I am having problems trying to understand a statement by Howe in his paper "On the role of the Heisenberg group in harmonic analysis". Here is the setting:
Howe defined the (reduced) Heisenber group as the set $\mathbb{H} = \mathbb{R^n} \times \mathbb{R^n} \times \mathbb{T}$ with the operation
$$ (y_1,x_1,z_1) \cdot (y_2,x_2,z_1) = (y_1 + y_2, x_1 + x_2, e^{-2\pi i y_2 \cdot x_1}z_1z_2). $$
He defines a unitary representation of $\mathbb{H}$ on $L^2(\mathbb{R}^n)$ by
$$ \rho(x,y,z) = M_yT_xS_z $$
where
$$ [M_y f](x') = e^{2\pi i y\cdot x'}f(x') $$ $$ [T_x f](x') = f(x' - x)$$ $$ [S_z f](x') = zf(x'). $$
Now this is where I am having difficulty understanding. Howe goes to explain (on page 824):
Let $\mathfrak{g}$ be the Lie algebra of $\mathbb{H}$. Let $e: \mathfrak{g} \to \mathbb{H}$ by the exponential map. We can choose a basis $X_j, Y_j, T$ for $\mathfrak{g}$ such that
$$e \left(\sum y_j Y_j + \sum x_j X_j + tT \right) = \left( y, x, e^{2\pi i(t - x\cdot y/2)} \right).$$
We find easily that
$$ \rho(X_j) = -\partial_{x_j},\, \rho(Y_j) = 2\pi i x_j,\, \rho(t) = 2\pi i. $$
I'm not sure how to interpret this last statement. Where do these last identities come from?
Let $\rho$ be a representation of $G$ on $V$ and $\dot{\rho}$ the induced representation of $\mathfrak{g}$. Then for $X \in \mathfrak{g}$, we have $$ \rho(\exp(X)) = \exp( \dot{\rho}(X) )$$ or in other words, a natural square commutes. This falls out of the definition of $\dot{\rho}$: any $X \in \mathfrak{g}$ generates a one-parameter subgroup of $G$, namely $t \mapsto \exp(tX)$, which we can then push through $\rho$ to get a one-parameter subgroup $t \mapsto \rho(\exp(tX))$ of $\mathrm{GL}(V)$, which (by standard Lie theory) determines a unique infinitesimal generator, $Y \in \mathfrak{gl}(V)$. Define $\dot{\rho}(X) := Y$.