4D models

What does a 4-dimensional shape look like? How can you draw/make one in 3 dimensions?

To understand what they look like, it's easiest to consider what happens when you go from a point to a line to a square to a cube and... beyond.

Suppose that when you move a point, it traces a line. As Paul Klee said: "A line is a dot that went for a walk!"

If you extend a point in the x-direction (giving it 1 dimension), you create a line. You can think of this as having 2 ends (vertices) and 1 edge.

If you slide this line in the y-direction (2 dimensions) you create a unit square, with 4 vertices and 4 edges (and 1 face).

Now if you slide this square in the z-direction you create a unit cube, with 8 vertices, 12 edges (2x4 for the two squares + 4 for the vertex traces) and 6 faces (2 for the two squares and 4 for the edge traces).

Here it is easy to see that if you slide this cube in the 4th dimension, you create a unit 'hypercube' with:

16 vertices
32 edges (2x12 for the two cubes + 8 for the vertex traces)
24 faces (2x6 for the two cubes + 12 for the edge traces)
8 cubes (2 for the two cubes and 6 for the face traces) - called 'cells'.

You can keep going! Can you work out how many of everything a 5D cube has? Here is a picture of one I made below (distorted to be demonstrable in 3 dimensions). The yellow lines represent the vertex traces into the 5th dimension.

working out V, E, F, C, etc...

You can work out the number of vertices etc for each model in a different way. For example, consider the cube. It has 2^3 = 8 vertices. Each vertex has 3 edges but each edge is shared by 2 vertices, so the number of edges = 8 x 3 / 2 = 12. Each vertex is the meeting of 3 faces with each face shared by 4 vertices, so the number of faces = 8 x 3 / 4 = 6.

The hypercube has 2^4 = 16 vertices. Each vertex has 4 edges with each edge shared by 2 vertices, so the number of edges = 16 x 4 / 2 = 32 edges. Each vertex is the meeting of 6 faces shared by 4 vertices, so the number of faces = 16 x 6 / 4 = 24 faces. 4 cube cells meet at each vertex, with each cube cell shared by 8 vertices, giving 16 x 4 / 8 = 8 cube cells.

If you do a similar calculation for the 5D cube, you will notice that there is a pattern to the number of each type of cell shared at each vertex - the binomial expansion...

shadow representations

You can represent higher dimensional shapes in a different way by considering what they look like from a particular view.

To see how, again consider a line (1 dimension). If you look at it along the x-axis (ie along its length) it will look like a point. It looks like it has only one vertex from this direction, but in fact it has 2 (one is hidden).

Consider the square. If you look at it along the y-direction (ie side on), it looks like a line - 1 edge with 2 vertices - but of course you know it has 4 vertices (double those you can see) and 4 edges (double those you can see + one for each vertex trace).

Now if we look at the cube face-on, we see a square. But you know there is another face directly behind it and 4 other faces from the edge traces, giving 6 faces.

So.. consider the hypercube viewed 'cube-on'... You will just see a cube! But you know there is one 'behind it' and 6 others from the face traces, giving 8 cube 'cells'. You could use this to analyse the number of vertices and edges too.

So the 5D cube will look like a hypercube, but with one hidden and 8 more hypercubes from the cube traces giving 10 hypercube cells... I think you get the idea!!

Other 4-D shapes

There are 6 4-dimensional polytopes (the general name for shapes) of which the hypercube is one. One of them is based on the dodecahedron called the 120-cell, so-called because it comprises of 120 dodecahedron cells, with 4 cells meeting at each vertex in the same way as the hypercube. As there are 20 vertices for each cell, there are 120 x 20 / 4 = 600 vertices!

You can demonstrate it in 3 dimensions using the shadow idea above. Here is one made by my year 10 students!