5
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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 $$ \nabla (e_1\otimes e^1 + \cdots + e_r\otimes e^r) = 0. $$

$\endgroup$
  • $\begingroup$ 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 $\nabla e_i$. $\endgroup$ – Asaf Shachar Mar 4 at 12:23
  • $\begingroup$ @AsafShachar: You are right. I was thinking of the trace rather than the actual $g$. I'll fix that. $\endgroup$ – Robert Bryant Mar 4 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.