# Difference between SU(N)xU(1) and U(N)?

1. Sep 15, 2009

### RedX

What's the difference between SU(N)xU(1) and U(N)? They seem to have the same # of generators: indeed, their generators seem to be exactly the same. The Lie algebra of their generators seem to be the same too.

But I'm guessing they're not the same, since we call the Standard Model (before spontaneous symmetry breaking) SU(2)xU(1) and not U(2). But I don't know how they're different mathematically.

2. Sep 15, 2009

### Hurkyl

Staff Emeritus
This sounds like a fun exercise. Let's try to relate them!

There is an obvious map SU(N) --> U(N). The determinant gives an obvious map U(N) --> U(1).

In fact, this seems to be a group extension -- it expresses U(1) as the quotient of U(N) by SU(N).

Now, what about a splitting? It's easy enough to find maps U(1) --> U(N) such that U(1) --> U(N) --> U(1) is the identity homomorphism. (e.g. extend the 1x1 matrix into an NxN by putting ones on the diagonal and zeroes everywhere else)

Oh, but this doesn't work to give an isomorphism SU(N)xU(1) with U(N) -- one thing that fails is that we need to stick U(1) into the center of U(N).

I believe... the center of U(N) is the set of all multiples of the identity matrix such that the multiplier has norm 1?

Well, now we have a problem. The obvious map U(1)-->U(N) that sends [a] to aI isn't a splitting of the determinant: the composite U(1)-->U(N)-->U(1) is the N-th power map.

The following is certainly true: there exists a map
SU(N) x U(1) --> U(N)​
defined by
(M, [a]) --> aM​
which is surjective and N-to-1.

This would imply that su(N)xu(1) = u(N).

I furthermore suspect this is the "closest" you can make these two groups.

3. Sep 15, 2009

### Avodyne

Well my math skills are not good enough to follow Hurkyl, but here's my pedestrian understanding:

An element of U(N) is an NxN unitary matrix $U$.

An element of SU(N) is an NxN unitary matrix with determinant one $u$.

An element of U(1) is a phase factor (a complex number with magnitude one) $e^{i\phi}$.

Given an element of U(N), can we uniquely decompose it into an element of SU(N) and an element of U(1)? That is, given $U$, is there a unique $e^{i\phi}$ and a unique $u$ such that $U=e^{i\phi}u$? The answer is no, because I can multiply $u$ by $e^{2\pi i n/N}$, where $n=0,1,\ldots,N{-}1$, and the new $u$ will still have determinant one, and I can simultaneously multiply $e^{i\phi}$ by $e^{-2\pi i n/N}$ and it will still have magnitude one. So for every element of U(N), there are N elements of SU(N)xU(1).

Last edited: Sep 16, 2009
4. Sep 15, 2009

### RedX

I'll have to think about Hurkyl's explanation a bit more. I don't know much group theory other than what's utterly necessary for a basic book on QFT. Once I'm done studying measure theory I'll study some group theory and get back to you.

I got Avodyne's explanation. So although there are n-possible ways to decompose U(n) into SU(n)xU(1), what if you specify the generators of each? Then isn't there only one way to make the decomposition with a given set of generators? You can't get an SU(n) matrix multiplied by a phase factor just by changing the group angle.

5. Sep 15, 2009

### Ben Niehoff

6. Sep 16, 2009

### vanesch

Staff Emeritus
I think that the Lie algebras su(n) x u(1) and u(n) are equivalent, if that is what you mean.
(a bit like su(2) and so(3) are equivalent). Correct me if I'm wrong.

However, I find the reasoning a bit strange to demonstrate that U(N) is an n-fold cover of SU(N) x U(1).

After all, what obliges us to take the map (V,phi) --> exp(i phi) V from SU(N) x U(1) into U(N) (which is surjective). What's wrong with the map (V,phi) --> exp(i phi/N) V, which, I think, is bijective ? Or what am I missing ?

7. Sep 16, 2009

### George Jones

Staff Emeritus
What happens, for example, when phi = 2*pi?

8. Sep 16, 2009

### RedX

That's one thing that's confused me. SU(2) and SO(3) are isomorphic which is a bit stronger than a homomorphism. To me, I don't see any difference between SU(2) and SO(3). They have the same commutation relations. Therefore, their group elements multiply the same way ( just use the Baker-Campbell-Hausdorff identity - since the commutator is the same, the group elements multiply the same). Yet SU(2) is associated with spin, and SO(3) with orbital angular momentum? Is this ad hoc? Couldn't SO(3) be associated with spin, and SU(2) with orbital angular momentum?

9. Sep 16, 2009

### George Jones

Staff Emeritus
The groups SU(2) and SO(3) are not isomorpic; there is a two-to-one homomorphism from SU(2) to SO(3), and SU(2)/{-1,1} and SO(3) are isomorphic. The Lie groups do, however, have isomorphic Lie Algebras.
Since, the groups SU(2) and SO(3) are different, they have different representations. In fact the representations of SO(3) is isomorphic to a representation of SU(2), but not every (spinor) representation of SU(2) is isomorphic to representation of SO(3).

10. Sep 16, 2009

### vanesch

Staff Emeritus
Mmm, right. My reasoning was that (V, 2 pi) would be mapped onto exp(i (2 pi/n) V, which is ALSO an element of SU(N), but I see now what's wrong: there's a "cut" in the map, and it is not continuous anymore across the phi = 0 and phi = 2 pi - epsilon boundary.
In other words, this map gives a kind of Riemann surface in (SU(N) x U(1)).
And the map I propose is not a homomorphism because it is not continuous in 2 Pi.

11. Sep 16, 2009

### haushofer

I'm not very strong with this kind of things, but maybe I can contribute something :)

I think the important thing here to remember is the following:

We have a Lie-group, which is basically a group which consists of elements connected to the identity in a continuous way. And we have the Lie-algebra of this group. The connection between the two is the exponential map: exponentiate the Lie algebra to obtain the Lie group.

Now, you can proof the following. Suppose we have a Lie group homomorphism between two Lie groups G and H, then this Lie group homomorphism "induces" a unique real linear map between the Lie algebras of H and G.

However, the converse is not necessarily true! So, such a real linear map between the Lie algebras of G and H does NOT necessarily induce a Lie group homomorphism between H and G!

You can show that the converse is only true if G is simply connected. But SO(3) is not simply connected. So the fact that the Lie algebras of SU(2) and SO(3) look the same doesn't proof that the associated Lie groups are "the same". Instead, SU(2) is a double covering of SO(3): SO(3) is isomorphic to SU(2)/{I,-I}.

Maybe this clearifies some things.

12. Sep 16, 2009

### Hurkyl

Staff Emeritus
Oh, I didn't connect these dots until this morning.

Let's assume my claim earlier that the center of U(N) is what I said it was. (Wikipedia confirms, BTW)

Then the center of U(N) contains exactly N distinct N-th roots of the identity matrix. They are elements of SU(N) as well.

U(1), of course, also has N distinct N-th roots of the identity.

Therefore, the center of SU(N)xU(1) must have N² N-th roots of the identity matrix. (pick any N-th root lying in SU(N) and pair it with any N-th root lying in U(1))

So, any surjective homomorphism SU(N)xU(1) --> U(N) must, at best, be N-to-1.

Last edited: Sep 16, 2009