DG, Part 4: Thank You, Herr Euler!
Jan. 22nd, 2006 09:42 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowThis post won't involve any calculus.
It will, however, involve topology.
But don't fret; the only real math we'll be doing is some simple arithmetic.
Picture a polyhedron - a cube, say. Count the vertices; call that v. Do the same for the edges (e) and the faces (f). Compute the quantity v-e+f. (In the case of a cube, v=8, e=12, and f=6. 8-12+6=2.) Try again with another polyhedron, say a square pyramid. It has 5 vertices - the peak, and the four vertices of the square base. There are 8 edges - the edges of the base, plus the lines connecting the vertices of the base to the peak. Finally, there are 5 faces - the base, plus four triangular faces surrounding the peak. 5-8+5=2 again. Prisms, octahedrons, cubes with pyramids on top or pyramids on every face - they'll all give you the same result.
This is usually known as Euler's Formula, after the Swiss mathematician Leonhard Euler, who was apparently the first to prove it. René Descartes seems to have noticed it first, though.
There's a bit of a catch, though; it doesn't work for every polyhedron. Take a cube and drill a square hole from the middle of one face through to the opposite face. Now, in order to treat this as a polyhedron, we have to make all the faces polygons, and (as matters stand) the two faces we just drilled aren't polygons - they've got holes in the middle. To correct for that, we draw in some new edges, connecting each vertex of a drilled face to a vertex of the hole. Now, there are 16 vertices - the original 8, plus 8 for the hole. There are 32 edges - the original 12, 12 more for the hole, and the 8 new edges we just drew in. Finally, there are 16 faces. (Four of the original faces are undamaged. The two drilled faces have each been replaced by four new faces. The hole contributes four more.) This time, we get 16-32+16=0.
The difference is that this new polyhedron has a hole in it. To put it another way: suppose your polyhedron were made of thin rubber, like a balloon, and inflate it. A cube, or any of the polyhedra I mentioned before, blows up into a sphere, or something spherelike. The drilled cube, though, inflates into a "torus" - a doughnut-shaped object. You can similarly conceive of polyhedra that look like pretzels - with three holes - and so forth.
It turns out that Euler's Formula can be modified to fit any polyhedron, thus: if the polyhedron, on being inflated, has g holes, then v-e+f=2-2g. This last number, 2-2g, is the "Euler characteristic" of the polyhedron. The thing to notice is that it is a topological quantity; it's unaffected by deformations like blowing it up, or indeed anything that doesn't involve tearing. From the point of view of differential geometry - of calculus, that is - a sphere and, say, a football (American or European, either one) are very different objects. But they have the same Euler characteristic. An American football has two vertices (the points), four edges (the seams), and four faces, for example. Nonetheless, there turn out to be interesting connections between the Euler characteristic and differential geometry.
By the way, the Euler characteristic and formula can be generalized to objects with more or less than three dimensions. In one dimension, the formula says that v=2 - a line segment has two endpoints. In two, it says that v-e=0 - a polygon has just as many edges as vertices. (Fascinating, no?) In four dimensions... well, consider a tesseract. It has vertices (16 of them), edges (32), faces (24), and "hyperfaces" (8), and 16-32+24-8=0. Other four-dimensional "polytopes" will give the same result, as long as they don't have holes in them - in other words, as long as they are topologically equivalent to hyperspheres.
Next time, I have to lay out a little groundwork, but then we'll be able to tie this stuff in with differential geometry.
It will, however, involve topology.
But don't fret; the only real math we'll be doing is some simple arithmetic.
Picture a polyhedron - a cube, say. Count the vertices; call that v. Do the same for the edges (e) and the faces (f). Compute the quantity v-e+f. (In the case of a cube, v=8, e=12, and f=6. 8-12+6=2.) Try again with another polyhedron, say a square pyramid. It has 5 vertices - the peak, and the four vertices of the square base. There are 8 edges - the edges of the base, plus the lines connecting the vertices of the base to the peak. Finally, there are 5 faces - the base, plus four triangular faces surrounding the peak. 5-8+5=2 again. Prisms, octahedrons, cubes with pyramids on top or pyramids on every face - they'll all give you the same result.
This is usually known as Euler's Formula, after the Swiss mathematician Leonhard Euler, who was apparently the first to prove it. René Descartes seems to have noticed it first, though.
There's a bit of a catch, though; it doesn't work for every polyhedron. Take a cube and drill a square hole from the middle of one face through to the opposite face. Now, in order to treat this as a polyhedron, we have to make all the faces polygons, and (as matters stand) the two faces we just drilled aren't polygons - they've got holes in the middle. To correct for that, we draw in some new edges, connecting each vertex of a drilled face to a vertex of the hole. Now, there are 16 vertices - the original 8, plus 8 for the hole. There are 32 edges - the original 12, 12 more for the hole, and the 8 new edges we just drew in. Finally, there are 16 faces. (Four of the original faces are undamaged. The two drilled faces have each been replaced by four new faces. The hole contributes four more.) This time, we get 16-32+16=0.
The difference is that this new polyhedron has a hole in it. To put it another way: suppose your polyhedron were made of thin rubber, like a balloon, and inflate it. A cube, or any of the polyhedra I mentioned before, blows up into a sphere, or something spherelike. The drilled cube, though, inflates into a "torus" - a doughnut-shaped object. You can similarly conceive of polyhedra that look like pretzels - with three holes - and so forth.
It turns out that Euler's Formula can be modified to fit any polyhedron, thus: if the polyhedron, on being inflated, has g holes, then v-e+f=2-2g. This last number, 2-2g, is the "Euler characteristic" of the polyhedron. The thing to notice is that it is a topological quantity; it's unaffected by deformations like blowing it up, or indeed anything that doesn't involve tearing. From the point of view of differential geometry - of calculus, that is - a sphere and, say, a football (American or European, either one) are very different objects. But they have the same Euler characteristic. An American football has two vertices (the points), four edges (the seams), and four faces, for example. Nonetheless, there turn out to be interesting connections between the Euler characteristic and differential geometry.
By the way, the Euler characteristic and formula can be generalized to objects with more or less than three dimensions. In one dimension, the formula says that v=2 - a line segment has two endpoints. In two, it says that v-e=0 - a polygon has just as many edges as vertices. (Fascinating, no?) In four dimensions... well, consider a tesseract. It has vertices (16 of them), edges (32), faces (24), and "hyperfaces" (8), and 16-32+24-8=0. Other four-dimensional "polytopes" will give the same result, as long as they don't have holes in them - in other words, as long as they are topologically equivalent to hyperspheres.
Next time, I have to lay out a little groundwork, but then we'll be able to tie this stuff in with differential geometry.
![[identity profile]](https://www.dreamwidth.org/img/silk/identity/openid.png){kind=small}
no subject
Date: 2006-01-22 05:46 pm (UTC)What is a six-dimensional object called?
no subject
Date: 2006-01-22 05:57 pm (UTC)no subject
Date: 2006-01-23 07:38 pm (UTC)Insufficient to make a superstring.