Does the shape operator always commute with an almost complex structure.

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Suppose $(M,g)$ be an odd dimensional manifold and suppose it has an extra algebraic structure which induces an almost complex structure on oriented hypersurfaces of $M$. (The example I have in my mind is that of $M$ having a Vector cross product which induces an almost complex structure).



If $N$ is a hypersurface of $M$ with almost complex structure, say $J$, then is it necessary that $J$ should commute with the Shape operator $A$ of $N$ in $M$, i.e., does



$AJ(X)=JA(X)$ holds for all $Xin TN$?



I was doing some calculations and was stuck at this point and couldn't think of a possible reasoning that this might always be true and couldn't come up with a counterexample too.



Thanks










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  • Doesn't your cross product example in $mathbb R^3$ provide a counterexample? Just make sure your surface has some non-umbilic points, e.g. a cylinder.
    – Anthony Carapetis
    Sep 8 at 22:58














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Suppose $(M,g)$ be an odd dimensional manifold and suppose it has an extra algebraic structure which induces an almost complex structure on oriented hypersurfaces of $M$. (The example I have in my mind is that of $M$ having a Vector cross product which induces an almost complex structure).



If $N$ is a hypersurface of $M$ with almost complex structure, say $J$, then is it necessary that $J$ should commute with the Shape operator $A$ of $N$ in $M$, i.e., does



$AJ(X)=JA(X)$ holds for all $Xin TN$?



I was doing some calculations and was stuck at this point and couldn't think of a possible reasoning that this might always be true and couldn't come up with a counterexample too.



Thanks










share|cite|improve this question





















  • Doesn't your cross product example in $mathbb R^3$ provide a counterexample? Just make sure your surface has some non-umbilic points, e.g. a cylinder.
    – Anthony Carapetis
    Sep 8 at 22:58












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Suppose $(M,g)$ be an odd dimensional manifold and suppose it has an extra algebraic structure which induces an almost complex structure on oriented hypersurfaces of $M$. (The example I have in my mind is that of $M$ having a Vector cross product which induces an almost complex structure).



If $N$ is a hypersurface of $M$ with almost complex structure, say $J$, then is it necessary that $J$ should commute with the Shape operator $A$ of $N$ in $M$, i.e., does



$AJ(X)=JA(X)$ holds for all $Xin TN$?



I was doing some calculations and was stuck at this point and couldn't think of a possible reasoning that this might always be true and couldn't come up with a counterexample too.



Thanks










share|cite|improve this question













Suppose $(M,g)$ be an odd dimensional manifold and suppose it has an extra algebraic structure which induces an almost complex structure on oriented hypersurfaces of $M$. (The example I have in my mind is that of $M$ having a Vector cross product which induces an almost complex structure).



If $N$ is a hypersurface of $M$ with almost complex structure, say $J$, then is it necessary that $J$ should commute with the Shape operator $A$ of $N$ in $M$, i.e., does



$AJ(X)=JA(X)$ holds for all $Xin TN$?



I was doing some calculations and was stuck at this point and couldn't think of a possible reasoning that this might always be true and couldn't come up with a counterexample too.



Thanks







differential-geometry manifolds riemannian-geometry submanifold global-analysis






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asked Sep 8 at 13:15









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  • Doesn't your cross product example in $mathbb R^3$ provide a counterexample? Just make sure your surface has some non-umbilic points, e.g. a cylinder.
    – Anthony Carapetis
    Sep 8 at 22:58
















  • Doesn't your cross product example in $mathbb R^3$ provide a counterexample? Just make sure your surface has some non-umbilic points, e.g. a cylinder.
    – Anthony Carapetis
    Sep 8 at 22:58















Doesn't your cross product example in $mathbb R^3$ provide a counterexample? Just make sure your surface has some non-umbilic points, e.g. a cylinder.
– Anthony Carapetis
Sep 8 at 22:58




Doesn't your cross product example in $mathbb R^3$ provide a counterexample? Just make sure your surface has some non-umbilic points, e.g. a cylinder.
– Anthony Carapetis
Sep 8 at 22:58















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