The Lumenary · Early undergraduate

Surfaces, polyhedra and the Euler characteristic

Name: ______________________

Date: ______________________

You will learn to

Prove Euler's formula $V - E + F = 2$ for connected plane graphs, and deduce that there are exactly five Platonic solids
Use Euler's formula to prove that $K_5$ and $K_{3,3}$ are not planar and that a football-shaped polyhedron has exactly twelve pentagons
Build the sphere, torus, Klein bottle and projective plane by gluing the edges of a square, and decide which are orientable
Compute the Euler characteristic of surfaces given as polygons with edge identifications, and identify them using the classification theorem
Explain why the Euler characteristic is a topological invariant and how it is tied to curvature (Descartes, Gauss–Bonnet)

1.
A convex polyhedron has $30$ edges and $12$ faces. How many vertices does it have?
2.
A football is stitched from $12$ pentagons and $20$ hexagons, with three faces meeting at every vertex. How many vertices does it have?
3.
A plane graph consists of three separate triangles, drawn side by side without touching. Compute $V - E + F$ (remember the outer face).
4.
Glue the sides of an octagon according to the word $a\,b\,c\,d\,a^{-1}b^{-1}c^{-1}d^{-1}$ . All eight corners turn out to be glued to a single vertex. The surface is orientable. What is its genus $g$ ?
5.
Which of these closed surfaces are orientable? (Choose all that apply.)
- A.The torus $aba^{-1}b^{-1}$
- B.The Klein bottle $aba^{-1}b$
- C.The sphere $aa^{-1}bb^{-1}$
- D.The projective plane $abab$
- E.The surface $abcdABCD$ (capitals for inverses)
6.
In the proof that there are only five Platonic solids we reach $(p - 2)(q - 2) < 4$ . Why does this force $p \le 5$ ?
- A.Because $q \ge 3$ , so $q - 2 \ge 1$ , hence $p - 2 < 4$
- B.Because $V - E + F = 2$ implies $p \le 5$ directly
- C.Because regular polygons with more than five sides do not exist
- D.Because $pF = 2E$ forces $p$ to be even
7.
In the Lean theorem platonic_solids, which hypothesis encodes every face is a $p$ -gon (with $F$ faces and $E$ edges)?
- A.p * F = 2 * E
- B.F = p * E
- C.p * E = 2 * F
- D.q * V = 2 * E
8.
The hexagon word $a\,b\,a^{-1}b^{-1}c\,c$ describes the connected sum of a torus and a projective plane. By the classification theorem, which surface is it?
- A. $N_3 = \R P^2 \mathbin{\#} \R P^2 \mathbin{\#} \R P^2$ (Dyck's surface)
- B.The torus
- C.The Klein bottle
- D.The orientable surface of genus 2
9.
Write euler_char(word) for a polygon word such as 'abAB' (lower case: side traversed along its arrow; upper case: against it; each letter used twice). Return $\chi = V - E + 1$ , where $V$ is the number of corner classes after gluing. Corner $i$ is the start of side $i$ ; a side running along its arrow goes from corner $i$ to corner $i + 1$ (mod the number of sides), one running against it goes the other way, and gluing two sides identifies their tails and their heads.
Computational — Practice

Answer key — Surfaces, polyhedra and the Euler characteristic

1.
$V = 2 + E - F = 2 + 30 - 12 = 20$ . (These are the numbers of the dodecahedron: $20$ vertices, $30$ edges, $12$ pentagonal faces.)
2.
The faces have $5 \cdot 12 + 6 \cdot 20 = 180$ corners in total, and each vertex is a corner of three faces, so $V = 60$ . Similarly $E = 180/2 = 90$ , and $V - E + F = 60 - 90 + 32 = 2$ . This is the truncated icosahedron, the shape of $\mathrm{C}_{60}$ .
3.
$V = 9$ , $E = 9$ , $F = 4$ , so $V - E + F = 4$ . In general a plane graph with $k$ connected components has $V - E + F = k + 1$ : joining the components by $k - 1$ new edges makes it connected without creating faces. The connectedness hypothesis in Euler's formula is necessary.
4.
$\chi = 1 - 4 + 1 = -2$ , and since the surface is orientable (each letter appears once forwards and once backwards), $2 - 2g = -2$ gives $g = 2$ : the double torus.
5.
Orientable: the torus, the sphere and the genus-2 surface. Non-orientable: the Klein bottle and the projective plane, each of which has a pair of sides glued 'with a twist'.
6.
Since $q \ge 3$ , $q - 2 \ge 1$ , so $(p - 2) \le (p - 2)(q - 2) < 4$ and $p \le 5$ ; symmetrically $q \le 5$ . Without the hypothesis $q \ge 3$ (e.g. $q = 2$ , a 'polyhedron' made of two $p$ -gons glued along their boundary) every $p$ would be allowed.
7.
Each face has $p$ edges, giving $pF$ pairs; each edge lies on two faces, giving $2E$ pairs. So p * F = 2 * E. The companion hypothesis q * V = 2 * E counts (vertex, edge) pairs.
8.
Non-orientable with $\chi = -1$ , so it is $N_k$ with $2 - k = -1$ : $k = 3$ . This is Dyck's theorem $T \mathbin{\#} \R P^2 \cong \R P^2 \mathbin{\#} \R P^2 \mathbin{\#} \R P^2$ — in the presence of a cross-cap, a handle can be traded for two more cross-caps. It also explains why the classification lists no 'mixed' surfaces.

def euler_char(word):
    n = len(word)
    parent = list(range(n))

    def find(x):
        while parent[x] != x:
            x = parent[x]
        return x

    ends = {}
    for i, ch in enumerate(word):
        tail, head = i, (i + 1) % n
        if ch.isupper():
            tail, head = head, tail
        ends.setdefault(ch.lower(), []).append((tail, head))
    for (t1, h1), (t2, h2) in ends.values():
        parent[find(t1)] = find(t2)
        parent[find(h1)] = find(h2)
    v = len({find(x) for x in range(n)})
    return v - n // 2 + 1

For 'abab' the corners fall into two classes, so $\chi = 2 - 2 + 1 = 1$ : the projective plane. Corner-chasing by hand is exactly this union–find computation.