Why does this Jacobian have full rank?

Question

I am doing a basic exercises and I have to show that the set of $n\times n$ orthogonal matrices form a manifold.

Naturally, I have defined a function $f(X) = X^TX-I$ and am considering $f^{-1}(0)$.

For any orthogonal matrix $U$ (and small $H$), I know that $$f(U+H) - f(U) = U^TH + H^TU + O(H)$$

Out of here I get that $J(U) = U^TH + H^TU$ for any $U$.

What I don't get is why the Jacobian necessarily has full rank. It is possible that I am misinterpreting something but I have calculated the Jacobian and don't know where to go with it.

Even at a particular point, $U = I$, I do not see why anyone would claim that $H + H^T$ has full rank. I would not know where to begin in assigning a rank to this value.

Thank you.

Answer 1

The Jacobian doesn't have full rank (for $n > 1$), since the orthogonal group $O(n)$ has positive dimension then.

The rank of the derivative is the codimension of $O(n)$. And the rank is not the rank of the matrix $U^TH + H^TU$, if that is confusing you, but the rank of the linear map $H \mapsto U^TH + H^TU$ from $\mathbb{R}^{n\times n}$ to $\mathbb{R}^{n\times n}$.

To determine the rank, and the kernel of $Df(U)$, write $H = U\cdot K$. (Since $U$ is invertible, every matrix $H$ can be written thus.)

Answer 2

Let $f(A)=A^TA$. Then $D_Af(H)=A^TH+H^TA$ maps onto the space of symmetric matrices $S$ of dimension $\tfrac{1}{2}n(n+1)$. Indeed $\forall C\in S$ let $H=\tfrac{1}{2}AC$, then $D_Af(H)=C$. Thus we have also found the rank of $D_Af$ to be $\tfrac{1}{2}n(n+1)$.

By the regularity theorem, the space of orthogonal matrices $O_n=\{A\in GL_n\mid A^TA=I_n\}=f^{-1}(I_n)$ is a smooth submanifold (or just manifold) of $\mathbb{R}^{n^2}$ of dimension $n^2-\tfrac{1}{2}n(n+1)=\tfrac{1}{2}n(n-1).$