# How does the constancy of an operator’s eigenvalues imply the integrability of its eigenvector distribution?

I am reading a paper, Yano and Ishihara, “Submanifolds with Parallel Mean Curvature Vector” (MSN), where the authors have constructed a linear operator, say $$A$$, on vector fields. They claim that this operator having locally constant eigenvalues is enough to imply that the distribution generated by the eigenvectors of $$A$$ is integrable. I am having trouble understanding this.

In order to have integrability, I believe we must show that $$[v,w]$$ is an eigenvector for any eigenvectors $$v,w$$ of $$A$$. Say that $$\lambda_1,\lambda_2$$ are the eigenvalues associated to $$v,w$$, respectively. The constant eigenvalue condition seems to imply that $$\nabla(Av) = A(\nabla v) = \lambda_1\nabla v,$$ but I do not understand how this implies the distribution is closed under Lie brackets. In particular, it is clear from the above that $$[Av,Aw] = \lambda_1\lambda_2[v,w]$$, and it is also clear that the bracket is $$A$$-linear in both arguments separately. With that said, why is it that $$[v,w]$$ is also an eigenvector of $$A$$? I do not see it.

EDIT: The operator $$A$$ is (as I understand it) not a priori a homomorphism of Lie algebras. It is something related to the shape operator of an immersed submanifold. The precise statement of the problem above can be found in the final paragraph on pg 108 of Yano and Ishihara, “Submanifolds with Parallel Mean Curvature Vector”.

• you will probably want to disclose which paper it is, for a helpful answer... – Carlo Beenakker Oct 20 '19 at 7:09
• Sure. The paper in question is “submanifolds with parallel mean curvature vector” by Yano and Ishihara. What I am confused about is the final paragraph on page 108, which is part of the discussion immediately preceding Lemma 2.8. – MathIsArt Oct 20 '19 at 14:19

After much deliberation, I believe my problem is that I have misunderstood the definition of integrability. From https://www.encyclopediaofmath.org/index.php/Distribution_of_tangent_subspaces, it looks like the bracket $$[v,w]$$ does not have to be itself an eigenvector, but only a linear combination of eigenvectors.
This seems to be true from the following computation: $$A([v,w]) = A(\nabla_vw - \nabla_wv) = \nabla_v(Aw) - \nabla_w(Av) = \lambda_2\nabla_vw - \lambda_1\nabla_wv.$$