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A hypercube in n dimensions, or an n cube, is the n dimensional analog of a cube.  It has two hyperfaces on each axis; the hyperfaces are n-1 dimensional hypercubes.

In general, we call the volume enclosed by a hypercube an n-volume.  Ordinary "volume" (measured in things like quarts and liters) is 3-volume.  Area (measured in things like acres and square meters) is 2-volume.  The magnitude of the volume of an n-cube with edge length r is rn.  We can form such an n-cube by "sweeping" an (n-1)-cube with edge length r in a direction perpendicular to itself, through a distance of r units.  The rest of this page just expands on, explains, and attempts to justify those statements.

Some examples of "low-dimension" n-cubes:
Figure 1:  Sweeping out a cube
Sweeping out a cube
To construct an n cube, we start with an n-1 cube, and sweep it through space perpendicular to the hyperplane in which it lies.   In figure 1 we show this operation for the first three dimensions.  We start with a point, at the origin, which sweeps out a line segment; the line segment then moves perpendicularly to its length, and sweeps out a square; the square then moves perpendicularly to its surface, and sweeps out a cube.  We have a hard time drawing anything higher than 3 dimensional figures so we stopped there.

By paying attention to what happens as we do this, we can learn what we need to know about how an n cube looks. For clarity, we'll usually call 2-dimensional faces "2-faces", and n-1 dimensional faces, "hyperfaces".  We want to know how many vertices, edges, and hyperfaces an n-cube has; each of those is described by a "recurrence relation", based on the number of vertices, edges, and hyperfaces an (n-1)-cube has, and by paying attention to the details as we build a cube we can see what those relations must be.



Reflective Symmetry Along the Axes

There is an additional item here worth emphasizing, which is that it does not make any difference what axis we start with.  We can sweep a point along any axis, then sweep the resulting line segment along any of the remaining axes, and so forth.

In other words, the hypercube which results is symmetric in swaps of the axes.

Furthermore -- and related -- it is symmetric in reflections through a hyperplane which passes through the center of the cube, and which is perpendicular to one of the axes.  Less obscurely, the cube is r units on an edge and we formed it by sweeping along each axis in turn, then if we reflect the cube across the plane xi = r/2, for any coordinate xi, it will be unchanged.

To see this, consider first the last axis, along which we swept an (n-1)-cube to form the n-cube.  We have two images of the (n-1)-cube "lying on" that axis:  One at 0, and one at r.  But think about what happens on the other axes during the final sweep:  Each of the faces of the (n-1)-cube, which is, itself, an (n-2)-cube, will sweep out a new (n-1)-cube, which will become a face of the newly formed n-cube.   But because of the way we performed the earlier sweeps, there will be a pair of these along each axis, and we'll end up with a pair of faces along each axis.

This is, of course, directly related to the fact that there are 2n faces on an n-cube.

This argument could be formalized into an induction proof fairly easily, but we won't be depending on it in any formal proofs so I won't go through that here.

Coordinates of the Vertices

Let's consider the process of "sweeping" which we used to construct the hypercubes.

We start with a single point in a 0 dimensional space, which has no coordinates.

We then introduce a single coordinate (the x coordinate), in which our point is at the origin; when we "sweep" it along the newly introduced axis, we create a second coordinate r units from the origin.  So, we have the vertices:
   (0), (r)

Next we introduce a second coordinate (the y coordinate), and place each existing vertex at 0 on the y axis:

  (0, 0), (r, 0)

That's still a line segment lying on the x axis.  When we sweep this segment r units along the y axis, we introduce additional vertices which have y values of r.  We now have four vertices:

    (0,0), (0, r), (r,0), (rr)

And so it goes.  With each "sweep" operation, we add a new "0" coordinate to each existing vertex and we add new vertices which have the same old coordinate values as the old vertices, but have "r" for the new coordinate value instead of 0.  For the cube, we thus obtain the set:

    (0,0,0), (0,r,0), (r,0,0),  (r,r,0), 
    (0,0,r), (0,r,r), (r,0,r),  (r,r,r)

If we do this again, the vertices of the 4 cube will be:

    (0,0,0,0), (0,r,0,0), (r,0,0,0),  (r,r,0,0),
    (0,0,r,0), (0,r,r,0), (r,0,r,0),  (r,r,r,0),
    (0,0,0,r), (0,r,0,r), (r,0,0,r),  (r,r,0,r),
    (0,0,r,r), (0,r,r,r), (r,0,r,r),  (r,r,r,r)

In general, we see (and we can prove by induction) that every possible combination of "r" and "0" is a vertex.

There are two possible values for each coordinate, and n coordinates, so the number of vertices must be 2n, which matches the value we already determined, above.

The Set Enclosed Within an n-Cube

Each time we "sweep" along an axis as we construct the cube, we start at 0 and move r units.  Thus, the length spanned by the cube on each axis is r units, and it includes all points in that region.

Since we sweep along each axis, all the points along each axis which fall between 0 and r must be included in the cube.

So, the set of points inside the cube must be all those points for which all coordinates are less than or equal to r. (This, by the way, shows that the faces must be where one or more coordinates take on the values r, and from that, we can in turn deduce that the coordinates of the vertices must all be 0 or r, as we already pointed out in the last section.)

Formally, we can say:


Or, if we wish to include the "skin", we could say the volume occupied by the cube is the set:


Page first posted on 11/04/2007