Greatest possible symmetry group of Lagrangian I have two very specific questions regarding the symmetry group of a lagrangian.

*

*N Complex Scalars, massive, non-interacting
Do we have a global $U(N)$ symmetry, or a global $O(2N)$ symmetry?


*1 Fermion, Massless, non-interacting
I have the idea that the symmetry group should be $U(1) \times U(1)_{chiral}$, but my teacher says its $U(2)$. Which one is correct? If it's the second one, an explicit proof would be very helpful!
 A: In the scalar field case, the Lagrangian is given by:
$$\mathcal{L} = \sum_{i=1}^N  \frac{1}{2}\partial_{\mu} \bar{\Phi}_i\partial^{\mu} \Phi_i + m^2 \bar{\Phi}_i \Phi_i $$
It has a $U(N)$ symmetry under the transformation:
$$\Phi'_i = \sum_{i=1}^N  U_{ij} \Phi_j, \quad U\in U(N)$$
If we write the scalar field in terms of their real and imaginary components
$$\Phi_i  = \phi_{i} + i \phi_{i+N}$$
The Lagrangian becomes of a $2N$-dimensional real scalar field
$$\mathcal{L} = \sum_{i=1}^{2N}  \frac{1}{2}\partial_{\mu} \phi_i\partial^{\mu} \phi_i + m^2 \phi_i \phi_i $$
The Lagrangian has an $O(2N)$ symmetry under:
$$\phi'_i = \sum_{i=1}^{2N}  O_{ij} \phi_j, \quad O\in O(2N)$$
Since it is the same Lagrangian, then it must have both symmetries. The difference between the two cases, is that the first transformation does not mix between the real and imaginary components of the fields, while the second does. Clearly, the second case includes the first: $U(N)\subset O(2N)$. In fact the $U(N)$ matrix can be written in the $SO(2N)$ basis as:
$$O_U = \begin{pmatrix}\operatorname{Re}U & -\operatorname{Im}U\\ 
\operatorname{Im}U & \operatorname{Re}U \end{pmatrix}$$
Therefore the answer to the first question is $SO(2N)$.
The second question
A single specie Dirac theory has no internal symmetries (It has Lorentz symmetry and the massless case, a conformal symmetry, but these are space time symmetries.). However at a fixed momentum, it does have an internal symmetry. The symmetry for a massive fermion is $U(2)\times U(2)$. The symmetry is due to the fact that there the spectrum includes two degenerate positive eigenvalues and two negative ones. However, in the massless case, the symmetry is indeed reduced to $U(1) \times U(1) $, because there are no right handed negative energies and left handed positive energy solutions of the Weyl equation.
Details
The Dirac Hamiltonian:
$$H(\mathbf{p})= c \mathbf{\alpha} \cdot \mathbf{ p} + \beta m c^2$$
Can be exactly diagonalized for a given $\mathbf{p}$ by means of the unitary transformation
$$U_{\mathbf{ p} } =  \exp \left ( \frac{\mathbf{\alpha} \cdot \mathbf{ p} }{| \mathbf{ p}|} \tan^{-1} \frac{| \mathbf{ p}|}{mc}\right )$$
Thus
$$ H(\mathbf{p})= U_{\mathbf{ p} } \begin{pmatrix}E_{|\mathbf{ p}|}  & & & \\ & E_{|\mathbf{ p}|}  & & \\ & & - E_{|\mathbf{ p}|} &\\ &  & & - E_{|\mathbf{ p}|} \end{pmatrix} U_{\mathbf{ p} }$$
With
$$ E_{|\mathbf{ p}|} = \sqrt{| \mathbf{ p}|^2 + m^2 c^4}$$
Thus the Hamiltonian is invariant under the unitary similarity transformation given in the block form
$$ U = U_{\mathbf{ p} } \begin{pmatrix}A & 0 \\0 & B\end{pmatrix}U_{\mathbf{ p} }^{-1}$$
With both $A,B \in U(2)$. This the symmetry is $U(2)\times U(2)$
In the massless case, the eigenvectors corresponding to left and right chiralities decouple and we are left with $U(1)\times U(1)$ each.
