# Is this a metric on the Grassmannian Manifold?

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.

• It is obvious that when n=1. Sep 7, 2013 at 14:20
• Minor observation: You are building a metric on the oriented Grassmannian, not the Grassmannian. If $S$ is an $n \times n$ orthogonal matrix with determinant $-1$, then $A$ and $AS$ represent the same point of the Grassmannian, but $\det(A^T (AS)) = \det(S) = -1$, so your formula gives $\sqrt{2}$, not $0$. Sep 7, 2013 at 15:05
• @David: to get a distance function for the grassmanian one has to add an absolute value, i.e. $$d(A,B)=\sqrt{1-|\det(A^TB)|}.$$ Sep 8, 2013 at 13:54
• @user35593: but to show this to be a metric (with the abs) value, requires slightly more work than just an invocation of CB.... Sep 10, 2013 at 15:38
• @DavidESpeyer Can you please give a reference, why this is a correct distance for oriented Grassmannian? Thanks! Oct 17, 2019 at 21:10

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

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.

• After seeing the Cauchy-Binet formula I found a proof without Schoenberg' theorem: Let $a,b,c \in \mathbb{R}^{\binom{m}{n}}$ be definded by terms from the Cauchy-Binet formula, i.e. $\det(A_{S,[n]})$. Then $\|a\|_2^2=\|b\|_2^2=\|c\|_2^2=1$. The triangle inequality for $\mathbb{R}^{\binom{m}{n}}$ gives $$\sqrt{\langle a-c,a-c \rangle}\leq \sqrt{\langle a-b,a-b \rangle}+\sqrt{\langle b-c,b-c \rangle}$$ which boils down to the desired inequality. Sep 7, 2013 at 20:40
• That is exactly the way to go! so Cauchy-Binet suffices, very nice, I should have thought about the problem a bit before shooting off my answer! Sep 7, 2013 at 20:51

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.

• The statement with the sinuses is wrong for e.g. $\alpha=\gamma=44^{\circ}$ and $\beta=90^{\circ}$. The squareroot is important. However thanks for your contribution. For the case $n\geq 2$ how do you define an angle between two subspaces? Sep 7, 2013 at 17:23
• Just consider a triangular pyramidal with three line a,b,c. the corresponded angle $$\alpha,\beta,\gamma$$, we will have:$$\alpha+\beta>\gamma$$, and $$\alpha+\gamma>\beta$$, and $$\beta+\alpha>\gamma$$. I f we did not use the simple inequality $$\sqrt{x+y}\leq \sqrt{x}+\sqrt{y}$$, we need to prove $$\sin\frac{\gamma}{2}\leq \sin\frac{\beta}{2}+\sin\frac{\alpha}{2}$$ Sep 8, 2013 at 0:33