Integration by parts for the index form

On one piece, choose a local frame and let $u,w$ be the coefficient columns of fields $U,W$, with $H$ and $B$ as in [F4]–[F5]. Then $g(U,W)=u^{\mathsf{T}}Hw$. Differentiating this expression and using $H^{\mathrm{\prime }}=B^{\mathsf{T}}H+HB$ gives $\frac{d}{dt}g(U,W)=g(D_{t}U,W)+g(U,D_{t}W)\mathrm{.}$ Apply this identity with $U=D_{t}V$. Since $V$ is $C^2$ and $W$ is $C^1$ on the piece, the function $G(t):=g(D_{t}V,W)$ is continuous up to its one-sided endpoints and satisfies $G^{\mathrm{\prime }}=g(D_t^{2}V,W)+g(D_{t}V,D_{t}W)$ in the interior. The identity is local in $t$, so no family of frames is selected.

On each piece, set $Q(t):=g(D_t^{2}V+R(V,\dot{\gamma })\dot{\gamma },W)\mathrm{.}$ The functions $Q$ and $G^{\mathrm{\prime }}$ extend continuously to that piece's endpoints, so they are integrable by [F7]. The index-form integrand from [F1] obeys $g(D_{t}V,D_{t}W)-g(R(V,\dot{\gamma })\dot{\gamma },W)=G^{\mathrm{\prime }}(t)-Q(t)$ by step 1.1 and bilinearity of $g$. Integral linearity [F8] and Newton–Leibniz [F6] therefore give, on $[t_{k-1},t_k]$, ${\int }_{t_{k-1}}^{t_k}(g(D_{t}V,D_{t}W)-g(R(V,\dot{\gamma })\dot{\gamma },W))dt=g(D_{t}V(t_k^{-}),W(t_k))-g(D_{t}V(t_{k-1}^{+}),W(t_{k-1}))-{\int }_{t_{k-1}}^{t_k}Q(t)dt\mathrm{.}$

Sum the identity of step 2.1 over the fixed finite subdivision. At an interior breakpoint $t_j$, continuity of $W$ makes its two boundary contributions $g(D_{t}V(t_j^{-}),W(t_j))-g(D_{t}V(t_j^{+}),W(t_j))=-g({\mathrm{\Delta }}_j{D}_{t}V,W(t_j))\mathrm{.}$ The two outer contributions are exactly $g(D_{t}V(b^{-}),W(b))-g(D_{t}V(a^{+}),W(a))$. Substituting the piecewise-sum definition [F1] for the supplied geodesic [F10] proves the displayed formula. Since it equals the subdivision independent $I_{\gamma }(V,W)$ for every common subdivision, the right-hand expression is independent of the chosen one as well.

If $W(a)=W(b)=0$, the outer endpoint term vanishes; with moving endpoints it must be retained. If $V=0$, the covariant-derivative terms vanish and [F9] makes the curvature term zero; if $W=0$, every term vanishes by bilinearity. The condition $a<b$ excludes a degenerate singleton interval, and all endpoint and breakpoint derivatives are one-sided. If $M$ is empty no supplied geodesic exists; in dimension zero all fields and both sides are zero. Dimension one requires no separate argument because no division by dimension or curvature symmetry is used in the local identity. The assumption [A1] is carried only through the declared index-form interface; this finite partition proof itself uses no countable selection and assumes no full Axiom of Choice. This is an equality, not an iff claim. [A1, F1, F2, F9, step 3.1] $\mathrm{\square }$