# I SU 2 and SU 3

1. Jul 25, 2016

### Josh1079

Hi, I'm recently reading some text on particle physics and there is a section on symmetries and group theory. It gave the definition of SU2 as the group of unitary 2*2 matrices and that SU3 is the group of unitary 3*3 matrices. However, it kind of confuses me when it mentioned representations of higher orders. What's the difference between a 3*3 representation of SU2 and SU3? Also, I don't really understand what it means when it mentioned something like "invariant under SU2 transformations", can anyone give an example of a vector that's invariant under SU2 transformations?

Thanks!

2. Jul 25, 2016

### Staff: Mentor

A representation is a mapping from the given group into a group of (regular, invertible) transformations of a vector space. In mathematical terms: A representation $(G,V,φ)$ of a group $G$ is a vector space $V$ together with a group homomorphism $φ: G \longrightarrow GL(V).$
So the number you mentioned, "$3 \times 3$ representation" refers to the dimension of the vector space (here $3$), not to the group! Thus it has nothing to do with whether you consider $SU(2)$ or $SU(3)$. An invariant vector $v$ under $SU(2)$ transformation means, that $φ(X)(v) = v$ for all $X \in SU(2)$, here unitary $2 \times 2$-matrices with determinant $1$. The mapping $φ$ in this context is often omitted and the equation is noted $X.v = v$ or $v^X = v$. Things become a bit messy if the vector space $V$ itself is a vector space of (not necessarily regular, since $0 \in V$) matrices.

An example for a representation of $SU(n)$ would be $φ: SU(n) \longrightarrow GL(\mathfrak{su}(n))$ where $φ: u \longmapsto uAu^{-1}$ for $u \in SU(n) \, , \, A \in \mathfrak{su}(n).$
It shouldn't be too difficult to find invariant vectors here or in a simplier representation $V$.

One last remark: A representation $(G,V,φ)$ is often simply called by "$G$ operates on $V$".

3. Jul 25, 2016

### Josh1079

Thanks a lot! So I guess I've mixed up the definitions.

But for the invariant vector question, I think that's what I thought initially until I saw a line stating that (1, 1, 1) is invariant under SU3 transformations. Actually, it stated η = (u(ubar) + d(dbar) + s(sbar))/√3 is invariant under SU3. This really confuses me.

4. Jul 25, 2016

### Staff: Mentor

I'm not sure here, what $u,d,s$ are. Unipotent, diagonal, symmetric matrices? And I haven't generators of $SU(3)$ in mind to verify that $(1,1,1)$ is invariant under $SU(3)$ by its natural representation (matrix multiplication / application on $\mathbb{C}^3$) $u.(1,1,1) = (u_{1i},u_{2i},u_{3i})$, i.e. all the row sums of $u$ should be equal to $1$. Seems wrong to me, so either it's another representation on $\mathbb{C}^3$ or the diagonal matrix $\mathbb{1} = (1,1,1)$ is meant, which is of course invariant under $SU(3)$.

5. Jul 25, 2016

### Josh1079

Actually, since I'm reading a particle physics text, the u d s refers to quarks. Maybe I should raise this on the physics section.