# I Help me understand gauge theory?

1. Apr 20, 2017

### quasar987

Hi everyone,

So I recently read a chapter in a math book that vaguely describe how connections on bundles occur in particle physics, but they are very cryptic about the physics part and I just want to know a little bit more about it. So I'll tell you what I read and then follow up with some questions I have.

The way the text is structured, they start by making the observation that by defining a real-valued differential 2-form $F$ on R^4 in a certain way in terms of the components of the E and B fields, Maxwell's equations take the elegant form $dF=d*F=0$. Then they remark that if one writes $F=dA$ by virtue of Poincaré's lemma, then the "potential" $A$ is defined up to the addition of an exact 1-form $df$ which we may write $d(ln(g))=g^{-1}dg$ for $g=e^f$. In particular, by commutativity, the "gauge transformation" $A\mapsto A+df$ can be written $A\mapsto g^{-1}Ag+g^{-1}dg$ which we recognize as the way the connection 1-forms of a connection on a line bundle transform under a change of local trivialisation with transition function $g:\mathbb{R}^4\rightarrow \mathbb{K}^*$. As to $F$, it can be written $F=dA+A\wedge A$ since the exterior product of real-valued 1-forms vanish. As such, we recognize $F$ as the curvature 2-form of the connection.

Here, the line bundle could be real, or it could be complex with structure group $U(1)$. In this case, we admit that $A$ and $F$ are really $i\mathbb{R}$-valued.

In the more general context of a connection on a vector bundle with (matrix) structure group $G$ on a space-time 4-manifold $M$, the Maxwell equation $dF=0$ becomes $dF = F\wedge A - A\wedge F$, or, regarding $F$ as the curvature form of the corresponding principal $G$-bundle, $DF=0$, where $D$ stands for exterior covariant differentiation. The other Maxwell equation $d*F=0$ generalizes to $D*F=0$ and this is called the Yang-Mills equation.

I also read on wikipedia that in the standard model, the structure group $G$ is $U(1)\times SU(2)\times SU(3)$ where each one of these 3 groups is a symetry group for some "internal structure" that a particle may have.

K so what is the internal structure that is related to U(1) exactly?

How do particles enter the picture? Are particles assumed to be represented by a wave function $\psi:M\rightarrow\mathbb{C}$ as in QM? Is U(1) related to the fact that the probability density $|\psi|^2$ is invariant under phase transformation $\psi\mapsto e^{i\phi}\psi$ ?? But then what is the relationship with the EM field?!

PS I only have an undergraduate background in physics.

2. Apr 20, 2017

### dextercioby

Which is the book? This is a technical subject not easily found. I also look for a clear account of fiber bundle formulation of classical field theory.

Last edited: Apr 20, 2017
3. Apr 20, 2017

### quasar987

Postnikov's Leçons de géométrie différentielle. I am not sure the text exists other than in russian and french.

4. Apr 21, 2017

### RockyMarciano

U(1) may refer to global phase or to local phase, but in this context of "internal structure" in physics it usually refers to the local phase and it is the structure group that corresponds to the fiber in the principal bundle that has $A$ as principal connection.
When switching to QFT things are more involved, there is a step from wavefunctions to operators.

Yes, but note that $\phi$ is not constant here but dependent on spacetime position, this is what makes the phase local, just as the wave function is invariant under phase changes independently of spacetime point, the connection $A$(substitute the scalar and vector potentials in the case of the Schrodinger equation in the presence of an EM field) must also be invariant to the change of local phase, thus called gauge connection.