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?
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.
