Complex structure : Existence of Local frame











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We have a real vector bundle $pi : Erightarrow M$ of rank $2n$ over smooth manifold $M$ and $Jin Gamma(E^*otimes E)$(set of smooth sections of vector the bundle) is a complex structure on $E$. And suppose $E$ is trivial over an open set $Usubseteq M$. Show that there is a local frame $(e_1,f_1,dots, e_n, f_n)$ of $E$ over $U$ where $f_i=J(e_i)$.




I have been trying to show this but I have not gotten any advance after I start to think that $e_i$ takes real part and $f_i$ takes imaginary part. I really need help from here. I thank in advance for any help :)




For reference, for any $pin M$, $J$ should satisfy $J(p)^2=-Id_{E_p}$ where $E_p=pi^{-1}({p}).$











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  • Can you see why there is an inner product on $E$ compatible with $J$ (over $U$)?
    – user10354138
    Nov 14 at 14:10










  • @user10354138 Yes. Actually I know that $E$ admits Riemannian metric.
    – LeB
    Nov 14 at 14:59










  • @user10354138 But I don't know if that is compatible with $J$
    – LeB
    Nov 14 at 15:34















up vote
1
down vote

favorite













We have a real vector bundle $pi : Erightarrow M$ of rank $2n$ over smooth manifold $M$ and $Jin Gamma(E^*otimes E)$(set of smooth sections of vector the bundle) is a complex structure on $E$. And suppose $E$ is trivial over an open set $Usubseteq M$. Show that there is a local frame $(e_1,f_1,dots, e_n, f_n)$ of $E$ over $U$ where $f_i=J(e_i)$.




I have been trying to show this but I have not gotten any advance after I start to think that $e_i$ takes real part and $f_i$ takes imaginary part. I really need help from here. I thank in advance for any help :)




For reference, for any $pin M$, $J$ should satisfy $J(p)^2=-Id_{E_p}$ where $E_p=pi^{-1}({p}).$











share|cite|improve this question






















  • Can you see why there is an inner product on $E$ compatible with $J$ (over $U$)?
    – user10354138
    Nov 14 at 14:10










  • @user10354138 Yes. Actually I know that $E$ admits Riemannian metric.
    – LeB
    Nov 14 at 14:59










  • @user10354138 But I don't know if that is compatible with $J$
    – LeB
    Nov 14 at 15:34













up vote
1
down vote

favorite









up vote
1
down vote

favorite












We have a real vector bundle $pi : Erightarrow M$ of rank $2n$ over smooth manifold $M$ and $Jin Gamma(E^*otimes E)$(set of smooth sections of vector the bundle) is a complex structure on $E$. And suppose $E$ is trivial over an open set $Usubseteq M$. Show that there is a local frame $(e_1,f_1,dots, e_n, f_n)$ of $E$ over $U$ where $f_i=J(e_i)$.




I have been trying to show this but I have not gotten any advance after I start to think that $e_i$ takes real part and $f_i$ takes imaginary part. I really need help from here. I thank in advance for any help :)




For reference, for any $pin M$, $J$ should satisfy $J(p)^2=-Id_{E_p}$ where $E_p=pi^{-1}({p}).$











share|cite|improve this question














We have a real vector bundle $pi : Erightarrow M$ of rank $2n$ over smooth manifold $M$ and $Jin Gamma(E^*otimes E)$(set of smooth sections of vector the bundle) is a complex structure on $E$. And suppose $E$ is trivial over an open set $Usubseteq M$. Show that there is a local frame $(e_1,f_1,dots, e_n, f_n)$ of $E$ over $U$ where $f_i=J(e_i)$.




I have been trying to show this but I have not gotten any advance after I start to think that $e_i$ takes real part and $f_i$ takes imaginary part. I really need help from here. I thank in advance for any help :)




For reference, for any $pin M$, $J$ should satisfy $J(p)^2=-Id_{E_p}$ where $E_p=pi^{-1}({p}).$








differential-geometry vector-bundles






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asked Nov 14 at 12:51









LeB

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  • Can you see why there is an inner product on $E$ compatible with $J$ (over $U$)?
    – user10354138
    Nov 14 at 14:10










  • @user10354138 Yes. Actually I know that $E$ admits Riemannian metric.
    – LeB
    Nov 14 at 14:59










  • @user10354138 But I don't know if that is compatible with $J$
    – LeB
    Nov 14 at 15:34


















  • Can you see why there is an inner product on $E$ compatible with $J$ (over $U$)?
    – user10354138
    Nov 14 at 14:10










  • @user10354138 Yes. Actually I know that $E$ admits Riemannian metric.
    – LeB
    Nov 14 at 14:59










  • @user10354138 But I don't know if that is compatible with $J$
    – LeB
    Nov 14 at 15:34
















Can you see why there is an inner product on $E$ compatible with $J$ (over $U$)?
– user10354138
Nov 14 at 14:10




Can you see why there is an inner product on $E$ compatible with $J$ (over $U$)?
– user10354138
Nov 14 at 14:10












@user10354138 Yes. Actually I know that $E$ admits Riemannian metric.
– LeB
Nov 14 at 14:59




@user10354138 Yes. Actually I know that $E$ admits Riemannian metric.
– LeB
Nov 14 at 14:59












@user10354138 But I don't know if that is compatible with $J$
– LeB
Nov 14 at 15:34




@user10354138 But I don't know if that is compatible with $J$
– LeB
Nov 14 at 15:34















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