2
$\begingroup$

I know that the inner automorphisms of the binary icosahedral group is given by the quotient $2I/\{\pm 1\}$ where $\{\pm 1\}$ is the center of $2I$. I also know the elements of the binary icosahedral group expressed in terms of quaternions.

Using this quaternionic representation (or otherwise), is there a simple way to show that $2I/\{\pm 1\} \cong A_5$?

$\endgroup$
7
  • $\begingroup$ en.wikipedia.org/wiki/Icosahedral_symmetry#Group_structure — briefly, it acts in the natural way on various compounds of five solids inscribed in the dodecahedron. $\endgroup$ May 7 '20 at 22:34
  • $\begingroup$ Or the five choices of three golden rectangles in an icosahedron. $\endgroup$
    – runway44
    May 8 '20 at 2:26
  • $\begingroup$ @StevenStadnicki That link discusses the icosahedral group only, while I'm asking about the binary icosahedral group $\endgroup$
    – Sky
    May 8 '20 at 6:58
  • $\begingroup$ But the quotient of the binary icosahedral group is the icosahedral group... $\endgroup$
    – runway44
    May 8 '20 at 8:16
  • $\begingroup$ @runway44 Yeah, but that's what I'm looking for an explanation of. Why is the quotient isomorphic to the icosahedral group? $\endgroup$
    – Sky
    May 8 '20 at 11:34
2
$\begingroup$

I assume you're familiar with quaternions. The key idea here is how they model 3D rotations. Given a unit vector $\mathbf{u}$ there is Euler's formula $p=\exp(\theta\mathbf{u})=\cos(\theta)+\sin(\theta)\mathbf{u}$, and the conjugate $p\mathbf{v}p^{-1}$ is $\mathbf{v}$ rotated around the oriented axis in the $\mathbf{u}$-direction by the angle $2\theta$. Often we speak of the quaternion $p$ and the corresponding rotation (conjugation-by-$p$) interchangeably, although $\pm p$ give the same rotation and are distinct quaternions.

The map $S^3\to\mathrm{SO}(3)$ has kernel $\{\pm1\}$ (not difficult to check). If $I$ is the icosahedral group (the rotational symmetry group of the regular icosahedron) then its preimage $2I$ has twice the size. By the orbit-stabilizer theorem, $I$ has size $|I|=5\cdot12=2\cdot30=3\cdot20=60$ (you can have $I$ act on $V$, $E$ or $F$). Once we have $120$ elements of $2I$ we have them all.

A regular icosahedron can be constructed from three perpendicular golden rectangles:

$\hskip 1in$ icosahedron

The vertices are the $2^2\cdot3=12$ cyclic permutations of $(0,\pm1,\pm\varphi)$ where $\varphi$ is the golden ratio. The $0$ coordinates are related to the fact three golden rectangles are chosen on coordinate planes, and the $2^2$ signs, corresponding to the four vertices of each rectangle, correspond to different coordinate plane reflections. To check this is a regular polyhedron it suffices to check the rectangles' smaller side length matches the distance between close corners of different rectangles (any two, by symmetry). The golden ratio $\varphi$ can be checked as the unique proportion that makes this true, using the distance formula.

Given a vertex, the five adjacent vertices can be found by either changing the sign of the $\pm1$, cycling the coordinates, or simultaneously cycling and changing the sign the coordinate where $0$ was. (I got this by checking the neighbors of $(0,1,\varphi)$ and generalizing.)

Any compound of three inscribed perpendicular golden rectangles is determined by a single edge of the icosahedron, so these compounds may be identified with the edges adjacent to a single vertex, hence there are five such compounds. (Personally, I pick the closest top corner of the purple rectangle in the above picture.) The stabilizer of the coordinate rectangle compound is $V_4\rtimes C_3$, where the Klein-four group $V_4$ is comprised of diagonal matrices $(\pm1,\pm1,\pm1)$ with determinant $+1$ and $C_3$ is comprised of cycle permutation matrices. Acting on the four other edges adjacent to the chosen vertex, the $V_4$ and $C_3$ act in the usual way $A_4$ (itself a semidirect product $V_4\rtimes C_3$) does on four points. From this we conclude that $I$ acts as the alternating group $A_5$ on the five golden rectangle compounds.

So far we know $2I/\{\pm1\}\cong A_5$. Let's find the elements of $2I$.

The aforementioned stabilizer has a double cover of size $24$ in $2I$. The kernel $\{\pm1\}$ is automatically present, the $180^{\circ}$ axis rotations correspond to $\pm\mathbf{i},\pm\mathbf{j},\pm\mathbf{k}$; together these form the quaternion group $Q_8$. A cyclic rotation around a diagonal line in the $\mathbf{u}$-direction ($\mathbf{u}$ normalized) by $\theta$ corresponds in $S^3$ to the unit quaternions

$$ \pm\exp\left(\frac{\theta}{2}\mathbf{u}\right)=\pm\exp\left(\frac{1}{2}\frac{2\pi}{3}\frac{\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k}}{\sqrt{3}}\right)$$

$$=\pm\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\frac{\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k}}{\sqrt{3}}\right)=\frac{1}{2}(\pm1\pm\mathbf{i}\pm\mathbf{j}\pm\mathbf{k}). $$

Combined with $Q_8$, these form the vertices of the $24$-cell. These elements have order $3$ or $6$ (corresponding to real parts $\mp1/2$ resp.) in $2I$, which in the quotient $I$ have order $3$ and thus correspond to face rotations. We can also pick an edge not parallel to a coordinate axis and find a different $180^{\circ}$ rotation. Since $\varphi^{-1}=\varphi-1$, let's pick $(1,\varphi,0)$ and $(0,-1,\varphi)$ whose midpoint is $\frac{1}{2}(1,\varphi^{-1},\varphi)$ and the corresponding quaternion in $2I$ is the pure imaginary $\frac{1}{2}(\mathbf{i}+\varphi^{-1}\mathbf{j}+\varphi\mathbf{k})$ (check this is a unit vector; this can be swift if you already know the formula $\varphi^n=F_n\varphi+F_{n-1}$ and how to find $F_n$ for negative $n$).

By conjugating by any of $\mathbf{i},\mathbf{j},\mathbf{k}$ we can change two signs at a time, hence get all possible signs, and by multiplying by them on the left or right we swap two pairs of components (including real) while also changing signs, so starting from the three cyclic permutations of $(0,1,\varphi)$ we get all even permutations of the quaternions $\frac{1}{2}(\pm\mathbf{i}\pm\varphi^{-1}\mathbf{j}\pm\varphi\mathbf{k})$. Of these, the ones with real parts $0$ are $180^{\circ}$ rotations, with real parts $\pm\frac{1}{2}$ are $120^{\circ}$ rotations, with real parts $\pm\frac{1}{2}\varphi$ must be the $72^{\circ}$ face rotations and with real parts $\pm\frac{1}{2}\varphi^{-1}$ must be the $144^{\circ}$ face rotations. (Recall with Euler's formula, bigger real part means smaller angle.)

Now we have $2I/\{\pm1\}\cong I$ by definition, $I\cong A_5$ by considering the action on five triple golden rectangle compounds, and have determined all $120$ elements of $2I$ in $S^3$ explicitly in terms of $\varphi$.

I have tried to make all the logic here self-contained, if terse; there shouldn't be anything you need to take on faith. I can clarify any statement or step on request.


We can also take the elements of $2I$, turn them into $2\times2$ complex matrices in the standard way (since $S^3\cong\mathrm{SU}(2)$) and then mod by $5$ to get elements of $\mathrm{SL}_2(\mathbb{F}_5)$. Since $-1=2^2$ in $\mathbb{F}_3$, the $i$s in the matrices become $2$, and the determinant is preserved by modding. (Note $5$ is the smallest prime we can mod by in which $0,\pm1,\pm i$ are all distinct.) Also the $\sqrt{5}$s in the golden ratio just drop out. By checking sizes and kernel we see that $\overline{2T}=\mathrm{SL}_2(\mathbb{F}_5)$. Since the kernel of $2T\to T$ (modded out) matches the kernel of $\mathrm{SL}_2(\mathbb{F}_5)\to\mathrm{PSL}_2(\mathbb{F}_5)$, we may also conclude $I\cong\mathrm{PSL}_2(\mathbb{F}_5)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.