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.