# intrinsic proof that the grassmannian is a manifold

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.

• 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? Oct 31, 2012 at 0:22
• I think this should answer the question. math.stackexchange.com/questions/1461066/… Dec 4, 2015 at 16:19

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.

• ♦ Can you briefly explain for a fix $d$ dim subspace $V$, how to find a unique $\phi\in Hom(A, B)$, such that $V=\{a+\phi(a):a\in A\}$ Jan 30, 2017 at 22:07
• @user136592, you are mixing some letters? Jan 30, 2017 at 22:45
• How do you prove that the Grassmannian is Hausdorff without appealing to the Plücker embedding? Jun 1, 2019 at 19:16
• Let $A$ be one-dimensional, $n =3$. How is $B_{\phi}$ ever of dimension 2? Jun 18, 2019 at 8:02
• @AlexanderGeldhof if $A$ is $1$-D then since since $\phi$ is linear, send the basis vector $v$ by $\phi$ and $\phi(v)$ will be $1$-D and thus $B_{\phi}=span\{v+\phi(v)\}$ as defined above by Mariano is $2$-D (in direct sums you add the dimensions) Nov 15, 2020 at 17:50