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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 Straightening theorem for vector fields ： ウィキペディア英語版 Straightening theorem for vector fields In differential calculus, the domain-straightening theorem states that, given a vector field $X$ on a manifold, there exist local coordinates $y\_1,\; \backslash dots,\; y\_n$ such that $X\; =\; \backslash partial\; /\; \backslash partial\; y\_1$ in a neighborhood of a point where $X$ is nonzero. The theorem is also known as straightening out of a vector field. The Frobenius theorem in differential geometry can be considered as a higher-dimensional generalization of this theorem. == Proof == It is clear that we only have to find such coordinates at 0 in $\backslash mathbb^n$. First we write $X\; =\; \backslash sum\_j\; f\_j(x)$ where $x$ is some coordinate system at $0$. Let $f\; =\; (f\_1,\; \backslash dots,\; f\_n)$. By linear change of coordinates, we can assume $f(0)\; =\; (1,\; 0,\; \backslash dots,\; 0).$ Let $\backslash Phi(t,\; p)$ be the solution of the initial value problem $\backslash dot\; x\; =\; f(x),\; x(0)\; =\; p$ and let :$\backslash psi(x\_1,\; \backslash dots,\; x\_n)\; =\; \backslash Phi(x\_1,\; (0,\; x\_2,\; \backslash dots,\; x\_n)).$ $\backslash Phi$ (and thus $\backslash psi$) is smooth by smooth dependence on initial conditions in ordinary differential equations. It follows that :$\backslash psi(x)\; =\; f(\backslash psi(x))$, and, since $\backslash psi(0,\; x\_2,\; \backslash dots,\; x\_n)\; =\; \backslash Phi(0,\; (0,\; x\_2,\; \backslash dots,\; x\_n))\; =\; (0,\; x\_2,\; \backslash dots,\; x\_n)$, the differential $d\backslash psi$ is the identity at $0$. Thus, $y\; =\; \backslash psi^(x)$ is a coordinate system at $0$. Finally, since $x\; =\; \backslash psi(y)$, we have: $=\; f\_j(\backslash psi(y))\; =\; f\_j(x)$ and so $=\; X$ as required. 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「Straightening theorem for vector fields」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.