## Rational forms of simple Lie algebras

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I am more or less familiar with the classification of real forms of complex semisimple Lie algebras. But as soon as I wander off into the domain of very-non-algebraically closed fields, things seem to become considerably more complex.

Is there a classification of the rational forms of complex semisimple finite-dimensional Lie algebras?

Of course, there is such a classification, and one can in principle carry it out using standard ideas involving Galois cohomology &c. The question is, rather, has it been written down explicitly and with a nice parametrization?

In particular:

How many rational forms does the complex algebra of type $G_2$ have?

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## 3 Answers

I think the answer to the first question is clearcut: yes, there is a (sort of) classification. On the other hand, it gets extremely complicated when you start over `$\mathbb{Q}$` rather than `$\mathbb{R}$`. As far as I know, all detailed treatments require (beyond an early stage) case-by-case examination of the simple types over `$\mathbb{C}$`. As you note, Galois cohomology gets implicated here, along with the careful study of inner and outer automorphism groups of each simple Lie algebra. The final chapter of Jacobson's 1962 book Lie Algebras lays out the algebraic program, which at that time hadn't disposed of all the exceptional types (some of my fellow graduate students were still occupied with that).

The main problem is that you get involved with the classification of various types of associative algebras along the way. This is easier to control over local fields, but there are infinitely many of those fields in the background of the study of Lie algebras over `$\mathbb{Q}$`. Not having spent time with this problem for many decades, I can't comment in more detail on how well satisfied one might be with the answers in the existing literature. But beyond Jacobson's book, there is a somewhat more "rational" approach developed in considerable detail by one of his former students George Seligman, which I reviewed for the AMS Bulletin here.

The exceptional Lie algebra of type `$G_2$` was apparently first studied in depth by Jacobson himself in Duke Math. J. 5 (1939); here the relevant (non-associative) algebra is the 8-dimensional Cayley (octonion) algebra whose forms over the ground field have to be identified.

None of this literature makes for easy reading, but in retrospect my impression (possibly false) is that infinitely many forms can exist over `$\mathbb{Q}$`. I'd be curious as to whether any recent literature simplifies the whole matter at all.

P.S. George McNinch has addressed some of this more precisely from the viewpoint of Galois cohomology. Maybe I should just add that the big obstacle in general to finding all forms of a simple Lie algebra over a given field is the possible existence of many anisotropic forms. Over `$\mathbb{R}$` one is lucky enough to find unique anisotropic (=compact) forms, but the classification of such forms is highly sensitive to the nature of the field. Classification methods for algebraic groups (Tits) or Lie algebras (discussed above) usually just reduce the problem in a unified way to the anisotropic case.

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Sorry: I was being a dim-wit.

(My answer now reflects a fair amount of editing -- sorry.)

There are only finitely many isomorphism classes of $\mathbf{Q}$-forms of simple groups of type $G_2$. I expect that means there are only finitely many iso classes of $\mathbf{Q}$-forms of simple Lie algebras of type $G_2$ though I confess I didn't carefully think through the transition from algebraic groups to Lie algebras.

Indeed, over any field $k$, $k$-forms of a simple algebraic group of type $G_2$ are classified by the cohomology set $H^1(k,H)$ where $H$ is a split group of type $G_2$. [We use here that $H$ is simple -- i.e. adjoint -- and that every $k$-automorphism of $H$ is inner].

Now, one knows that $H^1(\mathbf{Q}_p,H)$ is trivial for all primes p (since $\mathbf{Q}_p$ is a local field and $G$ is simply connected). And $H^1(\mathbf{R},H)$ is finite [there are only finitely many (two, I think?) real forms]. Hence the Hasse principle implies that $H^1(\mathbf{Q},H)$ is finite.

For a general field $k$, note that the cohomology set $H^1(k,G)$ identifies with (1) the set of isomorphism classes of octonion algebras over $k$, and (2) the set of isomorphism classes of certain quadratic forms known as 3-Pfister forms. For all this, see e.g. Serre's part of the book [Garibaldi, Merkurjev, Serre "Cohomological Invariants and Galois cohomology".

At least if the characteristic of $k$ is 0 and $H_1$ and $H_2$ are two $k$-forms of the group $G_2$, I hope that $H_1 \not \simeq H_2$ should imply that $\operatorname{Lie}(H_1) \not \simeq \operatorname{Lie}(H_2)$.

So to decide if there are infinitely many $k$-forms of the simple Lie algebra of type $G_2$, it is enough to exhibit infinitely many 3-Pfister forms over $k$ which are not isometric.

Now, Theorem 18.1 in [Serre, loc. cit.] calculates the group of cohomological invariants Inv$_k($Pfister$_3,\mathbf{Z}/2\mathbf{Z})$; it is a free module of rank 2 over the (infinite) cohomology ring $H^\bullet(k,\mathbf{Z}/2\mathbf{Z})$. And there is an invariant e such that for $\alpha_1,\alpha_2,\alpha_3 \in k^\times$, the value of e on the $3$-Pfister form $Q_\alpha = \langle \langle \alpha_1$, $\alpha_2$, $\alpha_3 \rangle \rangle$ is the cup-product $(\alpha_1)\cup (\alpha_2) \cup (\alpha_3)$ in $H^3(k,\mathbf{Z}/2\mathbf{Z})$ of the classes in $H^1(k,\mathbf{Z}/2\mathbf{Z}) = k^\times/k^{\times 2}$ determined by the $\alpha_i$. In fact, any "normalized invariant" is a $H^\bullet(k,\mathbf{Z}/2\mathbf{Z})$-multiple of $e$.

So to use these invariants to find lots of non-isometric 3-Pfister forms, you'd need at least that $H^3(k,\mathbf{Z}/2\mathbf{Z})$ is non-zero.

Now, $H^3$ is non-zero for $k=\mathbf{Q}((T))$ (or even $\mathbf{Q}_p((T))$) and I believe I would expect there to be infinitely many isometry classes of Pfister forms in those cases (but I'd be interested in seeing an argument...)

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If I understand you correctly: the Hasse principle implies that for $G$ split over $\mathbb Q$ we have $H^1(\mathbb Q,G)=H^1(\mathbb R,G)$ and, taking $G$ of split type $G_2$ this allows us to conclude that there are two rational forms of $G_2$, giving after extension of scalars to $\mathbb R$ the compact form and the one other form shown in the tables of, say, [Onishchik, Vinberg, Lie Groups and Lie Algebras III]. (Since this second real form is not compact, their rational Lie algebras cannot be isomorphic)  – Mariano Suárez-Alvarez Mar 8 2012 at 23:26
@Mario: Yes, for $G$ split of type $G_2$, and more generally for $G$ simply connected, $H^1(k,G)$ vanishes for $k = \mathbf{Q}_p$ so the Hasse principle implies what you say. When $G$ is no longer simply connected, that vanishing may fail. And the Hasse principle may fail. The issue about Lie algebras that gave me pause was this: does every $\mathbf{Q}$ form of a simple Lie algebra come from a form of the corresponding algebraic group? I guess the answer is "yes" since the automorphism group of the Lie algebra and algebraic group should coincide; but I didn't think too carefully about it – George McNinch Mar 9 2012 at 2:13

(Disclaimer: I am not an expert and easily may overlook something).

Results about forms of exceptional types are technically difficult and scattered over the literature. For example, the only known to me full description of forms of $D_4$ is buried (somewhat implicitly) inside the book: Knus, Merkurjev, Rost, Tignol, The Book of Involutions, AMS Colloq. Publ, Vol. 44., 1998, http://www.mathematik.uni-bielefeld.de/~rost/BoI.html . The works of Skip Garibaldi (using the language of algebraic groups) are also highly relevant.

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Thanks for the reference! – Mariano Suárez-Alvarez Apr 4 2012 at 2:49
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