Euler characteristics, platonic solids, and doughnuts

ellen

On 12 March two young mathematical researchers, Aditi Kar from Oxford University and Ellen Powell from Cambridge University, came to CoLA to tell us a little about the sort of algebra that Emmy Noether, the “mother of modern algebra”, did, and a little about real-life maths, beyond school maths. Aditi’s notes from the session are available here; and there are more notes here (read on).

aditi

They focused on the idea of invariants. At the simplest, when you count objects, the number you end up with is invariant (the same) whatever order you count them in, whatever size and shape and colour the objects are, etc.

Here are some other examples they didn’t have time to mention. If you translate (move across) a triangle, rotate it, or reflect it, its position and alignment change, but not its sides and its angles: they are invariant. If you draw the triangle with a different-coloured pen, it looks different, but its sides and angles are invariant. That’s what we mean by congruent triangles.

If you apply a linear transformation (a matrix) to something, it may be expanded or shrunk; or expanded in one direction and shrunk in another; or rotated; or reflected; or sheared. Lots of things change. But what was a line remains a line. What was a triangle remains a triangle. What was the origin remains the origin. Those are invariant.

If you rotate the axes relative to which you are measuring, then vectors and matrices change. But the dot product and cross product of vectors is invariant. The determinant and the trace (the total of the numbers down the leading diagonal) of matrices are invariant. If you take an algebraic expression like ax2+bxy+cy2, then b2−4ac is invariant when you rotate the x and y axes…

In the session, they looked at topological invariants. These are properties which remain invariant when you stretch and distort, as long you don’t rip.

If you wash a jumper at the wrong temperature, it may go way out of shape. But the number of holes in it – four: one at the neck, one at the end of each arm, and one at the bottom – remains invariant.

In the same way, if you had an infinitely stretchy coffee mug, you could stretch and squeeze it to look like a ring doughnut. Or stretch and squeeze the doughnut to look like a coffee mug. One topological property remains invariant: the number of holes in the shape.

Mug_and_Torus_morph

Polyhedra

A polyhedron is like a polygon, but in 3D. A polyhedron is a solid shape with lots of faces, while a polygon is a flat shape with lots of sides.

There are infinitely many regular polygons, but only five regular polyhedra (i.e. polyhedra where every face is the same, and every face is a regular polygon).

platonic

Proving that there are only five regular polyhedra

Proof 1

The total of angles at each vertex must be less than 2π

And that total must = a multiple of the internal angle of some regular polygon (π/3 for an equilateral triangle, π/2 for a square, 3π/5 for a pentagon, 2π/3 for a hexagon, etc.) And a multiple ≥ 3 – there must be at least three angles at every vertex.

But only 3, 4, or 5 times π/3 makes less than 2π (6 times is too big)

Only 3 times π/2 makes less than 2π (4 times is too big)

Only 3 times 3π/5 makes less than 2π (4 times is too big)

Even 3 times 2π/3 (or the angle of any regular polygon with more sides than a hexagon) ≥ 2π

Therefore the regular polyhedron must be:

3 triangles meeting at each vertex (tetrahedron)

4 triangles meeting at each vertex (octahedron)

5 triangles meeting at each vertex (icosahedron)

3 squares meeting at each vertex (cube); or

3 pentagons meeting at each vertex (dodecahedron)

Proof 2 uses the rule faces+vertices−edges=2, or f+v−e=2

Every face is an identical regular polygon. Let n be the number of sides of that polygon (3 for a triangle, 4 for a square, 5 for a pentagon, etc.)

And let k be the number of edges meeting at each vertex (what in D1 maths we call the degree or valency of the vertex).

Then, since each edge has two ends (is counted twice if we count the total number of edges vertex by vertex), and every edge has faces on two sides (is counted twice if we count the total number of edges face by face);

kv = 2e

and

nf = 2e

In other words:

e = kv/2

f = kv/n

Since f+v−e=2

kv/n + v − kv/2 = 2

Multiply both sides of the equation by 2n

v(2n + 2k − nk) = 4n

But n and v are positive, so in order for this to be true, 2n + 2k − nk > 0, or (n − 2)(k − 2) < 4

Since n must be at least 3 (you can’t have a polygon with only 2 sides), and k must also be at least 3, there are only five possibilities for (n, k):

(3, 3) = tetrahedron
(3, 4) = cube
(4, 3) = octahedron
(3, 5) = dodecahedron
(5, 3) = icosahedron

Euler characteristic

The rule f+v−e=2 is true not just for regular polyhedra, but for any connected network of faces, vertices, and edges which we could draw on the surface of a sphere. Example: a football. 32 faces, 60 vertices, and 90 edges. The number 2 is a defining property of the surface of a sphere, called its Euler characteristic.

Football

The rule f+v−e=2 is also true for any network which we could draw on a flat surface, if we reckon the “outside” of the network there as a face.

Any infinitely-stretchy network drawn on a sphere can be turned into one drawn on a flat surface by puncturing one of the faces and then stretching the sphere out until it lies flat on a surface. Any infinitely stretchy-and-squeezable network drawn on a flat surface can be turned into one drawn on a sphere by wrapping the flat surface onto a sphere, squeezing together all the outside fringes of the surface (not the edges drawn in the network, which remain distinct though distorted), and gluing the whole outside together so that you can’t see the join.

The stretching and squeezing changes lots of things, but edges remain edges (not necessarily straight), vertices remain vertices, and faces remain faces, and f+v−e remains the same.

Any stretchy regular polyhedron can be turned into a network of faces, vertices, and edges drawn on a sphere by making a small hole, blowing it up until it is spherical, and then closing the hole again.

To prove f+v−e=2 for a network of faces, vertices, and edges

Begin with a spanning tree T of the network. (You learn about spanning trees in D1. A tree is a network without “loops”, and a spanning tree is a tree which reaches every vertex. In D1 you learn how to find a spanning tree for every connected network. In D1 you also learn to call networks “graphs”. Nothing to do with the sort of graphs where you draw x and y axes).

For a tree, e = v−1, and f = 1 so f+v−e=2 is true

Now G can be obtained from T by adding edges.

Each time we add an edge, we also create a new face. So the value of e goes up by 1, but so does the value of f; and so f+v−e remains unchanged, at 2.

But if we draw a network on the surface of a torus (ring doughnut), and remember that something counts as a face only if it could be stretched out flat, we find

f+v−e=0

for a torus.

torus4

For a pretzel (with two holes)

DoubleTorus_800

f+v−e=−2

The Euler characteristic of a network (graph) remains the same (invariant) however you stretch it in different ways. So it is called a topological invariant. Topology is the mathematical study of properties which remain invariant after stretching.

And it turns out we can describe that topology by an algebraic equation:

f+v−e= Euler characteristic of a surface.