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Liviu Nicolaescu
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I think that you need to state very clearly what you mean by triangulation of the manifold, and what do you mean by "angles". (On a manifold that would require a choice of metric.) If by triangulation you mean the affine realizeation ofba finite simplicial complex inside some Euclidean space, and you measure the angles using the induced metric then the answer is no.

Consider the case $n=4$. The affine simplicial complex with $4$ vertices and homeomorphic to a manifold is given by the boundary of a $3$-dimensional simplex. In this case the curvature can only be positive, and clearly one can produce metrics on the $2$-sphere that are negative somewhere.

On the other hand, if you allow the faces of this tetrahedron to be curved then the answer could be positive. Here is one plausible solution.

On $S^2$ fix four points $p_1,\dotsc, p_4$ and a metric whose curvature at $p_i$ is $f(p_i)$. suchSuch a metric exists by Kazhdan-Warner. Next connect the points by geodesic arcs, and assume you can do this so that these arcs do not intersect in the interior. You now have a curved triangulation with the properties you desire.

I think that the question needs to be formulated more carefully.

I think that you need to state very clearly what you mean by triangulation of the manifold, and what do you mean by "angles". (On a manifold that would require a choice of metric.) If by triangulation you mean the affine realizeation ofba finite simplicial complex inside some Euclidean space, and you measure the angles using the induced metric then the answer is no.

Consider the case $n=4$. The affine simplicial complex with $4$ vertices and homeomorphic to a manifold is given by the boundary of a $3$-dimensional simplex. In this case the curvature can only be positive, and clearly one can produce metrics on the $2$-sphere that are negative somewhere.

On the other hand, if you allow the faces of this tetrahedron to be curved then the answer could be positive. Here is one plausible solution.

On $S^2$ fix four points $p_1,\dotsc, p_4$ and a metric whose curvature at $p_i$ is $f(p_i)$. such a metric exists by Kazhdan-Warner. Next connect the points by geodesic arcs, and assume you can do this so that these arcs do not intersect in the interior. You now have a curved triangulation with the properties you desire.

I think that the question needs to be formulated more carefully.

I think that you need to state very clearly what you mean by triangulation of the manifold, and what do you mean by "angles". (On a manifold that would require a choice of metric.) If by triangulation you mean the affine realizeation ofba finite simplicial complex inside some Euclidean space, and you measure the angles using the induced metric then the answer is no.

Consider the case $n=4$. The affine simplicial complex with $4$ vertices and homeomorphic to a manifold is given by the boundary of a $3$-dimensional simplex. In this case the curvature can only be positive, and clearly one can produce metrics on the $2$-sphere that are negative somewhere.

On the other hand, if you allow the faces of this tetrahedron to be curved then the answer could be positive. Here is one plausible solution.

On $S^2$ fix four points $p_1,\dotsc, p_4$ and a metric whose curvature at $p_i$ is $f(p_i)$. Such a metric exists by Kazhdan-Warner. Next connect the points by geodesic arcs, and assume you can do this so that these arcs do not intersect in the interior. You now have a curved triangulation with the properties you desire.

I think that the question needs to be formulated more carefully.

Source Link
Liviu Nicolaescu
  • 34.7k
  • 2
  • 91
  • 165

I think that you need to state very clearly what you mean by triangulation of the manifold, and what do you mean by "angles". (On a manifold that would require a choice of metric.) If by triangulation you mean the affine realizeation ofba finite simplicial complex inside some Euclidean space, and you measure the angles using the induced metric then the answer is no.

Consider the case $n=4$. The affine simplicial complex with $4$ vertices and homeomorphic to a manifold is given by the boundary of a $3$-dimensional simplex. In this case the curvature can only be positive, and clearly one can produce metrics on the $2$-sphere that are negative somewhere.

On the other hand, if you allow the faces of this tetrahedron to be curved then the answer could be positive. Here is one plausible solution.

On $S^2$ fix four points $p_1,\dotsc, p_4$ and a metric whose curvature at $p_i$ is $f(p_i)$. such a metric exists by Kazhdan-Warner. Next connect the points by geodesic arcs, and assume you can do this so that these arcs do not intersect in the interior. You now have a curved triangulation with the properties you desire.

I think that the question needs to be formulated more carefully.