# $\mathfrak{sl}(2,\mathbb C)$ real v. complex

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.

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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.

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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