4
$\begingroup$

In 1985, S.~Halperin conjectured in the topological context of maximal free torus actions on topological manifolds, that:

If $X$ is a topological space, then $$\dim H^*(X;\mathbb Q)\geq 2^{rk(X)}.$$

Where $rk(X):=\max\{n\in\mathbb{N};\text{ such that }\mathbb{T}^n\text{ acts almost freely on } X\}$ is the toral rank of $X$.

My first question: TRC is obvious for $rk(X)=0$ or 1. Do we know for which values of $rk(X)\geq 2$ the conjecture holds?.

Any comments and references are welcome

$\endgroup$

1 Answer 1

4
$\begingroup$

Volker Puppe has proved the conjecture for $rk_0(X)\le 3$, see the introduction here with a reference to the article of Puppe. There seems to be no general result for $rk(X)\ge 2$. The Lie algebra version, for nilpotent Lie algebras, has been proved for many classes of nilpotent Lie algebras (e.g., $2$-step nilpotent, or low dimension, see Proposition $2.2.7$ here).

$\endgroup$
6
  • $\begingroup$ Thanks Dietrich. Just note that following both USTIONOVSKY introduction and Puppe Theorem 1.1, the rational TRC is resolved for $rk_0(X)\geq 3$. $\endgroup$
    – MyIsmail
    Mar 27, 2015 at 20:47
  • $\begingroup$ Yes, thank you. Ustionovsky says $m\le 3$, right ? $\endgroup$ Mar 27, 2015 at 21:05
  • $\begingroup$ Yes, TRC is ok for $rk_0(X)\leq 3$. For $rk_0(X)\geq 3$, Puppe bounded the rational cohomological dimension by 2(r+1). Is there any other best approximation? $\endgroup$
    – MyIsmail
    Mar 28, 2015 at 8:03
  • $\begingroup$ I'v found the following: Aman showed in 2012 see arXiv:1204.6276 that $$\dim H^*(X;\mathbb Q)\geq 2(r+[r/3]).$$ 20 years before, Hilali proved in 1990 (see Theorem D) that $$\dim H^*(X;\mathbb Q)\geq 2(r^2+r).$$ $\endgroup$
    – MyIsmail
    Mar 31, 2015 at 10:08
  • $\begingroup$ @MyIsmail: I think the $2$ in the second bound should be $1/2$. $\endgroup$
    – Mark Grant
    Apr 3, 2015 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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