Return to Question

$\DeclareMathOperator\GL{GL}$Suppose that we have a Zariski-dense subgroup $\Gamma$ of $\GL(d,\mathbb{R})$. Let $\delta$ be the abscissa of convergence of the series $$ \sum_{x \in \Gamma} e^{-s \|x\|} $$ and suppose that this series diverges at $s = \delta$. If we have a subsemigroup $N$ of $\Gamma$ such that $$ \sum_{x \in N} e^{-\delta \|x\|} = \infty $$ 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.

$\DeclareMathOperator\GL{GL}$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.

Suppose that we have a Zariski dense subgroup $\Gamma$ of $GL(d,\mathbb{R})$. Let $\delta$ be the abscissa of convergence of the series $$ \sum_{x \in \Gamma} e^{-s \|x\|} $$ and suppose that this series diverges at $s = \delta$. If we have a subsemigroup $N$ of $\Gamma$ such that $$ \sum_{x \in N} e^{-\delta \|x\|} = \infty $$ 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.

$\DeclareMathOperator\GL{GL}$Suppose that we have a Zariski-dense subgroup $\Gamma$ of $\GL(d,\mathbb{R})$. Let $\delta$ be the abscissa of convergence of the series $$ \sum_{x \in \Gamma} e^{-s \|x\|} $$ and suppose that this series diverges at $s = \delta$. If we have a subsemigroup $N$ of $\Gamma$ such that $$ \sum_{x \in N} e^{-\delta \|x\|} = \infty $$ 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.

Suppose that we have a Zariski dense subgroup $\Gamma$ of $GL(d,\mathbb{R})$. Let $\delta$ be the abscissa of convergence of the series $$ \sum_{x \in \Gamma} e^{-s \|x\|} $$ and suppose that this series diverges at $s = \delta$. If we have a subsemigroup $N$ of $\Gamma$ such that $$ \sum_{x \in N} e^{-\delta \|x\|} = \infty $$ can we conclude that $N$ is Zariski dense in $GL(d,\mathbb{R})$?

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

Suppose that we have a Zariski dense subgroup $\Gamma$ of $GL(d,\mathbb{R})$. Let $\delta$ be the abscissa of convergence of the series $$ \sum_{x \in \Gamma} e^{-s \|x\|} $$ and suppose that this series diverges at $s = \delta$. If we have a subsemigroup $N$ of $\Gamma$ such that $$ \sum_{x \in N} e^{-\delta \|x\|} = \infty $$ 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.