Structure of derivative of exponential map

I am trying to calculate the derivative of the exponential map, and I am getting stuck on a few points.

Given a Lie group $$G$$, the Lie algebra $$\mathfrak{g}$$ can either be thought of as left-invariant vector fields of $$G$$ (which I will denote $$\mathcal{Lie}(G)$$) or the tangent space at the identity (denoted $$T_eG$$). Given a left-invariant vector field, there exists a flow $$\phi_X(t)$$ such that $$\phi_X(0)=e$$ and $$\dot{\phi}_X(t)=X_{\phi(t)}$$. The exponential map

$$\text{exp}:g \rightarrow G \\ X \mapsto \phi_X(1)$$ then satisfies certain properties, and in particular $$\text{exp}(tX) = \phi_X(t).$$

I want to show that the derivative of exp at the identity is given by the identity. It seems to follow immediately from looking at

$$D_0\text{exp}(X) = \left.\frac{d}{dt}\right|_{t=0} \text{exp}(tX) = \left.\frac{d}{dt}\right|_{t=0}\phi_X(t) = X_e,$$

where I understand the first equality to hold since we are looking at the derivative at the identity, i.e. with $$X=0$$ (is this correct?).

However, I want to understand the spaces in which everything is defined a bit better. It seems be the case that everything is defined 'up to isomorphism' in a sense, but I would like to clarify is this is correct.

To begin, the Lie algebra is a vector space, and so there is clearly an isomorphism between the Lie algebra $$\mathfrak{g}$$ and the tangent space at the identity of the Lie algebra $$T_0\mathfrak{g}$$. Thus, the above map begins by using this isomorphism so that $$D_0\text{exp}$$ is a map from the Lie algebra $$\mathcal{Lie}(G)$$ (which we consider in terms of vector fields). The map is then to the Lie algebra $$T_eG$$, now in terms of the tangent space. Hence, the result is the identity up to the isomorphism between them.

So to summarise:

Is the first step in my proof correct?

Is my understanding of the map correct?

It is helpful to first understand the definition of the derivatives in the abstract.

I will use the definition that for a smooth manifold $$M$$, $$T_pM$$ is the set of equivalence classes of curves $$\gamma: (-1,1)\to M$$ with $$\gamma(0)=p$$, where two curves are equivalent if their derivative in any chart is equal, i.e. $$\gamma\sim\gamma' \iff \frac{d}{dt}\bigg\rvert_{t=0}\phi\circ \gamma=\frac{d}{dt}\bigg\rvert_{t=0}\phi\circ \gamma'$$ for a chart $$\phi: U\to \mathbb{R}^n$$ a chart containing $$p$$. We say that the derivative of a curve $$\gamma$$ at time $$\tau$$ is the equivalence class of $$\gamma$$, i.e. $$$$\frac{d}{dt}\bigg\rvert_{t=\tau}\gamma:= [\gamma(t-\tau)]\quad (\ast)$$$$

The defining equation for the flow of a vector field is as follows:Given a time dependent diffeomorphism $$\phi: \mathbb{R}\times M\to M$$ and a point $$p\in M$$, we can associate a curve $$\gamma_p: I\to M$$ by $$\gamma_p(t)=\phi_t(p)$$. A time dependent diffeomorphism is said to be the flow for a vector field $$X\in \mathfrak{X}(M)$$ if for each $$p\in M$$ $$[\gamma_p]=X(p)$$ or more suggestively $$\frac{d}{dt}\bigg\rvert_{t=0}\phi_{t}(p)=X(p)$$

We then can easily define the derivative of a map $$f: M\to N$$ as $$D_pf: T_pM\to T_{f(p)}N$$ by $$D_pf[\gamma]=[f\circ \gamma]$$

Now we will look at the specific case of $$f=\exp: \mathfrak{g}\to G$$. If we define $$\exp(tX)$$ to be the flow of the left invariant vector field corresponding to $$X$$, $$\phi_X(t)$$ must satisfy $$\frac{d}{dt}\bigg\rvert_{t=0}\phi_X(t)(g)=X(g)$$ for all $$g\in G$$. Recall that $$D_0\exp: T_0 \mathfrak{g}\to T_{\exp(0X)}G=T_eG$$ and since $$\mathfrak{g}$$ is a vector space we have a canonical isomorphism $$\mathfrak{g}\cong T_0\mathfrak{g}$$. This means that the map $$D_0\exp$$ is really an endomorphism $$\mathfrak{g}\to \mathfrak{g}$$.

$$D_0\exp(X):=\frac{d}{dt}\bigg\rvert_{t=0} \exp(\gamma(t))$$ where $$\gamma(t)$$ is a curve with $$\frac{d}{dt}\bigg\rvert_{t=0} \gamma(t)=X$$, so in this case we can take $$\gamma(t)=tX$$ (under the canonical isomorphism $$T_0\mathfrak{g}\cong \mathfrak{g}$$.)

EDIT: The definition of $$(D_0\exp)(X)$$ is as follows, if $$\gamma: I\to \mathfrak{g}$$ with $$[\gamma]=X$$ then $$D_0\exp(X):=[\exp\circ \gamma]$$. Since $$\frac{d}{dt}\bigg\rvert_{t=0} (tX)=\left(\frac{d}{dt}\rvert_{t=0} t\right)X=X$$ then $$[tX]=X$$ and $$D_0\exp(X)=[\exp(tX)]=\frac{d}{dt}\bigg\rvert_{t=0} \exp(tX)$$ (this last equation comes from the notation given in equation $$(\ast)$$).

• Thanks for your answer. Could you please clarify a few details from the last paragraph. I don't see where your definition $D_0\text{exp}(X):=\left.\frac{d}{dt}\right|_{t=0} \text{exp}(\gamma(t))$ comes from (this seems to imply $X = \gamma(t)$, or have I misunderstood? And in this case, how can we take $\gamma(t)=tX$?). Commented Nov 25, 2022 at 22:10
• I have edited my answer to hopefully address your questions. Commented Nov 25, 2022 at 22:52

Your proof appears to be correct, but here is a short summary.

I also prefer and use below the following notation: Given a smooth map $$F: M \rightarrow N$$, the directional derivative of $$F$$ at $$p \in M$$ in the direction $$V \in T_pM$$ will be denoted $$D_VF(p) \in T_{F(p)}N.$$ By definition, if $$c$$ is a curve such that $$c(0) =p$$ and $$c'(0) = V$$, then $$D_VF(p) = \left.\frac{d}{dt}\right|_{t=0}F(c(t)).$$

Let $$\mathfrak{g} = T_eG$$ and $$\mathfrak{g}_L$$ denote the space of left invariant vector fields. The existence and uniqueness of ODEs implies that the map \begin{align*} \mathfrak{g}_L &\rightarrow \mathfrak{g}\\ X &\mapsto X(e) \end{align*} is a linear isomorphism. The existence and uniqueness of ODEs also leads to the definition of the exponential map as the map \begin{align*} \exp: \mathfrak{g} &\rightarrow G\\ X &\mapsto \phi_X(1), \end{align*} where $$\phi$$ is the solution to the ODE $$$$\tag{*} \frac{d}{dt}\phi_X(t) = X(\phi_X(t)).$$$$ The uniqueness of the solution also implies that for any $$t \in \mathbb{R}$$, $$\exp(tX) = \phi_X(t).$$

Evaluating (*) at $$t=0$$ yields $$\left.\frac{d}{dt}\right|_{t=0}\phi_X(t) = X(\phi_X(0)) = X(e) = X.$$ By this and the definition of the directional derivative, \begin{align*} D_X\exp(0) &= \left.\frac{d}{dt}\right|_{t=0}\exp(tX)\\ &= \left.\frac{d}{dt}\right|_{t=0}\phi_X(t)\\ &= X. \end{align*}

• Thanks for you answer. In the final step, we have that $\phi_X(0)=e$ (the $0$-vector field in $\mathfrak{g}_L$). Additionally, $\phi_X'(0) = X$ in your second last equation. Thus by your original notation, $D_X\text{exp}(e) = \left.\frac{d}{dt}\right|_{t=0}\text{exp}(\phi_X(t))$. But in your third last equation, $\phi_X(t) = \text{exp}(tX)$, and so then $D_X\text{exp}(e) = \left.\frac{d}{dt}\right|_{t=0}\text{exp}(\phi_X(t)) = \left.\frac{d}{dt}\right|_{t=0}\text{exp}(\text{exp}(tX))$, which isn't what you got. Can you tell me what I have misunderstood here? Commented Nov 26, 2022 at 0:31
• Note that the range of $\phi_X$ is $G$ and not $\mathfrak{g}_L$. In particular, $$\phi_X(0) = e \in G,$$ where $e$ is the identity element in $G$. Given $X \in \mathfrak{g}$, the definition of the directional derivative of $\exp: \mathfrak{g} \rightarrow G$ in the direction $X$ is defined to be $$D_X\exp(e) = \left.\frac{d}{dt}\right|_{t=0}\exp(c(t)),$$ where $c$ is any curve in $\mathfrak{g}$ such that $c(0) = 0$ and $c'(0) = X$. Here, I have set $$c(t) = tX.$$ Commented Nov 26, 2022 at 4:34
• Apologies if I am going in circles, so with your definitions $D_X \text{exp}:G\rightarrow \mathfrak{g}_L$? Rather than a map from $T_0\mathfrak{g} \cong \mathfrak{g}$? Commented Nov 26, 2022 at 16:44
• Sorry. I wrote it incorrectly. In the first line of the last display, it should say $D_X\exp(0)$ (I've fixed this). In your notation, this is $D_0\exp(X)$. You can verify this by looking at the definition of the directional derivative provided in my answer. Commented Nov 26, 2022 at 17:02
• Thanks for clarifying, I'm happy with that now! Commented Nov 26, 2022 at 17:08