## The Fourth Dimension, III

Last week we counted vertices, edges, faces, and cells on a hypercube.  Mainly, we thought by analogy — but how could we be sure our intuition was correct?  Today, we look at another say to make these same counts, and obtain the same results.  This suggests that maybe we were right all along….

So we begin by looking at the square and cube in a different way, and thinking (again) by analogy from there.

Let’s start with the number of edges on a square this time. We’ll start with one dimension less than the object we’re examining — later we’ll use the squares on a cube, and the cubes on a hypercube.  Now to count vertices, we note that each edge has two vertices — for 4 x 2 = 8 vertices.  But when we join the edges at vertices, two vertices merge, so we’ve overcounted by a factor of 2.  Thus, there are 8 / 2 = 4 vertices on a square (as we know).

What happens when we look at the cube in the same way?  Let’s start with the six squares.  To count edges, we see that each square on the cube has 4 edges, for a total of 6 x 4 = 24 edges.  But as with the vertices on the square, when we join two squares at their edges, two edges merge, so we’ve overcounted by a factor of 2.  This means there are 24 / 2 = 12 edges on the cube.

For vertices, we note that three squares meet at each vertex.  So there are  6 x 4 = 24 vertices on the 6 squares — but when we join the squares together, three vertices merge to one.  This means we’ve overcounted by a factor of 3, so there are 24 / 3 = 8 vertices on a cube.

Now begin with the eight cubes on a hypercube.  We note that the “8” comes from the sequence 2, 4, 6, 8,… for 2 vertices on a segment, 4 edges on a square, 6 squares on a cube, etc.  On the square, we saw that 2 vertices merged.  On the cube, we observed that 2 edges merged, then 3 vertices.  By analogy, on the hypercube, we should have 2 squares merging, 3 edges, and 4 vertices.  This pattern continues into higher dimensions as well.

Let’s use this analogy to check that we’ve counted correctly.  First, we count squares.  With 8 cubes, we have 8 x 6 = 48 squares.  But since the cubes meet square-to-square, we’ve overcounted by a factor of 2, so that there are just 48 / 2 = 24 squares on a hypercube.

Now on to the edges.  Since 8 cubes contribute 12 edges each, there are 96 edges in total.  But three cubes meet at each edge, so we’ve overcounted by a factor of 3.  This implies that there are in fact 96 / 3 = 32 edges on a hypercube.

Finally, we count the vertices.  With 8 vertices on each of 8 cubes, we have 64 vertices all together.  But four cubes meet at each vertex, so we have in fact overcounted by a factor of 4.  This results in just 64 / 4 = 16 vertices on a hypercube.

So we counted correctly!  Well, almost….  You might have noticed that we haven’t exactly defined what a hypercube is in a rigorous mathematical way.  Until we do, the best we can say is that we’ve counted something, although we don’t know precisely what they something is at the moment.  That will be a subject for a later post.

As you might expect, there are other ways to represent a hypercube besides the figure shown in last week’s post.  We chose that particular representation because of the way we were thinking by analogy.

We’ll look at two additional ways.  The first is analogous to looking at a cube face on, as shown below.

Of course the inner and outer squares are the same size on a cube — but we can’t help distorting faces of the cube when we make a two-dimensional sketch.  Four of the squares are distorted into trapezoids here.

Here is the analogous representation of the hypercube.

In this figure, the inner cube (in green with black edges) is in fact the same size as the outer cube (with blue edges); these cubes are directly opposite each other on the hypercube.  This is directly analogous to the inner and outer squares we saw in the earlier figure.

Further, note how the other six cubes are distorted into frustums of square pyramids — you can easily see the trapezoidal faces, which we also saw in the earlier figure.  If you look carefully, you can count the 16 vertices and 32 edges.  The 24 squares are a bit trickier — but begin with the 12 squares on the inner and outer cubes.  Each of the other 12 squares contains exactly one black edge and exactly one blue edge — they are going “out” from the black edges to the blue edges.  The perspective is different, but all the squares are there.

And of course there are the eight cubes — the inner cube, the outer cube, and the six frustums of square pyramids.  All the elements of the hypercube are indeed present, but again, in a different perspective.

I did save the best for last — my favorite representation of a hypercube, shown below.

I just love the symmetry of this image — the octagram inside the octagon.  If you look at the left image, you’ll see one of the eight cubes highlighted in blue.  When each of the eight cubes is transparently colored in the right image, you’ll see an interesting overlap of colors.

Now let’s count the vertices, edges, faces, and cells in this figure.  The 16 vertices are readily apparent in the inner octagram and the outer octagon.  The 32 edges can be seen by counting eight edges from both the octagram and octagon, and two additional edges connecting each vertex of the octagon to two vertices of the octagram.

The squares are a bit trickier here as well — but eight are easily visible as undistorted squares sharing one edge of the octagon.  But if you look carefully, you’ll also see 16 additional squares as rhombi in the figure.  Eight of these rhombi each share two edges with the octagon, and the eight others each share two edges with the octagram.  It might take staring for a minute, but they are all there.

And finally, there are the eight cubes. As seen in the left image, three consecutive edges of the octagon are enough to determine one of the cubes. Since we can take three consecutive edges of an octagon in exactly eight ways, we have found the eight cells on the hypercube.

So again, all the elements of a hypercube are present — it just takes looking from the right perspective to see them all.

Still remaining is to obtain the same counts by considering a rigorous definition of a hypercube.  That for a later post….

## 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!

## The Fourth Dimension, I

I have found that the topics of infinity and the fourth dimension really do always pique students’ interest.  When I had a few moments left at the end of a class period, I could sometimes casually remark about one of these topics, and I immediately had the attention of the entire classroom.

One of my projects has been writing an introductory book on polyhedra, and I’m in the middle of a proposal to have the book published.  A draft chapter was on the fourth spatial dimension, so I thought I’d share it here.  As much as I love the topic, I rather surprised myself by looking at my blog index and finding I’d never talked about it before….

What follows is the draft chapter, slightly edited as a stand-alone series.  Enjoy!

“What is the fourth dimension?

No, it’s not time.

Well, maybe it is if you’re studying physics.  Even then, we have certain intuitive ideas about how time works.

For example, if I imagine that it’s 9:17 a.m. in San Francisco, then I know what time it is in any other city in the world.  In New York City, it must be 12:17 p.m., since I add three hours to convert to Eastern Standard Time.

This assumption is just fine for getting along in daily life — and as far as most people are concerned, this way of thinking about time is right.  But in fact, it works only because in our daily lives, we move around fairly slowly — at least compared to the speed of light.

To understand what happens when particles do move close to the speed of light, you need to study special relativity — and here, our ordinary intuitions about time are no longer valid.

But we’re interested in a fourth spatial dimension.  How is this possible?  Where is it?  We are so used to living in a three-dimensional world, the idea of a fourth spatial dimension seems rather fantastic.

In 1884, Edwin Abbott’s delightful novella Flatland was published (Abbott, Edwin A. Flatland: A Romance in Many Dimensions. New York: Dover Thrift Edition. 1992).  The protagonist was none other than A Square, an inquisitive four-sided being living in a purely two-dimensional world called Flatland.

He was chosen as the Flatlander to receive a visit from a Sphere on the eve of the Third Millenium.  This was an unnerving visit for A Square, since the Sphere kept suggesting he consider the direction upward, but there was no upward for A Square.  There was North, South, East, and West, but A Square just couldn’t fathom this direction, “upward.”

Finally, the Sphere lifted A Square out of his two-dimensional world to show him the glory of Space.  Quite a revelation!

We are in the same predicament as A Square when it comes to contemplating a fourth spatial dimension.  Yes, we can look forward, backward, to our left and right, up and down, but nowhere else.  It doesn’t seem that there is enough room for a fourth dimension.  Where would it be?

In a typical high school geometry class, you would likely have been introduced to points, lines, and planes — and perhaps were told that points are 0-dimensional, lines are 1-dimensional, planes are 2-dimensional, and that all objects of these types live in a 3-dimensional space.  But while there were infinitely many points, lines, and planes, there was only one 3-dimensional space.

And while we do not encounter a fourth spatial dimension on a daily basis, we don’t actually encounter points, lines, or planes, either.  Can we actually see a point if it has no length?  How could we possibly see a line if it has no width?  It would be invisible.  These geometrical ideas are in fact mathematical abstractions — and once we enter the world of mathematical abstraction, our universe becomes almost unimaginably vast. A. R. Forsyth wrote almost one hundred years ago (A. R. Forsyth, Geometry of Four Dimensions, Cambridge University Press, New York, 1930, p. vii.):

Mathematically, there is no impassable bar against adventure into spaces of more than three dimensions of experience.

To get a handle on how to imagine the fourth dimension, we’ll look at one of the most popular and well-know denizens of that rarefied world — the hypercube, also known as a tesseract.”

What follows is an extended example of “thinking by analogy,” which would make this post much longer than I usually allow myself.  So that’s where we’ll start next week!