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By linearity of $f^*$, $$f^*\left(\sum_I a_I \, dx_I\right) = \sum_I f^* (a_I \, dx_I)$$

And if I want $df_I$, I would have to use the formula $f^*dx_i = df_i$. So $f^*$ disappears when introduced $df_i$ according to my computation.

$$f^*\left(\sum_I a_I \, dx_I\right) = \sum_I a_I (f^*dx_I) = \sum_I a_I \, df_I.$$

But according to the book, it is still there: $$f^*\left(\sum_I a_I \, dx_I\right) = \sum_I (f^*a_I) \, df_I.$$

Can anyone demystify it for me?

1LiterTears
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    $dx_I$ isn't even defined in general on/above the domain of $f$. You need to pull back the $dx_I$ to get a form on that. – Daniel Fischer Aug 06 '13 at 19:22
  • Hi Daniel, I figured probably I should consider $\sum a_I dx_I$ a $p$-form in $\mathbb{R^n}$ like: $2 dx_1 \wedge dx_2 + 4 dx_2 \wedge dx_3$ when $p=2, n = 3$? Then how can I get the RHS? – 1LiterTears Aug 06 '13 at 19:32
  • I'm not sure what your question is. The pull-back is defined by $f^\ast(\varphi) = \varphi\circ f$ for $0$-forms (functions) $\varphi$, $f^\ast(dx_i) = df_i$, and $f^\ast (\omega \wedge \psi) = \bigl(f^\ast(\omega)\bigr) \wedge \bigl(f^\ast(\psi)\bigr)$. Since for $0$-forms, $\wedge$ is just ordinary multiplication, you get $f^\ast(\sum \alpha_I, dx_I) = \sum \alpha_I \circ f df_I$. If you want to get $\sum \alpha_I\circ f dx_I$ (assuming that the domain and codomain of $f$ are the same, so the expression makes sense), that is not a natural operation, I don't know if it's useful for anything. – Daniel Fischer Aug 06 '13 at 19:40
  • What is the book's definition of pulling back forms? Just work directly from that. Maybe the confusion is that where you say "by linearity" isn't the definition of linearity. You should have $f^(\sum a_I dx_I)=\sum f^(a_I dx_I)$. – Matt Aug 06 '13 at 19:41
  • Hi Matt, the book's definition is $f^* \omega (x) = (df_x)^* \omega[f(x)]$, and for linear function $A$, $A^*T(v_1, \dots, v_p) = T(Av_1, \dots, Av_p)$. – 1LiterTears Aug 06 '13 at 19:45
  • Hi Daniel, I used the formula $f^dx_i = df_i$. So $f^$ disappears when introduced $df_i$ according to my computation. But according to the book, it is still there... – 1LiterTears Aug 06 '13 at 19:55
  • Hi @Matt, thanks. I was wrong on the brackets and I corrected it. But my problem is $f^*$ shall go away after introducing $df_i$. – 1LiterTears Aug 06 '13 at 19:58
  • Ah. I think I know the problem. The $a_I$ are functions, not constants. This is why they don't just move out. You're right in the example you give in the second comment because they are only constants there. You could have a two form that looks like $x_3^2 dx_1\wedge dx_2$ or something. – Matt Aug 06 '13 at 20:01
  • Right, $a_I$s are functions. But we can only distribute $f^$ over wedges. So $f^(x_3^2 dx_1 \wedge dx_2) = ?$ – 1LiterTears Aug 06 '13 at 20:03
  • @DanielFischer, I am really confused why $f^$ is still there after producing $df_I$. It looks like $f^(x_3^2 dx_1 \wedge dx_2) = f^(x_3^2) f^(dx_1 \wedge dx_2)$? – 1LiterTears Aug 06 '13 at 20:13
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    @Jellyfish In the book's definition $f^\ast\omega(x)= (df_x)^\ast \omega[f(x)]$, the $[f(x)]$ indicates the point at which $\omega$ has to be evaluated. Perhaps it'd be better to indicate that by a subscript $\omega_{f(x)}$ like for $df$. Then you have the fundamental fact that $dx_I(df_x v_1,,\ldots,,df_x v_p) = df_I(v_1,,\ldots,, v_p)$, thus $f^\ast(dx_I) = df_I$. – Daniel Fischer Aug 06 '13 at 20:19
  • @Jellyfish Yes. Note, for example, that it must be something like that, because if it were what you originally wrote, you would get $x_3^2 f^(dx_1\wedge dx_2)$ and the function $x_3^2$ wouldn't be on the right manifold. This case is somewhat confusing because you identify the manifolds since they are the same. So it is harder to see that $a_I$ wouldn't make sense in a more general situation. But this term of $f^a_I$ is just pulling back a function, and hence just composing $a_I\circ f$. – Matt Aug 06 '13 at 20:27
  • Thanks @Matt, I think I got it. – 1LiterTears Aug 06 '13 at 20:35
  • Thank you @DanielFischer, though it is frustrating that I never heard of the fundamental fact. GP gives very little information and floods a lot, apparently assumed some prerequisites, like algebraic topology or multilinear algebra or calculus on manifolds... – 1LiterTears Aug 06 '13 at 20:37
  • That fact is pure multilinear algebra. Ugly but elementary (not easy, you can make a lot of sign errors etc.) to verify by calculation, if one wants to. – Daniel Fischer Aug 06 '13 at 20:41
  • Yeah, exactly @DanielFischer! I am quite upset that nowhere (to my very ignorant knowledge) offers multilinear algebra course! And even more upsetting, every course assumes you know that kind of stuff. – 1LiterTears Aug 06 '13 at 20:42
  • Isn't that part of linear algebra courses anymore? You can't talk much about determinants without talking about alternating forms. – Daniel Fischer Aug 06 '13 at 20:47
  • Yes. But only determinants, not the notion of dual space, nor forms in general. @DanielFischer – 1LiterTears Aug 06 '13 at 20:51

1 Answers1

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By linearity of $f^*$, $$f^*\left(\sum_I a_I \, dx_I\right) = \sum_I f^* (a_I \, dx_I)$$

And if I want $df_I$, I would have to use the formula $f^*dx_i = df_i$.

$$f^*\left(\sum_I a_I \, dx_I\right) = \sum_I (f^* a_I) (f^*dx_I) = \sum_I (f^* a_I) \, df_I.$$

Note: $f^*$ distribute on $a$ as well, this is where the question arises.

1LiterTears
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