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In these notes of Pezzini and the obvious question arises, is a sphere, a "spherical variety". I'm interested in $L^2(X)$ for the space: $$ X = \{ x^2 + y^2 + z^2 = 1\} \subseteq \mathbb{R}^3 $$ In order to become a spherical variety I need a couple of things:

  • $G = $ reductive connected linear algebraic group. I'm guessing $G = \text{SO}_3$.
  • The Borel subgroup $B$ of orthogonal matrices would be the upper triangular orthogonal matrices.
  • The maximal torus would be $T \simeq \{ (e^{i\theta_1}, e^{i\theta_2}, e^{i\theta_3}): \theta_1 + \theta_2 + \theta_3 = 0 \}$ except those are complex. And I'm getting that $B = T$.
  • Unsure about the correct choice of $H$. It just says that $X$ is some $G$-variety, possible that $X = G/H$.

Wikipedia also has discussion of Homogeneous spaces. It has that:

  • $S^2 = \text{O}(3)/\text{O}(2)$
  • $\text{Gr}(k,n) = \text{O}(n) / \text{O}(k) \times \text{O}(n-k) $

Then maybe I'd have that $L^2(X)$ is just the spherical haromnics, even if I pass into positive characteristic.


As a complex space, I believe $S^2 \simeq \widehat{\mathbb{C}}= \mathbb{C} \cup \{ \infty\}$ is called the Riemann sphere or $\mathbb{C}P^1$ with a standard identification via Möbius transformations. If we identify $\mathbb{C}\simeq \mathbb{R}^2$ and remember the complex structure, we might be OK.

As a complex projective space we might consider the Hopf fibration $\mathbb{C}P^1 = S^3/S^1$ or possibly $U(2)/\big( U(1) \times U(1) \big)$, perhaps leading to appropriate choices of $G$ and $H$.

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    $\begingroup$ I am afraid you are confused. The notes you are quoting discuss complex manifolds. How do you define a real spherical variety? $\endgroup$
    – abx
    Commented Oct 14, 2017 at 18:55
  • $\begingroup$ @abx the Euclidean 2-sphere $S^2 \subset \mathbb{R}^3$ is a Riemann surface with it's complex structure. And the question makes sense. $\endgroup$ Commented Oct 14, 2017 at 19:34
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    $\begingroup$ I don't understand where you are heading to. As you observe, the sphere $S^2$ carries a complex structure, unique up to diffeomorphism, giving $\mathbb{CP}^1$; it is trivially a spherical variety. So what? $\endgroup$
    – abx
    Commented Oct 14, 2017 at 19:36

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Yes, $S^2$ is $PSL_2\mathbb{C}/B$ where $B$ is the Borel subgroup, so the Riemann sphere is spherical in that sense. Spherical varieties are called spherical because the Riemann sphere is one, with action of the rotation group: see here: Why are they called Spherical Varieties?

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  • $\begingroup$ One of the answers mentions $S^2 = SO(3)/SO(2)$ and says it's a "spherical homogeneous space". $\endgroup$ Commented Oct 14, 2017 at 19:42

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