Euler's formula for polytopes

Remarks

With $f_{-1} = 1$ for the empty face and $f_d = 1$ for $P$, the relation reads $\sum_{i=-1}^{d} (-1)^i f_i = 0$. One proof is topological: the boundary of $P$ is homeomorphic to the sphere $S^{d-1}$, and its faces form a cell decomposition. So the alternating sum is the Euler characteristic $1 + (-1)^{d-1}$ of the sphere. A combinatorial proof builds the boundary facet by facet in a shelling order and checks that the alternating sum changes correctly at each step.

For $d = 3$ the formula is Euler's polyhedron formula $f_0 - f_1 + f_2 = 2$. For $d = 4$ the alternating sum is $0$; the $4$-cube has $16 - 32 + 24 - 8 = 0$. For a polygon with $n$ vertices, $n - n = 0 = 1 - (-1)^2$.

It is the only linear equation satisfied by the f-vectors of all $d$-polytopes, up to scaling. For simplicial polytopes there are further linear relations, the Dehn–Sommerville equations.

Going deeper

For a student

The relation $\sum (-1)^i f_i = 1 - (-1)^d$ is the Euler–Poincaré relation. It is the first constraint any candidate f-vector must meet, and in dimension $3$ together with the inequalities $f_2 \le 2 f_0 - 4$ and $f_0 \le 2 f_2 - 4$ it characterises the f-vectors of $3$-polytopes. Its proof by shellings was historically delicate: early arguments assumed that the boundary of a polytope can be shelled, which was proved only later by Bruggesser and Mani. The same alternating sum, for general cell complexes, is the Euler characteristic of algebraic topology.