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May 6, 2012 at 21:51 history edited Renato G. Bettiol CC BY-SA 3.0
Corrected a problem, expanded Anton's comment and emphasized a few points
May 6, 2012 at 21:40 comment added Renato G. Bettiol @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!
May 6, 2012 at 21:38 comment added Renato G. Bettiol @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 $S^2\times_f S^2$?
May 6, 2012 at 20:52 vote accept Hee Kwon Lee
May 6, 2012 at 20:22 comment added Anton Petrunin 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.
May 6, 2012 at 19:59 vote accept Hee Kwon Lee
May 6, 2012 at 20:52
May 6, 2012 at 17:17 history edited Renato G. Bettiol CC BY-SA 3.0
Mention Robert Haslhofer's comment and improved formatting
May 6, 2012 at 17:04 history answered Renato G. Bettiol CC BY-SA 3.0