# Transformation of self-dual and anti-self-dual tensors and irreducibility of representations

I am working out exercise 2.5 of Maggiore's book. Part of the exercise is the following:

Verify that self-dual and anti-self-dual tensors are irreducible representations of (real) dimension three of the Euclidean group $SO(4)$, and verify that the six-dimensional representation $A^{\mu\nu}$ of $SO(4)$ decomposes into its self-dual and anti-self-dual parts.

Consider an antisymmetry tensor $A^{\mu\nu}$. The self-dual tensors are tensors that satisfy:

$$\tilde A^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}\tilde A_{\rho\sigma}$$

while the anti-self-dual is,

$$\bar A^{\mu\nu}=-\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}\bar A_{\rho\sigma}$$

I could show that (if someone is interested I can put the calculation):

$$\tilde A'^{\mu\nu}=\frac{1}{2}\epsilon^{\mu\nu\rho\sigma}\tilde A_{\rho\sigma}'$$

i.e., the tensor $\tilde A'^{\mu\nu}$ transforms in the same way $\tilde A^{\mu\nu}$ a $SO(4)$ transformation. (Similarly for the anti-self-dual). In other words, a self-dual tensor in a given frame will be a (anti) self-dual tensor in a "rotated" frame. The author, in the solution, then claims:

This means that self-dual and anti-self-dual tensors are irreducible representations of $SO(4)$, and that in Euclidean space a six-dimensional real antisymmetric tensor $A^{\mu\nu}$ decomposes into its self-dual and anti-self-dual parts.

My question is: why is that true? For me, showing that tensors transforms in the same way just means that if they were irreducible representations before the transformation, they continue to be irreducible representations after the transformation.

If I am right, how to show explicitly that $\bar A^{\mu\nu}$ and $\tilde A^{\mu\nu}$ do provide irreducible representations?

Thanks.

The result you described says that the projection of antisymmetric rank 2 $SO(4)$ tensors onto self-dual and anti-self-dual subspaces commutes with the action of $SO(4)$. This just implies that the space of antisymmetric rank 2 tensors of $SO(4)$ is reducible.
To show that the self-dual and anti-self-dual subspaces are themselves irreducible, think of these rank 2 tensors as separate Lie algebras, and look for isomorphisms with $\mathfrak{su}(2)$ (recall that $\wedge^2V$ can be identified with $\mathfrak{so}(4)$, which decomposes into a direct sum).
• Thank you very much. One more question, please: do you know a direct way to show the irreducibility of the subspaces ? PS: what does $\wedge^2V$ mean? – aprendiz Feb 8 '16 at 5:49
• Well, irreducibility is equivalent in this case to the fact that $\mathfrak{su}(2)$ is a simple Lie algebra. So you can either use this fact, or prove it again by mapping whatever proof you like on the simplicity of $\mathfrak{su}(2)$ to the irreducibility of the self/anti-self-dual irreps of $SO(4)$. Also, $\wedge^2V=V\wedge V$ is just a name for the space of antisymmetric rank 2 tensors written in terms of the wedge product (used, for example, in differential forms). – TotallyRhombus Feb 8 '16 at 14:39