Figuring of locally product Riemannian metric. If $S^1\times M$ where $M$ is a simply connected compact manifold has a metric $g$ with nonnegative sectional curvature, then its universal cover ${\bf R}\times M$ has a product metric by splitting theorem, which clearly has nonnegative sectional curvature. So $g$ is locally product. 
That is I think that $g$ will be obtained by cutting and pasting 
from product metric. Please, give me an example for $g$ ? 
 A: Recall that the flat torus is obtained by tiling $\mathbb{R}^2$ by rectangles and then quotienting out appropriately.  Modify this so that instead tiling by rectangles, you tile by parallellograms with no angle equal to $\pi/2$.  For definiteness, I'm using the parallelogram with corners at $(0,0)$, $(1,1)$, $(\sqrt{3},0)$, and $(1+\sqrt{3},1)$.  The short edge, then, is given by the line $y=x$.
The quotient defines a different flat metric $g$ on the torus, which is locally a product because it's a quotient of the canonical flat (product) metric on $\mathbb{R}^2$.
I claim this metric is not a product metric.
Proof:  Assume for a contradiction that the metric is a product, say $(S^1\times S^1, g_a + g_b)$.  By swapping the two factors, we assume that the length $l_a$ of $S^1\times \{p\}$ is smaller than or equal to the length $l_b$ of $\{p\}\times S^1$.
Now, we can actually recover $l_a$ from the isometry type of $S^1\times S^1$:  It's simply the length of the shortest closed geodesic.  On our parallelogram-torus, the shortest closed geodesic is the edge of length $\sqrt{2}$, so if we can decompose our parallelogram-torus as a product metric, one of the factors must be the $S^1$ along the shortest edge.
What does this tell us about the other factor?  Well, in a product metric, the two factors are perpendicular to each other, so our only other possible choice is the perpendicular direction from our shortest edge.  Since our parallelogram-torus is still homogeneous, it's enough to figure out what is perpendicular to the short edge emanating from $(0,0)$.
However, I claim that the geodesic emanating from $(0,0)$ perpendicular to $y=x$ (i.e., $y=-x$) never closes up, so is not an $S^1$, giving a contradiction.
To see this, note that the lattice defining the torus is the integral span of the two vectors $(1,1)$ and $(\sqrt{3},0)$, so what we're asking whether the line $y=-x$ every intersects a point of the form $a(1,1)+b(\sqrt{3},0)$ for $a,b\in\mathbb{Z}$.
One can easily solve $(x,-x) = (a+b\sqrt{2}, a)$ to get $a=-x$ and $b=\frac{2}{\sqrt{3}}{x}$.  But these can't both be integers unless $x=0$, so it never closes up.
