Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.