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?



Irreducible representations of U(n) and probability of being close to having fixed points
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.


