What is a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$?
up vote
1
down vote
favorite
I have been trying to find a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$ by considering the free group $Fr({x_i| iin mathbb Q})$ under a homomorphism $f$ into $GL(2, mathbb Q)$ by
$f(x_i):=begin{bmatrix}i&b\&{1over i}end{bmatrix}$ or $f(x_i):=begin{bmatrix}i&b\& {i^2} end{bmatrix}$ but I haven't been able to finagle it enough to get it to work. Is this at all the right approach? How exactly do I go about finding a presentation in the proper way? Thanks.
group-theory group-presentation combinatorial-group-theory
edited Nov 30 at 2:48
Shaun
8,015113577
asked Nov 19 at 20:46
mnewman
264
add a comment |
up vote
1
down vote
favorite
I have been trying to find a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$ by considering the free group $Fr({x_i| iin mathbb Q})$ under a homomorphism $f$ into $GL(2, mathbb Q)$ by
$f(x_i):=begin{bmatrix}i&b\&{1over i}end{bmatrix}$ or $f(x_i):=begin{bmatrix}i&b\& {i^2} end{bmatrix}$ but I haven't been able to finagle it enough to get it to work. Is this at all the right approach? How exactly do I go about finding a presentation in the proper way? Thanks.
group-theory group-presentation combinatorial-group-theory
edited Nov 30 at 2:48
Shaun
8,015113577
asked Nov 19 at 20:46
mnewman
264
2
Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
– Daniel Schepler
Nov 19 at 21:15
@DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
– mnewman
Nov 19 at 21:23
Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
– Daniel Schepler
Nov 19 at 21:35
2
Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
– reuns
Nov 19 at 21:36
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been trying to find a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$ by considering the free group $Fr({x_i| iin mathbb Q})$ under a homomorphism $f$ into $GL(2, mathbb Q)$ by
$f(x_i):=begin{bmatrix}i&b\&{1over i}end{bmatrix}$ or $f(x_i):=begin{bmatrix}i&b\& {i^2} end{bmatrix}$ but I haven't been able to finagle it enough to get it to work. Is this at all the right approach? How exactly do I go about finding a presentation in the proper way? Thanks.
group-theory group-presentation combinatorial-group-theory
edited Nov 30 at 2:48
Shaun
8,015113577
asked Nov 19 at 20:46
mnewman
264
I have been trying to find a presentation of the upper triangular subgroup of $GL(2, mathbb Q)$ by considering the free group $Fr({x_i| iin mathbb Q})$ under a homomorphism $f$ into $GL(2, mathbb Q)$ by
$f(x_i):=begin{bmatrix}i&b\&{1over i}end{bmatrix}$ or $f(x_i):=begin{bmatrix}i&b\& {i^2} end{bmatrix}$ but I haven't been able to finagle it enough to get it to work. Is this at all the right approach? How exactly do I go about finding a presentation in the proper way? Thanks.
group-theory group-presentation combinatorial-group-theory
group-theory group-presentation combinatorial-group-theory
edited Nov 30 at 2:48
Shaun
8,015113577
asked Nov 19 at 20:46
mnewman
264
edited Nov 30 at 2:48
Shaun
8,015113577
asked Nov 19 at 20:46
mnewman
264
edited Nov 30 at 2:48
Shaun
8,015113577
edited Nov 30 at 2:48
Shaun
8,015113577
edited Nov 30 at 2:48
Shaun
8,015113577
8,015113577
asked Nov 19 at 20:46
mnewman
264
asked Nov 19 at 20:46
mnewman
264
asked Nov 19 at 20:46
mnewman
264
264
2
Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
– Daniel Schepler
Nov 19 at 21:15
@DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
– mnewman
Nov 19 at 21:23
Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
– Daniel Schepler
Nov 19 at 21:35
2
Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
– reuns
Nov 19 at 21:36
add a comment |
2
Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
– Daniel Schepler
Nov 19 at 21:15
@DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
– mnewman
Nov 19 at 21:23
Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
– Daniel Schepler
Nov 19 at 21:35
2
Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
– reuns
Nov 19 at 21:36
2
2
Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
– Daniel Schepler
Nov 19 at 21:15
Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
– Daniel Schepler
Nov 19 at 21:15
@DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
– mnewman
Nov 19 at 21:23
@DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
– mnewman
Nov 19 at 21:23
Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
– Daniel Schepler
Nov 19 at 21:35
Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
– Daniel Schepler
Nov 19 at 21:35
2
2
Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
– reuns
Nov 19 at 21:36
Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
– reuns
Nov 19 at 21:36
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005503%2fwhat-is-a-presentation-of-the-upper-triangular-subgroup-of-gl2-mathbb-q%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
Personally, I would probably go with generators of the form $begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix}$, $begin{bmatrix} 1 & 0 \ 0 & b end{bmatrix}$, and $begin{bmatrix} 1 & c \ 0 & 1 end{bmatrix}$.
– Daniel Schepler
Nov 19 at 21:15
@DanielSchepler Thanks. I guess I was just approaching the problem wrongly. When is is that I should be looking to establish isomorphisms from the quotient of the free group with the kernel of some homomorphism like I was above, as compared to just trying stuff with generators?
– mnewman
Nov 19 at 21:23
Well, for example, you could write a "normal form" representation such as $begin{bmatrix} a & b \ 0 & c end{bmatrix} = begin{bmatrix} 1 & c^{-1} b \ 0 & 1 end{bmatrix} begin{bmatrix} a & 0 \ 0 & 1 end{bmatrix} begin{bmatrix} 1 & 0 \ 0 & c end{bmatrix}$ and then come up with enough relations to be able to write a product of two such expressions in the same form, and possibly some relations for inverses as well.
– Daniel Schepler
Nov 19 at 21:35
2
Otherwise, "send" $begin{bmatrix} a & c \ 0 & b end{bmatrix}, a in mathbb{Q}^*, b in mathbb{Q}^*,c in mathbb{Q}$ to the element $x_{a,b,c}$ of the free group generated by the elements of $mathbb{Q}^*times mathbb{Q}^* times mathbb{Q}$, then define the multiplication in term of elements of that free group.
– reuns
Nov 19 at 21:36