Two faces of a polytope: hull and intersection
From the previous guide you already trust that a compact convex body is rebuilt from its extreme points by Krein-Milman. A polytope is what you get when that set of corners is finite. There are two equivalent definitions, and the whole subject runs on the fact that they agree. The V-description: a polytope is the convex hull of finitely many points in R^n. The H-description: it is a bounded intersection of finitely many closed half-spaces. The Weyl-Minkowski theorem says these define exactly the same objects — every bounded H-polyhedron is the hull of its vertices, and vice versa.
Why insist on both? Because they make opposite things easy. From the V-side, projecting or taking a hull is trivial but reading off facets is work; from the H-side, slicing by an inequality is trivial but listing vertices is work. The cube in R^3 is the cleanest illustration: as a hull it is conv{(+/-1, +/-1, +/-1)}, eight points; as an intersection it is {x : -1 <= x_i <= 1}, six inequalities. Same solid, two account books. Converting between them — vertex enumeration vs. facet enumeration — is the computational heart of the theory, and it can blow up exponentially.
Faces are flat pieces cut by supporting hyperplanes
The structure of a polytope lives in its faces, and the clean definition uses the tool from guide 1. A supporting hyperplane H touches the polytope P without cutting through it; the intersection F = P ∩ H is a face of P. By convention P itself and the empty set are faces too (the 'improper' ones). Each face is again a polytope, of some dimension d. We name them by dimension: a 0-face is a vertex, a 1-face an edge, a (dim P - 1)-face a facet, and a (dim P - 2)-face a ridge.
Order the faces by inclusion and you get the face lattice, a finite graded poset with a bottom (the empty face) and a top (P). This lattice is the combinatorial soul of the polytope: two polytopes are 'combinatorially the same' precisely when their face lattices are isomorphic, even if one is a stretched, sheared version of the other. The cube and a generic box have the same face lattice; the cube and the octahedron do not — but watch this — they are dual, meaning one's face lattice is the other's turned upside down. Vertices of the cube (8) match facets of the octahedron (8); facets of the cube (6) match vertices of the octahedron (6).
Two structural facts earn their keep constantly. First, every face of a face is a face of P, and the intersection of two faces is a face — so the face lattice really is a lattice. Second, a d-polytope has faces of every dimension from 0 up to d, and (this is the engine of the next section) you can count them. We collect those counts into the f-vector f = (f_0, f_1, ..., f_{d-1}), where f_k is the number of k-dimensional faces. For the cube in R^3: f = (8, 12, 6) — eight vertices, twelve edges, six facets.
Euler's relation and its generalization
Now the first miracle. For any convex 3-polytope, the face numbers are not free — they satisfy V - E + F = 2. The cube: 8 - 12 + 6 = 2. The tetrahedron: 4 - 6 + 4 = 2. The octahedron: 6 - 12 + 8 = 2. The same '2' every time, regardless of how lopsided the polytope is. This is Euler's relation, and the Euler polyhedron formula is the statement that this alternating sum is a topological invariant: the boundary of a convex 3-polytope is a sphere S^2, and 2 is its Euler characteristic.
The clean way to see it is to recognize that the formula is not about the convex shape but about the cell structure on its boundary. The boundary of a convex polytope is homeomorphic to a sphere, and a sphere triangulated (or cell-divided) any way at all gives the same alternating count. So Euler's relation is really the Euler characteristic of S^(d-1) showing through. In general dimension d this becomes the Euler-Poincare relation: f_0 - f_1 + f_2 - ... + (-1)^(d-1) f_{d-1} = 1 - (-1)^d. For d = 3 the right side is 1 - (-1) = 2, recovering V - E + F = 2; for d = 4 it gives 0, so f_0 - f_1 + f_2 - f_3 = 0.
d = 3 : f_0 - f_1 + f_2 = 1 - (-1)^3 = 2 d = 4 : f_0 - f_1 + f_2 - f_3 = 1 - (-1)^4 = 0 d = 5 : f_0 - f_1 + f_2 - f_3 + f_4 = 1 - (-1)^5 = 2 cube (d=3): 8 - 12 + 6 = 2 ok 24-cell(d=4): 24 - 96 + 96 - 24 = 0 ok
Simplicial polytopes and the Dehn-Sommerville equations
Euler is one equation. For a special, very common class of polytopes there are many more. A polytope is simplicial if every facet is a simplex — in R^3 that means every face is a triangle, like the octahedron or the icosahedron, not the cube. Simplicial polytopes are the generic case (perturb the vertices a little and facets become simplices) and they are the right setting for the deepest counting laws. The dual notion, where every vertex sits in exactly d facets, is simple; the cube is simple, the octahedron is simplicial, and they are dual to each other.
Here is the second miracle. For a simplicial d-polytope, the f-vector satisfies not one but roughly d/2 independent linear equations — the Dehn-Sommerville equations. The slick way to state them uses the h-vector, an invertible repackaging of the f-vector defined by sum over i of f_{i-1}(t-1)^(d-i) = sum over i of h_i t^(d-i). In those coordinates the Dehn-Sommerville relations collapse to a single breathtaking statement: the h-vector is palindromic, h_i = h_{d-i} for all i. Symmetry you would never have guessed from the raw face counts.
- Take the octahedron, a simplicial 3-polytope, with f = (f_0, f_1, f_2) = (6, 12, 8).
- Compute the h-vector from sum f_{i-1}(t-1)^(d-i) = (t-1)^3 + 6(t-1)^2 + 12(t-1) + 8, expanding to t^3 + 3t^2 + 3t + 1.
- Read off h = (h_0, h_1, h_2, h_3) = (1, 3, 3, 1) — and there it is, palindromic: h_0 = h_3 and h_1 = h_2.
- Check against Euler: h_0 = 1 and the alternating sum of f matches as before — Dehn-Sommerville contains Euler as its h_0 = h_d clause.
How many faces can a polytope have? The Upper Bound Theorem
Dehn-Sommerville says which f-vectors are even arithmetically possible for simplicial polytopes; the next question is how large they can grow. Fix d and the number of vertices n. Which simplicial d-polytope on n vertices has the most facets? The answer is the cyclic polytope C(n, d), the convex hull of n points on the moment curve t -> (t, t^2, ..., t^d). Cyclic polytopes are 'neighborly': every subset of up to d/2 vertices forms a face, which is the maximum possible, and that maximality is exactly what pumps up the facet count.
The Upper Bound Theorem, conjectured by Motzkin and proved by McMullen in 1970, says cyclic polytopes are the champions: among all simplicial d-polytopes with n vertices, C(n, d) simultaneously maximizes f_k for every k. The Upper Bound Theorem is not a curiosity — it is the binding constraint in the worst-case analysis of geometric algorithms. When you bound the running time of a convex hull or a Voronoi computation in R^d, the face-count ceiling it provides (roughly n^(d/2)) is exactly the term that appears.
Be honest about what is and is not known here. The Upper Bound Theorem (a ceiling) has a companion Lower Bound Theorem (a floor, Barnette), and together with Dehn-Sommerville they were folded into the celebrated g-theorem of Billera-Lee and Stanley, which gives a complete characterization of exactly which integer vectors arise as f-vectors of simplicial polytopes. That is a genuinely deep result — Stanley's half used the hard Lefschetz theorem on a toric variety, not elementary combinatorics. The analogous full classification for general (nonsimplicial) polytopes in high dimension is far less complete, a reminder that 'count the faces' opens onto live research, not a closed book.