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$\newcommand{\R}{\mathbf R}$ Let $V$ be an $n$-dimensional vector space and $k$ be an integer less than $n$. A $k$-frame in $V$ is an injective linear map $T:\R^k\to V$. Let the set of all the $k$-frames in $V$ be denoted by $F_k(V)$. It is clear that $F_k(V)$ is an open subset of $L(\R^k, V)$.

Define a relation $\sim$ on $F_k(V)$ as follows: We write $S\sim T$ for two members $S$ and $T$ in $F_k(V)$ if and only if $\text{span }T=\text{span }S$.

It can be easily seen that $S\sim T$ if and only if there is a $\tau\in GL_k(\R)$ such that $T=S\circ \tau$.

The Grassmannian manifold $GR_k(V)$ is defined as the quotient space $F_k(V)/\sim$.

I know that the projection map $\pi:F_k(V)\to GR_k(\R)$ is an open map.

I am trying to prove that $GR_k(V)$ is a Hausdorff space.

I proved the above by noting that the above statement is just this: Given linearly independent lists $(u_1, \ldots, u_k)$ and $(v_1, \ldots, v_k)$ in $V$ which do not span the same subspace, there are neighborhoods $U_i$'s of $u_i$'s and $V_j$'s of $v_j$'s such that whenever $(u_1',\ldots, u_k')\in U_1\times \cdots\times U_k$ and $(v_1',\ldots, v_k')\in V_1\times \cdots\times V_k$, the lists $(u_1', \ldots, u_k')$ and $(v_1', \ldots, v_k')$ are linearly independent and do not span the same subspace.

I had a rather long proof of this. Basically I established that given hyperplanes $H$ and $K$ in $V$, there is a hyperplane $P$ in $V$ such that $P$ is "between" $H$ and $K$.

I am looking for a more direct approach than recasting the problem in the aformentioned way.

edit: Also, I am trying to avoid the use of matrices and coordinates as much as possible.

Thanks.

EDIT.

I finally was able to put down the kind of proof I was looking for. Here it is.

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  • $\begingroup$ This answer - math.stackexchange.com/a/225711/59575 - outlines why the Grassmannian is a manifold. In particular, it shows the Grassmannian is locally Euclidean, hence Hausdorff. $\endgroup$ – Zach Jun 3 '15 at 11:49
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    $\begingroup$ Are you suggesting that any locally Euclidean space is Hausdorff? $\endgroup$ – caffeinemachine Jun 3 '15 at 12:06
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    $\begingroup$ I was, which is a mistake. This approach would need some additional work. $\endgroup$ – Zach Jun 3 '15 at 12:08
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    $\begingroup$ To show $G_{k}(V)$ is Hausdorff, it may be easiest to equip $V$ with an inner product and argue (using Gram-Schmidt, for example) that every subspace has an orthonormal basis. Then you can view the Grassmannian as the quotient space $O(n)/O(k) \times O(n - k)$, and use compactness of the groups to your advantage. $\endgroup$ – Andrew D. Hwang Jun 3 '15 at 13:04
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Here's an alternative way to deal with the topology : work with orthonormal $k$-frames instead! That is, pick an inner product $\langle-|-\rangle$ on $V$ and study the by the Stieffel manifold of orthonormal $k$-frames $$V_k(V)=\lbrace (v_1,\dots,v_k)\in V^k\mid\langle v_i\mid v_j\rangle=\delta_{ij}\rbrace$$ The equivalence relation induced by that on $F_k(V)$ now coincides with the one coming from the natural action of the matrix group $O(k)$. The advantage of this approach is that the quotient is automatically Hausdorff:

Lemma. Suppose $G$ is a compact group (Hausdorff not needed), and $X$ is a Hausdorff space with a continuous group action $\rho:G\times X\to X$. Then $X/G$ is Hausdorff.

Proof. Suppose $x,y\in X$ belong to different $G$-orbits. Our hope is to find two disjoint open neighborhoods of $x$ and $y$ that are unions of orbits. For every $x'\in G.y$, there exists disjoint open sets $V_{x'}\ni x'$ and $V_y^{x'}\ni y$. By continuity of the action, for every $g\in G$ there exist open sets $O_g\ni g$ and $V_{x}^g\ni x$ such that $\rho(O_g\times V_{x}^g)\subset V_{g.x}$. By compactness of $G$, there is a finite open covering $G=\cup_{i=1}^NO_{g_i}$. Set $W_x=\cap_{i=1}^NV_{x}^{g_i}$ and $W_y=\cap_{i=1}^NV_{y}^{g_i.x}$, the claim is that $\rho(G\times W_x)$ and $\rho(G\times W_y)$ are disjoint open sets, and it's easily checked.


Thus the grassmannian of $k$-planes, defined as the quotient the Stieffel-manifold under the action of the matrix group $O(k)$ is Hausdorff. But does this answer your question? It does!

First of all note that the grassmannian, as you define it, is compact. This is because there is a contiuous surjection $V_k(V)\hookrightarrow Fr_k(V)\rightarrow Gr_k(V)$ : every subspace admits an orthonormal basis. Also, the Gram-Schmidtification map $$\mathrm{GS}:Fr_k(V)\to V_k(V)$$ is continuous, and compatible with the equivalence relations, and induces a continuous bijection between your definition of the Grassmannian and mine $$\overbrace{\widetilde{\mathrm{GS}}:\underbrace{Fr_k(V)/\sim}_{\text{compact}}\;\longrightarrow\;\underbrace{V_k(V)/\sim}_{\text{Hausdorff}}}^{\text{continuous bijection}}$$ Hence $\widetilde{\mathrm{GS}}$ is a homeomorphism, and your grassmannian is Hausdorff.

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  • $\begingroup$ Thanks for this answer. I will need some time to study it because a lot of things are new to me. I have on question. When you write "the equivalence relation by that no $F_k(V)$ coincides with.." do you mean that the equivalence relation on $V_k(V)$ gotten by restricting $\sim$ on $V_k(V)$ coincides with the natural action of $O_k(V)$? $\endgroup$ – caffeinemachine Jun 3 '15 at 16:40
  • $\begingroup$ Yes. ${}{}{}{}$ $\endgroup$ – Olivier Bégassat Jun 3 '15 at 16:41
  • $\begingroup$ Looks great on first look. Wasn't as difficult as I thought. Disturbingly neat! $\endgroup$ – caffeinemachine Jun 3 '15 at 16:49
  • $\begingroup$ "Disturbingly neat" ^^ oh really? But you're right, it's pretty mundane stuff. $\endgroup$ – Olivier Bégassat Jun 3 '15 at 16:53
  • $\begingroup$ I cannot call it mundane! From where I am this seems quite masterful. How do achieve this kind of awesomeness? $\endgroup$ – caffeinemachine Jun 4 '15 at 0:36
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One simple, clean way to show that the Grassmannian is Hausdorff is to define a metric on it. Here are the steps in the proof:

  1. Let $S$ be the unit sphere in $V$, and let $d$ be a metric on $S$. If $A\subset S$ is closed and $p\in S$, define $$ d(p,A) \;=\; \min\{d(p,a) \mid a\in A\}. $$ It is easy to show that $d(p,A)$ is a continuous function of $p$.

  2. Now, if $A,B\subset S$ are closed, define $$ d_H(A,B) \;=\; \max\biggl(\max_{a\in A} \,d(a,B),\, \max_{b\in B}\,d(b,A)\biggr) $$ Note that $d_H(A,B) = 0$ if and only if $A=B$.

  3. Finally, if $H$ and $K$ are $k$-dimensional subspaces of $V$, define $$ \rho(H,K) \;=\; d_H(H\cap S,K\cap S) $$ Then it suffices to prove that $\rho\colon \textit{GR}_k(V)\times \textit{GR}_k(V) \to\mathbb{R}$ is continuous. Since $\textit{GR}_k(V)$ is defined as a quotient space, this is the same as proving that the composition $$ F_k(V) \times F_k(V) \to \textit{GR}_k(V)\times \textit{GR}_k(V) \to\mathbb{R} $$ is continuous, which should be relatively quick. (The details depend on how you have defined the topology on $L(\textbf{R}^k,V)$.)

Of course, $d_H$ is actually a metric on the collection of closed subsets of $S$ (the Hausdorff distance), so $\rho$ is actually a metric on $\textit{GR}_k(V)$. Indeed, if you use the geodesic metric $d$ on $S$, then $\rho$ is precisely the angle between $k$-dimensional subspaces. However, we don't need to prove this if the goal is just to establish that $\textit{GR}_k(V)$ is Hausdorff.

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There are at least two approaches:

  1. Any $(n - k)$-plane $A \in Gr_{n - k}(V)$ determines a natural chart on the (open) subset $U_A \subset Gr_k(V)$ consisting of planes $B \in Gr_k(V)$ transverse to $A$ (i.e. such that $A \oplus B = V$): We get a map $$\Phi_A: \text{Hom}(A, B) \cong B \otimes A^* \to U_A$$ that sends any linear map $T: A \to B$ to its graph, $$\text{graph } T := \{(x, T(x)) : x \in A\} \subset A \oplus B = V.$$ If we fix bases of $A, B$, we can identify $\text{Hom}(A, B)$ with $M((n - k) \times k, \Bbb R) \cong \Bbb R^{(n - k) k}$, furnishing coordinates on $U_A$. At least when $k \neq 0, n$, for any $k$-planes $B, C \in Gr_k(V)$ there is some $(n - k)$-plane $A \in Gr_{n - k}(V)$ transverse to both, and hence $B$ and $C$ are both in $U_A$, in which case (because Euclidean space is Hausdorff) the topology separates $B$ and $C$. (Of course, there are more details to check, namely that the charts $(U_A, \Phi_A^{-1})$ are compatible with the smooth structure on $Gr_k(V)$ implicit in the frame construction of that space.)

  2. It's easy to see by choosing bases that $GL(V)$ acts transitively on $Gr_k(V)$, and with a little more work one can show that this action is smooth. If we choose a basis $(E_1, \ldots, E_n)$ of $V$, then the stabilizer of the $k$-plane $\langle E_1, \ldots, E_k \rangle \in Gr_k(V)$ is the Lie subgroup $H$ consisting of block upper triangular matrices of the form $$\pmatrix{\ast & \ast \\ 0 & \ast}$$ in $GL(n, \Bbb R) \cong GL(V)$, where the lower-left zero block has size $(n - k) \times k$. So, we can identify $$Gr_k(V) \cong GL(V) / H ;$$ in particular, $H$ is closed, and so $Gr_k(V)$ admits a unique smooth manifold structure such that the natural quotient map $GL(V) \to Gr_k(V)$ is a smooth submersion.

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  • $\begingroup$ Thanks fro the generous answer. I have one question. Can you argue that $U_A$ is open in $GR_k(V)$ under the definitions I am using for the Grassmannian. $\endgroup$ – caffeinemachine Jun 3 '15 at 13:50
  • $\begingroup$ You're welcome, I hope you found it useful. And yes, that shouldn't be too hard; one could, say, show that the quotient map defined by the relation $\sim$ is open and then show that the preimage of $U_A$ under that quotient is open. $\endgroup$ – Travis Willse Jun 3 '15 at 14:08
  • $\begingroup$ Okay then let me try doing that later. Am really sleepy. Will get back in a few hours. $\endgroup$ – caffeinemachine Jun 3 '15 at 14:10
  • $\begingroup$ I am facing the same kind of difficulty proving that $U_A$ is open in $GR_k(V)$ which I was facing in proving Hausdorffness of $GR_k(V)$. My argument would be on the same lines as what I have mentioned after the block quote in my question. Do you have an easy argument. I have a short proof of the fact that the projection map $\pi:F_k(V)\to GR_k(V)$ is an open map. $\endgroup$ – caffeinemachine Jun 3 '15 at 16:26
  • $\begingroup$ Since $\pi$ is open, it suffices to show that $\pi^{-1}(U_A)$ is open in $F_k(V)$, that is, that the set of $k$-frames spanning a plane transverse to $A$ is open: Pick a frame $(X_1,\ldots,X_{n - k})$ of $A$; for any $k$-frame $(Y_a)$, $A$ and $\langle Y_1,\ldots,Y_k\rangle$ are transverse iff (w.r.t. any, equivalently, every basis) the matrix $$Q[(Y_a)] := \pmatrix{X_1&\cdots &X_{n-k}&Y_1&\cdots Y_k}$$ has nonzero determinant. So, $\pi^{-1}(U_A)$ is the preimage of the open set $\Bbb R - \{0\}$ under the continuous the map $F_k(V)\to\Bbb R$ defined by $(Y_a)\mapsto\det Q[(Y_a)]$. $\endgroup$ – Travis Willse Jun 3 '15 at 17:42
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Another 'neat' proof that has not been mentioned already (I hope) goes as follows:

You can view $Gr(k,V)$ as equivalence classes of $n\times k$ matrices of full rank. Trivialised sets $U_\alpha$ are those matrices $X$ which give invertible $k\times k$ matrices $X_\alpha$ by selecting the rows $\alpha$. Here $\alpha$ is a multi-index of $k$ integers between $1$ and $n$. You would map these matrices to $\mathbb{R}^{k(n-k)}$ by changing coordinates to make the $\alpha$ submatrix the identity. It is 'easy' to see this is a cover with polynomial transition functions.

To show this space is Hausdorff, consider two points $A$, $B$ in $U_\alpha$, $U_\beta$ respectively, neither one being in the intersection. Define functions $h_{\alpha\beta}:U_\alpha \rightarrow\mathbb{R}$ and $h_{\beta\alpha}:U_\beta \rightarrow\mathbb{R}$ by $h_{\alpha\beta}(X):=det(X_{\beta}X_\alpha^{-1})$ and $h_{\beta\alpha}(X):=det(X_{\alpha}X_\beta^{-1})$ respectively. These are well defined and multiply to give $1$ on $U_\alpha\cap U_\beta$.

Since neither $A$ nor $B$ is in the intersection, they respectively map to $0$ by the functions. The claim is that $h_{\alpha\beta}^{-1}(-1,1)$ and $h_{\beta\alpha}^{-1}(-1,1)$ are disjoint open neighbourhoods for $A$ and $B$. To see they are disjoint, use the fact that $h_{\alpha\beta}$ and $h_{\beta\alpha}$ multiply to give $1$ on $U_\alpha \cap U _\beta$.

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  • $\begingroup$ Thanks for the response. It's probably because I was trying to avoid the use of matrices that this answer was not posted. Also, I had found the kind of proof I was looking for, which I linked in the question later. $\endgroup$ – caffeinemachine Dec 23 '16 at 1:36
  • $\begingroup$ Ah, I should of read the question entirely. $\endgroup$ – Dr. Thomas Underhill Dec 23 '16 at 16:24

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