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I'm trying to do a pullback of a differential form.

Where $\omega=\left( x_4- x_2 x_4\right)\frac{d}{\text{dx}_1}+\left( x_4+x_5\right)\frac{d}{\text{dx}_2}+\left( x_5-x_2\text{ }x_5\right)\frac{d}{\text{dx}_3}-\left(x_2\text{ }x_5-x_1 x_3\text{ }x_5\right)\frac{d}{\text{dx}_4}+x_1 x_2 x_3 x_4 \frac{d}{\text{dx}_5}$

is a differential form on $\mathbb{R}^{5}$

since $f : \mathbb{R}^{5} \to \mathbb{R}$ $$f=x_1 x_4+x_2 x_4-x_1 x_2 x_4+x_2 x_5+x_3 x_5-x_2 x_3 x_5-x_2 x_4 x_5-x_1 x_3 x_4 x_5+x_1 x_2 x_3 x_4 x_5$$

How I can calculate the pullback. I was try to do a pullback by definition.But I did not get a result because it is general in $\mathbb{R}^{5}$ and I am starting on this subject

How I take a general vector and calculate the form?

Can this be done? Can an assistant find a solution or mention the steps

Please help and thank you anyway

Emad kareem
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  • A pullback by ... what map? –  Jun 09 '17 at 01:40
  • @Bye_World see new update – Emad kareem Jun 09 '17 at 01:51
  • OK. Two more issues. (1) You've defined $\omega$ as a vector, not a differential form and (2) if I were to just mentally convert the $\frac{d}{dx_i}$'s to $dx_i$'s, you'd still need the codomain of $f$ to be $\Bbb R^5$. As of right now $\omega$ and $f$ are incompatible. –  Jun 09 '17 at 02:19
  • Here's an example of how to perform a pullback on a differential form. –  Jun 09 '17 at 02:21

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