Why is the exp map in the case of the Heisenberg group just the identity map? I would like to know why, in the case of  the 3-dimensional Heisenberg group $H^3$, We can identify $H^3$ with its Lie algebra $h_3$ through the exponential function $$\exp= id : h_3 \to H^3 .$$
i.e.; why $\exp= id$ ?
 A: I would say that $\exp$ is not id. However, the exp map is a bijection.
The Lie algebra $\frak h_3$ consists of matrices
$$\pmatrix{0&a&b\\0&0&c\\0&0&0}$$
and the group $H_3$ consists of matrices
$$\pmatrix{1&r&s\\0&1&t\\0&0&1}.$$
The exponential map takes
$$\pmatrix{0&a&b\\0&0&c\\0&0&0}\mapsto\pmatrix{1&a&b+ac/2\\0&1&c\\0&0&1}.$$
From this we see that $\exp$ is a bijection, but an identity map? Not likely!
A: It depends on which coordinates you use! As Angina Seng says above, if you think of the Heisenberg group and its Lie algebra as subsets of $M_3(\mathbb{R})$, then the map will not be the identity. (In fact you can see this just by noting that the subsets are disjoint, by looking at the diagonal of the matrix).
However, since exp is a diffeomorphism, you can define a group law on $\mathbb{R}^3$ and a Lie algebra structure on $\mathbb{R}^3$ such that the resulting group is isomorphic to $H^3$, the resulting Lie algebra is isomorphic to $h_3$, and the exponential map between the two is the identity.
This is called "using exponential coordinates", and it can be done whenever exp is a diffeomorphism (for instance, whenever the group is simply connected and nilpotent).
Explicitly, set $X = (1,0,0), Y = (0,1,0), Z=(0,0,1) \in \mathbb{R}^3$,
and define the Lie structure by $[X,Y]=Z, [X,Z]=[Y,Z]=0$.
By linearity, this means the Lie structure in these coordinates is
$$ [(a,b,c),(a',b',c')] = (0,0, ab' - ba'). $$
(So the Lie algebra is isomorphic to $h_3$).
Now, we use the Campbell-Baker-Hausdorff formula, which relates the Lie group law to its Lie algebra; since all nested commutators of depth three or more vanish (e.g. $[[X,Y],Y] = 0$), we get a group law
$$ A*B = A + B + \frac{1}{2}[A,B] $$
which translates in our coordinates to
$$ (a,b,c)*(a',b',c') = (a+a',b+b',c+c'+\frac{1}{2}(ab'-ba')).$$
Note that this is a different group law than what you get if you identify $H^3$ with $\mathbb{R}^3$ by just taking the upper triangular entries of the matrix. So the exponential coordinates are not necessarily the first coordinates you would have thought to choose. But again, the point is that under these definitions of $H^3$ and $h_3$, exp is the identity map.
Of course, this is an instance of a much more general principle, that is, whenever you have a bijection between two objects, you can think of them as having the same underlying set and go from there. So you can turn any bijection into the identity map by redefining things a little, and it's usually not deeper than that.
