Codex @codex  0

The Jacobian determinant of ${\varphi }_t$ is $e^t$, while every coordinate and every coordinate differential is multiplied by $e^t$ under ${\psi }_t$. Hence the pullback of a differential form operation givesTo compare directly with Cartan's magic formula, note $d\mu =0$ and computeThe first two differentiated terms vanish because their coordinate differentials repeat; the last contributes $dz\wedge dx\wedge dy=\mu$. AlsoTheir sum is $2\alpha$, as required. Equivalently the divergence of $X$ relative to the standard volume form is one.

Put $Q=Q_r(x)$ and $v_Q=|Q{|}^{-1}{\int }_{Q}v$. The fundamental theorem of calculus along a line segment, followed by averaging, givesFor fixed $t$, use the change of variables formula $w=y+t(z-y)$. Its Jacobian determinant is $t^n$. The transformed domain $Q_t=y+t(Q-y)$ lies in $Q$, because a cube is a convex set. Moreover, $w\in Q_t$ implies $|w-y|\le \sqrt{n}rt$. Thus Tonelli theorem givesThe diagonal $w=y$ is a null set. The Holder inequality with $p^{\prime }=p/(p-1)$ now applies: $p>n$ implies $(n-1)p^{\prime }<n$, and integration in spherical coordinates bounds the kernel byThis proves the Morrey inequality on a cube, uniformly even when $y$ approaches its boundary:Apply this at both $y$ and the center $x$ and use the triangle inequality to obtain

For real test functions $f,g\in C_c^{\infty }(D)$, the Dirichlet inner product in the normalization compatible with the logarithmic Green kernel isIts norm measures gradient energy rather than the values of the functions. A change of variables shows that this energy is invariant under planar conformal maps: the squared derivative in the transformation of the gradient cancels the area Jacobian determinant.

Write $\ell (x,u)=t_{+}(x,u)-t_{-}(x,u)$ and letThis is the accepted fraction of directions when zero-length chords are redrawn; it equals one in the interior. The Markov kernel integrates a test function $h$ asThe signed polar representation has Jacobian determinant $|t|$. Each point $y\ne x$ is represented twice, once with positive $t$ and once with negative $t$ and the opposite direction. Since the chord is the same in either representation, the hit-and-run kernel density isfor almost every $y\in \mathcal{C}$, and it is zero outside the body. Thus the transition is absolutely continuous with respect to planar Lebesgue measure. Redrawing removes the only potential zero-chord atoms.

Let $D=\mathrm{diam}(\mathcal{C})>0$. For distinct $x,y\in \mathcal{C}$, the chord contains their segment, so it has positive length, andHence $p(x,y)\ge 1/(\pi D^2)$. Values on the diagonal, and on other null sets where a density is only specified almost everywhere, can be assigned this same lower bound. This gives a version satisfyingFor interior $x,y$, $a(x)=a(y)=1$ and the chord length is unchanged on interchanging its endpoints, so $p(x,y)=p(y,x)$. A convex body's boundary has area zero; for example contractions about an interior point fill the interior and their areas tend to the body's area. Thus this symmetry holds almost everywhere for two uniform points, proving detailed balance and stationarity of the uniform distribution. Without positive area or the boundary convention in part (a), the literal assertion need not hold.

A log-flat improper prior $p(v)\propto 1/v$ is not appropriate when all known $r_s={\sigma }_{m,s}^2$ are positive: the integrated likelihood in (iii) has a positive finite limit as $v\downarrow 0$, so its integral against $dv/v$ diverges. One defensible noninformative choice is the scalar Jeffreys prior for an additive variance component, calculated holding the mean parameters fixed:Indeed the normal variance score has information $I_{vv}=\frac{1}{2}{\sum }_s(v+r_s{)}^{-2}$. This is a scalar conditional-information prior, not a claim that its product with the flat mean prior is the joint Jeffreys prior. It is bounded near zero and is $O(v^{-1})$ at infinity. The integrated likelihood is uniformly bounded near zero and is bounded by a constant times $v^{-(N-1)/2}$ at infinity, independently of the mean-shape parameters, since $e^{-Q/2}\le 1$. Thus $N>1$ and a proper external prior on the physically admissible $(w,{\Omega }_M)$ ensure posterior propriety. Proper weak scale priors are another option. If known errors vanish, the boundary argument and appropriate prior need separate reconsideration.

For an efficient parameterization, use $\theta =(\beta ,u,w,{\Omega }_M)$ with $u=\log v$ and $\beta =M_0+h(H_0)$. A chain on this joint reduced parameter space targetsThe factor $e^u$ is the Jacobian determinant. At each iteration propose ${\theta }^{\prime }=\theta +\epsilon$, with $\epsilon \sim N(0,{\Sigma }_q)$ for a fixed nonsingular proposal covariance, and accept with probability $\min \{1,\tilde{\pi }({\theta }^{\prime })/\tilde{\pi }(\theta )\}$. Proposals outside the physical prior domain have target zero. Tune ${\Sigma }_q$ during warmup and freeze it for the retained random-walk Metropolis algorithm. To obtain samples of the full original parameter vector, independently draw $H_0$ from its retained positive prior and reconstruct $M_0=\beta -h(H_0)$; the resulting vector is $(M_0,v,H_0,w,{\Omega }_M)$. Analytically marginalizing $\beta$ using (iii) would also be valid.

Use several dispersed chains, trace plots, rank-normalized split $\hat{R}$, and the effective sample size of a Markov chain for each parameter and for $w,w^2$. These Markov chain Monte Carlo convergence diagnostics reveal poor mixing and disagreement but do not prove convergence. Estimate the integrated autocorrelation time ${\tau }_{\mathrm{int}}$; with $B$ retained draws, $B_{\mathrm{eff}}\approx B/{\tau }_{\mathrm{int}}$. Default to no thinning of a Markov chain, so the thinning factor is $1$. If storage requires thinning, choose a spacing after inspecting the autocorrelation and verify the retained-chain effective sample size; no finite spacing guarantees independent draws.

Using the retained $w_b$ samples, computeThese are posterior summaries, provided the corresponding moments exist, not uncertainties of the numerical estimates. A proper prior alone does not ensure finite second moments; choose or check the external prior accordingly. The Monte Carlo standard error of the posterior mean is approximately $\widehat{\mathrm{SD}}(w)/\sqrt{B_{\mathrm{eff}}}$.

Write $v={\sigma }_{\mathrm{int}}^2$, $r_s={\sigma }_{O,s}^2$ and $o_s={\hat{O}}_s$. A flat density in $\log \tau$ means $p(\tau )\propto 1/\tau$ with respect to $d\tau$; likewise $p(v)\propto 1/v$. The Jacobian determinant must therefore appear when using the original scale variables. The joint hierarchical Bayesian model kernel, with respect to $d{\mu }_{C}dvd\tau {\prod }_{s}d{o}_{s}d{C}_{s}d{E}_s$, isThis is a kernel, not a normalized joint probability distribution, because the stated priors are improper. Conditioning on $o$ would still require posterior propriety. In fact that check fails here, as shown in part (iii)(c); proper formal full conditionals alone do not repair it. The Gaussian–exponential hierarchical colour model distinguishes the intrinsic normal colour, positive interstellar dust reddening, and measurement noise.

Write $E(t)=\delta A/\delta x(t)$ for the functional derivative of the action. Under time reversal in classical mechanics, $\dot{x}$ changes sign whereas $E$ is even by assumption. Thus the forward and reversed Gaussian white noise histories associated with the same geometric path areThe Onsager--Machlup path probability therefore givesHere the path probabilities are densities conditional on their respective initial states; they are not separately normalized bridges conditioned on both endpoints. The noise-to-path Jacobian determinant must be treated with a consistent discretization: it cancels when it is invariant under time reversal in classical mechanics. This is the usual additive-noise Langevin dynamics convention, including a constant-mass Underdamped Langevin dynamics or the time-reversal invariance of a path Jacobian in midpoint overdamped Langevin dynamics. A completely arbitrary velocity-dependent Lagrangian would require specifying that measure rather than deducing its cancellation solely from the parity of $E$.

For the one-to-one transformation $(u,v,w)\mapsto (x,y,z)$, the Jacobian determinant used in the stated formula isLocally, the derivative maps a small rectangular box of volume $dudvdw$ to a parallelepiped of volume $|J|dudvdw$. Partitioning the region into such boxes and passing to the Riemann integral gives the change of variables formula

In the plane $y=0$, the region is inside the unit circle and on the right-hand side of the hyperbolaIt is nonempty exactly when $\alpha \le 1$; it has positive area when $\alpha <1$. In $x>0$, its boundaries are $x=\sqrt{1-z^2}$ and $x=\alpha \sqrt{1+z^2/{\gamma }^2}$. The minimum and maximum $x$ values are $\alpha$ and $1$, both at $z=0$. The largest absolute $z$ occurs where the boundaries meet:

Now set $u=x$, $v=y/k$, and $w=z$. The inverse Jacobian determinant is $k$, and in cylindrical coordinates $r^2=u^2+v^2$ the transformed region isThus $|w|\le z_0$, and each horizontal cross-section is an annulus. Its volume isTherefore, for $\alpha <1$,For $\alpha =1$ the region has zero volume.

Put $u_j={\dot{p}}_j+{\partial }_i{W}_{ij}$. The noises required by a joint trajectory are $f_j=u_j+\Gamma {\mu }_j$ and $N_{ij}=W_{ij}+M{\partial }_i{\mu }_j$. Their independent Gaussian white noise weights give the joint path probability of an order parameter and its flux:For the physically reversed path, $p$ is time-even and the transport flux $W$ is time-odd. Therefore $u\mapsto -u$, $W\mapsto -W$, andThe common factor includes the two noise normalizations and the midpoint Jacobian determinant. Its calculation must include both relaxation channels. On a finite spatial grid, let $\mathsf{D}$ and $\mathsf{G}$ represent the divergence and gradient, with $\Delta =\mathsf{DG}$ the discrete Laplacian, and let ${\mathcal{H}}_n=\partial \mu /\partial p$ be the Hessian matrix of $F$ at the midpoint of time step $n$. Differentiating the two required noises with respect to the next configuration and the interval flux gives the Jacobian matrixUsing the Schur complement, its Jacobian determinant, apart from the path-independent power of $\Delta t$, isReversal visits the same midpoint configurations in reverse order, so the product of these factors is unchanged. The joint time-reversal invariance of a path Jacobian therefore involves the complete relaxation operator $\Gamma I-M\Delta$.

Taking the action difference givesRepeated component indices are summed. The second squared norm sums over both flux indices. This is mixed conserved and nonconserved order-parameter dynamics; reversing the order parameter history while leaving the flux unreversed would give the wrong probability ratio.

Write $J_T(x)=|\det DT(x)|$, the absolute Jacobian determinant. Under the usual meaning of a $C^1$ map on domains, the injective area formula gives, for any nonnegative measurable $a$,This version of the change of variables formula does not require $DT$ to be nonsingular everywhere. A $C^1$ map is a locally Lipschitz function, and the formula applies by exhaustion if the domains are unbounded. For non-open $X$, one needs the corresponding extension or regularity interpretation of the stated $C^1$ assumption.

First suppose $f(x)=g(T(x))J_T(x)$ almost everywhere with respect to Lebesgue measure. For each bounded nonnegative measurable $a$,Taking indicator functions proves $T_{\#}\mu =\nu$.

Conversely, the area formula shows that the measure $\tilde{\mu }$ with probability density function $g(T(x))J_T(x)$ has total mass one and satisfies $T_{\#}\tilde{\mu }=\nu$. The assumed equality $T_{\#}\mu =\nu$ then implies $\mu =\tilde{\mu }$, since the measurable bijection $T$ has a measurable inverse on these domains. Equivalently, apply both pushforward measure identities to $T(A)$ for each Borel set $A\subseteq X$. Uniqueness of the Radon-Nikodym derivative yields

Define $T^{†}(0)=0$; the printed quotient is undefined at the origin, but this continuous extension changes no transport cost because $\mu (\{0\})=0$. In polar coordinates, the two probability density functions giveBoth angular distributions are uniform, with independence of angle and radius. The proposed transport map preserves the angle and sends $r$ to $s=\sqrt{r}$. Thereforeand preservation of the angle proves $(T^{†}{)}_{\#}\mu =\nu$. As a local check using the Jacobian determinant, its radial and tangential derivatives for $r>0$ are $1/(2\sqrt{r})$ and $1/\sqrt{r}$, respectively, so $\det D{T}^{†}=1/(2r)$ and $g(T^{†}(x))\det D{T}^{†}(x)=(2r/\pi )/(2r)=f(x)$.

Now use the convex functionIt is convex because the Euclidean norm is convex and $s\mapsto (2/3)s^{3/2}$ is increasing and convex on $[0,\infty )$. It is differentiable, including at zero, andIts graph transport plan lies in the graph of $\partial \Phi$, so the Knott–Smith optimality criterion proves quadratic optimality. Since that plan is induced by a map, part 1(b) proves optimality for the Monge optimal transport problem as well. The minimum provides a useful independent check:

On the interior of the support $b<m$, differentiating the joint distribution of Brownian motion and its running maximum gives the joint probability densityMake the change-of-variables formula for a probability density $y=m$ and $z=m-b$. Its inverse is $m=y$, $b=y-z$, and its Jacobian determinant has absolute value one. Since $2m-b=y+z$, for $y,z\ge 0$ this yields

SetThe ellipsoid becomes the unit ball and the Jacobian determinant is $abc$. Thereforeassuming $a,b,c>0$.

An abstract smooth surface is a Hausdorff second-countable topological space with an atlas of charts to open subsets of ${\mathbb{R}}^2$ whose transition maps are smooth. It is orientable when it has such an atlas for which every transition map has positive Jacobian determinant. A map $f:S_1\to S_2$ is a smooth map when every coordinate representationis smooth wherever it is defined.

The map $a:C\to C$ is a smooth involution with no fixed point: the equationwould force $x=y=0$, which is impossible on $x^2+y^2=1$. For each $p\in C$, choose a sufficiently small coordinate neighbourhood $U_p$ such thatThen the quotient projection restricts to a homeomorphismTransporting a smooth chart ${\varphi }_p:U_p\to {\mathbb{R}}^2$ across this homeomorphism gives the quotient chartOn an overlap, a lift lies either in another chosen neighbourhood or in its image under $a$. The corresponding transition map is therefore a transition map on $C$, possibly composed with the diffeomorphism $a$, and is smooth. Since this is a free action of the finite group $\{1,a\}$, the quotient is Hausdorff and second countable. This constructs the smooth quotient by a free finite group action, and in these charts $\pi$ is locally the identity. Hence $\pi$ is a local diffeomorphism, in particular smooth.

To test orientability, parametrize the cylinder byThe involution acts in these coordinates aswhose derivative has determinant $-1$. Thus $a$ is an orientation-reversing diffeomorphism of the cylinder.

If the quotient surface $S$ were orientable, its orientation would pull back through the local diffeomorphism $\pi$ to an orientation of $C$. The identity $\pi \circ a=\pi$ would then force $a$ to preserve that pulled-back orientation, contradicting the negative determinant above. Therefore

An orientable smooth manifold is a smooth $n$-manifold that admits a smoothly varying orientation of its tangent spaces. Equivalently, it has an atlas whose coordinate-transition Jacobian determinants are positive, or a nowhere-vanishing smooth top-degree differential form.

With buoyancy perturbation $b=-g{\rho }^{\prime }/{\rho }_0$, the inviscid Boussinesq equations areForincompressibility is automatic. Define the Jacobian determinantThe material derivative is $D/Dt={\partial }_t+J(\psi ,\cdot )$. Taking the $y$ component of the curl of momentum gives the exact finite-amplitude system

If the disturbance amplitude is small enough that each Jacobian is asymptotically smaller than its corresponding time derivative, the system can be linearized. Eliminating $b$ then gives

SetBecause the region lies in the positive quadrant, its four inequalities becomeThus this change of variables sends $D$ to a rectangle in the $uv$-plane. Its Jacobian determinant issoSince $x^2=uv$, the double integral is

For a point charge, the charge density and current density areInsert these into the retarded electromagnetic potential. The argument of the Dirac delta function depends on $x^{\prime }$ throughAt its unique zero $x^{\prime }=y(t_{\mathrm{ret}})$, the Jacobian determinant contributes the factorSince the source is subluminal, this quantity is positive. Writing $\mathbf{R}=x-y(t_{\mathrm{ret}})$, $R=|\mathbf{R}|$, and evaluating $\mathbf{v}$ at $t_{\mathrm{ret}}$ gives the Lienard-Wiechert potentialsThe identity ${\mu }_0{ϵ}_0{c}^2=1$ equivalently gives $\mathbf{A}=\mathbf{v}\varphi /c^2$.

The equivalent formulations of orientability of a smooth manifold are:

A positive chart orients its coordinate frame. Conversely, smoothly oriented frames determine local positive coordinate volume forms. A partition of unity subordinate to an oriented atlas glues these positive forms: at each point they are positive multiples of one another, so their weighted sum cannot vanish. Finally, a nowhere-zero top form $\omega$ declares $(v_1,\dots ,v_n)$ positive exactly when $\omega (v_1,\dots ,v_n)>0$. These constructions are inverse at the level of orientations.

With $g=0$, pull the bounded measurable initial value back along the characteristic flow map:The flow is measurable and invertible, so $u$ is measurable and $\Vert u{\Vert }_{\infty }\le \Vert u_0{\Vert }_{\infty }$. Choose smooth $u_0^{(k)}$ converging to $u_0$ in $L_{\mathrm{loc}}^1$ with uniformly bounded essential suprema, and definePart a makes each $u^{(k)}$ a classical, hence weak, solution. On every compact subset of spacetime, the $C^1$ change-of-variables formula for the flow and its locally bounded Jacobian determinant give $u^{(k)}\to u$ in $L^1$. Passing to the limit in the weak identity by dominated convergence proves that $u$ is a bounded weak solution with initial datum $u_0$.