# SU(3) representation mystified need help

1. Aug 9, 2009

### jacobrhcp

Hi,

I'm currently reading a book on particle physics, which tells me this about SU(3):

"...The generators may be taken to be any 3x3-1=8 linearly independent traceless matrices. Since it possible to have only two of these traceless matrices diagonal, this is the maximum number of commuting generators..."

These few lines make close to no sense to me. I have some questions:

1. Why 3x3-1? A 3x3 matrix has 9 entries, so that might be the 9, and the 1 must somehow come from the unitarian and determinant=1 conditions. But how?

2. Why do they need to be traceless?

3. Most importantly: Why do you have a maximum of 2 matrices diagonal, and what does this have to do with the number of commuting generators?

Then he goes on being sensible again, but this I don't get.
If anyone can help me through these questions, I'd be very grateful.

Jacob

2. Aug 9, 2009

### Hurkyl

Staff Emeritus
When they say generators, they are referring to the Lie algebra su(3). Since all matrices in SU(3) have determinant 1, all matrices in su(3) must have trace 0.

3. Aug 9, 2009

### jacobrhcp

Hmm ... I must admit to know not much about Lie albegras and the sort, and your well-intended help (sincerely thank you) is of no use at all! :( This books claims to be able to explain these things about SU(3) on the physics level of rigor.
Are you sure there is no more straightforward way to understand this? As in, in terms of group and representation theory, and linear algebra (which I did in class).

4. Aug 9, 2009

### Ravid

Lie algebras are obtained by differentiating Lie groups at the identity. A formal introduction is not straightforward, but I can briefly explain this example: any element of SU(3) has determinant one and thus preserves $$e_1\wedge e_2\wedge e_3$$. Differentiating along a path $$X_t$$ we get

$$\left.\frac{d}{dt}\right|_{t=0}(X_t(e_1\wedge e_2\wedge e_3))=0$$

which by the product rule gives

$$0=X_0'(e_1)\wedge e_2\wedge e_3 + e_1\wedge X_0'(e_2)\wedge e_3 + e_1\wedge e_2\wedge X_0'(e_3) =\mathrm{tr}(X_0')e_1\wedge e_2\wedge e_3$$

Thus elements of the Lie algebra $$X_0'\in\mathfrak{su}(3)$$ must have trace 0. The generators are the nine entries of the matrix minus the one dependent element (each entry on the diagonal must be minus the sum of the other two); for the same reason the Cartan subalgebra $$\mathfrak h$$ of diagonal matrices is generated by two elements corresponding to any two diagonal entries of the matrix. Elements of this subalgebra commute, and moreover fail to commute with any other elements of $$\mathfrak{su}(3)$$, so it is a maximal abelian subalgebra.

5. Aug 9, 2009

### Hurkyl

Staff Emeritus
I'm afraid this works out to be an application of differential geometry as applied to groups. The following might not make immediate sense, but it will hopefully give you the idea of where things are going as you study.

Fortunately, matrix groups are fairly concrete -- M(3), the set of all 3x3 matrices, is isomorphic to R9. SU(3) is an eight-dimensional subspace1. To the identity, there is an eight-dimensional, affine tangent space. su(3) can be identified with the "displacement from the identity" vectors within that affine space.

It is convenient to find a way to name tangent vectors to other points of SU(3) via elements of su(3).

Through integral (or some other variety of) calculus, one can construct elements of SU(3) out of elements of su(3) -- this, I presume, is the etymology of physicsts calling them "(infinitessimal) generators".

(The infinitessimal part is that tangent vectors provide a pretty good model for the notion of an infinitessimal displacement)

1: Space here meant in the sense of "topological space" not "vector space"

6. Aug 10, 2009

### jacobrhcp

*tries hard to make an effort to understand some diff.geometry*

how is this:

GL(3,C) has dimension 18, right? The set of all hermitian 3x3 matrices is then clearly 9-dimensional as a vector space. Then by virtue of the determinant we can scrape of the last one to come down to 8.

So we can generate it by any set of 8 linearly independent matrices. Now it turns out that 8 linearly indepent matrices of SU(3) are always traceless which is due to something complicated you just told me.

Then ofcourse the diagonal matrices commute, and the others do not in the same representation, because then they are not simultaneously diagonalizable.

But this does not explain why none of the other matrices do not commute with the diagonal matrices, or why there can be only 2 diagonal matrices! Something I'd still very much like to understand.

And as for the term 'infinitesimal generator', i thought it comes from the idea that physicists like to express rotations using complex exponentials, and for small rotations they can make the first order taylor approximation to calculate with.

7. Aug 10, 2009

### aziz113

1. When they speak of generators they are referring to the Lie algebra of SU(3). This is the Lie algebra of traceless 3 X 3 matrices. This forms a vector space of dimension 8: 2 diagonal matrices E11-E22, E22-E33, and 6 non-diagonal matrices E12, E13, E21, E23, E31, E32 where Eij is the matrix with 1 in position ij and zero elsewhere.

2. Brain Hall's book "Lie groups, Lie algebras, and representations" gives a clear description of how you can calculate the Lie algebra of a matrix Lie group such as SU(3).

3. See "1".

Note: The representations of SU(3) are fairly complicated. You may want to start with SU(2).

Last edited: Aug 11, 2009