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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 :-)

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    $\begingroup$ I'm not sure what exactly "enhanced" means but there are no nontrivial Lie group maps between $SU(3)$ and $SO(5)$. $\endgroup$
    – algori
    Sep 19, 2011 at 17:38
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    $\begingroup$ That's a mind bending connection between the two groups. The geometry of their natural homogenous spaces are very different though. Just goes to show that the same space can have very different geometric structures. $\endgroup$
    – Charlie Frohman
    Sep 19, 2011 at 18:50
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    $\begingroup$ I think that the standard terminology is "squashed 7-sphere" and it is what geometers call the canonical variation. You exhibit the 7-sphere as a Hopf-like fibration $S^3 \to S^7 \to S^4$ and then you rescale the fibre metric. There are two values of the rescaling parameters for which the resulting geometry is Einstein: the round metric and the squashed metric. I am not sure what sort of relationship you expect between these different compactifications, though. Perhaps you could elaborate? $\endgroup$
    – José Figueroa-O'Farrill
    Sep 19, 2011 at 19:44
  • $\begingroup$ @José I expect to get some understanding of why Nature has choosen the CP^2xS3 way, which produces the L-R symmetric version of the standard model, instead of the more elegant, or obvious, S7. Of course part of the answer could be not lying here, but in upper dimensions, with CP4 (so SU(5)), S5xS3 (so Pati-Salam) or even S9 (so SO(10)). But the initial root of the evil seems to be in this S4 vs CP2 variation. $\endgroup$
    – arivero
    Sep 19, 2011 at 23:22
  • $\begingroup$ Hmm, looking now at the whole scheme, it seems that the involved groups have the same number of nodes in the dynkin diagram but different number of edges. Of course to change the number of edges is to change all the group structure, so it has not sense. But still, compare SO(8) to SU(3)xSU(2)xSU(2), which is the symmetry of CP2xS3. And also SO(5) and SU(3) differ in an edge, do them? Furthermore, the diagram with four isolated nodes is SU(2)^4, again one of the known families. $\endgroup$
    – arivero
    Sep 19, 2011 at 23:24

2 Answers 2

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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.

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  • $\begingroup$ As for the why, it is on physics, or metaphysics ground... I will add some context. $\endgroup$
    – arivero
    Sep 19, 2011 at 18:29
  • $\begingroup$ I think it is as good an answer as it can be done. It is fascinating how a discrete symmetry can transmute a group into another. I mean, naively one could have expected that the quotient manifold has only SO(3) symmetry, but instead it makes some magic and enhances up to SO(5). Something similar happens with the torus and the 2-sphere. $\endgroup$
    – arivero
    Sep 20, 2011 at 19:26
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$SU(3)$ has center of order 3 and $SO(5)$ has center reduced to the identity. In fact they are not even locally isomorphic: $SU(3)$ is of Cartan type $A_2$ and $SO(5)$ is of type $B_2$.

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