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Geometry and Dimensions. - Part 3

Updated: Apr 17, 2021

Circles, the tenth dimension and tesseracts: They're all here!


Another page from my notebook


Keywords: Flatland, fourth dimension, 3-D cross section, 4-D projections, tesseract, 3Blue1Brown, 4-D nets, simplex


Welcome to this blog post! Here we will meet our fellow Flatlanders, Spacelanders, Fourth-Dimensionlanders and we will go into the tenth dimension. Let’s go!

Higher Dimensions (continued…)


To consider a Fourth spatial dimension, we should first consider the realms of Flatland, our Spaceland and then consider Fourth-Dimensionland.

First of all, could the fourth dimension even exist?


To consider this question, we could put it to a Flatlander, a fictional 2-D character created by Edwin Abbott Abbott in his famous book, Flatland.


We could ask a Flatlander whether the third dimension exists. He would say no, since his body and his mind are accustomed to living in two dimensions and he just can’t grasp the third dimension. But contrary to his answer, you know there is a third dimension by experience. Using the same analogy, even if we disagree with the existence of the fourth dimension, it is possible that there is a dimension that our bodies are not accustomed to.

Also, if a Fourth-Dimensionlander passes through our Spaceland, what would it look like?

Well, we can try imagining what a Spacelander passing through Flatland would look like to a Flatlander.

In fact, the book “Flatland” addresses this question. The protagonist Flatlander, A. Square, sees a Spacelander, A.Sphere, go through the plane of Flatland, and here’s what it looks like:


A snippet from Ted-Ed’s reproduction of “Flatland”

This can be clearly reasoned as follows: A. Sphere is made up of 2-D circular cross-sections, and each of those show up in Flatland as he goes through the plane.


If we apply this same intuition to a hypersphere, then we can deduce that each of the 3-D spherical cross-sections of the hypersphere show up in our space, starting with a point to a sphere with the same diameter as the hypersphere and then to a point again as shown below.


A snippet from Ted-Ed’s reproduction of “Flatland”

We’ve just looked at an obtrusion of a 4-D shape into our space. Now let’s look at a projection of a 4-D shape onto our space.


First let’s consider a studio in Spaceland with a light bulb, a cube with opaque edges and a transparent filling and a planar screen where the projection of the cube will be cast upon. It would look something like this:


A projection of a cube: A square inside a square, or a hypotesseract from math.union.edu

This is because the square closer to the lightbulb would be shown to be larger because of perspective and the square farthest from the lightbulb would be shown smaller.

Now for a projection of a hypercube onto our space: the projection of a cube on a plane was a square in a square, connected with diagonal edges; the projection of a hypercube on our space would be a cube inside a cube, connected with diagonal edges as shown below:


A tesseract from en.wikipedia.org

This looks very weird. What looks even weirder is when a 4-D hypercube rotates while still being projected:



A rotating tesseract from simple.wikipedia.org

Why is this?

It would be easier to first consider a rotating cube projection and then consider the hypercube’s projection. This is what happens when a cube rotates:


From math.union.org

For this cube, the rightmost edge of the projection moves across from right to left whilst everything else moves to the right in one 90-degree rotation.


For the hypercube, therefore, the rightmost square (plane) of the projection moves across from front to back whilst everything else moves to the front in one 90-degree rotation.

So, we’ve got an idea of how projections and obtrusions of Fourth-Dimensionlanders on our space look like. But there is another concept I would like to share with you. I learnt this on a channel called 3Blue1Brown and I think that even if you don’t follow 3Blue1Brown, this thing would be fun to conceptualise:

Imagine a Flatland structure like this:



There is a square of side length 2 units, 4 blue circles with radius 1 unit and with centres positioned on the vertices of the square and a green circle with its centre on the centre of the square and with a radius such that the circle is tangent to each of the 4 blue circles.

What is the green circle’s 🟢 radius?

Answer: (sqrt(2) – 1) units

This can be easily derived since the distance from the centre of the green circle to a vertex of the square is defined to be sqrt(2) units. This distance is also defined to be the sum of the radius of the green and blue circles. Since the radius of the blue circles is 1 unit, the radius of the green circle is (sqrt(2) – 1) units.

Now let’s replicate this in three dimensions:

Photo from 3Blue1Brown

There is a cube of side length 2 units, 8 blue spheres with radius 1 unit and with centres positioned on the vertices of the cube and a green sphere with its centre on the centre of the cube and with a radius such that the sphere is tangent to each of the 8 blue spheres.

What is the green sphere’s radius?

Answer: (sqrt(3) – 1) units.

This can be easily derived since the distance from the centre of the green sphere to a vertex of the cube is defined to be sqrt(3) units. This distance is also defined to be the sum of the radius of the green and blue spheres. Since the radius of the blue spheres is 1 unit, the radius of the green sphere is (sqrt(3) – 1) units.

What will this be in Fourth-Dimensionland?

There will be a hypercube of side length 2 units, 16 blue hyperspheres with radius 1 unit and with centres positioned on the vertices of the hypercube and a green hypersphere with its centre on the centre of the hypercube and with a radius such that the hypersphere is tangent to each of the 16 blue hyperspheres.

What is the green hypersphere’s radius?

The answer is amazingly, 1 unit!

This can be easily derived since the distance from the centre of the green hypersphere to a vertex of the hypercube is defined to be sqrt(4) = 2 units. This distance is also defined to be the sum of the radius of the green and blue hyperspheres. Since the radius of the blue hyperspheres is 1 unit, the radius of the green hypersphere is sqrt(4) – 1 = 2 – 1 = 1 unit!

This is easy to prove, but let’s just think about this for a while: Not only is the green hypersphere the same size as the other hyperspheres and touching the other hyperspheres, but it is also touching the hypercube!

For even higher dimensions, one can see that the green n-sphere’s radius is actually greater than 1! The green n-sphere is reaching out of the n-cube! For example, in the tenth dimension, the green hyperhyperhyperhyperhyperhyperhypersphere’s (phew!) radius is sqrt(10) – 1 = 2.16 units (2 d.p.). That results in a 10-D volume of 5697.48 units^10 (2 d.p.), even larger than the 10-D cube’s 10-D volume (1024 units^10)!

If you are feeling very uneasy right now, I could empathise with you when I first learnt this!

But don’t worry, it’s just a concept. I’m not even sure the 10th dimension exists!

But brace yourself, because in Part 4, we will look at non-integer and even negative (?) dimensions!


4-D nets.


We have all cut out cube nets (from 2-D paper) and folded them to form the humble cube. How would a 4-D person cut out a hypercube net (out of 3-D paper) and fold the paper to make a hypercube? Well, from part 1 we know that there are 8 solids in the hypercube and that there are 3 cubes sharing each edge. Let's first describe a cube net in terms of the 6 squares in it:




There is a central square (orange) with 4 squares on its sides and on one of the outside squares, there is an extra square (red).


What would the 3-D equivalent of this be?


There would be a central cube with 6 cubes on its faces and on one of the outside cubes, there is an extra cube. If my calculations are correct, that is 8 cubes in total:


This is a lot to absorb. But how does the net fold into a hypercube?


Now, we know how to fold a cube net into a cube, but we simply do not have the perspective and intuition to fold 8 cubes into a hypercube that we cannot even see. What we can see, however, is the projection of the folding of the 8 cubes. This can be seen in the GIF below.

From Uncle Wikipedia.


I find it easier to first look at the 2-D net on the right and to realise what is happening before looking at the hypercube net.


This is great! But what about other 4-D shape nets?

Let's consider the "simplex", the 4D version of a tetrahedron. For the record, the 3-D projection of the simplex is:


To follow an analogy into higher dimensions, here is the 2-D net of the tetrahedron:


There is a central (orange) triangle surrounded by a triangle for each of its sides.


The 3-D equivalent is a central tetrahedron surrounded by a tetrahedron for each of its faces.


It would look like this:

Also, the process by which this folds into the projection is simple. The outmost corners of the outside tetrahedra move towards the centre of the central tetrahedron and meet at the centre. The points drag the edges with them.


These sorts of analogies also hold true in higher dimensions. As I said before, we will look at non-integer dimensions and even negative dimensions in Part 4. Please feel free to share your comments and thoughts below.


C u soon 😃!

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