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 $\{dx_1,...,dx_n \}$.


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 $\det\left(\frac{\partial (y_{i}\circ f\circ\Phi_{U}^{-1})}{\partial x_{j}}\right)_{i,j}.$

  • $\begingroup$ 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. $\endgroup$ – Algebrus Sep 14 '16 at 19:29
  • $\begingroup$ But it's also my Intention that we cannot eliminate $f$ from the determinant. $\endgroup$ – Algebrus Sep 14 '16 at 19:42
  • 1
    $\begingroup$ 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. $\endgroup$ – studiosus Sep 15 '16 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.