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エマルションは局所的に埋め込まれます。 📂幾何学

エマルションは局所的に埋め込まれます。

3173) イマージョンは局所的に埋め込まれる。

定理1

証明

ϕ\phiをイマージョンと仮定すると、dϕpd\phi{p}は単射である。埋め込みを示すには、ϕV\phi|_{V}(ϕV)1(\phi|_{V})^{-1}が全単射でなければならないため、最大でRm\mathbb{R}^{m}の座標まで考慮される [y1x1y1x2y1xny2x1y2x2y2xnymx1ymx2ymxn] \begin{bmatrix} \dfrac{\partial y_{1}}{\partial x_{1}} & \dfrac{\partial y_{1}}{\partial x_{2}} & \dots & \dfrac{\partial y_{1}}{\partial x_{n}} \\[1em] \dfrac{\partial y_{2}}{\partial x_{1}} & \dfrac{\partial y_{2}}{\partial x_{2}} & \dots & \dfrac{\partial y_{2}}{\partial x_{n}} \\[1ex] \vdots & \vdots & \ddots & \vdots \\[1ex] \dfrac{\partial y_{m}}{\partial x_{1}} & \dfrac{\partial y_{m}}{\partial x_{2}} & \dots & \dfrac{\partial y_{m}}{\partial x_{n}} \end{bmatrix}

逆関数定理を適用するために、次のマッピングを考えよう。

φ:(U1×Rmn=k)RmRm \varphi : (U_{1} \times \mathbb{R}^{m-n=k})\subset \mathbb{R}^{m} \to \mathbb{R}^{m}

 φ(x1,,xn,t1,,tk)= (y1(x1,,xn),,yn(x1,,xn),yn+1(x1,,xn)+t1,,yn+k(x1,,xn)+tk) \begin{align*} &\ \varphi ( x_{1}, \dots, x_{n}, t_{1}, \dots, t_{k} ) \\ =&\ \Big( y_{1}(x_{1},\dots,x_{n}), \dots, y_{n}(x_{1},\dots,x_{n}), y_{n+1}(x_{1},\dots,x_{n}) + t_{1}, \dots, y_{n+k}(x_{1},\dots,x_{n}) + t_{k} \Big) \end{align*}

det(dφq)= (y1,,yn,yn+1,,yn+k)(x1,,xn,t1,,tk)= [y1x1y1x2y1xny2x1y2x2y2xnymx1ymx2ymxn]=  \begin{align*} \det( d \varphi _{q}) =&\ \dfrac{\partial (y_{1}, \dots, y_{n}, y_{n+1}, \dots, y_{n+k})}{\partial (x_{1}, \dots, x_{n}, t_{1}, \dots, t_{k})} \\ =&\ \begin{bmatrix} \dfrac{\partial y_{1}}{\partial x_{1}} & \dfrac{\partial y_{1}}{\partial x_{2}} & \dots & \dfrac{\partial y_{1}}{\partial x_{n}} \\[1em] \dfrac{\partial y_{2}}{\partial x_{1}} & \dfrac{\partial y_{2}}{\partial x_{2}} & \dots & \dfrac{\partial y_{2}}{\partial x_{n}} \\[1ex] \vdots & \vdots & \ddots & \vdots \\[1ex] \dfrac{\partial y_{m}}{\partial x_{1}} & \dfrac{\partial y_{m}}{\partial x_{2}} & \dots & \dfrac{\partial y_{m}}{\partial x_{n}} \end{bmatrix} \\ =&\ \end{align*}


  1. Manfredo P. Do Carmo, Riemannian Geometry (Eng Edition, 1992), p13-14 ↩︎