A generalization of projective spaces are the Grassmann manifolds. Consider $G(m,n)$ as the set of all $m$-dimensional subspaces of $\mathbb{R}^n$. The topology on $G(m,n)$ should then arise analogously to the topology of the projective space: the space $\text{Mat}(m \times n, \mathbb{R})$ carries the Euclidean topology. Let $X$ be the subset of $m \times n$ matrices of rank $n$. Then the topology of $G(m,n)$ is the final topology of $f: X \rightarrow G(m,n)$ with $f(A) = \text{im}(A)$, where $\text{im}(A)$ is the image of $A$, that is, the subspace spanned by the columns of $A$.
I would like to know how to show that $G(m,n)$ is locally Euclidean of dimension $m(n-m)$? The dimension is something I read in a more advanced book on differentiable manifolds.
(too long for a comment, but maybe not a full answer either...)
Let $M\in X$, $$ M=\begin{pmatrix} v_1^1 & \cdots & v_1^n \\ \vdots & \ddots & \vdots \\ v_m^1 & \cdots & v_m^n \end{pmatrix} $$ Since the matrix is full rank by definition of $X$, there exist indices $I=\{i_1,...,i_m\}$, with $$ 1\leq i_1<\cdots<i_m\leq n, $$ such that the $m\times m$ square matrix $M_I$ obtained from $M$ retaining only the rows with indices $i_1,...,i_m$ is invertible.
The matrix $\widetilde M:=(M_I)^{-1}M$ satisfies $f(M)=f(\widetilde M)$. The matrix $\widetilde M$ moreover has only $m(n-m)$ "nontrivial" entries (the submatrix $\widetilde M_I$ is the $m\times m$ identity matrix).
The game to play now would be to show that:
the locus in $G(m,n)$ where $\det M_I\not =0$ is open (for any choice of indices $I$),
on this locus the coordinates given by the $m(n-m)$ nontrivial entries of $\widetilde M_I$ provide a local homeomorphism with $\mathbb R^{m(n-m)}$, and
changes of coordinates are smooth.