Cross-sections: a 4D body passing through our world

A resident of Flatland can never see a ball whole. When a ball passes through their plane, they see a point that swells into a circle, grows, shrinks and vanishes. We are in the same position with respect to four-dimensional bodies. If a tesseract passed through our space, we would see a three-dimensional solid appear, change shape and disappear. This film of slices is the most honest way to “look” at 4D: unlike a projection, a cross-section squashes nothing — it is a genuine piece of the four-dimensional body.

Cutting with a hyperplane

Take all points of four-dimensional space whose fourth coordinate equals a fixed number $h$. This is the hyperplane $w = h$: a three-dimensional space, an exact copy of ours, hung at height $h$ along the fourth axis. Its intersection with a four-dimensional body is the cross-section. Changing $h$ is like pulling the body through our world along the $w$ axis.

For a convex polytope the section has a simple structure. Every cell is a convex polyhedron, and the hyperplane cuts it along a convex polygon. These polygons make up the surface of the section, which is a convex polyhedron. If the hyperplane misses the vertices, the bookkeeping is transparent:

• every edge it crosses gives a vertex of the section;
• every 2-face it crosses gives an edge of the section;
• every cell it crosses gives a face of the section.

The section depends not only on $h$ but also on how the body is turned. It is convenient to name the orientation after what meets the hyperplane first: cell-first, face-first, edge-first or vertex-first. This means that the $w$ axis points from the centre of the polytope to the centre of a cell, of a 2-face, of an edge, or to a vertex. For each orientation the widget below computes the rotation that turns that direction to the $w$ axis, then cuts the body with the hyperplane $w = h$.

The tesseract in four orientations

Take the tesseract with edge 1 centred at the origin: $|x|, |y|, |z|, |w| \le \tfrac12$. Let the direction of travel be a unit vector $n$; the section is the set of points of the tesseract with $n\cdot p = h$.

Cell-first, $n = (0,0,0,1)$. The condition $w = h$ simply drops the fourth coordinate, and for every $|h| < \tfrac12$ the section is the unit cube. The cube appears all at once, stays unchanged and vanishes all at once, like a square passing flat through Flatland.

Face-first, $n = (1,1,0,0)/\sqrt2$, towards the centre $(\tfrac12, \tfrac12, 0, 0)$ of a square face. The coordinates $z$ and $w$ do not enter the condition, so the section is the $1\times1$ square in $z$ and $w$ times the segment that the line $x + y = \sqrt2\,h$ cuts out of the square in $x$ and $y$. That segment has length $\sqrt2 - 2|h|$. The result is a box $1 \times 1 \times (\sqrt2 - 2|h|)$: it grows out of a square, lengthens to $\sqrt2$ at $h = 0$ and shrinks back to a square at $|h| = 1/\sqrt2$.

Edge-first, $n = (1,1,1,0)/\sqrt3$, towards the midpoint $(\tfrac12, \tfrac12, \tfrac12, 0)$ of an edge. Now only $w$ is missing from the condition, and the section is the section of the cube by the plane $x + y + z = \sqrt3\,h$ times a segment of length 1. A plane perpendicular to the cube's diagonal cuts it in a triangle near a vertex and in a hexagon in the middle. So the tesseract edge-first is a triangular prism for $\tfrac{\sqrt3}{6} < |h| < \tfrac{\sqrt3}{2}$ and a hexagonal prism for $|h| < \tfrac{\sqrt3}{6} \approx 0.289$; at $h = 0$ the hexagon is regular.

Vertex-first, $n = (1,1,1,1)/2$. On the vertices of the tesseract $n\cdot p$ takes five values, $1, \tfrac12, 0, -\tfrac12, -1$, at one, four, six, four and one vertex. For $\tfrac12 < h < 1$ the hyperplane cuts a corner off the top vertex, and the section is a regular tetrahedron that grows to edge $\sqrt2$. For $0 < h < \tfrac12$ that tetrahedron has all four corners sliced off: a truncated tetrahedron (4 triangles and 4 hexagons), and at $h = \tfrac14$ all its edges equal $\tfrac{\sqrt2}{2}$ — the Archimedean solid. At $h = 0$ the section passes through six vertices of the tesseract and becomes a regular octahedron of edge $\sqrt2$. Then everything repeats in reverse, with the tetrahedra pointing the other way.

The counting rule is easy to check. At $h = \tfrac14$ vertex-first the hyperplane crosses 12 edges of the tesseract (the ones joining the levels $\tfrac12$ and $0$), and the truncated tetrahedron has 12 vertices. All 8 cubic cells are crossed, and it has 8 faces.

The other regular polytopes

The same method gives each of the other regular polytopes its own series. In the widget they are all inscribed in the unit sphere.

The 5-cell (the simplex) is a pyramid over a tetrahedron. Vertex-first its sections are tetrahedra growing from a point to the opposite cell. Cell-first gives the same in reverse: in a simplex a cell lies opposite each vertex. Edge-first the hyperplane separates two vertices from three and crosses $2 \cdot 3 = 6$ edges, and the section is a triangular prism: its length falls from the length of the edge to zero while the triangle at its base grows to the opposite 2-face.

The 16-cell, with vertices $\pm e_1, \dots, \pm e_4$, is a double pyramid over an octahedron. Vertex-first (along $e_4$) its sections are octahedra growing from a point to the octahedron with vertices $\pm e_1, \pm e_2, \pm e_3$ and then shrinking again. Cell-first is more interesting: the tetrahedral cell first acquires bevels, and the section is made of 4 large triangles, 4 small ones and 6 rectangles. By the middle the large and small triangles have become equal, the rectangles have become squares, and at $h = 0$ the section is a cuboctahedron.

The 24-cell cell-first starts as an octahedron, turns into a truncated octahedron (at $h = \pm\tfrac{\sqrt2}{4}$ all its edges are equal: the Archimedean solid) and is a cuboctahedron in the middle. Vertex-first it starts as a cube — the vertex figure of the 24-cell is a cube. Then the cube acquires hexagonal bevels (6 squares and 12 hexagons), and at the “equator” the section is a rhombic dodecahedron.

The 600-cell vertex-first is arranged in layers. Its vertices lie at the levels $\cos 0^\circ$, $\cos 36^\circ$, $\cos 60^\circ$, $\cos 72^\circ$, $\cos 90^\circ$ and symmetrically below: 1, 12, 20, 12, 30, 12, 20, 12, 1 vertices. The second layer forms an icosahedron, the third a dodecahedron, the fifth an icosidodecahedron. But the section at the height of a layer is not always the convex hull of that layer: edges joining every other layer also pierce the hyperplane. Counting by the rule above shows that at height $\cos 36^\circ \approx 0.809$ the section is exactly an icosahedron with the 600-cell's own edge, at $h = \tfrac12$ it is a dodecahedron with low pentagonal pyramids on its faces (32 vertices, 60 triangles), and at $h = 0$ an icosidodecahedron with pyramids on its twelve pentagons (42 vertices, 80 triangles).

Smooth bodies

Sections of smooth four-dimensional bodies are even easier: just put $w = h$ into their equations.

The hypersphere (strictly, the 4-ball) $x^2 + y^2 + z^2 + w^2 \le r^2$. With $w = h$ what remains is $x^2 + y^2 + z^2 \le r^2 - h^2$, a ball of radius $\sqrt{r^2 - h^2}$. A three-dimensional observer sees a point that swells into a ball of radius $r$ and shrinks back to a point: the direct analogue of the circle in Flatland. Turning the body changes nothing, since all directions are alike for the hypersphere.

The duocylinder is the product of two discs: $x^2 + y^2 \le r^2$ and $z^2 + w^2 \le r^2$. With $w = h$ the second condition becomes $|z| \le \sqrt{r^2 - h^2}$, and the section is a cylinder of radius $r$ and height $2\sqrt{r^2 - h^2}$: a round washer thickens into a cylinder of height $2r$ and flattens again. The duocylinder has no three-dimensional analogue: it can only exist in 4D, where there is room for two completely orthogonal planes.

The spherinder, or spherical cylinder, is a ball times a segment: $x^2 + y^2 + z^2 \le r^2$, $|w| \le a$. Cut along the segment, it always gives the same ball, which appears and vanishes abruptly, like the tesseract's cube cell-first. Cut across, $x = h$, it gives a cylinder of length $2a$ and radius $\sqrt{r^2 - h^2}$ that thins down to a segment. The cubinder is a disc times a square: along one side of the square its sections are identical cylinders; across the disc they are boxes with one side $2\sqrt{r^2 - h^2}$ that changes.

The tiger is the set of points within distance $r$ of the Clifford torus: $\bigl(\sqrt{x^2 + y^2} - R\bigr)^2 + \bigl(\sqrt{z^2 + w^2} - R\bigr)^2 \le r^2$. The name comes from the Japanese tora, “tiger”, which happens to sound like “torus”. With $w = h$ the second term contains $\sqrt{z^2 + h^2}$, and for each point at distance $\rho$ from the $z$ axis the allowed values of $z$ form two intervals, symmetric about zero, as long as $|h| < R - r$. So the section is two coaxial solid tori, shifted along the $z$ axis in opposite directions. At $|h| = R - r$ they touch, beyond that they merge into a single torus, which thins and vanishes at $|h| = R + r$. The ditorus is the set of points within distance $\rho$ of the surface of an ordinary torus; its section is a hollow torus, a thick-walled inner tube whose wall thins to nothing at $|h| = \rho$.

Cavalieri's principle and the volume of the 4-ball

The Italian mathematician Bonaventura Cavalieri stated in his Geometry of Indivisibles (1635) that if all sections of two solids by parallel planes have equal areas, the solids have equal volumes. In the fifth century the Chinese mathematicians Zu Chongzhi and his son Zu Gengzhi used the same idea to find the volume of a ball. In four dimensions the rule looks the same, with areas replaced by the volumes of the sections.

The most beautiful application is the volume of the four-dimensional ball. The section at height $h$ is a ball of radius $\sqrt{r^2 - h^2}$ with volume $\tfrac43\pi(r^2 - h^2)^{3/2}$. Substitute $h = r\sin t$; then $dh = r\cos t\,dt$ and $(r^2 - h^2)^{3/2} = r^3\cos^3 t$: $$\mathrm{Vol}_4 = \int_{-r}^{r} \frac43\pi\,(r^2 - h^2)^{3/2}\,dh = \frac43\pi r^4 \int_{-\pi/2}^{\pi/2} \cos^4 t\,dt = \frac43\pi r^4\cdot\frac{3\pi}{8} = \frac{\pi^2 r^4}{2}.$$ Here $\int_{-\pi/2}^{\pi/2}\cos^4 t\,dt = \tfrac{3\pi}{8}$, because $\cos^4 t = \tfrac18(3 + 4\cos 2t + \cos 4t)$ and the cosines of the multiple angles integrate to zero over this interval. The unit 4-ball has volume $\pi^2/2 \approx 4.935$. It fills $\pi^2/32 \approx 30.8\%$ of the circumscribed tesseract $[-1, 1]^4$, while an ordinary ball fills $\pi/6 \approx 52.4\%$ of its cube. The share keeps falling as the dimension grows; see the chapter on high dimensions.

Let us check the principle on the tesseract: its volume is 1 in every orientation, while the graphs of its sections are completely different. Cell-first, $\int_{-1/2}^{1/2} 1\,dh = 1$. Face-first, $$\int_{-1/\sqrt2}^{1/\sqrt2} \bigl(\sqrt2 - 2|h|\bigr)\,dh = 2\Bigl(\sqrt2\cdot\tfrac{1}{\sqrt2} - \tfrac12\Bigr) = 1.$$ Vertex-first, the tetrahedron of edge $2\sqrt2\,(1 - |h|)$ has volume $\tfrac83(1 - |h|)^3$, and the truncated tetrahedron is such a tetrahedron minus four corners of volume $\tfrac83(\tfrac12 - |h|)^3$ each. Integrating, the tetrahedra contribute $2\cdot\tfrac83\cdot\tfrac{1}{64} = \tfrac1{12}$ and the truncated tetrahedra $\tfrac{11}{12}$, once again $1$ in total.

The same principle shows that a four-dimensional pyramid has a quarter of the base volume times the height: the section at distance $t$ from the apex is the base scaled down by $H/t$, its volume is $V\,(t/H)^3$, and $\int_0^H V (t/H)^3\,dt = VH/4$. The 16-cell is two such pyramids of height 1 over an octahedron of volume $\tfrac43$, so its volume is $2\cdot\tfrac14\cdot\tfrac43 = \tfrac23$. In the widget the area under the graph is computed numerically: for the 5-, 8-, 16-, 24- and 600-cell inscribed in the unit sphere it comes out as $0.146$, $1$, $0.667$, $2$ and $3.863$ — exactly their four-dimensional volumes.

Alicia Boole Stott

Series of sections of the regular polytopes were studied in depth by Alicia Boole Stott (1860–1940). She was born on 8 June 1860 in Cork, Ireland, to the logician George Boole and Mary Everest Boole. She never went to university: she learned mathematics mostly from her mother and from the first books of Euclid's Elements. At seventeen she was introduced to four-dimensional geometry by Charles Howard Hinton, her sister's husband, who had devised wooden cubes for training four-dimensional intuition. Alicia developed a rare ability to see the fourth dimension and found on her own that there are exactly six regular polytopes.

Her method is the one used in this chapter: successive three-dimensional sections parallel to one of the cells. She constructed them by purely geometric means and made cardboard models of the sections. In 1900 her paper “On certain series of sections of the regular four-dimensional hypersolids” appeared in Amsterdam, with diagrams of sections of the 600-cell and the 120-cell. In it Boole Stott brought the word polytope into English, anglicising the German Polytop coined by Reinhold Hoppe in 1882.

From 1895 she worked for almost twenty years with the Dutch geometer Pieter Hendrik Schoute of Groningen; in 1914 the University of Groningen gave her an honorary doctorate. In the 1930s she worked with the young Harold Coxeter, and they presented a joint paper in Cambridge. The cardboard models of the sections of the 120- and 600-cell that she left with Schoute are still on display at the University of Groningen; another set of her models is kept at the Faulkes Institute for Geometry in Cambridge.

Summary

• The section of a four-dimensional body by the hyperplane $w = h$ is a three-dimensional solid; as $h$ changes, a three-dimensional observer sees a “film” of sections.
• For a convex polytope, crossed edges give the vertices of the section, 2-faces its edges, cells its faces.
• The tesseract cell-first is a motionless cube; face-first a box $1\times1\times(\sqrt2 - 2|h|)$; edge-first a triangular, then a hexagonal prism; vertex-first a tetrahedron, a truncated tetrahedron and an octahedron in the middle.
• The hypersphere's section is a ball of radius $\sqrt{r^2 - h^2}$, the duocylinder's a cylinder, the tiger's two tori that merge into one.
• Cavalieri's principle: the four-dimensional volume is $\int V(h)\,dh$. Hence the volume of the 4-ball, $\pi^2 r^4/2$, and the 4D pyramid rule “base × height / 4”.
• Alicia Boole Stott published series of sections of the regular polytopes in 1900 and brought the word polytope into English.

Next come the six regular polytopes: why there are exactly six and how they are built.

Sources

• A. Boole Stott. On certain series of sections of the regular four-dimensional hypersolids. Verhandelingen der Koninklijke Akademie van Wetenschappen te Amsterdam, 1900.
• H. S. M. Coxeter. Regular Polytopes. 3rd ed. Dover, 1973 — sections and projections of the regular polytopes.
• Alicia Boole Stott — biography in the MacTutor History of Mathematics archive (University of St Andrews).
• Alicia Boole Stott — Wikipedia: dates, Hinton, Schoute, the 1914 honorary doctorate, Coxeter, the word polytope.
• T. Phillips. The Princess of Polytopia: Alicia Boole Stott and the 120-cell — AMS Feature Column: sections parallel to a cell, the models in Groningen and Cambridge.
• I. Polo-Blanco. Alicia Boole Stott, a geometer in higher dimension. Historia Mathematica 35 (2008).
• Polytope — Wikipedia: R. Hoppe's Polytop (1882).
• Cavalieri's principle — Wikipedia: Cavalieri (1635), Zu Chongzhi and Zu Gengzhi.
• Tiger — Polytope Wiki: the definition of the tiger and the origin of its name.
• Duocylinder — Wikipedia.