In our DiffGeo-lectures we had a theorem that the differential of the local exponential map can be expressed through Jacobi-Fields:

Let $M$ be a Riemannian manifold, $p\in M, X\in T_pM$ such that $exp(X)$ is defined. Then for $W\in T_pM\simeq T_XT_pM$ it holds that $d_X exp_p(W)=Y(1)$ where $Y$ is the Jacobi-Field along $\gamma:[0,1]\to M, t\mapsto \gamma(t)=\exp(tX)$ with $Y(0)=0, Y'(0)=W$.

In the proof we said $d_X \exp_p(W)=\frac{d}{dt}\big|_{t=0} \exp(X+tW)$ and we chose the geodesic variation $\alpha: [1,0] \times ]-\varepsilon, \varepsilon[ \to M, (s,t)\mapsto \exp(s(X+tW))$.

Then we defined $Y\in\Gamma(\gamma^*TM)$ is defined by $Y(s)=\frac{d}{dt}\big|_{t=0}\alpha(s,t)$ and we concluded $d_X\exp(W)=Y(1)$.

I checked all this equations to really understand what we did there:

$Y(s)=\frac{d}{dt}\big|_{t=0}\alpha(s,t) = \frac{d}{dt}\big|_{t=0} \exp(s(X+tW)) = sW\exp(sX). \\d_X \exp_p(W)=\frac{d}{dt}\big|_{t=0} \exp(X+tW) = W\exp(X).$

So obviously $Y(1)=d_X\exp_p(W)$ and $Y(0)=0_X \in T_XT_pM$.

So this is all fine but here is what I don't get:

$Y'(s) = \frac{d}{ds} sW\exp(sX) = W\exp(sX)+sWX\exp(sX)$, hence $Y'(0) = W\exp(0)$.

But isn't $exp(0_p)=p$ as the geodesic for $0_p$ is just $\mu:[0,1]\to M, t\mapsto p$, hence $exp(0_p)=\mu(1)=p$ so why does it hold that $Y'(0)=W$?

I guess I'm also confused when we do things in $T_pM$ and when we do things in $T_XT_pM$.

I would like to get some explanations on this.


1 Answer 1


First, let me highlight your last question about the tangent space of $T_pM$. Let $\exp_p : T_p M \to M$ be the exponential map. It is a smooth map away from the cut locus of $p$. If $v$ is a point where $\exp_p$ is differentiable, then its differential at $v$ is a map $$ \mathrm{d}\exp_p(v) : T_vT_pM \to T_{\exp_p(v)}M. $$ Now, remember that $T_pM$ is a vector space, hence, as a manifold, its tangent bundle is trivial (which is more than trivializable!): there is a canonical identification $T_vT_pM \simeq T_pM$. We always use this identification in Riemannian geometry, and we say that the differential of $\exp_p $ at $v$ is a linear map: $$ \mathrm{d}\exp_p(v) : T_pM \to T_{\exp_p(v)}M. $$ Now, let us answer you main question. Let $w \in T_pM$ and let's try to identify what $\mathrm{d}\exp_p(v)\cdot w$ is. Consider the geodesic segment $\gamma : [0,1] \to M$ defined by $\gamma(t) = \exp_p(tv)$. In the following, I will keep $t$ as the parameter of the geodesic $\gamma$.

Let $s$ be a parameter in $(-\varepsilon,\varepsilon)$ with $\varepsilon >0$ and define the function $f(s,t) = \exp_p\left(t(v + sw)\right)$. It follows from the definition that $f(0,\cdot) = \gamma$, and that $\gamma_{s} = f(s,\cdot)$ is a geodesic. Hence, by a classical result of Riemannian geometry, the vector field $Y(t) = \frac{\partial \gamma_{s}(t)}{\partial s}|_{s=0}$ is a Jacobi field along $\gamma$. Moreover, by the formula we defined, it follows that $$ Y(t) = \left.\frac{\mathrm{d}}{\mathrm{d}s}\right|_{s=0} \exp_p\left(t(v+sw)\right) = \mathrm{d}\exp_{p}(t(v+sw))|_{s=0}\cdot \left.\frac{\mathrm{d}(t(v+sw)}{\mathrm{d}s}\right|_{s=0} = \mathrm{d}\exp_p(tv)\cdot(tw). $$ This is basically the chain-rule: $(f\circ g)'(t) = f'(g(t))\cdot g'(t)$. At $t=0$, this field satisfies $Y(0) = 0$ and at $t=1$, it satisfies $Y(1) = \mathrm{d}\exp_p(v)\cdot w$, which is what we wanted to compute.

Now, as a Jacobi field is determined by $Y(0)$ and $Y'(0)$, we have to indentify $Y'(0)$ to completely know $Y$, and thus, $Y(1) = \mathrm{d}\exp_p(v)\cdot w$. But if $\frac{\nabla}{\mathrm{d}t}$ denotes the covariant derivative along $\gamma$, then: \begin{align} Y(0) &= \left.\frac{\nabla}{\mathrm{d}t}\right|_{t=0}Y\\ &= \left.\frac{\nabla}{\mathrm{d}t}\right|_{t=0} \left.\frac{\partial}{\partial s}\right|_{s=0} f \\ &= \left.\frac{\nabla}{\mathrm{d}s}\right|_{s=0} \left.\frac{\partial}{\partial t}\right|_{t=0}f ~~~ \text{by a lemma of Riemannian geometry}\\ &= \left.\frac{\nabla}{\mathrm{d}s}\right|_{s=0} \left(v+sw \right)\\ &= w \end{align} because $\partial/\partial t |_{t=0} f(s,t)$ is the tangent vector at $0$ of the geodesic $\gamma_s$, that is, $v+sw$. The result follows.


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