Continuous homomorphisms $U(n) \to \mathbb{T}$.

Question

Let $U(n)$ be the group of $n \times n$ unitary matrices, and let $\mathbb{T} = \{ z\in \mathbb{C} : |z|=1\}$ be the circle group. I would like to say that every continuous homomorphism $U(n) \to \mathbb{T}$ is given by $u \mapsto \det(u)^k$ for some $k \in \mathbb{Z}$. Assuming this is even true, can we prove this without appealing to some general description of the irreducible representations of $U(n)$?

Answer 1

The derived subgroup of $U(n)$ is $SU(n)$, so any homomorphism into $S^1$ factors through the projection $U(n)\to U(n)/SU(n)$, whose codomain is isomorphic to $S^1$. Your question then amounts to asking what are the continuous homomorphisms $S^1\to S^1$. These are parametrized by the integers. In particular, your guess is correct.

Answer 2

Notation: Let $diag(z_1,z_2,\cdots,z_n)$ denote the diagonal matrix of order $n\times n$ with diagonal entries $z_1,z_2,z_3, \cdots, z_n$ respectively.