Let $\gamma:[0,\ell]\to M$ be a unit-speed <geodesic>, let $F(s,t)$ be a smooth variation with fixed endpoints, and let
$$
T=\dot\gamma,
\qquad
V=\left.\frac{\partial F}{\partial s}\right|_{s=0}
$$
be its <variation vector field>. Then $V(0)=V(\ell)=0$. If $V^\perp=V-\langle V,T\rangle T$ is its component normal to $\gamma$, the <second variation of Riemannian arc length> is
$$
\left.\frac{d^2}{ds^2}L(F(s,\cdot))\right|_{s=0}
=I(V^\perp,V^\perp)
=\int_0^\ell\left(
|D_tV^\perp|^2-
\langle R(V^\perp,T)T,V^\perp\rangle
\right)dt.
$$
Here $D_t=\nabla_T$ is the <covariant derivative> along $\gamma$, $R$ is the <Riemann curvature tensor>, and $I$ is the <Riemannian index form>. Fixed endpoints remove the boundary term. The normal projection removes a tangential change of parametrization, which does not change length to second order.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 131 / 2 / a / Solution

{title2=$V$}
{wiki=Variation_(geometry)}

For a variation $F(s,t)$ of a curve $\gamma(t)=F(0,t)$, the variation vector field is $V(t)=\partial_sF(0,t)$. A fixed-endpoint variation has $V=0$ at both endpoints.


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