Just to put things in perspective, recall that the Hopf Conjecture asks whether $S^2\times S^2$ admits a metric of positive sectional curvature. By the work of Hsiang-Kleiner, it is known that, if $S^2\times S^2$ admits such a metric, then its isometry group cannot contain a circle, and is hence finite.

Q: If $S^2\times S^2$ admits a metric with $sec>0$, what is known about its isometry group $G$?

The only results I know of are:

  1. [Edit suggested by Misha] The diagonal antipodal action of $\mathbb Z_2$ on $S^2\times S^2$, i.e., $\pm 1\cdot(x,y)=(\pm x,\pm y)$, cannot be isometric if $S^2\times S^2$ is equipped with a metric of positive curvature. By Weinstein's Thm, an orientation-preserving isometry of an even-dimensional positively curved manifold has a fixed point (and the antipodal map does not). Equivalently, it would induce a positively curved metric on the $2$-fold orientable cover of $\mathbb R P^2\times \mathbb R P^2$, hence on $\mathbb R P^2\times \mathbb R P^2$, but this contradicts Synge's Thm.

  2. From Wilking's thesis (Prop 4.2), any simple subgroup of $G$ is either cyclic or isomorphic to a group in a finite list $F_1,\dots,F_k$ of simple groups. (This is actually true for any finitely generated subgroups of isometries of a manifold with $Ric≥0$).

  3. From Fang's paper (Thm 1.2), $G$ cannot have a subgroup of sufficiently large odd order (but this lower bound is huge, since it is estimated with Gromov's universal constant for the total Betti number).

Apart from these, are there other known restrictions on what $G$ can be like?

  • 2
    $\begingroup$ You should also include the classical result that $RP^2\times RP^2$ does not admit a metric of positive sectional curvature. This gives one more restriction for $S^2\times S^2$. $\endgroup$
    – Misha
    Mar 11, 2012 at 23:47
  • $\begingroup$ @Misha: I believe there is, however, one little detail about using that $RP^2\times RP^2$ cannot have $sec>0$ to imply that the standard $Z_2\oplus Z_2$ action on $S^2\times S^2$ cannot be isometric w.r.t. a positively curved metric. By Weinstein's Theorem, any (orientation-preserving) isometry of a compact even-dimensional positively curved manifold always has a fixed point. Thus, the $Z_2\oplus Z_2$ action cannot be free on a positively curved $S^2\times S^2$, so the quotient wouldn't be $RP^2\times RP^2$, right? $\endgroup$ Mar 12, 2012 at 3:37
  • 1
    $\begingroup$ That's right. The manifold $RP^2\times RP^2$ is, of course, non-orientable, but it has an orientable 2-fold cover, whose fundamental group acts on $S^2\times S^2$ freely preserving orientation. $\endgroup$
    – Misha
    Mar 12, 2012 at 4:44
  • 1
    $\begingroup$ There must be some restrictions on smooth finite group actions on $S^2\times S^2$ regardless of curvature, e.g. groups that acts pseudo-freely are exactly those that cats linearly, see arxiv.org//abs/math/9906159, as shown by Michael McCooey who may tell you more on the subject. $\endgroup$ Mar 12, 2012 at 12:44

1 Answer 1


You should look at Andrew Hick's thesis,

Andrew Hicks, Group actions and the topology of nonnegatively curved $4$-manifolds, Illinois Journal of Mathematics. Volume 41, Issue 3 (1997), 421-437.

A corollary to his Theorem 2 shows that for a positively curved metric on $S^2\times S^2$ with $\delta$ pinching, the size of the isometry group is bounded above by a constant related to the pinching.

There are a few other interesting results about symmetries of nonnegatively curved 4-manifolds there.

  • 1
    $\begingroup$ @Luis: Thanks for the reference. I'm not sure I have read it before, I'll definitely check it out. By the way, welcome to MathOverflow -- it's great to see more differential geometers around here! $\endgroup$ May 2, 2013 at 2:00

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