Differences in mean curvatures for permuted coordinates of a manifold

Question

Below is an example demonstrating the differences in mean curvatures and one other quantity, and below is my question. Please comment if you need me to explain in more detail or clarify notation, or if there are any mistakes.

Example 1 Consider the manifold with coordinates given by $$(x,y,x+y, x^2).$$ At each point, there are two unit normal vectors $$n_1 = \frac{\sqrt{3}}{3} [1, 1, -1, 0]^T$$ and $$n_2 = \frac{\sqrt{3}}{3\sqrt{8x^2+3}} [4x,-2x,2x,-3].$$ The vector of mean curvatures in the directions of each of these normals is given by $$c=\left[0, -\frac{2\sqrt{3}}{(8x^2+3)^{3/2}}\right]^T.$$ Further, if we let $q$ be the vector with components $q^r = \operatorname{Tr}{N (Dn_r) N^T}$ where $Dn_r$ is the Jacobian derivative of the vector $n_r$, $r=1,2$ and $N$ is the matrix with $r$-th row equal to $n_r$, then we get that $q=\vec{0}$.

Brownian motion on this manifold has drift $\mu =N^T(q+c)$, i.e. $$\mu = \frac{1}{(8x^2+3)^2}[-8x, 4x, -4x, 6]^T.$$

Example 2 Now, permute the last two coordinates so that instead we have $$(x,y,x^2,x+y).$$ Now, we can obtain unit normal vectors $$n_1 = \frac{1}{\sqrt{4x^2+1}} [2x,0,-1,0]^T,$$ $$n_2 = \frac{1}{\sqrt{(4x^2+1)(8x^2+3)}}\left[1, 4x^2+1, 2x, -(4x^2+1)\right]^T,$$ mean curvature vector $c$, $$c=\left[-\frac{1}{(4x^2+1)^{3/2}}, -\frac{2x(16x^2+5)}{(4x^2+1)^{3/2}(8x^2+3)^{3/2}}\right]^T$$ and now a non-zero $q$, as defined above, $$q=\frac{1}{\sqrt{(4x^2+1)(8x^2+3)}}\left[\frac{1}{(4x^2+1)\sqrt{8x^2+3}}, -\frac{2x}{4x^2+1}\right]^T$$ and drift of a Brownian motion $$\mu = \frac{1}{(8x^2+3)^2}[-8x, 4x, -4x, 6]^T,$$ again from $\mu=N^T(q+c)$.

My questions:

1. Why are the mean curvatures different for these two permutations of coordinates of, essentially, the same manifold? (From a probabilist perspective, I get the same SDE for Brownian motion on both manifolds--this might be a blasphemous viewpoint to differential geometers, so please excuse my ignorance!)

2. What is the geometric interpretation of $q^r = \operatorname{Tr}(N (D n_r)N^T)$ and why does it change depending on the permutation of coordinates (or rather on the order of orthonormalizing the rows of $Df$ where $f(x,y,z,w)=0$ defines the manifold, via component definitions e.g. $f^1(x,y,z,w)=x+y-z$ and and $f^2(x,y,z,w)=x^2-w$) for the first presentation.

Answer 1

Question 1: The mean curvature measures the average deviation of the surface from its tangent planes, and it depends on the normal vectors and their derivatives. In Example 1, the normal vectors have a specific form that is different from Example 2, and this difference in normal vectors leads to different mean curvature vectors. Essentially, the way the surface deviates from its tangent planes is different for the two examples due to the different normal vectors.

Provided your probabilistic perspective, the SDE for Brownian motion only depends on the Riemannian metric of the manifold, which is the same for both examples. However, the drift term of the SDE, which depends on the mean curvature, is different for the two examples due to the difference in normal vectors and mean curvature vectors.

Answer 2

Here's the way I compute this as a differential geometer. I differentiate the parametrization, not the normal vector, to arrive at the second fundamental form. With the parametrization $$g(x,y)=(x,y,x+y,x^2),$$ I get $g_x = (1,0,1,2x)$ and $g_y = (0,1,1,0)$ spanning the tangent space, with corresponding orthonormal normal vectors $$N_1 = \frac1{\sqrt3}(1,1,-1,0) \quad\text{and}\quad N_2 = \frac1{\sqrt{9+24x^2}}(4x,-2x,2x,-3).$$ You and I have chosen the same normal vectors. The corresponding second fundamental forms are $$\text{II}_j = \begin{bmatrix} g_{xx}\cdot N_j & g_{xy}\cdot N_j \\ g_{yx}\cdot N_j & g_{yy}\cdot N_j\end{bmatrix},$$ and so the second-fundamental form is the vector-valued $2$-tensor $$\text{II}_1\otimes N_1 + \text{II}_2\otimes N_2 = \begin{bmatrix} 0 & 0\\0 & 0\end{bmatrix} \otimes N_1 + \frac{-6}{\sqrt{9+24x^2}}\begin{bmatrix} 1 & 0\\0 & 0\end{bmatrix} \otimes N_2.$$ The mean-curvature vector will be $$H=\text{tr}(\text{II}_1)N_1 + \text{tr}(\text{II}_2)N_2 = \frac{-6}{\sqrt{9+24x^2}}N_2 = \frac{-2}{3+8x^2}(4x,-2x,2x,-3).$$ This seems to agree with your result.

If I change the surface, as you did, by switching the final two coordinates, those two coordinates switch in everything I've computed and nothing else changes; I end up with $H'=\dfrac{-2}{3+8x^2}(4x,-2x,-3,2x)$. I continue not to understand what you've done. Why did you choose such arcane normal vectors now? Why didn't you just choose the corresponding normal vectors in the first case and compare the mean-curvature vectors with the two different orthonormal bases?