5
$\begingroup$

In 2002, by using Floer theory, Froyshov defined the $h$-invariant for intergal homology 3-spheres, which is a surjective group homomorphism $\Theta^3_{\Bbb Z}\to \Bbb Z$, where $\Theta^3_{\Bbb Z}$ is the group of homology cobordism classes of integral homology 3-spheres with connected sum as operation. (https://arxiv.org/pdf/math/9903083.pdf)

For Seifert homology spheres $\Sigma(a_1,\dots,a_n)$, there are some examples whose $h$-invariants are known. But is there a theorem or algorithm or formula concerning about a way to compute the $h$-invariant for a general Seifert homology sphere $\Sigma(a_1,\dots,a_n)$?

For the Ozsvath-Szabo $d$-invariant, a group homomorphism $\Theta^3_{\Bbb Z}\to 2\Bbb Z$ (which is also defined using Floer theory), there are some ways to calculate it for Seifert homology spheres (e.g. https://arxiv.org/pdf/math/0310083.pdf), but I cannot find any results for computing the $h$-invariant for Seifert homology spheres.

$\endgroup$
1
  • $\begingroup$ Do you want this for something in particular? There may be tricks in specific cases to get around the lack of a nice algorithm. $\endgroup$
    – mme
    Commented Oct 22, 2022 at 0:43

1 Answer 1

4
$\begingroup$

No, not yet. The state of the art in computing instanton homology for Seifert spaces is in this paper. This is like knowing how to compute $\widehat{HF}(\Sigma)$, which is not enough: you also want to understand $\widehat{HF}_{red}(\Sigma)$. More or less equivalently, you want comparable results for equivariant versions along the lines of the computation of $HF^+(\Sigma)$ for all Seifert spaces.

Such a computation does not exist in the literature nor do I believe it is forthcoming in the immediate future. You might get in touch with one of the authors of that paper. I know John Baldwin has put some thought into the equivariant versions of instanton homology and may be able to say whether or not the necessary calculations exist in his head.

$\endgroup$
1
  • 2
    $\begingroup$ It is expected but not known that d=2h. $\endgroup$
    – mme
    Commented Oct 19, 2022 at 16:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .