# General question about gauge invariance of Lagrangian. Illustration with $U(1)$

I am learning gauge theories and I am a little confused with things.

I would like to take the simplest example : the $U(1)$ invariance to check if I understood the basics.

When we work with electromagnetism, we know that we have a gauge invariance :

$$A_\mu \rightarrow A_\mu + \partial_\mu \alpha$$

The E.O.M are unchanged if we change like this the potential vector.

So, what we want is a Lagrangian invariant under such transformation. In practice, the E.M lagrangian :

$$\mathcal{L}_{EM}=F_{\mu \nu}F^{\mu \nu}$$

is invariant under thoose gauge transformations.

Now, we want to couple this field to the matter, so the most general Lagrangian we write is :

$$\mathcal{L}=F_{\mu \nu}F^{\mu \nu}+\partial_\mu \phi^* \partial^\mu \phi + \mathcal{L}_{int}(\phi, A)$$

As the electromagnetism is gauge invariant, the total Lagrangian must also follow this symmetry. Thus $\mathcal{L}_{int}(\phi, A)$ must be gauge invariant.

Finally, the simplest Lagrangian that follows the gauge invariance is :

$$\mathcal{L}=F_{\mu \nu}F^{\mu \nu}+D_\mu \phi^* D^\mu \phi$$

Where $D^{(A)}_{\mu}=\partial_\mu + ig A_\mu$.

And we know that under a group transformation $g \in U(1)$, we have by construction :

$$D^{(A')}_{\mu} (g \phi) = g D^{(A)}_{\mu}(\phi)$$.

First question : Link between $U(1)$ symmetry and vector potential.

If we say that we have $U(1)$ symmetry here, it is because to change the vector potential $A_\mu \rightarrow A_\mu + \partial_\mu \alpha$ is equivalent to act on $\phi$ with an $U(1)$ element : $e^{-ig \alpha(x)}$.

So finally, to talk about $U(1)$ invariance of this Lagrangian, or gauge invariance of the potential vector $A_\mu$ is just the same thing said in two different ways.

Also, it is by construction a local gauge symmetry because $\alpha$ can of course depend on space.

Am I right ?

Second question :

If I am right with what is written above, to try to write an $U(1)$ gauge invariant Lagrangian makes sense because it is linked to the gauge symmetry of electromagnetism (that we know well with maxwell equations).

But in gauge theories we also look for $SU(N)$ symmetries. How do we know from a physical point of view that it is the "good" group of symmetries ? For $U(1)$ it can be seen (if I am right with what I wrote) because of the gauge invariance of electromagnetism that we know from maxwell equation. But for QCD for example what is the intuition behind knowing it is the $SU(3)$ group of symmetry ?

Did theoretician just got their inspiration for electromagnetism and wondered what would happen if we write lagrangian following more general symmetries. They tried with $SU(2)$ and $SU(3)$ and saw that it indeed fitted to some experiments ? (I'm almost sure it wasn't like that but this paragraph is to try to explain better my second question).

• You might want to look up the calculation of overall color factor in QCD. There are certain experimentally-observable variables that allow us to tell the difference between $SU(3)$ and $SU(N)$ for some other $N$. As far as why the special unitary group is special on its own, I imagine that has to do with the fact that observables have to be unitary anyway. – probably_someone Dec 11 '17 at 18:01
• Chapter 7 of damtp.cam.ac.uk/user/examples/3P2Lb.pdf gives a perspective on Q1, the introduction + 6.5 + beginnings of chapters such as ch. 6 helps with Q2 – bolbteppa Dec 11 '17 at 18:14
• Just a note to say that the description you're using reflects taking a local section of a principal bundle with a U(1) gauge group; if you're interested in seeing how this works it is described in the last couple of chapters of Ishams Modern Differential Geometry for Physicsts. – Mozibur Ullah Dec 11 '17 at 19:06
• G is acting on different objects: on A as additive term and on phi as multiplication. This difference reflected fact that one is group U(1) while second is Lie algebra of this group. That's why one is represented by multiplication and another by addition. Other groups, as SU(3) etc are just theoretical generalisations of this concept, and gives famous gauge theories. From that zoo we choose this option which looks like promising to describe reality, by checking in experiment... – kakaz Dec 11 '17 at 19:09

I think there may be some confusion, or at least things may be made more precise. You say that,

So finally, to talk about $U(1)$ invariance of this Lagrangian, or gauge invariance of the potential vector $A_\mu$ is just the same thing said in two different ways.

The quantity of interest which we are checking for a symmetry (up to a total derivative) is the Lagrangian, such that the action is truly invariant. The Lagrangian itself depends on the potential $A_\mu$, so naturally if we say the Lagrangian has a $U(1)$ invariance, if we are to transform $\mathcal L$ then we have to define some way the group acts, which in this instance is on $A_\mu$.

It is totally analogous to for example, the $SU(2)$ symmetry of a function, $f(\mathbf x)$. The natural definition of symmetry would be to say that, $f(g^{-1}\mathbf x) = f(\mathbf x)$ for some $g \in SU(2)$, for example.

Also, it is by construction a local gauge symmetry because $\alpha$ can of course depend on space.

It is redundant to say, local gauge symmetry. By definition, if a transformation is local, i.e. depends on space-time coordinates, then it is a gauge transformation.

To say local gauge symmetry is like saying gauge gauge symmmetry. So, if a symmetry is global, it is simply a global symmetry and if it is local, it is a local symmetry or equivalently gauge symmetry.

Remember also that despite the name, a gauge symmetry should be thought of as a redundancy in our formulation of a theory, contrary to a global symmetry which leads to conservation laws.

Your second question should be posted separately, thought off the top of my head I believe there are related or duplicate questions to it.