# I SO(2n) representation on n complex fields

1. Jul 30, 2017

### hideelo

If I have a lagrangian which has terms of the form $\Psi^{\dagger}_\mu \Psi^\mu$ then I can decompose the n complex $\Psi$ fields into 2n real fields by $\Psi_\mu = \eta_{2\mu+1} + i\eta_{2\mu}$. When I look at the lagrangian now it seems to have SO(2n) symmetry from mixing the 2n real fields.

Is there any obvious choice of representing SO(2n) on the n complex $\Psi$ fields explicitly?

2. Jul 31, 2017

### vanhees71

The SU(N) is a $N^2-1$-dimensional Lie group, while SO(2N) is $2N(2N-1)/2=2N^2-N$ dimensional. So SO(2N) is in this sense larger than SU(N).

3. Jul 31, 2017

### hideelo

Youre not wrong, but I dont see how this helps

4. Jul 31, 2017

### Avodyne

Well we could work out the generators. The SO(2N) generator mixing $\eta_{2\mu+1}$ and $\eta_{2\mu}$ becomes a U(1) generator acting on $\Psi_\mu$, and SO(2N) generators mixing even indexed $\eta$'s have the same form when acting on $\Psi$, and similarly for odd indexed $\eta$'s. Then there are the SO(2N) generators that mix an even $\eta$ and an odd $\eta$; I suspect will we need the charge-conjugation operator (which exchanges $\Psi$ and $\Psi^\dagger$) to express those ...

5. Aug 1, 2017

### king vitamin

I worked on a field theory with the same enlarged symmetry a few years ago, see here: https://arxiv.org/abs/1603.05652

For my purposes, I was largely interested in how the SU(N) irreps fit inside the larger SO(2N) irreps, see section IV (page 21). I have a mapping between the creation operators in the SU(N) and SO(2N) notation in equations 55-56, but I don't think my mapping implies a simple SO(2N) action on the complex field itself. So I don't have a good answer for you, but maybe looking at what I did will help.

Also, chapter 23 of Georgi's textbook is entirely dedicated to $SU(N) \subset SO(2N)$.