The tangent plane of orthogonal group at identity.

Question

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 ~

Answer 1

The keyword is submersion.

The smooth map $F:M_n(\mathbb{R})\longrightarrow S_n(\mathbb{R})$ into the symmetric matrices has derivative $dF_P(H)=HP^T+PH^T$ surjective for every $P$ of the level set $F^{-1}(I_n)$. The submersion (or regular value) theorem says that $F^{-1}(I_n)$ (the orthogonal matrices) is a smooth submanifold of $M_n(\mathbb{R})$ of dimension $\dim M_n(\mathbb{R})-\dim S_n(\mathbb{R})=n^2-\frac{n(n+1)}{2}=\frac{n(n-1)}{2}$. Moreover, the tangent space at $P$ is given by $\ker dF_P$. In the case of $P=I_n$, this yields the antisymmetric matrices.

I believe this is covered in every text on differential geometry. So instead of a formal proof (essentially the implicit function theorem) that can be found everywhere, let us look at an easy example.

The smooth map $F:\mathbb{R}^2\longrightarrow \mathbb{R}$ defined by $F(x,y)=x^2+y^2$ has derivative $dF_{(x_0,y_0)}(h,k)=2x_0h+2y_0k$. This is surjective for every point $z_0=(x_0,y_0)$ of the unit circle $F^{-1}(1)$. So the circle is a smooth submanifold of $\mathbb{R}^2$. The tangent space at $z_0$ is the nullspace of $dF_{z_0}$, which is indeed the orthogonal of the gradient $(2x_0,2y_0)$, corresponding to our intuitive notion of tangent space to the circle at the point $z_0=(x_0,y_0)$ (modulo an affine translation).

Answer 2

$\exists$

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!