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The derivative of a map $F$ between manifolds $M$ and $N$ is defined by $$F_*X(f)= X(f \circ F)$$ where $X \in T_P(M)$, the tanget space at the point $P$.

We know that $$\left\{\frac{\partial}{\partial x^i}\bigg|_P\right\}_i$$ is a basis for $T_P(M)$. How to show that $$\left\{dx^i\bigg|_P\right\}_i$$ is a basis for the cotangent space $(T_P(M))^*$?

First, by $dx^i$, I guess we mean the derivative of the map $x^i$ as defined above, right? Is this map $x^i$ just picking out the ith coordinate? Secondly, to show that it is a basis, we need to show that $dx^i\left(\frac{\partial}{\partial x^j}\bigg|_P\right) = \delta^i_j.$ Where to go from here: $$\underbrace{(dx^i)_P}_{(\Phi_*)_P}\underbrace{\left(\frac{\partial}{\partial x^j}\bigg|_P\right)}_{X}f = \left(\frac{\partial}{\partial x^j}\bigg|_P\right)(f\circ x^i)?$$

I can use the chain rule but I am not sure exactly. Please help.

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up vote 5 down vote accepted

We work locally in a chart $(U,\phi)$ of $p$ on $M.$ Let $x_i:M\to \Bbb R$ denote the $i^{\rm th}$ coordinate function. Then $dx_i:M_p\to\Bbb R_{p_i}$ where $M_p,\Bbb R_{p_i}$ denote the tangent spaces at $p=(p_1,\ldots,p_n)$ and $p_i.$

By definition, $dx_i(v)(f)=v(f\circ x_i)$ for any tangent vector $v\in M_p$ and $C^\infty$-function $f$ at $p_i.$ In particular, choosing $v={\partial\over\partial x_j}|_{p},$ we would get $0$ unless $j=i,$ in which case we get $$dx_i({\partial\over\partial x_i}|_{p})(f)={\partial\over\partial x_i}|_{p}(f\circ x_i) \overset{\rm def}= {\partial(f\circ x_i\circ\phi^{-1})\over\partial r_i}|_{\phi(p)} = {\partial(f\circ (r_i\circ\phi)\circ\phi^{-1})\over\partial r_i}|_{\phi(p)} = {\partial(f\circ r_i)\over\partial r_i}|_{\phi(p)}$$

where $\phi: U\subseteq M\to \Bbb R^n$ is a chart of $p,$ and $r_i$ is a coordinate function on $\Bbb R^n.$

Now using calculus, ${\partial(f\circ r_i)\over\partial r_i}|_{\phi(p)} = {\partial r_i\over\partial r_i}(\phi(p))\times{\partial f\over\partial t}(p_i)={\partial f\over\partial t}(p_i),$ where $t$ is a coordinate on $\Bbb R.$ Thus, we have shown that $dx_i({\partial\over\partial x_i}|_{p}) = {\partial\over\partial t}|_{p_i}.$

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Thank you. Sorry for my denseness, but when you write the second equality ($\stackrel{def}{=}$) in the two lines of calculations in your answer, what is that? I don't see why that's the definition of that derivative. I know $x^i$ is defined on the manifold so we must take the chart back to $\mathbb{R}^n$ but I haven't come across that definition. – hopo2 Sep 9 '12 at 21:16
The definition I referenced is from Warner, p.15: ${\partial\over\partial x_i}|_p$ is the tangent vector at $p\in M$ defined by $f\mapsto {\partial(f\circ\phi^{-1})\over\partial r_i}|_{\phi(p)},$ with the notation from my answer, and $x_i=r_i\circ\phi.$ – Andrew Sep 9 '12 at 22:43
And where ${\partial\over\partial r_i}$ really means derivative on $\Bbb R^n.$ – Andrew Sep 9 '12 at 22:46

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