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Let $M$ be a simply connected closed Riemannian manifold. How does one find a necessary condition going both ways that may be imposed on $M$ (perhaps on the curvature of $M$ and on torsion) which guarantees that the rational cohomology ring $H^*(M;\mathbb{Q})$ needs at least two generators? That is, how does one force $M$ not to have rational cohomology that is the quotient of a polynomial ring?

Cross-posting on MSE: https://math.stackexchange.com/questions/2857279/show-that-the-rational-cohomology-ring-hm-mathbbq-needs-at-least-two-ge

Any help would be much appreciated. Thanks in advance!

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    $\begingroup$ I'm not sure about a necessary condition, but if your rational cohomology ring has at least two generators, then your manifold has infinitely many distinct closed geodesics. I don't know any conditions on curvature that can control the number of distinct closed geodesics however ... $\endgroup$
    – Tobias Shin
    Commented Jul 20, 2018 at 21:53
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    $\begingroup$ Presumably you're assuming that $dim H^k(M;\mathbb{Q})= 1$ for all $k \equiv 0 (\mod q)$ for some $1< q| dim M$? Also, I think you want to say that $H^*(M;\mathbb{Q})$ is not the quotient of a polynomial ring. $\endgroup$
    – Ian Agol
    Commented Jul 20, 2018 at 22:16
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    $\begingroup$ I don't think such results are known. Some examples of geometrically interesting manifolds with singly generated rational cohomology can be found in arxiv.org/abs/math/0210231 but there are many other examples and it is unclear how curvature could be relevant. $\endgroup$
    – Igor Belegradek
    Commented Jul 20, 2018 at 22:18
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    $\begingroup$ There is an equivalent condition in terms of rational homotopy groups: $H^*(M;\mathbb{Q})$ requires more than one generator iff $\pi_{\rm odd}(M)\otimes\mathbb{Q}$ is more than one-dimensional. $\endgroup$
    – Mark Grant
    Commented Jul 23, 2018 at 8:26
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    $\begingroup$ @Multivariablecalculus: well, it is a condition on the non-torsion part of the homotopy groups of $M$, since tensoring with $\mathbb{Q}$ killls torsion; see rational homotopy theory. (Torsion here has little to do with torsion in the sense of differential geometry.) $\endgroup$
    – Mark Grant
    Commented Jul 23, 2018 at 18:51

1 Answer 1

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If $M$ is simply-connected and has reducible holonomy, then a theorem of de Rham implies that $M$ is a product, and hence does not have homology generated by one element.

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  • $\begingroup$ Thanks! Could you elaborate a bit more on this, i.e. why does reducible holonomy imply this. The link you gave is quite terse. @IanAgol $\endgroup$
    – Sergio Charles
    Commented Jul 23, 2018 at 20:00
  • $\begingroup$ So to clarify, if $M$ is a simply-connected closed Riemannian manifold and has reducible holonomy then $H^*(M;\mathbb{Q})$ has at least two generators? I am asking as I am not familiar with this result. @IanAgol $\endgroup$
    – Sergio Charles
    Commented Jul 23, 2018 at 20:06
  • $\begingroup$ More precisely, Berger gave a complete classification of possible holonomy groups for the irreducible case; however, is such a complete classification know for the reducible case? $\endgroup$
    – Sergio Charles
    Commented Jul 23, 2018 at 20:11
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    $\begingroup$ @Multivariablecalculus: Coupling Berger's classification with de Rham's decomposition theorem, I suppose one does obtain a classification of reducible holonomy groups by requiring that each factor is one of the examples coming from Berger's list (actually, I'm not sure which on his list are realized by compact simply-connected manifolds). One can find metrics on product manifolds with irreducible holonomy, so the converse is false. $\endgroup$
    – Ian Agol
    Commented Jul 23, 2018 at 20:25
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    $\begingroup$ @Multivariablecalculus, I expect that a "generic" manifold in any reasonable sense will (a) have irreducible holonomy and (b) have homology that is not generated by a single element. So I expect the converse to be very false (but it's hard to quote precise results). $\endgroup$
    – Dylan Thurston
    Commented Jul 24, 2018 at 22:01

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