# GUT and Lie Algebra

1. Dec 4, 2008

### squarks

I hope this is the right place to ask this question:

Why does GUT Model has to have a rank of at least 4 (Such as Georgi-Glashaw Model of SU(5) )? In Georgi's Lie Algebras book it vaguely states that they correspond to the generators S, R, T3 and T8, where S and R are generators of U(1) and SU(2) respectively (No idea what T is -- Does it have something to do with colour or QCD?).

Can someone help me in the right direction? Thank you in advance

2. Dec 4, 2008

### BenTheMan

The standard model has rank 4, this means that anything which contains the standard model as a subgroup must have rank AT LEAST 4.

Note that SO(10) has rank 5 and E6 has rank 6, both of which people use to build GUTs.

3. Dec 4, 2008

### squarks

Thanks Ben

You answered my question from mathematical point of view (don't get me wrong, it was very helpful), but what does it mean physically? The U(1) generator gives spin, SU(2) generator gives rotation... I believe the the other two generators comes from SU(3), but not sure what they are.

or am I missing the whole point? The question that is being asked is "Why are rank 4 groups important in the context of describing the unification of strong, weak and electromagnetic interactions (from physics point of view)?"

4. Dec 4, 2008

### jambaugh

The issue of rank has to do with the total number of mutually commuting operators and thus total number of simultaneous observables. $$SU(3)_{\text{color}}$$ has rank 2 so we can (if we could observe color directly) see say redness and (blue+antigreen)ness simultaneously. These we assume are independent of
* total charge, $$Q$$ of the $$U(1)_{em}$$ gauge group
and
* the observable weak-isospin component $$I_3$$ of the $$SU(2)_{iso}$$which distinguishes e.g. the electron from its neutrino.

Now if we suppose we can distinquish two independent quark color components of the three, these plus the two definitely observable e-m and isospin charges yields 4 simultaneously observable gauge charges.

5. Dec 4, 2008

### humanino

Is there a reason behind your not mentioning redness + (blue+green)ness ?

6. Dec 5, 2008

### BenTheMan

No. The U(1) generator forms a linear combination with one of the SU(2) generators to give a photon. The orthogonal linear combination gives the Z boson.

Again, no. The SU(2) generators form linear combinations to make W+-, Z, and the photon.

I don't really understand what jambaugh's saying---so apologies if I'm repeating something (s)he wrote.

Let me attack this from another vantage point. You can take the generators of SU(N) and form what's called ladder operators---the example you're probably familiar with is SU(2) spin. There, you take the Pauli matrices and form linear combinations:

$$\sigma_{\pm} = \sigma_1 \pm i \sigma_2$$

The ladder operators work on a Hilbert space of spins, so that

$$\sigma_+ |\downarrow\rangle = \# |\uparrow\rangle$$,
$$\sigma_+ | \uparrow \rangle = 0$$,

where I've forgotten the numbers. Then the diagonal sigma matrix is left alone:

$$\sigma_3 |\uparrow\rangle = | \uparrow \rangle$$.

How does this work with W bosons? Well, SU(2) from the standard models isn't REALLY isospin, but is completely analogous to the SU(2) spin example I just showed. You can imagine the down quark as having "spin down" and the up quark as having "spin up". Then, schematically:

$$W^+ | down \rangle = |up\rangle$$, etc.

So the W+ and W- bosons change "upness" and "downness". The Z boson, on the other hand, leaves upness and downness alone.

We can generalize this. In SU(3), there are eight Gell-Mann matrices (generalization of Pauli Matrices). The diagonal Gell-Mann matrices (look them up on Wikipedia, usually $$\lambda_{3,8}$$) commute with each other. The other six form linear combinations which I forget, but look suspiciously like the $$\sigma_\pm$$ that I wrote down above. These six linear combinations of gluons change, say, a red up quark to an anti-blue up quark, for example. The diagonal generators don't change the color of the quark at all---they change red to anti-red or blue to anti-blue.

Now it turns out that these diagonal generators are very special: you can add or multiply two diagonal matrices and always get a diagonal matrix. Also, diagonal matrices A and B always obey AB - BA = 0. You can see they form sort of a subset of all of the generators which are floating around---technically, we call this the Cartan sub-Algebra. The number of diagonal generators of SU(N) turns out to be N-1, which is the rank of SU(N). In general, the rank of the gauge group is the number of diagonal generators of the algebra---so you can see that if you want to get the SM from some larger group, that larger group has to contain at least the diagonal bits to give you the proper mathematical structures that you need.

7. Dec 5, 2008

### jambaugh

I was picking "orthogonal" colors. The three color operators add to 0 = R+G+B.
I could have picked Redness and Greenness and simply pointed out the Blueness is B= -(R+G).

8. Dec 5, 2008

### squarks

Thanks for all of your help

I certainly understand more than I did, but not completely. I think it will become more clearer once I study field theory. For now, this is good enough.

Thanks again

9. Dec 5, 2008

### humanino

The reason I am wondering is, in deep inelastic scattering in the appropriate kinematics for factorisation theorems to apply, we "pull out" a single quark, leaving behind not an orthogonal component, but really the opposite color. Anyway, thanks for the answer

10. Dec 5, 2008

### jambaugh

I believe that this has to do with color conservation. You are leaving an orthogonal component but one in the vector representation of the su(3) color group('s Lie algebra).

The orthogonality I was using was in the context of the space of observables spanned by the commuting set of color operators. This is orthogonality in the adjoint representation of su(3) restricted to the particular choice of Cartan sub-algebra. This is the orthogonality in weight space. (or in its dual space to be precise).