Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 20 / 5 / i / Solution

For a rational homology three-sphere and any Spin-c structure, the infinity version of Heegaard Floer homology isIt is one Laurent tower, with one integral generator in each degree of a fixed parity and none in the opposite parity. Its absolute rational grading has an overall offset determined by $(Y,\mathfrak{s})$.

To compute the Euler characteristic of the hat group, tensor with $\mathbb{Q}$, which preserves integral ranks. Put $V=H{F}^{-}(Y,\mathfrak{s};\mathbb{Q})$. It is a finitely generated graded $\mathbb{Q}[U]$ module. Localizing at $U$ gives $H{F}^{\infty }$, so the structure theorem for finitely generated modules over a principal ideal domain gives one free summand and finitely many $U$-primary torsion summands. The torsion is $U$-primary because it vanishes after localization, or directly because the reduced graded group is supported in finitely many degrees.

The exact sequence of Heegaard Floer chain complexes given by multiplication by $U$ and passage to the hat quotient identifies the hat group, as a parity-graded vector space, with an extension of $V/UV$ by $\ker (U:V\to V)$ with the latter parity reversed. The extra absolute shift by two between $C{F}^{-}/UC{F}^{-}$ and $\widehat{CF}$ is even, so it does not affect this argument. ThusThe free summand contributes one vector to $V/UV$ and none to $\ker U$. A torsion summand $\mathbb{Q}[U]/U^r$, whose top generator has parity $ϵ$, contributes one vector of parity $ϵ$ to the quotient and one of the same parity to the kernel, since $U^{r-1}$ changes degree by an even number. Their contributions cancel. Only the single free summand remains, provingInteger torsion does not enter an Euler characteristic defined by ranks.