📂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

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

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。

$$ \varphi : (U_{1} \times \mathbb{R}^{m-n=k})\subset \mathbb{R}^{m} \to \mathbb{R}^{m} $$

$$ \begin{align*} &\ \varphi ( x_{1}, \dots, x_{n}, t_{1}, \dots, t_{k} ) \\ =&\ \Big( y_{1}(x_{1},\dots,x_{n}), \dots, y_{n}(x_{1},\dots,x_{n}), y_{n+1}(x_{1},\dots,x_{n}) + t_{1}, \dots, y_{n+k}(x_{1},\dots,x_{n}) + t_{k} \Big) \end{align*} $$

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