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Let $(M, g) = (\mathbb{R}^n, ds^2 = dr^2 + f^2(r) d\theta^2)$ be a Riemannian manifold, where $d\theta^2$ is the induced Riemannian metric on the unit sphere $S^{n-1}$. Find the sectional curvatures $K(\dot{\gamma}_t, v)$, where $v \in T_{\gamma(t)}M$ and $\gamma$ is a geodesic ray emanating from the pole $0 \in \mathbb{R}^n$.

The sectional curvature is given by $$K(\dot{\gamma}_t,v)= \frac{\langle R(\dot{\gamma}_t, v)v, \dot{\gamma}_t)\rangle}{|\dot{\gamma}_t \wedge v|^2}.$$ Now $$R(\dot{\gamma}_t, v)v= \nabla_{\dot{\gamma}_t}\nabla_v v - \nabla_v\nabla_{\dot{\gamma}_t}v - \nabla_{[\dot{\gamma}_t,v]}v$$ but I really don't know what can I do with this. Is it worth to consider a frame $(\partial_i)$ and express $v=v^i \partial_i$ and further expand the terms $\nabla_vv$? This feels like it's going to get very messy.

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1 Answer 1

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Here are some general facts about warped products $B\times_\phi F = (B\times F, \mathtt{g}^B\oplus \phi^2\mathtt{g}^F)$, where $(B,\mathtt{g}^B)$ and $(F,\mathtt{g}^F)$ are given.

Writing $H_{(x,y)} = T_xB\oplus \{0\}$ and $V_{(x,y)} = \{0\}\oplus T_yF$, we have that a $2$-plane $\Pi \subseteq T_{(x,y)}(B\times F)$ is horizontal, vertical, or mixed, according to whether $\Pi\subseteq H_{(x,y)}$, $\Pi\subseteq V_{(x,y)}$, or $\dim(\Pi \cap H_{(x,y)}) = \dim(\Pi \cap V_{(x,y)}) = 1$.

  1. $K(\Pi) = K^B(\Pi)$ for horizontal $\Pi$,
  2. $K(\Pi) = \phi(x)^{-2}K^F(\Pi) - \|{\rm d}(\log \phi)\|_x^2$ for vertical $\Pi$,
  3. $K(\Pi) = -\phi(x)^{-1} ({\rm Hess}\,\phi)_x(v,v)$ whenever $\Pi$ is mixed and $v\in \Pi \cap H_{(x,y)} $ is an unit vector.

This is a general consequence of curvature formulas for warped products, which are presented in O'Neill's Semi-Riemannian Geometry With Applications to Relativity.

Here, we have that $(M,\mathtt{g}) = (0,\infty) \times_f \Bbb S^{n-1}$. The velocity vector of a unit speed radial geodesic starting from the origin is simply $\partial_r$ and, in $(B,\mathtt{g}^B) = ((0,\infty), {\rm d}r^2)$ we have that $({\rm Hess}\,f)_r(\partial_r,\partial_r) = f''(r)$.

This means that taking $v = \partial_r$ in (3) above, and now letting $v\in T_{\gamma(t)}M$ be orthogonal to $\partial_r$, we have that $$K(\Bbb R \partial_r|_{\gamma(t)} \oplus \Bbb R v) = -\frac{f''(r)}{f(r)}.$$

Which should not be a surprise, if you ever computed the Gaussian curvature of a surface of revolution in $\Bbb R^3$.

(Note: writing this answer made me realize that the easy way to remember (3) is precisely to think of it as the natural generalization of the Gaussian curvature of a surface of revolution. I still don't know an interpretation for (2)).

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