## Fun with ECH capacities

I now want to talk about some new computations of ECH capacities, most of which are still conjectural.  These are inspired by the following question which Larry Guth asked me (although I might be remembering it wrong):

Question. Let $X$ be the union of a ball and a thin symplectic cylinder in ${\mathbb R}^4$.  In my notation for ellipsoids,

$X = E(1,1) \cup E(\infty,a)$

where $a<1$.  Note that this is the union in ${\mathbb R}^4$, not the disjoint union.  Now the question is, can we find obstructions to symplectically embedding an ellipsoid $E(b,c)$ into $X$, where $a and $bc>1$?

Note that $E(b,c)$ cannot symplectically embed into $E(1,1)$ alone (because $bc>1$ means that its volume is too big), or into $E(\infty,a)$ alone (because $c>a$, and apply your favorite symplectic capacity).  But it is harder to obstruct a symplectic embedding into all of $X$.  Maybe some classical symplectic capacities can say something about this, I don’t know.  In any case I am naturally motivated to ask:

Question. Compute the ECH capacities of $X$.

Based on evidence which I will describe below, I believe the following is true (and probably provable):

[UPDATE 5/23/13: Conjecture 1 is now proved in this post. Also, I corrected a typo in the statement of Conjecture 1 which mixed up $d$ and $d+1$.]

Conjecture 1. $X$ has the same ECH capacities as the disjoint union of $E(1,1)$ and $E(\infty,a)$.  More explicitly,

$c_k(X) = \max\left\{d + a\left(k-\frac{d(d+1)}{2}\right)\right\}$

where the maximum is over nonnegative integers $d$ such that $k- \frac{d(d+1)}{2}\ge 0$.

If this conjecture is true, then it does tell us something about Larry Guth’s question.  For example the conjecture implies that $c_k(X)$ is equal to $ak$ plus a constant when $k$ is sufficiently large.  In particular, if $c>a$ then $E(\infty,c)$ does not embed into $X$ (because it has ECH capacities $ck$), so $E(b,c)$ does not embed into $X$ if $b$ sufficiently large.

Now let’s talk about how to try computing the ECH capacities of $X$ and why I believe the above conjecture is true.  Recall that if $\Omega$ is a domain in the first quadrant of the plane, I define the notation

$X_\Omega = \{(z_1,z_2)\in{\mathbb C}^2 \mid (\pi|z_1|^2,\pi|z_2|^2) \in \Omega\}.$

In particular, the space $X$ of interest above is $X_\Omega$ where $\Omega$ is the union of the triangle with vertices $(0,0), (1,0), (0,1)$ with the strip bounded by the $x$ axis and the line $y=a$.  So let’s recall what we know about how to compute the ECH capacities of $X_\Omega$ for general $\Omega$.

Suppose first that $\Omega$ is a convex domain which does not touch the axes.  (By a limiting argument one can allow it to touch the axes in one or two points, and one may in fact be able to allow it to touch the axes arbitrarily.)  Then by a theorem from the paper “Quantitative ECH” which I described in the previous post, we have

$c_k(X_\Omega) = \min\{\ell(\Lambda)\}$

where the minimum is over convex polygonal paths $\Lambda$ in the plane that enclose at least $k+1$ lattice points (including lattice points on $\Lambda$ itself), and $\ell(\Lambda)$ is defined as follows.  Pick an arbitrary reference point $p$ in the plane.  If $e$ is a vector in the plane, let $q$ be a point on $\Lambda$ such that an outward normal to $\Lambda$ at $q$ is parallel to $e$ (if $\Lambda$ has a corner I regard every vector between the outward normals to the incident edges to be an outward normal at the corner), and define a new vector $v_e = q-p$.  Then

$\ell(\Lambda) = \sum_{e\in Edge(\Lambda)}e\cdot v_e$

where $Edge(\Lambda)$ denotes the set of edges of $\Lambda$, regarded as vectors, using the counterclockwise orientation of $\Lambda$.  Note that $\ell(\Lambda)$ does not depend on the choice of reference point $p$.  (Also $min\{\ell(\Lambda)\}$ would be the same if we used the clockwise orientation instead.)  This is a restatement of the formula in terms of dual norms which I described in the last post.

Now this formula for $c_k(X_\Omega)$, despite being computable in finite time, is a bit difficult to work with if you are trying to calculate by hand.  However we can simplify it somewhat when $\Omega$ is a polygon, which is our main case of interest.   The claim is that in this case, we can restrict attention to polygons $\Lambda$ such that the edges of $\Lambda$ are parallel to 90 degree rotations of the edges of $\Omega$.  To see this, look at the formula for $\ell(\Lambda)$.  Since $\Omega$ is a polygon, in the formula for $\ell(\Lambda)$, for each edge of $\Lambda$ one can take the corresponding point $q$ to be a corner of $\Omega$.  In particular, if we number the edges of $\Omega$ in cylic order as $E_1,\ldots,E_n$, then these determine corners $c_1,\ldots,c_n$ of $\Lambda$, such that the edges of $\Lambda$ between $c_i$ and $c_{i+1}$ have the point $q$ at the corner between $E_i$ and $E_{i+1}$, call this $q_i$.  This means that $\ell(\Lambda)$ depends only on the points $c_1,\ldots,c_n$, namely

$\ell(\Lambda) = \sum_{i=1}^n(c_{i+1}-c_i)\cdot(q_i-p).$

So given the points $c_1,\ldots,c_n$, we can assume without generality that the path $\Lambda$ is the largest path giving rise to these points.  And it is easy to see (or would be easy to see if we drew a picture) that this means exactly that the edges of $\Lambda$ are parallel to 90 degree rotations of the edges of $\Omega$.

Note that if $\Omega$ is a right triangle with legs parallel to the axes, or a rectangle with legs parallel to the axes, then this formula immediately recovers the ECH capacities of an ellipsoid or a polydisk.

When $\Omega$ is a triangle and $\Lambda$ is a 90 degree rotation of $\Omega$, scaled by $\lambda$, we can simplify a bit more;  in this case we have

$\ell(\Lambda) = 2A(\Omega)\lambda,$

where $A$ denotes area.  It is still somewhat unwieldy to minimize this over triangles that enclose at least $k+1$ lattice points, so later I will tell you a much more effective way to compute $c_k(X_\Omega)$ when $\Omega$ is a triangle.

Now what about when $\Omega$ is not convex? In this case the theorem that I used to compute the ECH capacities in the convex case does not apply (I will explain later why). However here are some general remarks which are useful.  To start, write $c_k(\Omega)=c_k(X_\Omega)$.  Now observe that $c_k(\Omega)$ is invariant under affine equivalence of $\Omega$.  Here I call two domains in the interior of the first quadrant “affine equivalent” if one can be obtained from the other by applying an element of $SL_2{\mathbb Z}$ and translating.  This is because the corresponding symplectic manifolds $X_\Omega$ are symplectomorphic (exercise).  Second, if we can decompose $\Omega$ as $\Omega_1\cup\Omega_2$, where $\Omega_1$ and $\Omega_2$ meet only along their boundaries, then it follows from the monotonicity and disjoint union axioms of the ECH capacities that we have a lower bound

$c_k(\Omega) \ge \max_{i+j=k}(c_i(\Omega_1) + c_j(\Omega_2)).$

Third, if $\Omega'$ is another domain which intersects $\Omega$ only along the boundary, then we likewise get an upper bound

$c_k(\Omega) \le \min_{l\ge 0}(c_{k+l}(\Omega\cup \Omega') - c_l(\Omega')).$

Now returning to the question at the top of this post, by a limiting argument it is enough to compute $c_k(\Omega)$ where $\Omega$ is a quadrilateral with vertices $(0,0), (A,0), (1-a,a), (0,1)$, where $a < 1 < A$.  To invoke the above bounds, we can write $\Omega=\Omega_1\cup \Omega_2$ where $\Omega_1$ is the triangle with vertices $(0,0), (1,0), (0,1)$, and $\Omega_2$ is the triangle with vertices $(1,0), (A,0), (1-a,a)$.  Also define $\Omega'$ to be the triangle with vertices $(A,0), (0,1), (1-a,a)$.  Then in the above bounds, $\Omega_1$ and $\Omega\cup\Omega'$ correspond to ellipsoids whose ECH capacities we can compute, while $\Omega_2$ is affine equivalent to a right triangle with legs parallel to the axes, so its ECH capacities can be computed the same way.  The ECH capacities of $\Omega'$ can be effectively computed using the special trick which I still need to tell you about (in another post) for dealing with more general triangles.  So we now have computable lower and upper bounds on $c_k(\Omega)$.

Now when I computed some examples, it miraculously seemed to be the case that the lower and upper bounds are equal!  In particular:

Conjecture 2. For the above $\Omega,\Omega_1,\Omega_2$ we have $c_k(X_\Omega) = c_k(X_{\Omega_1}\sqcup X_{\Omega_2})$.

Note that Conjecture 2 implies Conjecture 1 by taking the limit as $A\to\infty$.

Note also that it is certainly not true that we can get sharp lower bounds on $c_k$ of a general $\Omega$ by cutting it into pieces as above.  For example if you cut a square into four equal pieces, then the lower bound you get on $c_1$ of the square is too small by a factor of $2$.  Dusa McDuff’s paper “the Hofer conjecture on embedding symplectic ellipsoids” proves two combinatorial lemmas (2.4 and 2.6) describing cases where you do get sharp lower or upper bounds of the above form.  However it is unclear to me to what extent these lemmas can be generalized.  I tried to generalize them to prove Conjecture 2, didn’t quite get there, but expect that it is still doable, since Conjecture 2 is a combinatorial statement and should have a combinatorial proof.  In any case, we should try to find a clear conceptual reason why Conjecture 2 should be true.

If we don’t know why a conjecture should be true, then let’s make a more general conjecture which implies it!  In particular, here is a naive generalization of the theorem for computing ECH capacities of convex domains.

[UPDATE: I just realized that the Conjecture below is probably not what I want.  So the problem is now to formulate a better conjecture.]

Conjecture 3. Let $\Omega$ be any star-shaped (but not necessarily convex) domain in the first quadrant.  Then

$c_k(X_\Omega) = \min\{\ell(\Lambda,f)\}.$

Here the minimum is over pairs $(\Lambda,f)$, where $\Lambda$ is a closed polygonal path (not necessarily convex) which encloses (suitably interpreted if $\Lambda$ crosses itself) at least $k+1$ lattice points, and $q:Edge(\Lambda)\to\partial\Omega$ is a function which preserves the cyclic orderings, such that for each $e\in Edge(\Lambda)$, an outward normal vector to $\Omega$ at $q(e)$ is parallel to $e$.  Also

$\ell(\Lambda,f) = \sum_{e\in Edge(\Lambda)}e\cdot (q(e)-p)$

where $p$ is a reference point.

[UPDATE: The following paragraph is wrong, which is why I don’t think Conjecture 3 is right.]

Note that in the above minimum, if $\Omega$ is a polygon, then as before one can restrict attention to paths $\Lambda$ whose edges are parallel to 90 degree rotations of the edges of $\Omega$ (in the same cyclic order).  In particular, Conjecture 3 implies Conjecture 2 (because for the $\Omega$ in Conjecture 2, a minizing path as in Conjecture 3 encloses the union of minimizing triangles for the two subregions, and the costs add up).

Now here are two possible approaches to proving a formula for $c_k(\Omega)$ when $\Omega$ is not convex:

1) By a limiting argument, we can assume $\Omega$ is a polygon.  We can get lower bounds on $c_k(\Omega)$ by cutting $\Omega$ into convex pieces, and we can get upper bounds on $c_k(\Omega)$ by adding convex pieces to form something convex.  If we are very lucky, then we can prove combinatorially that the lower bounds and upper bounds agree (and equal the desired answer).  However this might be too much to hope for.

2) By a limiting argument, we can assume that the boundary of $\Omega$ is smooth.  Let’s also (with some loss of generality) restrict attention to the case where $\Omega$ does not touch the axes.  Since $\Omega$ is star-shaped, $X_\Omega$ is a Liouville domain, and we obtain a Morse-Bott contact form on its boundary (which is diffeomorphic to $T^3$).  We need to understand the ECH chain complex of this contact form.  This will require some generalization of my work with Michael Sullivan on the ECH of $T^3$. (My work with Michael Sullivan handles the case where $\Omega$ is convex).  But this post is already too long, so I’ll have to explain this some other time.