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Look at the following diagrams of Lie groups

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

Note that they are not subgroups of SO(8), but still it could be said that their roots remember SO(8) in some ways. For instance, they are also isometry groups of seven dimensional manifolds. My particular question is, what does happen with triality of representations of SO(8)? Is there some similar simmetry, or remmants of it, in the other groups?

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This question is vague, and I voted to close it. The triality of Lie algebra $\mathfrak{so}(8)$ refers to an isomorphism of its root system. This isomorphism has implications for simply connected Lie groups built from $\mathfrak{so}(8)$, and $SO(8)$ is not simply connected, but $Spin(8)$ is. Think of how you would mathematically define the remnant of triality, and refine the question. –  Sasha Sep 20 '11 at 14:10
    
First, note that $SO(8)$ doesn't have triality - you must pass to $Spin(8)$ to realize it. Second, what would a "remnant of symmetry" be? Third, in general, the symmetries of the Dynkin diagram are isomorphic to the outer autmorphism group of the simly connected Lie group corresponding to the Dynkin diagram. So, the second two have no notion of triality. –  Jason DeVito Sep 20 '11 at 14:11
    
So the answer is a plain "no, nothing survives"? –  arivero Sep 20 '11 at 14:20

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