Calculate the pullback $(F^*w)(p)(v_1,v_2)$

differential-geometrypullback

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

Best Answer

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, $$(F^*d\omega)_p(v_1,v_2)=(e^{p_1+2p_2}-e^{p_1+p_2})(dx(v_1)dy(v_2)-dx(v_2)dy(v_1))=1-e^{-1}.$$

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