Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For an abelian group $G$, one can give an explicit description of the homology ring $H_*(G, k)$ for e.g. $k=\mathbb{Q}, \mathbb{Z}_p$ or in general PIDs $k$ in which every natural number is invertible.

Is there some application or motivating example where explicit knowledge about the Pontryagin product structure of some (preferably not finitely generated) abelian group is useful?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

After conversing with Ken Brown, here is what I can say.

1) It is now known that the Thompson group $F$ is not a Kähler group, but before this was known, Brown had calculated its homology ring $H_*(F)$ using the Pontryagin product. The motivation was 2-fold, he loved that group and so wanted to determine all information about it, and there was an argument (which I cannot recall) using this ring to determine whether $F$ is Kähler. However, the argument ended up not working for some reason.

2) As something you might prefer: There are spectral sequence arguments in which the product structure lets you compute the differential just from a tiny bit of information about what it does on certain elements. (I have never seen these, but Brown recalls their existence, and if you stumble across such arguments it would be great to notify me about them!)

share|improve this answer
3  
I don't think 1) is true literally. For, $F$ isn't abelian, so there's no Pontryagin product on the homology of $F$. But as explained by Brown in math.cornell.edu/~kbrown/papers/homology.pdf, there's a hom. $F \times F \to F$ that induces a map $H_\ast(F) \otimes H_\ast(F) \to H_\ast(F)$ and the latter turns out to be helpful in the computation of the (co)homology of $F$. –  Ralph Dec 5 '12 at 10:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.