Gradient Like Vector Fields and Morse Trajectories

If ${\mathrm{grad}}_{g}f(x)=0$, then [F1] gives $d{f}_x(v)=g_x(0,v)=0$ for every $v$, so $d{f}_x=0$.

Conversely, if $d{f}_x=0$, then [F1] gives $g_x({\mathrm{grad}}_{g}f(x),v)=0$ for every $v$. Taking $v={\mathrm{grad}}_{g}f(x)$ and using positive definiteness gives ${\mathrm{grad}}_{g}f(x)=0$.

By [F2], the two implications say exactly that ${\mathrm{grad}}_{g}f(x)=0$ if and only if $x$ is critical.

By [F3], $(f\circ \gamma {)}^{\mathrm{\prime }}(t)=d{f}_{\gamma (t)}(\dot{\gamma }(t))$.

Substituting [F1] into step 1.1 and applying [F2] gives $(f\circ \gamma {)}^{\mathrm{\prime }}(t)=-g_{\gamma (t)}({\mathrm{grad}}_{g}f,{\mathrm{grad}}_{g}f)$.

The right-hand side in step 2.1 is $-\parallel {\mathrm{grad}}_{g}f(\gamma (t)){\parallel }_g^2$, proving the identity on $I$.

If $\gamma (t_0)$ were critical, [F2] would make the vector field $-{\mathrm{grad}}_{g}f$ vanish there, so the constant curve at $\gamma (t_0)$ is an integral curve through that point.

By [F3], that constant integral curve and $\gamma$ agree on $I$, contradicting the hypothesis that $\gamma$ is nonconstant. Hence $\gamma (t)$ is never critical.

By [F2] the gradient is nonzero at every $\gamma (t)$, and [F1] now gives $(f\circ \gamma {)}^{\mathrm{\prime }}(t)<0$ for every $t\in I$.

By [F1], choose pairwise disjoint Morse-coordinate neighbourhoods of the finitely many critical points, and smaller neighbourhoods inside them. On each smaller neighbourhood prescribe the local field in [F3].

On the complement of the smaller neighbourhoods, $df$ is nowhere zero. In each coordinate patch choose a vector $Y$ with $df(Y)<0$; a partition of unity and cutoffs that equal one on the smaller neighbourhoods patch these choices with the prescribed local fields to a smooth $X$ satisfying both clauses of [F3].

The resulting $X$ is smooth on the compact manifold $M$, so [F2] makes it complete. Therefore it is the required complete downward gradient-like field.

Each $K_T:=\overline{\gamma ([T,\mathrm{\infty }))}$ is a nonempty closed subset of $K$, the family is decreasing, and each $K_T$ is connected because it is the closure of the connected image of $[T,\mathrm{\infty })$.

If ${\bigcap }_{T\ge 0}{K}_T$ were empty, the open sets $K\setminus K_T$ would cover $K$; [F1] would give finitely many of them that cover. Since the $K_T$ decrease, one already covers, contradicting $K_T\mathrm{\ne }\mathrm{\varnothing }$. Thus $\omega (\gamma )$ is nonempty; it is closed in $K$, hence compact, and the nested connected-set argument makes it connected.

For fixed $s$ and any $T$, [F2] sends $\gamma ([T,\mathrm{\infty }))$ into $\gamma ([T+s,\mathrm{\infty }))$ after increasing $T$ if necessary. Continuity therefore sends $\omega (\gamma )$ into itself; applying the same argument to $-s$ gives equality.

Replacing $t$ by $-t$ gives the asserted nonempty compact connected invariant set $\alpha (\gamma )$ for a precompact negative tail.

On a precompact positive tail, $f\circ \gamma$ is decreasing by [F2] and bounded below because $f$ is continuous on its compact closure. It therefore has a finite limit $\mathrm{\ell }$; every point of $\omega (\gamma )$ is a limit of tail values and has $f=\mathrm{\ell }$. The same argument, with increasing time reversed, applies to $\alpha (\gamma )$.

Let $z$ lie in either limit set. If $z$ were noncritical, [F3] and [F2] would give a sufficiently short positive orbit segment from $z$ on which $f$ strictly decreases.

By [F1] the entire short segment in step 1.2 remains in the same limit set, whereas step 1.1 makes $f$ constant there. This contradiction proves that $z$ is critical.

By [F1], the negative-gradient field is complete, so $\gamma$ has domain $\mathbb{R}$. Both tails have compact closure because they lie in $M$.

By [F2] each of $\alpha (\gamma )$ and $\omega (\gamma )$ is nonempty and connected, and [F3] places it in $\mathrm{Crit}(f)$.

By [F4], $\mathrm{Crit}(f)$ is finite and hence discrete. A connected subset of a discrete finite set is one point, so both limit sets are single critical points.

A trajectory with singleton tail-limit set converges to that point: otherwise a sequence of tail times outside a fixed neighbourhood would have a limit point in the same tail-limit set. Thus the two displayed limits hold.

Integrating [F2] on $[-T,T]$ gives $f(\gamma (-T))-f(\gamma (T))={\int }_{-T}^T\parallel {\mathrm{grad}}_{g}f(\gamma (t)){\parallel }_g^{2}dt$.

Letting $T\to \mathrm{\infty }$ and using [F1] and continuity of $f$ yields the stated improper-integral equality.

By [F3], $f\circ \gamma$ is strictly decreasing, so $f(p)>f(q)$. The equality in step 2.1 therefore has strictly positive value.

By [F1], the coordinate flow solves $\dot{u}=2u$ and $\dot{v}=-2v$, hence $u(t)=e^{2t}u(0)$ and $v(t)=e^{-2t}v(0)$.

A point remains near $p$ and converges to it in forward time exactly when $u(0)=0$; in backward time exactly when $v(0)=0$. Thus the local stable disk is $\{u=0\}$ and the local unstable disk is $\{v=0\}$.

By [F2], the $u$-space has dimension $\lambda$ and is the negative Hessian space, while the $v$-space has dimension $n-\lambda$ and is the positive one. This gives the claimed disk dimensions and tangent spaces.

Let $D^u$ and $D^s$ be the local disks from [F1]. From the defining limits in [F2], every $x\in W^u(p)$ has ${\mathrm{\Phi }}_{-T}(x)\in D^u$ for some $T\ge 0$, and every $x\in W^s(p)$ has ${\mathrm{\Phi }}_T(x)\in D^s$ for some $T\ge 0$.

Hence $W^u(p)={\bigcup }_{T\ge 0}{\mathrm{\Phi }}_T(D^u)$ and $W^s(p)={\bigcup }_{T\ge 0}{\mathrm{\Phi }}_{-T}(D^s)$. By [F3], these are increasing compatible immersed-manifold charts transported from the disks.

The standard flow-exhaustion parametrization of these compatible disks identifies the unions with the corresponding Euclidean spaces; their dimensions are those in [F1]. Thus they are immersed submanifolds diffeomorphic to ${\mathbb{R}}^{\lambda }$ and ${\mathbb{R}}^{n-\lambda }$, with no embeddedness conclusion.

If $x\in W^s(p)$, then [F2] gives ${\mathrm{\Phi }}_t({\mathrm{\Phi }}_s(x))={\mathrm{\Phi }}_{t+s}(x)\to p$ as $t\to \mathrm{\infty }$, so ${\mathrm{\Phi }}_s(x)\in W^s(p)$ by [F1]. The same calculation with $t\to -\mathrm{\infty }$ proves inclusion for $W^u(p)$.

Applying step 1.1 with $-s$ and using the inverse in [F2] proves the reverse inclusions.

Therefore both stable and unstable sets are invariant under every fixed flow time.

If the orbit had period $T>0$, then $f(\gamma (T))=f(\gamma (0))$, contradicting [F1].

If ${\mathrm{\Phi }}_{t_k}(x)\to x$ with $t_k\to \mathrm{\infty }$, fix $s>0$. For all sufficiently large $k$, $t_k\ge s$, so [F1] gives $f({\mathrm{\Phi }}_{t_k}(x))\le f({\mathrm{\Phi }}_s(x))<f(x)$.

Continuity of $f$ makes the left side of step 1.2 tend to $f(x)$, a contradiction. Thus the orbit is not recurrent either.

Suppose the right endpoint $T$ of $I$ were finite. The local flow neighbourhoods supplied by [F3] cover the compact set $K$ in [F1], so finitely many suffice; their time radii have a positive minimum $\epsilon$.

Choose $t\in I$ with $T-t<\epsilon$. Since $\gamma (t)\in K$, the corresponding local solution from [F3] extends $\gamma$ beyond $T$; uniqueness identifies it with $\gamma$ on the overlap, contradicting [F4].

The same argument at the left endpoint proves $I=\mathbb{R}$. If endpoint levels lie in $[a,b]$, [F2] verifies the usual monotone trapping in that slab once the trajectory is known to remain between those levels.