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Take a smooth gauge transformation $g:{\mathbb{R}}^3\to SU(2)$. In the anti-Hermitian convention of the adjoint covariant derivative, setExpanding ${\partial }_j(g\Phi g^{-1})$ shows that its two differentiated-$g$ terms cancel with the extra terms in $[A_j^g,{\Phi }^g]$. Thus $D_j^g{\Phi }^g=g(D_j\Phi )g^{-1}$. The gauge field strength is $F_{jk}={\partial }_j{A}_k-{\partial }_k{A}_j+[A_j,A_k]$, and the same cancellation, or the commutator of covariant derivatives, gives $F_{jk}^g=g{F}_{jk}{g}^{-1}$.

The matrix trace is unchanged by conjugation, soEach field-strength norm, covariant-gradient norm and Higgs norm is therefore invariant pointwise. It follows that $V_{\lambda }(A^g,{\Phi }^g)=V_{\lambda }(A,\Phi )$. The displayed tensor made from these fields is also gauge invariant wherever $\Phi \ne 0$.

There is a normalization problem in the requested final identity. With the ordinary defining $2\times 2$ trace and the printed inner product, an explicit orthonormal basis isThe last identity follows from the Pauli matrix multiplication law. In particular it is not permissible to combine this orthonormality with unit structure constants of a Lie algebra.

Write $A_j=a_j^a{e}_a$ and $\Phi =\phi e_3$ with $\phi \ne 0$, and put $s=\mathrm{sgn}\phi$. Define $C_{jk}=a_j^1{a}_k^2-a_j^2{a}_k^1$. ThenOnly the bracket of the transverse terms contributes in the $e_3$ direction, givingHence the literal tensor printed in the question isThe transverse term does not cancel with this trace normalization of su(2).

A concrete counterexample to the printed closure assertion is $\Phi =e_3$, $A_1=e_1$, $A_2=x_3{e}_2$, $A_3=0$ on a ball. Directly, $F_{12}=\sqrt{2}{x}_3{e}_3$, $F_{23}=-e_2$ and $F_{31}=0$, while $D_1\Phi =-\sqrt{2}{e}_2$, $D_2\Phi =\sqrt{2}{x}_3{e}_1$ and $D_3\Phi =0$. It follows thatThe Higgs field never vanishes. If global finite energy is desired, multiply both nonzero gauge-potential components by a smooth compactly supported cutoff equal to one on this ball and keep $\Phi =e_3$ everywhere. The energy is finite and the same local counterexample remains. Thus the claimed closure is false with the ordinary trace normalization actually printed; it is not a consequence of gauge invariance alone.

The consistent 't Hooft electromagnetic tensor for the printed inner product is insteadIn the directional gauge, its commutator coefficient is $1/{\kappa }^2$, so the terms cancel exactly andOn a connected region where $\phi \ne 0$, $s$ is constant. Commutation of ordinary partial derivatives provesAt any nonzero-Higgs point, use the permitted local gauge transformation to align the field with a fixed internal direction. Since $\mathcal{F}$ is gauge invariant, the identity derived in that gauge holds in the original gauge on the same ball. These balls cover every nonzero-Higgs region, proving that the corrected tensor is a closed differential form there.

Alternatively, keeping the tensor's printed coefficient one but replacing the inner product by $-2\mathrm{tr}(XY)$ gives the orthonormal basis $e_a=-i{\sigma }_a/2$ with $\kappa =1$. The same calculation then gives $f_{jk}=s({\partial }_j{a}_k^3-{\partial }_k{a}_j^3)$ and the intended identity. Either repair makes the bracket and inner-product normalizations agree; they must not be silently mixed.

The supplied gamma matrices use the mostly-minus metric signature. The Pauli matrix multiplication law gives ${\sigma }^1{\sigma }^2{\sigma }^3=i{I}_2$, hence block multiplication yieldsThe off-diagonal blocks of every ${\gamma }^{\mu }$ connect opposite eigenspaces of this diagonal matrix. Explicitly,Thus $\{{\gamma }^5,{\gamma }^{\mu }\}=0$, and $({\gamma }^5{)}^2=I_4$. The chiral projectors $(1\mp {\gamma }^5)/2$ select the upper and lower two-component blocks, respectively.

Write the Dirac spinor as $\psi =({\chi }_L,{\chi }_R{)}^T$. Substituting the block matrices into the massless Dirac equation gives two independent Weyl spinor equations:They are decoupled because the mass term, which would join the two chirality components, is absent.

For a left-handed plane wave, use $p\cdot x=p^{0}t-\mathbf{p}\cdot \mathbf{x}$. Its coefficient $\xi$ obeysSince $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^2{I}_2$, the determinant is $(p^0{)}^2-|\mathbf{p}{|}^2$. A nonzero coefficient therefore requires the massless mass shell $\boxed{p^2=0}$. For positive energy and nonzero $\mathbf{p}$, put $P=|\mathbf{p}|=p^0$ and parameterize its direction by angles $\vartheta ,\phi$. ThenMultiplication gives $(\mathbf{\sigma }\cdot \hat{\mathbf{p}}){\chi }_{-}=-{\chi }_{-}$ and ${\chi }_{-}^{†}{\chi }_{-}=1$. Hence the left-handed massless plane-wave spinor can be writtenHere $N(p)$ is an arbitrary complex normalization; $|N|=\sqrt{2P}$ gives ${\lambda }_L^{†}{\lambda }_L=2P$. Away from the negative $z$ axis one can equivalently takeAt $\mathbf{p}=-P\hat{\mathbf{z}}$ choose $(1,0{)}^T$ instead. These are phase patches for the same one-dimensional kernel. The other frequency branch $p^0=-P$ uses the positive eigenvector ${\chi }_{+}=(\cos (\vartheta /2),e^{i\phi }\sin (\vartheta /2){)}^T$ for the direction of $\mathbf{p}$. At $p^{\mu }=0$, the algebraic equation vanishes and any two-component coefficient is possible, but no momentum direction or helicity is defined there.

A spatial rotation by $\alpha$ about $\hat{\mathbf{p}}$ acts on the two-component coefficient by $\exp (-i\alpha \mathbf{\sigma }\cdot \hat{\mathbf{p}}/2)$. The negative eigenvector therefore acquires the phase $e^{i\alpha /2}$. Since a state with helicity $h$ acquires $e^{-i\alpha h}$, the positive-energy left-handed particle has $\boxed{h=-1/2}$. There is only one such spin state at a fixed nonzero momentum, rather than two independent polarizations of the same chiral particle.

The helicity content of a quantized Weyl field also includes an antiparticle with opposite helicity. In the field expansion, the negative-frequency term multiplies an antiparticle fermionic creation operator, while the positive-frequency term multiplies a particle fermionic annihilation operator. Their adjoint roles give conjugate phases for the corresponding created states, so a left-handed field creates particles with $h=-1/2$ and antiparticles with $h=+1/2$. The right-handed Weyl spinor has the opposite assignments. Quantizing the full massless Dirac field, rather than one chiral component, therefore gives both particle helicities and both antiparticle helicities; every excitation has spin one-half, while chirality determines the helicity of each massless sector.

A Lie algebra representation is a linear map $\rho :\mathfrak{g}\to \mathrm{End}(V)$ preserving brackets: $\rho ([X,Y])=[\rho (X),\rho (Y)]$. Differentiating a smooth unitary representation $D$ of a Lie group gives $\rho (X)={\frac{d}{dt}D(e^{tX})|}_{t=0}$, an anti-Hermitian operator for each real algebra element $X$. In the physics convention one uses Hermitian generators $t_X=i\rho (X)$ and writes $D(e^{tX})=e^{-it{t}_X}$. Conversely, integration of a Lie-algebra representation gives a representation of the connected simply connected group. A representation of another group with the same algebra additionally has to be trivial on the kernel of its covering map.

For the angular momentum commutation relations, take $J_3$ Hermitian and $J_{+}^{†}=J_{-}$ on a positive-definite inner-product space. A highest-weight vector has $J_3|j,j⟩=j|j,j⟩$ and $J_{+}|j,j⟩=0$. Set $v_{\ell }=(J_{-}{)}^{\ell }|j,j⟩$. Commuting $J_3$ through each lowering operator gives $J_3{v}_{\ell }=(j-\ell )v_{\ell }$. Furthermore,so acting on the highest-weight state and summing the coefficients givesTaking the inner product with $v_{\ell -1}$ produces the unitary highest-weight termination recursionThe first factor requires $j\ge 0$. If $2j$ is not an integer, the first integer $\ell >2j+1$ would make the norm negative, while every preceding norm is positive. ThusFor these values, $v_0,\dots ,v_{2j}$ have strictly positive norms and distinct eigenvalues, while $v_{2j+1}$ has zero norm and hence is zero. They therefore span a space of dimension $2j+1$, invariant under all three generators. Its weights are $j,j-1,\dots ,-j$. This proves integer or half-integer spin and finite dimension $2j+1$. The positive-definite unitary assumption matters: an unrestricted algebraic highest-weight module need not terminate. Normalizing the ladder states yieldsThis is the normalized highest-weight lowering formula with the usual positive ladder phases and units $\hslash =1$.

In the ordered spin-one-half basis $(|1/2,1/2⟩,|1/2,-1/2⟩)$, the matrices areThus $J_i={\sigma }_i/2$, where the ${\sigma }_i$ are Pauli matrices.

Choose the active-rotation convention. The SU(2) lift of an axis-angle rotation isThe Pauli matrix multiplication law shows that conjugating $\mathbf{x}\cdot \mathbf{\sigma }$ by this matrix produces $(R\mathbf{x})\cdot \mathbf{\sigma }$, withThis is the ordinary three-dimensional rotation. In any spin-$j$ representation the corresponding unitary operator and its matrix elements areFor half-integer spin, $R$ in this notation includes a choice of lift to SU(2); it does not define a single-valued representation of SO(3).

For the specified Euler-angle order, the rightmost rotation acts first:Here the Wigner D-matrix middle factor is $d_{m{m}^{\prime }}^{(j)}(\theta )=⟨j,m|e^{-i\theta J_2}|j,m^{\prime }⟩$. It can be calculated explicitly by realizing spin $j$ as the symmetric product of $2j$ spin-one-half factors. Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The fundamental rotation sends $|+⟩$ to $c|+⟩+s|-⟩$ and $|-⟩$ to $-s|+⟩+c|-⟩$. Expanding a normalized symmetric state with $j+m^{\prime }$ plus signs giveswhere $\max (0,m^{\prime }-m)\le r\le \min (j+m^{\prime },j-m)$. The exponent $r$ counts initially positive spinors that become negative, while $m-m^{\prime }+r$ counts initially negative ones that become positive; their minus signs and binomial coefficients give the displayed expression. Every factorial argument is an integer even for half-integer spin.

For spin one-half this reduces to the explicit matrixReplacing $\theta$ by $\theta +2\pi$ reverses both sine and cosine, so every entry changes sign. In particular, a pure $2\pi$ rotation gives $D^{(1/2)}=-I$, while a $4\pi$ rotation gives $I$.

Finally, the Adjoint double cover from SU(2) to SO(3) is surjective by the axis-angle construction. Its kernel consists of matrices commuting with all Pauli matrices, hence scalar matrices; within SU(2) these are $\pm I$. ThusTheir Lie algebras are isomorphic, but their global topology and representations differ. In spin $j$, the central element $-I$ acts as $(-1{)}^{2j}I$, so exactly the integer-spin representations descend to ordinary SO(3) representations. This is the descent of an SU(2) representation to SO(3) criterion.

A SU(2) matrix has the form $\left(\begin{array}{cc}a& b\\ -\overline{b}& \overline{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$. Comparing this with the Pauli expansion givesConsequently all four Pauli coefficients must be real, and their norm condition isConversely, the Pauli matrix multiplication law gives $A{A}^{†}=(u_0^2+|\mathbf{u}{|}^2)I$ and $\det A=u_0^2+|\mathbf{u}{|}^2$, so these conditions imply unitarity and determinant one. For complex coefficients the determinant equation alone would not imply unitarity.

On either open hemisphere,Both signs are necessary globally. The two hemispheres meet along $u_0=0$, and the full four-coefficient constraint is exactly the unit sphere in ${\mathbb{R}}^4$. The coefficient map and its inverse are smooth, establishing SU(2) as the three-sphere as a manifold, not merely a dimension count.

For two real coefficient quadruples, expanding the matrix product with the Pauli matrix multiplication law givesFor an infinitesimal second factor, $v_0=1+O(|\mathbf{v}{|}^2)$. Thereforewhile $d{u}_0=-u_i{v}_i$. These variations are tangent to the unit sphere because $u_{0}d{u}_0+u_{i}d{u}_i=0$.

On a hemisphere, differentiation of the norm constraint gives ${\partial }_{u_i}{u}_0=-u_i/u_0$, and henceContracting with the Pauli-coordinate left-invariant vector fields on SU(2) and using ${\mu }_{ji}{u}_i=u_0{u}_j$ givesThe Pauli matrix multiplication law gives exactly the same expression for $-A{\sigma }_j$, soIt follows thatThis uses the Pauli matrix commutator identity. The commutator of two first-order vector fields is again a vector field, and the matrix entries of $A$ span the four coordinate functions $u_0,u_1,u_2,u_3$. Equality of their actions on these entries determines the tangent field. ThereforeThe real tangent field underneath $T_j$ is generated by the flow $A\mapsto A{e}^{it{\sigma }_j}$; left multiplication transports that tangent consistently, explaining its left invariance. The rescaled generators $-T_j/2$ have the usual positive-sign angular-momentum commutator.

To determine the Haar measure, compute the metric induced from the unit three-sphere. Since $d{u}_0=-u_{i}d{u}_i/u_0$, its line element isThe rank-one matrix determinant lemma givesThus the Pauli-coordinate Haar measure on SU(2) isup to an overall normalization. To check its group invariance, fix one unit coefficient quadruple and multiply by it. The product formulas define a linear transformation of the other quadruple. The norm of a product is the product of norms, as follows either from $A{A}^{†}$ or directly from the dot and cross products. Fixed multiplication by a unit quadruple is therefore an orthogonal transformation of ${\mathbb{R}}^4$. It preserves the round sphere metric and volume, on both the left and the right, proving that this is the invariant measure.

The same density can be checked directly against the vector fields. On a fixed hemisphere, $u_0/|u_0|$ is constant, andThe second term vanishes because its contractions involve either ${\delta }_{ik}$ or $u_i{u}_k$, both symmetric in $i,k$. Thus the density has zero divergence along every invariant tangent direction.

The hemisphere chart becomes singular at the equator, but its volume is finite:Both hemispheres give total volume $2{\pi }^2$, so probability-normalized Haar measure is $d^{3}u/(2{\pi }^2|u_0|)$ on each hemisphere. Finally, the three-sphere is closed and bounded in ${\mathbb{R}}^4$, hence compact; its smooth identification with SU(2) proves that $SU(2)$ is a compact group. The finite integral reflects this compactness, and the equatorial divergence is only a coordinate artefact.

Use signature $\eta =\mathrm{diag}(1,-1,-1,-1)$, the numerical matrices ${\sigma }_{\mu }=(I,{\sigma }_i)$ and ${\bar{\sigma }}_{\mu }=(I,-{\sigma }_i)$ given in the question, and raise genuine tensor indices with $\eta$. A real Minkowski spacetime vector corresponds to the Hermitian matrixFor SL(2,C) matrices, $X^{\prime }=AX{A}^{†}$ is Hermitian and has the same determinant. It therefore induces a real linear Lorentz transformation; composition of the congruence actions agrees with matrix multiplication. The kernel consists of $\pm I$: preservation of $X=I$ makes a kernel element unitary, and preservation of every Hermitian $X$ makes it scalar.

Polar decomposition of an invertible complex matrix connects every determinant-one complex matrix to its unitary factor, so the group is connected. The action gives the Lorentz spinor double cover of the proper orthochronous Lorentz group. It covers rotations via $A_R=e^{-i\vartheta \mathbf{n}\cdot \mathbf{\sigma }/2}$ and boosts via the positive Hermitian matrices below. Every proper orthochronous transformation is a boost followed by a rotation, since one can first match its image of the future unit time vector and then use its rotation stabilizer. The congruence construction does not cover spatial parity or time reversal. In particular parity has determinant $-1$ as a four-vector transformation; all transformations continuously produced from SL(2,C) have determinant $+1$.

Set $c_{\theta }=\cosh (\theta /2)$, $s_{\theta }=\sinh (\theta /2)$ and $N=\mathbf{n}\cdot \mathbf{\sigma }$. The Pauli matrix multiplication law gives $N^2=I$. The proposed boost matrix is $A_B=c_{\theta }I+s_{\theta }N$, with eigenvalues $e^{\pm \theta /2}$, determinant one and $A_B^{†}=A_B$. Split $\mathbf{x}=x_{\parallel }\mathbf{n}+{\mathbf{x}}_{\perp }$. The perpendicular Pauli part anticommutes with $N$, so multiplication givesThis is an active Lorentz boost. The image of a rest worldline has velocity $\mathbf{v}=\tanh \theta \mathbf{n}$, fixing the velocity sign convention.

For two nonzero boosts, write $A^{\prime }=c^{\prime }I+s^{\prime }{\mathbf{n}}^{\prime }\cdot \mathbf{\sigma }$ and $A=cI+s\mathbf{n}\cdot \mathbf{\sigma }$. Their product isThe last term is anti-Hermitian. Thus $A^{\prime }A$ is not Hermitian unless the axes are collinear; a pure boost has Hermitian lifts $\pm A_B$, so it cannot give the same Lorentz transformation. The noncollinear boost obstruction from Pauli products is thereforeThat rotation is a Wigner rotation. A zero-rapidity factor is the trivial exception, regardless of the arbitrary axis assigned to it.

Define $J_i=\frac{1}{2}{ϵ}_{ijk}{M}_{jk}$ and $K_i=M_{0i}$. The printed Lorentz algebra brackets giveSet $L_i=(J_i+i{K}_i)/2$, $R_i=(J_i-i{K}_i)/2$. ThenThis is the chiral decomposition of the complex Lorentz algebra. The two copies are the complexifications of the SU(2) algebras conventionally labelled left and right. They are not two independent compact real subalgebras of the real Lorentz algebra: $L,R$ are complex linear combinations of the real generators. The distinction is needed for noncompact boosts.

To verify the infinitesimal two-component action, putIts trace vanishes by antisymmetry, so $\det A=1+O({\omega }^2)$. The Pauli matrix multiplication law impliesSince $a^{†}=\frac{1}{4}{\omega }^{\mu \nu }{\bar{\sigma }}_{\nu }{\sigma }_{\mu }$, antisymmetrizing this identity yieldsConsequently $AX{A}^{†}=X+{\sigma }_{\mu }{\omega }^{\mu }{}_{\nu }{x}^{\nu }+O({\omega }^2)$, exactly the claimed infinitesimal coordinate transformation with ${\omega }^{\mu }{}_{\nu }={\omega }^{\mu \rho }{\eta }_{\rho \nu }$.

The two Weyl spinor representations transform asBoth maps preserve group multiplication. For $\Lambda =e^{\omega }$, choose its spin lift and write the finite matrices asFor rotations these coincide as the SU(2) doublet; for boosts their generator signs are opposite. Any complex-linear intertwiner would commute with all rotations and hence be scalar by the Schur lemma, but a nonzero scalar cannot intertwine the opposite boost matrices. They are therefore inequivalent. Infinitesimally their $K_i$ matrices are $-i{\sigma }_i/2$ and $+i{\sigma }_i/2$, while their $J_i$ matrices are both ${\sigma }_i/2$. Thus they have chiral labels $(1/2,0)$ and $(0,1/2)$. The finite matrices are representations of the spin cover; choosing $A$ or $-A$ matters for spinors even though their four-vector transformations coincide.

Using the supplied block gamma matrices, the generator of the Dirac spinor isThe antisymmetry of $\omega$ removes the symmetric Clifford part. Hence

There is a conjugation-order error in the last displayed gamma identity in the PDF. The Clifford algebra gives, with ${\gamma }^{\rho }={\eta }^{\rho \sigma }{\gamma }_{\sigma }$,Exponentiating this linear commutator action proves the inverse Lorentz action on gamma matricesThe second identity is the order required for the requested bilinears when $\psi \mapsto S\psi$. An explicit countercheck to the printed order is a positive boost along the third axis: ${\omega }^{03}=-\theta$ giveswhereas the printed right side has a plus sign. This cannot be repaired by dropping index raising; the same raising convention is needed in the preceding coordinate transformation.

The adjoint relation supplied in the question gives ${\Omega }^{†}{\gamma }^0=-{\gamma }^0\Omega$, hence the Dirac spinor pseudo-unitarity identity $S^{†}{\gamma }^{0}S={\gamma }^0$. The Dirac adjoint therefore transforms as $\bar{\psi }\mapsto \bar{\psi }{S}^{-1}$. It now follows that the Dirac scalar bilinear isthe vector isand the antisymmetric second-rank tensor isThese are respectively a Lorentz scalar, Lorentz four-vector and Lorentz tensor. The source's inconsistent gamma identity is replaced by its correct inverse/order pair; all three transformation laws then follow with the stated spinor transformation.

Use anti-Hermitian gauge potentials and the convention $D_i={\partial }_i+A_i$ for the gauge covariant derivative. Commuting these operators defines the gauge field strength:It is antisymmetric in its two spacetime indices, so $F_{xx}=F_{yy}=0$ and $F_{yx}=-F_{xy}$. In two dimensions there is therefore only one independent component, although that component is itself Lie-algebra valued.

Put $B_x=({\partial }_{x}g)g^{-1}$ and $B_y=({\partial }_{y}g)g^{-1}$. Differentiating $g^{-1}$ gives ${\partial }_i{g}^{-1}=-g^{-1}({\partial }_{i}g)g^{-1}$. Equality of mixed derivatives then gives the right Maurer-Cartan equationConsequently the scaled right Maurer-Cartan gauge potential has curvatureThus $\alpha =0$ and $\alpha =-1$ give zero curvature for every smooth $g$. At zero the potential is zero. At minus one it is a pure gauge potential: transforming the zero connection by $g$ gives $A=-dg{g}^{-1}$. If the gauge Lie algebra is abelian, the commutator vanishes for every $\alpha$. If it contains $X,Y$ with $[X,Y]\ne 0$, choose $g(x,y)=e^{xX}{e}^{yY}$; at the origin $B_x=X$ and $B_y=Y$. Thus in that case the two displayed values are the only choices flat for every $g$. If the potential is instead defined using $D=\partial -A$ and field components ${\partial }_x{A}_y-{\partial }_y{A}_x-[A_x,A_y]$, the same calculation reads $\alpha (1-\alpha )[B_x,B_y]$ and the nonzero pure-gauge value is $+1$. The sign convention must be specified.

For the specified SU(2) exponential, let $r=\sqrt{x^2+y^2}$ and, away from the origin, $n=(x/r,y/r,0)$. The Pauli matrix multiplication law gives $(n\cdot \sigma {)}^2=I$, so summing the exponential series yieldsThe continuous extension at the origin is $I$. For $r>0$, the second term is zero precisely when $\sin (r/2)=0$, and henceThese are infinitely many distinct circles.

At the origin, differentiating the exponential at zero gives $B_x=t_1$ and $B_y=t_2$. Since $[t_1,t_2]=t_3=-i{\sigma }_3/2$, the gauge field strength there isTo evaluate it on the circles, use polar coordinates. On $r=2\pi k$ the angular derivative of $g$ vanishes, while $B_r=-i(n\cdot \sigma )/2$. Therefore $B_x=(x/r)B_r$ and $B_y=(y/r)B_r$ commute, and $F_{xy}=0$ everywhere on every such circle, for every $\alpha$.

One can also see these circular curvature zeros for a planar SU2 exponential from a formula valid away from the origin. Write $n=(\cos \vartheta ,\sin \vartheta ,0)$, $m=(-\sin \vartheta ,\cos \vartheta ,0)$, and $t_n=-i(n\cdot \sigma )/2$, $t_m=-i(m\cdot \sigma )/2$. Direct differentiation givesand henceBoth coefficients vanish at the positive circle radii, and the expression tends to $t_3$ at the origin, agreeing with the direct calculation.