## Lagrangian capacities and Ekeland-Hofer capacities

References for this post:

[CM] K. Cieliebak and K. Mohnke, Punctured holomorphic curves and Lagrangian embeddings

[HN] M. Hutchings and J. Nelson, Cylindrical contact homology for dynamically convex contact forms in three dimensions

[BEYOND] M. Hutchings, Beyond ECH capacities

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

First, let me mention that Chris Wendl has a new blog.

Now I would like to comment on one remark in the recent preprint [CM] (which has many more interesting things in it). In this paper, Cieliebak and Mohnke define the “Lagrangian capacity” of a symplectic manifold $(X,\omega)$ as follows: If $T\subset X$ is a Lagrangian torus, define $A_{min}(T)$ to be the infimum of $\int_D\omega$ where $D\in\pi_2(X,T)$ and $\int_D\omega>0$. Then define $c_L(X,\omega)$ to be the supremum of $A_{min}(T)$ over all embedded Lagrangian tori $T$. Cieliebak-Mohnke then ask:

Question. For which domains $X$ in ${\mathbb R}^{2n}$ is it true that

$c_L(X) = \lim_{k\to\infty}\frac{1}{k} c_k^{EH}(X)$

where $c_k^{EH}$ denotes the $k^{th}$ Ekeland-Hofer capacity?

They conjecture that this is true for ellipsoids and ask whether it is true for all convex domains.

I would now like to present some evidence (based on more conjectures) that the answer to the Question is YES for an interesting family of examples, namely convex toric domains in ${\mathbb R}^4$. More precisely, what I will do is the following:

• Recall how to use cylindrical contact homology to define an ersatz version of the Ekeland-Hofer capacities, denoted by $c_k^{CH}$, which are conjecturally equal to them.
• Compute $c_k^{CH}$ for convex toric domains in ${\mathbb R}^4$, modulo a conjectural description of the cylindrical contact homology differential which is probably not too hard to prove.
• Deduce from the above computation that $lim_{k\to\infty}\frac{1}{k}c_k^{CH}(X) \le c_L(X)$ whenever $X$ is a convex toric domain in ${\mathbb R}^4$.
• Briefly discuss strategy for trying to prove the reverse inequality.

1. Cylindrical contact homology capacities

Let $X$ be a (strictly) star-shaped domain in ${\mathbb R}^4$ with boundary $Y$. Recall that the Liouville form

$\lambda = \frac{1}{2}\sum_{i=1}^2(x_idy_i-y_idx_i)$

restricts to a contact form on $Y$. Let us perturb $Y$ if necessary to ensure that this contact form is nondegenerate, and let us further assume that $\lambda|_Y$ is dynamically convex (which holds for example when $X$ is convex). We can then define the cylindrical contact homology $CH(Y,\lambda)$, as explained in [HN]. (The proof of invariance of cylindrical contact homology in the dynamically convex case and construction of cobordism maps on it are to appear in a sequel.) With the usual grading convention, this cylindrical contact homology is ${\mathbb Q}$ in degree $2,4,\ldots$ and $0$ in all other degrees.

If $k$ is a positive integer, we now define $c_k^{CH}(X)$ to the minimum over $L$ such that the degree $2k$ class in $CH(Y,\lambda)$ can be represented by a linear combination of good Reeb orbits, each of which has action $\le L$. One can use cobordism maps to show that this number does not depend on the almost complex structure used to define $CH(Y,\lambda)$ and is monotone with respect to symplectic embeddings. Also, this definition extends to any convex domain $X$ (where the boundary might not be smooth or nondegenerate) by taking $C^0$ limits.

It is conjectured that $c_k^{CH}(X)=c_k^{EH}(X)$ when $X$ is a convex domain in ${\mathbb R}^4$, or more generally a star-shaped domain whose boundary is the limit of hypersurfaces which are nondegenerate and dynamically convex. I made (a more general version of) this conjecture in this previous post, based on calculations for ellipsoids and polydisks (which I will explain below), and other people have made similar conjectures.

2. CH capacities of convex toric domains in ${\mathbb R}^4$

Recall that if $\Omega$ is a domain in the first quadrant in the plane, we define the “toric domain”

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

I’ll use the not entirely satisfactory term “convex toric domain” to indicate a domain $X_\Omega$ for which

$\Omega = \{(x,y)\mid 0\le x\le a,\; 0\le y \le f(x)\}$

where $f:[0,a]\to[0,\infty)$ is a nonincreasing concave function. Let’s now compute $c_k^{CH}(X_\Omega)$ where $X_\Omega$ is a convex toric domain.

As explained in [BEYOND], the boundary $Y$ of $X_\Omega$ can pe perturbed so that the contact form is nondegenerate and, up to large symplectic action, the simple Reeb orbits consist of the following:

• Elliptic orbits $e_{1,0}$ and $e_{0,1}$. Here $e_{1,0}$ is the circle in $Y$ where $z_1=0$, and $e_{0,1}$ is the circle in $Y$ where $z_2=0$.
• An elliptic orbit $e_{a,b}$, and a hyperbolic orbit $h_{a,b}$, for each pair $(a,b)$ of relatively prime positive integers. These arise from points on the boundary of $\Omega$ where the slope of a tangent line to $\partial\Omega$ is $-b/a$.

If $(a,b)$ are nonnegative integers (not both zero), let $d$ denote their greatest common divisor, and let $a'=a/d$ and $b'=b/d$. Let $e_{a,b}$ denote the $d$-fold cover of $e_{a',b'}$, and let $h_{a,b}$ denote the $d$-fold cover of $h_{a',b'}$ (when $a$ and $b$ are both nonzero). The generators of the cylindrical contact homology then consist of the following:

• $e_{a,b}$ and $h_{a,b}$ where $a,b$ are positive integers.
• $e_{d,0}$ and $e_{0,d}$ where $d$ is a positive integer.

It follows from calculations in [BEYOND] that the gradings of these generators are given by

$|e_{a,b}| = 2(a+b),$

$|h_{a,b}|=2(a+b)-1.$

Based on ECH calculations, I think the following should not be too hard to prove:

Conjecture. For a suitable generic $J$, the differential on the cylindrical contact homology chain complex is given by

$\partial e_{a,b}=0,$

$\partial h_{a,b} = d(\pm e_{a-1,b} \pm e_{a,b-1})$

where $d$ denotes the greatest common divisor of $a$ and $b$.

If you believe this, then it follows that the degree $2k$ homology generator is represented by $e_{a,b}$ with $a+b=k$, and these are all homologous. Thus $c_k^{CH}(X_\Omega)$ is the minimum of the symplectic action of $e_{a,b}$ where $a+b=k$.

What is this symplectic action? The calculations in [BEYOND] show that, up to some small error coming from the perturbation, the symplectic action of $e_{a,b}$ is given by

$A_\Omega(e_{a,b}) = (a,-b)\times p_{\Omega,-b/a}$

where $p_{\Omega,-b/a}$ denotes a point on $\partial\Omega$ where a tangent line to $\partial\Omega$ has slope $-b/a$, and $\times$ denotes the cross product. An equivalent way to say this, which is a bit more convenient for the present calculation, is

$A_\Omega(e_{a,b}) = \max\{ bx+ay \mid (x,y)\in\Omega\}.$

We conclude that

$c_k^{CH}(X_\Omega) = \min_{a+b=k}\max\{bx+ay\mid (x,y)\in\Omega\}$

where the minimum is over nonnegative integers $a,b$.

3. Examples of CH capacities

To become more comfortable with the above formula, let us compute some examples of CH capacities and check that they agree with the known formulas for ECH capacities.

First suppose that $\Omega$ is the rectangle with vertices $(0,0), (c,0), (0,d), (c,d)$ so that $X_\Omega$ is the polydisk $P(c,d)$. Then

$c_k^{CH}(P(c,d)) = \min_{a+b=k}(bc+ad) = k\max(c,d).$

This agrees with Ekeland-Hofer.

Next suppose that $\Omega$ is the triangle with vertices $(0,0), (c,0), (0,c)$ so that $X_\Omega$ is the ball $B(c)$. Then

$c_k^{CH}(B(c)) = \min_{a+b=k}\max(ac,bc) = c\lceil k/2 \rceil$

which also agrees with Ekeland-Hofer. Finally, one can generalize this to compute $c_k^{CH}$ of an ellipsoid and check that it agrees with Ekeland-Hofer. (The Ekeland-Hofer capacities of the ellipsoid $E(c,d)$ consist of the positive integer multiples of $c$ and $d$, arranged in nondecreasing order.) But I’ll skip this since it is an unnecessarily complicated way to compute the CH capacities of an ellipsoid. (It is much easier to just take a standard irrational ellipsoid with exactly two simple Reeb orbits.)

4. Comparison with the Lagrangian capacity

Let $t_0$ denote the largest positive real number $t$ such that $(t,t)\in\Omega$. I claim that

$c_L(X_\Omega)\ge t_0$

and

$\lim_{k\to\infty}\frac{1}{k}c_k^{CH}(X_\Omega)=t_0$.

The first claim is easy, because if $(t,t)\in\Omega$, then the torus $T=(\pi|z_1|^2 = \pi|z_2|^2=t)$ is a Lagrangian torus in $X_\Omega$ such that $A_{min}(T)=t$.

To prove the second claim, note that by Part 2, we have

$\lim_{k\to\infty}\frac{1}{k} c_k^{CH}(X_\Omega) = \min_{a+b=1}\max\{bx+ay\mid (x,y)\in\Omega\}.$

Here $a,b$ are now nonnegative real numbers instead of integers.

If $a+b=1$, then taking $(x,y)=(t_0,t_0)$ shows that $\max\{bx+ay\mid (x,y)\in\Omega\}\ge t_0$, and thus

$\lim_{k\to\infty}\frac{1}{k}c_k^{CH}(X_\Omega) \ge t_0$.

To prove the reverse inequality, consider a tangent line to $\partial\Omega$ through the point $(t_0,t_0)$. We can uniquely write this line in the form $bx+ay=L$ where $a+b=1$. Since this line is tangent to $\partial\Omega$, we have $\max\{bx+ay\mid(x,y)\in\Omega\}= L$, and thus

$\lim_{k\to\infty}\frac{1}{k}c_k^{CH}(X_\Omega)\le L$.

On the other hand, since the line goes through the point $(t_0,t_0)$, we have

$L = bt_0 + at_0 = t_0$.

It follows that

$\lim_{k\to\infty}\frac{1}{k}c_k^{CH}(X_\Omega) \le t_0$.

This completes the proof of the claims. We conclude that

$\lim_{k\to\infty}\frac{1}{k}c_k^{CH}(X_\Omega) \le c_L(X_\Omega)$.

5. How to prove the reverse inequality?

Now we would like to prove the reverse inequality

$\lim_{k\to\infty}\frac{1}{k}c_k^{CH}(X) \ge c_L(X)$.

where $X=X_\Omega$ (and here it is maybe not so important that $X$ is a convex toric domain). To do so, let $T\subset X$ be an embedded Lagrangian torus. We want to prove that there exists $D\in\pi_2(X,T)$ such that

$0 < \int_D\omega \le \lim_{k\to\infty}\frac{1}{k}c_k^{CH}(X)$.

(Actually, in this case, since $\pi_2(X,T)=H_2(X,T)$, we could allow $D$ to be any (not necessarily embedded) compact oriented surface in $X$ with boundary on $T$.)
I haven’t thought this through, but maybe one prove this using the methods in [HL]. Or maybe these methods will just prove the following weaker upper bound?

Namely, [CM,Cor. 1.3] and monotonicity of the Lagrangian capacity imply the upper bound

$c_L(X_\Omega) \le \frac{1}{2}\max\{x+y\mid (x,y)\in\Omega)\}$.

This agrees with our trivial lower bound $c_L(X_\Omega)\le t_0$ if and only if a tangent line to $\partial\Omega$ through $(t_0,t_0)$ has slope $-1$; or equivalently, $P(c,c)\subset X_\Omega \subset B(2c)$ for some $c$.

Posted in Contact homology | 1 Comment

## Version 2 of “Beyond ECH capacities”

I just posted a revised version of the preprint “Beyond ECH capacities” to the arXiv. It should appear on Monday, but you can view it here first. The new version corrects some embarrassing/horrifying typos, clarifies a few things, and includes the new application to sharpness of symplectic folding from the previous blog post.

There are still tons of calculations to do using the ideas in this preprint, if anyone is interested. Here are some examples of things to try:

• Study symplectic embeddings of the polydisk $P(a,1)$ into a ball when $a>12/5$, improving Theorem 1.3 and/or extending Theorem 1.4 in the preprint.
• Study symplectic embeddings of the polydisk $P(a,1)$ into the ellipsoid $P(bc,c)$ when $b$ is an integer and $a>2$, or when $b$ is not an integer, extending Theorem 1.5 in the preprint.
• Prove Conjecture A.3 in the preprint (regarding the ECH differential on the boundary of a suitably perturbed convex toric domain), which would allow Theorem 1.6 to be improved as explained in Remark 1.8.
• Study symplectic embeddings of the disjoint union of two polydisks into an ellipsoid or polydisk.
• Study symplectic embeddings of concave or convex toric domains into a concave toric domain.

In general, one would like to identify more cases when the obvious inclusion map, or the folding-type constructions in Felix Schlenk’s book, give optimal symplectic embeddings.

## Symplectic folding is sometimes optimal

Reference: [BEYOND] = “Beyond ECH capacities”

I have played with a few more calculations using the methods in [BEYOND]. Here is the most interesting thing I have found so far.

In [BEYOND, Thm. 1.2], it was shown, among other things, that if the polydisk $P(a,1)$ symplectically embeds into the four-dimensional ball $B(c)$, and if $2 \le a \le 4$, then $c\ge (10+a)/4$. On the other hand, Felix Schlenk showed using symplectic folding that if $2\le a\le 4$, then $P(a,1)$ symplectically embeds into $B(c)$ whenever $c> (4+a)/2$. These bounds agree for $a=2$ and disagree for $a>2$. One might ask whether one can improve one or both of these bounds to get them to agree. In fact, it turns out that symplectic folding is sometimes optimal, in the following sense:

Theorem. If $2\le a\le 12/5$, and if $P(a,1)$ symplectically embeds into $B(c)$, then $c\ge (4+a)/2$.

This is a direct application of [BEYOND, Thm. 1.18], and I will assume the statement of the latter theorem below. (In other words, the following is basically an addendum to be added in the next version of [BEYOND], unless I discover some way to improve it first.)

Proof of Theorem. Suppose that $2\le a\le 12/5$, that $P(a,1)$ symplectically embeds into $B(c)$, and that $c < (4+a)/2$. We will obtain a contradiction in four steps. Below, the symbol $\le$ between convex generators means $\le_{P(a,1),B(c)}$.

Step 1. We first show that if $\Lambda \le e_{1,1}^d$ with $d\le 9$, then $y(\Lambda) \le 1$.

If $y(\Lambda)\ge 2$, then as in Step 1 of the proof of [BEYOND, Thm. 1.2], we have

$3d-3+2a \le dc$.

Combining this with our assumption that $c < (4+a)/2$ gives

$(d-4)a > 2d-6$.

If $d<4$ then it follows that $a < 4/3$; if $d=4$ then it follows that $2<0$; and if $5\le d\le 9$ then it follows that $a>12/5$. Either way this contradicts our hypothesis that $2\le a \le 12/5$.

Step 2. We now show that if $\Lambda\le e_{d,d}$, and if $y(\Lambda)\le 1$, then $\Lambda$ includes a factor of $e_{1,0}$.

If not, then the only possibility for $\Lambda$ with the correct ECH index is

$\Lambda = e_{(d^2+3d-2)/2,1}$.

The action inequality in the definition of $\le_{P(a,1),B(c)}$ then implies that

$(d^2+3d-2)/2 + a \le dc.$

Combining this with our assumption that $c < (4+a)/2$ gives

$(d-2)a > d^2-d-2$.

If $d=1$ then it follows that $a<2$; if $d=2$ then it follows that $0<0$; and if $d\ge 3$ then it follows that $a > d+1$. Either way this contradicts our hypothesis that $2 \le a \le 12/5$.

Step 3. We now show that there does not exist any convex generator $\Lambda$ with $\Lambda \le e_{1,1}^9$.

If $\Lambda$ is such a generator, then we know from Step 1 that $y(\Lambda) \le 1$.

If $y(\Lambda) = 0$, then the only possibility for $\Lambda$ with the correct ECH index is $\Lambda = e_{1,0}^{54}$. Then $54 \le 9c$, which combined with our assumption that $c < (4+a)/2$ implies that $a > 8$, contradicting our hypotheses.

If $y(\Lambda) = 1$, then $x(\Lambda) \ge 27$, or else we would have $I(\Lambda) \le 106$, contradicting the fact that $I(\Lambda) = 108$. Since $x(\Lambda) \ge 27$, it follows that

$27 + a \le 9c$.

Combining this with our assumption that $c < (4+a)/2$ gives $a > 18/7$, contradicting our hypothesis that $a \le 12/5$.

Step 4. We now apply [BEYOND, Thm. 1.18] to $\Lambda'=e_{1,1}^9$, to obtain a convex generator $\Lambda$, and factorizations $\Lambda=\Lambda_1\cdots\Lambda_n$ and $\Lambda'=\Lambda_1'\cdots\Lambda_n'$, satisfying the three bullet points in [BEYOND, Thm. 1.18].

By Step 3 and the first bullet point, we must have $n>1$.

By Step 2 and the first two bullet points, all of the $\Lambda_i$ must be equal, and all of the $\Lambda_i'$ must be equal. Thus either $n=9$ and $\Lambda_i'=e_{1,1}$ for each $i$, or $n=3$ and $\Lambda_i'=e_{3,3}$ for each $i$.

If $n=9$, then by Steps 1 and 2, we have $\Lambda=e_{1,0}^2$ for each $i$. But then $I(\Lambda)=36$, contradicting the fact that $I(\Lambda)=108$.

If $n=3$, then by Step 1, and the facts that $I(\Lambda_i)=18$ and $x(\Lambda_i) + y(\Lambda_i) \ge 8$, the only possibilities are that $\Lambda_i = e_{1,0}^9$ for each $i$, or $\Lambda_i = e_{1,0}e_{6,1}$ for each $i$. In the former case we have $I(\Lambda) = 54$, and in the latter case we have $I(\Lambda)=102$. Either way, this contradicts the fact that $I(\Lambda)=108$.

QED

Remark. It is conceivable that with more work, the hypothesis $a\le 12/5$ could be weakened to $a \le (\sqrt{7}-1)/(\sqrt{7}-2) = 2.54858\cdots$. The significance of the latter number is that if $a$ is less than it, and if $d$ is sufficiently large with respect to $a$, then there does not exist any convex generator $\Lambda$ with $\Lambda \le e_{1,1}^d$. We might then be able to use arguments similar to the above to get a contradiction.

More generally, one can maybe get more information by considering all of the holomorphic curves that exist in the cobordism coming from a symplectic embedding. We know that certain curves must exist in order to give a chain map on ECH satisfying the required properties. However the existence of certain curves excludes the existence of others, for example when their intersection number would be negative. When you write down all of these conditions, it is like a giant logic puzzle, and the challenge is to extract significant information from it using a manageable amount of computation.

## Comparison with Ekeland-Hofer

Helmut Hofer asked me how the symplectic embedding obstructions in “Beyond ECH capacities” compare to the obstructions given by Ekeland-Hofer capacities, for symplectic embeddings between ellipsoids and polydisks. I happened to know the answer to this, using something which I chose to leave out of the paper (too many things going on, needed to try to stay focused!), so I would like to explain it here. First, the answer is the following:

• For symplectic embeddings of (four-dimensional) ellipsoids into ellipsoids or polydisks, ECH capacities give sharp obstructions (shown by McDuff), while Ekeland-Hofer capacities are often weaker.
• For a polydisk $P(a,b)$ into another polydisk $P(a',b')$ where $a\ge b$ and $a'\ge b'$, Ekeland-Hofer only tells us that $b\le b'$. ECH capacities say a bit more but are not very good. “Beyond ECH capacities” gives sharp obstructions in some cases.
• For a polydisk $P(a,b)$ into an ellipsoid $E(c,d)$, ECH capacities are not very good, and sometimes weaker than Ekeland-Hofer capacities. Ekeland-Hofer gives a sharp obstruction when $a=b$ (and a less good obstruction when $a\neq b$). “Beyond ECH capacities” can recover this (the sharp obstruction when $a=b$; I haven’t checked that I can recover all information given by all Ekeland-Hofer capacities when $a\neq b$).

I would now like to explain this last point. First let’s change the notation: given $a,b\ge 1$, we would like to find the infimum of $c$ such that the polydisk $P(a,1)$ symplectically embeds into the ellipsoid $E(bc,c)$. Observe that $P(a,1)$ trivially embeds into $E(bc,c)$ by inclusion if $c\ge 1+a/b$. When $a=1$, the converse is true:

Theorem 1. If $b\ge 1$ and $P(1,1)$ symplectically embeds into $E(bc,c)$, then $c\ge 1+1/b$.

Here’s how to prove this using Ekeland-Hofer capacities. Let’s denote the $k^{th}$ Ekeland-Hofer capacity by $c_k$. (Usually I use this symbol to denote ECH capacities, but we won’t be talking about ECH capacities in this post.) The Ekeland-Hofer capacities of $P(a,1)$ for $a\ge 1$ are given by

$c_k(P(a,1)) = k$.

On the other hand, $c_k(E(bc,c))$ is the $k^{th}$ entry in the list of all positive integer multiples of $bc$ or $c$, written in increasing order with repetitions. It follows that

$c_{k + \lfloor k/b \rfloor}(E(bc,c)) = kc.$

So if $P(1,1)$ symplectically embeds into $E(bc,c)$, then by the monotonicity property of Ekeland-Hofer capacities, for every positive integer $k$ we have

$k + \lfloor k/b \rfloor \le kc$.

Taking $k$ large gives $1 + 1/b \le c$ as desired. Now how do we recover this from “Beyond ECH capacities”? Doing something rather crude with the methods in that paper, which I will explain below, one obtains the following:

Theorem 2. Suppose $P(a,1)$ symplectically embeds into $P(bc,c)$ where $a,b\ge 1$. Suppose also that

$\sqrt{a/2} + \sqrt{1/(2a)} \le \sqrt{b} + 1/\sqrt{b}$.

Then

$2c \ge 1 + b^{-1} + \sqrt{1 + b^{-2}} + a(1 + b^{-1} - \sqrt{1 + b^{-2}}).$

For example, if $a=1$, then we obtain $c \ge 1 + b^{-1}$, recovering Theorem 1.  Another example is that if $b=1$, then we obtain

$c \ge (1 + 1/\sqrt{2}) + (1-1/\sqrt{2})a$

for $a\le 3+2\sqrt{2}$. This is nontrivial, but weaker than Theorem 1.2 in “Beyond ECH capacities” when $a>1$.

Now I will explain the proof of Theorem 2, assuming as a prerequisite the statement of Theorem 1.18 in “Beyond ECH capacities”. We will need the following immediate corollary of the latter theorem:

Theorem 3. Let $X_\Omega$ and $X_{\Omega'}$ be convex toric tomains, and suppose that $X_\Omega$ symplectically embeds into $X_{\Omega'}$. Let $\Lambda'$ be a convex generator which is minimal for $X_{\Omega'}$. Then there exists a convex generator $\Lambda$ such that

$I(\Lambda) = I(\Lambda')$,

$A_{\Omega}(\Lambda) \le A_{\Omega'}(\Lambda')$,

and

$x(\Lambda) + y(\Lambda) \ge x(\Lambda') + y(\Lambda')$.

(To address a question of Dan C-G: You can also say that $I - x - y$ defines a filtration on the ECH chain complex which is preserved by the cobordism map. However I’m not sure if the cobordism map necessarily induces an isomorphism on the homology of the associated graded, because I don’t know if this filtration will be preserved by the relevant chain homotopies. There are also variants of this filtration to play with coming from $J_0$ and $J_+$. An interesting topic to think about later.)

Here is how to deduce Theorem 2 from Theorem 3. Let $\Lambda'$ be a convex generator which is minimal for $E(bc,c)$ and has very large ECH index. Let us rescale this so that the area under the curve is $b/2$. Then the rescaled $\Lambda'$ is approximately a straight line from $(0,1)$ to $(b,0)$. Now the convex generator $\Lambda$ provided by Theorem 2, after rescaling by the same factor, is a curve from $(0,y)$ to $(x,0)$, for some positive real numbers $x$ and $y$, which is the graph of a nonincreasing concave function, and the area under this curve is approximately $b/2$. It follows that

$xy/2 \le b/2 \le xy$,

up to some error which can be made arbitrarily small by taking the ECH index of $\Lambda$ to be sufficiently large. The other two inequalities in Theorem 2 then tell us that, also up to a small error, we have

$x+ay \le bc$

and

$x+y \ge b+1$.

It follows that

$bc\ge \min\{x+ay | xy/2\le b/2 \le xy, x+y\ge b+1\}$.

It is now an exercise in undergraduate multivariable calculus to compute the minimum on the right hand side. If

$\sqrt{a/2} + \sqrt{1/(2a)} \ge \sqrt{b} + 1/\sqrt{b}$,

then the minimum is $\sqrt{2ab}$. Thus we conclude that $c\ge \sqrt{2a/b}$. This is just the volume constraint $vol(P(a,1)) \ge vol(E(bc,c))$. On the other hand, if

$\sqrt{a/2} + 1/\sqrt{1/(2a)} \ge \sqrt{b} + 1/\sqrt{b}$,

then the minimum is $(b+1+\sqrt{b^2+1})/2 + a(b+1-\sqrt{b^2+1})/2$, which proves Theorem 3.

One last remark: the asymptotics of the symplectic embedding obstructions coming from ECH capacities for large ECH index just recover the volume constraint. The example above shows that the asymptotics of the obstruction in Theorem 2 for large ECH index (and also the Ekeland-Hofer capacities for large $k$) sometimes say more.

## Beyond ECH capacities

In case the last couple of postings were confusing, I have now posted a preprint explaining the story. Here is the abstract:

“ECH (embedded contact homology) capacities give obstructions to symplectically embedding one four-dimensional symplectic manifold with boundary into another. These obstructions are known to be sharp when the domain and target are ellipsoids (proved by McDuff), and more generally when the domain is a “concave toric domain” and the target is a “convex toric domain” (proved by Cristofaro-Gardiner). However ECH capacities often do not give sharp obstructions, for example in many cases when the domain is a polydisk. This paper uses more refined information from ECH to give stronger symplectic embedding obstructions when the domain is a polydisk, or more generally a convex toric domain. We use these new obstructions to reprove a result of Hind-Lisi on symplectic embeddings of a polydisk into a ball, and generalize this to obstruct some symplectic embeddings of a polydisk into an ellipsoid. We also obtain a new obstruction to symplectically embedding one polydisk into another, in particular proving the four-dimensional case of a conjecture of Schlenk.”

There are lots more calculations to do, to try to use the techniques in this paper to obstruct more symplectic embeddings. If anyone is interested in working on some, please feel free to discuss this with me.

## Symplectic embeddings of polydisks into polydisks

If I am not mistaken, the methods in the previous post (plus a conjecture about the ECH chain complex of perturbed boundaries of convex toric domains) can be used to show that if $a,b,c$ are real numbers with $a,b\ge 1$ and $c>0$, and if $P(a,1)$ symplectically embeds into $P(bc,c)$, and if $a\le 2b$, then $a\le bc$. In other words, if you include one four-dimensional polydisk into another, such that the long sides are the same, and the short side of the domain is at least half the short side of the target, then this symplectic embedding is optimal.

The conjecture needed is that in the ECH chain complex of the (perturbed) boundary of a convex toric domain, a generator with only elliptic orbits represents a nontrivial homology class. (This would follow from a conjectural description of the differential in terms of rounding corners.) Without this conjecture, one can still prove a version of the above theorem in which the hypothesis $a\le 2b$ is strengthened somewhat. (When $b=1$ you can still just assume $a\le 2$.)

I’m working on writing this up cleanly.

## Hind-Lisi and more

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 $T^3$

[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 $P(2,1)$ symplectically embeds into the ball $B(b)$ then $b\ge 3$. (I’ll review what this notation means below.) This bound is optimal, because the polydisk $P(a,1)$ symplectically embeds into the ball $B(1+a)$ by inclusion. Also, the $2$ is as large as possible for this sort of result, because whenever $a>2$, according to Schlenk’s book, one can use “symplectic folding” to symplectically embed $P(a,1)$ into $B(b)$ for certain $b < 1+a$.

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 $a\in[1,8]$ and suppose that $P(a,1)$ symplectically embeds into $B(b)$. Then:

• If $1\le a\le 2$ then $b\ge 1+a$.
• If $2\le a\le 4$ then $b\ge (10+a)/4$.
• If $4\le a \le 9/2$ then $b\ge 7/2$.
• If $9/2 \le a \le 7$ then $b\ge (13+a)/5$.
• If $7\le a\le 8$ then $b\ge 4$.

Of course the third and fifth lines follow trivially from the lines above them. I have written the fifth line because the inequality $b\ge 4$ is significant until $a=8$, at which point it ties the volume constraint $b\ge\sqrt{2a}$.

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

Theorem 2. Let $a\in[1,2]$ and let $b$ be a positive integer. Then $P(a,1)$ symplectically embeds into $E(bc,c)$ if and only if $bc \le a + b$.

Note that $P(a,1)$ includes into $E(bc,c)$ whenever $bc \ge a + b$. Thus Theorem 2 asserts that the inclusion is sharp when $b$ is an integer and $1\le a\le 2$. It is unclear why integrality of $b$ 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 $\Omega$ is a domain in the (closed) first quadrant of the plane, we define the “toric domain”

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

Let us now define a “convex toric domain” to be a domain $X_\Omega$ where $\Omega=\{(x,y)\mid 0\le x \le A, 0 \le y \le f(x)\}$ where $f$ is a convex function such that $f(0)=A$ and $f'(0)\le 0$. (Sorry for the confusing terminology, but the region below the graph of a concave function is convex!) For example, the polydisk

$P(a,b) = \{z\in{\mathbb C}^2\mid \pi|z_1|^2\le a, \pi|z_2|^2\le b\}$

is a convex toric domain where $A=a$ and $f\equiv b$ so that $\Omega$ is a rectangle. Also, the ellipsoid

$E(a,b) = \{z\in{\mathbb C}^2\mid \pi|z_1|^2/a + \pi|z_2|^2/b\}$

is a convex toric domain where $A=a$ and $f$ is linear so that $\Omega$ is a triangle.

Convex toric domains should be contrasted with the “concave toric domains” considered in [CONCAVE]. For a concave toric domain, $f$ is convex and $f(A)=0$. 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 $P(2,1)$ symplectically embeds into $B(b)=E(b,b)$ then $b\ge 2$, although we actually want $b\ge 3$.

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 $J_0$ (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 $X_\Omega$ be a convex toric domain and let $Y$ denote its boundary. If $Y$ is smooth, then it has a natural contact form, given by the restriction of

$\lambda=\frac{1}{2}\sum_{i=1}^2(x_idy_i - y_idx_i).$

The claim is now that one can perturb $X_\Omega$ so that $Y$ is smooth, $\lambda$ 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 $\Lambda$ in the plane such that:

• $\Lambda$ starts at $(0,y)$ and ends at $(x,0)$ where $x$ and $y$ are nonnegative integers.
• The vertices of $\Lambda$ (the points where it changes direction) are at lattice points.
• The edges of $\Lambda$ (i.e. line segments between vertices) have nonpositive slope (possibly $-\infty$), and as one moves to the right, the slope of each edge is less than that of the preceding edge.
• Each edge of $\Lambda$ is labeled e’ or h’.
• Horizontal and vertical edges can only be labeled e’.

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

$I(\Lambda) = 2(L(\Lambda)-1) - h(\Lambda)$

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

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

$A_\Omega(\Lambda) = \sum_{e\in Edges(\Lambda)}v_e\times p_e.$

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

The more precise statement is that for any $L>0$ large and $\epsilon>0$ small, one can perturb $Y$ by a perturbation of size less than $\epsilon$ in an appropriate sense, such that there is a bijection between ECH generators of (actual) action less than $L$ with convex generators of (combinatorial) action less than $L$, such that the actual ECH index agrees with the combinatorial ECH index, and the action agrees with the combinatorial action up to $\epsilon$ 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 $\Omega$ 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 ${\mathbb Z}/2$ 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 $\Lambda$ is any path in which all edges are labeled e’, then $\Lambda$ is a cycle which represents the homology generator of grading $I(\Lambda)$. 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 $\Omega$ is the triangle with vertices $(0,0)$, $(a,0)$, and $(0,b)$, so that $X_\Omega = E(a,b)$. Let $P$ be a line in the plane of slope $-b/a$ which passes through at least one lattice point in the first quadrant. Let $\Lambda$ be the maximal convex generator to the left of $P$, with all edges labeled e’. Then $\Lambda$ is a cycle which represents a nontrivial homology class in the ECH of (perturbed) $\partial E(a,b)$.

Proof. Let $k=I(\Lambda)/2$. One can check that $\Lambda$ uniquely minimizes the combinatorial action among generators of index $2k$, and its combinatorial action agrees with the kth ECH capacity of $E(a,b)$. This is similar to Example 1.23 in [CONCAVE]. The claim follows immediately (because the degree $2k$ 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 $\Lambda$ is the only such generator.)

5. The cobordism map

Suppose now that a convex toric domain $X_\Omega$ with boundary $Y$ symplectically embeds into (the interior of) another convex toric domain $X_{\Omega'}$ with boundary $Y'$. Then $X_\Omega$ minus the interior of the image of $X_{\Omega'}$ is a strong symplectic cobordism from $Y'$ to $Y$. We further get a strong symplectic cobordism between the perturbed contact forms as above. This induces a map $ECH(Y')\to ECH(Y)$, which is an isomorphism, because the cobordism is diffeomorphic to the product $[0,1]\times S^3$. We know from my work with Taubes [CC2] using Seiberg-Witten theory that given an appropriate almost complex structure $J$ 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) $J$-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 $L$) can be chosen so that it only counts $I=0$ unbroken $J$-holomophic currents, such that the components are disjoint and have $I=0$, and the somewhere injective curve underlying each component is embedded and also has $I=0$.

To prove this, one needs to do a calculation, using the special nature of the contact forms in question, to show that if $J$ 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 $J$-holomorphic currents with ECH index index $I=0$. To do so, we use the quantity $J_0$ (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, $J_0$ can be computed combinatorially as follows. If $\Lambda$ is a convex generator, going from $(0,y)$ to $(x,0)$, then

$J_0(\Lambda) = I(\Lambda) - 2x - 2y - e(\Lambda).$

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

Now the significance of $J_0$ is that if $C$ is an $I=0$ holomophic curve from $\Lambda'$ to $\Lambda$ which does not have any multiply covered components, then

$J_0(\Lambda') - J_0(\Lambda) = -\chi(C) + O(C).$

Here $O(C)$ is the sum, over all Reeb orbits in $\Lambda$ or $\Lambda'$, of the number of ends of $C$ 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 $C$ are at simple orbits, and $C$ can have at most one negative end at covers of any given orbit. To make use of this fact, let $n(\Lambda)$ denote the total number of Reeb orbits in $\Lambda$, namely the number of edges, plus the number of h’ edges with at least one interior lattice point. Let $m(\Lambda)$ denote the total multiplicity of all Reeb orbits in $\Lambda$, namely the sum over all edges of the number of interior lattice points minus one. Then

$O(C) = m(\Lambda') - n(\Lambda').$

Also, if $C$ is connected with genus $g$, then

$\chi(C) = 2 - 2g - m(\Lambda') - n(\Lambda)$.

Putting the above together, we get

$2x(\Lambda) + 2y(\Lambda) = 2g - 2 + 2x(\Lambda') + 2y(\Lambda') + 2m(\Lambda') + h(\Lambda) - h(\Lambda').$

Since $g\ge 0$ and $h(\Lambda)\ge 0$, we conclude the following:

Lemma 6. Suppose there exists an $I=0$, connected, embedded curve from $\Lambda'$ to $\Lambda$. Suppose that all edges of $\Lambda'$ are labeled e’. Then

$x(\Lambda) + y(\Lambda) \ge x(\Lambda') + y(\Lambda') + m(\Lambda') - 1.$

(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 $P(a,1)$ symplectically embeds into $B(b)$, and if $1\le a \le 2$, then $b\ge 1 + a$.

Assume that $P(a,1)$ symplectically embeds into $B(b)$, assume that $a\ge 1$, and assume that $b < 1+a$. We will show that $b\ge 3$.

Step 1. Consider the convex generator $\Lambda'$ consisting of the straight line from $(0,1)$ to $(1,0)$, labeled e’. Then $I(\Lambda') = 4$ and $A_{B(b)}(\Lambda') = b$. By Lemmas 4 and 5, there is an $I=0$ current from $\Lambda'$ to some convex generator $\Lambda$ with $I(\Lambda)=4$ and $A_{P(a,1)}(\Lambda)\le b$.

There are in fact only three convex generators $\Lambda$ with $I(\Lambda)=4$: the horizontal line from $(0,0)$ to $(2,0)$, labeled e’, which has action $2$; the line from $(0,1)$ to $(1,0)$, labeled e’, which has action $1+a$; and the vertical line from $(0,2)$ to $(0,0)$, labeled e’, which has action $2a$. If there is a holomorphic curve from $\Lambda'$ to either of the latter two generators, then we immediately get $b\ge 1+a$ or $b\ge 2a$, which contradicts our assumption that $b< 1+a$. Thus $I=0$ curves from $\Lambda'$ can only go to the horizontal line.

Step 2. Consider the convex generator $\Lambda'$ consisting of the straight line from $(0,2)$ to $(2,0)$, labeled e’. Then $I(\Lambda')=10$ and $A_{B(b)}(\Lambda')=2b$. By Lemmas 4 and 5, there is an $I=0$ current $C$ from $\Lambda'$ to some convex generator $\Lambda$ with $I(\Lambda)=10$ and $A_{P(a,1)}(\Lambda)\le 2b$.

Now the current $C$ must actually be a connected, embedded holomorphic curve. Why? Well, if $C$ were disconnected, then each component would have $I=0$, and so by the end of Step 1, each component can only go to the horizontal line of length 2. Hence $\Lambda$ is a horizontal line of length 4. But this only has $I=8$ which is not big enough. Likewise, if $C$ is multiply covered, then again by the end of Step 1, the embedded curve underlying $C$ can only map to the horizontal line of length 2, so again $\Lambda$ is the horizontal line of length 4 which is impossible.

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

$x(\Lambda) + y(\Lambda) \ge 5$.

Now the action of $\Lambda$ is given by

$A_{P(a,1)}(\Lambda) = x(\Lambda) + a y(\Lambda).$

If $y(\Lambda)>0$, then $A_{P(a,1)}(\Lambda)\ge 4+a$, so $4+a\le 2b$. Combining this with our assumption $b < 1+a$ then gives $b > 3$.

Assume now that $2b < 4+a$. Then $\Lambda'$ can only be the horizontal line of length 5, labeled e’. In particular this gives $b\ge 5/2$. We need one more step to obtain $b\ge 3$.

Step 3. We are now assuming $b<1+a$ and $2b<4+a$. Consider the convex generator $\Lambda'$ given by the straight line from $(0,3)$ to $(3,0)$, labeled e’. This has $I=18$ and $A_{B(b)}=3b$. By Lemmas 4 and 5, there is a holomorphic current $C$ from $\Lambda'$ to a convex generator $\Lambda$ with $I(\Lambda) = 18$ and $A_{P(a,1)}(\Lambda) \le 3b$.

By Steps 1 and 2, $C$ is connected and embedded. (Otherwise $\Lambda$ is a horizontal line of length at most $7$, which only has $I=14$, which is too small.) Thus Lemma 6 applies to give

$x(\Lambda) + y(\Lambda) \ge 8$.

If $y(\Lambda)>0$, then $A_{P(a,1)}(\Lambda)\ge 7+a$, so $7+a\le 3b$. Combining this with our assumption that $b<1+a$ then gives $b>3$.

If $y(\Lambda)=0$, then $\Lambda$ must be a horizontal line of length $9$, since its ECH index is $18$. Thus $A_{P(a,1)}(\Lambda)=9$, so $b\ge 3$.

QED

Which holomorphic curve is giving the obstruction? If you analyze the above proof, what it is saying is that if $a\ge 1$ and if $P(a,1)$ symplectically embeds into $B(b)$, then at least one of the following holomorphic curves must exist:

• A pair of pants from the “diagonal” Reeb orbit of action $b$ to the “horizontal” and “vertical” simple orbits for the polydisk, implying that $b\ge 1+a$.
• A cylinder from $\Lambda'$ as in Step 1 to the vertical line of length 2, implying that $b\ge 2a$.
• A holomorphic curve with two positive ends at the “diagonal” orbit of action $b$, and negative ends with total action at least $4+a$, implying that $2b\ge 4+a$.
• A holomorphic curve with three positive ends at the “diagonal” orbit of action $b$, and either one negative end with action $9$, or with negative ends of total action at least $7+a$.

Either way, we obtain $b\ge 1+a$ if $a\le 2$. In any case, this looks quite different from the Hind-Lisi argument, which studies curves of “degree” $d$ and takes the limit as $d\to\infty$.

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 $a\ge 2$, $b<1+a$, $2b<4+a$, and $3b<7+a$. (We can make this last assumption without loss of generality because $(7+a)/3 > (10+a)/4$ when $a\ge 2$.) Let $\Lambda'$ be the straight line from $(0,4)$ to $(4,0)$, labeled e’. This has $I=28$ and $A_{B(b)}=4b$. Then there is a holomorphic current $C$ from $\Lambda'$ to a convex generator $\Lambda$ with $I(\Lambda)=28$ and $A_{P(a,1)}(\Lambda) \le 4b$. The above assumptions imply that $C$ is connected and embedded, so we can apply Lemma 6 to deduce that $x(\Lambda) + y(\Lambda) \ge 11$. Thus

$4b\ge \min(10+a,14).$

This gives the lower bound on $b$ in Theorem 1 for $2\le a \le 4$.

Step 5. Now assume $a\ge 4$, $b<1+a$, $2b<4+a$, $3b<7+a$, and $4b<10+a$. Let $\Lambda'$ be the straight line from $(0,5)$ to $(5,0)$. Then we similarly obtain

$5b\ge \min(13+a,20).$

This gives the lower bound on $b$ in Theorem 1 for $9/2\le a \le 7$.

QED.

9. Proof of Theorem 2.

We now prove that if $1\le a\le 2$ and $b$ is a positive integer, and if $P(a,1)$ symplectically embeds into $E(bc,c)$, then $bc \le a + b$.

Theorem 1 covered the case where $b=1$.

To prove the rest, suppose that $b>1$ is an integer and that $P(a,1)$ symplectically embeds into $E(bc,c)$. Assume that $a\ge 1$ and $a+b>bc$. We need to show that $a>2$. This is actually a little simpler than the proof of Theorem 1 and only requires two steps.

Step 1. Let $\Lambda'$ be the convex generator given by the straight line from $(0,1)$ to $(b,0)$, labeled e’. Then $I(\Lambda') = 2b$ and $A_{E(bc,c)}(\Lambda') = bc$. By Lemmas 4 and 5, there is a holomorphic current $C$ from $\Lambda'$ to a convex generator $\Lambda$ with $I(\Lambda) = 2b$ and $A_{P(a,1)}(\Lambda) \le bc$. Now $C$ must be connected and embedded, since $\Lambda'$ consists only of a single, simple Reeb orbit. Thus Lemma 6 implies that $x(\Lambda) + y(\Lambda) \ge b+1$. If $y(\Lambda)>0$ then $A_{P(a,1)}(\Lambda) = x(\Lambda) + ay(\Lambda) \ge a+b$, contradicting our assumption that $a+b>bc$. The conclusion is that $\Lambda$ can only be the horizontal line of length $b+1$, labeled e’.

Step 2. Now let $\Lambda'$ be the convex generator given by the straight line from $(0,2)$ to $(2b,0)$, labeled `e’. Then $I(\Lambda')= 6b+4$ and $A_{E(bc,c)}(\Lambda')=2bc$. By Lemmas 4 and 5, there is a holomorphic current $C$ from $\Lambda'$ to a convex generator $\Lambda$ with $I(\Lambda)= 6b+4$ and $A_{P(a,1)}(\Lambda)\le 2bc$. Now $C$ must be connected and embedded, since otherwise, by Step 1, $\Lambda$ is a horizontal line of length $2b+2$, which has index $4b+4<6b+4$. Thus Lemma 6 applies to give

$x(\Lambda) + y(\Lambda) \ge 2b+3$.

If $y(\Lambda)>0$ then $A_{P(a,1)}(\Lambda) = x(\Lambda) + a y(\Lambda) \ge 2b+2+a$. Thus $2b+2+a\le 2bc$. By assumption $2bc < 2a+2b$, so $2 < a$ and we are done.

If on the other hand $y(\Lambda)=0$, then $x(\Lambda)=3b+2$, so $3b+2\le 2bc$. By assumption $2bc<2a+2b$, so we get $b+2 < 2a$. Since we are assuming $b\ge 2$, it follows that $a>2$ so we are done again.

QED

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