Is there a known set $S=\{x \in G: x^2=1, x\ne1\}$ of elements of a simple compact Lie group $G$ ? By simple compact Lie group I consider $SO_n$, $SU_n$, $Sp_n$, $G_2$, $F_4$, $E_6$, $E_7$, $E_8$. (Feel free to extend this definition in case you find it useful). What are the conjugacy classes in this set and their dimensions ? By "conjugacy class" I mean any set $\{gxg^{-1}\}_{g\in G}$ for $x$ in above set $S$.

For example in $SO_8$ the dimensions are 0, 12, 16, 12. In $O_8$ we should add 7, 15, 15, 7. These are dimensions of grassmanians $G_{k,8-k}$. For spin groups it would be oriented grassmanians. For given grassmanian element $K$ we have the map in $O_8$ which reverse $K$ and fix perpendicular space.

I am especially interested in the answer for exceptional Lie groups $F_4, E_6, E_7, E_8$. Elements which square to one I would call "octonion reflections" there. The conjecture is that the conjugacy classes of involutions correspond to Riemanian symmetric spaces.

I would add later more information on how to define such elements in spin groups.

By another occasion I am also interested in set $\{x^2=-1\}$. Element $-1$ in the group can be defined as order 2 element in the center. This question is only valid for groups having $-1$. This should be another question, I suppose. For mentioned group $SO_8$ this set is 12-dimensional, equal to complex structures on $\mathbb R^8$. By looking at Clifford algebra $C_6$ we can prove that it is equal to oriented grassmanian $G_{2,6}^+$.

EDIT: I am mainly interested in answer for groups $SO_n$, $SU_n$, $Sp_n$ and exceptional ones listed above. The "conjecture" is that for three families there are subsets of dimensions $n, 2n, 4n$ corresponding to projective spaces $\mathbb RP^n, \mathbb CP^n, \mathbb HP^n$. For exceptional ones the sets correspond to exceptional symmetric spaces embedded in the group (all of them or some of them ?).

EDIT 2: By looking at this wikipedia list of all Riemannian symmetric spaces we can check whether each such space can be embedded in proper Lie group and what are the properties of this set in the group. For example let's count dimension of quaternion structures on $\mathbb C^4$. Fix the point $v_0$ in $\mathbb CP^3$. Quaternion structure is the element $j$ in $U_4$ which square to $-1$ and map each complex plane to perpendicular one. We have 4 dimensions to select $w$ in $\mathbb CP^2$ perpendicular to $v_0$. Additional $5^{th}$ dimension is allowed to rotate image of $v_0$ in $w$. In space $\mathbb C^2$ perpendicular to $<v_0,w>$ map $j$ has also one more dimension of freedom, but we should subtract one dimension, because the circle $e^{i\alpha}j$ gives the same quaternion structure (is it clear ?). This gives $AII$ symmetric space $SU_4/Sp_2$.

Here is link to my previous question in this subject.

Regards, Marek

  • $\begingroup$ Any restriction on the compact Lie group? otherwise this includes all finite groups. On the other hand restricting to connected compact Lie groups discards natural examples such as $O(n)$. $\endgroup$
    – YCor
    Aug 31, 2016 at 18:33
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    $\begingroup$ Anyway, a comment is that this set (for an arbitrary compact Lie group) consists of finitely many conjugacy classes (see D.H. Lee, T.S. Wu. On conjugacy of homomorphisms of topological groups. Illinois J. Math. 13 1969 694–699; projecteuclid.org/euclid.ijm/1256053429). $\endgroup$
    – YCor
    Aug 31, 2016 at 18:39
  • $\begingroup$ Two minor comments: 1) It probably helps to define "simple" more precisely. 2) I guess you don't intend to regard the identity element of the group as an involution even though it belongs to the set you define. $\endgroup$ Sep 1, 2016 at 13:45
  • $\begingroup$ I tried to clarify although I do not guarantee it makes you happy. Please use definition of "simple Lie group" whatever is more suitable for you and whatever interesting can be said about set $\{x^2=1\}$ for such a group. $\endgroup$
    – user21230
    Sep 1, 2016 at 14:23

1 Answer 1


For the case of a connected semisimple compact Lie group $G$, see this preprint, Section 3, where, following ideas of Kac and Vinberg, we describe set of conjugacy classes of $n$-th roots of a given central element $z$ of $G$. You should take $n=2$ and $z=1$ (or $z=-1$ in your sense).

EDIT: I can prove your conjecture that the conjugacy class of an involution $x$ in a connected, semisimple, compact Lie group $G$ is a Riemannian symmetric space of $G$. It suffices to assume that $x^2\in Z(G)$ but $x\notin Z(G)$, where $Z(G)$ denotes the center of $G$.

The conjugacy class $X$ of $x$ in $G$ is clearly $G/H$, where $H$ is the centralizer of $x$ in $G$. Consider the automorphism $i_x$ of $G$ defined by $$ i_x(g)=xgx^{-1}.$$ Our conditions on $x$ guarantee that that $(i_x)^2=1$, but $i_x\neq 1$. Clearly $$ H=\{g\in G\ |\ i_x(g)=g\}. $$ Thus $H$ is the subgroup of fixed points of an involutive automorphism of $G$, which means that $X=G/H$ is a Riemannian symmetric space of $G$.

In this way we obtain all Riemannian symmetric spaces of $G$ of inner type, i.e., the homogeneous spaces $X=G/H$, where $H$ is .the group of fixed points of some inner involutive automorphism of $G.$ See Helgason's book "Differential geometry, Lie groups, and symmetric spaces" for a list of inner and outer involutive automorphisms and the corresponding symmetric spaces. Note that if G is one of the exceptional simple groups $E_7,\, E_8,\, F_4,\, G_2$, then there are no outer involutive automorphisms of $G$ (because the Dynkin diagram of $G$ has no involutive automorphisms), and therefore, in this way we obtain all Riemannian symmetric spaces of $G$. The same refers to ${\rm Sp}_n$ and ${\rm SO}_{2n+1}$.

  • $\begingroup$ Your question is related to real Galois cohomology, see the arXiv preprint by Borovoi and Evenor. $\endgroup$ Sep 1, 2016 at 4:27
  • $\begingroup$ In the case when your group is of adjoint type, the question is related to classification of inner automorhisms of finite order of simple Lie algebras, see results of Kac in a book of Helgason. $\endgroup$ Sep 1, 2016 at 4:32
  • $\begingroup$ Magnificent ! How about E6 ? Which symmetric spaces can be obtained in this way ? I obtained $EIII$ 32-dim. Products of two commuting elements there land either in EIII or in ... other one. I don't have my notes now. $\endgroup$
    – user21230
    Sep 2, 2016 at 5:41
  • $\begingroup$ In this way you obtain $EII$ and $EIII$. $\endgroup$ Sep 2, 2016 at 6:59
  • $\begingroup$ This agree with my notes. Product of two commuting.elements in EIII is either in EIII or in EII. For a in EIII what is the dimension of such b that ab is in EII - I try to figure out this. In FII =OP2 I conjecture that product of some four elements gives FI element. I am behind in constructing E7 and E8. $\endgroup$
    – user21230
    Sep 2, 2016 at 7:42

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