Interior product (contraction) ι_X

Let $M$ be a smooth manifold and let $X$ be a vector field on $M$. For a differential form $\omega\in\Omega^k(M)$, the interior product (or contraction) of $\omega$ by $X$ is the $(k-1)$-form $\iota_X\omega\in\Omega^{k-1}(M)$ defined by

By convention, if $k=0$ then $\iota_X\omega:=0$.

The contraction satisfies the following standard identities.

• $C^\infty(M)$-linearity in the vector field: for $f\in C^\infty(M)$,

  $$\iota_{fX}\omega = f\,\iota_X\omega.$$

• Graded derivation rule for the wedge product : if $\alpha\in\Omega^p(M)$ and $\beta\in\Omega^q(M)$, then

  $$\iota_X(\alpha\wedge\beta) = (\iota_X\alpha)\wedge\beta + (-1)^p\,\alpha\wedge(\iota_X\beta).$$

• Cartan’s formula (interaction with differentiation): the Lie derivative of forms along $X$ is given by

  $$\mathcal{L}_X\omega = d(\iota_X\omega)+\iota_X(d\omega),$$

  where $d$ is the exterior derivative .

Examples

1. If $\alpha\in\Omega^1(M)$ is a $1$-form, then $\iota_X\alpha$ is the smooth function $\alpha(X)\in C^\infty(M)$ obtained by evaluating $\alpha$ on the vector field $X$.

2. On $M=\mathbb{R}^2$ with coordinates $(x,y)$, let $\omega=dx\wedge dy$ and $X=\frac{\partial}{\partial x}$. Then

   $$\iota_X\omega = dy.$$

3. On $M=\mathbb{R}^3$ with coordinates $(x,y,z)$, let $\omega=dx\wedge dy\wedge dz$ and $X=\frac{\partial}{\partial x}+2\frac{\partial}{\partial y}$. Then

   $$\iota_X\omega = dy\wedge dz - 2\,dx\wedge dz.$$