Skip to main content
added 53 characters in body
Source Link
M. Winter
  • 13.6k
  • 3
  • 29
  • 70

Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge graph, edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects. (it was pointed out to me by Ivan Izmestiev that this would not be as straight-forward because we don't know the edge graph from the metric)

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge graph, edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects. (it was pointed out to me by Ivan Izmestiev that this would not be as straight-forward)

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge graph, edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects. (it was pointed out to me by Ivan Izmestiev that this would not be as straight-forward because we don't know the edge graph from the metric)

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

added 99 characters in body
Source Link
M. Winter
  • 13.6k
  • 3
  • 29
  • 70

Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge graph, edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theoremIf Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects. ("a convex polyhedron is determinedit was pointed out to me by the metric on its surface"Ivan Izmestiev that this would not be as straight-forward), because from the metric we can read off the edge lengths and angular defects.

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge graph, edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects.

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge graph, edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects. (it was pointed out to me by Ivan Izmestiev that this would not be as straight-forward)

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

added 12 characters in body
Source Link
M. Winter
  • 13.6k
  • 3
  • 29
  • 70

Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge graph, edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects.

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects.

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

Let's consider 3-dimensional convex polyhedra $P\subset\Bbb R^3$. The angular defect at a vertex $v$ is $2\pi$ minus the sum of the interior angles of the incident faces at $v$.

Question:
Is a convex polyhedron determined (up to isometry) by its edge graph, edge lengths and angular defects?

If Yes, this would imply Alexandrov's uniqueness theorem ("a convex polyhedron is determined by the metric on its surface"), because from the metric we can read off the edge lengths and angular defects.

A potential approach is to reduce it to Alexandrov's theorem. Can we reconstruct the metric from the edge lengths and angular defects?

deleted 19 characters in body
Source Link
M. Winter
  • 13.6k
  • 3
  • 29
  • 70
Loading
deleted 19 characters in body
Source Link
M. Winter
  • 13.6k
  • 3
  • 29
  • 70
Loading
added 145 characters in body
Source Link
M. Winter
  • 13.6k
  • 3
  • 29
  • 70
Loading
Source Link
M. Winter
  • 13.6k
  • 3
  • 29
  • 70
Loading