Question about gauge symmetry confronted in Schwartz‘s book

Question

This picture is from Schwartz book on QFT on page 131. I cannot understand that:

1. What does the orange underlined sentences mean?

2. How is Equation 8.108 derived?

Could anyone kindly make some further explanation on this?

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By @Nikita ‘s answer, I think I’ve understand all the rest except this little calculation...

I could get that

$(A_\mu+\partial_\mu \pi)^2=A_\mu A^\mu-A_\mu \frac{2}{\square} \partial_\mu \partial_\nu A^\nu+\frac{1}{\square}\partial_\mu \partial_\nu A^\nu \frac{1}{\square}\partial^\mu \partial_\rho A^\rho$

While

$\frac{1}{\square}\partial_\mu \partial_\nu A^\nu \frac{1}{\square}\partial^\mu \partial_\rho A^\rho =\frac{1}{\square}\partial_\mu \partial_\nu A^\nu \frac{1}{\square}\partial_\rho A^\rho-\frac{\square}{\square}\partial_\nu A^\nu \frac{1}{\square}\partial_\rho A^\rho$

Why does $\frac{1}{\square}\partial_\mu \partial_\nu A^\nu \frac{1}{\square}\partial_\rho A^\rho$ vanish?

How to get $\frac{1}{\square} (\partial^\nu A^\mu)\partial_\mu A_\nu$?

Answer 1

1) Lagrangian (8.107) have following fields:

1) $A_\mu$ - four degrees of freedom

2) $\pi$ - one degree of freedom

Totally we have 5 d.o.f. One can use gauge symmetry to reduce degrees to four, fixing gauge for $A_\mu$, for example using Coulomb gauge $\partial_\mu A^{\mu}=0$. Using equation of motion for $A_\mu$ one can reduce d.o.f. to three: 2 in $A_\mu$ and 1 in $\pi$.

2) Equation (8.108) is obtained by solving equation of motion for $\pi$:

$$ \Box \pi + \partial_\mu A^{\mu} = 0 $$ $$ \pi = - \frac{1}{\Box}\partial_\mu A^{\mu} $$

And substitution them back to action (I present my calculations):

3) Obtained theory is not local due to therm $\frac{1}{\Box}\propto r^2$. This therm really mean:

$$ F_{\mu\nu}(x) \frac{1}{\Box} F^{\mu\nu}(x) =\int d^4y\; F_{\mu\nu}(x) G(x-y) F^{\mu\nu}(y) $$

Where $G(x-y) = \frac{1}{\Box}$ defined by equation:

$$ \Box G(x-y) = -\delta^{(4)}(x-y) $$

$$ G(x-y) = \int d^4p \frac{e^{ip(x-y)}}{p^2} = \frac{1}{(x-y)^2} $$

I believe that now meaning of nonlocality is more clear...