Show $-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}$ does not depend on $z$.
I'm trying to show $-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}$ does not depend on $zinmathbb{H}$ where $A,Bin SL_2{(mathbb{R})}$ and $j_A(z)=cz+d$, $A=
begin{bmatrix}
a & b \
c & d
end{bmatrix}$ and the branch cut is made at $pi$.
The book says it follows from the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ but I don't see why the statement above follows from the identity. I can see that the identity implies
$$-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}equiv0mod{(-pi,pi]}.$$
And I can also see that $-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}$ takes only 3 possible values, $-2pi, 0, 2pi$. However, I really don't see why this expression always takes the same value no matter what $z$ I choose. Can anybody give me some insights?
complex-analysis modular-forms
asked Jan 4 at 1:10
J.ShimJ.Shim
496
add a comment |
I'm trying to show $-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}$ does not depend on $zinmathbb{H}$ where $A,Bin SL_2{(mathbb{R})}$ and $j_A(z)=cz+d$, $A=
begin{bmatrix}
a & b \
c & d
end{bmatrix}$ and the branch cut is made at $pi$.
The book says it follows from the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ but I don't see why the statement above follows from the identity. I can see that the identity implies
$$-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}equiv0mod{(-pi,pi]}.$$
And I can also see that $-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}$ takes only 3 possible values, $-2pi, 0, 2pi$. However, I really don't see why this expression always takes the same value no matter what $z$ I choose. Can anybody give me some insights?
complex-analysis modular-forms
asked Jan 4 at 1:10
J.ShimJ.Shim
496
$A(z) = frac{az+b}{cz+d}$ and $ad-bc=1 implies A'(z) = frac{1}{(cz+d)^2}$ so $(A(B(z)))' = A'(B(z))B'(z)$ and $j_{AB}(z)= pm j_A(B(z))j_B(z)$. Then find the sign assuming $c > 0$ (for both $A,B$) so $Im(cz+d) > 0$ and pick a branch of $log$ sending $r e^{it}, t in (0,pi),r > 0$ to $r+it$. Extend to the case $c=0$ by continuity, finally look at the case $c < 0$.
– reuns
Jan 4 at 1:47
I can show the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ without much problem. What I meant is it's not obvious why the fact $-arg{j_{AB}(z)}+arg{j_A(Bz)}+arg{j_B(z)}$ does not depend on $z$ follows from the identity.
– J.Shim
Jan 4 at 6:41
1
With the conditions I mentioned $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z))$ is analytic in $z, Im(z) > 0$ and it takes its values in $2i pi mathbb{Z}$, thus it suffices to show that $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z)) = 0$ for one $z$ for it to be true for every $z, Im(z) > 0$
– reuns
Jan 4 at 7:02
Ah, it was so simple.. Thank you!
– J.Shim
2 days ago
add a comment |
I'm trying to show $-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}$ does not depend on $zinmathbb{H}$ where $A,Bin SL_2{(mathbb{R})}$ and $j_A(z)=cz+d$, $A=
begin{bmatrix}
a & b \
c & d
end{bmatrix}$ and the branch cut is made at $pi$.
The book says it follows from the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ but I don't see why the statement above follows from the identity. I can see that the identity implies
$$-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}equiv0mod{(-pi,pi]}.$$
And I can also see that $-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}$ takes only 3 possible values, $-2pi, 0, 2pi$. However, I really don't see why this expression always takes the same value no matter what $z$ I choose. Can anybody give me some insights?
complex-analysis modular-forms
asked Jan 4 at 1:10
J.ShimJ.Shim
496
I'm trying to show $-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}$ does not depend on $zinmathbb{H}$ where $A,Bin SL_2{(mathbb{R})}$ and $j_A(z)=cz+d$, $A=
begin{bmatrix}
a & b \
c & d
end{bmatrix}$ and the branch cut is made at $pi$.
The book says it follows from the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ but I don't see why the statement above follows from the identity. I can see that the identity implies
$$-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}equiv0mod{(-pi,pi]}.$$
And I can also see that $-arg{j_{AB}(z)}+arg{j_{A}(Bz)}+arg{j_{B}(z)}$ takes only 3 possible values, $-2pi, 0, 2pi$. However, I really don't see why this expression always takes the same value no matter what $z$ I choose. Can anybody give me some insights?
complex-analysis modular-forms
complex-analysis modular-forms
asked Jan 4 at 1:10
J.ShimJ.Shim
496
asked Jan 4 at 1:10
J.ShimJ.Shim
496
edited Jan 4 at 6:42
J.Shim
asked Jan 4 at 1:10
J.ShimJ.Shim
496
asked Jan 4 at 1:10
J.ShimJ.Shim
496
asked Jan 4 at 1:10
J.ShimJ.Shim
496
496
$A(z) = frac{az+b}{cz+d}$ and $ad-bc=1 implies A'(z) = frac{1}{(cz+d)^2}$ so $(A(B(z)))' = A'(B(z))B'(z)$ and $j_{AB}(z)= pm j_A(B(z))j_B(z)$. Then find the sign assuming $c > 0$ (for both $A,B$) so $Im(cz+d) > 0$ and pick a branch of $log$ sending $r e^{it}, t in (0,pi),r > 0$ to $r+it$. Extend to the case $c=0$ by continuity, finally look at the case $c < 0$.
– reuns
Jan 4 at 1:47
I can show the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ without much problem. What I meant is it's not obvious why the fact $-arg{j_{AB}(z)}+arg{j_A(Bz)}+arg{j_B(z)}$ does not depend on $z$ follows from the identity.
– J.Shim
Jan 4 at 6:41
1
With the conditions I mentioned $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z))$ is analytic in $z, Im(z) > 0$ and it takes its values in $2i pi mathbb{Z}$, thus it suffices to show that $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z)) = 0$ for one $z$ for it to be true for every $z, Im(z) > 0$
– reuns
Jan 4 at 7:02
Ah, it was so simple.. Thank you!
– J.Shim
2 days ago
add a comment |
$A(z) = frac{az+b}{cz+d}$ and $ad-bc=1 implies A'(z) = frac{1}{(cz+d)^2}$ so $(A(B(z)))' = A'(B(z))B'(z)$ and $j_{AB}(z)= pm j_A(B(z))j_B(z)$. Then find the sign assuming $c > 0$ (for both $A,B$) so $Im(cz+d) > 0$ and pick a branch of $log$ sending $r e^{it}, t in (0,pi),r > 0$ to $r+it$. Extend to the case $c=0$ by continuity, finally look at the case $c < 0$.
– reuns
Jan 4 at 1:47
I can show the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ without much problem. What I meant is it's not obvious why the fact $-arg{j_{AB}(z)}+arg{j_A(Bz)}+arg{j_B(z)}$ does not depend on $z$ follows from the identity.
– J.Shim
Jan 4 at 6:41
1
With the conditions I mentioned $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z))$ is analytic in $z, Im(z) > 0$ and it takes its values in $2i pi mathbb{Z}$, thus it suffices to show that $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z)) = 0$ for one $z$ for it to be true for every $z, Im(z) > 0$
– reuns
Jan 4 at 7:02
Ah, it was so simple.. Thank you!
– J.Shim
2 days ago
$A(z) = frac{az+b}{cz+d}$ and $ad-bc=1 implies A'(z) = frac{1}{(cz+d)^2}$ so $(A(B(z)))' = A'(B(z))B'(z)$ and $j_{AB}(z)= pm j_A(B(z))j_B(z)$. Then find the sign assuming $c > 0$ (for both $A,B$) so $Im(cz+d) > 0$ and pick a branch of $log$ sending $r e^{it}, t in (0,pi),r > 0$ to $r+it$. Extend to the case $c=0$ by continuity, finally look at the case $c < 0$.
– reuns
Jan 4 at 1:47
$A(z) = frac{az+b}{cz+d}$ and $ad-bc=1 implies A'(z) = frac{1}{(cz+d)^2}$ so $(A(B(z)))' = A'(B(z))B'(z)$ and $j_{AB}(z)= pm j_A(B(z))j_B(z)$. Then find the sign assuming $c > 0$ (for both $A,B$) so $Im(cz+d) > 0$ and pick a branch of $log$ sending $r e^{it}, t in (0,pi),r > 0$ to $r+it$. Extend to the case $c=0$ by continuity, finally look at the case $c < 0$.
– reuns
Jan 4 at 1:47
I can show the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ without much problem. What I meant is it's not obvious why the fact $-arg{j_{AB}(z)}+arg{j_A(Bz)}+arg{j_B(z)}$ does not depend on $z$ follows from the identity.
– J.Shim
Jan 4 at 6:41
I can show the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ without much problem. What I meant is it's not obvious why the fact $-arg{j_{AB}(z)}+arg{j_A(Bz)}+arg{j_B(z)}$ does not depend on $z$ follows from the identity.
– J.Shim
Jan 4 at 6:41
1
1
With the conditions I mentioned $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z))$ is analytic in $z, Im(z) > 0$ and it takes its values in $2i pi mathbb{Z}$, thus it suffices to show that $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z)) = 0$ for one $z$ for it to be true for every $z, Im(z) > 0$
– reuns
Jan 4 at 7:02
With the conditions I mentioned $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z))$ is analytic in $z, Im(z) > 0$ and it takes its values in $2i pi mathbb{Z}$, thus it suffices to show that $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z)) = 0$ for one $z$ for it to be true for every $z, Im(z) > 0$
– reuns
Jan 4 at 7:02
Ah, it was so simple.. Thank you!
– J.Shim
2 days ago
Ah, it was so simple.. Thank you!
– J.Shim
2 days ago
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
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%2f3061213%2fshow-argj-abz-argj-abz-argj-bz-does-not-depend-on-z%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
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%2f3061213%2fshow-argj-abz-argj-abz-argj-bz-does-not-depend-on-z%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
$A(z) = frac{az+b}{cz+d}$ and $ad-bc=1 implies A'(z) = frac{1}{(cz+d)^2}$ so $(A(B(z)))' = A'(B(z))B'(z)$ and $j_{AB}(z)= pm j_A(B(z))j_B(z)$. Then find the sign assuming $c > 0$ (for both $A,B$) so $Im(cz+d) > 0$ and pick a branch of $log$ sending $r e^{it}, t in (0,pi),r > 0$ to $r+it$. Extend to the case $c=0$ by continuity, finally look at the case $c < 0$.
– reuns
Jan 4 at 1:47
I can show the identity $j_{AB}(z)=j_A(Bz)j_B(z)$ without much problem. What I meant is it's not obvious why the fact $-arg{j_{AB}(z)}+arg{j_A(Bz)}+arg{j_B(z)}$ does not depend on $z$ follows from the identity.
– J.Shim
Jan 4 at 6:41
1
With the conditions I mentioned $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z))$ is analytic in $z, Im(z) > 0$ and it takes its values in $2i pi mathbb{Z}$, thus it suffices to show that $-log(j_{AB}(z))+log(j_A(Bz))+log(j_B(z)) = 0$ for one $z$ for it to be true for every $z, Im(z) > 0$
– reuns
Jan 4 at 7:02
Ah, it was so simple.. Thank you!
– J.Shim
2 days ago