Does this special vector field affect on sectional curvature?

Question

We have a nowhere vanishing vector field $X$ on a Reimannian manifold $(M,g)$, such that $\mathcal{L}_X g=2\alpha X^\flat \otimes X^\flat$ for a constant non zero $\alpha$. So, for every vector fields $Y,Z$ orthogonal to $X$, we have $\mathcal{L}_X g(Y,Z)=0$. Also, for every two arbitrary commutitive vector fields $Y,Z$ orthogonal to $X$, we can write $dX^\flat (Y,Z)=0$.

Can we deduce that $K(X,Y)=0$ for every vector field $Y$ orthogonal to $X$?

Personally, I think, we can. Because in this case we can deduce that $M$ locally has to be as a warped product manifold $M=I\times_{f(s)} F$ with $X=\mu \dfrac{\partial}{\partial s}$. Where $\mu$ is a smooth function on $I$, and $I$ is an interval. Now, the condition $\mathcal{L}_X g=2\alpha X^\flat \otimes X^\flat$ yields that, $f(s)$ has to be a constant.

Am I right?

Answer 1

This is not necessarily true. As a counter example, take $M=\mathbb{S}^3$ and consider the Killing field induced by some rotation. Then the Lie derivative of the metric will vanish (i.e., $\alpha=0$), but the sectional curvature is constant and positive.

Answer 2

On a Riemannian manifold without boundary it must hold that $\alpha=0$ --- i.e., $X$ is a killing field, like @Gabe K said.

In general, it is easy to establish that for a general vector field $X\in\mathfrak{X}(M)$,

$$ \operatorname{tr}_{g}\mathcal{L}_{X}g=2\operatorname{div}{X}.$$

On the other hand,

$$ \operatorname{tr}_{g}(2\alpha X^{\flat}\otimes X^{\flat})=2\alpha |X|^{2}_{g}.$$

So by comparing,

$$ \operatorname{div}{X}= \alpha |X|^{2}_{g}.$$

But by the divergence theorem,

$$\int_{M} \operatorname{div}{X}\, \operatorname{Vol}_{g}=0.$$

Which is to say,

$$ \alpha \int_{M} |X|^{2}_{g}\, \operatorname{Vol}_{g}=0.$$

Hence either $\alpha=0$ or $X=0$ identically.