Do the rotations in a single plane, and a single other rotation, generate all rotations?

Question

In $n>2$ dimensions, suppose we have a (possibly non-planar) rotation matrix $A\in SO(n)$, and a 1-parameter family of Givens rotations

$$G(\theta)=G_{21}(\theta)=\begin{bmatrix}\cos\theta&-\sin\theta&0&\cdots&0\\\sin\theta&\cos\theta&0&\cdots&0\\0&0&1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&1\end{bmatrix},\qquad\theta\in\mathbb R.$$

I expect that these generate the special orthogonal group (as a monoid), so any other rotation matrix $X\in SO(n)$ can be written as

$$X=G(\theta_k)\,A\,G(\theta_{k-1})\,\cdots\,A\,G(\theta_3)\,A\,G(\theta_2)\,A\,G(\theta_1)$$

for some angles $\theta_1,\cdots,\theta_k$.

Of course it won't always work, e.g. if $A$ is the identity, or if $A$ fixes (setwise) the plane $\mathbb R^2\subset\mathbb R^n$ or fixes a non-zero subspace orthogonal to that plane. But it should "almost always" work, I think. Can we make this more precise? Is there a simple condition on $A$ that guarantees generation of all of $SO(n)$?

In particular, does it work if $A$ doesn't fix any non-zero subspace orthogonal to $\mathbb R^2$? (We can focus on 1D and 2D subspaces, since if $A$ fixes a $k$D subspace with $k>2$ then it acts like $A\in O(k)$ which always has a 2D fixed subspace.)

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As pointed out by @DavidK, we can take $A$ to cycle the axes:

$$A=\begin{bmatrix}0&0&0&0&\cdots&0&(-1)^{n-1}\\1&0&0&0&\cdots&0&0\\0&1&0&0&\cdots&0&0\\0&0&1&0&\cdots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\cdots&0&0\\0&0&0&0&\cdots&1&0\end{bmatrix}$$

Then conjugation with $A$ produces other Givens rotations, such as $A\,G_{21}(\theta)\,A^{-1}=G_{32}(\theta)$. (Note that $A^{-1}=A^{2n-1}$ is in the monoid generated by $A$.) And these can be combined to make any rotation $X$, using Euler angles.

However, that is just one particular $A$, not "almost any" $A$.

Answer 1

This is true even in a bit greater generality. Suppose that $G$ is a connected compact simple Lie group (and $SO(n)$ satisfies all these properties). Let $H< G$ be a 1-parameter subgroup. Then for generic elements $g\in G$, the group $G$ is generated by $H$ and $g$. Here a subset of $G$ is "generic" if it is nonempty and its complement is given by a finite system of polynomial equations (actually, one equation suffices). In particular, a generic subset is open, dense and has full measure in $G$.

Here is one way to prove this. Let ${\mathfrak g}$ denote the Lie algebra of $G$ (for you, this is the space of skew-symmetric $n\times n$ matrices equipped with the matrix commutator $XY-YX$ serving as the Lie bracket). Let $\xi\in {\mathfrak g}$ denote a generator of the (1-dimensional) Lie algebra of $H$. (Physicists usually say that $\xi$ is a "generator of $H$" but this is an abuse of terminology.) The group $G$ acts on its Lie algebra via the adjoint representation, for you, this simply means: $$ Ad_g(\eta)= g \eta g^{-1}, g\in G, \eta\in {\mathfrak g}. $$ The key is that given any nonzero vector $\eta\in {\mathfrak g}$, for generic elements $g\in G$, the $Ad_{\langle g\rangle}$-orbit of $\eta$ spans ${\mathfrak g}$ as a vector space: $$ Ad_{\langle g\rangle}(\eta)=\{Ad_{g^k}(\eta): k\in \mathbb Z\}. $$
(This is where I use the assumption that $G$ is a simple Lie group: The adjoint representation is irreducible.) For $\eta=\xi$ as above, the elements $Ad_{g^k}(\eta)$ are generators of Lie algebras of conjugate subgroups $$ H_k:=g^k H g^{-k}< G. $$

What is this good for? Suppose that $\eta_0,...,\eta_{N-1}$ are basis elements of ${\mathfrak g}$, where $N=\dim(G)$. Consider the map $$ F: \mathbb R^N\to G, F(t_0,...,t_{N-1})= \exp(t_0\eta_0)\cdots \exp(t_{N-1}\eta_{N-1}). $$ The differential of this map at $\mathbf 0$ sends a vector $(t_0,...,t_{N-1})$ to the linear combination $$ \sum_{i=0}^{N-1} t_i \eta_i. $$ Thus, the differential is a linear isomorphism. By the Inverse Function Theorem, it follows that the image of $F$ contains a neighborhood $W$ of $1\in G$. In our situation, we take $\eta_i=Ad_{g^i}(\xi)$, $i=0,...,N-1$. Then the image of $F$ is the image of the product map $$ H_0\times ...\times H_{N-1}\to G, $$ $(h_0,...,h_{N-1})\mapsto h_0....h_{N-1}$.

Since $G$ is connected, the subset $W$ generates the entire group $G$. Hence, the subgroups $H_0,...,H_{N-1}$ generate $G$. But these subgroups are conjugates of $H$ under powers of $g$. Hence, $H$ and $g$ generate $G$.

Answer 2

Let $Q$ be the $90^\circ$ rotation in $yz$-plane taking $(0,0,1)$ to $(0,1,0)$. Then $Q$ together with $SO(2)$, rotations in the $xy$-plane generate all of $SO(3)$. In what follows $G(3)$ denotes the subgroup generated by $Q$ and $SO(2)$,

First of all any rotation in the $xz$-plane can be obtained by conjugating a rotation in $SO(2)$ by $Q$. Now let $v$ be a point on the unit sphere with spherical coordinates $(1,\theta,\phi)$. Then a rotation through angle $\phi$ in the $xz$-plane takes $(0,0,1)$ to the point with spherical coordinates $(1,0,\phi)$. Then the rotation through angle $\theta$ in $SO(2)$ takes $(1,0,\phi)$ to $v$. Thus we have shown that there is a rotation $R$ in $G(3)$ which takes $(0,0,1)$ to any point $v$ in the unit sphere. Then we can obtain any rotation with radial axis $v$ by conjugating elements of $SO(2)$ by $R$. Hence $G(3)=SO(3)$.

If $n>4$ it is impossible for the set of all rotations within a given plane $P_1$ together with a single rotation in another plane $P_2$ to generate all of $SO(n)$. The reason is that any element of the resulting subgroup $G(n)$ has to fix all vectors in the orthogonal complement of $P_1+P_2$. Since we can find elements of $SO(n)$ which move any nonzero vector, we must have $G(n)\subsetneqq SO(n)$. I am not sure whether it may be true for $n=4$.