# Periodicity trick for Kerr Black Holes

I am slightly confused concerning the euclidean section of a Kerr black hole. In page 5 of the following paper https://arxiv.org/abs/hep-th/9908022 it is said that in order to get the euclidean section, we need to set $$t \to i \tau$$ and $$a \to i a$$. (They consider general Kerr-Newman-AdS black holes but I am simply interested in Kerr asymptotically flat.) This makes sense because we want to keep the $$dt \otimes d\phi$$ components of the euclidean metric real. What confuses me is that if we do the analysis of the conical singularities as they mention, we will get the following periodicity for $$\tau$$ and $$\phi$$

$$$$\tau \sim \tau +\beta$$$$ $$$$\phi\ \sim \phi+i\beta\Omega_H$$$$ with $$\beta$$ the inverse temperature and $$\Omega_H$$ the angular velocity of the event horizon, namely $$$$\Omega_H=\frac{a}{r_{+}^2+a^2}$$$$ where $$r_{+}$$ is the event horizon and $$a$$ is the rotation parameter of the black hole. What is strange to me is that if we take $$a \to 0$$ in Boyer-Lindquist coordinates, we get that $$$$\phi \sim \phi$$$$ because $$\Omega_H$$ vanishes. This becomes a trivial identification and it does not tell us anything about the periodicity of the $$\phi$$ coordinate. However, we also know that if we take the $$a \to 0$$ limit, we get the Schwarzschild black hole in Schwarzschild coordinates. In Schwarzschild Euclidean, we should take the $$\phi$$ coordinate to have period $$$$\phi \sim \phi+2\pi$$$$ and even though the Boyer-Lindquist $$\phi$$ is different than the $$\phi$$ in Schwarzschild, they match in the limit I am considering $$a \to 0$$. What does this imply? Does this mean that even though Kerr goes to Schwarzschild in the limit $$a \to 0$$ as a lorentzian geometry, their euclidean sections are not connected continuously somehow?

Edit1: I also have the notion that in lorentzian Kerr, the $$\phi$$ coordinate has periodicity $$2\pi$$. When we go to Euclidean, we seem to get this other periodicity: but shouldn't the periodicity of $$2\pi$$ be preserved as well? At least that is what happens in Schwarzschild. So we would have both $$$$\phi\ \sim \phi+i\beta\Omega_H$$$$ $$$$\phi\ \sim \phi + 2\pi$$$$ It also confuses me that this manipulations are usually done based on the coordinate systems and therefore it is harder to get a notion of what it means to 'euclideanize' in a coordinate invariant way. If someone has a coordinate invariant way to talk about this analytic continuation, I would like to hear it.

Edit2: If we see what really is the expression in the identification of $$\phi$$, we get $$$$i\beta \Omega_H=i4\pi \frac{r_{+}a}{r_{+}^2\left(1-\frac{a^2}{r_{+}^2}\right)}$$$$ By doing the analytic continuation $$a \to ia$$, we have $$$$i\beta \Omega_H=-4\pi \frac{r_{+}a}{r_{+}^2\left(1+\frac{a^2}{r_{+}^2}\right)}$$$$ we see that it is alway less then $$2\pi$$ because $$$$r_{+}=a+\sqrt{2}a$$$$ defines extremality assuming the fact that we set $$a \to ia$$. So it seems to make the $$\phi$$ direction smaller in general. But if I try to compute the action on-shell $$$$I=\int_{\partial \mathcal{M}}K-K_0$$$$ I have to integrate from $$0$$ to $$2\pi$$ along $$\phi$$ to get the right result mentioned in https://doi.org/10.1103/PhysRevD.15.2752 because since we are sending the boundary to infinity only the leading order of $$1/r$$ matters which is the same as in Schwarzschild. So I am confused what kind of geometry we have along $$\phi$$.

The picture above is a torus in the complex plane where $$z = x + i y$$. The torus is defined by two equivelence relations, \begin{align} z \sim z + 1 \\ z \sim z + \tau. \end{align} If we write $$\tau = \tau_x + i \tau_y$$ then we can write the equivalence relations as \begin{align} (x, y) &\sim (x + 1, y) \\ (x, y) &\sim (x + \tau_x, y + \tau_y). \end{align} There is a "shift" in $$x$$ as $$y$$ goes from $$0$$ to $$\tau_y$$. Note that the range of x is $$1$$ and the range of $$y$$ is $$\tau_y$$. $$\tau_x$$ gives the shift.
This is exactly what is happening in the Euclidean Kerr metric. If one defines the variable $$\Phi \equiv \phi - \Omega t$$ where $$t$$ is euclidean time, then the equivalence relations are \begin{align} (t, \Phi) &\sim (t + \beta, \Phi) \\ (t, \Phi) &\sim (t, \Phi + 2 \pi ). \end{align} If one uses $$\phi = \Phi + \Omega t$$, then this becomes \begin{align} (t, \phi) &\sim (t + \beta, \phi + \beta \Omega) \\ (t, \phi) &\sim (t, \phi + 2 \pi ). \end{align} $$\beta \Omega$$ is therefore not giving the periodicity of $$\phi$$ but rather the "shift" in $$\phi$$ as one makes a full cycle in $$t$$.