First variation of volume for a normal variation

Fix $p\in M$ and extend a $g$-orthonormal basis $(e_1,\dots ,e_m)$ of $T_{p}M$ to local fields on $M$, lifted independently of $t$ to the product. Write $E_i(t)=d{F}_t(e_i)$ and $G_{ij}(t)=\bar{g}(E_i(t),E_j(t))$. In ambient coordinates, equality of mixed partials and the symmetric Christoffel symbols in [F5] give $D_t{E}_i{\mid }_{t=0}=D_{e_i}(dF({\mathrm{\partial }}_t)){\mid }_{t=0}={\bar{\mathrm{\nabla }}}_{e_i}V\mathrm{.}$

Metric compatibility and step 1.1 give ${\dot{G}}_{ij}(0)=\bar{g}({\bar{\mathrm{\nabla }}}_{e_i}V,e_j)+\bar{g}(e_i,{\bar{\mathrm{\nabla }}}_{e_j}V)\mathrm{.}$ At $p$, $G(0)=I$. Applying [F3]–[F4] therefore yields the pointwise density derivative ${\frac{d}{dt}\mid }_0{\mu }_{g_t}=\frac{1}{2}\mathrm{tr}(\dot{G}(0)){\mu }_g={\sum }_{i=1}^m\bar{g}({\bar{\mathrm{\nabla }}}_{e_i}V,e_i){\mu }_g\mathrm{.}$

Because $V$ is normal, [F6] converts each summand in step 2.1 to $-\bar{g}(V,{\mathrm{I}\mathrm{I}}_f(e_i,e_i))$. The trace formula [F1] now gives the global density identity ${\frac{d}{dt}\mid }_0{\mu }_{g_t}=-\bar{g}\text{\hspace{0.17em}⁣}(V,{\sum }_i{\mathrm{I}\mathrm{I}}_f(e_i,e_i)){\mu }_g=-m\bar{g}(V,{\mathbf{H}}_f){\mu }_g\mathrm{.}$ Both sides are intrinsic, so the pointwise calculation in an arbitrary orthonormal basis patches over $M$.

Write ${\mu }_{g_t}=J(t,p){\mu }_g$ on $K$. By [F2]–[F3], $J$ is smooth and positive near $\{0\}\times K$. Compactness supplies a closed parameter interval on which $\mid {\mathrm{\partial }}_{t}J\mid$ is bounded; the constant bound is integrable because [F8] gives $K$ finite volume. Thus [F7] passes the derivative through the fixed-domain integral, and step 3.1 gives $A_K^{\mathrm{\prime }}(0)={\int }_K{\mathrm{\partial }}_{t}J(0,p){\mu }_g=-m{\int }_K\bar{g}(V,{\mathbf{H}}_f){\mu }_g\mathrm{.}$

The last integrand in step 4.1 vanishes outside $\mathrm{supp}V$, which lies in ${\mathrm{int}}_{M}K$. Its extension by zero is therefore a smooth compactly supported density on $M$, and locality of the intrinsic integral in [F8] gives ${\int }_K\bar{g}(V,{\mathbf{H}}_f){\mu }_g={\int }_M\bar{g}(V,{\mathbf{H}}_f){\mu }_g\mathrm{.}$ The same equality holds for every eligible $K$, proving domain independence.

Now assume $M$ is boundaryless and drop the normality hypothesis. Decompose $V=V^{\mathrm{\top }}+V^{\perp }$. The tangent field corresponding to $V^{\mathrm{\top }}$ is compactly supported inside ${\mathrm{int}}_{M}K$, so [F9] supplies its flow ${\varphi }_t$, with ${\varphi }_t$ equal to the identity near $\mathrm{\partial }K$. Put ${\tilde{F}}_t=F_t\circ {\varphi }_t^{-1}$. Its variation field is $V-df(V^{\mathrm{\top }})=V^{\perp }$, while ${\varphi }_t(K)=K$ and [F10] gives ${\mathrm{Vol}}_{{\tilde{F}}_t^{\ast }\bar{g}}(K)={\mathrm{Vol}}_{F_t^{\ast }\bar{g}}(K)$. Applying steps 3.1–5.1 to $\tilde{F}$ proves the same formula for the original variation, since $\bar{g}(V^{\perp },{\mathbf{H}}_f)=\bar{g}(V,{\mathbf{H}}_f)$.

If $M$ is compact, it is itself an eligible compact domain (its interior relative to itself is $M$), so step 5.1 gives the total-volume formula.

For empty $M$ (of fixed positive dimension), both integrals and the derivative are zero. Dimension zero is excluded because the averaged vector contains $1\mathrm{/}m$; in dimension one the proof is the single diagonal Gram calculation. A zero variation field gives zero pointwise. Immersivity in [F2] excludes a degenerate pullback metric. The parameter value $0$ is interior to $(-\epsilon ,\epsilon )$, and the density argument applies to the boundary of $K$ without orientation or integration by parts. The all-variation clause is restricted to boundaryless $M$ because [F9]'s two-sided reparametrizing flow need not exist for a field transverse to a manifold boundary. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F1], [F6], and [F8]; [F9]–[F10], the pointwise finite-basis calculation, and one compactness bound add no choice.