We have the following setting:

We let $$ X = Gr_k(\mathbb{R}^n) = \{k\text{-dimensional linear subgroups of $\mathbb{R}^n$}\}$$ To topologize this space, we pick a fixed $V \in Gr_k(\mathbb{R}^n)$. Then any $k$-dimensional subspace can be obtained by applying some linear map to $V$. So we obtain a surjection \begin{align*} GL_n(\mathbb{R}) &\to Gr_k(\mathbb{R}^n)\\ M &\mapsto M(V). \end{align*} So we can given $Gr_k(\mathbb{R}^n)$ the quotient (final) topology. For example, $$ Gr_1(\mathbb{R}^{n + 1}) = \mathbb{RP}^n.$$ Now to construct a vector bundle, we need to assign a vector space to each point in $X$. But a point in $Gr_k(\mathbb{R}^n)$ is a vector space, so we have an obvious definition $$ E = \{(V, v) \in Gr_k(\mathbb{R}^n) \times \mathbb{R}^n: v \in V\}. $$ This has the evident projection $\pi: E \to X$ given by the first projection. We then have $ E_V = V.$ To see that this is a vector bundle, we have to check local triviality. We fix a $V \in Gr_k(\mathbb{R}^n)$, and let $$ U = \{W \in Gr_k(\mathbb{R}^n): W \cap V^\perp = \{0\}\}. $$ We now construct a map $\varphi: E|_U \to U \times V \cong U \times \mathbb{R}^k$ by mapping $(W, w)$ to $(W, pr_V(w))$, where $pr_V: \mathbb{R}^n \to V$ is the orthogonal projection.

Now if $W \in U$, then $pr_V(w) \not= 0$ since $W \cap V^\perp = \{0\}$. So $\varphi$ is a homeomorphism. We call this bundle $\gamma_{k, n}^\mathbb{R} \to Gr_k(\mathbb{R}^n)$.

I have some questions about that:

1) Could you please give me some intuition about how the topology we define "works"? I can understand the formal definition, but I cannot see what exactly we are doing.

2) Why is it $Gr_1(\mathbb{R}^{n + 1}) = \mathbb{RP}^n$? I assume that this might be a direct consequence of 1).

3)Why $U$ is an open neighborhood of V? I can see that $V\in U$ and I have an idea that goes like that: $W\in U$ means that for every $w\in W$ we have that $pr_V(w)\neq 0$. This holds for all the elements of $V$, and if we remain in a small neighborhood of $V$ then this will still hold for all elements of the spaces in this neighborhood, due to continuity. However, I cannot make that rigorous.

4) Why is $\phi$ a homeomorphism? I can see that it is bijection and continuous but not that its inverse is also continuous.

Thank you in advance

ps1: I used the source code of Dexter from here for the setting

ps2: I don't know if I should have split this post to several others. Please, let me know.

  • $\begingroup$ (1) -- when should two lines in $\mathbb{R}^2$ be close? $\mathbb{R}^3$? Can you define a distance between lines? Is the action of $GL_n$ continuous in this sense? Open? $\endgroup$ Apr 2 '17 at 3:14
  • $\begingroup$ [1/2]Well, in $\mathbb{R}^2$ we could say that two lines are close if their angle is small enough. In $\mathbb{R}^3$, we could do something similar, by considering the plane they span and measure their angle there. Regarding distance, the only idea that comes up to me is to take the $\mathbb{S}^1$ and $\mathbb{S}^2$ respectively and define the distance of two lines to be the distance between the points of intersection with the sphere at the upper hemisphere. This, of course, could give as another measure of "closeness". $\endgroup$
    – perlman
    Apr 2 '17 at 13:22
  • $\begingroup$ [2/2] For the questions regarding actions, I don't have any good idea. Actually, my main problem is that I cannot understand how does this action work. $\endgroup$
    – perlman
    Apr 2 '17 at 13:23
  • $\begingroup$ The action comes from taking a basis for the subspace, and applying the transformation to each basis vector, and taking the span of the result. It needs to be checked that the choice of basis doesn't affect the resulting subspace, but that's true. $\endgroup$ Apr 2 '17 at 14:32
  • $\begingroup$ So, if we take the coordinate representation of a basis vector of the subspace (with respect to the standard basis of $\mathbb{R}^n$) and take a matrix "near" the identity, the result of the action will be a vector that has coordinate representation close to the first one (because matrix multiplication is continuous). Doing the same for every basis vector of the subspace, we will end up with a subspace that has a basis for which each element is "close" to the initial one, so this subspace will be "close" to the initial subspace. Is that right? $\endgroup$
    – perlman
    Apr 2 '17 at 15:20

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