I hope to know the classification of real forms of complex simple Lie algebras of types $A$, $D$, $E$ up to inner automorphisms.

Let $\mathfrak{g}_1$ and $\mathfrak{g}_2$ be real forms of a complex simple Lie algebra $\mathfrak{g}$. We say that they are equivalent if there is an isomorphism $\mathfrak{g}_1 \to \mathfrak{g}_2$ which extends to an inner automorphism of $\mathfrak{g}$. However, there may be real forms which are isomorphic but not equivalent.

Where can we find the classification of such equivalence classes of real forms?

  • $\begingroup$ @O'Farrill: Thanks. But that is precisely my question: I want to distinguish real forms which not related by inner automorphism. $\endgroup$ – sunny Dec 14 '10 at 15:38
  • $\begingroup$ So are you after a refinement of the usual classification of real forms where you are allowed to use only inner automorphisms and not the whole automorphism group? $\endgroup$ – José Figueroa-O'Farrill Dec 14 '10 at 17:48
  • $\begingroup$ @O'Farrill: That is correct. $\endgroup$ – sunny Dec 14 '10 at 17:57

The one case (for $\frak g$ simple) where the two definitions of real forms (up to $Aut(\frak g)$ versus $Int(\frak g)$) don't agree is the following.

There is a real form of $\frak g=\frak s\mathfrak o(2n,\mathbb C)$ denoted $\frak s\frak o^*(2n)$.

For $n\ge4$ even $\frak g$ has two subalgebras isomorphic to $\frak s\frak o^*(2n)$ which are related by $Aut(\frak g)$, but not by $Int(\frak g)$. If $n\ge 3$ is odd there is only one such algebra up to $Int(\frak g)$. The difference is because the center of the simply connected group is $\mathbb Z/2\times\mathbb Z/2$ if $n$ is even, and $\mathbb Z/4$ if $n$ is odd.

$n=4$ is particularly interesting: the triality automorphism of $D_4$ interchanges these two copies of $\mathfrak s\mathfrak o*(2n)$, as well as $\frak s\frak o(6,2)\simeq \frak s\frak o^*(2n)$.

  • $\begingroup$ Thank you. Do you know if these are the only examples, or if there is a table listed somewhere? $\endgroup$ – sunny Dec 15 '10 at 14:28
  • $\begingroup$ I believe that is the only case. I don't know of such a table, but this comment reduces the table to the classical one, for example in Helgason, Tits, or Onischik-Vinberg. $\endgroup$ – Jeffrey Adams Dec 15 '10 at 19:20
  • $\begingroup$ @jda: Thank you. This seems like a situation where the experts know the classification (i.e. only the above two extra classes of isomorphic but not equivalent real forms), but is not written anywhere. I look up Helgason and Onischik-Vinberg, but it is not there. $\endgroup$ – sunny Dec 16 '10 at 13:15

I'd recommend looking at the article "Strong real forms and the Kac classification", by Jeffrey Adams -- it's an expository paper from 2005 which answers your question.

For more details, Adams is careful to distinguish between three notions which he calls "traditional real forms", "real forms", and "strong real forms". He works at the level of groups rather than Lie algebras, but let me repeat his definitions here.

Let $\mathfrak{g}$ be a simple complex Lie algebra. Let $G$ be the associated adjoint algebraic group over the complex numbers, so we identify $\mathfrak{g} = Lie(G)$ and $Int(\mathfrak{g}) = Int(G)$ and the homomorphism $G \rightarrow Int(G)$ is an isomorphism. There is a well-known short exact sequence: $$1 \rightarrow Int(G) \rightarrow Aut(G) \rightarrow Out(G) \rightarrow 1.$$ Adams defines:

  • A traditional real form of $G$ is an equivalence class of involutions in $Aut(G)$, where equivalence is given by conjugation by $Aut(G)$: $\iota \sim \alpha \iota \alpha^{-1}$ for any $\alpha \in Aut(G)$.

  • A real form of $G$ is an equivalence class of involutions in $Aut(G)$, where equivalence is given by conjugation by $Int(G)$: $\iota \sim Int(g) \iota Int(g^{-1})$ for all $g \in G$.

Associated to an involution $\iota$, there is a real algebraic group $G_R$, whose complexification coincides with $G$, and which has a maximal compact subgroup $K_R$ with complexification $K_C = G^\iota$.

Now, for the Lie algebra, let's make the analogous definitions:

  • A traditional real form of $\mathfrak{g}$ is an equivalence class of involutions in $Aut(\mathfrak{g})$, where equivalence is given by conjugation by $Aut(\mathfrak{g})$.

  • A real form of $\mathfrak{g}$ is an equivalence class of involutions in $Aut(\mathfrak{g})$, where equivalence is given by conjugation in $Int(\mathfrak{g})$.

Since $G$ is chosen to be the adjoint group, $Aut(G) = Aut(\mathfrak{g})$ and $Int(G) = Int(\mathfrak{g})$. When $\iota$ is an involution in $Aut(\mathfrak{g})$, there exists an antiholomorphic involution $\theta$ of $\mathfrak{g}$, such that $\mathfrak{g}^\theta$ is a compact real form, and $\iota \theta = \theta \iota$. In this way, $\iota$ yields a real Lie algebra $\mathfrak{g}^{\iota \theta} = \mathfrak{g}^{\theta \iota}$.

Furthermore, such a $\theta$ is unique up to $Int(\mathfrak{g}^\iota )$, so for any $\theta' = \gamma \theta \gamma^{-1}$ another such involution, we find that $$\mathfrak{g}^{\iota \theta'} = \mathfrak{g}^{\iota \gamma \theta \gamma^{-1}} = \mathfrak{g}^{\gamma \iota \theta \gamma^{-1}} = \gamma \left( \mathfrak{g}^{\iota \theta} \right).$$

If we consider an inner conjugate of $\iota$: $\iota' = \delta \iota \delta^{-1}$ for $\delta \in Int(\mathfrak{g})$, then one may choose a commuting Cartan involution $\theta' = \delta \theta \delta^{-1}$. In this way, we find $$\mathfrak{g}^{\iota' \theta'} = \delta \left( \mathfrak{g}^{\iota \theta} \right).$$

What this boils down to is that the suggested notion of inner-equivalence classes of real forms of $\mathfrak{g}$ coincides precisely with Adams notion of a real form of $G$, or my adaptation above to a real form of $\mathfrak{g}$. Hence the answer is given by classifying the equivalence classes of involutions in $Aut(G)$, where equivalence is given by conjugation in $Int(G)$.

Now, to come to the Kac classification that Adams discusses. We refer back to Adams for the details (or Adams-Barbasch-Vogan, Chapter 2).

Fix $G$, and also an involution in $\gamma \in Out(G)$; this fixes an "inner class" of real forms. Let $c$ be the order of $\gamma$: $c \in \{ 1,2 \}$. Let $\tilde \Delta_\gamma$ denote the affine twisted Dynkin diagram associated to the pair $(\mathfrak{g}, \gamma)$, with vertices $\{ \alpha_0, \ldots, \alpha_\ell \}$. (Fold the usual Dynkin diagram of $\mathfrak{g}$ by the automorphism $c$, and extend it by the lowest root). Let $n_i$ denote the usualy numbering of this Dynkin diagram, based on multiplicities in the highest root, for example.

A Kac marking of the Dynkin diagram will mean a subset $S \subset \{0, \ldots, n \}$ of the vertices of $\tilde \Delta_\gamma$, such that $$\sum_{i \in S} n_i = 2/c.$$ Thus at most two vertices are marked, and $n_i \leq 2$ for all marked vertices $i \in S$.

$Z(G_{sc})$ acts on $\tilde \Delta_\gamma$, and the orbits on the set of Kac markings parameterize the real forms within the inner class given by $\gamma$.

I think this is the best you'll get -- there aren't too many cases, but I don't know where to find a table of them.

  • $\begingroup$ +1. A very informative answer. $\endgroup$ – José Figueroa-O'Farrill Dec 14 '10 at 22:15
  • 1
    $\begingroup$ The following webpage seems to have a couple of papers by Jeffrey Adams, including the one mentioned in this answer: liegroups.org/papers $\endgroup$ – José Figueroa-O'Farrill Dec 14 '10 at 22:21
  • $\begingroup$ @Marty: Thanks for your comments. How does Z(G_{sc}) act on the the set of Kac markings? We need to know how it acts in order to obtain the orbits within the set of Kac markings. Anyway, for the statement that "The orbits of Z(G_{sc}) on the set of Kac markings parametrizes the real forms within inner class", do you know the reference? $\endgroup$ – sunny Dec 15 '10 at 0:45
  • $\begingroup$ That's in the paper of Jeff Adams, cited above. I thought about discussing this action too -- but I don't want to rewrite the whole paper of Adams in my answer here. $\endgroup$ – Marty Dec 15 '10 at 1:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.