Show that $SO_{3}$ contains the free group of rank 2 as a subgroup.

Question

I'm trying to prove that $SO_{3}$ contains the free group of rank 2, $F_{2}$, as a subgroup, by showing there are two rotations in $SO_{3}$ that are independent and hence generate $F_{2}$. I came across the following proof of this online:

$$ A(x,y,z) := (\frac{3}{5}x + \frac{4}{5} y, -\frac{4}{5} x + \frac{3}{5} y, z ); \quad B(x,y,z) := (x, \frac{3}{5}y + \frac{4}{5} z, -\frac{4}{5} y + \frac{3}{5} z ).$$ These are easily seen to be rotation matrices with inverses $$ A^{-1}(x,y,z) := (\frac{3}{5}x - \frac{4}{5} y, \frac{4}{5} x + \frac{3}{5} y, z ); \quad B^{-1}(x,y,z) := (x, \frac{3}{5}y - \frac{4}{5} z, \frac{4}{5} y + \frac{3}{5} z ).$$ Now we claim that no non-trivial composition of $A, B, A^{-1}, B^{-1}$ gives the identity. It suffices to show that no non-trivial composition of the operators $5A$, $5B$, $5A^{-1}$, $5B^{-1}$ gives a linear operator whose coefficients are all divisible by 5. We now work in the finite field geometry $F_5^3$, where $F_5 = \mathbb{Z}/5\mathbb{Z}$ is the field of order 5. Then we have $$ 5A(x,y,z) := (3x + 4y, -4x + 3y, 0); \quad 5B(x,y,z) := (0, 3y + 4z, -4y + 3z)$$ and $$ 5A^{-1}(x,y,z) := (3x - 4y, 4x + 3y, 0); \quad 5B^{-1}(x,y,z) := (0, 3y - 4z, 4y + 3z).$$ Each of these operators are rank one operators in $F_5^3$: \begin{align*} \text{range}(5A) &= \text{span}( (3,-4,0) ) = \ker(5A^{-1})^\perp\\ \text{range}(5A^{-1}) &= \text{span}( (3,4,0) ) = \ker(5A)^\perp\\ \text{range}(5B) &= \text{span}( (0,3,-4) ) = \ker(5B^{-1})^\perp\\ \text{range}(5B^{-1}) &= \text{span}( (0,3,4) ) = \ker(5B)^\perp. \end{align*} From this we see that any non-trivial combination of $5A$, $5A^{-1}$, $5B$, $5B^{-1}$ (in which $5A$ and $5A^{-1}$ are never adjacent, and $5B$ and $5B^{-1}$ are never adjacent) will always be a non-zero operator, as desired, because the ranges and kernels are skew.

I can't quite understand the proof from 'Each of these operators are rank one operators in $F_5^3$' onwards.

I would really appreciate it if someone could explain the proof in a way that is relatively easy to understand. Thanks for any help.

Answer 1

1. Before moving on to the actual answer I'll start with a quible about notation. As indicated by the use of the ":=" rather than "=" signs in the quoted proof, the operators $5A, 5A^{-1}, 5B, 5B^{-1}$ on $\mathbb{F}_5^3$ are not in a meaningful way 'the same' as the operators $5A$, $5B$, $5A^{-1}$ and $5B^{-1}$ on $\mathbb{R}^3$ that we started with. Also the operator $5A$ acting on $\mathbb{F}_5^3$ is not in any way the scalar 5 multiplied with another, invertible, operator named $A$ acting on $\mathbb{F}_5^3$ (since multiplying such an operator with 5 would result in the zero matrix, given that $5 = 0$ in this world) let alone that this non-existent operator would have an inverse $A^{-1}$ which, after multiplying with 5 (hence 0) would yield the matrix acting on $\mathbb{F}_5^3$ that is here denoted $5A^{-1}$. In other words the notation $5A, 5A^{-1}, 5B, 5B^{-1}$ on $\mathbb{F}_5^3$ is a bit misleading. To make the distinction clearer I will write $5A, 5A^{-1}, 5B, 5B^{-1}$ for the operators on $\mathbb{R}^3$ and $[5A], [5A^{-1}], [5B], [5B^{-1}]$ for the operators on $\mathbb{F}_5^3$. Note that by contrast $5A$ actually is 5 times the operator $A$ on $\mathbb{R}^3$.

2. The reason for moving to $\mathbb{F}_5^3$ is that we have:

Statement 1: suppose that operator $X$ on $\mathbb{R}^3$ is a product of $5A$, $5B$, $5A^{-1}$ and $5B^{-1}$ all whose matrix coefficient are divisible by 5, then the corresponding product of $[5A]$, $[5B]$, $[5A^{-1}]$ and $[5B^{-1}]$ is the zero-operator on $\mathbb{F}_5^3$.

(Since $5A, 5B, 5A^{-1}, 5B^{-1}$ have integer coefficients so does $X$ and hence the question of whether these coefficients are divisible by 5 makes sense.) This falls under the part of the proof you said you understood so I won't say much more about it, but please let me know if something here is unclear.

3. The objective of the part of the proof that you say you don't understand is to prove the following statement:

Statement 2. No product of $[5A], [5A^{-1}], [5B], [5B^{-1}]$, no matter how long or complicated will equal the zero operator on $\mathbb{F}_5^3$ as long as we take care that in our product $[5A]$ is never directly adjacent to $[5A^{-1}]$ and $[5B]$ is never directly adjacent to $[5B^{-1}]$.

Before explaining the proof of statement 2 let's see why it is useful. Combining it with statement 1 gives us:

Statement 3. No product of $5A, 5A^{-1}, 5B, 5B^{-1}$ (which is an operator on $\mathbb{R}^3$ with integer coefficients), no matter how long or complicated will have all its coefficients divisible by 5, as long as we take care that in our product $5A$ is never directly adjacent to $5A^{-1}$ and $5B$ is never directly adjacent to $5B^{-1}$.

Now our goal was to prove:

Statement 4a. No product of $A, A^{-1}, B, B^{-1}$, no matter how long or complicated will equal the identity operator on $\mathbb{R}^3$, as long as we take care that in our product $A$ is never directly adjacent to $A^{-1}$ and $B$ is never directly adjacent to $B^{-1}$.

Which is equivalent to

Statement 4b. No product of $n$ elements from $5A, 5A^{-1}, 5B, 5B^{-1}$, no matter how complicated will equal $5^n$ times the identity operator on $\mathbb{R}^3$, as long as we take care that in our product $5A$ is never directly adjacent to $5A^{-1}$ and $5B$ is never directly adjacent to $5B^{-1}$.

Now it is clear that statement 3 implies statement 4b and hence 4a. In fact it seems that statement 3 is a bit of an overkill, giving us much more than we need. Since statement 2 implies statement 3 we can say that statement 2 is much much stronger than our desired statement 4. At the same time it also sounds much much easier than statement 4, talking only about finite stuff and working with products (not) being zero rather than not being the identity. This paradox is what I like about the proof. (Of course sounding easy is subjective, our opinions on this may differ)

4. The proof of statement 2, part 1.

The text you quote claims

Statement 5a. Let $\mathbf{x} := (x, y, z) \in \mathbb{F}_5^3$ be an arbitrary vector. Then, no matter how cleverly you chose $x, y$ and $z$ the vector $[5A]\mathbf{x}$ will always lie in the one dimensional subspace spanned by the vector $(3, -4, 0) \in \mathbb{F}_5^3$, in other words, there will always be some scalar $p \in \mathbb{F}_5$ such that $[5A]\mathbf{x} = (3p, -4p, 0)$.

There are also statements $5b, 5c, 5d$ making similar claims about $[5A^{-1}], [5B], [5B^{-1}]$.

Let's see if we can understand the proof of this. In $\mathbb{F}_5$, dividing by 3 is the same as multiplying by 2. So from looking at the first coefficient of $[5A]\mathbf{x}$, which equals $3x - 4y$, I conclude that if a number $p$ as in statement 5a exists then it must be equal to $6x - 8y = x - 3y$. So we can take the flight forward and make a stronger claim:

Statement 5a'. Let $\mathbf{x} := (x, y, z) \in \mathbb{F}_5^3$ be an arbitrary vector. Then, no matter how cleverly you chose $x, y$ and $z$ the vector $[5A]\mathbf{x}$ will always equal the vector $(x - 3y)(3, -4, 0)$ in $\mathbb{F}_5^3$.

Of course statement 5a' still needs proof, but now it is easy. We can just compute the vector $[5A]\mathbf{x}$ from the definition and see that statement 5a' is correct.

In a similar way we can prove $5b, 5c, 5d$ telling us that $[5A], [5A^{-1}], [5B], [5B^{-1}]$ each have a teeny tiny one-dimensional subspace into which they map every vector they can lay their hands on.

6. Proof of statement 2, part 2.

We want to show that a product of elements from $[5A], [5B], [5A^{-1}], [5B^{-1}]$ of the right form is not the zero operator, i.e. that there always is some vector $\mathbb{x} \in \mathbb{F}_5^3$ that is not mapped to zero. We can work with induction on the length of our product. There are only four products of length 1 and for each of them you can check by hand that there are at least some $(x, y, z)$ not mapped to zero.

Now let's deal with products $X$ of length $n \geq 2$. It can always written in one of the following four forms:

$$Y[5A], Y[5A^{-1}], Y[5B], Y[5B^{-1}]$$

where $Y$ is a product of our operators of length $n - 1$. The nice thing (looking just at the first case where $X = Y[5A]$) now is that $X\mathbf{x} = Yp(3, -4, 0)$ for some scalar $p$. Had we chosen $\mathbf{x}$ in such a way that $[5A]\mathbf{x}$ is non-zero (and we know we could have done that) then we even get that $X\mathbf{x} = Yp(3, -4, 0) = pY(3, -4, 0)$ for some non-zero scalar $p$. We would conclude that $X\mathbf{x}$ is non-zero as soon as we had some magical way (i.e. some sort of induction hypothesis) that told us that $Y(3, -4, 0)$ is non-zero. Well now it is clear what we need to do. Instead of the single statement 2 we prove the four more complicated statements 6a, 6b, 6c, 6d below which have the pleasant property that they can be proven using induction:

Statement 6a. Let $Y$ be a product of $n$ terms from $[5A], [5B], [5A^{-1}], [5B^{-1}]$ the rightmost of which is not $[5A^{-1}]$. Then $Y(3, -4, 0)$ is non-zero.

Statement 6b. Let $Y$ be a product of $n$ terms from $[5A], [5B], [5A^{-1}], [5B^{-1}]$ the rightmost of which is not $[5A]$. Then $Y(3, 4, 0)$ is non-zero.

Statement 6c. Let $Y$ be a product of $n$ terms from $[5A], [5B], [5A^{-1}], [5B^{-1}]$ the rightmost of which is not $[5B^{-1}]$. Then $Y(0, 3, -4)$ is non-zero.

Statement 6d. Let $Y$ be a product of $n$ terms from $[5A], [5B], [5A^{-1}], [5B^{-1}]$ the rightmost of which is not $[5A]$. Then $Y(0, 3, 4)$ is non-zero.

The proof is simple but slightly annoying. To show 6a we need to distinguish 3 cases: $Y$ ends in $[5A]$, $Y$ ends in $[5B]$ and $Y$ ends in $[5B^{-1}]$. I'll demonstrate the second of these cases and leave the remaining 2 (or rather 11) to you. 'Ending in $[5B]$' means that $Y = Z[5B]$ for some product $Z$ of length $n-1$ (length 0 means $Z$ is the identity operator). Now $Y(3, -4, 0) = Z[5B](3, -4, 0) = Z(0, 9, -12) = 3Z(0, 3, -4)$. Now apply induction hypothesis 6c with $Z$ in the role of $Y$ and $n-1$ in the role of $n$.

(We need of course to first check the $n=1$ cases of these statements as well, our proof of the $n=1$-case of statement 2 above might not be sufficient)

Thus having proven statements 6abcd of all $n$ it follows as above that we have proven the more pleasant sounding statement 2 for all $n$ as well.