{ "schema_version": 1, "kind": "ConceptView", "producer": "concept-view-1", "concept_id": "urn:uuid:3ca9ee3c-aa64-49a8-8eac-e4e9fdde1f33", "labels": { "zh-CN": "Clebsch–Gordan 分解", "en": "Clebsch–Gordan decomposition" }, "alternative_labels": [], "scope_note": "", "identity_status": "active", "identity_revision": "f5d8adbd602ea3f8d129f349d8539fa89b6ba731d643f77ba2f5c990ee72ccb8", "occurrences": [ { "concept_id": "urn:uuid:3ca9ee3c-aa64-49a8-8eac-e4e9fdde1f33", "book_id": "woit-qtgr", "edition_id": "2025-10-20", "lang": "zh-CN", "route": "books/woit-qtgr/2025-10-20/zh/chapter-09/04-tensor-products-of-representations", "anchor": "first-bee1103babcc35ba870b-zh", "chapter": "9", "section": "9.4", "offset": 1461, "range": [ 8545, 9441 ], "source_revision": "1d4dd7dd9c43a9dd38d15e6377795a8ce87e81c38569e13b0b5f9485d63303a1", "source_category": "machine_draft", "review_status": "machine-draft-not-reviewed", "role": "theorem", "role_basis": { "kind": "registered-source-environment", "registered_position": { "route": "books/woit-qtgr/2025-10-20/zh/chapter-09/04-tensor-products-of-representations", "lang": "zh-CN", "section": "9.4", "signal": "named-theorem", "label": "Clebsch–Gordan 分解", "english": "", "offset": 1320 } }, "label": "> [!theorem] 定理 9.1（Clebsch–Gordan 分解）", "context": { "range": [ 8545, 9441 ], "quote": "\n> [!theorem] 定理 9.1（Clebsch–Gordan 分解）\n> 张量积表示分解为不可约表示的直和：\n>\n> $$\n> \\begin{aligned}\n> (\\pi_{V^{n_1}\\otimes V^{n_2}},V^{n_1}\\otimes V^{n_2})\\cong{}&\n> (\\pi_{n_1+n_2},V^{n_1+n_2})\\oplus(\\pi_{n_1+n_2-2},V^{n_1+n_2-2})\\\\\n> &\\oplus\\cdots\\oplus(\\pi_{|n_1-n_2|},V^{|n_1-n_2|})\n> \\end{aligned}\n> $$\n\n> [!proof]- 证明\n> 证明这一结果的一种方法是使用最高权理论、升算符与降算符以及 Casimir 算符公式。下一节将用特征标给出一个更简单的论证。这里先概述一种证明思路；更多细节可参见例如 [71] 的第 5.2 节。\n>\n> 设$v_{n_1}\\in V_{n_1}$和$v_{n_2}\\in V_{n_2}$是两个表示的最高权向量。那么，$v_{n_1}\\otimes v_{n_2}$是张量积表示中的最高权向量：它被$\\pi^{\\prime}_{n_1+n_2}(S_+)$湮灭，权为$n_1+n_2$。因此，$(\\pi_{n_1+n_2},V^{n_1+n_2})$会出现在分解中。将$\\pi^{\\prime}_{n_1+n_2}(S_-)$作用于$v_{n_1}\\otimes v_{n_2}$，可得到该表示中其余向量的一组基。在权$n_1+n_2-2$处，可以找到另一个与$(\\pi_{n_1+n_2},V^{n_1+n_2})$中的向量正交的最高权向量。对它作用下降算符，就得到$(\\pi_{n_1+n_2-2},V^{n_1+n_2-2})$。同理，在权$n_1+n_2-4$处还能找到另一个正交的最高权向量，并得到另一个表示。如此继续，直到权$|n_1-n_2|$为止。\n", "quote_sha256": "e92252c6fa85c907521406298b7f9df34048bb3689ea267561fa4471012539f9", "environment": "theorem", "excerpt_omitted": false }, "parallel_readings": [ { "route": "books/woit-qtgr/2025-10-20/en/chapter-09/04-tensor-products-of-representations", "lang": "en", "source_category": "machine_draft", "basis": "registered-book-edition-section-language-context; not-concept-equivalence" } ], "id": "occ-f76c49f853d356f9680a499f" }, { "concept_id": "urn:uuid:3ca9ee3c-aa64-49a8-8eac-e4e9fdde1f33", "book_id": "woit-qtgr", "edition_id": "2025-10-20", "lang": "en", "route": "books/woit-qtgr/2025-10-20/en/chapter-09/04-tensor-products-of-representations", "anchor": "first-243cb6d1e32d4cc7f484-en", "chapter": "9", "section": "9.4", "offset": 2029, "range": [ 8275, 9816 ], "source_revision": "4aa1a1d7f3fcb427daecdc46e53ab72991dfbeec261c3e7e58a5d923efb74b3f", "source_category": "machine_draft", "review_status": "machine-draft-not-reviewed", "role": "theorem", "role_basis": { "kind": "registered-source-environment", "registered_position": { "route": "books/woit-qtgr/2025-10-20/en/chapter-09/04-tensor-products-of-representations", "lang": "en", "section": "9.4", "signal": "named-theorem", "label": "Clebsch–Gordan decomposition", "english": "Clebsch–Gordan decomposition", "offset": 1888 } }, "label": "> [!theorem] Theorem 9.1 (Clebsch–Gordan decomposition)", "context": { "range": [ 8275, 9816 ], "quote": "\n> [!theorem] Theorem 9.1 (Clebsch–Gordan decomposition)\n> The tensor-product representation decomposes into irreducibles as\n>\n> $$\n> \\begin{aligned}\n> (\\pi_{V^{n_1}\\otimes V^{n_2}},V^{n_1}\\otimes V^{n_2})\\cong{}&\n> (\\pi_{n_1+n_2},V^{n_1+n_2})\\oplus(\\pi_{n_1+n_2-2},V^{n_1+n_2-2})\\\\\n> &\\oplus\\cdots\\oplus(\\pi_{|n_1-n_2|},V^{|n_1-n_2|}).\n> \\end{aligned}\n> $$\n\n> [!proof]- Proof\n> One proof uses highest-weight theory, the raising and lowering operators, and the formula for the Casimir operator. The next section gives a simpler argument using characters. Here is an outline; for more detail, see, for example, §5.2 of [71].\n>\n> Let $v_{n_1}\\in V_{n_1}$ and $v_{n_2}\\in V_{n_2}$ be highest-weight vectors for the two representations. Then $v_{n_1}\\otimes v_{n_2}$ is a highest-weight vector in their tensor-product representation: it is annihilated by $\\pi^{\\prime}_{n_1+n_2}(S_+)$ and has weight $n_1+n_2$. Thus, $(\\pi_{n_1+n_2},V^{n_1+n_2})$ occurs in the decomposition. Applying $\\pi^{\\prime}_{n_1+n_2}(S_-)$ to $v_{n_1}\\otimes v_{n_2}$ gives a basis for the remaining vectors in this representation. At weight $n_1+n_2-2$, one can find another highest-weight vector, orthogonal to the vectors in $(\\pi_{n_1+n_2},V^{n_1+n_2})$. Applying the lowering operator to it gives $(\\pi_{n_1+n_2-2},V^{n_1+n_2-2})$. Continuing in this way, one finds another orthogonal highest-weight vector at weight $n_1+n_2-4$, and obtains another representation. 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