I want to visualize Ricci flow solution on the following sphere 

Let $r> 0$

$L = \{ (x cos \theta, x sin \theta, x) | r < x < R \}$

$S$ : $(z-\sqrt{2} r)^2 + x^2 + y^2 = r^2$

$T$ : $ (z- \sqrt{2} R)^2 + x^2 + y^2 = R^2$

If $R$ is sufficiently large, then we have a two dimensional sphere $U$ enclosed by $L$, $S$, and $T$.

The Ricci flow solution on $U$ shrinks fastly around region covered by $S$, but the region covered by $L$ remains unchanged, since the Gaussian curvature is 0.

This confuses me. Where is wrong ?

MOTIVATION : I want to know the Ricci flow on a orbifold $O$ which is smooth except one point. In generally, is there a solution on $O$ ? Around a singuler point, the curvature is very large so that Ricci flow solution on $O$ shrinks to the singular point fastly.

If we consider normalized Ricci flow on $O$, the solution goes to a "canonical" orbifold ?

What I say is that if $O$ is a two sphere with exactly one point singularity, then the solution goes to $lim_{r \rightarrow 0} U$

Anything related with my opinion is welcome.

I have a second question. First notice the following. We can define $L_c = \{ (x cos \theta, x sin \theta, cx) | r < x < R \}$ so that we have $U_c$. In addition $lim_{r \rightarrow 0} U_c$ is an orbifold for some $c$

Question : Is any orbifold is a limit of manifolds in Gromov-Hausdorff sense ?

Question 2: In the following paper, a space with curvature $ \geq k$ is defined.

M. Gromov Y. Burago and G. Perelman, A.d. alexandrov spaces with curvature bounded below, Uspekhi Mat. Nauk 47 (2) (1992), 3–51.

Is a space with curvature $ \geq k$ is a limit of manifolds ?

Thank you for your attention.

I want to visualize Ricci flow solution on the following sphere 

Let $r> 0$

$L = \{ (x cos \theta, x sin \theta, x) | r < x < R \}$

$S$ : $(z-\sqrt{2} r)^2 + x^2 + y^2 = r^2$

$T$ : $ (z- \sqrt{2} R)^2 + x^2 + y^2 = R^2$

If $R$ is sufficiently large, then we have a two dimensional sphere $U$ enclosed by $L$, $S$, and $T$.

The Ricci flow solution on $U$ shrinks fastly around region covered by $S$, but the region covered by $L$ remains unchanged, since the Gaussian curvature is 0.

This confuses me. Where is wrong ?

MOTIVATION : I want to know the Ricci flow on a orbifold $O$ which is smooth except one point. In generally, is there a solution on $O$ ? Around a singuler point, the curvature is very large so that Ricci flow solution on $O$ shrinks to the singular point fastly.

If we consider normalized Ricci flow on $O$, the solution goes to a "canonical" orbifold ?

What I say is that if $O$ is a two sphere with exactly one point singularity, then the solution goes to $lim_{r \rightarrow 0} U$

Anything related with my opinion is welcome.

I want to visualize Ricci flow solution on the following sphere 

Let $r> 0$

$L = \{ (x cos \theta, x sin \theta, x) | r < x < R \}$

$S$ : $(z-\sqrt{2} r)^2 + x^2 + y^2 = r^2$

$T$ : $ (z- \sqrt{2} R)^2 + x^2 + y^2 = R^2$

If $R$ is sufficiently large, then we have a two dimensional sphere $U$ enclosed by $L$, $S$, and $T$.

The Ricci flow solution on $U$ shrinks fastly around region covered by $S$, but the region covered by $L$ remains unchanged, since the Gaussian curvature is 0.

This confuses me. Where is wrong ?

MOTIVATION : I want to know the Ricci flow on a orbifold $O$ which is smooth except one point. In generally, is there a solution on $O$ ? Around a singuler point, the curvature is very large so that Ricci flow solution on $O$ shrinks to the singular point fastly.

If we consider normalized Ricci flow on $O$, the solution goes to a "canonical" orbifold ?

What I say is that if $O$ is a two sphere with exactly one point singularity, then the solution goes to $lim_{r \rightarrow 0} U$

Anything related with my opinion is welcome.

I want to visualize Ricci flow solution on the following sphere 

Let $r> 0$

$L = \{ (x cos \theta, x sin \theta, x) | r < x < R \}$

$S$ : $(z-\sqrt{2} r)^2 + x^2 + y^2 = r^2$

$T$ : $ (z- \sqrt{2} R)^2 + x^2 + y^2 = R^2$

If $R$ is sufficiently large, then we have a two dimensional sphere $U$ enclosed by $L$, $S$, and $T$.

The Ricci flow solution on $U$ shrinks fastly around region covered by $S$, but the region covered by $L$ remains unchanged, since the Gaussian curvature is 0.

This confuses me. Where is wrong ?

MOTIVATION : I want to know the Ricci flow on a orbifold $O$ which is smooth except one point. In generally, is there a solution on $O$ ? Around a singuler point, the curvature is very large so that Ricci flow solution on $O$ shrinks to the singular point fastly.

If we consider normalized Ricci flow on $O$, the solution goes to a "canonical" orbifold ?

What I say is that if $O$ is a two sphere with exactly one point singularity, then the solution goes to $lim_{r \rightarrow 0} U$

Anything related with my opinion is welcome.

I have a second question. First notice the following. We can define $L_c = \{ (x cos \theta, x sin \theta, cx) | r < x < R \}$ so that we have $U_c$. In addition $lim_{r \rightarrow 0} U_c$ is an orbifold for some $c$

Question : Is any orbifold is a limit of manifolds in Gromov-Hausdorff sense ?

Question 2: In the following paper, a space with curvature $ \geq k$ is defined.

M. Gromov Y. Burago and G. Perelman, A.d. alexandrov spaces with curvature bounded below, Uspekhi Mat. Nauk 47 (2) (1992), 3–51.

Is a space with curvature $ \geq k$ is a limit of manifolds ?

Thank you for your attention.