Grassmannian as smooth manifold

[Korybut]

TL;DR

Grassmannian from projectors point of view

Hello!

There is a proof that Grassmannian is indeed a smooth manifold provided in Nicolaescu textbook on differential geometry. Screenshots are below

There are some troubles with signs in the formulas please ignore them they are not relevant. My questions are the following:
1. After (1.2.5) there is a matrix block form decomposition for projector and the line that puzzles me
"where for every subspace $K\rightarrow V$ we denote $I_K :K\rightarrow V$ the canonical inclusion, then $U=\Gamma_S$. This last "then $U=\Gamma_S$" causes troubles since it was already shown that subspaces can be identified with images of homomorphism. Why author put it here?
2. To show that bijection between projector and homomorphism is continuous some sort of metric in the space of homomorphism is required to induce topology. How topology on there homomorphism is built? (Perhaps here I am asking something every undergraduate students knows. Sorry for that)
3. Proof ends on showing that corresponding subspaces are isomorphic with homomorphism but in manifold theory I need a map to $R^n$ or $C^n$. How one approaches the latter?

Many thanks in advance

 

[jbergman]

TL;DR Summary: Grassmannian from projectors point of view

Hello!

There is a proof that Grassmannian is indeed a smooth manifold provided in Nicolaescu textbook on differential geometry. Screenshots are below
View attachment 346060View attachment 346061View attachment 346062View attachment 346063
There are some troubles with signs in the formulas please ignore them they are not relevant. My questions are the following:
1. After (1.2.5) there is a matrix block form decomposition for projector and the line that puzzles me
"where for every subspace $K\rightarrow V$ we denote $I_K :K\rightarrow V$ the canonical inclusion, then $U=\Gamma_S$. This last "then $U=\Gamma_S$" causes troubles since it was already shown that subspaces can be identified with images of homomorphism. Why author put it here?
2. To show that bijection between projector and homomorphism is continuous some sort of metric in the space of homomorphism is required to induce topology. How topology on there homomorphism is built? (Perhaps here I am asking something every undergraduate students knows. Sorry for that)
3. Proof ends on showing that corresponding subspaces are isomorphic with homomorphism but in manifold theory I need a map to $R^n$ or $C^n$. How one approaches the latter?

Many thanks in advance

I'm really having trouble seeing these screenshots on my phone. I will try and look on a desktop and reply.

 

[Ibix]

I'm really having trouble seeing these screenshots on my phone. I will try and look on a desktop and reply.

PF compresses images quite a lot so there is a lot of jpg artifact in them. I doubt the clarity will improve much on a large screen, I'm afraid.

OP - if the textbook is available online, which it looks like it is, a link would be better.

 

[Korybut]

Sorry, Indeed book is available online
https://www3.nd.edu/~lnicolae/Lectures.pdf

My screenshots are dealing with example 1.2.22 on page 15

 

[jbergman]

TL;DR Summary: Grassmannian from projectors point of view

Hello!

There is a proof that Grassmannian is indeed a smooth manifold provided in Nicolaescu textbook on differential geometry. Screenshots are below
View attachment 346060View attachment 346061View attachment 346062View attachment 346063
There are some troubles with signs in the formulas please ignore them they are not relevant. My questions are the following:
1. After (1.2.5) there is a matrix block form decomposition for projector and the line that puzzles me
"where for every subspace $K\rightarrow V$ we denote $I_K :K\rightarrow V$ the canonical inclusion, then $U=\Gamma_S$. This last "then $U=\Gamma_S$" causes troubles since it was already shown that subspaces can be identified with images of homomorphism. Why author put it here?
2. To show that bijection between projector and homomorphism is continuous some sort of metric in the space of homomorphism is required to induce topology. How topology on there homomorphism is built? (Perhaps here I am asking something every undergraduate students knows. Sorry for that)
3. Proof ends on showing that corresponding subspaces are isomorphic with homomorphism but in manifold theory I need a map to $R^n$ or $C^n$. How one approaches the latter?

Many thanks in advance

This is a messy proof and I would have to spend more time to look at it more precisely to answer your specific questions. I am more familiar with Lee's proof in his book Introduction to Smooth Manifolds which you can find in example 1.36 here.

The basic idea is that by choosing basis for $L$ and $L_{\perp}$, $\hom(L,L_{\perp})$ can be thought of an $M(n-k,k)$ matrix which is essentially $\mathbb{R}^{k(n-k)}$. $\Gamma_S$ is a map from $\hom(L,L_{\perp}) \rightarrow Gr_k(V, L)$. Hence, $\Gamma^{-1}_S: Gr_k(V, L) \rightarrow \hom(L,L_{\perp})$ or with the identification mentioned above by choosing a basis, $\Gamma^{-1}_S: Gr_k(V, L) \rightarrow M(n-k,k) \cong \mathbb{R}^{k(n-k)}$, which can serve as a chart for the open set $Gr_k(V, L)$ .

 

[mathwonk]

here are a couple of remarks,. not totally precise, but hopefully intuitively helpful.

1). since given two k diml subspaces of n-space, there is a linear automorphism of n-space taking one subspace to the other, the set of all such subspaces is "homogeneous", i.e. it looks the same near every point. consequently this set is a manifold, i.e. since any algebraically defined space is smooth at some point, if it looks the same at all points, then all points are smooth.

ok, so arguably it is a manifold, but of what dimension?
2. if you are familiar with reduced row echelon form of a rank k matrix with n columns, you know that the first k rows of such a matrix, give a distinguished basis of the k diml row space, a general k diml subspace of n space. for a general k diml subspace, it projects isomorphically onto the subspace of n-space spanned by the first k standard basis vectors, i.e. in general, the first k columns are pivots. then the particular such k diml row space is determined by the non pivot entries in the first k rows. there are exactly n-k such non pivot entries in each of the first k rows, and these give coordinates for the set of k diml subspaces of n-space that project isomorphically onto the subspace spanned by the first k standard basis vectors. thus this open set of the grassmannian is diffeomorphic to k(n-k) space, so that is the dimension of the homogeneous grassmannian.

 

[Korybut]

I am very thankful to everyone for clarifying the subject to me. Proof in the Lee's book is indeed consistent and there are no questions left