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Use the Einstein summation convention and the contraction of two Levi-Civita symbols. For the first cross product term,The last term cancels $(F\cdot \nabla )G_i$. Interchanging $F,G$ gives the analogous cancellation with $(G\cdot \nabla )F_i$. The surviving terms are $F_j{\partial }_i{G}_j+G_j{\partial }_i{F}_j={\partial }_i(F_j{G}_j)$ by the product rule. This proves the dot-product gradient identityThe proof only requires continuously differentiable vector fields; neither field needs to be an irrotational vector field.

For $a=(a_1,a_2,a_3)$ the map has the explicit formIts diagonal entries sum to zero and its conjugate transpose is its negative, because the three coordinates are real. Hence it is traceless and a skew-Hermitian matrix.

Direct multiplication of the given basis matrices yieldsThe reverse products give the negative commutators, and equal-index commutators vanish. Bilinearity then givesThus the cross product is represented by the commutator:This realizes the cross-product model of su(2) with exactly the normalization and cyclic orientation used here.

Use the positive graph $z=g(x,y)=\sqrt{1+x^2+y^2}$. The printed PDF bounds are $\sqrt{2}\le z\le \sqrt{5}$, so its projection is the annulus $1\le r\le 2$. The surface is the corresponding band of the upper sheet of a two-sheeted hyperboloid, bounded by circles of radii $1$ and $2$ at heights $\sqrt{2}$ and $\sqrt{5}$.

For the parametrization $\mathbf{R}(x,y)=(x,y,g(x,y))$, its tangent vectors are $(1,0,x/z)$ and $(0,1,y/z)$. Their cross product gives the upward vector surface element of a graph:Its magnitude gives the scalar surface area of a graph:Both formulas apply on $1\le \sqrt{x^2+y^2}\le 2$. Reversing the orientation changes only the sign of the vector element.

Let $R$ be rotation through $2\pi /3$ about $z$, and let $S$ be rotation through $\pi$ about $x$. These generate the order-six dihedral group with relationsnamely $\{I,R,R^2,S,RS,R^{2}S\}$. We specify its order explicitly because dihedral notation has two conventions. It is a proper finite subgroup of the full axial rotation group.

To see that it is sufficient, write a general matrix in blocks $T=\left(\begin{array}{cc}B& u\\ v^T& c\end{array}\right)$. Commutation with $R$ forces $u=v=0$, since a nontrivial planar $120$-degree rotation fixes no nonzero planar vector. Its planar part is $\cos (2\pi /3)I+\sin (2\pi /3)J$, with nonzero sine, so $B$ must commute with $J$ and has the form $aI+bJ$. Commutation with $S$ again kills $b$. Hence this six-element subgroup has precisely the required invariant tensors.

It is also smallest by order. Groups of orders $2,3,5$ are cyclic: a nonidentity element has order dividing the prime group order by Lagrange theorem. For a cyclic rotation subgroup, the nonzero antisymmetric tensor representing cross product with its rotation-axis direction is invariant, so it cannot force the symmetric form above. The trivial subgroup plainly imposes no constraint. A group of order four is either cyclic or has three nonidentity elements of order two; in the latter case they commute, since $(AB{)}^{-1}=AB$ and also equals $BA$. Commuting distinct half-turns in three dimensions have perpendicular axes: conjugation by one half-turn must preserve the other's axis, and a distinct preserved axis lies in its perpendicular plane. In the specified axial group, their three axes are consequently $z$ and two perpendicular horizontal directions. In coordinates along them every diagonal matrix is invariant, including one with unequal horizontal entries. Such a group again does not force the required form.

Thus the smallest subgroup has six elements, generated by the axial $120$-degree rotation and one horizontal half-turn. It is the smallest axial rotation group forcing second-rank transverse isotropy; conjugating the horizontal axis gives equally valid choices.

For a Lie bracket over $\mathbb{R}$ or $\mathbb{C}$, antisymmetry of a Lie bracket means $[x,y]=-[y,x]$. In particular $[x,x]=0$ in these characteristic-zero fields. The Jacobi identity isIt expresses compatibility of the bracket with its own adjoint action. Bilinearity must hold over the chosen base field, and the bracket must take its values in the same vector space.

For the matrix commutator, bilinearity and antisymmetry follow directly from distributivity. Associativity of matrix multiplication givesadding the two cyclic permutations cancels every monomial. Thus the Jacobi identity holds in the entire matrix algebra. For a specified linear subspace, the only additional bracket condition is closure: $AB-BA$ must belong to the subspace whenever $A,B$ do. It is unnecessary to require closure under the separate products $AB$ and $BA$.

To determine the special unitary Lie algebra, let $g(t)$ be a differentiable curve in the special unitary group with $g(0)=I$ and $A=g^{\prime }(0)$. Differentiating $g(t{)}^{†}g(t)=I$ gives $A^{†}+A=0$. Differentiating $\det g(t)=1$ at the identity gives $\mathrm{tr}A=0$. Conversely a traceless skew-Hermitian matrix has $e^{tA}$ unitary and $\det e^{tA}=e^{t\mathrm{tr}A}=1$, so it really is a tangent vector. ThereforeThis is a real Lie algebra of complex matrices: multiplication by $i$ generally leaves this real subspace. Its complexification is ${\mathfrak{sl}}_n(\mathbb{C})$, not the compact algebra itself. For $A,B\in \mathfrak{su}(n)$,Thus closure holds, and the already verified commutator identities establish all the Lie algebra axioms.

The cross-product Lie algebra on ${\mathbb{R}}^3$ is bilinear, antisymmetric and closed because the cross product has those properties. Its Jacobi identity follows from the vector triple-product identity:In the cyclic sum, the coefficients cancel by symmetry of the scalar product.

For the explicit relation to the SU(2) Lie algebra, take the three Pauli matrices and defineThese matrices are traceless and Skew-Hermitian, and the three images of the standard basis form a real basis of $\mathfrak{su}(2)$. Using the Pauli matrix commutator identity,Hence $F$ is a real Lie algebra isomorphism. The factor and sign $-i/2$ are essential for preserving the unscaled cross-product bracket. At the group level there is an Adjoint double cover from SU(2) to SO(3); the isomorphism of their tangent algebras does not identify the two global groups.

Differentiate the unit tangent vector with $d/ds=v^{-1}d/du$ and $v^{\prime }=({\mathbf{x}}^{\prime }\cdot {\mathbf{x}}^{\prime \prime })/v$. This givesThe curvature of a space curve is $\kappa =|d\mathbf{t}/ds|$. Squaring the displayed expression and simplifying givesThe last equality is the cross product norm identity. Equivalently, the hinted vector triple product rewrites $d\mathbf{t}/ds=-{\mathbf{x}}^{\prime }\times ({\mathbf{x}}^{\prime }\times {\mathbf{x}}^{\prime \prime })/v^4$.

For a differentiable curve $R(t)$ in the SO(3) group with $R(0)=I$, differentiating $R(t{)}^{T}R(t)=I$ gives $X^T+X=0$, where $X=R^{\prime }(0)$. Conversely, if $X$ is a real skew-symmetric matrix, $e^{tX}$ is an orthogonal matrix and has determinant $e^{t\mathrm{tr}X}=1$. HenceThe displayed generators of the SO(3) Lie algebra areThey are linearly independent and span the three independent entries of a real skew-symmetric matrix. Equivalently, $T_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The vector triple-product identity then givesThus the structure constants of a Lie algebra in this basis are

Use a real, antisymmetric convention throughout the SO(3) vector Higgs model. With gauge coupling $g$, the gauge covariant derivative and gauge field strength areThere is no factor of $i$ in these formulas because $T_a$ are real antisymmetric generators. For a local non-Abelian gauge transformation ${\Phi }^{\prime }=R\Phi$, with $R(x)$ in the SO(3) group, the compatible transformation of the gauge potential isIt gives $D_{\mu }^{\prime }{\Phi }^{\prime }=R{D}_{\mu }\Phi$ and $F_{\mu \nu }^{\prime }=R{F}_{\mu \nu }{R}^{-1}$. These covariance laws fix the relative signs in the gauge covariant derivative and gauge field strength.

An SO(3) invariant quartic scalar potential isSince $(R\Phi {)}^T(R\Phi )={\Phi }^T\Phi$, this scalar potential is gauge-invariant. Its minimum is the sphere ${\Phi }^T\Phi =v^2$, rather than a preferred direction in the internal vector space.

Writing ${\mathbf{A}}_{\mu }=(A_{\mu 1},A_{\mu 2},A_{\mu 3})$, the cross product form of the gauge covariant derivative is $D_{\mu }\Phi ={\partial }_{\mu }\Phi +g{\mathbf{A}}_{\mu }\times \Phi$. In components this isCorrespondingly, the gauge field strength has components

A covariantly constant vector Higgs field has constant norm, becauseThe last equality uses antisymmetry of the gauge potential. On a connected region write this norm as $\varphi \ge 0$. If $\varphi =0$, the vector Higgs field is identically zero. If $\varphi >0$, the SO(3) group acts transitively on unit vectors, so choose a smooth local non-Abelian gauge transformation rotating $\Phi /\varphi$ to ${\mathbf{e}}_3$. In this unitary gauge, $\Phi =(0,0,\varphi {)}^T$ with $\varphi$ constant. This is a local construction, and is global on a trivial contractible region; a nontrivial bundle can require more than one gauge patch.

For nonzero $\varphi$, substituting the constant aligned vector Higgs field into $D_{\mu }\Phi =0$ gives $g\varphi (A_{\mu 2},-A_{\mu 1},0{)}^T=0$. For nonzero gauge coupling, the most general compatible gauge potential and gauge field strength arewhere $a_{\mu }$ is an arbitrary real one-form. Thus the surviving gauge group is the $SO(2)$ stabilizer subgroup of ${\mathbf{e}}_3$. A covariantly constant vector Higgs field does not force the gauge field strength to vanish: it only requires the curvature to annihilate that field.

For the Higgs mechanism, choose a vacuum of the SO(3) invariant quartic scalar potential, so the constant norm is $v$. The canonically normalized kinetic term then containsThe two broken-direction gauge bosons have mass $gv$ (or $|g|v$ if the sign of $g$ is unrestricted), while the $T_3$ gauge boson remains massless. The two angular Goldstone bosons supply the massive vectors' longitudinal gauge-boson polarizations. The radial fluctuation remains a physical scalar field, with $m_h^2=2\lambda v^2$ for this normalization. The condition $D_{\mu }\Phi =0$ selects backgrounds without broken-direction gauge potentials; fluctuations about such a vacuum exhibit the two massive modes. Constancy of the norm alone does not require it to minimize the scalar potential.