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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.

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.