I am a beginner at Manifold. I am trying the following exercise:
Given a map $f: O(n, \mathbb{R}) \to GL(n, \mathbb{R})$ by $f(A)=AA^{T}$. I want to show that $O(n,\mathbb{R})$ is the inverse image of a regular value.
My attempt:
Since $f^{-1}(I)=O(n, \mathbb{R})$, so my guess is to show that $I$ is a regular value. That is to show for any $A \in O(n, \mathbb{R})$, the differential map $f_{*,A}:T_A(O(n, \mathbb{R})) \to T_{f(A)}(GL(n, \mathbb{R}))$ is surjective. Now for any $X \in T_A(O(n, \mathbb{R})$. Using a result I know that there is a curve $c(t)$ in $O(n, \mathbb{R})$ whith $c(0)=A$ and $c'(0)=X$ then,
$f_{*,A}(X)= \frac{d}{dt}f(c(t)) |_{t=0}=\frac{d}{dt}c(t)c(t)^{T} |_{t=0}=c'(t)c(t)^{T}+c(t)(c'(t))^{T}|_{t=0}=XA^{T}+AX^{T}$.
Now let for any matrix $Y\in T_{f(A)}(GL(n, \mathbb{R}))= T_{I}(GL(n, \mathbb{R}))=T_{I}(\mathbb{R^{n\times n}})= \mathbb R^{n\times n} $.
Where I got stuck:
I want to show there is some $X\in T_A(O(n, \mathbb{R})$ such that $XA^{T}+AX^{T}=Y$. I can't figure out how to proceed next.
Thanks for any insights!!