Consider the homogeneous space $X:=GL_2\times GL_2\times GL_2/ H$ where $H=GL_2$ is diagonally embedded into $GL_2\times GL_2\times GL_2$. My question is why $X$ is spherical (i.e., there is a Borel subgroup $B$ of $GL_2\times GL_2\times GL_2$ which has an open orbit on $X$).
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$\begingroup$ Using the obvious Isomorphism to $GL_2\times GL_2$, you can rule that out by dimension count. $\endgroup$– user130903Commented Oct 10, 2021 at 6:51
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1$\begingroup$ It is a calculation. Your homogeneous space $X$ has dimension $12-4=8$. You should try to construct a Borel subgroup $B$ such that $B\cap H$ has dimension one. Since $B$ is of dimension $9$, you will see that the $B$-orbit $B\cdot 1H\subset G/H$ is of dimension 8, hence dense (and open). $\endgroup$– Mikhail BorovoiCommented Oct 10, 2021 at 11:29
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