# How can we apply the exponential map to an element of a Lie group?

As far as I know, the exponential map is defined as a map $exp: \mathfrak{g} \to G$ from Lie algebra of the Lie group (or tangent space) to the Lie group itself. Now I have a problem in which I am asked to find an exponential map for a matrix Lie group. Well, I just use the definition of the exponential map for matrix via series. But I still cannot understand what it means to use the exponential map in such a way. Is it just a map $exp: G \to Mat(\mathbb{R}, N)$ from a Lie group to the space of all matrices?

The problem sounds: $\text{Write explicitly exponential map for the Lie group of matrices} \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix}, \text{if } a>0 \text{ and } b\in \mathbb{R}$

• You are right, $\exp$ is defined on a Lie algebra, even if Lie groups and Lie algebras may consist both of matrices. Sep 25, 2016 at 20:32

If $G$ is a matrix Lie group (i.e., a Lie subgroup of $GL(n,\mathbb R)$ for some $n$), then its Lie algebra $\mathfrak g$ can be canonically identified with a Lie subalgebra of the algebra of all $n\times n$ matrices. (See Prop. 8.41, Thm. 8.46, and Example 8.47 in my Introduction to Smooth Manifolds, 2nd ed.) So the exponential map of $G$ will take a matrix $X\in \mathfrak g$ and yield a matrix $\exp(X)\in G$.
For the explicit problem that you posted, your job is (1) to figure out what the Lie algebra of that group is, considered as a subalgebra of the Lie algebra $\frak g\frak l(n,\mathbb R)$ of all $n\times n$ matrices; and (2) to compute the exponential of an arbitrary element of that Lie algebra.
• You write that I shall take $X \in \mathfrak{g}$ and map it to $\exp(X)\in G$. But my $X \not\in \mathfrak{g}$: it is in $G$, Lie group itself. Sep 25, 2016 at 20:40
• @NickTerziev: I think you need to give more context before I can say anything useful. But notice that both $\mathfrak g$ and $G$ are subsets of the set of all $n\times n$ real matrices, so it's perfectly possible for a specific matrix to be both in $\mathfrak g$ and in $G$. Sep 25, 2016 at 20:46