Noether's 2nd Theorem and Local Gauge Identities

Question

I am trying to derive the so called Gauge Identities:
begin{equation}
D_nufrac{delta S}{deltaphi} = 0
end{equation}
Where $D_nu$ is an operator involving derivatives and $frac{delta S}{deltaphi}$ are the usual Euler-Lagrange equations.

So far I have taken the following local field transformation:
begin{equation}
bar{delta}phi(x) = varphi_nulambda^nu(x)simeq varphi_nulambda^nu + varphi^mu_nupartial_mulambda^nu
end{equation}

And varied the action:
begin{align}
bar{delta}S &= int d^4x~left(frac{partialmathcal{L}}{partialphi}bar{delta}phi + frac{partialmathcal{L}}{partialpartial_muphi}partial_mubar{delta}phiright)
&= int d^4x~left(frac{partialmathcal{L}}{partialphi}left(varphi_nu^0lambda^nu + varphi_nu^rhopartial_rholambda^nuright) + frac{partialmathcal{L}}{partialpartial_muphi}partial_muleft(varphi_nu^0lambda^nu + varphi_nu^rhopartial_rholambda^nuright)right)
&text{Integrate the second term by parts to get}
&= int d^4x~left(frac{partialmathcal{L}}{partialphi}left(varphi_nu^0lambda^nu + varphi_nu^rhopartial_rholambda^nuright) - partial_mufrac{partialmathcal{L}}{partialpartial_muphi}left(varphi_nu^0lambda^nu + varphi_nu^rhopartial_rholambda^nuright)right)
&= int d^4x~left[frac{partialmathcal{L}}{partialphi} - partial_muleft(frac{partialmathcal{L}}{partialpartial_muphi}right)right]varphi_nu^0lambda^nu + left[frac{partialmathcal{L}}{partialphi} - partial_muleft(frac{partialmathcal{L}}{partialpartial_muphi}right)right]varphi_nu^rhopartial_rholambda^nu
&text{Again integrate the second term by parts to get}
&= int d^4x~left[frac{partialmathcal{L}}{partialphi} - partial_muleft(frac{partialmathcal{L}}{partialpartial_muphi}right)right]varphi_nu^0lambda^nu - partial_rholeft(left[frac{partialmathcal{L}}{partialphi} - partial_muleft(frac{partialmathcal{L}}{partialpartial_muphi}right)right]varphi_nu^rhoright)lambda^nu
&=int d^4x~partial_mu mathcal{J}^mu(lambda)
end{align}

Recognising this stuff in the integral as the operator $D_nu$, I get the following:

begin{equation}
int d^4x ~left(partial_mu mathcal{J}^mu(lambda) - lambda^nu D_nufrac{delta S}{deltaphi}right) = 0
end{equation}

What I don't understand is how to now see that $D_nufrac{delta S}{deltaphi} = 0$ for arbitrary $lambda$.

What if I choose a parameter that doesn't make $mathcal{J}$ vanish on the surface for example?

Ali Seraj · Answer

Lets denote the spacetime manifold by $M$. For any localized function $lambda$ without support at the boundaries of spacetime, the first term in your last equation is vanishing, since it can be recast into a boundary integral, i.e. 
$$int_M partial_mu mathcal{J}^mu(lambda)=int_{partial M}dSigma_mu mathcal{J}^mu(lambda)=0$$
Now take a closed subregion $Nsubset M$ and consider functions $lambda$ on $M$ such that they are arbitrary on $N$ while vanishing outside. Using the above argument in your last equation implies that 
begin{equation}
int_N d^4x ~lambda^nu D_nufrac{delta S}{deltaphi}= 0
end{equation}
Since $lambda$ is arbitrary on $N$, we find that $D_nufrac{delta S}{deltaphi}= 0$ on $N$. Repeating this argument for a sequence of patches $N_i$, which altogether cover $M$, we conclude that $D_nufrac{delta S}{deltaphi}= 0$ on the whole manifold $M$. Would you agree?

Noether's 2nd Theorem and Local Gauge Identities

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