[proofplan]
We verify directly that $\delta_r$ preserves the Heisenberg multiplication by substituting the two dilated points into the group law. The only non-linear term is the central coordinate, and it scales correctly because $\operatorname{Im}((rz)\cdot\overline{rw})=r^2\operatorname{Im}(z\cdot\bar w)$. We then define the reciprocal dilation and check that it is a two-sided inverse, so the homomorphism is an automorphism. Finally, we compute the pushforward on smooth test functions using the chain rule and identify the resulting differential operators with $rX_j$, $rY_j$, and $r^2T$ at the dilated point.
[/proofplan]
[step:Check that the dilation preserves the Heisenberg product]
Let $p=(z,t)\in\mathbb H^n$ and $q=(w,s)\in\mathbb H^n$, where $z,w\in\mathbb C^n$ and $t,s\in\mathbb R$. By the group law,
\begin{align*}
pq=\left(z+w,t+s+2\operatorname{Im}(z\cdot\bar w)\right).
\end{align*}
Applying $\delta_r$ gives
\begin{align*}
\delta_r(pq)
=
\left(r(z+w),r^2\left(t+s+2\operatorname{Im}(z\cdot\bar w)\right)\right).
\end{align*}
On the other hand,
\begin{align*}
\delta_r(p)\delta_r(q)
=
(rz,r^2t)(rw,r^2s).
\end{align*}
Using the group law again,
\begin{align*}
\delta_r(p)\delta_r(q)
=
\left(rz+rw,r^2t+r^2s+2\operatorname{Im}((rz)\cdot\overline{rw})\right).
\end{align*}
Since $r$ is real,
\begin{align*}
(rz)\cdot\overline{rw}=r^2(z\cdot\bar w),
\end{align*}
and therefore
\begin{align*}
\operatorname{Im}((rz)\cdot\overline{rw})=r^2\operatorname{Im}(z\cdot\bar w).
\end{align*}
Thus
\begin{align*}
\delta_r(p)\delta_r(q)
=
\left(r(z+w),r^2\left(t+s+2\operatorname{Im}(z\cdot\bar w)\right)\right)
=
\delta_r(pq).
\end{align*}
Hence $\delta_r$ is a [group homomorphism](/page/Group%20Homomorphism).
[guided]
We need to prove that $\delta_r$ respects the group operation, so we compare the two possible orders of operations: first multiply and then dilate, or first dilate and then multiply.
Let $p=(z,t)\in\mathbb H^n$ and $q=(w,s)\in\mathbb H^n$, with $z,w\in\mathbb C^n$ and $t,s\in\mathbb R$. The Heisenberg product is
\begin{align*}
pq=\left(z+w,t+s+2\operatorname{Im}(z\cdot\bar w)\right).
\end{align*}
Applying the dilation map to this product gives
\begin{align*}
\delta_r(pq)
=
\left(r(z+w),r^2\left(t+s+2\operatorname{Im}(z\cdot\bar w)\right)\right).
\end{align*}
Now compute the other expression. We have
\begin{align*}
\delta_r(p)=(rz,r^2t)
\end{align*}
and
\begin{align*}
\delta_r(q)=(rw,r^2s).
\end{align*}
Multiplying these two points in $\mathbb H^n$ gives
\begin{align*}
\delta_r(p)\delta_r(q)
=
\left(rz+rw,r^2t+r^2s+2\operatorname{Im}((rz)\cdot\overline{rw})\right).
\end{align*}
The horizontal coordinate already agrees with $r(z+w)$. The central coordinate is the only place where the non-commutative part of the group law appears. Since $r$ is a real scalar,
\begin{align*}
(rz)\cdot\overline{rw}=r^2(z\cdot\bar w).
\end{align*}
Taking imaginary parts gives
\begin{align*}
\operatorname{Im}((rz)\cdot\overline{rw})=r^2\operatorname{Im}(z\cdot\bar w).
\end{align*}
Substituting this into the product formula yields
\begin{align*}
\delta_r(p)\delta_r(q)
=
\left(r(z+w),r^2\left(t+s+2\operatorname{Im}(z\cdot\bar w)\right)\right).
\end{align*}
This is exactly $\delta_r(pq)$, so $\delta_r$ is a group homomorphism.
[/guided]
[/step]
[step:Use the reciprocal dilation as the inverse automorphism]
Define the reciprocal dilation
\begin{align*}
\delta_{1/r}:\mathbb H^n\to\mathbb H^n,\qquad (z,t)\mapsto \left(\frac{1}{r}z,\frac{1}{r^2}t\right).
\end{align*}
For every $(z,t)\in\mathbb H^n$,
\begin{align*}
\delta_{1/r}(\delta_r(z,t))=(z,t)
\end{align*}
and
\begin{align*}
\delta_r(\delta_{1/r}(z,t))=(z,t).
\end{align*}
Thus $\delta_{1/r}$ is the inverse map of $\delta_r$. Since $\delta_r$ is a bijective group homomorphism, $\delta_r$ is a group automorphism.
[/step]
[step:Compute the pushforward on horizontal and central vector fields]
Let $p=(z,t)\in\mathbb H^n$, write $z_j=x_j+iy_j$, and set $q=\delta_r(p)=(rz,r^2t)$. Let $f:\mathbb H^n\to\mathbb C$ be a smooth function. By definition of pushforward,
\begin{align*}
((\delta_r)_*X_j)_q f=X_j(f\circ\delta_r)(p).
\end{align*}
Since
\begin{align*}
X_j=\partial_{x_j}+2y_j\partial_t,
\end{align*}
the chain rule gives
\begin{align*}
X_j(f\circ\delta_r)(p)
=
r(\partial_{x_j}f)(q)+2y_jr^2(\partial_tf)(q).
\end{align*}
Because the $y_j$-coordinate of $q$ is $ry_j$, this becomes
\begin{align*}
X_j(f\circ\delta_r)(p)
=
r\left((\partial_{x_j}f)(q)+2(ry_j)(\partial_tf)(q)\right)
=
(rX_jf)(q).
\end{align*}
Hence
\begin{align*}
(\delta_r)_*X_j=rX_j.
\end{align*}
Similarly, using
\begin{align*}
Y_j=\partial_{y_j}-2x_j\partial_t,
\end{align*}
the chain rule gives
\begin{align*}
Y_j(f\circ\delta_r)(p)
=
r(\partial_{y_j}f)(q)-2x_jr^2(\partial_tf)(q).
\end{align*}
Since the $x_j$-coordinate of $q$ is $rx_j$,
\begin{align*}
Y_j(f\circ\delta_r)(p)
=
r\left((\partial_{y_j}f)(q)-2(rx_j)(\partial_tf)(q)\right)
=
(rY_jf)(q).
\end{align*}
Thus
\begin{align*}
(\delta_r)_*Y_j=rY_j.
\end{align*}
Finally,
\begin{align*}
((\delta_r)_*T)_q f=T(f\circ\delta_r)(p).
\end{align*}
Since $T=\partial_t$, the chain rule gives
\begin{align*}
T(f\circ\delta_r)(p)=r^2(\partial_tf)(q)=(r^2Tf)(q).
\end{align*}
Therefore
\begin{align*}
(\delta_r)_*T=r^2T.
\end{align*}
The three pushforward identities hold for every $p\in\mathbb H^n$, every smooth function $f:\mathbb H^n\to\mathbb C$, and every $1\le j\le n$, so the asserted vector field identities follow.
[/step]
