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 $$ h(z,t)=\left(\frac{-z}{|z|^2-it},\frac{-t}{|z|^4+t^2}\right)=\left(-z\overline{A^{-1}},-t A^{-1} \overline{A^{-1}}\right).$$

I would like to show that: $$\begin{equation} (f\circ h)*Z=-\overline{A^{-1}}\left|(f*Z)\circ h+2\overline{z}\, (f*E_{z})\circ h\right|; \quad E_{z}:= zZ. \quad (2) \end{equation}$$ For this I need to prove: $$\begin{equation} z*Z=1,\,\,\, \overline{z}*Z=0,\,\,\, t*Z=i\overline{z}. \qquad\qquad\qquad (3.1)\\ A*Z=0,\quad \overline{A}*Z=2\overline{z}.\qquad\qquad\qquad\qquad\qquad \qquad (3.2)\\ (z\circ h)*Z=\overline{A^{-2}} \, (2|z|^2-\overline{A}), \quad (t\circ h)*Z=-i\overline{z}\, \overline{A^{-2}}. \quad (3.3) \end{equation}$$

By a simple calculation I found the equations (3.1) and (3.2), but I am not understood the notations $(z\circ h)$ and $(t\circ h)$ ??, so that I can check the equation (3.3)? Finally I need a helping hand to prove (2).

$\triangle$ I found these formulas, in the paper of: Korányi, Kelvin transforms and harmonic polynomials on the Heisenberg group. JFA 1982.

$\Rightarrow$ 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

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

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 $H^3$. If D is a left-invariant differential operator on $H^3$, 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 $H^3$ given by $A=|z|^2+it$, we define the map $h$ on $H^3-\left\{e\right\}$ by $$ h(z,t)=\left(\frac{-z}{|z|^2-it},\frac{-t}{|z|^4+t^2}\right)=\left(-z\overline{A^{-1}},-t A^{-1} \overline{A^{-1}}\right).$$

I would like to show that: $$\begin{equation} (f\circ h)*Z=-\overline{A^{-1}}\left|(f*Z)\circ h+2\overline{z}\, (f*E_{z})\circ h\right|; \quad E_{z}:= zZ. \quad (2) \end{equation}$$ For this I need to prove: $$\begin{equation} z*Z=1,\,\,\, \overline{z}*Z=0,\,\,\, t*Z=i\overline{z}. \qquad\qquad\qquad (3.1)\\ A*Z=0,\quad \overline{A}*Z=2\overline{z}.\qquad\qquad\qquad\qquad\qquad \qquad (3.2)\\ (z\circ h)*Z=\overline{A^{-2}} \, (2|z|^2-\overline{A}), \quad (t\circ h)*Z=-i\overline{z}\, \overline{A^{-2}}. \quad (3.3) \end{equation}$$

By a simple calculation I found the equations (3.1) and (3.2), but I am not understood the notations $(z\circ h)$ and $(t\circ h)$ ??, so that I can check the equation (3.3)? Finally I need a helping hand to prove (2).

$\triangle$ I found this formulas, in the paper of: Korányi, Kelvin transforms and harmonic polynomials on the Heisenberg group. JFA 1982.

$\Rightarrow$ 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

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 $$ h(z,t)=\left(\frac{-z}{|z|^2-it},\frac{-t}{|z|^4+t^2}\right)=\left(-z\overline{A^{-1}},-t A^{-1} \overline{A^{-1}}\right).$$

I would like to show that: $$\begin{equation} (f\circ h)*Z=-\overline{A^{-1}}\left|(f*Z)\circ h+2\overline{z}\, (f*E_{z})\circ h\right|; \quad E_{z}:= zZ. \quad (2) \end{equation}$$ For this I need to prove: $$\begin{equation} z*Z=1,\,\,\, \overline{z}*Z=0,\,\,\, t*Z=i\overline{z}. \qquad\qquad\qquad (3.1)\\ A*Z=0,\quad \overline{A}*Z=2\overline{z}.\qquad\qquad\qquad\qquad\qquad \qquad (3.2)\\ (z\circ h)*Z=\overline{A^{-2}} \, (2|z|^2-\overline{A}), \quad (t\circ h)*Z=-i\overline{z}\, \overline{A^{-2}}. \quad (3.3) \end{equation}$$

By a simple calculation I found the equations (3.1) and (3.2), but I am not understood the notations $(z\circ h)$ and $(t\circ h)$ ??, so that I can check the equation (3.3)? Finally I need a helping hand to prove (2).

$\triangle$ I found these formulas, in the paper of: Korányi, Kelvin transforms and harmonic polynomials on the Heisenberg group. JFA 1982.

$\Rightarrow$ 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

The Heisenberg group $H^3$ 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 $H^3$. If D is a left-invariant differential operator on $H^3$, 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 $H^3$ given by $A=|z|^2+it$, we define the map $h$ on $H^3-\left\{e\right\}$ by $$ h(z,t)=\left(\frac{-z}{|z|^2-it},\frac{-t}{|z|^4+t^2}\right)=\left(-z\overline{A^{-1}},-t A^{-1} \overline{A^{-1}}\right).$$

I would like to show that: $$\begin{equation} (f\circ h)*Z=-\overline{A^{-1}}\left|(f*Z)\circ h+2\overline{z}\, (f*E_{z})\circ h\right|; \quad E_{z}:= zZ. \quad (2) \end{equation}$$ For this I need to prove: $$\begin{equation} z*Z=1,\,\,\, \overline{z}*Z=0,\,\,\, t*Z=i\overline{z}. \qquad\qquad\qquad (3.1)\\ A*Z=0,\quad \overline{A}*Z=2\overline{z}.\qquad\qquad\qquad\qquad\qquad \qquad (3.2)\\ (z\circ h)*Z=\overline{A^{-2}} \, (2|z|^2-\overline{A}), \quad (t\circ h)*Z=-i\overline{z}\, \overline{A^{-2}}. \quad (3.3) \end{equation}$$

By a simple calculation I found the equations (3.1) and (3.2), but I am not understood the notations $(z\circ h)$ and $(t\circ h)$ ??, so that I can check the equation (3.3)? Finally I need a helping hand to prove (2).

$\triangle$ I found this formulas, in the paper of: Korányi, Kelvin transforms and harmonic polynomials on the Heisenberg group. JFA 1982.

$\Rightarrow$ 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 $H^3=\mathbb C\times \mathbb R$ (n=1).

Thank you in advance

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 $H^3$. If D is a left-invariant differential operator on $H^3$, 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 $H^3$ given by $A=|z|^2+it$, we define the map $h$ on $H^3-\left\{e\right\}$ by $$ h(z,t)=\left(\frac{-z}{|z|^2-it},\frac{-t}{|z|^4+t^2}\right)=\left(-z\overline{A^{-1}},-t A^{-1} \overline{A^{-1}}\right).$$

I would like to show that: $$\begin{equation} (f\circ h)*Z=-\overline{A^{-1}}\left|(f*Z)\circ h+2\overline{z}\, (f*E_{z})\circ h\right|; \quad E_{z}:= zZ. \quad (2) \end{equation}$$ For this I need to prove: $$\begin{equation} z*Z=1,\,\,\, \overline{z}*Z=0,\,\,\, t*Z=i\overline{z}. \qquad\qquad\qquad (3.1)\\ A*Z=0,\quad \overline{A}*Z=2\overline{z}.\qquad\qquad\qquad\qquad\qquad \qquad (3.2)\\ (z\circ h)*Z=\overline{A^{-2}} \, (2|z|^2-\overline{A}), \quad (t\circ h)*Z=-i\overline{z}\, \overline{A^{-2}}. \quad (3.3) \end{equation}$$

By a simple calculation I found the equations (3.1) and (3.2), but I am not understood the notations $(z\circ h)$ and $(t\circ h)$ ??, so that I can check the equation (3.3)? Finally I need a helping hand to prove (2).

$\triangle$ I found this formulas, in the paper of: Korányi, Kelvin transforms and harmonic polynomials on the Heisenberg group. JFA 1982.

$\Rightarrow$ 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