I am understanding the Heisenberg group interpreted as a group of operators. At the moment, I have been able to understand the following:
The set of matrices \begin{align} A=\begin{bmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{bmatrix} \end{align} is the Heisenberg group ($H$). If $\mathrm{e}$ is the exponential (matrix), then $\mathrm{e}^{X}=A$ with $X=\begin{bmatrix} 0 & a & b\\ 0 & 0 & c\\ 0 & 0 & 0 \end{bmatrix}$. The set of matrices $X$ is the Lie Algebra of the Lie Group $H$.If $\Pi$ is a representation, i.e., $\Pi:H\to \text{GL}(V)$ then, for each $A\in H$,
\begin{align}\Pi(A=\mathrm{e}^{X}):V\to V\end{align} and \begin{align} (\Pi(A)f)(x)=(\mathrm{e}^{\pi(X)}f)(x),\quad f\in V,\, \end{align} with $\pi(X)=\left.\frac{d}{dt}\right|_{t=0}\Pi(\mathrm{e}^{tX})$
But, I understand that the Heisenberg group of operators is $\Pi(A)=\mathrm{e}^{a\partial_x+bx+c}$, Given this, my questions arise:
Question 1. The representation is $\Pi(A)=\mathrm{e}^{a\partial_x+bx+c}$?
Question 2 . What is the vector space $V$? Can $V$ be the space $L^2(\mathbb{R})$?
Question 3.The exponential is $\mathrm{e}^{a\partial_x+bx+c}=\sum_{k=0}^{\infty}\frac{(a\partial_x+bx+c)^k}{k!}$?converges? in what sense?
Thanks!
Partial answer.
The Lie algebra of the Heisenberg group is generated by the trace--free and nilpotent matrices $$ X=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\,,\quad Y=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}\,,\quad Z=\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix} $$ which satisfy $$ XY=Z\,,\quad YX=XZ=ZX=YZ=ZY=0 $$ and therefore the commutation relations \begin{align} [X,Y]=Z\,,\quad [X,Z]=0\,,\quad [Y,Z]=0\,. \end{align} Three vector fields (operators) that satisfy the same relations and therefore generate the same Lie algebra are \begin{align*} \boldsymbol{X}&=\partial_x\,, &\boldsymbol{Y}&=x\,\partial_z+\partial_y\,,&\boldsymbol{Z}&=\partial_z\,. \end{align*}