As we all know that the irreducible representation for Heisenberg group can be classified easily when the group is over a finite field $\mathbb{F}_q$, where $q=p^n$ and $p$ is a prime greater than $2$. The Heisenberg group is defined to be $$\left\{\left.\begin{pmatrix}1 & \mathbf{x}& t\\0 & I_n &\mathbf{y} \\ 0 & 0 & 1\end{pmatrix}\right|t\in\mathbb{F}_q,\mathbf{x},\mathbf{y}\in\mathbb{F}_q^n \right\}$$ with matrix multiplication. The way to classify them is to use the isomorphism $$\varphi:\begin{pmatrix}1 & \mathbf{x}& t\\0 & I_n &\mathbf{y} \\ 0 & 0 & 1\end{pmatrix} \to (t-\frac{1}{2}\mathbf{x}\cdot \mathbf{y},\mathbf{x},\mathbf{y}^T).$$ But in characteristic $2$, the $1/2$ does not make any sense. There's a way to treat the case here, but it seems hard to classify irreducible representation for the group. So my question is:

Are there any references for the classification of irreducible representation of characteristic $2$? Or can you give me some hints of how to do so?

Thanks for your help!


1 Answer 1


Heisenberg groups over a finite field $\mathbb{F_q}$ with $q=2^m$ are abelian and its representations are all one-dimentional, i.e., characters. The classification of irreducible representations is given (among other references) in section $3$ of the article The Representations of the Heisenberg Group over a Finite Field by M. Misaghian. See section $2$ for the definition of the Heisenberg groups, also in characteristic $2$.

  • 6
    $\begingroup$ Are there different definitions of the Heisenberg group around? The group defined above is not abelian when $q$ is even and $n \ge 1$, and this agrees with the definition on the Wikipedia page, which says that the group is $D_8$ when $n=1$ and $q=2$. $\endgroup$
    – Derek Holt
    Aug 26, 2014 at 15:16
  • 1
    $\begingroup$ The author of the article define the Heisenberg group to be $(x,a)(x',a')=(x+x',a+a'+\langle x,x'\rangle)$, which differs my definition when $p=2$. But anyway, this definition is still valid and gives a different group. $\endgroup$ Aug 28, 2014 at 4:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.