Let $n\in\mathbb{N}$. Regard the real [orthogonal group](/page/Orthogonal%20Group) as

\begin{align*} O(n)=\{A\in M_n(\mathbb{R}): A^\top A=I_n\}, \end{align*}

and let

\begin{align*} \mathfrak{so}(n)=\{X\in M_n(\mathbb{R}): X^\top+X=0\}. \end{align*}

Then the tangent space to $O(n)$ at $I_n$ is $\mathfrak{so}(n)$. Equivalently, if $\varepsilon>0$ and $A:(-\varepsilon,\varepsilon)\to O(n)$ is differentiable with $A(0)=I_n$, then $A'(0)\in\mathfrak{so}(n)$; conversely, for every $X\in\mathfrak{so}(n)$ there exist $\varepsilon>0$ and a differentiable map $A:(-\varepsilon,\varepsilon)\to O(n)$ such that $A(0)=I_n$ and $A'(0)=X$.