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For the three exceptional compact Lie groups $E_8, E_7, E_6$ we have the inclusions $$ E_6 \subseteq E_7 \subseteq E_8. $$ What can we say about the the homogeneous spaces $$ E_8/E_7, ~~~~ E_7/E_6? $$ Do they look anything like spheres? I am motivated by the case of the homogeneous spaces $SU_n/SU_{n-1}$, $SO(n)/SO(n-1)$, and $Sp(n)/Sp(n-1)$, that are all spheres of some dimension.

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    $\begingroup$ There is a paper devoted to $E_8/E_7$: "On the homogeneous space $E_8/E_7$". By. Ichiro YOKOTA, Takao Imai and Osami YASUKURA. J. Math. Kyoto Univ. (JMKYAZ) 23-3 (1983) 467-473. $\endgroup$
    – Nick L
    Commented Sep 23, 2021 at 15:39
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    $\begingroup$ @Alain: Are you talking about real groups or complex groups? $\endgroup$
    – Sasha
    Commented Sep 23, 2021 at 17:45
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    $\begingroup$ @Sasha: he says compact, so he means real and compact. $\endgroup$
    – Ben McKay
    Commented Sep 23, 2021 at 18:27
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    $\begingroup$ See the answer of José Figueroa-O'Farrill to mathoverflow.net/questions/75525/… to see that they are not spheres. $\endgroup$
    – Ben McKay
    Commented Sep 23, 2021 at 18:29
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    $\begingroup$ @BenMcKay: The answer is the same. The list of compact groups acting transitively on the $n$-sphere was actually determined in this paper: Montgomery, Deane; Samelson, Hans Transformation groups of spheres. Ann. of Math. (2) 44 (1943), 454–470. A posteriori, it follows that all such groups act preserving a constant curvature metric. Of course, the point is that $E_8$ and $E_7$ don't appear on this list, so, in particular, no homogeneous space of the form $E_8/K$ or $E_7/K$ is homeomorphic to a sphere. $\endgroup$
    – Robert Bryant
    Commented Sep 24, 2021 at 17:28

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