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This problem is in Spivak's Differential Geometry (Ch.9 #37), and he gives a sketch of a proof which I have been unable to finish.

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So let's compute $\frac{dL(\overline{\alpha}(u))}{du}\mid_{u=0}$ where $L(\overline{\alpha}(u))=L_{0}^{t_{1}}(\gamma)+\int_{t_{1}}^{t_{0}+u}F(\,,\,)dt+\int_{t_{0}+u}^{1}F(\,,\,)dt$ , and of course $F(\alpha(u,t),\frac{\partial\alpha}{\partial t}(u,t))=\sqrt{\underset{i,j}{\sum}g_{ij}(\alpha)\frac{\partial\alpha^{i}}{\partial t}\frac{\partial\alpha^{j}}{\partial t}}$ (all the "$(u,t)$" omitted).

Well, $\frac{dL(\overline{\alpha}(u))}{du}=\frac{d}{du}\int_{t_{1}}^{t_{0}+u}F(\,,\,)dt+\frac{d}{du}\int_{t_{0}+u}^{1}F(\,,\,)dt$ . Now I apply the Leibniz integral rule, and the terms become

  • $F(\alpha(u,t_{0}+u),\frac{\partial\alpha}{\partial t}(u,t_{0}+u))+\int_{t_{1}}^{t_{0}+u}\frac{\partial}{\partial u}F(\alpha(u,t),\frac{\partial\alpha}{\partial t}(u,t))dt$

and

  • $\int_{t_{0}+u}^{1}\frac{\partial}{\partial u}F(\alpha(u,t),\frac{\partial\alpha}{\partial t}(u,t))dt-F(\alpha(u,t_{0}+u),\frac{\partial\alpha}{\partial t}(u,t_{0}+u))$

, respectively. Evaluating their sum at $u=0$ , we just get $\int_{t_{1}}^{1}\frac{\partial}{\partial u}F(\alpha(0,t),\frac{\partial\alpha}{\partial t}(0,t))dt$ .

Note that $\alpha$ is not exactly a variation on $\gamma$ since $\alpha(0,t)$ has a piece of $\gamma$ replaced by a geodesic (*). But anyway, $\alpha$ is a variation on the piecewise smooth curve $\alpha(0,t)$ , and the integral obtained yields the First Variation Formula for Length of $\alpha(0,t)\mid_{[t_{1},1]} .$ . For the moment ignore *, and assume this is the thing we want to show $\neq0$ . The integral term in the First Variation Formula will disappear, leaving

$\left\langle \frac{\partial\alpha}{\partial u}(0,t_{0}),\frac{\partial\alpha}{\partial t}(0,t_{0}^{+})-\frac{\partial\alpha}{\partial t}(0,t_{0}^{-})\right\rangle$ .

This is sort of like $\left\langle \frac{\partial\alpha}{\partial u}(0,t_{0}),\Delta_{t_{0}}\frac{d\gamma}{dt}\right\rangle$ , where we know $\Delta_{t_{0}}\frac{d\gamma}{dt}\neq0$ .

Assuming everything was correct so far, I have two questions:

1) Why (where) do we need $t_{1}$ to be sufficiently close to $t_{0}$, as hinted by Spivak ?

2) How can I conclude that the inner product term is not 0?

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2 Answers 2

If the metric changes sufficiently smoothly at the corner point ($C^1$ or something like that) then geodesics between points that approach the corner will closely approximate geodesics in the tangent space, where the answer follows from Minkowski or Cauchy inequalities applied to the tangent metric. This is not affected by the infinitesimal amount of sewing at the ends needed to change the geodesics to variations.

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So in essence there are vectors in $M_{\gamma(t_0)}$: a) representing the distance of the curve from $t_1$ to $t_0$, b)from $t_0$ to $t_0$+u, and c) connecting the ends of (a) and (b), representing the length of the geodesic. Then I know the change in (a)=(b) vs change in (c) as $u\to0$ is not equal to 0, provided the directions (a) and (b) are linearly independent. Can you state the inequalities you refer to? I don't doubt your method is ok, but I should be able to show it using the calculus of variations... –  Forever Mozart Jan 31 '13 at 5:55
    
You want that two sides of a triangle in the tangent space have length greater than the third, $|u| + |v| > |u+v|$, with equality only when collinear, and length measured by $|x|=\sqrt{g(x,x)}$. This is Minkowski's inequality. It can be proved by squaring and using the Cauchy-Schwarz inequality, which is true with the usual proof for any positive definite bilinear form. –  zyx Jan 31 '13 at 6:35
    
Ok. Now regarding your comment on approximating the corner geodesics with straight lines in $M_{\gamma(t_0)}$. Here is an earlier exercise I need to prove that should show this... If $v,w\in M_{p}$ then $\underset{v,w\to0}{\mbox{lim}}\frac{d(exp(v),exp(w))}{\left\Vert v-w\right\Vert }=1$ . That should do it. Still a little frustrated that the calculus of variations did not work. –  Forever Mozart Jan 31 '13 at 6:45
    
It probably does work, in a way that parallels the calculation proving Cauchy's inequality. Also, for the result to be true you might need some $C^1$ or other regularity assumption on the metric. –  zyx Jan 31 '13 at 6:50

Some more thoughts...

Can we just define $\alpha(0,t)$ to be $\gamma(t)$; in other words, let the geodesic "short-cuts" begin only when u>0? Would this destroy the required $C^\infty$ properties of $\alpha$? The geodesics are parametrized arclength, while the piece of $\alpha$ they approach can have an arbitrary parametrization.

Another note: By moving $t_1$ toward $t_0$ we can change the length of $\frac{\partial\alpha}{\partial t}(0,t_{0}^{-})$, since the geodesic portion is parametrized by arclength. This may be useful when we need to show the vectors in the inner product are not orthogonal.

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