Isomorphism ${1}times mathbb C^*to {1}times mathbb C^*$
$begingroup$
I am reading a text where I have trouble understanding an argument:
Let $f: mathbb C^*times mathbb C^*to mathbb C^*times mathbb C^*$ an isomorphism, such that $f({1}times mathbb C^*)= {1}times mathbb C^*$.
Further, let $|x|neq 1$.
The following argument is made:
$f_{{1}times mathbb C^*}: {1}times mathbb C^*to {1}times mathbb C^*$ is an isomorphism of linear groups. Therefore $f(1,|x|^2)in {(1,|x|^2),(1,|x|^{-2}}$.
Why should this follow? As far as I see it, the condition $f({1}times mathbb C^*)= {1}times mathbb C^*$ means that $f_{{1}times mathbb C^*}$ comes from an isomorphism $mathbb C^* to mathbb C^*$. But these could map $|x|^2$ to any value. For example the homomorphism could be a rotation.
abstract-algebra group-theory abelian-groups group-homomorphism linear-groups
asked Nov 23 '18 at 18:26
J.DoeJ.Doe
214
$endgroup$
add a comment |
$begingroup$
I am reading a text where I have trouble understanding an argument:
Let $f: mathbb C^*times mathbb C^*to mathbb C^*times mathbb C^*$ an isomorphism, such that $f({1}times mathbb C^*)= {1}times mathbb C^*$.
Further, let $|x|neq 1$.
The following argument is made:
$f_{{1}times mathbb C^*}: {1}times mathbb C^*to {1}times mathbb C^*$ is an isomorphism of linear groups. Therefore $f(1,|x|^2)in {(1,|x|^2),(1,|x|^{-2}}$.
Why should this follow? As far as I see it, the condition $f({1}times mathbb C^*)= {1}times mathbb C^*$ means that $f_{{1}times mathbb C^*}$ comes from an isomorphism $mathbb C^* to mathbb C^*$. But these could map $|x|^2$ to any value. For example the homomorphism could be a rotation.
abstract-algebra group-theory abelian-groups group-homomorphism linear-groups
asked Nov 23 '18 at 18:26
J.DoeJ.Doe
214
$endgroup$
1
$begingroup$
A rotation is not an endomorphism (homomorphism from a group to itself) of the multiplicative group of non-zero complex numbers. Consider that $1$ has to map to $1$, for instance.
$endgroup$
– Arthur
Nov 23 '18 at 18:35
$begingroup$
@Arthur: Oh, you are right. But what about $zmapsto z^k$? If $|k|neq 1$, then stil the claim in the question does not hold?
$endgroup$
– J.Doe
Nov 23 '18 at 18:45
$begingroup$
ah...nevermind, if $|k|neq 1$, then the homomorphism is not injective. But how can I be sure that there is no other possibility?
$endgroup$
– J.Doe
Nov 23 '18 at 18:47
add a comment |
$begingroup$
I am reading a text where I have trouble understanding an argument:
Let $f: mathbb C^*times mathbb C^*to mathbb C^*times mathbb C^*$ an isomorphism, such that $f({1}times mathbb C^*)= {1}times mathbb C^*$.
Further, let $|x|neq 1$.
The following argument is made:
$f_{{1}times mathbb C^*}: {1}times mathbb C^*to {1}times mathbb C^*$ is an isomorphism of linear groups. Therefore $f(1,|x|^2)in {(1,|x|^2),(1,|x|^{-2}}$.
Why should this follow? As far as I see it, the condition $f({1}times mathbb C^*)= {1}times mathbb C^*$ means that $f_{{1}times mathbb C^*}$ comes from an isomorphism $mathbb C^* to mathbb C^*$. But these could map $|x|^2$ to any value. For example the homomorphism could be a rotation.
abstract-algebra group-theory abelian-groups group-homomorphism linear-groups
asked Nov 23 '18 at 18:26
J.DoeJ.Doe
214
$endgroup$
I am reading a text where I have trouble understanding an argument:
Let $f: mathbb C^*times mathbb C^*to mathbb C^*times mathbb C^*$ an isomorphism, such that $f({1}times mathbb C^*)= {1}times mathbb C^*$.
Further, let $|x|neq 1$.
The following argument is made:
$f_{{1}times mathbb C^*}: {1}times mathbb C^*to {1}times mathbb C^*$ is an isomorphism of linear groups. Therefore $f(1,|x|^2)in {(1,|x|^2),(1,|x|^{-2}}$.
Why should this follow? As far as I see it, the condition $f({1}times mathbb C^*)= {1}times mathbb C^*$ means that $f_{{1}times mathbb C^*}$ comes from an isomorphism $mathbb C^* to mathbb C^*$. But these could map $|x|^2$ to any value. For example the homomorphism could be a rotation.
abstract-algebra group-theory abelian-groups group-homomorphism linear-groups
abstract-algebra group-theory abelian-groups group-homomorphism linear-groups
asked Nov 23 '18 at 18:26
J.DoeJ.Doe
214
asked Nov 23 '18 at 18:26
J.DoeJ.Doe
214
asked Nov 23 '18 at 18:26
J.DoeJ.Doe
214
asked Nov 23 '18 at 18:26
J.DoeJ.Doe
214
asked Nov 23 '18 at 18:26
J.DoeJ.Doe
214
214
1
$begingroup$
A rotation is not an endomorphism (homomorphism from a group to itself) of the multiplicative group of non-zero complex numbers. Consider that $1$ has to map to $1$, for instance.
$endgroup$
– Arthur
Nov 23 '18 at 18:35
$begingroup$
@Arthur: Oh, you are right. But what about $zmapsto z^k$? If $|k|neq 1$, then stil the claim in the question does not hold?
$endgroup$
– J.Doe
Nov 23 '18 at 18:45
$begingroup$
ah...nevermind, if $|k|neq 1$, then the homomorphism is not injective. But how can I be sure that there is no other possibility?
$endgroup$
– J.Doe
Nov 23 '18 at 18:47
add a comment |
1
$begingroup$
A rotation is not an endomorphism (homomorphism from a group to itself) of the multiplicative group of non-zero complex numbers. Consider that $1$ has to map to $1$, for instance.
$endgroup$
– Arthur
Nov 23 '18 at 18:35
$begingroup$
@Arthur: Oh, you are right. But what about $zmapsto z^k$? If $|k|neq 1$, then stil the claim in the question does not hold?
$endgroup$
– J.Doe
Nov 23 '18 at 18:45
$begingroup$
ah...nevermind, if $|k|neq 1$, then the homomorphism is not injective. But how can I be sure that there is no other possibility?
$endgroup$
– J.Doe
Nov 23 '18 at 18:47
1
1
$begingroup$
A rotation is not an endomorphism (homomorphism from a group to itself) of the multiplicative group of non-zero complex numbers. Consider that $1$ has to map to $1$, for instance.
$endgroup$
– Arthur
Nov 23 '18 at 18:35
$begingroup$
A rotation is not an endomorphism (homomorphism from a group to itself) of the multiplicative group of non-zero complex numbers. Consider that $1$ has to map to $1$, for instance.
$endgroup$
– Arthur
Nov 23 '18 at 18:35
$begingroup$
@Arthur: Oh, you are right. But what about $zmapsto z^k$? If $|k|neq 1$, then stil the claim in the question does not hold?
$endgroup$
– J.Doe
Nov 23 '18 at 18:45
$begingroup$
@Arthur: Oh, you are right. But what about $zmapsto z^k$? If $|k|neq 1$, then stil the claim in the question does not hold?
$endgroup$
– J.Doe
Nov 23 '18 at 18:45
$begingroup$
ah...nevermind, if $|k|neq 1$, then the homomorphism is not injective. But how can I be sure that there is no other possibility?
$endgroup$
– J.Doe
Nov 23 '18 at 18:47
$begingroup$
ah...nevermind, if $|k|neq 1$, then the homomorphism is not injective. But how can I be sure that there is no other possibility?
$endgroup$
– J.Doe
Nov 23 '18 at 18:47
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%2f3010681%2fisomorphism-1-times-mathbb-c-to-1-times-mathbb-c%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.
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%2f3010681%2fisomorphism-1-times-mathbb-c-to-1-times-mathbb-c%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
1
$begingroup$
A rotation is not an endomorphism (homomorphism from a group to itself) of the multiplicative group of non-zero complex numbers. Consider that $1$ has to map to $1$, for instance.
$endgroup$
– Arthur
Nov 23 '18 at 18:35
$begingroup$
@Arthur: Oh, you are right. But what about $zmapsto z^k$? If $|k|neq 1$, then stil the claim in the question does not hold?
$endgroup$
– J.Doe
Nov 23 '18 at 18:45
$begingroup$
ah...nevermind, if $|k|neq 1$, then the homomorphism is not injective. But how can I be sure that there is no other possibility?
$endgroup$
– J.Doe
Nov 23 '18 at 18:47