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There is a small domain issue in the printed notation. If ${\mathbb{R}}^3/{\mathbb{R}}^2$ means a quotient vector space by the horizontal plane, the displayed formula does not descend to that quotient vector space: $(0,0,0)$ and $(1,0,0)$ represent the same class but give different values. The intended construction is stereographic projection, restricted to the unit sphere with the north pole removed. Interpreting the slash as removal of the plane $z=1$ also supplies a suitable ambient domain. The holomorphic stereographic atlas of the sphere is obtained as follows.

Write $N=(0,0,1)$ and $S=(0,0,-1)$. On $U_N=S^2\setminus \{N\}$ use $\zeta =(x+iy)/(1-z)$. Its inverse, with $\zeta =u+iv$, isThese formulas give a smooth manifold chart from $U_N$ onto $\mathbb{C}$. On $U_S=S^2\setminus \{S\}$ choose the second manifold chartIts inverse is $x=2\mathrm{Re}\eta /(1+|\eta {|}^2)$, $y=-2\mathrm{Im}\eta /(1+|\eta {|}^2)$ and $z=(1-|\eta {|}^2)/(1+|\eta {|}^2)$, so this is also a smooth manifold chart onto $\mathbb{C}$. The conjugation in this second stereographic projection is essential. On the overlap, $x^2+y^2=1-z^2$, soBoth directions of this transition are holomorphic maps on ${\mathbb{C}}^{\times }$, with nonzero derivative. The two manifold charts cover the sphere, hence define a holomorphic atlas, giving precisely the Riemann sphere. If one instead used $x+iy$ in both manifold charts, the transition would be $1/\bar{\zeta }$ and would not be holomorphic.

Orient the sphere by this holomorphic atlas. The given volume form is smooth at infinity: replacing $\zeta$ by $1/\eta$ in its exterior product gives the same expressionMoreover, $id\zeta \wedge d\bar{\zeta }=2du\wedge dv$, so the normalization of this volume form isThus the printed volume form has half the area of the standard round unit sphere; replacing its integral by $4\pi$ would introduce an erroneous factor of two.

For $k\ge 1$, the holomorphic map $f(\zeta )={\zeta }^k$ extends over infinity, since the target reciprocal coordinate is ${\eta }^k$ when the source reciprocal coordinate is $\eta$. Its pullback of a differential form isUsing $s=r^{2k}$ givesThe degree of a map between oriented manifolds is thereforeFor $k=0$, the formula on the finite manifold chart is the constant $1$. Its unique continuous extension is also $1$ at infinity, rather than an undefined expression ${\infty }^0$. Its pullback of a differential form is zero and its degree of a map between oriented manifolds is zero.

For the preimage calculation when $k\ge 1$, choose a regular value $w\in {\mathbb{C}}^{\times }$. There are exactly $k$ distinct roots of ${\zeta }^k=w$. At each root the real Jacobian determinant is $|k{\zeta }^{k-1}{|}^2>0$, so every local contribution to the degree of a map between oriented manifolds is $+1$. Their sum is $k$, agreeing with the integral. The exceptional values $0$ and infinity are avoided because they are branch values when $k>1$. For the constant map, any $w\ne 1$ is a regular value with no preimages, giving the same answer zero. This establishes the degree of a power map of the Riemann sphere for every allowed $k$.

Differentiating the explicit characteristic flow map gives the block triangular Jacobian matrixIts Jacobian determinant is the product of its diagonal block determinants. Therefore

A Grassmann variable is an odd generator of a Grassmann algebra: ${\eta }_i{\eta }_j=-{\eta }_j{\eta }_i$, so ${\eta }_i^2=0$. For one generator, any function is $f(\eta )=a+b\eta$. The Berezin integral is the linear operationThus integration extracts a coefficient, rather than assigning a length or volume. For many generators it extracts the coefficient of the highest-degree monomial with the sign fixed by the order of the measure. Odd coefficients and Grassmann derivatives must retain their order; exchanging two odd objects changes the sign.

This operation is invariant under odd translations, because a translation only changes terms of lower degree. For an invertible ordinary matrix $S$ and $\eta =S\xi$, the Grassmann change-of-variables formula isThe inverse Jacobian determinant, rather than the ordinary commuting-variable Jacobian, compensates for the factor $\det S$ multiplying the top monomial. Integration agrees with the appropriate ordered Grassmann derivatives, but the orientation must be specified when combining barred and unbarred variables.

Let $E=L^1({\mathbb{R}}_x^d\times {\mathbb{R}}_v^d)$, and denote the free-transport semigroup by $(U_{t}g)(x,v)=g(x-tv,v)$. For fixed $v$, the spatial translation has unit Jacobian determinant, so the Tonelli theorem givesThus the given free term is an isometry on $E$. The operators form a strongly continuous semigroup: continuity first holds for smooth compactly supported functions by dominated convergence, and density plus the isometry extends it to every $L^1$ function.

Write $F=F_N$ and integrate the Kac master equation over $v_2,\dots ,v_N$. For a pair $i,j\ge 2$, the rotation acts only on integrated variables. Its unit Jacobian determinant makes the integrated gain identical to the integrated loss, so all those pairs cancel.

The only remaining pairs are $(1,j)$, $j=2,\dots ,N$. For such a pair, first integrate over every variable except $v_1$ and $v_j$. This yields the corresponding two-coordinate marginal distribution evaluated at the rotated pair. Permutation symmetry of $F_N$ makes all $N-1$ resulting integrals identical to the one for $(1,2)$. The coefficient isConsequently the Kac marginal evolution equation isThe time argument has been suppressed on the right. The loss is consistent with normalization, since $\int {\Pi }_2(v_1,v_2)d{v}_2={\Pi }_1(v_1)$. Under the printed definition $k<N$, this use of ${\Pi }_2$ requires $N\ge 3$. For $N=2$ the same formula holds with the natural extension ${\Pi }_N(F_N)=F_N$.

This identity is exact and generally unclosed. Replacing the two-coordinate marginal distribution by the product of one-coordinate marginals would produce the quadratic collision equation associated with Kac chaos. Permutation symmetry alone does not imply that product approximation.

Put $\mathbf{x}=r\mathbf{\omega }$, with $|\mathbf{\omega }|=1$. The Jacobian determinant in spherical coordinates gives $dS=r^{2}d\omega =r^2\sin \theta d\theta d\phi$, soDifferentiation under this fixed-domain integral yieldsThe last step is the divergence theorem and the Laplace equation. Hence $f$ is constant on radii for which the closed ball lies inside $V$. Continuity at the origin gives ${\lim }_{r\to 0}f(r)=\varphi (0)$, proving the mean value property for harmonic functions:

In cylindrical coordinates, $\rho =(x^2+y^2{)}^{1/2}$ and $\mathbf{F}=(f(\rho )x/\rho ,f(\rho )y/\rho ,0)$. The chain rule gives ${\partial }_x\rho =x/\rho$ and ${\partial }_y\rho =y/\rho$. ThereforeThis is cylindrically radial divergence, valid for $\rho >0$.

The bounded domain is outside the unit cylinder, below the inverse-square surface and above the lower plane:The upper plane touches only the rim $\rho =1,z=1$; it is not an extra positive-area top face. Revolving the shaded meridional region below around the $z$ axis gives the three-dimensional domain.

With the Jacobian determinant $dV=\rho d\rho d\varphi dz$ of cylindrical coordinates,Zero divergence gives $(\rho f{)}^{\prime }=0$, hence the most general solution on the positive radial interval isAlthough this vector field is singular on the axis if $A\ne 0$, the entire closure of $V$ lies at $\rho \ge 1$, so it is smooth on a neighbourhood of $V$ and the divergence theorem applies.

Its volume integral of divergence is zero. Compute the outward surface integral directly. On the inner cylinder, $\mathbf{n}=-{\mathbf{e}}_{\rho }$, $dS=d\varphi dz$, and the flux is $-2\pi A(1-1/4)=-3\pi A/2$. On the curved roof $z=g(\rho )={\rho }^{-2}$, the outward vector surface element of a graph isTherefore its flux isThe horizontal bottom has zero flux and the top rim has zero surface area. Summing giveswhich verifies the divergence theorem with every boundary contribution accounted for.

The Box-Muller transform uses independent uniforms to setThe radius has the unit Rayleigh distribution, with density $r{e}^{-r^2/2}$ for $r>0$, and the angle is independently uniform on $[0,2\pi )$. Their joint density is $(2\pi {)}^{-1}r{e}^{-r^2/2}$. The Cartesian change of variables has absolute Jacobian determinant $r$, so the joint density of $(X_1,X_2)$ isThe factorization proves that the outputs are independent random variables, each with a standard normal distribution.

The normalized Killing one-form $\eta =F^{-1}{V}_{\mu }d{x}^{\mu }$ is a closed differential form. The Poincare lemma gives a local scalar potential $\varphi$ with $\eta =d\varphi$, hence $V_{\mu }=F{\partial }_{\mu }\varphi$. Choose the adapted chart of part (i), so $V={\partial }_0$. ThenThe closed differential form condition supplies the integrability of these spatial derivatives. For a fixed $g$ and $V$, $f$ is determined locally up to an additive constant; its derivatives cannot be chosen freely despite the wording in the paper.

Use ${\bar{x}}^0=\varphi$, ${\bar{x}}^i=x^i$. The Jacobian determinant is one, so this is a local coordinate change. Holding ${\bar{x}}^i$ fixed gives $V={\partial }_{\bar{0}}$, while its metric-dual one-form is $v=Fd{\bar{x}}^0$. Thus ${\bar{g}}_{00}=F<0$ and ${\bar{g}}_{0i}=0$. The Killing equation still holds, so ${\partial }_{\bar{0}}{\bar{g}}_{\mu \nu }=0$. The metric is locally static:A closed differential form need not be globally exact; neither a global potential nor a global static chart follows without additional topological assumptions.

Set $U=X/(X+Y)$ and $V=X+Y$. Their inverse change of variables is $X=UV$, $Y=(1-U)V$, with domain $0<U<1$, $V>0$. The absolute Jacobian determinant of this inverse transformation isUse the independence of random variables to multiply the two original Gamma distribution densities. The change of variables formula then gives the joint probability densityFactor this into two normalized densities:The support is a product domain, and the factors are precisely a Beta distribution density and a Gamma distribution density. ThereforeThis beta-gamma independence uses the common rate $\theta$; it is a rate parameter, not the scale parameter $1/\theta$.

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.

Use the open strip $S=\{x+iy:0<y<\pi \}$; the printed closed strip denotes the region together with its exit boundary. Starting at $z=x+iy$, the conformal bijection $e^z$ sends $S$ onto the complex upper half-plane and the bottom edge $a\in \mathbb{R}$ onto $v=e^a>0$. The upper edge maps to the negative half-axis. The transformed initial point is $e^x\cos y+i{e}^x\sin y$.

By conformal invariance of planar Brownian motion, transform the given Poisson kernel for the upper half-plane and include the Jacobian determinant $dv/da=e^a$. The bottom-boundary harmonic measure of a strip isThis is the unconditional exit density on the bottom edge, not the density conditional on using that edge. Its total mass isIndeed substitution $v=e^{a-x}$ gives the positive-half-axis mass of the Poisson kernel, equal to $\frac{1}{2}+{\pi }^{-1}\arctan (\cot y)=1-y/\pi$ for $0<y<\pi$. To obtain the conditional bottom-edge density, divide by $1-y/\pi$. The top-edge density is $\sin y/[2\pi (\cosh (a-x)+\cos y)]$ and has mass $y/\pi$, a useful normalization check.

Use the Green function of killed planar Brownian motion, normalized as the density of expected occupation with respect to area:for nonnegative measurable $f$. Equivalently, $G_D(z,w)={\int }_0^{\infty }{p}_D(t,z,w)dt$, where $p_D$ is the killed Brownian transition density. With generator $\frac{1}{2}\Delta$, the distributional normalization is $-\frac{1}{2}{\Delta }_w{G}_D(z,w)={\delta }_z(w)$, and the singularity is $-{\pi }^{-1}\log |w-z|$ plus a locally harmonic function. This fixes the normalization of the Dirichlet Green function explicitly.

By conformal invariance of planar Brownian motion and its conformal Brownian clock,The Jacobian determinant of a conformal map is $|{\varphi }^{\prime }(w){|}^2$. Changing the area variable to $w^{\prime }=\varphi (w)$ givesUniqueness of the occupation density proves the desired equality almost everywhere. Both functions are continuous and harmonic away from their pole, so it holds at every $w\ne z$. The Green function is conformally invariant:If the Dirichlet Green function is instead normalized for $-\Delta$, both kernels are divided by two and the invariance statement is unchanged.

The Box–Cox transformation searches over transformations of the positive response,The plotted horizontal coordinate is $\lambda$ and the vertical coordinate is its maximized profile likelihood on the log scale. The likelihood includes the transformation Jacobian determinant; omitting it would compare response scales incorrectly. For the Gaussian regression it is, up to constants,The maximum occurs around $\lambda =2$. The horizontal 95% line is approximately ${\ell }_p(\hat{\lambda })-{\chi }_{1,0.95}^2/2$, and its intersections give an approximate interval roughly $1.2$ to $2.8$, read visually. In particular, $\lambda =1$ is below the line, and $\lambda =0$ is far below it.

Square the response: a convenient transformation is $Y_i^2$, equivalent up to affine rescaling to $T_2(Y_i)=(Y_i^2-1)/2$. Transform energy, not the sunny-days or latitude predictors. Refit and check the normal linear model on the transformed scale; the plot is evidence for a transformation, not proof that the transformed residual assumptions hold. Inference on transformed means cannot be inverted naively to obtain original-scale conditional means.

Fix the measure conventions before evaluating this zero-dimensional supersymmetric field theory. Take $dzd\bar{z}$ to mean ordinary positive area measure $d^{2}z=dxdy$, and choose Berezin integration orientations with $\int d^2\theta {\theta }_1{\theta }_2=\int d^2\overline{\theta }{\overline{\theta }}_1{\overline{\theta }}_2=1$. A different overall measure normalization multiplies all the unnormalized answers by the same constant. Assume $r=\mathrm{deg}P=\mathrm{deg}W-1\ge 1$, so $|P(z)|\to \infty$ at infinity and all polynomial insertions converge. An affine or constant $W$ does not give this confining integral and is outside this convergence hypothesis.

The Grassmann variables truncate the fermionic exponential. The coefficient of ${\theta }_1{\theta }_2{\overline{\theta }}_1{\overline{\theta }}_2$ is $P^{\prime }(z)\overline{P^{\prime }(z)}$, so the partition function becomesThe Jacobian determinant of the holomorphic map $w=P(z)$ is $|P^{\prime }(z){|}^2$. Away from its finitely many critical values, a polynomial of polynomial degree $r$ has $r$ preimages counted with multiplicity. The change of variables formula therefore givesCritical values form a set of area zero and do not affect the integral. This partition function measures the covering degree, even when some zeros of $P$ are repeated.

Orient both spheres in the standard way and normalize their area forms to total area $4\pi$. The degree of a map between oriented manifolds can be obtained in two ways. For a regular value $y$, use the degree as a sum of local degrees:Each inverse image is isolated by the inverse function theorem, and compactness makes the set finite. The determinant is computed in consistently oriented local coordinates. A second method is spherical degree by area pullback:where the last formula represents $g$ as a unit vector in ${\mathbb{R}}^3$. The pullback of a differential form already contains the signed Jacobian determinant; no extra $\sin \theta$ is to be inserted in that last coordinate expression.

To relate the methods, replace $\omega$ by a smooth top-degree differential form with the same total integral supported in a small neighbourhood of a regular value. Two such top-degree forms with equal integral differ by an exact differential form on $S^2$, by its top-degree de Rham cohomology. Their pullbacks therefore have the same integral by Stokes theorem. Over the chosen neighbourhood, $g$ splits into local inverse branches; the change of variables formula makes the contribution of each branch its orientation sign times $4\pi$. Their sum is precisely the first formula. Thus the area integral is an integer and agrees with the signed inverse-image count.

For a nonconstant rational map, first use common-factor reduction of a rational map so $p$ and $q$ are coprime. Write $k=\max (\mathrm{deg}p,\mathrm{deg}q)$ for these reduced polynomials. A generic finite target value $w$ has inverse images at the roots of $p-wq$: avoiding exceptional values makes its degree $k$ and its roots simple. The fundamental theorem of algebra supplies $k$ roots. A holomorphic map has positive real Jacobian determinant $|R^{\prime }{|}^2$ at a regular point, so every local sign is $+1$. HenceThe source leaves coprimality implicit. In an unreduced representation the answer is $\max (\mathrm{deg}p,\mathrm{deg}q)-\mathrm{deg}\gcd (p,q)$, including degree zero for a constant reduced map. For example $(z^2-1)/(z-1)$ extends to $z+1$ and has degree one, although the unreduced maximum degree is two. Exceptional inverse images at infinity or multiple roots do not change the degree of a rational map of the Riemann sphere.

For the rational map approximation for Skyrmions, use stereographic projection $z=\tan (\theta /2)e^{i\phi }$ and the unit target vectorCombine this rational map with a radial profile to form a special unitary group field:where $\mathbf{\sigma }$ are the Pauli matrices. The endpoint values make $U(0)=-1$ independent of angle and $U(\infty )=1$. Appropriate radial behaviour gives an admissible finite-energy field configuration. With $L_i=U^{†}{\partial }_{i}U$, choose the topological baryon number in the Skyrme model conventionSeparating the radial and angular factors givesThus the degree of a rational map of the Riemann sphere supplies the Skyrmion charge.

In conventional dimensionless massless Skyrme model units, its static energy reduces towith the angular Jacobian of a rational mapThe Cauchy-Schwarz inequality gives $\mathcal{I}\ge k^2$. These formulas follow from the radial strain $|f^{\prime }|$ and the two equal angular strains $\sin f\sqrt{J_R}/r$: the quadratic energy sums their squares and the quartic Skyrme term sums their pairwise products of squares. Minimize the angular integral in the rational map approximation over degree-$k$ maps, then minimize the remaining radial energy with the stated endpoints. This replaces a three-dimensional field minimization by finitely many map coefficients and an ordinary differential equation for $f$.

The method constructs a charge-$k$ variational approximation, with topology built in and with rotational symmetry of a rational map translated into combined spatial and isospin rotations. It is efficient for identifying shapes and providing initial data for unrestricted numerical relaxation. Its restrictions are equally concrete: it uses one radial profile and a holomorphic angular map independent of radius, so it cannot represent arbitrary radial-angular correlations, separated clusters, or all deformations. Apart from the degree-one Skyrmion hedgehog ansatz, it generally does not solve the full field equation exactly. Massive-pion terms can be included in the radial functional but do not remove these restrictions, and multi-shell or unrestricted fields may be needed for larger charges. Approximate energy minima and a final collective-coordinate quantization are distinct steps.

Choose $X=x-2y$ and $T=x$, so $(\alpha ,\beta ,\gamma ,\delta )=(1,-2,1,0)$. The Jacobian determinant of the coordinate change is two, so it is an invertible linear map. By the chain rule, ${\partial }_x={\partial }_X+{\partial }_T$ and ${\partial }_y=-2{\partial }_X$. ConsequentlyThis is the wave equation with $c=1$. The D'Alembert formula is $F(X,T)=U(X+T)+V(X-T)$, with arbitrary twice differentiable functions $U,V$. Returning to the old coordinates and absorbing factors of two into the arbitrary functions givesDirect differentiation checks the cancellation. Equivalently, first integrating ${\partial }_x(f_x+f_y)=0$ produces the same two-function family.

The boundary ray has $y/x=3/5$, and its intersection with the hyperbola is $(5/4,3/4)$. Under the given change of variables, $x^2-y^2=r^2$ and $y/x=\tanh \theta$. Thus the region becomesThe hyperbolic functions satisfy ${\cosh }^2\theta -{\sinh }^2\theta =1$, so the Jacobian determinant isIt is positive in the interior. The coordinate degeneracy at $r=0$ is a boundary set of area zero and does not affect the change of variables formula. ThereforeHere $\cosh (\log 2)=5/4$. The PDF specifies two line segments and one hyperbolic arc; the duplicated $y=0$ line in the TeX is not an extra boundary condition.