Definition: ([Hall] Definition 3.18) Let $G$ be a matrix Lie group. The Lie algebra of $G$, denoted $\mathfrak{g}$, is the set of all matrices $X$ such that $e^{tX}$ is in G for all real numbers $t$.

This Lie algebra is defined over $\mathbb{R}$. If I choose a basis $\{X_a\}_{a=1}^d$ in $\mathfrak{g}$ (here $d \equiv \dim \mathfrak{g}$), then any element $g$ in the identity component of $G$ can be expressed as

$$ g = \exp(\sum_{a=1}^d \theta_a X_a), \quad \theta_a \in \mathbb{R} $$

In the complexified Lie algebra $\mathfrak{g}_\mathbb{C}$, an arbitrary element can be written as

$$ X = \sum_{a=1}^d \theta_a X_a, \quad \theta_a \in \mathbb{C} $$

Note that now $\theta_a$ can be complex. Then we can construct the identity component of another matrix Lie group $G'$ as

$$ G' = \{e^X \mid X \in \mathfrak{g}_\mathbb{C}\} $$

The difference between $G$ and $G'$ is whether the parameters $\{\theta_a\}$ can be complex. In general, what is the relation between (the identity components of) $G$ and $G'$? When are they guaranteed to be the same?

[Hall]: Brian C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction (Second Edition)

Edit: Removed an erroneous statement according to Qiaochu Yuan's answer.

  • $\begingroup$ You cannot write every element in the identity component of $G$ as an exponential. This works for example when $G$ is compact or nilpotent but not in general. Instead every element is a product of exponentials. $\endgroup$
    – Callum
    Commented Aug 6, 2022 at 13:45
  • $\begingroup$ @Callum Could you please provide an example of matrix Lie group such that its identity component cannot be written as $\exp(\cdots)$ but only $\prod \exp(\cdots)$? $\endgroup$ Commented Aug 7, 2022 at 0:34
  • 1
    $\begingroup$ $SL(2,\mathbb{C})$ (or $SL(2,\mathbb{R})$) is connected but it contains $\begin{pmatrix}-1 & 1\\0&-1\end{pmatrix}$ which is not the exponential of a tracefree matrix $\endgroup$
    – Callum
    Commented Aug 7, 2022 at 0:49

1 Answer 1


They are never the same (unless $G$ is zero-dimensional). Even if $G$ is already a complex Lie group you make no use of that fact in the construction; you're choosing a real basis of its Lie algebra, so when you complexify the Lie algebra you always get another Lie algebra and hence another Lie group of twice the (real) dimension.

For example, sticking only to the level of Lie algebras for simplicity, the complexification of $\mathfrak{sl}_2(\mathbb{R})$ is $\mathfrak{sl}_2(\mathbb{C})$ as you might expect. But $\mathfrak{sl}_2(\mathbb{C})$ can itself be regarded as a real Lie algebra (sitting inside $M_4(\mathbb{R})$) and then complexified again; the result is $\mathfrak{sl}_2(\mathbb{C}) \times \overline{\mathfrak{sl}_2(\mathbb{C})}$, where the $\overline{(-)}$ indicates that scalar multiplication has been modified by a complex conjugate. However this turns out not to matter in this case because we have $\overline{\mathfrak{sl}_2(\mathbb{C})} \cong \mathfrak{sl}_2(\mathbb{C})$.


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