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I have a question about derivation of fonction $f:M\longrightarrow N$ where $M$ and $N$ are smooth manifold of dimension $n$. In my course, we try to compute $$\mathrm d_p f\left(\frac{\partial }{\partial x_i}\right)$$ where $\frac{\partial }{\partial x^i}$ is a derivation at $p$. They do as follow:

Let $(x^1,...,x^n)$ the coordinate on $p\in M$ and $(y^1,...,y^n)$ the coordinate on $q\in N$ and let $y^i=f^i(x^1,...,x^n)$. Therefore, for all $u=u(y^1,...,y^n)\in \mathcal C^\infty (N)$, we have $$\mathrm d _pf\left(\frac{\partial }{\partial x_i}\right)(u)=\frac{\partial }{\partial x^i}(u\circ f)=\sum_{j=1}^n\frac{\partial u}{\partial y^j}\frac{\partial f^j}{\partial x^i}$$ and thus $$\mathrm d _pf\left(\frac{\partial }{\partial x_i}\right)=\sum_{j=1}^n\frac{\partial f^j}{\partial x^i}\frac{\partial }{\partial y^j}.$$

My problems

1) How can we write $f^i(x^1,...,x^n)$ since $x^i\in\mathbb R$ and $f:M\longrightarrow N$ ?

2) How can we write $u(y^1,...,u^n)$ whereas $y^i\in \mathbb R$ and $u:M\longrightarrow \mathbb R$ ?

3) How can we write $\frac{\partial f^j}{\partial x^i}$ whereas $\frac{\partial }{\partial x^i}:\mathcal C^1(\mathbb R^n)\longrightarrow \mathbb R$ and here $f:M\longrightarrow N$.

4) How can we write $\frac{\partial u}{\partial y^j}$ whereas $\frac{\partial u}{\partial y^j}:\mathcal C^1(\mathbb R^n)\longrightarrow \mathbb R$ and that $u:M\longrightarrow \mathbb R$.

Actually they do the same in the book Introduction to smooth manifold of John Lee. So my problem is principally in the compatibility of those operations.

user301068
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1 Answers1

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Let $(U,\varphi)$ is the chart that gives $(x^1,...,x^n)$. Then $f(x^1,...,x^n)$ refer to $f$ read in the chard $(U,\varphi)$. In other word $f(x_1,...,x_n)$ mean $f\circ \varphi^{-1}(x^1,...,x^n)$ and thus, $\frac{\partial f^j}{\partial x^i}$ means $$\frac{\partial }{\partial x^i}f^j\circ \varphi^{-1}(x^1,...,x^n).$$

Now, let $(V,\psi)$ the chart that gives $(y^1,...,y^n)$. I let you adapt this argument to $u$ to give sense to all your questions.

idm
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