Stable Unstable Manifolds and Morse Smale Transversality

The Morse--Smale condition says that the two immersion maps from $W^u(p)$ and $W^s(q)$ are transverse. Thus [F2] makes their fibre product a smooth manifold. Its evaluation image is precisely $W^u(p)\cap W^s(q)$, hence, by the point-marked convention, $\tilde{\mathcal{M}}(p,q)$.

The dimension of that transverse fibre product is $\lambda (p)+(n-\lambda (q))-n=\lambda (p)-\lambda (q)$ by [F1].

Equality $s\cdot \gamma =\gamma$ says $\gamma (t+s)=\gamma (t)$ for every $t$, so if $s\mathrm{\ne }0$ then $\gamma$ is periodic with nonzero period $\mathrm{\mid }s\mathrm{\mid }$.

By [F2] the orbit is nonconstant, while [F1] makes $f\circ \gamma$ strictly decreasing; it therefore cannot be periodic. Hence $s=0$, which is exactly freeness.

Continuity and the endpoint limits give a time $t_0$ with $f(\gamma (t_0))=c$; strict decrease in [F1] makes $t_0$ unique. Translating by $t_0$ therefore gives exactly one representative in the stated slice.

Conversely, two slice representatives in one time orbit differ by a translation, and step 1.1 forces that translation to be zero. The slice lies in the embedded hypersurface of [F2], giving the asserted identification.

Choose a regular $c$ strictly between $f(q)$ and $f(p)$. The map $f$ restricted to $\tilde{\mathcal{M}}(p,q)$ has nonzero derivative along the flow direction, since $df(X)<0$ away from critical points. Hence its $c$-level is a codimension-one smooth submanifold.

By [F2], that level is bijective to $\mathcal{M}(p,q)$; transport its smooth structure across this bijection. Its dimension is $\lambda (p)-\lambda (q)-1$ by [F1] and step 1.1.

If $\mathcal{M}(p,q)$ were nonempty, [F1] would give it dimension $\lambda (p)-\lambda (q)-1\le -1$.

A nonempty smooth manifold has a nonnegative integer dimension, contradicting step 1.1. Thus the moduli space is empty.

Differentiating $\dot{\gamma }+{\mathrm{grad}}_{g}f(\gamma )=0$ in a decaying variation gives the displayed operator, by [F1]. Its kernel is the tangent space of the point-marked solution set.

The hyperbolic Hessians at $p$ and $q$ give exponential dichotomies at the two ends. The standard first-order Fredholm theorem therefore gives index $\lambda (p)-\lambda (q)$. Its adjoint solvability condition identifies the dual cokernel with the annihilator of $T_{\gamma (0)}{W}^u(p)+T_{\gamma (0)}{W}^s(q)$, rather than canonically identifying the cokernel itself with a tangent-space quotient.

That annihilator vanishes exactly when the two tangent spaces span $T_{\gamma (0)}M$, which is transversality. Since a Fredholm operator is onto exactly when its cokernel vanishes, this proves the biconditional.

At each $x$, the finite-dimensional kernel and cokernel in [F1] permit the standard local Lyapunov--Schmidt reduction of $P$ to a $C^h$ map between finite-dimensional spaces whose source-target dimension difference is $m$.

Fredholm maps are locally proper on suitable closed neighbourhoods. Using the countable bases, choose countably many such neighbourhoods $N_i$ covering $\mathcal{X}$. On each $N_i$, Lyapunov--Schmidt reduction and [F3] show that the image of the critical set has empty interior; properness makes that image closed, hence nowhere dense.

Every critical value belongs to one of these countably many nowhere-dense images. Their union is meagre, so [F2] makes its complement residual; every point of that complement is a regular value of $P$.

Metric variations supported where $df\mathrm{\ne }0$ supply the missing cokernel directions of the fixed-metric operator in [F1]; consequently the universal section linearization is onto.

The Banach implicit-function theorem therefore makes its zero set a Banach manifold. Eliminating the trajectory variable leaves a Fredholm projection to the metric parameter, in the sense of [F2].

The asymptotic Morse splitting computes the parametrized projection's index as $\lambda (p)-\lambda (q)$. Time translation is free and contributes the one-dimensional kernel direction; passing to the quotient therefore lowers the induced projection's index to $\lambda (p)-\lambda (q)-1$.

At a zero $(g,\gamma )$ of the universal section, write its surjective linearization as $L(\delta g,\xi )=A(\delta g)+D_{\gamma }\xi \mathrm{.}$ Then the tangent space to the universal zero set is $\ker L$, and the differential of its metric projection is $(\delta g,\xi )\mapsto \delta g$. If $D_{\gamma }$ is onto, this projection is onto by solving $D_{\gamma }\xi =-A(\delta g)$. Conversely, if the projection is onto, write any target vector as $A(\delta g)+D_{\gamma }\xi$ using surjectivity of $L$, choose $(\delta g,\eta )\in \ker L$, and subtract to obtain $D_{\gamma }(\xi -\eta )$ equal to that target. Hence a metric is a regular value of the projection exactly when every corresponding $D_{\gamma }$ is onto, which by [F4] is exactly the required stable--unstable transversality.

If $p=q$, the stable and unstable tangent spaces at $p$ are complementary eigenspaces of the hyperbolic gradient linearization; strict descent excludes any other intersection. If $p\mathrm{\ne }q$ and $f(p)\le f(q)$, strict descent makes the intersection empty. These cases hold for every metric. Hence suppose $f(p)>f(q)$.

Near each metric $g_{\ast }$ choose compact local unstable and stable disks at $p,q$, with parametrizations depending continuously in $C^1$ on $g$ in a $C^2$ neighbourhood $U$. Here is the parameter dependence needed: in coordinates near either zero fix its hyperbolic linearization $L$ at $g_{\ast }$ and write $X_g(x)=Lx+R_g(x)$. After a cutoff on a sufficiently small ball, $R_g$ has uniformly small Lipschitz constant. For prescribed small stable coordinate $z$, the integral equation $x(t)=e^{tL}z+{\int }_0^t{e}^{(t-s)L}{P}_s{R}_g(x(s))ds-{\int }_t^{\mathrm{\infty }}{e}^{(t-s)L}{P}_u{R}_g(x(s))ds$ is a uniform contraction on a small exponentially weighted space of paths on $[0,\mathrm{\infty })$. Its fixed point and its derivative with respect to $z$ depend continuously on $g$; the derivative solves a linear contraction equation. Evaluation at zero gives the stable disk. Reversing time gives the unstable disk. Restrict to smaller closed parameter balls so both disks extend beyond their boundaries. Finite-time flows also depend continuously in $C^1$ on $g$.

For integers $j,k\ge 0$, let $T_{jk}\subset U$ require transversality of the disk maps ${\mathrm{\Phi }}_j^g{D}_g^u(p)$ and ${\mathrm{\Phi }}_{-k}^g{D}_g^s(q)$ at every coincidence of points in their compact parameter domains. Tangent spaces here are those of the extended disks, including at boundary points. This is an open condition: if failing metrics converged in $C^2$ to a metric satisfying it, compactness would give convergent coincidence parameters, and the closed rank-deficiency condition would contradict transversality at their limit. Every point of a global stable or unstable manifold eventually flows into the interior of the corresponding local disk. Thus all $T_{jk}$ together are equivalent to the desired global transversality on $U$.

Each $T_{jk}\cap {\mathcal{G}}^{\mathrm{\infty }}$ is dense in $U\cap {\mathcal{G}}^{\mathrm{\infty }}$. Indeed, start at any smooth $g_0$ in a specified basic smooth neighbourhood controlling derivatives through order $r$. Choose finite $H\ge \max \{r,2\}$ above the Sard--Smale threshold. Fix $g_0$ on small disjoint critical neighbourhoods. Every trajectory between distinct critical points leaves their union, so metric variations outside them are the variations allowed in [F1]. By [F1], [F2], step 1.1 and Baire, there are arbitrarily $C^H$-close metrics in this affine Banach parameter space for which the pair is transverse. Choose one, $g_1$, still in $U$ and within the prescribed derivative bounds; it satisfies $T_{jk}$.

Approximate $g_1$ by a smooth symmetric tensor using convolution in a finite coordinate cover and a smooth partition of unity. This converges in $C^H$ because $H$ is finite and $M$ is compact. Sufficiently close approximants remain positive definite, remain in the prescribed neighbourhood, and still satisfy $T_{jk}$ by its $C^2$ openness. Only this compact-disk condition needs to survive smoothing. Since $g_0$ was arbitrary, this proves the claimed smooth density without fixing any common critical-neighbourhood data throughout $U$.

The smooth symmetric tensors on compact $M$ form a separable complete metrizable space under their countable derivative seminorms; positivity is an open condition. Hence ${\mathcal{G}}^{\mathrm{\infty }}$ is second countable and completely metrizable (an open subset admits an equivalent complete metric). Choose countably many neighbourhoods $U_i$ as above and closed sets $V_i\subset U_i\cap {\mathcal{G}}^{\mathrm{\infty }}$ whose interiors cover ${\mathcal{G}}^{\mathrm{\infty }}$. This is obtained by taking sufficiently small closed balls from a countable metric basis. For each $i,j,k$, set $O_{ijk}=({\mathcal{G}}^{\mathrm{\infty }}\setminus V_i)\cup (T_{i,jk}\cap {\mathcal{G}}^{\mathrm{\infty }})\mathrm{.}$ These sets are open and dense by steps 2.1 and 4.1.

By [F3] and dependent choice, ${\bigcap }_{i,j,k}{O}_{ijk}$ is dense and residual. Any metric in this intersection belongs to some $V_i$, hence satisfies every $T_{i,jk}$ and therefore the global pair transversality by step 2.1. The complement of the desired set is consequently contained in a countable union of closed nowhere dense sets, so the desired set itself is residual. Together with step 1.2 this covers all ordered pairs.

By [F2], there are only finitely many ordered pairs $(p,q)$ of critical points. For each pair, take the residual set of metrics supplied by [F1].

Their finite intersection is residual and consists exactly of metrics for which every $W^u(p)$ is transverse to every $W^s(q)$, namely the Morse--Smale metrics.

Thus the conclusion is residual in the smooth metric space. No openness or simultaneous statement about varying functions or continuation families has been used.

Work componentwise. A compact manifold has finitely many connected components, and [F1] gives finitely many critical points on each. On a fixed component, order that component's critical values. Its continuous image under $f$ is an interval, so every value chosen strictly between consecutive critical values is attained; it is regular because there is no critical value between them. Thus its fibre is nonempty, and [F3] supplies the smooth level hypersurface. These finitely many levels cut the components of $M$ into finitely many bands.

On a component of positive dimension, write its critical points as $c_1,\dots ,c_N$ with ${\alpha }_1>\cdots >{\alpha }_N$, where ${\alpha }_j=f(c_j)$. Induct on the destination: after stage $j$, require $W^s(c_h)⋔W^u(c_i)$ for every $h\le j$ and every $i$. The assertion for $h=1$ is automatic, since a maximum has stable manifold consisting only of itself. For a general stage $j$, the case of a local maximum is automatic for the same reason. Otherwise the stable disk at $c_j$ has a compact sphere section $Q$ of dimension $\dim M-\lambda (c_j)-1$ at a level just above ${\alpha }_j$. Choose this section outside the fixed ball but inside its larger chart. A short flow collar of it lies above the fixed ball's maximum height, below ${\alpha }_{j-1}$, and outside every other fixed ball, by their disjoint height windows.

Let $L$ be the collar's upper regular level. The stable section $Q\subset L$ has a tubular product $Q\times D^{\lambda (c_j)}$, furnished by the unstable coordinates in the Morse chart and transport by the flow. Write $Q_w=Q\times \{w\}$ for small $w$. The evaluation $(q,w)\mapsto (q,w)\in L$ is a submersion. For each higher critical point $c_i$, the unstable slice $W^u(c_i)\cap L$ is an immersed manifold: local unstable disks are transported by finite-time flow, and the flow direction is transverse to $L$. Cover each such slice by countably many embedded immersion charts. Apply [F2] to the family $Q_w$ and each chart image. The union of the exceptional null sets is null, so there are arbitrarily small $w$ for which $Q_w$ is transverse to every higher unstable slice. For $\lambda (c_j)=0$, the stable manifold is open and this transversality already holds without a perturbation.

Realize the chosen displacement by a field perturbation in the collar. In product flow coordinates $(q,u,z)$ with $X=-{\mathrm{\partial }}_z$, bottom $z=0$, and top $z=T$, take a smooth function $\beta$ supported in $(0,T)$ with ${\int }_0^T\beta (z)dz=1$, and a cutoff $\chi (u)$ equal to one near zero and zero near the boundary of the normal disk. Put $X_w=-{\mathrm{\partial }}_z-\beta (z)\chi (u){\sum }_{\mathrm{\ell }}{w}_{\mathrm{\ell }}{\mathrm{\partial }}_{u_{\mathrm{\ell }}}\mathrm{.}$ For small $w$, backward flow from $(q,0,0)$ stays where $\chi =1$ and reaches $(q,w,T)$. Thus the stable sphere at the upper section becomes precisely $Q_w$. The perturbation is zero near the entire collar boundary, extends smoothly by $X$, and tends to zero in $C^1$ with $w$. It preserves the fixed critical balls.

All higher unstable slices at $L$ are unchanged: their backward trajectories lie above $L$, whereas the perturbation is below $L$. Hence step 3.1 gives the desired transversality for destination $c_j$. Every trajectory from a higher critical point to $c_j$ crosses $L$, and flow transports this tangent-space condition along it. For each earlier destination $c_h$, $h<j$, its stable manifold lies in $\{f\ge {\alpha }_h\}$, entirely above the perturbation. At any intersection there, the backward trajectory defining the unstable manifold also stays in this unchanged superlevel region. Thus both tangent manifolds at every previously established intersection are unchanged. This proves the induction invariant of step 2.1.

There are finitely many stages and components, so choose each parameter within a finite $C^1$ error budget. On the compact collar supports $df(X)<0$ has a strict margin, so sufficiently small choices retain $df(X_w)<0$; on the fixed balls the exact local normal form is untouched. Closedness gives completeness. Self-intersections at a critical point have complementary stable/unstable tangent spaces, and strict descent excludes other identical-endpoint or reversed-value connections. Cross-component intersections are empty. Zero-dimensional components require no perturbation. Thus the final field is arbitrarily $C^1$-close, agrees on all fixed balls, and is Morse--Smale.

Substitution in [F1] gives $\dim \mathcal{M}(p,q)=1-1=0$.

A zero-dimensional smooth manifold is discrete, proving the claim without any compactness assertion.

Substitution in [F1] gives $\dim \mathcal{M}(p,q)=2-1=1$.

Therefore the moduli space is a one-dimensional smooth manifold, with no assertion about its ends.

Integrating the strict decrease in [F1] along each component gives $f(p_i)>f(p_{i+1})$.

Since each component exists, the contrapositive of [F2] gives $\lambda (p_i)>\lambda (p_{i+1})$.

Each strict index drop is at least one and indices are nonnegative, so there can be at most $\lambda (p_0)$ components.