# Quantifying the angle metric on the Grassmannian in terms of the norm on the exterior power

Let $$V$$ be a finite-dimensional Hilbert space and $$Gr_k(V)$$ the Grassmannian of $$k$$-dimensional subspaces of $$V$$. The $$k$$th exterior power $$\bigwedge^k(V)$$ can be equipped with a scalar product by extending $$\langle v_1\wedge \cdots \wedge v_k, w_1\wedge \cdots \wedge w_k\rangle_{\bigwedge^k(V)} := \det{\left(\langle v_i,w_j\rangle_{V}\,_{i,j}\right)}$$ bilinearly to all $$k$$-vectors.

The Grassmannian can be equipped with a metric which measures the angle (more exactly I think the sine of the angle) between two $$k$$-dimensional subspaces by $$d(U,W) := \left\| P_U-P_W \right\|$$ where $$P_U:V\to U$$ is the projection onto U and the norm is the operator norm (see for example this discussion and Equation (3), p. 3428 in the article linked therein).

I know that there is the Plücker embedding which maps $$G_k(V)$$ to the projectivisation of the $$k$$th exterior power. In fact, the image of the Plücker embedding corresponds to the simple $$k$$-vectors.

For example, if $$U$$ and $$W$$ coincide then the $$k$$-wedge of two distinct bases are off by a scalar factor (the determinant of the change of basis matrix). So it would be nice to have something like $$d(U,W) \leq \left\|u_1\wedge\cdots\wedge u_k - w_1\wedge\cdots\wedge w_k\right\|_{\bigwedge^k(V)}$$ for any bases $$(u_i)$$ and $$(w_i)$$ of $$U$$ and $$W$$.

Is there a way of estimating the angle metric on the Grassmannian in terms of the norm on the exterior power?

$$\newcommand{\R}{\mathbb{R}} \newcommand{\U}{\mathbf{U}} \newcommand{\W}{\mathbf{W}} \newcommand{\A}{\mathbf{A}} \newcommand{\tr}{\mathrm{tr}} \newcommand{\rank}{\mathrm{rank}}$$ Choosing an arbitrary basis, we may assume $$V=\R^n$$. We will show that the mapping \begin{align} \mathbb{S}\big(\bigwedge\nolimits^k_s(\R^n)\big)&\to Gr_k(\R^n)\\ \sigma=u_1\wedge\ldots\wedge u_k &\mapsto\operatorname{span}\{u_1,\ldots,u_k\}=\{v\in\R^n\colon v\wedge\sigma=0\} \end{align} is continuous w.r.t. the angle metric on the Grassmannian and induced inner product metric on the exterior power. The following lemma implies this assertion. Here $$\mathbb{S}(X)$$ denotes the set of elements in $$X$$ of unit length and $$\bigwedge\nolimits^k_s(\R^n)$$ is the set of simple $$k$$-vector.

Let $$U,W\in Gr_k(\R^n)$$ and choose orthonormal bases $$u_1,\ldots,u_k$$ and $$w_1,\ldots,w_k$$ of $$U$$ and $$W$$, respectively. Let further $$\U,\W\in\R^{n\times k}$$ be the matrices with columns $$(u_i)$$ and $$(w_i)$$, respectively. Finally, define the unit $$k$$-vectors $$\sigma=u_1\wedge\cdots\wedge u_k$$ and $$\tau=w_1\wedge\cdots\wedge w_k$$.

Lemma \begin{align} d(U,W) \leq \sqrt{k}\left\|\sigma-\tau\right\|_{\bigwedge^k(\R^n)} \end{align}

Proof First, note that by the Cauchy-Schwartz inequality \begin{align} \det(\U^T\W)=\langle\sigma,\tau\rangle_{\bigwedge^k(\R^n)} \leq \|\sigma\|_{\bigwedge^k(\R^n)}\|\tau\|_{\bigwedge^k(\R^n)}=1. \tag{1}\label{eq:CS_ineq} \end{align} Secondly, the Euclidean operator norm of a linear operator $$A$$ is bounded by its Frobenius norm, i.e. $$\|A\| \leq \|A\|_F$$. This follows from the fact that the Euclidean norm and the Frobenius norm of a vector coincide and the submultiplicativity of the Frobenius norm. Therefore, \begin{align} d(U,W)^2 &= \left\| P_U-P_W \right\|^2 \leq \left\| P_U-P_W \right\|_F^2 =\left\| P_U\right\|_F^2 +\left\|P_W \right\|_F^2 -2\left\langle P_U,P_W \right\rangle_F. \end{align} Using that $$P_U=\U^T\U$$ and $$P_W=\W^T\W$$ are idempotent, we get that $$\left\| P_U\right\|_F=\rank(P_U)=k$$ and also $$\left\| P_W\right\|_F=k$$. Thus, \begin{align} \left\| P_U-P_W \right\|_F^2 &=2k-2\tr(\U\U^T\W\W^T) =k\big(2-\tfrac2k \tr((\U^T\W)^T\U^T\W)\big). \end{align} Due to the inequality of the arithmetic and geometric mean, it holds that \begin{align} \sqrt[k]{\det(\A)} \leq \tfrac1k \tr(\A) \end{align} for any matrix $$\A\in\R^{k\times k}$$ (because the trace is the sum and the determinant the product of the eigenvalues of $$\A$$). If $$\det(\A)\leq1$$ then for any $$k\geq2$$ \begin{align} \sqrt[2]{\det(\A)} \leq \tfrac1k \tr(\A). \tag{2}\label{eq:AGM} \end{align} Therefore we can estimate using \eqref{eq:CS_ineq} and \eqref{eq:AGM} for $$\A=(\U^T\W)^T\U^T\W$$ \begin{align} \left\| P_U-P_W \right\|_F^2 &\leq k\big(2-2\sqrt{\det((\U^T\W)^T\U^T\W)}\big) =k\big(2-2\det(\U^T\W)\big) \end{align} Finally, direct computation shows that \begin{align} \|\sigma-\tau\|_{\bigwedge^k(\R^n)}^2=\|\sigma\|_{\bigwedge^k(\R^n)}^2+\|\tau\|_{\bigwedge^k(\R^n)}^2 - 2 \langle\sigma,\tau\rangle_{\bigwedge^k(\R^n)}^2 = 2-2\det(\U^T\W) \end{align} which concludes the proof.

The next statement is not really needed but just included here for completion.

Corollary For arbitrary $$\sigma,\tau$$ arbitrary $$k$$-vectors such that they span the subspaces $$U$$ and $$W$$, it holds that \begin{align} d(U,W) \leq \sqrt{k}\left\|\tfrac{\sigma}{\left\|\sigma\right\|_{\bigwedge^k(\R^n)}}- \tfrac{\tau}{\left\|\tau\right\|_{\bigwedge^k(\R^n)}}\right\|_{\bigwedge^k(\R^n)} \leq \sqrt{k}\frac{\left\|\sigma-\tau\right\|_{\bigwedge^k(\R^n)}}{\left\|\sigma\right\|_{\bigwedge^k(\R^n)}\left\|\tau\right\|_{\bigwedge^k(\R^n)}} \end{align}

Proof If $$\sigma=a_1\wedge\cdots\wedge a_k$$ for an arbitrary basis $$(a_1,\ldots,a_k)$$ of $$U$$ then we can change basis to an orthonormal basis (u_1,\ldots,u_k), e.g. via QR-decomposition of $$\A=\U\mathbf{R}$$. Then $$\sigma=\det(\mathbf{R})u_1\wedge\cdots\wedge u_k$$ and $$\left\|\sigma\right\|_{\bigwedge^k(\R^n)}=\det(\mathbf{R})$$. Then use the lemma for $$\sigma/\left\|\sigma\right\|_{\bigwedge^k(\R^n)}$$ and $$\tau/\left\|\tau\right\|_{\bigwedge^k(\R^n)}$$. The second inequality follows from the fact that $$2ab\leq a^2+b^2$$ for any real numbers $$a,b$$.