I'm reading sec 2.2 from this article Arxiv:2201.10768 and there is part that I don't understand
Some context: Let $A(t) \in GL(n)$ and consider its polar decomposition $A(t)=U(t)P(t)$, where $U(t) \in O(n)$ and $P(t)$ is a positive-definite symmetric matrix. From this we can define a projector operator from $GL(n)$ to $O(n)$ via $\mathbb{P}(A):=U$. From orthogonality $U(t)U^T(t)=I$ follows that $\frac{dU}{dt} = U\Omega $, where $\Omega$ is a skew-symmetric matrix (Indeed, $\Omega =-U\frac{d}{dt} U^T U$).
Why the differential map of the projection operator is $d\mathbb{P}_{A}(\dot{A})=\Omega$ ?
I tried by doing the following: Let consider a curve $A(t)$ with $A(0)=A$ and then
$$d\mathbb{P}_{A}(\dot{A})= \frac{d}{dt} (\mathbb{P} \circ A(t))|_{t=0} = \frac{d}{dt} (\mathbb{P}(A(t)))|_{t=0} = \frac{d}{dt} U(t) |_{t=0}= U(0)\Omega(0) $$
Of course my calculations are not coherent with $d\mathbb{P}_{A}(\dot{A})=\Omega$. What I'm doing wrong?