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  What is the difference between the groups $PSU(N)$ and $SU(N)$?

+ 4 like - 0 dislike

What is the difference between the groups $PSU(N)$ and $SU(N)$? For example how is $PSU(2,2|4)$ different than $SU(2,2|4)$?

This post imported from StackExchange Physics at 2015-02-03 12:59 (UTC), posted by SE-user Marion
asked Feb 2, 2015 in Theoretical Physics by Marion Edualdo (250 points) [ no revision ]
retagged Feb 3, 2015
What do you mean "what's the difference?"? By definition, the projective (special) unitary group is the [right] quotient of the (special) unitary group by its center, so the difference is rather obvious.

This post imported from StackExchange Physics at 2015-02-03 12:59 (UTC), posted by SE-user ACuriousMind
Yes, could you give me more details about this or a specific reference??? To me what you said is not obvious at all. Thanks a lot!

This post imported from StackExchange Physics at 2015-02-03 12:59 (UTC), posted by SE-user Marion
The Wikipedia article covers the most important facts, in particular, that $\mathrm{SU}(N)$ is an $N$-fold cover of its projective version, and that the projective versions of the unitary and special unitary group agree.

This post imported from StackExchange Physics at 2015-02-03 12:59 (UTC), posted by SE-user ACuriousMind
Thanks. But can you provide some physical intuition? People understand $SU(N)$ because of the familiarity with QCD, but not the case of $PSU(N)$.

This post imported from StackExchange Physics at 2015-02-03 12:59 (UTC), posted by SE-user Marion
You can understand the difference already for $N=2$, namely $SU(2)$ v.s. $SO(3)$. $SU(2)$ is simply connected and is the 2-fold cover of $SO(3)$. They look the same locally (i.e. the same Lie algebra), but have different global topology.

1 Answer

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Simply put $PSU(N,N| \mathcal{N} )=U(N,N)\times U(\mathcal{N})/U(1)$.  More generally a superalgbera of the form $SU(A|B)$ has a bosonic sub-super-algebra $SU(A) \times SU(B) \times U(1)$. The $U(1)$ phase decouples from the rest of the subalgebra for the case of $A=B$, e.g. and this is denoted by putting the $P$ letter in front of $SU(2,2|4)$ by which we denote the projective group. Thus, the correct way to say what the full global superalgebra of the $\mathcal{N}=4$ theory is the $PSU(2,2|4)$. I think a good reference is Beisert's review and various stringy textbooks like Polchinski, BBS, etc. 

answered Feb 3, 2015 by conformal_gk (3,625 points) [ revision history ]
edited Feb 4, 2015 by conformal_gk
It is not completely correct to write $SU(N)/U(1)$ because there is no $U(1)$ in $SU(N)$. The $U(1)$ you are referring to is probably the diagonal $U(1)$ in $U(N)$ and so the correct thing to do is to take the quotient of $SU(N)$ by the intersection of this $U(1)$ with $SU(n)$, i.e. $PSU(n)=SU(n)/(\mathbb{Z}/n\mathbb{Z})$ where $\mathbb{Z}/n\mathbb{Z}$ is the cyclic subgroup of $n$-roots of unity.
You are right, I wanted to write something else. I will correct it now. Let me know if I have written it correctly now (and maybe you could also provide some nice references if you have any).

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