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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 $<s>$ is dense in G.

$<s>$ denotes the subgroup of G generated by S.

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  • $\begingroup$ My guess is $n=3$. $\endgroup$ – Gae. S. Jan 10 '20 at 1:12
  • $\begingroup$ what is your intuition for this? $\endgroup$ – Mathew Jan 10 '20 at 1:12
  • $\begingroup$ 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. $\endgroup$ – Gae. S. Jan 10 '20 at 1:14
  • $\begingroup$ I highly doubt it to be possible with less than $3$, but I've been wrong before. $\endgroup$ – Gae. S. Jan 10 '20 at 1:18
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    $\begingroup$ Interesting problem. Got me curious and I found this: heldermann-verlag.de/jlt/jlt02/MORRPL.PDF . So, the answer is 2. $\endgroup$ – Robert Bell Jan 10 '20 at 2:43
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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: enter image description here

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.

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