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5.12: Cubes in higher dimensions

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    81434
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    This final section on elementary geometry seeks to explore fresh territory by going beyond three dimensions. Whenever we try to jump up to a new level, it can help to first take a step back and 'take a longer run up'. So please be patient if we initially take a step or two backwards.

    We all know what a unit 3D-cube is. And - going backwards - it is not hard to guess what is meant by a unit “2D-cube": a unit 2D-cube is just another name for a unit square. It is then not hard to notice that a unit 3D-cube can be constructed from two unit 2D-cubes as follows:

    • first position two unit 2D-cubes 1 unit apart in 3D space, with one directly above the other;
    • make sure that each vertex of the lower 2D-cube is directly beneath a vertex of the upper 2D-cube;
    • then join each vertex of the upper 2D-cube to the corresponding vertex below it.

    Perhaps a unit 2D-cube can be constructed in a similar way from “unit lD-cubes"! This idea suggests that a unit “1D-cube” is just another name for a unit line segment.

    Take the unit lD-cube to be the line segment from 0 to 1:

    • position two such 1D-cubes in 2D (e.g. one joining (0,0) to (1, 0), and the other joining (0,1) to (1,1));
    • check that each vertex of the lower 1D-cube is directly beneath a vertex of the upper 1D-cube;
    • then join corresponding pairs of vertices - one from the upper 1D-cube and one from the lower 1D-cube ((0, 0) to (0,1), and (1,0) to (1,1)) - to obtain a unit 2D-cube.

    Having taken a step back, we repeat (and reformulate) the previous construction of a 3D-cube: o

    • position two such unit 2D-cubes in 3D: with one 2D-cube joining (0, 0, 0) to (1,0,0), then to (1,1, 0), then to (0,1, 0) and back to (0, 0, 0), and the other 2D-cube joining (0,0,1) to (1, 0,1), then to (1,1,1), then to (0,1,1), and back to (1,0, 0);
    • with one 2D-cube directly above the other,
    • then join corresponding pairs of vertices - one from the upper 2D-cube and one from the lower 2D-cube ((0,0,0) to (0,0,1), and (1, 0,0) to (1, 0,1), and (1,1, 0) to (1,1,1), and (0,1, 0) to (0,1,1)) - to obtain a unit 3D-cube.

    To sum up: a unit cube in 1D, or in 2D, or in 3D:

    • has as “vertices” all points whose coordinates are all “0s or 1s” (in 1D, or 2D, or 3D)
    • has as “edges” all the unit segments (or unit 1D-cubes) joining vertices whose coordinates differ in exactly one place
    • and a unit 3D-cube has as “faces” all the unit 2D-cubes spanned by vertices with a constant value (0 or 1) in one of the three coordinate places (that is, for the unit 3D-cube: the four vertices with x = 0, or the four vertices with x = 1; or the four vertices with y = 0, or the four vertices with y = 1; or the four vertices with z = 0, or the four vertices with z = 1).

    A 3D-cube is surrounded by six 2D-cubes (or faces), and a 2D-cube is surrounded by four 1D-cubes (or faces). So it is natural to interpret the two end vertices of a 1D-cube as being '0D-cubes'. We can then see that a cube in any dimension is made up from cubes of smaller dimensions. We can also begin to make a reasonable guess as to what we might expect to find in a '4D-cuhe'.

    Problem 225

    (a) (i) How many vertices (i.e. 0D-cubes) are there in a 1D-cube?

    (ii) How many edges (i.e. 1D-cubes) are there in a 1D-cube?

    (b) (i) How many vertices (or 0D-cubes) are there in a 2D-cube?

    (ii) How many “faces” (i.e. 2D-cubes) are there in a 2D-cube?

    (iii) How many edges (i.e. 1D-cubes) are there in a 2D-cube?

    (c) (i) How many vertices (or 0D-cubes) are there in a 3D-cube?

    (ii) How many 3D-cubes are there in a 3D-cube?

    (iii) How many edges (i.e. 1D-cubes) are there in a 3D-cube?

    (iv) How many “faces” (i.e. 2D-cubes) are there in a 3D-cube?

    (d) (i) How many vertices (or 0D-cubes) do you expect to find in a 4D-cube?

    (ii) How many 4D-cubes do you expect to find in a 4D-cube?

    (iii) How many edges (i.e. lD-cubes) do you expect to find in a 4D-cube?

    (iv) How many “faces” (i.e. 2D-cubes) do you expect to find in a 4D-cube?

    (v) How many 3D-cubes do you expect to find in a 4D-cube?

    Problem 226

    (a) (i) Sketch a unit 2D-cube as follows. Starting with two unit lD-cubes - one directly above the other. Then join up each vertex in the upper lD-cube to the vertex it corresponds to in the lower lD-cube (directly beneath it).

    (ii) Label each vertex of your sketch with coordinates (x, y) (x,y = 0 or 1) so that the lower 2D-cube has the equation “y = 0” and the upper 2D-cube has the equation “y = 1”.

    (b) (i) Sketch a unit 3D-cube, starting with two unit 2D-cubes - one directly above the other. Then join up each vertex in the upper 2D-cube to the vertex it corresponds to in the lower 2D-cube (directly beneath it).

    (ii) Label each vertex of your sketch with coordinates (x, y, z) (where each x,y,z = 0 or 1) so that the lower 2D-cube has the equation “z = 0” and the upper 2D-cube has the equation “z = 1”.

    (c) (i) Now sketch a unit 4D-cube in the same way - starting with two unit 3D-cubes, one “directly above” the other.

    [Hint: In part (b) your sketch was a projection of a 3D-cube onto 2D paper, and this forced you to represent the lower and upper 2D-cubes as rhombuses rather than genuine 2D-cubes (unit squares). In part (c) you face the even more difficult task of representing a 4D-cube on 2D paper; so you must be prepared for other “distortions”. In particular, it is almost impossible to see what is going on if you try to physically position one 3D-cube “directly above” the other on 2D paper. So start with the “upper” unit 3D-cube towards the top right of your paper, and then position the “lower” unit 3D-cube not directly below it on the paper, but below and slightly to the left, before pairing off and joining up each vertex of the upper 3D-cube with the corresponding vertex in the lower 3D-cube.]

    (ii) Label each vertex of your sketch with coordinates (w, x, y, z) (where each w,x,y, z = 0 or 1) so that the lower 3D-cube has the equation “z = 0” and the upper 3D-cube has the equation “z = 1”.

    Problem 227 The only possible path along the edges of a 2D-cube uses each vertex once and returns to the start after visiting all four vertices.

    (a) (i) Draw a path along the edges of a 3D-cube that visits each vertex exactly once and returns to the start.

    (ii) Look at the sequence of coordinate triples as you follow your path. What do you notice?

    (b) (i) Draw a path along the edges of a 4D-cube that visits each vertex exactly once and returns to the start.

    (ii) Look at the sequence of coordinate 4-tuples as you follow your path. What do you notice?


    This page titled 5.12: Cubes in higher dimensions is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Alexandre Borovik & Tony Gardiner (Open Book Publishers) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.