# Real forms of the general linear Lie superalgebra

I'm interested in a classification of the real forms of the general linear Lie superalgebra $$\mathfrak{gl}_{m|m}(\mathbb{C})$$.

The real forms of the simple complex Lie superalgebras were classified by Serganova in the paper Classification of real simple Lie superalgebras and symmetric spaces. For $$\mathfrak{gl}_{m|n}(\mathbb{C})$$, with $$m \ne n$$, we have a decomposition of Lie superalgebras $$\mathfrak{gl}_{m|n}(\mathbb{C}) = \mathfrak{sl}_{m|n}(\mathbb{C}) \oplus \mathbb{C}$$. Since $$\mathfrak{sl}_{m|n}(\mathbb{C})$$ is simple, and the only real form of the abelian complex Lie algebra $$\mathbb{C}$$ is the abelian real Lie algebra $$\mathbb{R}$$, we can use Serganova's classification to get all the real forms of $$\mathfrak{gl}_{m|n}(\mathbb{C})$$.

The situation for $$\mathfrak{gl}_{m|m}(\mathbb{C})$$ is more subtle, since it does not decompose as a direct sum of the corresponding simple Lie superalgebra $$\mathfrak{psl}_{m|m}(\mathbb{C})$$ and a center. There are natural enlargements of the real forms of $$\mathfrak{psl}_{m|m}(\mathbb{C})$$ to real forms of $$\mathfrak{gl}_{m|m}(\mathbb{C})$$. But it's not clear to me if one obtains all real forms of $$\mathfrak{gl}_{m|m}(\mathbb{C})$$ in this way.

This seems a natural enough question that I expected to find an answer in the literature that somehow bootstraps off the known classification of real forms for the simple Lie superalgebras. However, I haven't been able to locate it. I'm also interested in the analogous question for the Lie superalgebras of types $$P$$ and $$Q$$.

• As @YCor has hinted by his edit, this kind of problems are solved using Galois cohomology. You can read about that in Serre's book "Galois Cohomology", Sections III.1 and I.5. The case of the field $\Bbb R$ is covered also in Section 3 of this preprint. Dec 15, 2022 at 19:09
• If necessary, I can help you. Edit your your question: add a calculation of the automorphism group ${\rm Aut}({\mathfrak g})$ for your Lie algebra $\mathfrak{g}=\mathfrak{gl}_{m|m}(\mathbb{C})$. I am interested in even automorphisms only (preserving the grading). Then I will compute the real forms of $\mathfrak g$. Dec 15, 2022 at 19:18

It turns out that one does get all real forms of $$\mathfrak{gl}_{m|m}(\mathbb{C})$$ in the way described in the question. To see this, one can show that a real form of $$\mathfrak{gl}_{m|m}(\mathbb{C})$$ is uniquely determined by the induced real form of $$\mathfrak{psl}_{m|m}(\mathbb{C})$$. This same method also works for the Lie superalgebras of type $$P$$ and $$Q$$. I've written up the details in Appendix B of this paper.