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Probability of satisfying a word in a compact group
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This question is inspired by Probability of commutation in a compact group, which asked whether $P(xyx^-1y^-1 = 1)$ could take values strictly between $0$ and $1$ on a compact connected group. That question turned out to have a rather easy answer, but one that was quite particular to the word $xyx^-1y^-1$.
Question: Let $w$ be an arbitrary word in $k$ letters, and let $G$ be a compact connected group. Let $x_1, dots, x_k$ be drawn uniformly from $G$. Can $P(w(x_1, dots, x_k)=1)$ be strictly between $0$ and $1$?
By the Peter--Weyl theorem we may assume that $G$ is a compact connected Lie group.
There is some low-hanging fruit, like $w = [[x,y],z]$, but that's still very specialized. Admittedly I am not even clear on the words $w = x^n$.
gr.group-theory lie-groups
asked 2 hours ago
Sean Eberhard
3,3551327
add a comment |Â
up vote
6
down vote
favorite
This question is inspired by Probability of commutation in a compact group, which asked whether $P(xyx^-1y^-1 = 1)$ could take values strictly between $0$ and $1$ on a compact connected group. That question turned out to have a rather easy answer, but one that was quite particular to the word $xyx^-1y^-1$.
Question: Let $w$ be an arbitrary word in $k$ letters, and let $G$ be a compact connected group. Let $x_1, dots, x_k$ be drawn uniformly from $G$. Can $P(w(x_1, dots, x_k)=1)$ be strictly between $0$ and $1$?
By the Peter--Weyl theorem we may assume that $G$ is a compact connected Lie group.
There is some low-hanging fruit, like $w = [[x,y],z]$, but that's still very specialized. Admittedly I am not even clear on the words $w = x^n$.
gr.group-theory lie-groups
asked 2 hours ago
Sean Eberhard
3,3551327
Particularly in Lie groups, there's lots of other measure-theoretic and potential-theoretic notions of "smallness" that one could ask about. If it does have measure zero, what's its Hausdorff dimension? Is it a polar set? Et cetera.
– Nate Eldredge
2 hours ago
2
the solution space for a $k$ letters word will be a closed subvariety of $G^k$, and as such it will have either measure 0 or 1. Morover, if $G$ is not commutative then it contains a free group hence this subvariety will be proper, thus of 0 meaure.
– Uri Bader
2 hours ago
@UriBader Why does full dimension imply measure 1? I'm thinking of for example the solution space to $x^2=1$ in $O(2)$. Of course $O(2)$ is not connected, so it's not a counterexample.
– Sean Eberhard
1 hour ago
Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity.
– Uri Bader
33 mins ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
This question is inspired by Probability of commutation in a compact group, which asked whether $P(xyx^-1y^-1 = 1)$ could take values strictly between $0$ and $1$ on a compact connected group. That question turned out to have a rather easy answer, but one that was quite particular to the word $xyx^-1y^-1$.
Question: Let $w$ be an arbitrary word in $k$ letters, and let $G$ be a compact connected group. Let $x_1, dots, x_k$ be drawn uniformly from $G$. Can $P(w(x_1, dots, x_k)=1)$ be strictly between $0$ and $1$?
By the Peter--Weyl theorem we may assume that $G$ is a compact connected Lie group.
There is some low-hanging fruit, like $w = [[x,y],z]$, but that's still very specialized. Admittedly I am not even clear on the words $w = x^n$.
gr.group-theory lie-groups
asked 2 hours ago
Sean Eberhard
3,3551327
This question is inspired by Probability of commutation in a compact group, which asked whether $P(xyx^-1y^-1 = 1)$ could take values strictly between $0$ and $1$ on a compact connected group. That question turned out to have a rather easy answer, but one that was quite particular to the word $xyx^-1y^-1$.
Question: Let $w$ be an arbitrary word in $k$ letters, and let $G$ be a compact connected group. Let $x_1, dots, x_k$ be drawn uniformly from $G$. Can $P(w(x_1, dots, x_k)=1)$ be strictly between $0$ and $1$?
By the Peter--Weyl theorem we may assume that $G$ is a compact connected Lie group.
There is some low-hanging fruit, like $w = [[x,y],z]$, but that's still very specialized. Admittedly I am not even clear on the words $w = x^n$.
gr.group-theory lie-groups
gr.group-theory lie-groups
asked 2 hours ago
Sean Eberhard
3,3551327
asked 2 hours ago
Sean Eberhard
3,3551327
asked 2 hours ago
Sean Eberhard
3,3551327
asked 2 hours ago
Sean Eberhard
3,3551327
asked 2 hours ago
Sean Eberhard
3,3551327
3,3551327
Particularly in Lie groups, there's lots of other measure-theoretic and potential-theoretic notions of "smallness" that one could ask about. If it does have measure zero, what's its Hausdorff dimension? Is it a polar set? Et cetera.
– Nate Eldredge
2 hours ago
2
the solution space for a $k$ letters word will be a closed subvariety of $G^k$, and as such it will have either measure 0 or 1. Morover, if $G$ is not commutative then it contains a free group hence this subvariety will be proper, thus of 0 meaure.
– Uri Bader
2 hours ago
@UriBader Why does full dimension imply measure 1? I'm thinking of for example the solution space to $x^2=1$ in $O(2)$. Of course $O(2)$ is not connected, so it's not a counterexample.
– Sean Eberhard
1 hour ago
Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity.
– Uri Bader
33 mins ago
add a comment |Â
Particularly in Lie groups, there's lots of other measure-theoretic and potential-theoretic notions of "smallness" that one could ask about. If it does have measure zero, what's its Hausdorff dimension? Is it a polar set? Et cetera.
– Nate Eldredge
2 hours ago
2
the solution space for a $k$ letters word will be a closed subvariety of $G^k$, and as such it will have either measure 0 or 1. Morover, if $G$ is not commutative then it contains a free group hence this subvariety will be proper, thus of 0 meaure.
– Uri Bader
2 hours ago
@UriBader Why does full dimension imply measure 1? I'm thinking of for example the solution space to $x^2=1$ in $O(2)$. Of course $O(2)$ is not connected, so it's not a counterexample.
– Sean Eberhard
1 hour ago
Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity.
– Uri Bader
33 mins ago
Particularly in Lie groups, there's lots of other measure-theoretic and potential-theoretic notions of "smallness" that one could ask about. If it does have measure zero, what's its Hausdorff dimension? Is it a polar set? Et cetera.
– Nate Eldredge
2 hours ago
Particularly in Lie groups, there's lots of other measure-theoretic and potential-theoretic notions of "smallness" that one could ask about. If it does have measure zero, what's its Hausdorff dimension? Is it a polar set? Et cetera.
– Nate Eldredge
2 hours ago
2
2
the solution space for a $k$ letters word will be a closed subvariety of $G^k$, and as such it will have either measure 0 or 1. Morover, if $G$ is not commutative then it contains a free group hence this subvariety will be proper, thus of 0 meaure.
– Uri Bader
2 hours ago
the solution space for a $k$ letters word will be a closed subvariety of $G^k$, and as such it will have either measure 0 or 1. Morover, if $G$ is not commutative then it contains a free group hence this subvariety will be proper, thus of 0 meaure.
– Uri Bader
2 hours ago
@UriBader Why does full dimension imply measure 1? I'm thinking of for example the solution space to $x^2=1$ in $O(2)$. Of course $O(2)$ is not connected, so it's not a counterexample.
– Sean Eberhard
1 hour ago
@UriBader Why does full dimension imply measure 1? I'm thinking of for example the solution space to $x^2=1$ in $O(2)$. Of course $O(2)$ is not connected, so it's not a counterexample.
– Sean Eberhard
1 hour ago
Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity.
– Uri Bader
33 mins ago
Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity.
– Uri Bader
33 mins ago
add a comment |Â
1 Answer
1
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up vote
4
down vote
The negative answer follows easily from a very useful fact that should be better known, so I am writing it explicitly.
Fact (Tannaka, Chevalley):
Every (connected) compact Lie group is isomorphic as a topological group to the group of real points of a (connected) reductive affine real algebraic group.
Moreover, the Haar measure on such a group is given by a volume form on the corrseponding variety.
We thus may view (for a given word $w$ in the rank $k$ free group $F_k$) the word map $w:G^kto G$ as a morphism of real varieties and its solution space (the fiber over the identity element) as a the real points of a closed real subvariety. Since $G$ is connected, this subvariety will be either $G^k$ itself, or a lower dimensional one. Its measure, accordingly, will be either 0 or 1.
We could be more precise:
If $G$ is trivial then the measure is 1.
If $G$ is non-trivial abelian then it is isomprphic to a torus $textSO(2)^n$, and it is easy to see that the measure is 1 iff $w$ is in the commutator group of $F_k$, otherwise the measure is 0.
If $G$ is non-abelian then it contains a free group on $k$ generators (by Tits alternative, if you want to hit it with a hammer), thus the corresponding subvariety is proper and its measure is necessarily 0.
answered 35 mins ago
Uri Bader
5,99911531
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The negative answer follows easily from a very useful fact that should be better known, so I am writing it explicitly.
Fact (Tannaka, Chevalley):
Every (connected) compact Lie group is isomorphic as a topological group to the group of real points of a (connected) reductive affine real algebraic group.
Moreover, the Haar measure on such a group is given by a volume form on the corrseponding variety.
We thus may view (for a given word $w$ in the rank $k$ free group $F_k$) the word map $w:G^kto G$ as a morphism of real varieties and its solution space (the fiber over the identity element) as a the real points of a closed real subvariety. Since $G$ is connected, this subvariety will be either $G^k$ itself, or a lower dimensional one. Its measure, accordingly, will be either 0 or 1.
We could be more precise:
If $G$ is trivial then the measure is 1.
If $G$ is non-trivial abelian then it is isomprphic to a torus $textSO(2)^n$, and it is easy to see that the measure is 1 iff $w$ is in the commutator group of $F_k$, otherwise the measure is 0.
If $G$ is non-abelian then it contains a free group on $k$ generators (by Tits alternative, if you want to hit it with a hammer), thus the corresponding subvariety is proper and its measure is necessarily 0.
answered 35 mins ago
Uri Bader
5,99911531
add a comment |Â
up vote
4
down vote
The negative answer follows easily from a very useful fact that should be better known, so I am writing it explicitly.
Fact (Tannaka, Chevalley):
Every (connected) compact Lie group is isomorphic as a topological group to the group of real points of a (connected) reductive affine real algebraic group.
Moreover, the Haar measure on such a group is given by a volume form on the corrseponding variety.
We thus may view (for a given word $w$ in the rank $k$ free group $F_k$) the word map $w:G^kto G$ as a morphism of real varieties and its solution space (the fiber over the identity element) as a the real points of a closed real subvariety. Since $G$ is connected, this subvariety will be either $G^k$ itself, or a lower dimensional one. Its measure, accordingly, will be either 0 or 1.
We could be more precise:
If $G$ is trivial then the measure is 1.
If $G$ is non-trivial abelian then it is isomprphic to a torus $textSO(2)^n$, and it is easy to see that the measure is 1 iff $w$ is in the commutator group of $F_k$, otherwise the measure is 0.
If $G$ is non-abelian then it contains a free group on $k$ generators (by Tits alternative, if you want to hit it with a hammer), thus the corresponding subvariety is proper and its measure is necessarily 0.
answered 35 mins ago
Uri Bader
5,99911531
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The negative answer follows easily from a very useful fact that should be better known, so I am writing it explicitly.
Fact (Tannaka, Chevalley):
Every (connected) compact Lie group is isomorphic as a topological group to the group of real points of a (connected) reductive affine real algebraic group.
Moreover, the Haar measure on such a group is given by a volume form on the corrseponding variety.
We thus may view (for a given word $w$ in the rank $k$ free group $F_k$) the word map $w:G^kto G$ as a morphism of real varieties and its solution space (the fiber over the identity element) as a the real points of a closed real subvariety. Since $G$ is connected, this subvariety will be either $G^k$ itself, or a lower dimensional one. Its measure, accordingly, will be either 0 or 1.
We could be more precise:
If $G$ is trivial then the measure is 1.
If $G$ is non-trivial abelian then it is isomprphic to a torus $textSO(2)^n$, and it is easy to see that the measure is 1 iff $w$ is in the commutator group of $F_k$, otherwise the measure is 0.
If $G$ is non-abelian then it contains a free group on $k$ generators (by Tits alternative, if you want to hit it with a hammer), thus the corresponding subvariety is proper and its measure is necessarily 0.
answered 35 mins ago
Uri Bader
5,99911531
The negative answer follows easily from a very useful fact that should be better known, so I am writing it explicitly.
Fact (Tannaka, Chevalley):
Every (connected) compact Lie group is isomorphic as a topological group to the group of real points of a (connected) reductive affine real algebraic group.
Moreover, the Haar measure on such a group is given by a volume form on the corrseponding variety.
We thus may view (for a given word $w$ in the rank $k$ free group $F_k$) the word map $w:G^kto G$ as a morphism of real varieties and its solution space (the fiber over the identity element) as a the real points of a closed real subvariety. Since $G$ is connected, this subvariety will be either $G^k$ itself, or a lower dimensional one. Its measure, accordingly, will be either 0 or 1.
We could be more precise:
If $G$ is trivial then the measure is 1.
If $G$ is non-trivial abelian then it is isomprphic to a torus $textSO(2)^n$, and it is easy to see that the measure is 1 iff $w$ is in the commutator group of $F_k$, otherwise the measure is 0.
If $G$ is non-abelian then it contains a free group on $k$ generators (by Tits alternative, if you want to hit it with a hammer), thus the corresponding subvariety is proper and its measure is necessarily 0.
answered 35 mins ago
Uri Bader
5,99911531
edited 17 mins ago
answered 35 mins ago
Uri Bader
5,99911531
answered 35 mins ago
Uri Bader
5,99911531
answered 35 mins ago
Uri Bader
5,99911531
5,99911531
add a comment |Â
add a comment |Â
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Particularly in Lie groups, there's lots of other measure-theoretic and potential-theoretic notions of "smallness" that one could ask about. If it does have measure zero, what's its Hausdorff dimension? Is it a polar set? Et cetera.
– Nate Eldredge
2 hours ago
2
the solution space for a $k$ letters word will be a closed subvariety of $G^k$, and as such it will have either measure 0 or 1. Morover, if $G$ is not commutative then it contains a free group hence this subvariety will be proper, thus of 0 meaure.
– Uri Bader
2 hours ago
@UriBader Why does full dimension imply measure 1? I'm thinking of for example the solution space to $x^2=1$ in $O(2)$. Of course $O(2)$ is not connected, so it's not a counterexample.
– Sean Eberhard
1 hour ago
Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity.
– Uri Bader
33 mins ago