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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Lie group–Lie algebra correspondence ： ウィキペディア英語版 Lie group–Lie algebra correspondence In mathematics, Lie group–Lie algebra correspondence allows one to study Lie groups, which are geometric objects, in terms of Lie algebras, which are linear objects. In this article, a Lie group refers to a real Lie group. For the complex and ''p''-adic cases, see complex Lie group and ''p''-adic Lie group. In this article, manifolds (in particular Lie groups) are assumed to be second countable; in particular, they have at most countably many connected components. == Basics == Given a Lie group ''G'', let $\backslash operatorname(G)$ be its associated Lie algebra—the Lie subalgebra of the Lie algebra of vector fields on ''G'' that consists of vector fields ''X'' that are invariant under left translations; that is, for any ''g'', ''h'' in ''G'', :$(dl\_g)\_h(X\_h)\; =\; X\_$ where $l\_g:\; G\; \backslash to\; G,\; x\; \backslash mapsto\; gx$ and $(dl\_g)\_h:\; T\_h\; G\; \backslash to\; T\_\; G$ is the differential of $l\_g$ between tangent spaces.〔In other words, a vector field on ''G'' is left-invariant if it is $l\_g$-related to itself for any ''g'' in ''G''.〕 The left-invariance amounts to the fact that the vector bundle map over ''G'' :$\backslash phi:\; G\; \backslash times\; \backslash operatorname(G)\; \backslash to\; TG$ given by $\backslash phi\_g(X)\; =\; X\_g$ is an isomorphism (note φ makes $TG$ a Lie group). It follows that the canonical map :$\backslash operatorname(G)\; \backslash to\; T\_e\; G,\; \backslash ,\; X\; \backslash mapsto\; X\_e$ is an isomorphism of vector spaces and one usually identifies $\backslash operatorname(G)$ with $T\_e\; G$. (There is also a third incarnation of $\backslash operatorname(G)$; see "related constructions" below.) In particular, the dimension of ''G'' as a real manifold is the dimension of the vector space $\backslash operatorname(G)$, and $\backslash operatorname(G)\; =\; \backslash operatorname(G^0)$ where $G^0$ is the connected component of the identity element. If :$f:\; G\; \backslash to\; H$ is a Lie group homomorphism, then its differential at the identity element :$df\; =\; df\_e:\; \backslash operatorname(G)\; \backslash to\; \backslash operatorname(H)$ is a Lie algebra homomorphism (brackets go to brackets), which has the following properties: *$\backslash operatorname(\backslash operatorname(f))\; =\; \backslash operatorname(df)$.〔More generally, if ''H''' is a closed subgroup of ''H'', then $\backslash operatorname(f^(H\text{'}))\; =\; (df)^(\backslash operatorname(H\text{'})).$〕 *If the image of ''f'' is closed,〔This requirement cannot be omitted; see also http://math.stackexchange.com/questions/329753/image-of-homomorphism-of-lie-groups〕 then $\backslash operatorname(\backslash operatorname(f))\; =\; \backslash operatorname(df)$ and the first isomorphism theorem holds: ''f'' induces the isomorphism of Lie groups: ::$G/\backslash operatorname(f)\; \backslash to\; \backslash operatorname(f)$. *The product rule holds: if $f,\; g:\; G\; \backslash to\; H$ are Lie group homomorphisms, then $d(fg)(X)\; =\; df(X)\; +\; dg(X).$ *The chain rule holds: if $f:\; G\; \backslash to\; H$ and $g:\; H\; \backslash to\; K$ are Lie group homomorphisms, then $d(g\; \backslash circ\; f)\; =\; (dg)\; \backslash circ\; (df).$ *The differential of the (multiplicative) inverse is the additive inverse: $d(g\; \backslash mapsto\; g^)(X)\; =\; -X.$ In particular, if ''H'' is a Lie subgroup (i.e., a closed subgroup) of a Lie group ''G'', then $\backslash operatorname(H)$ is a Lie subalgebra of $\backslash operatorname(G)$. Also, if ''f'' is injective, then ''f'' is an immersion and so ''G'' is said to be an immersed (Lie) subgroup of ''H''. For example, $G/\backslash operatorname(f)$ is an immersed subgroup of ''H''. If ''f'' is surjective, then ''f'' is a submersion and if, in addition, ''G'' is compact, then ''f'' is a principal bundle with the structure group its kernel. (Ehresmann's lemma) Let $G\; =\; G\_1\; \backslash times\; \backslash cdots\; \backslash times\; G\_r$ be a direct product of Lie groups and $p\_i:\; G\; \backslash to\; G\_i$ projections. Then the differentials $dp\_i:\; \backslash operatorname(G)\; \backslash to\; \backslash operatorname(G\_i)$ give the canonical identification: :$\backslash operatorname(G\_1\; \backslash times\; \backslash cdots\; \backslash times\; G\_r)\; =\; \backslash operatorname(G\_1)\; \backslash oplus\; \backslash cdots\; \backslash oplus\; \backslash operatorname(G\_r)$. If $H,\; H\text{'}$ are Lie subgroups of a Lie group, then $\backslash operatorname(H\; \backslash cap\; H\text{'})\; =\; \backslash operatorname(H)\; \backslash cap\; \backslash operatorname(H\text{'}).$ Let ''G'' be a connected Lie group. If ''H'' is a Lie group, then any Lie group homomorphism $f:\; G\; \backslash to\; H$ is uniquely determined by its differential $df$. Precisely, there is the exponential map $\backslash operatorname:\; \backslash operatorname(G)\; \backslash to\; G$ (and one for ''H'') such that $f(\backslash operatorname(X))\; =\; \backslash operatorname(df(X))$ and, since ''G'' is connected, this determines ''f'' uniquely.〔In general, if ''U'' is a neighborhood of the identity element in a connected topological group ''G'', then $\backslash bigcup\_\; U^n$ coincides with ''G'', since the former is an open (hence closed) subgroup. Now, $\backslash operatorname:\; \backslash operatorname(G)\; \backslash to\; G$ defines a local homeomorphism from a neighborhood of the zero vector to the neighborhood of the identity element.〕 For example, if ''G'' is the Lie group of invertible real square matrices of size ''n'' (general linear group), then $\backslash operatorname(G)$ is the Lie algebra of real square matrices of size ''n'' and $\backslash displaystyle\; \backslash exp(X)\; =\; e^X\; =\; \backslash sum\_0^\backslash infty$. The next criterion is frequently used to compute the Lie algebra of a given Lie group. Let ''G'' be a Lie group and ''H'' an immersed subgroup. Then :$\backslash operatorname(H)\; =\; \backslash (tX)\; \backslash in\; H\; \backslash text\; t\; \backslash text\; \backslash mathbb.$ For example, one can use the criterion to establish the correspondence for classical compact groups (cf. the table in "compact Lie groups" below.) $\backslash operatorname$ defines a functor from the category of Lie groups to the category of finite-dimensional real Lie algebras. Lie's third theorem states that $\backslash operatorname$ defines an equivalence from the subcategory of simply connected Lie groups to the category of finite-dimensional real Lie algebras. Explicitly, the theorem contains the following two statements: *Every finite-dimensional real Lie algebra is the Lie algebra of some simply connected Lie group. *If $\backslash phi:\; \backslash operatorname(G)\; \backslash to\; \backslash operatorname(H)$ is a Lie algebra homomorphism and if ''G'' is simply connected, then there exists a (unique) Lie group homomorphism $f:\; G\; \backslash to\; H$ such that $\backslash phi\; =\; df$. Of two, the first one is fundamental since the second follows from the first applied to the graph of a Lie algebra homomorphism. Perhaps, the most elegant proof of the theorem uses Ado's theorem, which says any finite-dimensional Lie algebra (over a field of any characteristic) is a Lie subalgebra of the Lie algebra $\backslash mathfrak\_n$ of square matrices. The proof goes as follows: by Ado's theorem, we assume $\backslash mathfrak\; \backslash subset\; \backslash mathfrak\_n(\backslash mathbb)\; =\; \backslash operatorname(GL\_n(\backslash mathbb))$ is a Lie subalgebra. Let ''G'' be the subgroup of $GL\_n(\backslash mathbb)$ generated by $e^$ be a simply connected covering of ''G''; it is not hard to show that $\backslash widetilde$ is a Lie group and that the covering map is a Lie group homomorphism. Since $T\_e\; \backslash widetilde\; =\; T\_e\; G\; =\; \backslash mathfrak$, this completes the proof. Example: Each element ''X'' in the Lie algebra $\backslash mathfrak\; =\; \backslash operatorname(G)$ gives rise to the Lie algebra homomorphism :$\backslash mathbb\; \backslash to\; \backslash mathfrak,\; \backslash ,\; t\; \backslash mapsto\; tX.$ By Lie's third theorem, as $\backslash operatorname(\backslash mathbb)\; =\; T\_0\; \backslash mathbb\; =\; \backslash mathbb$ and exp for it is the identity, this homomorphism is the differential of the Lie group homomorphism $\backslash mathbb\; \backslash to\; H$ for some immersed subgroup ''H'' of ''G''. This Lie group homomorphism, called the one-parameter subgroup generated by ''X'', is precisely the exponential map $t\; \backslash mapsto\; \backslash operatorname(tX)$ and ''H'' its image. The preceding can be summarized to saying that there is a canonical bijective correspondence between $\backslash mathfrak$ and the set of one-parameter subgroups of ''G''. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Lie group–Lie algebra correspondence」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.