Linear endomorphisms of sl(2,C) after forgetting its bracket

Let $\mathfrak g=\mathfrak{sl}_2(\mathbb C)$ , the complex vector space of traceless $2\times2$ matrices. As a complex vector space it has dimension $3$ , and as an underlying real vector space it has dimension $6$ . Consequently, after choosing bases,

\begin{aligned} \operatorname{End}_{\mathbb C\text{-}\mathbf{Vect}}(\mathfrak g)&\cong M_3(\mathbb C),& \operatorname{Aut}_{\mathbb C\text{-}\mathbf{Vect}}(\mathfrak g)&\cong GL_3(\mathbb C),\\ \operatorname{End}_{\mathbb R\text{-}\mathbf{Vect}}(\mathfrak g)&\cong M_6(\mathbb R),& \operatorname{Aut}_{\mathbb R\text{-}\mathbf{Vect}}(\mathfrak g)&\cong GL_6(\mathbb R). \end{aligned}

These are vector-space morphisms, without the requirement of preserving the Lie bracket.

Bases and scalar restriction

A complex basis is

E=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad F=\begin{pmatrix}0&0\\1&0\end{pmatrix},\quad H=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.

A real basis is $(E,F,H,iE,iF,iH)$ . Complex-linear maps form the real-linear maps commuting with the complex structure $X\mapsto iX$ . Entrywise conjugation is real-linear and invertible but does not commute with this complex structure.

Retaining the bracket

The Lie-algebra categories additionally impose $T([X,Y])=[T(X),T(Y)]$ . For example $T=2\operatorname{id}$ is an invertible linear map but is not a Lie-algebra homomorphism: the two sides are $2[X,Y]$ and $4[X,Y]$ . This distinguishes linear automorphisms from Lie-algebra automorphisms.

The group $SL(2,\mathbb C)$ itself is a different carrier and is not a vector space; see its real and complex group maps.