singular metric (with essential singularity)

Question

Working on some $Q$-curvature equation in dimension $4$, I have been faced with singular metric of the form $(\mathbb{B}, e^{-1/\vert x\vert ^2} \vert dx\vert)$. I try to figure out to what those metrics look like. In order to do so, I refer to the well known $2$-dimensional where we have conical metric $(\mathbb{D}, \vert x\vert^{\beta-1} \vert dx\vert)$ which are isometric to a cone of angle $2\pi \beta$. If $\beta=n$ is an integer, I can think about it as the neighborhood of $0$ of and $n$-fold branched unit disc. My question is: is there is a similar interpretation for $(\mathbb{D}, e^{-1/\vert x\vert ^2} \vert dx\vert)$ (in dimension 2)? Can we think about this metric as the limit of some more classical degenerate metric? Any reference will be welcome. Thx in advance

Edit: Thanks to the post of @ubik, I can reformulate the question as follows. Let $g=e^{-\frac{d}{r^2}}(dr^2 +r^2 d\theta^2)$, we set $s=\int_0^r e^{-\frac{d}{2 t^2}}\, dt$, then we get $$ s= \int_0^r \frac{t^3}{d} \frac{d}{dt}\left( e^{-\frac{d}{2t^2}}\right)\, dt\sim \frac{r^3}{d} e^{-\frac{d}{2r^2}}$$ Hence $$re^{-\frac{d}{2r^2}} \sim s\frac{d}{r^2}$$ $$\log(s)\sim -\frac{d}{2r^2}$$ which gives $$g=ds^2 +J(s)^2d\theta^2$$ where $$J(s)\sim 2s\log\left(\frac{1}{s}\right)$$ My question what: is the interpretation of this metric, is it a cusp? is there is a standard model?

Answer 1

When $\beta>0$, the metric $$g=e^{-\frac{1}{|z|^\beta}}|dz|^2$$ has a conical singularity of angle $(\beta+2)(\beta+1)\pi$ at $z=0$ and non positive Gaussian curvature. Using polar coordinate $(r,\theta)\in (0,+\infty)\times \mathbb{S}^1$ we have $$g=e^{-\frac{1}{r^\beta\,}}\left((dr)^2+r^2(d\theta)^2\right).$$ Using the change of variable $$s=\int_0^r e^{-\frac{1}{2\,t^\beta}}dt=\int_0^r \frac{2t^{\beta+1}}{\beta+1} \frac{d}{dt}e^{-\frac{1}{2\,t^\beta}} dt=\frac{2r}{(\beta+2)(\beta+1)} e^{-\frac{1}{2\,r^\beta}}-\int_0^r \frac{2t^{\beta+2}}{(\beta+2)(\beta+1)}e^{-\frac{1}{2\,t^\beta}} dt$$ we get $$g=(ds)^2+J(s)^2(d\theta)^2$$ where $\lim_{s\to 0+}J(s)/s=\frac{(\beta+2)(\beta+1)}{2}$ hence the result.

The bad singularity at $z=0$ is an artefact of the coordinates that are not suitable to understand the geometry of $g$.

Sorry for my very stupid mistake ! I hope that this other computation is OK ! We have : $$s=\int_0^r e^{-\frac{1}{2\,t^\beta}}dt=\int_0^r \frac{2t^{\beta+1}}{\beta+1} \frac{d}{dt}e^{-\frac{1}{2\,t^\beta}} dt=\frac{2r^{\beta+1}}{\beta+1} e^{-\frac{1}{2\,r^\beta}}-\int_0^r 2t^{\beta}e^{-\frac{1}{2\,t^\beta}} dt.$$ Hence $$s=\frac{2r^{\beta+1}}{\beta+1} e^{-\frac{1}{2\,r^\beta}}\left(1+\mathcal{O}\left(r^\beta\right)\right).$$ Hence we have $$\log(s)=-\frac{1}{2r^{\beta}}+(\beta+1)\log(r)+\log\left(2/(\beta+1)\right)+ \mathcal{O}\left(r^\beta\right),$$ And $$J(s)=re^{-\frac{1}{2\,r^\beta}}=\frac{\beta+1}{2} s r^\beta\left(1+\mathcal{O}\left(r^\beta\right)\right)$$

Hence $$J(s)\simeq_{s\to 0+}\frac{\beta+1}{2} s \frac{1}{2\log(1/s)}.$$