Isotropic covariance functions on a lattice in $R^n$

Question

In many spatial statistics references, isotropic covariance functions are usually defined in terms of Euclidean distance. But does this generalize to different norms on $R^n$? For example, if we took the exponential covariance function,

$$C(\textbf{s},\textbf{t}) = \exp(-||\textbf{s}-\textbf{t}||)$$

does, $C(\cdot,\cdot)$ remain isotropic on $R^n$ if $||\cdot||$ was for example, the $\ell_\infty$ norm, $||\textbf{s}-\textbf{t}|| = \max\limits_{1\leq i \leq n}|s_i-t_i|$? My original guess was yes, because all norms are equivalent on $R^n$. However, I might be misinterpreting what isotropy means here. I believe the official definition requires that $C(\textbf{s},\textbf{t})$ is invariant under rotations, but I'm not sure how that would apply to the $\ell_\infty$ norm.

Answer 1

The short answer: For $l_p$ norms where $p\neq 2$, the only linear transformations that preserve the norm are signed permutation matrices. It's only for $l_2$ that orthogonal matrices (and therefore rotations) preserve the norm. So the covariance function $$ C(\mathbf{s},\mathbf{t}) = \exp(-\lVert\mathbf{s}-\mathbf{t} \rVert_p) \neq C(\mathbf{R}\mathbf{s},\mathbf{R}\mathbf{t})$$ is not isotropic, as rotations $\mathbf{R}$ change the covariance.

The long answer: The question is quite frequent in linear algebra, see for example this mathoverflow post.