The Heisenberg group $H_1$ is the set $\mathbb C\times \mathbb R$ endowed with the group law $$ (z,t)\cdot(w,s) =\left (z+w, \,t+s+\tfrac{1}{2}\Im m(z \bar{w})\right); \quad \forall z,w \in \mathbb C\,\text{and}\, \,t,s\in \mathbb R.$$

Let $\delta$ be the Dirac distribution supported at $e=(0,0)$ the identity element on $H_1$. If D is a left-invariant differential operator on $H_1$, we have $Df = D(f *\delta) = f*D\delta$; by the usual abuse of language we write this as $f * D$. Let $Z$ the left-invariant vector field given by $$\begin{equation} f*Z=\left(\frac{\partial}{\partial z}+i\overline{z}\frac{\partial}{\partial t}\right)f. \qquad (1) \end{equation}$$

Using the function $A$ on $H_1$ given by $A=|z|^2+it$, we define the map $h$ on $H_1-\left\{e\right\}$ by $$ h(z,t)=\left(\frac{-z}{|z|^2-it},\frac{-t}{|z|^4+t^2}\right)=\left(-z\overline{A^{-1}},-t A^{-1} \overline{A^{-1}}\right).$$

By the use of the following identities $$\begin{equation} z*Z=1,\,\,\, \overline{z}*Z=0,\,\,\, t*Z=i\overline{z}. \qquad\qquad\qquad (1.1)\\ A*Z=0,\quad \overline{A}*Z=2\overline{z}.\qquad\qquad\qquad\qquad\qquad \qquad (1.2)\\ (z\circ h)*Z=\overline{A^{-2}} \, (2|z|^2-\overline{A}), \quad (t\circ h)*Z=-i\overline{z}\, \overline{A^{-2}}. \quad (1.3) \end{equation}$$ By a simple calculation we found the equations (1.1) and (1.2), and for (1.3) see the answer of @F Zaldivar below.

Finally, I ask if someone can help me to check the following
$$\begin{equation} (f\circ h)*Z=-\overline{A^{-1}}\left|(f*Z)\circ h+2\overline{z}\, (f*E_{z})\circ h\right|; \quad E_{z}:= zZ. \quad (2) \end{equation}$$

$\triangle$ I found these formulas, in the paper of: Korányi, Kelvin transforms and harmonic polynomials on the Heisenberg group. JFA 1982. The result is given in the case of $H_n=\mathbb C^n\times \mathbb R$, and I tried to do it again in the case $H_1=\mathbb C\times \mathbb R$ (n=1).

Thank you in advance

  • 1
    $\begingroup$ If you “don’t understand” the notation $\circ$, then you might want to say where it’s from? $\endgroup$ Mar 28 at 1:43
  • $\begingroup$ @Francois Ziegler, ahh yes, sorry, I found this in the paper of: Korányi, Kelvin transforms and harmonic polynomials on the Heisenberg group. JFA 1982. $\endgroup$
    – Z. Alfata
    Mar 28 at 5:29

1 Answer 1


First, the reference is: https://www.sciencedirect.com/science/article/pii/0022123682900787

Second, in the notation of the cited paper $H_n={\mathbb C}^n\times {\mathbb R}$. Hence, you must decide if you want $H_3={\mathbb C}^3\times {\mathbb R}$ or $H_1={\mathbb C}^1\times {\mathbb R}$. For the former, its elements are of the form $(z,t)$, where $z=(z_1,z_2,z_3)$ and the compositions $z_k\circ h$ and $t\circ h$ just take the corresponding (complex or real) coordinate of $h(z,t)$. For $H_1$ one just has a single coordinate for $z$.

For the calculation of $(t\circ h)*Z$ first you write $A=z\overline{z}+it$ and $\overline{A}=z\overline{z}-it$. Then, $A\overline{A}=(z\overline{z})^2+t^2$ and thus $A^{-1}\overline{A^{-1}}=\frac{1}{|z|^4+t^2}$. Next,

\begin{align*} (t\circ h)*Z &=\Big(\frac{\partial}{\partial z}+i\overline{z}\frac{\partial}{\partial t}\Big)\Big(-tA^{-1}\overline{A^{-1}}\Big)\\ &=\Big(\frac{2t\overline{z}}{(|z|^4+t^2)(|z|^2-it)}\Big)-i\overline{z}A^{-1}\overline{A^{-1}}\\ &=\frac{\overline{z}}{|z|^4+t^2}\Big(\frac{2t}{|z|^2-it}-i\Big)\\ &=-i\overline{z}\left(\frac{(|z|^2+it)^2}{(|z|^4+t^2)(|z|^2-it)}\right)\\ &=-i\overline{z}\overline{A^{-2}}. \end{align*}

The other calculations are similar.

  • $\begingroup$ @Zaldivar, No, I want $H_1= \mathbb C\times \mathbb R$. I idit my question. $\endgroup$
    – Z. Alfata
    Mar 28 at 18:09
  • $\begingroup$ @F Zaldivar, Thank you for your indication. If you can help me to check the equation (2) I would be very grateful $\endgroup$
    – Z. Alfata
    Mar 29 at 14:04

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