The famous “cube inside a cube” on the covers of books about the fourth dimension is not a tesseract. It is the tesseract’s shadow. The tesseract itself is made of eight identical cubes, yet in the picture one cube is big, one is small, and the other six look like truncated pyramids. A shadow depends on how the light falls, and one solid can cast very different shadows. In this chapter we look at the kinds of “light” available for four-dimensional solids, derive a formula for each, and see what each shadow hides.
A shadow is a projection
A shadow works simply: each point of a solid goes to the spot where the ray of light through it hits the floor. Mathematicians call such a map a projection. Two kinds of light matter.
The sun is so far away that its rays are parallel. Let them fall vertically, let the floor be the plane $y = 0$ and the height of a point its $y$-coordinate. Then the shadow of the point $(x, y, z)$ is $(x, z)$: the height is simply forgotten. This is parallel, or orthographic, projection. The size of the shadow does not depend on how high the object hangs.
A lamp is a point. Hang it at height $d$ above the origin, at $L = (0, d, 0)$. The ray from the lamp through the point $P = (x, y, z)$ consists of the points $L + t\,(P - L)$. Its height is $d + t\,(y - d)$, which vanishes at $t = d/(d - y)$. So the shadow of the point is $$\bigl(X, Z\bigr) = \frac{d}{d-y}\,\bigl(x, z\bigr).$$ This is central, or perspective, projection. The higher the point, the closer it is to the lamp and the more its shadow is stretched: the factor $d/(d-y)$ grows with height.
Now hang a cube under the lamp with its faces horizontal. The top face is nearer the lamp, and its shadow is a big square. The bottom face is farther, and its shadow is a smaller square inside the big one. The four vertical edges give four segments joining the corners of the squares. The result is a “square inside a square”. With $d = 3$ and the faces at heights 1.7 and 0.7, the factors are $3/1.3 \approx 2.31$ and $3/2.3 \approx 1.30$.
Raise the lamp to $d = 12$: the squares nearly coincide, because a distant lamp shines almost like the sun, $d/(d-y) \to 1$. Switch to “Sun” and they coincide exactly. Turn the cube vertex on to the light: the outline of the shadow becomes a hexagon. Edge colour shows height: amber is nearer the lamp, blue is farther away.
That is the whole secret of the “cube inside a cube”. Stand a tesseract with one cell facing a four-dimensional “lamp”: the near cell gives a big cube, the far cell a small one inside it, and the eight edges running along the fourth axis join their vertices. The six side cells turn into truncated pyramids between them, just as the side faces of the cube turned into trapezoids.
From four dimensions to three
Let’s carry both formulas one floor up. The “floor” is now our space $w = 0$, and “height” is the fourth coordinate $w$.
Parallel projection simply forgets $w$: $$(x, y, z, w) \longmapsto (x, y, z).$$
Perspective projection. Put an eye at the point $E = (0, 0, 0, d)$ on the $w$-axis. The line from the eye through a point $p = (x, y, z, w)$ is $E + t\,(p - E)$; its $w$-coordinate $d + t\,(w - d)$ vanishes at $t = d/(d - w)$. Hence $$\mathbf X = \frac{d}{d - w}\,(x, y, z) = \frac{(x, y, z)}{1 - k\,w}, \qquad k = \frac1d.$$ The second form is handier: a single parameter $k$ turns one projection smoothly into the other. At $k = 0$ (the eye infinitely far away) we get the parallel shadow, and for $0 < k < 1$ a perspective view. This is exactly how the lenses on the showcase work. The formula holds as long as the eye is outside the solid: points with $w > d$ would be behind the eye.
Let’s do the numbers for the tesseract with edge 1 centred at the origin: its vertices are the points $(\pm\frac12, \pm\frac12, \pm\frac12, \pm\frac12)$, and two of its cells lie in the hyperplanes $w = \pm\frac12$. With $d = 2$ the near cell is stretched by $2/1.5 = 4/3$ and the far one shrunk to $2/2.5 = 4/5$, so the outer cube is $5/3 \approx 1.67$ times the inner one. Each of the other six cells has four vertices at $w = \frac12$ and four at $w = -\frac12$: one face is stretched, the opposite one shrunk, and the cube becomes a truncated pyramid. All eight cells are identical cubes, yet only two of them look like cubes in the picture.
The perspective divide and homogeneous coordinates
Everything in the perspective formula is linear except the division by $1 - kw$. That division can be put off until the very end. Append a 1 to the point, $(x, y, z, w, 1)$, and multiply by a matrix: $$\begin{pmatrix} X_1 \\ X_2 \\ X_3 \\ q \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -k & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \\ 1 \end{pmatrix},$$ $$\mathbf X = \frac{(X_1, X_2, X_3)}{q}.$$ The quadruple $(X_1 : X_2 : X_3 : q)$ gives the homogeneous coordinates of a point of three-dimensional space: proportional quadruples stand for the same point, and the point itself is recovered by dividing by the last number. Homogeneous coordinates were invented independently by August Möbius (Der barycentrische Calcul, 1827), Karl Wilhelm Feuerbach (1827) and Julius Plücker (1830). Their charm is that perspective becomes a matrix multiplication, so rotations and perspective can be combined into a single matrix.
Every graphics card does the same thing. A vertex shader outputs four numbers, the “clip coordinates” $(x, y, z, w)$, and the hardware divides the first three by the last. That step is known as the perspective divide. The clash of names is unfortunate: for the graphics card, $w$ is just a bookkeeping divisor, not a fourth dimension. The showcase on this site divides twice in a row: first the four-dimensional point is projected into our space by $\mathbf X = (x,y,z)/(1-kw)$, then an ordinary three-dimensional camera divides once more to put the picture on the screen.
The Schlegel diagram
What if we move the eye right up to one cell, so that it is just outside that cell’s hyperplane but still “below” the hyperplanes of all the other cells? Then only that one cell is visible from the eye. It becomes a frame, and all the other cells project inside it, filling it without overlaps like a honeycomb. Such a picture is called a Schlegel diagram.
One floor down this is familiar: look at a cube with your nose almost against one face, and you see a square inside a square, with the four side faces as trapezoids in between. A dodecahedron gives a pentagon divided into eleven smaller pentagons. For the tesseract, the Schlegel diagram is the same “cube inside a cube”, but now we know what it means: the outer cube is the near cell, the inner one the far cell, the six truncated pyramids are the side cells, and no cell overlaps another. The whole structure of the polytope, which cells border which, is on view without overlaps.
The diagram is named after the German mathematician Victor Schlegel (1843–1905). In 1883 he published a long treatise, “Theorie der homogen zusammengesetzten Raumgebilde”, in the journal of the Leopoldina academy, and in the same year he made models of the projections of all six regular four-dimensional polytopes out of brass wire and silk thread. They were sold as teaching models, first by Ludwig Brill in Darmstadt and later by his successor Martin Schilling in Leipzig. In 1886 he published “Ueber Projectionsmodelle der regelmässigen vier-dimensionalen Körper” (on projection models of the regular four-dimensional solids).
Stereographic projection
All the vertices of the regular polytopes lie on the three-dimensional sphere, the “surface” of a four-dimensional ball. For solids on a sphere there is a particularly beautiful shadow. Let’s start, as usual, one dimension down.
Take the unit circle $x^2 + y^2 = 1$, its “north pole” $N = (0, 1)$ and the line $y = 0$. Draw a ray from $N$ through a point $P = (x, y)$ of the circle until it meets the line. By similar triangles, the meeting point is $$X = \frac{x}{1 - y}.$$ Every point of the circle except the pole itself lands on exactly one point of the line, and every point of the line is the image of exactly one point of the circle; the inverse is $x = \dfrac{2X}{1+X^2}$, $y = \dfrac{X^2 - 1}{X^2 + 1}$. The south pole goes to zero, the points $(\pm1, 0)$ stay put, and as $P$ approaches the pole, $X$ runs off to infinity. The pole itself goes to “the point at infinity”. For the sphere it is the same, from the pole $N = (0, 0, 1)$ onto the plane $z = 0$: $$(X, Y) = \frac{(x, y)}{1 - z}.$$ This is stereographic projection.
Move the point $P$ towards the pole and watch $X$: 1.92 at 35°, 3.73 at 60°, 11.4 at 80°, 114.6 at 89°. The coloured dots on the circle are spaced evenly, 15° apart, but their images on the line spread out more and more. The green arcs mark two angles that are always equal: that is the key to the proof below. In the “Sphere → plane” mode, compare the angle between the circles on the sphere with the angle between their images; the latter is computed afresh from the projection formulas, not copied.
It is one of the oldest map projections. Hipparchus may have known it in the 2nd century BC (Synesius hinted as much, somewhat vaguely, around 400 AD), and the earliest surviving description is Ptolemy’s Planisphaerium, 2nd century AD. The astrolabe is built on it: its plate is a stereographic map of the celestial sphere. The name “stereographic” was given by François d’Aguilon in 1613. That the projection takes every circle on the sphere to a circle was proved in general by al-Farghānī in the 9th century. That it preserves angles was proved by Thomas Harriot in notes he never published, and the first published proof came from Edmond Halley in the Philosophical Transactions of 1695–96. Halley frankly said that he had learned the fact from de Moivre and that Hooke had already shown it to the Royal Society; only the proof was his own.
Circles go to circles
Stereographic projection takes every circle on the sphere to a circle in the plane if the circle does not pass through the pole $N$, and to a straight line if it does.
Every circle on the sphere is the intersection of the sphere with a plane $ax + by + cz = e$. Express a point of the sphere through its image $(X, Y)$ by the inverse formulas: with $R^2 = X^2 + Y^2$, $$\begin{gathered} x = \frac{2X}{1 + R^2}, \quad y = \frac{2Y}{1 + R^2}, \\ z = \frac{R^2 - 1}{R^2 + 1}. \end{gathered}$$ (Check: $x^2 + y^2 + z^2 = \frac{4R^2 + (R^2-1)^2}{(R^2+1)^2} = 1$ and $1 - z = \frac{2}{R^2+1}$, so $x/(1-z) = X$.) Substitute into the equation of the plane and multiply by $1 + R^2$: $$\begin{gathered} (c - e)(X^2 + Y^2) + 2aX + 2bY \\ {} - (c + e) = 0. \end{gathered}$$ If $c \ne e$, this is a circle with centre $\bigl(-\frac{a}{c-e}, -\frac{b}{c-e}\bigr)$ and radius $\frac{\sqrt{a^2+b^2+c^2-e^2}}{|c-e|}$; the expression under the root is positive because the plane cuts the sphere in a circle, which means it is closer than 1 to the centre: $e^2 < a^2 + b^2 + c^2$. If $c = e$, the plane passes through $N = (0, 0, 1)$ and the equation becomes linear, $aX + bY = c$: a straight line. Since the projection is one-to-one away from the pole, the image of the circle is the whole of that circle or the whole of that line. ∎
Angles are preserved
A map that preserves the angles between curves is called conformal. Stereographic projection is conformal: two curves on the sphere cross at the same angle as their images in the plane.
If two curves on the sphere cross at a point $P \ne N$ at an angle $\alpha$, their stereographic images cross at the point $P'$ at the same angle $\alpha$.
The angle between curves is the angle between their tangent lines, and the tangent lines at $P$ lie in the tangent plane $T_P$. A tangent line $\ell$ at $P$ and the line $NP$ span a plane $\Pi_\ell$; projection from $N$ takes $\ell$ to the line $\Pi_\ell \cap \{z = 0\}$, which is the tangent to the image curve at $P'$. Now reflect the whole of space in the plane $\sigma$ that is perpendicular to the segment $NP$ and bisects it. The sphere goes to itself (its centre lies in $\sigma$), and $N$ and $P$ swap places, so the tangent plane $T_P$ goes to the tangent plane $T_N$. Every plane $\Pi_\ell$ contains the line $NP$, which is perpendicular to $\sigma$, and therefore goes to itself. So the line $\ell = \Pi_\ell \cap T_P$ goes to the line $\Pi_\ell \cap T_N$, and the angle between two tangent lines in $T_P$ equals the angle between the corresponding lines in $T_N$ (a reflection preserves angles). Finally, $T_N$ is the plane $z = 1$, parallel to the plane $z = 0$, and one plane $\Pi_\ell$ cuts two parallel planes along parallel lines. So the angle in $T_N$ equals the angle between the images in the plane $z = 0$. ∎
In the flat case the proof can be seen in the widget: the green arcs are the angle between the ray and the tangent at $P$, and the angle between the ray and the line at $X$. They are equal because the tangents to a circle at the two ends of the chord $NP$ make equal angles with it, and the tangent at $N$ is parallel to the line.
Angles are preserved; lengths are not. For the circle parametrised by angle, $x = \cos t$, $y = \sin t$, the derivative of the image is $$\begin{aligned} \frac{dX}{dt} &= \frac{d}{dt}\,\frac{\cos t}{1 - \sin t} \\ &= \frac{-\sin t\,(1 - \sin t) + \cos^2 t}{(1 - \sin t)^2} \\ &= \frac{1 - \sin t}{(1 - \sin t)^2} = \frac{1}{1 - y}. \end{aligned}$$ So a small arc near a point at height $y$ is stretched by $1/(1 - y)$: halved at the south pole, unchanged at the equator, stretched without limit near the north pole. For a sphere of any dimension the factor is the same, $1/(1 - w)$, where $w$ is the “height” of the point above the equator.
The three-sphere in our space
Up one floor. The three-sphere $S^3$ is the set of points $(x, y, z, w)$ with $x^2 + y^2 + z^2 + w^2 = 1$; the pole is $(0, 0, 0, 1)$, the “equator” is our space $w = 0$, and $$\mathbf X = \frac{(x, y, z)}{1 - w}.$$ This is the perspective formula again with $k = 1$: the eye sits right on the sphere. Both theorems carry over word for word, with one extra coordinate: two-spheres on $S^3$ go to spheres or planes, circles go to circles or lines, and angles are preserved.
The vertices of the regular polytopes on the showcase lie on the unit $S^3$, but their edges are straight segments, chords, and do not lie on the sphere. So the stereographic lens first “inflates” the polytope: it replaces every edge by the arc of a great circle between the same vertices, and every cell by a piece of the sphere. Only then does it project. Arcs of great circles go to arcs of circles, which is where the round edges and pillow-shaped cells of stereographic pictures come from.
The scale factor $1/(1-w)$ also explains the main impression these pictures make. A cell near the “south pole” $w = -1$ is shrunk by half, while a cell centred at $w = 0.9$ is magnified ten times: a twenty-fold difference, although in four dimensions all the cells are identical. The cell that contains the pole itself is turned inside out and becomes the “outside” of all the others: its boundary appears as a huge sphere, and its interior runs off to infinity. All of $\R^3$ ends up filled with cells, with no gaps and no overlaps. See what this looks like for the 120-cell: the showcase, stereographic lens.
Three lenses side by side
Below, one solid is shown three ways at once, and all three panels turn together. The shadow ($k = 0$) keeps parallel lines parallel, but everything overlaps. Perspective ($k = 1/d$) pulls near and far apart. Stereographic projection ($k = 1$, edges as arcs) allows no overlaps and keeps angles, but distorts sizes heavily.
Press “Schlegel diagram”: the tesseract turns one cell towards the eye. In perspective you see the frame with a cube inside, in the stereographic view the “pillows”, and in the shadow the two end cells merge into one cube while the sixteen vertices fold into eight points. Drag on any panel to push the solid into the fourth dimension; the colour of each point shows its $w$. Try the 5-cell: its Schlegel diagram is a tetrahedron split into four tetrahedra.
What a shadow loses, and how to get it back
A projection from four dimensions to three throws away one number out of four. What does that cost?
- Overlaps. All the points on one ray land on one point of the shadow. Edges that never meet in four dimensions cross in the picture; cells pass through each other; you cannot tell what is nearer. In the parallel shadow of a tesseract turned cell first, two cells coincide and six are squashed into squares.
- Lengths and angles. All 24 faces of the tesseract are identical squares, but in perspective many of them look like trapezoids. Stereographic projection keeps angles but not lengths; parallel projection keeps parallel lines parallel but not angles.
- Convexity and straightness. In stereographic projection straight edges become arcs and flat faces become pieces of spheres.
The lost coordinate can be brought back by other means. The first is colour: on the showcase and in every widget of this guide, points with negative $w$ (far from the four-dimensional eye) are blue, points with positive $w$ (near) are amber, with violet and pink in between. The second is motion: when a solid rotates in four dimensions, the way its parts slide through each other lets the brain guess at depth, just as turning a wire frame reveals its three-dimensional shape. The third is several projections at once, as in the widget above. The fourth is sections instead of shadows: they lose nothing, but show only a piece of the solid; they have a chapter of their own.
All three projections are one formula, $\mathbf X = (x, y, z)/(1 - kw)$, with different $k$: $0$ for the parallel shadow, $1/d$ for perspective with the eye at distance $d$, $1$ for stereographic projection of solids on the unit sphere. Choosing $k$ means choosing what to sacrifice.
Summary
- A shadow is a projection: parallel light forgets one coordinate, a point light divides by the distance to the source: $\mathbf X = \frac{d}{d-w}(x, y, z)$.
- The “cube inside a cube” is the perspective shadow of a tesseract turned cell first towards the eye, exactly like the “square inside a square” of a cube under a lamp.
- The division in perspective can be put off: in homogeneous coordinates perspective is linear, and graphics cards perform this “perspective divide” for every vertex.
- A Schlegel diagram is a perspective view with the eye right next to one cell: that cell becomes the frame, and the others fit inside it without overlaps.
- Stereographic projection takes circles to circles or lines and preserves angles; its local scale factor is $1/(1-w)$. For $S^3$ it is the same formula with $k = 1$.
The pictures on the showcase move because the solids rotate. How rotation works when there are no longer any axes is the subject of the next chapter.
Sources
- H. S. M. Coxeter. Regular Polytopes. 3rd ed. New York: Dover, 1973: projections and Schlegel diagrams.
- Schlegel diagram and Victor Schlegel on Wikipedia.
- G. W. Hart. 4D Polytope Projection Models by 3D Printing: on Schlegel’s models.
- Stereographic projection on Wikipedia (history and properties).
- H. A. Kastrup. On the Advancements of Conformal Transformations and their Associated Symmetries in Geometry and Theoretical Physics, Annalen der Physik 17 (2008), 631–690: the history of stereographic projection.
- E. Halley. An Easie Demonstration of the Analogy of the Logarithmick Tangents to the Meridian Line…, Phil. Trans. 19 (1695–1697), 202–214.
- A. F. Möbius. Der barycentrische Calcul. Leipzig, 1827; J. Plücker. Über ein neues Coordinatensystem, J. reine angew. Math. 5 (1830), 1–36.
- August Möbius, MacTutor History of Mathematics.