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May 3, 2020 at 16:26 comment added skd Yep, thanks! (I also just added a construction of a candidate for Y.)
May 3, 2020 at 16:25 history edited skd CC BY-SA 4.0
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May 3, 2020 at 11:04 comment added Robert Bryant You are quite right that $\mathrm{G}_2/\mathrm{SU}(2)$ is not a topological group. Since, as you point out, there is a fibration $S^5\to \mathrm{G}_2/\mathrm{SU}(2)\to S^6$, all the homotopy groups $\pi_i(\mathrm{G}_2/\mathrm{SU}(2))$ for $i =0,1,2,3$ are trivial, so it can't be a compact Lie group. Also, because $\pi_3\bigl(\mathrm{Spin}(9)/\mathrm{G}_2\bigr)$ is finite (probably trivial, but I haven't checked to be sure), it can't be a compact Lie group. For similar reasons, your $Y$, if it exists, cannot be a topological group either.
May 2, 2020 at 20:41 history edited skd CC BY-SA 4.0
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May 2, 2020 at 19:59 history asked skd CC BY-SA 4.0