I have some trouble with the derivation of the second variation formula in do Carmo's famous "Riemannian Geometry" (p. 197f.). The proposition is the following:
2.8 Proposition Let $(M,\langle\cdot,\cdot\rangle)$ be a Riemannian manifold and let $\gamma:[0,a]\to M$ be a geodesic. Assume that $f:(-\epsilon,\epsilon)\times[0,a]\to M$ is a proper variation of $\gamma$. Let $E:(-\epsilon,\epsilon)\to\mathbb{R}$ be the energy function associated to $f$, i.e.: $E(s):=\int_{0}^{a}\left\langle\frac{\partial f}{\partial s}(s,t),\frac{\partial f}{\partial s}(s,t)\right\rangle\operatorname{d}\!t$ Then: $\frac{1}{2}E''(0)=-\int_{0}^{a}\left\langle V(t),\frac{D^{2}V}{dt}+R\left(\dot{\gamma},V\right)\dot{\gamma}\right\rangle\operatorname{d}\!t-\sum_{i=1}^{k}\left\langle V(t_{i}),\frac{DV}{dt}(t_{i}^{+})-\frac{DV}{dt}(t_{i}^{-})\right\rangle$ where $R$ is the curvature on $M$ and $V:[0,a]\to TM$ is the variational field given by $V(t):=\frac{\partial f}{\partial s}(0,t)$ - is this well-defined everywhere? - and: $\frac{DV}{dt}(t_{i}^{+}):=\lim_{t\downarrow t_{i}}\frac{DV}{dt}(t)\quad \frac{DV}{dt}(t_{i}^{-}):=\lim_{t\uparrow t_{i}}\frac{DV}{dt}(t)$
I know that the proposition still is not well-defined as it is not clear what the $t_{i}$ stand for and so on. Let me start with a few comments on notation: $\frac{D}{dt}$ denotes the covariant derivative along a curve, or if $f:(-\epsilon,\epsilon)\times [0,a]\to M$ is a parametrized surface, then by convention $\frac{D}{\partial t}V(s_{0},t_{0})$ is the covariant derivative of the field $V:(-\epsilon,\epsilon)\times [0,a]\to TM$ along the curve defined by $t\mapsto f(s_{0},t)$ at the point $t_{0}$ and similarly for the operator $\frac{D}{\partial s}$.
Now I give you the definition of a variation as it appears in do Carmo's book: Let $c:[0,a]\to M$ be a piecewise differentiable curve. A function $f:(-\epsilon,\epsilon)\times[0,a]\to M$ is a variation of $c$ iff:
- $f(0,t)=c(t)$ for all $t\in[0,a]$
- There exists a partition $0=t_{0}<\cdots
such that $f\big|_{(-\epsilon,\epsilon)\times[t_{i},t_{i+1}]}$ is differentiable for all $0\leq i\leq k$. - $f$ is a proper variation of $c$ if $f(s,0)=c(0)$ and $f(s,a)=c(a)$ for all $s\in(-\epsilon,\epsilon)$.
At the time it is already clear that: $\frac{1}{2}E'(s)=\sum_{i=0}^{k}\left.\left\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle\right|_{t_{i}}^{t_{i+1}}-\int_{0}^{a}\left\langle\frac{\partial f}{\partial s},\frac{D}{\partial t}\frac{\partial f}{\partial t}\right\rangle\operatorname{d}\!t$ and hence differentiation with respect to $s$ yields: $\frac{1}{2}E''(s)=\sum_{i=0}^{k}\left.\frac{d}{ds}\left\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle\right|_{t_{i}}^{t_{i+1}}-\int_{0}^{a}\frac{d}{ds}\left\langle\frac{\partial f}{\partial s},\frac{D}{\partial t}\frac{\partial f}{\partial t}\right\rangle\operatorname{d}\!t$ Using standard properties of the Levi-Civita connection, this yields: $\frac{1}{2}E''(s)=\sum_{i=0}^{k}\left.\left\langle\frac{D}{\partial s}\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle\right|_{t_{i}}^{t_{i+1}}+\sum_{i=0}^{k}\left.\left\langle\frac{\partial f}{\partial s},\frac{D}{\partial s}\frac{\partial f}{\partial t}\right\rangle\right|_{t_{i}}^{t_{i+1}}-\int_{0}^{a}\left\langle\frac{D}{\partial s}\frac{\partial f}{\partial s},\frac{D}{\partial t}\frac{\partial f}{\partial t}\right\rangle\operatorname{d}\!t-\int_{0}^{a}\left\langle\frac{\partial f}{\partial s},\frac{D}{\partial s}\frac{D}{\partial t}\frac{\partial f}{\partial t}\right\rangle\operatorname{d}\!t$ So far so good. The issue now is the following: "Putting $s=0$ in the expression above, we obtain that the first [...] $<$term is$>$ zero, since $f$ is proper and $\gamma$ is a geodesic."
I do not see the reason for this. There are two obvious options: it clearly holds if for all $i$ the map $s\mapsto f(s,t_{i})$ is a geodesic or if $\frac{\partial f}{\partial t}(0,t)\equiv 0$. The latter is false by assumption (somewhere in the beginning of the book: geodesics are by definition non-trivial). The second is a rather strong assumptionand would be mentioned somewhere. A third possibility would be that the map $t\mapsto\frac{D}{\partial s}\frac{\partial f}{\partial s}(s,t)$ is continuous and indeed I assume that this is the case. I do not see why this follows from the definition of a variation. Does anybody have an idea?