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Equiped with the law $(a,b,c)\circ (a',b',c') = (a + a', b + b', c + c' + ab'),$ the matrix representation of the Heisenberg group $H^3$ is given by $$ \begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix}; \quad a,b,c\in \mathbb{R}. $$ My question, what the matrix representation of $H^3$, if the law in $H^3$ is given by $$(x,y,z) . (x',y',z') = (x + x', y + y', z + z' + xy'-yx').$$ thank you in advance

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    $\begingroup$ There is no "the" matrix representation, but many faithful matrix representations. Your alternative law for $H^3$ also admits a faithful 3-dimensional representation, which can be found using Lie algebra coordinates. $\endgroup$ – YCor Mar 5 '16 at 19:47
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I think the answer is as follows, we have:

  • $$ \begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix} \begin{pmatrix} 1 & a' & c'\\ 0 & 1 & b'\\ 0 & 0 & 1\\ \end{pmatrix}= \begin{pmatrix} 1 & a+a' & c+c'+ab'\\ 0 & 1 & b+b'\\ 0 & 0 & 1\\ \end{pmatrix} $$ is corresponds to $(a,b,c)\circ (a',b',c') = (a + a', b + b', c + c' + ab')$.

. And $$ \begin{pmatrix} 1 & 0 & 0 & y \\ x & 1 & b & z \\ 0 & 0 & 1 & - x\\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & y' \\ x' & 1 & y' & z' \\ 0 & 0 & 1 & - x'\\ 0 & 0 & 0 & 1 \\ \end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 & y+y' \\ x+x' & 1 & y+y' & z + z' + xy'-yx' \\ 0 & 0 & 1 & - x-x'\\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$ is corresponds to $(x,y,z).(x',y',z')=(x + x', y + y', z + z' + xy'-yx')$.

It follows that, the matrix representation of $H^3$, if the law in $H^3$ is $(x,y,z) . (x',y',z') = (x + x', y + y', z + z' + xy'-yx')$, is given by $$\begin{pmatrix} 1 & 0 & 0 & y \\ x & 1 & y & z \\ 0 & 0 & 1 & - x\\ 0 & 0 & 0 & 1 \\ \end{pmatrix}, \quad x,y,z \in \mathbb R$$ thank you for any remark or comment.

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