# Does $\nabla g=\omega(\cdot) g$ imply $\nabla$ is metric w.r.t a conformal rescaling of $g$?

This is a cross-post.

Let $$E$$ be a smooth vector bundle over a manifold $$M$$, where $$\text{rank}(E) > 1,\dim M > 1$$. Suppose that $$E$$ is equipped with a metric $$g$$ and an affine connection $$\nabla$$, such that $$\nabla_X g=\omega (X) g$$ for every $$X \in \Gamma(TM)$$. (Here $$\omega$$ is a one form).

Must $$\omega$$ be closed?

Clearly, $$\nabla$$ is metric-compatible ($$\nabla g=0$$) iff $$\omega=0$$. Moreover, $$\omega=d\phi$$ is exact if and only if $$\nabla s=0$$ where $$s=e^{-\phi}g$$, i.e. $$\nabla$$ is metric w.r.t a positive conformal rescaling of $$g$$. So, an alternative formulation of the question is the following:

Suppose that $$\nabla g=\omega (\cdot) g$$ for some $$\omega \in \Omega^1(M)$$. Must $$\nabla$$ be metric w.r.t a local conformal rescalings of $$g$$?

Differentiating $$\nabla g=\omega (\cdot) g$$, we get $$R(X,Y)g=d\omega(X,Y)g$$, so if $$\nabla$$ is flat then $$\omega$$ is closed.

I required $$\text{rank}(E) > 1$$, since if the rank is $$1$$, $$\nabla g$$ can always be written as $$\omega (\cdot) g$$ for a suitable $$\omega$$, so the assumption always holds, but I think that $$d\omega=0$$ does not always hold. Maybe this can be used to construct a counter example of higher rank by taking a direct sum of line bundles.

The answer is 'no'. For example, just take $$M$$ to be $$\mathbb{R}^n$$ (for $$n>1$$), and $$E = M\times \mathbb{R}^r$$ for some $$r>1$$. Let $$\omega$$ be any $$1$$-form on $$M$$, and define a connection $$\nabla$$ on $$E$$ by setting $$\nabla e_i = -\tfrac{1}{2} \omega\otimes e_i$$ where $$e_i$$ for $$1\le i\le r$$ is some basis for the sections of $$E$$ over $$M$$. Then, if $$e^i$$ are the dual basis of $$E^*$$, the metric $$g = (e^1)^2 + \cdots + (e^r)^2$$ satisfies $$\nabla g = \omega\otimes g$$.
Note: We extend the connection $$\nabla$$ as a connection on $$E^*$$ by the usual rule: I.e., we require that $$\nabla (e_1\otimes e^1 + \cdots + e_r\otimes e^r) = 0.$$
• Thank you. I think that the factor $\frac{1}{r}$ is superfluous, that is it should be $\frac{1}{2}$ instead of $\frac{1}{2r}$ in the definition of $\nabla e_i$. Mar 4, 2019 at 12:23
• @AsafShachar: You are right. I was thinking of the trace rather than the actual $g$. I'll fix that. Mar 4, 2019 at 12:29