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Let $\mathfrak{g}$ be a finite-dimensional complex Lie algebra and let $R \subset \mathbb{C}$ be a subring. Say that $\mathfrak{g}$ is defined over $R$ if there exists a basis $x_1, ... x_n$ for $\mathfrak{g}$ such that the structure constants $c_{ijk}$ of the bracket $$[x_i, x_j] = \sum_k c_{ijk} x_k$$ all lie in $R$. It is classical that all semisimple $\mathfrak{g}$ are defined over $\mathbb{Z}$. But this is also true for some non-semisimple $\mathfrak{g}$ such as the Lie algebra of $n \times n$ upper triangular or strictly upper triangular matrices.

In fact, I don't know an example of such a $\mathfrak{g}$ which isn't defined over $\mathbb{Z}$ although I would be surprised if they didn't exist. Can someone construct one or prove that they don't exist? If they do exist, is a weaker statement true? For example, are all such $\mathfrak{g}$ defined over a number field?

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up vote 6 down vote accepted

Three dimensional Lie algebras (over $\mathbb{C}$, say) vary in moduli: see for instance this paper for a description. (I am not conversant with the details here...) In particular, one of the connected components of the moduli space has dimension one, so the generic point of this moduli space cannot be defined over any algebraic extension of $\mathbb{Q}$.

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Ah, there it is. In one of Pete Clark's answers he links to a paper giving a continuous family $L_a^3$ of pairwise non-isomorphic solvable $3$-dimensional Lie algebras over any field. Explicitly, these are spanned by $x_1, x_2, x_3$ satisfying $$[x_3, x_1] = x_2, [x_3, x_2] = ax_1 + x_2$$

for a parameter $a$ (and, I suppose, $[x_1, x_2] = 0$). Since $\mathbb{C}$ is uncountable the conclusion follows.

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Right -- this is similar to but more direct than my answer (which requires some understanding / good faith about properties of moduli spaces). – Pete L. Clark Jan 1 '12 at 8:20
Upon further reflection, I think we need to be a little careful with these considerations, since in the setup of your answer / my previous answer we are looking at $\mathbb{C}$-isomorphisms, but that's not what you're asking for here. In particular, every Lie algebra can be defined over a countable field, and -- if I am not mistaken -- the finite dimensional $\mathbb{C}$-Lie algebras fall into countably many isomorphism classes as Lie algebras over $\mathbb{Z}$. – Pete L. Clark Jan 1 '12 at 8:28
@Pete: I'm not sure what you're saying. Having an integral form is a property of an isomorphism class of Lie algebras; moreover, the integral form determines the isomorphism class of the Lie algebra, so only countably many isomorphism classes of Lie algebras can have integral forms. Am I missing something here? – Qiaochu Yuan Jan 1 '12 at 8:54
It depends whether you are considering $\mathbb{C}$-algebra isomorphisms or $\mathbb{Z}$-algebra isomorphisms. Consider for instance the case of elliptic curves. Any two elliptic curves with transcendental $j$-invariant are isomorphic as $\mathbb{Z}$-schemes, including for instance the elliptic curves with $j$-invariants $e$ and $\pi$, but these are not isomorphic as varieties over $\mathbb{C}$. – Pete L. Clark Jan 1 '12 at 15:43
Can you explicit an $a$ such that $L_a^3$ is not defined over $Z$ ? – user10676 Jan 1 '12 at 16:38

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