[step:Decompose $Y$ into normal and tangential parts and identify the corrections]Write $Y = Y_n + g(Y, \dot\gamma) \dot\gamma$ where $Y_n$ is the pointwise orthogonal projection of $Y$ onto $\dot\gamma^\perp$. Set $\phi(t) := g(Y(t), \dot\gamma(t))$, so $Y(t) = Y_n(t) + \phi(t) \dot\gamma(t)$. Differentiate this decomposition along $\gamma$:
\begin{align*}
Y' = \nabla_t Y = \nabla_t Y_n + \phi'(t) \dot\gamma(t) + \phi(t) \nabla_t \dot\gamma = \nabla_t Y_n + \phi'(t) \dot\gamma(t),
\end{align*}
since $\nabla_t \dot\gamma = 0$. Pairing with $\dot\gamma$ and using $\nabla_t \dot\gamma = 0$ together with metric compatibility:
\begin{align*}
g(Y', \dot\gamma) = g(\nabla_t Y_n, \dot\gamma) + \phi'(t).
\end{align*}
Now $Y_n \perp \dot\gamma$ at every $t$, so $g(Y_n, \dot\gamma) \equiv 0$. Differentiating this identity in $t$ via (M):
\begin{align*}
0 = \frac{d}{dt} g(Y_n, \dot\gamma) = g(\nabla_t Y_n, \dot\gamma) + g(Y_n, \nabla_t \dot\gamma) = g(\nabla_t Y_n, \dot\gamma),
\end{align*}
so $g(Y', \dot\gamma) = \phi'(t)$. Squaring and integrating:
\begin{align*}
\int_0^T g(Y', \dot\gamma)^2 \, d\mathcal{L}^1(t) = \int_0^T \phi'(t)^2 \, d\mathcal{L}^1(t).
\end{align*}
The integrand $|Y'|_g^2$ in Part 1 splits using the same orthogonal decomposition. Since $\nabla_t Y_n \perp \dot\gamma$ from above and $Y' = \nabla_t Y_n + \phi'(t) \dot\gamma$:
\begin{align*}
|Y'|_g^2 = |\nabla_t Y_n|_g^2 + \phi'(t)^2 \cdot |\dot\gamma|_g^2 = |Y'_n|_g^2 + \phi'(t)^2,
\end{align*}
where $Y'_n := \nabla_t Y_n$ and $|\dot\gamma|_g = 1$.
For the curvature term, observe $R(Y, \dot\gamma)Y = R(Y_n + \phi \dot\gamma, \dot\gamma)(Y_n + \phi \dot\gamma)$. By the antisymmetry $R(\dot\gamma, \dot\gamma) = 0$ in the first slot and $R(\cdot, \cdot)\dot\gamma$ pairs with $\dot\gamma$ to vanish (using $g(R(X, Y)\dot\gamma, \dot\gamma) = -g(R(X,Y)\dot\gamma, \dot\gamma)$ from skew-symmetry in the last pair, see [Symmetries of the Riemann Curvature Tensor](/theorems/2704)), every term involving the tangential factor $\phi \dot\gamma$ contributes zero to $R(Y, \dot\gamma, Y, \dot\gamma) = g(R(Y, \dot\gamma)Y, \dot\gamma)$. Concretely:
\begin{align*}
R(Y, \dot\gamma, Y, \dot\gamma) = R(Y_n, \dot\gamma, Y_n, \dot\gamma).
\end{align*}
For the boundary term, $\frac{\nabla Y}{ds}(t, 0)$ refers to the variation; under the same decomposition the tangential and normal parts contribute separately. The normal piece corresponds to $\frac{\nabla Y_n}{ds}(t, 0)$ in the formula stated in Part 2.
Combining: the second variation of $\ell$ becomes
\begin{align*}
\frac{d^2}{ds^2} \ell(\gamma_s)\Big|_{s = 0} &= \frac{d^2}{ds^2} E(\gamma_s)\Big|_{s = 0} - \int_0^T \phi'(t)^2 \, d\mathcal{L}^1(t) \\
&= g\!\left(\frac{\nabla Y}{ds}(t, 0), \dot\gamma\right)\Big|_0^T + \int_0^T \Big[|Y'_n|_g^2 + \phi'(t)^2 - R(Y_n, \dot\gamma, Y_n, \dot\gamma)\Big] d\mathcal{L}^1(t) - \int_0^T \phi'(t)^2 \, d\mathcal{L}^1(t).
\end{align*}
The two $\phi'(t)^2$ integrals cancel exactly. We now show the boundary term reduces to $g(\nabla Y_n / ds, \dot\gamma)\big|_0^T$ rigorously.
Define $\Phi(t, s) := g(Y(t, s), \partial_t H(t, s) / |\partial_t H(t, s)|_g)$ on a neighbourhood of $s = 0$ where $\partial_t H$ is non-vanishing. At $s = 0$, $\Phi(t, 0) = \phi(t)$. The endpoint-fixing hypothesis gives $\partial_s H(0, s) = Y(0, s) \equiv 0$ and $\partial_s H(T, s) = Y(T, s) \equiv 0$ for all $s$, hence $\Phi(0, s) \equiv 0$ and $\Phi(T, s) \equiv 0$ as functions of $s$. Differentiating in $s$:
\begin{align*}
\partial_s \Phi(0, 0) = 0, \qquad \partial_s \Phi(T, 0) = 0,
\end{align*}
and also $\Phi(0, 0) = \phi(0) = 0$, $\Phi(T, 0) = \phi(T) = 0$ (since $Y(0) = Y(T) = 0$).
The $\nabla_s$-derivative of the tangential summand $\Phi(\cdot, s)\, \partial_t H / |\partial_t H|_g$ at $s = 0$, $t \in \{0, T\}$, expands as
\begin{align*}
\partial_s \Phi(t, 0) \cdot \dot\gamma(t) + \Phi(t, 0) \cdot \nabla_s\!\left[\frac{\partial_t H}{|\partial_t H|_g}\right]\bigg|_{s = 0}.
\end{align*}
Both $\partial_s \Phi(t, 0) = 0$ and $\Phi(t, 0) = 0$ at $t \in \{0, T\}$, so the entire tangential contribution to the boundary term vanishes. Hence at the endpoints, $\nabla_s Y$ and $\nabla_s Y_n$ have the same inner product with $\dot\gamma$:
\begin{align*}
g(\nabla_s Y, \dot\gamma)\big|_{t \in \{0, T\}} = g(\nabla_s Y_n, \dot\gamma)\big|_{t \in \{0, T\}}.
\end{align*}
We obtain
\begin{align*}
\frac{d^2}{ds^2} \ell(\gamma_s)\Big|_{s = 0} = g\!\left(\frac{\nabla Y_n}{ds}(t, 0),\, \dot\gamma(t)\right)\Big|_0^T + \int_0^T \Big[|Y'_n|_g^2 - R(Y_n, \dot\gamma, Y_n, \dot\gamma)\Big] d\mathcal{L}^1(t),
\end{align*}
which is the statement of Part 2.[/step]
