[Math] Why SU(3) is not equal to SO(5)

ct.category-theorydg.differential-geometrygr.group-theoryrt.representation-theory

I am asking in the sense of isometry groups of a manifold. SU(3) is the group of isometries of CP2, and SO(5) is the group of isometries of the 4-sphere. Now, it happens that both manifolds are related by Arnold-Kuiper-Massey theorem: $\mathbb{CP}^2/conj \approx S^4$; one is a branched covering of the other, the quotient being via complex conjugation.

Now, for the case of a manifold and a lower dimensional submanifold, it is not rare to find that the corresponding isometry groups are subgroups one of the other. But here, which is the equivalent result? is SO(5) an "enhanced SU(3)" in some way?

The context of the question comes from 11D Kaluza Klein, more particularly from the classification of Einstein metrics in compact 7-manifolds. It is easy to produce from the 7-sphere metric a "squeezed sphere" whose isometry group is, instead of SO(8), just the one of $S^4 \times S^3$. But it is not known if there is some relationship between the 7-sphere and the "Witten manifolds" of the kind $CP^2 \times S^3$.

EDIT: to add more context, some dynkin diagrams.

o====o  SO(5), isometries of the sphere S4
o----o  SU(3) are the isometries of CP2
o    o  SU(2)xSU(2), isometries of  S2xS2. Also SO(4), so isometries of S3

So it seems that the quotient under conjugation has implied, or is compensated by, some change in the angles between roots, but not in the number of roots.

For isometries of 7-manifolds we have also some similarities.

        o                  o                         o
       /                                        
      /                                          
o----o    SO(8)    o----o     SU(3)xSO(4)    o====o     SO(5)xSO(4)
      \
       \
        o                  o                         o

where the first diagram is the [isometry group of] the seven sphere, the last is the squashed sphere, and the intermediate is the one I am intrigued about, as it contains the physicists standard model gauge group.

By the way, the last drawing makes one to ask about how triality survives in the representations of these product groups, but that is other question 🙂

Best Answer

Why would you think that they are related?

The map $\mathbb{CP}^2/\text{conjugation} \to S^4$ is only $SO(3)$-equivariant, where $SO(3) \subset SU(3)$ consists of the real matrices and $SO(3) \subset SO(5)$ is the maximal subgroup acting irreducibly on the 5-dimensional vector representation of $SO(5)$.

Not sure if this answers your question, though.

Related Question