# Clarification: Why the gauge symmetry of pure Yang-Mills is $PU(n)$ and not $SU(n)$? [closed]

I am quoting the following from the Wikipedia article on the projective unitary group:

In the pure Yang–Mills $SU(n)$ gauge theory, which is a gauge theory with only gluons and no fundamental matter, all fields transform in the adjoint of the gauge group $SU(n)$. The $Z/n$ center of $SU(n)$ commutes, being in the center, with $SU(n)$-valued fields and so the adjoint action of the center is trivial. Therefore the gauge symmetry is the quotient of $SU(n)$ by $Z/n$, which is $PU(n)$ and it acts on fields using the adjoint action described above.

In this context, the distinction between $SU(n)$ and $PU(n)$ has an important physical consequence. $SU(n)$ is simply connected, but the fundamental group of $PU(n)$ is $Z/n$, the cyclic group of order $n$. Therefore a $PU(n)$ gauge theory with adjoint scalars will have nontrivial codimension 2 vortices in which the expectation values of the scalars wind around $PU(n)$'s nontrivial cycle as one encircles the vortex. These vortices, therefore, also have charges in $Z/n$, which implies that they attract each other and when $n$ come into contact they annihilate. An example of such a vortex is the Douglas–Shenker string in $SU(n)$ Seiberg–Witten gauge theories.

1. What is the center of $SU(n)$?
2. What does it mean the adjoint action to be trivial? Which action are they talking about?
3. I struggle to understand why the resulting gauge symmetry is $PU(n)$.
4. What are the "codimension 2 vertices" of the scalars?
5. Does this apply for pure $SU(3)$ QCD?
6. References?

## closed as too broad by Danu, user10851, John Rennie, ACuriousMind♦, JamalSMar 22 '15 at 16:54

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• You would probably do well to split up this question into multiple smaller ones. It seems awfully broad as of right now. – Danu Mar 21 '15 at 22:54
• Why is it broad? I specifically mention where I am confused. – Marion Mar 21 '15 at 23:01
• The fact that there are 6 subquestions seemed indicative to me, but maybe I'm wrong. Someone with a better understanding than me may be able to properly answer it! Please don't take the vote-to-close personally, by the way :) – Danu Mar 21 '15 at 23:03
• Did you vote to close? You prefer it closed than answered? – Marion Mar 22 '15 at 0:24
• 1. and 2. are trivial math questions, 3. is answered in your quote, 4. is actually a physics question, and for 5. you have to ask yourself the question if the center acts trivially on all objects in QCD. I agree that this is too broad, and would advise you to actually only ask 4. – ACuriousMind Mar 22 '15 at 13:50

$SU(N)$ is the $N$-fold cover of $PSU(N)$. They share the same Lie algebra, so the Yang-Mills action would look identical locally. The center of $SU(N)$ is just $Z_N$. At the level of representations, the fundamental representation of $SU(N)$ is a projective representation of $PU(N)$, and only the adjoint ones are linear representations of $PU(N)$.
If the matter fields all transform in the adjoint representation, then it makes sense to say that the gauge group is actually $PU(N)$. A simple explanation is that by taking tensor product of adjoint representations you never get the fundamental ones, so the Hilbert space is restricted.
Because $PSU(N)=SU(N)/Z_N$, the global topology of $PU(N)$ is nontrivial. For example, the fundamental group $\pi_1(PU(N))=Z_N$, so there are nontrivial "vortex lines" in the scalar matter field, around which you pick up a holonomy in the center $Z_N$. These topological excitations themselves are one-dimensional objects, and have "codimension" 2.
Quarks in $SU(3)$ QCD transform as the fundamental representation.