Residual gauge invariance

Question

I am trying to understand residual gauge invariance in context of Yang-Mills theory and Einstein's gravity.

In Yang-Mills theory, we know that the transformation $$A_\mu\rightarrow A'_\mu\ =\ A_\mu\ +\ \partial_\mu \Lambda$$ leaves the field strength $F_{\mu\nu}$ and hence, the equations of motion invariant. (I have supressed the gauge index on $A_{\mu}$ for brevity.)

Similarly, in GR under the transformation $$x^\mu \rightarrow x'^\mu\ =\ x^\mu\ +\ \xi^\mu\ , $$ the metric transforms as $$g'^{\mu\nu}(x')\ =\ \frac{\partial x'^\mu}{\partial x^\rho}\frac{\partial x'^\nu}{\partial x^\sigma}\ g^{\rho\sigma}(x) $$ Now, for linearised gravity, $$g_{\mu\nu}\ =\ \eta_{\mu\nu}\ +\ h_{\mu\nu}$$ So, the perturbation $h_{\mu\nu}$ transforms as $$h'_{\mu\nu}\ =\ h_{\mu\nu}\ -\ \partial_\mu \xi_{\nu}\ -\ \partial_\nu \xi_{\mu}$$ assuming that $\frac{\partial \xi^\mu}{\partial x^{\nu}}$ is very small. This transformation leaves the field equations invariant. This is what we mean by gauge invariance in GR. (Please correct me if I am mistaken).

After one fixes the gauge, we are left with some residual gauge symmetry. I don't understand what this precisely means for both the cases. Does it amount to putting some additional conditions on the gauge parameters, $\Lambda$ and $\xi^\mu$ ? How can I work this out explicitly?

Also, I've been trying to find some good material on this. So, please add any link, which might be relevant.

Answer 1

This statement is incorrect:

In Yang-Mills theory, we know that the transformation $$A_\mu \rightarrow A'_\mu = A_\mu + \partial_\mu \Lambda$$ leaves the field strength $F_{\mu\nu}$ and hence, the equations of motion invariant.

For starters, the transformation law for non-abelian gauge theories is given by: $$A_\mu \rightarrow A'_\mu = G A_\mu G^{-1} + \frac{i}{g} \left(\partial_\mu G\right) G^{-1},$$ where $G$ is the gauge transformation operator. In an abelian theory, $G = \exp\left(ig \Lambda(x)\right)$. For $\mathrm{SO}(N)$, $$G = \exp\left(g \Lambda(x)\right),$$ where $\Lambda^T = -\Lambda$ is a real operator that is anti-symmetric under interchange of gauge indices (and the factor of $i$ in the gauge transformation rule is removed). Similarly, for $\mathrm{SU}(N)$ we use $$G = \exp\left(-ig\Lambda(x)\right),$$ with $\Lambda$ Hermitian.

With that transformation rule for $A_\mu$, you can verify that the field strength becomes gauge covariant, $$F_{\mu\nu} \rightarrow F'_{\mu\nu} = G F_{\mu\nu} G^{-1}.$$ That's why the Lagrangian density has a trace over the gauge indices on the $F_{\mu\nu} F^{\mu\nu}$ term – otherwise, it would not be gauge invariant.

Now, for residual gauge symmetries we just look for gauge transformations that leave the gauge fixing condition invariant. In the Lorenz gauge in electromagnetism, for example, the fixing condition is: $$\partial^\mu A_\mu = 0.$$ Let's plug in the gauge transformed field to get: $$\partial^\mu A'_\mu + \partial^\mu \partial_\mu \Lambda = 0.$$ Clearly, the necessary and sufficient condition for $A'_\mu$ to satisfy the gauge fixing condition is $$\partial^\mu \partial_\mu \Lambda = 0.$$ Since there are functions, $\Lambda$, that satisfy our requirement there must be some gauge degrees of freedom that the Lorenz condition does not fix. Ergo, there are some residual degrees of freedom in our gauge group.

I'm not as adept with gravity as I am with gauge theories, though, so I'll leave that part for someone else to explain, but I imagine that it comes down to some coordinate freedoms that are left open, depending on how the "gravitational gauge" is fixed.

Answer 2

In few words:

When you're looking to Residual Gauge transformation you're already assuming that your object (say Lagrangian) is invariant under a Gauge transformation "h". You fix the gauge, and than you ask if there exists some "h_1" related to "h" by a gauge transformation such that "h_1" satisfies the same gauge condition. Then you have a residual gauge trasformation.