$\mathcal{J}^\mu(\epsilon) = J^\mu(\epsilon) + E_\alpha Y^{\alpha,\mu} \epsilon$
at constant $\epsilon$ : $\mathcal{J}^\mu = (J^\mu + E_\alpha Y^{\alpha,\mu})\epsilon$
and so $\mathcal{Q}(\epsilon) = \epsilon \int_V (J^\mu + E_\alpha Y^{\alpha,\mu})\mathrm{d}V$
and with (14): $\mathcal{Q}(\epsilon) = \epsilon\int_V\mathrm{d}_\mu J^{\mu,\nu}\mathrm{d}V = 0$, if $J^{\mu,\nu}$ vanishes on $\partial V$.
Now, think about what the $J^{\mu,\nu} = \pi^\mu_\alpha Y^{\alpha,\nu}$ are: They are the coefficients for the field transformations where the field transforms non-trivially with the derivative …

In the degenerate case I am confused as to what we do because:
In the eigenraum of the degenerate eigenvalue you would diagonalize the perturbation operator, find the eigenvalues and the corresponding eigenvectors. The eigenvectors can, if I stand correct, be expressed as a linear combination of the basis kets of the eigenraum. This is as far as I go.
Now, in the Hilbert space of system, one, substitutes, the unperturebed eigenvectors of the degenerate eigenvalue, with the eigenvectors of the perturbed operator in the eigenraum of the eigenvalue. But what can you say about the perturbed eig…