The universal metric--trajectory projection is Fredholm

The affine space of $C^h$ symmetric tensors vanishing on the specified critical neighbourhoods is a closed Banach subspace $K$ of the $C^h$ tensor space on compact $M$. Positivity is open, giving ${\mathcal{G}}^h$. Near a connecting trajectory $\gamma$, use a fixed smooth background exponential map to represent nearby curves as small sections of $E=C_0^1(\mathbb{R},{\gamma }^{\ast }TM)$, and parallel transport to identify their equation values with $F=C_0^0(\mathbb{R},{\gamma }^{\ast }TM)$. The resulting universal section is $\mathrm{\Psi }(g,\eta )=\dot{\eta }+{\mathrm{grad}}_{g}f(\eta )$ in these charts. It is $C^{h-1}$: on the compact target $M$, derivatives of the finite-dimensional exponential and gradient maps through order $h-1$ are uniformly bounded, so their Taylor remainders are uniform in the supremum norms; tensor variations vanish on the critical neighbourhoods and curve variations decay at both ends, so every derivative takes values in $F$.

At a zero $(g,\gamma )$, differentiation gives $L(k,\xi )=Ak+D_{\gamma }\xi$, where $D_{\gamma }$ is [F1] and $Ak=-g^{-1}k({\mathrm{grad}}_{g}f,\cdot )$ along $\gamma$. The tensor term follows by differentiating $g^{-1}$: $\delta (g^{-1})=-g^{-1}k{g}^{-1}$. Since $D_{\gamma }$ is Fredholm, its range $R$ is closed and its quotient $Q=F\mathrm{/}R$ is finite-dimensional. To prove $L$ onto it suffices to prove that the image of $A:K\to F$ spans $Q$.

Suppose $A(K)$ did not span $Q$. A nonzero functional on the finite-dimensional quotient annihilating its image would lift to a nonzero bounded $\mathrm{\ell }\in F^{\ast }$ with $\mathrm{\ell }{D}_{\gamma }=0$ and $\mathrm{\ell }A=0$. In a parallel frame write $D_{\gamma }z=z^{\mathrm{\prime }}+H(t)z$, and let ${\mathrm{\Phi }}^{\mathrm{\prime }}=-H\mathrm{\Phi }$, $\mathrm{\Phi }(0)=I$. By [F3], $\mathrm{\Phi }(t)$ is invertible for every finite $t$. For any compactly supported continuous vector function $\psi$ with $\int \psi =0$, its compactly supported $C^1$ primitive $v(t)={\int }_{-\mathrm{\infty }}^t\psi (s)ds$ satisfies $D_{\gamma }(\mathrm{\Phi }v)=\mathrm{\Phi }\psi$, so $\mathrm{\ell }(\mathrm{\Phi }\psi )=0$. Subtracting a fixed compactly supported scalar function of integral one shows that $\mathrm{\ell }(\mathrm{\Phi }\psi )=c\cdot \int \psi$ for a fixed vector $c$. Hence on compactly supported $w$, $\mathrm{\ell }(w)=\int a(t)\cdot w(t)dt$ with $a(t)=\mathrm{\Phi }(t{)}^{-T}c$. Compactly supported functions are dense in $F$, so $c\mathrm{\ne }0$; thus $a(t)$ is continuous and nowhere zero.

Since $p\mathrm{\ne }q$, strict descent makes $\gamma$ injective and gives a point $x=\gamma (t_0)$ outside the fixed critical neighbourhoods with $v={\mathrm{grad}}_{g}f(x)\mathrm{\ne }0$. A small regular-value slab about $f(x)$ meets this orbit only in a short compact time interval. For any target vector $b$ a symmetric endomorphism sending $v$ to $b$ is $B=\frac{b\otimes v+v\otimes b}{\mathrm{\mid }v{\mathrm{\mid }}^2}-\frac{⟨b,v⟩}{\mathrm{\mid }v{\mathrm{\mid }}^4}v\otimes v\mathrm{.}$ Choose a tensor $k_0$ at $x$ for which $a(t_0)\cdot A{k}_0>0$, extend it smoothly, and multiply by a nonnegative bump supported in a small ball inside the slab. Shrinking the ball preserves nonnegative pairing along the entire short orbit segment and positive pairing near $t_0$. The resulting $k\in K$ has $Ak$ compactly supported along $\gamma$ and $\mathrm{\ell }(Ak)=\int a\cdot Ak>0$, contradicting $\mathrm{\ell }A=0$. Thus $A(K)$ spans $Q$ and $L$ is onto.

Choose finitely many tensors whose $A$-images form a basis of $Q$, giving a bounded lift $J:Q\to K$ with ${\pi }_{Q}AJ=I_Q$. By [F4] split $E=\ker D_{\gamma }\oplus E_1$, and let $B_R:R\to E_1$ be the bounded inverse of $D_{\gamma }{\mathrm{\mid }}_{E_1}$. Then $S(w)=(J{\pi }_{Q}w,B_R(w-AJ{\pi }_{Q}w))$ is a bounded right inverse of $L$. Consequently $I-SL$ is a bounded projection onto $\ker L$, and [F5] makes the local universal zero set a $C^{h-1}$ Banach manifold with tangent $\ker L$. For the projection $\pi (g,\gamma )=g$, its tangent kernel is $\ker D_{\gamma }$ and its image is $\ker ({\pi }_{Q}A)$, a closed subspace of codimension $\dim Q$. Therefore $d\pi$ is Fredholm with index $\dim \ker D_{\gamma }-\dim Q=\lambda (p)-\lambda (q)$ by [F1].

Time translation of a nonconstant trajectory is free because $f\circ \gamma$ is strictly decreasing. Fix any $c$ with $f(q)<c<f(p)$. Each connecting trajectory crosses $f=c$ exactly once, and at that crossing $\frac{d}{dt}f(\gamma (t))=-\mathrm{\mid }{\mathrm{grad}}_{g}f{\mathrm{\mid }}_g^2<0$. Evaluation $\gamma \mapsto f(\gamma (0))$ is $C^{h-1}$ in the trajectory charts. The ordinary implicit-function theorem therefore gives a local $C^{h-1}$ slice $f(\gamma (0))=c$ transverse to the translation direction $\dot{\gamma }\in \ker D_{\gamma }$. Since solutions of the $C^{h-1}$ gradient ODE depend smoothly on initial data and time on compact intervals, translating solutions to their unique crossing varies smoothly in these zero-set charts. The slices are compatible and identify the free translation quotient with a Banach manifold. Its projection to ${\mathcal{G}}^h$ is the restriction of $\pi$ to a codimension-one tangent hyperplane that removes the one-dimensional translation line from $\ker d\pi$ while leaving the image of $d\pi$ unchanged. Its Fredholm index is therefore $\lambda (p)-\lambda (q)-1$.