---- > [!definition] Definition. ([[Riemannian exponential map]]) > Let $(M,g)$ be a [[Riemannian manifold]], $p \in M$ and $D$ the [[Levi-Civita connection]] on $M$. Let $p \in M$. > The map $\begin{align} \exp_{p}: T_{p}M &\to M \\ a & \mapsto \gamma_{p}(1, a) \end{align}$ is called the **(Riemannian) exponential map at $p \in M$**. It is [[well-defined]] and smooth on $B_{g}(0, \varepsilon) \subset T_{p}M$ for $\varepsilon>0$ small enough. >![[Pasted image 20250521111321.png|500]] [^1] > [!definition] Definition. (Logarithmic map) > Since $\exp_{p}$ is smooth around the origin (viewing $T_{p}M$ as $(\mathbb{R}^{n}, g)$), can look at the [[differential of a smooth map between smooth manifolds|differential]] $(d \exp_{p}) |_{a=0}:T_{0}(T_{p}M) \to T_{p}M$ there. After canonically identifying $T_{0}(T_{p}M)$ with $T_{p}M$, one can calculate that $(d \exp_{p}) |_{a=0}$ is just the [[identity map]] on $T_{p}M$.[^1] By the [[inverse function theorem]], there exists a ball of radius (say) $r_{0}$ about $0 \in T_{p}M$ such that $\exp_{p}:B_{g}(0, r_{0}) \to M$ is a [[diffeomorphism]] onto its image $U \subset M$. The inverse map $\log_{p}=\exp_{p} ^{-1}:U \to \mathbb{R}^{n}$ is called the **logarithmic map around $p$**. ^definition [^1]: Using the calculation of [[differential of a smooth map between smooth manifolds|differential]] via the curve $tX$ passing through $X$... $\begin{align} (d\operatorname{exp}_{p} |_{0})X &=\frac{d}{dt} |_{t=0} \operatorname{exp} (tX) \\ &= \frac{d}{dt} |_{t=0} \gamma_{p}(1, tX) \\ &= \frac{d}{dt} |_{t=0} \gamma_{p}(t, X) \\ & = \dot{\gamma} _{p}(0) \\ & = X. \end{align}$ > [!basicproperties] > For points in $T_{p}M$ about which $\exp_{p}$ is defined but fails to be an [[smooth immersion|immersion]], i.e. those points $v \in T_{p}M$ for which $d \exp_{p} |_{v}X=0$ for some $X \neq 0$, we have that the the geodesic $\exp_{p}(tv)$ has [[conjugate points|conjugate endpoints]]. (Picture: sphere) > The [[variation]] witnessing this is $f(s,t)=\exp_{p}(t (v+sX))$, with $J(t)=\frac{ \partial }{ \partial s } |_{s=0} \exp_{p}(t(v+sX))$. Indeed, since $f(s, 0) \equiv p$, $J(0)=0$. And $\begin{align} J(1)&= \frac{ \partial }{ \partial s } |_{s=0} \exp_{p}( v+sX) \\ &= d \exp_{p} |_{v} X \\ &=0. \end{align}$ ^proposition ---- #### [^1]: Image taken from: https://www.researchgate.net/figure/The-exponential-map-allow-us-to-locally-carry-the-Euclidean-structure-of-the-tangent_fig5_311807546 ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```