# Strake's splitting theorem for complex sectional curvature

Strake's splitting theorem: Let $(M^n,g)$ be an open manifold with sectional curvature $K\ge 0$ and soul $\Sigma^k$. If the normal holonomy group of $\Sigma$ is trivial, $M$ splits isometrically into $\Sigma \times \mathbb{R}^{n-k}$, where $\mathbb{R}^{n-k}$ carries a complete metric with $K\ge 0$.

Since nonnegative complex sectional curvature $K_\mathbb{C}\ge0$ implies $K\ge0$ but the converse does not hold for $n\ge 4$, my question is: Does an open manifold with $n\ge 4$, $K_\mathbb{C}\ge0$ and trivial normal holonomy of $\Sigma$ split into $\Sigma \times \mathbb{R}^{n-k}$, where $\mathbb{R}^{n-k}$ carries a complete metric with $K_\mathbb{C}\ge 0$?

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I am not sure if I am wrong, but does this not follow from each of the fibers $\{p\}\times \mathbb{R}^{n-k}$ being totally geodesic (Strake's splitting being a metric splitting)? – Luis Guijarro Feb 3 at 13:45