Skip to main content
MathOverflow

Return to Question

added 14 characters in body
Source Link
Kikiriku
Kikiriku
  • 101
  • 1
  • 4

Let $M$ be a smooth oriented manifold. Does there exist a smooth measure $m$ on $M$ which is not induced by the volume form of some Riemannian metric $g$ on $M$? I would say that the set of volume forms induced by Riemannian metrics is strictly contained in the set of all smooth measures on $M$...My interest would be to have some criteria for deciding whether a given measure on $M$ is induced by a Riemannian metric or not

Let $M$ be a smooth oriented manifold. Does there exist a measure $m$ on $M$ which is not induced by the volume form of some Riemannian metric $g$ on $M$? I would say that the set of volume forms induced by Riemannian metrics is strictly contained in the set of all measures on $M$...My interest would be to have some criteria for deciding whether a given measure on $M$ is induced by a Riemannian metric or not

Let $M$ be a smooth oriented manifold. Does there exist a smooth measure $m$ on $M$ which is not induced by the volume form of some Riemannian metric $g$ on $M$? I would say that the set of volume forms induced by Riemannian metrics is strictly contained in the set of all smooth measures on $M$...My interest would be to have some criteria for deciding whether a given measure on $M$ is induced by a Riemannian metric or not

Source Link
Kikiriku
Kikiriku
  • 101
  • 1
  • 4

Measures on Riemannian manifolds which are not induced by the volume form of some Riemannian metric

Let $M$ be a smooth oriented manifold. Does there exist a measure $m$ on $M$ which is not induced by the volume form of some Riemannian metric $g$ on $M$? I would say that the set of volume forms induced by Riemannian metrics is strictly contained in the set of all measures on $M$...My interest would be to have some criteria for deciding whether a given measure on $M$ is induced by a Riemannian metric or not