# Is $SL_2(\mathbb{O})$ a Lie group?

In JOhn Baez's article The Octonions, he writes the lie algebra isomorphism $\mathfrak sl_2(\mathbb {O}) \space \tilde = \space \mathfrak so(9, 1)$. Why is the former a Lie algbra, even though $SL_2(\mathbb O)$ is not a Lie group, because it is clearly non-associative?

• Check the definitions carefully; they don't mean the obvious things. – Qiaochu Yuan Oct 8 '17 at 8:53
• If there is such a thing, it is not likely to be associative. – Lord Shark the Unknown Oct 8 '17 at 9:30
• @LordSharktheUnknown It is. – anon Oct 11 '17 at 13:45

## 1 Answer

$\mathfrak{sl}(2,\mathbb{O})$ is the lie subalgebra of $\mathfrak{gl}(\mathbb{O}^2)$ (treating $\mathbb{O}^2\cong\mathbb{R}^{16}$ as real vector spaces) generated by the linear operators expressible as multiplication by a traceless octonionic $2\times 2$ matrix. And then the Lie group $\mathrm{SL}(2,\mathbb{O})$ within $\mathrm{GL}(16,\mathbb{R})$ is generated by the image of $\mathfrak{sl}(2,\mathbb{O})$ under $\exp$. This is more or less said on page $28$ of Baez's article.

Note $\mathrm{SL}(2,\mathbb{O})$ is associative because composition is associative; it just usually doesn't match up with $2\times2$ octonionic matrix multiplication. If $A$ and $B$ are $2\times 2$ octonionic matrices and $L_A$ is used to denote the left multiplication map $L_A(X)=AX$, then $L_A\circ L_B=L_{AB}$ iff $A,B$ are both $2\times2$ real matrices. If you instead represent $A,B$ by $16\times16$ real matrices $C,D$, then the $16\times16$ real matrix associated with the composition $L_A\circ L_B$ will be the matrix product $CD$.