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I'm trying to learn a bit about differential forms to supplement my study in analysis, but I'm having a hard time with some of the basic manipulations.

Anyway, suppose $\Omega$ is an open set in $\mathbb{C}$, and $\varphi:\Omega\to\mathbb{C}$ a smooth map. For a function $f$, I have the definition $\varphi^*f=f\circ\phi$, (when this makes sense for $f$ of course).

I also have the definitions $$ \varphi^*\,dx=\frac{\partial\varphi_1}{\partial x}\,dx+\frac{\partial\varphi_1}{\partial y}\,dy, \qquad \varphi^*dy=\frac{\partial\varphi_2}{\partial x}\,dx+\frac{\partial\varphi_2}{\partial y}\,dy, $$ where $\varphi_1$ is the $x$ component of $\varphi$ and $\varphi_2$ is the $y$ component. For a $1$-form $h=f\,dx+g\,dy$, $$ \varphi^*h=(\varphi^*f)\varphi^*\,dx+(\varphi^*g)\varphi^*\,dy. $$

What I don't get is why how $d(\varphi^* f)=\varphi^*df$.

I thought $d(\varphi^*f)=d(f\circ \phi)=f\,d\varphi+\varphi \,df$. I think my big problem here is I don't understand how to apply $d$ to a composition of functions. I also think $$ \varphi^*(df)=\varphi^*\left(\frac{\partial f}{\partial x}\,dx+\frac{\partial f}{\partial y}\,dy\right) $$ but I don't know how to take this further. They don't look equal to me. How does equality follow here? Thank you.

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The expression for $d(f \circ \phi)$ looks off to me. You want to use a chain rule, not a product rule. –  Dylan Moreland Apr 19 '12 at 0:22

1 Answer 1

up vote 4 down vote accepted

As you have written, we have $$\tag{1}\varphi^*(df)=\varphi^*\left(\frac{\partial f}{\partial u}\,du+\frac{\partial f}{\partial v}\,dv\right).$$ Let $\varphi(x,y)=(u(x,y),v(x,y))=(\varphi_1(x,y),\varphi_2(x,y))$. Again, by what you have written, i.e. $\varphi^*(fdu+gdv)=(\varphi^*f)\varphi^*\,du+(\varphi^*g)\varphi^*\,dv$, $(1)$ can be written as $$\tag{2}\varphi^*(df)=\varphi^*\left(\frac{\partial f}{\partial u}\right)\,\varphi^*du+\varphi^*\left(\frac{\partial f}{\partial v}\right)\,\varphi^*dv.$$ By definition you have, i.e. $$\varphi^*\,du=\frac{\partial u}{\partial x}\,dx+\frac{\partial u}{\partial y}\,dy, \qquad \varphi^*dy=\frac{\partial v}{\partial x}\,dx+\frac{\partial v}{\partial y}\,dy,$$ (I wrote $\varphi_1$ as $u$ and $\varphi_2$ as $v$) and $$\tag{3} \varphi^*F=F\circ\varphi\mbox{ for a function }F,$$ $(2)$ can be written as $$\varphi^*(df)=\frac{\partial f}{\partial u}\circ\varphi\left(\frac{\partial u}{\partial x}\,dx+\frac{\partial u}{\partial y}\,dy\right)+\frac{\partial f}{\partial v}\circ\varphi\left(\frac{\partial v}{\partial x}\,dx+\frac{\partial v}{\partial y}\,dy\right)$$ $$=\left(\frac{\partial f}{\partial u}\circ\varphi\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\circ\varphi\frac{\partial v}{\partial x}\right)dx+\left(\frac{\partial f}{\partial u}\circ\varphi\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\circ\varphi\frac{\partial v}{\partial y}\right)dy$$ $$=\frac{\partial (f\circ\varphi)}{\partial x}dx+\frac{\partial (f\circ\varphi)}{\partial y}dy=d(f\circ \varphi)=d(\varphi^*f)$$ where the third equality follows from chain rule, and the last inequality follows from $(3)$.

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Thanks Paul. I really appreciate the thoroughness of your explanation. –  Hana Bailey Apr 19 '12 at 4:53
    
You are welcome. Good that it helps. –  Paul Apr 19 '12 at 7:12

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