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added comments from Greg Lupton and link to notes of Vicente Munoz

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edited Jun 29, 2015 at 9:28

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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

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).

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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created Jun 19, 2015 at 7:40

35.9k
8
95
198

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).