Def: A section of ${\bigwedge }^k(T^{\ast }M)$ is called a differential $k$-form, or just a $k$-form; this is a continuous tensor field whose value at each point is an alternating tensor. The integer $k$ is called the degree of the form. We denote the space of Smooth $k$-forms by $$\Omega^k(M) := \Gamma\left({\textstyle \bigwedge}^{! k}( T^* M)\right). $$
The wedge product of two differential forms is defined pointwise: $(\omega \wedge \eta {)}_p:={\omega }_p\wedge {\eta }_p$. Thus, the wedge product of a $k$-form with an $l$-form is a $(k+l)$-form. If $f$ is a $0$-form and $\eta$ is a $k$-form, we interpret the wedge product $f\wedge \eta$ to mean the ordinary product $f\eta$. If we define $$\Omega^*(M) := \bigoplus_{k = 0}^n\Omega^k(M),$$then the wedge product turns ${\mathrm{\Omega }}^{\ast }(M)$ into an associative, anticommutative graded algebra.

In any smooth chart, a $k$-form $\omega$ can be written locally as $$\omega = \omega_ I; dx^{i_1}\wedge\dots \wedge dx^{i_k} = \omega_I ; dx^I,$$where the coefficients ${\omega }_I$ are continuous functions defined on the coordinate domain, and use $d{x}^I$ as the abbreviation for $d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}$. We are also extending Einstein's summation convention to differential forms, where it is is understood that we are summing over all increasing multi-indices.

Obs: A differential form $\omega$ is smooth on $U$ iff if the component functions ${\omega }_I$ are smooth.

Obs: We can see that $d{x}^I$ are the elementary differential forms, and satisfy the following identity $$dx^{i_1}\wedge\dots \wedge dx^{i_k}\left(\frac{\partial}{\partial x^{j_1}}, \dots,\frac{\partial}{\partial x^{j_k}} \right) = \delta^I_J. $$Thus the component functions ${\omega }_I$ of $\omega$ are determined by $$\omega_I = \omega \left(\frac{\partial}{\partial x^{i_1}}, \dots,\frac{\partial}{\partial x^{i_k}} \right)$$

Def: If $F:M\to N$ is a smooth map and $\omega$ is a differential form on $N$, the pullback $F^{\ast }\omega$ is a differential form on $M$, defined as for any covariant tensor field: $$(F^*\omega)p(v_1,\dots, v_k) = \omega(dF_p (v_1),\dots, dF_p(v_k)). $$

Lemma: Suppose $F:M\to N$ is smooth.

• $F^{\ast }:{\mathrm{\Omega }}^k(N)\to {\mathrm{\Omega }}^k(M)$ is linear over $\mathbb{R}$.
• $F^{\ast }(\omega \wedge \eta )=(F^{\ast }\omega )\wedge (F^{\ast }\eta )$.
• In any smooth chart, $$F^*(\omega_I ; dy^{i_1}\wedge\dots\wedge dy^{i_k}) = (\omega_I \circ F); d(y^{i_1}\circ F)\wedge \dots \wedge d(y^{i_1}\circ F).$$

Pullback Formula for Top-Degree Forms: Let $F:M\to N$ be a smooth map between $n$-manifolds with or without boundary. If $(x^i)$ and $(y^i)$ are smooth coordinates on open subsets $U\subseteq M$ and $V\subseteq N$, respectively, and $u$ is a continuous real-valued function on $V,$then the following golds on $U\cap F^{-1}[V]$: $$F^*(u; dy^1\wedge\dots \wedge dy^n) = (u \circ F)\det\left(\frac{\partial F_i}{\partial x^j}\right) ; dx^1\wedge \dots \wedge dx^n.$$
Cor: If $(U,(x^i))$ and $(V,(y^i))$ are overlapping smooth coordinate charts on $M$. then the following identity holds on $U\cap V$: $$dy^1\wedge \dots \wedge dy^n = \det\left(\frac{\partial y^i}{\partial x^j}\right) ; dx^1\wedge\dots \wedge dx^n.$$
Def: Interior multiplication also extends naturally to vector fields and differential forms, simply by letting it act pointwise: if $X\in \mathfrak{X}(M)$ and $\omega \in {\mathrm{\Omega }}^k(M)$, define a $(k-1)$-form $X⌟\omega =i_X\omega$ by $$(X; \lrcorner; \omega)_p := X_p;\lrcorner; \omega_p. $$
Properties of the Interior Product: Let $X$ be a smooth vector field on $M$.

• If $\omega$ is a smooth differential form, the $i_X\omega$ is smooth.
• $i_X:{\mathrm{\Omega }}^k(M)\to {\mathrm{\Omega }}^{k-1}(M)$ is linear over ${\mathcal{C}}^{\mathrm{\infty }}(M)$ and therefore corresponds to a smooth bundle homomorphism $i_X:{\bigwedge }^k(T^{\ast }M)\to {\bigwedge }^{k-1}(T^{\ast }M)$.

An important operation to differential forms is the exterior derivative

Prop: Suppose $M$ is a smooth manifold and $X\in \mathfrak{X}(M)$. Then the interior multiplication $i_X:{\mathrm{\Omega }}^{\ast }(M)\to {\mathrm{\Omega }}^{\ast }(M)$ is an antiderivation of degree $-1$ whose square is $0$.

Cartan's Lemma: Let $M$ be a smooth $n$-manifold with or without boundary, and let $({\omega }^1,\dots ,{\omega }^k)$ be an ordered $k$-tuple of smooth $1$-forms on an open subset $U\subseteq M$ such that $({\omega }^1{|}_p,\dots ,{\omega }^k{|}_p)$ is linearly independent for each $p\in U$. Given smooth $1$-forms ${\alpha }^1,\dots ,{\alpha }^k$ on $U$ such that $$\sum_{i = 1}^j \alpha^i \wedge \omega^i = 0,$$then each ${\alpha }^i$ can be written as a linear combination of ${\omega }^1,\dots ,{\omega }^k$ with smooth coefficients.

Prop: For each nonnegative integer $k$, there is a contravariant functor ${\mathrm{\Omega }}^k:\mathsf{D}\mathsf{i}\mathsf{f}\mathsf{f}\to \mathsf{V}\mathsf{e}{\mathsf{c}}_{\mathbb{R}}$, which to each smooth manifold $M$ assigns the vector space ${\mathrm{\Omega }}^k(M)$ and to each smooth $F$ the pullback $F^{\ast }$. We see that the exterior derivative is a natural transformation from ${\mathrm{\Omega }}^k$ to ${\mathrm{\Omega }}^{k+1}$.

On Riemannian Manifolds

Let $(M,g)$ be an oriented Riemannian $n$-manifold.

For each $k=1,\cdots n$, $g$ determines a unique inner product on ${\bigwedge }^k(T_p^{\ast }M)$, denoted by $⟨\cdot ,\cdot {⟩}_g$ just like the inner product $T_{p}M$ satisfying $$\langle\omega^1\wedge \dots \wedge \omega^k,\eta\wedge\dots \wedge\eta^k \rangle_k := \det(\langle\omega^i, \eta^j\rangle_g) $$whenever ${\omega }^1,\dots ,{\omega }^k,{\eta }^1,\dots ,{\eta }^k$ are covectors at $k$. We are using The Tangent-Cotangent Bundle Isomorphism for Riemannian Manifolds.

We see that the Riemannian volume form $d{V}_g$ is the unique positively oriented $n$-form that has unit norm with respect this inner product.

Prop: For each $k=0,\dots ,n$ there is a unique smooth bundle homomorphism $\star :{\bigwedge }^k(T^{\ast }M)\to {\bigwedge }^{n-k}(T^{\ast }M)$ satisfying $$\omega\wedge (\star\eta) =\langle \omega, \eta\rangle_g ; dV_g $$for all smooth $k$-forms $\omega$, $\eta$.

Def: The map from the above proposition is called the Hodge star operator.

In any smooth local coordinates $(x^i)$, we can calculate the Hodge dual of a basic $k$-form $$\star(dx^{i_1}\wedge\dots \wedge dx^{i_k}) = \sqrt{\det g} ;g^{i_1 j_1}\cdots g^{i_k j_k} \varepsilon_{j_1,\dots, j_k} ; dx^{j_{k+1}}\wedge \dots\wedge dx^{j_n},$$where ${\epsilon }_{j_1,\dots ,j_n}$ is the Levi-Civita Symbol with ${\epsilon }_{1,\dots ,n}=1$. This we can extend this to general differential forms. Let $\alpha ={\alpha }_{i_1,\dots ,i_k}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}$, $$\star\alpha = \sqrt{\det g} ;g^{i_1 j_1}\cdots g^{i_k j_k} \varepsilon_{j_1,\dots, j_k}; \alpha_{i_1,\dots, i_k} ; dx^{j_{k+1}}\wedge \dots\wedge dx^{j_n}. $$

Prop:

• We see that $\star :{\bigwedge }^0(T^{\ast }M)\to {\bigwedge }^n(T^{\ast }M)$ is given by $\star f=fd{V}_g$.
• We see that $\star (\star \omega )=(-1{)}^{k(n-k)}\omega$ if $\omega$ is a $k$-form.