Given an adjacency matrix $M$. Are there any simple properties that $M$ must have such that it represents the triangular mesh of a sphere. (examples are icosohedron, or geodesic spheres). What about for a torus?
One thing I can think of is that every line must belong to two and only two triangles. I think you would write this as $M_{ij}=1 \implies (M^2)_{ij}=2$. Are there any other formulae?
Can we express this some way in terms of perhaps the cube $M^3$ of the adjacency matrix?
We need $(M^3)_{ii}=2(M^2)_{ii}$ for a "locally planar" triangular mesh (for every edge incident with $i$, thee are two triangles with vertex $i$), but by itself this is weaker than your condition.
To distinguish between sphere and torus, we should look for the Euler characteristic $v+f-e$. We have $v$ vertices from the dimension of the matrix, $e=\frac 12\operatorname {Tr}M^2$ edges, and $f=\frac 16\operatorname {Tr}M^3$ triangular faces. By the condition in the first paragraph, $\operatorname {Tr}M^3=2\operatorname {Tr}M^2$, or (unsurprisingly) $3f=2e$., so that (given that we have a triangular mesh in the first place), the Euler characteristic is $$\dim V-\frac 16\operatorname{Tr}M^2. $$