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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 Adjoint representation of a Lie algebra ： ウィキペディア英語版 Adjoint representation of a Lie algebra In mathematics, the adjoint endomorphism or adjoint action is a homomorphism of Lie algebras that plays a fundamental role in the development of the theory of Lie algebras. Given an element of a Lie algebra $\backslash mathfrak$, one defines the adjoint action of on $\backslash mathfrak$ as the map :$\backslash operatorname\_x\; :\backslash mathfrak\backslash to\; \backslash mathfrak\; \backslash qquad\; \backslash text\; \backslash qquad\; \backslash operatorname\_x\; (y)\; =\; ()$ for all in $\backslash mathfrak$. The concept generates the adjoint representation of a Lie group Ad. In fact, ad is the differential of Ad at the identity element of the group. == Adjoint representation == Let $\backslash mathfrak$ be a Lie algebra over a field . Then the linear mapping :$\backslash operatorname:\backslash mathfrak\; \backslash to\; \backslash operatorname(\backslash mathfrak)$ given by is a representation of a Lie algebra and is called the adjoint representation of the algebra. (Its image actually lies in Der$(\backslash mathfrak)$. See below.) Within End$(\backslash mathfrak)$, the Lie bracket is, by definition, given by the commutator of the two operators: :$()=\backslash operatorname\_x\; \backslash circ\; \backslash operatorname\_y\; -\; \backslash operatorname\_y\; \backslash circ\; \backslash operatorname\_x$ where ○ denotes composition of linear maps. If $\backslash mathfrak$ is finite-dimensional, then End$(\backslash mathfrak)$ is isomorphic to $\backslash mathfrak(\backslash mathfrak)$, the Lie algebra of the general linear group over the vector space $\backslash mathfrak$ and if a basis for it is chosen, the composition corresponds to matrix multiplication. Using the above definition of the Lie bracket, the Jacobi identity :$]+\backslash right)(z)\; =\; \backslash left(\backslash operatorname\_\backslash right)(z)$ where , , and are arbitrary elements of $\backslash mathfrak$. This last identity says that ad really is a Lie algebra homomorphism; i.e., a linear mapping that takes brackets to brackets. In a more module-theoretic language, the construction simply says that $\backslash mathfrak$ is a module over itself. The kernel of ad is, by definition, the center of $\backslash mathfrak$. Next, we consider the image of ad. Recall that a derivation on a Lie algebra is a linear map $\backslash delta:\backslash mathfrak\backslash rightarrow\; \backslash mathfrak$ that obeys the Leibniz' law, that is, :$\backslash delta\; (())\; =\; ()\; +\; (\backslash delta(y))$ for all and in the algebra. That ad''x'' is a derivation is a consequence of the Jacobi identity. This implies that the image of $\backslash mathfrak$ under ad is a subalgebra of Der$(\backslash mathfrak)$, the space of all derivations of $\backslash mathfrak$. 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「Adjoint representation of a Lie algebra」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.