Representations of the Lorentz Group

Derivation

Representations of the Lorentz Group

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Statement

The complexified Lie algebra of the proper orthochronous Lorentz group decomposes as \(\mathfrak{so}(1,3)_{\mathbb C}\cong\mathfrak{su}(2)\oplus\mathfrak{su}(2)\). Consequently every finite-dimensional irreducible representation is labelled by an ordered pair of half-integers \((j_1,j_2)\), has dimension \((2j_1+1)(2j_2+1)\), and carries rotational spin content \(j=|j_1-j_2|,\,|j_1-j_2|+1,\dots,j_1+j_2\). This single classification delivers the scalar \((0,0)\), the two Weyl spinors \((\tfrac12,0)\) and \((0,\tfrac12)\), the Dirac bispinor \((\tfrac12,0)\oplus(0,\tfrac12)\), and the four-vector \((\tfrac12,\tfrac12)\) as the elementary carriers of Lorentz-covariant field theory.

Why it matters

Special relativity demands that a physical field transform in a definite way under Lorentz transformations, but it does not tell you which fields are allowed. The representation theory of the Lorentz algebra answers exactly that question: it enumerates, once and for all, the possible spins and index structures a local field may carry. Every Lagrangian in the Standard Model is built from these blocks — scalars for the Higgs, Weyl and Dirac spinors for the fermions, four-vectors for the gauge bosons.

The decomposition into two commuting \(\mathfrak{su}(2)\) factors also explains why spinors come in two distinct handednesses and why parity exchanges them. It is the algebraic origin of chirality, and hence of the entire structure of the electroweak interaction.

Assumptions

The Lorentz generators satisfy the standard \(\mathfrak{so}(1,3)\) commutation relations.If the bracket structure were different we would be classifying representations of a different group; the whole spin classification would change. We restrict to the proper orthochronous component \(SO^+(1,3)\) connected to the identity.Discrete parity \(P\) and time reversal \(T\) are handled separately; dropping this lets \(P\) mix \((j_1,j_2)\) with \((j_2,j_1)\), forcing reducible reps such as the Dirac spinor. We work with the complexified algebra \(\mathfrak{so}(1,3)_{\mathbb C}\) when forming the combinations \(A_i,B_i\).The real form \(\mathfrak{so}(1,3)\) is not the direct sum of two real \(\mathfrak{su}(2)\)'s (boosts are non-compact); the isomorphism holds only after complexification, and \(A_i^\dagger=B_i\) rather than \(A_i^\dagger=A_i\). Representations are finite-dimensional.Unitary representations of the non-compact Lorentz group are infinite-dimensional; the finite-dimensional \((j_1,j_2)\) reps used for fields are non-unitary, which is physically acceptable because unitarity is imposed on the Hilbert space of states, not on the field components.

Derivation

\[ [M^{\mu\nu},M^{\rho\sigma}] = i\left(\eta^{\nu\rho}M^{\mu\sigma}-\eta^{\mu\rho}M^{\nu\sigma}-\eta^{\nu\sigma}M^{\mu\rho}+\eta^{\mu\sigma}M^{\nu\rho}\right) \]

The six antisymmetric generators \(M^{\mu\nu}=-M^{\nu\mu}\) of \(SO^+(1,3)\) obey this algebra, fixed by the requirement that they preserve the Minkowski metric \(\eta=\mathrm{diag}(+,-,-,-)\). B

\[ J_i \equiv \tfrac12\,\epsilon_{ijk}M^{jk},\qquad K_i \equiv M^{0i} \]

Repackage the six generators into three rotations \(J_i\) (spatial-spatial) and three boosts \(K_i\) (time-space). This is a change of basis, legal because it is invertible. A

\[ [J_i,J_j]=i\epsilon_{ijk}J_k,\quad [J_i,K_j]=i\epsilon_{ijk}K_k,\quad [K_i,K_j]=-i\epsilon_{ijk}J_k \]

Substitute the definitions of Step 2 into the master bracket of Step 1 and contract the metric factors. The first relation shows \(J_i\) generate \(\mathfrak{su}(2)\) (rotations); the second shows \(K_i\) is a vector under rotations; the crucial minus sign in the third reflects the non-compact, hyperbolic nature of boosts. C

\[ A_i \equiv \tfrac12\left(J_i + iK_i\right),\qquad B_i \equiv \tfrac12\left(J_i - iK_i\right) \]

Form complex linear combinations. This step requires complexifying the algebra: \(A_i,B_i\) are genuine elements only of \(\mathfrak{so}(1,3)_{\mathbb C}\). Note \(J_i=A_i+B_i\) and \(K_i=-i(A_i-B_i)\), so the map is invertible. B

\[ [A_i,A_j]=i\epsilon_{ijk}A_k,\qquad [B_i,B_j]=i\epsilon_{ijk}B_k,\qquad [A_i,B_j]=0 \]

Expand each bracket bilinearly using Step 3. For example \([A_i,A_j]=\tfrac14\big([J_i,J_j]+i[J_i,K_j]+i[K_i,J_j]-[K_i,K_j]\big)=\tfrac14\big(i\epsilon_{ijk}J_k+i\cdot i\epsilon_{ijk}K_k+i\cdot i\epsilon_{ijk}K_k+i\epsilon_{ijk}J_k\big)=i\epsilon_{ijk}A_k\). The cross-term \([A_i,B_j]\) cancels identically. C

\[ \mathfrak{so}(1,3)_{\mathbb C}\;\cong\;\mathfrak{su}(2)_A\,\oplus\,\mathfrak{su}(2)_B \]

Step 5 exhibits two mutually commuting copies of the \(\mathfrak{su}(2)\) algebra. Since finite-dimensional irreps of a direct sum are tensor products of irreps of each factor, each factor is labelled by a Casimir eigenvalue \(j_1(j_1+1)\) and \(j_2(j_2+1)\) with \(j_1,j_2\in\{0,\tfrac12,1,\dots\}\). C

\[ \dim(j_1,j_2)=(2j_1+1)(2j_2+1),\qquad \vec J=\vec A+\vec B \]

The physical rotation generator is the sum of the two \(\mathfrak{su}(2)\) angular momenta, so ordinary spin follows the Clebsch–Gordan rule \(j=|j_1-j_2|,\dots,j_1+j_2\). Reading off the spin content assigns each \((j_1,j_2)\) to a familiar field. B

Result

\[ \boxed{\;\mathfrak{so}(1,3)_{\mathbb C}\cong\mathfrak{su}(2)\oplus\mathfrak{su}(2)\;\Longrightarrow\; \text{irreps }(j_1,j_2),\ \dim=(2j_1+1)(2j_2+1)\;}\]

\[ \begin{array}{lll} (0,0) & \text{spin }0 & \text{scalar }\phi\\ (\tfrac12,0) & \text{spin }\tfrac12 & \text{left Weyl }\psi_L\\ (0,\tfrac12) & \text{spin }\tfrac12 & \text{right Weyl }\psi_R\\ (\tfrac12,0)\oplus(0,\tfrac12) & \text{spin }\tfrac12 & \text{Dirac }\Psi\\ (\tfrac12,\tfrac12) & \text{spin }0\oplus 1 & \text{four-vector }A^\mu \end{array} \]

Reading. The Lorentz group is, up to complexification, two independent rotation groups. A field is fixed by how it spins under each: \((0,0)\) does not spin at all (scalar); \((\tfrac12,0)\) and \((0,\tfrac12)\) spin under one factor only (the two chiralities of a spin-\(\tfrac12\) particle); \((\tfrac12,\tfrac12)\) spins under both and reassembles into a spin-0 (time component) plus spin-1 (space components) four-vector. Parity swaps the two labels, which is why a parity-invariant fermion must combine both Weyl pieces into a Dirac field.

Units check. The classification is purely algebraic and dimensionless: \(j_1,j_2\) are half-integers, dimensions are pure counts. The generators \(J_i,K_i\) are dimensionless (angular momentum measured in units of \(\hbar\); rapidity is dimensionless), and every commutator \([\,\cdot\,,\cdot\,]\) returns an object of the same kind, so both sides of each bracket carry identical (null) dimension. Mass dimension enters only later, through the fields themselves in a Lagrangian, not through this group-theoretic labelling.

Limiting cases

Set \(j_2=0\): the \(B\)-factor is trivial and the rep reduces to a single \(\mathfrak{su}(2)_A\) rep of spin \(j_1\) — a purely left-handed object.
Set \(j_1=j_2=j\): the diagonal \((j,j)\) contains a scalar (spin 0) at the bottom of its Clebsch–Gordan tower; \((\tfrac12,\tfrac12)\) is the vector, \((1,1)\) contains the traceless symmetric rank-2 tensor.
Take the non-relativistic limit \(K_i\to 0\): boosts decouple, \(A_i\) and \(B_i\) collapse onto the single physical rotation \(J_i\), and \((j_1,j_2)\) reduces to ordinary \(SU(2)\) spin \(j=j_1+j_2\) content.
Antisymmetric tensor \(F^{\mu\nu}\): the six components form \((1,0)\oplus(0,1)\), the self-dual and anti-self-dual parts, each spin 1 — the algebraic home of the electromagnetic field strength.

Breaks when

Unitary field representations are demanded. The finite-dimensional \((j_1,j_2)\) reps are non-unitary because boosts are represented by non-Hermitian matrices (\(K_i\) anti-Hermitian in these reps). For unitary representations of the full Poincaré group one must use Wigner's infinite-dimensional little-group construction instead; the \((j_1,j_2)\) labels then describe field components, not physical states.
Parity, charge conjugation, or time reversal are included. The decomposition classifies only the connected component \(SO^+(1,3)\). Discrete symmetries relate distinct \((j_1,j_2)\): parity maps \((j_1,j_2)\to(j_2,j_1)\), so a parity-covariant theory of massive fermions cannot use a single Weyl rep and is forced to the reducible Dirac sum.
The spacetime dimension is not four. The accidental isomorphism \(\mathfrak{so}(1,3)_{\mathbb C}\cong\mathfrak{su}(2)\oplus\mathfrak{su}(2)\) is special to four dimensions. In \(D\ne4\) the spin group is different (e.g. \(\mathrm{Spin}(1,2)\), \(\mathrm{Spin}(1,4)\)) and the two-label classification does not hold.
Spacetime is curved or the algebra is deformed. On a generic curved manifold there is no global Lorentz symmetry; the classification survives only in the tangent-space (vierbein) frame. A deformed / quantum-group symmetry likewise invalidates the ordinary Lie-algebra representation theory.

Failure modes

Treating \(A_i,B_i\) as Hermitian. Because \(K_i\) is non-Hermitian in finite-dimensional reps, \(A_i^\dagger=B_i\), not \(A_i^\dagger=A_i\). Forgetting this leads to the false conclusion that the reps are unitary.
Claiming the real isomorphism \(\mathfrak{so}(1,3)=\mathfrak{su}(2)\oplus\mathfrak{su}(2)\). The equality holds only after complexification; the real Lorentz algebra is \(\mathfrak{sl}(2,\mathbb C)\) (six real dimensions), whose maximal compact subalgebra is a single \(\mathfrak{su}(2)\).
Adding the labels the wrong way. The spin content is the Clebsch–Gordan tensor product \(j_1\otimes j_2\), giving \(|j_1-j_2|\le j\le j_1+j_2\), not the single value \(j_1+j_2\). Students frequently write "\((\tfrac12,\tfrac12)\Rightarrow\) spin 1" and lose the spin-0 (time) component.
Confusing \((\tfrac12,0)\oplus(0,\tfrac12)\) with \((\tfrac12,\tfrac12)\). The direct sum is the four-component Dirac spinor (two chiralities); the tensor-labelled \((\tfrac12,\tfrac12)\) is the four-vector. Both are four-dimensional but transform completely differently.
Assigning a definite chirality to a massive Dirac field. A Dirac mass term \(m\bar\Psi\Psi=m(\bar\psi_L\psi_R+\bar\psi_R\psi_L)\) couples the two Weyl pieces, so chirality is not conserved; only for \(m=0\) do \(\psi_L,\psi_R\) decouple.

Discussion

The heart of the result is an accidental Lie-algebra isomorphism unique to four spacetime dimensions. Because \(\mathfrak{so}(1,3)_{\mathbb C}\) splits into two commuting angular-momentum algebras, a Lorentz field is completely characterised by a pair of "spins", one for each factor. The physical rotation subgroup sits on the diagonal, \(\vec J=\vec A+\vec B\), so ordinary spin is a Clebsch–Gordan sum of the two. This is why the vector representation \((\tfrac12,\tfrac12)\) contains both a spin-0 and a spin-1 piece — the timelike and spacelike parts of a four-vector — and why the antisymmetric field strength \(F_{\mu\nu}\) splits into self-dual \((1,0)\) and anti-self-dual \((0,1)\) halves that Maxwell's equations treat symmetrically.

The two \(\mathfrak{su}(2)\) factors are exchanged by parity. This is the deep reason chirality exists: a left-handed Weyl field \((\tfrac12,0)\) and a right-handed one \((0,\tfrac12)\) are inequivalent representations that no proper Lorentz transformation can interconvert. Nature exploits this in the weak interaction, which couples only to left-handed fermions — a fact impossible to state without the two-factor decomposition. A Dirac fermion is the parity-symmetric completion \((\tfrac12,0)\oplus(0,\tfrac12)\), and its mass term is precisely the invariant that pairs the two halves.

At the group level the covering group of \(SO^+(1,3)\) is \(SL(2,\mathbb C)\), the double cover, whose two inequivalent two-dimensional representations — the fundamental \(2\) acting on \(\psi_L\) and its complex conjugate \(\bar 2\) acting on \(\psi_R\) — are exactly \((\tfrac12,0)\) and \((0,\tfrac12)\). Spinor indices are then dotted or undotted according to which factor acts, and the invariant \(\epsilon_{\alpha\beta}\) tensors of each \(SL(2,\mathbb C)\) factor are the machinery for building Lorentz scalars from spinor bilinears. The non-unitarity of the finite-dimensional reps is not a defect: field components need only transform covariantly, while unitarity is a statement about the Hilbert-space inner product secured separately by Wigner's classification of Poincaré reps by mass and little-group spin.

Common misconceptions. The isomorphism is often quoted without the essential word "complexified"; the real Lorentz algebra is \(\mathfrak{sl}(2,\mathbb C)\), a simple algebra, and does not split into two real \(\mathfrak{su}(2)\)'s. Relatedly, the two \(\mathfrak{su}(2)\) factors are not spin and isospin, nor two independent physical angular momenta — only their diagonal sum is observable as spin. Finally, the labels \((j_1,j_2)\) are tensor factors, so the dimension multiplies but the spin content adds via Clebsch–Gordan.

Worked examples

Dimension and spin content of the rep \((1,\tfrac12)\).

Apply \(\dim=(2j_1+1)(2j_2+1)\) and the Clebsch–Gordan rule \(j=|j_1-j_2|,\dots,j_1+j_2\).

\[ \dim=(2\cdot 1+1)(2\cdot\tfrac12+1)=3\times 2=6 \]

Multiply the two factor dimensions.

\[ j=\left|1-\tfrac12\right|,\dots,1+\tfrac12 \;=\;\tfrac12,\ \tfrac32 \]

Tensor the two angular momenta: \(1\otimes\tfrac12=\tfrac12\oplus\tfrac32\). Check the count: \(2+4=6\), matching the dimension.

\[ (1,\tfrac12):\quad \dim=6,\quad \text{spin }\tfrac12\oplus\tfrac32 \]

Reading. This six-component object is the Lorentz content of the Rarita–Schwinger vector-spinor \(\psi_\mu\) building block (before imposing constraints), which is used to describe spin-\(\tfrac32\) fields. Units check. Dimensionless counts; \(2+4=6\) confirms consistency.

Verify \([A_1,A_2]=iA_3\) explicitly from the boost/rotation brackets, with numbers.

Use \(A_i=\tfrac12(J_i+iK_i)\) and the Step-3 relations \([J_1,J_2]=iJ_3\), \([J_1,K_2]=iK_3\), \([K_1,J_2]=iK_3\), \([K_1,K_2]=-iJ_3\).

\[ [A_1,A_2]=\tfrac14\big([J_1,J_2]+i[J_1,K_2]+i[K_1,J_2]-[K_1,K_2]\big) \]

Expand bilinearly; the \(i^2=-1\) turns the \([K_1,K_2]\) term's sign.

\[ =\tfrac14\big(iJ_3 + i(iK_3) + i(iK_3) - (-iJ_3)\big)=\tfrac14\big(iJ_3 - K_3 - K_3 + iJ_3\big) \]

Substitute the four numerical brackets from Step 1 and simplify each \(i\cdot i=-1\).

\[ =\tfrac14\big(2iJ_3 - 2K_3\big)=\tfrac{i}{2}\big(J_3+iK_3\big)\cdot 1 = i\cdot\tfrac12(J_3+iK_3)=iA_3 \]

Group real and imaginary parts: \(2iJ_3-2K_3=2i(J_3+iK_3)\), then divide by 4.

\[ [A_1,A_2]=iA_3 \]

Reading. The \(A\)-generators close on themselves into a clean \(\mathfrak{su}(2)\), with the boost contributions reassembling exactly into \(A_3\). The identical computation with \(-i\) gives \([B_1,B_2]=iB_3\), and the mixed bracket \([A_1,B_2]=\tfrac14(iJ_3+iK_3-iK_3-iJ_3)=0\) confirms the two factors commute. Units check. Every term is a dimensionless generator; both sides are single generators, dimensionally identical.

Problems

Compute the dimension and spin content of \((1,1)\). Identify which familiar tensor sits at the top spin.
Solution
\(\dim=(2\cdot1+1)^2=9\). Spin content \(1\otimes1=0\oplus1\oplus2\), with dimensions \(1+3+5=9\) ✓. The top spin-2 piece is the traceless symmetric rank-2 tensor \(h_{\mu\nu}\) (linearised graviton); the spin-0 is its trace and spin-1 the antisymmetric remainder within the reducible tensor product of two vectors.
Show that the antisymmetric tensor representation carried by \(F_{\mu\nu}\) is \((1,0)\oplus(0,1)\), and give the dimension count.
Solution
\(F_{\mu\nu}\) is antisymmetric with \(\binom{4}{2}=6\) independent components. Under \(SO^+(1,3)\) it decomposes into self-dual \(F^+_{\mu\nu}=\tfrac12(F_{\mu\nu}+\tfrac{i}{2}\epsilon_{\mu\nu\rho\sigma}F^{\rho\sigma})\) and anti-self-dual \(F^-\) parts, each with 3 complex components. These are \((1,0)\) and \((0,1)\): \(\dim=3+3=6\) ✓. Both are spin 1, as expected for the electromagnetic field strength.
A field transforms in \((\tfrac12,\tfrac12)\). Show by counting that it has the right number of components to be a four-vector, and give its spin decomposition.
Solution
\(\dim=(2\cdot\tfrac12+1)^2=2\times2=4\), matching the four components of \(A^\mu\). Spin content \(\tfrac12\otimes\tfrac12=0\oplus1\): the spin-0 part is the time component \(A^0\) (a rotational scalar) and the spin-1 part is the spatial vector \(\vec A\). Dimensions \(1+3=4\) ✓.
Using \(J_i=A_i+B_i\), determine the eigenvalue of the rotational Casimir \(\vec J^2\) that can appear in the rep \((\tfrac12,0)\), and confirm it is a pure spin-\(\tfrac12\) object.
Solution
For \((\tfrac12,0)\), the \(B\)-factor is trivial (\(j_2=0\), \(\vec B=0\)), so \(\vec J=\vec A\) and \(\vec J^2=\vec A^2=j_1(j_1+1)=\tfrac12\cdot\tfrac32=\tfrac34\). The only allowed spin is \(j=|{\tfrac12}-0|=\tfrac12\). Hence \((\tfrac12,0)\) is a single spin-\(\tfrac12\) (left-handed Weyl) field, dimension \(2\).
Explain, using the parity map \((j_1,j_2)\to(j_2,j_1)\), why a massive spin-\(\tfrac12\) fermion in a parity-invariant theory must be described by four (not two) components, and write the corresponding rep.
Solution
A single Weyl field \((\tfrac12,0)\) is mapped by parity to \((0,\tfrac12)\), a different (inequivalent) representation. A parity-invariant theory must contain both, so the smallest allowed object is the reducible direct sum \((\tfrac12,0)\oplus(0,\tfrac12)\), of dimension \(2+2=4\) — the Dirac bispinor \(\Psi=\begin{pmatrix}\psi_L\\ \psi_R\end{pmatrix}\). The Dirac mass term \(m\bar\Psi\Psi=m(\bar\psi_L\psi_R+\text{h.c.})\) couples the two halves and is the unique Lorentz- and parity-invariant mass, requiring both chiralities to be present.