Volume via Jacobi fields: Proof of Lemma 5.4 of Sakai's book "Riemannian Geometry" I am self-teaching Riemannian geometry and am stuck with the proof of Lemma 5.4 on page 65 of Takashi Sakai's book "Riemannian Geometry". The lines where I am stuck with are highlighted in the attached picture. In the picture, the $$\Vert \cdot \Vert$$ denotes the (absolute value of the signed) volume of the parallelepiped spanned by the vectors it embraces (and so please ignore the wedge $\wedge$ operator. For example, $\Vert v \wedge w \Vert$ for two m-dimensional vectors $v$ and $w$ is the (absolute value of the signed) area of the parallelogram spanned by $v$ and $w$).
I got why there is factor $t^{1-m}$ there but did not get why $$\Vert Y_{1}(t) \wedge \cdots \wedge Y_{m-1}(t)\wedge \dot{\gamma}(t)\Vert = \Vert Y_{1}(t) \wedge \cdots \wedge Y_{m-1}(t)\Vert,$$ where $\gamma_{u}(t)$ is a unit speed geodesic with initial speed vector $u=e_m$ (when $t=0$). Note that the Jacobi fields are not necessarily orthogonal along $\gamma_u(t)$.
I am not able to see how $\dot{\gamma}(t)$ and $\Vert \dot{\gamma}(t) \Vert \equiv 1$ plays their roles in the height-times-base formula for the above identity. Could someone please explain why the above identity is true? Any help or hint is appreciated.
Added: I think I got it. By Gauss Lemma, $\dot{\gamma}_u(t)$ is orthogonal to each $Y_{i}(t)$ for $i=1,\cdots,m-1$. Since $\Vert \dot{\gamma}(t) \Vert \equiv 1$, the height of the parallelepiped is $1$. So, the base area, $\Vert Y_{1}(t) \wedge \cdots \wedge Y_{m-1}(t)\Vert$, equals the volume.

 A: If $u$ is a unit tangent vector to $p \in M$, then $\gamma(t) = \exp_p(tu)$ is a unit speed geodesic. If $v \in T_pM$ is orthogonal to $u$, the Gauss lemma shows that
$$
g_{\gamma(t)}\left(\mathrm{d}\exp_p(tu)\cdot u, \mathrm{d}\exp_p(tu)\cdot v\right) = g_p(u,v)
$$
and then, $\mathrm{d}\exp_p(tu)\cdot v$ is orthogonal to $\mathrm{d}\exp_p(tu)\cdot u = \gamma'(t)$.
If $Y_i(t) = t\mathrm{d}\exp_p(tu)\cdot e_i$, then $Y_i(t)$ is colinear to $\mathrm{d}\exp_p(tu)e_i$, and if $e_i$ is orthogonal to $u$, then $Y_i(t)$ is orthogonal to $\gamma'(t)$. thus, if $Y_1(t),\ldots,Y_{m-1}(t)$ are all orthogonal to $\gamma'(t)$, one has
\begin{align}
\left\|Y_1(t)\wedge\cdots\wedge Y_{m-1}(t)\wedge \gamma'(t) \right\| &= \left\| Y_1(t) \wedge \cdots \wedge Y_{m-1}(t)\right\| \left\|\gamma'(t) \right\| \\
&= \left\| Y_1(t) \wedge \cdots \wedge Y_{m-1}(t)\right\|
\end{align}
because $\gamma$ has unit speed.
The first equality may seem a bit obscur. Here is a reason why. If $(e_1,\ldots,e_{m-1},u)$ is an orthonormal basis of $T_pM$, one can transport it as a parallel frame along $\gamma(t)$, say $(E_1(t),\ldots,E_{m-1}(t),\gamma'(t))$. It is an orthonormal basis of $T_{\gamma(t)}M$. The norm $\|Y_1(t)\wedge \cdots \wedge Y_{m-1}(t) \wedge \gamma'(t)\|$ is the absolute value of the determinant $\det\left(Y_1(t),\ldots,Y_{m-1}(t),\gamma'(t)\right)$ in any orthonormal basis. Thus, in the orthonormal basis $E_1(t),\ldots,E_{m-1}(t),\gamma'(t)$, this determinant have the form
$$
\begin{vmatrix}
g(Y_1(t),E_1(t)) & \cdots & g(Y_{m-1}(t),E_{1}(t)) & 0 \\
\vdots &  & \vdots & 0 \\
g(Y_1(t),E_{m-1}(t)) & \cdots & g(Y_{m-1}(t),E_{m-1}(t))\\
0 & 0 & 0 & \|\gamma'(t)\|
\end{vmatrix}
$$
and the formula is thus straightforward.
In the proof you are showing, the author takes $(e_1,\ldots,e_{m-1},u)$ to be and orthonormal basis. Thus, it is clear that $e_i$ and $u$ are orthogonal.
Remark: You can prove that $Y_i$ are all orthogonal to $\gamma'$ without the Gauss lemma, if you know that $V_i(t)=\mathrm{d}\exp_p(tu)\cdot e_i$ are Jacobi fields. Knowing they are Jacobi fields, they satisfy
$$
V_i'' = -R(\gamma',V_i)\gamma'
$$
Let $f(t) = g_{\gamma(t)}\left(V_i(t),\gamma'(t)\right)$. It is a smooth function, and one easily shows that
$$
f''(t) = g_{\gamma(t)}\left( V_i''(t),\gamma'(t)\right) = -R\left(\gamma'(t),V_i(t),\gamma'(t),\gamma'(t)\right) = 0
$$
the last equality being a direct consequence of the fact that the Riemann tensor is skew symmetric in the two last variable. Thus, $f$ is an affine function, of the form $f(t) = f'(0)t + f(0)$. The initial data saying that $V_i(0)$ and $V'_i(0)$ are orthogonal to $\gamma'(0)$ show that $f(0)=f'(0)=0$, thus $f=0$ and $V_i$ is orthogonal to $\gamma'(t)$.
Comment: there is a general use of these expressions for studying the volume outside a submanifold in a riemannian manifold. They are named Heintze-Kärcher equalities / inequalities, named after Heintze and Kärcher. You can find online (for example here) how they seemed to have first appeared.
