This is a repost (lightly edited) of a post that I wrote in February 2024 that got lost in the long dark teatime of the soul that ensued in the interim.

Spaces of constant curvature

In this post, I want to kind of bring together the theme of curvature and the theme of affine or projective geometry. My goal is to introduce the spaces of constant curvature in a unified way as subspaces of Euclidean space and explore their symmetries as groups of matrices.

The $n$-sphere in $\mathbb{R}^{n+1}$

Do you remember the equation for the unit circle?

$x^2 + y^2 = 1$

This is a $1$-dimensional object living in the plane $\mathbb{R}^2$, also affectionately known as $\mathbb{R}^{1+1}$, and we can easily generalize this equation to $\mathbb{R}^{n+1}$ as

$x_0^2 + x_1^2 + \cdots + x_n^2 = 1.$

The set of points $(x_0,\ldots,x_n)$ satisfying this equation make up the $n$-sphere sitting inside $\mathbb{R}^{n+1}$. For like, precisely two values of $n$, we know what to expect: the unit circle in $\mathbb{R}^2$, and the unit sphere in $\mathbb{R}^3$. It’s a joy to imagine what the unit $n$-sphere feels like for other values of $n$. For example, there ought to be a “$0$-sphere” sitting in $\mathbb{R}^{0+1}$, and indeed there is: the two points $x = \pm 1$ such that $x^2 = 1$.

Or the $3$-sphere, which is somehow like and unlike $3$-dimensional space. I won’t push you too hard here, but if you want a brain-twister, try this on: If you take two disks (2-dimensional filled-in circles), say made out of plastic or silly putty, and imagined gluing or sewing them together along their boundary circle and then inflating them, you’d end up with a 2-dimensional sphere, or at least, something which you could reasonably expect to mold into the right shape. This actually generalizes: to go down a dimension, you’d be gluing two intervals along their boundary points to obtain the circle $S^1$. To go up a dimension and get $S^3$, you’d take two balls (3-dimensional filled-in spheres) and somehow glue their outsides together. You can’t do this in reality, but it sure makes for something fun to try and think about correctly!

Ah, the above was what I wrote just now in 2026, but I appear to have written still another funny brain-bender back in 2024, so here goes:

If you’ll permit me an aside, here is my favorite way to think of the $3$-sphere; actually it works for $n$ arbitrary, but for 3 I can really picture something if I wrack my brain a little. Let’s go back to the circle: pick a point on the circle, like say the point $(0,1)$, and throw it away. What you’re left with, if you kind of sit with it for a moment, is kind of like an open interval in $\mathbb{R}$, just curled up funny. Actually, you can do something pretty neat with these leftover points, called stereographic projection. Imagine you’re a little surveyor standing at the point $(1,0)$ in $\mathbb{R}^2$, and for each other point $(p,q)$ on the circle, it’s your job to draw the straight line from $(1,0)$ to $(p,q)$ and extend it from a line segment to a whole ray. And then second, note down where that line hits the $y$-axis.

An equation for an arbitrary point on this line is $(tp + (1-t), tq)$, so the point $(0,y)$ on the line will be such that $y = tq$ and $tp + (1-t) = 0$. This latter equation implies that $t = \frac{1}{1 - p}$, so $y = \frac{q}{1 - p}$.

In $n$ dimensions, the equation for the line becomes

$t \mapsto (tx_0 + (1-t), tx_1,\ldots,tx_n) = t\mathbf{x} + (1-t)\mathbf{e}_0,$

and the point on the hyperplane $x_0 = 0$ is

$\frac{\displaystyle (0, x_1, \ldots, x_n)}{1 - x_0}.$

Anyway, it’s clear that we can reverse this process: for any point $\mathbf{p}$ on the hyperplane $x_0 = 0$, there is a unique point $\mathbf{q}$ on the unit $n$-sphere such that stereographic projection from $(1,0,\ldots,0) = \mathbf{e}_0$ yields $\mathbf{p}$. As the point $\mathbf{q}$ gets closer and closer to $\mathbf{e}_0$, the point $\mathbf{p}$ goes “off to infinity”. Thus, one way to think of the $n$-sphere is like $n$-dimensional space with one extra point at infinity that “wraps it all up” just like the circle is the real line (that open interval) wrapped up with one extra point.

Linear transformations and isometries

One conceit of this post is that we love matrix multiplication. Consider a linear transformation of $\mathbb{R}^{n+1}$, or equivalently an $(n+1)\times(n+1)$ matrix. We want two things to happen: first our transformation (or matrix $M$) should be invertible in the sense that we can undo it. If you remember linear algebra, you might know that this comes down to asking that the determinant of $M$ is nonzero.

The other thing we want to have happen is that the matrix sends the unit sphere to itself. For this it helps to realize that the equation $x_0^2 + cdots + x_n^2 = 1$ has a vector interpretation: if $\mathbf{x} = (x_0,\ldots,x_n)$, we have that the coordinates of $\mathbf{x}$ satisfy the equation for the unit sphere if and only if the dot product satisfies

$\mathbf{x} \cdot \mathbf{x} = x_0^2 + \cdots + x_n^2 = 1.$

This dot product may also be expressed as a matrix product $\mathbf{x}^T \mathbf{x} = 1$, where we think of $\mathbf{x}$ as an $(n+1)\times 1$ column vector and its transpose $\mathbf{x}^T$ as the corresponding $1 \times (n + 1)$ row vector.

For later, I want to point out the sneaky fact that we can put in any $(n+1)\times(n+1)$ between $\mathbf{x}^T$ and $\mathbf{x}$ and we’ll still end up with a number. We’ll stick with the identity matrix for now.

Anyway, this actually helps us figure out what we want: $M\mathbf{x}$ is also a column vector, and so for it to belong to the unit sphere, we want

$(M\mathbf{x})\cdot(M\mathbf{x}) = (M\mathbf{x})^T(M\mathbf{x}) = \mathbf{x}^TM^T M\mathbf{x} = \mathbf{x}^T\mathbf{x} = 1.$

This turns out to happen for every choice of $\mathbf{x}$ just when $M^TM=I$, where $I$ is the identity matrix with 1s on the main diagonal and 0s elsewhere. Actually this implies that $M$ is invertible, and furthermore that $M M^T = 1$ as well, so that $M^T$ is the inverse of $M$.

Let’s suppose we’re working with the unit circle in $\mathbb{R}^2$ and that we have a symmetry $M$ which sends the point $(1,0)$ to $(p,q)$. A neat linear algebra fact is that this means that the first column of $M$ is $(p,q)$! Since we need $M^TM = I$, if the other column is $(c,d)$, we compute

$M^TM = \begin{pmatrix}p & q \\c & d\end{pmatrix}\cdot\begin{pmatrix}p & c \\q & d\end{pmatrix} = \begin{pmatrix} p^2 + q^2 & pc + qd \\ cp + dq & c^2 + d^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Since $p^2 + q^2 = 1$, one of $p$ or $q$ is nonzero, so we can divide by it to see that (supposing it is $p$)

$c = -\frac{qd}{p}.$

Then $c^2 + d^2 = 1$ implies $\frac{q^2d^2}{p^2} + d^2 = 1$, so (doing some tricksy algebra) $d^2 = p^2$. So either $d = p$, in which case $c = -q$, or $d = -p$, in which case $d = q$. That is, there are exactly two choices for $M$,

$\begin{pmatrix}p & -q \\ q & p\end{pmatrix}\quad\text{ and } \quad\begin{pmatrix}p & q \\ q & -p\end{pmatrix}.$

These are distinguished by the sign of their determinant or equivalently by whether they preserve or reverse orientation, meaning that one of them is more like a rotation, while the other is like a rotation followed by a flip.

Metrics

There are two ways to think of the $n$-sphere as a metric space. The first is to restrict the metric from Euclidean space to the sphere. This works fine, but is a little weird: the distance on the circle from $(1,0)$ to $(0,1)$ would be their Euclidean distance $\sqrt{2}$, but the actual shortest path on the circle between these points is longer; in fact it has length equal to the arc length of the quarter circle it traces out. Since we “know” that the perimeter of a circle is $2\pi$ times its radius, we should expect the distance on the circle between $(1,0)$ and $(0,1)$ to be $\frac{pi}{2}$.

This latter distance, measuring the angle between $\mathbf{p}$ and $\mathbf{q}$ is connected to their dot product. For the unit sphere in general, we have that this angle $\theta$ satisfies

$\cos\theta = \mathbf{p} \cdot \mathbf{q}.$

(The formula holds in general if you additionally divide out the norms of $\mathbf{p}$ and $\mathbf{q}$, which is unnecessary for our unit vectors.) Anyway, this distance,

$d(\mathbf{p},\mathbf{q}) = \arccos (\mathbf{p}\cdot\mathbf{q})$

is called the round metric on the unit sphere. In a post which has yet to be restored to this blog, we computed that the sectional curvature of this metric is constant and equal to 1.

Euclidean space

In previous posts which have yet to be restored to this blog, we saw that affine subspaces of $\mathbb{R}^n$ may be productively thought of as linear subspaces of $\mathbb{R}^{n+1}$. If $\{ \mathbf{p}_1, \ldots, \mathbf{p}_k \}$ are affinely independent points in $\mathbb{R}^n$, they are a basis for the linear subspace of $\mathbb{R}^{n+1}$ spanned by the vectors $\{\mathbf{v}_1,\ldots,\mathbf{v}_k\}$, where $\mathbf{v}_i = (1,\mathbf{p}_i)$; that is, the 0th coordinate of $\mathbf{v}_i$ is 1 and the others are what they are in $\mathbf{p}_i$. One reason for setting the first coordinate to something nonzero is that if one of the $\mathbf{p}_i$ is the origin, thinking of it as a vector by adding a 0 is unproductive, since the zero vector is never part of a linearly independent set.

Thus the line in $\mathbb{R}^2$ through the points $(3,7)$ and $(4,5)$ corresponds to the linear subspace (a plane) in $\mathbb{R}^3$ spanned by the vectors $(1,3,7)$ and $(1,4,5)$. Indeed, when you look at this plane in $\mathbb{R}^3$, the places where it slices through the affine plane $x_0 = 1$ forms a line which passes through our given points.

If we take a column vector $\mathbf{x} = (x_0,x_1,x_2)$ and a $3\times3$ matrix $M$, the first coordinate of the vector $M\mathbf{x}$ is the dot product of $\mathbf{x}$ and the first row of $M$ (interpreted as a column vector for you sticklers). In other words, if the first row of $M$ is $(m_{00}, m_{01}, m_{02})$, we have that the last coordinate of $M\mathbf{x}$ is

$m_{00}x_0 + m_{01}x_1 + m_{02}x_2.$

Supposing that $\mathbf{x}$ determines a point on the affine plane $x_0 = 1$. For $M$ to preserve this property, we see that we must have $m_{01} = m_{02} = 0$ while $m_{00} = 1$. In other words, we can write $M$ in block form as

$M = \begin{pmatrix} 1 & \mathbf{0} \\ \mathbf{y} & A \end{pmatrix}$

where $A$ is a $2\times2$ matrix, $mathbf{y}$ is a column vector, and $\mathbf{0}$ is the all-zeroes row vector. Invertible matrices of this form are part of the affine group. This happens if and only if $A$ is invertible.

To be a rigid motion of Euclidean space, we want distances to be preserved. Recall that the distance between vectors $\mathbf{x}$ and $\mathbf{y}$ is

$d(\mathbf{x},\mathbf{y}) = \|\mathbf{x} - \mathbf{y}\| = \sqrt{(\mathbf{x} - \mathbf{y}) \cdot (\mathbf{x} - \mathbf{y})}.$

For vectors on the affine plane $x_0 = 1$, the distance between vectors corresponds to the usual Euclidean distance between points, since the subtraction of vectors means the presence or absence of $x_0$ does not change the calculation.

Doing a little matrix multiplication, we see that in $n$ dimensions, the image of the vector $(1,\mathbf{x})$ under $M$ as notated above is $(1,A\mathbf{x}+\mathbf{y}$. Therefore the matrix $M$ will preserve distances on the hyperplane $x_0 = 1$ just when the matrix $A$ satisfies

$A\mathbf{v} \cdot A\mathbf{w} = \mathbf{v} \cdot \mathbf{w}$

for all $n$-dimensional column vectors $\mathbf{v}$ and $\mathbf{w}$. In other words, the matrix $A$ turns out to be exactly one of the matrices from the previous section! That’s pretty neat.

Before going on, I want to point out one more thing: since $A^T A = A A^T = I$, doing block matrix multiplication, we see that

$M^T\begin{pmatrix}0 & \mathbf{0} \\ \mathbf{0} & I \end{pmatrix}M = \begin{pmatrix}1 & \mathbf{y}^T \\ \mathbf{0} & A^T\end{pmatrix}\begin{pmatrix}0 & \mathbf{0}\\ \mathbf{0} & A \end{pmatrix} = \begin{pmatrix} 0 & \mathbf{0} \\ \mathbf{0} & I \end{pmatrix}.$

So a matrix $M$ preserving the affine hyperplane $x_0 = 1$ is a rigid motion of Euclidean space if and only if it satisfies $M^T g M = g$, where $g$ is that funky block matrix in the middle above.

Hyperbolic space

Okay, this one is slightly weird but very very cool. Earlier, we stipulated that an $(n+1)times(n+1)$ matrix belonged to the symmetry group of $S^n$ when it satisfied the equation $M^T I M = I$. Crucially for the analogy I am making, the identity matrix has first diagonal entry 1. Just now, we saw that if we replace $I$ with the matrix $g$ which is diagonal with a 0 in the first diagonal entry and 1s otherwise, we get the rigid motions of Euclidean space thought of as an affine hyperplane (provided we also preserve the hyperplane). The curvature of Euclidean space is 0. It’s extremely tempting, therefore, to see what happens when we let the first diagonal entry be $-1$!

So let $Q$ be the diagonal matrix with first entry $-1$ and all other diagonal entries $1$, and suppose that $M$ satisfies $M^T Q M = Q$. Notice that $M$ preserves the (one-sheeted) hyperboloid comprising those points in $\mathbb{R}^{n+1}$ satisfying

$-x_0^2 + x_1^2 + \cdots + x_n^2 = 1;$

these are the vectors $\mathbf{x}$ for which $\mathbf{x}^T Q \mathbf{x} = 1$. But it actually also preserves the two-sheeted hyperboloid

$\mathbf{x}^T Q \mathbf{x} = -x_0^2 + x_1^2 + \cdots + x_n^2 = -1.$

One reason to prefer the two-sheeted hyperboloid is that it doesn’t have a “hole”, and another is that the two-sheeted hyperboloid contains the point $(1,\mathbf{0})$, just as the sphere and the affine plane do.

Since the matrix $Q$ is invertible, the equation $M^T Q M = Q$ implies that $M$ is invertible (this was not automatic for $g$ above). The group of matrices satisfying this equation is kind of curious: an element of this group may swap the two sheets of the hyperboloid, or it may fix them. Let’s suppose it fixes them, so that $\mathbf{x}_0 > 0$ implies $M\mathbf{x}_0 > 0$.

Just as we found the round metric on the $n$-sphere in $\mathbb{R}^{n+1}$, it would be nice to have a formula for the distance between two points on the hyperboloid. It’s a little tough to know precisely what we’re asking for here, but one reasonable thing to ask would be for linear subspaces of $\mathbb{R}^{n+1}$ to intersect one sheet of the hyperboloid in a convex set.

Let’s start in $\mathbb{R}^2$. Then the hyperbola is the set of points $(x,y)$ satisfying $x^2 - y^2 = 1$ and $x > 0$. There are many ways to parametrize the hyperbola, but I’d like to argue that the best one is maybe the weirdest:

$t \mapsto (\cosh t, \sinh t).$

The functions $\cosh$ and $\sinh$ are pronounced “kosh” (like as in “Oshkosh”) and “cinch” (like as in “it’ll be a cinch!” or if you say “sin” and follow it with a “ch” to make it rhyme with “clinch”), and you maybe met them once upon a time in Calculus. They’re defined as

$\cosh x = \frac{e^x + e^{-x}}{2} \quad\text{ and }\quad \frac{e^x - e^{-x}}{2}.$

Let’s do a little math, shall we? Notice that first of all if we write $a = e^x$, we can compute

$\cosh^2 x - \sinh^2 x = \frac{(a + \frac{1}{a})^2}{4} - \frac{(a - \frac{1}{a})}{4} = \frac{a^2 + 2 + \frac{1}{a^2}}{4} - \frac{a^2 - 2 + \frac{1}{a^2}}{4} = \frac{4}{4} = 1.$

This, and the calculation $\cosh 0 = 1$ shows that we really do parametrize the upper sheet of the hyperboloid.

The reason I prefer this parametrization is that the derivative of $\sinh t$ is (one checks) $\cosh t$ and vice versa. Thus the velocity vector of the curve $(\cosh t, \sinh t)$ is the unit vector $(\sinh t,\cosh t)$ when we use $\|\mathbf{x}\| = \sqrt{\mathbf{x}^TQ\mathbf{x}}$ as our funky way of measuring norms.

This is significant, because it says that with respect to this Minkowski norm, the curve $(\cosh t,\sinh t)$ is parametrized by arc length, so it makes sense to define the distance between two points $(\cosh s,\sinh s)$ and $(\cosh t,\sinh t)$ on the hyperbola as $|s - t|$!

Doing a little computation, note that we have

$\cosh s \cosh t - \sinh s \sinh t = \frac{e^{t - s} + e^{s - t}}{2} = \cosh (t - s).$

In fact this generalizes: the distance between $\mathbf{x}$ and $\mathbf{y}$ is $\operatorname{arccosh} |\mathbf{x}^T Q \mathbf{y}|$.

As it turns out, this metric on the hyperboloid turns it into a Riemannian manifold of dimension $n$ and constant sectional curvature $-1$. Since the embedding of the hyperboloid into $\mathbb{R}^{n+1}$ is not isometric (in the sense of Riemannian geometry, when $\mathbb{R}^{n+1}$ has the ordinary Riemannian metric), we can’t compute this directly from the definition of the hyperboloid, so the curvature tensor turns out to be actually useful.

There’s so much to say here (for one thing, I didn’t even sketch the computation above), but this post is long in the tooth as it is, so I’ll sign off here.