Compactness of the Grassmannian

Let $V$ be a finite-dimensional inner product space. For $0 \leq d \leq \text{dim}(V)$, define the Grassmannian $G(V, d)$ to be the set of all $d$-dimensional linear subspaces of $V$, equipped with the metric $d(X, Y) = \Vert P_X - P_Y \Vert$, where $P_X, P_Y$ is the orthogonal projector onto $X, Y$, respectively.

I want to use the following fact (to prove other things): The Grassmannian $G(V, d)$ is compact with respect to the metric topology.

I am a physicist. I am quite interested in mathematical questions, but I am not good enough at mathematics to see immediately that the bold statement above is true, so I tried to look it up in the literature. However, I only found this statement either just stated but unproved or the compactness was inferred from from some kind of structures or techniques which I don't know. (I've seen several definitions of the Grassmannian, and I think they are equivalent, but I cannot see this directly.)

Can someone tell me where I can look up a proof of the bold statement above which is understandable with basic knowledge in topological and metric spaces? If the proof is easy enough, I am also happy with a presentation of the proof (idea) here, instead of a reference.

Alternatively, I would also be happy if one could show me a proof for the compactness of $G(V,d)$ in the case where $V$ is a finite-dimensional normed space and $$d(X, Y) = \sup\limits_{x \in X, \Vert x \Vert = 1} \inf \{ \Vert x - y \Vert: y \in Y\} \,.$$

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The easiest explanation that I can think of (at the moment) is the following:

The Grassman manifold $G_{n}(m)$ consisting of all subspaces of $\mathbb{R}^m$ of dimension $n$ is a homogeneous space obtained by considering the natural action of the orthogonal group $O(m)$ on the Stiefel manifold $V_{n}(m)$. The Lie group $O(m)$ is compact and we conclude that $G_{n}(m)$ is a compact space.

I think that there is probably an easier explanation but the above constitutes the conceptual reason as to why the Grassman manifold is compact. In any case, I think that Grassman manifolds are conceptually simple to understand if you think of them as homogeneous spaces.

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Of course, one needs to know that the Stiefel manifold is compact to apply this. And note, there are competing definitions of Stiefel manifold and not all of them are compact! I've seen the Stiefel manifold described as the set of $k$ frames in $\mathbb{R}^n$ (i.e, ordered $k$ tuples of independent vectors), which is not compact, and I've also seen the Stiefel manifold described as the set of orthonormal $k$ frames in $\mathbb{R}^n$. This set is compact. –  Jason DeVito May 24 '12 at 0:39

At the risk of introducing one more a priori different topology on the Grassmannian, here's an easy way to see it's compact. I write $\mathbb{R}$ in what follows, though this applies equally well to $\mathbb{C}$.

We can realize $G(V,d)$ as a quotient of something compact, so it's definitely compact. Namely, write $S := \{v \in V: \|v\| = 1 \}$, picking an isomorphism of $(V, \langle\ \rangle) \simeq (\mathbb{R}^n, \cdot)$, this is just $S^{n-1} \subseteq \mathbb{R}^n$.

Now, consider the collection $C$ of $d$ tuples of orthogonal vectors, in $S^d$, $S^d$ is compact, and this collection is the 0 locus of $$S^d \rightarrow \mathbb{R}^{\binom{d}{2}} \quad (v_1, \ldots, v_d) \mapsto (v_i \cdot v_j)_{i< j}$$ So $C$ is the pullback of a point, hence closed, hence with the induced topology from $S^d$ it's compact; to get the Grassmannian, quotient $C$ by saying two $d$ tuples are the same, if their span gives the same $d$ plane.

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It should be reasonable to see this topology is the same as either of the induced ones you suggested...I'll think about it. In the mean time, I hope this helps. –  uncookedfalcon May 22 '12 at 6:53
Thanks for your answer! Although it took me a moment, I can follow your proof that the Grassmannian is compact with respect to the topology induced from $S^d$. As you mentioned, however, the most important thing remains unclear to me: I don't see that this topology coincides with the one generated by $d(X,Y)$ defined above. If we manage to see why this must be the case, this would answer my question, although a more direct approach working with the metric structure only would be more satisfying. –  Tom Jonathan May 22 '12 at 7:29
Dear Tom, how would you use a metric structure to prove that a space is compact? –  Georges Elencwajg May 22 '12 at 8:40
Dear Georges, your comment somehow sounds like you have a useful hint for me, but you want me to hit on it rather than telling me directly. However, I can't see it, so can you explain it to me? –  Tom Jonathan May 23 '12 at 4:06

This can be seen directly. I use $d(X,Y) = \|P_X - P_Y\|$. To show $G(V,d)$ is compact, it suffices to know that $P := \{P_X\} \subseteq End(V)$ is compact, where $P$ is the collection of all orthogonal projections onto $d$-planes.

(That this suffices is arguably the definition: take a sequence of $d$ planes $X_i$, consider the $P_{X_i}$; extract a convergent subsequence $P_{X_j} \rightarrow P_Y$, by definition, $X_j \rightarrow Y$ as desired)

To show compactness, we need that $P$ is closed and bounded, as choosing a basis for $V$ gives a homeomorphism $End(V) \simeq \mathbb{R}^{n^2}$.

I use the following convention: let $b_i$ be an orthonormal basis for $V$, for $T \in End(V)$, set $$\|T\| := \sum_i \|T(b_i)\| = \sum_i \langle T(b_i), T(b_i) \rangle$$One checks (i) this is indeed a norm, and (ii) this norm is invariant under orthogonal change of basis.

Bounded: I claim we have the trivial bound $\|P_X\| \leqslant n$. Why? Each column has norm at most 1, as projection only ever drops something's norm, and the $b_i$ have norm 1. (To see the thing about projection only dropping norm, we can write any $v$ uniquely as $w + w^\perp, w \in X, w^\perp \in X^\perp$, then $$\|P(v)\| = \sqrt{\langle w,w\rangle} \quad \|v\| = \sqrt{\langle w + w^\perp, w + w^\perp\rangle} = \sqrt{\langle w,w\rangle + \langle w^\perp, w^\perp\rangle}$$by bilinearity)

Closed: This is slightly more involved, I first outline the steps in the proof, then fill in the details below.

1. The first observation is that $d(X,Y)$ really does measure "distance" between $d$ planes. To be precise, if $d(X,Y) = \epsilon$, there exists orthonormal bases $x_i$ for $X$, $y_i$ for $Y$, such that $d(x_i, y_i) \leqslant f(\epsilon)$, where $\lim_{\epsilon \downarrow 0} f(\epsilon) = 0$.

2. If $P_{X_i}$ is a Cauchy sequence of projection operators, possibly after passing to a subsequence one can construct a sequence of orthonormal bases $x_{i,j}$ for $X_i$, $1 \leqslant j \leqslant n$, such that for fixed $j$, $x_{i,j}$ is Cauchy. Informally, we have a Cauchy sequence of not single vectors but orthonormal bases.

3. The $x_{i,j} \rightarrow y_j$, a collection of $k$ orthonormal vectors, writing their span as $Y$, we have that $P_{X_i} \rightarrow P_Y$, as desired.

Right, now the proofs. To be honest, they're kinda ugly:

(1) Informally, one takes an orthonormal basis for $X$, projects it down to $Y$, tweaks it to make them perpendicular, and then normalizes the resulting vectors.

Let $x_1, \ldots x_d$ be an orthonormal basis for $X$. Write $x_1 = P_Y(x_1) + P_Y(x_1)^\perp$, we have $$d(x_1, P_Y(x_1)) = \| P_Y(x_1)^\perp \| \leqslant \| P_X - P_Y \| = \epsilon$$Hence by the reverse triangle inequality, $\| P_Y(x_1) \| \geqslant 1 - \epsilon$, thus setting $y_1 := \frac{P_Y(x_1)}{\| P_Y(x_1) \|}$, we have $$d(x_1, y_1) \leqslant d(x_1, P_Y(x_1)) + d(P_Y(x_1), y_1) \leqslant 2 \epsilon$$

Suppose we have defined $y_1, \ldots y_{k-1}$, satisfying $d(x_i, y_i) \rightarrow 0$ as $\epsilon \downarrow 0$, which I denote by $d(x_i, y_i) \sim v(\epsilon)$ ("vanishes" with $\epsilon$). By abuse of notation, I use $v(\epsilon)$ to also denote vectors, whose length vanishes with $\epsilon$, e.g. $x_i - y_i = v(\epsilon)$.

Consider $P_Y(x_k)$. We have $$\langle P_Y(x_k), y_i \rangle = \langle x_k + v(\epsilon), x_i + v(\epsilon) \rangle$$$$= \langle x_k, v(\epsilon) \rangle + \langle x_i, v(\epsilon) \rangle + \langle v(\epsilon), v(\epsilon) \rangle = v(\epsilon)$$by bilinearity and Cauchy Schwartz. Hence, subtracting of the component of $P_Y(x_k)$ along $y_1, \ldots y_{k-1}$ results in a vector $P_Y(x_k)'$ of length $1 - v(\epsilon)$, and setting $y_k := \frac{P_Y(x_k)'}{\| P_Y(x_k)' \|}$, we have $$d(x_k, y_k) \leqslant d(x_k, P_Y(x_k)) + d( P_Y(x_k), P_Y(x_k)') + d(P_Y(x_k)', y_k) = v(\epsilon)$$as desired.

(2) Let $P_{X_i}$ be a Cauchy sequence, construct a subsequence $P_{Y_j}$ as follows:

Inductively, suppose we have chosen (i) $Y_{j-1} = X_{i}$, such that for all $k \geqslant i$, for any orthonormal basis $x_i$ for $X_i$, we can pick an orthonormal basis $y_i$ for $X_k$ such that $d(x_i, y_i) \leqslant \frac{1}{2^{j-1}}$ (we can do ths by 1), and (ii) an orthonormal basis for $Y_{j-1}$, $x_{j-1,i}$.

Pick $Y_j$ to satisfy for (i) blah blah $\leqslant \frac{1}{2^{j}}$, and (ii) an orthonormal basis $x_{j,i}$ for $Y_j$ satisfying $d(x_{j-1,i}, x_{j,i}) \leqslant \frac{1}{2^{j-1}}$

Why does this work? Well, by construction, for fixed $i$, the $x_{j,i}$ satisfy the distance between consecutive terms decreases geometrically, that they form a Cauchy sequence follows immediately from (i) the triangle inequality, and (ii) geometric series converge. Namely, for any $\epsilon$, we can pick $j$, such that $\sum_{k \geqslant j} 2^{-k} < \epsilon$, hence for any $n,m \geqslant j$, we have $$d(x_{n,i}, x_{m,i}) \leqslant \sum_{k=n}^{m-1} d(x_{k, i}, x_{k+1, i}) \leqslant \sum_{k = j}^\infty d(x_{k,i}, x_{k+1, i}) \leqslant \sum_{k \geqslant j} 2^{-k} < \epsilon$$

(3) By Cauchyness $x_{j,i} \rightarrow y_i$, by the (tautological) continuity of $\langle \rangle: V \times V \rightarrow V$, and $\| \; \|: V \rightarrow \mathbb{R}$, we deduce the $y_i$ are orthonormal. I claim $P_{Y_i} \rightarrow P_Y$, where $Y$'s just the span of the $y_i$. The proof is the "reverse" of 1, and proceeds similarly: namely, if $P_X, P_Y$ are projections, such that $X,Y$ admit orthonormal bases $x_i, y_i$ satisfying $d(x_i, y_i) \leqslant \epsilon$, then $d(P_X, P_Y) = v(\epsilon)$.

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For general knowledge, in terms of spacing \| gives a better result for a norm, compare: $$\|v\|\neq ||v||$$ –  Asaf Karagila May 23 '12 at 23:32
haha cheers yeah that looks worlds better –  uncookedfalcon May 24 '12 at 4:46