0

SU(3) is a subgroup of O(6). Therefor we can say it can be represented in 6 dimensional space. It has one invariant which is:

$$x^2+y^2+z^2+w^2+u^2+v^2$$

So far the group compatible with this is O(6). To get the subgroup SU(3) I can introduce the invariant:

$$xy-yx + zw-wz +uv-vw$$

Since this is an invariant of Sp(3) and combining these two invariants would give SU(3).

However, the second invariant is zero unless we specify that the coordinates are non-commutative.

Therefor (apart from specifying a complex struture which I say is equivalent to doing the above e.g. expand out $(x+iy)(x-iy)$ you get the same in terms of real an imaginary parts), is there no way to specify the group SU(3) in terms of invariants using commuting variables? In terms of the Lie groups I can only work out how to do $O(n)$ and $F_4$ in this way.

(BTW I'm not sure if the above gives SU(3) or U(3))

zooby
  • 4,649
  • 1
    Not at all sure if that's equivalent to your question, but the two non-isomorphic fundamental representations of $SU(3)$ on $\mathbb C^3$ are conjugate to each other, so they have no real structure and somehow "need" complex entries, even if you view $\mathbb C^3$ as $\mathbb R^6$. – Torsten Schoeneberg Nov 25 '20 at 19:02
  • @Torsten Thanks, Do you know what lie groups have a real structure? (Maybe there is a non-fundamental representation of SU(3) too) – zooby Nov 26 '20 at 12:24
  • For compact Lie groups (like $SU(N)$), Bourbaki gives us criteria to decide which representations have a real structure: https://math.stackexchange.com/a/2774741/96384. For example, the adjoint representation always does (obviously). In the non-compact case, it gets trickier, cf. https://math.stackexchange.com/a/3298058/96384. – Torsten Schoeneberg Nov 26 '20 at 16:11

0 Answers0