A Riemannian metric on $S^2 \times S^2$ of nonnegative curvature that is not a product

Question

Good afternoon,

There is an example of a Riemannian metric on $S^2 \times S^2$ of nonnegative sectional curvature that is not a product metric. I know there is one; however, I cannot find a specific reference. Any suggestions?

Answer 1

A discussion of nonnegatively curved metrics on $S^2\times S^2$ can be found in the survey by B. Wilking, see page 26 and last paragraph on page 25. In particular, there is a one parametric family of metrics of nonnegative curvature on $S^2\times S^2$ which are not moved by any diffeomorphism to a product metric.

Answer 2

One has to be careful with perturbations of metrics of nonnegative curvature, because that may introduce negative curvature.

Here's another approach which gives you a nonnegatively curved metric.

Start with $S^3\times S^2$ with the product of round metrics. (Note that the round metric on S^3 is its biinvariant metric).

Consider the $S^1$ action on $S^3 \times S^2$ where it acts as the Hopf action on $S^3$ and simultaneously rotates the $S^2$ factor $2k$ times around for some integer $k$.

To make it explicit, thinking of $S^2$ as the unit sphere in the imaginary quaternions, the action can be described as $z*(p,q) = (zp, z^k q \overline{z}^k)$.

The action is clearly free and the quotient is diffeomorphic to $S^2 \times S^2$. Since the circle is acting isometrically, there is an induced submersion metric on $S^2\times S^2$. By the O'Neill formulas for a submersion, this metric has nonnegative curvature. When k = 0, one gets the usual product of round metrics, but when $k\neq 0$ the metric is, in general, not a product.

Edit I'm now no longer certain that for $k\neq 0$, the metric is not a product. I am confident that for $k\neq 0$, the metric is not a product of round metrics, but I don't see any reason they can't be a product of two nonnegatively curved metrics.

However, here is an example (sorry for doubling the length of my post!): Let $G = S^3\times S^3$. Let $g_0$ denote a biinvariant metric on $G$. Writing $\mathfrak{g}$ for the Lie algebra of $G$, set $\mathfrak{p}$ to be the Lie algebra of the diagonal $S^3$ and choose $\mathfrak{q}$ to be $g_0$-orthogonal to $\mathfrak{p}$.

Fix a positive real number $t$ and define a new inner product $g_1 = g_0|_{\mathfrak{q}} + \frac{t}{t+1}g_0|_{\mathfrak{p}}$ and left translate it around $G$ to give a left invariant, right $\Delta S^3$ invariant metric. Such a metric is called a Cheeger deformation of $g_0$ and it is known that $g_1$ has nonnegative sectional curvature.

Give $G\times G$ the product metric $g_0+g_1$ and consider the space $\Delta S^3 \backslash G\times G/ T^2$ where the $T^2$ acts on $G\times G$ as $(z,w)*(p,q,r,s) = (pz^{-1}, q, rw^{-1},sw^{-1})$.

(The map $G\times G\rightarrow G$ sending $(p,q,r,s)$ to $(r^{-1}p, s^{-1}q)$, or something like it if I've made a mistake, induces a diffeomorphism between $\Delta S^3\backslash G\times G/T^2$ and $G/T^2 = S^2\times S^2$, where the $G/T^2$ is referring to the action of $T^2$ on $G$ spelled out before the edit with k=1).

As above, there is an induced submersion metric of nonnegative sectional curvature by the O'Neill formulas. Finally, to prove that this is NOT a product metric, one observes that at generic points, there is a unique plane with 0 sectional curvature, while for a product metric, there should be infinitely many planes of 0 curvature.

The observation comes from

P.Müter, Krümmungserhöhende Deformationen mittels Gruppenaktionen, Ph.D. thesis, University of Münster, 1987.