Pearson Correlation of the Principal Curvatures

Question

Consider the set of all possible smooth 3D surface patches. Now, let's say that I draw $K$ random points from that set, where each point is parametrized by its principal curvatures $\kappa_1$ and $\kappa_2$. So, I have $K$ points, where each is embedded as a vector in $\mathbb{R}^2$, and I stack all of them into a matrix $A$ of size $K \times 2$. By calculating the Pearson correlation coefficient of the two columns of $A$, I can estimate the correlation between the two principal curvatures.

Intuitively, I would expect the correlation to approach $0$ as $K \to \infty$, since both $\kappa_1$ and $\kappa_2$ can take any value, regardless of the other. In other words, in my intuition, picking a random point from the family of smooth manifolds is equivalent (at least for me) to drawing two random real numbers, one after the other, which represent the principal curvatures.

However, I cannot show it through numeric experimentation. In my experiments, where I follow the exact same process as described above for $K > 10^6$, I get that the correlation is around $0.5$, which is counter-intuitive, at least for me.

Is it possible to prove that the correlation should be $0$? Or is my intuition wrong?

Answer 1

As stated in the comments, you need to propose a clear and unequivocal definition of the parametrization and the measure you want to use to define the random patches. Nevertheless, I think it is possible to shed some light on the question and maybe also on your intuition by proposing a simple approach to randomisation.

Let us use Monge patches around the pount $(u,v)=(0,0)$ and let's specialize to quadratic surfaces, and choose coordinates such that they are orthogonal in $(0,0)$. To make these random in a well-defined fashion, write: $$ h(u,v) = \frac X 2 u^2 + \frac Y 2 v^2$$ where $X, Y$ are random variables with finite first and second moments. Each random instance of $X,Y$ will provide you with a random patch of graph $(u,v,h(u,v))$ around $(0,0).$ The principal curvatures are then two random variables. Their correlation is $$ \text{Corr}(\kappa_1,\kappa_2) = \frac{\text{Cov}(\kappa_1,\kappa_2)}{\sqrt{\text{Var}(\kappa_1)\text{Var}(\kappa_2)}}.$$

To go further, one needs to fix the measure, i.e. define the joint distribution of $(X,Y).$ There is basically no right or wrong distribution, which one to use is just a matter of choice. Of course, to obtain closed formulas, one needs simple assumptions. Following the preferences of the OP those will be:

1. $X$ and $Y$ are independent.
2. $X$ and $Y$ are uniform on the interval $[0,L]$ for an $L>0.$

Using the formulas cited above one can conclude that Gauss and mean curvature at the point $(0,0)$ are $$ K = XY\quad\text{ and }\quad H=\frac 1 2 (X+Y).$$

Furthermore, the principal curvatures can be expressed in terms of $H$ and $K$ as $$ \kappa_1=H + \sqrt{H^2-K}\quad\text{ and }\quad \kappa_2=H - \sqrt{H^2-K}.$$

Using those formulas conclude: $$ \sqrt{H^2-K} = \frac{1}{2} \lvert X - Y \rvert$$ $$ \kappa_1 = H + \sqrt{H^2-K} = \max\{X,Y\}$$ $$ \kappa_2 = H - \sqrt{H^2-K} = \min\{X,Y\}.$$

Now plug those into the definition of covariance: \begin{align} \text{Cov}(\kappa_1,\kappa_2) &=\mathbb{E}[\kappa_1 \kappa_2] - \mathbb{E}[\kappa_1] \mathbb{E}[\kappa_2]\\ & = \mathbb{E}[K] - \big( \mathbb{E}[H]^2 - \mathbb{E}[\sqrt{H^2 - K}]^2 \big)\\ &= \mathbb{E}\Big[\frac{1}{2} \lvert X - Y \rvert\Big]^2 \geq 0. \end{align} For the last line use $\mathbb{E}[K] = \mathbb{E}[X] \mathbb{E}[Y] = \mathbb{E}[\frac 1 2 (X+Y)]^2 = \mathbb{E}[H]^2$ due to independence assumption. This shows already the problem with the OP's intuition: The principal curvatures will always have non-negative correlation. Furthermore, since they are max and mins of a common pair of variables, they will be independent only in trivial cases!

Finally obtain by straightforward integration: $$ \text{Cov}(\kappa_1,\kappa_2) = \mathbb{E}\Big[\frac{1}{2} \lvert X - Y \rvert\Big]^2 = \frac 1 4 (\frac 1 3 L)^2 = \frac 1 {36} L^2$$ and $$ \text{Var}[\kappa_1] = \text{Var}[\kappa_2] = \frac 1 {18} L^2$$ to arrive ultimately at $$ \text{Corr}(\kappa_1,\kappa_2) = \frac{\frac 1 {36} L^2}{\frac 1{18}L^2}=\frac 1 2.$$ The result seems to be in full agreement with the observations of the OP.

As a quick quality check, verify the results by simulation:

n = 1e6

f_sample = function(L){ 
  X = L * runif(n)
  Y = L * runif(n)
  
  K = X * Y
  H = (X + Y) / 2
  
  kappa_1 = H + sqrt(H^2-K)
  kappa_2 = H - sqrt(H^2-K)
  
  cor(kappa_1, kappa_2)
}

tt = sapply(c(1,10,100,1000), f_sample)
print(tt)
[1] 0.4991383 0.5012970 0.5004247 0.5005081

As a final remark, observe that the distributions are not invariant under change of coordinates. It may be more aligned with the geometric spirit to specify distributions for the invariants instead, and then work backwards to coordinates or to properties of the patch. But this seems to me much harder.