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        "quote": "> [!exercise] <a class=\"object-permalink\" href=\"#problem-19-18\">习题 19.18</a> \n>\n> (a) 证明一个李代数 $\\mathcal{A}$ 的所有李代数自同构构成的群是 $\\operatorname { A u t } ( { \\mathcal { A } } ) \\subseteq G L ( { \\mathcal { A } } )$ 的一个李子群\n>\n> (b) 若线性算符 $D:\\mathcal{A}\\to\\mathcal{A}$ 满足 $D [ X , Y ] = [ D X , Y ] + [ X , D Y ]$，则称其为 $\\mathcal{A}$ 上的 <u id=\"concept-19-derivation\" class=\"concept-term\" data-concept-identity=\"urn:uuid:d15da6db-7ba5-435d-9eec-3740a703c137\" data-concept-route=\"/concept-identities#concept-d15da6db-7ba5-435d-9eec-3740a703c137\">导子(derivation)</u>。证明 $\\mathcal{A}$ 的所有导子构成的集合形成一个李代数，$\\partial(\\mathcal{A})$，即 $\\operatorname{Aut}(\\mathcal{A})$ 的李代数。",
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