Root system of fixed point Lie sub-algebra

Question

It is known that a non-simply laced simple root system can be constructed from the simply-laced root system by folding the Dynkin diagram and hence the corresponding non-simply-laced Lie algebra can be constructed by taking the fixed points of a non-trivial diagram automorphism (outer automorphism).

Now let $\theta$ be an inner automorphism of order $2$ of a simple Lie algebra $\mathfrak{g}$ over $\mathbb C$ and let $\mathfrak{g}= \mathfrak{g}_0 \oplus \mathfrak{g}_1$ be the eigendecomposition. The fixed point subalgebra $\mathfrak{g}_0$ is reductive. Now is there a way to get the root system of $\mathfrak{g}_0$ from the root system of $\mathfrak{g}$?

Answer 1

Let ${\frak g}$ be a simple Lie algebra over $\Bbb C$, and let $\theta$ be an inner involution of ${\frak g}$, that is, an inner automorphism of ${\frak g}$ of order dividing 2. Such automorphisms are classified by Kac labelings of the extended Dynkin diagram ${\widetilde D}={\widetilde D}({\frak g})$.

We fix a Cartan subalgebra ${\frak t}\subset{\frak g}$ and a Borel subalgebra ${\frak b}\supset {\frak t}$ and consider the Dynkin diagram $D({\frak g})=D({\frak g},{\frak t},{\frak b})$, whose vertices are the simple roots $\alpha_1,\dots,\alpha_\ell$. We consider also the extended Dynkin diagram ${\widetilde D}$ whose vertices are $\alpha_1,\dots,\alpha_\ell$ and $\alpha_0$, where $\alpha_0$ is the lowest root (the opposite to the highest root). There is a unique linear relation $$m_0\alpha_0+m_1\alpha_1+\dotsb+m_\ell\alpha_\ell=0$$ normalized such that $m_0=1$. It is easy to see that the numbers $m_i$ are positive integers; we write them near the vertices of the extended Dynkin diagram. See Table 6 in Lie groups and algebraic groups by Onishchik and Vinberg, or Table 1 in Section 9 of the paper Galois cohomology of real semisimple groups via Kac labelings by Borovoi and Timashev.

Example: Here I give the coefficients $m_i$ for the extended Dynkin diagram of type ${\sf E}_7$. The extreme right-hand vertex corresponds to the lowest root $\alpha_0$.

A Kac labeling of ${\widetilde D}$ is a family of nonnegative integers ${\bf q}=(q_0,q_1,\dots,q_\ell)$ satisfying $$ m_0q_0+m_1q_1+\cdots+m_\ell q_\ell=2.$$ Clearly we have $q_i\le 2$ for all $i$, and there can be either one vertex (type I) $\alpha_i$ with nonzero $q_i$, or two such vertices (type II). See Table 7, Types I and II, in the book by Onishchik and Vinberg, where the list of all possible Kac labelings is given. For a Kac labeling $\bf q$, the vertices with $q_i\neq 0$ are painted in black in this table (this determines uniquely the values of $q_i$ for all vertices $\alpha_i$).

According to Victor Kac (1969), the inner involutions $\theta$ of ${\frak g}$ (up to conjugation by automorphisms of ${\frak g}$) correspond bijectively to the Kac labelings of ${\widetilde D}({\frak g})$ (up to automorphisms of ${\widetilde D}({\frak g})$). The fixed subalgebra ${\frak g}^\theta$ for $\theta=\theta({\bf q})$ is reductive. It is semisimple for Kac labelings ${\bf q}$ of type I, and has one-dimensional center for ${\bf q}$ of type II. The Dynkin diagram of the derived subalgebra of ${\frak g}$ is obtained from the extended Dynkin diagram ${\widetilde D}$ by removing the black vertices. See Table 7 in the book by Onishchik and Vinberg.

Example 1. Here $\theta$ corresponds to the real form $\sf EVI$ of ${\sf E}_7$. We have ${\frak g}^\theta={\sf A}_1\oplus{\sf D}_6$.

Example 2. Here $\theta$ corresponds to the real form $\sf EVII$ of ${\sf E}_7$. We have ${\frak g}^\theta={\sf E}_6\oplus {\Bbb C}$.

Kac classified all automorphisms of ${\frak g}$ (inner and outer) of any finite order $r$. See Chapter 3, Section 3 in: V. V. Gorbatsevich, A. L. Onishchik, and E. B. Vinberg, Structure of Lie groups and Lie algebras, Lie Groups and Lie Algebras III, Encyclopaedia of Mathematical Sciences, Vol. 41, Springer–Verlag, Berlin, 1994.

For an outer involution $\theta$ of ${\frak g}$, the Lie algebra ${\frak g}^\theta$ is semisimple. The Dynkin diagram of ${\frak g}^\theta$ is obtained by removing the black vertex corresponding to a Kac labeling from the affine Dynkin diagram of $({\frak g},\theta)$; see Table 7, Type III in the book by Onishchik and Vinberg. In particular, it is not true that for all outer involutions $\theta$, the Dynkin diagram of ${\frak g}^\theta$ is obtained by folding the Dynkin diagram of ${\frak g}$.

Acknowledgements. To draw the extended Dynkin diagrams for this answer, the dynkin-diagrams package of @BenMcKay was used.

EDIT. I describe an inner automorphism $\nu_{\bf q}$ of $\frak g$ corresponding to a Kac labeling ${\bf q}=(q_0,q_1,\dots,q_\ell)$ of $\widetilde D$. Let $G$ be the connected semisimple $\Bbb C$-group of adjoint type with Lie algebra $\frak g$ (we may take for $G$ the identity component of ${\rm Aut}\, \frak g$). Let $T\subset G$ be the maximal torus of $G$ with Lie algebra $\frak t$. Let $t\in T$ be the element such that $\alpha_i(t)=(-1)^{q_i}$ for all simple roots $\alpha_i\colon T\to{\Bbb C}^\times$. Then we associate to $\bf q$ the conjugacy class of the inner automorphism $\nu_{\bf q}:={\rm Ad}(t)$ of $\frak g$.

I describe the action of $\nu_{\bf q}$ on $\frak g$. We have the root decomposition $${\frak g}={\frak t}\oplus\bigoplus_{\beta\in R} {\frak g}_\beta\, ,$$ where $R=R(G,T)$ denotes the root system. Then $\nu_{\bf q}$ acts on $\frak t$ trivially, and it acts on the root subspace ${\frak g}_\beta$ as $(-1)^{s_\beta}$, where $$ s_\beta =c_1 q_1+\dots+c_\ell q_\ell\quad\text{for} \quad \beta=c_1\alpha_1+\dots+c_\ell\alpha_\ell$$ with $c_i\in {\Bbb Z}$.

Concerning proofs: see, for instance, Borovoi and Timashev, 2015. This is much easier to read than our 2021 Transformations Group paper referred to above.

Answer 2

I haven't yet got an idea of what sort of answer would be satisfactory; it probably depends on how you are thinking of the automorphisms. Here's one attempt, just to have something written; it is entirely elementary, so probably unsatisfactory, but you can let me know how it falls short of the goal, and we can see if it can be fixed.

You probably think of a diagram automorphism as completely determined by the permutation it induces on the simple roots, and then the roots of the fixed-point Lie algebra are in bijection with the orbits on the full root datum. (To be clear, one cannot simply arbitrarily permute the simple roots, but knowing what a diagram automorphism does on simple roots determines its behaviour on the whole Lie algebra.)

An inner automorphism does not permute roots at all (at least if you measure roots with respect to a torus pointwise fixed by the inner automorphism). If you're looking for a way of parameterising such an automorphism similar to the outer-automorphism description, then you might think of it as completely specified by a scalar for each simple root—namely, the scalar by which the inner automorphism acts on the corresponding root space. These scalars can be specified arbitrarily, and they determine the behaviour of the automorphism on every root space—namely, having attached the scalars $(c_\alpha)_{\alpha \in \Delta}$ to the simple roots, we attach to the root $r = \sum n_\alpha\alpha$ the scalar $c_r = \prod c_\alpha^{n_\alpha}$. For you, the various $c_\alpha$ will be $\pm1$. Then the roots of the fixed-point Lie algebra are precisely those for which the associated scalar is $1$.

Answer 3

This is covered in detail in Endomorphisms of Algebraic Groups by Robert Steinberg, AMS Memoirs #80, 1968. What you are getting is a quotient of the root system (obtained by restricting roots to $H^\theta$, or equivalently looking at $\theta$-orbits of roots). The book by Onischik and Vinberg (Lie Groups and Algebraic Groups, 1990) has a nice treatment (see Table 6 in the references).

In particular you get non-simply laced root systems from outer automorphisms of simply laced ones, for example $D_4\rightarrow G_2$, $E_6\rightarrow F_4$, $D_n\rightarrow B_{n-1}$. The last one is easy to see: $SO(2n-1)$ is the (identity component of the) fixed points of an outer automorphism of $SO(2n)$.