Deriving the Discrete Heisenberg Group generators.
Multi tool use
$begingroup$
How can we derive the generators of the Discrete Heisnberg Group?
Everyone seems to just state this as a given and never actually derive it from scratch.
I'm looking for a (somewhat) elementary derivation
group-theory group-presentation
edited Dec 21 '18 at 17:36
Shaun
9,789113684
$endgroup$
add a comment |
$begingroup$
How can we derive the generators of the Discrete Heisnberg Group?
Everyone seems to just state this as a given and never actually derive it from scratch.
I'm looking for a (somewhat) elementary derivation
group-theory group-presentation
edited Dec 21 '18 at 17:36
Shaun
9,789113684
$endgroup$
2
$begingroup$
The Wikipedia page gives an explicit formula for an arbitrary element in terms of the generators x and y (z can be written in terms of x and y also, Wikipedia gives the computation for that). This is under the section "discrete Heisenberg group". You can try to verify these formulas by computation.
$endgroup$
– Lorenzo
Dec 18 '18 at 15:52
$begingroup$
I have no idea how i missed that, but still a derivation from scratch would be nice
$endgroup$
– user371732
Dec 18 '18 at 16:13
add a comment |
$begingroup$
How can we derive the generators of the Discrete Heisnberg Group?
Everyone seems to just state this as a given and never actually derive it from scratch.
I'm looking for a (somewhat) elementary derivation
group-theory group-presentation
edited Dec 21 '18 at 17:36
Shaun
9,789113684
$endgroup$
How can we derive the generators of the Discrete Heisnberg Group?
Everyone seems to just state this as a given and never actually derive it from scratch.
I'm looking for a (somewhat) elementary derivation
group-theory group-presentation
group-theory group-presentation
edited Dec 21 '18 at 17:36
Shaun
9,789113684
edited Dec 21 '18 at 17:36
Shaun
9,789113684
edited Dec 21 '18 at 17:36
Shaun
9,789113684
edited Dec 21 '18 at 17:36
Shaun
9,789113684
edited Dec 21 '18 at 17:36
Shaun
9,789113684
9,789113684
asked Dec 18 '18 at 15:45
user371732
2
$begingroup$
The Wikipedia page gives an explicit formula for an arbitrary element in terms of the generators x and y (z can be written in terms of x and y also, Wikipedia gives the computation for that). This is under the section "discrete Heisenberg group". You can try to verify these formulas by computation.
$endgroup$
– Lorenzo
Dec 18 '18 at 15:52
$begingroup$
I have no idea how i missed that, but still a derivation from scratch would be nice
$endgroup$
– user371732
Dec 18 '18 at 16:13
add a comment |
2
$begingroup$
The Wikipedia page gives an explicit formula for an arbitrary element in terms of the generators x and y (z can be written in terms of x and y also, Wikipedia gives the computation for that). This is under the section "discrete Heisenberg group". You can try to verify these formulas by computation.
$endgroup$
– Lorenzo
Dec 18 '18 at 15:52
$begingroup$
I have no idea how i missed that, but still a derivation from scratch would be nice
$endgroup$
– user371732
Dec 18 '18 at 16:13
2
2
$begingroup$
The Wikipedia page gives an explicit formula for an arbitrary element in terms of the generators x and y (z can be written in terms of x and y also, Wikipedia gives the computation for that). This is under the section "discrete Heisenberg group". You can try to verify these formulas by computation.
$endgroup$
– Lorenzo
Dec 18 '18 at 15:52
$begingroup$
The Wikipedia page gives an explicit formula for an arbitrary element in terms of the generators x and y (z can be written in terms of x and y also, Wikipedia gives the computation for that). This is under the section "discrete Heisenberg group". You can try to verify these formulas by computation.
$endgroup$
– Lorenzo
Dec 18 '18 at 15:52
$begingroup$
I have no idea how i missed that, but still a derivation from scratch would be nice
$endgroup$
– user371732
Dec 18 '18 at 16:13
$begingroup$
I have no idea how i missed that, but still a derivation from scratch would be nice
$endgroup$
– user371732
Dec 18 '18 at 16:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since the discrete Heisenberg group is defined to be the subgroup of $GL_3(Bbb{Z})$ consisiting of upper-unitriangular matrices, it is clear that the generators are given by $x,y,z$, where
$$
begin{pmatrix} 1 & a & c\ 0 & 1 & b\ 0 & 0 & 1\ end{pmatrix}=y^bz^cx^a, $$
see Wikipedia. Here it is enough to consider $x$ and $y$ since $z=[x,y]$ by matrix multiplication.
answered Dec 18 '18 at 16:14
Dietrich BurdeDietrich Burde
81.4k648106
$endgroup$
add a comment |
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$begingroup$
Since the discrete Heisenberg group is defined to be the subgroup of $GL_3(Bbb{Z})$ consisiting of upper-unitriangular matrices, it is clear that the generators are given by $x,y,z$, where
$$
begin{pmatrix} 1 & a & c\ 0 & 1 & b\ 0 & 0 & 1\ end{pmatrix}=y^bz^cx^a, $$
see Wikipedia. Here it is enough to consider $x$ and $y$ since $z=[x,y]$ by matrix multiplication.
answered Dec 18 '18 at 16:14
Dietrich BurdeDietrich Burde
81.4k648106
$endgroup$
add a comment |
$begingroup$
Since the discrete Heisenberg group is defined to be the subgroup of $GL_3(Bbb{Z})$ consisiting of upper-unitriangular matrices, it is clear that the generators are given by $x,y,z$, where
$$
begin{pmatrix} 1 & a & c\ 0 & 1 & b\ 0 & 0 & 1\ end{pmatrix}=y^bz^cx^a, $$
see Wikipedia. Here it is enough to consider $x$ and $y$ since $z=[x,y]$ by matrix multiplication.
answered Dec 18 '18 at 16:14
Dietrich BurdeDietrich Burde
81.4k648106
$endgroup$
add a comment |
$begingroup$
Since the discrete Heisenberg group is defined to be the subgroup of $GL_3(Bbb{Z})$ consisiting of upper-unitriangular matrices, it is clear that the generators are given by $x,y,z$, where
$$
begin{pmatrix} 1 & a & c\ 0 & 1 & b\ 0 & 0 & 1\ end{pmatrix}=y^bz^cx^a, $$
see Wikipedia. Here it is enough to consider $x$ and $y$ since $z=[x,y]$ by matrix multiplication.
answered Dec 18 '18 at 16:14
Dietrich BurdeDietrich Burde
81.4k648106
$endgroup$
Since the discrete Heisenberg group is defined to be the subgroup of $GL_3(Bbb{Z})$ consisiting of upper-unitriangular matrices, it is clear that the generators are given by $x,y,z$, where
$$
begin{pmatrix} 1 & a & c\ 0 & 1 & b\ 0 & 0 & 1\ end{pmatrix}=y^bz^cx^a, $$
see Wikipedia. Here it is enough to consider $x$ and $y$ since $z=[x,y]$ by matrix multiplication.
answered Dec 18 '18 at 16:14
Dietrich BurdeDietrich Burde
81.4k648106
answered Dec 18 '18 at 16:14
Dietrich BurdeDietrich Burde
81.4k648106
answered Dec 18 '18 at 16:14
Dietrich BurdeDietrich Burde
81.4k648106
answered Dec 18 '18 at 16:14
Dietrich BurdeDietrich Burde
81.4k648106
81.4k648106
add a comment |
add a comment |
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2
$begingroup$
The Wikipedia page gives an explicit formula for an arbitrary element in terms of the generators x and y (z can be written in terms of x and y also, Wikipedia gives the computation for that). This is under the section "discrete Heisenberg group". You can try to verify these formulas by computation.
$endgroup$
– Lorenzo
Dec 18 '18 at 15:52
$begingroup$
I have no idea how i missed that, but still a derivation from scratch would be nice
$endgroup$
– user371732
Dec 18 '18 at 16:13