The operation "dilation" on the Heisenberg group?

Question

I would like to know why "dilation" on the 3-dimensional Heisenberg group $H^3$ is given by the following operation

$\begin{align} \varphi_{\lambda} : & \qquad H^3 \longrightarrow H^3\\ &(x,y,t)\to \varphi_{\lambda} (x,y,t)= (\lambda x, \lambda y, \lambda^2 t) \end{align}\qquad\mbox{for}\, \lambda\geq0.$

Where the Heisenberg group $H^3$ is the set $\mathbb R^2\times \mathbb R$ endowed with the group law $$(x,x,t)\bullet(x',y',t') =\left (x+x',y+y',t+t'+\frac{1}{2}(x y'-x' y)\right).$$

Thank you in advance

Answer 1

My best guess: it plays well with the group operation on $\mathbb{H}^3$. Notice that there are quadratic terms in the third component for the group operation. The Heisenberg group is kind of a warping of the usual group structure on $\Bbb R^3-$which is just vector addition of course. If you dilate a sum, it should be the sum of the dilations. We would want the same here since we are just warping the usual structure on $\Bbb R^3$, that is

$$\mathcal{D}_{\lambda}((x,y,t)\cdot(x',y',t')) = \mathcal{D}_{\lambda}(x,y,t)\cdot\mathcal{D}_{\lambda}(x',y',t').$$

Written out slightly,

$$\mathcal{D}_{\lambda}\left(x+x',y+y',t+t' + \frac{1}{2}\left(xy'-x'y\right)\right) = \mathcal{D}_{\lambda}(x,y,t)\cdot\mathcal{D}_{\lambda}(x',y',t').$$

Notice the appearance of quadratic terms $xy'$ and $x'y$ in the third component of the group law. If you dilate the $x$, $y$ linearly (which you might as well, or change variables to make the dilation linear), you have to dilate the $t$ quadratically in order for this equation to hold.

In algebraic terms, it's a specific kind of automorphism.

Answer 2

Because then you have that the dilation is a group homomorphism, i.e. $$\delta_r (x*y)=\delta_r(x)*\delta_r(y)$$ And the 1-parameter property: $$\delta_{rr'}(x)=\delta_r(\delta_{r'}(x)).$$ (Well, in fact this would hold with the usual scalar multiplication in $R^3$ as well.)

It also is compatible with the measure and Hausdorff dimension of the group.

As others answered correctly, there is good reason why one would have been able to guess/arrive at this formula.

Answer 3

Also, in slightly different coordinates, the Heisenberg group is the collection of matrices $\pmatrix{1 & x & t \cr 0 & 1 & y \cr 0 & 0 & 1}$, with matrix multiplication, and the dilation is the conjugation by $\pmatrix{\lambda & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & \lambda^{-1}}$.

(The relation between these coordinates and the ones you gave amounts to using "exponential coordinates" for matrices, meaning to exponentiate the Lie algebra $\pmatrix{0 & x & t \cr 0 & 0 & y \cr 0 & 0 & 0}$, or similar.)