## What is the probability $\Pr(||U-I||_{op}<\varepsilon)$ of a Haar-random unitary being close to the identity?

7

3

If one generates an $$n\times n$$ Haar random unitary $$U$$, then clearly $$\Pr(U=I)=0$$. However, for every $$\epsilon>0$$, the probability $$\Pr(\|U-I\|_{op}<\varepsilon)$$ should be positive. How can this quantity be computed?

Exatly? Approximately? Upper/lower bounds? Analytically? Numerically? – Norbert Schuch – 2020-08-14T11:20:07.917

3https://arxiv.org/abs/1506.07259 could contain some useful information or techniques (they compute said quantity, but for a different distance measure). – Norbert Schuch – 2020-08-14T11:27:08.827

Let's say we want an analytical lower-bound? – Calvin Liu – 2020-08-15T15:08:10.603

Are you saying this randomly, or is this what you really care about? Also, do you care about some specific $n$, any $n$, or maybe the behavior for large $n$? (I think if there is a motivation behind this question, it would be helpful to know it.) – Norbert Schuch – 2020-08-15T15:26:13.313

Note that the operator norm distance is smaller than the Frobenius distance, to the corresponding epsilon-ball is larger. So an exact number or lower bound derived in the paper I quoted above will also give lower bounds for the operator norm distance. – Norbert Schuch – 2020-08-15T15:27:41.940

I specifically care about an analytical lower-bound for n=4. – Calvin Liu – 2020-08-15T15:39:51.427

Ah! Then why don't you say that? -- Have you checked if said paper gives bounds/results for n=4? -- I guess you care about small $\varepsilon$? – Norbert Schuch – 2020-08-15T16:41:14.347

What values of $\epsilon$ are you interested in? On average $U$ and $I$ are going to be almost maximally far apart, i.e. the expected value of the operator norm will be almost maximal $\mathbb{E}U[|U-I|\infty]\approx 2$, and Haar random unitaries exponentially concentrate around the average. You can write down lower bounds on the probability you're interested in, they'll just be extremely small for values of $\epsilon$ a little less than 2. – 4xion – 2020-08-25T01:38:54.313

1@CalvinLiu I'll leave this as a comment having not carefully worked out the details, but you can think of this probability as computing the ratio of the volume of an $\epsilon$-ball around $I$ to the volume of the unitary group (with respect to the operator norm), equivalently you can think about this as the 1/size of an $\epsilon$-net for $U(n)$. Very roughly, this is going to be ${\rm Pr}(|U-I|_\infty \leq \epsilon) \sim (n/\epsilon^2)^{-n^2}$. – 4xion – 2020-08-25T15:36:59.707

1If you need a rigorous lower bound you should be able to compute this more precisely (look up refs related to volumes of balls in the unitary group and $\epsilon$-nets) – 4xion – 2020-08-25T15:38:30.523