an n cube
, is the n
dimensional analog of a cube. It has two hyperfaces
on each axis; the hyperfaces are n
In general, we call the volume enclosed by a hypercube an n-volume
Ordinary "volume" (measured in things like quarts and liters)
. 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
We can form such an n
-cube by "sweeping"
)-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
- A 0-cube is a point,
and has no defined volume.
- A 1-cube is a line
segment; its 1-volume is its length.
- A 2-cube is a square;
its 2-volume is its area.
- A 3-cube is an
- A 4-cube is a tesseract,
and is occasionally just called a "hypercube".
1: Sweeping out a cube
To construct an n
start with an n
-1 cube, and sweep
it through space perpendicular to the hyperplane in which it lies.
In figure 1
show this operation for the first three
dimensions. We start with a point, at the origin, which
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
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.
- We started with a
point (0 dimensions). A point has 0 edges, 0 "2-faces", and 1
vertex. It also has zero hyperfaces. It has no
We sweep the point through space r
units to form a line segment, or 1-cube,
with length r. It has 1 edge, zero
"2-faces", and 2 vertices:
- The single vertex of the point was
replaced with two new vertices
when it swept along: one (old) vertex at its starting
and one (new) vertex where it ended. We observe that the
vertex was doubled as a result. In
general, we will see that an n-dimensional hypercube will have 2n
general, just as the single point formed two images
of itself -- one
where it started sweeping, one where it finished sweeping -- we would
expect any hypercube to form two images of itself
if it were swept
through space perpendicular to the plane in which it lies.
each edge will form two "image edges", each face will form two "image
faces", and so forth.
- The edge was created as the vertex
swept along. It seems clear that every vertex will
form an edge when it's swept through space.
- The 1-volume of the line segment is r,
the distance which was "swept out" by the point.
- The line segment has two "hyperfaces", which are its
vertices; they lie on the axis along which the point swept.
We sweep the line segment through space
to form a square, or 2-cube.
The square has four vertices, four edges, and four
"hyperfaces" (the edges).
- The two vertices on the line segment each "doubled",
giving us four vertices. The number of vertices is,
- The line segment "doubles", leaving an edge where it
started, and forming another where it ended.
of the two vertices of the original line segment has swept out a new
edge. Thus, we have four edges altogether: two for
edge in the 1-cube, plus one for each vertex in the 1-cube.
- The 2-volume of the square is the distance swept
times the "1-volume" of the 1-cube, or r2.
- The square has four "hyperfaces", which are its edges.
The line segment moved along the y
axis, and left two images of itself along that axis. The two vertices
of the segment each swept out new edges, and they lie on the x
see that in general, (n-1)-cube forms two images of itself, which
become two hyperfaces of the n-cube. In other words, the
n-cube is the volume swept out by one of its hyperfaces.
In addition, each hyperface of the (n-1)-cube sweeps out a new
(n-1)-cube, which in turns becomes a face of the n-cube.
So, we acquired one hyperface for each hyperface of the (n-1)-cube, plus
two new hyperfaces:
we're "sweeping" along each axis in turn, this means we have one pair
of hyperfaces along each axis; the total number of hyperfaces is
We sweep the square through space to
form a cube, or 3-cube.
The 3-cube has eight vertices, 10 edges, and six "hyperfaces"
(in the case of a 3-cube, they're the ordinary 2-faces).
|Figure 4: "3-cube"|
- The four vertices on the square each "doubled", giving us
eight vertices. Again, we have 2n
- Again, each edge of the square "doubled", giving us two
new edges in place of each old one.
- In addition, each vertex on the square swept out a new
edge as it moved. Thus, the number of edges on the 3-cube is
equal to the number of vertices on the 2-cube plus twice
the number of edges on the 2-cube:
- The 3-volume of the 3-cube is the area of the square, times
the distance the square moved, or r3.
square formed two faces for our cube: one where it started, and one
where it ended. In addition, each edge (hyperface) of the
formed a new square. So the number of hyperfaces on the
6, which is, as we expected, 2n.
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.
-- 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
/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
-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
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
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:
Next we introduce a second coordinate (the y
coordinate), and place each existing vertex at 0
on the y
(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
(0,0), (0, r
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
" for the new coordinate value instead
of 0. For the cube, we thus obtain the set:
If we do this again, the vertices of the 4
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