Let $GL(n,R)$ be general linear group and $S(n,R)$ be set of symmetric matrices. Both viewed as manifolds. Let $f: GL(n,R)->S(n,R), f(A)=A^TA$
I want to prove that $f$ is smooth map. How do I do it?
I wanted to consider coordinate representation of $f$, but I don't know what is smooth structure for $GL(n,R)$ and $S(n,R)$.
I know that smooth structure for vector space is just one chart: the whole space with the map that after the base is chosen maps vectors to its coordinates in this basis. (*)
I know that if we have an open subset $U$ of manifold $M$ then we can get an atlas for $U$ (thought of as a manifold on its own) by taking charts $(V,\varphi)$ of $M$ such that $V \subset U$ (**)
In my case set of matrices $M(n,R)$ is manifold and a vector space so it has smooth structure (*), $GL(n,R)$ is open subset of it, but I can't use atlas (**) because there is no such chart.