# Clarification for the terminology "complex representation" of "real" Lie algebra

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?

• @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 $W$. Apr 8, 2021 at 22:37
• 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. Apr 8, 2021 at 22:53
• @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. Apr 9, 2021 at 4:28
• >> 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? Aug 5, 2021 at 1:54

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.
• 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? Apr 8, 2021 at 9:55
• 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. Apr 8, 2021 at 13:00
• 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 $x \in \mathfrak g$. Apr 8, 2021 at 22:26