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I am asked to calculate $(F^*w)(p)(v_1,v_2)$ given that $w=xydx+zdy+xdz$, $p=(1,-1,0)$, $v_1=(0,1,1),v_2=(1,0,1)$ and $F(x,y,z)=(e^y,e^x,e^y)$

As far as I am concerned, this should be an easy enough mechanical problem, however I just started studying differential forms and I am not sure I understand...

I know that $(F^*w)(p)(v_1,v_2)$ is defined as: $$(F^*w)(p)(v_1,v_2)=w(F(x))(dF_pv_1,dF_pv_2)$$ Therefore $w(F(x))=e^ye^xdx+e^ydy+e^ydz$ and $dF_p=\begin{bmatrix}0&e^{-1}&0\\e&0&0\\0&e^{-1}&0\end{bmatrix}$. Therefore $dF_pv_1=(e^{-1},0,e^{-1})^t,dF_pv_2=(0,e,0)^t$.

Now how do I follow?

Edit: Screenshot of the problem: enter image description here Translation: Consider the differential form ... of $R^3$ and the application ... given by

Calculate ... where

John Keeper
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    $\omega$ — and hence $F^*\omega$ — is a $1$-form. Why are you evaluating on a pair of vectors $v_1,v_2$? – Ted Shifrin Jun 07 '19 at 00:25
  • I don't understand your comment, do you know where I can find examples of computations of 1-forms like this? I have studied the theory again and again but I can't seem to understand what differential forms are overall. – John Keeper Jun 07 '19 at 00:36
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    A $k$-form wants to eat $k$ vectors; hence a $1$-form wants to eat a single vector. I don't know what source you're learning from. You might be interested in checking out my YouTube videos — start with 3510 day 24 (the complete version of that lecture is at the bottom of the list); then proceed ... – Ted Shifrin Jun 07 '19 at 00:41
  • @JohnKeeper You may like this link: https://math.stackexchange.com/questions/576638/how-to-calculate-the-pullback-of-a-k-form-explicitly – cmk Jun 07 '19 at 00:46

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Perhaps, it's best to show what the general pullback is. Since the pullback acts by substitution, $$F^*(\omega)=(e^xe^y)d(e^y)+e^yd(e^x)+e^yd(e^y)=e^{x}e^ydx+(e^{x}e^{2y}+e^{2y})dy.$$ In general, if $\omega=\sum\limits_{j} a_jdx^j$ in a chart, then $$F^*\omega=\sum\limits_{i,j} (a_i\circ F)\frac{\partial F^i}{\partial x^j}dx^j.$$ As Ted's comment states, this is a one-form, so it cannot act on two vectors.

It can, however, act on a single tangent vector. If $p=(p_1,p_2,p_3),$ and $u=(u_1,u_2,u_3)$ is a tangent vector at $p$, then $$(F^*\omega)_p(v)=e^{p_1+p_2}u_1+(e^{p_1+2p_2}+e^{p_2})u_2.$$

EDIT: The question has been corrected to calculate $(F^*d\omega)_p(v_1,v_2).$ I will put a spoiler below on how to do it, but you should try to see if you can do it yourself first.

Since $d$ commutes with $F^*$, we have $$F^*(d\omega)=d(F^*\omega)=(e^{x}e^ydx+e^{x}e^ydy)\wedge dx+(e^xe^{2y}dx+2e^xe^{2y}dy+2e^{2y}dy)\wedge dy=e^xe^ydy\wedge dx+e^xe^{2y}dx\wedge dy=(e^xe^{2y}-e^xe^y)dx\wedge dy.$$ Evaluating, $$(F^*d\omega)_p(v_1,v_2)=(e^{p_1+2p_2}-e^{p_1+p_2})(dx(v_1)dy(v_2)-dx(v_2)dy(v_1))=1-e^{-1}.$$

cmk
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  • No, what the OP had was correct. He never wrote down $F^*\omega$. – Ted Shifrin Jun 07 '19 at 00:54
  • In the end, I'm not entirely sure what they were doing, but it should be beneficial for them to know what $F^*\omega$ is and how it's computed. – cmk Jun 07 '19 at 00:58
  • I don't disagree with you about the ultimate goal, but you're not reading carefully. Perhaps the use of "Therefore" should be criticized, but not the math. – Ted Shifrin Jun 07 '19 at 00:59
  • Okay, I read it again, and you're correct, they had not pulled back the forms themselves, yet. I'll rephrase my response. – cmk Jun 07 '19 at 01:02
  • @JohnKeeper I've edited my post to include a similar computation to what you asked for (yours was not possible, but this is, arguably, the next closest thing). – cmk Jun 07 '19 at 01:47
  • @cmk thank you for your time. You mean the problem could not be solved as it was stated? I double checked the statement of the problem and it is exactly what I writed. So for the problem to be answerable, $v_1$ and $v_2$ must be components of a vector and not vectors? – John Keeper Jun 07 '19 at 01:52
  • Was it $d\omega$ possibly? 1-forms can only act on one vector, and 2-forms act on pairs. Is there any chance that you could attach a screenshot or reference? I will respond in the morning! – cmk Jun 07 '19 at 01:55
  • @cmk I edited the post with a screenshot. I am from catalonia so it is in catalan... I translated it although there is not much to translate – John Keeper Jun 07 '19 at 02:00
  • Note that they gave you specific vectors $v_1$ and $v_2$. Was the problem written specifically as you typed it, or did they ask, in fact, for $F^\omega(v_1)$ and $F^\omega(v_2)$, as cmk correctly surmises? – Ted Shifrin Jun 07 '19 at 03:18
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    @JohnKeeper I just read it again, and it asks you for to calculate the pullback of $d\omega$, which is a 2-form. Can you adapt my answer to this case? – cmk Jun 07 '19 at 11:21
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    @TedShifren I just read it again, and it asks for the pullback of $d\omega$ evaluated on that pair, which makes more sense! – cmk Jun 07 '19 at 11:22
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    @JohnKeeper I added a spoiler to check if you so desire. – cmk Jun 07 '19 at 12:10
  • @cmk thank you so much. I am ashamed I did not see that $d$ in $dw$ before.. how? Anyway thank you for your answer I will accept it. However I just realised that you wrote $F^(\omega)=(e^xe^y)d(e^x)+e^yd(e^y)+e^yd(e^y)$ when it should be $F^(\omega)=(e^xe^y)d(e^y)+e^yd(e^x)+e^yd(e^y)$, right? – John Keeper Jun 07 '19 at 14:04
  • It appears so, thank you; I will correct it when I have a moment in an hour! – cmk Jun 07 '19 at 14:08
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    @JohnKeeper I did a quick edit, let me know if you find another mistake! – cmk Jun 07 '19 at 14:24