Wedge product of differential forms

Let $M$ be a smooth manifold . The wedge product is the canonical bilinear product on the graded algebra of differential forms .

Definition

For $\alpha \in \Omega^k(M)$ and $\beta \in \Omega^\ell(M)$, their wedge product $\alpha\wedge\beta \in \Omega^{k+\ell}(M)$ is defined pointwise by the alternating product of multilinear forms on each tangent space :

for all $p\in M$ and $v_1,\dots,v_{k+\ell}\in T_pM$.

Equivalently, $(\alpha\wedge\beta)_p$ is the alternation of the tensor product $\alpha_p\otimes \beta_p$.

Key properties

If $\alpha\in\Omega^k(M)$, $\beta\in\Omega^\ell(M)$, and $\gamma\in\Omega^m(M)$, then:

• Bilinearity: $\wedge$ is $\mathbb{R}$-bilinear in each argument.
• Associativity: $(\alpha\wedge\beta)\wedge\gamma=\alpha\wedge(\beta\wedge\gamma)$.
• Graded-commutativity: $\alpha\wedge\beta = (-1)^{k\ell}\,\beta\wedge\alpha.$ In particular, if $k$ is odd then $\alpha\wedge\alpha=0$.
• Compatibility with pullback: for any smooth map $F:M\to N$, the pullback of forms satisfies $F^*(\alpha\wedge\beta)=F^*\alpha\wedge F^*\beta$.
• The wedge product is the product appearing in the graded Leibniz rule for the exterior derivative .

Examples

1. Coordinate 1-forms on $\mathbb{R}^3$.
   With standard coordinates $(x,y,z)$, the 2-form $dx\wedge dy$ satisfies $(dx\wedge dy)(\partial_x,\partial_y)=1$ and changes sign when the vectors are swapped: $(dx\wedge dy)(\partial_y,\partial_x)=-1$.

2. A wedge computation with functions.
   Let $\alpha = f\,dx + g\,dy$ and $\beta = h\,dz$ on $\mathbb{R}^3$, where $f,g,h$ are smooth functions. Then

   $$\alpha\wedge\beta = fh\,dx\wedge dz + gh\,dy\wedge dz.$$

3. Wedge of a 1-form with itself is zero (odd degree).
   On any manifold, $dx\wedge dx=0$. More generally, if $\eta$ is any 1-form then $\eta\wedge\eta=0$ by graded-commutativity with $k=\ell=1$.