Vanishing of the Nijenhuis tensor
$begingroup$
The Nijenhuis tensor is defined to be:
$$(1):quad
N_J(X,Y)equiv[X,Y]+J[JX,Y]+J[X,JY]-[JX,JY],
$$
for vector fields $X$ and $Y$ on the manifold $M$, equipped with Almost Complex Structure $J$:
$$(2):quad
J:TMrightarrow TMquad|quad J^2=-I_{TM}.
$$
The requirement is to show that given $J$ is integrable, $N_J$ is vanishing. I am having an issue with the following computation which gives a vanishing $N_J$, regardless of whether $J$ is integrable or not:
$$
N_J(X,Y)equiv[X,Y]+J[JX,Y]+J[X,JY]-[JX,JY]
$$
$$
=[X,Y]+J^2XY-JYJX+JXJY-J^2YX-[JX,JY]
$$
$$
=[X,Y]+J^2[X,Y]
$$
$$
=[X,Y]-[X,Y]=0quadtext{(by (2) alone)}.
$$
I find this very strange because this makes absolutely no reference at all to $J$ being integrable or not, rather this follows solely from the definition of $J$ in $(2)$. Suggesting that if $J$ is an almost complex structure, then $N_J$ is always vanishing. Is it that I can write $(2)$ only for integrable $J$ ? If someone could please explain what is it that I am not doing right or do not understand correctly ?...
differential-geometry complex-manifolds almost-complex
$endgroup$
add a comment |
$begingroup$
The Nijenhuis tensor is defined to be:
$$(1):quad
N_J(X,Y)equiv[X,Y]+J[JX,Y]+J[X,JY]-[JX,JY],
$$
for vector fields $X$ and $Y$ on the manifold $M$, equipped with Almost Complex Structure $J$:
$$(2):quad
J:TMrightarrow TMquad|quad J^2=-I_{TM}.
$$
The requirement is to show that given $J$ is integrable, $N_J$ is vanishing. I am having an issue with the following computation which gives a vanishing $N_J$, regardless of whether $J$ is integrable or not:
$$
N_J(X,Y)equiv[X,Y]+J[JX,Y]+J[X,JY]-[JX,JY]
$$
$$
=[X,Y]+J^2XY-JYJX+JXJY-J^2YX-[JX,JY]
$$
$$
=[X,Y]+J^2[X,Y]
$$
$$
=[X,Y]-[X,Y]=0quadtext{(by (2) alone)}.
$$
I find this very strange because this makes absolutely no reference at all to $J$ being integrable or not, rather this follows solely from the definition of $J$ in $(2)$. Suggesting that if $J$ is an almost complex structure, then $N_J$ is always vanishing. Is it that I can write $(2)$ only for integrable $J$ ? If someone could please explain what is it that I am not doing right or do not understand correctly ?...
differential-geometry complex-manifolds almost-complex
$endgroup$
1
$begingroup$
The condition $J^2 = - I$ is definitely not true just for integrable $J$. For example, $S^6$ has an almost complex structure $J$ coming from thinking of it as the unit octonians, but this almost complex structure is not integrable. In your computation, it's not clear to me that $J[JX,Y] = J^2XY - JYJX$, and similarly for the third term. I agree that $J(V+W) = JV + JW$ for any vectors $V$ and $W$, but when you write $[JX,Y] = JXY - YJX$, you have to remember that the terms $JXY$ and $YJX$ are not actually vector fields themselves, so there is no reason for $J$ to distribute over them....
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:28
$begingroup$
Perfect ... that answers it! you got me spot on :)
$endgroup$
– Kong
Nov 30 '18 at 19:32
$begingroup$
Well, I wish I knew how to actually do the computation!
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:35
1
$begingroup$
Hint: If $X$ is any tangent vector, then $X-iJX$ is an $i$-eigenvector of $J$, and $X+iJX$ is a $(-i)$-eigenvector.
$endgroup$
– Jack Lee
Nov 30 '18 at 22:56
add a comment |
$begingroup$
The Nijenhuis tensor is defined to be:
$$(1):quad
N_J(X,Y)equiv[X,Y]+J[JX,Y]+J[X,JY]-[JX,JY],
$$
for vector fields $X$ and $Y$ on the manifold $M$, equipped with Almost Complex Structure $J$:
$$(2):quad
J:TMrightarrow TMquad|quad J^2=-I_{TM}.
$$
The requirement is to show that given $J$ is integrable, $N_J$ is vanishing. I am having an issue with the following computation which gives a vanishing $N_J$, regardless of whether $J$ is integrable or not:
$$
N_J(X,Y)equiv[X,Y]+J[JX,Y]+J[X,JY]-[JX,JY]
$$
$$
=[X,Y]+J^2XY-JYJX+JXJY-J^2YX-[JX,JY]
$$
$$
=[X,Y]+J^2[X,Y]
$$
$$
=[X,Y]-[X,Y]=0quadtext{(by (2) alone)}.
$$
I find this very strange because this makes absolutely no reference at all to $J$ being integrable or not, rather this follows solely from the definition of $J$ in $(2)$. Suggesting that if $J$ is an almost complex structure, then $N_J$ is always vanishing. Is it that I can write $(2)$ only for integrable $J$ ? If someone could please explain what is it that I am not doing right or do not understand correctly ?...
differential-geometry complex-manifolds almost-complex
$endgroup$
The Nijenhuis tensor is defined to be:
$$(1):quad
N_J(X,Y)equiv[X,Y]+J[JX,Y]+J[X,JY]-[JX,JY],
$$
for vector fields $X$ and $Y$ on the manifold $M$, equipped with Almost Complex Structure $J$:
$$(2):quad
J:TMrightarrow TMquad|quad J^2=-I_{TM}.
$$
The requirement is to show that given $J$ is integrable, $N_J$ is vanishing. I am having an issue with the following computation which gives a vanishing $N_J$, regardless of whether $J$ is integrable or not:
$$
N_J(X,Y)equiv[X,Y]+J[JX,Y]+J[X,JY]-[JX,JY]
$$
$$
=[X,Y]+J^2XY-JYJX+JXJY-J^2YX-[JX,JY]
$$
$$
=[X,Y]+J^2[X,Y]
$$
$$
=[X,Y]-[X,Y]=0quadtext{(by (2) alone)}.
$$
I find this very strange because this makes absolutely no reference at all to $J$ being integrable or not, rather this follows solely from the definition of $J$ in $(2)$. Suggesting that if $J$ is an almost complex structure, then $N_J$ is always vanishing. Is it that I can write $(2)$ only for integrable $J$ ? If someone could please explain what is it that I am not doing right or do not understand correctly ?...
differential-geometry complex-manifolds almost-complex
differential-geometry complex-manifolds almost-complex
edited Nov 30 '18 at 17:11
Kong
asked Nov 30 '18 at 17:06
KongKong
315
315
1
$begingroup$
The condition $J^2 = - I$ is definitely not true just for integrable $J$. For example, $S^6$ has an almost complex structure $J$ coming from thinking of it as the unit octonians, but this almost complex structure is not integrable. In your computation, it's not clear to me that $J[JX,Y] = J^2XY - JYJX$, and similarly for the third term. I agree that $J(V+W) = JV + JW$ for any vectors $V$ and $W$, but when you write $[JX,Y] = JXY - YJX$, you have to remember that the terms $JXY$ and $YJX$ are not actually vector fields themselves, so there is no reason for $J$ to distribute over them....
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:28
$begingroup$
Perfect ... that answers it! you got me spot on :)
$endgroup$
– Kong
Nov 30 '18 at 19:32
$begingroup$
Well, I wish I knew how to actually do the computation!
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:35
1
$begingroup$
Hint: If $X$ is any tangent vector, then $X-iJX$ is an $i$-eigenvector of $J$, and $X+iJX$ is a $(-i)$-eigenvector.
$endgroup$
– Jack Lee
Nov 30 '18 at 22:56
add a comment |
1
$begingroup$
The condition $J^2 = - I$ is definitely not true just for integrable $J$. For example, $S^6$ has an almost complex structure $J$ coming from thinking of it as the unit octonians, but this almost complex structure is not integrable. In your computation, it's not clear to me that $J[JX,Y] = J^2XY - JYJX$, and similarly for the third term. I agree that $J(V+W) = JV + JW$ for any vectors $V$ and $W$, but when you write $[JX,Y] = JXY - YJX$, you have to remember that the terms $JXY$ and $YJX$ are not actually vector fields themselves, so there is no reason for $J$ to distribute over them....
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:28
$begingroup$
Perfect ... that answers it! you got me spot on :)
$endgroup$
– Kong
Nov 30 '18 at 19:32
$begingroup$
Well, I wish I knew how to actually do the computation!
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:35
1
$begingroup$
Hint: If $X$ is any tangent vector, then $X-iJX$ is an $i$-eigenvector of $J$, and $X+iJX$ is a $(-i)$-eigenvector.
$endgroup$
– Jack Lee
Nov 30 '18 at 22:56
1
1
$begingroup$
The condition $J^2 = - I$ is definitely not true just for integrable $J$. For example, $S^6$ has an almost complex structure $J$ coming from thinking of it as the unit octonians, but this almost complex structure is not integrable. In your computation, it's not clear to me that $J[JX,Y] = J^2XY - JYJX$, and similarly for the third term. I agree that $J(V+W) = JV + JW$ for any vectors $V$ and $W$, but when you write $[JX,Y] = JXY - YJX$, you have to remember that the terms $JXY$ and $YJX$ are not actually vector fields themselves, so there is no reason for $J$ to distribute over them....
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:28
$begingroup$
The condition $J^2 = - I$ is definitely not true just for integrable $J$. For example, $S^6$ has an almost complex structure $J$ coming from thinking of it as the unit octonians, but this almost complex structure is not integrable. In your computation, it's not clear to me that $J[JX,Y] = J^2XY - JYJX$, and similarly for the third term. I agree that $J(V+W) = JV + JW$ for any vectors $V$ and $W$, but when you write $[JX,Y] = JXY - YJX$, you have to remember that the terms $JXY$ and $YJX$ are not actually vector fields themselves, so there is no reason for $J$ to distribute over them....
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:28
$begingroup$
Perfect ... that answers it! you got me spot on :)
$endgroup$
– Kong
Nov 30 '18 at 19:32
$begingroup$
Perfect ... that answers it! you got me spot on :)
$endgroup$
– Kong
Nov 30 '18 at 19:32
$begingroup$
Well, I wish I knew how to actually do the computation!
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:35
$begingroup$
Well, I wish I knew how to actually do the computation!
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:35
1
1
$begingroup$
Hint: If $X$ is any tangent vector, then $X-iJX$ is an $i$-eigenvector of $J$, and $X+iJX$ is a $(-i)$-eigenvector.
$endgroup$
– Jack Lee
Nov 30 '18 at 22:56
$begingroup$
Hint: If $X$ is any tangent vector, then $X-iJX$ is an $i$-eigenvector of $J$, and $X+iJX$ is a $(-i)$-eigenvector.
$endgroup$
– Jack Lee
Nov 30 '18 at 22:56
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%2f3020336%2fvanishing-of-the-nijenhuis-tensor%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%2f3020336%2fvanishing-of-the-nijenhuis-tensor%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
$begingroup$
The condition $J^2 = - I$ is definitely not true just for integrable $J$. For example, $S^6$ has an almost complex structure $J$ coming from thinking of it as the unit octonians, but this almost complex structure is not integrable. In your computation, it's not clear to me that $J[JX,Y] = J^2XY - JYJX$, and similarly for the third term. I agree that $J(V+W) = JV + JW$ for any vectors $V$ and $W$, but when you write $[JX,Y] = JXY - YJX$, you have to remember that the terms $JXY$ and $YJX$ are not actually vector fields themselves, so there is no reason for $J$ to distribute over them....
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:28
$begingroup$
Perfect ... that answers it! you got me spot on :)
$endgroup$
– Kong
Nov 30 '18 at 19:32
$begingroup$
Well, I wish I knew how to actually do the computation!
$endgroup$
– Jason DeVito
Nov 30 '18 at 19:35
1
$begingroup$
Hint: If $X$ is any tangent vector, then $X-iJX$ is an $i$-eigenvector of $J$, and $X+iJX$ is a $(-i)$-eigenvector.
$endgroup$
– Jack Lee
Nov 30 '18 at 22:56