Convexity of energy of geodesic

Question

In 34 page of Jost's Riemannian geometry and geometric analysis. How to compute the $\frac{d^2}{dt^2}E$ ? Seemly, there are some wrong in the below calculate according this question.But I fail to get the results.

Answer 1

Professor Jost is obviously a big fan of coordinate calculations. If you like a coordinate free calculation: Note the heat flow is

$$\tag{1} \frac{\partial }{\partial t} = \nabla_{\frac{\partial }{\partial s}} \frac{\partial }{\partial s}, $$

then for $E(u_t) = \frac 12\int_{\mathbb S^1} \left\| \frac{\partial}{\partial s} \right\|^2 ds $, we have

$$\begin{split} \frac{d}{dt} E(u_t) &= \frac{1}{2}\frac{d}{dt}\int_{\mathbb S^1} \left\| \frac{\partial}{\partial s} \right\|^2 ds \\ &= \int_{\mathbb S^1} \left\langle \nabla_{\frac{\partial}{\partial t}}\frac{\partial}{\partial s}, \frac{\partial}{\partial s} \right\rangle ds\\ &= \int_{\mathbb S^1} \left\langle \nabla_{\frac{\partial}{\partial s}}\frac{\partial}{\partial t}, \frac{\partial}{\partial s} \right\rangle ds\\ &= -\int_{\mathbb S^1} \left\langle \frac{\partial}{\partial t}, \nabla_{\frac{\partial}{\partial s}}\frac{\partial}{\partial s} \right\rangle ds. \end{split}$$

(Note that the term $\int_{\mathbb S^1}\frac{\partial}{\partial s}\left\langle\frac{\partial}{\partial t}, \frac{\partial}{\partial s} \right\rangle ds$ vanishes by the Fundamental Theorem of Calculus). Using the heat flow equation $(1)$, we have

$$ \frac{d}{dt} E(u_t) = - \int_{\mathbb S^1} \left\langle \frac{\partial}{\partial t}, \frac{\partial }{\partial t}\right\rangle ds.$$

Then

$$\begin{split} \frac{d^2}{dt^2} E(u_t) &= -\frac{d}{dt} \int_{\mathbb S^1} \left\langle \frac{\partial}{\partial t}, \frac{\partial }{\partial t}\right\rangle ds \\ &=-2\int_{\mathbb S^1} \left\langle \nabla_{\frac{\partial}{\partial t}}\frac{\partial}{\partial t}, \frac{\partial }{\partial t}\right\rangle ds \\ &= -2\int_{\mathbb S^1} \left\langle \nabla_{\frac{\partial}{\partial t}}\nabla_{\frac{\partial}{\partial s}}\frac{\partial}{\partial s}, \frac{\partial }{\partial t}\right\rangle ds \ \ \ \ \ \ (1) \text{ again} \\ &= -2 \int_{\mathbb S^1} \left(\left\langle \nabla_{\frac{\partial}{\partial s}}\nabla_{\frac{\partial}{\partial t}}\frac{\partial}{\partial s}, \frac{\partial }{\partial t}\right\rangle + \left\langle R\left( \frac{\partial}{\partial s} , \frac{\partial}{\partial t} \right)\frac{\partial}{\partial s}, \frac{\partial }{\partial t}\right\rangle\right)ds \ \ \ \ \ \ \ (*)\\ &= 2\int_{\mathbb S^1} \left\langle \nabla_{\frac{\partial}{\partial t}}\frac{\partial}{\partial s}, \nabla_{\frac{\partial}{\partial s}}\frac{\partial }{\partial t}\right\rangle ds-2 \int_{\mathbb S^1}\left\langle R\left( \frac{\partial}{\partial s} , \frac{\partial}{\partial t} \right)\frac{\partial}{\partial s}, \frac{\partial }{\partial t}\right\rangle ds \\ &= 2\int_{\mathbb S^1} \left\| \nabla_{\frac{\partial}{\partial s}}\frac{\partial }{\partial t}\right\|^2 ds -2 \int_{\mathbb S^1}\left\langle R\left( \frac{\partial}{\partial s} , \frac{\partial}{\partial t} \right)\frac{\partial}{\partial s}, \frac{\partial }{\partial t}\right\rangle ds \end{split}$$

where in $(*)$ we used the definition of curvature tensor and the fact that $\left[ \frac{\partial }{\partial s}, \frac{\partial }{\partial t}\right] =0$. Thus this term does not really have a sign, unless the space has nonpositive sectional curvature.