Elements close to the identity in real Lie groups
I am having trouble with the following.
Let $G=mathrm{SL}(2,mathbb{R})$. Let $Gamma<G$ be a non co-compact (arithmetic) lattice. Is it true that there exists $epsilon>0$ such that if ${u_1,dots,u_n}subset Gamma$ are such that $|u_i-1|<epsilon$ for all $1leq ileq n$, then the $u_i$'s are unipotent? The Zassenhaus lemma implies that if $epsilon>0$ is small enough, then the $u_i$'s generate a nilpotent subgroup. But must this subgroup be unipotent (i.e. all eigenvalues are actually $1$)?
Thanks!
lie-groups
asked Nov 28 '18 at 19:27
Math-user
29617
add a comment |
I am having trouble with the following.
Let $G=mathrm{SL}(2,mathbb{R})$. Let $Gamma<G$ be a non co-compact (arithmetic) lattice. Is it true that there exists $epsilon>0$ such that if ${u_1,dots,u_n}subset Gamma$ are such that $|u_i-1|<epsilon$ for all $1leq ileq n$, then the $u_i$'s are unipotent? The Zassenhaus lemma implies that if $epsilon>0$ is small enough, then the $u_i$'s generate a nilpotent subgroup. But must this subgroup be unipotent (i.e. all eigenvalues are actually $1$)?
Thanks!
lie-groups
asked Nov 28 '18 at 19:27
Math-user
29617
1
do you mean ${u_1,dots,u_n}subsetGamma$ ? If not, what is the role of $Gamma$? Also, what is $n$? (as it stands, the question doesn't change if you set $n=1$)
– user8268
Nov 28 '18 at 20:09
yes sorry, I meant ${u_1,dots,u_n}subsetGamma$. $n$ is any integer, so yes if $n=1$, this should hold too. Thanks for the comment!
– Math-user
Nov 28 '18 at 20:28
1
I am still confused: $Gamma$ is discrete in $G$, right? But that would mean that for a sufficiently small $epsilon>0$ the condition $|u-1|<epsilon$, $uinGamma$, implies $u=1$.
– user8268
Nov 28 '18 at 20:38
Yes. When the real rank of $G$ is $geq 2$, by Margulis' arithmeticity theorem, there is a neighborhood of the identity $W$ such that for any non co-compact lattice, the intersection with this neighborhood contains only unipotent elements. But this does not apply to $mathrm{SL}(2,mathbb{R})$.
– Math-user
Nov 28 '18 at 20:55
add a comment |
I am having trouble with the following.
Let $G=mathrm{SL}(2,mathbb{R})$. Let $Gamma<G$ be a non co-compact (arithmetic) lattice. Is it true that there exists $epsilon>0$ such that if ${u_1,dots,u_n}subset Gamma$ are such that $|u_i-1|<epsilon$ for all $1leq ileq n$, then the $u_i$'s are unipotent? The Zassenhaus lemma implies that if $epsilon>0$ is small enough, then the $u_i$'s generate a nilpotent subgroup. But must this subgroup be unipotent (i.e. all eigenvalues are actually $1$)?
Thanks!
lie-groups
asked Nov 28 '18 at 19:27
Math-user
29617
I am having trouble with the following.
Let $G=mathrm{SL}(2,mathbb{R})$. Let $Gamma<G$ be a non co-compact (arithmetic) lattice. Is it true that there exists $epsilon>0$ such that if ${u_1,dots,u_n}subset Gamma$ are such that $|u_i-1|<epsilon$ for all $1leq ileq n$, then the $u_i$'s are unipotent? The Zassenhaus lemma implies that if $epsilon>0$ is small enough, then the $u_i$'s generate a nilpotent subgroup. But must this subgroup be unipotent (i.e. all eigenvalues are actually $1$)?
Thanks!
lie-groups
lie-groups
asked Nov 28 '18 at 19:27
Math-user
29617
asked Nov 28 '18 at 19:27
Math-user
29617
edited Nov 28 '18 at 20:28
asked Nov 28 '18 at 19:27
Math-user
29617
asked Nov 28 '18 at 19:27
Math-user
29617
asked Nov 28 '18 at 19:27
Math-user
29617
29617
1
do you mean ${u_1,dots,u_n}subsetGamma$ ? If not, what is the role of $Gamma$? Also, what is $n$? (as it stands, the question doesn't change if you set $n=1$)
– user8268
Nov 28 '18 at 20:09
yes sorry, I meant ${u_1,dots,u_n}subsetGamma$. $n$ is any integer, so yes if $n=1$, this should hold too. Thanks for the comment!
– Math-user
Nov 28 '18 at 20:28
1
I am still confused: $Gamma$ is discrete in $G$, right? But that would mean that for a sufficiently small $epsilon>0$ the condition $|u-1|<epsilon$, $uinGamma$, implies $u=1$.
– user8268
Nov 28 '18 at 20:38
Yes. When the real rank of $G$ is $geq 2$, by Margulis' arithmeticity theorem, there is a neighborhood of the identity $W$ such that for any non co-compact lattice, the intersection with this neighborhood contains only unipotent elements. But this does not apply to $mathrm{SL}(2,mathbb{R})$.
– Math-user
Nov 28 '18 at 20:55
add a comment |
1
do you mean ${u_1,dots,u_n}subsetGamma$ ? If not, what is the role of $Gamma$? Also, what is $n$? (as it stands, the question doesn't change if you set $n=1$)
– user8268
Nov 28 '18 at 20:09
yes sorry, I meant ${u_1,dots,u_n}subsetGamma$. $n$ is any integer, so yes if $n=1$, this should hold too. Thanks for the comment!
– Math-user
Nov 28 '18 at 20:28
1
I am still confused: $Gamma$ is discrete in $G$, right? But that would mean that for a sufficiently small $epsilon>0$ the condition $|u-1|<epsilon$, $uinGamma$, implies $u=1$.
– user8268
Nov 28 '18 at 20:38
Yes. When the real rank of $G$ is $geq 2$, by Margulis' arithmeticity theorem, there is a neighborhood of the identity $W$ such that for any non co-compact lattice, the intersection with this neighborhood contains only unipotent elements. But this does not apply to $mathrm{SL}(2,mathbb{R})$.
– Math-user
Nov 28 '18 at 20:55
1
1
do you mean ${u_1,dots,u_n}subsetGamma$ ? If not, what is the role of $Gamma$? Also, what is $n$? (as it stands, the question doesn't change if you set $n=1$)
– user8268
Nov 28 '18 at 20:09
do you mean ${u_1,dots,u_n}subsetGamma$ ? If not, what is the role of $Gamma$? Also, what is $n$? (as it stands, the question doesn't change if you set $n=1$)
– user8268
Nov 28 '18 at 20:09
yes sorry, I meant ${u_1,dots,u_n}subsetGamma$. $n$ is any integer, so yes if $n=1$, this should hold too. Thanks for the comment!
– Math-user
Nov 28 '18 at 20:28
yes sorry, I meant ${u_1,dots,u_n}subsetGamma$. $n$ is any integer, so yes if $n=1$, this should hold too. Thanks for the comment!
– Math-user
Nov 28 '18 at 20:28
1
1
I am still confused: $Gamma$ is discrete in $G$, right? But that would mean that for a sufficiently small $epsilon>0$ the condition $|u-1|<epsilon$, $uinGamma$, implies $u=1$.
– user8268
Nov 28 '18 at 20:38
I am still confused: $Gamma$ is discrete in $G$, right? But that would mean that for a sufficiently small $epsilon>0$ the condition $|u-1|<epsilon$, $uinGamma$, implies $u=1$.
– user8268
Nov 28 '18 at 20:38
Yes. When the real rank of $G$ is $geq 2$, by Margulis' arithmeticity theorem, there is a neighborhood of the identity $W$ such that for any non co-compact lattice, the intersection with this neighborhood contains only unipotent elements. But this does not apply to $mathrm{SL}(2,mathbb{R})$.
– Math-user
Nov 28 '18 at 20:55
Yes. When the real rank of $G$ is $geq 2$, by Margulis' arithmeticity theorem, there is a neighborhood of the identity $W$ such that for any non co-compact lattice, the intersection with this neighborhood contains only unipotent elements. But this does not apply to $mathrm{SL}(2,mathbb{R})$.
– Math-user
Nov 28 '18 at 20:55
add a comment |
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1
do you mean ${u_1,dots,u_n}subsetGamma$ ? If not, what is the role of $Gamma$? Also, what is $n$? (as it stands, the question doesn't change if you set $n=1$)
– user8268
Nov 28 '18 at 20:09
yes sorry, I meant ${u_1,dots,u_n}subsetGamma$. $n$ is any integer, so yes if $n=1$, this should hold too. Thanks for the comment!
– Math-user
Nov 28 '18 at 20:28
1
I am still confused: $Gamma$ is discrete in $G$, right? But that would mean that for a sufficiently small $epsilon>0$ the condition $|u-1|<epsilon$, $uinGamma$, implies $u=1$.
– user8268
Nov 28 '18 at 20:38
Yes. When the real rank of $G$ is $geq 2$, by Margulis' arithmeticity theorem, there is a neighborhood of the identity $W$ such that for any non co-compact lattice, the intersection with this neighborhood contains only unipotent elements. But this does not apply to $mathrm{SL}(2,mathbb{R})$.
– Math-user
Nov 28 '18 at 20:55