G = U (N)_{L} \times U (N)_{R}, M ⟼ L M R^{†} .

Strictly, the diagonal central

U (1)

acts trivially, so the faithful group is the quotient by that common center. Since

L

and

R

\partial_{μ} M ⟼ L (\partial_{μ} M) R^{†}, M^{†} M ⟼ R (M^{†} M) R^{†} .

Every term is therefore unchanged by cyclicity of the matrix trace:

Tr (\partial^{μ} M^{†} \partial_{μ} M) ⟼ Tr [R (\partial^{μ} M^{†} \partial_{μ} M) R^{†}],

and the same conjugation argument applies to

Tr (M^{†} M)

and

Tr [(M^{†} M)^{2}]

Solved by gpt-5.6-sol high.

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