# A question about pullback of the K-form

Let $M$ be an oriented $m$-dimensional manifold. Suppose the support of $\omega$ is in an open subset $U$ of $M$, and $\phi \colon U \to R^m$, $\psi \colon U \to R^m$ are two different charts on $M$ defining the same orientation.

In coordinates {$r_1,r_2,...r_m$} on U, we can see $\omega$ as $f dr_1\wedge dr_2\wedge \cdots \wedge dr_m$. Then the pull back $({\phi}^{-1})^{*}\omega = (f{\phi}^{-1}) dr_1{\phi}^{-1}\wedge \cdots \wedge dr_m{\phi}^{-1}$, say $(f{\phi}^{-1})dx_1\wedge dx_2\wedge \cdots \wedge dx_m,$ where define $dr_i{\phi}^{-1}$ to be $dx_i$

Similarly, $({\psi}^{-1})^{*}\omega = (f{\psi}^{-1}) dr_1{\psi}^{-1}\wedge \cdots \wedge dr_m{\psi}^{-1}$, say $(f{\psi}^{-1})dy_1\wedge dy_2\wedge \cdots \wedge dy_m$.

As we know $$\int_{\phi(U)}({\phi}^{-1})^{*}\omega = \int_{\psi(U)}({\psi}^{-1})^{*}\omega.$$ Note $$\int_{\phi(U)}({\phi}^{-1})^{*}\omega = \int_{\phi(U)}f({\phi}^{-1})(x)dx_1\wedge dx_2\wedge \cdots \wedge dx_m := \int_{\phi(U)}(f{\phi}^{-1})(x)dx,$$ and similarly $$\int_{\psi(U)}({\psi}^{-1})^{*}\omega = \int_{\psi(U)}(f{\psi}^{-1})(y)dy.$$ And $\int_{\psi(U)}(f{\psi}^{-1})(y)dy$ may not be equal to $\int_{\phi(U)}(f{\phi}^{-1})(x)dx$ with the reason in the picture below:

So it seems that it leads to a contraction, that is $\int_{\phi(U)}({\phi}^{-1})^{*}\omega = \int_{\psi(U)}({\psi}^{-1})^{*}\omega$ first and then they are not equal. Can you tell me where I went wrong? Thanks.

I think I see the issue.

Let $\omega$ be an $m$-form on $M$, supported in $U \subset M$. Let $\tau \colon U \to \mathbb{R}^m$ be an oriented chart, with coordinates $(r_1, \ldots, r_m)$. On $U \subset M$, we can write $\omega = f\,dr_1 \wedge \cdots \wedge dr_m$.

Let $\phi, \psi \colon U \to \mathbb{R}^m$ be oriented charts, also on $U \subset M$. It is true that $$\int_{\phi(U)} (\phi^{-1})^*\omega = \int_U \omega = \int_{\psi(U)} (\psi^{-1})^*\omega.$$ Let $(z_1, \ldots, z_m)$ be the standard coordinates on $\mathbb{R}^m$. It is also true that \begin{align*} \int_{\phi(U)} (\phi^{-1})^*\omega & = \int_{\phi(U)} (\phi^{-1})^*(f dr_1 \wedge \cdots \wedge dr_m) \\ & = \int_{\phi(U)} (f \circ \phi^{-1}) \cdot (\phi^{-1})^*(dr_1 \wedge \cdots \wedge dr_m) \\ & = \int_{\phi(U)} (f \circ \phi^{-1}) \cdot (d(r_1 \circ \phi^{-1}) \wedge \cdots \wedge d(r_m \circ \phi^{-1})) \\ & = \int_{\phi(U)} (f \circ \phi^{-1}) \cdot (d(z_1 \circ \tau \circ \phi^{-1}) \wedge \cdots \wedge d(z_m \circ \tau \circ \phi^{-1})) \end{align*} and similarly, \begin{align*} \int_{\psi(U)} (\psi^{-1})^*\omega & = \int_{\psi(U)} (\psi^{-1})^*(f dr_1 \wedge \cdots \wedge dr_m) \\ & = \int_{\psi(U)} (f \circ \psi^{-1}) \cdot (\psi^{-1})^*(dr_1 \wedge \cdots \wedge dr_m) \\ & = \int_{\psi(U)} (f \circ \psi^{-1}) \cdot (d(r_1 \circ \psi^{-1}) \wedge \cdots \wedge d(r_m \circ \psi^{-1})) \\ & = \int_{\psi(U)} (f \circ \psi^{-1}) \cdot (d(z_1 \circ \tau \circ \psi^{-1}) \wedge \cdots \wedge d(z_m \circ \tau \circ \psi^{-1})) \\ \end{align*}

The point is now that $d(z_1 \circ \tau \circ \phi^{-1}) \wedge \cdots \wedge d(z_m \circ \tau \circ \phi^{-1})$ is not necessarily equal to the euclidean volume form $dz_1 \wedge \cdots \wedge dz_m$. Similarly, $d(z_1 \circ \tau \circ \psi^{-1}) \wedge \cdots \wedge d(z_m \circ \tau \circ \psi^{-1})$ is not necessarily equal to $dz_1 \wedge \cdots \wedge dz_m$, either. That's the issue.

Addendum: Now, I have a feeling that this might leave you unsatisfied. That is, you might say something like: Well, can't I just define $x_i = r_i \circ \phi^{-1} = z_i \circ \tau \circ \phi^{-1}$ and $y_i = r_i \circ \psi^{-1} = z_i \circ \tau \circ \psi^{-1}$? Because if I do that, then I will get $$\int_{\phi(U)} (\phi^{-1})^*\omega = \int_{\phi(U)} (f \circ \phi^{-1})\,dx_1 \wedge \cdots \wedge dx_m$$ and $$\int_{\psi(U)} (\psi^{-1})^*\omega = \int_{\psi(U)} (f \circ \psi^{-1})\,dy_1 \wedge \cdots \wedge dy_m,$$ and indeed you do. However, that's not what your book means by by $dx$ and $dy$. What your book means is that $dx$ and $dy$ are both the euclidean volume form $dz_1 \wedge \cdots \wedge dz_m$.

The euclidean volume form is defined to be the $m$-form $dz_1 \wedge \cdots \wedge dz_m$ on $\mathbb{R}^m$, where $(z_1, \ldots, z_m)$ are the standard Cartesian coordinates. So, for example, in $\mathbb{R}^2$, the euclidean volume form is $$dz_1 \wedge dz_2 = r\,dr \wedge d\theta.$$ In $\mathbb{R}^3$, it is $$dz_1 \wedge dz_2 \wedge dz_3 = r\,dr \wedge d\theta \wedge dz_3 = \rho^2\sin\phi \,d\rho \wedge d\theta \wedge d\phi.$$

• Thank you, that's a very good answer! – 6666 Jun 16 '16 at 21:54