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Let $M$ be a simply connected closed Riemannian manifold. How does one find a condition 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!

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!

Let $M$ be a simply connected closed Riemannian manifold. How does one find a condition that may be imposed on $M$ (say on the curvature of $M$) 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!

Let $M$ be a simply connected closed Riemannian manifold. How does one find a condition 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!

Let $M$ be a simply connected closed Riemannian manifold. How does one find a condition that may be imposed on $M$ (say on the curvature of $M$) 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 polynomial cohomology 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!

Let $M$ be a simply connected closed Riemannian manifold. How does one find a condition that may be imposed on $M$ (say on the curvature of $M$) 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!