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Write $q_1={\bar{Q}}^{\dot{1}}$ and $q_2={\bar{Q}}^{\dot{2}}$. In the Jacobi-consistent convention just fixed,Applying the commutator product rule twice givesHere ${\sum }_i{\sigma }_i^2=3I$ produces the constant term; moving the $J_i$ to the right accounts for it. ConsequentlyThus the physically consistent numerical coefficients areIf one mechanically retains the literal positive-sign column transformation instead, the coefficient tuple would be $(\frac{1}{2},\frac{3}{4},1,-i,1,\frac{3}{4},1,i,-1)$. It cannot describe a unitary multiplet with the supplied rotation algebra, by the explicit Jacobi counterexample in part (a).

Take the spin-zero Clifford vacuum $|\Omega ⟩$, normalized to one and annihilated by the two annihilation charges. This is a lowest Fock state of a positive-mass representation, not a zero-energy vacuum annihilated by every charge. For the first one-charge state,The ladder commutators are $[J_{+},q_1]=-q_2$, $[J_{-},q_1]=0$, $[J_{+},q_2]=0$, $[J_{-},q_2]=-q_1$. Therefore these two states are the $j=1/2$ pair. A phase choice $|\frac{1}{2},+\frac{1}{2}⟩=-N{q}_2|\Omega ⟩$ makes the usual positive ladder coefficient hold.

Raising the dotted index only changes the oscillator basis by the unitary antisymmetric epsilon matrix, so $\{q_a,q_b^{†}\}=2m{\delta }_{ab}$. Since $q_a^{†}|\Omega ⟩=0$,The positive convention would formally label this state with $j_3=+1/2$ while $J_{+}$ acts nontrivially on it, contradicting the highest-weight ladder rule. This supplies an independent state-level check of the needed sign correction.

The two-charge state is a rotation singlet. Indeed, using $q_1^2=q_2^2=0$ and $q_2{q}_1=-q_1{q}_2$, the product rule gives $[J_i,q_1{q}_2]=0$ for every $i$: only the sum of the two diagonal spinor coefficients survives, and their trace is zero. Hence the requested commutators and quantum numbers areIts squared norm is $(2m{)}^2$ before multiplication by $N^4$, by the two oscillator anticommutators.

The spin-zero massive N=1 supermultiplet therefore consists of $|\Omega ⟩$, $N{q}_1|\Omega ⟩$, $N{q}_2|\Omega ⟩$ and $N^2{q}_1{q}_2|\Omega ⟩$. Their spins are $0,1/2,1/2,0$, all with the same mass $m$. Starting with an even vacuum, the two spin-zero states are bosonic and the two one-charge states fermionic. They are the two real scalar degrees of freedom and the two on-shell fermion polarizations of a massive chiral multiplet. Any product of three creation charges repeats an index and vanishes; annihilation charges reduce to these four states by the canonical anticommutation relations. There are exactly four independent states.

First use the actual PDF boost bracket $[K_i,K_j]=-i{ϵ}_{ijk}{J}_k$; replacing the final $J_k$ by $K_k$, as the TeX transcription does, would be a different and incorrect algebra. Expanding the three brackets yieldsThusThe two sets have SU(2)-type brackets. Their compact real form exponentiates to $SU(2)\times SU(2)$, whereas their full complexification exponentiates to $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$. This does not turn the physical real Lorentz group into a compact product. With Hermitian $J,K$, $A^{†}=B$, and the two factors are tied by the Lorentz reality condition; the connected physical spin cover is $SL(2,\mathbb{C})$.

For the right-handed spinor representation, the supplied sigma formula gives ${\bar{\sigma }}^{ij}={ϵ}_{ijk}{\sigma }_k/2$ and ${\bar{\sigma }}^{0i}=i{\sigma }_i/2$. ThereforeIt is the $(0,1/2)$ representation in this definition of $A,B$. These are finite-dimensional representation matrices; the boosts are not Hermitian for a positive-definite spinor inner product, as expected for a noncompact group. They are distinct from the operator-component commutator coefficients discussed in Question 1.

For the group map, encode a real Minkowski vector by the Hermitian matrixThe trace identity $\frac{1}{2}\mathrm{Tr}({\bar{\sigma }}^{\mu }{\sigma }_{\nu })={\delta }^{\mu }{}_{\nu }$ recovers its contravariant components. For $N\in SL(2,\mathbb{C})$, $X^{\prime }=NX{N}^{†}$ is Hermitian and has the same determinant. Expanding $X^{\prime }$ in that basis givesThe coefficients are real because the trace of a product of Hermitian matrices is real. Equality of determinants for all $x$ proves ${\Lambda }^T\eta \Lambda =\eta$. The action for $N_1{N}_2$ is the composition of the two actions, proving the homomorphism law.

Since $SL(2,\mathbb{C})$ is connected and the identity maps to the identity, its image has determinant $+1$ and preserves time orientation. More directly, positive-definite matrices representing future timelike vectors remain positive definite under $X\mapsto NX{N}^{†}$. The kernel is $\{I,-I\}$: if every Hermitian $X$ is fixed, $X=I$ first makes $N$ unitary, and then commutation with all Hermitian matrices makes it a scalar with determinant one.

Conversely, $SU(2)$ gives every spatial rotation, and positive Hermitian determinant-one matrices $\exp (\mathbf{b}\cdot \mathbf{\sigma }/2)$ give every pure boost. Rotation-boost decomposition yields every proper orthochronous Lorentz transformation. Thus the formula gives the Lorentz spinor double coverIt is a map into $SO(1,3)$ as asked, with precisely its identity component as image; it does not produce disconnected time-reversing transformations.

Set $\hslash =1$ and let $N=2I\in {\mathbb{Z}}_{\ge 0}$. A concrete irreducible spin representation is the symmetric powerStart with $N$ copies of the defining SU(2) doublet and restrict their tensor product representation to the completely symmetric subspace. Its orthonormal basis consists of symmetric states with $I+m$ up components and $I-m$ down components, where $m=I,I-1,\dots ,-I$. There are $N+1=2I+1$ such states. The total generators are $J_i={\sum }_{a=1}^N{\sigma }_i^{(a)}/2$, acting on this subspace, where ${\sigma }_i$ are the Pauli matrices. This also constructs the trivial representation when $I=0$.

The spin ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfyThe Casimir operator $J^2={\sum }_i{J}_i^2$ commutes with every generator. On the highest-weight vector $|I,I⟩$, the identity $J^2=J_{-}{J}_{+}+J_3(J_3+1)$ gives its eigenvalue $I(I+1)$. With phases chosen to make the lowering coefficients positive,The squared norm of the lowered state follows from $J_{+}{J}_{-}=J^2-J_3(J_3-1)$. The ladder stops exactly at $m=-I$. Every weight is connected to every other by these operators, and $J_3$ has distinct eigenvalues. Hence any invariant subspace contains a weight vector and then the whole ladder: the representation is irreducible.

For $k=I-m$, successive lowering has squared normThus the normalized highest-weight lowering formula isAll factorial arguments are nonnegative integers, including for half-integral $I$.

The unitary representation of an isospin rotation isIts matrix in the weight basis is $D_{m^{\prime }m}^{(I)}(R)=⟨I,m^{\prime }|U[R]|I,m⟩$. Insert the completeness relation between two operators to obtainThese verify the group representation and unitarity properties, and the ladder argument gives irreducibility. Here rotations carry their SU(2) lifts: $U(2\pi )=(-1{)}^{2I}I$. Integer $I$ descends to the ordinary SO(3) group; half-integral $I$ is a representation of its double cover, not a single-valued representation of $SO(3)$.

For $I=1$, order the basis as $(|1,1⟩,|1,0⟩,|1,-1⟩)$. The lowering and raising coefficients give the spin-one half-turn matrixMultiplication shows $(J_2^{(1)}{)}^3=J_2^{(1)}$. Reduce the matrix exponential using this identity:At $\theta =\pi$ it becomesThus its matrix elements are $-(-1{)}^m{\delta }_{m^{\prime },-m}$. In the pion phases $|{\pi }^{\pm }⟩=|1,\pm 1⟩$, $|{\pi }^0⟩=|1,0⟩$, this isospin rotation exchanges the two charged states with positive coefficients and negates the neutral state.

Charge conjugation is linear and unitary here. Similarity preserves commutators, soConsequently the conjugated generators obey the same Lie algebra. The specified signs also give $\mathcal{C}{J}_{\pm }{\mathcal{C}}^{-1}=-J_{\mp }$. Since $|{\pi }^{\pm }⟩=J_{\pm }|{\pi }^0⟩/\sqrt{2}$ and the neutral state has positive charge conjugation eigenvalue,The signs depend on the chosen charged-state phases, which have been fixed by the ladder convention.

Conjugation by $R_{\pi }=e^{-i\pi J_2}$ changes $(J_1,J_2,J_3)$ to $(-J_1,J_2,-J_3)$, exactly as charge conjugation does. Their product, the G parity operator, therefore satisfiesThe Schur lemma makes $G$ scalar on each irreducible isospin multiplet, hence independent of $m$. On the neutral pion, $G|{\pi }^0⟩=-\mathcal{C}|{\pi }^0⟩=-|{\pi }^0⟩$. Therefore the pion G-parity is

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.

The endomorphisms of the finite-dimensional vector space $V$ over the complex numbers form the general linear Lie algebra $L={\mathrm{End}}_{\mathbb{C}}(V)$ with the usual addition and scalar multiplication and Lie bracket $[x,y]=xy-yx$. This commutator is bilinear and antisymmetric, and expanding the six products verifies the Jacobi identity.

For a Lie subalgebra $\mathfrak{g}=L_1$, being an abelian Lie algebra means $[\mathfrak{g},\mathfrak{g}]=0$. A nilpotent Lie algebra has ${\gamma }_1=\mathfrak{g}$, ${\gamma }_{j+1}=[\mathfrak{g},{\gamma }_j]$ eventually zero; a solvable Lie algebra has ${\mathfrak{g}}^{(0)}=\mathfrak{g}$, ${\mathfrak{g}}^{(j+1)}=[{\mathfrak{g}}^{(j)},{\mathfrak{g}}^{(j)}]$ eventually zero. These are the lower central series of a Lie algebra and derived series of a Lie algebra, respectively. Nilpotence is a condition on the Lie bracket, and does not require every member to be a nilpotent endomorphism: a nonzero scalar multiple of the identity spans an abelian Lie algebra.

A flag of a vector space is an increasing chain of vector subspaces. The flag we construct is a complete flag, $0=V_0\subset V_1\subset \cdots \subset V_d=V$, where $\dim V_i=i$, and each $V_i$ is an invariant subspace for $\mathfrak{g}$. We first prove the common-eigenvector assertion in the Lie theorem, by induction on $\dim \mathfrak{g}$; the zero algebra is immediate. For nonzero solvable $\mathfrak{g}$, its derived algebra is proper, so there is a codimension-one ideal of a Lie algebra $\mathfrak{k}$ containing $[\mathfrak{g},\mathfrak{g}]$. Write $\mathfrak{g}=\mathfrak{k}\oplus \mathbb{C}x$. By induction there are $v\ne 0$ and a linear functional $\lambda$ on $\mathfrak{k}$ such that $kv=\lambda (k)v$ for every $k\in \mathfrak{k}$.

Let $W$ be the cyclic subspace spanned by $v,xv,x^{2}v,\dots$. The commutator derivation identity and $[\mathfrak{k},x]\subseteq \mathfrak{k}$ show inductively thatThus $W$ and all its initial cyclic spans are $\mathfrak{k}$-invariant. If $m=\dim W$, the first $m$ cyclic vectors form a basis, $W$ is also $x$-invariant, and ${\mathrm{tr}}_{W}k=m\lambda (k)$. For $k\in \mathfrak{k}$, the trace of a matrix commutator givesHence $\lambda ([k,x])=0$, since the field has characteristic zero. The nonzero common weight spaceis $x$-invariant: $k(xw)=x(kw)+[k,x]w=\lambda (k)xw$. The restriction of $x$ to $V_{\lambda }$ has an eigenvector, because $\mathbb{C}$ is an algebraically closed field. This is a common eigenvector for $\mathfrak{g}$. Its line is invariant, and repeating the argument on the quotient vector space gives the complete invariant flag. Equivalently, this proves simultaneous triangularization of a Lie algebra representation.

In a basis adapted to this complete flag, every member of $\mathfrak{g}$ is upper triangular, so every member of its derived algebra is strictly upper triangular. Products of $d$ strictly upper triangular matrices vanish, and each iterated Lie bracket of $d$ such matrices is a sum of these products. Consequently the derived algebra is a nilpotent Lie algebra. We may therefore take $L_2=[L_1,L_1]$: it is an ideal, and the quotient Lie algebra $L_1/L_2$ is abelian. This also covers $\dim V=0$.

The SU(2) group consists of complex $2\times 2$ matrices $U$ with $U^{†}U=I$ and $\det U=1$. The SO(3) group consists of real $3\times 3$ matrices $R$ with $R^{T}R=I$ and $\det R=1$. Differentiate these equations along a path through the identity. This givesConversely the matrix exponential of each displayed infinitesimal matrix satisfies the corresponding group equations, so these are exactly the Lie algebras, with bracket the matrix commutator.

Using the Pauli matrices, put $t_a=-i{\sigma }_a/2$. They form a real basis of the SU(2) Lie algebra. The Pauli matrix commutator identity gives $[t_a,t_b]={ϵ}_{abc}{t}_c$. Define $J_a$ on ${\mathbb{R}}^3$ by $J_{a}v=e_a\times v$. These are a basis of the SO(3) Lie algebra; the vector triple-product identity gives $[J_a,J_b]={ϵ}_{abc}{J}_c$. Thusis a real linear bijection preserving the bracket. This is the SU(2)-SO(3) Lie algebra isomorphism. It is a Lie-algebra isomorphism, not a group isomorphism: the Adjoint double cover from SU(2) to SO(3) has kernel $\{I,-I\}$.

The SU(3) group is defined similarly by $U^{†}U=I$ and $\det U=1$ on complex $3\times 3$ matrices. One SU(2) subgroup is $\{\mathrm{diag}(V,1):V\in SU(2)\}$. The real orthogonal matrices with determinant one form an SO(3) group subgroup, since real orthogonality is also complex unitarity.

Under the defining group action, the group orbit of $e_1$ is exactly the unit sphere in ${\mathbb{C}}^3$:Unitarity proves containment. Conversely, extend a unit vector $z$ to an orthonormal basis and use it as the first column of a unitary matrix. Multiplying the last column by the inverse of its determinant makes that determinant one without changing the first column. The isotropy group of $e_1$ must also preserve its orthogonal complement, and therefore isThis proves the unit sphere orbit of the defining special unitary action, including $S^5\cong SU(3)/SU(2)$.

For the complex quadric, write $z=x+iy$. Its defining relation separates intoThe Hermitian norm is then $z^{†}z=1+2|y{|}^2$, which is not constant on this set. For example, $e_1$ and $\sqrt{2}{e}_1+i{e}_2$ both satisfy the complex bilinear relation but have Hermitian norms one and three. Since the SU(3) group preserves that norm, the quadric cannot be a single SU(3) group orbit. In fact it is not even invariant under the whole group: $\mathrm{diag}(e^{it},e^{-it},1)e_1$ fails the bilinear relation when $e^{2it}\ne 1$.

The real SO(3) group does preserve the quadric, acting simultaneously on $x$ and $y$. Its group orbits are classified completely by $r=|y|\ge 0$. For $r=0$, $y=0$ and $x$ is a real unit vector, giving the group orbit $S^2$ and stabilizer subgroup $SO(2)$. For $r>0$, the vectors $x/\sqrt{1+r^2}$ and $y/r$ are orthonormal. Adjoining their cross product gives an oriented orthonormal frame. There is a unique rotation taking the standard frame to this one; hence the action is transitive at fixed $r$ and the stabilizer subgroup is trivial. Thus the real rotation orbits on a complex unit quadric areIn particular the real subgroup does not act transitively on the whole quadric. The parametrization $x=\sqrt{1+|y{|}^2}n$ with $n\in S^2$ and $y\perp n$ also identifies the quadric, as a real manifold, with the tangent bundle of $S^2$.

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.

A weight of a representation of $SU(3)$ is a joint eigenvalue of two commuting generators spanning a Cartan subalgebra. Equivalently it describes the character by which the diagonal maximal torus acts on a weight vector. Use the Hermitian generatorsTheir multiples by $i$ are in the SU(3) Lie algebra. A vector with weight $(a,b)$ acquires the phase $e^{i(a\theta +b\phi )}$ under $\exp (i\theta I_3+i\phi Y)$. The conventional generator $H_8$ is related by $Y=2H_8/\sqrt{3}$; using $H_8$ instead just rescales the vertical weight coordinate.

In the defining group representation, the coordinate vectors $e_u,e_d,e_s$ are joint eigenvectors. The weight diagram of the defining SU(3) representation therefore hasComplex conjugation reverses all torus phases, so the weights of $\overline{3}$ are $-{\mu }_u,-{\mu }_d,-{\mu }_s$, each with multiplicity one. The conjugation bar is essential: the tensor product here is $3\otimes \overline{3}$.

Weights add in a tensor product of group representations. Thus the nine weights of this product are ${\mu }_i-{\mu }_j$. There are three zero weights from $i=j$, and the remaining six areTo turn this weight calculation into a direct-sum decomposition, identify the tensor product with $\mathrm{End}({\mathbb{C}}^3)$ by $v\otimes \overline{w}\mapsto v{w}^{†}$. The group acts by $M\mapsto UM{U}^{-1}$. This gives two invariant spaces,The scalar line has weight zero and is the trivial group representation $1$. The traceless space has the six nonzero weights just computed, represented by the off-diagonal matrix units $E_{ij}$, and two zero weights, represented by traceless diagonal matrices. Its dimension is eight and it is the complex Adjoint representation $8$. ThereforeThis proves the octet and singlet in a fundamental SU3 tensor product with the correct central weight multiplicities.