# Meaning of X as in SU(3) X SU(2) X U(1)

1. Jul 20, 2011

### arlesterc

I am wondering what the meaning of X is in formulations such as SU(3) X SU(2) X U(1). The symbol is used a lot but I've never seen it explained. I'm assuming it's not any kind of multiplication but ... Clarification would be appreciated.

2. Jul 20, 2011

### George Jones

Staff Emeritus
3. Jul 20, 2011

### vanhees71

It means not only the Cartesian product of the sets but also the direct product of groups. If $G_1$ and $G_2$ are two groups, then one can build another group by considering ordered pairs $(g_1,g_2) \in G_1 \times G_2$ (here, it's the Cartesian product of sets!) with the group product defined by

$$(g_{11},g_{21}) \cdot (g_{12},g_{22})=(g_{11} \cdot g_{12},g_{21}\cdot g_{22}).$$

It is very easy to prove that this definition indeed leads to a group. Together with this group operation the Cartesian product becomes the direct product of groups, i.e., in addition to being just the set of ordered pairs of group elements it also has an algebraic structure making it another group.

4. Jul 20, 2011

### tom.stoer

Think about a "vector" or a "group of numbers" with indices i and a; no index for U(1). You can write this as fai; now let the SU(3) matrices act on a which means a=1,2,3 (the 3*3 matrices have two indices a,b) and let the SU(2) matrices act on i which means i=1,2 (the 2*2 matrices have two indices i,k).

The product of groups means that one group is blind for the indices of the other group and vice versa.

You find something like that quite frequntly, e.g. a quark field has three indices from SU(3), SU(2) and from the Lorentz group (a spinor index with 1,2,3,4); all these groups do commute with each other.

5. Jul 20, 2011

### arlesterc

Thanks so much for the quick and informative responses.

If I am understanding correctly, SU(3) is a series of 3*3 matrices, SU(2) is a series of 2*2 matrices and U(1) is a single number. SU(3) X SU(2) X U(1) is therefore some kind of set made up of all the possible combinations of these - X is the 'combining/grouping/cartesian product' operation.

My problem was understanding how how a 3*3 matrix gets 'combined' with a 2*2 matrix. While I now get the idea that it's possible, an example might help how one particular 3*3 gets combined with a 2*2.

Example

z
a b c
d e f
g h i

y
j k
l m

z X y= ?????

Actual unitary matrices example would be most helpful.

I definitely appreciate your time and and any further time on this and that this forum is here. This is the kind of thing that is well understood among physicists but for 'interested in physics' types like myself can, despite endless googling, sit on the brain for far too long without a resource like this forum.

Finally if someone can point me to a good Webcast of the 'Eightfold Way' theory - an actual lecture as opposed to a 'wonderful world of physics' type show - I would appreciate it as well. As far as my level of understanding I have managed to get through - with some heavy lifting and a lot of 'brow knitting'- all of Leonard Susskind's lectures with about 75% comprehension so I don't mind 'might be a bit difficult/includes higher math' - a university 'first level' introduction would be the right fit.

6. Jul 20, 2011

### tom.stoer

It's rather simple. Let's assume we have two such matrices Aab and Bik and a vector w.r.t both groups, i.e. fbk. Then the for the two matrices acting on f, i.e.

A * B * f

one finds the indx notation

f'ai = Aab Bik fbk

7. Jul 21, 2011

### arlesterc

In your example, are the matrices A and B of the same order e.g. 3 * 3 or are they of different orders? I don't have any problem understanding the multiplication of a 3 * 3 with a 3 *3 - it's a 3*3 with a 2*2 that I don't quite understand. As well, a worked example of one case would make it easier to understand what the X (cartesian product) actually does.

e.g.
1 1 1 2 2
1 1 1 X 2 2 = ?????
1 1 1

I also see another possibility for the meaning of X from the previous explanations where let us say that there are:
a (running from 1 ... g) 3*3 matrices e.g. 20 3*3 matrics
b (running from 1 ...h) 2*2 matrices e.g. 15 2*2 matrices
and then a X b would mean combinations such as the a1 3*3 matrix and the b1 2*2 matrix, or the a1 3*3 matrix and the b2 2*2 matrix ... basically the pairing of all possible 3*3 with all possible 2*2 matrices each such pairing being a member of the set a X b.

8. Jul 21, 2011

### tom.stoer

For SU(2) * SU(3) or anything like that you don't multiply the matrices!

Think about my f as a collection of numbers faip... with indices a,i,p,... and a=1...A, i=1...I, p=1...P which could be visualized in several dimensions. For aip f is a three-dimensional object with A*I*P entries.

If you want to calculate the action of a matrix M, e.g. Mba acting on the index a you construct A-tupels as follows: chose rows (columns, ...) in this multi-dimensional f.

set i=1, p=1 and let a=1...A; this is you first A-tupel fa11
set i=1, p=2 and let a=1...A; this is you second A-tupel fa12
...
set i=I, p=P and let a=1...A; this is you A-tupel faIP

Now for each such A-tupel (with indices i,p fixed) you calculate the product "matrix M times A-tupel" which means calculate

Mba fa11
Mba fa12
...
Mba faIP
(sum over a=1...A)

Then you put all these new A-tupels back into your multi-dimensional f.

So by adding another factor to SU(2) * SU(3) you add another dimension to f and you get another index.

9. Jul 23, 2011

### naima

We have here proper subgroups: SU(2) is a subgroup of SU(3).
is this related to the fact that quarks cans exhange W and Z
unlike elctrons with gluons?

10. Jul 23, 2011

### tom.stoer

The SU(2) for the weak force and the SU(3) for the strong force have nothing to do with each other; therefore this SU(2)weak is not the subgroup of the SU(3)strong b/c they act on different spaces = on different indices as explained above.

You can see this by considering QCD with only two colors i.e. SU(2)strong. That does not mean that the weak and the strong force become identical; they would still be different, unrelated forces.

11. Jul 23, 2011

### naima

Hi Tom
I agree with you.

In fact I'd like to find a mathematical structure (and a sub one) describing that the bosons
of the weak force act on a greater number of particle than those of SU(3)C

12. Jul 23, 2011

### tom.stoer

What do you mean by "on a greater number".

The bosons of the weak force act on ALL fermions b/c all fermions carry a weak index i=1,2; some of them carry in addition a strong index a=1..3. So strictly speaking what you want to achive has already been achieve.

If you look for a unification of strong and weak force then you need a structure which unifies SU(2) and SU(3) i.e. contains both of them as subgroups and allowes for a spontaneous symmetry breaking seperating them. This can be achieved via SU(5), SO(10), E(6) and other (larger) groups. They introduce additional gauge bosons which become massive due to a new Higgs mechanism.

You can calculate the gauge bosons für SU(N) in the following way:

# (gauge bosons) = dim (adjoin rep) SU(N) = N²-1

for SU(2) you get # =3 (which agrees with W+, W-, Z0)
for SU(3) you get # = 8 which agrees with the eight gluons.

So in SU(5) you get # = 24 = 3 +8 + 13

I guess one of 13 is the photon so there are 12 new gauge bosons.

These new bosons carry new interactions which can e.g. cause proton decay. That's the reason why most ogf these simple GUTs are ruled out experimentally (protons would decay too fast)

13. Jul 23, 2011

### naima

Thank you for the time you spend on this forum.
I am not looking for unification. I just want to stay in the SM.
Of course all have been achieved (just have to look to the lagrangian)
My question is here: photons act on L,R electrons and quarks, W and Z only act on left electrons and quarks and gluons just on quarks. It must be difficult to act on everything and just THE photon succeeds
and the great family of the gluons only succeeds with the quarks! can this be justified in a mathematicall
thing?
Well this is just for Aesthetics!

14. Jul 23, 2011

### Fredrik

Staff Emeritus
You wouldn't multiply z with y. The members of the direct product group are ordered pairs (z,y), so you would be multiplying such pairs with each other. This multiplication is defined by (z,y)(u,v)=(zu,yv).

15. Jul 25, 2011

### arlesterc

Thanks for all the help. It took a few rounds of reading but I started to get the idea
and then this Wiki article and the simple example in it finally nailed it for me:

http://en.wikipedia.org/wiki/Cartesian_product

"For example, the Cartesian product of the 13-element set of standard playing card ranks {Ace, King, Queen, Jack, 10, 9, 8, 7, 6, 5, 4, 3, 2} and the four-element set of card suits {♠, ♥, ♦, ♣} is the 52-element set of all possible playing cards: ranks × suits = {(Ace, ♠), (King, ♠), ..., (2, ♠), (Ace, ♥), ..., (3, ♣), (2, ♣)}. The corresponding Cartesian product has 52 = 13 × 4 elements. The Cartesian product of the suits × ranks would still be the 52 pairings, but in the opposite order {(♠, Ace), (♠, King), ...}. Ordered pairs (a kind of tuple) have order, but sets are unordered. The order in which the elements of a set are listed is irrelevant; you can shuffle the deck and it's still the same set of cards."

Thanks again for getting the grey cells rattled.