# I SU(3) as a gauge group

1. Jan 10, 2017

### CAF123

The lagrangian of a non interacting quark is made to be invariant under local SU(3) transformations by introduction of a new field, the gauge field, giving rise to the gluon. This gives us a locally gauge invariant lagrangian for the quark field and together with the construction of a locally gauge invariant kinetic term gives us the classical QCD lagrangian (ignoring the redundancy in the large degeneracy of gauge transformations resulting in additional ghost terms etc)

This gauging of the local symmetry results in the transformation law of the gauge field to be $A_{\mu} \rightarrow U A_{\mu}U^{-1} - i/g (\partial_{\mu} U) U^{-1}$. It is said that the gluon field transforms in the adjoint representation of the gauge group SU(3) which contains all group elements U such that $U = \exp(i \theta^a t^a)$ with $\theta^a$ constant.

My question is,
even though QCD is invariant under local gauge transformations, that is transformations of the form $\exp(i \theta^a(x) t^a)$ with $\theta^a$ now a function of x, we write the gauge group of the SM as $SU(3)_c \otimes SU(2)_L \otimes U(1)_Y$. But SU(3)_c only contains transformations where theta is restricted to be a constant. So why are the local transformations not contained in the gauge group of the SM? Is it because the charge of the gauged SU(3) symmetry is colour and since this is an internal degree of freedom, we restrict to global transformations independent of spacetime?

Thanks!

2. Jan 10, 2017

### MathematicalPhysicist

Have you read Srednicki cover to cover?
I remember some of your notation from his book, even if he doesn't discuss your particular problem, you can check his references.

I can't help more than this, I cannot remember everything I read from it.

3. Jan 10, 2017

### CAF123

I have used Srednicki in the past but I didn't like his presentation of a few things so sought other books. I don't have access to his book at the moment. I'll see if any QFT experts on the forum can provide any insight - certainly in the resources I have been using I couldn't find an answer to my question specifically.

4. Jan 10, 2017

### MathematicalPhysicist

The answer to your questions should appear on chapters 84-86 of Srednicki, or so I think.

Check it out in case no one replies, my 2 cents.

5. Jan 10, 2017

### CAF123

I found his book on the net but couldn't see anything relevant in those chapters - I checked his chapter on QCD too and on group representations but found nothing to help. I've done courses on this stuff before but more thinking on the matter is leading to more questions haha - I'll wait to see if anyone has an idea.

6. Jan 10, 2017

### Dr.AbeNikIanEdL

Could you maybe state again what the actual question is?

7. Jan 10, 2017

### CAF123

Sure. Basically the question stemmed from the transformation law for the gluon field - it's said everywhere that the quarks transform in the fundamental rep of the gauge group SU(3) and the gluons in the adjoint rep of the gauge group SU(3). The latter statement is only clear for global transformations, i.e make U independent of x then the $\partial_{\mu} U$ term disappears and we have the usual $A_{\mu} \rightarrow UA_{\mu}U^{-1}$ transformation law for objects transforming in the adjoint rep. The gauge group is also called $SU(3)_c$ because the degree of freedom that is mixed under these transformations is colour. This gauge group is what appears in the SM gauge group. But why? The full symmetry of QCD contains local transformations too (by construction) so why is this not inherent in the SM gauge group?

it feels like a misnomer to call $SU(3)_c$ a gauge group because it is restricted to global transformations and we gauged a local symmetry.

8. Jan 10, 2017

### Orodruin

Staff Emeritus
SU(3) is not restricted to global symmetries. If it was there would be no gluons. You would just have a global symmetry and say "so what?"

9. Jan 10, 2017

### CAF123

Ok, so why is it then said that the gluons transform under the adjoint representation of SU(3)? Under local SU(3) transformations (ie the gauged symmetry), $A_{\mu} \rightarrow UA_{\mu}U^{-1} - i/g (\partial_{\mu}U)U^{-1}$ which is not the transformation law for something transforming in the adjoint representation.

Thanks!

10. Jan 10, 2017

### Orodruin

Staff Emeritus
The field tensor (which contains the physical degrees of freedom) transforms under the adjoint representation. It is not clear what you really mean by saying "gluons transform".

11. Jan 10, 2017

### CAF123

I also made a discussion here

The physical degrees of freedom are just the $A_{\mu}^a$ no? In that post, I do a small computation showing $A_{\mu}^a t^a$ transforms as $U A_{\mu}^a t^a U^{-1}$ which infinitesimally looks like $A_{\mu}^c \rightarrow A_{\mu}^c + \theta^a f^{acb}A_{\mu}^b$ which is just the transformation law for the components of a 8x1 real vector (c.f infinitesimal cross product rule in 3D Euclidean with $\epsilon_{ijk}$) It's just the reconcilation of the result of my computation and the transformation law $A_{\mu} \rightarrow U A_{\mu}U^{-1} - i/g (\partial_{\mu}U)U^{-1}$ which I'm confused about.

Thanks :)

12. Jan 10, 2017

### Orodruin

Staff Emeritus
No. There are unphysical degrees of freedom that you need to fix using gauge conditions. Just like you need to fix a gauge in normal electrodynamics. If $A^\mu$ was completely physical you would not need to bother with gauge fixings and ghosts.

13. Jan 10, 2017

### samalkhaiat

You are dealing with two different groups here. The global compact group $G$, and the infinite-dimensional local gauge group $G(M)$ ($M$ being a contractible space-time manifold). With respect to $G$, the gauge field behaves like a genuine tensor transforming by the adjoint representation: $$\mathbb{A}^{g}_{\mu}(x) = g^{-1}\mathbb{A}_{\mu}(x)g ,$$ where $\mathbb{A}_{\mu} (x) = A^{a}_{\mu}(x) T_{a}$. Locally, i.e., with respect to $G(M)$, and exactly because of the inhomogeneous (second) term in its transformation law: $$\mathbb{A}^{g}_{\mu}(x) = g^{-1}(x) (\mathbb{A}_{\mu}(x) + \partial_{\mu}) g(x) ,$$ the gauge field is, therefore, interpreted as a $\mathcal{L}(G)$-valued connection, $\Gamma(x) = (A^{a}_{\mu}(x) dx^{\mu})T_{a}$, on the trivial principal bundle $\pi (G,M)$.

We have similar story in GR. The Christoffel symbol $$\Gamma (x) = \left( \Gamma^{\rho}_{\mu\nu}(x) dx^{\mu} \right) T^{\nu}{}_{\rho} ,$$$$[T^{\mu}{}_{\nu} , T^{\rho}{}_{\sigma}] = \delta^{\rho}_{\nu} T^{\mu}{}_{\sigma} - \delta^{\mu}_{\sigma} T^{\rho}{}_{\nu} ,$$ is interpreted as a $gl(n,R)$-valued connection on $\mathscr{F}_{M}\left(GL(n,R),M\right)$, the frame bundle of $M$. The $T^{\mu}{}_{\nu}$ is the natural basis of the Lie algebra $gl(n,R) \sim T_{e}\left(G(n,R) \right)$. Now let $g$ be the transition function mappings $g \ : \ U \cap \bar{U} \to GL(n,R)$, where $U, \bar{U} \subset M$ such that $U \cap \bar{U}$ is not the empty set . Then you have $$\bar{\Gamma}(x) = g^{-1}(x) \left( \Gamma (x) + d \right) g(x) .$$ Again this transformation law consists of two part, the first one is tensorial, but the second one is not.

14. Jan 11, 2017

### CAF123

I see, so just to check, it is the former that is called SU(3) and since the global transformations therein mix the colour charge of QCD, it is denoted as SU(3)c? The latter are then transformations in the image of the map $\mathbb{R}^4 \simeq \mathcal{M} \rightarrow \mathrm{SU}(3), x \mapsto \exp(i \theta^a(x) t^a)$, where $\mathcal M$ denotes Minkowski space. (I.e all local SU(3) transformations are in the latter group and the global transformations in the former)...That all fine?

I didn't understand the abstract mathematical terminology in terms of bundles but would be interested to read more about it all from this abstract setting - certainly I have never come across a treatment done in this way in any standard QFT text - could you refer me to a reference?

Why is it then that in the SM gauge group it is $\mathrm{SU}(3)_c$ that is appearing and not this infinite dimensional local gauge group? I thought (as pointed out in OP) that it maybe due to colour being an internal quantum number so transforms under transformations independent of spacetime but maybe there is another argument.

I see. So all of this is the underlying reason why we say the Christoffel symbols don't transform like tensors?

Thanks!

15. Jan 11, 2017

### samalkhaiat

To make that statement correct, you must add the phrase “up to inhomogeneous term”. Or, as I explained before, another correct statement is “it transforms in the adjoint representation of the global group $SU(3)$”. But the absolutely correct statement is of course “under the local (gauge) group $SU(3)$ the gluon field transforms like a connection or compensating field”
By the way, when we write $SU(3)_{c}$, we mean that this (local and/or global) group acts on the colour space of each quark-flavour, for example, $d = (d_{1}, d_{2} , d_{3})^{t}$. And we do that just to distinguish it from the flavour group $SU(3)_{f}$ which acts on the flavour space of quarks $q = (u,d,s)^{t}$.

No, we don’t. It is meaningless to use the tensor product sign $\otimes$ between groups.

Who said that? The gauge group of the SM is $$G_{SM} = SU(3)_{c} \times \left( SU(2) \times U(1) \right)_{ew} ,$$ and all factor groups in $G_{SM}$ are understood to be local gauge groups.

One can not give an answer to a wrong question: In QCD, the group $SU(3)_{c}$ is the local gauge group.

Answering a wrong question, almost always, leads to gibberish: What do you mean by “the charge of the gauged SU(3) symmetry is colour”? In field theories, the symmetry charges are the observables of the theory. So, don’t confuse the symmetry charges with the internal index space of the fields.

16. Jan 13, 2017

### CAF123

1) So, as Orodruin already mentioned, the field tensor transforms under an adjoint representation too - is this now the adjoint representation of the local gauge group rather than the gluon which is the adjoint representation for the global gauge group (= subgroup of local gauge group with theta restricted to be constant)

2) In my small computation I mentioned in the other thread which I linked from this thread (given again below), I derived the adjoint transformation law $A_{\mu} \rightarrow U A_{\mu}U^{-1}$ - where did I use the fact that I was working in the global gauge group in this derivation?
$(U t_a U^{-1})_{ij} = D_{ab}(t_b)_{ij}$. The U's and the t_a can be in any representation carrying the indices i,j but I don't see where I am using the fact I'm restricting to the global gauge group to obtain the adjoint rep transformation law.

Thanks!

17. Jan 14, 2017

### samalkhaiat

Again, you are asking a wrong question. The word representation always refers to the global (i.e., finite-dimensional) groups, i.e., not to the local (i.e., infinite-dimensional) groups. This is because we (mathematicians) know very little about the representation theory of infinite-dimensional groups on space-time of dimension >2.

The equation $$\mbox{Ad}_{g} X_{a} \equiv g X_{a} g^{-1} = D_{a}{}^{b}(g) X_{b} ,$$ defines the finite-dimensional (matrix) representation, $D(gh) = D(g)D(h)$, of the finite-dimensional (global) group.

18. Jan 21, 2017

### CAF123

Ok thanks! It's clear now - I wanted to ask a few final things about the mappings $\text{ad}$ and $\rho$ that are usually used in the literature:

The adjoint map at the level of the lie group is such that $$\text{ad}: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak g),$$ taking $\lambda_a \mapsto \text{ad}_{\lambda_a}$ where $$\text{ad}_{\lambda_a}: \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\,\text{with}\,\,\, \lambda_b \mapsto [\lambda_a, \lambda_b].$$

So, this means that $$\lambda_b \rightarrow \text{ad}_{\lambda_a} \lambda_b = [\lambda_a, \lambda_b] = if_{abc}\lambda_c.$$

Equivalently, $\rho: \mathfrak{g} \rightarrow \mathfrak{gl}(\mathfrak{g})$ is also defined as the map such that $\rho(\lambda_a)_{bc}\lambda_c = [\lambda_a, \lambda_b] = if_{abc}\lambda_c$. I'm just trying to get all the indices to make sense so what is the relationship between $\text{ad}$ and the $\rho?$

Thanks!

19. Jan 21, 2017

### samalkhaiat

For matrix Lie groups, they are the same thing. Both define the adjoint representation of (matrix) Lie algebra (or Lie algebra homomorphism) on $gl(\mathscr{L})$, the Lie algebra of the general linear group over the “vector space” $\mathscr{L}(G)$.

I believe, I went through this with you before. The representation of the group $G$ on its Lie algebra $\mathscr{L}(G)$, i.e., the pair $\left(\mbox{Ad} , \mathscr{L}(G) \right)$ is obtained by differentiating the conjugation map (automorphism) $\varphi_{g}(h) = ghg^{-1}$ at the identity, $$\mbox{Ad}_{g} = \mbox{d} (\varphi_{g})|_{e} : \ \mathscr{L}(G) \to \mathscr{L}(G) ,$$ So, by writing $h = e^{tX}$ for $X \in \mathscr{L}(G)$, you get $$\mbox{Ad}_{g}(X) = g \ \left( \frac{d}{dt} e^{tX} \right)_{t = 0} \ g^{-1} = g \ X \ g^{-1} .$$ Similarly, $\mbox{ad}_{X}$ is obtained by differentiating $\mbox{Ad}_{g}$ at the identity: $$\mbox{ad}_{X} = \mbox{d} (\mbox{Ad}_{g})|_{e} : \ \mathscr{L}(G) \to \mathscr{L}(G) .$$ Then, for $X,Y \in \mathscr{L}(G)$, write $g=e^{tX}$ and get $$\mbox{ad}_{X} (Y) = \frac{d}{dt}\left( \mbox{Ad}_{(e^{tX})} (Y)\right)_{t = 0} = \frac{d}{dt}\left( e^{tX}Ye^{-tX}\right)_{t=0} = [X , Y] .$$

20. Jan 22, 2017

### CAF123

Thanks! Indeed it was mentioned before but I want to check the indices on the $\rho$. E.g for the mapping $\text{ad}_X(Y) = [X,Y],$ if we let $X \rightarrow \lambda^a, Y \rightarrow \lambda^b$ then $\text{ad}_{\lambda^a}(\lambda^b) = [\lambda^a, \lambda^b]$. Or we can formulate this as $$\text{ad}_X: \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\text{with}\,\,\,\,Y \mapsto [X,Y].$$

Now what are the equivalent statements for $\rho$? $\rho$ was defined as $\rho (\lambda_a)_{bc}(\lambda_c) = [\lambda_a,\lambda_b]$ but here in this case, in the notation of X and Y used above, the X would be $\lambda_a$ and the Y would be $\lambda_c$ yes? So this seems to suggest I would write something like $$\rho_{(\lambda_a)_{bc}} : \mathfrak{g} \rightarrow \mathfrak{g}\,\,\,\,\,\text{with}\,\,\,\, \lambda_c \mapsto [\lambda_a, \lambda_b]$$

So in one case the index $b$ is used to specify an entry in the matrix representation $\rho$ (as in $\rho(\lambda_a)_{\mathbf{b}c}$) and in the other case (first case with the map $\text{ad}$) it is an index on the $\lambda^b$. Is this the right way to think about it?

Thanks again!

EDIT: Well actually, I've also seen the equation $\rho(\lambda_a)(\lambda_b)= [\lambda_a, \lambda_b]$ used so I think all the equation $\rho(\lambda_a)_{bc} \lambda_c = [\lambda_a, \lambda_b]$ is saying is that the basis vectors {\lambda} of the lie algebra mix between each other under the matrix representation of the adjoint action.

Last edited: Jan 22, 2017