Vectorfields satisfying SO(3) algebra 2 dimensional?

Question

Let $M$ be an $n$-dimensional smooth connected manifold and $V_1$, $V_2$ and $V_3$ vectorfields whose commutators satisfy the $SO(3)$ algebra: \begin{equation} [V_1,V_2]=V_3 \end{equation} \begin{equation} [V_2,V_3]=V_1 \end{equation} \begin{equation} [V_3,V_1]=V_2 \end{equation} Is the vectorspace $V_x = \langle V_1(x),V_2(x),V_3(x)\rangle \subset T_x M$ at most two-dimensional $\forall x \in M$? If so, is any compact submanifold $N \subset M$ that is an integral manifold of $\langle V_1,V_2,V_3 \rangle$ homeomorphic to a sphere?

See e.g. https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html to appreciate the relevance of this question.

EDIT: The answer from Andrew can in fact also be used to negatively answer the same question as before but now with $(M,g)$ a three dimensional (semi)-Riemannian manifold of our choice where $V_1$, $V_2$ and $V_3$ are required to be Killing vectorfields. We can endow $\mathbb{R}^4$ with the metric $ds^2=dx^2+dy^2+dz^2+dv^2$ for which the vectorfields $V_i$ indeed generate isometries (Killing vectorfields). The three vectorfields are also tangent to $S^3\subset \mathbb{R}^4$. Hence, taking $(M,g)=(S^3,\gamma)$ where $\gamma$ is the induced metric and where we take the induced vectorfields $W_1=V_1\left.\right|_{S^3},W_2=V_2\left.\right|_{S^3},W_3=V_3\left.\right|_{S^3}$. $W_1$, $W_2$ and $W_3$ are Killing vectorfields on $(M,g)$ which span the full 3-dimensional tangent space of $M$ at every point $x$. This follows from \begin{equation} \det((r\partial_r,V_1,V_2,V_3)^t)=\det\begin{pmatrix} x & y & z & v \\ -y & x & v & -z \\ -z & -v & x & y \\ -v & z & -y & x\end{pmatrix}=(x^2+y^2+z^2+v^2)^2. \end{equation}

Answer 1

$\newcommand{\Lie}[1]{#1}\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}\newcommand{\Cpx}{\mathbf{C}}$The Lie algebra $\Lie{su}(2)$ is isomorphic to the Lie algebra $\Lie{so}(3)$ because $SU(2)$ double-covers $SO(3)$. The left action of $SU(2)$ on $\Cpx^{2} \simeq \Reals^{4}$ has three-dimensional orbits, which (assuming I understand your question) shows that $V_{x}$ can be three-dimensional.

A concrete example of three linear independent vectorfields obtained in this way is \begin{equation} V_1:\mathbb{R}^4 \to \mathbb{R}^4:(x,y,z,v)\mapsto\frac{1}{2}(-y,x,v,-z) \end{equation} \begin{equation} V_2:(x,y,z,v)\mapsto\frac{1}{2}(-z,-v,x,y) \end{equation} \begin{equation} V_3:(x,y,z,v)\mapsto\frac{1}{2}(-v,z,-y,x). \end{equation}

Answer 2

Even when $n=3$, the system \begin{equation} \begin{cases} & [V_1,V_2]=V_3 \\ & [V_2,V_3]=V_1 \\ & [V_3,V_1]=V_2 \end{cases} \end{equation} is way too undetermined to have $\dim \langle V_1(x),V_2(x),V_3(x)\rangle <3$ everywhere. Take for instance the manifold $M=\mathbb{R}^3\setminus L$, where $L$ is the vertical line determined by $x^2+y^2=0$, and the following three vectorfields (which are globally defined): \begin{equation} \begin{cases} & V_1(x,y,z)=-y\partial_x+x\partial_y = \partial_{\varphi} \\ & V_2(x,y,z)=(-z\partial_y+y\partial_z)+\lambda \frac{y}{(x^2+y^2)^{\frac{1}{2}}}(x\partial_x+y\partial_y+z\partial_z) = (-z\partial_y+y\partial_z) + \lambda \sin(\varphi)r\partial_r\\ & V_3(x,y,z)=(-x\partial_z+z\partial_x )+\lambda \frac{x}{(x^2+y^2)^{\frac{1}{2}}}(x\partial_x+y\partial_y+z\partial_z) = (-x\partial_z+z\partial_x) + \lambda \cos(\varphi)r\partial_r. \end{cases} \end{equation} where $\lambda \in \mathbb{R}$ is arbitrary.

If we fix the value $\lambda=\sqrt{2}$ and we evaluate the vectorfields in the point (x,y,z)=(1,1,1) we get \begin{equation} \begin{cases} & V_1(1,1,1)=(-1,1,0) \\ & V_2(1,1,1)=(0,-1,1)+(1,1,1)=(1,0,2)\\ & V_3(1,1,1)=(1,0,-1)+(1,1,1) = (2,1,0). \end{cases} \end{equation} So clearly $\dim \langle V_1(1,1,1) , V_2(1,1,1) , V_3(1,1,1)\rangle=3$.