
#1
Aug411, 08:45 AM

P: 27

Roger Penrose wrote in "The Road to Reality" that the symmetry to describe the weak force is the "group SU(2) x U(1) or more correctly, U(2)". (page 641)
And "The group might be expressed as SU(2) x U(1)/Z2, where the '/Z2' means 'factor out by a Z2 subgroup'. However, there is more than one such subgroup, so this notation is not fully explicit. The notation 'U(2)' automatically picks out the correct one." (page 654) I like to understand what the group U(2) is. The group U(1) corresponds to the circle group. wiki: "U(1) is the multiplicative group of all complex numbers with absolute value 1, i.e., the unit circle in the complex plane." But to what corresponds the group U(2)? 



#2
Aug411, 04:29 PM

Sci Advisor
HW Helper
P: 11,863

Unitary 2*2 matrices. Search the internet/books for a general definition of U(n), then take n=2.




#3
Aug411, 04:51 PM

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PF Gold
P: 4,768

Recall that O(n) is the group of matrices (which we identify with the linear maps they represent with respect to the standard basis in R^n) wich preserves the euclidean inner product. That is, those matrices O such that <u,v> = <Ou,Ov>.
In C^n, there is a natural inner product (more precisely, a hermitian product because it is not symetric, but rather "conjugate symetric") given by [tex](w,z)=\sum_{i=1}^{n}w_i\overline{z_i}[/tex] U(n) is then to ( , ) what O(n) is to < , >. That is, U(n) = complex nxn matrices U such that (Uw,Uz) = (w,z) for all w,z in C^n. In the special case n=1, U(1)=1x1 matrices (i.e. element of C!) u such that [tex]uw\overline{uz}=uw\overline{u}\overline{z}=u^2w\overline{z}=w\overlin e{z}[/tex] These are precisely the element u with u=1; i.e. those lying on the unit cirlce in C. Hence U(1)=S^1. 



#4
Aug611, 11:31 AM

Sci Advisor
P: 1,716

What is Unitary group U(2) 



#5
Aug611, 08:35 PM

P: 491

The nearest analogue in the sense you describe is that SU(2) is the group of unit quarernions.




#6
Aug1111, 10:13 AM

P: 27

I am not realy helped by the answers. The only remark that is a bit helpfull is Lavinia
Is there someone else that can clearify U(2)? 



#7
Aug1111, 12:13 PM

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P: 11,863

Quasar gave you the definition. What's not clear to you ?




#8
Aug1211, 02:59 PM

P: 491

How to understand the symmetry group for the weak interaction? Well, there's two W bosons and a Z boson. There's a problem with this distinction, however: if we decide to rotate our gauge at each point, the laws of physics are unchanged, so we want to understand the gauge symmetry behind it. What symmetries are out there? Is any gauge acceptable? Certainly not all of them: there's a welldefined notion of field strength, and we also want our particles to be "orthogonal" in a certain sense. So we can think (at the moment) of our symmetry group being rotation matrices that interchange the particles. We want the field strength to be preserved, and we also want orthogonality (and orientation) to be preserved, so the first guess would be matrices that do precisely that: this is the group SO(3). There's a problem with this, however: if SO(3) is our symmetry group, the gauge transformation that takes W+ to W+ and then W to W is the same as the gauge transformation that does it in the reverse order, i.e. those define the same gauge if we were right. Problem is, experiments contradict this  this is the idea of spin. So we need to try again and look for a symmetry group where those gauge transformations are allowable, but don't define the same gauge. Mathematicians have worked this out such a group: it's called SU(2). Now for the electroweak interaction, we throw in the photon, which has symmetry group U(1), the circle. Again, the naive guess would be that the symmetry group would be to take the product of these two groups, i.e. allowable gauge transformations would be rotations of the W and Z bosons, followed by rotations in the photon, in any order. This would yield the symmetry group SU(2) x U(1). Again, however, there's a problem (which I don't quite understand because I'm not a physicist), which I believe is called the Higgs effect. This arises because there's an interaction between the weak and electric fields, hence gauges which don't appear to be equal are actually equal. Thus we have to shrink our symmetry group to account for this equality of gauges. Physicists have worked out how to do this: it turns out that if we view this as SU(2) x U(1), for each gauge, there is precisely one other gauge which defines the same gauge. The math behind this is that to obtain our true symmetry group, we need to divide out by this symmetry, yielding (SU(2) x U(1))/Z_2, which is the group U(2). 


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