A question about two parallel integrable distributions Give a Riemannian manifolds $(M,g)$,$\nabla$ is its connection.
Suppose we have two  distributions $E$ and $F$ on $(M,g)$,that are orthogonal complements of each other in $TM$.In addition,assume that the distributions are parallel,i.e.,if two vector fields $X$,$Y\in E$,then $ \nabla_{X}Y \in E$.
Show that around any point $M$ there is a product neighborhood $U=V_{E} \times V_{F}$ such that $(U,g)=(V_{E} \times V_{F},g|_{E}+g|_{F})$,where $g|_{E}$ and $g|_{F}$ are the restrictions of $g$ to the two distributions.
I have tried the special situation which $dimF=1$ .
Suppose   locally $E$ consists of a orthonormal frame  $e_{i} 1\leq i \leq n$ ,$F$ consists of a unit vector field $e_{n+1}$,the coframe donated by $w^i,w^{n+1}$.
Suppose $\nabla e_{n+1} = w_{n+1} ^{j}e_j$, then 
$$ w_{n+1} ^{j}(e_{i})=g(\nabla_{e_i} e_{n+1},e_{j})
                        =-g(\nabla_{e_i} e_{j},e_{n+1})=0 $$
the last equation because the condition $E$ is parallel.
Thus we got:
$$w_{n+1} ^{j}=0$$
By the torsion-free equation,we get :
$$ dw^{n+1}=w^{i} \wedge w_{i} ^{n+1}=w^{i} \wedge -w_{n+1} ^{i}=0$$
By the  Poincaré's lemma : Locally there is a function $f$,such that
$$df=w^{n+1}$$
Because 
$$\nabla_{e_i} e_{j}-\nabla_{e_j} e_{i}=[e_{i},{e_j}]$$
Hence $E$ is integrable distribution,so suppose in a local coordinate $x^{i},x^{n+1}$,$$E=span\{\frac{\partial}{\partial x^{1}}...,\frac{\partial}{\partial x^{n}}\}$$
then$\partial_{x^{n+1}} f \neq 0$,
we construct a transformation :
$$T:(x^{1},...,x^{n},x^{n+1}) \rightarrow (y^{1},...,y^{n},y^{n+1})$$
$$(x^{1},...,x^{n},x^{n+1})\mapsto (x^{1},...,x^{n},f)$$
By the check of Theorem of Inverse Function,this is a diffeomorphism ,and $T_{\star} e^{n+1}=\frac{\partial}{\partial y^{n+1}}$,$T_{\star} e^{i}=\frac{\partial}{\partial y^{i}}$
Pick $V_{E}=\{y^{n+1}=0\}$,$V_{F}=\{y^{i}=0\}$,we get the answer.
But I can't go to the general situation which $dimV\neq 1$.
If you can give me some suggestions,I will appreciate your help
 A: I had the answer.In the below,assume $1\leq i ,j\leq n$,$n+1\leq \alpha,\beta\leq n+m$.
Because $E$,$F$ are integrable distributions,we can take coordinates $x^{i},x^{\alpha}$;$y^{j},y^{\beta}$,such that:
$$E=span\{\frac{\partial}{\partial x^{1}}...,\frac{\partial}{\partial x^{n}}\}$$
$$F=span\{\frac{\partial}{\partial y^{n+1}}...,\frac{\partial}{\partial y^{n+m}}\}$$
So by the little check,we claim $(x^{\beta},y^{\i})$ is local coordinates.
The only we need to show is that $g\mid _{E}$(or respect to $g\mid _{F}$) is independent  of $y^{\beta}$(or respect to $x^{i}$).
This is equivalent  to show  
$$ \partial_{\alpha}g_{i j}=0$$
But due to $E$,$F$ are orthogonal complement of each other ,we get :
$$\Gamma_{ i j \alpha}=\frac{1}{2} (\partial_{\alpha}g_{i j}+\partial_{i}g_{j \alpha}-\partial_{j}g_{i \alpha})=\frac{1}{2} \partial_{\alpha}g_{i j}   $$
On the other side :
$$\Gamma_{ i j \alpha}=g(\nabla_{\frac{\partial}{\partial x^{j}}} \frac{\partial}{\partial x^{i}},\frac{\partial}{\partial y^{\alpha}})=0$$
The last equation by the virtue of parallel condition of $E$.
Then we proved our question.
