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.


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 $dx_i({\partial\over\partial x_i}|_{p}) = {\partial\over\partial t}|_{p_i}.$

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – hopo2 Sep 9 '12 at 21:16
  • $\begingroup$ 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 $x_i=r_i\circ\phi.$ $\endgroup$ – Andrew Sep 9 '12 at 22:43
  • $\begingroup$ And where ${\partial\over\partial r_i}$ really means derivative on $\Bbb R^n.$ $\endgroup$ – Andrew Sep 9 '12 at 22:46
  • $\begingroup$ I thought we wanted to show that $dx_i({\partial\over\partial x_i}|_{p}) = \delta_{ij}$, instead of $dx_i({\partial\over\partial x_i}|_{p}) = {\partial\over\partial t}|_{p_i}.$ $\endgroup$ – Soap Jul 9 '17 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.