# Is this measure employed in the Faddeev-Popov procedure related to the Haar measure?

In the Faddeev-Popov procedure one defines the Faddeev-Popov determinant through the formula $$\int {\mathcal{D}\alpha \ } \delta\big[G(A^\alpha)\big]\Delta[A]=1,\tag{1}$$

where $$G(A^\alpha)$$ is the gauge-fixing condition and $$A^\alpha$$ is the gauge field $$A$$ transformed by a finite gauge transformation obtained exponentiating $$\alpha$$:

$$(A^\alpha)^a_\mu T^a=e^{i\alpha^a T^a}A^b_\mu T^b e^{-i \alpha^c T^c}+\dfrac{i}{g}e^{i\alpha^a T^a}\partial_\mu e^{-i\alpha^b T^b}.\tag{2}$$

Question: Is this $$\cal D \alpha$$ measure related to the Haar measure on the gauge group $${\cal G}$$? In other words, is (1) exactly the same as $${\int \mathcal{D}g \ }\delta\big[G(A^g)\big]\Delta[A]=1\tag{3}$$

where now $$A^g=g Ag^{-1}+\dfrac{i}{e} g \partial_\mu g^{-1}\tag{4}$$

where in (4) the measure $${\cal D}g$$ is the Haar measure? What is the precise relation between $$\cal D\alpha$$ and $$\cal D g$$?

I mean, heuristically (1) and (3) are doing the exact same thing, since by varying all $$\alpha$$ we vary all $$g$$ and since $$A^\alpha = A^{g(\alpha)}$$ where $$g(\alpha)=e^{i\alpha^a T^a}$$, but it is not clear to me if the measure $${\cal D}\alpha$$ is related (and how it is related) to the Haar measure $${\cal D}g$$.

My intuition here is that this is analogous to what we do with the Lebesgue measure $$d^nx$$ in $$\mathbb{R}^n$$ when we express it in various coordinate systems. We can express it in cartesian, spherical or cylindrical coordinates but it is always the same measure. My intuition is that here we are using the functions $$\alpha$$ as the "coordinates" and in (1) we are just writing down the Haar measure in these coordinates. Is that correct?

Your intuition is correct. One way to understand this is as follows. Let $$\mathcal{C}$$ be the space of all gauge connections $$A$$ and let $$G$$ be the gauge group. Then the space $$\hat{\mathcal{C}} = \mathcal{C} /G$$ can be understood as the space of gauge inequivalent connections.
To make this construction explicit, we can think of the action of $$G$$ on a single $$A$$, namely the $$A^g$$ you mention, as defining some orbit of gauge-equivalent connections in $$\mathcal{C}$$. The object $$\hat{\mathcal{C}}$$ is then the space in which every connection within a single orbit is identified as a single point -- we "collapse" the orbits down. A good simple visual for how this works would be to consider the space $$\mathbb{R}^2$$ and the orbits traced out by the group of rotations about the origin, namely all concentric circles centered at the origin. If we are dealing with a rotationally symmetric problem, we can equivalently "collapse" all of the circles down to the positive real axis.
To form an integral over all connections, which is what the end goal is, we need to integrate over all the gauge inequivalent connections (which is what the path integral was over to begin with) and then also over gauge orbit for each inequivalent connection. This is what the procedure is really doing. Since the orbit is explicitly parametrized by the gauge group (to form the $$A^g$$), the volume of the orbit may be taken to be that of the gauge group itself. Hence the measure introduced in the Faddeev-Popov method is indeed the Haar measure.
Now for the final point. Observe that for gauge transformations which are close to the identity, $$g(\alpha)=1-iT^a\alpha_a$$ and so the measure $$\mathcal{D}g\propto \mathcal{D}\alpha_a$$ up to an overall constant which doesn't matter in the path integral. For transformations which do not have small $$\alpha$$, we may use the left-invariance of the Haar measure to translate back to the region near the identity so this result is, in fact, globally applicable.