I am reading the book "Lie Groups, Lie Algebras and Representations: An elementary introduction" by B. C. Hall. There, the author defines what is meant by (finite-dimensional) representations of Lie groups and Lie algebras.

Particularly, he states:

Definition: If $\mathfrak{g}$ is a real or complex Lie algebra, then a finite-dimensional complex representation of $\mathfrak{g}$ is a a Lie algebra homomorphism $\pi: \mathfrak{g} \rightarrow \mathfrak{gl} \left( V \right)$, where $V$ is a finite dimensional complex vector space.

Here, $\mathfrak{gl} \left( V \right)$, is identified with the Lie algebra of matrices (by choosing a basis on $V$).

Now, the definition is clear when $\mathfrak{g}$ is a complex Lie algebra. However, if $\mathfrak{g}$ is a real Lie algebra, how does this definition work? For $\pi: \mathfrak{g} \rightarrow \mathfrak{gl} \left( V \right)$ to be a Lie algebra homomorphism, it is required that $\pi$ be a linear transformation first. But since $V$ is complex, $\mathfrak{gl} \left( V \right)$ is complex and therefore, $\pi$ cannot be a linear transformation (since the domain and the codomain are defined over different fields).

To make sense of this definition, I have two things in my mind:

(1) We consider the vector space $V$ to be real but with double dimension. However, this does not really suit the needs. For then, it coincides with the definition of a real representation.

(2) We only require $\pi$ to be real-linear, instead of a (truly) linear transformation. This would make more sense since then there would be a difference between real representations and complex representations of the same real Lie algebra.

Am I on the right path to understand the concept?

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    $\begingroup$ @AndreasCap has given the right answer, it is option 2. I would rephrase it like this: If you have a real Lie algebra $\mathfrak g$ and define a representation as a map $\pi: \mathfrak g \rightarrow End(V)$, then the map $\pi$ is only real-linear either way, because its domain does not even have complex scalars acting on it. But in a complex rep, for all $x \in \mathfrak g$, $\pi(x) \in End_{\mathbb C}(V)$ is complex-linear on a complex space $V$; whereas in a real rep, for all $x \in \mathfrak g$, $\pi(x) \in End_{\mathbb R}(W)$ is real-linear on a real space $W$. $\endgroup$ Apr 8, 2021 at 22:37
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    $\begingroup$ That being said, I think one should warn you already that later in the theory, among these complex representations of a real Lie algebra you are getting a grip on here, one can make a further threefold distinction into "("truly") complex", "pseudoreal" ("quaternionic") and "real" representations. The last ones are complex in the sense of your question, but "come from" real ones via tensoring -- and are somewhat rare among the complex ones ... The nomenclature is truly unfortunate and confusing. If you encounter that, see e.g. the intro here math.stackexchange.com/a/4026224/96384. $\endgroup$ Apr 8, 2021 at 22:53
  • $\begingroup$ @TorstenSchoeneberg: Thanks! I am sure I will require this additional nomenclature somewhere in future. Also, I get it now, the meaning of a complex representation of a real Lie algebra. $\endgroup$ Apr 9, 2021 at 4:28
  • $\begingroup$ >> The last ones are complex in the sense of your question, but "come from" real ones via tensoring --> Torsten - Do you mean that the "pseudoreal" ("quaternionic") and "real" representations among these complex representations of a real Lie algebra? $\endgroup$ Aug 5, 2021 at 1:54

1 Answer 1


The point here is what you mean by $\mathfrak{gl}(V)$. You don't really have to work with matrices here, since you can view $\mathfrak{gl}(V)$ as the space of linear maps from $V$ to itself with the commutator as a bracket, i.e. $[f,g]=f\circ g-g\circ f$. When talking about complex representations, you always require that $V$ is a complex vector space and that $\mathfrak{gl}(V)$ is interpreted as the space $\mathfrak{gl}_{\mathbb C}(V)$ of complex linear maps from $V$ to itself.

Now $\mathfrak{gl}_{\mathbb C}(V)$ is a complex Lie algebra and thus also a real Lie algebra. The standard convention then is that by a complex representation of a real Lie algebra $\mathfrak g$ one means a homomorphism $\mathfrak g\to \mathfrak{gl}_{\mathbb C}(V)$ of real Lie algebras. For a complex Lie algebra $\mathfrak g$, one defines a complex representation as a a homomorphism $\mathfrak g\to \mathfrak{gl}_{\mathbb C}(V)$ of complex Lie algebras, i.e. one in addition requires that the map $\mathfrak g\to \mathfrak{gl}_{\mathbb C}(V)$ is linear over the complex numbers. (So there is a difference between complex representations of a complex Lie algebra and complex representations of the underlying real Lie algebra.)

This can also be nicely expressed in the picutre of a representation via an action $\mathfrak g\times V\to V$. If $\mathfrak g$ is real, one requires that the action map is complex linear in the second varible, whereas if $\mathfrak g$ is complex, one requires the map to be complex bilinear.

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    $\begingroup$ What do we mean by $\mathfrak{gl}_{\mathbb{C}} \left( V \right)$ is a real Lie algebra? Won't it mean that as a vector space we are considering it to be defined on $\mathbb{R}$? If that is the case, we automatically imply that we are also considering $V$ to be a real vector space. Am I right? $\endgroup$ Apr 8, 2021 at 9:55
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    $\begingroup$ I don't see the need to consider $V$ as a real vector space. But since any complex vector space is a real vector space and any complex bilinear map is real bilinear any complex Lie algebra is a real Lie algebra. The space $L_{\mathbb C}(V,V)$ has an addition and a multiplication by real scalars, so it is a real vector space. And the commutator is bilinear over $\mathbb R$, skew symmetric and satisfies the Jacobi identity. So this is a real Lie algebra and no "interpretation" is needed. $\endgroup$ Apr 8, 2021 at 13:00
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    $\begingroup$ OP: Say $V$ has complex dimension $n$. If we considered it as a real vector space, then it would have real dimension $2n$, and we could consider the space $End_{\color{red}{\mathbb R}}(V)$ which would have real dimension $(2n)^2=4n^2$. But that is not what is meant here. Instead, we keep $V$ complex, we consider the space $End_{\color{red}{\mathbb C}}(V)$ which has complex dimension $n^2$, and consider that one as a real vector space: It has real dimension $2n^2$. In other words, the map $\pi$ is only real-linear, but $\pi(x) \in End(V)$ is complex-linear for every $x \in \mathfrak g$. $\endgroup$ Apr 8, 2021 at 22:26
  • $\begingroup$ @Torsten Schoeneberg - could you write your answer on this? I am confused by this answer.. $\endgroup$ Aug 5, 2021 at 1:24
  • $\begingroup$ How about the case where 𝑉 has only a real odd dimension 𝑛? $\endgroup$ Aug 5, 2021 at 1:30

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