This ties together several ideas at once: the Lorentz algebra, its complexification, and how irreducible representations are labeled by \((j_+, j_-)\).
Let us go step by step and explicitly show why
\[ (1,0)\oplus(0,1) \]is the adjoint representation of \(SO(3,1)\).
The Lorentz algebra is generated by rotations \(J_i\) and boosts \(K_i\), with commutation relations
\[ [J_i, J_j] = i \epsilon_{ijk} J_k, \qquad [J_i, K_j] = i \epsilon_{ijk} K_k, \qquad [K_i, K_j] = -i \epsilon_{ijk} J_k. \]So there are 6 generators total:
\[ \{J_i, K_i\}, \qquad i=1,2,3. \]This already tells us that the adjoint representation must be 6-dimensional.
Define
\[ J_i^{\pm} = \frac{1}{2}(J_i \pm i K_i). \]Now compute their commutators. One finds
\[ [J_i^+, J_j^+] = i \epsilon_{ijk} J_k^+, \qquad [J_i^-, J_j^-] = i \epsilon_{ijk} J_k^-, \qquad [J_i^+, J_j^-] = 0. \]This shows that the complexified Lorentz algebra splits as
\[ \mathfrak{so}(3,1)_{\mathbb C} \cong \mathfrak{su}(2)_+ \oplus \mathfrak{su}(2)_-. \]So \(J_i^+\) generate one \(SU(2)\), \(J_i^-\) generate another, and the two sets commute with each other.
Each set \(\{J_i^+\}\) and \(\{J_i^-\}\) transforms like angular momentum generators.
So:
For a representation \((j_+,j_-)\), the dimension is
\[ (2j_+ + 1)(2j_- + 1). \]Therefore:
\[ \dim(1,0) = (2\cdot 1 + 1)(2\cdot 0 + 1) = 3, \qquad \dim(0,1) = (2\cdot 0 + 1)(2\cdot 1 + 1) = 3. \]So
\[ (1,0)\oplus(0,1) \]has dimension
\[ 3+3=6, \]which exactly matches the number of Lorentz generators.
The adjoint representation is defined by how generators transform under commutation:
\[ X \mapsto [T_a, X]. \]So we ask: how do the generators \(J_i\) and \(K_i\) transform under the Lorentz algebra itself?
From the commutation relations above, the set \(\{J_i, K_i\}\) is closed under commutation. Therefore the six generators furnish the adjoint representation.
Now rewrite this 6-dimensional space in the chiral basis:
\[ \mathrm{Span}\{J_i, K_i\} = \mathrm{Span}\{J_i^+\} \oplus \mathrm{Span}\{J_i^-\}. \]Each piece transforms independently:
\[ \mathrm{Span}\{J_i^+\} \sim (1,0), \qquad \mathrm{Span}\{J_i^-\} \sim (0,1). \]Hence the adjoint representation decomposes as
\[ \boxed{\mathrm{Adjoint\ of\ } SO(3,1) = (1,0)\oplus(0,1).} \]This is also exactly the decomposition of an antisymmetric rank-2 Lorentz tensor \(F_{\mu\nu}\), which has 6 independent components. Over the complex numbers, it splits into self-dual and anti-self-dual parts:
\[ F_{\mu\nu} \longleftrightarrow (1,0)\oplus(0,1). \]So the same 6-dimensional Lorentz object can be viewed either as:
The Lorentz algebra has 6 generators. These split into two commuting \(SU(2)\) algebras. Each gives a 3-dimensional spin-1 representation, so the adjoint representation is
\[ 3 \oplus 3 = (1,0)\oplus(0,1). \]This is additional material to complement page 477 of Group Theory in a Nutshell by Anthony Zee.
Created with assistance from ChatGPT.