Interior product - Knowlpedia

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

(\iota_X\omega)_p(v_1,\dots,v_{k-1}) = \omega_p\bigl(X_p,v_1,\dots,v_{k-1}\bigr) \qquad (p\in M).

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

Identities

The operator $\iota_X$ is $\mathbb{R}$ -linear in $X$ and is a graded derivation of degree $-1$ on the exterior algebra:

\iota_X(\alpha\wedge\beta)=(\iota_X\alpha)\wedge\beta+(-1)^{\deg\alpha}\alpha\wedge(\iota_X\beta).

It is linked to the Lie derivative by Cartan’s identity

\mathcal{L}_X=\mathrm{d}\circ\iota_X+\iota_X\circ\mathrm{d},

where $\mathrm{d}$ is the exterior derivative.

Examples

A basic contraction in $\mathbb{R}^2$ . With $\omega=\mathrm{d}x\wedge \mathrm{d}y$ and $X=\partial_x$ , one has $\iota_X\omega=\mathrm{d}y$ .
Volume form in $\mathbb{R}^3$ . For $\omega=\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$ and $X=\partial_z$ , one gets $\iota_X\omega=\mathrm{d}x\wedge\mathrm{d}y$ .
Symplectic geometry viewpoint. On a symplectic manifold $(M,\omega)$ , the assignment $X\mapsto \iota_X\omega$ identifies vector fields with 1-forms when $\omega$ is nondegenerate; Hamiltonian vector fields are characterized by $\iota_X\omega=\mathrm{d}H$ for some function $H$ .