# Invariance of Functional Integration Measure

Let us consider the functional integral: $$\int \mathcal{D} A e^{iS[A]}$$ where $S[A]$ is the action for $U(1)$ gauge field and $$\mathcal{D}A\equiv \mathcal{D}A_0 \mathcal{D}A_1 \mathcal{D}A_2 \mathcal{D}A_3; \\ \mathcal{D}A_i = \prod_x dA_i(x).$$

Now I have two questions:

1. How to show that the integration measure $\mathcal{D} A$ is invariant under the gauge transformation: $$A_\mu (x) \to A_\mu (x) + \frac{1}{e}\partial_\mu \alpha(x)$$ 2. How to show that the integration measure $\mathcal{D} A$ is invariant under the Lorentz transformation?

• Hints: 1. $\alpha(x)$ is a set function, it isn't being varied over like $A_\mu(x)$ is. 2. How does a measure change under a change of coordinates in regular integration?
– Will
Jun 30 '13 at 16:41
• Doesn't the answer to this depend on the regulator? In particular, couldn't one concoct some terrible regulator that would render the integration measure non-gauge-invariant? Jun 30 '13 at 18:18
• @joshphysics: You can certainly break symmetries with a non-covariant regulator, but it is almost universally assumed that one is using a gauge-covariant regulator since the consequence of gauge non-covariance are so dire. Jul 1 '13 at 22:29
• @Will: Can u elaborate the hint for question no. 2 ? Jul 2 '13 at 5:59
• Lorentz transformations take $A_\mu \rightarrow A^\prime_{\nu} = \Lambda_{\nu}^{\mu}A_{\mu}$, right? Now, consider the process you go through when changing coordinates in regular integration of several variables. Hopefully this helps :)
– Will
Jul 2 '13 at 6:13