Dtjytyk

Is it true that "carbohydrates are of no use for the basal metabolic need"?

How do I stop a creek from eroding my steep embankment?

How does debian/ubuntu knows a package has a updated version

Why do we bend a book to keep it straight?

Why are Kinder Surprise Eggs illegal in the USA?

Why is my conclusion inconsistent with the van't Hoff equation?

Why did the Falcon Heavy center core fall off the ASDS OCISLY barge?

Fundamental Solution of the Pell Equation

What does the "x" in "x86" represent?

Should I use a zero-interest credit card for a large one-time purchase?

Can a non-EU citizen traveling with me come with me through the EU passport line?

Can a USB port passively 'listen only'?

What's the meaning of 間時肆拾貳 at a car parking sign

How to bypass password on Windows XP account?

Bete Noir -- no dairy

What does "fit" mean in this sentence?

Extract all GPU name, model and GPU ram

Should I discuss the type of campaign with my players?

How to run gsettings for another user Ubuntu 18.04.2 LTS

String `!23` is replaced with `docker` in command line

What does an IRS interview request entail when called in to verify expenses for a sole proprietor small business?

Dating a Former Employee

Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?

How to answer "Have you ever been terminated?"

Generating a particular element of $mathrm{SL}_2(Bbb{Z})$ from generators

2

$begingroup$

This should be easy, I hope.

$mathrm{SL}_2(mathbb Z) = { begin{bmatrix} a & b \ c & d end{bmatrix} : a,b,c,d in mathbb Z, ad-bc=1}$

A book I was reading said that the elements of $mathrm{SL}_2(mathbb Z)$ could be generated by:

$ begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} $ and $ begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} $.

I have one element that I think should be in $mathrm{SL}_2(mathbb Z)$ that I can't quite figure out how to generate from these. That's this one:

$ begin{bmatrix} 1 & 0 \ 1 & 1 end{bmatrix} $

How do I form this matrix from the others?

Thanks.

---

Edit: In Diamond and Shurman's A First Course in Modular Forms $mathrm{SL}_2(mathbb Z)$ is called the modular group (on page 1). I am supposing that this is a group under matrix multiplication, since the additive matrix identity has determinant 0 and thus wouldn't be in the group.

Confirming with a second source that the matrices I've listed are generators for $mathrm{SL}_2(mathbb Z)$, here's some notes by Conrad that mention this fact immediately.

matrices group-theory modular-forms

share|cite|improve this question

edited Mar 24 at 8:45

Brahadeesh

6,53642365

asked Jan 1 '17 at 6:32

ctesta01ctesta01

23018

$endgroup$

add a comment | 

2

$begingroup$

This should be easy, I hope.

$mathrm{SL}_2(mathbb Z) = { begin{bmatrix} a & b \ c & d end{bmatrix} : a,b,c,d in mathbb Z, ad-bc=1}$

A book I was reading said that the elements of $mathrm{SL}_2(mathbb Z)$ could be generated by:

$ begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} $ and $ begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} $.

I have one element that I think should be in $mathrm{SL}_2(mathbb Z)$ that I can't quite figure out how to generate from these. That's this one:

$ begin{bmatrix} 1 & 0 \ 1 & 1 end{bmatrix} $

How do I form this matrix from the others?

Thanks.

---

Edit: In Diamond and Shurman's A First Course in Modular Forms $mathrm{SL}_2(mathbb Z)$ is called the modular group (on page 1). I am supposing that this is a group under matrix multiplication, since the additive matrix identity has determinant 0 and thus wouldn't be in the group.

Confirming with a second source that the matrices I've listed are generators for $mathrm{SL}_2(mathbb Z)$, here's some notes by Conrad that mention this fact immediately.

matrices group-theory modular-forms

share|cite|improve this question

edited Mar 24 at 8:45

Brahadeesh

6,53642365

asked Jan 1 '17 at 6:32

ctesta01ctesta01

23018

$endgroup$

add a comment | 

$begingroup$

This should be easy, I hope.

$mathrm{SL}_2(mathbb Z) = { begin{bmatrix} a & b \ c & d end{bmatrix} : a,b,c,d in mathbb Z, ad-bc=1}$

A book I was reading said that the elements of $mathrm{SL}_2(mathbb Z)$ could be generated by:

$ begin{bmatrix} 1 & 1 \ 0 & 1 end{bmatrix} $ and $ begin{bmatrix} 0 & -1 \ 1 & 0 end{bmatrix} $.

I have one element that I think should be in $mathrm{SL}_2(mathbb Z)$ that I can't quite figure out how to generate from these. That's this one:

$ begin{bmatrix} 1 & 0 \ 1 & 1 end{bmatrix} $

How do I form this matrix from the others?

Thanks.

---

Edit: In Diamond and Shurman's A First Course in Modular Forms $mathrm{SL}_2(mathbb Z)$ is called the modular group (on page 1). I am supposing that this is a group under matrix multiplication, since the additive matrix identity has determinant 0 and thus wouldn't be in the group.

Confirming with a second source that the matrices I've listed are generators for $mathrm{SL}_2(mathbb Z)$, here's some notes by Conrad that mention this fact immediately.

matrices group-theory modular-forms

share|cite|improve this question

edited Mar 24 at 8:45

Brahadeesh

6,53642365

asked Jan 1 '17 at 6:32

ctesta01ctesta01

23018

$endgroup$

share|cite|improve this question

edited Mar 24 at 8:45

Brahadeesh

6,53642365

asked Jan 1 '17 at 6:32

ctesta01ctesta01

23018

share|cite|improve this question

edited Mar 24 at 8:45

Brahadeesh

6,53642365

asked Jan 1 '17 at 6:32

ctesta01ctesta01

23018

edited Mar 24 at 8:45

Brahadeesh

6,53642365

edited Mar 24 at 8:45

Brahadeesh

6,53642365

asked Jan 1 '17 at 6:32

ctesta01ctesta01

23018

asked Jan 1 '17 at 6:32

ctesta01ctesta01

23018

1 Answer
1

active

oldest

votes

3

$begingroup$

Well, let's look at what multiplying by each generator does.

We have that: $$Ae_1 = begin{pmatrix} a & b \ c & d end{pmatrix}begin{pmatrix} 1 & 1 \ 0 & 1end{pmatrix} = begin{pmatrix}a & a+b \ c & c+dend{pmatrix}$$
We have that:
$$Ae_2 = begin{pmatrix} a & b \ c & d end{pmatrix}begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix} = begin{pmatrix}b & -a \ d & -cend{pmatrix}$$
These are the "moves" we have to work with. Now, lets try to build your matrix.

It looks like we'll want to start with the first generator. Just for fun, lets try applying $e_2$ first:
$$e_1e_2 = begin{pmatrix}1 & -1 \ 1 & 0end{pmatrix}$$
Now, it's worth mentioning that $e_2^2 = -I$, where $I$ is the identity matrix. As we weren't just trying to multiply by $-1$, it makes sense to try applying $e_1$ now:
$$e_1e_2e_1 = begin{pmatrix} 1 & 0 \ 1 & 1end{pmatrix}$$
This is exactly what we wanted, so we're done.

Another way we could have worked with this is by computing $e_1^{-1}$ and $e_2^{-1}$, and tried "working backwards" from the matrix we want.

share|cite|improve this answer

answered Jan 1 '17 at 7:03

MarkMark

5,846727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It was simple after all! Thanks a bunch!
  $endgroup$
  – ctesta01
  Jan 1 '17 at 7:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ctesta01 note that Conrad's notes gives an algorithimic way to solve this. That's how he proves those are the generators, given some generic matrix he shows how you can iteratively find the generators.
  $endgroup$
  – Mark
  Jan 1 '17 at 7:21
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2079111%2fgenerating-a-particular-element-of-mathrmsl-2-bbbz-from-generators%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

Well, let's look at what multiplying by each generator does.

We have that: $$Ae_1 = begin{pmatrix} a & b \ c & d end{pmatrix}begin{pmatrix} 1 & 1 \ 0 & 1end{pmatrix} = begin{pmatrix}a & a+b \ c & c+dend{pmatrix}$$
We have that:
$$Ae_2 = begin{pmatrix} a & b \ c & d end{pmatrix}begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix} = begin{pmatrix}b & -a \ d & -cend{pmatrix}$$
These are the "moves" we have to work with. Now, lets try to build your matrix.

It looks like we'll want to start with the first generator. Just for fun, lets try applying $e_2$ first:
$$e_1e_2 = begin{pmatrix}1 & -1 \ 1 & 0end{pmatrix}$$
Now, it's worth mentioning that $e_2^2 = -I$, where $I$ is the identity matrix. As we weren't just trying to multiply by $-1$, it makes sense to try applying $e_1$ now:
$$e_1e_2e_1 = begin{pmatrix} 1 & 0 \ 1 & 1end{pmatrix}$$
This is exactly what we wanted, so we're done.

Another way we could have worked with this is by computing $e_1^{-1}$ and $e_2^{-1}$, and tried "working backwards" from the matrix we want.

share|cite|improve this answer

answered Jan 1 '17 at 7:03

MarkMark

5,846727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It was simple after all! Thanks a bunch!
  $endgroup$
  – ctesta01
  Jan 1 '17 at 7:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ctesta01 note that Conrad's notes gives an algorithimic way to solve this. That's how he proves those are the generators, given some generic matrix he shows how you can iteratively find the generators.
  $endgroup$
  – Mark
  Jan 1 '17 at 7:21
  

  
  
  
  

add a comment | 

3

$begingroup$

Well, let's look at what multiplying by each generator does.

We have that: $$Ae_1 = begin{pmatrix} a & b \ c & d end{pmatrix}begin{pmatrix} 1 & 1 \ 0 & 1end{pmatrix} = begin{pmatrix}a & a+b \ c & c+dend{pmatrix}$$
We have that:
$$Ae_2 = begin{pmatrix} a & b \ c & d end{pmatrix}begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix} = begin{pmatrix}b & -a \ d & -cend{pmatrix}$$
These are the "moves" we have to work with. Now, lets try to build your matrix.

It looks like we'll want to start with the first generator. Just for fun, lets try applying $e_2$ first:
$$e_1e_2 = begin{pmatrix}1 & -1 \ 1 & 0end{pmatrix}$$
Now, it's worth mentioning that $e_2^2 = -I$, where $I$ is the identity matrix. As we weren't just trying to multiply by $-1$, it makes sense to try applying $e_1$ now:
$$e_1e_2e_1 = begin{pmatrix} 1 & 0 \ 1 & 1end{pmatrix}$$
This is exactly what we wanted, so we're done.

Another way we could have worked with this is by computing $e_1^{-1}$ and $e_2^{-1}$, and tried "working backwards" from the matrix we want.

share|cite|improve this answer

answered Jan 1 '17 at 7:03

MarkMark

5,846727

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It was simple after all! Thanks a bunch!
  $endgroup$
  – ctesta01
  Jan 1 '17 at 7:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @ctesta01 note that Conrad's notes gives an algorithimic way to solve this. That's how he proves those are the generators, given some generic matrix he shows how you can iteratively find the generators.
  $endgroup$
  – Mark
  Jan 1 '17 at 7:21
  

  
  
  
  

add a comment | 

$begingroup$

Well, let's look at what multiplying by each generator does.

We have that: $$Ae_1 = begin{pmatrix} a & b \ c & d end{pmatrix}begin{pmatrix} 1 & 1 \ 0 & 1end{pmatrix} = begin{pmatrix}a & a+b \ c & c+dend{pmatrix}$$
We have that:
$$Ae_2 = begin{pmatrix} a & b \ c & d end{pmatrix}begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix} = begin{pmatrix}b & -a \ d & -cend{pmatrix}$$
These are the "moves" we have to work with. Now, lets try to build your matrix.

It looks like we'll want to start with the first generator. Just for fun, lets try applying $e_2$ first:
$$e_1e_2 = begin{pmatrix}1 & -1 \ 1 & 0end{pmatrix}$$
Now, it's worth mentioning that $e_2^2 = -I$, where $I$ is the identity matrix. As we weren't just trying to multiply by $-1$, it makes sense to try applying $e_1$ now:
$$e_1e_2e_1 = begin{pmatrix} 1 & 0 \ 1 & 1end{pmatrix}$$
This is exactly what we wanted, so we're done.

Another way we could have worked with this is by computing $e_1^{-1}$ and $e_2^{-1}$, and tried "working backwards" from the matrix we want.

share|cite|improve this answer

answered Jan 1 '17 at 7:03

MarkMark

5,846727

$endgroup$

share|cite|improve this answer

answered Jan 1 '17 at 7:03

MarkMark

5,846727

answered Jan 1 '17 at 7:03

MarkMark

5,846727

answered Jan 1 '17 at 7:03

MarkMark

5,846727