# The Fourth Dimension, II

Last week, I introduced the idea of a fourth spatial dimension.  The typical question students ask is simply, “Where is it?”  These doesn’t seem to anywhere it can go — and in fact, since we live (insofar as we know it) in a three-dimensional world, there really isn’t anywhere it can go.

So the fourth spatial dimension must be, for us, an abstraction.  To think about it, we must somehow relate it to things we already know — in particular, geometry in dimensions zero, one, two, and three.  Therefore, to think about the tesseract — the four-dimensional analogue of the cube — we must somehow think by analogy.

In particular, we can think about how we go from a point (zero dimensions) to a line segment (one dimension), and then from a line segment to a square, a square to a cube, and then make the analogous leap from a three-dimensional cube to a four-dimensional tesseract, or hypercube.  We’ll look one way to do this in today’s post, and next week, we’ll see a different way.  It’s a good idea to look at this “new” fourth dimensional from different perspectives to make sure our thinking by analogy is accurate.

To create a line segment from a point, we think of it moving one unit (we need to be specific) along a “new” first dimension to create another point — these two points are vertices of the line segment, and one edge is created.

Now move this segment one unit along a second dimension which is perpendicular to the line segment, as shown below.

Now let’s count.  We have two vertices and one edge for each of the two segments (the one at the bottom in black, and the one at the top in blue), and each vertex creates another line segment (shown in red above) as it moves from the bottom segment to the top segment.  This gives a total of four vertices and four edges on a square — as expected.

Thinking by analogy, we now imagine a square moving up along a perpendicular dimension, as shown below.

Counting vertices, edges, and faces, we have 8 vertices, 8 edges, and 2 faces from the bottom and top squares, shown in black and blue, respectively.  Now each vertex of the bottom square creates an edge as it moves up (creating 4 more edges, shown in red), and each edge of the bottom square creates a new face (creating 4 new faces, shown in pale yellow).  This gives a total of 8 vertices, 12 edges, and 6 faces.  Notice the strategy: count the bottom and top figures, and then notice what is created by vertices and edges as they move along a perpendicular dimension.

Now it’s time to extend this strategy into the fourth dimension.  To do so, we need to imagine a cube moving out along a fourth spatial dimension — and of course, it is difficult to imagine because we are so used to our three-dimensional world.

Look at the above figure.  Let’s think of the cube outlined in black as the base cube.  Move this cube out along a fourth spatial dimension — so that each vertex creates an edge (shown in red) as it moves to the top cube (shown in blue).

Before we start counting, we need to introduce a little terminology.  The four-dimensional analogue of a polyhedron is called a polytope, and in addition to vertices, edges, and faces, we have three-dimensional cells on a polytope.

Now let’s count the vertices, edges, faces, and cells on a hypercube in just the same way as in the previous two examples. For vertices, we have 16 total — 8 from the base cube, and 8 from the top cube.  For the edges, we have 12 each for the base and top cubes, and each vertex of the base cube creates a new edge (shown in red).  This gives a total of 12 + 12 + 8 = 32 edges on the hypercube.

What about faces?  Well, we have 6 faces each for the base and top cubes, and each edge of the base cube creates another square face (as it did when we created the cube).  This gives us 6 + 6 + 12 = 24 faces.  Finally, to count the cells, we have the base and top cubes, and each square face of the base cube moves out to create another cube (such as the one shown in yellow, with the square from the base cube shown in darker yellow).  This gives 2 + 6 = 8 cells (cubes) on the hypercube.

How can we be sure this is correct?  Next week, we’ll look at a different (though certainly related) way to count the number of vertices, edges, face, and cells on a hypercube.  We will see that we do in fact obtain the same counts.

As a final remark, we briefly look at Euler’s Formula in four dimensions.  Recall that $V-E+F=2,$ where $V, E,$ and $F$ represent the numbers of vertices, edges, and faces on a convex polyhedron.  Now if $C$ represents the number of cells on a polytope, the four-dimensional analogue of Euler’s Formula is

$V-E+F-C=0.$

In our case, we have the true statement

$16-32+24-8=0.$

We won’t go into more depth here, as that would take us quite a bit further afield.  But if you’re interested to know more, you can always look up Euler’s Formula in higher dimensions, and extensions which have to do with the topology of geometrical objects, which is fairly straightforward when the objects are convex.  When they’re nonconvex, the situation is decidedly more difficult.

Stay tuned for next week’s post, where we look at yet another way to count the numbers of vertices, edges, faces, and cells on our friendly tesseract!

### Vince Matsko

Mathematician, educator, consultant, artist, puzzle designer, programmer, blogger, etc., etc. @cre8math