 Solution

A smooth differential form of degree $p$ is a smooth section of ${\Lambda }^p{T}^{\ast }M$, soIt is equivalently a smooth alternating covariant $p$-tensor field. In a coordinate chart $x^1,\dots ,x^n$ it has a unique expressionwith smooth coefficients. At $p=0$ these are the smooth functions, and for $p>n$ the space is zero.

Define the exterior derivative in a chart byTo check that it is well-defined, let $y^1,\dots ,y^n$ be other smooth coordinates. The expression $d{y}^a={\sum }_i({\partial }_i{y}^a)d{x}^i$ satisfiesbecause the coefficients are symmetric and the wedge products antisymmetric. The coordinate formula also satisfies the graded Leibniz rule, and for a function $b$ the chain rule gives $d_{x}b={\sum }_a(\partial b/\partial y^a)d{y}^a=d_{y}b$. Applying these two facts to $\omega ={\sum }_I{b}_{I}d{y}^I$ yields $d_x\omega ={\sum }_I{d}_y{b}_I\wedge d{y}^I=d_y\omega$. Thus the local operators agree on overlaps and define a global linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$.

A second application in any chart givesagain by symmetry of mixed derivatives and antisymmetry of the wedge product of differential forms. Consequently every exact differential form is closed. Put ${\Omega }^{-1}(M)=0$ and define the de Rham cohomology byThe identity $d^2=0$ is exactly what puts the denominator inside the numerator, making this quotient meaningful. Since ${\Omega }^p(M)=0$ for $p>\dim M$, $H_{\mathrm{dR}}^p(M)=0$ in those degrees.