# Understanding: Differental of Riemannian exponential expressed through Jacobi-Fields

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.

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.