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The first variation formula says that $$\left.\frac{\mathrm d }{\mathrm d s}\right|_{s=0}\ell(\varphi_s)=\frac{1}{c}\left[\left<\frac{\partial \varphi}{\partial s}(0,t),\frac{\partial \varphi}{\partial t}(0,t)\right>\right]_{t=a}^b-\frac{1}{c}\int_a^b\left<\frac{\partial \varphi}{\partial s}(0,t),\nabla _{\dot\gamma }\dot\gamma \right>\mathrm d t.$$

where $\gamma :[a,b]\longrightarrow M$ is $\mathcal C^2$, $\|\dot \gamma \|=c$, $$\varphi:]-\varepsilon,\varepsilon[\times [a,b]\longrightarrow M,$$ is a variation of $\gamma $ s.t.$$\varphi=\varphi(s,t),\quad \varphi(s,a)=\gamma (a),\quad \varphi(s,b)=\gamma (b)\quad \text{and}\quad \varphi(0,t)=\gamma (t).$$

Now, I have a corollary that says that if $\gamma $ minimise then length between $\gamma (a)$ and $\gamma (b)$, then $\gamma $ is a geodesic. The proof goes like that : Let $\varphi$ like above. Since $\gamma $ minimise the length, $$0=\left.\frac{\mathrm d }{\mathrm d s}\right|_{s=0}\ell(\varphi_s)\underset{(*)}{=}-\frac{1}{c}\int_a^b\left<\frac{\partial \varphi}{\partial s}(0,t),\nabla _{\dot\gamma }\dot\gamma \right>\mathrm d t,$$ and thus, by the fondamental theorem of variation calculus, $\nabla _{\dot\gamma }\dot\gamma =0$, and thus $\gamma $ is a geodesic.

I dont understand the equality $(*)$, i.e. why $$\left[\left<\frac{\partial \varphi}{\partial s}(0,t),\frac{\partial \varphi}{\partial t}(0,t)\right>\right]_{t=a}^b=0.$$

user330587
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1 Answers1

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Because $\dfrac{\partial\phi}{\partial s}(0,a) = \dfrac{\partial\phi}{\partial s}(0,b) = 0$! As you yourself wrote, the endpoints stay fixed during the variation. Indeed, $\dfrac{\partial\phi}{\partial s}(s,a) = \dfrac{\partial\phi}{\partial s}(s,b) = 0$ for all $s$.

Ted Shifrin
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  • Dear Ted, I'm sorry but I don't see why $\frac{\partial \partial }{\partial s}(s,a)=\frac{\partial \varphi}{\partial s}(s,b)$ for all $s$. It indeed stay fix, but I don't see in what it implies that the derivate are nulle. – user330587 Jun 26 '16 at 16:23
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    What does the partial derivative mean? You fix $t=a$ and you ask how $\phi(s,a)=\gamma(a)$ is varying as you vary $s$. Well, it's not! – Ted Shifrin Jun 26 '16 at 16:29