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2
votes
Is a linear vector field a geodesible vector field?
This observation seems to be very easy,
but it takes care of many examples.
Suppose that $A$ corresponds to a contraction
(that is, all eigenvalues are $< 1$ in absolute value).
Decompose ${\Bbb R}^n …
9
votes
2
answers
361
views
Smooth rank one foliations with closed leaves
When the foliation is (transversally) Riemannian, this is proven in the book "Riemannian Foliations" by P. Molino. …
5
votes
Smooth rank one foliations with closed leaves
Wadsley, Geodesic foliations by circles,
J. Differential Geometry {\bf 10} (1975), no. 4, 541--549.) …
3
votes
Foliations by holomorphic curves on complex surfaces
This is a question which is related to Lagrangian foliations on hyperkaehler manifolds, but much of the results are conjectures, or unpublished. … The tangent bundle of a general non-algebraic K3 has no rank 1
coherent subsheaves, hence this K3 has no holomorphic foliations. …
4
votes
0
answers
152
views
Calabi–Yau theorem and complex Monge–Ampère equation for transversally Kähler manifolds
For taut foliations, one has also
Poincaré duality on the basic cohomology, and the
identification between the basic cohomology and the basic
harmonic forms if a transversal Riemannian structure
is given …