I have a quadratic form $f(x) = x^t A x$ where A is 3x3 real symmetric and $f$ satisfies $f(x) = f(-x)$ and now restricted to $||x|| = 1$ this is a well defined map on the projective plane (when antipodal points of sphere are identified). I'm trying to find the critical points of this map $f: \mathbb{R}P^2 \to \mathbb{R}$.
I understand that to find the critical points of a map on a surface we need to look at the local form of the map and then determine when the gradient of this is zero. I.e. for a map $F: S \to \mathbb{R}$ from a surface S, then $F$ has a critical point at $p$ iff $f_{loc}(x,y) = F(\sigma(x,y))$ for some parametrisation $\sigma$.
Using local coordinates on the sphere $(x,y) \to (x, y, (1-x^2 -y^2)^{1/2})$ and then locally we have $$f_{loc} (x,y) = (x,y, (1-x^2 -y^2)^{1/2})^t A (x, y, (1-x^2 -y^2)^{1/2})$$
I'm struggling to find the critical points of this. I'd quite like to diagonalise $A$ but I'm not sure what coordinates to use to do this when $f$ is written locally?
Thanks
Suppose the eigenvalues of $A$ are $\lambda_1\leq\dots\leq\lambda_n$, and $e_1,\dots,e_n$ are corresponding orthonormal eigen vectors. Any $x\in\mathbb{S}^{n-1}$ can be written as $x=\sum x_ie_i$. With frame {$e_i$}, $A$ is diagonalized, and $xAx^t=\sum\lambda_ix_i^2$. Let $x\in\mathbb{S}^{n-1}$, and $v$ be any vector tangent to $\mathbb{S}^{n-1}$. Take $\gamma(t)\subset\mathbb{S}^{n-1}$ such that $\gamma(0)=x$, and $\gamma'(0)=v$. Then $\frac{d}{dt}\gamma(t)A\gamma(t)^t=2\sum\lambda_ix_iv_i$. In order to make $x$ a critical point, we need to have $(\lambda_1x_1,\dots,\lambda_nx_n)$ is perpendicular to any $v$, I.e., parallel to $x$. Particularly, all the eigen vectors are critical points.