Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm a Lie theory novice, so please bear with me.

My understanding is that the Lie algebra $\mathfrak g$ of a matrix Lie group $G$ is the pair $(V, [\cdot, \cdot ])$ where $V$ is the real vector space over the set of all matrices $X$ for which $e^{tX}\in G$ for all $t\in\mathbb R$, and $[\cdot, \cdot]$ is the matrix commutator.

This leads me to believe that the Lie algebra $\mathfrak{sl}(2,\mathbb C)$ of the matrix Lie group $\mathrm{SL}(2, \mathbb C)$ is the pair $(V, [\cdot, \cdot])$ where $V$ is a real vector space over the set of traceless, $2\times 2$ complex matrices with matrix commutator.

However, it seems to me common that the symbol $\mathfrak{sl}(2,\mathbb C)$ is used to refer to a complex Lie algebra. Is it common to simply extend the field to $\mathbb C$ and call the resulting Lie algebra $\mathfrak{sl}(2,\mathbb C)$? Does the terminology depend on the context?

Thanks for the help.

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

First: The definition of Lie groups and a Lie algebras can vary depending on who you ask.

Given a (matrix) Lie group $G$ (so $G\subseteq GL_n(\mathbb{C})$), the Lie algebra of $G$ is the set $$ \mathfrak{g} = \{X \in G \mid e^{tX} \in G \; \forall\; t\in \mathbb{R}\}. $$ This is clearly then a real vector space. But it isn't necessarily a complex vector space. We always get a real Lie algebra. We only get a complex Lie algebra if $iX\in \mathfrak{g}$ for all $X\in \mathfrak{g}$. So just because the entries are complex, doesn't mean that the Lie algebra is complex.

Specifically considering $\mathfrak{sl}_2(\mathbb{C})$ you get the set of $2\times 2$ matrices with complex entries of trace zero. Again this is automatically a real vector space, but we can try to check if it is a complex vector space. We check that $$ i\pmatrix{a & b \\ c & -a} = \pmatrix{ia & ib \\ ic & -ia}. $$ This again has trace zero, so indeed $\mathfrak{sl}_2(\mathbb{C})$ is a complex Lie algebra. We usually then call the Lie group complex if the Lie algebra turns out to be a complex Lie algebra. Other complex Lie groups are $GL_n(\mathbb{C}), SL_n(\mathbb{C})$, $SO_n(\mathbb{C})$, and $Sp_n(\mathbb{C})$.

Another example: Consider the Lie group $SU(n)$ of all $n\times n$ unitary matrices (with entries from $\mathbb{C}$) with determinant $1$. In this case you can find that the Lie algebra $\mathfrak{su}(n)$ is the space of all $n\times n$ complex matrices $X$ where $X^* = -X$ ($*$ being complex conjugate transposed) and with trace $0$. This is not a complex Lie algebra, but only a real Lie algebra. So $SU(n)$ is not a complex Lie group.

Hopefully I didn't say anything wrong.

share|improve this answer
Thanks Thomas. Just to be sure, what I gather is that when we write the symbol $\mathfrak{sl}(2,\mathbb C)$, we usually consider the underlying vector space to be $\mathbb C$ as a matter of convention since the constraint that characterizes the set is not spoiled by extending to complex multiplicaton even though the original construction from the group resulted in a real vector space, and this is the general convention for other Lie algebras arising from matrix Lie groups. Do you know of a reference that discusses these conventions btw? Thanks again. –  joshphysics Feb 18 '13 at 19:37
@joshphysics: Yes, I would say that $\mathfrak{sl}(2,\mathbb{C})$ is considered as a complex vector space. As for a reference. There are several good books out there. One I can think of is Brian Hall: "Lie Groups, Lie Algebras, and Representations", but I can't remember how much it discusses real vs. complex Lie groups. –  Thomas Feb 18 '13 at 19:41
Ok yeah I have actually been using Hall which I like quite a bit, but it's not the most descriptive on this point (as far as I have been able to tell). Thanks once again. –  joshphysics Feb 18 '13 at 19:43
Yet another aspect that adds to the terminological confusion is that representations of real Lie groups/algebras are often on complex vector spaces... so then it can be convenient to "complexify" the Lie algebra/group. Thus, ${\frak s}{\frak l}(2,\mathbb R)$ becomes ${\frak s}{\frak l}(2,\mathbb C)$, and complexifying the "real" Lie algebra ${\frak s}{\frak l}(2,\mathbb C)$ produces two_copies of itself. –  paul garrett Feb 18 '13 at 20:42
@paulgarrett Thanks for that! I was actually wondering yesterday what would happen if one were to complexify $\mathfrak{sl}(2,\mathbb C)$ considered as a Lie algebra over the reals. Then I stopped thinking about it because I heard my adviser's voice saying "you're a physicist, stop thinking about this and do research" –  joshphysics Feb 18 '13 at 22:33
show 3 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.