In the book Brian C Hall, Lie Groups, Lie Algebras, and Representations, a complex matrix Lie group is a matrix Lie group $G \subset \mathrm{GL}_n(\mathbb{C})$ whose Lie algebra $\mathfrak{g} \subset \mathfrak{gl}_n(\mathbb{C})$ is a subspace of $\mathfrak{gl}_n(\mathbb{C})$ over $\mathbb{C}$, not only over $\mathbb{R}$.

Question: we know that a Lie group homomorphism $\phi : G \to H$ gives a Lie algebra homomorphism $\bar{\phi} : \mathfrak{g} \to \mathfrak{h}$ over $\mathbb{R}$. Is it true that $\bar{\phi}$ is linear over $\mathbb{C}$ if $G, H$ are complex matrix Lie groups?

If it is not true in general, then for a complex representation $\rho : G \to \mathrm{GL}_n(\mathbb{C})$ of a complex matrix Lie group $G$, $\bar{\rho} : \mathfrak{g} \to \mathfrak{gl}_n(\mathbb{C})$ is not a Lie algebra representation over $\mathbb{C}$. If we want it over $\mathbb{C}$, then we need to introduce another imaginary unit $j$ to construct $\mathfrak{g}_{\mathbb{C}}$ and $\bar{\rho} : \mathfrak{g}_{\mathbb{C}} \to \mathfrak{gl}_n(\mathbb{C})$. However, I don't think such a discussion is not in the book so I am confused (see Propositions 3.38 and 4.6).


2 Answers 2


It is unfortunately true that the introductory literature on Lie groups flips back and forth between complex Lie groups (and Lie algebras), and real ditto. Further, there is "restriction of scalars" which is a "functor" that "forgets" that a complex Lie group is a complex manifold, etc., and only "remembers" that it is a real manifold, etc. And, oppositely, there is "complexification" of a real Lie group/algebra.

(A possibly counter-intuitive aspect to the situation is that a complex Lie group/algebra, with scalars restricted to $\mathbb R$, and then complexified, is not at all the original thing! These two functors are adjoints, but not "inverses".)

It depends on the context to know whether complex Lie groups/algebras are meant to be thought of (by restriction of scalars) as real Lie groups/algebras. Sometimes this happens without any explicit mention!

But in the context of the first/basic classification result on simple Lie algebras, surely everything is meant to be complex, and the maps are complex-linear/holomorphic...

  • $\begingroup$ Thank you for your answer, paul! So a group hom should be holomorphic if I desire to obtain a Lie algebra hom over C from it. Otherwise, there is a counterexample tkf suggested below. $\endgroup$ Jan 21, 2022 at 8:31
  • $\begingroup$ @MathDrifter Right-o. :) $\endgroup$ Jan 21, 2022 at 18:09

We have a Lie group homomorphism $\phi\colon GL_1(\mathbb{C})\to GL_1(\mathbb{C})$ sending $$\phi\colon (z)\mapsto (z/|z|).$$

Then $\bar\phi$ is not linear over $\mathbb{C}$. Identifying the Lie algebra with $\mathbb{C}$ in the natural way, we have: $$\phi\colon \begin{array}{ccc}1&\mapsto& 0 \\ i&\mapsto& i\end{array}$$


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