Saturday, 4 September 2010

reference request - A Riemannian metric on S^2 times S^2 of nonnegative curvature that is not a product

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 $S^3times S^2$ 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 $S^2times S^2$. 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 $kneq 0$ the metric is, in general, not a product.



Edit I'm now no longer certain that for $kneq 0$, the metric is not a product. I am confident that for $kneq 0$, 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 $G = S^3times S^3$. 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 $Gtimes G$ the product metric $g_0+g_1$ and consider the space $Delta S^3 backslash Gtimes G/ T^2$ where the $T^2$ acts on $Gtimes G$ as $(z,w)*(p,q,r,s) = (pz^{-1}, q, rw^{-1},sw^{-1})$.



(The map $Gtimes Grightarrow G$ 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 $Delta S^3backslash Gtimes G/T^2$ and $G/T^2 = S^2times S^2$, 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.

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