$$$$

17  Foundations

After a rather deep dive into the foundations and history of plane geometry, we are ready to leave the familiar behind and explore other worlds! The first new geometry we will consider is…..well…..actually also familiar: its the sphere. We’ve even met this geometry in our discussion of $\pi$, where we noted that using analogous arguments to what we did in the plane, the distance formula in three dimensions is a natural generalization of the pythagorean theorem, which provides an equation for the sphere.

Definition 17.1 (The Sphere (Points)) The (unit) sphere is the set of points $(x,y,z)\in {\mathbb{R}}^3$ lying at distance $1$ from the origin. $${\mathbb{S}}^2=\{(x,y,z)\in {\mathbb{E}}^3\mid x^2+y^2+z^2=1\}$$

It’s important to remember that by sphere mathematicians usually mean the surface, not the interior (we will call the interior of the sphere the ball). Thus, ${\mathbb{S}}^2$ is two dimensional, which is why we denote it this way, and call it the two-sphere.

The sphere has been studied since ancient times: we came across it most recently while analyzing the work of Archimedes, but it became of particular importance outside of mathematics around the same time, when Eratosthenes calculated the circumference of the Earth (quite accurately). But in both of these contexts we are picturing the sphere extrinsically, from the perspective of three-dimensional beings that could hold it in their hands.

Remark 17.1. Centuries earlier around 450BCE, the pre-Socratic philosopher Anaxagoras correctly postulated that the earth was a sphere, floating freely in the vacuum. But no one knew its size!

The big change in perspective here is that we are going to think of the sphere as a geometry all on its own, just like we did for the plane! We will work with coordinates in three dimensions to make our lives easier, but the surrounding 3-dimensional space is of no interest or consequence to us: the only space that is “real” is the surface of the sphere itself.

In some sense we are very used to this: as this is how we actually live our lives! Since evolution did not grace the great apes with wings, we humans spend almost all of our time walking around on the surface of a large sphere, unable to meaningfully interact with the totality of the 3-dimensional space it is embedded in. However, this isn’t totally helpful, as our two main ways of sensing the world around us, sight and sound depend on the physics of 3-dimensional space, and are not constrained to the sphere.

For me it is helpful to think about spherical geometry as the geometry a mathematically gifted-ant would discover if it lived its entire life on an orange weak and downward pointed eyes only able to perceive its immediate vicinity on the peel. What curves on the orange would the ant call lines? How would the ant measure angles and distances? Does the ant’s mathematics contain the pythagorean theorem?

Remark 17.2. The first treatments of spherical geometry as a true intrinsic geometry in its own right come not from silly thought experiments about ants of course, but rather from navigation using the stars, where the celestial sphere modeled the sky, and spherical trigonometry was first developed.

17.1 Calculus on ${\mathbb{S}}^2$

Having put all the work into understanding a modern, calculus-based approach to geometry in the plane, we will reap significant benefits here by seeing how many of the ideas remain conceptually the same on the sphere. Our infinitesimal foundations all rely on being able to take derivatives, so the first thing we should wonder is what is the derivative of a curve on the sphere? Happily, because the sphere lives in ${\mathbb{E}}^3$ and we understand Euclidean calculus well, we can directly borrow that notion:

Definition 17.2 (Calculus on the Sphere) The sphere inherits its notion of calculus from the 3-dimensional space it lives in: if $\gamma$ is a curve on the sphere then $\gamma (t)=(x(t),y(t),z(t))$ and ${\gamma }^{\prime }=⟨x^{\prime },y^{\prime },z^{\prime }⟩$.

To really get things moving, we need to define a notion of tangent space to each point on the sphere. This space should be the set of all infinitesimal tangent vectors to curves to the sphere. Here we need to put a little more thought in than we did for the plane, where we just noted that the derivative to a planar curve was also a 2-dimensional vector, so the tangent space at each point should be another copy of the plane. Why? Well here we have represented points on the sphere with three coordinates, and so tangent vectors also have three coordinates. But this doesn’t mean the tangent space at each point is three dimensional! Indeed, there are many three dimensional vectors at each point which are not tangent to any curve on the sphere.

Tangent vectors to ${\mathbb{S}}^2$ at a point are the derivatives of curves passing through that point.

Proposition 17.1 (Tangents to Curves on the Sphere) If $\gamma$ is a curve lying on the surface of the sphere passing through a point $\gamma (t)=p$, then its tangent vector ${\gamma }^{\prime }(t)$ is orthogonal to $p$ in ${\mathbb{E}}^3$.

Tangent vectors to a curve on the sphere through $p\in {\mathbb{S}}^2$ are orthogonal to the point $p$, thought of as a vector from the origin in ${\mathbb{E}}^3$.

Proof.

This argument relied on the observation that the dot product has its own product rule, which is a straightforward algebraic computation from its definition.

Exercise 17.1 (Product Rule for Dot Product) Let $f(t)=⟨f_1(t),f_2(t),f_3(t)⟩$ and $g(t)=⟨g_1(t),g_2(t),g_3(t)⟩$ be two vector functions. Prove that the dot product satisfies the product rule: $$\frac{d}{dt}(f(t)\cdot g(t))=f^{\prime }(t)\cdot g(t)+f(t)\cdot g^{\prime }(t)$$

Definition 17.3 (The Sphere (Tangent Vectors)) If $p\in {\mathbb{S}}^2$, then the tangent space $T_p{\mathbb{S}}^2$ is the set of all vectors in ${\mathbb{E}}^3$ which are orthogonal to $p$: $$T_p{\mathbb{S}}^2=\{q\in {\mathbb{E}}^3\mid p\cdot q=0\}$$ In coordinates, if $p=(p_1,p_2,p_3)$,these are the points $⟨x,y,z⟩$ such that $p_{1}x+p_{2}y+p_{3}z=0$.

Tangent spaces to the sphere are linear subspaces of ${\mathbb{E}}^3$ containing all vectors perpendicular to the position.

17.2 Geometry on ${\mathbb{S}}^2$

Now that we have points and tangent vectors, we need to bring the actual geometry into the picture. In our original development of ${\mathbb{E}}^2$ we encoded all of geometry via the notion of an infinitesimal length. We then went on to develop all the higher level concepts like lengths of curves, and eventually angles - before discovering that the we could measure angles easily with the dot product! But since we can also measure infinitesimal lengths with the dot product, we saw that we could alternatively take this as the basis of all of geometry. We will take this bold new approach here with the sphere.

Definition 17.4 (The Sphere’s Dot Product) If $v=⟨v_1,v_2,v_3⟩$ and $w=⟨w_1,w_2,w_3⟩$ are two tangent vectors on the sphere based at a point $p$ their dot product is computed using the standard dot product on ${\mathbb{E}}^3$: $$v\cdot w:=v_1{w}_1+v_2{w}_2+v_3{w}_3$$

This gives rise immediately to our notion of infinitesimal length:

Definition 17.5 (Infinitesimal Length on ${\mathbb{S}}^2$) Given a vector $v=⟨v_1,v_2,v_3⟩$ in the tangent space $T_p{\mathbb{S}}^2$, the infinitesimal length of $v$ is the square root of its dot product with itself:

$$\parallel v\parallel =\sqrt{v\cdot v}=\sqrt{v_1^2+v_2^2+v_3^2}$$

Thus, each tangent space comes with an infinitesimal version of the pythagorean theorem, just like we had for ${\mathbb{E}}^2$! Remember what tangent spaces are all about: they’re encoding the result of a limiting process of infinite zoom: the fact that we see the pythagorean theorem here on the tangent space is just the statement that zooming in on a point, a sphere appears to be flat! This we are quite used to from living on the surface of the earth. The magic we will see soon is that in fact all of spherical geometry can be recovered from this infinitesimal flatness.

Remark 17.3. Some people are too impressed by this fact, and have the mistaken impression that the earth actually is flat!

To define the length of a curve on ${\mathbb{S}}^2$ we follow the exact approach from ${\mathbb{E}}^2$, and use integration to promote thse infinitesimal lengths to finite ones.

Definition 17.6 (Lengths of Curves on ${\mathbb{S}}^2$) Length of a curve is the integral of its infinitesimal lengths: $$\mathrm{length}(\gamma )={\int }_I\parallel {\gamma }^{\prime }(t)\parallel d$$

Now for angles, our new foundations make everything much easier! Instead of working hard (to define an angle as the arclength of the unit circle in the tangent space, spanned by two tangent vectors at a point), we instead note that we already know how this is related to the dot product in Euclidean space, and we know the tangent space IS euclidean (Definition 17.5). Thus, we can take the relation to the dot product as our definition:

Definition 17.7 (Angles on ${\mathbb{S}}^2$) The angle between two vectors on the sphere is defined using the inner product: $$\sphericalangle (v,w)=\arccos \left(\frac{v\cdot w}{\parallel v\parallel \parallel w\parallel }\right)$$

Where here $\arccos$ can be calculated by the integral expression we derived in Proposition 14.2 (or by your calculator, which does this faster!)

Exercise 17.2 Consider the curves $\alpha (t)=(\cos t,\sin t,0)$ (the equator of the sphere), and $\beta (t)=(0,\sin (t),\cos (t))$ (a line of longitude). Prove that they

• Intersect each other at the $t=\pi /2$
• Form a right angle at their point of intersection.

17.3 Isometries of ${\mathbb{S}}^2$

Our fundamental tool for working with Euclidean space was isometries. In our development of the geometry, we tried to seek out as many isometries early on as we coould, and then continually used them to make our lives easier: moving points to the origin, lines to the $x$-axis, and so on.

The same approach will prove benificial on the sphere: it’ll be nice to be able to move points to the north pole, or circles to the equator when we desire. So, let’s track down some isometries! But first - what is an isometry here? We defined an isometry before as a function which preserved infinitesimal lengths, but that was because infinitesimal lengths were the foundation of our geometry. Now we’ve decided to take the dot product as our foundations so, perhaps we should change our definition of isometry here too?

Definition 17.8 (Isometries on ${\mathbb{S}}^2$) An isometry of ${\mathbb{S}}^2$ is a function $\varphi :{\mathbb{S}}^2\to {\mathbb{S}}^2$ which preserves the dot product. Precisely, this means that if $p\in {\mathbb{S}}^2$ is a point and $v,w\in T_p{\mathbb{S}}^2$ are tangent vectors, then

$$v\cdot w=(D{\varphi }_{p}v)\cdot (D{\varphi }_{p}w)$$

However, it doesn’t actually matter which we take as our definition (preserving infinitesimal length, or the dot product) they pick out precisely the same class of maps! In practice, when we want to prove something is an isometry, we will either show it preserves the dot product, or that it preserves infinitesimal lengths, whichever is easier. This perhaps surprising claim is justified by a result:

Theorem 17.1 A function $f:{\mathbb{S}}^2\to {\mathbb{S}}^2$ (or ${\mathbb{E}}^2\to {\mathbb{E}}^2$, or ${\mathbb{E}}^3\to {\mathbb{E}}^3$…) preserves all infinitesimal lengths if and only if it preserves the dot product.

One direction of this theorem is straightforward: if a map $\varphi$ preserves the dot product, then it certainly preserves infinitesimal lengths! After all, preserving the dot product means that for any vector $v$, we have $$v\cdot v=(D{\varphi }_{p}v)\cdot (D{\varphi }_{p}v)$$

But length is just the square root of this expression, so this immediately implies $\parallel v\parallel =\parallel D{\varphi }_{p}v\parallel$. The perhaps more surprising direction is the reverse: if a map preserves all infinitesimal lengths, then it actually preserves the dot product. The trick here is to show that it’s actually possible to compute the dot product of two vectors using infinitesimal lengths (the reverse of what we did above!)

Exercise 17.3 (Dot Products from Lengths) Prove that if $v,w$ are two vectors then the following equation is true: $$\parallel v+w{\parallel }^2=\parallel v{\parallel }^2+\parallel w{\parallel }^2+2⟨v,w⟩$$

Solve this for the dot product (moving all the other terms to the other side of the equation), and then prove the following fact: if $\parallel u\parallel =\parallel D{\varphi }_{p}u\parallel$ for all vectors, then $v\cdot w=(D{\varphi }_{p}v)\cdot (D{\varphi }_{p}w)$ (Hint: apply $D\varphi$ to the equation!)

Because isometries are defined using the same basic machinery here as Euclidean space (preserving infinitesimal quantities) the theorems we proved there about their composition and inversion carry over without any change:

Theorem 17.2 The composition of any two isometries of the sphere is an isometry, and the inverse of any isometry of the sphere is an isometry.

So to find isometries of the sphere we just need to track down functions on ${\mathbb{E}}^3$ that preserve the dot product. But in linear algebra at least, such functions already have a name!

Definition 17.9 If $A$ is a linear map ${\mathbb{E}}^n\to {\mathbb{E}}^n$ such that preserves the dot product $(Av)\cdot (Aw)=v\cdot w$, then $A$ is called an orthogonal matrix.

Example 17.1 The linear map $(x,y,z)\mapsto (x,y,-z)$ is represented by an orthogonal matrix: $$\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right)$$

The isometry $(x,y,z)\to (x,y,-z)$ on the sphere.

Exercise 17.4 A permutation matrix is a square matrix where every row and column has exactly one “1”, and the other entries are zero. Prove the following permutation matrix is an orthogonal matrix:

$$\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 1\end{array}\right)$$

These maps preserve the dot product on ${\mathbb{E}}^3$, but we need a little more than that to be sure they are isometries! Isometries of the sphere need to actually be maps ${\mathbb{S}}^2\to {\mathbb{S}}^2$.

Corollary 17.1 If $\varphi (x)=Ax$ is an orthogonal transformation of ${\mathbb{E}}^3$, then $\varphi$ sends the unit sphere to the unit sphere.

Proof. Since $A$ is orthogonal it preserves the dot product. THus it preserves infinitesimal lengths, and so it preserves distances in ${\mathbb{E}}^3$. This means if $p\in {\mathbb{S}}^2$ (so that $p$ is distance 1 from the origin $O$) then $\varphi (p)$ is also on the sphere (its distance 1 from $\varphi (O)$, but $\varphi$ sends the origin to itself, because its a linear map).

Putting these facts together gives the following powerful theorem telling us tons of isometries of the sphere! (In fact, these are all the isometries of the sphere. But we don’t need that here)

Theorem 17.3 If $A$ is an orthogonal matrix, then the function $p\mapsto Ap$ is an isometry of the sphere.

This theorem gives us access to tons of isometries: all we need to do is track down orthogonal $3\times 3$ matrices. We’ve already seen a couple explicit examples above (the reflection $(x,y,z)\mapsto (x,y,-z)$ and the permutation matrices in the examples), but it will prove useful to dig a little deeper and try to figure out what kind of matrices are orthogonal. The following theorem of Linear Algebra gives us a complete classification:

Theorem 17.4 A matrix is an orthogonal matrix if and only if all of its columns are unit vectors, and each column is orthogonal (hence the name) to every other.

Now that we know the algebraic description of isometries ($3\times 3$ number squares where all the columns are orthonormal) we turn to the geometry: what do isometries of the sphere do?

The two most useful properties of isometries by far in ${\mathbb{E}}^2$ were the ability to move points around, and the ability to rotate any tangent vector to any other: these were the properties we called homogenity and isotropy. It was these two properties that gave the plane its incredible symmetry.

The sphere is of course very symmetric looking as well, and we are used to from our experience in day-to-day life with the ability to rotate a sphere any which way we like. But now we should prove it:

Any point can be sent to the north pole of the sphere, or equivalently, you can send any point to the north pole via an isometry.

Proposition 17.2 (Any Point Moves to the North Pole) Let $N=(0,0,1)$ denote the north pole of the sphere, and $p$ and arbitrary point on the sphere. Then there is an isometry of ${\mathbb{S}}^2$ which moves $N$ to $p$. (And thus its inverse moves $p$ to $N$).

Proof. We will find an orthogonal matrix $A$ so that the isometry $\varphi (x)=Ax$ takes $N$ to $p$. Since $N=(0,0,1)$, applying a linear map $A$ to the vector $N$ gives us the third colum of $A$. So, to begin to assemble such a map we will make its third column be $p$: $$A=\left(\begin{array}{ccc}\ast & \ast & p_1\\ \ast & \ast & p_2\\ \ast & \ast & p_3\end{array}\right)$$ Now we just need to find values for six missing entires so that the all columns are orthogonal and unit length. In fact, there are many ways to do this! And we don’t need any explicit solution we just need to know of their existence. So we will work column-by-column.

Call the second column of this matrix $u=(u_1,u_2,u_3)$. We know this must be orthogonal to $p$, so we have an equation this must satisfy: $$u\cdot p=u_1{p}_1+u_2{p}_2+u_3{p}_3=0$$ This is a single linear equation in three variables, and so it has many solutions (a two-dimensional space of solutions, in fact)! Taking any solution, we can rescale it to unit length, and use that as our second column.

Now for the first column, we have three unknowns (its three entires): but we have two equations - it must dot product with both the second and third column to zero. This still has an infinite number of solutions (in linear-algebra-speak, there’s one ‘free variable’), and choosing any solution and normalizing it gives a viable first column.

Remark 17.4. The argument I give here is a soft or qualitative argument: we prove the existence of something without actually computing it. If you would like to actually compute a specific matrix that takes $N$ to $p$ (which is often useful in real-world applications of spherical geometry), you can do so by starting with any two vectors $u,v$ where that $\{u,v,p\}$ is linearly independent, and apply the Gram-Schmidt process.

Theorem 17.5 (The Sphere is Homogeneous) Given any two points $p$ and $q$ on the sphere, there is an isometry taking $p$ to $q$:

Proof. Let $N$ be the north pole of the sphere. Then by Proposition 17.2, we can find an isometry $|phi$ taking $N$ to $p$, and another isometry $\psi$ taking $N$ to $q$. We will apply our by-now-standard trick, and compose one of these with the inverse of the other!

Specifically, the map ${\varphi }^{-1}$ is an isometry which takes $p$ to $N$, and $\psi$ takes $N$ to $q$ so the composition $\psi \circ {\varphi }^{-1}$ takes $p$ to $q$, as desired.

Next, we wish to see the sphere is also isotropic. We will do this in two parts (just like we did for ${\mathbb{E}}^2$)! First, we show that you can rotate the sphere about some specific point, and then we use homogenity to show we can actually do this at any point.

Rotating the sphere about a point.

Proposition 17.3 Let $N$ be the north pole, and $v$ be any unit vector in $T_N{\mathbb{S}}^2$. Then there exists an isometry $\varphi$ of the sphere which fixes $N$ and takes $⟨1,0,0⟩\in T_N{\mathbb{S}}^2$ to $v$.

Proof. First, what sort of a vector is $v$? Its a unit vector in $T_N{\mathbb{S}}^2$, but what set of vectors is this? By Definition 17.3, its the set of vectors orthogonal to $N=(0,0,1)$. That is, the vectors $⟨v_1,v_2,0⟩$: its a horizontal Euclidean plane! So, $⟨v_1,v_2⟩$ is a unit vector in this plane, and we want to rotate $⟨1,0⟩$ to this vector, and we know a matrix in the plane (from Euclidean geometry!) that does this:

$$\left(\begin{array}{cc}{v}_1& -v_2\\ v_2& v_1\end{array}\right)$$

How can we write down a transformation of ${\mathbb{E}}^3$ which does this to the horizontal plane and fixes the vertical direction (thus fixing $N$)? We can just insert it as the top $2\times 2$ block of the matrix:

$$A=\left(\begin{array}{ccc}{v}_1& -v_2& 0\\ v_2& v_1& 0\\ 0& 0& 1\end{array}\right)$$

Its easy to see that this takes $⟨1,0,0⟩$ to $v$: as $v$ is the first column of this matrix! So all we need to see is that this is actually an isometry: that $A$ is an orthogonal matrix.

But this is likewise straightforward: we can take the dot product of any two columns and see we get zero (try it!) and, each column is unit length (because $v$ was by hypothesis, and $(0,0,1)$ is).

Exercise 17.5 Use Proposition 3 and Theorem 1 to show the sphere is isotropic: that given any point $p\in {\mathbb{S}}^2$ and any two unit vectors $v,w\in T_p{\mathbb{S}}^2$, there exists an isometry of ${\mathbb{S}}^2$ fixing $p$ and taking $v$ to $w$.

(Hint: first show you can do this when $p$ is the north pole! Then use homogenity and a conjugation)

Now we have access to isometries that can move any point to any other point, and also rotate any vector to any other vector. This prepares us to prove the analog of the Euclidean theorem Exercise 21.

Exercise 17.6 Let $p,q$ be any two points on the sphere, and $v$ a unit vector at $p$ and $w$ a tangent vector at $q$. Then there is an isometry of ${\mathbb{S}}^2$ taking $(p,v)$ to $(q,w)$.

The sphere is homogeneous and isotropic.

These are essentially all the facts that we will need about isometries of the sphere! But we would be remiss to not mention one very useful dichotomy between isometries of the sphere: the familiar groups of rotations vs reflections. Like everything else we’ve studied in this section, this concept is also captured infinitesimally (ie with linear algebra).

Definition 17.10 An isometry of the sphere $\varphi (x)=Ax$ is a reflection if the $detA=-1$ and is a rotation if $detA=1$.

This lets us see computationally that the matrix in Example 17.1 and ?exm-permutation-orthogonal are both reflections, whereas the matrix we created in ?prp-sphere-homogeneous-step is a rotation.

Exercise 17.7 Prove that you can find a rotation which takes $N$ to any point $p$ of the sphere.

(Hint: our earlier construction produces an isometry, but we don’t know if its a rotation or reflection. If it is a reflection, can you modify it somehow so that it becomes a rotation, without changing the fact that it sends $N$ to $p$?)