[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 is a domain in the first quadrant in , then we define the “toric domain”
For example if is a right triangle with legs on the axes then is an ellipsoid; and if is a rectangle with two sides on the axes then is a polydisk. In general it is interesting to compute the ECH capacities of . When is convex and does not touch the axes, there is a combinatorial formula for the ECH capacities of , which is also valid in some (and conjecturally all) cases when 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 when is star-shaped but not convex. Here is one bit of progress:
Theorem. Let and let . That is, where is the union of the triangle and the horizontal strip . Then the ECH capacities of are given by
Here is an integer. (This was conjectured in this earlier post, except that there I made a typo mixing up and .) This gives some obstructions to symplectic embeddings into which we can explore later. But first, here is the proof of the theorem:
Proof. Step 1: setup. By the definition of ECH capacities, where is a Liouville domain which can be symplectically embedded into the interior of . Given , let . Since the are nested domains whose union is , it follows from the monotonicity axiom for ECH capacities that
So we need to show that
To do so, fix and assume that is small with respect to .
Step 2: the lower bound. We first show that
By the monotonicity and disjoint union axioms for ECH capacities, we know that
From the computation of the ECH capacities of an ellipsoid, we know that where is the unique nonnegative integer such that
Also, is affine equivalent to a right triangle of height , so it symplectically embeds into an ellipsoid with axis , and in fact has the same ECH capacities (see Exercise 4.16(b) in the ECH lecture notes), which means that , provided that and is sufficiently small with respect to . In computing the above maximum, we can restrict attention to the case where for some nonnegative integer (since otherwise we can decrease for free and profitably increase ). The desired lower bound on follows.
Step 3: the upper bound. We now prove that is less than or equal to the claimed value. To calculate , we need to approximate by a Liouville domain for which the contact form on the boundary is nondegenerate. We have for a certain domain . Let be obtained from by slightly increasing the slope of the edge with slope , and rounding the concave corner (preserving concavity). Let be obtained from by perturbing so that the Morse-Bott circles of Reeb orbits on the boundary (up to some large symplectic action ) split into elliptic and hyperbolic Reeb orbits. To complete the proof, we will show that if is an ECH generator for with , then the symplectic action of satisfies
up to some small error depending on the size of the perturbation from to .
The embedded Reeb orbits in the boundary of , up to action , are given as follows. There is an elliptic orbit corresponding to the upper left corner of the domain ; we denote this orbit by , and it has symplectic action approximately . For every pair of relatively prime positive integers with , there is an elliptic orbit and a hyperbolic orbit . These arise from the point on the boundary of where a tangent vector to the boundary of is parallel to the vector . They both have symplectic action approximately . (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 be an ECH generator. This is a formal product of orbits and where no factor may be repeated. Let denote the sum over all factors of the subscript, and let denote the sum over all factors of the subscript. Write where is the exponent of . Then by the previous paragraph, the symplectic action of is given (up to some small error) by
To describe the ECH index of , let denote the following polygonal path in the plane. The path starts at the point , and the first edge goes to . After that, for each or factor in with , there is an edge in with edge vector . These edges are arranged in order of increasing slope. These edges take us to the point . The path then goes horizontally to , and finally vertically back to the starting point .
Let denote the number of lattice points enclosed by , not including lattice points on the “upper boundary”, namely the part of the boundary from to . I claim that the ECH index of is given by
where denotes the number of hyperbolic factors in . 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 .
We need to show that if then
We will in fact show that
with . (The next paragraph will show that .)
To see this, observe that , not including the upper boundary, encloses all of the lattice points in the triangle with vertices , and , and there are of these. In addition, the line segment from to , and the line segment from to , include an additional lattice points which are enclosed by and not on the upper boundary. This gives a lower bound on the lattice point count , from which it follows that
In the inequality that we want to prove, , so it is enough to show that
By our choice of and our previous calculation of , this inequality is
This says that if then , and if then . This follows immediately from the definition of , and so we have proved the desired inequality.
The above calculation shows that an ECH generator of index with maximum symplectic action has the form with action , or with which has action . The choice of that gives the maximum symplectic action depends on .