# Bound between distance between Rotation Matrices

Let $$\|\cdot\|_F$$ denote the Fröbenius norm on the set of $$d\times d$$ matrices. By restriction this induces a metric on $$SO(n)$$.

Let's make an observation. Since $$X\in SO(n)$$ is a rotation matrix then it is an isometry hence if $$\lambda$$ is an eigenvalue of $$A$$ with corresponding eigenevector $$x$$ we have that $$\|x\|=\|Ax\|=\|\lambda x\|= |\lambda| \|x\| \,\Rightarrow\, |\lambda|=1.$$ Therefore, we get the crude bound \begin{aligned} \sup_{X, Y \in SO(n)} \|X-Y\|_F \leq & \sup_{X,Y \in SO(n)} \|X\|_F + \|Y\|_F \\= & \sup_{X, Y \in SO(n)} \sqrt{ \sum_{i=1}^n \lambda_i(X) } + \sqrt{ \sum_{i=1}^n \lambda_i(Y) } \\= & 2\sqrt{n} , \end{aligned} where I use $$\lambda_i(X)$$ to emphasize the $$i^{th}$$ eigenvalue of $$X$$.

However, here are my two issues with this bound:

1. It is not specific to $$SO(n)$$ and applies to any set of linear isometries of $$\mathbb{R}^n$$,
2. It is clearly crude since it entirely disregards the distance between $$X$$ and $$Y$$ and only looks at their "norm" individually..

Is a sharp(er?) estimate for $$\sup_{X,Y \in SO(n)} \|X-Y\|_F,$$ known? Specifically, can we bound this quantity by $$1$$?

• What is $\|\cdot\|_F$?
– YCor
May 13, 2020 at 10:22
• @YCor I added a host of details...
– ABIM
May 13, 2020 at 10:46
• What is the Frobenius norm? May 13, 2020 at 10:52
• Well, but the bound is achieved on $Y=-X$ if $n$ is even... May 13, 2020 at 12:58
• And for odd $n$ we can use that the inequality is invariant w.r.t simultaneous left or right multiplication of X and Y and then reduce the problem to the even case. If follows that the bound for $n=2m+1$ is the same as for $n=2m$. Mar 2, 2021 at 10:28

First Point: The bound isn't sharp

Consider the case where $$n=2$$. The every matrix in $$SO(n)$$ is of the form $$A_{\theta} \triangleq \begin{pmatrix} cos(\theta) & -sin(\theta)\\ sin(\theta) & cos(\theta), \end{pmatrix}$$ for some $$\theta \in [0,2\pi]$$ (note: fun easy proof of compactness of $$SO(2)$$). In particular, $$\|A_0 - A_{\frac{\pi}{2}}\|_F = \left\|\begin{pmatrix} 1 & -1\\ 1 & 1, \end{pmatrix}\right\|_F= \sqrt{4} = 2.$$ So $$2\sqrt{2}$$ is not sharpe.

Second point: $$1$$ cannot be achieved This also shows that $$1$$ cannot be achieved if $$SO(n)$$ is metrized by the Fröbenius norm, since we just got $$2$$...

Third point/question: $$1$$ can maybe be Achieved if we instead consider the Spectral Norm Recall that the spectral (or Operator norm) of a $$n\times n$$ matrix is given by $$\|X\|_{\infty} = \max_{i=1,\dots,n} |\sigma_i(A)|.$$ Therefore, by your remark on the eigenvalues of any $$A \in SO(n)$$ we have that $$\sup_{X,Y \in So(n)}\, \|X-Y\|_{\infty} \leq 2.$$ However, a quick computation shows that $$\|A_0 - A_{\frac{\pi}{2}}\|_{\infty} = \sqrt{2}>1.$$ So $$1$$ cannot be achied..

Suggestion: If you're willing to take any metric induced by a norm on the set of $$n\times n$$ matrices then I would just use $$\|X-Y\|_n' := \frac1{2\sqrt{n}} \|X-Y\|_F.$$ Note that it generates the same topology on $$SO(n)$$ since all norms are equivalent on finite-dimensional normed spaces... So, if you can use this, then your bound will give you a metric induced by a norm which is uniformly bounded by $$1$$ on $$SO(n)$$!

Hopefully this works for you.

• I encourage you to follow-up and find out if the spectral norm can achieve value $1$, this would be interesting?
– ABIM
May 13, 2020 at 11:11