Mixing
$f,g$ 2 linearly independent weight 12 entire modular form give rise to isomorphism between $SL_2(Z)backslash...
Let $M_{12}$ be weight 12(integral weighted) entire modular form(i.e. holomorphic at cusps). So over $dim_C(M_{12})=2$. Now pick out any $f,gin M_{12}$ linearly independent.
$textbf{Q:}$ The book says $[f,g]:SL_2(Z)backslash Hto CP_1$ is isomorphism where $H$ is the upper half plane union all cusps. Does the book mean identify $SL_2(Z)backslash H$ not as an orbifold rather than simply a sphere without orbifold structure?(I could not see why this has to be biholomorphic if it is equipped with orbifold structure as there will be locally branched covering.) I can see this is obviously bijection but I do not see this as homeomorphism.
Ref. Zagier 1-2-3 Modular Forms pg 11 Cor 2.
complex-analysis algebraic-geometry
asked Nov 21 '18 at 1:36
user45765
2,5952722
add a comment |
Let $M_{12}$ be weight 12(integral weighted) entire modular form(i.e. holomorphic at cusps). So over $dim_C(M_{12})=2$. Now pick out any $f,gin M_{12}$ linearly independent.
$textbf{Q:}$ The book says $[f,g]:SL_2(Z)backslash Hto CP_1$ is isomorphism where $H$ is the upper half plane union all cusps. Does the book mean identify $SL_2(Z)backslash H$ not as an orbifold rather than simply a sphere without orbifold structure?(I could not see why this has to be biholomorphic if it is equipped with orbifold structure as there will be locally branched covering.) I can see this is obviously bijection but I do not see this as homeomorphism.
Ref. Zagier 1-2-3 Modular Forms pg 11 Cor 2.
complex-analysis algebraic-geometry
asked Nov 21 '18 at 1:36
user45765
2,5952722
I think Zagier is just being imprecise. On p. 9, he remarks that the two elliptic points become singularities on $overline{Gamma(1) backslash mathfrak{h}}$, and since $mathbb{P}^1(mathbb{C})$ is nonsingular this means they can't be isomorphic. What he probably actually means is that once we resolve these singularities, the resulting nonsingular Riemann surface is isomorphic to $mathbb{P}^1(mathbb{C})$. One can resolve the singularities by defining specific charts that take into account these stabilizers: this is done for instance in $S2.2$ of Diamond and Shurman or $S1.8$ of Miyake.
– André 3000
Nov 22 '18 at 0:45
add a comment |
Let $M_{12}$ be weight 12(integral weighted) entire modular form(i.e. holomorphic at cusps). So over $dim_C(M_{12})=2$. Now pick out any $f,gin M_{12}$ linearly independent.
$textbf{Q:}$ The book says $[f,g]:SL_2(Z)backslash Hto CP_1$ is isomorphism where $H$ is the upper half plane union all cusps. Does the book mean identify $SL_2(Z)backslash H$ not as an orbifold rather than simply a sphere without orbifold structure?(I could not see why this has to be biholomorphic if it is equipped with orbifold structure as there will be locally branched covering.) I can see this is obviously bijection but I do not see this as homeomorphism.
Ref. Zagier 1-2-3 Modular Forms pg 11 Cor 2.
complex-analysis algebraic-geometry
asked Nov 21 '18 at 1:36
user45765
2,5952722
Let $M_{12}$ be weight 12(integral weighted) entire modular form(i.e. holomorphic at cusps). So over $dim_C(M_{12})=2$. Now pick out any $f,gin M_{12}$ linearly independent.
$textbf{Q:}$ The book says $[f,g]:SL_2(Z)backslash Hto CP_1$ is isomorphism where $H$ is the upper half plane union all cusps. Does the book mean identify $SL_2(Z)backslash H$ not as an orbifold rather than simply a sphere without orbifold structure?(I could not see why this has to be biholomorphic if it is equipped with orbifold structure as there will be locally branched covering.) I can see this is obviously bijection but I do not see this as homeomorphism.
Ref. Zagier 1-2-3 Modular Forms pg 11 Cor 2.
complex-analysis algebraic-geometry
complex-analysis algebraic-geometry
asked Nov 21 '18 at 1:36
user45765
2,5952722
asked Nov 21 '18 at 1:36
user45765
2,5952722
asked Nov 21 '18 at 1:36
user45765
2,5952722
asked Nov 21 '18 at 1:36
user45765
2,5952722
asked Nov 21 '18 at 1:36
user45765
2,5952722
2,5952722
I think Zagier is just being imprecise. On p. 9, he remarks that the two elliptic points become singularities on $overline{Gamma(1) backslash mathfrak{h}}$, and since $mathbb{P}^1(mathbb{C})$ is nonsingular this means they can't be isomorphic. What he probably actually means is that once we resolve these singularities, the resulting nonsingular Riemann surface is isomorphic to $mathbb{P}^1(mathbb{C})$. One can resolve the singularities by defining specific charts that take into account these stabilizers: this is done for instance in $S2.2$ of Diamond and Shurman or $S1.8$ of Miyake.
– André 3000
Nov 22 '18 at 0:45
add a comment |
I think Zagier is just being imprecise. On p. 9, he remarks that the two elliptic points become singularities on $overline{Gamma(1) backslash mathfrak{h}}$, and since $mathbb{P}^1(mathbb{C})$ is nonsingular this means they can't be isomorphic. What he probably actually means is that once we resolve these singularities, the resulting nonsingular Riemann surface is isomorphic to $mathbb{P}^1(mathbb{C})$. One can resolve the singularities by defining specific charts that take into account these stabilizers: this is done for instance in $S2.2$ of Diamond and Shurman or $S1.8$ of Miyake.
– André 3000
Nov 22 '18 at 0:45
I think Zagier is just being imprecise. On p. 9, he remarks that the two elliptic points become singularities on $overline{Gamma(1) backslash mathfrak{h}}$, and since $mathbb{P}^1(mathbb{C})$ is nonsingular this means they can't be isomorphic. What he probably actually means is that once we resolve these singularities, the resulting nonsingular Riemann surface is isomorphic to $mathbb{P}^1(mathbb{C})$. One can resolve the singularities by defining specific charts that take into account these stabilizers: this is done for instance in $S2.2$ of Diamond and Shurman or $S1.8$ of Miyake.
– André 3000
Nov 22 '18 at 0:45
I think Zagier is just being imprecise. On p. 9, he remarks that the two elliptic points become singularities on $overline{Gamma(1) backslash mathfrak{h}}$, and since $mathbb{P}^1(mathbb{C})$ is nonsingular this means they can't be isomorphic. What he probably actually means is that once we resolve these singularities, the resulting nonsingular Riemann surface is isomorphic to $mathbb{P}^1(mathbb{C})$. One can resolve the singularities by defining specific charts that take into account these stabilizers: this is done for instance in $S2.2$ of Diamond and Shurman or $S1.8$ of Miyake.
– André 3000
Nov 22 '18 at 0:45
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%2f3007140%2ff-g-2-linearly-independent-weight-12-entire-modular-form-give-rise-to-isomorph%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%2f3007140%2ff-g-2-linearly-independent-weight-12-entire-modular-form-give-rise-to-isomorph%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
I think Zagier is just being imprecise. On p. 9, he remarks that the two elliptic points become singularities on $overline{Gamma(1) backslash mathfrak{h}}$, and since $mathbb{P}^1(mathbb{C})$ is nonsingular this means they can't be isomorphic. What he probably actually means is that once we resolve these singularities, the resulting nonsingular Riemann surface is isomorphic to $mathbb{P}^1(mathbb{C})$. One can resolve the singularities by defining specific charts that take into account these stabilizers: this is done for instance in $S2.2$ of Diamond and Shurman or $S1.8$ of Miyake.
– André 3000
Nov 22 '18 at 0:45