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For the unheaded continuation, write the vector triple product asThe Levi-Civita symbol contraction is ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$. One way to check its sign is to fix distinct $i,j$: the only nonzero term has $k$ the remaining index, and the pairs $(l,m)=(i,j),(j,i)$ contribute $+1,-1$ respectively. When $i=j$, both sides vanish. Contracting with the vectors now gives both requested notations:If $\mathbf{a}$ and $\mathbf{b}$ are orthogonal, then $\mathbf{a}\cdot \mathbf{b}=0$. With the specified $\mathbf{c}$, the perpendicularity of the cross product gives $\mathbf{a}\cdot \mathbf{c}=|\mathbf{a}{|}^2$. ThereforeThe same answer covers the degenerate case in which one of these two vectors is zero.

Expanding the cross product component by component gives its cross-product matrix:It is a real skew-symmetric matrix and annihilates $\mathbf{n}$. By the vector triple product,so $A^2=\mathbf{n}{\mathbf{n}}^T-I$ for a unit vector $\mathbf{n}$. Choose a unit vector $\mathbf{u}\perp \mathbf{n}$ and put $\mathbf{v}=\mathbf{n}\times \mathbf{u}$. Then $(\mathbf{u},\mathbf{v},\mathbf{n})$ is an orthonormal basis and $A\mathbf{u}=\mathbf{v}$, $A\mathbf{v}=-\mathbf{u}$, $A\mathbf{n}=0$. Thus the characteristic polynomial is $\lambda ({\lambda }^2+1)$ and the cross-product matrix spectrum isOver the complex numbers, corresponding eigenvectors are $\mathbf{n}$, $\mathbf{u}-i\mathbf{v}$, and $\mathbf{u}+i\mathbf{v}$. Only the zero eigenvalue has real eigenvectors.

For the specified third coordinate axis, direct multiplication givesIts zero eigenspace is $\mathrm{span}\{(0,0,1)\}$, and its $-1$ eigenspace is the whole plane $\mathrm{span}\{(1,0,0),(0,1,0)\}$. Hence the eigenvalues of $A^2$ are $-1,-1,0$, with every nonzero vector in the appropriate eigenspace an eigenvector.

Use the resistive induction equation with constant magnetic diffusivity,Seek ${\mathbf{B}}_{\mathrm{tot}}=B_0{\mathbf{e}}_z+B(s,z,t){\mathbf{e}}_{\varphi }$. The axisymmetric velocity $\mathbf{u}=s\omega (s,z){\mathbf{e}}_{\varphi }$ is incompressible, and the proposed field is solenoidal. The toroidal magnetic field is parallel to the velocity, so it does not contribute to their cross product. The remaining term isThe uniform axial field has zero vector Laplacian. For the azimuthal part, differentiation of the cylindrical basis gives ${\partial }_{\varphi }^2{\mathbf{e}}_{\varphi }=-{\mathbf{e}}_{\varphi }$. ConsequentlyThere is no source for a radial or axial perturbation, so zero initial values of those components remain zero. Equating the toroidal components provesThe last Laplacian is the scalar axisymmetric Laplacian. The additional $-B/s^2$ term is essential for a vector component in a rotating basis. The source is the Omega effect: rotation that varies along an axial field line winds it into a toroidal magnetic field.

If $u\cdot v=0$, use the positively oriented orthonormal basis $(u,v\times u,v)$. The cross product identities give $Au=u+v\times u$, $A(v\times u)=-u$ and $Av=0$. In this basis,Its characteristic polynomial is $\lambda ({\lambda }^2-\lambda +1)$, so

Let $[v{]}_{\times }$ denote the cross product matrix, with $[v{]}_{\times }x=v\times x$. ThenThe two eigenvalue calculations use an orthonormal basis adapted to $u,v$.

For $v=(v_1,v_2,v_3)$ define the skew-symmetric matrixThen $\hat{v}x=v\times x$. This is a linear isomorphism ${\mathbb{R}}^3\to \mathfrak{so}(3)$, with inverse $\varphi (\xi )=({\xi }_{32},{\xi }_{13},{\xi }_{21})$. Since rotations preserve the cross product,Thus $\varphi$ intertwines the Adjoint representation with the standard representation. The vector triple-product identity gives, for every $x$,Consequently $[\hat{v},\hat{w}]=\widehat{v\times w}$ and the chosen normalization satisfiesNo undetermined scale remains in this explicit matrix convention.

Identify $SU(2)$ with the unit quaternions usingThe relations $jz=\overline{z}j$ and $j^2=-1$ verify multiplication, and every matrix in $SU(2)$ has this form. If $v$ is imaginary, then $\overline{v}=-v$. Quaternionic conjugation reverses product order and $q^{-1}=\overline{q}$, soThus conjugation preserves $\mathrm{Im}\mathbb{H}\cong {\mathbb{R}}^3$. Multiplicativity of the quaternion norm shows that this action preserves the inner product. The unit quaternions form the connected three-sphere, so its determinant, equal to one at the identity, is always one. We obtain a homomorphism $C:SU(2)\to SO(3)$.

Its group kernel consists of unit quaternions commuting with every imaginary quaternion. Commuting with both $i$ and $j$ forces a quaternion to be real, so the group kernel is exactly $\{\pm 1\}$. To prove surjectivity explicitly, let $u$ be a unit imaginary quaternion and put $q=\cos (\theta /2)+u\sin (\theta /2)$. For imaginary $u,v$, multiplication satisfies $uv=-u\cdot v+u\times v$. Expanding $qv{q}^{-1}$ therefore givesThis is the Rodrigues rotation formula about axis $u$. Every element of $SO(3)$ has such an axis-angle description: an odd-dimensional orthogonal matrix of determinant one has an eigenvector of eigenvalue one, and its action on the perpendicular plane is a plane rotation. Hence $C$ is onto, andThis is the quaternionic form of the Adjoint double cover from SU(2) to SO(3). Its differential sends an imaginary quaternion $u$ to $2\hat{u}$, because $[u,v]=2u\times v$, consistent with the bracket normalization used above.

Use the equal-time points to estimate velocity and acceleration with centered finite differences:At the endpoints use one-sided formulas of the desired order. A differentiated path and its time parametrization can instead supply these quantities. Let $g_k$ denote the gravitational acceleration, directed downwards. In the accelerating camera, the effective downward vector is $g_k-a_k$, so the locally felt upward vector is $U_k=a_k-g_k$. For a stationary camera this gives $-g_k$, fixing the sign.

The apparent-gravity camera frame is obtained by orthogonal projection and a cross product:When $W_k\ne 0$, $X_k,Z_k$ are perpendicular unit vectors, and $Y_k$ completes a right-handed orthonormal basis: $X_k\times Y_k=Z_k$. Also $U_k=(U_k\cdot X_k)X_k+\Vert W_k\Vert Z_k$, so the felt-up vector lies in the $XZ$ plane, with its transverse component pointing towards positive $Z$.

The stated exclusion of $a_k=g_k$ ensures $U_k\ne 0$, but it does not ensure $W_k\ne 0$. If felt-up is parallel to travel, the roll angle is undetermined rather than the construction being impossible. Every transverse orthonormal basis then has the required plane property. For continuity, project the previous $Z$ into $X_k^{\perp }$ and normalize it, using $Y_k=Z_k\times X_k$. If this projection is also too small, choose a coordinate unit vector least aligned with $X_k$ and project that vector instead. Switch to this transported choice near the parallel case to avoid numerical amplification. A stopped vehicle likewise needs a chosen limiting travel tangent vector, since its zero velocity itself has no direction. None of this requires a Frenet frame, which can fail where curvature vanishes.

Let $⟨\cdot ⟩$ denote the spatial average. Assume periodicity, or an equivalent homogeneous averaging limit in which total derivatives average to zero and integration by parts is valid. Fix the induced field to have zero mean. Expand the solution of the resistive induction equation in small magnetic Reynolds number:The uniform test magnetic field commutes with differentiation. Since $\Delta \mathbf{u}=-\mathbf{u}$, the first equation has solution ${\mathbf{b}}_1=(\mathbf{B}\cdot \nabla )\mathbf{u}$.

The quadratic second-order forcing need not be monochromatic, so one must not simply replace its inverse Laplacian by multiplication by $-1$. Instead use the self-adjointness of the Laplacian under the average:This proves the monochromatic small-Reynolds-number mean electromotive force expansion

For the symmetry of the first-order monochromatic alpha tensor, writeIts antisymmetric part is determined by contraction with the Levi-Civita symbol. The contraction identity givesThe second term vanishes by incompressibility; the first is $⟨{\partial }_j(u_j{u}_p)⟩=0$. Since every three-dimensional antisymmetric second-rank tensor is equivalent to its contracted axial vector, ${\alpha }^{(1)}$ is symmetric. No Fourier series or Fourier transform is required.

To derive the next identity, set $\mathbf{C}=\mathbf{u}\times (\mathbf{B}\cdot \nabla )\mathbf{u}$. Expanding the curl and cross product in components,The average of the second term is zero: integration by parts makes it $-⟨C_i{\partial }_j{u}_j⟩$. Integrating the first term by parts givesExtracting the coefficient of $B_j$ now yieldsInterchanging $i$ and $j$ reverses the cross product, proving the antisymmetry of the second-order monochromatic alpha tensor:The second coefficient of the alpha tensor can therefore contribute turbulent magnetic pumping, while its symmetric part vanishes at this order.

Use suffix notation, summing repeated indices, and the Levi-Civita symbol convention $(A\times B{)}_i={ϵ}_{ijk}{A}_j{B}_k$. Contracting the outer symbol with the symbol from the second cross product gives the epsilon-delta identitywhere $\delta$ is the Kronecker delta. ThusThe second term is zero because $A_p{A}_{\ell }$ is symmetric in $p,\ell$ and the Levi-Civita symbol is antisymmetric. The cyclic rearrangement of the remaining symbol has positive sign. Its contraction is the scalar triple product $A\cdot (B\times C)$. HenceThis cross products with a common vector identity requires no independence assumption, so it also covers zero or parallel vectors.

Write $\mathbf{x}=s\mathbf{n}+\mathbf{v}$, with $\mathbf{v}\cdot \mathbf{n}=0$. The linear map sends $s\mathbf{n}$ to $as\mathbf{n}$ and sends $\mathbf{v}$ to $a\mathbf{v}+b(\mathbf{n}\times \mathbf{v})$. On the perpendicular plane, $\mathbf{v}$ and $\mathbf{n}\times \mathbf{v}$ are perpendicular vectors of equal length. ThusThe expression in parentheses is a rotation in three dimensions by $\vartheta$ about the oriented $\mathbf{n}$ axis, with the positive direction fixed by the cross product. Therefore the map scales the axial component by $a$, scales the perpendicular component by $\sqrt{a^2+b^2}$, and rotates the perpendicular plane by $\arctan (b/a)$. Since $a,b>0$, the angle lies between zero and $\pi /2$. This is an axis-angle decomposition, with different axial and transverse scale factors.

Use suffix notation, with repeated indices summed from one to three. The cross product, divergence, and curl have components $(\mathbf{A}\times \mathbf{B}{)}_i={ϵ}_{ijk}{A}_j{B}_k$, $\nabla \cdot \mathbf{A}={\partial }_i{A}_i$, and $(\nabla \times \mathbf{A}{)}_i={ϵ}_{ijk}{\partial }_j{A}_k$. Here $ϵ$ is the Levi-Civita symbol.

The product rule givesUsing the cyclic identity ${ϵ}_{ijk}={ϵ}_{kij}$ in the first term and ${ϵ}_{ijk}=-{ϵ}_{jik}$ in the second identifies these terms as $B_k(\nabla \times \mathbf{A}{)}_k$ and $-A_j(\nabla \times \mathbf{B}{)}_j$. ThereforeFor the second identity, contract two Levi-Civita symbols, using ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$:The four component expressions translate intoBoth derivations require only continuously differentiable vector fields; no additional divergence-free assumptions are used.

For a regular parametrized surface $\sigma (u,v)$, the first fundamental form is the induced metricChoose the unit normal $N=({\sigma }_u\times {\sigma }_v)/|{\sigma }_u\times {\sigma }_v|$. With the convention $I{I}_{ij}=N\cdot {\sigma }_{ij}$, the second fundamental form isThe Gaussian curvature, the product of the principal curvatures, isFor a compact smooth embedded surface without boundary, the Gauss-Bonnet theorem giveswhere $e(S)$ is the Euler characteristic. The no-boundary condition is needed for this form; a smooth embedded surface with boundary has the additional boundary geodesic curvature integral.

For the surface of revolution, useThe parameters are periodic, with local open parameter rectangles providing the coordinate charts. Put $R=r+a\sin u$ and $D=\sqrt{a^2{\cos }^{2}u+b^2{\sin }^{2}u}$. Since $r>a>0$ and $b>0$, both $R$ and $D$ are positive. The tangent vectors areTheir inner products are $E=D^2$, $F=0$, and $G=R^2$. ThusTheir cross product is $R(b\sin u\cos v,b\sin u\sin v,a\cos u)$, so its normalization gives the outward unit normalNext,Taking their inner products with $N$ yields $e=-ab/D$, $f=0$, and $g=-bR\sin u/D$. Hence the fundamental forms of an elliptic ring torus giveChoosing the inward unit normal changes both displayed coefficients of $II$ to their negatives and leaves $I$ and $K$ unchanged. As a consistency check, $K=a{b}^2\sin u/(R{D}^4)$; its integral over the torus is zero, since $KdA=a{b}^2\sin ududv/D^3$ cancels under $u\mapsto u+\pi$. This agrees with $e(S)=0$ in the Gauss-Bonnet theorem.

There is a sign inconsistency in the printed model. For the ordinary cross product let $J(a)z=a\times z$. The vector triple-product identity gives $[J(a),J(b)]=J(a\times b)$. Consequently the printed $D_{\mu }={\partial }_{\mu }+eJ(A_{\mu })$ has gauge curvatureIts gauge curvature cannot instead contain the printed negative quadratic term. For a noncommuting pure gauge potential for this gauge covariant derivative, $F^{+}=0$ whereas $F^{-}=-2e{A}_{\mu }\times A_{\nu }\ne 0$. Thus the printed gauge kinetic term assigns a nonzero value to a connection obtained by applying the gauge-field transformation law to the zero connection. Literally the printed action has global rotations but not the requested local triplet gauge symmetry. The cross-product convention for an adjoint covariant derivative makes the minimal repair explicit: change the gauge field strength quadratic term to a plus sign, retaining the printed gauge covariant derivative. Changing the sign in the gauge covariant derivative instead would be another consistent convention. The remaining calculation uses the first repair.

The corrected theory has local SO(3) gauge symmetry, equivalently the adjoint-field realization of the SU(2) Lie algebra; its adjoint scalar fields and gauge fields are insensitive to the center of a group of SU(2). Explicitly, $\Phi \mapsto R(x)\Phi$ andBoth $D_{\mu }\Phi$ and $F_{\mu \nu }$ transform as triplets, so their dot products and the scalar potential are invariant. Assume $\lambda >0$, $e\ne 0$ and take $v\ge 0$ without loss of generality. For $v>0$ the minima have ${\Phi }^2=v^2$. Choose the scalar-field vacuum ${\Phi }_0=v(0,0,1)$, whose stabilizer consists of rotations around the third axis: the residual gauge symmetry is SO(2), locally U(1).

In unitary gauge write $\Phi =(0,0,v+h)$, $A_{\mu }=A_{\mu }^3$ and $W_{\mu }^{\pm }=(A_{\mu }^1\mp i{A}_{\mu }^2)/\sqrt{2}$. These are the neutral massless gauge field, two charged massive gauge fields and a real radial scalar field. DefineSubstitution gives $F_{\mu \nu }^3=f_{\mu \nu }-ie{C}_{\mu \nu }$ and $(F_{\mu \nu }^1\mp i{F}_{\mu \nu }^2)/\sqrt{2}=G_{\mu \nu }^{\pm }$. The entire physical-field Lagrangian density is thereforeIn particular $D_{\mu }\Phi =(e(v+h)A_{\mu }^2,-e(v+h)A_{\mu }^1,{\partial }_{\mu }h)$; this directly verifies the scalar field kinetic and vector mass terms. Expanding the scalar potential gives $\lambda v^2{h}^2/2+\lambda v{h}^3/2+\lambda h^4/8$. Comparing the quadratic terms with standard real scalar field and charged-vector mass terms yields the adjoint triplet Higgs spectrum,This also rewrites the purely vector interactions, which are already contained in the displayed gauge field strengths rather than omitted from the theory.

The original symmetry remains unbroken at $v=0$. The unique minimum is then $\Phi =0$, with three massless gauge fields and three massless real scalar field components; the quartic scalar field interaction remains. The physical degrees of freedom number $3\times 2+3=9$. At $v>0$ the two broken generators supply two Goldstone bosons, eaten to give longitudinal gauge-boson polarizations of the charged vectors. The physical count is $2\times 3+2+1=9$: two massive vectors, one massless vector and one radial scalar field. The scalar components eaten in unitary gauge are not additional physical particles.

For the broken phase all interactions containing the physical scalar field areMultiplying interaction coefficients by $i$ and differentiating with respect to the external fields includes the identical-field factorials. The scalar vertices of an adjoint triplet Higgs model areFor example, the $h^3$ rule contains $3!$, whereas the $hh{W}^{+}{W}^{-}$ rule contains only $2!$ from its two identical scalar fields. There are no $hAA$ or $hhAA$ vertices: the radial scalar field is neutral under the residual U(1). There is no derivative scalar-vector vertex in this physical spectrum in unitary gauge.

For completeness, if the unbroken phase is chosen, its physical scalar fields are the three components ${\varphi }_a$, not a single radial $h$. Expanding its gauge-covariant kinetic term givesWith all momenta incoming and Fourier transform convention $e^{-ipx}$, the three unbroken-phase rules areThe first rule follows by differentiating once on each possible scalar: for $a=3,b=1,c=2$, the interaction is $e{A}_{\mu }^3({\varphi }_1{\partial }^{\mu }{\varphi }_2-{\varphi }_2{\partial }^{\mu }{\varphi }_1)$, which fixes the momentum sign. The remaining rules follow from the quadratic gauge interaction and $-\lambda ({\varphi }_a{\varphi }_a{)}^2/8$. There are no unbroken-phase cubic scalar field Feynman vertices.

Use a steady axisymmetric flow governed by ideal magnetohydrodynamics and take the Newtonian gravitational potential to be axisymmetric. Work locally on connected regular magnetic flux surfaces with nonzero poloidal flow. As usual for a wind without an imposed toroidal loop voltage, $E_{\varphi }=0$; regularity on an included rotation axis guarantees this condition. It is needed to exclude steady cross-field flow in an annular domain.

Since $\nabla \varphi ={\mathbf{e}}_{\varphi }/R$, the poloidal magnetic flux function givesThe poloidal magnetic field is divergence-free. The zero toroidal electric field, with $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, gives ${\mathbf{u}}_p\parallel {\mathbf{B}}_p$. Define $k$ by $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$. Steady mass conservation then givesThus $k=k(\psi )$ is the magnetohydrodynamic mass loading, the mass flux per unit poloidal magnetic flux. Subtracting $(k/\rho )\mathbf{B}$ from the full velocity leaves only an azimuthal component. The cross product isThe steady ideal magnetohydrodynamic induction equation requires $0=\nabla \times (\omega \nabla \psi )=\nabla \omega \times \nabla \psi$. Consequently the field-line angular velocity is $\omega =\omega (\psi )$ and

Neither gas pressure nor axisymmetric gravity exerts an azimuthal torque. Multiplying the azimuthal momentum equation by $R$ combines the cylindrical curvature terms into angular-momentum transport:Insert $\rho {\mathbf{u}}_p=k{\mathbf{B}}_p$ and use ${\mathbf{B}}_p\cdot \nabla k=0$. It follows thatso the magnetohydrodynamic angular-momentum invariant isThe two terms account respectively for matter and Maxwell stress tensor transport of angular momentum.

Finally, subtracting $\gamma$ times the logarithmic mass density equation from the logarithmic pressure equation gives $D\ln (p/{\rho }^{\gamma })/Dt=0$. The specific entropy of a fixed-composition ideal gas is a function of $p/{\rho }^{\gamma }$. Steadiness and axisymmetry reduce its advection to ${\mathbf{u}}_p\cdot \nabla s=0$. HenceThese flux-label functions are local on connected regular surfaces; disconnected components with the same numerical label need not share constants without an additional matching condition.

Use the Levi-Civita symbol contraction ${ϵ}_{ijk}{ϵ}_{i\ell m}={\delta }_{j\ell }{\delta }_{km}-{\delta }_{jm}{\delta }_{k\ell }$ in suffix notation:Thus the cross product contraction expands as

The parametric equation of a straight line through $a$ with unit direction $\hat{t}$ isIf the two lines meet, then $a_1+s{\hat{t}}_1=a_2+t{\hat{t}}_2$ for some real $s,t$, so $a_1-a_2$ lies in the linear span of their direction vectors. Its dot product with their cross product is consequently zero:This is not sufficient without excluding parallel lines. For instance $a_1=(0,0,0)$, $a_2=(0,1,0)$ and ${\hat{t}}_1={\hat{t}}_2=(1,0,0)$ satisfy it, but the lines are distinct and never meet. For nonparallel lines the condition is sufficient, because the two directions span the plane perpendicular to their nonzero cross product.

For nonparallel skew lines, put $n=({\hat{t}}_1\times {\hat{t}}_2)/|{\hat{t}}_1\times {\hat{t}}_2|$. Every joining vector is $a_1-a_2+s{\hat{t}}_1-t{\hat{t}}_2$, with fixed component $(a_1-a_2)\cdot n$ along $n$. Its length is at least the absolute value of that component. Its perpendicular component can be canceled by a unique choice of $s,t$, since the two independent directions span $n^{\perp }$. This lower bound is attained, proving the distance between skew lines:

The derivative matrixis symmetric in $j,k$. Contracting it with the antisymmetric Levi-Civita symbol gives the curlDirect differentiation of the scalar potential givesso $\mathbf{F}=\nabla \varphi$. To identify the level sets, use the cross product identityIf $\mathbf{\omega }\ne 0$, write $\mathbf{x}={\mathbf{x}}_{\parallel }+{\mathbf{x}}_{\perp }$ relative to its direction. Then $\varphi =-|\mathbf{\omega }{|}^2|{\mathbf{x}}_{\perp }{|}^2/2$. Thus each negative level $\varphi =c$ is a circular cylinder with axis $\mathbb{R}\mathbf{\omega }$ and radius $\sqrt{-2c}/|\mathbf{\omega }|$. The zero level is the axis itself; positive levels are empty. If $\mathbf{\omega }=0$, the scalar potential is identically zero, so its zero level is all of space and every other level is empty.