- At this point you may like to click here. A new window will open where you can find my definition of the 4th dimension. This links to a simple explanation of a 4D cube ... how it relates to a point, a line, a square and an ordinary cube ... and how to draw it ... and how it rolls.
In our previous post (31 October 2014), I hope I proved to you that the 4D hypersphere exists.
One other path for groping our way towards the true nature of the
4D hypersphere is to bear in mind that there must be straight-edged shapes that
fit inside it, with their corners on its surface. That brings us back to
the question: how many regular solids are there in 4D?
In 3D, as we know, there are five regular Platonic solids (not
counting the sphere) and any one of the five would fit precisely inside a 3D
sphere of the appropriate size. The 4 corners of the tetrahedron would
precisely touch the inner surface of the sphere. Ditto (assuming the
sizes were correct) the 6 corners of an octahedron, or the 8 corners of a cube,
or the 12 corners of an icosahedron (which has 20 triangular faces), or finally
the 20 corners of a dodecahedron (which has 12 pentagonal faces). If you
were explaining to a Flatlander person what a sphere was, it might help you to
show them, say, the dodecahedron (not that they can see more than one face at a
time) and to explain that it would fit precisely inside such a sphere.
So indeed would the cube, which is perhaps easier to explain.
In our attempts towards a mental picture of a 4D hypersphere
therefore, it may help to look at any regular hypersolids there may be and to
reflect that all of them, if correctly sized (and if of course they do
"really" exist), would fit precisely inside such a hypersphere.
So: what 4D hypersolids might possibly exist?
To work out which 3D regular solids might exist, we considered
what 2D regular polygons might be used as building blocks to construct the
solids. We then considered how many of the polygons could be
fitted round a corner of the eventual 3D shape. E.g. the internal angle
of a square is 90 degrees. 4 x 90 = 360, hence there isn't room for four
squares to meet at a corner, because the four of them fill up the whole plane.
There is room, however, for three squares to meet at a corner.
And of course three squares do indeed meet at the corner of a cube.
Therefore squares are certainly a possible building block to construct a
3D solid with. There are other qualifying polygons of course -
triangles, pentagons - but not hexagons. Why not? Because to make
a corner you need at least three of them to meet and the internal angle of a
regular hexagon is 120 degrees. 3 x 120 = 360 so 3 hexagons meeting
would fill up the entire plane.
We could try the same approach in 4D. In the 1880s, many
did. It's difficult though, and much confusion resulted. Instead,
let's try something slightly different. Rather than worrying about how
many building blocks can fit round a point (a corner or vertex), let's try
worrying about how many can fit round an edge.
What building blocks shall we try? Obviously if the
eventual 4D hypersolid is to be regular, it must be built from 3D building
blocks which are themselves regular. So there are only 5 choices which
are possible (tetrahedron, cube, octahedron, dodecahedron and icosahedron) for
us to use as building blocks. Will all (or indeed any) of these
fit round an edge of the eventual hypersolid?
One of them we can eliminate straight away. Consider the
regular icosahedron. What is the dihedral angle between two adjacent
faces? What indeed is meant by a dihedral angle? The only time I
have ever came across this phrase before was in a description of Hurricane and
Spitfire fighter planes in the Battle of Britain in World War Two. The
wings of these aircraft did not stick straight out horizontally from the
fuselage, but were inclined slightly upwards. The dihedral angle was the
angle between the surfaces of the two wings of the aircraft.
In just the same way the dihedral angle of an icosahedron (or any
other solid figure) is the angle that one of its surfaces makes with an
adjacent surface when the two of them meet. In the case of an
icosahedron, wherever two surfaces meet along an edge, this dihedral angle is
slightly more than 120 degrees. 3 x this angle is more than 360 degrees.
Hence you cannot assemble three icosahedrons together round an edge.
They won't fit. The icosahedron cannot be used as one of our
building blocks.
What about the cube as a building block? The dihedral angle
between adjacent faces of a cube is 90 degrees. So 3 of them could
fit round an edge perfectly well. But not 4. Hence there is one
possibility for a regular 4D hypersolid made out of 3D cubes. This does
not prove that such a thing exists; all we have done is to show that it
is not impossible.
What about the tetrahedron, the simplest
possible 3D shape, as a building block? Because it is the simplest, the
tetrahedron is sometimes known as a "simplex", or 3-simplex to show
that it's in 3 dimensions. So a 2D triangle would be a 2-simplex, the
tetrahedron is a 3-simplex and the 4D equivalent (if there is one) would be a
4-simplex. Could we configure a 4-simplex? Yes we could, provided
that we could fit 3 tetrahedra round an edge. What is the dihedral
angle? It's more than 60 degrees but not much more. So there's
plenty of room to fit three round an edge. So that's another possible
regular 4D shape, using the tetrahedron as a building block and with 3 of them
meeting at each edge.
Let's stay with the tetrahedron as our building block for the
moment. Could 4 of them meet at an edge? What's the dihedral
angle? It's less than a right angle, so yes they could, there's plenty
of room. That's another possible 4D shape therefore.
What about 5 tetrahedra meeting at an edge? Possible?
Yes. But only just. Five times the dihedral angle of a
tetrahedron is just less that 360 degrees. So, in theory at least,
that's another possible 4D shape, which we could just about manage to
construct, using the tetrahedron as a building block.
6 at an edge? Not possible. 6 times the dihedral angle
is more than 360 degrees. So that possibility is ruled out. So
much, therefore, for the tetrahedron as a building block. It seems to
have done us pretty well.
