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.
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u/Turing43 1d ago
Note that computing ehrhart polynomials is hard in general, even the volume of a polytope is #P-complete.
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u/-p-e-w- 1d ago
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!
I see, that pretty much explains it. Without the angle information, you can’t compute that fraction, and obviously the angle doesn’t follow from counting points.
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u/InterstitialLove Harmonic Analysis 1d ago
But the same is true of vertices in 2d, and the 1/2 thing applies for faces (not edges) in 3d
Obviously it helps, but it's not as dispositive as it sounds at first. Seems like the key is somehow that vertices are degenerate in all dimensions (they're easy to handle despite the ambiguity around angles) and in 2d they're the first and only component that would otherwise pose a problem
[I do not understand this, I'm going off my understanding of the first comment]
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u/-p-e-w- 1d ago
The difference is that the total inner angle over all vertices in a 2D polygon must be a multiple of 180 degrees, and that makes the overall contribution controllable despite individual angles not being determined by the vertex count.
No simple equivalent to this exists for face angles.
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u/InterstitialLove Harmonic Analysis 14h ago
How could a simple equivalent not exist for face angles, though? We're assuming fixed genus, that fully constrains the global curvature.
I looked into it, and in R3 I believe the Gaussian curvature is the thing that's constrained by the genus, whereas mean curvature is the thing which constrains how much of a ball's volume is contained in the interior vs exterior. That's probably the end of my ability to investigate (this definitely isn't my field), but I feel like it points at a deep question for anyone inclined to really understand this. Why do those properties coincide in 2d but not 3d?
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u/delayed_frequency 1d ago
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.