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"quote": "后续章节将假定读者具备扎实的线性代数基础，我们也会用到该学科中的若干具体结论。其中有些内容在常见的线性代数课程中并不受重视，例如[对偶空间(dual space)](../../../../../concepts/linear-algebra/dual-space)所起的作用，以及把不同类别的可逆矩阵作为群来研究。目前讨论的[向量空间(vector space)](../../../../../concepts/linear-algebra/vector-space)都是有限维的。后文还会涉及无限维状态空间及其引出的各种问题。",
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"quote": "注意，这些对易关系与李代数 $\\mathfrak{su}(2)$ 的基向量 $X_1,X_2,X_3$ 所满足的[方程 3.5](../chapter-03/02-commutation-relations-for-pauli-matrices#equation-3-5) 完全相同，因此 $\\mathfrak{so}(3)$ 与 $\\mathfrak{su}(2)$ 是同构的李代数。二者都是带有相同[李括号(Lie bracket)](../../../../../concepts/lie-theory/lie-bracket)运算的[向量空间(vector space)](../../../../../concepts/linear-algebra/vector-space) $\\mathbf R^3$。这种运算在另一语境中也很熟悉：它就是标准基向量的[叉积(cross product)](../../../../../concepts/linear-algebra/vector-cross-product)：",
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"quote": "作为 $\\mathbb R$ 上的[向量空间(vector space)](../../../../../concepts/linear-algebra/vector-space)，$\\mathbb H$ 与 $\\mathbb R^4$ 同构。",
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"quote": "\n> [!proof]- 证明\n> 映射\n>\n> $$\n> L\\longmapsto\\mu_L\n> $$\n>\n> 显然给出了矩阵空间与二次多项式空间之间的[向量空间(vector space)](../../../../../concepts/linear-algebra/vector-space)同构。为证明它也是李代数同构，可利用泊松括号的雅可比恒等式：\n>\n> $$\n> \\begin{aligned}\n> &\\{\\mu_L,\\{\\mu_{L'},\\mathbf c_q\\cdot\\mathbf q+\\mathbf c_p\\cdot\\mathbf p\\}\\}\n> -\\{\\mu_{L'},\\{\\mu_L,\\mathbf c_q\\cdot\\mathbf q+\\mathbf c_p\\cdot\\mathbf p\\}\\}\\\\\n> &\\qquad=\\{\\{\\mu_L,\\mu_{L'}\\},\\mathbf c_q\\cdot\\mathbf q+\\mathbf c_p\\cdot\\mathbf p\\}\n> \\end{aligned}\n> $$\n>\n> 该式左侧为 $\\mathbf c_q''\\cdot\\mathbf q+\\mathbf c_p''\\cdot\\mathbf p$，其中\n>\n> $$\n> \\binom{\\mathbf c_q''}{\\mathbf c_p''}=(LL'-L'L)\\binom{\\mathbf c_q}{\\mathbf c_p}\n> $$\n>\n> 因此，右侧的线性映射由下式给出：\n>\n> $$\n> \\{\\mu_L,\\mu_{L'}\\}=\\mu_{[L,L']}\n> $$\n>\n> $\\square$\n",
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"quote": "\n> [!proof]- Proof\n> The map\n>\n> $$\n> L\\longmapsto\\mu_L\n> $$\n>\n> is a [vector space](../../../../../concepts/linear-algebra/vector-space) isomorphism from the matrices to the quadratic polynomials. To show that it is a Lie-algebra isomorphism, use the Jacobi identity for the Poisson bracket:\n>\n> $$\n> \\begin{aligned}\n> &\\{\\mu_L,\\{\\mu_{L'},\\mathbf c_q\\cdot\\mathbf q+\\mathbf c_p\\cdot\\mathbf p\\}\\}\n> -\\{\\mu_{L'},\\{\\mu_L,\\mathbf c_q\\cdot\\mathbf q+\\mathbf c_p\\cdot\\mathbf p\\}\\}\\\\\n> &\\qquad=\\{\\{\\mu_L,\\mu_{L'}\\},\\mathbf c_q\\cdot\\mathbf q+\\mathbf c_p\\cdot\\mathbf p\\}.\n> \\end{aligned}\n> $$\n>\n> The left-hand side is $\\mathbf c_q''\\cdot\\mathbf q+\\mathbf c_p''\\cdot\\mathbf p$, where\n>\n> $$\n> \\binom{\\mathbf c_q''}{\\mathbf c_p''}=(LL'-L'L)\\binom{\\mathbf c_q}{\\mathbf c_p}.\n> $$\n>\n> Therefore, the linear map on the right is given by\n>\n> $$\n> \\{\\mu_L,\\mu_{L'}\\}=\\mu_{[L,L']}.\n> $$\n>\n> $\\square$\n",
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