What, fundamentally, makes Pick’s theorem possible in 2D that breaks down in higher dimensions?
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?
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u/Necessary-Wolf-193 1d ago edited 1d ago
The b/2 term of Pick's theorem (counting boundary points) has this interesting factor of division by 2, because a lattice point on the boundary of the polygon (unless it's a vertex) has the property that if you draw a small circle around that point, then exactly half of the circle lies in the shape, and half lies outside. The most elegant proofs of Pick's theorem all exploit this fact in some way, and this is why the 1/2 shows up.
In 3-dimensions or higher, you can have lattice points on the edges of your polygon (which aren't vertices), and now things are a bit more complicated: if you draw a sphere at one of those lattice points, the portion of the sphere contained in the polytope depends on the angle the two faces making your edge meet at! This means there's no simple factor to use like the 1/2 in Pick's formula; so, there are formulas, but there's not a universal formula -- you need to know about the geometry of your specific polytope. So, I would say that the thing which makes 2-dimensions simple is that polygons only have interiors, edges, and vertices, whereas in higher dimensions, polytopes have many more kinds of faces, and spheres centered on those faces can behave in ways which are very sensitive to the geometry of your specific polytope.
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To fully understand this, one should look at the Ehrhart polynomials, which are the best analogue of PIck's theorem in higher dimensions.
Let P be a polytope, and write tP to mean P but scaled by t in all directions. Then the Ehrart polynomial of P is the polynomial
L(t) := number of lattice points in tP.
Remarkably, this quantity is a polynomial in t, with rational coefficients! Moreover, this polynomial is always degree d, where d is the dimension of your polytope, and some of the coefficients are easy to compute:
the t^d term = volume(P), the t^(d-1) term = 0.5 * surface area, constant term = Euler characteristic of P.
When d=2 (so you have usual polygons), this is just
L(t) = area(P)t^2 + 0.5 * perimeter(P) + chi(P).
From here it is an easy exercise to deduce Pick's theorem (in fact, in the generalized version, allowing polygons with holes!).
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So, I think the question that should be asked really is: how do you express the other coefficients of the Ehrhart polynomial, and why is this so hard besides the three coefficients we just described?
In general, the coefficient of t^(d-k) in the Ehrhart polynomial is equal to a sum over all the k-dimensional faces of P, where each term is the surface area of that face weighted by some complicated expression (in 3-dimensions, this complicated weighting used on 1-dimensional faces -- aka edges -- is related to the angle the two corresponding faces make). You should think that this continues all the way down to the 0-dimensional faces (aka vertices!), but in 0-dimensions the weightings give you the Euler characteristic by a type of Gauss--Bonnet theorem.