Relation of commutators $[A_{\xi},A_{\eta}]$ with the normal bundle

81 Views Asked by At

Let $(\overline{M}^{n+m}, \newcommand\pair[1]{\left\langle #1 \right\rangle} \pair{\cdot,\cdot})$ be a pseudo-Riemannian manifold and $M^n \subseteq \overline{M}$ a non-degenerate submanifold. We have the Ricci $\renewcommand\vec[1]{{\boldsymbol #1}}$equation $$\pair{R^\perp(\vec{X},\vec{Y})\vec{\xi},\vec{\eta}} = \overline{R}(\vec{X},\vec{Y},\vec{\xi},\vec{\eta}) + \pair{[A_{\vec{\xi}},A_{\vec{\eta}}](\vec{X}),\vec{Y}},$$for all $\vec{X},\vec{Y} \in \mathfrak{X}(M)$ and $\vec{\xi},\vec{\eta} \in \mathfrak{X}^\perp(M)$. Do Carmo says that if $\overline{M}$ has constant sectional curvature, then the Ricci equation reads $$\pair{R^\perp(\vec{X},\vec{Y})\vec{\xi},\vec{\eta}} = \pair{[A_{\vec{\xi}},A_{\vec{\eta}}](\vec{X}),\vec{Y}},$$and so the normal bundle of $M$ is flat (i.e., $R^\perp = \vec{0}$) if and only if $[A_{\vec{\xi}},A_{\vec{\eta}}] = \vec{0}$ for all $\vec{\xi}, \vec{\eta} \in \mathfrak{X}^\perp(M)$.

I'm sure I'm missing something obvious here, but I think that the sectional curvature of the ambient should not only be constant, but be identically null. And it seems that a few papers (e.g. this one) use that result.

How can we make that conclusion if the sectional curvature is a non-zero constant?

1

There are 1 best solutions below

0
On BEST ANSWER

Stupid me forgot that if the sectional curvature is a $\newcommand\pair[1]{\left\langle #1\right\rangle}$constant$\renewcommand\vec[1]{{\boldsymbol #1}}$ $K_0$, then $$\overline{R}(\vec{X},\vec{Y})\vec{\xi} = K_0\left(\pair{\vec{Y},\vec{\xi}}\vec{X} - \pair{\vec{X},\vec{\xi}}\vec{Y} \right) = \vec{0}.$$Oh well.