Let $M$ a Riemannian manifold and let $D$ be a smooth distribution on $M$. Let $D^{\perp}$ denote the distribution on $M$ that assigns to each $x \in M^{n}$ the orthogonal complement of $D(x)$ in $T_xM$. According to the orthogonal splitting $TM=D\oplus D^{\perp}$, we write $$ X=X^{v}+X^{h}$$ for any $X \in \Gamma(TM)$. We denote $$ \nabla_{X}^{h}Y=(\nabla_{X}Y)^{h},$$ for all $X,Y \in \Gamma(TM).$
The splitting tensor $C$ of $D$ is the map $C:\Gamma(D)\times \Gamma(D^{\perp})\rightarrow \Gamma(D^{\perp})$ defined by $$C_{T}X:= C(T,X)=-\nabla_{X}^{h}T.$$
Follows from properties of levi-civita connection that $C$ is $C^{\infty}$(M)-bilinear.
Here we have that $D^{\perp}$ is integrable iff $C_T$ is self-adjoint for all $T \in \Gamma(D)$, because \begin{align*} \langle C_TX,Y\rangle-\langle X, C_{T}Y\rangle&=-\langle \nabla_{X}^{h}T,Y\rangle+\langle X,\nabla_{Y}^{h}T\rangle\\ &=\langle \nabla_{X}Y-\nabla_{Y}X,T\rangle\\ &=\langle[X,Y],T\rangle \end{align*} for all $X,Y \in \Gamma(D^{\perp})$ and $T\in \Gamma(D).$
My question is:
If $D^{\perp}$ has rank 2 and it is integrable then at any point of $M$ either there exists $T_0 \in \Gamma(D)$ such that $C_T=\langle T,T_0\rangle I$ for all $T \in \Gamma(D)$ or there exists $T_1 \in \Gamma(D)$ such that $C_{T_1}$ (is symmetric and) has two distinct real eigenvalues. Here $I$ is the identity tensor of $D^{\perp}$.
How can I prove that? Thank you.