Consider the projective model of $HP^{n}$ it is an open subset of the projective space, as an open subset we can restrict the FS form of $\mathbb{C}P^{n}$, given by the hermitian inner product of signature $(1,n)$ in $\mathbb{C}^{n+1}$.
What I want to prove is that this form restricted to the hyperbolic is invariant under the action of $U(1,n)$.
I read on Epstein's Analytical and Geometric aspects of hyperbolic space that the tangent space of a point in the hyperbolic is isomorphic to the ortogonal space of that point, considering any lift. Is this because you can see the projective space as a quotient of the unit sphere? I was thinking to use this representation to try proving the invariance.
Or if anyone knows another path it would be more that welcome.
This should just be a simple calculation in multivariable calculus. The calculation takes place entirely in the indefinite inner product space, not in the projective space.
Let me state what happens in the real hyperbolic setting as opposed to the complex hyperbolic setting. The only differences arise in calculus and linear algebra over $\mathbb{C}$ rather than over $\mathbb{R}$, but the notation is simpler over $\mathbb{R}$.
Consider $\mathbb{R}^{n+1}$ with coordinates $x_1,...,x_{n+1}$, and use the bilinear pairing $$x \cdot y = x_1 y_1 - x_2 y_2 - ... - x_{n+1} y_{n+1} $$ with associated indefinite norm of signature $(1,n)$ given by $$|x|^2 = |x_1|^2 - |x_2|^2 - ... - |x_{n+1}|^2 $$ We are considering one sheet of the two-sheeted hyperboloid, namely $$\mathbb{RH}^n = \{x \in \mathbb{R}^{n+1} \mid |x|^2=1, \, x_1 > 0\} $$ Given $p \in \mathbb{RH}^n$, the fact that the tangent space of $\mathbb{RH}^n$ at $p$ is the orthogonal subspace at $p$ should be a simple multivariable calculus problem. By definition $\mathbb{RH}^n$ is a level set of $|x|^2$ passing through $p$. And a simple calculation shows that the tangent space of that level set at $p$ is equal (up to a constant factor of $2$) to the orthogonal subspace of the vector $p$, where "orthogonality" is determined using the indefinite inner product.
The courage to believe this comes from the fact that it's all obvious using the standard Euclidean norm on $\mathbb{R}^{n+1}$, using the unit sphere $S^n$ as a level set of the norm: at a point $p \in S^n$, a calculation shows that $p$ equals the gradient of the norm at $p$ (up to a factor of $2$); and the tangent space of $S^n$ at $p$ equals the orthogonal subspace of the gradient vector.