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A tangent vector at $p$ is a derivation at a point: a real-linear map $v:C_p^{\infty }(M)\to \mathbb{R}$ on germs of smooth functions satisfyingAddition and scalar multiplication preserve this identity, defining the tangent space $T_{p}M$ as a vector space. A derivation annihilates constants. In a manifold chart $x=(x^1,\dots ,x^n)$ centered near $p$, the local factorization $f-f(p)={\sum }_i(x^i-x^i(p))f_i$, with $f_i(p)={\partial }_{i}f(p)$, gives $v(f)={\sum }_{i}v(x^i){\partial }_{i}f(p)$. The factorization follows by integrating the first derivatives along a straight segment in a sufficiently small coordinate ball. ThereforeConversely these coordinate derivatives satisfy the derivation identity, so they really give a basis rather than merely a spanning family.

The cotangent space is the dual vector space $T_p^{\ast }M=(T_{p}M{)}^{\ast }$. Its basis $d{x}^i{|}_p$ is characterized by $d{x}^i({\partial }_j)={\delta }_j^i$. If $\alpha =a_{i}d{x}^i=b_{j}d{y}^j$, the chain rule gives the cotangent coordinate transitionThe cotangent bundle is the disjoint union $T^{\ast }M={\coprod }_{p\in M}{T}_p^{\ast }M$, with projection $\pi (p,\alpha )=p$. For each base manifold chart $(U,x)$, defineIts inverse sends $(z,a)$ to $(x^{-1}(z),a_{i}d{x}^i{|}_{x^{-1}(z)})$. On overlaps the transition is $(x,a)\mapsto (y(x),b)$, with the displayed fibre-linear transformation. Its coefficients and inverse are smooth, so these maps give a smooth atlas in dimension $2n$. Give the total space the topology obtained by transporting the product topology through these charts. Different base points are separated by disjoint base neighborhoods; distinct points over the same base point are separated within one bundle chart. A countable base smooth atlas and countable product bases give second countability. Thus the total space is a Hausdorff, second-countable smooth manifold. The same charts are vector bundle trivializations, since $\pi$ is product projection and their fibre changes are invertible linear maps.

The intrinsic canonical one-form on a cotangent bundle isIn the bundle chart it is $\lambda ={\sum }_i{a}_{i}d{x}^i$. This definition is coordinate independent, hence its exterior derivative is the globally defined smooth differential two-formThis sign follows the fibre-first order in this problem; the equally common position-first symplectic form is $-d\lambda$. The volume formnever vanishes. A smooth manifold is orientable exactly when it admits a nowhere-zero top-degree form; its positive ordered bases determine a consistent orientation. Consequently $T^{\ast }M$ is an orientable smooth manifold, even when $M$ is not. This is cotangent bundle orientation. In dimension zero the same conclusion uses the nowhere-zero zero-form $1$.

Any nonsingular holomorphic function $f({\Phi }_1,\dots ,{\Phi }_n)$ of chiral superfields is itself a chiral superfield. The graded Leibniz rule and the chain rule giveThus sums, products, convergent power series and inverses on patches where the denominator is nonzero preserve chirality. An ordinary function involving ${\Phi }^{†}$ is generally not chiral, because $\bar{D}{\Phi }^{†}$ need not vanish. This is why a superpotential is holomorphic. Spacetime derivatives also commute with $\bar{D}$, so derivatives of chiral superfields remain chiral; chirality by itself does not impose renormalizability.

The characteristic curves solve $\dot{X}=X$, hence $X(t,\xi )=e^t\xi$. By the chain rule, $d[u(t,X(t,\xi ))]/dt=0$. Pulling back along the characteristic flow map givesFor $u_0\in C^1$, this is a classical solution: $u_t=-e^{-t}x{u}_0^{\prime }(e^{-t}x)$ and $u_x=e^{-t}{u}_0^{\prime }(e^{-t}x)$ cancel in the transport equation, and at $t=0$ the initial condition holds. Conversely, constancy along every characteristic curve forces this formula.

The tangent-line inequality for a differentiable concave function is $f(r)\le f(s)+f^{\prime }(s)(r-s)$. Taking $r=u=U_x$ and using the Hamilton-Jacobi equation yieldsAlong $x(\tau )=x_0+\tau f^{\prime }(s)$, the chain rule identifies the left side with $dU(\tau ,x(\tau ))/d\tau$. Integrating givesStrict concavity makes equality possible exactly when $u=s$ along the line, which will select the maximizing characteristic curve.

The original PDF assumes $u\ge 0$ at the beginning of this question; the TeX conversion omits that hypothesis. First replace a smooth nonnegative solution by $v=u+\epsilon >0$, which solves the same homogeneous divergence-form elliptic equation. For $w=v^{q/2}$, the chain rule gives the favorable termin the distributional sense. Thus $w$ is a weak subsolution. Testing its subsolution inequality with ${\zeta }^{2}w$ yields the same Caccioppoli inequality as before. More precisely, testing the equation for $v$ with ${\zeta }^2{v}^{q-1}$ givesIndeed the main term is at least $(q-1)\lambda \int {\zeta }^2{v}^{q-2}|\nabla v{|}^2$, and the cross term is bounded by twice $\Lambda$ times the square roots of this weighted gradient integral and $\int v^q|\nabla \zeta {|}^2$. Let $\epsilon \downarrow 0$ to obtain the stated power energy estimate.

Apply the fixed-domain Sobolev embedding theorem to $\zeta u^{q/2}$. The crucial constant stays uniform before taking the $2/q$ power:where $D_{\ell }\ge 1$ is independent of $q$. This is uniform power gain for elliptic solutions. The exponent is $q\alpha$, not the TeX conversion's $q^{\alpha }$. The PDF's weaker prefactor $C_{\ell }$ outside the $2/q$ power follows from this stronger estimate, but that weaker form alone cannot be iterated infinitely.

For completeness, the estimate also extends to nonnegative locally $H^1$ weak solutions without assuming they are already bounded. For $s\ge 0$ define $F_M(s)=s^{q/2}$ up to $M$ and continue it linearly with slope $(q/2)M^{q/2-1}$ above $M$; set ${\Psi }_M(s)={\int }_0^s{F}_M^{\prime }(r{)}^{2}dr$. Both have bounded derivatives for fixed $M$. Direct calculation gives ${\Psi }_M/F_M^{\prime }\le F_M$ and $F_M(s)\le (q/2)s^{q/2}$. Test the weak equation with ${\zeta }^2{\Psi }_M(u)$, using smooth approximations to these Sobolev chain rule functions if needed. The resulting weighted energy bound is $\int {\zeta }^2|\nabla F_M(u){|}^2\le 4R^2\int F_M(u{)}^2|\nabla \zeta {|}^2$. If $u\in L^q(B_{r_2})$, dominated convergence on the right and the Fatou lemma on the Sobolev left give the boxed gain as $M\to \infty$, with the same uniform constant. Starting with $q=2$, this justifies every subsequent iteration step.

The stellar adiabatic exponents are fixed-composition, constant-specific entropy derivatives:The chain rule gives ${\Gamma }_1({\Gamma }_2-1)/{\Gamma }_2={\Gamma }_3-1$. Define the pressure derivatives ${\chi }_{\rho }=(\partial \log P/\partial \log \rho {)}_T$ and ${\chi }_T=(\partial \log P/\partial \log T{)}_{\rho }$. The first law of thermodynamics, with $du=Tds+Pd\rho /{\rho }^2$, givesCombining this with $d\log P={\chi }_{\rho }d\log \rho +{\chi }_{T}d\log T$ on an adiabat givesThus the specific-heat ratio is not generally equal to the three stellar adiabatic exponents.

For the mixture, ${\chi }_{\rho }=\beta$ and ${\chi }_T=4-3\beta$. The specific heat capacity at constant volume, obtained by differentiating $u$ at fixed mass density, isTherefore the adiabatic exponents of a monatomic gas-radiation mixture and its specific-heat ratio areIf a relation involving only the exponents is wanted, eliminate $\beta$ from ${\Gamma }_1=\beta +(4-3\beta )({\Gamma }_3-1)$:For a pure monatomic perfect gas, $\beta =1$ and $\gamma ={\Gamma }_1={\Gamma }_2={\Gamma }_3=5/3$. For any calorically perfect gas with constant heat capacities, the same equality holds with its own ${\gamma }_g$. In a genuine gas-radiation mixture, $0<\beta <1$, $\gamma ={\Gamma }_1/\beta$ while the exponents are given separately above.

In the radiation limit, ${\Gamma }_1={\Gamma }_2={\Gamma }_3=4/3$ and the adiabatic temperature gradient is $1/4$. This follows independently from photon entropy: a comoving volume $V$ has $S\propto V{T}^3$, so an adiabat obeys $T\propto V^{-1/3}$ and $P\propto V^{-4/3}$. However, $c_P/c_V$ is singular in the pure-radiation limit, not $4/3$. This is the pure-radiation constant-pressure heat-capacity singularity. The equation $P=a{T}^4/3$ fixes temperature whenever pressure is fixed, so an ordinary constant-pressure temperature derivative is not available; along the mixture limit $\gamma \to \infty$ and $\beta \gamma \to 4/3$. For photons alone, mass-specific quantities additionally require a material mass label. The often quoted radiation index $4/3$ is its pressure-density adiabatic exponent, not a finite constant-pressure/constant-volume heat-capacity ratio.

For an autonomous Hamiltonian function, the chain rule and Hamilton's equations giveThus the energy is constant along each characteristic:The time-independent hypothesis is implicit in the displayed expression in this subpart. For the time-dependent Hamiltonian function allowed earlier, the correct Hamiltonian energy balance is insteadFor example $H(t,x,v)=t+|v{|}^2/2$ has a unique global Hamiltonian flow, but its value along a curve increases at unit rate. Thus conservation cannot be claimed for arbitrary time-dependent $H$.

Use the inverse change of variables $u=(x+y)/2$ and $v=(x-y)/2$. The chain rule gives the differential operatorsThe mixed partial derivatives cancel, assuming the twice continuously differentiable solution appropriate here. Consequently $f_{xy}=(g_{uu}-g_{vv})/4$.

The transformed equation isThe factors of $1/2$ belong to the inverse Jacobian matrix; using the forward transformation instead would give an incorrect factor.

For a smooth map $\mathbf{u}$, the Jacobian determinant is the determinant of its Jacobian matrix $A_{ai}={\partial }_i{u}_a$. In Einstein summation convention,Differentiate the supplied expression for $V_i$ by the product rule. The term differentiating ${\partial }_j{u}_a$ vanishes because ${\partial }_i{\partial }_j{u}_a$ is symmetric in $i,j$, while ${ϵ}_{ijk}$ is antisymmetric; the term differentiating ${\partial }_k{u}_b$ vanishes similarly. Only differentiation of $u_c$ remains:The last equality follows by the cyclic relabelling $(i,j,k)\mapsto (j,k,i)$, which has positive sign. This proves the Jacobian determinant as a divergence identity.

For a composition $\mathbf{w}=\mathbf{u}\circ \mathbf{v}$, the chain rule givesTaking determinants yieldsFor the first-order perturbation, use spatial derivative control in perturbations of the identity map: the usual smooth-family interpretation gives $D{\mathbf{u}}_t=I+tD\mathbf{F}+o(t)$ locally. Multilinearity of the determinant shows that its linear term comes from choosing one column of $tD\mathbf{F}$ and all other columns from $I$. This gives the trace:The regularity qualification matters: a pointwise $o(t)$ remainder for maps does not by itself give an $o(t)$ remainder for spatial derivatives. For instance $(x+t^2\sin (x/t^2),y,z)$ has a uniform $o(t)$ displacement and $\mathbf{F}=0$, but its Jacobian determinant at $x=0$ is $2$ for $t\ne 0$. Thus the first expansion needs the remainder in $C^1$ locally, or an equivalent smoothness assumption on the family.

Under the subsequent group law, ${\mathbf{u}}_0$ is the identity and ${\mathbf{u}}_{-t}$ is the inverse of ${\mathbf{u}}_t$. Write $j(t,\mathbf{x})=J[{\mathbf{u}}_t](\mathbf{x})$. Applying the composition formula to ${\mathbf{u}}_{t+h}={\mathbf{u}}_h\circ {\mathbf{u}}_t$ and the short-time expansion gives the Jacobian evolution of a smooth flow:HenceIf only the pointwise generator expansion is assumed initially, the group law still gives ${\partial }_t{\mathbf{u}}_t(\mathbf{x})=\mathbf{F}({\mathbf{u}}_t(\mathbf{x}))$. Uniqueness and smooth dependence for this smooth ordinary differential equation make it a smooth flow map, thereby justifying the differentiated expansion used above. Thus a divergence-free generator hasBy the change of variables formula, these maps preserve ordinary Euclidean volume and orientation. This is the Euclidean instance of a volume-preserving vector field.

Away from the origin, the given vector field is the gradient of the globally single-valued radius:The sphere has centre $(5,3,2)$, at distance $\sqrt{38}$ from the origin, and radius one. Its intersection circle is therefore wholly away from the origin, since every point on the sphere has distance at least $\sqrt{38}-1>0$. Along the closed curve the chain rule gives $\mathbf{F}\cdot d\mathbf{x}=dr$. Consequently, independently of orientation,The intersection is nondegenerate: the centre's distance from the plane is $4/\sqrt{35}<1$. No parametrization of the resulting circle is needed because $r$ is a potential of a conservative vector field throughout the punctured space.

The chain rule along a smooth curve isDifferentiating the circular parametrization gives the tangent vectorThus the chain rule turns the angular derivative of $u$ into $-y{u}_x+x{u}_y$. The given partial differential equation makes this equal to $u$, so along every portion of the circle inside the upper half-plane, $U(t)=u(\mathbf{x}(t))$ satisfies the ordinary differential equation $U^{\prime }=U$. Its solution is $U(t)=U(t_0)e^{t-t_0}$.

For the two specified points take $a=\sqrt{2}$. The initial point corresponds to $t_0=\pi /4$ and the terminal point to $t_1=3\pi /4$. The entire intervening arc lies in $y>0$, so no continuation outside the domain is used. The requested value isThese circular characteristic curves also show locally that the general solution has the polar-coordinate form $u(r,\theta )=C(r)e^{\theta }$, with $0<\theta <\pi$; there is no requirement of angular periodicity on the upper half-plane.

A conservative vector field on ${\mathbb{R}}^n$ is a vector field $\mathbf{V}=\nabla F$ for a globally defined smooth potential of a conservative vector field $F$. The chain rule givesso its line integral is path independent.

Green theorem states that, for a bounded planar region $D$ with piecewise smooth boundary and continuously differentiable $P,Q$ on a neighbourhood of its closure,The boundary is positively oriented: the region lies on the left while it is traversed, so outer components run counterclockwise and hole boundaries clockwise.

For the proposed potential of a conservative vector field, the fundamental theorem of calculus and differentiation under the integral sign giveSince $Q_x=P_y$, the last integral equals $P(x,y)-P(x,0)$, including negative $y$ with the usual signed-integral convention. Hence the proposed potential is global andThe fact that the vector field is defined on all of ${\mathbb{R}}^2$ ensures both straight integration segments stay inside its domain; a hole in the domain could obstruct a global potential.

Set $s=\log x$ and $Y(s)=y(e^s)$, so $0\le s\le 1$. By the chain rule, the equation becomes $Y^{\prime \prime }+2Y^{\prime }+(1+\lambda )Y=0$. The substitution $Y=e^{-s}v$ reduces it further toMultiplication by $\overline{v}$ and integration by parts gives $\lambda {\int }_0^1|v{|}^2={\int }_0^1|v^{\prime }{|}^2>0$ for a nonzero solution, ruling out zero and negative eigenvalues. For $\lambda =k^2>0$, the first boundary condition gives $v=A\sin (ks)$, and the second gives $k=n\pi$, $n=1,2,\dots$. Thus all eigenpairs arewith arbitrary nonzero scalar multiples of each eigenfunction. These are the complete list since all solutions of the constant-coefficient differential equation have been considered; the transformed sine system is also complete in the corresponding weighted Hilbert space.

Parametrize a streamline by $\mathbf{r}(s)$ with $d\mathbf{r}/ds=\mathbf{u}(\mathbf{r}(s))$. The chain rule givesTherefore $H$ is constant along every streamline. A different regular parametrization merely multiplies this derivative by a scalar and gives the same conclusion. For time-dependent flow, the claim concerns streamlines at a fixed time, rather than Lagrangian trajectories.

A smooth vector field is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle, so $X_p\in T_{p}M$. In a manifold chart, $X=X^i{\partial }_i$ with smooth coefficients. It acts on a smooth function by $Xf=X^i{\partial }_{i}f$. For a diffeomorphism $\alpha :M\to N$, the pushforward of a vector field isThe Lie bracket of vector fields is the commutator of their actions on smooth functions:The second derivatives of $f$ cancel, so this is again a vector field. The corresponding Jacobi identity is $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

A local flow of $X$ is a smooth map $(t,p)\mapsto {\varphi }_t(p)$ on an open neighbourhood of $\{0\}\times M$ satisfying ${\varphi }_0(p)=p$ and ${\partial }_t{\varphi }_t(p)=X_{{\varphi }_t(p)}$. On a sufficiently small neighbourhood and time interval, each ${\varphi }_t$ is a diffeomorphism onto its image, with inverse ${\varphi }_{-t}$. Uniqueness of integral curves of a vector field gives ${\varphi }_{s+t}={\varphi }_s\circ {\varphi }_t$ wherever both sides are defined. These are local statements; no assumption of a complete vector field is required.

A smooth tensor field of type $(r,s)$ is a smooth section of $(TM{)}^{\otimes r}\otimes (T^{\ast }M{)}^{\otimes s}$, where the factors are the tangent bundle and cotangent bundle. The flow definition of the Lie derivative of a tensor field isHere pullback applies $d{\varphi }_t^{-1}$ to each vector factor and the dual of $d{\varphi }_t$ to each covector factor, evaluating $T$ at ${\varphi }_t(p)$. Thus all tensors being differentiated lie in the same fibre over $p$.

For a smooth function, ${\varphi }_t^{\ast }f=f\circ {\varphi }_t$, and the chain rule givesFor a vector field $Y$, in local coordinates the expansions ${\varphi }_t(x)=x+tX(x)+O(t^2)$ and $d{\varphi }_t(x)=I+td{X}_x+O(t^2)$ giveConsequentlyPullback preserves tensor products and tensor contractions, so differentiation makes this definition a tensor derivation. It therefore agrees on every tensor field with the Lie derivative of a tensor field determined by these two formulas.

Now put ${\psi }_t=\alpha \circ {\varphi }_t\circ {\alpha }^{-1}$. The chain rule givesThus ${\psi }_t$ is the local flow of ${\alpha }_{\ast }X$. In the case $M=N$, if ${\alpha }_{\ast }X=X$, uniqueness of integral curves of a vector field makes ${\psi }_t={\varphi }_t$ on their common domains, or $\alpha \circ {\varphi }_t={\varphi }_t\circ \alpha$. Conversely, differentiating this commuting identity at zero gives $d{\alpha }_p{X}_p=X_{\alpha (p)}$, hence ${\alpha }_{\ast }X=X$. This proves that diffeomorphism invariance of a vector field is equivalent to commuting with its local flow, with every identity understood on the domain where its compositions exist.

In a manifold chart, write a smooth differential form as $\eta ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$, with $I$ increasing. Define its exterior derivative byTo check that this definition is global, change coordinates to $y$. The chain rule expresses $d{x}^i$ as ${\sum }_j({\partial }_j{x}^i)d{y}^j$. Differentiating it gives zero because the second partial derivatives of $x^i$ are symmetric and $d{y}^k\wedge d{y}^j$ is antisymmetric. The coordinate definition satisfies the graded Leibniz rule, so differentiating the transformed expression for $\eta$ leaves exactly ${\sum }_{I}d{a}_I\wedge d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. This agrees with the original formula and proves independence of the manifold chart. The same cancellation of mixed derivatives gives $d^2=0$.

The de Rham cohomology is thereforethe vector space of closed differential forms modulo exact differential forms.

Here is a Green-theorem proof of exactness on a disc. Identify the disc with a round disc centred at zero, and let $\eta =Pdx+Qdy$ be a closed differential form. Set $f(x)={\int }_{[0,x]}\eta$. For $x$ and $x+h$ in the disc, Green theorem on the oriented triangle with vertices $0,x,x+h$ gives zero boundary integral, since ${\partial }_{x}Q-{\partial }_{y}P=0$. HenceTaking $h=\epsilon v$ and dividing by $\epsilon$ yields $d{f}_x(v)={\eta }_x(v)$. The radial integral defining $f$ is smooth, so $\eta =df$ andDegenerate triangles follow by continuity. The same proof works on any star-shaped plane domain, including the whole plane.

Integration requires an orientation: take $M$ to be an oriented compact $n$-dimensional smooth manifold without boundary. The Generalized Stokes theorem says ${\int }_{M}d\beta =0$ for every smooth $(n-1)$-form $\beta$. Every $n$-form is closed by dimension, and changing it by an exact differential form leaves its integral unchanged. Thusis a well-defined linear map $H_{\mathrm{dR}}^n(M;\mathbb{R})\to \mathbb{R}$. Compactness alone does not supply the orientation omitted from the printed integration statement.

For an $n$-dimensional Lie group, choose a nonzero $\lambda \in {\Lambda }^n{T}_e^{\ast }G$ and define the invariant volume form on a Lie group byEach $d{L}_{g^{-1}}$ is an isomorphism, so $\omega$ is nowhere zero and defines an orientation. The composition identity $L_{(hg{)}^{-1}}\circ L_h=L_{g^{-1}}$ gives $L_h^{\ast }\omega =\omega$. Moreover, every left-invariant differential form of degree $n$ is determined by its value at $e$, so the space of such forms is one-dimensional.

Under the supplied invariant-representative assumption, all top-degree de Rham cohomology classes are multiples of $[\omega ]$. For compact $G$, orient it so $\omega$ is positive; then ${\int }_G\omega >0$, whereas the Generalized Stokes theorem makes every exact top form have integral zero. Therefore $[\omega ]\ne 0$ andunder that assumption. For actual compact Lie groups, this argument applies when $G$ is connected. The supplied assumption fails for disconnected groups: on $S^1\times \mathbb{Z}/2$, the two components have independent degree-one classes, while globally left-invariant top forms form only a one-dimensional space. In general the top-degree de Rham cohomology of a compact Lie group is ${\mathbb{R}}^c$, where $c$ is its number of connected components.

For $S^2$, use $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with the plane. The disc argument gives $\eta {|}_U=d{f}_U$ and $\eta {|}_V=d{f}_V$ for any closed differential one-form $\eta$. The overlap $U\cap V$ is connected, so $d(f_U-f_V)=0$ makes the difference a constant. Adjust one potential by that constant; they glue to a global smooth function $f$ with $df=\eta$. ThusFinally, $S^3$ is a compact connected Lie group: the identificationidentifies it smoothly with the special unitary group $SU(2)$. Matrix multiplication and inversion supply its smooth group operations. The preceding invariant-volume argument therefore gives

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.

All derivatives below are left Grassmann derivatives. Decompose the supersymmetric covariant derivatives into odd piecesWhen these operators act on an arbitrary superfield, the graded Leibniz rule givesThe anticommutator of the two pure Grassmann derivatives vanishes. Also $\{A_{\alpha },B_{\dot{\beta }}\}=0$: the spacetime derivatives commute, while the $\theta$ and $\bar{\theta }$ coefficients anticommute. Adding these terms proves the supercovariant derivative algebra with left derivatives,For two unbarred derivatives, neither ${\partial }_{\alpha }$ differentiates the $\bar{\theta }$ coefficient of the other operator. Their coefficient products anticommute, so $\boxed{\{D_{\alpha },D_{\beta }\}=0}$. The same argument with barred and unbarred coordinates exchanged gives $\boxed{\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0}$. These signs follow the derivatives printed in this paper; a convention replacing $\bar{D}$ by $-\bar{D}$ reverses the mixed sign as well.