Past exam of the groups and representations course of the University of Oxford / 2018 / 3

Differentiate the defining relation at $\Lambda =I+tX+O(t^2)$. The linear term gives the Lie algebraSolving this equation between matrices yieldswhere a convenient basis isThe first two generate Lorentz boosts and the last a spatial rotation. Direct commutators giveThere are three independent real coefficients, so ${\dim }_{\mathbb{R}}\mathfrak{g}=3$. A Cartan subalgebra is $\mathbb{R}J$, and the rank of a semisimple Lie algebra is one. Explicitly, if $X=a{K}_1+b{K}_2+cJ$ and $Y=a^{\prime }{K}_1+b^{\prime }{K}_2+c^{\prime }J$, their commutator vanishes exactly whenFor nonzero $X$, these equations force $Y$ to be proportional to $X$, so there is no two-dimensional commuting subalgebra. The split real rank is also one, realized by the semisimple Lorentz boost generator $K_1$. ThusFor this Lorentz algebra in three spacetime dimensions, disconnected components of the full Lorentz group do not change its tangent Lie algebra at the identity.

In the complexification of a Lie algebra, defineThe relations in a becomeThe special unitary Lie algebra $\mathfrak{su}(2)$ has real basis $S_i=-i{\sigma }_i/2$, where the Pauli matrices satisfy $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$. Its commutators are the same: $[S_i,S_j]={ϵ}_{ijk}{S}_k$. The complex-linear map $L_i\mapsto S_i$ therefore preserves the bracket and maps a basis to a basis. ConsequentlyThe real Lie algebras are not isomorphic: multiplying the boosts by $i$ changes their real form. For example, the Adjoint representation of a Lie algebra evaluated at $K_1$ has real eigenvalues $0,1,-1$, whereas the corresponding nonzero compact SU(2) generators have purely imaginary adjoint eigenvalues.

Every complex Lie algebra representation of the real algebra extends uniquely by complex linearity to its complexification of a Lie algebra. The finite-dimensional complex irreducible representations are therefore classified by a nonnegative integer highest weight $m$, just as for SU(2). There is one equivalence class for eachThese are statements about Lie algebra representations. They do not assert positive-definite unitarity for the noncompact real form, or automatic descent to the full Lorentz group. In particular, half-integer $j$ gives a sign under a $2\pi$ spatial rotation, so its integration needs a covering group.

Choose the following Pauli matrix realization, with its spinor basis fixed for the later Dirac adjoint:The Pauli matrix multiplication rule verifiesExtending linearly therefore gives a two-dimensional Lie algebra representation. For an irreducible representation, it suffices to rule out an invariant line. Any such line would have to be an eigenline of $r(K_1)$, whose two eigenvalues are distinct. Those two coordinate lines are interchanged by $r(K_2)$, so neither is invariant. There is no proper nonzero invariant subspace.

This is a spinorial representation rather than a single-valued two-dimensional representation of the identity component of the specified orthogonal group:The mismatch explicitly exhibits the need for the spin double cover. It causes no obstruction to the requested infinitesimal construction.

The Clifford algebra condition for signature $(-,+,+)$ isEach Pauli matrix squares to $I_2$, and distinct Pauli matrices anticommute. Therefore the given choice obeysThese check all diagonal and off-diagonal anticommutators, provingThus they are viable three-dimensional gamma matrices. Their Hermitian adjoints satisfy ${\gamma }_0^{†}=-{\gamma }_0$ and ${\gamma }_a^{†}={\gamma }_a$ for $a=1,2$. Two components are minimal: one-dimensional nonzero scalar matrices cannot anticommute while all having nonzero squares.

The spinor basis in c is compatible with the supplied gamma matrices. The rotation generator $r(J)$ is anti-Hermitian and commutes with ${\gamma }_0=-i{\sigma }_1$. The two boost generators are Hermitian matrices and anticommute with ${\gamma }_0$. It follows separately for all three generators, and hence for every real $T\in \mathfrak{so}(1,2)$, thatFor the infinitesimal Lorentz transformation, $\delta \psi =r(T)\psi$ and $\delta {\psi }^{†}={\psi }^{†}r(T{)}^{†}$. Hence the Dirac adjoint transforms contragrediently:The two contributions to the scalar variation cancel:This proves the spinor invariant in three spacetime dimensions has infinitesimal Lorentz invariance. Real transformation parameters matter because the Hermitian adjoint conjugates coefficients; this is not an invariance under arbitrary complexified parameters. For ordinary commuting spinor components, the bilinear is purely imaginary in this mostly-plus convention, and multiplying it by $i$ gives a real invariant. Its reality is separate from its invariance.