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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 $M$ 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?

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?

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 $M$ 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.

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 $M$ 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?

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 $\nabla_Y X$=0 for every vector field $Y$ orthogonal to $X$.

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 $M$ 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.