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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?

Many thanks for your help!

glS
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Edward Hughes
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  • @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
  • And thanks for pointing me to the second definition... I can't believe I missed it... – rschwieb Feb 21 '13 at 17:25
  • @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
  • What do you mean "No"? That's exactly what my point is :P I think perhaps I need to see how you are using these two terms though... it sounds like we might be using the words in different ways. – rschwieb Feb 21 '13 at 20:30
  • 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
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    OK, i had been taking "R/C" representation to mean that the corresponding homomorphism was also R/C linear, but now I think I see you mean it to just indicate that it goes into $M_n(R)$ or $M_n(C)$ respectively. Thanks for your patience. Nixing the obsolete comments. – rschwieb Feb 21 '13 at 20:43
  • Yes, exactly that. – Edward Hughes Feb 21 '13 at 20:43

1 Answers1

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  1. Well, the main point is that that the finite-dimensional representations of the complex Lie group $SL(2,\mathbb{C})$ (and its corresponding complex Lie algebra $sl(2,\mathbb{C})$) by definition have to be complex manifolds. In particular, the representation maps should be holomorphic maps between complex manifolds. This rules out the complex conjugate representation $\bar{d}$ from the very beginning.

  2. Yes, $e$ and $\bar{e}$ are inequivalent representations.

Qmechanic
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