# Pullback of a differential form over arbitraty complete field $K$

Let $K$ be a complete field and let $X, Y$ be $K$-analytic manifolds of Dimension $n$. If $(U,\Phi_U), (V,\Psi_V)$ are charts on $X, Y$ with $\Phi_U = (x_1,...,x_n), \Psi_V(y) = (y_1,...,y_n)$, then any $K$-analytic differential form $\beta$ of degree $n$ on $Y$ has a local representation $$\beta(y) = g(y) dy_1 \wedge ... \wedge dy_n$$ on $V$, where $g \colon V \to K$ is a $K$-analytic function. Now let $f \colon X \to Y$ be a $K$-analytic map and $U' = U \cap f^{-1}(V) \neq \emptyset$, then the pullback $f^*(\beta)$ has a local representation $$f^*(\beta)(x) = g(f(x)) d(y_1 \circ f) \wedge ... \wedge d(y_n \circ f)$$ on $U'$ which can be simplified to $$f^*(\beta)(x) = g(f(x)) \frac{\partial (y_1,...,y_n)}{\partial(x_1,..,x_n)} dx_1 \wedge ... \wedge dx_n,$$ where $\frac{\partial (y_1,...,y_n)}{\partial(x_1,..,x_n)}$ denotes the determinant of the $n \times n$-matrix with $\frac{\partial y_i}{\partial x_j}$ as its $(i,j)$-entry and where $\frac{\partial y_i}{\partial x_j}$ simply denotes $\frac{\partial (y_i \circ \Phi_U^{-1})}{\partial x_j}$.

My problem is that I don't understand this last step. Does anyone have an idea how it works? Especially I do not know how I can represent $d(y_1 \circ f)$ as a linear combination of the Basis $\{dx_1,...,dx_n \}$.

## 1 Answer

We have $$d(y_{i}\circ f)=\frac{\partial (y_{i}\circ f\circ\Phi_{U}^{-1})}{\partial x_{1}}dx_{1}+\cdots+\frac{\partial (y_{i}\circ f\circ\Phi_{U}^{-1})}{\partial x_{n}}dx_{n}.$$

Therefore $d(y_{1}\circ f)\wedge \cdots\wedge d(y_{n}\circ f)$ equals

$\left(\frac{\partial (y_{1}\circ f\circ\Phi_{U}^{-1})}{\partial x_{1}}dx_{1}+\cdots+\frac{\partial (y_{1}\circ f\circ\Phi_{U}^{-1})}{\partial x_{n}}dx_{n}\right)\wedge\cdots\wedge \left(\frac{\partial (y_{n}\circ f\circ\Phi_{U}^{-1})}{\partial x_{1}}dx_{1}+\cdots+\frac{\partial (y_{n}\circ f\circ\Phi_{U}^{-1})}{\partial x_{n}}dx_{n}\right).$

We multiply out and get $$\sum_{(i_{1},\ldots,i_{n})\in\{1,\ldots,n\}^{n}}\frac{\partial (y_{1}\circ f\circ\Phi_{U}^{-1})}{\partial x_{i_{1}}}dx_{i_{1}}\wedge\cdots\wedge\frac{\partial (y_{n}\circ f\circ\Phi_{U}^{-1})}{\partial x_{i_{n}}}dx_{i_{n}}.$$

Since the wedge product is antisymmetric, terms for which any two of $i_{1},\ldots,i_{n}$ are equal will vanish. Hence the terms that survive are exactly the terms for which $(i_{1},\ldots,i_{n})$ is a permutation of $(1,\ldots,n)$. We thus get

$$\sum_{\sigma\in S_{n}}\frac{\partial (y_{1}\circ f\circ\Phi_{U}^{-1})}{\partial x_{\sigma(1)}}\cdots\frac{\partial (y_{n}\circ f\circ\Phi_{U}^{-1})}{\partial x_{\sigma(n)}}dx_{\sigma(1)}\wedge\cdots\wedge dx_{\sigma(n)}$$

Again by antisymmetry of the wedge product, this equals $$\left(\sum_{\sigma\in S_{n}}sign(\sigma)\frac{\partial (y_{1}\circ f\circ\Phi_{U}^{-1})}{\partial x_{\sigma(1)}}\cdots\frac{\partial (y_{n}\circ f\circ\Phi_{U}^{-1})}{\partial x_{\sigma(n)}}\right)dx_{1}\wedge\cdots\wedge dx_{n}$$

And the expression between brackets is exactly $\det\left(\frac{\partial (y_{i}\circ f\circ\Phi_{U}^{-1})}{\partial x_{j}}\right)_{i,j}.$

• Thank you for your answer! Do you have any idea why $\frac{\partial (y_{i}\circ f\circ\Phi_{U}^{-1})}{\partial x_j} = \frac{\partial (y_{i} \circ \Phi_{U}^{-1})}{\partial x_j}$? In other words: $f$ should not stay in the determinant. – Algebrus Sep 14 '16 at 19:29
• But it's also my Intention that we cannot eliminate $f$ from the determinant. – Algebrus Sep 14 '16 at 19:42
• The composition $y_{i}\circ\Phi_{U}^{-1}$ seems weird to me since $\Phi_{U}^{-1}:\Phi_{U}(U)\subset\mathbb{R}^{n}\rightarrow X$ and $y_{i}:Y\rightarrow\mathbb{R}$. I guess that it is a typo in your text book. – studiosus Sep 15 '16 at 8:38