A locally CAT(0) metric on length space means that every point in it has a geodesically convex neighborhood such that every triangle in it is slimmer than the comparison triangle in the Euclidean plane. For example, the Riemannian metric with nonpositive sectional curvature is a locally CAT(0) metric.

Let $M$ be a closed manifold such that $M\times S^1$ admits a locally CAT(0) metric. Does $M$ also admit a locally CAT(0) metric?

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    $\begingroup$ A suggestion: There should be a version of the splitting theorem for groups with infinite center. For Riemannian locally CAT(0) metrics it can be found in Schroeder's Inventiones splitting theorem paper. In this version the universal cover of a closed nonpositively curved manifold with $\pi_1\cong H_1\times H_2$ splits as the product of three factors $X_0\times X_1\times X_2$ where $X_0$ is Euclidean and each $H_i$ acts trivially on $X_{3-i}$ and by translation on $X_0$. Try to deform the $H_i$-action to the trivial one through translations by keeping $H_1\times H_2$ discrete. $\endgroup$ Jul 30, 2019 at 20:48
  • $\begingroup$ It seems that I can show that $M\times \mathbb{S}^1$ splits isometrically, but it does not imply the existence of a CAT(0)-metric on $M$ --- we only get a CAT(0)-metric on a manifold $M'$ that is bordant to $M$. $\endgroup$ Nov 12, 2022 at 16:12
  • $\begingroup$ + If $M\times \mathbb{S}^1$ is a torus, then the answer is yes, but it is not trivial; see mathoverflow.net/questions/403202 $\endgroup$ Nov 15, 2022 at 20:51

1 Answer 1


Here is a partial answer, I will prove the following:

If $M\times \mathbb{S}^1$ equipped with locally $\mathrm{CAT}(0)$ length metric, then there is isometric $\mathbb{S}^1$-action on $M\times \mathbb{S}^1$ with parallel geodesic orbits.

Denote by $N$ the universal cover of $M\times \mathbb{S}^1$; it is a $\mathrm{CAT}(0)$ length metric. Let $\gamma$ be a shortest circle in $M\times \mathbb{S}^1$ that is homotopic to $p\times \mathbb{S}^1$. Denote by $\tilde\gamma_\alpha$ lifts of $\gamma$ in $N$; each $\tilde\gamma_\alpha$ is a line in $N$. Denote by $b_\alpha$ the Busemann function associated to $\gamma_\alpha$.

Note that projection of $\gamma_\alpha$ to $\gamma_\beta$ is isometric for any $\alpha$ and $\beta$. It follows that $b_\alpha-b_\beta$ is constant for any $\alpha$ and $\beta$. Therefore, the gradient $\nabla b_\alpha$ is independent of $\alpha$ and invariant with respect to the action of deck transformations on $N$.

It follows that $M\times \mathbb{S}^1$ admits a vector field $v$ such that $\nabla b_\alpha$ is a lift of $v$ for any $\alpha$. (Gradient flow on singular spaces is discussed in our book.)

Note that the flow $\Phi^t$ along $v$ is distance-noncontracting. Since it is defined on a closed manifold $M\times \mathbb{S}^1$, $\Phi^t$ is isometry for any $t$. It follows that $b_\alpha$ is affine (it is convex and concave at the same time). By the line strip theorem, $N$ splits isometrically as $L\times \mathbb{R}$. Passing back to $M\times \mathbb{S}^1$, we get the statement.


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