Euler characteristic

Removing one face of a convex polyhedron and stretching the remaining surface flat onto a plane produces a connected planar graph whose bounded faces correspond to the remaining $F - 1$ faces of the polyhedron, while the unbounded outer region represents the removed face.

Following Cauchy (1813), project the polyhedron onto a plane as a connected planar graph with $V$ vertices, $E$ edges, and $F$ faces (including the outer face). If the graph contains any cycle, deleting one edge on that cycle merges two adjacent faces into one, reducing both $E$ and $F$ by $1$ while leaving $V$ and hence $V - E + F$ unchanged. Repeat until no cycles remain; the resulting connected acyclic graph is a tree, which has $F = 1$ (only the outer face) and satisfies $E = V - 1$. Therefore $V - E + F = V - (V - 1) + 1 = 2$.

This theorem makes the Euler characteristic a complete topological fingerprint for closed 2-manifolds (once orientability is known): to decide what surface a complicated polygon-gluing produces, you merely compute $V - E + F$ and check whether the gluing reverses orientation!

Triangulate the compact connected surface $S$ into finitely many triangles and glue them edge-to-edge; cutting the triangulated surface open along a spanning tree of its dual graph produces a single polygon $P$ with $2n$ directed boundary edges, grouped into $n$ identified pairs. Repeatedly zipping together an adjacent pair of the form $a a^{-1}$ (which removes a fold with no effect on the surface) and cutting-and-pasting along a diagonal whenever two like-labelled edges are separated, one reduces $P$ to one of two canonical normal forms: the orientable word $a_1 b_1 a_1^{-1} b_1^{-1} \cdots a_g b_g a_g^{-1} b_g^{-1}$, producing the surface $\Sigma_g$ of genus $g$; or, if an orientation-reversing pair appears, the non-orientable word $c_1 c_1 \cdots c_k c_k$, producing $N_k$, the connected sum of $k$ projective planes.

In the orientable normal form, all $4g$ corners of the polygon are glued to a single point, so $V = 1$; the $4g$ boundary edges pair up into $E = 2g$; and the interior of the polygon remains one face, $F = 1$. Hence $\chi(\Sigma_g) = V - E + F = 1 - 2g + 1 = 2 - 2g$. In the non-orientable normal form, all $2k$ corners again meet at one vertex ($V = 1$), the $2k$ edges pair into $E = k$, and again $F = 1$, so $\chi(N_k) = 1 - k + 1 = 2 - k$. Since $g \mapsto 2 - 2g$ is strictly decreasing and $k \mapsto 2 - k$ is strictly decreasing on the non-negative integers, each is injective; therefore knowing whether $S$ is orientable together with its Euler characteristic $\chi(S)$ determines $g$ or $k$ uniquely, and hence determines the homeomorphism type of $S$ completely.

The left-hand side $\int_M K\,dA$ is purely differential-geometric (changing from point to point as you dent or stretch the surface), while the right-hand side $2\pi\,\chi(M)$ is a discrete topological integer times $2\pi$. If you dent a sphere, regions of positive curvature increase only at the exact expense of new saddle regions of negative curvature, keeping the total integral locked at $4\pi$; on a torus ($\chi = 0$), positive outer curvature and negative inner curvature always cancel to $0$.

Choose a smooth geodesic triangulation of the closed orientable surface $M$, with $V$ vertices, $E$ edges, and $F$ triangular faces $T_1, \dots, T_F$, fine enough that each triangle lies inside a single coordinate chart. Applying Green's theorem to the position vector along the boundary of a small triangle gives the local Gauss–Bonnet identity: for each face $T_f$ with interior angles $\alpha_{f,1}, \alpha_{f,2}, \alpha_{f,3}$ and geodesic curvature $\kappa_g$ along its boundary $\partial T_f$, $\displaystyle\iint_{T_f} K\,dA + \int_{\partial T_f} \kappa_g\,ds = (\alpha_{f,1}+\alpha_{f,2}+\alpha_{f,3}) - \pi$.

Summing this identity over all $F$ triangles, the geodesic-curvature line integral along each interior edge is traversed exactly twice, once from each adjacent triangle and in opposite directions, so these boundary terms cancel completely and $\displaystyle\iint_M K\,dA = \sum_{f=1}^F (\alpha_{f,1}+\alpha_{f,2}+\alpha_{f,3}) - F\pi$. The angles surrounding each of the $V$ vertices sum to a full turn, so the grand total of all interior angles equals $2\pi V$. Because every triangle has $3$ sides and every edge borders exactly $2$ triangles, $3F = 2E$, i.e. $F\pi = (2E - 2F)\pi$. Substituting both facts into the summed identity gives $\displaystyle\iint_M K\,dA = 2\pi V - 2\pi E + 2\pi F = 2\pi(V-E+F) = 2\pi\,\chi(M)$.

This explains why a steady wind pattern on Earth ($S^2$, $\chi = 2$) must always have at least one calm eye (a cyclone or anticyclone of total index $2$), whereas a torus ($T^2$, $\chi = 0$) can be combed completely flat by a nowhere-vanishing vector field running along its longitudes!

To prove the Poincaré–Hopf index formula $\sum_{v(p)=0} \operatorname{ind}_p(v) = \chi(M)$, replace $v$ by a homotopic vector field built from a Morse function on a fine triangulation of $M$: a source of index $+1$ at each of the $V$ vertices, a saddle of index $-1$ at the midpoint of each of the $E$ edges, and a sink of index $+1$ at the centre of each of the $F$ faces. Because the index of a vector field restricted to the boundary of a small disk around each zero is a homotopy invariant, and $v$ can be continuously deformed into this canonical cellular field without ever crossing zero outside the disks already accounted for, the sum of the indices of $v$ equals the sum of the indices of the canonical field.

Adding up the canonical indices gives $V(+1) + E(-1) + F(+1) = V - E + F = \chi(M)$, so $\sum_{v(p)=0} \operatorname{ind}_p(v) = \chi(M)$ holds for every closed surface $M$. Specialising to the sphere, $\chi(S^2) = 2 \neq 0$. If a continuous tangent vector field on $S^2$ had no zero anywhere, the left-hand side of the index formula would be the empty sum $0$, contradicting $\chi(S^2) = 2$. Therefore every continuous tangent vector field on $S^2$ must vanish at some point $p \in S^2$ — you cannot comb a hairy ball flat without a cowlick.