do Carmo: Second Variation Formula 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 $0=t_{0}<\cdots<t_{k+1}=a$ 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?
 A: Let $0\leq i\leq k$ and $\alpha,\beta>0$ such that the image of $f\big|_{(-\alpha,\alpha)\times(t_{i}-\beta,t_{i}+\beta)}$ is contained in $U\subseteq M$ where $(U,\rho)$ is a coordinate neighbourhood of $f(0,t_{i})$. Then there exist continuous $\lambda_{1},\ldots,\lambda_{n}:(-\alpha,\alpha)\times(t_{i}-\beta,t_{i}+\beta)\to\mathbb{R}$ such that $$\frac{\partial f}{\partial s}(s,t)=\sum_{j=1}^{n}\lambda_{j}(s,t)\frac{\partial}{\partial x_{j}}\bigg|_{f(s,t)}$$ It follows from the definition of the variation that $\lambda_{j}\big|_{(-\alpha,\alpha)\times[t_{i},t_{i}+\beta)}$ and $\lambda_{j}\big|_{(-\alpha,\alpha)\times(t_{i}-\beta,t_{i}]}$ are differentiable. Furthermore it is a defining property of the covariant derivation that if well defined, we obtain:
$$\frac{D}{\partial s}\bigg|_{(s,t)}\frac{\partial f}{\partial s}=\sum_{j=1}^{n}\frac{\partial \lambda_{j}}{\partial s}(s,t)\frac{\partial}{\partial x_{j}}\bigg|_{f(s,t)}+\sum_{j=1}^{n}\lambda_{j}(s,t)\frac{D}{\partial s}\bigg|_{(s,t)}\frac{\partial}{\partial x_{j}}\\
=\sum_{j=1}^{n}\frac{\partial \lambda_{j}}{\partial s}(s,t)\frac{\partial}{\partial x_{j}}\bigg|_{f(s,t)}+\sum_{j,m=1}^{n}\lambda_{j}(s,t)\lambda_{m}(s,t)\left(\nabla_{\frac{\partial}{\partial x_{m}}}\frac{\partial}{\partial x_{j}}\right)_{f(s,t)}$$
So all there is to check is that $\frac{\partial \lambda_{j}}{\partial s}(s,t_{i}^{+})=\frac{\partial \lambda_{j}}{\partial s}(s,t_{i}^{-})$. But this follows from the definition of $\frac{\partial \lambda_{j}}{\partial s}(s,t_{i})$. More formally, we can define $\lambda_{j}^{1}:=\lambda_{j}\big|_{(-\alpha,\alpha)\times(t_{i}-\delta,t_{i}]}$ and $\lambda_{j}^{2}:=\lambda_{j}\big|_{(-\alpha,\alpha)\times[t_{i},t_{i}+\delta)}$. By definition we have: 
$$\frac{\partial \lambda_{j}}{\partial s}(s,t_{i}^{+})=\lim_{t\downarrow t_{i}}\frac{\partial\lambda_{j}}{\partial s}(s,t)=\lim_{t\downarrow t_{i}}\frac{\partial\lambda_{j}^{2}}{\partial s}(s,t)$$ 
and: 
$$\frac{\partial \lambda}{\partial s}(s,t_{i}^{-})=\lim_{t\uparrow t_{i}}\frac{\partial\lambda_{j}}{\partial s}(s,t)=\lim_{t\uparrow t_{i}}\frac{\partial\lambda_{j}^{1}}{\partial s}(s,t)$$ 
As the restrictions are continuously differentiable by assumption, we obtain $\frac{\partial \lambda_{j}}{\partial s}(s,t_{i}^{+})=\frac{\partial \lambda_{j}^{2}}{\partial s}(s,t_{i})$ and $\frac{\partial \lambda_{j}}{\partial s}(s,t_{i}^{-})=\frac{\partial \lambda_{j}^{1}}{\partial s}(s,t_{i})$. Explicit calculation now yields:
$$\frac{\partial\lambda_{j}}{\partial s}(s,t_{j}^{+})=\lim_{h\to 0}\frac{\lambda_{j}^{2}(s+h,t_{i})-\lambda_{j}^{2}(s,t_{i})}{h}=\lim_{h\to 0}\frac{\lambda_{j}(s+h,t_{i})-\lambda_{j}(s,t_{i})}{h}\\ \frac{\partial\lambda_{j}}{\partial s}(s,t_{j}^{-})=\lim_{h\to 0}\frac{\lambda_{j}^{1}(s+h,t_{i})-\lambda_{j}^{1}(s,t_{i})}{h}=\lim_{h\to 0}\frac{\lambda_{j}(s+h,t_{i})-\lambda_{j}(s,t_{i})}{h}$$
So indeed this is independent of whether we take the derivative from below $t_{i}$ or from above $t_{i}$.
As a synopsis: as expected this is nothing but the fact that at $t_{i}$ the definition of a variation leads to the smooth curve $s\mapsto f(s,t_{i})$. Hence all your favourite operations can be applied to this curve and as $t\mapsto f(s,t)$ is continuous, also the derivatives are continuous in the parameter. 
P.s.:@Jason: I am sure that this just boils down to the solutions of a differential equation being continuous with respect to the initial conditions. The formal discussion just covers a special case and indeed the proof is nothing but the fact that continuity means continuity from both sides. Thank you for your patience anyway.
A: I don't see any problem with defining the variational field. By assumption $f$ is smooth on each half-closed rectangle $(-\epsilon,\epsilon)\times [t_i,t_{i+1}]$. So, as long as we fix $t\in [0,a]$ and vary $s$, we have a smooth map $f(\cdot,t)$
into $M$ even if $t$ is one of the break points. The variational field is found by taking its derivative at the point where $s=0$. It is continuous, but may be  nonsmooth at the break points.
Now, why is $E'(0)=0$? First, we have a telescoping sum $$\sum_{i=0}^{k} \left\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle\bigg|_{t_{i}}^{t_{i+1}} = 
 \left\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle\bigg|_{0}^{a} =0$$
  because the $s$-derivative vanishes at the endpoints. 
Next, inside of the integral we have 
$$
\frac{D}{dt}\frac{\partial f}{\partial t}\bigg|_{s=0} = \frac{D}{dt} \dot{\gamma}(t) =0
$$ 
  because the velocity field of a geodesic is "constant" (i.e., parallel). 
