Being a non-mathematician, excuses if I make errors in formulating this question.
Consider a general quadric surface of the implicit form: $$ f(x,y,z)=Ax^2+ By^2 + Cz^2 + 2Dxy + 2Exz + 2Fyz + 2Gx + 2Hy + 2Iz + J = 0 $$ and in the vectorial notation: $\mathbf{a}^T\mathbf{q} = 0$, where: $$ \begin{align} %\vec{q} &= [A B C D E F G H I J]\\ \vec{q} &= \begin{bmatrix} A & B & C & D & E & F & G & H & I & J \end{bmatrix}^T \\ \vec{a} &= \begin{bmatrix} x^2 & y^2 & z^2 & 2xy & 2xz & 2yz & 2x & 2y & 2z & 1 \end{bmatrix}^T \end{align} $$
It is also common to re-arrange the terms of $\mathbf{q}$ to a matrix form $Q$ such that $\mathbf{x}^TQ\mathbf{x}=0$.
Now, in essence, given a set of similar quadrics $S=\{\mathbf{q}_1..\mathbf{q}_n\}$, I would like to compute a mean quadric, which resembles all the shapes in the best way. Or I would also like to interpolate them from some 'shortest path'. Of course this is not really an easy task because averaging the coefficients doesn't guarantee a similar surface (or the mean surface), because:
The parameters have highly non-linear dependencies: Adding a perturbation of $\delta$ to a sphere-like ellipsoid doesn't cause the same shape change as perturbing a thin ellipsoid.
The geodesic distance on the manifold of the quadric coefficients (if such a thing exists) is probably not the same as a simple distance.
At this point, I would like to raise the further questions:
- Is there a manifold of these coefficients?
- If so, is it somewhat Grassmannian or Riemannian? I know that the matrix is not an SPD, but still the question remains.
- Is it possible to find a manifold distance for similarity of quadratic surfaces? If not, is it provable?