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Let $F\twoheadrightarrow G$ be an epimorphism of groups, $F$ being finitely generated and free. Let $H$ be its kernel. Consider a lifting $i:G\hookrightarrow F$ of the epimorphism.

Every element of $C[F]$ is of the form $a=\sum_{g\in G} a(g)\,i(g)$, where $a(g)\in C[H]$. For any character $\chi$ of $H$ it is easy to see that $$\|a\|_{\max}^2\;\ge\;\max\left\{\sum_{g\in G}\left|\,\chi(a(g))\right|^2\, ,\,\sum_{g\in G} \left|\,\chi\!\left(g^{-1}(a(g))\right)\right|^2\right\}$$ (here the action of $G$ is induced by the adjoint action of $F$).
Proof: Consider the action of $a$ and $a^*$ in the representation induced by the character.

Question: Can this estimate be improved?
For example, could the following be true: Fix an integer $p$. Let $w(g)$ be the partial word of length $p$ in $i(g)$. Then $$\|a\|^2_{\max}\ge C_p\sum\left|\,\chi\left(w(g)^{-1}(a(g))\right)\right|^2$$ with $C_p$ polynomial in $p\,$?

What should I read in order even to approach proving/disproving such things?

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  • $\begingroup$ You mean "every element of $C[F]$", right? Could you clarify the "it is easy to see" computation. It seems like in the case where $a = a(g)$ for a single $g$ you are saying that the operator norm of a matrix must exceed its trace ... I must be confused about something. $\endgroup$
    – Nik Weaver
    Jun 24, 2015 at 16:18

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