Let $m\ge 2$, and let $G={\rm SO}^*(4m)$ denote the "quaternionic" real form of the special orthogonal group ${\rm SO}(4m,\mathbb C)$ of type ${\sf D}_{2m}$. Let $\tau\in{\rm Aut}_{\Bbb R}(G)$ be a real automorphism of $G$, that is, an automorphism defined over $\Bbb R$. My Galois-cohomological calculations suggest that then $\tau$ is an inner automorphism.

Question. Is it true that, although $G$ does have complex outer automorphisms, it has no real outer automorphisms?

Clarification: I regard $G={\rm SO}^*(4m)$ as an algebraic group over $\Bbb R$. By a complex inner automorphism of $G$ I mean an element of the group ${\rm Inn}(G)(\Bbb C)$, where ${\rm Inn}(G)=G/Z(G)$. By a real inner automorphism of $G$ I mean an element of ${\rm Inn}(G)(\Bbb R)$.

  • $\begingroup$ This has been systematically looked in "Gundogan, The component group of the automorphism group of a simple Lie algebra and the splitting of the corresponding short exact sequence", J. Lie Theory 20 (2010), no. 4, 709-737. But I'm not sure what the conclusion is. Apparently, the automorphism group of the real Lie algebra, as Lie group, has 2 components. But I don't see if the resulting homomorphism of the component group to the complex Out is an isomorphism or is the trivial map; this might come from a more careful reading. $\endgroup$ – YCor May 11 '18 at 10:21
  • $\begingroup$ Just to clarify the notation, $G$ is a real Lie algebra here, right? Otherwise I do not know how to make sense of a Lie-group homomorphism being defined 'over a field'. But maybe there is a definition I don't know about? $\endgroup$ – Vincent May 11 '18 at 11:07
  • 2
    $\begingroup$ @Vincent: Abusing notation, I meant the real algebraic group $G$ over $\Bbb R$ with the group of real points $G(\Bbb R)={\rm SO}^*(4m)$. However, if you answer for the Lie algebra, I will be quite happy. $\endgroup$ – Mikhail Borovoi May 11 '18 at 11:23
  • $\begingroup$ One has to be careful about the meaning of outer automorphism. Let $G$ be the corresponding adjoint (connected) group. One can consider the group $G(\mathbf{R})$ and say that its elements are inner automorphisms. But one can say that inner automorphisms are only those in the unit component of $G(\mathbf{R})$ in the real topology. Unlike in the complex case these can be distinct. From the question it seems that one uses the first definition to define the group of outer automorphisms. But I'm clarifying, especially if one talks of the outer automorphism group of the Lie algebra... $\endgroup$ – YCor May 11 '18 at 15:01
  • 1
    $\begingroup$ Just to illustrate my previous comment, consider the Lie algebra $\mathfrak{sl}_2(\mathbf{R})$. Then conjugation by a matrix with determinant $-1$ is not an inner automorphism (but it's an inner automorphism in the adjoint group $\mathrm{PGL}_2(\mathbf{R})$, which is connected in the Zariski sense but has 2 real (Lie) components. $\endgroup$ – YCor May 11 '18 at 18:55

No, there are no real outer automorphisms of $G = SO^*(4m)$. Suppose, for sake of contradiction, that one exists and call it $\phi$. One of the half-spin representations $\rho \!: G \to GL(V)$ is real, and the composition $\rho \phi$ provides an irreducible representation defined over $\mathbb{R}$ of $G$ that is (by examining highest weights) the other half-spin representation. But the other half-spin representation is quaternionic, a contradiction.

(The fact that one half-spin rep of $SO^*(4m)$ is real and the other is quaternionic also played a role in this related MO question "The Tits classes of simply connected simple real groups".)

Generalizing, we can replace $k = \mathbb{R}$ with any field and $G$ with a semisimple $k$-group. The general statement is that the Tits algebras of $G$ provide an obstruction to the existence of outer $k$-automorphisms of $G$. See Theorem 11 in my paper Outer automorphisms of algebraic groups and determining groups by their maximal tori (Michigan Mathematical Journal 61 #2 (2012), 227-237). The specific example of groups of type $^1D_n$ as in this question is at the top of page 233.

| cite | improve this answer | |

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.