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The Toral Rank conjecture in its original form runs something along the lines of: suppose we have an almost free action of the $n$-torus $T^n$ on a nice topological space $X$. (Say a closed CW complex, or a closed manifold.) Then the sum of the Betti numbers of $X$ is at least $2^n$.

I have been thinking specifically about the case when the action of $T^n$ is free (rather than almost free) in the smooth manifold case. However I do not seem to be able to find any references in the literature which deal specifically with free torus actions (there is a version of the conjecture for $\mathbb{Z}/p$ coefficients, but that is not what I want). Do any experts on the conjecture know if the free case is a folklore result? Are the free and almost-free cases equivalent?

Thanks!

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The conjecture is most commonly stated with rational coefficients: if $X$ is a simply-connected finite CW complex with an almost-free action of $T^n$, then $$ \operatorname{rk} H^\ast(X;\mathbb{Q})\ge 2^n. $$ In the rational case the almost-free and free cases are equivalent. The reason is that if $X$ admits an almost-free $T^n$ action then it is rationally homotopy equivalent to a finite complex $X'$ with a free $T^n$ action, and of course $H^\ast(X;\mathbb{Q})\cong H^\ast(X';\mathbb{Q})$. This is originally due to Halperin; you'll find a nice discussion in the book Algebraic Models in Geometry by Félix, Oprea and Tanré, Section 7.3 (in particular Proposition 7.17).

Edit: I emailed Greg Lupton about this. He said that the version of the TRC for smooth, free $T^n$-actions on manifolds is strictly more specialized than the usual form of the TRC, which is a rational homotopy statement that doesn't need the space to be a manifold (or the action to be smooth). For example, it includes spaces that needn't satisfy Poincare duality over the rationals. He mentioned that Steve Halperin (in his paper "Rational Homotopy and Torus Actions") gives an example of something of the rational homotopy type of a wedge of spheres that admits a free circle action.

However, it appears that your version of the conjecture is still unknown and interesting. See these notes of Vicente Munoz, in particular from page 13 onwards:

http://www-ma2.upc.edu/sxd/FME/VicenteMunoz-ToralRankConjecture.pdf

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  • $\begingroup$ Thanks for the enlightening answer! I have looked briefly at the book, although unfortunately I am completely new to minimal models. I had a question about Proposition 7.17, however. The proposition mentions being able to choose X' such that it is a compact manifold in certain cases. Am I correct in understanding that the free torus action in these instances can be chosen to be smooth as well as free? (After all, the original torus action is only a continuous action.) I looked for the reference provided in the proposition but found nothing. $\endgroup$
    – anon271
    Commented Jun 29, 2015 at 3:39
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    $\begingroup$ Basically, suppose that I can show: if $X$ is a compact manifold with a smooth, free action of $T^n$, then the sum of the Betti numbers of $X$ is at least $2^n$. Will this suffice to show the general conjecture? Is it considered to be an interesting case? (Sorry for the repeated questions!) $\endgroup$
    – anon271
    Commented Jun 29, 2015 at 3:47
  • $\begingroup$ Looking at the proof of 7.17, it seems that if $X'$ can be chosen to be a compact manifold, then the action is smooth and free by construction ($X'$ is the total space of a smooth principal $T^n$-bundle). However, I'm not sure we can always find a manifold in the rational homotopy type. There is a well-developed rational surgery theory for answering this type of question (see Theorem 3.2 in the book, or here: mathoverflow.net/questions/115911/…) $\endgroup$
    – Mark Grant
    Commented Jun 29, 2015 at 8:01
  • $\begingroup$ BTW, I'm not really an expert in the TRC, but I know a couple...I'll email them to alert them of your question. $\endgroup$
    – Mark Grant
    Commented Jun 29, 2015 at 8:02

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