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<a<1 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<n/m<1, 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.

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