I'm currently reading the book of Do Carmo but I don't understand something in the proof of Hadamard's theorem. He first proves two lemmas which I understand completely but in the actual proof of the theorem, he states that $\exp_p : T_pM \to M$ is a local isometry because it is a local diffeomorphism. However, he then says that the induced metric on $T_pM$ is complete because the geodesics that pass through the origin are straight lines. I don't really see how this follows and I was wondering if someone could help me understand.
Proof of Hadamard’s theorem in book of Do Carmo
differential-geometrygeodesicriemannian-geometry
Related Solutions
Regarding your comment above: yes, if $M$ is complete and connected with non-postiive curvature, then the universal cover $\widetilde{M}$ of $M$ inherits a complete connected non-positive curvature metric (just pull-back the Riemannian metric from $M$), and so satisfies the conditions (and so also the conclusions!) of the Cartan--Hadamard theorem. One then finds that $\widetilde{M}$ is homeomorphic to the tangent space of any point $\widetilde{m}$ of $\widetilde{M}$ (via $\exp_{\widetilde{m}}$. If $m$ is the image of $\widetilde{m}$ in $M$, then the tangent space to $\widetilde{m}$ is naturally identified with the tangent space to $m$, and the exponential map $\exp_m$ is naturally identified with the composite of $\exp_{\widetilde{m}}$ and the projection $\widetilde{M} \to M$.
Thus indeed, one finds that $\exp_m$ is a covering map.
This circle of ideas is frequently applied, e.g. in the context of hyperbolic manifolds. If $M$ is a compact connected hyperbolic manifold, then one finds by Cartan--Hadamard that $\widetilde{M}$ is isometric to hyperbolic space of the appropriate dimension, and so $M$ is isometric to a quotient of hyperbolic space by a discrete cocompact group of isometries. (This is why hyperbolic manifold theory interacts so tightly with certain parts of group theory.)
Also, one concludes that if $M$ is compact and connected (and positive dimensional --- i.e. not a single point!) with non-positive curvature then $\pi_1(M)$ is infinite (because, writing $M$ as a quotient of $T_m M$ via $\exp_m$, we see that to get something compact, we have to quotient out by an infinite group of diffeos).
Partial answer
Let $M$ be a complete Riemannian manifold. Then every two points can be joined by a length minimizing geodesic ($\gamma$ is called length minimizing between $p$, $q$ if for any piece-wise smooth curves $\eta$ joining $p$ and $q$ we have $L(\eta)\ge L(\gamma)$). The following are equivalent:
All geodesics in $M$ are length-minimizing.
All points in $M$ are joined by a unique geodesic.
The exponential map $\exp_p : T_pM \to M$ is a diffeomorphism for all $p \in M$.
$(1)\Rightarrow (2)$: If not, let $p, q\in M$ and $\gamma_1, \gamma_2 : [0,1] \to M$ be two geodesics joining $p, q$. Then both $\gamma_1, \gamma_2$ cannot be length minimizing when $t>1$.
$(2)\Rightarrow (1)$ is obvious, since length minimizing curve must be a geodesic and every two points can be joined by at least one geodesic as $M$ is complete.
$(2)\Rightarrow (3)$: By completeness, $\exp_p$ is surjective. If it is not injective, then there are two geodesics which starts at $p$ and interest at some points. Indeed it has to be a diffeomorphism. If not, then there are two points that are conjugate to each other. Thus there are (yet another) two points $p, q$ so that they are joined by a geodesic $\gamma$ which is not length minimizing. By completeness, $p$ and $q$ are also joined by another geodesic $\eta$, thus it contradicts $(2)$.
$(3)\Rightarrow (2)$ is also obvious.
So there are huge topological constraint given by $(3)$. Not completely sure about the curvature constraint, except the one you mentioned.
Best Answer
They are using the following fact:
If $F: (M, g)\to (N, h)$ is a local isometry, then a curve $\gamma$ in $M$ is a geodesic if and only if $F(\gamma)$ is a geodesic in $N$.
In our situation, the straight lines $\{ tw : t\in \mathbb R\}$ passing through the origin is a geodesic in $T_pM$ with the induced metric, since $\exp_p(tw)$ is a geodesic in $M$. Then for all $w\in T_pM \cong T_0(T_pM)$, then geodesic passing through $0$ with vector $w$ is just $t\mapsto tw$ and thus is defined for all $t\in \mathbb R$. Thus implies that $T_pM$ is complete with the induced metric.