Action of the Haar measure on the Heisenberg group

Question

The Heisenberg group is $\mathbb{H}^N=\mathbb{R}^{2N+1} = \left\{ (x,y, \tau ) \right\}\in \mathbb{R}^N \times \mathbb{R}^N \times \mathbb{R}$ equipped with the group operation

\begin{equation} (x,y, \tau ) \circ (\tilde{x}, \tilde{y}, \tilde{ \tau } ) = (x+ \tilde{x}, y+\tilde{y}, \tau +\tilde{ \tau } + 2 \left( x \cdot \tilde{y} - \tilde{x} \cdot y\right)) \end{equation}

I have a question which is the following. Some materials I'm studying say without much detail that the Haar measure $d\eta$ in $\mathbb{H}^N$ is "similar" to the Lebesgue measure $dxdyd\tau$ in $\mathbb{R}^{2N+1}$. However, I don't know if when they say similar they coincide in any measurable Borel set. For example, if $f(x,y, \tau )=a(x)b(y)c(\tau)$ has separable variables with $a, b, c$ functions Lebesgue integrable, I don't know if I can solve $\int_{D} f(\eta)d\eta$ (where $D$ is a ball in the Heisenberg group for example) it in the same way as in $\mathbb{R}^N$, separating into three integrals depending on $x$, $y$ and $\tau$ or if I can estimate above or below by exchanging Haar's measure for Lebesgue's measure. Thank you in advance if anyone can clarify this question for me.

Answer 1

I am not sure why they would use the word "similar". With $H^N$ identified with $\mathbb{R}^{2N+1}$ as you defined it, Lebesgue measure on $\mathbb{R}^{2N+1}$ literally is the Haar measure on $H^N$ (up to a constant scaling, of course). Therefore, any method you might use to compute an integral with respect to Lebesgue measure, is certainly just as valid for the integral with respect to Haar measure, since they are precisely the same thing.

One way to verify this is to write down the map $g \mapsto h \circ g$ in the coordinates you gave, and compute its Jacobian, which is easily seen to be a triangular matrix with $1$s on the diagonal. So its Jacobian is 1, which means it preserves Lebesgue measure. Since $h$ was arbitrary, we have shown that Lebesgue measure is a left-invariant measure on $H$, so it is a Haar measure. And we know from general theory that Haar measure is unique up to scaling.

If you want a more intrinsic statement, you can note that since $H^N$ is nilpotent, connected, and simply connected, the exponential map $\exp : \mathfrak{h}^N \to H^N$ is a diffeomorphism. Since $\mathfrak{h}^N$ is a finite-dimensional vector space, it admits Lebesgue measure $m$, i.e. a nontrivial translation-invariant measure, unique up to scaling. Now we look at the pushforward of $m$ under $\exp$, and using the Baker-Campbell-Hausdorff-Dynkin formula, we can verify much as before that this measure on $H^N$ is invariant under left (and right) translation.