How many hypersolids are there in 4 dimensions?

New readers:

• 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.

4D hypersolids and the 4D hypersphere

New readers:

• 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.

So  -  from our previous post (24 June 2014) it's fairly easy to see that the possible regular solids in 3 dimensions are 5 in number and that there can't be any more.   The  ancient Greeks had worked this out, together with vast amounts of other geometrical stuff (remember Euclid?), which is why the 5 solids in question are known as the "Platonic Solids".    Plato knew about them.   Arguably, if you include the sphere, there are 6 of the things, but the sphere is different in kind.

Now, what about in 4 dimensions?   Are there any hyper-Platonic hypersolids lurking there?   Can we use a similar approach to show, say, the maximum number that there might be?   How many would you guess that there are, if you were coming new to the subject?   And what do you think they would look like?   Some of them, we shall find, are truly remarkable objects, altogether splendid in their symmetry and complexity.   And, curiously enough, 100+ years ago or so, the race was on to find how many there were.   Disagreement and controversy ensued, as we'll see later, in spite of the fact that, as some would say, these objects can't "really" exist (whatever that means).

Let's see if they do exist.

Rather unconventionally, I'd like to start with the 4D hypersphere, successor to the 2D circle and the 3D sphere.   Does this exist and, more enigmatically, can you imagine it?

Imagine a charity prize auction, with special and unusual rules designed to raise huge amounts of money from  2 bidders, Mr X and Mr Y.   They both bid, say for example they stop bidding at  $x and $y respectively, and here are the rules.   Both their bids are winners and  between the two of them the amount they have to pay to share the prize is worked out thus:  take the square of Mr X's bid (that's x squared), add the square of Mr Y's bid (y squared) making x squared plus y squared so far, then take the square root of the total.   That gives us  the total amount of money which, between them, they have to pay to share the prize under these particular rather strange rules.

And by the way, if they were feeling contrary, they could bid negative amounts if they wanted to.   It wouldn't cost them any less in the end, because the square of minus x (or minus y) is of course the same as the square of plus x or plus y.

Now draw a graph of all this.   Say for the sake of argument that Mr X bids $300 and Mr Y bids $400.   So we plot a first limb of $300 horizontally along the x axis and a second limb of $400 perpendicularly up, parallel to the y axis, from the x = $300 point.   How much do they have to pay?   Draw a line from the origin O to the top end of that second limb.   That line represents what they have to pay, namely of course $500.   We have constructed a right angled triangle whose sides, in true Pythagorean style, are $300, $400, and $500 long.

Now draw a circle with its centre at the origin O and its radius $500.   Every point on that circle will represent a possible pair of bids from Mr X and Mr Y, bids which will result in $500 being owed by the pair of them.   Rather annoyingly, x = plus or minus $300 and y = plus or minus $400 (or the other way round, giving the $400 bid to Mr X and the $300 bid to Mr Y) are the only ones which come out as nice whole numbers.   But why worry about that?   Say Mr X bids $100 and Mr Y bids $(100 x 2 x root 6).   That would also mean they owed $500 between them.   And of course there are vast numbers of other x and y pairs whose squares add up to 250,000.   Each of these pairs gives a point on the circle.

Thus far, all this may seem a  needlessly complex and pointless way of slipping Pythagoras in by the back door.   But wait.   Now imagine there are 3 bidders, X, Y and Z.   Same sort of rules.   All three must bid and the amount they owe between them is given by the square root of x squared plus y squared plus z squared.   Draw a graph of this.   Or, if you don't have any 3D graph paper, imagine it, with a z axis perpendicular to the xy plane.   Instead of a 2D circle with its centre at the origin O, the result will now be a 3D sphere with its centre at O.

Say for instance that Mr X, the old skinflint, bids $100, Mr Y bids $200 and Mr Z also bids $200.   So we plot a first limb of $100 along the x axis, and from the end of the first limb we plot a second limb of $200 vertically up, parallel to the y axis.   Now from the end of that second limb we plot a third limb, again of $200,   This third limb sticks horizontally outwards, perpendicular to the paper, and is therefore at right angles to the plane containing the first and second limbs.   In other words the third limb runs parallel to the z axis.

How far away from the origin O is the remote end of that third limb?   We use Pythagoras.   The answer is the square root of (x squared plus y squared plus z squared).   Which, because I have chosen the figures that way, comes to $300.   ($300 is the square root of $10,000 plus $40,000 plus $40,000.)

Every point  on the surface of this nice 3D sphere will represent a trio of bids, respectively of $x, $y and $z.   And the total amount owing by the three bidders between them will be equal to the radius of the sphere, namely the square root of (x squared + y squared + z squared).   And, using the figures I've put in, the radius comes to $300.   So every point on the sphere's surface represents a trio of bids, a trio which results in Messrs X, Y and Z between them owing $300.

You will expect me now to wheel in a fourth bidder.   I would hate to disappoint you;  that is indeed the next step.   Enter Mr W, who bids $w.   We must now add a fourth axis, the w axis.   This w axis has to be at right angles not only to the x axis but also to the y axis and to the z axis.   Not possible?   Let's imagine it anyway and see what happens.   The four bidders between them must now pay, in dollars,  the square root of (x squared + y squared + z squared + w squared).   And that dollar figure will be the radius of a 4D hypersphere.   Every point on the hypersphere's hypersurface will represent a quartet of bids, respectively of $x, $y, $z and $w.

Say for instance, to make things easy, that the four bidders, X, Y, Z and W, by coincidence bid $100 each.   So to plot this on our imaginary 4D graph paper, we plot a first limb for $100 worth along the x axis.   From the end of the first limb we plot a second limb worth $100 vertically up and parallel to the y axis.   From the end of the second limb we plot a third limb worth $100 sticking horizontally out of the paper and parallel to the z axis.   And from the end of that third limb we plot a fourth limb, again worth $100, in some mysterious imaginary direction parallel to the w axis (and therefore at right angles to all three of the  other limbs).

Now consider the far end of that fourth limb.   How far distant is it from the origin O?   Once again, we use Pythagoras.   The distance we want is given by the square root of (x squared + y squared + z squared + w squared).   Which, in this case, in dollars, is the square root of (100 squared + 100 squared + 100 squared + 100 squared).   Equals the square root of $40,000.   Equals $200. 
  

So we have a 4D hypersphere, radius $200.   Every point on the surface of the hypersphere represents a quartet of bids, one each from Messrs, X, Y, Z and W.   And the amount which, between the four bidders, they owe is given by the radius of the hypersphere, namely $200.   It's strange, in a way, isn't it, that while all four of them were prepared to pay as much as $100, and said so in their bidding, they end up only having to find $50 each.   That's four dimensions for you.   (Of course the same sort of thing can be true in two, three, in fact in any number of dimensions.)

Now, it is very hard for us 3D people to imagine a 4D hypersphere.   I can't help much.   If you have a 2D circle with centre at O, you can spin it about the x axis (or the y axis) and if you then solidify the spinning surface you have a 3D sphere.   So try spinning this 3D sphere and see if that helps.   I have to admit it doesn't help me much.   Perhaps the reason is I'm not spinning it correctly.   In 2D you spin about a point.   In 3D you spin about a line (an axis).   In 4D you have to spin about a surface - the xy plane, say.

I shall return to this later.   For now, I hope I have proved to you that the 4D hypersphere exists.   It must exist, however hard it may be to imagine.   It would be ludicrous to suggest that you could not possibly have an auction as described (however silly the rules are) and that you could not possibly have 4 bidders at it.

Indeed, obviously you could have 5, 6, 7, any number of bidders.   Pythagoras works in any number of dimensions, up to infinity.   The 5D hypersphere exists (and is, no doubt, even harder to imagine).   So do corresponding sphere-derived shapes in all higher dimensions.   However many dimensions you consider, there will always be a nearly-spherely shape waiting there to greet you.

FOR MORE ON PYTHAGORAS, see Comment below.

Are there really only five Platonic Solids?

New readers:

• 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.

Any number of people have tried in vain to disprove the supposed impossibility of various geometrical items, such as trisecting an angle with ruler and compasses, and have published long screeds on such matters.   Why then are there no crackpots getting busy with their knives and raw potatoes and writing about 3D shapes?   The reason, I think, is that it's fairly easy to see, and convince oneself, that only the five listed in the 8 June 2014 post, the so-called "Platonic Solids", are possible.

Is this true however?   Are there really only five?   Arguably not, I suggest.   What about the sphere?   Just as a circle can be thought of as an infinite number of infinitely short straight lines arranged in circular formation, so a sphere can be thought of as an infinite number of infinitely small faces arranged in spherical formation.   And these faces, of course, if you make them small enough, can be thought of as being bounded by infinitely short straight lines.   All of the same length, since you ask, so that perfect regularity is guaranteed

What you must imagine therefore is a round ball whose surface consists of an infinite number of  infinitely small, straight-edged, regular-shaped faces.   In just the same way that an infinite number of infinitesimal straight lines can, and indeed do, combine to make a circle in two dimensions, so an infinite number of infinitesimally small surfaces can and do combine to make a sphere in three dimensions.   Now, you may complain that the supposed nearly-spherely shape that you're imagining in your mind's eye feels a bit rough and knobbly to you.   If that's your problem, the answer is simple.   You haven't made the faces small enough.   There's an infinite number of them, remember, and they're infinitely small.   If they really were infinite in number, you wouldn't feel the knobbles at all.   The result would be a perfectly smooth sphere.   Indeed, why do I say "would be"?   With an infinite number of the things, the result, like it or not, IS a sphere.

Let's now go into four dimensions.   When thinking of possible 3D solids, a useful approach is to think of a point in 3D space and work out how many of a given 2D shape could be fitted in round that one point.   Consider 2D squares for instance.   Three of them can readily converge at a point, as they do at the corner of an ordinary cube.   There isn't room for four squares to converge however.   If four squares do converge, 4 x 90 degrees = 360 degrees so they fill up the whole plane, leaving nowhere for the volume (of any possible 3D solid) to go.  So if you're building a 3D shape, you can have three squares meeting at a point, but not four or more.

So much for squares.   What about triangles?   There's plenty of room for three to meet at a point.   Think of the corner of a tetrahedron.   There's room for four to meet also, as at the corner of an octahedron.   What about five triangles meeting?   They're equilateral triangles and 5 x 60 degrees = 300 degrees, i.e. less than 360 degrees.   Just OK therefore.   And five triangles do indeed meet at the corner of an icosahedron.   6 x 60 degrees however = 360 degrees and there isn't room for any such corner to exist.   So when you're building a 3D shape out of equilateral triangles, you can't have six or more of the triangles meeting at a point.

So we've done squares and triangles, what about pentagons?   It turns out that there's room for three of them to meet at a point, but there isn't room for four.   Hexagons?   Put three hexagons together at a point, their internal angles are 120 degrees, 3 x 120 = 360, so there isn't room even for as few as three hexagons to form part of a solid shape.   Heptagons, octagons and so on?   Again, no room for them to meet at a point and form part of a 3D shape.

These arguments certainly do not prove that the five Platonic solids exist.   They do show clearly, however, that there can be no more than the five (not counting the sphere, which is a bit of a digression).  And in practice, as it happens, all five do exist.   The one we've not mentioned yet is the dodecahedron, which has three regular pentagons meeting at each corner and turns out to have twelve of these pentagonal faces in all.

All this goes some way to explaining why the people with their knives and their raw potatoes have not spent much time cutting.   They can go through the above thought processes and can see there's no point.

What about the equivalent in four dimensions?

Regular shapes in various numbers of dimensions

New readers:

• 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.

Here is a series of numbers, which are in this order for a reason.   (Some of them, I suppose, are arguable, in that you could claim they should be different, but these ones will do for me.)   Here they are. 

1;  1;  infinity;  5;  6;  3;  3;  3;  3;  and guess the next number.

Any ideas?   What they are supposed to be, and indeed what they are at least in my mind, is the numbers of convex regular shapes, bounded by straight lines, that are possible in various numbers of dimensions.   So the numbers of dimensions corresponding to the above are: 

0;  1;  2;  3;  4;  5;  6;  7;  8;  9;  ?

In zero dimensions the only shape possible is a point.   Hence the answer in zero dimensions is 1.   Now, this is what I mean by arguable.   Is a point, a simple dot on a piece of paper, truly a shape?   And  even if it is a shape, is it convex and is it bounded by straight lines?   Arguably not.   In which case the first number in my series should be 0, zero.

What about one dimension?   Here the only shape possible is a straight line.   Is that a convex shape bounded by straight lines?   I think arguably it is.   So for my part I'd leave the second number in the series as 1.   But you may feel differently.

In two dimensions we are relatively untrammelled.   We can draw an equilateral triangle, a square, a pentagon, a hexagon and so on.   Whatever number of edges we want, however big, it is theoretically possible to draw a regular polygon (that's what they're called) which is convex and has that number of edges, or straight line sides.   So the third number in the series is infinity.

Now, there is a special feature of infinity.   If we were in the business of drawing these dreaded regular polygons, we could draw them, or at any rate imagine them, as big or small as we like.   We could have a regular thousand-sided figure, or a million- or trillion-sided figure if we prefer, with each of its regular straight line sides a mile long, a millimetre long, or whatever we choose.   But what if we draw a circle?

It is perfectly permissible to think of a circle, say a circle one inch in diameter, as consisting of an infinite number of infinitely short straight lines all joined together in such a way as to make a circle. So, on that way of looking at it, a circle is in fact an example of a regular convex polygon with an infinite number of sides.   This may be relevant when we look at higher dimensions.

It is surprising and extraordinary in a way, is it not,  after the infinite pastures of two dimensions, to find that the regular convex shapes in three dimensions are, as Lewis Carroll put it, "provokingly few in number".   They consist of the tetrahedron, which has four triangular faces, the cube, with its six square faces, the octahedron, with its eight triangular faces, the dodecahedron, with its twelve pentagonal faces, and finally the icosahedron which has twenty triangular faces.    Surely someone could create some extra regular shapes if they put their mind to it and had a sharp knife and a raw potato?