Heisenberg group

Groups, Rings, Domains, Modules, etc, Galois theory
Post Reply
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Heisenberg group

#1

Post by Papapetros Vaggelis »

Let

\(\displaystyle{H=\left\{\begin{pmatrix}
1&a &b \\
0&1 &c \\
0&0 &1
\end{pmatrix}:a\,,b\,,c\in\mathbb{Z}\right\}\leq GL(3,\mathbb{Z})}\) .

Prove that the group \(\displaystyle{\left(H,\cdot\right)}\) is a nilpotent group.
dmsx
Posts: 9
Joined: Thu Jan 28, 2016 8:52 pm

Re: Heisenberg group

#2

Post by dmsx »

It is fairly easy to check (and maybe come up with) that the center of $H$ is $Z(H)=\left \{ \begin{pmatrix} 1 &0 &b \\ 0 &1 &0 \\ 0 &0 &1 \end{pmatrix}:b\in \mathbb{Z} \right \}$.

Define $f:H\rightarrow \mathbb{Z}\times \mathbb{Z}$, $\begin{pmatrix} 1 &a &b \\ 0 &1 &c \\ 0 &0 &1 \end{pmatrix} \overset{f}{\mapsto} (a,c)$. We observe that $f$ is an epimorphism (routine check), $kerf=Z(H)$ and therefore $H/Z(H)\cong \mathbb{Z}\times \mathbb{Z}$. That means that $H/Z(H)$ is abelian, so $Z(H/Z(H))=H/Z(H)=\zeta _2(H)/\zeta _1(H)$, which means $\zeta _2(H)=H$ (i am using the notation $\zeta _n(H)$ for the factors of the upper central ceries of $H$). Therefore, $1=\zeta _0(H)\vartriangleleft \zeta _1(H)=Z(H)\vartriangleleft \zeta _2(H)=H$ is a central series for $H$, so $H$ is nilpotent.

In general, if $R$ is a commutative ring with a unit $1_R(=1)$ and $G=GL(n,R)$ is the group of $n\times n$ invertible matrices, show that $H=\left \{ \begin{pmatrix} 1 &* &. &. &* \\ 0 &1 &* &. &* \\ . &. &. &. &. \\ . &. &. &. &* \\0 &. &. &0 &1 \\ \end{pmatrix}:*\in R \right \}\leq G$ is nilpotent. Furthermore, i leave as an exercise the following:

Give an example of two non-isomorphic, non-abelian, infinite, nilpotent groups. I will post the answer by Friday. Take your time.
dmsx
Posts: 9
Joined: Thu Jan 28, 2016 8:52 pm

Re: Heisenberg group

#3

Post by dmsx »

It's due time. Solutions are given for both exercises:

For the first exercise, denote this particular $H$ as $U_n(R)$. Facing an $n\times n$ matrix as a $3\times 3$ block matrix we write:

$U_n(R)=\left \{ \begin{pmatrix} 1 &A &b \\ 0_{n-2} &I_{n-2} &C \\ 0 &0_{n-2} &1 \end{pmatrix}:A\epsilon M_{1\times n-1}(R), b\in R, C\epsilon M_{n-1\times 1}(R) \right \}$

where $A$ is a row vector with $n-2$ columns, $C$ is a column vector with $n-2$ rows, $I_{n-2}$ is the identity matrix and $0_{n-2}$ denotes a zero vector (the one on the first column is a column vector and the one on the last row is a row vector). Dimensions are the appropriate ones (namely, they add up to $n$).

Since multiplication of matrices in block form follows the usual product rules of matrices, we can use the above example generalising appropriately (alternatively, we can use induction, which i do not encourage). So the general case is completely analogous to the case $n=3$.

Considering the above observations, we can reach the following result:

$\zeta _1(U_n(R))=Z(U_n(R))\left \{ \begin{pmatrix}1 &0 &. &. &. &. &0 &* \\ 0 &1 &0 &. &. &. &0 &0 \\ . & &. & & & & &. \\ . & & &. & & & &. \\ . & & & &. & & &. \\ . & & & & &. & &. \\ 0 & & & & & &1 &0 \\ 0 &0 &. &. &. &. &0 &1 \end{pmatrix}:*\epsilon R \right \}$,

$\zeta _2(U_n(R))=\left \{ \begin{pmatrix}1 &0 &. &. &. &0 &* &* \\ 0 &1 &0 &. &. &. &0 &* \\0 &. &. &. &. &. &. &0 \\ . & & &. & & & &. \\ . & & & &. & & &. \\ . & & & & &. & &. \\ 0 & & & & & &1 &0 \\ 0 &0 &. &. &. &. &0 &1 \end{pmatrix}:*\epsilon R \right \}$, $...$,

$\zeta _{n-2}(U_n(R))=\left \{ \begin{pmatrix}1 &0 &* &. &. &* &* &* \\ 0 &1 &0 &* &. &. &* &* \\ . & &. & &. & & &* \\ . & & &. & &. & &. \\ . & & & &. & &. &. \\ . & & & & &. & &* \\ 0 & & & & & &1 &0 \\ 0 &0 &. &. &. &. &0 &1 \end{pmatrix}:*\epsilon R \right \}$,

so $\zeta _{n-1}(U_n(R))=U_n(R)$ is nilpotent of class $n-1$, just like the previous post.


Corollary: There exist nilpotent groups of class $c$, for every $c\in \mathbb{N}$.


For the second exercise, consider the (discrete) Heisengroup group $H$ that is defined above and $H\times \mathbb{Z}$. We have:

* $H$ is nilpotent (showed above), non-abelian ($Z(H)\neq H$) and infinite (contains a copy of $\mathbb{Z}$)
* $H\times \mathbb{Z}$ is nilpotent (as a direct product of nilpotent groups), non-abelian ($Z(H\times \mathbb{Z})=Z(H)\times Z(\mathbb{Z})=Z(H)\times \mathbb{Z}\neq H\times \mathbb{Z}$) and infinite (contains a copy of $\mathbb{Z}$)

These two groups, however, are non-isomorphic: if they were, they would have isomorphic centers (if two groups are isomorphic, then restricting the isomorphism to the center induces an isomorphism between the two centers - it is very easy to check). However:

$Z(H\times \mathbb{Z})=Z(H)\times \mathbb{Z}\cong \mathbb{Z}\times \mathbb{Z} \ncong \mathbb{Z}=Z(H)$, they are non-isomorphic because: $\underset{n-times}{\underbrace{\mathbb{Z}\times ... \times \mathbb{Z}}}\cong \underset{m-times}{\underbrace{\mathbb{Z}\times ... \times \mathbb{Z}}}\Leftrightarrow n=m$.
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Heisenberg group

#4

Post by Papapetros Vaggelis »

:clap2: :clap2: :clap2: :clap2: :clap2: :clap2: :clap2: :clap2:

Thank you dmsx.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 6 guests