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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?

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2 Answers 2

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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. \begin{equation}\frac{d}{dt}\bigg\rvert_{t=\tau}\gamma:= [\gamma(t-\tau)]\quad (\ast)\end{equation}

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 $(\ast)$).

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  • $\begingroup$ 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 $\gamma(t)=tX$?). $\endgroup$
    – Bedge
    Commented Nov 25, 2022 at 22:10
  • $\begingroup$ I have edited my answer to hopefully address your questions. $\endgroup$
    – J.V.Gaiter
    Commented Nov 25, 2022 at 22:52
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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 \begin{equation}\tag{*} \frac{d}{dt}\phi_X(t) = X(\phi_X(t)). \end{equation} 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*}

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  • $\begingroup$ 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? $\endgroup$
    – Bedge
    Commented Nov 26, 2022 at 0:31
  • $\begingroup$ 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 $$c(t) = tX.$$ $\endgroup$
    – Deane
    Commented Nov 26, 2022 at 4:34
  • $\begingroup$ 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 $T_0\mathfrak{g} \cong \mathfrak{g}$? $\endgroup$
    – Bedge
    Commented Nov 26, 2022 at 16:44
  • $\begingroup$ 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. $\endgroup$
    – Deane
    Commented Nov 26, 2022 at 17:02
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    $\begingroup$ Thanks for clarifying, I'm happy with that now! $\endgroup$
    – Bedge
    Commented Nov 26, 2022 at 17:08

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