Is Heisenberg group Euclidean? I'm reading an article speaking about Heisenberg group $\mathbb H^n$ and some of its properties. 
Now, I have some questions to ask, hoping to be clear enought.
Reading the introduction I've understood that we can identify $\mathbb H^n$ with $\mathbb R^{2n+1}$ so we can denote points of Heisenberg group as $P=(x_1,\dots, x_n, y_1,\dots,y_n,t)$. On $\mathbb H^n$ we have an operation 
$\cdot: \mathbb H^n\times \mathbb H^n\to \mathbb H^n$ defined as $$P\cdot P'= (x_1+x'_1,\dots,y_n+y_n',t+t'+\frac{1}{2}\sum_{j=1}^n x_iy'_i-x'_iy_i),$$ which is not commutative. We can define dilatations $\delta_\lambda:\mathbb H^n\to\mathbb H^n$, $$\delta_{\lambda}(P)=(\lambda x_1,\dots,\lambda y_n,\lambda^2 t).$$
Now, we can equipe $\mathbb H^n$ with a norm $||\cdot||$ defined as $$||P||=max\{|(x_1,\dots,x_n,y_1\dots,y_n)|,|t|^{\frac{1}{2}}\},$$ turning $\mathbb H^n$ in a metric space with distance $d,\, d(P,Q)=||P^{-1}\cdot Q||.$
This said, I'm going to add some my considerations and I would like to know if they are a correct way of thinking or not.
Since I can identify $\mathbb H^n$ with $\mathbb R^{2n+1}$, I can think about $(\mathbb H^n,d)$ as $(\mathbb R^{2n+1},d)$.
Then, since all norms on $\mathbb R^{2n+1}$ are equivalent, so are the distance $d$ and the Euclidean distance $D$, hence $(\mathbb H^n,d)$ and $(\mathbb R^{2n+1},D)$ are topologically equivalent. So, if I think about $\mathbb H^n$ as a smooth manifold, it coincides with $\mathbb R^{2n+1}$, in the sense that I have a unique chart ($\mathbb H^n$ itself) and the homeomorphism is given by the identity. 
On the other hand, in the introduction there's also written that $\mathbb H^n$ is not Euclidean and this seems to contrast what I've said so far.
Could someone help me to understand? 
 A: Here are several comments that might help clarify things for you.
First, the "norm" $\|\cdot\|$ on $\mathbb H^n$ is not a norm in the sense of linear algebra, because it doesn't satisfy $\|cP\| = |c|\,\|P\|$ for $c\in \mathbb R$. So the theorem that "all norms are equivalent" doesn't apply to it. In particular, if we let $|\cdot|$ denote the Euclidean norm on $\mathbb R^{2n+1}$, then $\|P_\varepsilon\|/|P_\varepsilon|\to \infty$ as $\varepsilon\to 0$, where $P_\varepsilon = (0,\dots,0,\varepsilon)$. Thus the distance functions $d$ and $D$ are not uniformly equivalent.
However, the topology determined by $\|\cdot\|$ is the same as the Euclidean topology. You can see this by noting that $|P|\le \|P\| \le |P|^{1/2}$ when either $|P|$ or $\|P\|$ is less than $1$, so small balls in either topology are also open in the other topology. 
The statement in the article that "$\mathbb H^n$ is not Euclidean" might simply be referring to the fact that the distance function determined by $\|\cdot\|$ is not the Euclidean distance function, or even uniformly equivalent to it, so the geometry that it determines is not Euclidean geometry.
However, another way that this statement might be interpreted is in terms of Riemannian geometry. As a Lie group, $\mathbb H^n$ possesses a left-invariant Riemannian metric (many of them, in fact), so we can endow $\mathbb H^n$ with one of these metrics $g$ and ask whether the Riemannian manifold $(\mathbb H^n,g)$ is isometric to $\mathbb R^{2n+1}$ with its Euclidean metric. Because every left-invariant metric on $\mathbb H^n$ has nonzero curvature, the answer is no.
