2
$\begingroup$

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 $G\ne N\times F$?

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Fan, why $\pi'(x)y=\omega^2y$? $\endgroup$ Jan 9, 2017 at 21:11
  • $\begingroup$ @SergeiAkbarov, because $y x y^{-1}$ equals $x^{-1} = x^2$. $\endgroup$
    – LSpice
    Jan 9, 2017 at 21:25
  • $\begingroup$ I don't understand something. I thought, $y$ must be an element of ${\mathcal B}(L_2(F))$? $\endgroup$ Jan 9, 2017 at 21:31
  • $\begingroup$ Ah, yes, you mean the indicator function of the non-trivial element in $F$... OK. $\endgroup$ Jan 9, 2017 at 21:44

Your Answer

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

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