# Exponentiating a representation on the Heisenberg Lie algebra

We know that every representation $\rho_*: \mathfrak{h} \to GL(V)$ lifts to a representation $\rho: H \to GL(V)$, because the Heisenberg algebra $H$ of upper triangular matrices is simply connected. Given the basis $X = E_{12}$, $Y = E_{23}$, and $Z = E_{13}$, consider the representation of $\mathfrak{h}$ on $C^\infty(\mathbf{R})$ given by

$$(Xf)(x) = \pi i x f(x)\ \ \ (Yf)(x) = f'(x)\ \ \ (Zf)(x) = - \pi i f(x)$$

Exponentiating, we find that the induced representation on $H$ is given by

$$\begin{pmatrix} 1 & a & \\ & 1 & \\ & & 1 \end{pmatrix} f (x) = e^{\pi i a x} f(x)\ \ \ \begin{pmatrix} 1 & & b \\ & 1 & \\ & & 1 \end{pmatrix} f(x) = f(x + b)$$ $$\begin{pmatrix} 1 & & \\ & 1 & c \\ & & 1 \end{pmatrix} f(x) = e^{- \pi i c} f(x)$$

But we can write an arbitrary element of the Heisenberg group as

$$M = \begin{pmatrix} 1 & a & b \\ & 1 & c \\ & & 1 \end{pmatrix} = \begin{pmatrix} 1 & & \\ & 1 & c \\ & & 1 \end{pmatrix} \begin{pmatrix} 1 & a & \\ & 1 & \\ & & 1 \end{pmatrix} \begin{pmatrix} 1 & & b \\ & 1 & \\ & & 1 \end{pmatrix}$$

And so we can determine the action defined on the entire group as $(Mf)(x) = e^{\pi i(ax - c)} f(x + b)$. However, when I attempt to formally check that this is actually a representation of $H$, it seems this map isn't a representation at all, because $(M(Nf)) \neq (MN)(f)$. What's going wrong here? Is this a problem with infinite dimensional representations that I am not aware of?

• Page 3 of this paper may be useful. – Mark Apr 25 '17 at 5:00
• Almost perfect for what I was looking for! From a quick read, it seems the problem is that we don't necessarily have an exponential on $L^2(\mathbf{R})$. But can't we embed this space into $L^2(\mathbf{C})$ and consider the exponential defined by the Gelfand theory? – Jacob Denson Apr 25 '17 at 5:36
• There is the WP article. – Cosmas Zachos May 1 '17 at 21:42