# Three-dimensional simple Lie algebras over the rationals

Let $\mathfrak g$ be a three-dimensional $\mathbf Q$-vectorspace endowed with the structure of a simple Lie algebra. How many non-isomorphic such $\mathfrak g$ are there?

Over the complex numbers, the answer is 1, while over the reals there are two non-isomorphic ones.

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–  Mariano Suárez-Alvarez Oct 19 '11 at 16:28

There are infinitely many. As you know, there are two non-isomorphic three-dimensional Lie algebras over $\mathbb R$ that coincide over $\mathbb C$: $\mathfrak {sl}_2(\mathbb R)$ and $\mathfrak {su}(2)$. To understand what happens over $\mathbb Q$, note that $\mathfrak {su}(2)$ is the derived subalgebra of the quaternion algebra over $\mathbb R$ (which is four-dimensional with a one-dimensional center).
So how many non-isomorphic quaternion algebras are there over $\mathbb Q$? There are infinitely many, though this is non-trivial. For every even set $S$ of absolute values of $\mathbb Q$ (which are the real one and the $p$-adic absolute values), there exists a unique quaternion algebra $D$ over $\mathbb Q$ such that for every $\nu\in S$, $D\otimes_\mathbb Q\mathbb Q_\nu$ is the unique quaternion algebra over $\mathbb Q_\nu$ (here $\mathbb Q_\nu$ denotes the completion of $\mathbb Q$ with respect to the absolute value $\nu$) and for every $\nu\not\in S$, $D\otimes_\mathbb Q\mathbb Q_\nu\simeq M_2(\mathbb Q_\nu)$, and all such quaternion algebras arise in this way. This is essentially an application of quadratic reciprocity (interpreted in terms of the Hilbert symbol).
Of course, the Lie algebra associated to $D$ is four-dimensional (and has a one-dimensional center), so take either the derived subalgebra of $D$ or, equivalently, take elements in $D$ with "reduced" trace zero to get a three-dimensional simple algebra. I can't quite rule out the existence of other algebras (the above are the "inner forms" of $\mathfrak{sl}_2(\mathbb Q)$, there might be non-isomorphic "outer forms", but this is well outside my competency)
+1 Wow! I should have seen this, but failed to make the connection. Even I know that the different quaternion algebras you listed are pairwise non-isomorphic as $\mathbf{Q}$-algebras (they represent distinct classes in the Brauer group over the rationals). Is it obvious that this implies that the kernels of the reduced trace are then pairwise non-isomorphic as Lie algebras? –  Jyrki Lahtonen Oct 19 '11 at 16:29
Jyrki, let $\mathfrak {sl}_1(D)$ be the kernel of the reduced trace. Then $D\simeq \mathfrak {sl}_1(D)\oplus\mathbb Q$ as Lie algebras (since $\mathbb Q$ is the center of $D$ as an algebra). If $\mathfrak {sl}_1(D_1)\simeq \mathfrak {sl}_1(D_2)$, then $\mathfrak {sl}_1(D_1)\oplus\mathbb Q\simeq \mathfrak {sl}_1(D_2)\oplus\mathbb Q$, so $D_1\simeq D_2$. –  B R Oct 19 '11 at 20:13