Mixing
Meaning of $h_*$ complex map.
$begingroup$
I have come across the following:
Let $Omega$ be a convex domain in $mathbb{C}^n$ and let $h: Omega rightarrow mathbb{C}^n$ (or $mathbb{R}^n$) be a smooth mapping with $||h_*||= sup_x || h_*(x)|| < 1$, then the mapping $H$ taking $x mapsto x + h(x)$ is a diffeomorphism of $Omega$ onto $HOmega$.
What does $h_*$ stand for in this context?
complex-analysis dynamical-systems complex-geometry diffeomorphism
asked Jan 14 at 16:26
airiairi
233
$endgroup$
add a comment |
$begingroup$
I have come across the following:
Let $Omega$ be a convex domain in $mathbb{C}^n$ and let $h: Omega rightarrow mathbb{C}^n$ (or $mathbb{R}^n$) be a smooth mapping with $||h_*||= sup_x || h_*(x)|| < 1$, then the mapping $H$ taking $x mapsto x + h(x)$ is a diffeomorphism of $Omega$ onto $HOmega$.
What does $h_*$ stand for in this context?
complex-analysis dynamical-systems complex-geometry diffeomorphism
asked Jan 14 at 16:26
airiairi
233
$endgroup$
add a comment |
$begingroup$
I have come across the following:
Let $Omega$ be a convex domain in $mathbb{C}^n$ and let $h: Omega rightarrow mathbb{C}^n$ (or $mathbb{R}^n$) be a smooth mapping with $||h_*||= sup_x || h_*(x)|| < 1$, then the mapping $H$ taking $x mapsto x + h(x)$ is a diffeomorphism of $Omega$ onto $HOmega$.
What does $h_*$ stand for in this context?
complex-analysis dynamical-systems complex-geometry diffeomorphism
asked Jan 14 at 16:26
airiairi
233
$endgroup$
I have come across the following:
Let $Omega$ be a convex domain in $mathbb{C}^n$ and let $h: Omega rightarrow mathbb{C}^n$ (or $mathbb{R}^n$) be a smooth mapping with $||h_*||= sup_x || h_*(x)|| < 1$, then the mapping $H$ taking $x mapsto x + h(x)$ is a diffeomorphism of $Omega$ onto $HOmega$.
What does $h_*$ stand for in this context?
complex-analysis dynamical-systems complex-geometry diffeomorphism
complex-analysis dynamical-systems complex-geometry diffeomorphism
asked Jan 14 at 16:26
airiairi
233
asked Jan 14 at 16:26
airiairi
233
asked Jan 14 at 16:26
airiairi
233
asked Jan 14 at 16:26
airiairi
233
asked Jan 14 at 16:26
airiairi
233
233
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
From context it has to be the induced map of the tangent bundles, in other words, the Jacobian matrix $h_*=Dh=h'=J_h$. Then the claim is just a consequence of the inverse function theorem, as the derivative of $x+h(x)$, which is $I+h'(x)$, is invertible everywhere with $|(I+h'(x))^{-1}|le(1-|h_*|)^{-1}$.
answered Jan 14 at 16:37
LutzLLutzL
58.7k42055
$endgroup$
add a comment |
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%2f3073404%2fmeaning-of-h-complex-map%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From context it has to be the induced map of the tangent bundles, in other words, the Jacobian matrix $h_*=Dh=h'=J_h$. Then the claim is just a consequence of the inverse function theorem, as the derivative of $x+h(x)$, which is $I+h'(x)$, is invertible everywhere with $|(I+h'(x))^{-1}|le(1-|h_*|)^{-1}$.
answered Jan 14 at 16:37
LutzLLutzL
58.7k42055
$endgroup$
add a comment |
$begingroup$
From context it has to be the induced map of the tangent bundles, in other words, the Jacobian matrix $h_*=Dh=h'=J_h$. Then the claim is just a consequence of the inverse function theorem, as the derivative of $x+h(x)$, which is $I+h'(x)$, is invertible everywhere with $|(I+h'(x))^{-1}|le(1-|h_*|)^{-1}$.
answered Jan 14 at 16:37
LutzLLutzL
58.7k42055
$endgroup$
add a comment |
$begingroup$
From context it has to be the induced map of the tangent bundles, in other words, the Jacobian matrix $h_*=Dh=h'=J_h$. Then the claim is just a consequence of the inverse function theorem, as the derivative of $x+h(x)$, which is $I+h'(x)$, is invertible everywhere with $|(I+h'(x))^{-1}|le(1-|h_*|)^{-1}$.
answered Jan 14 at 16:37
LutzLLutzL
58.7k42055
$endgroup$
From context it has to be the induced map of the tangent bundles, in other words, the Jacobian matrix $h_*=Dh=h'=J_h$. Then the claim is just a consequence of the inverse function theorem, as the derivative of $x+h(x)$, which is $I+h'(x)$, is invertible everywhere with $|(I+h'(x))^{-1}|le(1-|h_*|)^{-1}$.
answered Jan 14 at 16:37
LutzLLutzL
58.7k42055
answered Jan 14 at 16:37
LutzLLutzL
58.7k42055
answered Jan 14 at 16:37
LutzLLutzL
58.7k42055
answered Jan 14 at 16:37
LutzLLutzL
58.7k42055
58.7k42055
add a comment |
add a comment |
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%2f3073404%2fmeaning-of-h-complex-map%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
