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For continuously differentiable vector fields, the gradient of their dot product expands as displayed. The contraction of two Levi-Civita symbols gives $[F\times (\nabla \times G){]}_i=F_j{\partial }_i{G}_j-F_j{\partial }_j{G}_i$; adding the directional derivative cancels the second term. Interchanging the fields and using the product rule proves the identity. It does not require either field to be irrotational.

For differentiable vector fields, expand the cross products using the contraction of two Levi-Civita symbols. The directional-derivative terms cancel the corresponding terms in the two cross products, leaving ${\partial }_i(F_j{G}_j)$. This proves the displayed gradient identity by the product rule.

Apply the contraction of two Levi-Civita symbols to the cross product:Thus the vector triple product identity is

Using the Levi-Civita symbol and the contraction of two Levi-Civita symbols, the $i$th component of the vector triple product isThus the triple-product identity is $a\times (b\times c)=(a\cdot c)b-(a\cdot b)c$.

Apply this with the unit vector $n$ to obtain $\Pi (x)=x-(n\cdot x)n$. Dot products and scalar multiplication are linear in $x$, so $\Pi (sx+ty)=s\Pi (x)+t\Pi (y)$. Its matrix representation isThe fixed-vector equation is equivalent to $(n\cdot x)n=0$. Since $n\ne 0$, all its solutions are $x\in n^{\perp }$. Also $\boxed{\Pi (n)=0}$. Every vector decomposes as $x=\Pi (x)+(n\cdot x)n$, with $\Pi (x)$ orthogonal to $n$. Hence $\Pi$ is the orthogonal projection onto the plane through the origin normal to $n$: it fixes that plane and removes the perpendicular component.

In Einstein summation convention, the curl of a cross product has componentsThe second line uses the contraction of two Levi-Civita symbols; the last uses the product rule. These components prove the curl of a cross product identity, including both divergence and directional-derivative terms.

Now apply that identity to $m\times v$. On the orientable surface, $m=n$, $m\cdot v=0$, and $|m|=1$. Differentiating $m\cdot m=1$ gives $m_i{\partial }_j{m}_i=0$. Taking a dot product with $m$ in the curl of a cross product formula therefore gives, on $S$,Only tangency on the surface is needed; no claim is made that the extension $v$ is tangent away from it. The right side is the surface divergence of the tangent vector field.

With the boundary oriented according to Stokes theorem, cyclic invariance of the scalar triple product giveswhere $u$ is the outward conormal of a surface boundary. Thus Stokes theorem provesThis is the surface divergence theorem for a tangent field.

For the disc in the $z=0$ plane, take $n=(0,0,1)$ and $v=\mathbf{r}=(x,y,z)$. The field is tangent on the disc, and ${\partial }_j{v}_i={\delta }_{ij}$. The contraction is $3-|n{|}^2=2$, so the surface integral is $2\pi R^2$. On the positively oriented circular boundary, $u=(\cos \theta ,\sin \theta ,0)$ and $v=Ru$, so the boundary flux is $R(2\pi R)=2\pi R^2$. Both sides agree.

Using Einstein summation convention and the contraction of two Levi-Civita symbols,This proves the vector triple product identity directly from the Levi-Civita symbol and Kronecker delta.

Define the scalar triple product by $[a,b,c]=a\cdot (b\times c)$. Its expression ${ϵ}_{ijk}{a}_i{b}_j{c}_k$ is unchanged by cyclically relabeling the three indices: a three-cycle has positive sign. ThusIn particular, let $\Delta =[e_1,e_2,e_3]$. Since the three vectors form a basis, their determinant $\Delta$ is nonzero, so their reciprocal basis vectors are well defined. Cyclic invariance and vanishing of a scalar triple product with repeated vectors giveTo calculate the new scalar triple product explicitly, apply the already proved vector triple product identity with $A=e_3\times e_1$:Consequentlyso the reciprocal vectors also form a basis. Taking the reciprocal basis of that basis now givesThe same cyclic calculation gives ${\hat{\hat{e}}}_2=e_2$ and ${\hat{\hat{e}}}_3=e_3$, proving reciprocal basis involution.

Finally, write $V=V_1{e}_1+V_2{e}_2+V_3{e}_3$. Taking a dot product with ${\hat{e}}_j$ isolates $V_j$. Applying the same argument to the reciprocal basis, whose reciprocal is the original basis, gives the two sets of coordinates:No orthonormality assumption is needed; the reciprocal basis is what makes these coordinate-extraction formulas work.

Use suffix notation, summing repeated indices. The contraction of two Levi-Civita symbols givesAdding the directional derivative $(F\cdot \nabla )G_i=F_j{\partial }_j{G}_i$ cancels the second term. Interchanging $F,G$ gives another pair whose sum is $G_j{\partial }_i{F}_j$. The total right-hand side is thereforethe $i$th component of the gradient of the dot product. Since this holds for each component, the dot-product gradient identity is proved:It holds for any continuously differentiable vector fields; no irrotationality assumption is used.

The contraction of two Levi-Civita symbols gives $[F\times (\nabla \times G){]}_i=F_j{\partial }_i{G}_j-F_j{\partial }_j{G}_i$. Similarly $[G\times (\nabla \times F){]}_i=G_j{\partial }_i{F}_j-G_j{\partial }_j{F}_i$. Add the two directional-derivative terms to cancel the last terms, leavingThe product rule has therefore proved the gradient of a dot product identity

Use the Levi-Civita symbol and sum over repeated indices. The contraction of two Levi-Civita symbols and the product rule giveReading the four terms as divergence and directional derivatives proves the curl of a cross product identity

Use suffix notation, summing repeated indices, and write ${\partial }_i=\partial /\partial x_i$. For a twice continuously differentiable vector field, mixed derivatives commute. Thus the divergence of its curl satisfiesbecause the derivatives are symmetric in $i,j$, while the Levi-Civita symbol is antisymmetric in those indices. HenceFor the second identity, use the contraction of two Levi-Civita symbols:Rearranging each component proves the vorticity cross-product identityThe second identity needs first derivatives; the first additionally uses equality of the mixed second derivatives.

Covariance of the fundamental gauge covariant derivative requires $D_{\mu }^{f\prime }(g\chi )=g{D}_{\mu }^f\chi$ for every column $\chi$. Expanding the left side and cancelling $g{\partial }_{\mu }\chi$ gives $({\partial }_{\mu }g)+A_{\mu }^{\prime }g=g{A}_{\mu }$. Therefore the necessary and sufficient Yang-Mills gauge transformation isIn operator notation $D_{\mu }^{f\prime }=g{D}_{\mu }^f{g}^{-1}$. Expanding its commutator on an arbitrary column shows $[D_{\mu }^f,D_{\nu }^f]\chi =F_{\mu \nu }\chi$: the second derivatives cancel, as do the terms containing a single derivative of $\chi$. Conjugating this identity proves the gauge field strength transformationThus the gauge field strength transforms homogeneously even though the potential has an inhomogeneous derivative term.

For the adjoint scalar field, write $B_{\mu }=({\partial }_{\mu }g)g^{-1}$ and ${\Phi }^{\prime }=g\Phi g^{-1}$. Differentiating the inverse matrix givesThe extra terms cancel, proving $D_{\mu }^{\prime }{\Phi }^{\prime }=g(D_{\mu }\Phi )g^{-1}$ for the adjoint covariant derivative. Its curvature is obtained by another direct expansion:where the Jacobi identity combines the last two terms.

For the equations of motion take compactly supported variations, and assume $e\ne 0$, as required by the Lagrangian coefficient. Invariance of the Killing form impliesIndeed, the two connection terms sum to $\kappa ([A_{\mu },X],Y)+\kappa (X,[A_{\mu },Y])=0$. This is the gauge-covariant integration by parts identity. The needed variations areAntisymmetry of the gauge field strength combines its two variation terms. Applying gauge-covariant integration by parts to the action $S=\int \mathcal{L}{d}^{4}x$ givesHere $\kappa ([\delta A_{\nu },\Phi ],D^{\nu }\Phi )=\kappa (\delta A_{\nu },[\Phi ,D^{\nu }\Phi ])$ fixes the sign of the scalar current. Since the Killing form is nondegenerate on a semisimple Lie algebra, independent variations give the coupled Yang-Mills equations and scalar equationEquivalently, $D_{\mu }{F}^{\mu \nu }=e^2[D^{\nu }\Phi ,\Phi ]$. These signs use the stated Minkowski metric, with upper spatial derivatives $D^i=-D_i$.

It remains to verify every equation for the static Bogomolny equations. The hypotheses imply $F_{0i}=0$ and $D_0\Phi =0$, so the temporal gauge equation is automatically zero. The gauge-theory Bianchi identityfollows, for example, from the Jacobi identity for the fundamental gauge covariant derivatives, using their commutator $F_{ij}$. Define $B_i=\frac{1}{2}{\epsilon }_{ijk}{F}_{jk}$. Contracting the gauge-theory Bianchi identity with ${\epsilon }_{ijk}$ gives $D_i{B}_i=0$. The Bogomolny equations say $B_i=e{D}_i\Phi$, so $e\ne 0$ implies $D_i{D}_i\Phi =0$. In the static configuration $D_{\mu }{D}^{\mu }\Phi =-D_i{D}_i\Phi$, verifying the scalar equation.

For the spatial gauge equation, use $F^{ij}=F_{ij}$ and $D^j\Phi =-D_j\Phi$. The Bogomolny equations, the curvature commutator proved above, and the contraction of two Levi-Civita symbols givesince ${\epsilon }_{ijk}{\epsilon }_{ik\ell }=-2{\delta }_{j\ell }$. This is exactly $D_i{F}^{ij}+e^2[\Phi ,D^j\Phi ]=0$. Thus the static first-order Bogomolny system satisfies all four gauge equations and the scalar equation, with no extra ansatz or boundary assumption needed for this local implication. This is the mechanism by which Bogomolny monopole equations imply Yang-Mills-Higgs equations.

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.

In Einstein summation convention, the curl isApply another curl and use the contraction of two Levi-Civita symbols:Smoothness allows the partial derivatives to commute. This proves the curl of the curl identity:

Insert ${\Omega }_{jk}=-{ϵ}_{jkl}{\varphi }^l$ and the result for $S^{jk}$ into the Spinor representation of the Lorentz group. The contraction of two Levi-Civita symbols ${\sum }_{j,k}{ϵ}_{jkl}{ϵ}^{jkm}=2{\delta }_l{}^m$ givesThe matrix exponential of this block diagonal matrix is the stated $O$, so ${\psi }^{\prime }(x)=O\psi ({\Lambda }^{-1}x)$. The plus sign in its exponent follows from both minus signs above; it is tied to the rotation-parameter convention in the PDF.

For a rotation through $2\pi$ about $z$, the Pauli matrix has eigenvalues $\pm 1$, hence $e^{i\pi {\sigma }^3}=-I_2$. The spacetime rotation is the identity, whileThis is the spinor sign under a full spatial rotation. A vector returns to itself already at $2\pi$. The comparison concerns the exponentiated Lorentz group representations: a Lie algebra by itself has no distinguished element called a $2\pi$ rotation. The sign records the nontrivial lift of this rotation loop to the Spin group.

Both sides are antisymmetric in $j,k$ and in $p,q$. If either pair has equal indices, both sides vanish. It therefore suffices to take $j\ne k$ and $p\ne q$. The sum over $i$ is nonzero exactly when $(p,q)=(j,k)$ or $(k,j)$; it is respectively $1$ or $-1$. This is exactlythe contraction of two Levi-Civita symbols.

First separate the symmetric tensor and antisymmetric tensor parts:The trace of the antisymmetric part vanishes. In three dimensions, remove one third of the trace from the symmetric part, and identify the remaining antisymmetric part with an axial vector. The scalar, symmetric-traceless and axial tensor decomposition isHere $S_{ji}=S_{ij}$ and $S_{ii}=P_{ii}-P_{kk}=0$. The contraction of two Levi-Civita symbols verifies the reconstruction:ThereforeThe pieces contain $1$, $5$ and $3$ independent components respectively, accounting for all nine entries. Taking the trace, then the traceless symmetric part, then the antisymmetric part shows that the decomposition is unique.

For a differentiable vector field in Cartesian coordinates, the divergence and curl arewith summation over repeated indices. In components the curl is $({\partial }_y{F}_z-{\partial }_z{F}_y,{\partial }_z{F}_x-{\partial }_x{F}_z,{\partial }_x{F}_y-{\partial }_y{F}_x)$. The contraction of two Levi-Civita symbols and the product rule giveRecognizing the directional derivatives gives the curl of a cross product identity

In Cartesian coordinates a rank-$r$ tensor is an array whose components transform under a change of orthonormal axes by one rotation matrix factor per index:An isotropic tensor has unchanged components under every proper rotation. The Levi-Civita symbol obeysfor $Q\in SO(3)$, proving rotation invariance of the Levi-Civita symbol. Under an orthogonal transformation of determinant $-1$ its sign reverses; including such transformations requires the pseudotensor distinction. Here isotropy is with respect to proper rotations.

Contract the determinant identity with ${ϵ}_{ijk}$. Since ${ϵ}_{ijk}{ϵ}_{ijk}=3!=6$, one obtainsEvery index on the right is contracted, so the tensor component transformation law leaves no free rotation factor: the determinant is a scalar. Equivalently, $A^{\prime }=QA{Q}^T$ gives $\det A^{\prime }=(\det Q{)}^2\det A=\det A$.

For the matrix inverse, defineThis is the cofactor matrix. To verify the inverse formula rather than just recognize it, contract with $A_{ib}$ and use the determinant identity again:Here the contraction of two Levi-Civita symbols gives ${ϵ}_{ajk}{ϵ}_{ijk}=2{\delta }_{ai}$. Therefore $A{C}^T=(\det A)I$ and, when $\det A\ne 0$,These are the Levi-Civita formulas for a three-dimensional determinant and inverse.

Now differentiate the cubic determinant formula with respect to the individual entry $A_{ab}$. The three differentiated factors give equal contributions: exchanging two paired indices changes the sign of both epsilon symbols, leaving their product unchanged. ThusSubstituting the inverse expression proves the entrywise derivative of the determinant:The cofactor version remains valid for singular matrices; only the inverse version requires invertibility.

For the time-dependent Jacobian matrix, the stated smooth expansion impliesApplying the chain rule and the entrywise determinant derivative gives Jacobi's formula near $t=0$:At zero, $C=I$, so ${{\partial }_t\det A|}_0={\sum }_a{\partial }_a{u}_a=\nabla \cdot u$. A Taylor expansion therefore yieldsThe Jacobian determinant measures local volume change, so this establishes volume preservation to first order. An arbitrary family specified only through its first-order expansion need not preserve volume exactly at higher orders.

Use the Levi-Civita symbol and the contraction of two Levi-Civita symbols in Cartesian coordinates. With repeated indices summed,The last line is the product rule. Reading its four terms in vector notation proves the curl of a cross product identity:

In Cartesian coordinates, write ${\partial }_i=\partial /\partial x_i$ and use the Levi-Civita symbol. The contraction of two Levi-Civita symbols givesHere the Kronecker delta selects the indicated indices, and continuous second partial derivatives allow their order to be interchanged. Identifying the gradient, divergence and componentwise Laplacian proves the curl of the curl identity: