## Why aren’t there three variants of ECH?

Frequently asked question: Heegaard Floer homology comes in three variants which fit into an exact triangle, namely $HF^+$, $HF^-$, and $HF^\infty$. Likewise for Seiberg-Witten Floer homology which has $\widehat{HM}$, $\check{HM}$, and $\overline{HM}$. So why is there only one ECH?

Answer: We can algebraically cook up two other variants of ECH if we really want to. Recall that ECH is the homology of a chain complex $C_*$ which has a degree $-2$ chain map $U$ on it. We could define an “infinity” of “bar” version of ECH to be the inverse limit of the $U$ map on this chain complex. For the algebraically challenged like me, this means that we define $C_*^\infty$ to be the set of sequences $\{x_k\}_{k\ge 0}$ where $x_k\in C_{*+2k}$ and $Ux_{k+1}=x_k$ for all $k\ge 0$. The differential on $C_*^\infty$ is defined simply by $\partial\{x_k\} = \{\partial x_k\}$. The homology of this complex, which we can denote by $ECH^\infty$, should agree with suitable versions of $HF^\infty$ and $\overline{HM}$.

There is also a chain map $C_*^\infty\to C_*$ sending $\{x_k\}\mapsto x_0$.  (In a previous version of this post I suggested that we could define a third variant of ECH to be the homology of the kernel of this map, and then get an exact triangle which would agree with suitable versions of the exact triangles in the other two Floer theories. But actually the map $C_*^\infty\to C_*$ is not surjective, so we don’t have a short exact sequence of chain complexes here. So I need to think more about how to get a third version of ECH with an exact triangle.)

Now the interesting thing is that $HF^\infty$ and $\overline{HM}$ are relatively simple, determined by the triple cup product on the three-manifold. It is very plausible that at least $\overline{HM}$ should have a “classical” description, since it is determined entirely by the reducibles which can be understood explicitly. I don’t know enough about Heegaard Floer theory to summarize why $HF^\infty$ has a simple description. In any case, it is not at all obvious from the above definition that this should be the case for $ECH^\infty$. If anyone can think of a direct argument for this, then it might lead to a proof of the Weinstein conjecture without using Seiberg-Witten theory.