Building $\mathfrak{so}(1,3)$ reps using $\mathfrak{so}(1,3)\cong \mathfrak{su}(2)\oplus \mathfrak{su}(2)$

Question

I'm going through the representation theory of $\mathfrak{so}(1,3)$, building Dirac/Weyl spinors and vectors, and I'm a bit confused on the mathematical definitions involved. We have $\mathfrak{so}(1,3)\cong \mathfrak{su}(2)\oplus \mathfrak{su}(2)$ as algebras. I'm good with the formalities of this isomorphism: if $J^i$ and $K^i$ are our elements of the six-dimensional algebra $\mathfrak{so}(1,3)$, we can always write any linear combination of $J^i$ and $K^i$ as $a_i J_+^i+b_jJ_-^j$ where $J_+$ satisfies its own $su(2)$ algebra, $J_-$ does too, and all $J_+$ and $J_-$ commute.

To build out a representation of $\mathfrak{so}(1,3)$, I see the notation $\left(n_1,n_2\right)$ used. So $\left(\frac{1}{2},0\right)$ can denote the left-handed Weyl spinors, $\left(0,\frac{1}{2}\right)$ the right-handed ones, $\left(\frac{1}{2},\frac{1}{2}\right)$ vectors, and $\left(\frac{1}{2},0\right)\oplus\left(0,\frac{1}{2}\right)$ Dirac spinors. I'm a bit confused about what this notation really means.

It can't be the tensor product: $\frac{1}{2}\otimes \frac{1}{2}\cong 1\oplus 0$ is still just a representation of the three-dimensional Lie algebra $su(2)$, with group action (for example) $J^i=(\frac{1}{2}\sigma^i)\otimes 1+ 1\otimes (\frac{1}{2}\sigma^i)$.

Is the following a good definition for $(n_1,n_2)$?

Take $V_1$ to be the vector space of the spin $n_1$ representation, and $V_2$ the vector space corresponding to $n_2$. Then $(n_1,n_2)$ is the representation of $su(2)\oplus su(2)$ which acts on the vector space $V_1\otimes V_2$, with action $(A,B)(v_1\otimes v_2)=(Av_1)\otimes(B v_2)$ (where $(A,B)\in su(2)\oplus su(2)$ and their actions on $V_i$ are determined by the spin $n_i$ representation).

I think this is correct, I just got very confused with the contrast between addition of angular momentum: in this case the vector space is still the tensor product, but the group action is different.

Answer 1

The definition for the $(n_1,n_2)$ representation in the question is almost correct, but not quite. $(A,B)(v_1\otimes v_2)=(Av_1)\otimes(B v_2)$ is not a linear action and probably isn't even well-defined (consider $A=A_1+A_2$ and $B=B_1+B_2$).

The answer lies in the different meanings for the tensor product of representations. There are two distinct ways to form tensor products of representations, and they are explained in Brian Hall's Lie Groups, Lie Algebras, and Representations, 2ed, definitions 4.19 and 4.20. Hall writes after these two definitions:

The notation is, unfortunately, ambiguous, since if $\Pi_1$ and $\Pi_2$ are representations of the same group $G$, we can regard $\Pi_1\otimes\Pi_2$ either as a representation of $G$ or as a representation of $G\times G$. We must, therefore, be careful to specify which way we are thinking about $\Pi_1\otimes\Pi_2$.

To clarify these two different ways we have to define them:

Definition 1. (Tensor product as a representation of $\mathfrak{g}\oplus\mathfrak{h}$). Let $\Pi_1$ and $\Pi_2$ be two representations of two Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$. Let $V_1$ and $V_2$ be the vector spaces on which they act. Then we can define a representation of $\mathfrak{g}\oplus\mathfrak{h}$ on $V_1\otimes V_2$, by defining the group action $(\pi_1,\pi_2)\cdot (v_1\otimes v_2)=(\pi_1\cdot v_1)\otimes v_2+v_1\otimes (\pi_2\cdot v_2)$. This representation is sometimes denoted $\Pi_1\otimes \Pi_2$. For our purposes let's call this $\Pi_1\otimes_1 \Pi_2$

Definition 2. (Tensor product as a representation of $\mathfrak{g}$). Let $\Pi_1$ and $\Pi_2$ be two representations of the same Lie algebra $\mathfrak{g}$. Let $V_1$ and $V_2$ be the vector spaces on which they act. Any $a\in \mathfrak{g}$ can act on an element of $V_1$ or an element of $V_2$ by virtue of these two representations. Then we can define a representation of $\mathfrak{g}$ on $V_1\otimes V_2$, by defining the group action $a\cdot (v_1\otimes v_2)=(a\cdot v_1)\otimes v_2+v_1\otimes (a\cdot v_2)$ for $a\in \mathfrak{g}$. This representation is, confusingly, also sometimes denoted $\Pi_1\otimes \Pi_2$. Let's call this $\Pi_1\otimes_2 \Pi_2$

In addition of angular momentum, we have, for example, $\mathbf{\frac{3}{2}}\otimes_2 \mathbf{\frac{3}{2}}\cong \mathbf{3}\oplus \mathbf{2}\oplus \mathbf{1}\oplus \mathbf{0}$, a reducible representation of $\mathfrak{su}(2)$. (Note there is no ambiguity for direct sum as defined by Hall defn. 4.12. Both sides are reps of $\mathfrak{su}(2)$).

In our case, we define $\left(\mathbf{n}_1,\mathbf{n}_2\right)=\mathbf{n}_1\otimes_1\mathbf{n}_2$, an irreducible representation of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$. I can guess that the parentheses notation is adopted instead of the tensor product notation to emphasize the difference between the two ways, but it is in fact a certain tensor product of two representations!

The confusion in the question was caused by asking, if $\left(\mathbf{n}_1,\mathbf{n}_2\right)$ is a tensor product, why don't we have $\left(\mathbf{\frac{1}{2}},\mathbf{\frac{1}{2}}\right)\cong\mathbf{1}\oplus\mathbf{0}$? The answer is that the righthand side is a representation of $\mathfrak{su}(2)$ while the lefthand side is a representation of $\mathfrak{su}(2)\oplus \mathfrak{su}(2)$, so the two sides couldn't possibly be isomorphic.

It's probably best to stick with definition 2 in physics, so that the meaning of tensor products in the addition of angular momentum remains unambiguous. Both of these representations indeed act on the tensor product of the two underlying vector spaces, but the actions are different and the algebras they represent are different.