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.

  • $\begingroup$ 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

1 Answer 1


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!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.