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Given a vector field $n$ on a Riemannian manifold $(M,g)$ we can define \begin{equation} \Delta_x=\{v\in T_xM : g(v,n_x)=0\} \end{equation}

I have to find the condition on $n$ such that this distribution is involutive.

I thought to use the definition of involutive, so $\Delta_x$ is involutive if we have: $v,w\in \Delta_x \Rightarrow [v,w]\in \Delta_x$. This leads to the condition $\forall v,w\in T_xM , g([v,w],n_x)=0$.

Is this true? Can I rewrite this condition in a better way?

Piera V
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    I guess distributions stands for a smooth distribution, meaning that one always has a local basis of smooth vector fields for $\Delta$? To rewrite it one can for example use the Leibniz rule for the Lie derivative and its relation with contraction. Or one can also look at the $1$-form $g(\cdot, n)$, see: https://math.stackexchange.com/questions/1943016/condition-for-integrability-foliation/3428929#3428929 – Florian Mar 07 '20 at 14:14
  • Yes, I meant smooth distribution. Thanks for the reference. – Piera V Mar 08 '20 at 18:58

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