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 $1$?

  • $\begingroup$ What is $\|\cdot\|_F$? $\endgroup$
    – YCor
    May 13, 2020 at 10:22
  • $\begingroup$ @YCor I added a host of details... $\endgroup$
    – ABIM
    May 13, 2020 at 10:46
  • $\begingroup$ What is the Frobenius norm? $\endgroup$ May 13, 2020 at 10:52
  • 1
    $\begingroup$ Well, but the bound is achieved on $Y=-X$ if $n$ is even... $\endgroup$ May 13, 2020 at 12:58
  • 1
    $\begingroup$ 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 $n=2m$. $\endgroup$
    – user35593
    Mar 2, 2021 at 10:28

1 Answer 1


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.

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

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