A simple result on homotopy in Riemannian manifolds

This is a simple result on the homotopy of two curves with the same endpoints in a Riemannian manifold {(M, g)}, which is raised in today’s Riemannian geometry lesson. Professor sketched the proof of the result but there’s one point which seems unclear (at least to me), and I also have some doubt about the statement of the result. I am posting it here in the hope that some of you can actually help me improve the statement of the following result.

Theorem 1 Let {\alpha: [0, 1]\rightarrow M} be a continuous curve, then there exists {\epsilon>0} (depending on {\alpha}) such that if {\beta:[0, 1]\rightarrow M} is another continuous curve with the same endpoints, satisfying

\displaystyle  \max_{t\in[0,1]}d(\alpha(t), \beta(t))<\epsilon. \ \ \ \ \ (1)

Then {\alpha} and {\beta} are homotopic relative to their endpoints. i.e. there exists continuous {\sigma=\sigma(s, t): [0, 1]\times [0,1]\rightarrow M} such that {\sigma(0, t)=\alpha(t)}, {\sigma(1, t)=\beta(t)}, {\sigma(s, 0)=\alpha(0)=\beta(0)} for all {s} and {\sigma(s, 1)=\alpha(1)=\beta(1)} for all {s}.

Remark 1 In the lecture, it was said that {\epsilon} can be chosen to be half of the injectivity radius of M (if it is positive). I am not able to show this, and my {\epsilon} cannot be explicitly written down (due to inverse function theorem argument, see below).

Proof: Let us define the map {E: U\subset T(M)\rightarrow M\times M} by {E(p, v)=(p, \exp_p(v))}, where {U} is an open set (which can be chosen to be maximal) containing the zero section {0} of {T(M)} such that the exponential map is well-defined. Then by inverse function theorem, {E} is a diffeomorphism in an open set {U'} of {U}, containing {0}. This is because {E} is {C^1} and the differential of {E} at a “point” {(p, 0)\in T(M)} is {\begin{pmatrix} I &I\\ 0 &I \end{pmatrix}} which is invertible. (I’ve omitted some details. )

Let {V=E(U')} which is an open set in {M\times M}, containing the diagonal {\{(p, p)\in M\times M: p\in M\}}. Let {\epsilon} be so small that if {\beta} satisfies (1) then {(\alpha(t), \beta(t))\subset V} and by shrinking {\epsilon} further, that {\beta(t)} lies in the convex normal neighborhood of {\alpha(t)} for all {t} (i.e. every two points in this neighborhood can be joined by a unique minimal geodesic which is contained in this neighborhood). Then for all fixed {t}, there is a unique minimal geodesic {\sigma_t=\sigma_t(s): [0, 1]\rightarrow M} such that {\sigma_t(0)=\alpha(t), \sigma_t(1)=\beta(t)}. Now we define {\sigma(s, t)=\sigma_t(s)}. Note that we then have {\sigma(0, t)=\alpha(t)}, {\sigma(1, t)=\beta(t)}, {\sigma(s, 0)=\alpha(0)=\beta(0)} for all {s} and {\sigma(s, 1)=\alpha(1)=\beta(1)} for all {s}.

It remains to show that {\sigma} is continuous (on the square {[0,1]\times [0, 1]}). By our construction, the inverse of {E|_{U'}} (which for convenience still denote by {E}) exists on {V}. So let {(\alpha(t), V(t))=E^{-1}(\alpha(t), \beta(t))}, where {V(t)\in T_{\alpha(t)}(M)}, {V(t)} will be continuous. It is then easy to see that

\displaystyle \sigma(s, t)=\exp_{\alpha (t)}(s V(t))

which is continuous. \Box

This entry was posted in Geometry. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s