Smooth approximation of polyhedral metric!
do someone have an idea, how to prove that :
For every polyhedral metric $(S,d)$ of curvature $leq -1$, (S=surface), there exists a sequence of smooth metrics, converging uniformly to $(S,d)$.
And why the metric in the neigbourhood of any singularity will be of the shape $dr^2+f^2(r)dphi^2$ with $f$ a suitable function.
Thnak you for any answer
semi-riemannian-geometry
asked Nov 26 at 11:06
kamerove
816
add a comment |
do someone have an idea, how to prove that :
For every polyhedral metric $(S,d)$ of curvature $leq -1$, (S=surface), there exists a sequence of smooth metrics, converging uniformly to $(S,d)$.
And why the metric in the neigbourhood of any singularity will be of the shape $dr^2+f^2(r)dphi^2$ with $f$ a suitable function.
Thnak you for any answer
semi-riemannian-geometry
asked Nov 26 at 11:06
kamerove
816
add a comment |
do someone have an idea, how to prove that :
For every polyhedral metric $(S,d)$ of curvature $leq -1$, (S=surface), there exists a sequence of smooth metrics, converging uniformly to $(S,d)$.
And why the metric in the neigbourhood of any singularity will be of the shape $dr^2+f^2(r)dphi^2$ with $f$ a suitable function.
Thnak you for any answer
semi-riemannian-geometry
asked Nov 26 at 11:06
kamerove
816
do someone have an idea, how to prove that :
For every polyhedral metric $(S,d)$ of curvature $leq -1$, (S=surface), there exists a sequence of smooth metrics, converging uniformly to $(S,d)$.
And why the metric in the neigbourhood of any singularity will be of the shape $dr^2+f^2(r)dphi^2$ with $f$ a suitable function.
Thnak you for any answer
semi-riemannian-geometry
semi-riemannian-geometry
asked Nov 26 at 11:06
kamerove
816
asked Nov 26 at 11:06
kamerove
816
asked Nov 26 at 11:06
kamerove
816
asked Nov 26 at 11:06
kamerove
816
asked Nov 26 at 11:06
kamerove
816
816
add a comment |
add a comment |
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