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As you know, the Hopf conjecture is about the existence of positively curved metric on $S^2\times S^2$. Hsiang-Kleiner have shown that there exists no positively curved metric admitting $S^1$-action on $S^2\times S^2$.

My question is simple. If $(S_1=S^2,g)$ and $(S_2=S^2,h)$ are positively curved, then for any positive function $f: S_1 \rightarrow \mathbb{R}$, is a warped metric $g+ fh$ not positively curved ? Or is this statement not proved ?

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The warped product admits an $S^1$ action. Thus it cannot be positively curved by the Hsiang-Kleiner result that you quoted... –  Robert Haslhofer May 6 '12 at 10:23
    
Thank you for yor correction. However I have another question. If $e: S_2 \rightarrow \mathbb{R}$ is a positive function, then is a doubly warped metric $eg+fh$ not positively curved ? –  Hee Kwon Lee May 6 '12 at 12:55
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@Robert, no in general warped products do not admit an $S^1$ action. –  Anton Petrunin May 6 '12 at 20:19
    
@Anton: oops, thanks for catching my mistake! (I haven't read the question carefully enough, and thought that $S^2$ carries the standard metric) –  Robert Haslhofer May 6 '12 at 21:12
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3 Answers

up vote 5 down vote accepted

As already observed above, if you consider a warped product metric $g_1 + f \cdot g_2$ on $S^2\times S^2$ obtained from a positively curved metric on each factor, it will not have positive curvature. This can be seen in the following way. The formula for the sectional curvature of "mixed planes" (i.e., spanned by vectors $X$ and $Y$, one from each factor) with respect to the warped metric is essentially (up to renormalization) the hessian of $f$ in the direction of $Y$, $\mathrm{Hess} \; f(Y,Y)$. Since the domain of $f$ is $S^2$ (which is compact), $f$ will have a minimum and a maximum, so its hessian cannot be always positive (or negative) definite. Thus, the curvature of mixed planes with respect to warped product metrics cannot be always $>0$ (or $<0$). You can find a slightly more precise description of this fact on a paper by Leysen and Verstraelen "On warped products and a conjecture of H. Hopf." Soochow J. Math. 13 (1987), no. 2, 175–178.

For a similar reason, a double-warped metric will also not have positive curvature, since the hessians of the warping functions are not definite. The Hopf conjecture is a nasty (but terribly intriguing) problem...


EDIT (due to Anton's comment): One comment above mentioned using Hsiang-Kleiner's result (a positively curved metric on $S^2\times S^2$ cannot have an isometric circle action) to answer the question. This indeed works if we are warping a positively curved metric on one $S^2$ that has a circle in its isometry group (e.g. the round metric) with another positively curved metric on the other $S^2$ (and the warping function is defined on this second factor). In this way, there is a circle acting isometrically in the warped metric and Hsiang-Kleiner's result applies. Nevertheless, in general, a positively curved metric on $S^2$ does not have an isometric circle action, so this reasoning cannot be used.

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Your second way is correct, BUT the first one is not. In general warped products do not admit an $S^1$ action AND Kleiner is not working here. –  Anton Petrunin May 6 '12 at 20:22
    
@Anton: Thanks for your comment! Now I am a little confused: isn't it true that if $\phi$ is an isometry of $(M,g_M)$ and $\psi$ is an isometry of $(N,g_N)$ such that $f\circ\psi=\psi$, then $(\phi,\psi)$ is an isometry of $(M\times N,g_M+f g_N)$? In this way, there is an inclusion $Iso(M,g_M)\subset Iso(M\times N,g_M +f g_N)$, so if the first $S^2$ has the round metric, there would be a circle in the isometry group of the warped product $S^2\times_f S^2$? –  Renato G Bettiol May 6 '12 at 21:38
    
@Anton: Oh, OK, I think I got it. The OP was asking about warping any two positively curved metrics on $S^2$ (not the round metric), and these in general don't have any symmetries. So the warped products will not have symmetries either. I am editing my answer to reflect this. Thanks! –  Renato G Bettiol May 6 '12 at 21:40
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I think the metrics you get might have non-negative curvature, but not positive curvature. If, for example, you just take $g$ and $h$ to be the standard metrics on $S^2$ and $e$ and $f$ to be $1$, then if you work out the sectional curvature on any plane with a tangent vector in each $S^2$ factor you get zero. Positive curvature is much harder than non-negative curvature.

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An alternative proof of this result is to realize that the injection into the first factor $i:S^2\to S^2\times S^2$ gives you a "soul" of the warped product. By that I mean that the composition of the Riemannian submersion (for the warped metric) $S^2\times S^2\to S^2, (p,q)\to p$ with $i$ provides you with an analog of the Sharafutdinov map for open nonnegatively curved manifolds. Then you can adapt the proof of Perelman's proof of the soul conjecture to rule out positive curvature, as well as gain some rigidity for nonnegative curvature. The details appear in the last section of (sorry for the shameless self-promotion)

V. Berestovskii and L. Guijarro, A metric characterization of Riemannian submersions, Annals of Global Analysis and Geometry 18 (2000), pp.577-588

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