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I am trying to make sense of pullbacks of forms as described in Guillemin and Pollack. If $f:X\rightarrow Y$ is a smooth map and $\omega$ is a $p$-form on $Y$, we define a $p$-form $f^*\omega$ on $X$ as follows. If $f(x)=y$, then $f$ induces a derivative map $df_x:T_x(X)\rightarrow T_y(Y)$. Since $\omega(y)$ is an alternating $p$-tensor on $T_y(Y)$, we can pull it back to $T_x(X)$ using the transpose $(df_x)^*$. We then define the pullback of $\omega$ by $f$ by $f^*\omega(x)=(df_x)^*\omega[f(x)]$.

(Note: In general, we define the transpose as $A^*T(v_1,...,v_p)=T(Av_1,...,Av_p)$)

I am having trouble actually applying this definition to real examples. Guillemin and Pollack give none.

For example, if $f:(\pi/2,\pi/2)\rightarrow \mathbb{R}$ is given by $f(t)=\sin t$ and $\omega=dx/\sqrt{1-x^2}$, then I begin as follows. We have that $df_t=f'(t)=\cos t$. Then $f^*\omega(t)=(\cos t)^*\omega$. Is this correct? How do I proceed?

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  • $\begingroup$ Your $\omega$ is defined only when $x\neq \pm1$, so it is not a one form on $\mathbb R$, but $\mathbb R\setminus\{\pm 1\}$. Is $g = f$? $\endgroup$ – user99914 Apr 24 '18 at 21:20
  • $\begingroup$ Yes, g should have been f $\endgroup$ – ponchan Apr 24 '18 at 21:22
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It may help to unravel the notations, let $M,N$ be manifolds, $\omega\in\Lambda^pT^*N$ and $f\colon M\rightarrow N$ be smooth, then: $$\forall x\in N,\forall (v_1,\ldots,v_p)\in T_xN,(f^*\omega)_x(v_1,\ldots,v_p):=\omega_{f(x)}(T_xf(v_1),\ldots,T_xf(v_p)),$$ this formula defines an element of $\Lambda^pT^*M$.

Assume that $M=(\pi/2,\pi/2)$, $N=\mathbb{R}$, $\omega:x\mapsto\frac{\mathrm{d}x}{\sqrt{1-x^2}}$ and $f\colon t\mapsto\sin(t)$, then one has: $$(f^*\omega)_t(v)=\omega_{f(t)}(T_tf(v))=\frac{\cos(t)v}{\sqrt{1-\sin(t)^2}}=\frac{\cos(t)v}{|\cos(t)|}=v,$$ since $T_tf\colon\mathbb{R}\rightarrow\mathbb{R}$ is the multiplication by $f'(t)$ and $\cos>0$ on $M$. Hence, $f^*\omega\equiv\mathrm{d}t$.

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  • $\begingroup$ Since $\omega$ is a $1$-form, shouldn't we be pulling back to a $1$-form in $M$? $\endgroup$ – ponchan Apr 24 '18 at 21:12
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    $\begingroup$ The first paragraph of my answer concerns the general case. In the second paragraph, $f^*\omega\colon t\mapsto\mathrm{d}t$ is a $1$-form on $M$, right? I have changed my $x$'s to $t$'s to be coherent with your notations. $\endgroup$ – C. Falcon Apr 24 '18 at 21:16
  • $\begingroup$ It looks like you are pulling back $\omega$ to the function, or $0$-form, $\frac{\cos(t)}{|\cos(t)|}$ $\endgroup$ – ponchan Apr 24 '18 at 21:20
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    $\begingroup$ Don't forget the $v$ hiding in the formula I have written! $\endgroup$ – C. Falcon Apr 24 '18 at 21:21
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    $\begingroup$ To be very precise probably you should write $v \frac{\partial}{\partial t}$ instead of $v$. $\endgroup$ – user99914 Apr 24 '18 at 21:24

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