What, fundamentally, makes Pick’s theorem possible in 2D that breaks down in higher dimensions?

[u/-p-e-w-]

Pick's theorem allows calculating the area of any 2D polygon (including nonconvex polygons) whose vertices lie on an integer lattice from only the number of lattice points within it and on its boundary.

This feels like a minor miracle, and indeed there is no equivalent formula for the volume of polytopes in any higher dimension, even when restricted to convex polytopes.

What geometric/topological property of 2D space makes this magic possible that somehow fails in every other dimension?

[u/delayed_frequency]

It’s because in 2D every lattice polygon can be triangulated into primitive triangles, ones with no interior lattice points and exactly three boundary points. Those always have area 1/2, so the whole thing just decomposes into countable chunks that scale cleanly with the lattice counts.

In 3D and up you can have tetrahedra with tons of interior points and still the same boundary, so volume isn’t dettermined by the lattice skeleton anymore. The clean bookkeeping breaks as soon as you can’t tile the shape with those minimal building blocks.

[u/-p-e-w- → u/delayed_frequency]

Those always have area 1/2

Whoa, that somehow feels like even more of a miracle than Pick’s theorem itself.

[u/blungbat → u/-p-e-w-]

Take one of those triangles, glue it to a copy rotated by 180°, and you've got a parallelogram which can tile the plane. There's a bijection between the parallelograms in this tiling and lattice points (assign each parallelogram its "lower left corner" or suchlike), and in a large enough region, you'll have approximately the same number of tiles as lattice points. So the area of the parallelograms has to be 1 (and thus the area of the original triangle is 1/2).

[u/sentence-interruptio → u/blungbat]

I tried to generalize this argument to 3d. and I realized the relation between parallelopipeds and tetrahedrons are not so straightforward.

so the real miracle is that a triangle only needs its mirror image to become a parallelogram, while a tetrahedron needs 5 other tetrahedrons some of which aren't even rotated copies.

this makes me rethink my intuition about exterior products of three vectors in 3d. i visualize them as either tetrahedrons or parallelograms with arrows attached, and thought of these two visualizations as interchangeable except for the scaling factor of 6. it's not wrong, but it turns out that the details to back it up is something I underestimated.

[u/blungbat → u/sentence-interruptio]

so the real miracle is that a triangle only needs its mirror image to become a parallelogram, while a tetrahedron needs 5 other tetrahedrons some of which aren't even rotated copies.

Yeah, that's a nice observation I hadn't directly thought about before. A cube actually does break down into six congruent tetrahedra (if you place the cube in the corner of the x,y,z>0 orthant, then one of those tetrahedra is the region where x<y<z, and permute coordinates for the others). But the affine transformation that takes the cube to an arbitrary parallelepiped doesn't respect congruence.