# Square root in complex reductive groups

Let $G$ be a connected complex reductive linear algebraic group. Does every $g\in G$ have a square root? (That is, some $a\in G$ such that $a^2=g$.)

• Using Jordan-decomposition it is easy to show that $\begin{pmatrix}-1&1\\0&-1\end{pmatrix}$ has no square root in $\mathrm{SL}(2,\mathbb{C})$.
– SHP
Aug 16, 2017 at 13:13
• The expression "complex reductive" may cause some confusion here. The term "reductive" was apparently introduced by Borel and Tits in their 1965 treatment of algebraic groups, where the unipotent radical is required to be trivial. But Jordan decomposition is not well-defined in an arbitrary real or complex Lie group. Aug 16, 2017 at 13:32
• @JimHumphreys I'm interested in connected complex linear algebraic groups for which every algebraic representation is isomorphic to a direct sum of irreducible subrepresentations.
– Pete
Aug 16, 2017 at 13:56
• @Pete: This just means reductive algebraic group over an algebraically closed field of characteristic 0 (but the complex numbers are unnecessary for this). Over any algebraically closed field, "(connected) reductive" translates structurally into the almost-direct product of a semisimple (connected) algebraic group and a central torus. But in prime characteristic the representation theory of a semisimple group is more complicated. Aug 16, 2017 at 15:27
• However, a weaker result is true. Let $k$ be a field, $G/k$ be a connected reductive linear algebraic group, $n$ be a positive integer. Then the power morphism $G\to G,x\mapsto x^n$ is dominant. That means the image is Zariski dense in $G$. Aug 18, 2022 at 12:53

As the comment shows the answer is negative in general. Perhaps it is worth to mention that for connected compact Lie groups the answer is yes, because its exponential map is surjective. In general, if the exponential map of a Lie group $G$ is surjective, then every group element $g \in G$ has a square root, i.e. an element $a \in G$ with $a^2 = g$, since $\exp(x)$ has $\exp(x/2)$ as a square root for any $x \in \mathfrak {g}=\operatorname{Lie}(G)$.
• Just stating the obvious for completeness sake: more generally, same argument above shows the existence of $n$-th roots in compact connected Lie groups, $n\in\mathbb{N}$.