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I am reading a paper in which the author has a group $G$ admitting a representation $\pi$ on a vector space $V$. Let $g \in G$ be a group element. The author refers to a so-called "commutator of $g$ on $V$". It is denoted $[g,V]$ in the paper.

Could someone help me in clarifying what is usually meant by this term?

My guess is that $\left[g,V\right]$ is the subspace of $V$ generated by the co-invariants $\pi(g)v -v$. This is the only meaning I can see that makes sense. However, the element $g$ appears only once in this formula, whereas it should appear twice in a commutator.

It is not being used immediately in the paper so it is hard to figure out from the context.

Thank you.

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    $\begingroup$ @The_Lar: Your interpretation is correct. Let $H=(V^+)G$ be the semidirect product where $V^+$ is the additive group of $V$, and $G$ acts on $V$ via $\pi$. The back story is that if we set up multiplicative notation in $H$, so that $\pi(g)v=gvg^{-1}$, then $\pi(g)v-v$ is $gvg^{-1}v^{-1}$ in multiplicative notation. $\endgroup$
    – Richard Lyons
    Dec 13 at 2:23
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    $\begingroup$ I tend to agree with Richard Lyons and the_lar (it's the given span), and not the quotient by the span as suggested by Andy Putman. the_lar, which paper are you quoting? this should clarify everything. $\endgroup$
    – YCor
    Dec 13 at 7:30
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    $\begingroup$ Among group theorists, your interpretation is definitely the one used, and $\pi(g)v-v$ may be interpreted as a commutator for the reason given by @RichardLyons . Note also that $g$ acts as the identity on the quotient space $V/[g,V].$ $\endgroup$
    – Geoff Robinson
    Dec 13 at 10:12
  • $\begingroup$ @RichardLyons Thank you. This clarifies matters. $\endgroup$
    – the_lar
    Dec 17 at 12:49
  • $\begingroup$ @GeoffRobinson I see, thank you. I would consider myself a group theorist.. Well at least I was able to make the right guess... $\endgroup$
    – the_lar
    Dec 17 at 12:50

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