I've been reading Manifolds, Tensor Analysis, and Applications recently, and have a question about how to construct a two dimensional integrable distribution on $SO(3)$.
Let $M$ be a manifold, the Local Frobenius Theorem says that a subbundle $E$ of $TM$ is involutive if and only if it is integrable. So I'm trying to construct two vector fields $ X, Y $ defined on open sets of $SO(3)$ such that $[X, Y]$ take values in the distribution generated by $X, Y$.
This two vector fields $X, Y$ can not be left-invariant at the same time, otherwise they can be moved to the identity, and $[X,Y]$ can be computed directly by the Lie-bracket of $\mathfrak{so(3)}$. Identifying $\mathfrak{so(3)}$ with $R^3$ by $$\left ( \begin{array} &0 & -c & b \\ c & 0 &-a \\ -b & a & 0\\ \end{array} \right ) \cong \left ( \begin{array} & a\\b\\c\\ \end{array} \right ) , $$ it can be verified that for any two linearly independent vectors $v_1, v_2$ in $R^3$, $v_1 \times v_2$ can not be in $span\{v_1, v_2 \}$, which means that $[X, Y]$ can not take values in the distribution generated by $X, Y$.
And I got stuck here. My questions are:
- For any given vector fields $ X, Y $, how to compute $[X, Y]$ directly, where $X, Y$ are not assumed to be left-invariant;
- Do such a two dimensional integrable distribution exist? If yes, how can I construct it? If not, how to prove?
Any hint or comment will be appreciate.
I believe that the answer is that there does exist such a distribution. To see this, recall that $S^3$ is the double cover of $\mathrm{SO}(3) \cong \mathbb{RP}^3$ and notice that the Reeb foliation of $S^3$ is stable under the antipodal map.