Lecture notes on ECH 8: The ellipsoid example

In general, most examples of ECH are difficult to compute directly. However the ECH of an ellipsoid with a certain standard contact form can be computed in a page or two. So let us now do that, as this should help with understanding some of the definition of ECH.

Contact type hypersurfaces

Before describing the example, let me review some basics. Let (X^{2n},\omega) be a symplectic manifold. An oriented hypersurface Y in X is said to be of contact type if there exists a contact form \lambda on Y (in 2n-1 dimensions this means that \lambda\wedge(d\lambda)^{n-1}>0) such that d\lambda = \omega|_Y. If there is a Liouville vector field V defined in a neighborhood of Y (the Liouville condition means that {\mathcal L}_V\omega=\omega) such that V is transverse to Y, then Y is of contact type, and we can take \lambda=(\imath_V\omega)|_Y. If Y is compact then the converse is true, i.e. if Y is of contact type then there exists a transverse Liouville vector field. (The proof is a bit trickier but we will not need this.) If Y is of contact type, then up to scaling by positive functions, the Reeb vector field does not depend on \lambda, and is given by the kernel of \omega|_Y.

Now let X={\mathbb R}^{2n} with the standard symplectic form \omega=\sum_{i=1}^ndx_idy_i. Then the radial vector field \rho=\frac{1}{2}\sum_{i=1}^n(x_i\partial_{x_i}+y_i\partial_{y_i}) is  Liouville vector field. Call a hypersurface Y\subset{\mathbb R}^{2n} star-shaped if it is transverse to the radial vector field \rho. Then by the above discusion, any star-shaped hypersurface is of contact type, and a corresponding contact form is given by the restriction of \imath_\rho\omega = \frac{1}{2}\sum_{i=1}^n(x_idy_y-y_idx_i).

Ellipsoids

Given positive real numbers a,b>0, define the ellipsoid

E(a,b) = \{(z_1,z_2)\in {\mathbb C}^2 \mid \frac{\pi|z_1|^2}{a} + \frac{\pi|z_2|^2}{b} \le 1\}.

This is a symplectic four-manifold with boundary, where the symplectic form is the restriction of the standard symplectic form on {\mathbb R}^4={\mathbb C}^2. By the above discussion, its boundary

Y=\partial E(a,b) = \{(z_1,z_2)\in{\mathbb C}^2 \mid \frac{\pi|z_1|^2}{a} + \frac{\pi|z_2|^2}{b} = 1\}

is a contact type surface in {\mathbb R}^4, and

\lambda = \frac{1}{2}(x_1dy_1-y_1dx_1+x_2dy_2-y_2dx_2)|_Y

is a corresponding contact form on Y. Let us now compute ECH(Y,\lambda). (We are omitting the argument \Gamma\in H_1(Y) from the notation since this is necessarily 0.)

The Reeb vector field

We first need to find the Reeb orbits. Even this is difficult for general contact manifolds, but for the present example it is an exercise to show the following:

(1) The Reeb vector field is given by

R = 2\pi\left(\frac{1}{a}\frac{\partial}{\partial\theta_1} + \frac{1}{b}{\partial}{\partial \theta_2}\right).

Here we are using polar coordinates r_1,\theta_1,r_2,\theta_2 on {\mathbb C}^2, so that z_k=r_ke^{i\theta_k}.

(2) \lambda is nondegenerate if and only if a/b is irrational. In this case there are only two embedded Reeb orbits, namely \gamma_1=(z_2=0) and \gamma_2=(z_1=0). Both of these orbits are elliptic. There is a canonical trivialization \tau of \xi over \gamma_1 and \gamma_2 such that \gamma_1 has rotation angle a/b, and \gamma_2 has rotation angle b/a (see installment 5 for the definition of rotation angle).

The chain complex

Assume that a/b is irrational. Since the Reeb orbits are \gamma_1,\gamma_2 are elliptic, an ECH generator corresponds to a pair of nonnegative integers m_1,m_2, where we take \gamma_i with multiplicity m_i. We will denote this generator by \gamma_1^{m_1}\gamma_2^{m_2}.

Since c_1(\xi)=0 is torsion and \Gamma=0, the chain complex has a canonical {\mathbb Z}-grading. Since all Reeb orbits are elliptic, it follows from the parity property of the ECH index (see the previous installment) that all generators have even grading. Since the differential decreases the grading by one (I should have said that in the previous installment but it is obvious), it follows that the differential is zero. So to complete the calculation of the ECH in this case, we just need to calculate the grading of each generator.

A calculation is usually easier if you know the answer in advance. In the present case, the calculation of the Seiberg-Witten Floer homology of S^3 is a (relatively) easy example in Kronheimer-Mrowka’s book. The result of this calculation together with Taubes’s isomorphism tell us (if we use {\mathbb Z} coefficients) that

ECH_*(Y,\lambda) = \left\{ \begin{array}{cl} {\mathbb Z}, & *=0,2,4,\ldots,\\ 0, & \mbox{otherwise}.\end{array}\right.

So let us try to calculate the ECH index and see if we can recover this result.

Calculation of the grading

What is the grading of the generator \alpha=\gamma_1^{m_1}\gamma_2^{m_2}? Since we have normalized the grading of the empty set (corresponding to \gamma_1=\gamma_2=0), the grading of \alpha is I(\alpha,\emptyset,Z)\in{\mathbb Z} where Z\in H_2(Y,\alpha,\emptyset) is unique. We can write this as

I(\alpha) = c_\tau(\alpha) + Q_\tau(\alpha) + CZ_\tau^I(\alpha)

where the notation is that c_\tau(\alpha)=c_\tau(\alpha,\emptyset,Z), Q_\tau(\alpha)=Q_\tau(Z), and CZ_\tau^I(\alpha)=CZ_\tau^I(\alpha,\emptyset).

By the above calculation of rotation angles, and the fact that rotation angle is multiplicative under iteration, we have

CZ_\tau^I(\alpha) = \sum_{k=1}^{m_1}(2\lfloor ka/b\rfloor + 1) + \sum_{k=1}^{m_2}(2\lfloor kb/a\rfloor + 1).

Exercise. c_\tau(\alpha) = m_1 + m_2.

Exercise. Q_\tau(\alpha) = 2m_1m_2.

Putting the above together, we get that

I(\alpha) = 2\bigg((m_1+1)(m_2+1)-1 + \sum_{k=1}^{m_1}\lfloor ka/b\rfloor + \sum_{k=1}^{m_2}\lfloor kb/a\rfloor \bigg).

So evidently I(\alpha) is always an even nonnegative integer. But why on earth should there be exactly one generator for each even nonnegative integer grading?

What we need to remember about ECH is that it is kind of a symplectic shadow of Seiberg-Witten theory. Things in ECH that are natural from a Seiberg-Witten perspective can appear bizarre when viewed from a purely symplectic perspective.  So if Seiberg-Witten theory suggests that we do something which looks crazy, then don’t panic, and it should work out.

Now let’s take a deep breath and stare at the above formula for I(\alpha). Observe that the right hand side can be interpreted as follows: let T be the triagle in the plane which is bounded by the coordinate axes together with the line through (m_1,m_2) with slope -a/b. Then \frac{1}{2}I(\alpha) is just the number of lattice points in the triangle T (including the boundary) minus 1. (To see this, divide the triangle T into a rectangle and two smaller triangles along the lines y=m_2 and x=m_1.) Now take a line with slope -a/b and move it up and to the right. Every time the line crosses a lattice point in the first quadrant, this gives a new ECH generator whose grading is two more than the grading of the previous one. QED. (Phew!)

 

 

 

 

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