Musings

$Spin(8)$ has a triality symmetry (an outer autmomorphism of the Lie algebra), which permutes the three 8-dimensional irreducible representations: $8_v$, $8_s$, and $8_c$. $\mathfrak{g}_2\subset \mathfrak{so}(8)$ is the invariant subalgebra. (I’ll conveniently pass back and forth between the complex form of the Lie algebra and the compact real form of the group, as both are of interest to us.) What I want to do is describe that triality symmetry very explicitly and, thereby, the realization of $\mathfrak{g}_2$.

Note that $Spin(8)$ contains an $({SU(2)}^4)/\mathbb{Z}_2$ subgroup, under which the adjoint decomposes as $$28 = (3,1,1,1)+(1,3,1,1)+(1,1,3,1)+(1,1,1,3)+(2,2,2,2)$$ Under this decomposition, the action of triality is easy to describe: pick one of the $\mathfrak{sl}(2)$ subalgebras to hold fixed, and consider all permutations of the other three (supplemented by the obvious action on the $(2,2,2,2)$).

That’s triality. Looked at this way, it seems absurdly simple. The above description gives a perfectly concrete action of triality, as permutations of the generators. And we can push a little harder, and really understand $\mathfrak{g}_2$, this way.

The subalgebra, invariant under the $S_3$ permutations, is $\mathfrak{g}_2\subset \mathfrak{so}(8)$, under which $$28 = 14 + 7 \otimes V$$ where $V$ is the 2-dimensional irreducible representation of $S_3$. In terms of our previous decomposition, $$G_2 \supset (SU(2)\times {SU(2)}_D)/\mathbb{Z}_2$$ where the first $SU(2)$ is the one you kept fixed, and ${SU(2)}_D$ is the diagonal $SU(2)$ of the three which are permuted by triality. Under this embedding, $$\begin{split} 14 &= (3,1)+(1,3)+(2,4)\\ 7 &= (1,3) + (2,2) \end{split}$$

An explicit basis of antisymmetric $8\times 8$ matrices which give this $\mathfrak{g}_2$ subalgebra is as follows. First, we embed ${\mathfrak{sl}(2)}^4$, by taking the $8\times 8$ matrix to be block-diagonal, with $4\times 4$ blocks containing ${\mathfrak{sl}(2)}^2$, as $$\begin{matrix} \begin{split} H_L &= \sigma_2 \otimes \mathbb{1}\\ X_L &= \tfrac{1}{2} (\sigma_3+i\sigma_1) \otimes \sigma_2\\ Y_L &= \tfrac{1}{2} (\sigma_3-i\sigma_1) \otimes \sigma_2 = X_L^\dagger \end{split}&,\qquad\qquad \begin{split} H_R &= \mathbb{1} \otimes \sigma_2\\ X_R &= \tfrac{1}{2} \sigma_2 \otimes (\sigma_3+i\sigma_1)\\ Y_R &= \tfrac{1}{2} \sigma_2 \otimes (\sigma_3-i\sigma_1) = X_R^\dagger \end{split} \end{matrix}$$ where we’ve chosen the normalization conventions $$\begin{split} [X,Y]&=H\\ [H,X]&=2X\\ [H,Y]&=-2Y \end{split}$$

We pick one of these (the ${\mathfrak{sl}(2)}_L$ in the upper left-hand block) to hold fixed, and embed our second $\mathfrak{sl}(2)$ diagonally in the other three: $$\begin{split} H_1 &= \tfrac{1}{2} (\mathbb{1}+\sigma_3)\otimes \sigma_2 \otimes 1\\ X_1 &= \tfrac{1}{4} (\mathbb{1}+\sigma_3)\otimes (\sigma_3+i\sigma_1)\otimes\sigma_2\\ Y_1 &= X_1^\dagger\\ H_2 &= \tfrac{1}{2} (\mathbb{1}-\sigma_3)\otimes \sigma_2 \otimes \mathbb{1} + \mathbb{1}\otimes \mathbb{1} \otimes \sigma_2\\ X_2 &= \tfrac{1}{4} (\mathbb{1}-\sigma_3)\otimes (\sigma_3+i\sigma_1)\otimes \sigma_2 + \tfrac{1}{2} 1\otimes \sigma_2\otimes (\sigma_3+i\sigma_1)\\ Y_2 &= X_2^\dagger \end{split}$$

The highest weight of the $(2,4)$ is $$S_{1,3} = \tfrac{1}{4} \sigma_2\otimes (\sigma_3+i\sigma_1)\otimes (\sigma_3+i\sigma_1)$$ The remaining ones, e.g., $S_{-1,3} = [Y_1, S_{1,3}]$, are obtained by acting with the lowering operators, $Y_{1,2}$. With this choice of Cartan, the simple roots of $\mathfrak{g}_2$ correspond to $X_2$ (short root) and $S_{1,-3}$ (long root).

$$X_2$$

This 8-dimensional representation of $G_2$, as it’s reducible, is not the most convenient one for studying the representation theory of $G_2$. But it’s tailor-made for our purpose, which is understanding the embedding in $Spin(8)$. With an explicit embedding in hand, we can manufacture a distinguished triple, $(H,X,Y)$ for each nilpotent orbit of $\mathfrak{g}_2$, and see how it sits in $\mathfrak{so}(8)$. But that, probably, holds little interest for the general reader, so I’ll end here.