Suppose $\pi:U(n)\rightarrow GL(V)$ is a positive-dimensional irreducible representation of the unitary group. Given $\varepsilon>0$, how could one rigorously show that the probability that $|\det(I-\pi(g))|<\varepsilon$ as $g$ is chosen in $U(n)$ (with respect to the Haar probability measure) is small (in a quantifiable way, hopefully)?
Edit: In light of David's answer, what if $n$ is large? Under what conditions do irreducible representations lead to the above phenomenon?
Best Answer
This isn't true. Let $V$ be the standard two dimensional representation of $U(2)$ and let $W = (\det V)^{-1} \otimes \mathrm{Sym}^2(V)$. If the eigenvalues of $g$ acting on $V$ are $z_1$ and $z_2$, then the eigenvalues of $g$ acting on $W$ are $z_1 z_2^{-1}$, $1$ and $z_1^{-1} z_2$. So $\det (\mathrm{Id} - \pi_W(g))=0$ for all $g$.
Okay, general nonsense about representation theory of $U(n)$ first. Let $T$ be the torus of diagonal matrices inside $U(n)$, and write $(z_1, \ldots, z_n)$ for coordinates on $T$. Irreducible representations of $U(n)$ are called $V_{\lambda}$ where $\lambda$ is a weakly decreasing $n$-tuples of integers: $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$. When we decompose $V_{\lambda}$ into characters of $T$, the character $\prod z_j^{\mu_j}$ occurs if and only if $(\mu_1, \ldots, \mu_n)$ is in the convex hull of the $n!$ permutations of $(\lambda_1, \ldots, \lambda_n)$.
The example I gave above is $\lambda = (1,-1)$, so $(0,0)$ is in the convex hull of $(1,-1)$ and $(-1, 1)$.
Lemma Let $(a_1, \ldots, a_n) \in \mathbb{R}^n$. Then $(0,0,\ldots,0)$ is in the convex hull of the $n!$ permutations of $(a_1, \ldots, a_n)$ if and only if $\sum a_i=0$.
Proof If $\sum a_i=0$, then the average of the permutation is the $0$-vector. If $\sum a_i = h \neq 0$, then all the permutations are in the affine hyperplane where the coordinates sum to $h$. $\square$
So you want to study representations where $\sum \lambda_i = h \neq 0$. The integer $h$ can be thought be described by saying that the central matrix $z \mathrm{Id}_n$ acts by $z^h$, so you want the center of $U(n)$ to act nontrivially.
Given this, what can we say? The multiplicity of the character $(z_1, \ldots, z_n) \mapsto \prod z_j^{\mu_j}$ in the representation $V_{\lambda}$ is called the Kotska number $K_{\lambda \mu}$. So
$$\det (1-\pi(g)) = \prod_{\mu} (1-\prod_j z_j^{\mu_j})^{K_{\lambda \mu}}.$$
Every conjugacy class of $U(n)$ meets the torus $T$, and conjugacy classes correspond to unordered $n$-tuples $\{ z_1, z_2, \ldots, z_n \}$. The volume of the conjugacy class through $(z_1, \ldots, z_n)$ is proportional to $\prod_{i \neq j} |z_i - z_j| = \prod_{i<j} |z_i-z_j|^2$ (remember that the $z_i$ are complex numbers of norm $1$).
So you want some way to control the integral of $\prod_{i<j} |z_i-z_j|^2$ over the region where $\prod_{\mu} (1-\prod_j z_j^{\mu_j})^{K_{\lambda \mu}}$ is small.
If you instead were looking the eigenvalue of $\pi(g)$ closest to $1$, so you wanted $\min_{\mu} |1-\prod_j z_j^{\mu_j}|$ to be large, I think you might have a chance. The assumption that $\sum \mu_j = h \neq 0$ would mean that multiplying $(z_1, \ldots, z_n)$ by $w$ would multiply $\prod_j z_j^{\mu_j}$ by $w^h$ while leaving the volume of the conjugacy class unchanged. The probability for an individual $\mu$ would just be the probability that a random element of $S^1$ wasn't too small, and then you could use a union bound.
But the different terms $(1-\prod_j z_j^{\mu_j})$ will be correlated in complicated ways, so I don't know how you'll control their product.