Let $E \to B$ be a Hermitian vector bundle. If $E$ has a projectively flat connection, then its total Chern character has the form $\mbox{ch}(E) = \mbox{rank} \cdot \exp(\mbox{slope})$. Is the converse true? In other words, can I deduce that a vector bundle has a projectively flat connection just by looking at its Chern character?

1$\begingroup$ It seems unlikely to me, because the existence of a projectively flat connection is equivalent to the traceless part of the Atiyah class vanishing, and that is more complicated than the Chern character, which can be computed from it up to the first Chen class. $\endgroup$ – Ben McKay Mar 1 at 20:42

2$\begingroup$ Hi Dimitri ! Isn't a particular case of your question asking whether a bundle with vanishing Chern classes admits a flat connection ? $\endgroup$ – Nicolas Tholozan Mar 1 at 23:11

$\begingroup$ Nicolas, I found an example of a vector bundle with vanishing Chern classes, that lacks a flat connection. It is not obvious to me that it also lacks a projectively flat connection. But maybe it's a lead. $\endgroup$ – Dimitri Zvonkine Mar 2 at 6:24
Let $E$ is a Hermitian vector bundle with vanishing Chern classes.
Proposition: If $E$ admits a projectively flat Hermitian connection, then it admits a flat Hermitian connection.
Proof: Let $\nabla$ be a projectively flat Hermitian connection. Then its curvature $F_\nabla$ has the form $$F_\nabla = i\omega \,\mathrm{Id}_E$$ for some closed real $2$form $\omega$.
Since $$c_1(E) = \left [\frac{i}{2\pi} \mathrm{Trace}(F_\nabla)\right ] = \left[ \frac{\mathrm{rank}(E)}{2\pi} \omega\right ] = 0~,$$ there exists a real $1$form $\alpha$ such that $\mathrm d \alpha = \omega$.
Now, the connection $\nabla  i \alpha\, \mathrm{Id}_E$ is another Hermitian connection with curvature $$F_\nabla i \nabla(\alpha\, \mathrm{Id}_E) = 0~.$$ QED
Now, there are examples of bundles with vanishing Chern classes that do not admit a flat connection. See for instance this question: Does bundle with torsion Chern classes admit flat connection?