# Find the number of elements needed to generate a dense subset of $SO_4(\mathbb{R})$

Consider the topological group $$G=SO_4(\mathbb{R})$$

Find the smallest $$n \in \mathbb{N}$$ such that $$\exists s=\{g_1,g_2...g_n\} \subset G$$ for which $$$$ is dense in G.

$$$$ denotes the subgroup of G generated by S.

• My guess is $n=3$. – Gae. S. Jan 10 '20 at 1:12
• what is your intuition for this? – Mathew Jan 10 '20 at 1:12
• It's $n=1$ for $SO(\Bbb R^2)$ because you need just one suitable rotation to get close to all rotations. $SO(\Bbb R^2)$ needs just the rotations along two different axis, and I guess it goes on. – Gae. S. Jan 10 '20 at 1:14
• I highly doubt it to be possible with less than $3$, but I've been wrong before. – Gae. S. Jan 10 '20 at 1:18
• Interesting problem. Got me curious and I found this: heldermann-verlag.de/jlt/jlt02/MORRPL.PDF . So, the answer is 2. – Robert Bell Jan 10 '20 at 2:43

To save the labor, I will write $$SO_4$$ for $$\operatorname{SO}_4$$ and write $$so_4$$ for $$\mathfrak{so}_4$$.

The answer is $$2$$.

We first note that a general element of $$SO_4(\Bbb R)$$ is a double rotation, i.e. if $$g \in SO_4(\Bbb R)$$, then there is an orthogonal decomposition $$\Bbb R^4 = V \oplus W$$, such that $$g$$ is the composition of a rotation on $$V$$ and a rotation on $$W$$.

This means that a single element $$g$$ cannot generate a dense subgroup of $$SO_4(\Bbb R)$$, because $$g^n(V) \subseteq V$$ for any $$n\in \Bbb Z$$, and (since $$V$$ is closed) hence $$h(V)\subseteq V$$ for any $$h$$ in the closure of the subgroup generated by $$g$$.

Now it suffices to show that two elements can generate a dense subgroup of $$SO_4(\Bbb R)$$.

It is easier to work with the Lie algebra. Recall that the Lie algebra $$so_4(\Bbb R)$$ is $$6$$-dimensional, with basis $$(A_i, B_i)_{i = 1, 2, 3}$$, as defined here. Below is a screen shot of the matrices:

The Lie bracket is given, for $$\{i, j, k\} = \{1, 2, 3\}$$, by:

• $$[A_i, A_i] = [A_i, B_i] = [B_i, B_i] = 0$$;
• $$[A_i, A_j] = \pm A_k$$;
• $$[A_i, B_j] = \pm B_k$$;
• $$[B_i, B_j] = \pm A_k$$.

The signs $$\pm$$ are not quite important for us.

Now pick any two numbers $$\lambda, \mu\in \Bbb R$$ such that $$2\pi, \lambda, \mu$$ are linearly independent over $$\Bbb Q$$. Let $$X$$ be the matrix $$\begin{pmatrix} \cos\mu & 0 & 0 & -\sin\mu\\ 0 & \cos\lambda & -\sin\lambda & 0\\ 0 & \sin\lambda & \cos\lambda & 0\\ \sin\mu & 0 & 0 & \cos\mu \end{pmatrix}.$$ In other words, $$X$$ is a double rotation with respect to the decomposition $$\Bbb R^4 = V\oplus W$$, where $$V = \Bbb Re_2 \oplus \Bbb Re_3$$ and $$W = \Bbb Re_1 \oplus \Bbb Re_4$$.

If we look at the group $$H$$ of all double rotations with respect to the decomposition $$V\oplus W$$, then it is clearly isomorphic to $$(\Bbb R/2\pi \Bbb Z)^2$$ with the obvious isomorphism, and our element $$X$$ corresponds to the element $$(\lambda, \mu)$$ under this isomorphism. By our choice of $$\lambda, \mu$$, we see that $$X$$ generates a dense subgroup of $$H$$.

Now pick another matrix $$Y$$ as follows $$\begin{pmatrix} \cos\lambda & 0 & -\sin\lambda & 0\\ 0 & 0 & 0 & 0\\ \sin\lambda & 0 & \cos\lambda & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}.$$ The number $$\lambda$$ could be the same as before, or could be any irrational real number. For similar reason, $$Y$$ generates a dense subgroup of $$K$$, which is all rotations on the plane $$\Bbb R_1 \oplus \Bbb R_3$$.

We now prove that $$X$$ and $$Y$$ generates a dense subgroup of $$SO_4(\Bbb R)$$.

Let $$G$$ be the closure of the subgroup generated by $$X$$ and $$Y$$. Then $$G$$ is a closed subgroup of $$SO_4(\Bbb R)$$, and hence its Lie algebra $$g$$ is a Lie subalgebra of $$so_4(\Bbb R)$$.

But we already see that $$G$$ contains $$H$$ and $$K$$, and hence $$g$$ contains the corresponding Lie subalgebras $$h$$ and $$k$$.

It is clear that $$h$$ has dimension $$2$$, with generators $$A_1$$ and $$B_1$$; and $$k$$ has dimension $$1$$, with generator $$A_2$$. From the Lie bracket formulas, we see that these three generate the whole $$so_4(\Bbb R)$$.

Hence $$g$$ is the whole $$so_4(\Bbb R)$$ and we are done.