In putting the sections together some of the outlines in them disappear. The line TW for instance is not wanted.
We notice that PQTW and TWRS are each the half of a hexagon. Now QV and VR lie in one straight line. Hence these two hexagons fit together, forming one hexagon, and the line TW is only wanted when we consider a section of the whole figure, we thus obtain the solid represented in the lower part of [fig. 74]. Equal repetitions of this figure, called a tetrakaidecagon, will fill up three-dimensional space.
To make the corresponding four-dimensional figure we have to take five axes mutually at right angles with five points on each. A catalogue of the positions determined in five-dimensional space can be found thus.
Fig. 75.
Take a cube with five points on each of its axes, the fifth point is at a distance of four units of length from the first on any one of the axes. And since the fourth dimension also stretches to a distance of four we shall need to represent the successive sets of points at distances 0, 1, 2, 3, 4, in the fourth dimensions, five cubes. Now all of these extend to no distance at all in the fifth dimension. To represent what lies in the fifth dimension we shall have to draw, starting from each of our cubes, five similar cubes to represent the four steps on in the fifth dimension. By this assemblage we get a catalogue of all the points shown in [fig. 75], in which L represents the fifth dimension.
Now, as we saw before, there is nothing to prevent us from putting all the cubes representing the different stages in the fourth dimension in one figure, if we take note when we look at it, whether we consider it as a 0h, a 1h, a 2h, etc., cube. Putting then the 0h, 1h, 2h, 3h, 4h cubes of each row in one, we have five cubes with the sides of each containing five positions, the first of these five cubes represents the 0l points, and has in it the i points from 0 to 4, the j points from 0 to 4, the k points from 0 to 4, while we have to specify with regard to any selection we make from it, whether we regard it as a 0h, a 1h, a 2h, a 3h, or a 4h figure. In [fig. 76] each cube is represented by two drawings, one of the front part, the other of the rear part.
Let then our five cubes be arranged before us and our selection be made according to the rule. Take the first figure in which all points are 0l points. We cannot have 0 with any other letter. Then, keeping in the first figure, which is that of the 0l positions, take first of all that selection which always contains 1h. We suppose, therefore, that the cube is a 1h cube, and in it we take i, j, k in combination with 4, 3, 2 according to the rule.
The figure we obtain is a hexagon, as shown, the one in front. The points on the right hand have the same figures as those on the left, with the first two numerals interchanged. Next keeping still to the 0l figure let us suppose that the cube before us represents a section at a distance of 2 in the h direction. Let all the points in it be considered as 2h points. We then have a 0l, 2h region, and have the sets ijk and 431 left over. We must then pick out in accordance with our rule all such points as 4i, 3j, 1k.
These are shown in the figure and we find that we can draw them without confusion, forming the second hexagon from the front. Going on in this way it will be seen that in each of the five figures a set of hexagons is picked out, which put together form a three-space figure something like the tetrakaidecagon.