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

+ 4 like - 0 dislike
175 views

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
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.

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.
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.
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.