Nonnegative Isotropic Curvature Conditions
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The complex sectional curvature of a manifold is defined by extending the curvature endomorphism to the complexified tangent bundle bilinearly.
We say that a Riemannian manifold $(M,g)$ is PIC (positive isotropic curvature), if the complex sectional curvature on all isotropic planes is positive, $rm{PIC1}$ is $rm{PIC}$ on $M times mathbb{R}$ and $rm{PIC2}$ is $rm{PIC}$ on $M times mathbb{R}^2$.
It is easy to prove that $rm{K} > 0$ (sectional)) implies $rm{Ric} > 0$ (Ricci) implies $rm{Scal} > 0$ (scalar). Also, it is easy to prove that $rm{PIC2}$ implies $rm{PIC1}$ implies $rm{PIC}$.
Are the following implications true:
$rm{PIC2}$ implies $rm{K} > 0$
$rm{PIC1}$ implies $rm{Ric} > 0$
$rm{PIC}$ implies $rm{Scal} > 0$
How would one go about proving these implications and are there any sources for this?
analysis differential-geometry curvature
asked Nov 16 at 14:06
AlexError
3816
add a comment |
up vote
0
down vote
favorite
The complex sectional curvature of a manifold is defined by extending the curvature endomorphism to the complexified tangent bundle bilinearly.
We say that a Riemannian manifold $(M,g)$ is PIC (positive isotropic curvature), if the complex sectional curvature on all isotropic planes is positive, $rm{PIC1}$ is $rm{PIC}$ on $M times mathbb{R}$ and $rm{PIC2}$ is $rm{PIC}$ on $M times mathbb{R}^2$.
It is easy to prove that $rm{K} > 0$ (sectional)) implies $rm{Ric} > 0$ (Ricci) implies $rm{Scal} > 0$ (scalar). Also, it is easy to prove that $rm{PIC2}$ implies $rm{PIC1}$ implies $rm{PIC}$.
Are the following implications true:
$rm{PIC2}$ implies $rm{K} > 0$
$rm{PIC1}$ implies $rm{Ric} > 0$
$rm{PIC}$ implies $rm{Scal} > 0$
How would one go about proving these implications and are there any sources for this?
analysis differential-geometry curvature
asked Nov 16 at 14:06
AlexError
3816
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The complex sectional curvature of a manifold is defined by extending the curvature endomorphism to the complexified tangent bundle bilinearly.
We say that a Riemannian manifold $(M,g)$ is PIC (positive isotropic curvature), if the complex sectional curvature on all isotropic planes is positive, $rm{PIC1}$ is $rm{PIC}$ on $M times mathbb{R}$ and $rm{PIC2}$ is $rm{PIC}$ on $M times mathbb{R}^2$.
It is easy to prove that $rm{K} > 0$ (sectional)) implies $rm{Ric} > 0$ (Ricci) implies $rm{Scal} > 0$ (scalar). Also, it is easy to prove that $rm{PIC2}$ implies $rm{PIC1}$ implies $rm{PIC}$.
Are the following implications true:
$rm{PIC2}$ implies $rm{K} > 0$
$rm{PIC1}$ implies $rm{Ric} > 0$
$rm{PIC}$ implies $rm{Scal} > 0$
How would one go about proving these implications and are there any sources for this?
analysis differential-geometry curvature
asked Nov 16 at 14:06
AlexError
3816
The complex sectional curvature of a manifold is defined by extending the curvature endomorphism to the complexified tangent bundle bilinearly.
We say that a Riemannian manifold $(M,g)$ is PIC (positive isotropic curvature), if the complex sectional curvature on all isotropic planes is positive, $rm{PIC1}$ is $rm{PIC}$ on $M times mathbb{R}$ and $rm{PIC2}$ is $rm{PIC}$ on $M times mathbb{R}^2$.
It is easy to prove that $rm{K} > 0$ (sectional)) implies $rm{Ric} > 0$ (Ricci) implies $rm{Scal} > 0$ (scalar). Also, it is easy to prove that $rm{PIC2}$ implies $rm{PIC1}$ implies $rm{PIC}$.
Are the following implications true:
$rm{PIC2}$ implies $rm{K} > 0$
$rm{PIC1}$ implies $rm{Ric} > 0$
$rm{PIC}$ implies $rm{Scal} > 0$
How would one go about proving these implications and are there any sources for this?
analysis differential-geometry curvature
analysis differential-geometry curvature
asked Nov 16 at 14:06
AlexError
3816
asked Nov 16 at 14:06
AlexError
3816
asked Nov 16 at 14:06
AlexError
3816
asked Nov 16 at 14:06
AlexError
3816
asked Nov 16 at 14:06
AlexError
3816
3816
add a comment |
add a comment |
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