## 3-Sphere notes

**Rotations**

A rotation in 2 dimensions leaves one single point stationary. A rotation in 3 dimensions leaves an axis stationary. A rotation in 4 dimensions leaves a plane stationary. So in 4d everything rotates around a plane. Now we can additionally choose to rotate that stationary plane, thereby doing 2 planar rotations at once. I believe this is what is referred to as an isoclinic or Clifford rotation, whereby you have a pair of rotating orthogonal 2 planes that intersect only at one point. [3]

**Hopf Circles, Hopf Fibrations, Hopf Links**

The intersection of a complex line in C² containing the origin, with S³ is a circle. This is true for every line passing through the origin whose equation is of the form **z2=a*z1** where **a** is a complex number and **z2** and **z1** are the coordinates of C². By ‘line’ we mean one-dimensional subspace which we can think of as a plane. Thus there is a circle in S³ for each complex number **a** (including the case where **a** is infinite). The sphere S³ is therefore filled with circles, one for each point of S², that is, for each complex number **a**. (The **a**s fill all of C¹ which we can identify with the Riemann Sphere). As Thurston says “exactly one Hopf circle passes through each point of S³” and “the three-sphere is a two-sphere’s worth of circles”.[4]

(Note: A complex line in C² is a one dimensional subspace thereof. We can visualize a complex line in C² by imagining a plane in R⁴, but not all planes in R⁴ are one dimensional subspaces of C². [1])

The Hopf fibration is this map from the unit 3-sphere to the unit 2-sphere. (This is different than stereographic project to R³!).

- The inverse image of each point on the 2-sphere is a circle. Thus, these images decompose the 3-sphere into a continuous family of circles, and each two distinct circles form a Hopf link. The linking is what makes the fibration a non-trivial fibration.[9]
- The inverse image of a circle of latitude on
*S*^{2}under the fiber map is a torus, and these project to nested toruses in R³ which fill space. The individual fibers link to Villarceau circles on these tori with the exception of the circle through the projection point and the one through its opposite point: the former maps to a straight line, the latter to a unit circle perpendicular to, and centered on, this line, which may be viewed as a degenerate torus whose radius has shrunken to zero. Every other fiber image encircles the line as well, and so, by symmetry, each circle is linked through*every*circle, both in R^{3}and in*S*^{3}. Two such linking circles form a Hopf link in R^{3}[10]

The n-dimensional complex project space CPⁿ is the set of all complex lines on Cⁿ⁺¹ *passing through the origin*. [2] Thus C² projects to CP¹, the Riemann sphere. Homogenous coordinates represent the same point in the projective space. As we saw above **a=z2/z1, **and each value of **a** specifies a different set of homogeneous coordinates, therefore a line through the origin of C² maps to a point in CP¹. CP¹ is the set of all complex one-dimensional subspaces of C². A fancier way to say it, as Wikipedia does is: “**CP**^{1} is the quotient of **C**^{2}\{0} by the equivalence relation which identifies (*z*_{0}, *z*_{1}) with (*λ* *z*_{0}, *λ* *z*_{1}) for any nonzero complex number *λ*. On any complex line in **C**^{2} there is a circle of unit norm, and so the restriction of the quotient map to the points of unit norm is a fibration of *S*^{3} over **CP**^{1}.”

**Stereographic Projection from S³ to R³**

So the Hopf fibration can be thought of as a map from S³ to **CP**^{1}, or to S². But that is not really a visualization tool at all, it is just a map. To visualize entities embedded in S³ we can stereographically project to R³. I’m not going to summarize how to do that, because it is in a million places on the web, e.g. [7].

Under stereographic projection of S³ to R³,

- Radial planes in R³ correspond to 2-spheres through N (the north pole, i.e. infinity in R³) in S³.
- The equator of S³ projects to a sphere , S², of radius 1 centered at the origin of R³.

I initially confused Hopf Circles with Great circles and C² with R⁴. Not every two dimensional plane in R⁴ is a one dimensional subspace of C² . Any plane in R⁴ that goes through the origin intersects S³ in a great circle. But Hopf circles are the intersections of complex lines with S³. Note that unlike for S², two great circles need not intersect in S³.

**Entities embedded in S³**

**Clifford Torus.**

This is a flat torus that lives in R⁴ and is embedded in S³. “Flat” means it has zero Gaussian curvature everywhere. By “embedded in S³”, we mean that all points on the surface are unit distance from the origin. So both of these statements are mind-bending but also strangely intuitive. If you use the standard formula for a Clifford Torus and perform stereographic projection:

// Clifford Torus Parameterization var f=1/Math.sqrt(2); var xx=f*Math.cos(u); var yy=f*Math.sin(u); var zz=f*Math.cos(v); var rr=f*Math.sin(v); // Stereographic projection: var x=xx/(1-rr); var y=yy/(1-rr); var z=zz/(1-rr);

you get… a donut:

However, if you rotate the projection point:

var xx = Math.cos(u+theta) *Math.cos(v+phi); var yy = Math.cos(u+theta) *Math.sin(v+phi); var zz = Math.sin(u+theta) *Math.cos(v+phi); var rr = Math.sin(u+theta) *Math.sin(v+phi);

you get something that looks like a plane with handles:

I got these formulas from “Sculptures in S3” by Saul Schleimer and Henry Segerman, which is a very interesting paper with many practical (as in fun) applications.

This plane with handles has more structure than it seem, so I left some gaps in the surface which brings out the structure more:

A Clifford torus divides the 3-sphere into 2 solid tori. This was hard for me to grasp initially. But if you consider the * surface* of the donut above (the Clifford torus is a surface), it in-fact divides R³ into 2 solid tori. It is clear that the inside of the donut is a solid torus, but the outside is also a solid torus in the pre-image of the projection, i.e. in S³. Perhaps a little sloppy, but that’s the intuition.

There are other fun parameterizations in the Segerman/Schleimer paper. Here are a few I coded up:

Inverted (3,11) Torus Knot from Robert Woodley on Vimeo.

Trefoil on S3 from Robert Woodley on Vimeo.

Trefoil on S3, II. from Robert Woodley on Vimeo.

[1] https://math.stackexchange.com/questions/1265551/how-to-verify-whether-r2-is-a-subspace-of-the-complex-vector-space-c2

[2] http://www.math.poly.edu/courses/projective_geometry/chapter_three/node1.html

[3] http://eusebeia.dyndns.org/4d/vis/10-rot-1

[4] Three-dimensional Geometry and Topology, Volume 1. page 103.

[5] http://emerald.tufts.edu/~gwalsh01/dissertationfinal.pdf

[6]Topology, Geometry, and Gauge Fields, Foundations – Gregory Naber, page 17.

[7] Sculptures in S3. Saul Schleimer and Henry Segerman. https://arxiv.org/pdf/1204.4952.pdf

[8] https://en.wikipedia.org/wiki/Villarceau_circles

[9] https://en.wikipedia.org/wiki/Hopf_link

[10] https://en.wikipedia.org/wiki/Hopf_fibration

## Intuition for Semi-Direct Products

### Example 1: C3⋊C2 = S3

You can’t do the complicated stuff with being clear on the simple stuff. So to clarify semi-direct products lets start with pretty much the simplest case.

C3xC2 = C6 is a direct product. Thus, C3 and C2 are normal subgroups of the cyclic group C6 which has order 6.

C3⋊C2 = S3 is a semi-direct product. Only C3 is a subgroup of the symmetric group S3 which also has order 6.

C6 and S3 are the only 2 groups of order 6.

C3 has 2 automorphisms, the identity automorphism: {e ↦ e, y ↦ y, y²^{ } ↦ y²^{ }} and a bijective one: {e ↦ e, y ↦y², y²^{ } ↦ y^{ }}

The identity automorphism is used in the first, direct product, case to build C6.

The bijective automorphism of C3 is used to build S3. The second Cayley diagram below for S3 shows how C2 is used to map the identity automorphism to the bijective one via a semi-direct product. The inner circle goes in the reverse direction of the outer circle.

Here is the Cayley diagram for C6 from the great free program Group Explorer:

And for S3:

Generators for S3 are r^{³}= f²= 1. And fr=r²f.

S3 is isomorphic to the dihedral group of same order, sometimes called D3, sometimes called D6.

So that’s the simple case. Let’s move up to the a larger example: A4.

### Example 2: (C2×C2)⋊C3 = A4

(C2×C2) is the Klein 4 group, aka V4. V4 is normal in A4, so the semi-direct product of v4⋊C3 will involve an automorphism of V4 by C3.

The automorphism group of V4 is isomorphic to S3:

This arrangement shows the isomorphism to S3 clearly:

So we’re looking for an intuition of the semi-direct product, V4⋊C3 = A4. S3 has C3 as a subgroup (as we saw in the first part of this post), so there are two homomorphisms from C3 to S3, those would be the outer ring and the inner ring in Figure 4, or the left and right columns respectively in Figure 3. Choosing the injective one (the inner ring) gives us A4 which might be portrayed something like this:

The point of Figure 4 is to provide an intuition showing that A4 is the semi-direct product of C3 and the injective automorphism of V4. I don’t find the wiring diagrams of the V4 automorphisms in Figure 3 very intuitive, but Figures 4 demonstrates that there are 2 possibilities when choosing the automorphisms, one choice (the outer ring in Figure 4) leads to a direct product, the other choice leads to the semi-direct product A4 as shown in Figure 5.

References:

[1] Algebra: Chapter 0, Part 0. By Paolo Aluffi. Page 232.

[2] http://www.weddslist.com/groups/building/sdp.html

[3] Visual Group Theory by Nathan Carter

[4] Slides by Matthew Macauley, Clemson University. Notes for Math 4120. http://www.math.clemson.edu/~macaule/classes/m17_math4120/index.html

## Intuitions for SU(2) and SO(3)

This article is about a single strange topic which doesn’t have a pithy title but which manifests in several areas:

- Gimbal Lock
- Orientation Entanglement
- Quantum Mechanics
- The theory of rotation groups

The underlying concepts that relate these topics comes from Group Theory/Abstract Algebra. I understood the algebra behind these concepts but had little or no intuition for them. When that happens I feel like the algebra is worthless, or a kind of trick. Intuition is key. So after grappling for a week with these concepts I have some intuitions which is what this article is about. I’m not sure a mathematician would agree with everything here, and I’d love to have such feedback, but by writing this I’m able to organize my thoughts in some sort of coherent manner.

Perhaps it is simplest to start with a topic that confronts anyone who does even a little graphics programming: Gimbal Lock.

When you first start working with a framework like three.js and you need to rotate an object, it is natural to turn to Euler angles, invented by the great man himself. Euler angles are a set of 3 angles that describe the orientation of a rigid body in a fixed coordinate system. The three angles are sometimes called yaw, pitch and roll. Or heading, elevation and bank. The first angle might rotate the object around the X-axis, the second around its (new) Y-axis and the third around its (new) Z-axis. There are several different ways to define these angles, but however you define them, there are always 3 angles. They are widely used in graphics programming to rotate objects.

Gimbals are mechanical devices. A single gimbal allows rotation around a single axis. A set of 3 orthogonally nested gimbals extends this to 3 separate axes. In an inertial system, the 3 gimbal system senses rotation about all axes of 3 dimensional space. They have been used in ships, airplanes and rockets. However, they suffer from a flaw called Gimbal Lock whereby in certain orientations, when 2 gimbals rotate around the same axis, the system loses one degree of freedom. The term Gimbal Lock comes from the world of mechanical engineering. Mathematicians refer to it as a singularity. What this means is that sometimes when you reach a certain point in your rotation, you can’t go any further by incrementing the Euler angles and you have to do a 180 degree flip. Here is an example of that 180 degree flip in a simple three.js animation. The viewer is doing a polar orbit and must execute a flip at each pole:Note: This is not a purely abstract problem of graphics programming. It had real world consequences: The threat of Gimbal lock famously occurred during both Apollo 11 and Apollo 13. The Apollo 11 spacecraft actually had to execute such a 180 degree flip to avoid it. It was an especially dangerous threat to Apollo 13 right after the initial explosion when the spacecraft was pitching about wildly.[3]

Back to programming. The usual advice if you want to avoid 180 degree flips in your computer game is to stop using Euler Angles and to start using Quaternions instead. Libraries like three.js offer both methods.

So why do Euler angles suffer from Gimbal lock and Quaternions do not? To answer that, let’s first try to get a deeper understanding of rotations in 3 dimensions.

There is a mathematical group called SU(2) defined as follows:

This group will be very relevant to this discussion because it turns out that SU(2) is isomorphic to the 3-sphere, S^{³}, as is demonstrated here in wikipedia. This is totally intuitive since SU(2) describes a sphere in C² space, i.e. in 2-dimensional complex space. What is 2-dimensional complex space? Well, one dimensional complex space (C^{¹}) is usually drawn as a plane with the imaginary axis being the vertical axis and the real axis being the horizontal axis. Mathematicians might say that C^{¹} is diffeomorphic to R², i.e. we can think of it as being 2 dimensional. So we can think of C² as being 4 dimensional, it is diffeomorphic to R⁴. The fact that SU(2) is a unitary matrix with determinant = 1 (|α|²+|β|² = 1, as stated above) means it maps to a sphere in 4 dimensional space, aka S^{³}. So that’s the intuition.

This is relevant because SU(2) is also isomorphic to the subgroup of quaternions of norm 1. They describe the same rotational space.

So SU(2) is isomorphic to S^{³}^{.} It is also isomporphic to the group of unit quaternions. So that’s why quaternions can describe all 3-D rotations without singularities and why SO(3) cannot.

So that’s why quaternions work, but why don’t Euler angles work? Why do they have a singularity?

Euler’s rotation theorem states that any rotation in 3 dimensions can be described by a rotation of some angle around some axis. We can specify the axis of such a rotation using 2 angles, and the radius of the axis vector specifies the angle of rotation around the axis. (You may remember a similar representation of the angular velocity vector from freshman physics.) The union of these vectors form a ball with radius of π. Negative rotations correspond to vectors of negative length. A vector of radius π represents the same Euler rotation as a vector of radius -π. So antipodal points on the surface of this solid 3d sphere are identified.

Euler angles are a map onto the group SO(3), the group of all rotations about the origin of three-dimensional Euclidean space R^{3} under the operation of composition. There are many ways to map onto SO(3), Euler angles are just one way. But all maps onto SO(3) suffer from a similar parameterization problem as Euler angles do. [4]

What is this SO(3) group then? It is a normal subgroup of O(3), the group of 3×3 orthogonal matrices, but with determinant equal to 1:

O(3) = {A ∈ GL(3) | A ^{-1}= A^{T} }

So, SO(3) = { A ∈ O(3) | det(A) = 1 }

Now what is the relation between SU(2) and SO(3)? SU(2) is a ‘double cover’ of SO(3). SU(2)/{±I} and SO(3) are isomorphic. That means that every point in SO(3) maps to 2 points in SU(2). The effect of the double cover is to identify antipodal points in SO(3).

Another way to look at it: A sphere (S²) can be seen as the union of 2 discs (D²) joined at the equator of the sphere. So a 3-sphere (S^{³}) can be seen as the union of 2 balls (D^{³}). The ball representing SO(3) described in the previous paragraph on Euler rotations is only one such ball. We need another ball.

I find SO(3) to be really strange. On the one hand it is the most intuitive space for us to work with. We intuitively understand 3-D rotations and Euler angles. A child could understand them. A (normal) child could not understand quaternions. Yet simple as it seems SO(3) is full of strange phenomena/singularities like Gimbal lock and exhibits weird rotation behavior like the plate trick or the Dirac belt trick:

Let us linger on the Dirac belt trick for a second. This is an example of orientation entanglement. If the object you are rotating in 3-d space is embedded in a fixed framework like a room, then after one rotation of 360 degrees, the orientation of your object is entangled with respect to the room. You have to turn the object an additional 360 degrees to untangle it.

This is because SO(3) is not simply connected. It is not simply connected because antipodal points are identified. SU(2) is simply connected. But since SO(3) = SU(2)/{±I}, antipodal points in SO(3) are identified and it is not simply connected. I think of it this way: SU(2) represents 2 Spheres: S2+ and S2-. Taking the quotient identifies these 2 spheres via the antipodes.

How can we get an intuition for this statement: The identification of the antipodes in SO(3) means SO(3) is not simply connected? This is discussed in many places, I quote one example here:

“in the ball with antipodal surface points identified, consider the path running from the “north pole” straight through the interior down to the south pole. This is a closed loop, since the north pole and the south pole are identified. This loop cannot be shrunk to a point, since no matter how you deform the loop, the start and end point have to remain antipodal, or else the loop will “break open”. In terms of rotations, this loop represents a continuous sequence of rotations about the *z*-axis starting and ending at the identity rotation (i.e. a series of rotation through an angle φ where φ runs from 0 to 2π)” https://en.wikipedia.org/wiki/Rotation_group_SO(3)

That article goes onto explain why with an additional rotation you get a closed loop.

Because SO(3) is not simply connected, problems will always occur when trying to parameterize this space of 3D rotations. Gimbal lock is an example of a problem caused by trying to parameterize SO(3).

The fact that SU(2) is a double cover of SO(3) has some deep applications to physics. “Quantum mechanical spin is not described by a vector as in classical angular momentum. It is described by a complex-valued vector with two components called a spinor. There are subtle differences between the behavior of spinors and vectors under coordinate rotations, stemming from the behavior of a vector space over a complex field.”[6] In other words the spinor Spin(3) is isomorphic to SU(2). Physicists use Pauli Matrices as generators of SU(2). If you “rotate” an electron through an angle of 2π it is not the same as what you started with. Fermions have spin of 1/2. This is related to the fact that electrons are described by representations of SU(2) and not SO(3). It takes a rotation of 720 degrees for a fermion to return to its original state.

I hope the intuitions described above are not too sloppy, and that I haven’t mixed geometry and topology in a way that is unacceptable.

Addendum 1:

Another way to state Euler’s Rotation Theorem: If A is an element of SO(3) where A ≠ I, then A has a one dimensional eigenspace. This is not to say that A does not also have a two-dimensional eigenspace but that A must have a one-dimensional eigenspace. It is this eigenspace that is known as the axis of rotation. https://en.wikipedia.org/wiki/Orientation_(geometry)

Addendum 2:

“A 2×2 matrix with complex entries represents an element of U(2) iff U*U=1. So the determinant detU is a complex number of modulus 1. The group U(1) consists of all complex numbers of modulus 1, which allows us to rephrase the previous finding as follows: Taking the determinant is a group homomorphism, det: U(2) → U(1). Indeed detUV = detUdetV and det1 = 1. The kernel of the group homomorphism det is a subgroup of U(2)”. https://en.wikipedia.org/wiki/Unitary_group

That subgroup is SU(2). So: U(2) = U(1) × SU(2). Or alternatively, U(2) = S^{3}×S^{¹}

U(1) is a circle. So we’re factoring out a whole circle to get SU(2).

It appears that U(2) can’t be identified with a manifold, see here: https://math.stackexchange.com/a/843892

Identifying SU(2) with S^{3} makes it much easier to think about. The fact that it is simply connected is obvious for instance.

References:

[1] By Euler2.gif: Juansempere derivative work: Xavax (This file was derived from Euler2.gif:) [CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0) or GFDL (http://www.gnu.org/copyleft/fdl.html)], via Wikimedia Commons. From: https://en.wikipedia.org/wiki/Euler_angles

[2] By Lookang many thanks to Fu-Kwun Hwang and author of Easy Java Simulation = Francisco Esquembre (Own work) [CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons. From: https://en.wikipedia.org/wiki/Gimbal

[3] “https://www.universetoday.com/119984/13-more-things-that-saved-apollo-13-part-9-avoiding-gimbal-lock/”

[4] https://en.wikipedia.org/wiki/Rotation_group_SO(3)

[5] By JasonHise (Own work) [CC0], via Wikimedia Commons. https://commons.wikimedia.org/wiki/File%3ABelt_Trick.gif

[6] https://en.wikipedia.org/wiki/Spin-%C2%BD#Complex_phase

Other references:

http://www.mat.univie.ac.at/~westra/so3su2.pdf

http://alistairsavage.ca/mat4144/notes/MAT4144-5158-LieGroups.pdf

## Tetrahedral symmetry

I was curious how one would go about rendering a triangle group using Möbius transforms. It took a while! I wanted to jot down the underlying theory so I wouldn’t forget it later. Perhaps these notes will be of interest to others as well. Here goes:

Given a triangle with angles *π/l, π/m, π/n*, you can tile a plane if *1/l + 1/m + 1/n = 1*. For instance if *l,m,n* = 2, 4, and 4 respectively, then the sum of the angles of the triangle is *π/2 + π/4 + π/4 = π*, and you can tile a plane with isosceles right triangles which would look something like this [1]:

That’s what a tiling of a Euclidean plane looks like. What does a tiling of a sphere look like?

The sum of the angles of a spherical triangle is always greater than π. *T*here are only a few values for *l,m,n* that form a pattern covering the sphere just once without any overlaps or gaps. The possible values for (*l,m,n*) are[2]:

(p p 1), (p 2 2), (3 3 2), (4 3 2), (5 3 2).

The sides of these triangles form great circles or geodesics.

Let’s look in detail at the (3 3 2) tiling. This fundamental region is a triangle with angles: π/3, π/3, π/2. This is another isosceles right triangle. The total sum of the angles is 7π/6. The area of a single triangle (for sphere radius = 1) is π/6. Since the total area of the unit sphere is 4π, it will take 24 of our triangles to the tile the sphere. If we did so, we’d get a result like this:

There are 6 points where 4 vertices (each of angle π/2) meet, and 8 points where 6 vertices (each of angle π/3) meet. The symmetries consist of rotations and reflections.

Rotations: there are 3 axes with order 2 rotation and 4 axes with order 3 rotation (antipodal points share an axis). And we count the identity (no rotation) as one of the rotations. So altogether:

- 1 identity rotation (no rotation)
- 4×2 rotations, (not 4×3 since we don’t want to double count the identity),
- 3×1 rotations, (not 3×2 since we don’t want to double count the identity),

1 + 4×2 + 3×1 = 12 rotational symmetries altogether.

Reflections: The illustration immediately below shows 3 possible reflection planes for our tiling. But this is not all of the possible planes of reflection of course. Each plane in the illustration corresponds to one of the great circles (aka geodesics) traced out by the sides of the triangles: each such great circle is a plane of rotation. There are 6 of these great circles in this tiling. To these 6 reflections are added 6 so-called roto-reflections, in which, using the same 6 planes, a rotation by π is followed by a reflection.

The group of 12 rotations and 12 reflections corresponds to the group S4, which is of order 4! = 24.

So what about tetrahedrons? At some point this article has to mention tetrahedrons, given the title.

Wikipedia says: “*A regular tetrahedron has 12 rotational (or orientation-preserving) symmetries, and a symmetry order of 24 including transformations that combine a reflection and a rotation.*

*The group of all symmetries is isomorphic to the group S _{4}, the symmetric group of permutations of four objects, since there is exactly one such symmetry for each permutation of the vertices of the tetrahedron.*”

And indeed the following diagram also from wikipedia [4] shows that the rotations and reflections of a tetrahedron correspond exactly to those of the (3,3,2) triangle group that we have been working with.

This tetrahedral group “arises if we inscribe a regular tetrahedron in a sphere and mark its vertices together with the projections of the centers of the faces and the midpoints of the edges on the the sphere, the center of projection being the center of the sphere.”[3] That article goes on to say that the defining relations of the group are:

*L²=M²=N²=1, (LM)³=(MN)³=(NL)²=1*

where *L, M, N* are the reflections in the side of the (3,3,2) triangle.

So to summarize, there are 24 symmetries of the (3,3,2) triangle group and these consist of 12 rotations and 12 reflections.

Now, rotations are orientation-preserving transformations, whereas reflections are orientation-reversing motions. Triangle groups are not examples of Keinian groups. Kleinian groups are groups of möbius transformations, which are orientation-preserving.[5]

Triangle groups were studied first in the 19th century and the study of Kleinian groups and eventually hyperbolic geometry grew out of those foundations. So lets pursue this analysis into the world of Kleinian groups.

“*Every group of motions, which is generated by reflections has a subgroup of index two. This subgroup consists of directly conformal motions, that is of, möbius transformations.*“[5]

If we throw out the reflections, we are left with a set of 12 rotations. This subgroup is isomorphic with A4, the alternating group of permutations of four symbols. Wikipedia says: “*The set of orientation-preserving symmetries forms a group referred to as the alternating subgroup A _{4} of S_{4″}.*

Generators for A4 satisfy these relations:

u³ = v³ = (uv)² = I

We can generate A4 with two three-cycles [6]:

U = (123), and V=(124).

It is interesting to work out all the permutations of these generators and see that there are indeed only 12 of them:

Generators | Result | Already Seen? |

I | 1234 | |

U | 2314 | |

V | 2431 | |

UU | 3124 | |

UV | 4321 | |

VU | 3412 | |

VV | 4132 | |

UUU | 1234 | I |

UUV | 3241 | |

UVU | 4132 | VV |

UVV | 1342 | |

VUU | 1423 | |

VUV | 3124 | UU |

VVU | 4213 | |

VVV | 1234 | I |

UVVU | 2143 | |

UVVV | UI=U | |

VUUU | VI=V | |

VUUV | 2143 | UVVU |

VUVU | 1234 | (VU)^2=I |

VUVV | 3241 | UUV |

VVUU | 4321 | UV |

VVUV | 1423 | VUU |

VVVU | IU=U | |

VVVV | IV=V |

So the elements of the A4 group in terms of our generators is: I, U, V, UU, UV, VU, VV, UUV, VVU, UVV, VUU, UVVU.

Nomenclature seems inconsistent across writers, but lets call the tetrahedral group that is isomorphic with S4, T*(3,3,2). And the tetrahedral group that is isomorphic with A4, we’ll call T(3,3,2). As above, the defining relations of T(3,3,2) are

u³ = v³ = (uv)² = 1,

where u = LM and v = MN and *L, M, N* are the reflections in the side of the (2,3,3) triangle [3].

From [3], we get some sample values for U and V:

Finding an appropriate fundamental domain was not easy, but a triangle with vertices at these corners works:

[0, -x + xi, x+xi]

where x= -0.366025393.

In spherical coordinates (phi, theta) these corners correspond to:

[(0,0), (-PI/4,len), (+PI/4,len)]

where len = acos((cos(PI/3.)+cos(PI/3.)*cos(PI/2.))/(sin(PI/3.)*sin(PI/2.))) = 0.955316618 rad.

Given this fundamental domain and the list of generators above, we can finally tile the Riemann Sphere with (2,3,3) tirangles using Möbius transformations! The fact that we are using only the orientation preserving symmetries means we get a triangle tiling that covers half of the sphere and leaves empty space elsewhere:

So to summarize, the triangle tiling was formed by repeatedly applying the 2 Möbius transformations to the given fundamental domain in the 12 combinations indicated by A4 symmetry. Implementation note: we are rotating this sphere not by rotating the camera or the mesh, but again by applying Möbius transformations to the texture as discussed in this article, which is why the equirectangular projection to the right in the picture above is morphing in that weird way.

**Addendum 1: The A4 Group**

The A4 group is of order 12. It has the following non-trivial subgroups:

- Three groups of order 2, isomorphic to Z2.
- Four groups of order 3, isomorphic to Z3.
- One group of order 4, isomorphic to the Klein 4 group: V4. Also isomorphic to the dihedral group D2.

Lets linger on the V4 group for a second. If you stretch a tetrahedron like this:

It no longer displays A4 symmetry. You’ll never find an axis that you can rotate by an angle of 2π/3 and have it look the same. But it still displays V4 symmetry, the symmetry of a baseball for instance [9]:

**Conjugacy Classes of A4**

There are 4 conjugacy classes of A4:

1. One class containing the identity.

2. One class containing the three 180 degree rotations through opposite edges.

3. One class for the four forward rotations of *π*/3 around each corner.

4. One class for the four reverse rotations of *π*/3 around each corner.

Note that classes 3 and 4 are merged in S4 but disjoint in A4 because A4 does not have a reflection.[8]

If the identity is { 1, 2, 3, 4},

Class 2 consists of these permutations:

{ (2 1 4 3), (3 4 1 2), (4 3 2 1) }

Class 3 consists of these permutations:

{ (2 3 1 4), (1 4 2 3), (3 2 4 1), (4 1 3 2) }

Class 4 consists of these permutations:

{ (3 1 2 4), (1 3 4 2), (4 2 1 3), (2 4 3 1) }

To visualize this, label the vertices of the tetrahedron 1,2,3 and 4 in any order. Then the numbers listed in the conjugacy class are the permutations of the vertices for that class.

You could also label the permutations using our generators:

The identity is * I*,

Class 2 consists of these permutations:

{ * UVVU, VU, UV* }

Class 3 consists of these permutations:

{ * U, VUU, UUV, VV*}

Class 4 consists of these permutations:

{ * UU, UVV, VVU, V* }

Note, the transformation of * U* is a rotation around vertex 4 and the

*transformation is a rotation around vertex 3.*

**V****Normal Subgroups:**

A normal subgroup must be a combination of conjugacy classes. These are our conjugacy classes (from above):

e – the identity element

x, y, z – 180 degree flips

a, b, c, d – 120-degree clockwise rotations

a², b², c², d² – 120-degree counterclockwise rotations.

The number of elements in each class are 1, 3, 4, and 4 respectively. By LaGrange’s theorem the order of a subgroup must divide the order of the group. The only non-trivial combination of classes that meets that constraint is {e, x, y, z} which is V4. But that on its own is not enough to determine normalcy. However this observation from Armstrong [10] helps: “If H is a subgroup of G, then each conjugate gHg^{-¹ }is also a subgroup of G and has and has the same order as H. Therefore, if G has no other subgroup of the same order as H, then H must be a *normal* subgroup of G.” So V4 is a normal subgroup of A4.

Since it is normal, we can take a quotient, i.e.:

C3 = A4/V4

A4 = V4⋊C3

This is a semi-direct product because C3 is not normal in A4. Details:

V4 is normal in A4. C3 is not normal in A4. Since |V4| and |C3| are relatively prime, we have V4 ∩ C3 = {e}. Lets refer to A4, V4 and C3 as G, N and H respectively. Every element in G is uniquely expressible in the form *nh. *The uniqueness follows from N ∩ H = {e}. Since N is normal in G, for each h ∈ H we have an automorphism of N given by *n ↦ hnh*^{-¹ }. In other words, the automorphism is the inner automorphism of *h* restricted to *N*.

I’m not sure why it is so important to mention the automorphism. The semi-direct product is a product of a normal group and a non-normal group – that seems to be the main point. * Update: *I got a deeper understanding of the role automorphism groups in semi-direct products and wrote about it here.

For the Klein 4 group V4 we have the relations *a² = b²*=1 and *ab=ba*. For C3 we have *c*³=1. Due to the inner automorphism, we can also state that *cac*^{-¹} = *ab* and *cbc*^{-¹} = *a*, for the appropriate choice of *a* and *b*.[11]

I am assured in many places that that last presentation represents a group that is isomorphic to A4 though I don’t really see it. But we already know we’re working with A4.

**Addendum 2:**

In the first chapter of the book *Finite Möbius Groups, Minimal Immersions of Spheres, and Moduli,* Gabor Toth works out a set of Möbius transforms corresponding to all 3 polyhedral symmetry groups. These are different from the ones I used above. Here I apply his tetrahedral rotations to a so-called 360 photo (taken on a Ricoh Theta):

And just for kicks, here is the same process applied to another photo using his octahedral symmetries:

Octahedral symmetries are double those of the tetrahedron which makes intuitive sense when you consider that you can fit 2 tetrahedrons in a cube, e.g. [7]:

[1] https://commons.wikimedia.org/wiki/File:Tile_V488_bicolor.svg#/media/File:Tile_V488_bicolor.svg

[2] Coxeter, H. S. M. (1974) *Regular Complex Polytopes*, Cambridge Unitvesity Press, London. Page 19.

[3] *Noneuclidean Tesselations and Their Groups. *Academic Press, Oct 18, 1974. Page 73 on.

[4] By I, Cronholm144, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=2295116

[5] *Geometry of Group Representations*: Proceedings of the AMS-IMS-SIAM Joint Summer Research Conference Held July 5-11, 1987 with Support from the National Science Foundation. William Mark Goldman, Andy R. Magid. American Mathematical Soc., 1988 – Mathematics. Page 4.

[6] http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/genset.pdf

[7] By Steelpillow (Own work) [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons

[8] http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/conjclass.pdf

[9] By Tage Olsin – Cropped from Image:Baseball.jpg by Tage Olsin, CC BY-SA 2.0, https://commons.wikimedia.org/w/index.php?curid=2345172

[10] M.A. Armstrong, Groups And Symmetry. Page 114.

[11] http://sierra.nmsu.edu/morandi/notes/semidirect.pdf, page 3.

## Congruence Group Γ(2)

Modular Group Γ(1) generated the hyperbolic tessellations seen previously. Its generators are:

Or in matrix form: [0,1;-1,0], [1,1;1,0].

The Congruence Group Γ(2) can be generated by these matrices:

[1,2;0,1], [1,0;-2,1]

The tiling looks like this:

You can view a version I wrote in Three.js/WebGl here. You can spin it and pan and zoom. It is clear in this version that the tessellation lives on the upper-half plane of the Reimann sphere.

As before the orange region is the fundamental domain. It and all of the tiles are 4 sided and touch the rim at 4 points. Here is another representation of the fundamental domain from Wolfram Alpha[1]:

The 4 corners of this fundamental domain are -1, 0, 1, and ∞. The red letters indicate the effect of the initial tiling of 1 letter words (e.g. T, T¯¹, S, S¯¹). These areas correspond to the yellow quadrilaterals in the tiling above.

The imaginary axis cutting the fundamental domain in two is an additional axis of symmetry we could impose on the tiling. Then it would look just like the Farey graph depicted in the previous post and the disk would be tiled by ideal triangles. Ideal triangles in hyperbolic space have interesting properties (to quote wikipedia)[3]:

- All ideal triangles are congruent to each other.
- The interior angles of an ideal triangle are all zero.
- An ideal triangle has infinite perimeter.
- An ideal triangle is the largest possible triangle in hyperbolic geometry.

**The Modular Necklace**

If you have read the book “Indra’s Pearls” you will be familiar with a similar tiling which they describe as “The modular necklace”:

“A modular necklace is a tangent chain of four circles in which adjacent disks are paired by two transformations *a* and *b*… The transformations *a, b, *and *ab* are all parabolic and S = Fix(a), Q = Fix(b), *a(P)=R, b(R) = P *so that* P = Fix(ba) and R = Fix(ab)… *The four points P, Q, R and S always lie on a circle (or line) which is the limit set of the group. The limit circle is perpendicular to all circles in the chain. Both inner and outer tiles have their sides matched in the same way and the surfaces made by gluing up these tiles are each spheres with three punctures or cusps.”[2]

Following their naming, P, Q, R and S are the tangency points of the four circles *a, A, b, *and* B *which are positioned where the letters T, T¯¹, S, S¯¹ are in the diagram above, respectively. The möbius transform *a *corresponds to T and the möbius transform *b* corresponds to S.

If we look at the tiling this way, then it makes no sense to restrict ourself to the upper half plane and the limit set is very clear:

Another point of view on the same tiling:

[1] Weisstein, Eric W. “Modular Group Lambda.” From *MathWorld*–A Wolfram Web Resource. http://mathworld.wolfram.com/ModularGroupLambda.html

[2] “Indra’s Pearls: The Vision of Felix Klein”. By David Mumford, Caroline Series, David Wright. Page 214.

[3] https://en.wikipedia.org/wiki/Ideal_triangle

## Ford Circles and Farey Graphs

Here is an image of a tessellation or tiling of the upper half-plane. Under some group of symmetries, an initial triangle (for instance the one in orange) covers the plane without overlaps. (The initial triangle is not ideal because only one of its angles is 0). The group of symmetries used are those of the modular group. See my previous post where I go into this in greater detail.

Note that each triangle has a vertex on the rim of the circle. The rim of the circle is the real projective number line.

The orange initial triangle, aka the fundamental domain, is touching the real number line at the point of infinity. The green triangle right below the orange triangle touches the real number line at the point 0. The green triangle = S(orange triangle), where S -> -1/z. Every other vertex touching the rim of the disc can be obtained by applying some combination of S and T. (T->z+1).

**Ford Circles**

Lets start with 2 circles, one with a radius=1 touching at 0. Another is a line, Im(z) = 1 (a circle with an infinite radius that is). These are the 2 large circles in the picture below. If we then apply S and T repeatedly we get the Ford circles.

The set of points where the circles touch the real number line is the set of rational numbers, Q. This is the same set of points where the vertexes of the tessellated initial triangles touch the rim.

From [4], I learned the following interesting facts:

- The circle touching the real axis at the reduced fraction a/c has radius 1/2c². This explains why the circles for reduced fraction a/c and a’/c have the same radius.
- The circles touching z = a/c and z = b/d > a/c are tangential to each other because ad − bc = 1. Such circles are images of the 2 initial circles touching z = 0 and z = ∞ under SL(2, Z) (determinant = 1).
- The circle between these tangential circles touches at (a+b)/(c+d). Because the latter circle is the image of the circle between the circles touching z = 0 and z = ∞, namely the circle touching z = 1, under

.

The formula (a+b)/(c+d) is the formula for calculating a mediant between 2 *Farey neighbors, *a/b and c/d.

“The Farey graph is the PSL(2,Z) images of the imaginary axis, and so contains all of Q* as its points at infinity. It is not hard to see that p/q and r/s are connected by an edge iff ps-qr = +/-1, so it is in some sense a geometric recording of the matrix group.” [1]

Here the images of the imaginary axis are highlighted and the Farey graph is evident:

[1] http://mduchin.math.tufts.edu/notes/hyp-groups-course.pdf

[2] “Indra’s Pearls”, pages 210-213.

[3] http://homepages.warwick.ac.uk/~masbb/HypGeomandCntdFractions-2.pdf

[4] http://www.maths.ed.ac.uk/~aar/Whittaker2012A

## Tessellation of the Hyperbolic Plane on the Riemann Sphere

I’m interested in learning more about the modular group and this article looking into hyperbolic tessellations represents my initial efforts to collect my thoughts on the topic.

We are familiar with hyperbolic tessellations from the artwork of M.C. Escher and from numerous renderings of the Poincare Disk (which was really invented by Beltrami, not Poincaré):

The disk model is not the only model of hyperbolic space. There is also the Klein model, the hyperboloid model, and the upper half plane model, and probably some others. Let’s consider the upper half plane model. One way to look at the Poincaré half plane model is as a view on that part of complex plane where the imaginary part is positive. So we’re only considering half of the complex plane, the upper half. Here is what a tiling of the upper half plane looks like:

###### By Fropuff (from en wikipedia) [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/)], via Wikimedia Commons

To quote wikipedia: “The projective linear group PGL(2,**C**) acts on the Riemann sphere by the Möbius transformations. The subgroup that maps the upper half-plane, **H**, onto itself is PSL(2,**R**)”. So Möbius transformations with real coefficients preserve the upper half-plane. That means that any such transformation will keep all points that are already on the upper half-plane on the upper half plane. (And incidentally, the real numbers are mapped to themselves).

If we look at the upper half plane model this way (as the upper half of the complex plane), then when we render it on the Riemann Sphere, it looks like this:

In this depiction, the tiled portion is the upper half plane. The red hemisphere is not the upper half plane, it is the lower half plane. The green axis points to (∞, ∞i), the blue axis intercepts the sphere at (-1,0i) and the red axis intercepts the sphere at (0,-i).

The tessellation depicted above is of the modular group, PSL(2,Z), where Z indicates the set of integers. So in other words the modular group Γ is those mobius transformations with integer coefficients and unit determinant. It is a subgroup of PSL(2,R) (but not a normal subgroup).

“The modular group can be shown to be generated by the two transformations

so that every element in the modular group can be represented (in a non-unique way) by the composition of powers of *S* and *T*. Geometrically, *S* represents inversion in the unit circle followed by reflection with respect to the imaginary axis, while *T* represents a unit translation to the right.” * z* is a complex number of course. And as mentioned above, S and T are Möbius transforms with integer coefficients and determinant = 1.

So we can use those 2 transformations and apply them over and over. What region should we start with as our initial tile (or to use the technical terminology: what * fundamental domain* should we choose?). A common choice is depicted in the grey region in the upper half plane figure above. It is that region where |z| is > 1 and |Re(z)| < 1/2. The orange triangle in the spherical tiling above is the very same region.

You can view my Three.js version of this tessellation here where you can spin and zoom it as you please.

Note that we are drawing this diagram using WebGL shaders. Pixel shaders (aka fragment shaders) are called once for each pixel. In our case a pixel maps to a complex coordinate. So our code is called repeatedly and handed a different complex coordinate on each call. The pixel shader has to decide what color to use for this given pixel/complex coordinate. So it has to do the Möbius transforms in reverse. (I discuss this issue in greater detail here). This shows how the tesselation works [3]:

Thus, in code, the logic is:

if |z| < 1 then apply S^{-1}else if Re(z) < -.5 apply T, else T^{-1}

(remember we’re doing everything backwards because we’re working with pixel shaders).

By a suitable placement of the camera you can get a view that looks just like the Poincaré disk:

This provides an intuition for believing that there is a simple mapping from the upper half plane model to the disk model, and indeed there is, it is called the Cayley transform.

**Rational Numbers, Q**

In case it wasn’t clear, the rim of the upper half plane is the real line, or the projective real line (the real line plus a point at infinity) to be precise. A cool thing about the modular group is that each point where a tessellation intersects the real line is a rational number. That makes sense because any combinations of powers of S and T can only generate a rational number.

In other words, the points where the tessellations of the fundamental domain hit the boundary are those points s ∈ R ∪ {∞} that are fixed by a parabolic element of Γ. The points are precisely Q ∪ {∞}, where Q stands for the set of rational numbers. The parabolic element is the transform T above. You can tell it is parabolic because the Trace of the 2×2 matrix corresponding to the Möbius transform for T is 2.

For z ∈ SL2(R), z is parabolic if z can be conjugated to T . Any parabolic element z generates an infinite discrete subgroup of SL2(R) consisting solely of parabolic elements; and z fixes no points of H and a unique point of P1(R). The parabolic points for SL2(Z) are P1(Q). [4]

The fundamental domain in orange above has one point on the real axis. Each transformation in the tessellation moves that point around the rim and generates a new rational number.

**Hyperbolic Tesselations**

So what does this have to do with hyperbolic tesselations? We’ve explained how a special set of Möbius transformations, those that generate the modular group, can tile the upper half plane. The upper half plane is a model of hyperbolic geometry. This was the famous insight that came to Poincaré in a flash:

*“At that moment which I put my foot on the step the idea came to me, without anything in my former thoughts seeming to have paved the way for it, that the transformations I had used to define the Fuscian functions were identical with those of non-Euclidean geometry”* L’Invention Mathématique, Henri Poincaré

Or as wikipedia says: “Möbius transformations are also isometries of the hyperbolic plane“. This seems a strange statement. Möbius transformations are conformal on the complex plane: they preserve angles, but the don’t preserve lengths. Isometry means ‘length preserving’. This is possible because in hyperbolic space, in marked contrast to Euclidean space, equal angles guarantee that corresponding side lengths are equal in the hyperbolic metric.[1] A hyperbolic triangle’s size is determined by its angles.

**Algebraic vs Geometric**

The matrices [0,1,-1,0] and [1,1,1,0] generate SL(2Z). Both matrices have determinant = 1. That means they are volume and orientation preserving.

The analogous Möbius transforms:

**And finally….**

As a final note, since we’re working with the Riemann sphere we can do all sorts of fun mappings from 360 video footage, as I detailed elsewhere. So piping video through the tessellation gives you:

Video 5 from Robert Woodley on Vimeo.

.

—————-

[1] http://www.mathematica-journal.com/issue/v9i3/contents/ModularGroup/ModularGroup.pdf

[2] https://math.dartmouth.edu/~m125x15/quat-book-chap29-072315.pdf

[3] http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/SL(2,Z).pdf

[4] https://homepages.warwick.ac.uk/~masiao/modforms/examples2.pdf

[5] http://www.math.cornell.edu/~hatcher/TN/TNch1.pdf

## Möbius Transformations on Spherical Photos and Videos

Spherical Video presents some interesting challenges, for instance, how do you zoom? Or how do you rotate on any arbitrary axis? As the mathematician Henry Segerman pointed out in a post for EleVR, you can achieve both of the above using Möbius transformations. The transformations are conformal (they preserve angles), and they map circles to circles (considering a line to be a circle of infinite radius). Professor Segerman’s work is the inspiration for the project I am describing here.

I wrote an implementation using Möbius transformations to manipulate spherical images in WebGL/Three.js. The implementation is here. Click on the ‘?’ for help or watch this video.

So what is going on here?

### The mathematical procedure:

Spherical cameras such as the Ricoh Theta save images in an equirectangular format. This is then wrapped around a sphere to give the spherical effect. (With three.js you create a material whose texture is the saved image from the Theta and then create a mesh using this material and a sphere geometry.) The X and Y of the equirectangular format map to the longitude and latitude of the sphere.

The Riemann sphere is a representation of the complex plane as a sphere using reverse stereographic projection. Möbius transformations are transformations of the complex plane. Our three.js sphere is a normal sphere in 3-d cartesian space (R3); each point on its surface has an (x,y,z) coordinate where x, y and z are real numbers. We can convert our x,y,z point to a point in the complex plane if we take our sphere to be a Riemann sphere. Or in jargon we can say equivalently: P¹(C) is diffeomorphic to the sphere S². The formulas are here (and elsewhere).

So for instance the south pole is (0,0,-1) in cartesian/R3 space and (0,0i) in complex space. The north pole is (0,0,1) in cartesian space/R3 and (∞, ∞i) in complex space. (The Reimann sphere is really the complex plane plus one point which is the point at the north pole, but I digress).

### The technical procedure:

To implement our transformations, we are going to move pixels around on the texture. We can only do this in the shader since doing it in javascript would be way to slow. Specifically we are going to do it in the fragment shader. Our vertex shader will be short and boring, doing only what is necessary to build the sphere’s vertices. Nothing special there. The fragment shader is where the work is done.

The fragment shader is called when the graphics layer needs to know what color to put at a given UV coordinate. We will pluck that pixel from our transformed texture. The steps to execute on each call to the fragment shader are:

1 – Get the UV coordinates. ‘UV’ is industry nomenclature for the X and Y coordinates of the texture that the graphics layer is trying to draw.

2 – We know that the rectangular texture must map to a sphere. The top of the rectangle maps to the north pole, the bottom to the south pole. So we can calculate the X, Y, and Z coordinate of the sphere corresponding to the UV coordinate. That is, we know which point of the sphere we are drawing.

3 – This sphere is also a Riemann sphere as mentioned above, so we calculate the point (a + bi) on the complex plane corresponding to the cartesian point (X,Y,Z) on the sphere, using the formulas mentioned above.

4 – We now know our location on the complex plane and we apply as many transformations as we’d like. When we are done we have our new complex point (c + di).

5 – We reverse the above steps and calculate the cartesian X,Y,Z corresponding to (c + di). Then we calculate the UV corresponding to this X,Y,Z and we use the color at that pixel, UV.

The only Möbius transformations I’ve implemented for now are those to do rotation and zoom. By default the fixed points are antipodal though you can explicitly set the fixed points using the Epsilon 1 and Epsilon 2 buttons. The video linked to above gives a good overview of what all the different buttons do.

This is a rotation around 2 non-antipodal fixed points:

In addition to Möbius transforms to do rotation and zoom, there are some other complex transformation options.

You may have noticed that we start with the point on the complex plane (a + bi) for which we need a pixel. We then transform (a + bi) into (c + di) to get the pixel we will draw. Thus the transformation is really a reverse transformation.

Infinities of volleyball players on North Avenue Beach (Works at 60FPS on video!):

This article by Henry Segerman gives a comprehensive overview of the math behind these transformations and includes more examples of other transformations you can do.

## Thought Bubbles

A window is a two-dimensional hole in a two-dimensional plane that allows you to see into a three-dimensional world. So what if we could make three-dimensional holes?

The original idea came from a three.js demo by altered qualia where he was demo’ing fresnel shaders:

I didn’t care much about fresnel shaders, but was intrigued by the bubbles. I realized that rendering the outer sphere was not needed to render the bubbles:

These bubbles are three-dimensional windows that can be taken anywhere. Spherical holograms.

To render the above image, you need 2 cameras, each one looking at a separate texture. Now look at this next image. If we flood Michigan avenue we need 2 cameras as well, one for the surrounding sphere, and one to capture the reflection as well.

We can view the scene of a flooded Michigan Avenue at home, via one of our holo-bubbles from above. The camera count is now 3, and you have to deal with the fact that while you want them all to turn in sync, some will need to be more wider angle than others.

We can make the inner camera have such a wide angle that it captures the zenith and the nadir, and the cursor both zooms and moves in space. It is a very sensitive. (Try it!).

This final image is one of the strangest I’ve made to date. Only one camera, with the sphere in the middle seemingly refracting the surroundings but even though the image is in the sphere is reversed it tracks the outer image as they rotate. How is this possible?

Most of the above images can and should be clicked to see the rendering in WebGL.

## Complex Surfaces

I wrote a separate version of formula toy that handles complex functions. So, you can type in a function like: f=sqrt(g). Both f and g are expected to be complex functions:

g = u +iv

and

f = w +ix

where u is the real component of g, v is the imaginary component of g, etc.

To draw this surface we need 4 axes, so formula toy uses a color gradient for the 4th axis. We name these as follows:

– fR – the real component of f.

– fI – the imaginary component of f.

– gR – the real component of g.

– gI – the imaginary component of g.

And you can choose which complex axis maps to which cartesian axis.

This flexible way of doing the mappings allows for simple multi-valued functions/Riemann surfaces:

Click on the images to open the surface in formula toy. More details here.

## Torus Knots

Formula Toy is a simple and free WebGL app I wrote that allows you to enter in 3-d formulas and see the resulting surface. Sort of like Desmos, but for 3D.

When I first wrote it you could express your formulas in 3 different coordinate systems: cartesian, spherical (polar), and cylindrical. I recently add toroidal which is not very useful except for drawing toruses:

radius=.5+sin(phi)/(15)*3

AND I added parametric surfaces. If you choose this option, the system is expecting formulas for X, Y, and Z. You are given U and V. For instance a helix could be expressed as:

u=u*(2*pi); v=v*(6*pi); x=u*cos(v); y=u*sin(v); z=v;

Parametric equations that are functions of a single variable (t) instead of 2 variables (u,v), don’t display because they have zero thickness. For instance the parametric formula for a trefoil knot is a function of 1 variable. To make this visible in formula toy, you need a tube geometry so that the knot doesn’t have zero thickness/volume. Three.js offers a tube geometry but you can also achieve the same effect using the parametric geometry, but with a little extra math. We can start with a torus, and make that our ‘tube’ that will be bent into a trefoil. Here is the parametric formula for a torus:

u=u*(2*pi); v=v*(2*pi); x=(4+cos(v))*cos(u); y=(4+cos(v))*sin(u); z=sin(v);

So to plot a trefoil:

// Setup u=u*(2*pi); v=v*(2*pi); phi=u;

// Parametric formula for a trefoil pp=2; // A trefoil is a (2,3) torus ring. qq=3; rr=cos(qq*phi)+2; // xx,yy,zz are the formula for the trefoil, a function of one variable (phi) xx=rr*cos(pp*phi); yy=rr*sin(pp*phi); zz=-3*sin(qq*phi);

// Modified Torus formula to bend a torus into a trefoil shape: x=(4*xx+cos(v)*xx/rr); y=(4*yy+cos(v)*yy/rr); z=sin(v)+zz/rr;

So that works, and is succinct, but you could also achieve the same thing in a more mathematically correct but longer way. The correct way is to take 2 normal vectors to the derivative of the trefoil formula with respect to phi. Pseudo-code would look something like this:

var dv = df(pp, qq, phi); var north = new THREE.Vector3(0,0,1); var v1 = cross(dv, north); v1.normalize(); var v2 = cross(dv, v1); v2.normalize(); x = xy + (cos(theta)*v1.x + sin(theta)*v2.x)*.2, y = yy + (cos(theta)*v1.y + sin(theta)*v2.y)*.2, z = zz + (cos(theta)*v1.z + sin(theta)*v2.z)*.2,

But that code uses three.js objects so is beyond what you can do in formula toy.

A trefoil is just on example of a torus knot. It is a (2,3) torus knot when means it winds 3 times around a circle in the interior of the torus, and 2 times around the torus’ axis of rotational symmetry. There is a whole family of torus knots (*p,q*). *p* and *q* correspond to **pp** and **qq** in the parametric formula above.

For instance, here is a (*3,7*) knot:

Below is another example which shows a (*5,6*) torus knot winding around a torus. This was not done in Formula Toy since it draws 2 surfaces (the knot and the torus), and Formula Toy only draws one surface. (but it was all done with three.js).

## Rotations, Transformations – Geometries and Meshes

I was driving myself batty trying to get all my rotations and transformations to behave correctly in three.js. So in these kind of cases one must always pare down to essentials. As follows:

Lets create a simple mesh and place it on the scene:

// Example 1 var geo = new THREE.BoxGeometry(5,5,20,32); _mesh = new THREE.Mesh(geo, new THREE.MeshNormalMaterial());

Then in the render loop, we rotate it around the Z axis (the blue axis):

_mesh.rotation.z = -_tick * Math.PI/256;

The result:

This just a screen grab, an animated gif. Hence the jerk when the animation restarts.

Now what happens if we rotate the mesh in the initial setup:

// Example 2 var geo = new THREE.BoxGeometry(5,5,20,32); _mesh = new THREE.Mesh(geo, new THREE.MeshNormalMaterial()); _mesh.rotateY(Math.PI/2);

Code in the render loop is the same as before:

_mesh.rotation.z = -_tick * Math.PI/256;

Now the block rotates along the X (red) axis, even though we told it to rotate it along the Z axis.

If we specify rotation order in the initial setup, then it will rotate around the Z axis:

// Example 3 _mesh.rotation.order = 'ZXY';

The result:

Now lets try some translation. Initial setup:

// Example 4 var geo = new THREE.BoxGeometry(5,5,20,32); _mesh = new THREE.Mesh(geo, new THREE.MeshNormalMaterial()); _mesh.rotation.order = 'ZXY'; _mesh.rotateY(Math.PI/2); _mesh.position.set(0,-6,0); _scene.add( _mesh);

Render loop is the same, rotate around Z axis. The result is that the translation is applied and the object rotates around the its new local axis which was also shifted downwards:

Three.js also supports the axis/angle method of specifying rotations:

// Example 5 var geo = new THREE.BoxGeometry(5,5,20,32); _mesh = new THREE.Mesh(geo, new THREE.MeshNormalMaterial()); // these have no effect because in render() we will directly modify the internal rotation matrix //_mesh.rotateY(Math.PI/4); //_mesh.rotation.order = 'ZXY';

Render loop:

var axis = new THREE.Vector3( 0, 0, 1 ); var angle = _tick * Math.PI / 256; // matrix is a THREE.Matrix4() _matrix.makeRotationAxis( axis.normalize(), angle ); _mesh.rotation.setFromRotationMatrix( _matrix );

The result is identical to what we achieved above in Example 1 with mesh.rotation. But often it is easier to conceptualize an axis of rotation rather than a succession of Euler angles.

We can mimic the Example 3 result using makeRotationY and applying it to the __geometry__.

// Example 6 var geo = new THREE.BoxGeometry(5,5,20,32); geo.applyMatrix( new THREE.Matrix4().makeRotationY( Math.PI/2 ) ); _mesh = new THREE.Mesh(geo, new THREE.MeshNormalMaterial()); _scene.add( _mesh);

Render loop is same as Example 5, with the desired result.

Now why did that work whereas mesh.rotationY is ignored? Because we rotated the geometry; the vertices were changed. This will be clearer if we translate the geometry:

// Example 7 var geo = new THREE.BoxGeometry(5,5,20,32); geo.applyMatrix( new THREE.Matrix4().makeRotationY( Math.PI/2 ) ); geo.applyMatrix( new THREE.Matrix4().makeTranslation(0,-6,0) ); _mesh = new THREE.Mesh(geo, new THREE.MeshNormalMaterial()); _scene.add( _mesh); // put sphere at mesh origin var sphere = new THREE.Mesh( new THREE.SphereGeometry(1,20,20), new THREE.MeshNormalMaterial()); sphere.position.set(0, 0, 0,); // we could put the sphere anywhere // and the box would rotate around it

Render loop code is the same as the previous 2 examples:

var axis = new THREE.Vector3( 0, 0, 1 ); var angle = -_tick * Math.PI / 256; // matrix is a THREE.Matrix4() _matrix.makeRotationAxis( axis.normalize(), angle ); _mesh.rotation.setFromRotationMatrix( _matrix );

The result:

The sphere represents the center of our geometry, its origin. The vertices have been rotated and translated from the local origin, where the sphere is.

If we’re rotating geometries, it is helpful to create a set of axes to show the local rotation axes of the geometry.

// Example 8: // put axes at mesh origin with mesh rotation // drawAxes is my routine which is based on THREE.AxisUtils(). See animation link below. _scene.add(drawAxes(10, _mesh.position, _mesh.rotation));

Since we’re just rotating the mesh around the Z (blue) axis, we can use this more compact syntax in the render loop:

_mesh.rotation.z = -_tick * Math.PI/128;

The result:

So to review: we wanted a geometry to rotate around a point that was external to the geometry. We did that by transforming the vertices of the geometry using applyMatrix.

There is another way to accomplish the same result: attach the box mesh to a parent mesh. Set the child’s position (that is to say, the box’s position) to be relative to the parent. And then place the parent where one wants, and rotate the parent if appropriate:

// Example 9: var geo = new THREE.BoxGeometry(5,5,20,32); _mesh = new THREE.Mesh(geo, new THREE.MeshNormalMaterial()); _mesh.rotation.y = -Math.PI/2; _mesh.position.set(0,-6,0); // _mesh will be the child _sphere = new THREE.Mesh( // _sphere will be the parent new THREE.SphereGeometry(1,20,20), new THREE.MeshNormalMaterial()); _sphere.position.set(0,-12,0); _sphere.rotation.x -= Math.PI/8; _sphere.add(_mesh); // add child to parent _scene.add(_sphere); // put axes at parent origin with parent rotation _scene.add(drawAxes(10, _sphere.position, _sphere.rotation));

And in the render loop, rotate the _sphere, not the _mesh.

When is this method (building a parent child relationship) preferable to changing the geometry? I couldn’t see much difference until I had to work with physi.js to apply physics to some of my meshes. Physi.js only works with meshes that have certain simple geometries (sphere, cone, box, etc). It will however work with compound meshes and thus with more complicated geometries, but in this case a parent/child relationship between the meshes is required.

Full three.js animation is here.

Using r69 of three.js.

## Drawing Pentatope Cross-Sections in three.js

A triangle is the simplest regular figure in 2 dimensions. Its 3 dimensional analogue is a tetrahedron. Its 4 dimensional analogue is the pentatope. Another term for these 3 geometries is: simplex. Many articles online explain these further.

I wanted to view what it would look like if a pentatope passed through our 3-dimensional space. There are many examples of what the projection of a pentatope onto 3-d space would look like (ie its shadow), but I didn’t want to do that because while projections can be pretty, they seem less intuitive than cross-sections.

To summarize, the plan was: take 3-dimensional cross-sections of the pentatope (a 4 dimensional shape). The actual drawing will be done using the 3-dimensional drawing toolkit called three.js.

Our equation for the 3-dimensional cross-section is:

ax+by+cz+K = w

This is a linear equation; we’re not dealing with curved cross-sections for now, though that would be fun. a,b, and c are constants. x,y,z, and w are the 4 axes. K is a constant that we change to get different cross-sections.

But why do we need 4 dimensions to describe a 3-dimensional space? Because we are locating that 3-d space in 4 dimensions. Just as the equation for a plane in 3-d space requires 3 dimensions (e.g. x+y = z describes a plane in 3-d space).

What about our equation for pentatope? Well we don’t have an equation, but we have a collection of points. Our pentatope has 5 corners. Lines connect each point to every other point, resulting in 10 lines. These 10 lines in turn form 10 faces (2-d planes). These 10 faces in turn from 5 3-d tetrahedrons.

Note that:

– The intersection of a 4-d line and a 3-d space is a point.

– The intersection of a 4-d plane and a 3-d space is a line.

– The intersection of a 4-d volume and a 3-d space is a plane.

Our cross-section is going to look like a wire-frame, in other words a collection of lines. So we will capture the intersection of the pentatope’s faces with our cross-sectional space.

Our pseudo-code would look like this:

foreach (face in faces) { // 10 of these // Each face is a triangle. Lets name the vertices p0, p1, p2. These points have 4-d coords. // point1 and 2 will be the points we use to draw a line in 3-d space using three.js. // they have 3-d coords of course. point1 = calculateIntersectionOfLineWithSpace(p0,p1); point2 = calculateIntersectionOfLineWithSpace(p0,p2); drawLineUsingThreeJS(point1, point2); }

So the interesting bit is in ‘calculateIntersectionOfLineWithSpace()’:

We must use parametric equations to work with lines in dimensions higher than 2. The parametric form for a line defined by the points p0 and p1 is:

p = p0 + t*v (1) where v = p1 - p0.

We need to calculate what the value of t is at the intersection with our cross-sectional space. (p0 and p1 are the vertices of one of the faces of our pentatope.) So we need to solve for t.

t = (Pt - P0)/V

Pt is the point of intersection of the line with the cross-sectional space. Given our formula above for our 3-d cross-sectional space, the equation for t is:

t = (W0-aX0-bY0-cZ0-K)/(aXv+bYv+cZv-Wv) Where P0 = (X0, Y0, Z0, W0) and V is (Xv, Yv, Zv, Wv).

Plugging t into equation (1) above we calculate the point of intersection of our 4-d line with our 3-d space. We return (x,y,z) from ‘calculateIntersectionOfLineWithSpace()’ above, discarding w.

And we simply change K to see different cross-sections. Here is the animated result:

http://rwoodley.org/MyContent/WIP/30-Simplex/index.html

A still:

## Minecraft Menger Sponge – STEAM project.

What has zero volume and an infinite surface area? A Menger Sponge of course. If your child is a Minecraft fan like my 7 year old, then this simple fractal can be used to make a good afternoon STEAM project that teaches math and programming concepts. (Note you should already know how to modify Minecraft code. That would take more than afternoon to get a grip on. And of course you should know Java).

First we built a Level 1 Menger cube with snap cubes. Then we came up with the X,Y, and Z coordinates for each of the 20 blocks that make up the Level 1 cube. This was a major goal of our project: increased fluency with 3-d coordinates.

Then it was not a difficult leap to see how our Java code used these 20 coordinates to place blocks.

public static void drawblock(World world, int x, int y, int z, int startx, int starty, int startz) { int metadata = world.getBlockMetadata(x, y, z); Block block = Block.getBlockById(35); boolean res = world.setBlock(startx + x, starty + y, startz + z+5, block, metadata, 3); System.out.println("Placing block at " + x + "," + y + ", res = " + res); } public boolean onItemUse(ItemStack par1ItemStack, EntityPlayer par2EntityPlayer, World world, int startx, int starty, int startz, int par7, float par8, float par9, float par10) { Menger1Item.drawLevel1Cube(world, startx, starty, startz); return true; } public static void drawLevel1Cube(World world, int startx, int starty, int startz) { drawblock(world, 0, 0, 0, startx, starty, startz); drawblock(world, 1, 0, 0, startx, starty, startz); drawblock(world, 2, 0, 0, startx, starty, startz); drawblock(world, 0, 1, 0, startx, starty, startz); drawblock(world, 2, 1, 0, startx, starty, startz); drawblock(world, 0, 2, 0, startx, starty, startz); drawblock(world, 1, 2, 0, startx, starty, startz); drawblock(world, 2, 2, 0, startx, starty, startz); drawblock(world, 0, 0, 1, startx, starty, startz); drawblock(world, 0, 2, 1, startx, starty, startz); drawblock(world, 2, 0, 1, startx, starty, startz); drawblock(world, 2, 2, 1, startx, starty, startz); drawblock(world, 0, 0, 2, startx, starty, startz); drawblock(world, 1, 0, 2, startx, starty, startz); drawblock(world, 2, 0, 2, startx, starty, startz); drawblock(world, 0, 1, 2, startx, starty, startz); drawblock(world, 2, 1, 2, startx, starty, startz); drawblock(world, 0, 2, 2, startx, starty, startz); drawblock(world, 1, 2, 2, startx, starty, startz); drawblock(world, 2, 2, 2, startx, starty, startz); }

Here is the Level 1 Cube in Minecraft:

At this point we wanted to make higher level cubes. A simple recursive algorithm was called for. This was too much for my son, not surprisingly. It is excerpted below if you want to do something similar. The main concept I tried to convey was that each level was a new power of 20.

Level 1 = 20^1 = 20. Length of side: 3 blocks.

Level 2 = 20^2 = 400. Length of side: 9 blocks.

Level 3 = 20^3 = 8000. Length of side: 27 blocks.

Level 4 = 20^4 = 160,000. Length of side: 81 blocks.

Level 5 = 20^5 = 3,200,000. Length of side 243 blocks.

At this point you could point out how the volume is shrinking with each level. E.g: 400/9^3 < 3200000/243^3. Over time the volume will approach zero! Ah, the mysteries of fractals and limits.
Back to the concrete: We started with a level 2 cube. Exploring this next level up was important to understand recursion. My son kindly illuminated the structure with torches as night fell:

We skipped right to Level 4 which was beautiful and impressive. When night fell in our Minecraft world, the local wild-life (mobs) colonized our structure and served to give a nice sense of scale and depth.

Take a tour of this level 4 Menger Cube:

Level 4 Menger Cube from Robert Woodley on Vimeo.

Could Minecraft handle a Level 5 cube? That was the question on our minds. 3,200,000 blocks. It took about 5 minutes and the laptop labored mightily. But it worked! There were nice glitch effects as Minecraft struggled to build out the structure. Also the top was truncated as Minecraft prevented us from going into outer space.

This incredible fractal structure is best understood through video:

Level 5 Menger Cube in Minecraft from Robert Woodley on Vimeo.

Complete Java class to build a level 5 Menger Cube:

package com.example.examplemod; import net.minecraft.block.Block; import net.minecraft.entity.player.EntityPlayer; import net.minecraft.item.Item; import net.minecraft.creativetab.CreativeTabs; import net.minecraft.item.ItemStack; import net.minecraft.world.World; public class Menger1Item extends Item { public Menger1Item() { setMaxStackSize(64); setCreativeTab(CreativeTabs.tabMisc); setUnlocalizedName("MengaLevel1Cube"); } public static void drawblock(World world, int x, int y, int z, int startx, int starty, int startz) { int metadata = world.getBlockMetadata(x, y, z); Block block = Block.getBlockById(35); boolean res = world.setBlock(startx + x, starty + y, startz + z+5, block, metadata, 3); System.out.println("Placing block at " + x + "," + y + ", res = " + res); } public boolean onItemUse(ItemStack par1ItemStack, EntityPlayer par2EntityPlayer, World world, int startx, int starty, int startz, int par7, float par8, float par9, float par10) { Menger1Item.drawLevel1Cube(world, startx, starty, startz); return true; } public static void drawLevel1Cube(World world, int startx, int starty, int startz) { int[][] i5coords = returnCoords(81); for (int i5 = 0; i5 < 20; i5++) { int[][] i4coords = returnCoords(27); for (int i4 = 0; i4 < 20; i4++) { int[][] i3coords = returnCoords(9); for (int i3 = 0; i3 < 20; i3++) { int[][] i2coords = returnCoords(3); for (int i2 = 0; i2 < 20; i2++) { int[][] coords = returnCoords(1); for (int i1 = 0; i1 < 20; i1++) { drawblock(world, i5coords[i5][0] + i4coords[i4][0] + i3coords[i3][0] + i2coords[i2][0] + coords[i1][0], i5coords[i5][1] + i4coords[i4][1] + i3coords[i3][1] + i2coords[i2][1] + coords[i1][1], i5coords[i5][2] + i4coords[i4][2] + i3coords[i3][2] + i2coords[i2][2] + coords[i1][2], startx, starty, startz); } } } } } } public static int[][] returnCoords(int level) { int[][] coords = { {0, 0, 0}, {1, 0, 0}, {2, 0, 0}, {0, 1, 0}, {2, 1, 0}, {0, 2, 0}, {1, 2, 0}, {2, 2, 0}, {0, 0, 1}, {0, 2, 1}, {2, 0, 1}, {2, 2, 1}, {0, 0, 2}, {1, 0, 2}, {2, 0, 2}, {0, 1, 2}, {2, 1, 2}, {0, 2, 2}, {1, 2, 2}, {2, 2, 2}, }; for (int i = 0; i < 20; i++) { for (int j = 0; j < 3; j++) coords[i][j] *= level; } return coords; } }

## CVision: Computer Vision Workbench

CVision: A computer vision WorkBench and Utilities written in WinForms that wraps OpenCV. I wrote it for my purposes, there is plenty to improve: https://github.com/rwoodley/CVision.

Features:

- Has all OpenCV color options and color maps.
- Histogram Equalization
- 4 Blurs
- Many Morph Modes:
- ERODE, DILATE, OPEN, CLOSE, GRADIENT, TOPHAT, BLACKHAT

- You can specify morph type (RECT, CROSS, ELLIPSE) as well as kernel size.

- Other transformations: NOT, SOBELX SOBELY, LAPLACIAN, SCHARRX1, SCHARRY1, CANNY, THRESHOLD
- Ability to group operations into ‘recipes’.
- Boolean operations.
- Intelligent contours and rotation.
- Ability to process in batch mode across a directory full of images.
- All intermediate transformations are saved.

## The Visual Representation of High Dimension Spaces

Our brains struggle visualizing spaces with more than 3 dimensions. This is a problem we tried to address in our FaceCloud and FaceField projects. We present here a possible solution using fractional dimensions to represent higher dimensions. The examples here are all based on the Eigenfaces face recognition algorithm where we were dealing with high dimension PCA spaces. But the visualization methodology is not specific to PCA.

The model we used for the Face Field project is a 60-dimensional ‘face space’. That is, we work with 60 Eigenvectors (or Eigenfaces) for all of our operations, be it classic face recognition, the anti-face calculation, or synthetic face generation.

We’ve repeatedly tried to create visualizations of this 60-dimensional space. Working with images makes this easier because one can see immediately what many of the Eigenvectors encode. Working with faces is even better because we’re so tuned to reading faces.

For instance the first Eigenface codes for both lighting and gender.

In this picture we see the ‘mean face’ or ‘average face’ in the center, and the results of shifting the first eigenface’s value (that is, its eigenvalue) from -1 to 1. Note that the face on the left is: female, white face on dark background. The face on the right is: male, dark face on light background.

The coordinates for the mean face are {0,0,0,0,0,…..,0} – zeroes for all 60 eigen values.

The coordinates for the face on the left are {-1,0,0,0,0,…..,0} – minus one for the first eigen value, 0 for the remaining 59.

The coordinates for the face on the right are {1,0,0,0,0,…..,0} – one for the first eigen value, 0 for the remaining 59.

—-

Lets take this a step further and look at the first 3 Eigenfaces. All of the faces below have Eigenvalues of 0 for all Eigenfaces beyond the 3rd one. The values for the first 3 are displayed on the diagram. We have just discussed the first Eigenface in our preceding paragraph. It is represented here by the axis that goes from the upper-left to the lower-right. The second Eigenface seems to encode for side lighting. The third Eigenface encodes for top vs bottom lighting which also incidentally encodes for hair.

But wait, you say, where is the face corresponding to these eigenvalues: {1,1,1}. or {1,1,0} or {-1,0,1}? These are indeed not displayed. One could imagine a cube with the X, Y and Z axes being the first 3 eigen vectors and these other missing faces being the vertices and edges of the cube. So go imagine that because I haven’t got a graphic handy that displays that. Even if I did, what would we have accomplished? Simply displaying 3 dimensions of data in 3 dimensions. The topic of this article concerns High Dimensional Spaces.

We can take the ‘Face Sextant’ above a step further and use Adelheid Mers’ fractal 3-line matrix to browse a 6 dimensional face space:

I only typed in the coordinates for a few faces. Hopefully you can easily divine what the other values would be. Once again we are missing the edges and vertices of the fractal cube (i.e. the face at {1,1,1,1,1,1} is not on this diagram).

So this process could be repeated over and over-again, adding 3 new dimensions to our visualization on each iteration. It is a loss-less method of representing higher dimensions using smaller and smaller fractal spaces.

In terms of actually making a usable visualization, the faces would get smaller of course so we would need a pan and zoom capability. Here is a visualization using this approach:

Face Cloud from Robert Woodley on Vimeo.

It displays faces with non-zero eigenvalue for the first 12 eigenfaces. In other words it is a visualization of 12 dimensional space. Note also it is using a tetrahedral approach as opposed to a cubic approach. That is, the coordinates for the first 4 faces to emanate from the Mean Face are: {1,1,1}, {-1,-1,1}, {1,1,-1}, {-1,-1,-1}. This reduces the number of faces, thereby reducing the visual clutter and giving the graphics card a chance to keep up with the calculations.

That was a video. The original was written in javascript using three.js and can be run here using chrome on a computer with a good graphics card.

—–

Separately, many of you have seen the Synthetic Face machine. It allows you to tweak all 60 eigen values to get any face you want from the face space, in real-time:

So it is not a way to visualize high dimension spaces, but it is useful for browsing those spaces and pulling faces at random from them. Indeed this was how we generated the faces on display at the “Enter The Matrix” show at the Chicago Cultural Center.

## Face Field Update

10 Synthetic Faces are now on display at the Chicago Cultural Center through August, as part of Adelheid Mers’ “Enter the Matrix” exhibit. They look great in large format and are quite powerful. Some photos below.

These are faces pulled at random from the 60-dimensional PCA space that we have been working with for some time now. You can create your own Synthetic Face at http://facefield.org/SynthFace.aspx. (Click on the face that appears for options.)

## Formula Toy

Back at the age of 17, thanks to the liberal access policies of Indiana University’s Wrubel Computing Center, I was able to write short little computer programs (on cards!) that would graph 3 dimensional surfaces using a pen plotter. The little nerd that was me was thrilled at the results, and my bedroom wall was plastered with graphs of various mathematical functions of my creation. And incidentally it was an extremely helpful way to get a visual understanding of mathematical geometries, including alternative coordinates systems – cartesian, spherical, and cylindrical.

In these days, there is no shortage of packages that will draw 3-D mathematical surfaces. However, nothing that I found was totally simple. All required a learning curve, or a download or a plugin. I wanted to build something whereby you would simply type in a formula, hit enter, and see the surface. Hence was born Formula Toy. It uses the amazing three.js library which is a wrapper around WebGL. The dependence on WebGL means that it won’t run on every single browser because WebGL is still an emerging standard. However it should work on most MACs and on most Window’s desktops, at least if you use a modern browser like Chrome.

You can either go directly to http://formulatoy.net or you can look at some examples and just click on ones there that intrigue you which will pop you straight into formula toy. There is some help text here.

## The Hairy Blob, 800 Ping Pong Balls, and a Mindstorms RoboCam

#### The Hairy Blob:

At the Hairy Blob exhibition at the Hyde Park Art Center last spring, visitors were invited to draw an image of time on a ping pong ball and toss it into a net that was suspended from the ceiling.

#### 800 Ping Pong Balls:

What to do with over 800 ping pong balls?

How to document 800 three dimensional objects in less than 5 years?

#### Mindstorms RoboCam:

Our ping pong cam was an NXT Mindstorms robot (which rotated the balls) driven by a laptop that was simultaneously taking pictures. Controlling the robot from an external device was suprisingly difficult. NHK.MindSqualls did the job, but just.

Ping Pong Robo Cam and Laptop setup:

We did the scanning over Thanksgiving weekend at the Roger Brown house in New Buffalo, MI.

#### The installation:

Intalled at the Mers Micro Museum, a Raspberry Pi drives the display. Some javascript randomly selects from the 800, and then starts a few of them spinning. First it shows a random batch of the day time balls and then a random batch of the night time balls. And so on, indefinitely.

You can also spin the balls online.

## Anti-Face Model Specification and Calculation Details

**Update 10/2013**: We have implemented this in a free iPhone app which is available here:

https://itunes.apple.com/us/app/anti-face/id690376775

Our site FaceField.org (and now the iPhone app) uses the EigenFaces methodology as implemented in Open CV to calculate a special kind of face that we have labelled an ‘Anti Face’.

Model specification:

– Over 1000 faces were used.

– The faces were all facing the camera straight-on. We used a specially designed Haar classifier to ensure that we excluded faces looking to the side.

– The faces that we fed into the PCA calculation are slightly larger than what the Haar classifier detected so that we didn’t cut off the chins.

– The faces were all subject to histogram equalization.

– The faces were all sized to 200×200 pixels.

– We took the first 60 eigenvectors.

The Antiface calculation is simple: we do a subspace projection of the uploaded face into the 60-dimensional EigenFace space. It is interesting to view this ‘reconstructed face’ as we call it. In a normal face recognition calculation you would then compute the nearest face to this reconstructed face. The anti-face is simply the reconstructed face except that every weight is multiplied by -1.

In other words:

Let Ω ∋ R⁶⁰ be the subspace projection of the uploaded face.

Ω=(w₁,w₂,..w₆₀) where wᵢ= uᵢᵀ(Γ-Ψ) for i=1…60.

u ∋ R⁶⁰ is the Eigenface. Γ is the uploaded face image and Ψ is the mean face.

Then the antiface, Ω’ is (-1*w₁,-1*w₂,..-1*w₆₀).

So prominent features in the reconstructed faces (with high weighting, meaning far from the mean) would be equally prominent, though opposite, in the anti-face.

Why 60 dimensions? Originally we tried higher numbers like 200 because at that point the reconstructed face looks indistinguishable from the original face. However the anti-face looks very little like a face and is highly distorted and muddied. It seems that it such high-dimension space not everything looks like a face, however in a lower dimension space like 60 you’re likely to get a face no matter where you land.

Two Face/Anti-Face pairs as examples: