Considering a non-rotating and non-charged 2+1 dimensional black hole, known as the BTZ black hole which obtained by adding a negative cosmological constant $\Lambda=-\frac{1}{l^2},l\ne0$ to the Einstein-Hilbert action, resulting the following metric:
$$ds^2=-\left(\frac{r^2}{l^2}-M\right)dt^2+\left(\frac{r^2}{l^2}-M\right)^{-1}dr^2+r^2d\theta^2,\quad \theta\in[0,2\pi]$$
Where $M$ is the black holes mass. There is one horizon to this black hole given as a singularity in the metric in the radius $$r=\sqrt{M}l$$ But this is not a real curvature singularity of the system as one can make a coordinate transformation in which the horizon is no longer constitute singularity as in Schwarzschild black hole. In order to find intrinsic curvature singularity one needs to appeal to the Kretschmann scalar of the metric given by: $$K = R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}$$ where $R_{\mu\nu\rho\sigma}$ is the Riemann tensor of the metric and see whenever it have a singularity, its singularity is in particular the real intrinsic curvature singularity of the metric, according to BTZ black hole the Kretschmann scalar is: $$K=-\frac{6}{l^4}$$ which doesn't depends on $r$ at all, thus didn't have a curvature singularity.
However, in many sources the BTZ black hole have another kind of singularity known as conical singularity, is there is some parallel test or operator which given a metric and in particular BTZ black hole, can reveal whenever the metric contain a conical singularity?
