I am trying to show that the map $\varphi : \operatorname{O}(3) \rightarrow \mathbb{R}^3$ defined by $$ \begin{pmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \end{pmatrix} \longmapsto (x_{12} , x_{13} , x_{23} )$$
is a local diffeomorphism near the identity, however I think that I may be overlooking something.
If we first regard the same mapping $\Phi : \operatorname{GL}_{3}(\mathbb{R}) \rightarrow \mathbb{R}^3$ defined by the same rule as $\varphi$, then we can compute the coordinate representation of $\Phi$ using the global coordinates $(x^{11}, x^{12} , x^{13} , x^{21}, x^{22} , x^{23}, x^{31}, x^{32}, x^{33})$ on $\operatorname{GL}_{3}(\mathbb{R})$ and $(y_{1} , y_{2} , y_{3})$ on $\mathbb{R}^3$
$$
\Phi(x^{11}, x^{12} , x^{13} , x^{21}, x^{22} , x^{23}, x^{31}, x^{32}, x^{33}) = (x_{12}, x_{13}, x_{23})
$$
from which we see that $\Phi$ is clearly smooth and so $\Phi|_{\operatorname{O}(3)} = \varphi$ is smooth as well. Further, for any $X = (x_{ij}) \in \operatorname{GL}_{3}(\mathbb{R})$, the Jacobian matrix of the differential is
$$
d_{X} \Phi = \begin{pmatrix}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
\end{pmatrix},
$$
which clearly has rank equal to 3. It is tempting to say that $d_{X} \varphi$ has rank 3 at this point, from which it follows that $d_{X} \varphi$ is invertible and so $\varphi$ is a local diffeomorphism, but I do not believe this is correct. How could I go about finishing this problem off?
EDIT: I think that I figured things out with the help of Tsemo's answer. If we let $\operatorname{Sym}_{3}(\mathbb{R})$ denote the space of all $3 \times 3$ symmetric matrices with real entries, then it is easy to see $\operatorname{O}(3) = F^{-1}(I)$, where $F : \operatorname{GL}_{3}(\mathbb{R}) \rightarrow \operatorname{Sym}_{3}(\mathbb{R})$ is defined by $A \mapsto AA^t$. It is not too hard to work out that $d_{A} F(X) = XA^t + AX^t$ for all $X \in T_{A} \operatorname{GL}_{3}(\mathbb{R})$ and so $d_{I} F = X + X^t$. It is not too difficult to show that $F$ is a submersion from which it follows that $T_{I} \operatorname{O}_{3} \simeq \ker d_{I} F$. Working in standard coordinates and identifying the operators $\partial/\partial x^{ij}$ with the matrices that are 1 at the ijth entry and 0 elsewhere, one can show that $\ker d_{I} F$ is spanned by
$$
\begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}.
$$
The claim then follows from the inverse function theorem.
Best Answer
The tangent space of the identity $\rm T_IO(3)$ is the vector space of antisymmetric matrices, you can extend $\varphi$ to a linear map $\Phi$ defined on the space of $3\times 3$ matrices, we deduce that $\Phi$ is equal to its differential and the restriction of $d\Phi$ to $\rm T_IO(3)$ is an isomorphism. The result follows from the local inversion theorem.