The tangent plane of orthogonal group at identity. Why the tangent plane of orthogonal group at identity is the kernel of $dF_I$, the derivative of $F$ at identity, where $F(A) = AA^T$?
Thank you ~

$\exists$

Proposition Let $Z$ be the preimage of a regular value $y\in Y$ under the smooth map $f: X \to Y$. Then the kernel of the derivative $df_x:T_x(X) \to T_y(Y)$ at any point $x \in Z$ is precisely the tangent space to $Z, T_x(Z).$

Proof:
Since $f$ is constant on $Z$, $df_x$ is zero on $T_x(Z)$. But $df_x: T_x(X)\to T_y(Y)$ is surjective, so the dimension of the kernel of $df_x$ must be
$$\dim T_x(X) - \dim T_y(Y) = \dim X - \dim Y = \dim Z.$$
Thus $T_x(Z)$ is a subspace of the kernel that has the same dimension as the complete kernel; hence $T_x(Z)$ must be the kernel.
Proposition on Page 24, Guillemin and Pollack, Differential Topology
Jellyfish, you should really read your textbook!
