Given a representation $(n, m)$ of the Lorentz group, is the little group representation just the tensor product $n \otimes m$?

Question

I've been reading Weinberg's QFT Vol 1. and more specifically section 5.6. I would like to know if my understanding is correct or if I missed something. He starts with the full Lorentz group $\mathrm{SO}(3, 1)$ and shows that we have two commuting copies of $\mathfrak{su}(2)$ generators $A_i$ and $B_i$. He then says that the 'rotation group' is generated by $J_i = A_i + B_i$. Is the 'rotation group' the same thing as the little group (for massive particles)?

These $J_i$ then form a representation of the tensor product $V_A \otimes V_B$ of the two $\mathfrak{su}(2)$ irreps $V_A, V_B$ (if I remember correctly). Is it correct to say then, that given some irrep of the Lorentz group $(A, B)$ the corresponding representation of the little group is nothing but $V_A \otimes V_B$? His examples seem to confirm this, e.g. a vector $(1/2, 1/2)$ would then be in the $1 \oplus 0$ representation of the (universal cover) of the little group $\mathrm{SO}(3)$, but he doesn't state it this way which makes me wary.

Answer 1

1. Consider the restricted Lorentz group $SO^+(3,1;\mathbb{R})$ and its complexification $$SO(3,1;\mathbb{C}).\tag{1}$$ Picking the COM frame the massive little group becomes the 3D rotation group $SO(3,\mathbb{R})$.

2. The double covers are isomorphic to $SL(2,\mathbb{C})$, $$G~:=~SL(2,\mathbb{C})_L\times SL(2,\mathbb{C})_R\tag{2}$$ and $H:=SU(2)$, respectively, cf. e.g. this Phys.SE post.

3. The (double cover of the) little group $H$ is a subgroup of $G$ via the diagonal imbedding $$H~\ni~h\quad\mapsto\quad(h,h)~\in~G.\tag{3}$$

4. If $\rho: G\to GL(V)$ is a group representation for $G$, it induces a restricted representation $\rho|_H:H\to GL(V)$ on the subgroup $H$.

5. For an irreducible $G$-representation $$(j_L,j_R)~=~j_L\otimes j_R, \qquad j_L, j_R~\in~ \frac{1}{2}\mathbb{N}_0,\tag{4}$$ the restricted $H$-representation is $$j_L\otimes j_R~\cong~\oplus_{j=|j_L-j_r|}^{j_L+j_R} j,\tag{5}$$ as mentioned in OP's title. Note that the notation in expression (4) and (5) are deceptively similar. However we stress that it is implicitly implied that expression (4) is a $G$-representation, while expression (5) is an $H$-representation.

Answer 2

I am not going to read Weinberg's book with you to your satisfaction, nor should I.

Indeed, the little group of a massive particle (go to its rest frame) is the rotation group SO(3), sharing a Lie algebra with SU(2).

Presumably, your text has reminded you, like (A2) of this, that for an so(1,3) irrep π labelled by integer of half-integer spins (n,m), the three coproduct operators $$ \pi_{(n,m)}(J_i) = J^{(m)}_i \otimes {\mathbb I}_{(2n+1)}+{\mathbb I}_{(2m+1)} \otimes J^{(n)}_i $$ manifestly satisfy the so(3)~su(2) Lie algebra commutation relations. (This is just the rotation subgroup of the Lorentz group, the little group.) Check it again.

These operators are huge matrices acting on $(2n+1)\times (2m+1)$-dimensional vectors, so the tensor product reducible representation you get by "adding" spins n and m in elementary QM, as you properly exemplify.

Moreover, your text should have exemplified that the likewise huge, $(2n+1)\times (2m+1)$ by $(2n+1)\times (2m+1)$ -dimensional, group matrices $$ e^{i\vec \theta \cdot (\vec J^{(m)} \otimes {\mathbb I}_{(2n+1)}+{\mathbb I}_{(2m+1)} \otimes \vec J^{(n)} )} = e^{i\vec \theta \cdot \vec J^{(m)} } \otimes e^{i\vec \theta \cdot \vec J^{(n)} } $$ provide the conventional reducible representation of the rotation group of QM.