# When can the basis of a Lie algebra always be made Hermitian?

In Zee's book Group Theory in a Nutshell for Physicists, Chapter VI.3, it is stated that (quoted contents are in italic):

Consider a general Lie algebra with $$n$$ generators defined by

$$[X^a, X^b] = i {f^{ab}}_c X^c \tag{1}$$

A footnote says that here we only discuss the algebras of compact Lie groups. On the next page, Zee argues that the structure constants should be real:

Hermiticity of $$X^a$$ implies that the $${f^{ab}}_c$$ are real.

A few pages later, after converting to the Cartan-Weyl basis $$H^i, E_\beta$$ for which

$$[H^i, E_\beta] = \beta^i E_\beta \tag{16}$$

Zee again assumes that $$H^i$$ are Hermitian to show that if $$\beta$$ is a root vector, so is $$-\beta$$. So my questions are:

Can the generators $$\{X^a\}$$ of a Lie algebra, or the $$\{H^i\}$$ in the Cartan-Weyl basis (or the representation matrices of them) always be made Hermitian? Are there any restrictions (e.g. compact, semi-simple, connected etc) on the Lie group to make this true? Please kindly provide references to related theorems.

(I am a physics student, and I apologize if my statements are not clear or rigorous enough.)

• After some search in my textbook library, I find Chapter 16 of Pal's A Physicist's Introduction to Algebraic Structures very useful, and maybe suitable for people with a high tolerance of "physicists' level of rigorosity". Theorem 16.6 there states that Every finite-dimensional representation of a compact Lie group is equivalent to a unitary representation (with a proof). Aug 3, 2022 at 0:57
• For the simplest case of $\mathfrak{su}_2$, cf. math.stackexchange.com/q/3318978/96384 Aug 9, 2022 at 23:56

To write things in mathematical language and notation here, we are considering an $$n$$-dimensional matrix Lie algebra $$\mathfrak{g} \subseteq \mathfrak{gl}_d(\mathbb{C})$$ (to a mathematician a "general Lie algebra" would not come equipped with such an embedding, and "$$n$$ generators" would mean something quite different). The question is when up to conjugacy such a Lie algebra can be chosen to lie in the unitary Lie algebra $$\mathfrak{u}(d)$$ of skew-Hermitian matrices (to get Hermitian matrices we need to multiply everything by $$i$$ all the time which is annoying and I don't recommend doing it; that's where the factor of $$i$$ in the first expression comes from).
The answer is that $$\mathfrak{g}$$ needs to be compact in the sense that it integrates to a compact subgroup $$G \subseteq GL_d(\mathbb{C})$$ (this is slightly different from what "compact Lie algebra" means abstractly; this is the precise meaning of Zee's footnote). When this is true we can apply the unitary trick, starting with the standard inner product on $$\mathbb{C}^d$$ and averaging it over $$G$$ with respect to Haar measure to get a $$G$$-invariant inner product, which (after finding an orthonormal basis of this new inner product) conjugates $$G$$ into the unitary group $$U(d)$$ and hence which conjugates its Lie algebra into the unitary Lie algebra.
• When you say "an $n$-dimensional matrix Lie algebra $\mathfrak{g} \subseteq \mathfrak{gl}_d(\mathbb{C})$", do you actually mean $\mathfrak{gl}_n(\mathbb{C})$?. Aug 3, 2022 at 0:42
• @Zhengyuan: no. $\mathfrak{g}$ is $n$-dimensional but that doesn't constrain the ambient $\mathfrak{gl}_d$ except that we need $d^2 \ge n$. Aug 3, 2022 at 9:32