The complex orthogonal group $O(n+1, \mathbb{C})$ acts transitively on the complex quadric $$ Q_{n-1} := \{[z_0:z_1: \cdots :z_n] : z_0^2 + \cdots z_n^2 = 0 \} \subset \mathbb{CP}^n. $$
What is known about the geometry of curves, surfaces, invariant tensors, etc. in this geometry?
If I'm not mistaken, the case $n = 2$, where the quadric is a conic in the projective plane, is just conformal geometry.
Consider a $(n+2)$-dimensional vector space $V$ over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Let $V'$ be the set of all non-zero vectors. Then $V'$ is an open (real or complex) submanifold of $V$. The projective space $\mathbb{P}:=V'/\sim$ (where $\sim$ is the equivalence relation on $V'$ denied by $v\sim w$ iff there exists some $\lambda\in\mathbb{K}$ such that $w=\lambda v$), is a compact manifold of dimension $n+1$ (over $\mathbb{K}$), such that $V'\to\mathbb{P}$ is a submersion. In particular, $\mathbb{P}$ consists of all the lines $[v]=\{\lambda v :\lambda\in\mathbb{K}\}$ through $v\in V'$. Now, $T\mathbb{P}$ is identified with $TV'/\sim$ where we appropriate extend $\sim$ in $TV'\cong V'\times V$. In particular, if $E=\mathbb{P}\times V\to \mathbb{P}$ is the trivial bundle over $\mathbb{P}$ and $$F:=\{([v], x) : x\in[v]\}$$ denotes the canonical line bundle over $\mathbb{P}$, then one can show that there is a natural isomorphism of vector bundles $T\mathbb{P}\cong {\rm Hom}(F, E/F)$.
Let us pass now to the definition of the quadric. Assume $n\geq 1$ and consider the light cone in $V'$, i.e. the $(n+1)$-submanifold (hypersurface) $\cal{C}$ of $V'$ defined by $$ {\cal{C}}=\{v\in V' : g(v, v)=0\}, $$ where $g$ is a scalar product on $V$ satisfying specific properties (e.g. for $\mathbb{K}=\mathbb{R}$ the signature of $g$ must be $(p+1, q+1)$ with $p, q\geq 0$). Then, the quadric $Q$ is a $n$-dimensional submanifold of $\mathbb{P}$ defined by $$ Q=\{[v]\in\mathbb{P} : v\in\cal{C}\subset V'\}. $$ In particular, the map $V'\to\mathbb{P}$ restricts to a submersion ${\cal{C}}\to Q$. Also the vector bundle $E\to\mathbb{P}$ restricts to a vector bundle over $Q$ and similarly for the line bundle $F\to\mathbb{P}$.
Now, the Lie group ${\rm SO}(g)$ acts transitively on $Q$ with kernel $A=\{\rm Id \}$ if $n$ is odd, and $A=\{{\rm Id}, -{\rm Id}\}$ if $n$ is even. The quotient ${\rm SO}(g)/A$ is the (effective) Möbius group of transformations on $Q$.
Exercise: Show that $TQ\cong {\rm Hom}(F, F^{\perp}/F)$, where $F^{\perp}$ is the orthogonal subbundle of $F$ relative to the fibre metric $g$ on $E$.
Mention that although $g$ induces a scalar product in the fibres of the bundle $F^{\perp}/F\to Q$, the quadric $Q$ inherits only a weaker conformal structure, which is preserved by the Möbius group.
Example: The complex quadric, say of (complex) dimension n, $$ Q_{n}=\{[z]\in\mathbb{C}P^{n+1} : (z,z)=0\} $$ is diffeomorphic to the Grassmannian ${\rm Gr}_{+}(2, n):={\rm SO}(n+2)/({\rm SO}(n)\times{\rm SO}(2))$ of oriented two planes in $\mathbb{R}^{n+2}$.
The complex quadric admits a holomorphic conformal structure induced by the quadratic form $(z, z)$ on $\mathbb{C}^{n+2}$. It admits also a Kähler structure, induced by the hermitian form $(z, \bar{z})$, which can be thought of as induced by the embedding of $Q_{n}$ in $\mathbb{C}P^{n+1}$. A complex quadric doesn't admit any smooth complex-bilinear Riemannian metric, although it admits a Kähler metric as we said above. Further details on complex quadrics can be found in the books of Kobayashi and Nomizu.
