Let $\Gamma:]-\epsilon,\epsilon[\times [a,b]\rightarrow M$ be a proper variation through geodesics, must the length of the transverse curves $\Gamma_s$ be constant as $s$ varies ?
I thought the ellipsoid could give a counterexample, because it has two equators of unequal length. However, I read on Wikipedia that the shorted path between two points on the equator of an ellipsoid need not live in the equator. Thus, it seems that I don't understand an ellipsoid's geodesics very well and therefore not sure if it is a counterexample.
Thank you
Answer is yes.
Since each $\Gamma_s$ is a geodesic with domain $[a,b]$, therefore each transverse curve will have length $(b-a)|(\Gamma_s)'(0)|=(b-a)|\partial_2\Gamma(s,0)=(b-a)\sqrt{<\partial_2\Gamma(s,0),\partial_2\Gamma(s,0)>}|$. It suffices to show that $<\partial_2\Gamma(s,0),\partial_2\Gamma(s,0)>$ is constant as $s$ varies.
$$\frac{d}{ds}<\partial_2\Gamma(s,0),\partial_2\Gamma(s,0)>=2<\nabla_1∂ _2 Γ(s,0), ∂ _2 Γ(s,0)>=2<\nabla_2∂ _1 Γ(s,0), ∂ _2 Γ(s,0)>=0 $$.
The rightmost term in the above line is $0$, because $\partial_1\Gamma$ is Jacobi field (as $\Gamma$ is variation through geodesics), and the tangential part of this Jacobi field is $0$ ( because it is a fixed variation). It follows that $\partial_1\Gamma$ is orthogonal to $\partial_2\Gamma$, thus differentiating $<\partial_1\Gamma, \partial_2\Gamma>=0$ with respect to the second argument gives:
$$0=<\partial_1\Gamma, \partial_2\Gamma>'=<\nabla_2\partial_1\Gamma,\partial_2 \Gamma>+<\partial_1\Gamma,\nabla_2\partial_2\Gamma>=<\nabla_2\partial_1\Gamma,\partial_2 \Gamma>$$
($\nabla_2\partial_2 \Gamma=0$ because $\Gamma$ is variation through geodesics).
An explanation to why the tangential part of the Jacobi field is 0: