# A 4x4 matrix representation of SU(3)?

Is it possible to find a representation of the infinitesimal generators of the special unitary group SU(3) that contains 4 by 4 matrices, by say taking a Kronecker product of its irreducible representation(s) with itself?

I know this is possible for SU(2), where one can express the three 4 by 4 matrices spanning the unit quaternion group in terms of some of the generators of SU(4), which are 4 by 4 matrices as well.

I am trying to do the same for SU(3). This question is motivated by the investigation of higher dimensional gauge theories. Thanks.

• Can you not just extend the matrices of size $3$ by a "trivial" block of size $1$ ? So the direct sum of the natural representation and the trivial one. – Dietrich Burde Oct 21 '14 at 18:14
• So basically adding a "1" entry to each of the Gell-Mann matrices? Would you say this is the only way of obtaining a 4 x 4 representation? – itsqualtime Oct 21 '14 at 20:03
• Actually not the Gell-Mann matrices since those are its infinitesimal generators, but rather its fundamental 3x3 rep – itsqualtime Oct 21 '14 at 21:10
• I don't think this is possible in a way other than what Dietrich Burde described. SU(2) is special in the sense that there is up to isomorphism a single irreducible rep of each dimension. With SU(3) the sequence of possible dimensions of irreducible reps goes like 1, 3, 3, 6, 8, 6, 10,... IIRC at least the 8-dimensional and 10-dimensional reps play a role in particle physics. – Jyrki Lahtonen Oct 22 '14 at 20:26
• Jyrki - that's very interesting and particularly relevant to what I'm doing. Do you know of any ref. that would provide more details on the number and dimensionality of the IRR of an arbitrary SU(n) group? – itsqualtime Oct 23 '14 at 19:56

You might have a look at Fulton's book Young tableaux for the combinatorics relevant for computing these dimensions. Here is the relevant fact: the irreducible representations of $SU(n)$ may be indexed by partitions with at most $n-1$ parts in such a way that the dimension of the irreducible $L(\lambda)$ corresponding to a partition $\lambda$ is the number of column-strict Young tableaux (in the alphabet $\{1,2,\dots,n\}$) on $\lambda$. This gives a practical way of computing the dimensions that is a bit faster than using Weyl's character formula (the classical formula that can be expressed in a root-system uniform fashion) or more recent tools such as the Littelmann path model. The point being, for $SU(n)$ you don't need to know about root systems to operate a machine that will compute the dimensions. (In fact, things are even better: Schur functions give you the characters of the irreps and not just their dimensions).

In your example, the irreps would therefore be indexed by the partitions (in roughly increasing order of size) $$(0),(1),(1,1),(2),(2,1),(3),(2,2),(3,1),(4),\dots$$ of dimensions $$1,3,3,6,8,10,6,15,15,...$$ as noted in Jyrki's comment above. The upshot for your problem is that a four dimensional representation is one of three things: the sum of four copies of the trivial irrep, or the sum of one copy of the trivial irrep with one of the two $3$ dimensional irreps.

As noted by the accepted answer and the comments, the answer is no.

Note, we can prove this without using Young Tableaux: we can check the possible dimensions as noted in Jyrki's comment above by directly using the known dimension formula for $$SU(3)$$:

The irreducible representations are indexed by a pair of non-negative integers $$(m_1,m_2)$$ and the dimension of the irreducible representation is given by

\begin{align} \dim(V_{m_1,m_2}) = \frac{1}{2}(m_1+1)(m_2+1)(m_1+m_2+2) \end{align}

One checks that there is no pair of non-negative integers that gives $$\dim(V_{m_1,m_2})=4$$ by testing the first few (and only) possibilities directly.