On the map $E: O\subset TM \to M \times M$ by $E(v)=(\pi(v), \exp(v) )$

Question

Let $\exp$ be the exponential map on the Riemannian manifold M and $O$ is its domain in $TM$. Consider the map $E: O\subset TM \to M \times M$ by $E(v)=(\pi(v), \exp(v) )$, where $\pi$ is the canonical map $TM \to M$. It is easy to show that $dE: T_{0_p}(TM) \to T_{(p,p)}(M\times M)$ is nonsingular and thus the inverse map theorem gives local diffeomorphisms via $E$ of neighborhoods of $0_p \in TM$ onto neighborhoods of $(p,p)$ for every $p\in M$. My question is:

How to show that $E$ can be extended to a diffeomorphism from a neighborhood of the zero section of $TM$ onto an open neighborhood of the diagonal in $M\times M$ ? (cf. the last paragraph of Page 131 of Riemannian Geometry (GTM171) written by Peter Petersen)

The book has a sketch of proof, but I cannot fully understand. The main difficulty may be how to guarantee this diffeomorphism is injective, thus I cannot simply fit those local diffeomorphisms together to give the desired diffeomorphism.

Please help, thanks!

Answer 1

If the map fails to be injective then $(\pi(v),\exp (v))=(\pi(w),\exp(w))$ for some $v\not=w$. But this implies in particular that $\pi(v)=\pi(w)$, i.e., both vectors are over the same point of $M$. Thus you are reduced to a local problem after all, namely that of the exponential map being a local diffeomorphism, which follows from a general fact about ordinary differential equations.

Answer 2

For each $x \in M$, let $D_x \subseteq T_x M$ be the largest set on which $\exp_x$ can be defined as a diffeomorphism (its boundary $\partial D_x$ is the cut locus of $x$ in $T_x M$). Notice that $D = \bigcup \limits _{x \in M} D_x$ is contained in $O$. For brevity, I shall denote the restriction $\exp \big| _D$ by $\exp$ again. Notice that $\exp_x D_x$ is an open subset of $M$, diffeomorphic to $D_x$ by construction (its complementary is the cut locus of $x$ in $M$, a closed subset of measure $0$). Clearly, then, $\exp D$ is an open tubular neighbourhood of the diagonal in $M \times M$ (the zero section of $TM$ being transported onto the diagonal itself).

Notice that $\exp$ is injective: if $\exp (u) = \exp (v)$ for some $u, v \in D$, there are two possibilities:

• if $\pi (u) \ne \pi (v)$, clearly $\exp (u) = \exp (v)$;
• if $\pi (u) = \pi (v) = x$, and $\exp_x (u) = \exp_x (v)$, then since $\exp_x$ is a diffeomorphism on $D_x$ (by construction of $D_x$) you get that $u=v$.

Notice that $\exp : D \to \exp(D)$ is also trivially surjective; it is, therefore, bijective, and has an inverse $\exp ^{-1}$.

The map $\exp : D \to \exp(D)$ is smooth (this is proved using facts about the dependence of ODEs on their initial conditions). Since $\textrm d (\exp)$ is invertible at the zero section of $TM$, you can restrict yourself to some open $U_x \subseteq D_x$ and $U = \bigcup \limits _{x \in M} U_x$ such that $\exp \big| _U$ is locally invertible, with a smooth local inverse. But on $\exp (U)$ this local inverse must coincide locally with $\exp^{-1} \big| _{\exp (U)}$ constructed above, so $\exp^{-1} \big| _{\exp (U)}$ must then be smooth too (because smoothness is a local property). This implies that $(\exp \big| _U)^{-1} = \exp^{-1} \big| _{\exp (U)}$ is a smooth inverse of $\exp \big| _U$, so $\exp \big| _U : U \to \exp(U)$ is a diffeomorphism as desired.

In fact, I believe that $U=D$ and that the restrictions in the above paragraph are merely pedantic, but I don't feel like trying to prove this (it is too technical and meticulous and I lack the necessary patience).