Skip to main content
1 of 2
user avatar
user avatar

Let $(M,g)$ be a (bad) Riemannian $2$-orbifold with one cone point $x_{1}$ of order $p\geq2$ and underlying space a topological $2$-sphere, i.e., a teardrop with Euler characteristic $\chi(M)=1+\frac{1}{p}$. For $\varepsilon>0$ small, we have length $\operatorname{L}(\partial B_{\varepsilon}(x_{1}))\approx2\pi p^{-1}\varepsilon$ and geodesic curvature $\kappa(\partial B_{\varepsilon }(x_{1}))\approx\varepsilon^{-1}$. The Gauss-Bonnet formula is, using $\chi(M-B_{\varepsilon}(x_{1}))=1$, $$ \int_{M}Rd\mu=\lim_{\varepsilon\rightarrow0}\int_{M-B_{\varepsilon}(x_{1} )}Rd\mu=2\pi+\lim_{\varepsilon\rightarrow0}\int_{\partial B_{\varepsilon }(x_{1})}\kappa ds=2\pi\chi(M). $$

A vector field $X$ is smooth on $M$ if it is smooth on $M-x_{1}$ and its $p$-fold lift via the orbifold chart $\pi:B\rightarrow U$, where $U$ is a neighborhood of $x_{1}$, is smooth (the same applies to functions). This implies that $\lim_{x\rightarrow x_{1}}X\left( x\right) =X\left( x_{1}\right) =0$. Moreover, the divergence theorem is true for $X$: $\int _{M}\operatorname{div}X\,d\mu=0$. In particular, if $u$ is a smooth function on $M$, then $\int_{M}\Delta u\,d\mu=0$.

Conversely, we may solve $\Delta f=h$ if (and only if) $\int_{M}hd\mu=0$. E.g., use the calculus of variations to obtain a minimizer $f\in W^{1,2}(M)$ of the functional $J\left( f\right) =\int_{M}(\frac{1}{2}\left\vert \nabla f\right\vert ^{2}+hf)d\mu$ with the constraint $\int_{M}fd\mu=0$ (so we can apply the Poincar'{e} inequality; the constraint anchors the function). By regularity theory, the minimizer $f$ is smooth and satisfies $-\Delta f+h=\operatorname{const}$, which must be zero. This works on an orbifold with isolated singularites.

Let $r$ be the average scalar curvature. Since $\int_{M}(r-R)d\mu=0$, there exists $f$ (unique up to an additive constant) such that $\Delta f=r-R$, called the curvature potential. Note that $\nabla f\left( x_{1}\right) =0$.

The only closed bad $2$-orbifolds are the teardrop and the spindle (i.e., a bad orbifold with two cone points, of orders $p,q\geq2$, where $p\neq q$ and $\chi=\frac{1}{p}+\frac{1}{q}$). On these, Lang-Fang Wu proved that for any initial smooth metric with positive curvature, the normalized Ricci flow (NRF) asymptotically aproaches a (nonconstant curvature) shrinking gradient Ricci soliton. More specifically, consider the NRF modified by diffeomorphisms, defined by $\frac{\partial}{\partial t}g=-2\left( \operatorname{Ric} +\nabla^{2}f-\frac{r}{2}g\right) $, where the curvature potential $f$ satisfies $\frac{\partial f}{\partial t}=\Delta f-\left\vert \nabla f\right\vert ^{2}+rf$. (The right side of the first equation is actually trace-free, which implies that the area form is pointwise constant under the flow). Then, as $t\rightarrow\infty$, $g\left( t\right) $ and $f\left( t\right) $ converge to a $C^{\infty}$ metric $g_{\infty}$ and a $C^{\infty}$ function $f_{\infty}$, each exponentially fast in all $C^{k}$-norms. They satisfy $\operatorname{Ric}(g_{\infty})+\nabla_{g_{\infty}}^{2}f_{\infty }-\frac{r}{2}g_{\infty}=0$ and $\Delta_{g_{\infty}}f_{\infty}-\left\vert \nabla f_{\infty}\right\vert _{g_{\infty}}^{2}+rf_{\infty}=0$.

So, Wu proved that under the NRF on a bad closed $2$-orbifold there exists a gradient Ricci soliton structure $(\tilde{g}_{\infty},\tilde{f}_{\infty})$ equivalent to $(g_{\infty},f_{\infty})$ such that $g\left( t\right) $ asymptotically approaches $\varphi_{t}^{\ast}\tilde{g}_{\infty}$, where $\{\varphi_{t}\}$ is generated by $\operatorname{grad}_{\tilde{g}_{\infty} }\tilde{f}_{\infty}$. If the initial metric yields a shrinking gradient Ricci soliton structure, then we have equality, i.e., $g\left( t\right) =\varphi_{t}^{\ast}g\left( 0\right) $.

Disclaimer: I haven't been thinking about orbifolds recently, so some of the above is what I think is true from memory without checking the details; please look out for errors.

user41263