Is this a metric on the Grassmannian Manifold?

Question

Let $m>n$ and consider the Set $$S_{m,n}=\{A \in \mathbb{R}^{m \times n}\lvert A^TA=I_n \}.$$ Does the function $d\colon S_{m,n} \times S_{m,n} \rightarrow \mathbb{R}$ defined by $$d(A,B)=\sqrt{1-\det(A^TB)}$$ define a pseudometric on $S_{m,n}$? (A pseudometric satisfies all conditions of a metric except that two elements can also have distance zero.) Consider the equivalence relation $A \sim B$ if there exist an orthogonal $Q \in \mathbb{R}^{n \times n}$ with $A=BQ$. The set $S_{m,n}$ together with the equivalence relation can be identified with the grassmannian manifold $Gr(n,\mathbb{R}^m)$. Does $d$ define a metric on $Gr(n,\mathbb{R}^m)$? This question interests me because im trying to approximate (interpolate) functions which take values in the grassmanian manifold and the above metric would open up a possibility for approximating such functions. The difficult part is the triangle-inequality, i.e. for all $A,B,C \in S_{m,n}$ we need to prove that $$\sqrt{1-\det(A^TC)}\leq \sqrt{1-\det(A^TB)}+\sqrt{1-\det(B^TC)}.$$

Thanks for any help in advance.

Answer 1

EDIT Actually, Cauchy-Binet suffices as the OP notices in the comments. I'll leave my overkill proof here for your amusement.

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The proof below appeals to a famous result of Schoenberg (I've simplified the statement a bit), and basic linear algebra.

Schoenberg's theorem (see e.g., [Prop. 3.2, 1]). Let $X$ be a nonempty set and $\psi: X \times X \mapsto \mathbb{R}$ be positive definite kernel. Then, there exists an RKHS $H$ and a map $\varphi : X \to H$ such that \begin{equation*} \|\varphi(x)-\varphi(y)\|_H^2 = \frac{1}{2}[\psi(x,x)+\psi(y,y)] - \psi(x,y). \end{equation*}

We show that the function $\psi(A,B) = \det(A^TB)$ is positive definite, which as a result of Schoenberg's theorem shows that \begin{equation*} 1-\det(A^TB) = \|\varphi(A)-\varphi(B)\|_H^2, \end{equation*} from which the triangle inequality is immediate.

To prove the positive definiteness of $\psi$, we show that it is an inner-product by invoking the Cauchy-Binet formula (using Wikipedia's notation, except that for us $A$ is $m \times n$): \begin{equation*} \det(A^TB) = \sum_{S \in \binom{[m]}{n}} \det(A^T_{[n],S})\det(B_{S,[n]}) = \sum_{S \in \binom{[m]}{n}} \det(A_{S,[n]})\det(B_{S,[n]}) = \langle \phi(A), \phi(B)\rangle. \end{equation*}

[1] C. Berg, J. P. R. Christensen, and P. Ressel. Harmonic Analysis on Semigroups: Theory of Positive Definite and Related Functions, Springer GTM 100, 1984.

Answer 2

I believe this is true, this inequality has interesting geometry. However it is not an answer:

Let us see m>n=1. take inner product $(a,b)=\cos \gamma, (b,c)=\cos \alpha, (a,c)=\cos b$, by Euclidean geometry it is obvious $\alpha+\gamma>\beta$

To prove : $$\sqrt{1-\det(A^TC)}\leq \sqrt{1-\det(A^TB)}+\sqrt{1-\det(B^TC)}.$$

$$\sqrt{1-\det(A^TB)}+\sqrt{1-\det(B^TC)}\geq \sqrt{2-\det(A^TB)-det(B^{T}C)}.$$

It is suffices to prove: $$2-\det(A^TB)-det(B^{T}C)\geq 1-det(A^TC)$$.

For n=1, it is equivalent to prove the following: $$\sin^{2}(\frac{\beta}{2})\leq \sin^{2}(\frac{\alpha}{2})+\sin^{2}(\frac{\gamma}{2})$$

Notice the fact that $\alpha+\gamma>\beta$, this is obvious.

When $n\geq 2$, I think it is better to consider the n-dimension subspace in $R^{m}$, $\det(A^TB)$ is just cosin angel of two subspaces.