Computing the pullback of an exterior differential of a vector field at a point given two linear independent vectors

Question

Given the $1$-differential form $\omega=xzdx+xdy+xdz$ and the vector field $F(x,y,z)=(e^y,e^z,e^x)$, I must compute $$(F^{*}d\omega)_{p}(v_1,v_2)$$ where $p=(1,-1,0), v_1=(1,1,0)$ and $v_2=(1,0,1)$. This is in essence (if I am right), the pullback of the exterior differential of the $1$-differential form $\omega$. I get stuck almost at the end of the process, let me explain.

First of all, I have to compute the exterior differential of $\omega$ $$d\omega=dx\wedge dy+(1-x)dx\wedge dz$$Now, for the pullback $$(F^{*}d\omega)=e^y dy \wedge e^z dz + (1-e^y)e^y dy\wedge e^x dx = e^{y+z}dy\wedge dz \wedge e^x dx=e^{x+y+z}dx\wedge dy\wedge dz$$ right?

The problem comes whenever I have to apply this to my points and respective vectors. I guess that the coefficient $$e^{x+y+z}|_{(x,y,z)=(1,-1,0)}=1$$ but how do I get the final result? Please help.

Answer 1

I would like to propose an answer that uses the fact that the pull-back and the exterior differential commute (a fact that is sometimes referred to as the "naturality" of the pull-back). I personally find this approach easier since it does not require to pull-back anything more than $1$-forms. I will give all the details to show how elementary the computations are.

Let us first compute $F^*\omega$. We have \begin{align} F^*\omega &= F^*(xzdx + xdy + xdz) \\ &= (x\circ F)(z\circ F)d(x\circ F) + (x\circ F)d(y\circ F) + (x\circ F)d(z\circ F)\\ &= e^ye^xd(e^y) + e^yd(e^z) + e^yd(e^x)\\ &= e^ye^xe^ydy + e^ye^zdz + e^ye^xdx\\ &= e^{x+y}dx + e^{x+2y}dy + e^{y+z}dz. \end{align} Hence, we have \begin{align} F^*(d\omega) &= d(F^*\omega) \\ &= d(e^{x+y})\wedge dx + d(e^{x+2y})\wedge dy + d(e^{y+z})\wedge dz \\ &= (e^{x+y}dx + e^{x+y}dy)\wedge dx + (e^{x+2y}dx + 2e^{x+2y}dy)\wedge dy + (e^{y+z}dy + e^{y+z}dz)\wedge dz\\ &= e^{x+y}dy\wedge dx + e^{x+2y}dx\wedge dy + e^{y+z}dy\wedge dz\\ &= (e^{x+2y} -e^{x+y})dx\wedge dy + e^{y+z}dy\wedge dz. \end{align} Therefore, at $p=(1,-1,0)$, we have $$ d(F^*\omega)_p = (e^{-1}-1)dx\wedge dy +e^{-1}dy\wedge dz. $$ If $v_1 = (1,1,0)$ and $v_2 = (1,0,1)$, then \begin{align} dx\wedge dy (v_1,v_2) &= dx(v_1)dy(v_2) - dy(v_1)dy(v_2) \\ &=1 \times 0 - 1\times 1 \\ &= -1,\\ \text{and} \quad dy\wedge dz (v_1,v_2) &= dy(v_1)dz(v_2)-dy(v_2)dz(v_1) \\ &= 1\times 1 - 0\times 0\\ &= 1. \end{align} Ultimately, one has $$ F^*(d\omega)_p(v_1,v_2) = (e^{-1}-1)\times -1 + e^{-1}\times 1 = 1. $$

Answer 2

The problem is in the last chain of equalities. You sum 2-forms and end up with a volume form. The correct pull-back, as you wrote, is:

$$ F^* d \omega = e^{y+z} dy \wedge dz - e^{x+y} (1- e^y) dx \wedge dy$$

Looking at $(v_1, v_2)$ as a 2-vector:

$$ v_1 \wedge v_2 = (e_1 + e_2 ) \wedge (e_1 + e_3)= e_1 \wedge e_3 - e_1 \wedge e_2 + e_2 \wedge e_3 .$$

And thus:

$$ (F^* d \omega)_{(1,-1,0)} (v_1, v_2 )=[e^{-1} dy \wedge dz - (1- e^{-1}) dx \wedge dy ](e_1 \wedge e_3 - e_1 \wedge e_2 + e_2 \wedge e_3)$$

which you can compute being 1. By the way, as noted in the coments pull back and differential commute, you can try doing $d F^* \omega$ and see that the form you get is the same.