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:

$$ (l_g)_* A (f) = A(f\circ l_g) = A^i \partial_i (f\circ l_g\circ x^{-1})(x(e)) = A^i \left(\frac{\partial f}{\partial x^i}\right)_g $$


$$ \dot{\gamma}^i = A^i $$

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?

  • 1
    $\begingroup$ An integral curve goes the other way around $\mathbb R\rightarrow G$ $\endgroup$
    – Alessandro
    Sep 1, 2022 at 19:06
  • 4
    $\begingroup$ 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. $\endgroup$ Sep 1, 2022 at 21:56

1 Answer 1


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)).

  • $\begingroup$ 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 $\text{GL}(n,\mathbb{R})$ $\endgroup$
    – John K
    Sep 5, 2022 at 17:38
  • $\begingroup$ 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? $\endgroup$
    – John K
    Sep 5, 2022 at 18:55
  • $\begingroup$ 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 $\endgroup$
    – usr0192
    Sep 10, 2022 at 23:53

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