## The ECH capacities of a ball union a cylinder

[UPDATE 2: I corrected some small mistakes pointed out by Vinicius. I’ll write a detailed explanation of the index calculation in another post.]

Recall that if $\Omega$ is a domain in the first quadrant in ${\mathbb R}^2$, then we define the “toric domain”

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

For example if $\Omega$ is a right triangle with legs on the axes then $X_\Omega$ is an ellipsoid; and if $\Omega$ is a rectangle with two sides on the axes then $X_\Omega$ is a polydisk. In general it is interesting to compute the ECH capacities of $X_\Omega$. When $\Omega$ is convex and does not touch the axes, there is a combinatorial formula for the ECH capacities of $X_\Omega$, which is also valid in some (and conjecturally all) cases when $\Omega$ is convex and does touch the axes. For details, see e.g. section 4.3 of the ECH lecture notes.

Dan, Vinicius, Keon, and I have been discussing how to compute the ECH capacities of $X_\Omega$ when $\Omega$ is star-shaped but not convex. Here is one bit of progress:

Theorem. Let $0 and let $X=B(1)\cup E(\infty,a)\subset {\mathbb R}^4$. That is, $X=X_\Omega$ where $\Omega$ is the union of the triangle $0\le \mu_1,\mu_2,\mu_1+\mu_2\le 1$ and the horizontal strip $0\le \mu_1; 0\le \mu_2\le a$. Then the ECH capacities of $X$ are given by

$c_k(X) = \max\left\{d+a\left(k-\frac{d(d+1)}{2}\right)\mid d\ge 0, d(d+1)\le 2k\right\}.$

Here $d$ is an integer. (This was conjectured in this earlier post, except that there I made a typo mixing up $d$ and $d+1$.) This gives some obstructions to symplectic embeddings into $X$ which we can explore later. But first, here is the proof of the theorem:

Proof. Step 1: setup. By the definition of ECH capacities, $c_k(X) = \sup\{c_k(X',\omega)\}$ where $(X',\omega)$ is a Liouville domain which can be symplectically embedded into the interior of $X$. Given $\epsilon>0$, let $X_\epsilon = B(1) \cup E(\epsilon^{-1},a)$. Since the $X_\epsilon$ are nested domains whose union is $X$, it follows from the monotonicity axiom for ECH capacities that

$c_k(X) = \lim_{\epsilon\to 0+} c_k(X_\epsilon).$

So we need to show that

$\lim_{\epsilon\to 0+} c_k(X_\epsilon) = \max\left\{d+a\left(k-\frac{d(d+1)}{2}\right)\mid d\ge 0, d(d+1)\le 2k \right\}.$

To do so, fix $k$ and assume that $\epsilon$ is small with respect to $k$.

Step 2: the lower bound. We first show that

$c_k(X_\epsilon) \ge \max\left\{d+a\left(k-\frac{d(d+1)}{2}\right)\mid d\ge 0, d(d+1)\le 2k \right\}.$

By the monotonicity and disjoint union axioms for ECH capacities, we know that

$c_k(X_\epsilon) \ge \max\{c_{k_1}(B(1)) + c_{k_2}(X_\epsilon\setminus B(1)) \mid k_1+k_2=k\}.$

From the computation of the ECH capacities of an ellipsoid, we know that $c_{k_1}(B(1)) = d$ where $d$ is the unique nonnegative integer such that

$\frac{d(d+1)}{2} \le k_1 \le \frac{d(d+3)}{2}.$

Also, $X_\epsilon\setminus B(1)$ is affine equivalent to a right triangle of height $a$, so it symplectically embeds into an ellipsoid with axis $a$, and in fact has the same ECH capacities (see Exercise 4.16(b) in the ECH lecture notes), which means that $c_{k_2}(X_\epsilon\setminus B(1))=ak_2$, provided that $k_2\le k$ and $\epsilon$ is sufficiently small with respect to $k$. In computing the above maximum, we can restrict attention to the case where $k_1=d(d+1)/2$ for some nonnegative integer $d$ (since otherwise we can decrease $k_1$ for free and profitably increase $k_2$). The desired lower bound on $c_k(X_\epsilon)$ follows.

Step 3: the upper bound. We now prove that $c_k(X_\epsilon)$ is less than or equal to the claimed value. To calculate $c_k(X_\epsilon)$, we need to approximate $X_\epsilon$ by a Liouville domain for which the contact form on the boundary is nondegenerate. We have $X_\epsilon = X_\Omega$ for a certain domain $X_\Omega$. Let $\Omega'$ be obtained from $\Omega$ by slightly increasing the slope of the edge with slope $-1$, and rounding the concave corner (preserving concavity). Let $X_\epsilon'$ be obtained from $X_{\Omega'}$ by perturbing so that the Morse-Bott circles of Reeb orbits on the boundary (up to some large symplectic action $L>k$) split into elliptic and hyperbolic Reeb orbits. To complete the proof, we will show that if $\alpha$ is an ECH generator for $\partial X_\epsilon'$ with $I(\alpha)=2k$, then the symplectic action of $\alpha$ satisfies

${\mathcal A}(\alpha) \le \max\left\{d+a\left(k-\frac{d(d+1)}{2}\right)\mid d\ge 0, d(d+1)\le 2k \right\},$

up to some small error depending on the size of the perturbation from $X_\epsilon$ to $X_\epsilon'$.

The embedded Reeb orbits in the boundary of $X_\epsilon'$, up to action $L$, are given as follows. There is an elliptic orbit corresponding to the upper left corner of the domain $\Omega'$; we denote this orbit by $e_{1,0}$, and it has symplectic action approximately $1$. For every pair of relatively prime positive integers $m,n$ with $0, there is an elliptic orbit $e_{m,n}$ and a hyperbolic orbit $h_{m,n}$. These arise from the point on the boundary of $\Omega'$ where a tangent vector to the boundary of $\Omega'$ is parallel to the vector $(m,-n)$. They both have symplectic action approximately $am+(1-a)n$. (For the calculation of the Reeb orbits and their symplectic actions in the boundaries of general toric domains, see Section 4.3 in the ECH lecture notes.)

Now let $\alpha$ be an ECH generator. This is a formal product of orbits $e_{m,n}$ and $h_{m,n}$ where no $h_{m,n}$ factor may be repeated. Let $M$ denote the sum over all factors of the $m$ subscript, and let $N$ denote the sum over all factors of the $n$ subscript. Write $M=M_0+M_1$ where $M_0$ is the exponent of $e_{1,0}$. Then by the previous paragraph, the symplectic action of $\alpha$ is given (up to some small error) by

${\mathcal A}(\alpha) = M_0 + aM_1 + (1-a)N.$

To describe the ECH index of $\alpha$, let $\Lambda$ denote the following polygonal path in the plane. The path $\Lambda$ starts at the point $(0,M_0+N)$, and the first edge goes to $(M_0,N)$. After that, for each $e_{m,n}$ or $h_{m,n}$ factor in $\alpha$ with $n>0$, there is an edge in $\Lambda$ with edge vector $(m,-n)$. These edges are arranged in order of increasing slope. These edges take us to the point $(M,0)$. The path $\Lambda$ then goes horizontally to $(0,0)$, and finally vertically back to the starting point $(0,M_0+N)$.

Let $L(\Lambda)$ denote the number of lattice points enclosed by $\Lambda$, not including lattice points on the “upper boundary”, namely the part of the boundary from $(0,M_0+N)$ to $(M,0)$. I claim that the ECH index of $\alpha$ is given by

$I(\alpha) = 2L(\Lambda)+ h$

where $h$ denotes the number of hyperbolic factors in $\alpha$. I will not prove this here because it would take a lot space, but it is similar to the calculation of the ECH index for the standard contact form on $T^3$.

We need to show that if $I(\alpha)=2k$ then

${\mathcal A}(\alpha) \le \max\left\{d+a\left(k-\frac{d(d+1)}{2}\right)\mid d\ge 0, d(d+1)\le 2k \right\}.$

We will in fact show that

${\mathcal A}(\alpha) \le d+a\left(k-\frac{d(d+1)}{2}\right)$

with $d=M_0+N$. (The next paragraph will show that $d(d+1)\le 2k$.)

To see this, observe that $\Lambda$, not including the upper boundary, encloses all of the lattice points in the triangle with vertices $(0,0),(d-1,0)$, and $(0,d-1)$, and there are $d(d+1)/2$ of these. In addition, the line segment from $(M_0,N)$ to $(d,0)$, and the line segment from $(d,0)$ to $(M,0)$, include an additional $\max\{M_1-1,0\}$ lattice points which are enclosed by $\Lambda$ and not on the upper boundary. This gives a lower bound on the lattice point count $L(\Lambda)$, from which it follows that

$I(\alpha)\ge d(d+1)+2\max\{0,M_1-1\}.$

In the inequality that we want to prove, $k=I(\alpha)/2$, so it is enough to show that

${\mathcal A}(\alpha) \le d+a\max\{0,M_1-1\}.$

By our choice of $d$ and our previous calculation of ${\mathcal A}(\alpha)$, this inequality is

$M_1-N\le \max\{0,M_1-1\}.$

This says that if $M_1=0$ then $N\ge 0$, and if $M_1>0$ then $N>0$. This follows immediately from the definition of $N$, and so we have proved the desired inequality.

The above calculation shows that an ECH generator of index $2k$ with maximum symplectic action has the form $e_{1,0}^d$ with action $d$, or $e_{1,0}^{d-1}e_{m,1}$ with $m=k-d(d+1)/2+1>0$ which has action $d+a(k-d(d+1)/2)$. The choice of $d$ that gives the maximum symplectic action depends on $a$.