I have a problem of analysis in computing some integrals, and I would like to share with you what I have done so far, hoping that someone will help me to get through this.
First of all, let me fix some notation.
We can think at the Heisenberg group $\mathbb H^n$ as $\mathbb C^n \times \mathbb R$, so we can denote its points $P=(z,t)$, where $z \in \mathbb C^n$ and $t\in \mathbb R.$
Now, we endow $\mathbb H^n$ with a metric $||\cdot||_\infty$ defined as in the following: $$||P||_\infty=\text{max}\{|z|,\sqrt{|t|}\},$$ the open balls w.r.t. $||\cdot||_\infty$ of centre $P$ and radius $r$ will be indicated with $U(P,r)$.
Having said this, I'm interested in calculate an integral of the form:
$$\int_{\partial U(0,r)} f(P)d\mathcal H^{2n}.$$
Now, notice that $\partial U(0,r)\subset \mathbb H^n=\mathbb R^{2n+1}=\mathbb C^n \times \mathbb R$ is a cylinder whose basis are $$B_1=\{|z|\leq r, t= r^2\} \text{ and } B_2=\{|z|\leq r, t= -r^2\}, $$ while the lateral surface is $$S=\{|z|=r,-r^2\leq t\leq r^2\}.$$
Thus $\int_{\partial U(0,r)}=\int_{B_1}+\int_{B_2}+\int_S.$ Please, tell me if I am missing something or there are any mistakes up to here.
Now, I know that over each of the two bases $f(P)=2|z|$, while over $S$ we have $f(P)=1$, so $$\int_S f(P) d\mathcal H^{2n}=\text{ area of } S$$ and $$ \int_{B_{1,2}}f(P)d\mathcal H^{2n}=2 \int_{B_{1,2}}|z|d\mathcal H^{2n}.$$
I already know the results: $$\int_{B_{1,2}}f(P)d\mathcal H^{2n}= \frac{2\omega_{2n}}{2n+1} r^{2n+1},\\ \int_{S}f(P)d\mathcal H^{2n}=2(2n-1)\omega_{2n}r^{2n+1},$$ where $\omega_m$ is the Lebesgue measure of the $m$- dimensional ball of $\mathbb R^m.$ I want to verify them.
The simplest case is when $n=1$, because $\mathbb H^1=\mathbb C \times \mathbb R=\mathbb R^3$; we denote its points as $(x,y,t)$, $x,y,t\in \mathbb R$, or $(z,t)$, $z\in \mathbb C,t\in \mathbb R$.
In this case, $B_1$ is the image of the map $\sigma: (\rho,\theta)\mapsto (\rho \cos \theta,\rho \sin \theta,r^2), \rho\in[0,r], \theta \in[0,2\pi].$ Moreover $$\frac{\partial \sigma}{\partial \rho}=(\cos \theta, \sin \theta, 0)\\ \frac{\partial \sigma}{\partial \theta}=(-\rho \sin \theta, \rho \cos \theta, 0),$$ so that $\frac{\partial \sigma}{\partial \rho}\wedge \frac{\partial \sigma}{\partial \theta}=(0,0,\rho).$
Thus $\int_{B_1}f(P)d\mathcal H^1=2 \int_{B_1}|z|d\mathcal H^{1}=2 \int_0^{2\pi}\int_0^r \rho \cdot \rho d\rho d\theta= \frac{4}{3}\pi r^3. $ I don't understand why, but this result seems to be wrong: according to the previous results it should be $\frac{2}{3}\pi r^3$.
On the other hand, we have that $\text{Area of } S = 2\pi r \cdot 2r^2= 4\pi r^3$ and also this result is not the one expected, it should be $2 \pi r^3$.
I know that my results are not so wrong (they differ by a factor 2), but I would like to know where I'm making mistakes.
Another thing I would like to ask is how to generalize this for $n>1.$ For example, for the integral over $S$ I have to calculate the $2n$-dimensional area of the cylinder, so I would say $\text{Area of }S= 2n \omega_{2n}\cdot 2r^2$, which is clearly not. Then, for the integral over the two bases, should I have always to use a parameterization of the bases, do all the computations hoping to do not make any mistake or is there a easier way to arrive to the expected result? I really hope so.
Any kind of help will be very appreciated and thanks for your patience in reading all the post.