Prove $SO(3)$ is a smooth regular surface in $\mathbb R^{3\times 3}$

Question

I proved this for $O(3)$. To do that, I defined $O(3)$ implicitly by $f(A) = AA^t - I = 0$. Identifying $\mathbb R^{3\times 3}$ with $\mathbb R^9$, I got $$f (a_{11}, ..., a_{33}) = \cases { a_{11}^2 + a_{12}^2 + a_{12}^2 - 1\\\\ a_{21}^2 + a_{22}^2 + a_{23}^2 - 1\\\\ a_{31}^2 + a_{32}^2 + a_{33}^2 - 1\\\\ a_{11}a_{21} + a_{12}a_{22} + a_{13}a_{23}\\\\ a_{11}a_{31} + a_{12}a_{32} + a_{13}a_{33}\\\\ a_{21}a_{31} + a_{22}a_{32} + a_{23}a_{33},}$$ and then showed that rank of $(\partial{f_i}/\partial{x^j})$ is maximal on $O(3)$. I tried to do it similarly for $SO(3)$. In this case $f(A) = \det A(AA^t) - I= 0$ should work, but then $(\partial{f_i}/\partial{x^j})$ seems to look like $$\small\begin{pmatrix} d_{11} + 2a_{11} & -d_{12} + 2a_{12} & d_{13} + 2a_{13} & -d_{21} & d_{22} & -d_{23} & d_{31} & -d_{32} & d_{33}\\ d_{11} & -d_{12} & d_{13} & -d_{21} + 2a_{21} & d_{22} + 2a_{22} & -d_{23} + 2a_{23} & d_{31} & -d_{32} & d_{33}\\ d_{11} & -d_{12} & d_{13} & -d_{21} & d_{22} & -d_{23} & d_{31} + 2a_{31} & -d_{32} + 2a_{32} & d_{33} + 2a_{33}\\ a_{21} & a_{22} & a_{23} & a_{11} & a_{12} & a_{13} & 0 & 0 & 0\\ a_{31} & a_{32} & a_{33} & 0 & 0 & 0 & a_{11} & a_{12} & a_{13}\\ 0 & 0 & 0 & a_{31} & a_{32} & a_{33} & a_{21} & a_{22} & a_{23}\\ \end{pmatrix},$$ where $d_{ij}$ are corresponding minors of $A$. I'm not sure how to show its rank is maximal. Can someone help with this, or should I use some other approach?

Answer 1

I would recommend that you not work in coordinates like this. Think of the space of $3\times 3$ matrices as a vector space $V$ and compute derivatives from the definition. In particular, since $f(A) = AA^\top$ is smooth, we know that for any $B\in V$, we have \begin{align*} df_A(B) &= \lim_{t\to 0}\frac{f(A+tB)-f(A)}t = \lim_{t\to 0}\frac{t(AB^\top + BA^\top +tBB^\top)}t \\ &= AB^\top + BA^\top. \end{align*} Now, you already noticed that $f$ maps $V$ to the vector space $W$ of symmetric $3\times 3$ matrices. We want to show that $df_A\colon V\to W$ is surjective for any $A\in O(n)$. So, for an arbitrary $C\in W$, your charge is to figure out $B\in V$ so that $df_A(B)=C$.