A reference for this material is Chapter 14 of John M. Lee. Introduction to smooth manifolds. Second edition. Grad. Texts in Math., 218. Springer, New York, 2013. xvi+708pp. ISBN: 978-1-4419-9981-8.

Solution:

Exercise. Do it at least for \(k =1\)!

Solution:

Since the vector fields \( \tfrac {\partial }{\partial x ^1}, \ldots , \tfrac {\partial }{\partial x ^n} \) commute (partial derivatives commute), when we plug them into an exterior derivative, only the terms in the first sum show up. For \(1 \leq j_0 < \ldots < j_k \leq n\), we have

\begin{align*} d \omega (\tfrac {\partial }{\partial x ^{j_0}}, & \ldots , \tfrac {\partial }{\partial x ^{j_k}} ) = \sum _{\ell = 0} ^k (-1 ) ^{\ell } \frac {\partial }{\partial x ^{j_{\ell }}} \omega (\tfrac {\partial }{\partial x ^{j_0}}, \ldots ,\hat {\tfrac {\partial }{\partial x ^{j_{\ell }}}} , \ldots , \tfrac {\partial }{\partial x ^{j_k}} ) \\ & = \sum _{\ell = 0} ^k (-1 ) ^{\ell } \frac {\partial f }{\partial x ^{j_{\ell }}} dx ^{i_1} \wedge \cdots \wedge dx ^{i_k} ( \tfrac {\partial }{\partial x ^{j_0}} , \ldots , \hat {\tfrac {\partial }{\partial x ^{j_{\ell }}}} , \ldots , \tfrac {\partial }{\partial x ^{j_k}} ) \end{align*}

This sum is zero unless

\[ \{ i_1, \ldots , i_k \} \subset \{ j_0, \ldots , j_k \} . \]

In that case, the only summand that is not zero is the one with

\[ \{ j _{\ell } \} = \{ j_0, \ldots , j_k \} \backslash \{ i_1, \ldots , i_k \} , \]

and that summand is

\[ (-1) ^{\ell } \frac {\partial f}{\partial x ^{j_{\ell }}} . \]

Plugging \((\tfrac {\partial }{\partial x ^{j_0}}, \ldots , \tfrac {\partial }{\partial x ^{j_k}} )\) into the right hand side of the expression of the proposition, we get zero unless

\[ \{ i_1, \ldots , i_k \} \subset \{ j_0, \ldots , j_k \} . \]

In that case, if we set

\[ \{j_{\ell } \} = \{ j_0, \ldots , j_k \} \backslash \{ i_1, \ldots , i_k \} , \]

the only summand that is not zero is

\[ \frac {\partial f}{\partial x ^{j_{\ell }}} d x ^{j_{\ell }} \wedge d x ^{i_1 } \wedge \cdots \wedge dx ^{i_k} ( \tfrac {\partial }{\partial x ^{j_0}}, \ldots , \tfrac {\partial }{\partial x ^{j_k}} ) = (-1 )^{\ell } \frac {\partial f}{\partial x ^{j_{\ell }}} . \]

In either case, the two expressions agree.

Solution:

By linearity, it is enough to check in coordinates. Namely,

\begin{align*} \omega & = f \, dx^{i_1} \wedge \cdots \wedge dx ^{i_k}\\ \eta & = g \, dx^{j_1} \wedge \cdots \wedge dx ^{j_{\ell }}. \end{align*}

In that case,

\[ \omega \wedge \eta = fg \, dx^{i_1} \wedge \cdots \wedge dx ^{i_k} \wedge dx^{j_1} \wedge \cdots \wedge dx ^{j_{\ell }}. \]

Taking the derivative,

\begin{align*} d (\omega \wedge \eta ) & = \sum _{a = 1 }^n \frac {\partial (fg)}{\partial x ^a} dx ^a \wedge dx^{i_1} \wedge \cdots \wedge dx ^{i_k} \wedge dx^{j_1} \wedge \cdots \wedge dx ^{j_{\ell }} \\ & = \sum _{a = 1 }^n \frac {\partial f }{\partial x ^a} g \, dx ^a \wedge dx^{i_1} \wedge \cdots \wedge dx ^{i_k} \wedge dx^{j_1} \wedge \cdots \wedge dx ^{j_{\ell }} \\ & \,\,\,\,\,\,\,\, + \sum _{a = 1 }^n f \frac {\partial g}{\partial x ^a} \, dx ^a \wedge dx^{i_1} \wedge \cdots \wedge dx ^{i_k} \wedge dx^{j_1} \wedge \cdots \wedge dx ^{j_{\ell }} \\ & = \left [ \sum _{a = 1}^n \frac {\partial f }{\partial x ^a} dx ^a \wedge dx^{i_1} \wedge \cdots \wedge dx ^{i_k} \right ] \wedge \left [ g dx^{j_1} \wedge \cdots \wedge dx ^{j_{\ell }} \right ]\\ & \,\,\,\,\,\, + (-1) ^k \left [ f dx^{i_1} \wedge \cdots \wedge dx ^{i_k} \right ] \wedge \left [ \sum _{a = 1 }^n \frac {\partial g}{\partial x ^a} \, dx ^a \wedge dx^{j_1} \wedge \cdots \wedge dx ^{j_{\ell }} \right ] \\ & = (d\omega ) \wedge \eta + (-1)^k \omega \wedge (d \eta ) . \end{align*}

Solution:

We first check it for functions.

\begin{align*} d^2 f (X,Y) & = X (df (Y) ) - Y (df (X)) - df ([X,Y]) \\ & = X Yf - YXf - [X,Y] f \\ & = 0 . \end{align*}

Then, working in local coordinates, if

\[ \omega = f dx^{i_1} \wedge \cdots \wedge dx ^{i_k} , \]

then

\begin{align*} d^2 \omega & = d\left [ df \wedge dx^{i_1} \wedge \cdots \wedge dx ^{i_k} \right ] \\ & = ( d^2 f ) \wedge dx^{i_1} \wedge \cdots \wedge dx ^{i_k} \\ & = 0 . \end{align*}