I am trying to do an exercise from Hirsch's Book , Differential Topology, that basically wants me to prove that $G_{n,k}$ is a manifold.
The Grassmanian manifold $G_{n,k}$ of $k$-dimensional subspaces or $k-$planes of $\mathbb{R}^n$ is given an atlas as follows. Let $E\subset \mathbb{R}^n$ be a $k-$plane , we can identify $\mathbb{R}^n\cong E\bigoplus E^\perp$ . Every $k$-plane near enough $E$ is the graph of a unique linear map $T:E\rightarrow E^{\perp}$.
Now I don't see why this statement is true and how this is going to give me an open set, can anyone provide me some hints ? Thanks in advance.
Hint: You can interpret "$F$ is close enough to $E$" as "$F\cap E^\perp=\{0\}$" and the subspaces with this property define the open neighborhood of $E$ that you are looking for. Assuming this, any elment of $\mathbb R^n$ can be uniquely written as the sum of an element of $F$ and an element of $E^\perp$. Doing this for elements of $E$ and then mapping to the $E^\perp$-component defines a linear map $E\to E^\perp$, whose graph is $F$. Since the space $L(E,E^\perp)$ of linear maps is a finite dimensional vector space, you can construct charts as having values in this space. (Alternatively, choose bases and interpret charts as having values in spaces of matrices.)