Why is the set of all bases of a vector space a manifold?

Question

Let $V$ be an $\mathbb{R}$ vector space. Then the collection of all ordered bases of $V$ denoted by $P(V)$ seemingly is manifold. How does one see this ? I suppose that second-countability is inhereted from some $\mathbb{R}^n$ diffeomorphic to $V$. But how to think about the local Euclidean property. Also, what's happening in the case where the dimension of $V$ is not finite.

Answer 1

The set of all bases of $V$ is in 'natural' bijection with $GL_{\dim(V)}(\mathbb{R})$ by assigning to a matrix $M$ the basis consisting of the vectors that form the columns of $M$ in some fixed basis $\mathcal{B}_0$ of $V$ (or the rows if you will). Hence, the natural manifold structure on $P(V)$ is that of $GL_{\dim(V)}(\mathbb{R})$ seen as an open subset of $M_{\dim(V)}(\mathbb{R})\cong \mathbb{R}^{\dim(V)^2}$.