# How curvature and field strength are exactly the same?

I am watching this lecture series by Fredric Schuller:

In this part he discusses the Lie algebra valued one and two forms on the principal bundle that are pulled back to the base manifold.

He shows the relationship between general relativity and electromagnetism in the classical theory.

He emphasize that for instance that the curvature of space-time is exactly the same as the non-abelian Yang-Mills field strength.

It is not clear to me in what sense they are "exactly" the same.

Isn't the physical electromagnetism filed strength tensor a real valued object and abelian?

• Are you familiar with the general notion of the curvature of a connection, taking values in the endomorphism bundle? May 25, 2018 at 9:55
• Yes, I am, the question is that the physical field strength is not Lie algebra valued, it is a real valued tensor.
– user56963
May 25, 2018 at 10:03

Simply, field strength tensor of a non-Abelian Yang-Mills reads as follows: $$F_{\mu\nu} \equiv \left[ \mathcal{D}_\mu, \mathcal{D}_\nu \right]$$ where $\mathcal{D}_\mu$ is the covariant derivative with respect to the potential one-form, $A_\mu \equiv A_\mu^i T_i$, and $T$'s are the generators of the Lie group in subject, while in gravity the Riemann curvature tensor reads as: $$R_{\mu\nu} \equiv \left[ \nabla_\mu, \nabla_\nu \right]$$ where $\nabla_\mu$ is the covariant derivative with respect to the connection, $\omega_\mu \equiv \omega_{\mu} {}^a {}_b \Lambda_a^b$, and $\Lambda$ is the generator of the group associated with the principle bundle (the group is the general linear group, $GL(\mathbb{R}^m)$ in the affine case). By the way here, $R_{\mu \nu} \equiv R_{\mu\nu} {}^a {}_b \Lambda_a^b$ is the Riemann tensor (not to be confused with Ricci tensor).

As a matter of fact, torsion tensor is also the same thing but for the soldering form: $$T_{\mu\nu} \equiv \left[ D_\mu, D_\nu \right]$$ where $D_\mu$ is the covariant derivative with respect to the tetrad (vielbein, soldering) field, $\theta_\mu \equiv \theta_{\mu}^a P_a$, and $P$ is the generator of translation (i.e. four-momentum).

As you can see, they are just the same thing, i.e. the commutator of the covariant derivative, except the fact that they are living with different group. But the same mathematical object.

• So is it right to say that in the case of electromagnetism the Lie group U(1) is different from the case of the general relativity Lie group which is GL? Otherwise, curvature and field strength are exactly the same?
– user56963
May 26, 2018 at 3:15
• Yes, exactly. (Actually, in Einstein's General Relativity, the group is Lorentz group, $SO(1,3)$, but it is just a subgroup of GL(1,3), so it is indeed true what you said). Jul 21, 2018 at 12:27

In electromagnetism, the gauge group is $\text{U}(1)$, whose Lie algebra can be identified with $i\mathbb R$, eg. the set of all imaginary numbers.

The gauge connection $\mathcal A$ is an $i\mathbb R$-valued 1-form, so we can write it as $$\mathcal A=iqA,$$ where $q$ is a coupling constant (charge), and $A$ is now an ordinary 1-form.

Calculating the curvature, we get $$\mathcal F=\mathrm d\mathcal A+\mathcal A\wedge\mathcal A=\mathrm d\mathcal A=iq\mathrm dA=iq F,$$ where $F$ is the usual field strength. As you can see, $F$ is a real-valued 2-form, while $\mathcal F$ is an imaginary, hence $\mathfrak u(1)\simeq i\mathbb R$-valued 2-form.

I would like to point out that the curvature of space-time and the Yang-Mills field strength is not "absolutely same" nontheless. They are objects of a similar type, but the curvature of space-time is the curvature of the Levi-Civita connection on $F(M)$ (or $O(M)$, or $TM$, however you'd like it) and the Yang Mills field strength is proportional to the curvature of a connection on a principal $\text{SU}(N)$ bundle.

• I think you mean "$\mathcal{F}$ is a [...] 2 form" May 25, 2018 at 13:01
• @Uldreth So is it right to say that in the case of electromagnetism the Lie group U(1) is different from the case of the general relativity Lie group which is GL? Otherwise, curvature and field strength are exactly the same?
– user56963
May 25, 2018 at 14:07

I have not watched the video but...

The physical differences in classical reality are as important as the mathematical similarities.

The curvature of space-time is NOT a force field, as in Electromagnetism. The Christoffel symbol, or connection coefficient, creates force on test particles in the classical theory. This has always been an issue and one of the things that lead first attempts at unification to add fields to the metric. Coincidentally the extra connection terms transform as a curvature tensor.

Quantum mechanics is, for better or worse, something we do to a classical system (with the possible exception of non-relativistic spin, but in the Dirac theory and in QFT the field is seen as a classical structure on the manifold). There are more differences to discuss but I think this is the biggest. The very nature of the force is different at a classical (and empirical) level.

As for EM being Abelian that is due to the nature of the gauge symmetry, U(1) is an Abelian group, SU(2), SU(3), etc are non-abelean groups. Just because the curvature is a tensor doesn't mean it isn't Lie Algebra Valued (it is). It carries 2 space-time indices and two lie algebra indices. For U(1) the internal indices are trivial. The group action is on an "internal degree of freedom" like charge, phase, iso-spin, etc. That feature does not get in the way of seeing the similarity. In GR the "internal indices" are space-time directions or vielbien directions in the local tangent space at each point in the manifold.

To unify GR and SM researchers have been trying to capitalize on any similarity they can find (and this is a smart move, it's not made up) but to really accept these one needs to change the way we look at F = ma in GR.

An other difference is that the field action is first order in curvature while gauge theories are second order, again for classical theory first order R --> second order Connection, which is the force field. A higher order GR theory would make things look more alike but produce new issues.

The curvature in GR produces a force gradient dF/dx, which relates to perturbations in orbital motion of test particles.