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Let $G$ be a finite group, $N$ its normal subgroup, and $\pi:N\to{\mathcal B}(X)$ a unitary representation of $N$ in a Hilbert space $X$. Consider the induced representation $\pi':G\to{\mathcal B}(L_2(F,X))$, where $F=G/N$.

Let us extend the representations $\pi$ and $\pi'$ to the group algebras: $$ \pi:{\mathbb C}[N]\to{\mathcal B}(X),\qquad \pi':{\mathbb C}[G]\to{\mathcal B}(L_2(F,X)). $$

Question:

is it true that for all elements $a\in {\mathbb C}[N]$ the norms of the operators $\pi(a)\in {\mathcal B}(X)$ and $\pi'(a)\in {\mathcal B}(L_2(F,X))$ coincide $$ ||\pi(a)||=||\pi'(a)||\quad ? $$

I was sure, that this is true for all $G$, $N$ and $\pi$ (and not only for finite groups), but recently I found unexpectedly, that I can prove this only for the case of $G=N\times F$. Is it possible that there exists a counterexample for $G\ne N\times F$?

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No. Let $G=S_3$ generated by 3-cycle $x$ and involution $y$. Let $N=A_3$. Let $ x$ act on $X=\mathbb C$ by multiplication by the third root of unity $\omega$. Let $a=1+ix$. Then $\pi(a)$ has norm $|1+i\omega|$ but $\pi'(a)y=(1+i\omega^2)y$ showing $\pi'(a)$ has larger norm.

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  • $\begingroup$ Fan, why $\pi'(x)y=\omega^2y$? $\endgroup$
    – Sergei Akbarov
    Commented Jan 9, 2017 at 21:11
  • $\begingroup$ @SergeiAkbarov, because $y x y^{-1}$ equals $x^{-1} = x^2$. $\endgroup$
    – LSpice
    Commented Jan 9, 2017 at 21:25
  • $\begingroup$ I don't understand something. I thought, $y$ must be an element of ${\mathcal B}(L_2(F))$? $\endgroup$
    – Sergei Akbarov
    Commented Jan 9, 2017 at 21:31
  • $\begingroup$ Ah, yes, you mean the indicator function of the non-trivial element in $F$... OK. $\endgroup$
    – Sergei Akbarov
    Commented Jan 9, 2017 at 21:44

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