Exterior derivative well-defined

Question

In a differential geometry book I am currently reading the exterior derivative for a k-form $\omega$ on a manifold is defined via coordinate-patches, that is given a chart $(U,x)$ and a coordinate representation of $\omega$ on $U$ $$\omega=\sum_{i_1<\cdots<i_k}a_{i_1...i_k}dx^{i_1}\wedge\cdots\wedge dx^{i_k}$$ the exterior derivative on $U$ is defined as $$d\omega=\sum_{i_1<\cdots<i_k}da_{i_1...i_k}\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}$$ I want to convince myself that this definition for $d\omega$ is well defined, i.e. for a pair $(U,x),(V,y)$ of overlapping charts: $$\sum_{i_1<\cdots<i_k}da_{i_1...i_k}\wedge dx^{i_1}\wedge\cdots\wedge dx^{i_k}=\sum_{j_1<\cdots<j_k}db_{j_1...j_k}\wedge dy^{j_1}\wedge\cdots\wedge dy^{j_k}$$ on $U\cap V$, where $a_{i_1...i_k}$ and $b_{j_1...j_k}$ are the corresponding coefficient functions of $\omega$ with respect to $(U,x)$ and $(V,y)$. I tried plugging in the following transformation rules, but it got too messy: $$b_{j_1...j_k}=\frac{\partial x^{i_1}}{\partial y_{j_1}}\cdots\frac{\partial x^{i_k}}{\partial y_{j_k}}a_{i_1...i_k}$$ $$dy^j=\frac{\partial y^{j}}{\partial x_{i}}dx^i$$ I would appreciate any help, thanks in advance!

Edit: I will try to outline my attempt so far: When I use $dy^j=\frac{\partial y^{j}}{\partial x_{i}}dx^i$ and $b_{j_1...j_k}=\frac{\partial x^{i_1}}{\partial y_{j_1}}\cdots\frac{\partial x^{i_k}}{\partial y_{j_k}}a_{i_1...i_k}$ on the right hand side and write $db_{j_1...j_k}=\frac{\partial}{\partial y_{j}}(b_{j_1...j_k})dy^j$, I would proceed by applying the product rule for k+1 factors. After that I am left with a lot of sums and I don't know how to go on. If it helps I could write out each step, but maybe someone knows an easier way than my brute force attempt?

Answer 1

This computation is messy. However, there is a way to avoid it! It is not hard to convince oneself that $d$, as you define it, is the unique operator that eats $k$-forms and returns $k+1$-forms, which satisfies the following (intrinsic) requirements:

$1)$ It is $\mathbb{R}$-linear. That is, $d\lambda\omega=\lambda d\omega$ for any differential form $\omega$ and $\lambda\in\mathbb{R}$.

$2)$ For any function $f$, $df$ is the differential of $f$. That is, the $1$-form $df$ is given by $$df(X)=X(f).$$

$3)$ It satisfies the Leibniz rule $$d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^k\alpha\wedge d\beta,$$ where $k$ is the degree of $\alpha$.

$4)$ It squares to zero. That is, for any form $\omega$ we have $$dd\omega=0.$$

In addition, $d$ is clearly local, in the sense that if $\omega=\omega'$ in a neighborhood of $p$, then $d\omega(p)=d\omega'(p)$.

Now, you know that $d$ is well-defined on any coordinate chart, and on any intersection of such charts. So, by uniqueness, you can conclude that the different definitions on an intersection of coordinate charts all agree with one another.

Answer 2

,Although messy, we can still do it by purely chart representation. Let $w=\Sigma a_I dx^I=\Sigma_J b_J dy^J$ and assume $w$ is $k$-form and the manifold has dim $n$. Then since they are equal , we can find:

$b_J=\Sigma_I \det (\partial x^I/\partial y^J) a_I $, where $\partial x^I/\partial y^J$ denotes the jacobian matrix of $x^{i_1},...,x^{i_k}$ for $i_1<...<i_k$ with respect to $y^{j_1},...,y^{j_k}$ where $j_1<...<j_k$ .

Then we take the exterior derivative definition, we want to prove the following: $$\sum\limits_I {\sum\limits_i {\frac{{\partial {a_I}}}{{\partial {x^i}}}d{x^i} \wedge d{x^I}} } = \sum\limits_J {\sum\limits_j {\frac{{\partial {b_J}}}{{\partial {y^j}}}d{y^i} \wedge d{y^J}} } $$.

Substitute the above relationship into the right hand side , we obtain :

\begin{array}{l} \sum\limits_J {\sum\limits_j {\frac{{\partial {b_J}}}{{\partial {y^j}}}d{y^i} \wedge d{y^J}} } = \sum\limits_J {\sum\limits_j {\sum\limits_I {((\frac{{\partial {a_I}}}{{\partial {y^j}}})} } } \det (\frac{{\partial {x^I}}}{{\partial {y^J}}}) + {a_I}\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {y^J}}})}}{{\partial {y^j}}})d{y^j} \wedge d{y^J}\\ = \sum\limits_J {\sum\limits_j {\sum\limits_I {\sum\limits_i {\frac{{\partial {a_I}}}{{\partial {x^i}}} \cdot \frac{{\partial {x^i}}}{{\partial {y^j}}}\det (\frac{{\partial {x^I}}}{{\partial {y^J}}})d{y^j} \wedge d{y^J}} } } } + \sum\limits_J {\sum\limits_j {\sum\limits_I {{a_I}\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {y^J}}})}}{{\partial {y^j}}}d{y^j} \wedge d{y^J}} } } \end{array}

By the chain rule and also interchanging the order of summation, we see the first term is just equal to $\sum\limits_I {\sum\limits_i {\frac{{\partial {a_I}}}{{\partial {x^i}}}d{x^i} \wedge d{x^I}} } $ and we only need to show the second term is zero.

Due to $a_I$ arbitrary, we shall prove for any fixed $I$ $$\sum\limits_J {\sum\limits_j {\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {x^J}}})}}{{\partial {y^j}}}d{y^j} \wedge d{y^J}} } = 0.$$

Since if $j$ lies in the ordered $J$, $dy^j\wedge dy^J$ is zero, so we can do it for every $U$ subset of $\{1,2,...,n\}$ where $card(U)=k+1$.

$$\sum\limits_J {\sum\limits_j {\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {y^J}}})}}{{\partial {y^j}}}d{y^j} \wedge d{y^J}} } = \sum\limits_{U \subseteq {\rm{ \{ 1,2,}}...{\rm{,n\} ,}}U{\rm{ has }}k + 1{\rm{ elements}}} {\sum\limits_{j \in U} {\frac{{\partial \det (\frac{{\partial {x^I}}}{{\partial {y^J}}})}}{{\partial {y^j}}}d{y^j} \wedge d{y^J}} } $$ where for every $U$ selected and for each $j\in U$ selected $J$ is uniquely determined by requiring $J\subset U$. And we can show for each summand $U$, it is zero.

Let us denote $I=(i_1,...,i_k)$ with $i_1<...<i_k$ and $U$=$\{j_1,...,j_k,j_{k+1}\}$ where $j_1<...<j_k<j_{k+1}$, for each pick of $j$, $J$ is uniquely determined. Then we can see by definition of determinant and the differentiation formula of determinant , the second order derivative term related to $i_1$ is given as the determinant of the following matrix when $j=j_1$ is picked: \begin{array}{*{20}{c}} {\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_2}}}\partial {y^{{j_1}}}}}}&{\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_3}}}\partial {y^{{j_1}}}}}}&{\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_4}}}\partial {y^{{j_1}}}}}}&{...}&{\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_{k + 1}}}}\partial {y^{{j_1}}}}}}\\ {\frac{{\partial {x^{{i_2}}}}}{{\partial {y^{{j_2}}}}}}&{\frac{{\partial {x^{{i_2}}}}}{{\partial {y^{{j_3}}}}}}&{\frac{{\partial {x^{{i_2}}}}}{{\partial {y^{{j_4}}}}}}&{...}&{\frac{{\partial {x^{{i_2}}}}}{{\partial {y^{{j_{k + 1}}}}}}}\\ {\frac{{\partial {x^{{i_3}}}}}{{\partial {y^{{j_2}}}}}}&{\frac{{\partial {x^{{i_3}}}}}{{\partial {y^{{j_3}}}}}}&{\frac{{\partial {x^{{i_3}}}}}{{\partial {y^{{j_4}}}}}}&{...}&{\frac{{\partial {x^{{i_3}}}}}{{\partial {y^{{j_{{\rm{k + 1}}}}}}}}}\\ \vdots & \vdots & \vdots & \vdots & \vdots \\ {\frac{{\partial {x^{{i_k}}}}}{{\partial {y^{{j_2}}}}}}&{\frac{{\partial {x^{{i_k}}}}}{{\partial {y^{{j_3}}}}}}&{\frac{{\partial {x^{{i_k}}}}}{{\partial {y^{{j_4}}}}}}& \cdots &{\frac{{\partial {x^{{i_k}}}}}{{\partial {y^{{j_{k + 1}}}}}}} \end{array} and we collect the terms for other value of $j$ similarly and note that by $dy^{j}\wedge dy^{J}$ it changes sign once when $j$ are incremented as $j_1,...,j_{k+1}$ in order.

Now let us observe the result by expanding of determinant along the first row using Laplace formula: we can easily find that the term related to $\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_p}}}\partial {y^{{j_q}}}}}$ can be represented as $${( - 1)^{p + 1}}\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_p}}}\partial {y^{{j_q}}}}} \cdot \det (\frac{{\partial ({x^{{i_2}}},{x^{{i_3}}},...,{x^{{i_k}}})}}{{\partial ({y^{{z_1}}},{y^{{z_2}}},...,{y^{{z_{k - 1}}}})}}) \cdot {( - 1)^r}$$ where the sign term $(-1)^{p+1}$ is due to $dy^{j}\wedge dy^{J}$ and the second sign change is due to the use of Laplace formula of determinant and it is not hard to find that we can have $r=q+1$ when $q<p$ and $r=q$ when $q>p$ and also $z_1,...,z_{k-1}$ are just indices in $\{j_1,...,j_{k+1}\}$ excluding $j_p,j_q$ and of course $z_1<....<z_{k-1}$ is required.

Now we can see that if we set $j=j_p$ and we can pick a $j_q$ in $J$ and then we can also set $j=j_q$ and pick $j_p$ for $J$ accordingly in the above summation for $U$ and they have different sign but since $\frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_p}}}\partial {y^{{j_q}}}}} = \frac{{{\partial ^2}{x^{{i_1}}}}}{{\partial {y^{{j_q}}}\partial {y^{{j_p}}}}}$, thus the summation is zero and thus there is no term related to the second derivative of $x^{i_1}$ and similar arguments can be applied to $x^{i_2},...,x^{i_k}$(easily by just interchanging the rows of the jacobian matrix).

Thus we conclude the proof.