The group operation of $\mathbb H$ is defined by: $$[x,y,t]\cdot [x',y',t']= [x+x',y+y',t+t'-2xy'+2x'y] $$ The subgroup of $GL_3 (\mathbb R)$ has the form $$G= \Bigg\{\begin{bmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{bmatrix}: a,b,c \in \mathbb R \Bigg\}$$
My guess is since $$ [x,y,t][x',y',t']=[x+x',y+y',t+t'+(x+y)(x'-y')-(x-y)(x'+y')]$$
Define the function \begin{align*} \phi: & \mathbb H^1 \rightarrow G \\ & [x,y,t] \rightarrow \begin{bmatrix} 1 & x+y & f(x,y,t) \\ 0 & 1 & x-y \\ 0 & 0 & 1 \end{bmatrix} \end{align*}
So $\phi([x,y,t]\cdot[x',y',t'])=\phi([x,y,t])\cdot\phi([x',y',t'])$. But I couldn't figure out $f$, can someone give a hint?
Thanks.