4
$\begingroup$

Let su(2) be the Lie algebra of SU(2), thought of as a representation of SU(2) by conjugation.

Let S be the rank-2 symmetric tensors over the complex numbers, acted on by SU(2) in the obvious way.

I have seen it asserted that the complexification of su(2) is isomorphic (as a representation) to S, and presumably there is some simple explicit isomorphism that should be obvious. Alas, to me it is not obvious. What is that isomorphism?

1 Answers 1

2

The map $C^2 \otimes C^2 \ni v \otimes w \mapsto v w^t$ allows us to identify the space of rank-2 symmetric tensors with the space of symmetric $2\times 2$ matrices, which I'll call $Sym$. The corresponding action of an element $A \in SU(2)$ on such a matrix $B$ is then $ABA^t$. The complexification of $su(2)$ is $sl(2,C)$, the space of traceless $2\times 2$ complex matrices. An isomorphism of these two reps is given by the map $ sl(2,C) \to Sym, B \mapsto B \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right). $ That this is a map of representations follows from the fact that $ A^{-1} \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right) = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right) A^t $ for $A \in SU(2)$. This fact is easily verified by using the fact that every element in $SU(2)$ is of the form $ \left(\begin{array}{cc} z & w \\ -\bar w & \bar z \end{array}\right) $ with $|z|^2 + |w|^2 = 1$.

I found this isomorphism by just playing around, and I think I just got lucky. There is probably another one that is less ad-hoc.

  • 0
    Eric: Thanks for this terrific answer. I would vote it up if I could figure out how to get registered; the process seems to keep failing. I hope others will vote it up for me.2011-01-03