I recently learned about Gaussian processes and the squared or non-squared exponential kernel, but also that there are finite metric spaces for which these functions do not constitute covariance matrices, i.e. $A_{ij}:=\exp(-d(i,j))$ resp. $A_{ij}:=\exp(-d(i,j)^2)$ are not positive semidefinite.
Is there a way to gain more intuition about why this approach fails sometimes and how metric spaces in general look like where it doesn't? Are there homeomorphisms $f:[0,\infty)\to(0,1]$ such that $A_{ij}:=f(d(i,j))$ is positive semidefinite for all metric spaces? Is there a special reason to use the squared or non-squared exponential kernel in Gaussian processes or are there some 'naturally occuring' processes, which give rise to these covariance matrices?
[EDIT] When I look at samples from Gaussian processes on $\mathbf{R}$ with squared resp. non-squared exponential kernel, they look smooth resp. continuous but very non-smooth, and it makes somehow sense that the squared exponential kernel does not work with the $1$-norm on $\mathbf{R}^2$, since this would give rise to a process that looks smooth along the horizontals and verticals, but very non-smooth along the diagonals.
Somehow I have the idea that positive semidefiniteness of a symmetric matrix with non-negative entries and ones on the diagonal means that I cannot weave positive and negative weights too closely together. If v is some non-zero weights vector, then $\sum_i\text{sign}(v_i)A_{ij} v_j$ denotes how strongly the node $i$ is 'influenced' by nodes of opposite sign, a positive value means that the total incluence does not outweigh the value of $i$. This sum is then weighted by $|v_i|$, so that we care more about nodes with large weights being outweighted by their surroundings. Does that make any sense? I cannot bring this together with covariance though.