Lifting curves from Riemannian submersions: reconciling claims from two books

Question

Let $\pi \colon \bar M \to M$ be a Riemannian submersion.

Let $c \colon I \to M$ be a smooth curve on $M$.

I am interested in lifting $c$ to a horizontal smooth curve $\bar c$ on $\bar M$, so that $c = \pi \circ \bar c$ and $\bar c'(t)$ is horizontal (that is, normal to the fiber at $\bar c(t)$) for all $t$.

The following are my understanding of two claims in two differential geometry books:

Lang (Fundamentals of differential geometry, 1999) says it is possible to lift $c$ for all $t$, that is, such that $\bar c$ is well defined on the whole interval $I$. -- see XIV, Prop. 3.5, p386.

Gallot, Hullin and Lafontaine (Riemannian geometry, 2004) say it can only be done locally in general. The counter-example they give involves a situation where $\pi$ is not surjective (so, while it is a Riemannian submersion, it is not a quotient map), and $c$ passes through points of $M$ that have no pre-image in $\bar M$. -- See Prop. 2.109 (counter-example in the proof).

I do not see where Lang assumes $\pi$ to be surjective. I have two questions:

1. Did I miss where Lang excludes the situation noted by Gallot et al.?

2. If we add the requirement that $\pi$ be surjective, is that enough to guarantee that we can indeed lift $c$ to a unique horizontal curve $\bar c$ for all $t$?

Answer 1

Lang's proposition is just wrong as stated. I don't see any place in that chapter where he assumes that $\pi$ is surjective (or anything else, other than a Riemannian submersion).

In fact, even surjectivity is not sufficient. For a counterexample, take $M=\mathbb S^1\subseteq\mathbb R^2$ and $\widetilde M = \mathbb (-1,1)\times\mathbb R\subseteq\mathbb R^2$, both with their standard Riemannian metrics, and let $\pi\colon \widetilde M\to M$ be the map $\pi(x,y) = (\cos 2\pi x,\sin 2\pi x)$. Then $\pi$ is a surjective Riemannian submersion. But if $\alpha\colon [0,2]\to \mathbb S^1$ is the curve $\alpha(t) = (\cos 2\pi t, \sin 2\pi t)$, then the unique horizontal lift $A$ of $\alpha$ satisfying $A(0) = (0,0)$ is $A(t) = (t,0)$, and this curve cannot be extended beyond the subinterval $[0,1)\subseteq[0,2]$.

It's not enough to assume that $\pi$ is a quotient map, either. The map $\pi$ defined above is a surjective smooth submersion, and therefore it's a quotient map. (See Proposition 4.28 in my Introduction to Smooth Manifolds, 2nd ed.)

There are several sufficient conditions that will guarantee that every curve has a horizontal lift defined on its entire domain. As Nicolas Bournal pointed out, Besse shows that one such condition is that $\widetilde M$ is complete. (It need not be connected, as long as each component is geodesically complete.)

Another sufficient condition is that $\pi$ be a proper map (meaning the preimage of every compact subset is compact). The idea of the proof is that if a lift is defined only on a proper subinterval, then its image lies in a compact subset of $\widetilde M$, and you can use local lifting together with local uniqueness to extend the lift a little farther.

Neither condition (completeness nor properness) is necessary, though. Here's another counterexample. Let $M=\mathbb S^1$ as before and $\widetilde M = \mathbb R\times (-1,1)\subseteq\mathbb R^2$, with $\pi(x,y) = (\cos 2\pi x,\sin 2\pi x)$. If $\alpha\colon [a,b]\to\mathbb S^1$ is any smooth curve, there is a global function $A_0\colon [a,b]\to \mathbb R$ such that $\alpha(t) = (\cos 2\pi A_0(t),\sin 2\pi A_0(t))$ (by the theory of covering spaces), and then the horizontal lifts of $\alpha$ to $\widetilde M$ are of the form $A(t) = (A_0(t), y_0)$ for $y_0\in (-1,1)$, all of which are defined on the whole domain of $\alpha$. In this case, $\widetilde M$ is not complete, and the map $\pi$ is not proper.

I don't know of any simple necessary and sufficient condition. Besse says that the horizontal distribution is called Ehresmann complete if every curve has a horizontal lift starting at any point in the initial fiber and defined on the whole interval.

Answer 2

Michor (Topics in differential geometry, 2008) gives a rather detailed treatment of this question for the case of fiber bundles. Fiber bundles equipped with a 'complete connection' (also called an 'Ehresmann connection') have the property that curves on (what in my case is the) quotient manifold can be lifted entirely.

Michor (pp204-206) gives sufficient conditions for a fiber bundle to have that property, and likewise there are known sufficient conditions for a quotient manifold to be a fiber bundle (see Lee's intro to smooth manifolds Pb 21-6 for example), but it seems that neither sets of sufficient conditions are necessary.

The absence of a stronger statement in these closely related treatments suggests that having a surjective Riemannian submersion may not be sufficient to be able to lift curves entirely (but I do not think these elements allow for a definitive conclusion).

From these references, we can extract the following special case:

If $G$ is a compact Lie group acting smoothly and freely on $\bar M$, then $M = \bar M / G$ is a quotient manifold, and it has the property that any smooth curve $c \colon I \to M$ can be lifted to a horizontal curve $\bar c \colon I \to \bar M$ (with the same domain $I$).

Another relevant reference is Kobayashi and Nomizu Vol. I, Prop. II.3.1 (p69) -- thanks to P.-A. Absil for pointing this out. I did not fully understand the claim and proof in that reference, but it seems that it might offer a stronger conclusion.

EDIT: another useful reference on this topic is: Arthur Besse, Einstein Manifolds, 1987 -- Section 9.E. There, one can read that if $\pi$ is a surjective Riemannian submersion, then it is sufficient (but not necessary) also to require the total space $\bar M$ to be connected and complete in order to have the property that all smooth curves on $M$ (with non-vanishing velocity) can be lifted horizontally on their entire domain. Thanks to Mario Lezcano-Casado and Estelle Massart for pointing this out.