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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Mar 15 at 18:33

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The full global group is SO(2N) which is much larger than the U(N), with N(2N-1) versus N² generators, for e=0. You already saw this with your N=2 example, which, however, alas! does not extend to all N, since SU(2) is pseudo real, and its conjugate doublet transforms as a doublet, which makes the right action feasible.

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!

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