# Check an equation on the Heisenberg group $H_1$

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 $$$$f*Z=\left(\frac{\partial}{\partial z}+i\overline{z}\frac{\partial}{\partial t}\right)f. \qquad (1)$$$$

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 $$$$z*Z=1,\,\,\, \overline{z}*Z=0,\,\,\, t*Z=i\overline{z}. \qquad\qquad\qquad (1.1)\\ A*Z=0,\quad \overline{A}*Z=2\overline{z}.\qquad\qquad\qquad\qquad\qquad \qquad (1.2)\\ (z\circ h)*Z=\overline{A^{-2}} \, (2|z|^2-\overline{A}), \quad (t\circ h)*Z=-i\overline{z}\, \overline{A^{-2}}. \quad (1.3)$$$$ 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
$$$$(f\circ h)*Z=-\overline{A^{-1}}\left|(f*Z)\circ h+2\overline{z}\, (f*E_{z})\circ h\right|; \quad E_{z}:= zZ. \quad (2)$$$$

$$\triangle$$ 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).

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

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*}
• @Zaldivar, No, I want $H_1= \mathbb C\times \mathbb R$. I idit my question. Mar 28 at 18:09