# Affine differential geometry. Is Calabi's hypersurface isotropic?

I am in the framework of (equi)affine differential geometry. Let $E$ be a centro-equiaffine space, that is a real vector space of dimension $n$, together with the special linear group $SL_n(R)$. Let $e_1, \cdots e_n$ be a system of linear coordinates for this space.

I am interested in Calabi's hypersurface $C$, given by:

$\prod_{i=1}^nx_i=1.$

I am interested in the isotropy group of $C$, fixing $(1, \cdots, 1)$ namely: $\{g\in SL_n(R)| g(C)=C\text{ and }g(1, \cdots, 1)=(1, \cdots, 1)\}$

My calculus gave me that this isotropy group is not continuous, in the sense that it is composed only of the permutations of the coordinates, and of the "diagonal transformations" of the type:

$g(x_1, \cdots, x_n)=(\lambda_1*x_1, \cdots, \lambda_n*x_n)$, with $\prod_{i=1}^n\lambda_i=1$.

First question: Could you please confirm my calculus of the isotropy group?

I was surprised by the result because the two fundamental structures that characterize a Blaschke hypersurface, namely : the affine metric $h$, and the shape operator $S$, have (in the case of Calabi's hypersurface) both isotropic properties, in the sense that both the scalar curvature (of the affine metric) and the affine curvature don't depend of the choice of a direction on the hypersurface.

Second question: (If my calculus concerning the first question were right) : How can we understand that an affine sphere could be non isotropic (in the sense that the group of isotropy is not transitive relatively to the linear directions), even if all its fundamental Blaschke structures (affine metric, shape operator) ARE isotropic?

Also, the 'diagonal' transformation that you write down fixes $(1,\ldots,1)$ only when it is the identity. The stabilizer of this point in the stabilizer of $C$ is actually discrete.