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?



Projectively flat connection
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?](https://mathoverflow.net/questions/91819/does-bundle-with-torsion-chern-classes-admit-flat-connection?rq=1)


