli space

2012-07-19T10:34:01Z

I am trying to assemble the answers to the question(s) that were hashed out in the comments (and also in a separate discussion with Jonny Evans). This answer is community wiki since it is the outcome of collaborative discussion. Please feel free to edit this.

Rephrasing of question: does there exist a non-constant holomorphic curve in any symplectic manifold? (presumably, for generic choice of compatible J) Can we say more in dimension 4? Are there conditions we can put on the symplectic manifold so that there exist curves?

A generic K3 surface (which is a symplectic 4-manifold) does not have any curves, so the answer to the question in complete generality is "no". We therefore reinterpret the question to be about finding a large class of 4-manifolds for which we can say something.
If we drop the non-constant condition, there are the trivial (constant) holomorphic curves. This is why we require non-constant holomorphic curves.
If we allow ourselves to find a $J$-holomorphic curve for a very special (not generic!) almost complex structure $J$, it suffices to find an embedded symplectic surface and then construct $J$ to make this surface $J$-holomorphic. In dimension 4, we can find a symplectic surface by finding a Donaldson divisor.
If there exists a $J$-holomorphic sphere in $N$, then there exist $J$-holomorphic maps from domains of all genus, by composing with a branched cover.
There are two obvious infinite families of examples for which we can find non-constant holomorphic curves. The first are products of symplectic manifolds with surfaces. The second family of examples is obtained by blowing up a symplectic 4-manifold.
Another family of examples come from 4-manifolds $(N, \omega)$ for which the Gromov-Taubes invariant is non-vanishing. For instance, if $c_1(TN) \ne 0$.

Answer by Sam Lisi for History Question: AUTObiography of Mathematicians

2012-07-19T01:00:49Z

Ulam : Adventures of a mathematician
My memory is that it is full of amusing Von Neumann stories.

strong contactomorphism group inside contactomorphism group

2012-04-12T15:06:59Z

Let $(M, \xi)$ be a closed contact manifold with co-oriented contact structure $\xi = \ker \alpha$. Let $\mathrm{Cont}(M, \alpha)$ be the group of diffeomorphisms that preserve the contact form $\alpha$, and let $\mathrm{Cont}^+(M, \xi)$ be the diffeomorphisms that preserve $\xi$ with its co-orientation (i.e. that pull $\alpha$ back to a positive multiple of $\alpha$).

Does the inclusion $\mathrm{Cont}(M, \alpha) \hookrightarrow \mathrm{Cont}^+(M, \xi)$ induce a weak homotopy equivalence?

My guess is that this is too much to hope for, but I don't have any candidate for a counter-example.

Motivation: This question came up in understanding the question of when a contact fibre bundle admits a global contact form with diffeomorphic Reeb dynamics on every fibre.

flexibility of almost contact ``Reeb'' vector fields

2012-03-22T12:21:40Z

New version of the question:

Given an odd dimensional manifold $V$, an almost contact structure is a pair of $(\alpha, \omega)$, where $\alpha$ is a non-vanishing 1-form and $\omega$ is a 2-form whose restriction to $\ker \alpha$ is non-degenerate. In particular, I don't ask that $\omega$ be closed. From this pair, I define a vector field $R$ with the properties that $\alpha(R) = 1$, and $\omega(R, \cdot) = 0$.

A class of examples come from hypersurfaces in a symplectic manifold. In those examples, the 2-form is the restriction of the ambient symplectic form (and is thus closed), and the 1-form is obtained by contracting the symplectic form with a normal vector field.

Suppose $V$ is a closed manifold. Can I deform the pair $(\alpha, \omega)$ so that the only minimal invariant sets of the resulting vector field are non-degenerate periodic orbits?

More generally, does this class of vector fields have any rigidity to it?

Explanation of the change: The first version of this question restricted attention to the class of examples of hypersurfaces in a symplectic manifold. From what Alvarez Paiva commented, it seems unlikely that this is doable. I am now allowing a larger family of deformations (since the 2-form is allowed to vary among maximally non-degenerate 2-forms, dropping the condition of being closed.)

Why I am asking this: I started thinking about this question in trying to understand some examples related to the question of putting a contact structure on an almost contact manifold. I don't think that my line of thinking is related to the original question anymore, but I am still curious to understand how soft almost contact is (compared to contact).

Answer by Sam Lisi for Persistence of boundary punctured holomorphic curves

2012-02-20T12:17:15Z

Yes, if the linearized operator is also surjective. If the linearized operator is not surjective (i.e. you don't have transversality), what I say is not applicable.

To prove this, you want to use an implicit function theorem argument. You can, for instance, set up the problem so that you are looking for sections of the normal bundle to your curve that solve a certain equation. In the case without boundary, this is done in Hofer-Wysocki-Zehnder Properties of pseudoholomorphic curves in symplectizations 3. I believe the case with boundary is done in Ekholm, Etnyre & Sullivan's two papers on contact homology -- they avoid some of the difficulties by considering a special class of contact manifolds, but the Fredholm theory is completely general. The deformations of Legendrian and of Reeb flow (i.e. deformations of the contact form) can be realized as small deformations of the equation. The surjectivity of the linearized operator is now precisely what you need to apply the implicit function theorem.

Note that this proof says that for sufficiently small deformations, there exists a solution. It's much more delicate to estimate how large a small deformation you are allowed.

Answer by Sam Lisi for Dimension of moduli space in Lagrangian Floer homology

2011-10-21T23:01:18Z

(1) The virtual dimension is only equal to $-i(\xi)$ once you have chosen enough trivializations and so forth that it works out that way. This formula hides that this choice of trivialization is what sees $\Lambda$. Take a look at, for instance, the Bourgeois-Ekholm-Eliashberg paper on the surgery formula for contact homology for a discussion of choice of trivializations. I will give a very fast summary. (It is also discussed in Eliashberg-Givental-Hofer's introduction to SFT, and in a number of other papers on the subject of contact homology etc.)

The Conley-Zehnder index is associated to a path of symplectic matrices starting at the identity and ending at a matrix without any eigenvalue equal to $1$. In order to obtain a path of matrices from a periodic orbit, you must trivialize $TM$ over $\xi$. In particular, the Conley-Zehnder index depends on trivialization.

The general index formula, however, has a Conley-Zehnder index term and a Maslov index term (and a relative Chern number term, if your boundary trivializations don't extend to a trivialization of $u^*TM$). The CZ index and the Maslov index each depend on trivialization. The difference/sum does not.

In this case, what you have implicitly done is trivialized $u^*TM$ so that the Lagrangian loop $u|_{{0}\times S^1}^*TL$ in $u|_{{0}\times S^1}^*TM$ has Maslov index $0$. Note that $u^*TM$ is trivial because $u$ is a surface with boundary. Your choice of trivialization on the boundary now extends to one of $u^*TM$. Finally, by the asymptotic convergence theorem, this induces a trivialization of $\xi^*TM$. You use this trivialization to compute the Conley-Zehnder index.

(2) Orbicular is correct. You can see the problem in a very simple toy problem. Consider finding holomorphic maps from the disk $D$ to $\mathbb{C}$. A totally real (actually Lagrangian) boundary condition is given by asking the boundary of the disk to map to the unit circle -- this has a 3 dimensional solution space. Let's take $P = \mathbb{C}$, say. Start with the standard map on the disk given by $u(z) = z$. Now take any simple closed curve in $\mathbb{C}$ that is close to the unit circle. Then, there exists a 3-parameter family of holomorphic disks with boundary on that curve. The space of such curves is infinite dimensional, showing that my linearized problem has an infinite dimensional kernel.

To see the problem for $P$ of lower dimension, we want to look in $\mathbb{C}^2$. Here, choose a generic circle for the boundary condition. For general choice of circle, you get a boundary value problem you cannot solve. You can work out by hand the infinite codimension condition you get on the circle that actually admits any solutions at all.

Answer by Sam Lisi for Elliptic estimates and regularity of the $\overline{\partial}$-operator with totally real boundary conditions in $W^{1,p},$ $1
2011-10-21T16:53:36Z

I'm almost certain that Abbas-Hofer does this in the appendix in which they prove the $L^p$ estimates. If I recall correctly, the proof shows the estimate in weak $L^1$ and for $L^2$, and then uses interpolation to get $1 < p \le 2$. You then obtain the $2 < p < \infty$ by duality. If you don't have access to it, I will try to find some references to the published literature.

The $p>2$ condition comes in to play once we want to show that a sequence holomorphic curves with gradient bounds have $C^\infty$ bounds. The induction uses that $W^{1,p}$ is a Banach algebra (notably to deal with the nonlinear almost complex structure $J$).

Answer by Sam Lisi for Why are "bad orbits" excluded from symplectic field theory, contact homology etc.?

2011-04-14T14:56:33Z

I think Eigenbunny's explanation cuts to the heart of the matter. Here are some more specific details in the case of symplectic field theory.

First of all, to get some intuition for what bad orbits look like, in the case of a 3 dimensional contact manifold, the bad orbits are precisely the hyperbolic orbits that have non-orientable stable and unstable manifolds. For simplicity, I will talk about cylinders, but the general case is very similar.

(1) The bad orbits must be thrown out because of an orientation problem. In SFT, the simple orbits are decorated with a choice of marked points on each one, called markers. Also, the domain curves are decorated with asymptotic markers. These are a choice of marked point in the circle at infinity for each of the cylindrical ends of the domain. Now consider a holomorphic half-cylinder asymptotic to the $m$-fold cover of a simple orbit $\gamma$. By the convergence result, the image of the asymptotic marker makes sense, and is a point on $\gamma$. We want the image of the asymptotic marker to match the marker on the orbit. Given an unmarked curve, we have $m$ possible locations for the asymptotic marker on the domain.
To understand how to glue, we want to think of how a level 2 building can arise as the limit of cylinders. Suppose we break at the $m$-fold cover of $\gamma$. The two ends asymptotic to this orbit don't obtain an asymptotic marker -- only a marker relative to each other. We can think of this as having $m$ possible choices of asymptotic marker on each end, and then identifying the ones obtained by a simultaneous rotation, giving us $m$ possible choices in the end. Now, for us to count these curves, we need the rotation of the markers to consistent with the orientations. In general it is, but for bad orbits, this is orientation reversing.

This business with markers is one of the places where we see the equivariant nature of SFT: we break the $S^1$ symmetry with these markers, and then quotient out by all possible choices to remove the dependence.

For more details, Bourgeois and Mohnke's Coherent orientations in symplectic field theory does a good job explaining this in detail. Also, see the Introduction to SFT paper by Eliashberg, Givental & Hofer (Section 1.8.4 and Remarks 1.9.2, 1.9.6)

(2) This is also because of the $\mathbb Z_m$ action. Consider the simplest example, where we look at the boundary of a 1-dimensional moduli space of cylinders connecting orbits a and a'. (By this, I mean 1 dimensional after the quotient by the translation.) Suppose one boundary component consists of a level 2 building with two cylinders, a to b and b to a', and b is a bad orbit. Suppose b has multiplicity $m=2k$. Then, the boundary building has $m$ different possible decorations by asymptotic markers. If you chase through the orientations, the end result is that these terms all cancel out, thus saving us from having problems with $d^2$.

(3) The answer to the most optimistic version of this question is no: here is no reason for the moduli space to be empty in general. In order to get a signed count of zero, one needs to make sense of orientations. The way orientations are done in SFT, these moduli spaces asymptotic to bad orbits don't even get one, so the signed count doesn't make sense. Arguably, this isn't the only way we might define orientations (to get the same theory). A related situation in which something like this is true is in the isomorphism between linearized contact homology and $S^1$-equivariant symplectic homology (actually the $SH^+$ part), due to Bourgeois and Oancea. Their non-equivariant contact homology involves decorating each Reeb orbit with a Morse function and imposing markers. The ``bad'' orbits end up dying (over $\mathbb Q$) because the differential of the maximum, instead of being $0$, is then twice the minimum.

Why do A_\infty functors form an A_\infty category?

2011-03-11T15:51:01Z

I am in a reading group studying Seidel's book (Fukaya Categories and Picard-Lefschetz Theory). All of the participants have backgrounds in symplectic topology/pseudoholomorphic curve methods. We are stuck in trying to understand the chapter presenting the algebraic background for Fukaya Categories.

Seidel makes the following claim: Non-unital $A_\infty$ functors $\mathcal F: \mathcal A \rightarrow \mathcal B$ are themselves the objects of a non-unital $A_\infty$ category. The morphisms $\mathrm{hom}( \mathcal F_0, \mathcal F_1)$ are something he calls (following Fukaya) pre-natural transformations. (The morphisms $T$ for which $\mu_1(T) = 0$ are the natural transformations.) Seidel then provides the formulae for the compositions $\mu_d$. (This is discussed in Section (1d) of the book [page 10].)

In our working group, we tried to check that these formulae for the compositions satisfied the $A_\infty$ associativity equations, but were unable to do so beyond $\mu_1$.

I have two questions (that may be the same question):

Why do these composition maps satisfy the $A_\infty$ associativity equations? Is there a way of understanding this geometrically?

Answer by Sam Lisi for Floer's space closed under products?

2010-11-03T05:42:56Z

Floer chooses $\epsilon_k$ so that this $\epsilon$ space is dense in $L^2$ (this should be equivalent to dense in $C^\infty$, since the latter is dense in $L^2$, and Floer's $\epsilon$ space sits in $C^\infty$).

To show that this subspace is dense in $L^2$, it suffices to show that one can approximate indicator functions.
For this, he needs the $\epsilon_k$ to go to zero very fast. His explicit construction in Lemma 5.1 (of the "unregularized gradient flow" paper) is to take a fixed cut-off function $\beta$, and approximate the characteristic function of a rectangle. The approximation to a characteristic function is going to be a product of terms that behave like $\beta(x/\delta)$ (with a better approximation as $\delta \rightarrow 0$). Thus, the behaviour of the $\epsilon$-norm is going to be roughly: \[ \sum_{k=0}^\infty \epsilon_k \delta^{-k} a_k, \] where $a_k = \sup | D^k \beta |$. We need this to converge for each $\delta > 0$. Floer takes $\epsilon_k = (a_k k^k)^{-1}$. In particular, then, these constants are going to $0$. Following this argument, it seems we can take the $\epsilon_k$ to be on the order of $1/k!$.

Note that if the sequence $\epsilon_k$ is not summable, we expect the space to be very small. In particular, consider this norm on a compact interval, say $[-\pi, \pi]$. Then, cos(x) is not in the space.

The Floer $\epsilon$ space forms a Banach algebra if $\epsilon_k$ decays faster than $1/k!$.
Then, \[ \sum \epsilon_k |D^{(k)}(fg)| \le \sum_{k=0}^\infty \sum_{l=0}^{k} \epsilon_k | D^{(l)} f| |D^{(k-l)} g | \binom{k}{l} = \sum_{l=0}^\infty |D^{(l)} f| \sum_{p=0}^{\infty} \binom{l+p}{p} \epsilon_{p+l} | D^{(p)}g| \] When $\epsilon_{p+l} \binom{l+p}{p} \le \epsilon_p \epsilon_l$ , we are then in business. In particular, this works for Floer's original construction.

Answer by Sam Lisi for Can the Darboux theorem be strengthened?

2010-10-17T04:57:14Z

(Sorry I don't have enough reputation to make this a comment.)

I don't fully understand your question, in particular since I don't know which formulations of Darboux's theorem you are concerned with. The version I would describe as the "classical" one is that each point of a symplectic manifold admits a neighbourhood diffeomorphic to a standard symplectic ball in $R^{2n}$. In the proof I know, you need to shrink the size of your neighbourhood twice: the first time, in order to have that $\omega_1$ and $\omega_0$ are connected through symplectic forms, and the second time, to guarantee that the flow $\mathbf{F}_t$ exists to time $t=1$.

Whatever ways you may have of sidestepping one or both of these, you definitely need to shrink the size the neighbourhood somehow. Indeed, the answer to the more global question (how big a neighbourhood can you embed?) is a very subtle one related to symplectic capacities and to symplectic packing problems. For instance, Gromov's non-squeezing theorem shows that in the case of the symplectic cylinder $D^2 \times \mathbb{R}^2$, you can't get a symplectic ball any bigger than the one of radius 1. You can, however, easily smoothly embed a ball of much bigger radius (even in a volume preserving way). When you pull back the symplectic form to this ball, you will not be able to deform it to the standard one, except in a neighbourhood of a given point.

Comment by Sam Lisi

2013-04-09T10:58:42Z

The letter from Wilder is also available on the 4th page of this pdf file from the Notices ams.org/notices/200311/comm-batterson.pdf

Comment by Sam Lisi

2013-04-07T21:07:05Z

@J.Martel : If the form on $M$ is exact and $\Sigma$ is closed, then the curve is constant. Perhaps you are thinking of punctured curves? @Hwang: This is probably a stupid comment, but what do you mean by the free part? I thought the splitting wasn't natural.

Comment by Sam Lisi

2013-01-22T09:06:36Z

Thank you for clearing up my confusion about real maps vs their halves.

Comment by Sam Lisi

2012-11-05T12:48:53Z

If one forgot about this Poisson red herring, one could consider symplectic manifolds admitting some kind of codimension 2 foliation, for instance symplectic fibre bundles over surfaces. Here's an alternative question that might be what you are looking to ask. Assume $(M, \omega)$ fibres symplectically. Does there exist a compatible almost complex structure $J$ for which the fibres are $J$ invariant? Does this question help you with what you want to do? (sorry for the pun earlier, I couldn't resist)

Comment by Sam Lisi

2012-10-03T22:29:28Z

Also take a look at this blog post by Joel Kamnitzer on the Secret Blogging Seminar: sbseminar.wordpress.com/2009/02/27/… He doesn't do the whole thing, but deals with a special case... well, I guess the gap is a large amount of what you are interested in.

Comment by Sam Lisi

2012-09-17T20:45:32Z

The second link is now dead. Do you have an updated link or a bibliographic reference we can search?

Comment by Sam Lisi

2012-09-16T10:53:53Z

The second symplectization, also $\mathbb{R}\times M$ is where a translation invariant $J$ should live -- this makes sense for a stable Hamiltonian structure, and should really be thought of as a kind of blown up or stretched out version of $(0,1) \times M$. If you consider e.g. the Hofer energy we use for pseudoholomorphic curves (or even better, the definition in the case of a stable Hamiltonian structure), you see that we take a sup over a family of forms that tame $J$, each of which have (uniformly bounded) finite volume. These comments may or may not be relevant to you.

Comment by Sam Lisi

2012-09-16T10:50:48Z

IMO (and I am perhaps a crank about this), there are two things we call "symplectization". The first should be thought of canonically as consisting of all elements of $T^*M$ whose kernel is the contact structure. If the contact structure is co-orientable, there are two connected components, each of which is naturally an $R^+$ bundle. Choosing one and taking the log gives you the symplectization you wrote with the symplectic form you wrote.

Comment by Sam Lisi

2012-08-31T07:07:49Z

Only somewhat related: Mike Usher has an interesting article about a converse of this (in finite dimensions) arxiv.org/abs/1207.0889

Comment by Sam Lisi

2012-08-24T09:26:13Z

Also, to emphasize something implicit in the examples of Weinstein and Hofer-Zehnder is that a small perturbation won't allow you to make them contact. (This is discussed somewhat in Hofer-Zehnder, if I am not mistaken.) This tells you that being of contact-type is not generic for a hypersurface in $\mathbb{R}^4$.

Comment by Sam Lisi

2012-08-24T09:09:19Z

I haven't read Weinstein's paper, so this example may be essentially the same, but Hofer and Zehnder give an example of a wine bottle $S^3$ in $\mathbb{R}^4$ in their book (at the end of section 4.3).

Comment by Sam Lisi

2012-07-16T10:43:54Z

Is this a new question, or have I misunderstood the original questions? If you want a sequence of manifolds $N_i$ for which there exist non-constant pseudoholomorphic curves of genus $g$, then take your favourite sequence of symplectic manifolds $X_i$ and let $N_i=X_i \times \Sigma_g$. On this, you can take a split almost complex structure, $A= [pt]\otimes [\Sigma_g]$ and the pseudoholomorphic curve will be $\{pt\} \times \Sigma_g$.

Comment by Sam Lisi

2012-07-15T14:28:03Z

Two comments: For question 1, the answer is trivially yes if you take $A = 0$ and constant holomorphic curves... is that a reasonable interpretation of the question? There is also a typo in the index formula. If you want the real dimension of $M$, the formula is $n(2-2g) + 2c_1(A)$.

Comment by Sam Lisi

2012-04-10T13:24:08Z

I somehow just saw this question. While you've already answered it yourself, you furthermore always have this inequality on a compact manifold $M$, since $|| \nabla f ||$ is continuous on $M$ and thus bounded, and $u'(s) = - \nabla f$.

Comment by Sam Lisi

2012-03-28T12:33:53Z

@BS, alvarezpaiva, sorry, I did mean minimal and the manifold is closed. I am editing the question to clarify this.