Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 3 / b / i

Expand the matrix trace explicitly:Write ${\Phi }_{ij}=(x_{ij}+i{y}_{ij})/\sqrt{2}$ and collect the $2N^2$ real fields into a vector $X$. ThenBoth quadratic forms are preserved by every constant orthogonal matrix $R$, acting as $X\mapsto RX$. Thus the full linear internal symmetry contains $O(2N^2)$, with connected special orthogonal group $SO(2N^2)$ as requested. The enlargement beyond unitary transformations is possible because the real and imaginary components have identical kinetic and mass terms and no interactions distinguishing them.

For a Hermitian matrix $\Phi$, there are $N$ real diagonal entries and $N(N-1)/2$ independent complex off-diagonal entries, hence $N^2$ real fields. Choose an orthonormal basis $T_A$ of Hermitian matrices for the trace inner product, $\mathrm{tr}(T_A{T}_B)={\delta }_{AB}$, and put $\Phi ={\sum }_A{X}_A{T}_A/\sqrt{2}$. The same calculation giveswhere these are the connected rotation groups; their full orthogonal extensions are $O(2N^2)$ and $O(N^2)$. A purely quadratic theory with $m^2<0$ would have an unstable scalar potential, but this algebraic symmetry statement still holds.