Here, https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#Examples, we can see a number of examples:
- The unit circle centered at $0$ in the complex plane is a Lie group (called the circle group) whose tangent space at $1$ can be identified with the imaginary line in the complex plane, $\{it:t\in\mathbb R\}$. The exponential map for this Lie group is given by
$$
it \mapsto \exp(it) = e^{it} = \cos(t) + i\sin(t),\,
$$
- In the quaternions $\mathbb {H}$ , the set of quaternions of unit length form a Lie group (isomorphic to the special unitary group SU(2)) whose tangent space at 1 can be identified with the space of purely imaginary quaternions, $\{it+ju + kv :t, u, v\in\mathbb R\}$. The exponential map for this Lie group is given by
$$
{\displaystyle \mathbf {w} :=(it+ju+kv)\mapsto \exp(it+ju+kv)=\cos(|\mathbf {w} |)1+\sin(|\mathbf {w} |){\frac {\mathbf {w} }{|\mathbf {w} |}}.\,}$$
But then, they give out this definition which they claim is the general definition:
- The Lie group–Lie algebra correspondence also gives the definition: for $X$ in
${\mathfrak {g}}t\mapsto \exp(tX)$ is the unique Lie group homomorphism corresponding to the Lie algebra homomorphism $t\mapsto tX$. (note: $\operatorname {Lie}({\mathbb {R}})={\mathbb {R}}$.)
I can see why the first example needs just one real number $t$ since there is only one term $i$ in the exponential. But, in the case of say the general linear group… how many real numbers to I need? Why is the general definition contains just $t$, and not a plurality of real numbers? Are we trying to generated a subgroup of $X$ by using a single parameter $t$ (the one-parameter group), but if we wanted to generate the full group we will need multiple real variables instead of just one? I am trying to understand what the exponential map does.
Best Answer
You can apply the exponential map to any element of the Lie algebra -- which gives a number of free parameters equal to the dimension of the Lie algebra (which is also the manifold dimension of the group).
However, when used that way the exponential map is generally not a homomorphism. It can't be in general because that would require all group elements in the image of the exponential map to commute with each other, and the original Lie group is not usually abelian.
On the other hand, if you restrict to an (arbitrary!) one-dimensional subspace of the Lie algebra, the exponential map does become a homomorphism from a subgroup of (the additive group of) the Lie algebra into the Lie group. This is why the one-parameter case has a special position in the theory, and some results only hold for that case.