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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 …

No. Take $g_2=1$, $g_1\notin T$, then $g_1 T\cap g_2 T=g_1T\cap T=\emptyset$. The differential of the map $$\phi\colon T\times T\to G,\quad (t_1,t_2)\mapsto g_1 t_1 g_2 t_2=g_1t_1t_2$$ at the point $( …

Yes, the toral component of a connected Lie group is equal to the toral component of its solvable radical. Let $G$ be a Lie group, $S$ its solvable radical, and $\mathrm{TC}(G)$ denote the toral co …

Concerning rational parametrizations of algebraic groups similar to the Cayley transform see this answer, which deals with Cayley maps, i.e., equivariant birational isomorphisms between an algebraic g …

The irreducible complex representations of the simply connected simple group $G=Sp_{r,{\mathbb C}}$ of type $C_r$, for $r>1$, of dimension $n<{\rm dim}\ G$ are listed in the paper of Andreev, Vinberg …

Consider the subgroup $N:=\langle k\rangle\subset \mathbb{Z}^n$. There exists a basis $f_1,\dots,f_n$ of $\mathbb{Z}^n$ such that $uf_1$ is a basis of $N$, where $u\in \mathbb{Z}$, $u>0$, see Vinberg, …

There is NO such example. Note that any semisimple algebraic ${\mathbb{R}}$-group $H$ of Hermitian type has a compact (anisotropic) maximal torus. Indeed, by a definition of a group of Hermitian t …

$\DeclareMathOperator{\Tr}{Tr} \DeclareMathOperator{\ad}{ad} \DeclareMathOperator{\Lie}{Lie} \newcommand{\g}{{\mathfrak g}} \newcommand{\z}{{\mathfrak z}} \newcommand{\s}{{\mathfrak s}} \newcommand{\O …

$ \newcommand{\g}{{\mathfrak g}} \newcommand{\h}{{\mathfrak h}} \newcommand{\t}{{\mathfrak t}} \newcommand{\C}{{\mathbb C}} \newcommand{\Ad}{{\rm Ad}} \newcommand{\ad}{{\rm ad}} $Theorem. Let $G$ be …

The abstract subgroup generated by $H$ and $K$ is closed. We may assume that $G$ is connected. The groups $G$, $H$, $K$ are the groups of real points of real algebraic groups $\mathbf{G}$, $\mathbf{H …

See Bourbaki, Groupes et algèbres de Lie, IV.2.7, Corollary of Theorem 5 (due to Tits). We assume that $K$ has at least four elements and that the Lie algebra is absolutely simple. We assume also th …

The answer is YES. It suffices to assume that $H_1$ and $H_2$ are conjugate over $\mathbb{C}$ or, what is the same, that they are conjugate over $\overline{\mathbb{Q}}$. Theorem 1. Let $G$ be a co …

Let $\psi\colon B\to C$ be a homomorphism of real Lie groups, where the group $C$ is connected. Let $B^0$ denote the identity component of $B$, and we set $\pi_0(B)=B/B^0$, then $\pi_0(B)$ is a discr …

Irreducible real representations of complex type of a compact group correspond to irreducible complex representations that do not admit an invariant bilinear form. Irreducible real representations of …

(I add details to my comments.) The answer depends on $n=4r$. Write $G=Sp(r)/\mu_2$. If $r=1$, then $G\simeq SO_3$, so $G$ admits a faithful 4-dimensional representation into $SO_4$. Similarly, if $r …