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$\DeclareMathOperator\GL{GL}$Suppose that we have a Zariski-dense subgroup $\Gamma$ of $\GL(d,\mathbb{R})$. Let $\delta$ be the abscissa of convergence of the series $$ \sum_{x \in \Gamma} e^{-s \log\|x\|} $$ and suppose that this series diverges at $s = \delta$. If we have a subsemigroup $N$ of $\Gamma$ such that $$ \sum_{x \in N} e^{-\delta \log\|x\|} = \infty $$ can we conclude that $N$ is Zariski dense in $\GL(d,\mathbb{R})$? (Edit: it has been pointed out in the comments that $N$ being Zariski dense is equivalent to the subgroup generated by $N$ being Zariski dense).

Intuitively this feels like it should be true - but I can't see a way in to prove it! Any suggestions would be appreciated.

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    $\begingroup$ The inverse of an element is in its Zariski closure so this is equivalent to whether the group generated by N is Zariski dense. $\endgroup$ May 20, 2021 at 16:17
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    $\begingroup$ In more detail, my answer to mathoverflow.net/questions/246974/… shows the Zariski closure of a subsemigroup is a group over an algebraically closed field using Ax-Grothendieck. But this is also valid over R by Białynicki-Birula, A., Rosenlicht, M.: Injective Morphisms of Real Algebraic Varieties. Maybe there is a more direct argument to deduce or from the complex case $\endgroup$ May 20, 2021 at 16:22
  • $\begingroup$ Thanks for the comment - I'll add a reference to this in the question. $\endgroup$ May 20, 2021 at 16:26
  • $\begingroup$ Actually reading the comments to my answer a Zariski closed submonoid M of GL_n is a subgroup just because any closed subspace is Noetherian and a direct algebraic argument $\endgroup$ May 20, 2021 at 16:27
  • $\begingroup$ Eventually if $\Lambda$ is the subgroup generated by $N$, then $\sum_{x\in\Lambda}e^{-\delta\|x\|}=\infty$. Hence, it is no restriction to assume that $N$ is a subgroup. $\endgroup$
    – YCor
    May 20, 2021 at 21:18

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