What is the $n$-dimensional matrix representation of an element $g\in U(1)$? Is it simply the $n$-dimensional identity matrix times an exponential factor $\text{e}^{\text{i}\alpha}$? This would fit the condition that its determinant is a complex number with norm one.
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3There is no such thing as the $n$-dimensional representation of $U(1)$. – user10354138 Nov 06 '18 at 03:08
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Hmm. I meant: for e.g. $U(2)=U(1)\otimes SU(2)$, what does the n-dimensional representation of the U(1)-part look like? – ersbygre1 Nov 06 '18 at 03:12
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Since $U(1)$ is compact, its (continuous, complex, finite-dimensional) representations are unitary and thus the direct sum of irreps by the Peter-Weyl theorem. By the Schur lemma, such irreps are all $1$-dimensional; that is, they're given by $\chi(t) = t^n$ (identifying $U(1)$ with the unit circle in $\mathbb{C}$) for some integer $n$. The representations of $U(1)$ are thus given by $t \to (t^{n_1}, \dots, t^{n_k})$ over some basis of $\mathbb{C}^k$ for $n_i\in \mathbb{Z}$.
anomaly
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I see. So generally, $n_i\neq n_j$? How can an element of U(1) be parametrized by $k$ numbers? – ersbygre1 Nov 06 '18 at 04:02
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The $n_i$ are arbitrary integers. It's the representation that's parametrized by $k$ integers, not $U(1)$ itself; the representation is generally faithful, but it's not a bijection onto $\mathbb{C}^n$ or $GL_n(\mathbb{C})$. – anomaly Nov 06 '18 at 04:58
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The argument of the character in question, identifying $U(1)$ with the unit circle in $\mathbb{C}$. – anomaly Nov 06 '18 at 07:07