# Conjugate Representations for $\mathfrak{sl}(2,\mathbb{C})$

Let $\mathfrak{sl}(2,\mathbb{C})$ be the complex Lie algebra of $SL(2,\mathbb{C})$ and $\mathfrak{sl}(2,\mathbb{C})_\mathbb{R}$ be its realification; that is $\mathfrak{sl}(2,\mathbb{C})_\mathbb{R}$ is $\mathfrak{sl}(2,\mathbb{C})$ considered as a real Lie algebra.

Let $d$ be an irrep of $\mathfrak{sl}(2,\mathbb{C})$ and $e$ an irrep of $\mathfrak{sl}(2,\mathbb{C})_\mathbb{R}$. Define the complex conjugate representations $\bar{d}$ and $\bar{e}$ in the usual way.

Am I right in thinking that $d$ and $\bar{d}$ are equivalent representations, which $e$ and $\bar{e}$ are inequivalent? My reasoning is as follows.

The irreps of $\mathfrak{sl}(2,\mathbb{C})$ are the spin-$j$ representations, unique in each dimension. The irreps of $\mathfrak{sl}(2,\mathbb{C})_\mathbb{R}$ are the restrictions of the irreps of $\mathfrak{sl}(2,\mathbb{C})\oplus \mathfrak{sl}(2,\mathbb{C})$, which are uniquely labelled by $(j_1,j_2)$, with the $(j_1,j_2)$ representation conjugate to the $(j_2,j_1)$ representation.

Further I assume that this reasoning can be extended to any complex (perhaps semisimple?) Lie algebra $\mathfrak{g}$. Would this be a fair conclusion?

@rschwieb: I didn't say that $e$ had to be a real representation. Real Lie algebras can happily have complex representations. One then uses the second definition of the three on the Wikipedia page. – Edward Hughes Feb 21 '13 at 16:06
@rschwieb: No - you can have complex representations that are only $\mathbb{R}$-linear; for example the $(j_1,j_2)$ representations of $\mathfrak{sl}(2,\mathbb{C})_\mathbb{R}$. – Edward Hughes Feb 21 '13 at 20:20
Yes - that's right. Sorry I was unclear! To my knowledge there's no better way to phrase it. I take the convention that a $k$-representation of $\mathfrak{g}$ is a representation of $\mathfrak{g}$ on some $k$-vector space. That is it's a $l$-linear homomorphism, where $l$ is the underlying field of the Lie algebra $\mathfrak{g}$. – Edward Hughes Feb 21 '13 at 20:39