Yesterday we were discussing the Heisenberg group in our weekly Geometric group theory workshop. The game of Mario came up! If you have not played Mario, then here is a the only thing you need to know. Mario can jump and hit floating bricks. These bricks may pop open to produce gold coins.
The method of simplifying products of elements in Heisenberg group is similar. First a few definitions.
Let \( \Gamma \) be a group generated by \( \{a, b, c\} \). Moreover set of relators contain \( ac = ca, bc = cb, ba = cab \). We say \(\Gamma\) is the Heisenberg group.
One way to think about this is: \( c\) commutes with everything, \(a\) and \( b\) do not commute but they are \( c\) away from actually commuting. Hence we do not have \( ba = ab \) but we have \( ba = c \cdot ab \).
We claim that every element of Heisenberg group looks like \( a^m b^n c^p \) where \( (m, n, p) \in \mathbb{Z}^3 \). We play a game of Mario with the elements of Heisenberg group to illustrate this.
Choose two elements \( a^m b^n c^p \) and \( a^r b^s c^t \) in \( \Gamma \). Lets multiply them and see what happens.
\[ a^m b^n c^p \cdot a^r b^s c^t \]
Since \( c \) commutes with everything, hence we can pull the \( c \)’s to the right and get the first simplification.
\[ a^m \cdot b^n \cdot a^r \cdot b^s \cdot c^t \cdot c^p \]
Now we focus on the \( b^n \cdot a^r \) in the middle of the expression. Think of each \( b\) as Mario and each of the \(a\)’s in \(a^r \) as a gold coin bearing brick. We will push the \( b\) to the right replacing \( b \cdot a \) by \(c \cdot a \cdot b\). The gold coin is \(c\). That is if we push the \(b\) to the right through one \(a\) then one \(c \) is produced and \(ba\) becomes \(ab\).
Now we can push that \( b \) through all the \( r \) copies of \(a\) so that it moves to the right and sits with \( b^s \). In this process \( r \) gold coins (\(c\)’s) will be produced. Hence we have \( c^r \) produced.
We will need to do this \(n \) times (for each of the Mario’s or \(b\)’s in \(b^n \)). Hence \( n \times r \) gold coins or \(c\)’s are produced in the process. After this is done we have the following expression.
\[ a^m \cdot a^r \cdot c^{nr} \cdot b^n \cdot b^s \cdot c^t \cdot c^p \]
Finally we push all the \( c\)’s to the right and get the final expression.
\[ a^{m+r} b^{n+s} c^{t + p + nr} \]
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