Computing the pull-back with exterior derivative

Question

I don't understand how to do the following question: Property 5 is $\phi$*d$\omega$=d$\phi$ *$\omega$

Thank you!

Answer 1

You have $$ F : \Bbb R^3 \to \Bbb R^6: (x, y, z) \mapsto (x, y, z, xyz, x^3 - y, 6x + 3z) $$ And you have a $2$-form, $\omega$ on $\Bbb R^6$. You'd like to "compute" $$ \eta = F^{*}\omega, $$ whatever that might mean, which should be a $2$-form on $\Bbb R^3$. That means that $\eta$ should be something that looks like $$ \eta(p, q, r) = f(p,q,r) dp \wedge dq + g(p,q,r) dq \wedge dr + h(p,q,r) dp \wedge dr $$ i.e., you're supposed to write down the functions $f, g, h$. If you prefer a notation with lots of superscripts, etc., then you want $$ \eta(x^1, x^2, x^3) = f(x^1, x^2, x^3) dx^1 \wedge dx^2 + \ldots, $$ but frankly, I think that's worse to read. To each their own.

I have no idea what the hint is getting at, but let's look at what $$ \eta(p,q,r) (e_1, e_2) $$ should be. On the one hand, it should be

\begin{align} \eta(p, q, r)(e_1, e_2) &= f(p,q,r) dp \wedge dq (e_1, e_2)+ g(p,q,r) dq \wedge dr (e_1, e_2) + h(p,q,r) dp \wedge dr(e_1, e_2)\\ &= f(p,q,r) dp \wedge dq (e_1, e_2)+ g(p,q,r)\cdot 0 + h(p,q,r) \cdot 0\\ &= f(p,q,r) \cdot 1 \end{align} Hold that thought.

On the other hand, it should also be

\begin{align} \eta(p, q, r)(e_1, e_2) &= (F^* \omega)(p,q,r) (e_1, e_2)\\ &= \omega(F(p,q,r))(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\ &= \omega((p, q, r, pqr, p^3 - r, 6p + 3r))(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\ &= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\ \end{align} Now we need to compute $F_{*}$, which is just the derivative of $F$, so at $(p,q,r)$, it's multiplication by the matrix $$ \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ qr & pr & pq \\ 3p^2& -1 & 0 \\ 6 & 0 & 3} $$

Multiplying this by $e_1$ gives the first column; multiplying by $e_2$ gives the second column. So

\begin{align} \eta(p, q, r)(e_1, e_2) &= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\ &= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(\pmatrix{1 \\ 0 \\ 0 \\ qr\\ 3p^2 \\ 6 }, \pmatrix{ 0 \\ 1 \\ 0 \\ pr \\ -1 \\ 0 })\\ \end{align}

So all we have to do is evaluate $dw^3 \wedge dw^4$ on that pair of vectors (we get $0$), $dw^6 \wedge dw^5$ (messier), and $dw^1 \wedge dw^2$. This last one gives $1 \cdot 1 - 0 \cdot 0 = 1$. So the middle one remains. Calling the two vectors $v_1$ and $v_2$, we need to compute \begin{align} s &= dw^6(v_1) dw^5(v_2) - dw^6(v2) dw^5(v_1) \\ &= 6 \cdot (-1) - 0 \cdot 3p^2 \\ &= -6. \end{align} So we end up with

\begin{align} \eta(p, q, r)(e_1, e_2) &= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(F_* (p,q,r) (e_1), F_* (p,q,r) (e_2))\\ &= ( (p+r)~dw^3 \wedge dw^4 + 2 ~dw^6 \wedge dw^5 + q ~dw^1 \wedge dw^2)(\pmatrix{1 \\ 0 \\ 0 \\ qr\\ 3p^2 \\ 6 }, \pmatrix{ 0 \\ 1 \\ 0 \\ pr \\ -1 \\ 0 })\\ &= ( (p+r)~\cdot 0 + 2 \cdot (-6) + q \cdot 1\\ &= -12 + q \end{align} from which we conclude that $f(p, q, r) = -12 + q$.

Now all you have to do is repeat that kind of computation to find the functions $g$ and $h$, and you're done.