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Let $G$ be a finite abelian group, and let $G\longrightarrow \hat{G}$, $g\mapsto \chi_g$ be a group isomorphism from $G$ to its dual (i.e. the abelian group of group morphisms from $G$ to $\textbf{C}^{\times}$). We define two endormorphisms of $V=\textbf{C}[G]$ as a $\textbf{C}$-vector space by their action on $g\in G$: $$\rho_S(g)=\frac{1}{\sqrt{|G|}}\sum_{h\in G}{\overline{\chi_h(g)}h},$$ $$\rho_{T^2}(g)=\chi_g(g)g.$$ I have several questions:

  1. If $\Gamma$ is the matrix group generated by $S=\left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$ and $T^2=\left(\begin{smallmatrix} 1 & 2 \\ 0 & 1 \end{smallmatrix}\right)$, does $\rho:\Gamma\longrightarrow \operatorname{GL}(V)$ defines a representation?

  2. Is this the so-called Weil representation attached to $G$?

  3. What do we know about the reducibility of $\rho$?

Many thanks!

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    $\begingroup$ The Weil representation attached to $G$ is a representation of the Heisenberg group of $G$, not of $G$ itself. The Heisenberg group of $G$ contains a copy of $G$ itself, which acts on the Weil representation by translation. A nice paper on this is Prasad - On character values ….. $\endgroup$
    – LSpice
    Aug 3, 2018 at 12:37
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    $\begingroup$ Ok, thanks! Actually my confusion comes from the following text web.stanford.edu/~tonyfeng/245C.pdf where definition 2.2.4 seems close to what I introduce in this question. $\endgroup$
    – Stabilo
    Aug 3, 2018 at 12:46
  • $\begingroup$ Yes, you're quite right! I'm not sure how that compares with the Weil representation as defined by Prasad. Note that the standard reduction of a Weil representation into the even and odd subspaces works here, too, so that your representation (if it is a representation) is certainly reducible if $G$ is not the trivial group. $\endgroup$
    – LSpice
    Aug 3, 2018 at 12:49
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    $\begingroup$ Isn't $S^2=-1$ the only relation of the group generated by $S$ and $T^2$? So in that case (1) is certainly true. $\endgroup$
    – Will Sawin
    Aug 3, 2018 at 13:37
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    $\begingroup$ @LSpice My notation $S^2=-1$ was intended to denote $S^4=1$ and $S^2$ is central, so $S^2 T^2= T^2 S^2$. (Actually I worked it out for the corresponding subgroup of $PSL_2(\mathbb Z)$ and then hastily added the minus sign.) $\endgroup$
    – Will Sawin
    Aug 3, 2018 at 15:00

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