I was trying to prove that the grassmannian is a manifold without picking bases, is that possible?
Here's what I've got, let's start from projective space.
Take $V$ a vector space of dimension n, and $P(V)$ its projective space. To imitate the standard open sets when you have a basis, consider a hyperplane $H$. We can form a (candidate open) subset $U_H$ consisting of those lines $L \in P(V)$ such that $L \oplus H = V$.
For the Grassmannian you can proceed similarly, say you want to construct $Gr(d,V)$. Take a subspace H of dimension $c = n – d$, and consider the set $U_H$ of those subspaces $W \in Gr(d,V)$ such that $W \oplus H = V$.
I'm not really sure how to proceed after this. Any hints? the main problem is that $U_H$ should be isomorphic to affine space but I can't seem to cook up the natural candidate for it.
Best Answer
let $n=dim V$.
Let $A$ be a $(n-d)$-dimensional subspace of $V$ and let $\mathcal U(A)$ be the subset of the Grassmanian of all subspaces $B$ of dimension $d$ such that $A\oplus B=V$. If $B$ is any element of $\mathcal (U)$, then there is a bijection $\hom(A,B)\to\mathcal U(A)$ such that the image of a linear map $\phi:A\to B$ is the subspace $B_\phi=\{a+\phi(a):a\in A\}$.
Then the set of all $\mathcal U(A)$ with those bijections, for all $A$, is an atlas for $G(d,V)$.
You do need to check that transition functions are smooth, though.