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

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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?
This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion
You would probably do well to split up this question into multiple smaller ones. It seems awfully broad as of right now.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Danu
Why is it broad? I specifically mention where I am confused.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion
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 :)

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Danu
Did you vote to close? You prefer it closed than answered?

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion
Like I explained above, I think that the question, in its current form, is a bit too broad. This does not mean that I don't want to see it answered.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Danu
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.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user ACuriousMind
I am not sure why $Z_n$ is the center of $SU(n)$ since, for example a $U(1)$ within $SU(n)$ should commute with all elements of $SU(n)$. Additionally, I am not sure how a $Z_n$ transformation acts on QCD fields. A $SU(3)$ transformation is just some matrix acting on quarks or gluons. But $Z_n$?

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion

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$SU(N)$ 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.

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Meng Cheng
answered Mar 22, 2015 by (550 points)
Hi and thanks. Can you provide me with some reference with the context you provide above (i.e. with focus on gauge theories)?

This post imported from StackExchange Physics at 2015-03-23 13:03 (UTC), posted by SE-user Marion

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