Intuition and existence for convex neighbourhoods

[center o bass]

Im trying to get some intuition for convex neighbourhoods which is neighbourhoods $U$ such that for any two points $p$ and $q$ in U there exists a unique geodesic connecting $p$ and $q$ staying within $U$.

QUESTION 1: These kind of neighbourhoods can be shown to always exist for Riemannian manifolds something i find a bit puzzling. Can someone offer some intuition on why such neighbourhoods must always exist?

QUESTION 2: Furhermore -- and more technical -- are their existence fundamentally related to the exponential map in the following way?

For any Riemannian manifold there always exists neighbourhoods $U$ of $p$ and $V$ of the origin of $T_pM$ such that $\exp_p: V \to U$ is a diffeomorphism (it is smoothly invertible). This means that for a point $q$ in $U$, we are able to get to the unique tangent vector $v \in T_pM$ with the property that the unique geodesic whose tangent vector at $p$ is $v$ satisfies $\gamma_v(1) =q$. In other words, there must exist a unique geodesic going through $p$ and $q$ -- for were it not unique, then we would not be able to pick out a unique vector $v$ with the above property and thus $\exp_p$ would not be a diffeomorphism within this region.

 

[wabbit]

For 1 : In short, locally a Riemannian manifold is just (an open ball in) $\mathbb{R}^n$. And of course you know the answer for flat space.

Longer form : You can always diagonalize your metric at a point, and also (not really needed) rescale to obtain the form $\sum dx_i^2$. Now around that point the metric won't have that form, but it is smooth, so it will be close to that. And by taking a small enough neighborhood you can make the deviations negligible, so you are left with something as close as you want to make it to an open ball in flat space with its usual metric.

For 2 : Yes, you can use that, and it's probably easier than to expand the argument above into a fully rigorous proof. It requires of course that you have already proven that the exponential map is a local diffeomorphism.

 

[lavinia]

QUESTION 2: Furhermore -- and more technical -- are their existence fundamentally related to the exponential map in the following way?

For any Riemannian manifold there always exists neighbourhoods $U$ of $p$ and $V$ of the origin of $T_pM$ such that $\exp_p: V \to U$ is a diffeomorphism (it is smoothly invertible). This means that for a point $q$ in $U$, we are able to get to the unique tangent vector $v \in T_pM$ with the property that the unique geodesic whose tangent vector at $p$ is $v$ satisfies $\gamma_v(1) =q$. In other words, there must exist a unique geodesic going through $p$ and $q$ -- for were it not unique, then we would not be able to pick out a unique vector $v$ with the above property and thus $\exp_p$ would not be a diffeomorphism within this region.

You need to do a little more work because in theory the length minimizing geodesic connecting two other points in U (two points other than p) may wander outside of U. If that happens U is not convex.

 

[center o bass]

You need to do a little more work because in theory the length minimizing geodesic connecting two other points in U (two points other than p) may wander outside of U. If that happens U is not convex.

How would I go ahead and proving that the geodesic will in fact not wander out of U? Furthermore, if that is proven, my argument only shows that we can get to map any other poing in $U$ back to $p$. How would I now extend this to any two points in this region -- not just $p$ and any other point -- being connected by a unique geodesic?

 

[lavinia]

How would I go ahead and proving that the geodesic will in fact not wander out of U? Furthermore, if that is proven, my argument only shows that we can get to map any other poing in $U$ back to $p$. How would I now extend this to any two points in this region -- not just $p$ and any other point -- being connected by a unique geodesic?

I don't know this proof in detail but Wabbit has given the right clue, that the rescaled metric using geodesic normal coordinates in a small enough region more or less approximates the flat metric.

BTW: The "rescaled metric" is the Euclidean distance in the tangent space at the point of origin of the geodesic polar coordinate system applied to the the inverse of the exponential map.

Intuitively, in the flat metric, a geodesic - i.e. a straight line - that does not start at the origin of a polar coordinate system has a minimum distance to the origin and at this point, it will be tangent to a sphere that is centered at the origin. This minimum is the only critical point of the distance function of the line to the origin. However a curve that leaves the coordinate system and then returns will on the contrary have a maximal distance to the origin and so can not be a geodesic. In the general case, one would show that in a small enough region the rescaled metric again can only have a critical point that is a minimum along a non-radial geodesic.

- Without proving convexity, one can always show that there is an open neighborhood of any point in the manifold so that any two points in the neighborhood are connected by a unique geodesic of length less than a small enough number, ε. The thing is that this geodesic might leave the neighborhood then return. To prove this weaker theorem one proves that the map,

$(x,v) -> (x,exp_{x}(tv))$

from the tangent bundle into the cartesian product of the manifold with itself, MxM, is non-singular at t = 0 so by the Implicit Function Theorem there is a neighborhood (in the tangent bundle) of any point,x, that is mapped diffeomorphically into a neighborhood in MxM. One can choose this neighborhood so that the tangent vectors are all of length less that some small,ε. Now in MxM choose a sub-neighborhood of the form, BxB. This can be done because the tangent bundle is locally trivial. Then each point in B is connected by a unique geodesic of length less than ε.

The exponential map, $exp_{x}(tv)$ itself, exists from general theorems about the existence of solutions to ordinary differential equations and the relation that for any constant, k, g(kt) is a geodesic whenever g(t) is a geodesic. This is easily seen from the form of the differential equation for a geodesic.

 

[wabbit]

Reading that last exchange I realize that my statement above "It would probably be easier" needs to be taken with a grain of salt (or two)