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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:

  1. $f(0,t)=c(t)$ for all $t\in[0,a]$
  2. 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 $0\leq i\leq k$.
  3. $f$ 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?

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2 Answers

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.

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I follow everything up until the final line. How does the computation you did tell you the left and right hand limits agree? (Also, if you type "@Jason" in your message, I'll receive a notification telling me to come back here. Under some circumstances, which I forget, the system automatically sends notifications, but it never hurts to purposefully do it anyway). –  Jason DeVito Jan 1 '13 at 18:26
    
@JasonDeVito : The final line is a bit cut short: if we look at $\lambda_{j}^{1}:=\lambda_{j}\big|_{(-\alpha,\alpha)\times(t_{i}-\delta,t_{i}]}$ and if we look at $\lambda_{j}^{2}:=\lambda_{j}\big|_{(-\alpha,\alpha)\times[t_{i},t_{i}+\delta)}$‌​, then because $\lambda_{j}^{1},\lambda_{j}^{2}$ are continuously differentiable: $$\frac{\partial\lambda_{j}}{\partial s}(s,t_{i}^{+})=\lim_{h\to 0,h>0}\frac{\lambda_{j}^{1}(s+h,t_{i})-\lambda_{j}^{1}(s,t_{i})}{h}=\lim_{h\to 0,h>0}\frac{\lambda_{j}(s+h,t_{i})-\lambda_{j}(s,t_{i})}{h}$$ Similarly from below. –  M. Luethi Jan 1 '13 at 18:53
    
In other words, I was just being dense ;-). The argument you just wrote is exactly what I used in my head to convince myself the $\lambda$s were continuous. I've voted you up now. –  Jason DeVito Jan 1 '13 at 19:37
    
Thank you. And yes, the answer is pretty straight forward and bot you and Pavel had good arguments (which is what I was thinking of as well), still I was keen to find out how to prove this. Do you know a nice argument why it should be obvious? –  M. Luethi Jan 1 '13 at 21:03
    
I may be misthinking, but doesn't everything boil down to the fact that a function on the real line is continuous iff it's continuous from the left and continuous from the right? –  Jason DeVito Jan 1 '13 at 23:43
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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).

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Unfortunately the previous answer and with it my comment was deleted. I am fine with $E'(0)=0$, no issues there. My problems arise with $$\sum_{i=0}^{k}\left.\left\langle\frac{D}{\partial s}\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle\right|_{t_{i}}^{t_{i+1}}=0$$ Given that the variational field need not be smooth at the endpoints, why do we know that $\frac{D}{\partial s}\frac{\partial f}{\partial s}(0,t_{i}^{+})=\frac{D}{\partial s}\frac{\partial f}{\partial s}(0,t_{i}^{-})$ whenever $1\leq i\leq k$? –  M. Luethi Jan 1 '13 at 15:53
    
@M.Luethi I see. You obviously put a ton of effort into writing the question, but stopped short of making the actual question unambigious. "the first [...] zero, since f is proper and γ is a geodesic. I do not see the reason for this." is not nearly as clear as "Why is $$\sum_{i=0}^{k}\left.\left\langle\frac{D}{\partial s}\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right\rangle\right|_{t_{i}}^{t_{i+1}}$$ equal to zero?" –  user53153 Jan 1 '13 at 17:01
    
The problem seems to be that text within $<$...$>$ is not displayed. It should have been "the first [...] $<$term is$>$ zero, since f is proper and γ is a geodesic." Sorry. Corrected this. Btw. I athink I almost got the answer in local coordinates, which is a bit tedious but should be true anyway. –  M. Luethi Jan 1 '13 at 17:40
    
@M.Luethi: I didn't mean to cause you difficulties by deleting my previous answer. I can undelete it if you want. –  Jason DeVito Jan 1 '13 at 17:43
    
No problem. If you want, you could check my answer :) –  M. Luethi Jan 1 '13 at 17:53
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