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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.

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perhaps the only solution is to use the plucker embedding? this would still leave the question to how to prove that P(V) is a manifold, although the candidate affine space for $U_H$ could be taken to be $H$ perhaps? – John Salvatierrez Oct 31 '12 at 0:22
I think this should answer the question.… – caffeinemachine Dec 4 '15 at 16:19
up vote 8 down vote accepted

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.

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