Orthogonal group is a regular submanifold of $GL(n,\Bbb R)$

Question

I want to show that $O(n)$ is a regular submanifold of $GL(n,\Bbb R)$. I think that I can use constant rank theorem but how? I am putting the picture that what I did. Please help me I want to learn.

Answer 1

Consider the function $f:A\mapsto A^tA$ defined on the vector space $V=M_n(\mathbb R)$ of $n\times n$ matrices.

Let $A_0\in O(n,\mathbb R)$.

Let $B\in V$ and let us compute the derivative of $f$ at $A_0$ in the direction of $B$, namely, $$D_{A_0}f(B):=\lim_{h\to0}\frac{f(A_0+hB)-f(A_0)}{h}.$$ Using the definition of $f$, we have \begin{align}\frac{f(A_0+hB)-f(A_0)}{h}&=\frac{(A_0+hB)^t(A_0+hB)-A_0^tA}{h}\\ &=\frac{A_0^tA_0+h(B^tA_0+A_0^tB)+h^2B^tB-A_0^tA_0}{h}\\ &=B^tA_0+A_0^tB+hB^tB, \end{align} so that obviously the limit above is $$D_{A_0}f(B)=B^tA_0+A_0^tB.$$ Can you compute the rank of the differential $D_{A_0}f:V\to V$ that we have just computed?

Notice that $D_{Id}f(A^tB)=D_Af(B)$. As long as $A$ is invertible, this implies that the maps $D_{Id}f$ and $D_Af$ have the same rank, so it is enough to compute the rank of $D_{Id}$.