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.
Best Answer
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 $$ \mathfrak{g} = \{X \in M_n(\mathbb{C}) \mid e^{tX} \in G \; \forall\; t\in \mathbb{R}\}. $$ 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.