Vanishing of the Nijenhuis tensor












2












$begingroup$


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 ?...










share|cite|improve this question











$endgroup$








  • 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
















2












$begingroup$


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 ?...










share|cite|improve this question











$endgroup$








  • 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














2












2








2


1



$begingroup$


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 ?...










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










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