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What is the unitary dual of the Heisenberg group over non-archimedean local fields k of characteristic two? This is well-known for the real Heisenberg group and in fact, when local fields have characteristic different from two, the unitary dual the same structure as for real Heisenberg group. See e.g. "On a notion of rank for unitary representations of the classical groups", Roger Howe

Why does this not work for non-archimedean local fields k of characteristic two? Can we not just use Mackey machine? see Page 307 Example 3 in "UNITARY REPRESENTATIONS OF GROUP EXTENSIONS. I" Acta 1958 BY GEORGE W. MACKEY

We define the Heisenberg group as a group of matrices (not using a symplectic form) $$ \pmatrix{1 & x & t \\ 0 & 1 & y \\ 0 & 0 & 1}_, $$ where $x,y,t\in k$.

After checking, I think the answer is Yes. The proof is identical with real case.

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    $\begingroup$ Note that, strictly speaking, the unitary dual is not "the same as for [the] real Heiseberg group", because the central character is a unitary character of $k$, and so depends on the field. (Likewise, the unitary representations with trivial central character are one-dimensional, hence are parametrized by the Pontryagin dual of $k^2$, and so depend on $k$). Perhaps, you can say that, for appropriate fields $k$, the unitary dual has the same structure, namely, there is a unique irreducible complex unitary representation for any non-trivial unitary character of $k$. $\endgroup$
    – Victor Protsak
    Commented Feb 5, 2017 at 2:59
  • $\begingroup$ @VictorProtsak: You are right. I guess the unitary dual has the same structure even for local fields of characteristic two. please see section 1.2 of V. Lafforgue's paper: "vlafforg.perso.math.cnrs.fr/haagerup-rem.pdf" for an evidence $\endgroup$
    – m07kl
    Commented Feb 5, 2017 at 10:21

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