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Completing squares reveals the geometry of the sheared quartic double-well gradient flow:The critical points satisfy $2(x+y)=0$ and $4x^3-2x+2y=0$, so they are $(0,0)$, $(1,-1)$ and $(-1,1)$. The Hessian matrix isAt the origin its determinant is $-8$, so $(0,0)$ is a saddle point, with $V=0$. At either other critical point, the determinant is $16$ and the upper-left entry is $10$, so it is a positive-definite matrix. They are strict local minima; the completed squares show that both are global minima, with $V=-1$.

A contour at level $c$ obeysThere are no contours for $c<-1$, two isolated minimum points for $c=-1$, and two separate closed ovals for $-1<c<0$. At $c=0$ the two lobes meet at the saddle point; near the origin their tangents are $y=(-1\pm \sqrt{2})x$. For $c>0$ one closed contour surrounds both minima.

Along the gradient flow, the gradient-flow dissipation identity isAt the stated initial point, $V(1,-1/2)=-3/4$. Its trajectory stays in the invariant sublevel set $V\le -3/4$. This set is compact because $V$ is a coercive function. It also excludes every point with $x=0$, where $V=y^2\ge 0$; continuity therefore keeps the trajectory in its positive-$x$ component.

To justify its limit rather than merely infer it from the sketch, integrate the dissipation identity: ${\int }_0^{\infty }\Vert \nabla V{\Vert }^{2}dt$ is finite. On this compact set the gradient and its time derivative are bounded, so $\Vert \nabla V{\Vert }^2$ is uniformly continuous. A nonnegative uniformly continuous function with finite integral tends to zero: otherwise separated intervals of a fixed positive height and width would force an infinite integral. Every accumulation point is consequently a critical point. In the positive-$x$ component below this energy level the only one is $(1,-1)$. Compactness then implies convergence, andThis is a compact gradient-flow trapping criterion; the drawn trajectory illustrates, rather than replaces, the convergence proof.

The Levi-Civita connection is the unique connection on a vector bundle $TM$ that is torsion-free and metric-compatible. These conditions areExistence and uniqueness follow from the Koszul formula; equivalently, these are the defining properties that characterize the connection.

Define the correction tensorand set $D_{X}Y={\nabla }_{X}Y+A(X,Y)$. All terms of $A$ are $C^{\infty }(M)$-linear in both $X$ and $Y$, so $D$ is a connection. The symmetry $A(X,Y)=A(Y,X)$ preserves vanishing of the torsion tensor.

To check compatibility with $\tilde{g}$, computeOn the other hand,The cross terms cancel, so their sum is $2X(\sigma )g(Y,Z)$. Multiplying by $e^{2\sigma }$ proves $X(\tilde{g}(Y,Z))=\tilde{g}(D_{X}Y,Z)+\tilde{g}(Y,D_{X}Z)$. Hence $D$ is torsion-free and compatible with $\tilde{g}$, so uniqueness of the Levi-Civita connection givesThis is the Levi-Civita connection under conformal rescaling.

On a nonempty compact manifold without boundary, $\sigma$ attains a maximum at some $p$. The derivative of a smooth function vanishes at an interior extremum, so $(d\sigma {)}_p=0$ and $(V_{\sigma }{)}_p=0$. The correction tensor therefore vanishes at $p$, givingThis is equality for all vector fields at the same point, not an assertion that either connection vanishes there. More generally, the two connections agree at any critical point of $\sigma$. Conversely, if their correction vanishes for all $X,Y$ at a point of positive dimension, evaluating $g(A(v,v),v)=g(v,v)d\sigma (v)$ for every $v$ shows that the point is critical. This is the critical-point criterion for equal conformal connections.

The boundaryless hypothesis is the usual convention for “manifold” here. If manifolds with boundary were included, compactness alone would not suffice: on $[0,1]$ with $g=d{x}^2$ and $\sigma =x$, one has ${\tilde{\nabla }}_{{\partial }_x}{\partial }_x={\partial }_x$ everywhere while ${\nabla }_{{\partial }_x}{\partial }_x=0$. An extremum at the boundary need not be a critical point.

Expand the function into separated terms: $f=g(x)+h(y)$, where $g=x^2-x^4$ and $h=y^8-y^4$. ThenPut $a=2^{-1/2}$ and $b=2^{-1/4}$. The nine critical points are all pairs from $x\in \{0,\pm a\}$ and $y\in \{0,\pm b\}$. The diagonal Hessian matrix has entries $2-12x^2$ and $56y^6-12y^2$.

At $(0,\pm b)$ both entries are positive, giving strict local minima of value $-1/4$. At $(\pm a,\pm b)$ they have opposite signs, giving four saddle points of value zero. At the three points on $y=0$ the Hessian is degenerate, so its sign alone does not classify them. Use the higher-order test for separated extrema: $h(y)=-y^4+O(y^8)$ is negative for small nonzero $y$. At $(0,0)$ the positive $x^2$ term and negative $y^4$ term give a saddle. At $(\pm a,0)$, $g-1/4=-(x^2-1/2{)}^2$, so every sufficiently small nonzero displacement decreases $f$ and these are strict local maxima.

For the level curves, the exact zero set consists of the two parabolas $x=\pm y^2$ and the oval $x^2+y^4=1$. They cross at the four nondegenerate saddle points. Small positive levels enclose each maximum; small negative levels enclose each minimum. The degenerate central saddle has quartic narrowing, locally $x^2-y^4=c$. Far along the $x$ axis the function is negative, while far along the $y$ axis it is positive. These signs and zero curves determine the connectivity shown in the sketch.

Use the Weinstein neighborhood theorem: a neighborhood of a compact Lagrangian submanifold is symplectomorphic to a neighborhood of the zero section of its cotangent bundle, with the identification equal to the identity on that submanifold. Take the canonical sign $-d\lambda$ in this identification. We also use C1 openness of diffeomorphisms: on a compact manifold, all smooth self-maps sufficiently close in the $C^1$ topology to a fixed diffeomorphism are themselves diffeomorphisms.

In $M\times M$ put $\Omega =-{\mathrm{pr}}_1^{\ast }\omega +{\mathrm{pr}}_2^{\ast }\omega$. The diagonal $\Delta$ is Lagrangian. For a symplectomorphism $\varphi$, its graph ${\Gamma }_{\varphi }$ is also Lagrangian, because its pullback of $\Omega$ is $-\omega +{\varphi }^{\ast }\omega =0$.

If $\varphi$ is sufficiently $C^1$-close to the identity, ${\Gamma }_{\varphi }$ lies in the fixed Weinstein neighborhood of $\Delta$. Its image in $T^{\ast }\Delta$ is transverse to the cotangent fibers and is a section: the projection of that image to $\Delta \cong M$ is $C^1$-close to the identity, hence is a diffeomorphism on compact $M$. Reparametrizing by this projection identifies the image with $\mathrm{graph}(\sigma )$ for a small differential one-form $\sigma$.

The preceding graph of a closed one-form is Lagrangian criterion gives $d\sigma =0$. Since $H_{\mathrm{dR}}^1(M)=0$, the de Rham cohomology definition gives $\sigma =df$. Intersections with the zero section are precisely the critical points of $f$; under the neighborhood identification these are the intersections ${\Gamma }_{\varphi }\cap \Delta$, hence the fixed points of $\varphi$.

On a nonempty compact manifold without boundary, $f$ has a maximum and a minimum. If it is nonconstant, these occur at distinct critical points. If it is constant, $df=0$ everywhere, so the whole graph is the diagonal and every point is fixed. For positive-dimensional $M$, there are at least two distinct fixed points. This is the nearby exact Lagrangian intersection lemma applied to the diagonal. No connectedness assumption is needed.

The usual positive-dimensional convention is necessary for the assertion: if zero-dimensional symplectic manifolds are allowed, a single-point $M$ has $H^1=0$ and only one fixed point. That is a literal exception to the printed statement.

For $ϵ>0$, dominant balance for algebraic roots gives two roots on the scale $\sqrt{ϵ}$ and a third on the scale ${ϵ}^{2/3}$. On the first scale put $t=\sqrt{ϵ}s$. The equation becomes $s^5-s^3+\sqrt{ϵ}=0$. The derivative of $s^5-s^3$ at either $s=1$ or $s=-1$ is two, so the first correction is $-\sqrt{ϵ}/2$. On the second scale put $t={ϵ}^{2/3}v$; then $1-v^3+{ϵ}^{1/3}{v}^5=0$. Writing $v=1+a{ϵ}^{1/3}+\cdots$ gives $a=1/3$. Thus all three real-root expansions for positive $ϵ$ areThese are fractional-power root expansions. To check that none is missing, the derivative of the polynomial is $t^2(5t^2-3ϵ)$. Its nonzero critical points are $\pm \sqrt{3ϵ/5}$; for sufficiently small positive $ϵ$ the polynomial is positive at the negative one and negative at the positive one. Its three monotone ranges therefore contain exactly these three real roots. The remaining two small roots correspond to the nonreal cube roots of unity in the second balance.

The printed limit does not specify the sign of $ϵ$. If a two-sided real limit is intended, also put $ϵ=-\eta$ with $\eta >0$. The polynomial $t^5+\eta t^3-{\eta }^3$ is strictly increasing and has just one real root. The same dominant balance for algebraic roots gives the negative-parameter branch

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.