Ever wanted to know what a 4 dimensional cube looks like? How can we even do that?

One explanation of what the 4th dimension is is time. This sort of works, thinking of a 3D object changing through time, as a single 4 dimensional entity. What about a 4th spatial dimension, what would that look like? It's very hard (maybe impossible?) to imagine because we're used to thinking in 3 dimensional space, however computers might be able to help...

To see how we're going to visualize a 4D cube, we have to first look at how we can visualize a 3D cube. The screen you're reading this on is a 2D surface-- there are two directions up-down and left-right. Any point on the screen can be specified with two numbers, two coordinates, that we can write as \((x,y)\) or \(\begin{bmatrix} x \\ y \end{bmatrix}\) or simply as \(\vec{p}\). That last notation denotes a vector which is technically not a point (think of it like a number with a direction) but that's for another post. Three dimensional points have a third direction, forward-back, which is usually denoted as \(z\), so any point in 3D space can be written as \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\). So how do we display a 3D object, a bunch of points \(\begin{bmatrix} x \\ y \\ z\end{bmatrix}\), on a 2D screen? The answer is in the Transformation.

There are all sorts of transformations that go from 3D to 2D but we'll use a simple one: \(\begin{bmatrix} x \\ y \\ z \end{bmatrix} \rightarrow \begin{bmatrix} x \\ y \end{bmatrix}\)

It'll be convenient so express this as a matrix multiplication. If you don't remember (or know) what a matrix is or know how matrix multiplication works, look at the following equation and notice the pattern and pay attention to columns. \[ P \vec{v} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \end{bmatrix} + z \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} x \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ y \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} \] If the above operation makes sense, you're good to go. Also notice this matrix \(P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\) takes a point \(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\) and maps it to the point \(\begin{bmatrix} x \\ y \end{bmatrix}\). This is exactly the transformation we were looking for!

Now if all we do is our simple transformation, we won't really be able to tell it's a 3D thing we're looking at-- we've
just thrown out all the depth information. What we'll need to do is move the object around, that way we can get multiple
perspectives and see the 3D object. Now we could do this by modelling a camera and moving that around, like a video game,
but that'll be overly complex and really only necessary if we want to do lighting and shadows. Instead we'll go with a simpler
solution: **rotation**

As you might've guessed (or seen on the computer screen), rotations are transformations that can be expressed through matrix multiplication (and a bit of trigonometry [maybe there will be a future page on this, just know that \(\sin\) and \(\cos\) take in some angle \(\theta\) and output numbers related to triangles]). In order to rotate a 2D point \(\begin{bmatrix} x \\ y \end{bmatrix}\) by some angle \(\theta\) we can use \(R = \begin{bmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{bmatrix} \).

However, if we rotate after we've mapped the point down to 2D, we'll already have lost the depth information so it'll be no good. Instead we need to rotate in 3D first, then map down to 2D. And in 3D there are actually 3 ways we can rotate: rotate about the x-axis, y-axis, or z-axis. These are all just variations on the 2D rotation \(R\) but we leave one of the coordinates unchanged. \[ R_x = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos{\theta} & -\sin{\theta} \\ 0 & \sin{\theta} & \cos{\theta} \end{bmatrix}, R_y = \begin{bmatrix} \cos{\theta} & 0 & -\sin{\theta} \\ 0 & 1 & 0 \\ \sin{\theta} & 0 & \cos{\theta} \end{bmatrix}, R_z = \begin{bmatrix} \cos{\theta} & -\sin{\theta} & 0 \\ \sin{\theta} & \cos{\theta} & 0 \\ 0 & 0 & 1 \end{bmatrix} \] Let's see how this works (and give another example of matrix multiplication): \[ R_y \vec{v} = \begin{bmatrix} \cos{\theta} & 0 & -\sin{\theta} \\ 0 & 1 & 0 \\ \sin{\theta} & 0 & \cos{\theta} \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = x \begin{bmatrix} \cos{\theta} \\ 0 \\ \sin{\theta} \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + z \begin{bmatrix} -\sin{\theta} \\ 0 \\ \cos{\theta} \end{bmatrix} = \begin{bmatrix} x \cos{\theta} \\ 0 \\ x \sin{\theta} \end{bmatrix} + \begin{bmatrix} 0 \\ y \\ 0 \end{bmatrix} + \begin{bmatrix} -z \sin{\theta} \\ 0 \\ z \cos{\theta} \end{bmatrix} = \begin{bmatrix} x \cos{\theta} - z \sin{\theta} \\ y \\ x \sin{\theta} + z\cos{\theta} \end{bmatrix} \] So we see that \(y\) is left unchanged and \(x\) and \(z\) are intermingled in a trigonometric way expressing the rotation.

Below you will find a 3D cube to play with. Try moving the sliders one at a time first, the 3D "effect" happens when you move more than one slider.

**Rotate about x axis:**

**Rotate about y axis:**

**Rotate about z axis:**

Now the question of how to render a 4 dimensional cube. To add a forth dimension we just add another number to our points like so: \( \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} \). Now the question is what are the verticies of a 4D cube? We can start by looking at what happens when we go from a square (a 2D "cube") to a cube (a 3D cube). For a square the points are: \( \begin{bmatrix} 0 \\ 0 \end{bmatrix} \), \( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \), \( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \), and \( \begin{bmatrix} 0 \\ 1 \end{bmatrix} \). For a cube the points are: \( \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \), \( \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \), \( \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \), \( \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \), \( \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \), \( \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \), \( \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \), and \( \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} \). Notice what we did to go from a square to a cube, we took all the square's points and added a \( 0 \) or a \( 1 \). So to get a 4D cube, we'll take all of the cube's points and add a \( 0 \) or a \( 1 \) to end up with 16 points.

Now we have the question of which verticies are connected. For a square and a cube we know where to draw the edges, we know not to connect the diagonals for instance, but how to we know which verticies to connect in the 4D case? We have to look for a pattern that doesn't depend on us being able to "visualize" it to know where to put the edges. In the case of a square we have \( \begin{bmatrix} 0 \\ 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 \\ 0 \end{bmatrix} \rightarrow \begin{bmatrix} 1 \\ 1 \end{bmatrix} \rightarrow \begin{bmatrix} 0 \\ 1 \end{bmatrix} \rightarrow \begin{bmatrix} 0 \\ 0 \end{bmatrix} \). Notice the pattern, there's an edge between the points if they only differ by 1. You'll see this is the same thing that happens for a cube too. So there's our rule, we can use this to know which 4D verticies to connect!

We'll use the same ideas of projecting and rotating that we did on the 3D cube to visualize the 4D cube. Here our projection matrix is similar but a little bigger: \[ P\vec{v} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \end{bmatrix} + z \begin{bmatrix} 0 \\ 0 \end{bmatrix} + w \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} x \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ y \end{bmatrix} + \vec{0} + \vec{0} = \begin{bmatrix} x \\ y \end{bmatrix} \] Also now we have 6 rotation matricies for all the ways that we could swap two components: \[ R_{\alpha} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos{\alpha} & -\sin{\alpha} \\ 0 & 0 & \sin{\alpha} & \cos{\alpha} \end{bmatrix}, R_{\beta} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos{\beta} & 0 & -\sin{\beta} \\ 0 & 0 & 1 & 0 \\ 0 & \sin{\beta} & 0 & \cos{\beta} \end{bmatrix}, R_{\gamma} = \begin{bmatrix} \cos{\gamma} & 0 & 0 & -\sin{\gamma} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \sin{\gamma} & 0 & 0 & \cos{\gamma} \end{bmatrix}, \] \[ R_{\delta} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos{\delta} & -\sin{\delta} & 0 \\ 0 & \sin{\delta} & \cos{\delta} & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, R_{\epsilon} = \begin{bmatrix} \cos{\epsilon} & 0 & -\sin{\epsilon} & 0 \\ 0 & 1 & 0 & 0 \\ \sin{\epsilon} & 0 & \cos{\epsilon} & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}, R_{\zeta} = \begin{bmatrix} \cos{\zeta} & -\sin{\zeta} & 0 & 0 \\ \sin{\zeta} & \cos{\zeta} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. \]

Below you'll find a demo of the 4D cube, again to really see it you need to rotate in several ways.

**\( \alpha \):**

**\( \beta \):**

**\( \gamma \):**

**\( \delta \):**

**\( \epsilon \):**

**\( \zeta \):**