By Detlef Gromoll
In the earlier 3 or 4 a long time, there was expanding recognition that metric foliations play a key function in knowing the constitution of Riemannian manifolds, quite people with optimistic or nonnegative sectional curvature. in reality, all recognized such areas are constituted of just a consultant handful through metric fibrations or deformations thereof.
This textual content is an try to record a few of these buildings, lots of that have basically seemed in magazine shape. The emphasis here's much less at the fibration itself and extra on the best way to use it to both build or comprehend a metric with curvature of mounted sign up a given space.
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Additional resources for Metric Foliations and Curvature
7. The Riccati equation for Jacobi ﬁelds 35 F = π −1 (π(c(0))). 1, if t0 ∈ I is a focal point of F along c, then t0 is a conjugate point of π ◦ c: in fact, the Jacobi ﬁeld J above is projectable at 0, hence everywhere by (i). π∗ J is then a Jacobi ﬁeld that vanishes at 0 and t0 . It turns out that the order of focal points of c and conjugate points of π ◦ c, as well as the indexes of these geodesics, are the same, see  for details and further results. 7 The Riccati equation for Jacobi ﬁelds When dealing with Jacobi ﬁelds, it is often useful to decompose the second-order Jacobi diﬀerential equation into two ﬁrst-order ones.
As we did for S, it is straightforward to check that A(t) is a well-deﬁned linear map. As usual, assume that Y (t0 ) = J(t0 ) for Y = J h as above, and consider orthonormal Dh -parallel ﬁelds X1 , . . Xd with X1 (t0 ) = J(t0 ). Notice that for Z ∈ V, J , Z (t0 ) = J, Z (t0 ) = J, AZ (t0 ), so that Similarly, J v (t0 ) = A∗ (t0 )J(t0 ). 6) imply and Xi h ≡ 0. 7. The Riccati equation for Jacobi ﬁelds and 39 Dh2 Y + Rh Y + 3AA∗ Y = 0. 4), we conclude S + S 2 + Rh + 3AA∗ = 0. 2 (Wilking, ). Let J denote an (n−1)-dimensional space of Jacobi ﬁelds orthogonal to a geodesic c : R → M n , with self-adjoint Riccati operator.
If J is a Jacobi ﬁeld along a geodesic c of M , then |J|2 = J, J = 2 J ,J = 2( J , J + J , J ) = 2(|J |2 − R(J, c) ˙ c, ˙ J ) ≥ 0. If F is totally geodesic, then the holonomy ﬁelds have constant norm, so that they must be parallel by the preceding inequality. 6) then implies that A∗c˙ J ≡ 0 for a holonomy ﬁeld along a horizontal geodesic c. Since c˙ and J are arbitrary, A ≡ 0, and M splits locally as a metric product. In particular, a negatively curved manifold admits no totally geodesic metric foliations.
Metric Foliations and Curvature by Detlef Gromoll