References for this post:

[HL] R. Hind and S. Lisi, Symplectic embeddings of polydisks

[CONCAVE] K. Choi, D. Cristofaro-Gardiner, D. Frenkel, M. Hutchings, and V. Ramos, Symplectic embeddings into four-dimensional concave toric domains

[T3] M. Hutchings and M. Sullivan, Rounding corners of polygons and the ECH of

[CC2] M. Hutchings and C. H. Taubes, Proof of the Arnold chord conjecture in three dimensions II

[BUDAPEST] M. Hutchings, Lecture notes on ECH

**0. Introduction**

Hind and Lisi [HL] recently proved that if the polydisk symplectically embeds into the ball then . (I’ll review what this notation means below.) This bound is optimal, because the polydisk symplectically embeds into the ball by inclusion. Also, the is as large as possible for this sort of result, because whenever , according to Schlenk’s book, one can use “symplectic folding” to symplectically embed into for certain .

So this result of Hind-Lisi is very nice. For a while I was wondering if one can reprove it using ECH methods; previously I couldn’t, but I was missing a couple of details, and I think I now know how to do this. Moreover the new method gives further obstructions to symplectically embedding polydisks into ellipsoids, and more generally symplectic embeddings of “convex toric domains”, which go beyond the obstructions coming from ECH capacities. Here are a couple of sample results.

First, we can extend the Hind-Lisi result to give some seemingly new obstructions to symplectic embeddings of skinnier polydisks into balls (I have no idea whether these are sharp).

**Theorem 1.** Let and suppose that symplectically embeds into . Then:

- If then .
- If then .
- If then .
- If then .
- If then .

Of course the third and fifth lines follow trivially from the lines above them. I have written the fifth line because the inequality is significant until , at which point it ties the volume constraint .

We can also extend the Hind-Lisi result in a different direction by replacing balls with integral ellipsoids:

**Theorem 2.** Let and let be a positive integer. Then symplectically embeds into if and only if .

Note that includes into whenever . Thus Theorem 2 asserts that the inclusion is sharp when is an integer and . It is unclear why integrality of should be relevant here, but the method I am using works better for rational ellipsoids and best for integral ellipsoids.

I would now like to give an introduction to all of this, with a more formal writeup to come later.

**1. Convex toric domains**

Recall that if is a domain in the (closed) first quadrant of the plane, we define the “toric domain”

Let us now define a “convex toric domain” to be a domain where where is a convex function such that and . (Sorry for the confusing terminology, but the region below the graph of a concave function is convex!) For example, the polydisk

is a convex toric domain where and so that is a rectangle. Also, the ellipsoid

is a convex toric domain where and is linear so that is a triangle.

Convex toric domains should be contrasted with the “concave toric domains” considered in [CONCAVE]. For a concave toric domain, is convex and . Note that an ellipsoid is both a convex toric domain and a concave toric domain; nothing else is.

Recall that ECH capacities give obstructions to symplectically embedding any symplectic four-manifold (typically with boundary) into another. McDuff showed that ECH capacities give a sharp obstruction to symplectically embedding one (four-dimensional) ellipsoid into another. More generally, at the Simons Center in June, Dan Cristofaro-Gardiner presented a proof of the remarkable result that ECH capacities give a sharp obstruction to symplectically embedding any concave toric domain into any convex toric domain.

On the other hand, ECH capacities do not give very good obstructions to symplectically embedding a convex toric domain into a concave toric domain, e.g. a polydisk into an ellipsoid. For example they only imply that if symplectically embeds into then , although we actually want .

**2. Introduction to the strategy.**

To improve on the obstructions given by ECH capacities, the idea is a follows. First of all, the obstruction given by ECH capacities works as follows: if one symplectic four-manifold (with contact boundary) embeds into another, then we get a strong symplectic cobordism between the two contact manifolds, which induces a cobordism map on ECH. Whenever this cobordism map is nontrivial, a (possibly broken) holomorphic curve in the (completed) cobordism must exist, and the fact that this holomorphic curve has positive symplectic area gives us an inequality.

Now in general if we do not know very much about the holomorphic curve that will exist, then we will not get a very sharp inequality. The idea is to get restrictions on which holomorphic curves can exist in the cobordism, and thus improve the inequality. To do so we will use two facts:

(a) For the symplectic cobordism given by a symplectic embedding of one convex toric domain into another (the target of the embedding can also be more general), the holomorphic currents that one wants to count to define the ECH cobordism map are better behaved than usual (in particular there are no negative ECH index multiple covers).

(b) The quantity (a variant on the ECH index) bounds the topological complexity of holomorphic curves (when they are not multiply covered) and thus can be used to show that certain kinds of holomorphic curves cannot contribute to the ECH cobordism map (e.g. because the genus of a holomophic curve cannot be negative).

That was a bit vague, so let us now explain some details.

**3. Convex ECH generators**

Let be a convex toric domain and let denote its boundary. If is smooth, then it has a natural contact form, given by the restriction of

The claim is now that one can perturb so that is smooth, is nondegenerate, and the generators of the ECH chain complex (up to large symplectic action) have the following combinatorial description.

**Definition 3.** A convex generator is a polygonal (i.e. comprised of finitely many line segments) path in the plane such that:

- starts at and ends at where and are nonnegative integers.
- The vertices of (the points where it changes direction) are at lattice points.
- The edges of (i.e. line segments between vertices) have nonpositive slope (possibly ), and as one moves to the right, the slope of each edge is less than that of the preceding edge.
- Each edge of is labeled `e’ or `h’.
- Horizontal and vertical edges can only be labeled `e’.

If is a convex generator, define its combinatorial ECH index by

where is the number of lattice points in the region bounded by and the axes (including lattice points on the boundary), and is the number of edges labeled `h’.

So far, the definition of a “convex generator” and its combinatorial ECH index do not depend on the convex domain . However there is also a combinatorial notion of symplectic action of , which does depend on ; it is given by

Here is the vector determined by , that is the difference between the final and initial endpoints; denotes the cross product of vectors; and denotes a point on such that a tangent line to at is parallel to . (If is a corner of , this means that is on the lower left of the line through parallel to .)

The more precise statement is that for any large and small, one can perturb by a perturbation of size less than in an appropriate sense, such that there is a bijection between ECH generators of (actual) action less than with convex generators of (combinatorial) action less than , such that the actual ECH index agrees with the combinatorial ECH index, and the action agrees with the combinatorial action up to error. This is analogous to Lemma 3.3 in [CONCAVE] and proved similarly. (If you actually read this proof and try to compare, note that here we are first perturbing so that its boundary is nearly horizontal at the beginning and nearly vertical at the end.)

**4. The chain complex**

Now let us consider the ECH of the boundary of a convex toric domain with coefficients. We know that the homology of the chain complex has one generator in each nonnegative even degree. What do the homology generators actually look like? And before discussing this question we should maybe first ask if there is a combinatorial formula for the differential.

I conjecture that if is any path in which all edges are labeled `e’, then is a cycle which represents the homology generator of grading . This would follow from a more general conjecture about the differential. Namely, extend each convex path to a closed polygon by appending “virtual” edges along the axes, with all of the virtual edges labeled `e’. The conjecture is then that the differential acts on these extended generators by rounding corners and locally losing one `h’ (possibly plus some relatively unimportant “double rounding” terms), just as in [T3]. Keon Choi may know how to prove this, related to his thesis work.

Anyway, to study symplectic embeddings into ellipsoids, we don’t need to know all of that. We just need to know the following fact, which can be proved more quickly:

**Lemma 4.** Suppose is the triangle with vertices , , and , so that . Let be a line in the plane of slope which passes through at least one lattice point in the first quadrant. Let be the maximal convex generator to the left of , with all edges labeled `e’. Then is a cycle which represents a nontrivial homology class in the ECH of (perturbed) .

**Proof.** Let . One can check that uniquely minimizes the combinatorial action among generators of index , and its combinatorial action agrees with the kth ECH capacity of . This is similar to Example 1.23 in [CONCAVE]. The claim follows immediately (because the degree homology class is represented by a cycle which is a sum of generators with action less than or equal to the kth ECH capacity, and is the only such generator.)

**5. The cobordism map**

Suppose now that a convex toric domain with boundary symplectically embeds into (the interior of) another convex toric domain with boundary . Then minus the interior of the image of is a strong symplectic cobordism from to . We further get a strong symplectic cobordism between the perturbed contact forms as above. This induces a map , which is an isomorphism, because the cobordism is diffeomorphic to the product . We know from my work with Taubes [CC2] using Seiberg-Witten theory that given an appropriate almost complex structure on the cobordism, this ECH cobordism map is induced by a chain map, such that whenever a coefficient of the chain map is nonzero, there is a (possibly broken) -holomorphic current with ECH index zero between the corresponding ECH generators.

In particular, the chain map decreases the symplectic action. All of our symplectic embedding obstructions come from this fact.

Now the key observation which gives us some control over this chain map is:

**Lemma 5.** The chain map (up to any given symplectic action ) can be chosen so that it only counts unbroken -holomophic currents, such that the components are disjoint and have , and the somewhere injective curve underlying each component is embedded and also has .

To prove this, one needs to do a calculation, using the special nature of the contact forms in question, to show that if is generic, then multiply covered curves with negative ECH index cannot arise here. I will explain this some other time. I will just remark for now that this lemma also holds for symplectic embeddings of convex or concave toric domains into concave toric domains. (It doesn’t work for concave toric domains into convex toric domains, but this is the case where Dan showed that ECH capacities already give a sharp symplectic embedding obstruction.)

**6. Controlling topological complexity**

For the above cobordism arising from a symplectic embedding of one convex toric domain into another, we now want to control the topological complexity of -holomorphic currents with ECH index index . To do so, we use the quantity (this is an integer similar to the ECH index, not an almost complex structure), see e.g. Section 5.2 of the lecture notes [BUDAPEST].

In our situation, can be computed combinatorially as follows. If is a convex generator, going from to , then

Here denotes the number of edges of that are labeled `e’, plus the number of edges that are labeled `h’ and contain at least one interior lattice point. (In other words, is the number of elliptic orbits in the orbit set corresponding to .)

Now the significance of is that if is an holomophic curve from to which does not have any multiply covered components, then

Here is the sum, over all Reeb orbits in or , of the number of ends of at covers of that orbit minus one. This is proved as in Exercise 5.9 of [BUDAPEST].

We can be a bit more specific as follows. By the partition conditions, we know that all positive ends of are at simple orbits, and can have at most one negative end at covers of any given orbit. To make use of this fact, let denote the total number of Reeb orbits in , namely the number of edges, plus the number of `h’ edges with at least one interior lattice point. Let denote the total multiplicity of all Reeb orbits in , namely the sum over all edges of the number of interior lattice points minus one. Then

Also, if is connected with genus , then

.

Putting the above together, we get

Since and , we conclude the following:

**Lemma 6.** Suppose there exists an , connected, embedded curve from to . Suppose that all edges of are labeled `e’. Then

(OK, I hate the writing style where the proof comes before the statement, sorry.)

Anyway, to prove Theorems 1 and 2, we will apply Lemma 6 repeatedly. Let’s see now how this works.

**7. Proof of the first part of Theorem 1.**

To start, let us prove the first part of Theorem 1, namely that if symplectically embeds into , and if , then .

Assume that symplectically embeds into , assume that , and assume that . We will show that .

*Step 1.* Consider the convex generator consisting of the straight line from to , labeled `e’. Then and . By Lemmas 4 and 5, there is an current from to some convex generator with and .

There are in fact only three convex generators with : the horizontal line from to , labeled `e’, which has action ; the line from to , labeled `e’, which has action ; and the vertical line from to , labeled `e’, which has action . If there is a holomorphic curve from to either of the latter two generators, then we immediately get or , which contradicts our assumption that . Thus curves from can only go to the horizontal line.

*Step 2.* Consider the convex generator consisting of the straight line from to , labeled `e’. Then and . By Lemmas 4 and 5, there is an current from to some convex generator with and .

Now the current must actually be a connected, embedded holomorphic curve. Why? Well, if were disconnected, then each component would have , and so by the end of Step 1, each component can only go to the horizontal line of length 2. Hence is a horizontal line of length 4. But this only has which is not big enough. Likewise, if is multiply covered, then again by the end of Step 1, the embedded curve underlying can only map to the horizontal line of length 2, so again is the horizontal line of length 4 which is impossible.

Since is connected and embedded, we can apply Lemma 6 to conclude that

.

Now the action of is given by

If , then , so . Combining this with our assumption then gives .

Assume now that . Then can only be the horizontal line of length 5, labeled `e’. In particular this gives . We need one more step to obtain .

*Step 3.* We are now assuming and . Consider the convex generator given by the straight line from to , labeled `e’. This has and . By Lemmas 4 and 5, there is a holomorphic current from to a convex generator with and .

By Steps 1 and 2, is connected and embedded. (Otherwise is a horizontal line of length at most , which only has , which is too small.) Thus Lemma 6 applies to give

.

If , then , so . Combining this with our assumption that then gives .

If , then must be a horizontal line of length , since its ECH index is . Thus , so .

QED

*Which holomorphic curve is giving the obstruction?* If you analyze the above proof, what it is saying is that if and if symplectically embeds into , then at least one of the following holomorphic curves must exist:

- A pair of pants from the “diagonal” Reeb orbit of action to the “horizontal” and “vertical” simple orbits for the polydisk, implying that .
- A cylinder from as in Step 1 to the vertical line of length 2, implying that .
- A holomorphic curve with two positive ends at the “diagonal” orbit of action , and negative ends with total action at least , implying that .
- A holomorphic curve with three positive ends at the “diagonal” orbit of action , and either one negative end with action , or with negative ends of total action at least .

Either way, we obtain if . In any case, this looks quite different from the Hind-Lisi argument, which studies curves of “degree” and takes the limit as .

**8. Proof of the rest of Theorem 1.**

To prove the rest of Theorem 1, we continue to play the above game. Namely:

*Step 4.* Assume , , , and . (We can make this last assumption without loss of generality because when .) Let be the straight line from to , labeled `e’. This has and . Then there is a holomorphic current from to a convex generator with and . The above assumptions imply that is connected and embedded, so we can apply Lemma 6 to deduce that . Thus

This gives the lower bound on in Theorem 1 for .

*Step 5.* Now assume , , , , and . Let be the straight line from to . Then we similarly obtain

This gives the lower bound on in Theorem 1 for .

QED.

**9. Proof of Theorem 2.**

We now prove that if and is a positive integer, and if symplectically embeds into , then .

Theorem 1 covered the case where .

To prove the rest, suppose that is an integer and that symplectically embeds into . Assume that and . We need to show that . This is actually a little simpler than the proof of Theorem 1 and only requires two steps.

*Step 1.* Let be the convex generator given by the straight line from to , labeled `e’. Then and . By Lemmas 4 and 5, there is a holomorphic current from to a convex generator with and . Now must be connected and embedded, since consists only of a single, simple Reeb orbit. Thus Lemma 6 implies that . If then , contradicting our assumption that . The conclusion is that can only be the horizontal line of length , labeled `e’.

*Step 2.* Now let be the convex generator given by the straight line from to , labeled `e’. Then and . By Lemmas 4 and 5, there is a holomorphic current from to a convex generator with and . Now must be connected and embedded, since otherwise, by Step 1, is a horizontal line of length , which has index . Thus Lemma 6 applies to give

.

If then . Thus . By assumption , so and we are done.

If on the other hand , then , so . By assumption , so we get . Since we are assuming , it follows that so we are done again.

QED