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Let $g_S$ be a Riemannian metric on the $n$-dimensional sphere $S^{n}$ and consider the space $M=(0,a)\times S^{n}$ with the warped metric $g=dt^2+f(t)^2g_S$, where $f\colon [0,a)\to \mathbb{R}$ is a smooth function such that $$ f(0)=0\,,\quad f(t) >0\quad t\in (0,a)\,. $$ I would like to understand which are the assumptions on $f$ in order to add a point $O$ to $M$ such that the geodesic balls centered at $O$ are the slices $\{t\}\times S^{n}$.

This fact occurs for instance if $M$ is a space form and $O$ is the origin.

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  • $\begingroup$ Are you requiring that $M\cup\{O\}$ have a smooth structure such that $g$ is smooth at $O$? $\endgroup$
    – Robert Bryant
    Commented Oct 21, 2015 at 8:50
  • $\begingroup$ Yes, I require that. My goal is to have a suitable generalization of space forms. $\endgroup$
    – pedro
    Commented Oct 21, 2015 at 9:49
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    $\begingroup$ Smoothness at $O$ is the only restriction, right? If I understand notation correctly, for any metric of this kind the shortest path from the origin to a point is a ray (proof is the same as in proving that line segments are shortest paths in Euclidean space). Thus the function $f$ does not matter so much, except for its behavior near $0$. Unless you provide more information about what you expect your 'generalized space form' to look like, I don't think it is possible to give a good answer. $\endgroup$
    – Benoît Kloeckner
    Commented Oct 21, 2015 at 11:55

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Since you assume that $M\cup \{O\}$ is smooth at $O$, the metric $g_S$ must be proportional to the standard metric on $\mathbb{S}^n$. Therefore, after rescaling $f$, we can assume that $g_S$ is the canonical metric. In this case, we should assume that $f'(0)=1$ --- this is a necessary and sufficient condition.

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