Demystifier said:
I am not sufficiently fluent in mathematician's terminology, perhaps
@samalkhaiat may help. From a physical perspective, I would advise you to ask yourself the following related question. In electrodynamics, is the covariant derivative
$$\nabla_{\mu}=\partial_{\mu}-ieA_{\mu}$$
metric compatible?
The covariant derivative of EM is compatible with the spacetime metric. This can be seen by the definition given by Nakahara in his textbook Geometry, Tpology and Physics, p.253(see image below). You can see that my definition and metric being covariantly constant are equal. The metric compatibility condition is not exclusive to the Levi-Civita connection. In the EM case, the covariant derivative is compatible with the spacetime metric, but it is not the Levi-Civita connection that performs parallel transports on the spacetime manifold. And the compatibility with the metric
need not refer to the Levi-Civita connection. It is a property that
any connection can have, either L-C or not.
Please correct me if I am misunderstanding something here.
Now, in the case on the Berry connection, since I am always using $$\partial_{\mu}(<n|m>)=<\partial_{\mu}n|m>+<n|\partial_{\mu}m>$$, I think that the Berry connection
is compatible with the metric(since the covariant derivative is given by $$\nabla_{\mu}=\partial_{\mu}-ieA_{\mu}$$ as you said), although not the spacetime metric. I suspect that the metric that is relevant to the Berry connection case is that which is used to define the inner product between states; I mean, since we do have a well-defined inner product between states, then there should be a metric behind it.
Lastly, note that there exists some kind of metric that comes from what is called the Quantum Geometric tensor.
@Demystifier you are correct in that I am not so fluent in the language of bundles, so I am effectively(though not purposely) trying to convert my understanding of the Berry language to the language used in Riemannian geometry(which I know well, at least at an introductory level). :)
@vanhees71 I am tagging you too so you can give an opinion on what I am saying(if you don't mind, that is) :)