Cartan-Hadmard Theorem

{\textbf{Thm: }} Suppose {(M,g)} is a complete manifold

\displaystyle \text{Sec}_M\leq 0

Then {\text{exp}_p:T_pM\rightarrow M} is a covering map. Moreover if {M} is simply connected, then {M} is diffeomorphic to {\mathbb{R}^n}.

Remark: {p} is the north pole of {\mathbb{S}^n}, {\text{exp}_p:\mathbb{R}^n\rightarrow \mathbb{S}^n} is the exponential map, then {\text{exp}_p^*g} will not be a metric on {\mathbb{R}^n} and {\text{exp}_p^{-1}(-p)} are concentric circles. Proof: Since {\text{Sec}_M\leq 0}, we have no conjugate points of {p}, {d\text{exp}_p} is nonsingular on {T_pM}. So {\tilde{g}=\text{exp}_p^*g} will be a metric on {M}. In fact {(T_pM,\tilde{g})} is a complete manifold. This is implied from Hopf-Rinow thm and the fact that lines through origin of {T_pM} are geodesics under {\tilde{g}} and they can be extended to infinity.

Since {\text{exp}_p} is a local isometry between {(T_pM,\tilde{g})} and {(M,g)}, then

\displaystyle \{V_i|\text{exp}_pV_i=q\}

is a discrete set. Choose {0<\delta<\text{injrad}(q)}. Denote {\text{exp}_p=\phi}

\displaystyle \phi^{-1}(B_\delta(q))=\cup_iB_{\delta}(V_i)

\displaystyle \phi: B_{\delta}(V_i)\rightarrow B_\delta(q)\text{ is homeomorphism}

\displaystyle B_i\cap B_j=\emptyset,\quad i\neq j

namely {\phi} will be the covering map.

For any {p'\in B_\delta(q)}, there exists a geodesic {\gamma} connects {p'} and {q}. For any {W_i\in \text{exp}^{-1}_pp'}, there exists a local lifting of {\gamma}, say {\tilde{\gamma}}, such that {\gamma=\phi\circ\tilde\gamma}. Since {\phi} is a local isometry, then it is length preserving. Since {\tilde{\gamma}} ends up at some {V_{i_0}}, then {W_i} must be in {B_\delta(V_{i_0})}.

Next {\phi:B_{\delta}(V_i)\rightarrow B_\delta(q)} is injective. Otherwise suppose {W_1, W_2\in B_{\delta}(V_i)} with {\phi(W_1)=\phi(W_2)=p'}. Suppose {\tilde\gamma_1} and {\tilde\gamma_2} are geodesics connecting {V_i} and {W_1} and {W_2} respectively, then both {\gamma_1=\phi\circ\tilde\gamma_1} and {\gamma_2=\phi\circ\tilde\gamma_2} will connect {q} and {p'}. {\gamma_1} and {\gamma_2} will be two different geodesics because {\phi} is a local isometry. This can not happen because {p'} is inside the injective domain.

{\phi:B_{\delta}(V_i)\rightarrow B_\delta(q)} is onto. Suppose {\text{exp}_qtv} be the minimal geodesic from {q} to {p'}. Then there exist {w\in T_{V_i}T_pM} such that {d\phi (w)=v}, then {\phi(\text{exp}_{V_i}tw)} will be a geodesic in {M} with initial speed {v}. Therefore it must concide with {\text{exp}_ptv}. Since {\phi} preserve the length, we have {p'\in (B_\delta(V_i))}. {\phi} is onto.

Suppose {w\in B_\delta(V_i)\cap B_\delta(V_j)}, {\tilde{\gamma}_i} and {\tilde\gamma_j} are two minimal geodesics from {w} to {V_i} and {V_j}. Then {\tilde\gamma_i} and {\tilde\gamma_j} are different near {w}, then {\gamma_i=\phi\circ\tilde{\gamma}_i} and {\gamma_j=\phi\circ\tilde{\gamma}_j} will be two different geodesics connecting {q} and {\phi(w)}. Contracdtion implies {B_\delta(V_i)} and {B_\delta(v_j)} are disjoint. \Box

Remark: If we change the restriction to {\text{Ric}_M\leq 0}, the theorem will not be true. Counter example comes from the Calabi-Yau manifold, which is compact, simply connected and has vanishing ricci curvature.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

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

You are commenting using your 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 )

Google+ photo

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

Connecting to %s

%d bloggers like this: