# Computing sectional curvature on a model Riemannian manifold

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.

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)).