# Why is the Lie group exponential map an exponential?

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?

• An integral curve goes the other way around $\mathbb R\rightarrow G$ Commented Sep 1, 2022 at 19:06
• It's called an exponential because it satisfies $\exp((t_1 + t_2) A) = \exp(t_1 A) \exp(t_2 A)$, and also because for $G = GL_n$ it reproduces the ordinary matrix exponential. Commented Sep 1, 2022 at 21:56

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)).
• Thank you for the detailed response. I see where I have forgotten the group action in the derivative. Will accept and look for the souped-up version of the argument for $\text{GL}(n,\mathbb{R})$ Commented Sep 5, 2022 at 17:38
• Actually how does one do differentiation with respect to the group operation? Is true that in general, for a group product of two $\gamma:\mathbb{R}\rightarrow G$ s.t. $\gamma(\lambda) = \gamma_1(\lambda)\cdot \gamma_2(\lambda)$, the product rule holds? Commented Sep 5, 2022 at 18:55
• Hmm, good question. For a subgroup of GL_n (and noting that GL_n is a subset of M_n nxn matrices), yes the product rule holds (by roughly the same proof as in calculus), and the + is the addition of nxn matrices (there's an identification of tangent spaces going on). See Brian Hall's book on Lie Groups. For abstract Lie group, I'm not sure exactly: the multiplication map m: G x G --> G is differentiable and gives a map $dm: TG_a \times TG_b \to TG_{ab}$ by (for a local function around ab) $dm(A, B)(f)|_{ab} = A(f(xb)) + B(f(ay))$ ... but I'm not sure Commented Sep 10, 2022 at 23:53