Let $G$ be a Lie Group with associated Lie Algebra defined over the tangent space at identity of $G$. An integral curve $\gamma^A: \mathbb{R}\rightarrow G$ associated to algebra element $A\in T_eG$ if for the vector field $X^A$ defined point-wise as:
$$
X^A_g = (l_g)_* (A)
$$
the relation
$$
X_{\gamma,\gamma(\lambda)} = X^A_{\gamma(\lambda)}
$$
holds. The pushforward of $l_g$ establishes $L(G)\cong T_eG$. Then the exponential map $\text{exp}:T_eG\rightarrow G$ is defined as:
$$
\text{exp}(\lambda A) = \gamma^A(\lambda)
$$
From this definition for the exponential map, it wasn't obvious to me why it is called an exponential map at all, or why I have seen it expanded as a usual exponential Maclaurin series:
$$
\text{exp}(\lambda A) = 1 + \lambda A + \frac{\lambda^2 A^2}{2!}+…
$$
To try to derive this, I explore the condition that defines the integral curves of A, i.e. $X_{\gamma,g}=X^A$. Let $\gamma(\lambda) = g$. The LHS is:
$$
X_{\gamma,g} (f) = \dot{\gamma}^i \left(\frac{\partial f}{\partial x^i}\right)_g
$$
The RHS:
$$
(l_g)_* A (f) = A(f\circ l_g) = A^i \partial_i (f\circ l_g\circ x^{-1})(x(e)) = A^i \left(\frac{\partial f}{\partial x^i}\right)_g
$$
Thus
$$
\dot{\gamma}^i = A^i
$$
I don't know how one would go from here to show that $\gamma$ can be written as a Maclaurin series. How does one solve this differential equation generally?
Moreover, how does one interpret the components $A^i$? Are these the individual numbers in the matrix representation of the Lie algebra element?
Best Answer
Nice job of carefully unwinding all the definitions. But I think at one of the last steps you have an error and so the conclusion that the diff eq is $$ \dot{\gamma}^i = A^i $$ isn't quite right (or maybe I don't understand the notation, e.g. $A^i$) - The RHS is independent of g, the LHS depends on g (it is the tangent vector to $\gamma$ at $g$). I think the error arises on the RHS in the equation $$ A^i \partial_i (f\circ l_g\circ x^{-1})(x(e)) = A^i \left(\frac{\partial f}{\partial x^i}\right)_g $$ ... on LHS you have the derivative of the group action, whereas on the RHS you are not - I'm not explaining that well and so a simple example will illustrate:
Consider two different Lie Groups $G_1=\mathbb{R}$ under addition and $G_2=\mathbb{R}^*$ under multiplication.
Then for $G_1$, $l_g(x)=g+x$, and if $A=a \frac{\partial}{\partial x}\vert_e$ for some $a \in \mathbb{R}$ and $g=\gamma(\lambda)$, then the diff eq is $$\frac{d}{dt}(f\circ \gamma)\Big\vert_{t=\lambda} = A(f\circ l_g)\Big\vert_e = a \frac{\partial}{\partial x}(f(g+x))\Big\vert_{x=0} $$ Taking $f=x$ (the usual coordinate chart of $\mathbb{R}$) we get $$\frac{d}{dt}\gamma(t) \Big\vert_{t=\lambda} = a \frac{\partial}{\partial x}(g+x)\Big\vert_{x=0}= a \cdot 1 $$, and so the DE is $$\frac{d\gamma}{d t } =a$$ which has the solution $$\gamma(t)=at$$ and the Lie Group 'exponential' for $(\mathbb{R}, +)$ is $$\text{exp}(at)=at$$in this case ... but there is no exponential function at all! So I never liked that this map was called $\text{exp}$.
Now for $G_2=\mathbb{R}^*=\text{GL}_1(\mathbb{R})$, $l_g(x)=gx$, and if $A=a \frac{\partial}{\partial x}\vert_e$ for some $a \in \mathbb{R}$ and $g=\gamma(\lambda)$, then the diff eq is $$\frac{d}{dt}(f\circ \gamma)\Big\vert_{t=\lambda} = A(f\circ l_g)\Big\vert_e = a \frac{\partial}{\partial x}(f(gx))\Big\vert_{x=1} $$ Taking $f=x$ (the usual coordinate chart of $\mathbb{R}$) we get $$\frac{d}{dt}\gamma(t) \Big\vert_{t=\lambda} = a \frac{\partial}{\partial x}(gx)\Big\vert_{x=1}= a \cdot g $$, and since $g=\gamma(t)$ so the DE is $$\frac{d\gamma}{d t } =a\gamma(t)$$ which has the solution $$\gamma^A(t)=e^{at}$$ and the Lie Group exponential for $\text{GL}_1(\mathbb{R})$ $$\text{exp}(at)=e^{at}$$in this case. As Qiaochu Yuan points out in the comments, for $GL_n$, we get the matrix exponential (for example by a souped-up version of this argument (which is the $n=1$ case)).