Cross-posted from MSE (https://math.stackexchange.com/questions/4886279/concentration-of-measure-on-spheres-with-respect-to-a-unitary-of-trace-approxima), where it hasn’t received any answer yet:

This question arose out of my attempt to understand how a unitary of trace approximately zero acts on the unit sphere of a $n$-dimensional Hilbert space. First, some context:

We note that, by concentration of measures for spheres, we have the following: Let $S^{n-1} = \{x \in \mathbb{R}^n: \|x\|_2 = 1\}$ denote the unit sphere of $\mathbb{R}^n$. Let $v_n \in S^{n-1}$ be randomly chosen according to the canonical probability measure on $S^{n-1}$. For any fixed $\epsilon > 0$, as $n \to \infty$, the probability that $v_n$ lies $\epsilon$-close to any given equator goes to $1$. See, for example, here.

Reframing this in terms of orthogonal matrices, we see that an equator is exactly the the intersection of the unit sphere with the invariant subspace of an orthogonal matrix whose eigenvalues are all $1$ except one $-1$ with multiplicity $1$. That is to say, we have the following: Let $v_n \in S^{n-1}$ be randomly chosen according to the canonical probability measure on $S^{n-1}$. Fix $\epsilon > 0$, and fix a sequence of matrices $T_n \in \mathbb{M}_n(\mathbb{R})$ s.t. $T_n$ is an orthogonal matrix with eigenvalue $1$ of multiplicity $n-1$ and eigenvalue $-1$ of multiplicity $1$ for all $n$. Then the probability that $\|T_nv_n - v_n\|_2 < \epsilon$ goes to $1$ as $n \to \infty$. By essentially the same proof, the complex version of this also holds, namely, the above result still holds if $v_n$ is instead randomly chosen on the unit sphere of $\mathbb{C}^n$ and $T_n \in \mathbb{M}_n(\mathbb{C})$ are unitary matrices with the same condition on eigenvalues.

Now, I’m interested in analogues of this for unitary matrices with (normalized) trace close to $0$. Drawing from the intuition from real Hilbert spaces and orthogonal matrices, it would seem that such matrices should have a large region on the unit sphere where the vectors in the region are sent to vectors approximately orthogonal to the original one. Then by a concentration of measure style argument, it should be the case that the probability that a uniformly random vector on the unit sphere will be sent to a vector orthogonal to it goes to $1$ as the dimension goes to $\infty$. I’ve been unable to prove it, however. The following is the precise question statement:

Let $v_n \in S^{2n-1} \subset \mathbb{C}^n$ be randomly chosen according to the canonical probability measure on $S^{2n-1}$. Fix $\epsilon > 0$, and fix a sequence of matrices $T_n \in \mathbb{M}_n(\mathbb{C})$ s.t. $T_n$ is a unitary matrix with $\frac{1}{n}\mathrm{Tr}(T_n) < \frac{\epsilon}{2}$ for all $n$. Is it the case then that the probability that $|\langle T_nv_n, v_n \rangle| < \epsilon$ goes to $1$ as $n \to \infty$?

Some thoughts on the matter: Again, this seems intuitively plausible. We also observe that the stronger condition that $\langle T_nv_n, v_n \rangle = 0$ is equivalent to (after diagonalizing $T_n$) $\sum_{i=1}^n \lambda_{ni}|(v_n)_i|^2 = 0$, where $\lambda_{ni}$ are the eigenvalues of $T_n$ and $(v_n)_i$ is the $i$-th coordinate of $v_n$. This is a complex co-dimension $1$ condition, same as the equator example in the complex case. Furthermore, this condition is certainly satisfiable. Indeed, $\frac{1}{n}\mathrm{Tr}(T_n) < \frac{\epsilon}{2}$ implies that any $v_n$ with $|(v_n)_i| = \frac{1}{\sqrt{n}}$ for all $i$ satisfies the desired condition. And furthermore any $v$ that is $\frac{\epsilon}{4}$-close to such an $v_n$ would satisfy the desired condition as well, again similar to the equator example.

However, I’ve been unable to make much progress on this as I’m not quite familiar with the intricacies of the concentration of measure, and the same method as in the equator example doesn’t exactly apply.

Any help on this is highly appreciated!