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"quote": "\n> [!theorem] 拓展\n>\n> *Digression. 严格说来，群 $U(n)$ 并不是乘积群 $SU(n)\\times U(1)$。若把 $U(1)$ 识别为 $U(n)$ 中由 $e^{i\\theta}\\mathbf{1}$ 构成的子群，那么矩阵*\n>\n> $$\n> e^{i\\frac{m}{n}2\\pi}\\mathbf{1}\n> $$\n>\n> *当 $m$ 为整数时同时属于 $SU(n)$ 和 $U(1)$。因此，$U(n)$ 并非这两个群的直积；它是一个[半直积群(semi-direct product group)](../chapter-18/02-semi-direct-product-groups#woit-semi-direct-product-group)的例子。第 18 章将讨论这类群。*\n",
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