Let $G$ be a connected linear algebraic group over the field of complex number ${\Bbb C}$. Let $G({\Bbb C})$ denote the complex Lie group of ${\Bbb C}$-points of $G$. Let $\sigma$ be an anti-holomorphic involution of $G({\Bbb C})$, that is, a an automorphism of the real Lie group $$\sigma\colon G({\Bbb C})\to G({\Bbb C})$$ such that $\sigma$ is anti-holomorphic and $\sigma^2={\rm id}$.

The anti-holomorphic involution $\sigma$ naturally acts on the ring of holomorphic function on $G({\Bbb C})$: $$({}^\sigma\!\! f)(g)=\overline{f(\sigma^{-1}(g))},$$ where the bar denotes complex conjugation (and, of course, $\sigma^{-1}=\sigma$).

We say that $\sigma$ as above is anti-regular, if, when acting on the ring of holomorphic functions on $G$, $\sigma$ preserves the subring of regular functions (recall that $G$ is an algebraic group).

Question. Are all anti-holomorphic involutions anti-regular in the following cases: (1) $G$ is a connected linear algebraic group; (2) $G$ is a (connected) reductive algebraic group; (3) $G$ is a (connected) semisimple algebraic group?

Remark. An anti-regular involution $\sigma$ of $G({\Bbb C})$ defines by Galois descent a real structure on $G$. Indeed, we may put $$ G_{\Bbb R}={\rm Spec}\,({\Bbb C}[G]^\sigma),$$ where ${\Bbb C}[G]^\sigma$ is the subring of fixed points of $\sigma$ in the ring of regular functions ${\Bbb C}[G]$ on $G$.

Conversely, an algebraic group $G_{\Bbb R}$ over ${\Bbb R}$ defines a complex algebraic group $G:=G_{\Bbb R}\times_{\Bbb R} {\Bbb C}$, and the complex conjugation on ${\Bbb C}$ induces by functoriality an anti-regular involution $\sigma$ on $G({\Bbb C})$.

  • $\begingroup$ Do you already know whether all bi-holomorphic group automorphisms are regular? do you know the answer to your question when $G$ is abelian? $\endgroup$
    – YCor
    Sep 23, 2019 at 19:00
  • $\begingroup$ @YCor: No, I do not know that. $\endgroup$ Sep 23, 2019 at 19:03
  • $\begingroup$ Actually, I work with anti-regular involutions. My working definition of a real algebraic group is: a pair $(G, \sigma)$, where $G$ is a complex algebraic group and $\sigma$ is an anti-regular involution. $\endgroup$ Sep 23, 2019 at 19:07
  • $\begingroup$ However, some other people (say, Dmitri Akhiezer), work with a pair $(G,\sigma)$, where $G$ is a complex semisimple group, and $\sigma$ is an anti-holomorphic involution. I would like to be sure that this is the same, at least for semisimple groups. $\endgroup$ Sep 23, 2019 at 19:11

1 Answer 1


(1): No; (2,3): Yes (and also for unipotent groups).

On the abelian group $\mathbb{G}_{\mathrm{a}}\times \mathbb{G}_{\mathrm{m}}=\mathbf{C}\times\mathbf{C}^*$, consider the anti-holomorphic involution $$(z,w)\mapsto (\bar{z},\exp(i\bar{z})\bar{w}):$$ it is not "anti-regular".

In the semisimple case, it's the same. One can reduce to $G$ simply connected, and in this case, the group of holomorphic automorphisms corresponds to the group of automorphisms of the Lie algebra, and this is the same as the algebraic automorphism group. Now, since there exists an algebraic real form, there exists at least one algebraic anti-regular automorphism, and hence the whole coset of anti-holomorphic automorphisms consists of anti-regular ones.

In the case of a torus $(\mathbf{C}^*)^d$, the answer is yes, and actually every (holomorphic or anti-holomorphic) endomorphism is regular or anti-regular. For this, it is enough to prove the case of $d=1$, and indeed every (anti)holomorphic endomorphism has the form $z\mapsto z^d$ or $z\mapsto \bar{z}^d$ for some $d\in\mathbf{Z}$.

The reductive case follows: every (holomorphic or anti-holomorphic) automorphism is regular or anti-regular (by acting on the derived subgroup on the one hand and the connected center on the other hand).

Also for $G$ unipotent, the (holomorphic or anti-holomorphic) automorphism group is the same as the automorphism group of the Lie algebra and hence acts (anti)-regularly.

  • $\begingroup$ Does the constructed anti-holomorphic involution on $({\bf C}^*)^2$ preserve the group structure? $\endgroup$ Sep 24, 2019 at 5:32
  • $\begingroup$ @MikhailBorovoi Good point, corrected (with conclusion changed). $\endgroup$
    – YCor
    Sep 24, 2019 at 9:33

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