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If we have $\Omega$ - domain in $R^{n}$ and standard basis of $R^{n}$: $\{x^{1},..,x^{n}\}$, and $\{x^{*}_{1},..,x^{*}_{n}\}$ the dual affine coordinate system on $R_{n}^{*}$. And if we define map $\iota$ from $\Omega$ to $R_{n}^{*}$ by $x_{i}^{*}\circ\iota=-\frac{\partial \phi}{\partial x^{i}}$, how do we obtain that $\iota_{*}(\frac{\partial}{\partial x^{i}})=-\sum(g_{ij}\circ\iota^{-1})\frac{\partial}{\partial x_{j}^{*}}$, knowing that $g_{ij}=\frac{\partial^{2}\phi}{\partial x^{i}\partial x^{j}}$.

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I assume this just means there exists some smooth function $\phi$ with this property? Also, the notation confuses me a bit. What is $x_i^*\circ \iota$? It seems $\iota$ takes a point of $\mathbb{R}^n$ and outputs a linear function $\mathbb{R}^n\to \mathbb{R}$. Since $x_i^*$ is also a linear function $\mathbb{R}^n\to \mathbb{R}$, I'm not sure how to compose them. – Matt Nov 18 '12 at 6:19
@Matt: I think the $x_i^*$ are meant to be coordinate functions on $(\mathbb{R^n})^*$ so that together they define a (global) chart. – Jesse Madnick Nov 18 '12 at 7:14
up vote 1 down vote accepted

I have to say, it took me a while to figure out your notation.

In general, if $F\colon \mathbb{R}^n \to N$ is a smooth map, $F(p) = q$, and $\{x^1, \ldots, x^n\}$ are coordinates on $\mathbb{R}^n$, and $\{y^j\}$ are coordinates on $N$, then

$$F_*\left( \left. \frac{\partial}{\partial x^i}\right|_p \right) = \sum_{j} \left.\frac{\partial(y^j \circ F)}{\partial x^i}(p)\frac{\partial}{\partial y^j}\right|_{q}$$

In your case, we take $F = \iota$ and $y^j = x^*_j$, so that

$$\begin{align*} \iota_*\left( \left.\frac{\partial}{\partial x^i}\right|_p \right) &= \sum_{j=1}^n \left.\frac{\partial(x^*_j \circ \iota)}{\partial x^i}(p)\frac{\partial}{\partial x^*_j}\right|_{q} \\ & = \sum_{j=1}^n \left.\frac{\partial}{\partial x^i}\left(-\frac{\partial \phi}{\partial x^j}\right)(p)\frac{\partial}{\partial x^*_j}\right|_{q} \\ & = -\sum_{j=1}^n \left.g_{ij}(p) \frac{\partial}{\partial x^*_j}\right|_{q} \\ \end{align*}$$ Now write $p = \iota^{-1}(q)$.

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