Post by 1dave on Sept 13, 2020 16:54:49 GMT -5
a sphere has a single surface.
A torus has a single surface and a single "hole," so . . .
www.johndcook.com/blog/2012/10/30/euler-characteristic/
Euler characteristic with dice
Posted on 30 October 2012 by John
For any convex solid, V – E + F = 2 where V is the number of vertices, E the number of edges, and F the number of faces. The number 2 in this formula is a topological invariant of a sphere, called its Euler characteristic.
But if you compute the Euler characteristic for a figure with a hole in it, you get a different value. For a torus (the surface of a doughnut) we get V – E + F = 0.
You can demonstrate this with eight 6-sided dice. A single die has 8 vertices, 12 edges, and 6 faces, and so V – E + F = 2. Next join two dice together along one face.
![1](https://www.johndcook.com/dice2.jpeg)
two dice joined together
Before joining, the two dice separately have 16 vertices, 24 edges, and 12 faces. But when we join them together, we have 4 fewer vertices since 4 pairs of edges are identified together. Similarly, 4 pairs of edges are identified, and 2 faces are identified. So the joined pair now has 12 vertices, 20 edges, and 10 faces, and once again V – E + F = 2.
We can keep on adding dice this way, and each time the Euler characteristic doesn’t change. Each new die adds 4 vertices, 8 edges, and 4 faces, so V – E + F doesn’t change.
![2](https://www.johndcook.com/dice7.jpeg)
seven dice in a U-shape
But when we join the dice into a circle, the Euler characteristic changes when we put the last die in place.
![3](https://www.johndcook.com/dice8.jpeg)
eight dice in a torus shape
The last die doesn’t change the total number of vertices, since all its vertices are identified with previous vertices.
The last die adds 4 edges. It adds a net of 2 faces: it adds 4 new faces, but it removes 2 existing faces.
Exercise: Use a similar procedure to find the Euler characteristic of a two-holed torus.
A torus has a single surface and a single "hole," so . . .
www.johndcook.com/blog/2012/10/30/euler-characteristic/
Euler characteristic with dice
Posted on 30 October 2012 by John
For any convex solid, V – E + F = 2 where V is the number of vertices, E the number of edges, and F the number of faces. The number 2 in this formula is a topological invariant of a sphere, called its Euler characteristic.
But if you compute the Euler characteristic for a figure with a hole in it, you get a different value. For a torus (the surface of a doughnut) we get V – E + F = 0.
You can demonstrate this with eight 6-sided dice. A single die has 8 vertices, 12 edges, and 6 faces, and so V – E + F = 2. Next join two dice together along one face.
![1](https://www.johndcook.com/dice2.jpeg)
two dice joined together
Before joining, the two dice separately have 16 vertices, 24 edges, and 12 faces. But when we join them together, we have 4 fewer vertices since 4 pairs of edges are identified together. Similarly, 4 pairs of edges are identified, and 2 faces are identified. So the joined pair now has 12 vertices, 20 edges, and 10 faces, and once again V – E + F = 2.
We can keep on adding dice this way, and each time the Euler characteristic doesn’t change. Each new die adds 4 vertices, 8 edges, and 4 faces, so V – E + F doesn’t change.
![2](https://www.johndcook.com/dice7.jpeg)
seven dice in a U-shape
But when we join the dice into a circle, the Euler characteristic changes when we put the last die in place.
![3](https://www.johndcook.com/dice8.jpeg)
eight dice in a torus shape
The last die doesn’t change the total number of vertices, since all its vertices are identified with previous vertices.
The last die adds 4 edges. It adds a net of 2 faces: it adds 4 new faces, but it removes 2 existing faces.
So the net change to the Euler characteristic is 0 – 4 + 2 = -2.
The last die lowers the Euler characteristic from 2 to 0.
The last die lowers the Euler characteristic from 2 to 0.
Exercise: Use a similar procedure to find the Euler characteristic of a two-holed torus.