# How to find the full global symmetry group of a Lagrangian of $N$ complex scalars?

I have the Lagrangian $$\mathcal{L} = \frac{1}{2}D_\mu \Phi^\dagger D^\mu \Phi - \frac{m^2}{2} \Phi^\dagger \Phi - \frac{\lambda}{4}(\Phi^\dagger \Phi)^2 - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ where $$\Phi = (\Phi_1, ..., \Phi_N)\in \mathbb{C}^N$$ is a vector of $$N$$ complex scalar fields and $$D_\mu \Phi = (\partial_\mu + ieA_\mu) \Phi.$$

I am supposed to find the global symmetry group $$G$$ of this Lagrangian. Obviously, $$\text{U}(N) \subseteq G,$$ but, since for the case that $$N=2$$ I know that there exists a larger $$\text{SU}(2)_L\times \text{SU}(2)_R$$ symmetry, I assume that for $$N\geq 3$$ the symmetry group $$G$$ is also larger than what I have written above.

For $$N=2$$, I know that the $$\text{SU}(2)_R$$ symmetry becomes apparent when rewriting the Lagrangian using $$\tilde\Phi = \begin{bmatrix} \Phi_2^* & \Phi_1 \\ -\Phi_1^* & \Phi_2\end{bmatrix}$$ where terms like $$\frac{1}{2}\text{Tr}[\tilde\Phi^\dagger \tilde\Phi]$$ appear. Here, the two distinct $$\text{SU}(2)$$ symmetries become visible due to the cyclicity of the trace.

However, I don't know how to generalise this idea to $$N$$ complex scalars, so I am unsure on how to proceed with finding more symmetries of the general Lagrangian with $$N$$ scalars.

• Isnt the $U(1)$ contained in the $SU(2)_R$? Also, isn't the $U(1)$ lost when moving to real scalar fields, as the rotations in the complex planes get turned into $SO(2)$ subgroups? Commented Mar 15 at 18:33

You may see this symmetry easiest by recasting your complex N-vectors as real 2N-vectors, $$\Phi_i \mapsto \begin{pmatrix} \Re \Phi_i \\ \Im \Phi_i\end{pmatrix}=\begin{pmatrix} \phi_{2i-1} \\ \phi_{2i}\end{pmatrix},$$ stacked on top of each other, as $$\Phi^\dagger \cdot \Phi= \vec \phi\cdot \vec\phi,$$ so invariant under SO(2N).
And likewise for all scalar bilinears, except those multiplied by just one power of $$A_\mu$$, $$\sum_j^N 2eA_\mu(\phi_{2j-1}\partial^\mu \phi_{2j}- \phi_{2j}\partial^\mu \phi_{2j-1} ),$$ where the same SO(2) orthogonal matrix keeps each of the N terms invariant! Stacking them in diagonal blocks yields an SO(2), the only invariance of N=1, and, as you all but pointed out in your comment, one generated by one of the 6 generators of SO(4) for N=2. (Considering $$su(2)\times su(2)\sim so(4)$$.)
I can see a larger SO(2)×SO(N) in this term, but not the full SO(2N) of the rest. Frankly, I don't see it even in your custodial $$\tilde \Phi$$ language for N=2. In the real 4-vector language of this answer, observe that the term is schematically of the form $$\vec \phi' A \vec \phi$$, where $$A=\begin{bmatrix}0&1&0&0\\ -1&0&0&0\\ 0&0&0&1\\ 0&0&-1&0\end{bmatrix}, \qquad T=\begin{bmatrix}0&0&1&0\\ 0&0&0&0\\ -1&0&0&0\\ 0&0&-0&0\end{bmatrix},$$ and I picked T to be another generator of SO(4). It is then evident this bilinear is not invariant under the rotation effected by T, since $$[A,T]\neq 0$$. There is another generator commenting with A: can you see which? So this bilinear is not fully SO(4) invariant, as you should be able to confirm in your language!