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:

- It is not specific to $SO(n)$ and applies to any set of linear isometries of $\mathbb{R}^n$,
- 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 $1$?

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