3 Heisenberg’s group
3.1 Construction
Given \(k\) a field, \(V\) a \(k\) vector space and \(V^*\) its dual vector space, we define the Heisenberg set associated to \(V\) by \(\mathcal{H}(V):=\{ (z,x,y) \in k\times V\times V^*\} \).
\(\mathcal{H}(V)\) is in bijection with \(k\times V\times V^*\).
Trivial.
We define an internal law on Heisenberg by the following formula : \((z_1,x_1,y_1)*(z_2,x_2,y_2) = (z_1+z_2+y_1(x_2),x_1+x_2,y_1+y_2)\) for every \((z_1,x_1,y_1),(z_2,x_2,y_2)\in \mathcal{H}(V)\).
The inverse of \((z,x,y)\in \mathcal{H}(V)\) is given by the formula \((-z- y(-x), - x ,- y)\).
Compute \(h*h^{-1}\).
We check the axioms of a group.
Under our identification of the bidual, the map \(\Phi : (z,x,y) \mapsto (z,y,x)\) defines a bijection between \(\mathcal{H}(V)\) and \(\mathcal{H}(V^*)\).
Compute \(\Phi \circ \Phi ^{-1}\) and \(\Phi ^{-1}\circ \Phi \).
Under our identification of the bidual, the map define in 85 is a group antiisomorphism from \(\mathcal{H}(V)\) to \(\mathcal{H}(V^*)\).
Compute that \(\Phi (h_1*h_2)=\Phi (h_2)*\Phi (h_1)\).
3.2 Center
We define the center of \(\mathcal{H}(V)\) by the set \(\mathcal{Z}_{\mathcal{H}(V)}:=\{ (z,0,0)\in \mathcal{H}(V),\ z\in k\} \)
The center of Heisenberg \(\mathcal{Z}_{\mathcal{H}(V)}\) is a subgroup of \(\mathcal{H}(V)\).
We check the axioms and compute.
The set define in 87 is indeed the center of \(\mathcal{H}(V)\).
3.3 Commutator and nilpotency
Let \(H_1:=(z_1,x_1,y_1)\) and \(H_2:=(z_2,x_2,y_2)\) be two elements of \(\mathcal{H}(V)\). The commutator of \([H_1,H_2]\) is \((y_1(x_2)-y_2(x_1),0,0)\).
We compute \(H_1*H_2*H_1^{-1}*H_2^{-1}\).
If \(V\) isn’t trivial, the subgroup of \(\mathcal{H}(V)\) generates by commutators isn’t trivial too.
By contradiction. If it was trivial, every element of \(\mathcal{H}(V)\) would belong to its center. Because \(V\) isn’t trivial, there exists \(x\in V\) such that \(x\ne 0\). Thus, \((0,x,0)\) would belong to the center. But because of the definition of the center, \(x = 0\). We get a contradiction.
If \(h:=(z,x,y)\) belongs to the commutator subgroup, then \(x=0\) and \(y=0\).
We compute.
If \(V\) isn’t trivial, \(\mathcal{H}(V)\) is a two step nilpotent group.
3.4 Short exact sequence
The map \(\varphi :z\mapsto (z,0,0)\) defines a homomorphism from \((k,+)\) to \(\mathcal{H}(V)\).
Trivial.
We suppose \(\varphi (x)=\varphi (y)\) and we show \(x=y\). No difficulties.
The map \(\psi :(z,x,y)\mapsto (x,y)\) defines a homomorphism from \(\mathcal{H}(V)\) to \(V\times V^*\).
Trivial.
Trivial.
We have a short exact sequence \(0\rightarrow k \stackrel{\varphi }{\rightarrow } \mathcal{H}(V) \stackrel{\psi }{\rightarrow } V\times V^* \rightarrow 0\).
We check that the kernel of \(\psi \) is exactly the image of \(\varphi \).
The pullback \(V\times \{ 0\} \) by \(\psi \) defines a subgroup of \(\mathcal{H}(V)\).
We check that it is a subgroup.
Check that \(h_1*h2=h2*h1\).
The subgroup defined in 99 is a normal subgroup of Heisenberg.
Check that \(g*h*g^{-1}\in \psi ^{-1}(V)\) for every \(g\in \mathcal{H}(V)\) and \(h\in \psi ^{-1}(V)\).
The subgroup defined in 99 is maximal among the commutative subgroups of \(\mathcal{H}(V)\).
By contradiction. If it’s not, then there exists \(Q\) a commutative subgroup such that \(\psi ^{-1}(V)\subset Q\) and \(Q\ne \psi ^{-1}(V)\). Let \(q:=(z,x,y)\in Q\backslash \psi ^{-1}(V)\). In particular, we have for every \(h=(a,b,0)\in \psi ^{-1}(V)\) that \(x*h=h*x\). We compute this equality and find out that for every \(b\in V\), \(y(b)=0\). Thus \(y=0\) and \(q\in \psi ^{-1}(V)\). Contradiction.
The pullback of \(\{ 0\} \times V^*\) by \(\psi \) defines a subgroup of \(\mathcal{H}(V)\).
We check the axioms.
Check that \(h_1*h2=h2*h1\).
The subgroup defined in 103 is ai normal subgroup of Heisenberg. .
Check that \(g*h*g^{-1}\in \psi ^{-1}(V^*)\) for every \(g\in \mathcal{H}(V)\) and \(h\in \psi ^{-1}(V^*)\).
The subgroup defined in 103 is maximal among the commutative subgroups of \(\mathcal{H}(V)\).
By contradiction. If it’s not, then there exists \(Q\) a commutative subgroup such that \(\psi ^{-1}(V^*)\subset Q\) and \(Q\ne \psi ^{-1}(V^*)\). Let \(q:=(z,x,y)\in Q\backslash \psi ^{-1}(V^*)\). In particular, we have for every \(h=(a,0,b)\in \psi ^{-1}(V^*)\) that \(x*h=h*x\). We compute this equality and find out that for every \(b\in V^*\), \(b(x)=0\). Thus \(x=0\) and \(q\in \psi ^{-1}(V^*)\). Contradiction.
3.5 Specifities of the case \(k=\mathbb {F}_q\)
If \(k\) is a finite field, then \(|\mathcal{H}(V)|=|k|\times |V|^2\).
With 81 we know \(\mathcal{H}(V)\cong k \times V \times V^*\). Because \(V\) is finite dimensional, \(|V|=|V^*|\), thus we get the result.
If \(k\) is a finite field, then \(|\mathcal{Z}_{\mathcal{H}(V)}|=|k|\).
Trivial, the center being isomorphic to \(k\).
If \(k\) is a finite field, then \([\mathcal{H}(V):\mathcal{Z}_{\mathcal{H}(V)}]=|V|^2\).