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Along the downward gradient flow, the gradient-flow dissipation identity, the chain rule and the defining property of the Riemannian gradient giveA smooth vector field on a compact manifold is complete, so the trajectory exists for every $t\ge 0$. In particular, its function value never increases.

If the trajectory is still in $C_{[a,b]}$ at time $t$, its monotone function value has stayed in $[a,b]$ throughout $[0,t]$. ConsequentlyThus a uniform exit bound isThe trajectory is outside the closed band for every $t>T$. Once it leaves below level $a$, monotonicity prevents its return. The argument also covers the zero-width band $a=b$.

For the total occupation-time bound for a gradient flow, let $m={\min }_{M}f$ and $F={\max }_{M}f$, which exist by compactness. For every $R>0$, the same gradient flow identity givesTaking $R\to \infty$, by monotone convergence theorem, gives the total occupation-time boundThis allows arbitrarily many visits to $C_{\epsilon }$; it bounds their total duration. Neither of these two estimates needs the Morse function hypothesis.

For exponential convergence of a Morse gradient flow, first observe that a limiting point $z=x_{\infty }$ is a critical point. Indeed, writing ${\varphi }_s$ for the flow and using continuous dependence on initial data,Differentiation at $s=0$ gives $\nabla f(z)=0$.

We use the Morse lemma: near a nondegenerate critical point of Morse index $\lambda$, there are coordinates $(u,v)\in {\mathbb{R}}^{\lambda }\times {\mathbb{R}}^{n-\lambda }$ in whichChoose a Riemannian metric equal to the Euclidean metric on a smaller ball in these coordinates. Such a global Riemannian metric exists by a partition of unity: patch this metric to any background metric outside a slightly larger ball. One can do this simultaneously at every critical point, since a Morse function on a compact manifold has only finitely many critical points. The metric is chosen before forming its gradient flow; changing the metric can change trajectories.

For this metric, any trajectory converging to $z$ eventually remains in the smaller coordinate ball. From some time $t_0$ onward its equations are exactlyConvergence forces $u(t_0)=0$, and thenThe straight radial segment stays in the coordinate ball, so its length bounds the Riemannian distance from above:Increase $C$ beyond both $e^{2t_0}|v(t_0)|$ and ${\max }_{0\le t\le t_0}{e}^{2t}d({\gamma }_x(t),z)$. Then the requested strict estimate holds for all $t\ge 0$:This proof uses the Morse lemma, existence and uniqueness for smooth ordinary differential equations, continuous dependence on initial data, and the existence of a partition of unity. The expanding coordinates explain why convergence to a saddle occurs only along its stable manifold.

For a counterexample, use an arc coordinate $x$ around a point of the circle, its flat local metric, and a globally smooth function that equals $x^4$ on this arc. Such a function can be made by a smooth cutoff, agreeing with a positive constant outside a larger arc. For $x_0>0$ sufficiently small the downward gradient flow remains in the arc and isIt converges to the degenerate critical point $0$, and its Riemannian distance from $0$ is $x(t)$ for sufficiently large $t$. For every $k>0$, $e^{kt}x(t)\to \infty$, so no exponential bound is possible. This counterexample is on a compact manifold as well.

Take the normalizationThe first variation of geodesic energy says that the critical points on the fixed-endpoint path space are exactly the affinely parametrized geodesics. The kernel of the second variation of geodesic energy, or Riemannian index form, is the space of Jacobi fields vanishing at both endpoints. Hence a geodesic is a nondegenerate critical point exactly when that space is zero, or equivalently when its terminal endpoint is not a conjugate point to its initial endpoint along the given geodesic.

The Morse index theorem states that the index is the sum of the multiplicities of all conjugate points strictly inside the parameter interval:The multiplicity at the terminal endpoint gives nullity rather than an additional contribution to the index. These statements use the usual Sobolev $H^1$ completion of the path space; smooth paths have the same homotopy type.

First suppose $p\ne q$ as well as $p\ne -q$, and put $\theta =d(p,q)\in (0,\pi )$. There is a unique great circle through these two points. Every critical path runs along it with constant speed, possibly passing around it extra times. Ifthen a complete enumeration without repetitions isIts length is $L=|\ell |$. Equivalently the two positive length lists areThere are no other critical paths, because a nonconstant geodesic of the round sphere is a constant-speed great circle and its plane must contain $p,q$.

To compute conjugate points and indices of round-sphere geodesics, the round unit sphere has constant sectional curvature $1$. Along a geodesic of speed $L$, a normal Jacobi field in a parallel direction satisfies $y^{\prime \prime }+L^{2}y=0$. A field with $J(0)=0$ is therefore a constant multiple of $\sin (Lt)$ in each of the $n-1$ normal directions. Its tangential component satisfies $y^{\prime \prime }=0$ and contributes no endpoint-vanishing field. Consequently the conjugate points occur at $Lt=j\pi$, each with multiplicity $n-1$. Since neither length list contains an integer multiple of $\pi$, all the displayed critical paths are nondegenerate, and the Morse index theorem gives

The permitted case $p=q$ requires separate treatment. There is a constant geodesic, which has index and nullity zero. All other critical paths areFor each $r$, this is a family parametrized by $S^{n-1}$; both directions of traversal are included through $v$ and $-v$. The interior conjugate points number $2r-1$, and the endpoint is also conjugate. ThusThese are Morse-Bott critical manifolds for $n>1$, rather than nondegenerate critical points. For $n=1$ the same formulas give zero index and nullity, with the two isolated directions comprising $S^0$.

For the requested homology calculation, choose distinct nonantipodal endpoints. Concatenation with a fixed path back to the basepoint gives a homotopy equivalence between this fixed-endpoint path space and the based loop space $\Omega S^n$. We use the Morse cell-attachment theorem for geodesic energy: on a complete compact Riemannian manifold, the fixed-endpoint energy, when all its critical points are nondegenerate, gives a CW complex of the same homotopy type with one cell of dimension equal to the index of each critical point. One can obtain this theorem from finite-dimensional broken-geodesic approximations and ordinary Morse theory; critical energy values tend to infinity here.

The two index lists interleave to give exactly one cell in every dimension $j(n-1)$, $j\ge 0$. If $n>2$, these dimensions are separated by at least two. The cellular homology groups therefore have zero boundary maps, since no occupied cell dimension has an occupied dimension one lower. With integer coefficients,This determines the graded abelian groups. The argument does not need an identification of the multiplication on loop-space homology.

Finally apply the Freudenthal suspension theorem: for an $(n-1)$-connected based CW complex $X$, $n\ge 2$, the suspension homomorphismis an isomorphism for $k\le 2n-2$ and a surjection for $k=2n-1$. A sphere $S^n$ is $(n-1)$-connected, and its reduced suspension of a topological space is homeomorphic to $S^{n+1}$. HenceThe isomorphism is the suspension map. Its stable range is also reflected by the Morse theory cell structure of $\Omega S^{n+1}$: after its bottom $n$-cell, the next positive-dimensional cell has dimension $2n$. For the stated homology problem $n>2$, all the connectivity hypotheses apply.

At any point, the linearly independent vectors $X_1,\dots ,X_n$ can be completed by $Y$ to a basis of $T_{p}M$. A nowhere-zero volume form is nonzero on every basis, soThus $\boxed{\eta \text{ is nowhere zero}}$. In particular, if it is an exact differential form $df$, its potential has no critical points. This is the role of pointwise independence in the boundary obstruction for commuting volume-preserving vector fields.

By the preceding sections, $\eta$ is a nowhere-zero closed differential form and $j^{\ast }\eta =0$. Injectivity of $j^{\ast }:H_{\mathrm{dR}}^1(M)\to H_{\mathrm{dR}}^1(\partial M)$ gives $[\eta ]=0$. The definition of de Rham cohomology then givesfor a smooth $f$, with no critical points. If the boundary were connected, $j^{\ast }df=0$ would make $f$ constant there. A nonconstant function on a compact manifold attains both extrema; a value differing from the common boundary value gives an interior extremum, contradicting $df\ne 0$. If $f$ were constant everywhere there would be the same contradiction. Empty boundary is also impossible, because an extremum would necessarily be interior. Thus $\boxed{\partial M\text{ has at least two connected components}}$, the boundary obstruction for commuting volume-preserving vector fields.

The solid torus $S^1\times D^2$ retracts onto its first circle. Its degree-one de Rham cohomology is generated by that circle's angular one-form, which restricts to a nonzero class on $S^1\times S^1$; hence the restriction map is injective. Its boundary is connected. Extensions of the two specified coordinate vector fields would be tangent there and would satisfy all the forbidden conditions. Therefore no such pair of extensions exists, for any volume form.

LetIts gradient isIf this vanishes, the first and third coordinates give $x=y=z$, and the middle coordinate then gives $y=0$. Thus the origin is the only critical point of $F$. It has been removed from $V$, so $0$ is a regular value of the restriction under consideration. The regular level set theorem shows thatis a smooth surface.

A critical point of a smooth map is a point where its derivative is not surjective; its image is a critical value. A regular value is a target value all of whose preimages have surjective derivative.

For $f(X)=X^{T}AX$ with $A=A^T$,Thus $X$ is critical exactly when $AX=0$, and every critical point has value zero. Hence zero is the only possible critical value; it is a critical value precisely when $A$ is singular. If $A$ is singular, every point of its kernel is critical, so there can be infinitely many critical points; if $A$ is invertible, only the origin is critical.

$dU/dt=\nabla U\cdot (-\nabla U)=-|\nabla U{|}^2\le 0$. The descending gradient flow trajectory from $(2,1)$ stays in its bounded contour basin and tends to the only critical point there, $(1,0)$.

$\nabla U=(3x^2-3+y^2,2y(x+1))$, so the critical points are $(\pm 1,0)$. The Hessian matrix at $(1,0)$ is $\mathrm{diag}(6,4)$, making it the local minimum; $(-1,0)$ is degenerate and not a minimum.

For a holomorphic phase $\varphi$, a saddle point is a critical point $z_0$ with ${\varphi }^{\prime }(z_0)=0$. At a simple saddle, ${\varphi }^{\prime \prime }(z_0)\ne 0$, and the quadratic Taylor term determines the local descent directions.