How can we compute the exact value of some rotational symmetric n-by-n convolution kernel

Question

The commonly used $3\times 3$ Laplacian convolution kernel $\frac{1}{6}\begin{bmatrix}1 & 4 & 1\\4 & -20 & 4\\1 & 4 & 1\end{bmatrix}$ is only an approximation of rotational symmetry, as mentioned in the Wikipedia article on the Discrete Laplace operator.

Also, I found the first order Sobel operator to have an exactly rotational symmetric form without any proof.

So my questions: How can we compute the exact value of a rotational symmetric $n\times n$ convolution kernel for

1. the $\nabla$ operator?
2. the Laplacian operator?

Answer 1

I don't think you can. The problem is that you represent a function $f(\mathbf x)$ on a Cartesian mesh and you apply the operator to that. If you rotate the function, i.e., you try $f(A\mathbf x)$ (where $A$ is a $2\times 2$ rotation matrix), then you will get different function values at the vertices of the mesh and different values if you apply the discrete Laplace to it. In other words, even though the Laplacian operator is rotationally invariant, the Laplace operator on a mesh is not and can not be.