# A Is the Berry connection compatible with the metric?

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1. Jan 3, 2018

### vanhees71

I don't understand the statement "In the EM case, the covariant derivative is compatible with the spacetime metric". I don't see what it has to do with the spacetime metric at all. In some hand-waving way one could say that the covariant derivative lives in the tangent space of the Lie group, but not in tangent space of the spacetime manifold. I guess mathematicians won't like this hand-wavy argument at all.

2. Jan 3, 2018

### Joker93

What I am saying is that we can talk about metric compatibility no matte what the metric or the connection is. I am just taking the definition given by Nakahara(see edited previous post). So, indeed it does not have to do only with the spacetime metric.
So, in essence, I am saying that we can say that a connection is compatible with a metric(not the metric) if the derivative of the inner product is given by the product rule as given by Nakahara.
I can't quite understand the language of bundles yet, so forgive me for pushing it a bit.

3. Jan 3, 2018

### Demystifier

The geometry of principal bundles is not Riemannian geometry. Or in physical words, the geometry of gauge theories and Berry connections is not Riemannian geometry. You are trying to do an impossible conversion.

4. Jan 3, 2018

### Joker93

Of course not, and that is the reason that I am trying to understand what you are saying, and sorry for pushing it.
What I am asking, in essence, is whether or not we can define metric compatibility outside of Riemannian geometry(where we are concerned with the tangent bundle). I thought that the definition of metric compatibility does not have to do with the bundle considered but with the relationship between the given connection with the given metric. I mean, the definition does not seem to be concerned with the bundle into consideration.

5. Jan 3, 2018

6. Jan 3, 2018

### vanhees71

Sure, that's the usual definition of a connection compatible with the metric of a Riemannian space, but there's no metric related with the gauge-covariant derivative it could be compatible with.

7. Jan 3, 2018

### Joker93

But, since there is the inner product between quantum states, then couldn't we find, at least in principle, the corresponding(to the inner product) metric?

8. Jan 3, 2018

### romsofia

I've, personally, never heard of the Berry connection. Are you asking if the Berry Connection is torsion free, or is not symmetric in the christoffel symbols?

It sounds like you're asking if this connection is coupled to Riemannian geometry? I'm also confused as the others because, in principle, to me you're asking something along the lines of the minimal coupling principle. So i'll outline the steps!

If you want to couple something to Riemannian geometry, essentially, what you do is:
1) Replace $n_{\mu\nu}$ with $g_{\mu\nu}$
2) Replace patrial/total derivatives with covariant derivaties
3) use $\sqrt{-g}$ to saturate all tensor densities to zero. If you're in an action integral, you've probably seen this as $d^4x$ becoming $d^4x\sqrt{-g}$

You can try this out with the E+M tensor $F_{\mu\nu} \rightarrow \nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}$ and you'll see the symmetry of the christoffel symbols here.

I'd also suggest learning about differential forms if you have not, it makes this whole concept of tangent spaces and all that easier to swallow (IMO). If I'm going down the right trail here, let me know and I'll try to add on more about this principle.

9. Jan 3, 2018

### Joker93

I am not talking about minimal coupling, nor does my question have to do with the spacetime metric in particular.
The Berry connection is like the 4-potential in EM and the Berry curvature is like the EM tensor, but you must replace the spacetime derivatives with derivatives eith respect to certain parameters. It is a problem in QM and does not have to do with Riemannian geometry(which has to do with the tangent bundle), but it has to do with the principal bundle(as others pointed out).
What I am asking though is about metric compatibility(which I think that a physicist sees for first time in General Relativity).
It is defined as the connection for which the following is true:
D(<m|n>)=<Dm|n>+<m|Dn> where <m|n> is the inner product between states and D is the covariant derivative associated with the connection that we are considering.
I the picture in post 20(Nakahara) you can see that it is equivalently defined as the vanishing of the covariant derivative of the metric that is associated with that inner product.
So, in essence, what I am asking is whether or not the Berry connection is compatible with the metric associated with the inner product between quantum states.
I hope I did not confuse you.
Thank you!

10. Jan 4, 2018

### Demystifier

Riemannian geometry can be defined without language of fiber bundles. But Riemannian geometry is not the most general geometry, and fiber bundles are introduced to describe more general geometries. In the language of fiber bundles, Riemannian geometry is a special case related to tangent bundles. As I said, you don't need to use any bundles at all to talk about Riemannian geometry, but then you cannot talk about non-Riemannian geometries such as geometry of Berry connection.

11. Jan 4, 2018

### Demystifier

Last edited: Jan 4, 2018
12. Jan 4, 2018

### Demystifier

If you look at do Carmo's Riemannian Geometry, page 50, Definition 2.1, you will see that he defines affine connection as a map from $\chi(M)\times\chi(M)$ to $\chi(M)$, where $\chi(M)$ is a set of vector fields. In a physics-friendly language, this means that the affine connection is an object with 3 indices, all of which belong to the same space $\chi(M)$. Indeed, the Christoffel connection $\Gamma^{\mu}_{\alpha\beta}$ has 3 such indices. On the other hand, a Yang-Mills connection $A^{\mu}_{ab}$ has one spacetime index ${\mu}$ and two gauge group indices $a,b$, so it is not an affine connection. The electromagnetic connection $A^{\mu}\equiv A^{\mu}_{11}$ is a special case of Yang-Mills, where the gauge-group indices are trivial because they take only one value 1 as the gauge group $U(1)$ is 1-dimensional. The Berry connection is very much like $A^{\mu}$ because both lack (non-trivial) lower indices. In other words, Berry connection is not an affine connection.

13. Jan 4, 2018

### Joker93

Thank you for all your remarks.
But, if you can check Nakahara on p.397, he defines a metric connection(metric compatible with a metric) as just a metric that preserves the inner product.
He does make a connection with Riemannian geometry through what he defines as Riemannian structure.
So, if I proceed on with this definition that has to do with fibre bundles, then the Berry connection is compatible with the metric that is used to take inner products between quantum states {|n>}, since the following is true:
$$\partial_\mu(<n|m>)=<\nabla_\mu n|m>+<n|\nabla_\mu m>$$ where $$\nabla_\mu=\partial_\mu+iA_\mu$$ where $A_\mu$ is the gauge field(or the Berry connection in this case).
I can't see anywhere in that section of Nakahara that restricts us to have affine connections in order for the definition he gives(for metric connections on fibre bundles) to be valid. Please correct me if I am wrong or I have missed something.

Also, on your post 31, I visited the link but I still can't figure out why we cannot say that the Berry connection is compatible with the the particular metric that has to do with the inner product between quantum states since the above is valid.

14. Jan 5, 2018

### Demystifier

Nakahara defines the Riemannian structure on $E$. What you are missing is to ask yourself - what is $E$ in your case? Is it the quantum Hilbert space, or is it the parameter space associated with Berry connection? The metric you are talking about is a metric on the quantum Hilbert space. By contrast, the Berry connection is a connection on the parameter space. Therefore the metric in (10.73) cannot be interpreted as the metric in the quantum Hilbert space. If you want to talk about metric compatibility of the Berry connection, then you need another metric, a metric on the parameter space.

Last edited: Jan 5, 2018
15. Jan 5, 2018

### Joker93

This really clears things up now. Thanks!
Berry, in the article I attach below, talks about a quantum geometric tensor which measures distances in parameter space. Might it be that this metric has any connection with what we are discussing?

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16. Jan 5, 2018

### Demystifier

This metric and Berry connection live in the same space, which is already a connection (pun alert!) between them. Whether Berry connection is compatible with that metric is at least a meaningful question.

17. Jan 5, 2018

### Joker93

18. Jan 6, 2018

### samalkhaiat

"I prefer to pass".