$$\rho^a_i (M^2) = -\tilde{\rho}^a_i (M^2)$$ which gives us $$\langle 0|[J^{a,\mu}_0(y), \phi_{i,0}(x)]|0\rangle =
i \frac{\partial}{\partial y_\mu} \int dM^2 \rho^a_i(M^2) (\Delta_+(y-x;M^2) -
\Delta_+(x-y;M^2))$$ Now, since the current $J^{a,\mu}_0$ is conserved, we can act by $\partial/\partial y_\mu$ to get $$0 = \int dM^2 M^2 \rho^a_i(M^2)(\Delta_+(y-x;M^2) - \Delta_+(x-y;M^2))$$ and thus we get $$M^2 \rho^a_i(M^2) = 0. \label{eq1}$$

Now consider the situation when the symmetry is broken. For $\mu =0,
x^0=y^0=t$, $$\langle 0|[J^{a,0}_0(\vec{y},t), \phi_{i,0}(\vec{x},t)]|0\rangle = i
\delta(\vec{y}-\vec{x}) \int dM^2 \rho^a_i(M^2).$$

Integrating and using (\[cQ\]), we get $$\sum_j T^a_{ij}< \phi_{j,0}(x)> = i\int \rho^a_i(M^2). \label{eq2}$$

Eqs. (\[eq1\]) and (\[eq2\]) are compatible only if $$\rho^a_i(M^2) = i \delta(M^2) \sum_j T^a_{ij}\langle 0|
\phi_{j,0}(0)|0 \rangle$$ As long as the symmetry is broken, the spectral density $\rho^a_i$ is proportional to $\delta(M^2)$. Since such a term can arise only in a theory with massless particles, we are forced to conclude that a broken symmetry with $T^a_{ij}\langle 0|\phi_{j,\theta}(0)|0\rangle~\neq 0$ requires the existence of a massless particle with the same quantum numbers as $J^{a,0}_\theta$. These are nothing but the Goldstone bosons.

Spontaneously Broken Local Symmetries & twisted standard model
--------------------------------------------------------------

Now given the map between the twisted fields and untwisted ones (eq. \[twistfield\]) and our earlier established rules for getting the correlation functions for 