# Interior product

In mathematics, the interior product (also known as interior derivative, interior multiplication, inner multiplication, inner derivative, insertion operator, or inner derivation) is a degree −1 (anti)derivation on the exterior algebra of differential forms on a smooth manifold. The interior product, named in opposition to the exterior product, should not be confused with an inner product. The interior product $\iota _{X}\omega$ is sometimes written as $X\mathbin {\lrcorner } \omega .$

## Definition

The interior product is defined to be the contraction of a differential form with a vector field. Thus if $X$ is a vector field on the manifold $M,$ then $\iota _{X}:\Omega ^{p}(M)\to \Omega ^{p-1}(M)$ is the map which sends a $p$-form $\omega$ to the $(p-1)$-form $\iota _{X}\omega$ defined by the property that $(\iota _{X}\omega )\left(X_{1},\ldots ,X_{p-1}\right)=\omega \left(X,X_{1},\ldots ,X_{p-1}\right)$ for any vector fields $X_{1},\ldots ,X_{p-1}.$
When $\omega$ is a scalar field (0-form), $\iota _{X}\omega =0$ by convention.
The interior product is the unique antiderivation of degree −1 on the exterior algebra such that on one-forms $\alpha$ $\displaystyle \iota _{X}\alpha =\alpha (X)=\langle \alpha ,X\rangle ,$ where $\langle \,\cdot ,\cdot \,\rangle$ is the duality pairing between $\alpha$ and the vector $X.$ Explicitly, if $\beta$ is a $p$-form and $\gamma$ is a $q$-form, then $\iota _{X}(\beta \wedge \gamma )=\left(\iota _{X}\beta \right)\wedge \gamma +(-1)^{p}\beta \wedge \left(\iota _{X}\gamma \right).$ The above relation says that the interior product obeys a graded Leibniz rule. An operation satisfying linearity and a Leibniz rule is called a derivation.

## Properties

If in local coordinates $(x_{1},...,x_{n})$ the vector field $X$ is given by
$X=f_{1}{\frac {\partial }{\partial x_{1}}}+\cdots +f_{n}{\frac {\partial }{\partial x_{n}}}$
then the interior product is given by $\iota _{X}(dx_{1}\wedge ...\wedge dx_{n})=\sum _{r=1}^{n}(-1)^{r-1}f_{r}dx_{1}\wedge ...\wedge {\widehat {dx_{r}}}\wedge ...\wedge dx_{n},$ where $dx_{1}\wedge ...\wedge {\widehat {dx_{r}}}\wedge ...\wedge dx_{n}$ is the form obtained by omitting $dx_{r}$ from $dx_{1}\wedge ...\wedge dx_{n}$.
By antisymmetry of forms, $\iota _{X}\iota _{Y}\omega =-\iota _{Y}\iota _{X}\omega ,$ and so $\iota _{X}\circ \iota _{X}=0.$ This may be compared to the exterior derivative $d,$ which has the property $d\circ d=0.$
The interior product with respect to the commutator of two vector fields $X,$ $Y$ satisfies the identity $\iota _{[X,Y]}=\left[{\mathcal {L}}_{X},\iota _{Y}\right]=\left[\iota _{X},{\mathcal {L}}_{Y}\right].$Proof. For any k-form $\Omega$, ${\mathcal {L}}_{X}(\iota _{Y}\Omega )-\iota _{Y}({\mathcal {L}}_{X}\Omega )=({\mathcal {L}}_{X}\Omega )(Y,-)+\Omega ({\mathcal {L}}_{X}Y,-)-({\mathcal {L}}_{X}\Omega )(Y,-)=\iota _{{\mathcal {L}}_{X}Y}\Omega =\iota _{[X,Y]}\Omega$and similarly for the other result.

## Cartan identity

The interior product relates the exterior derivative and Lie derivative of differential forms by the Cartan formula (also known as the Cartan identity, Cartan homotopy formula or Cartan magic formula): ${\mathcal {L}}_{X}\omega =d(\iota _{X}\omega )+\iota _{X}d\omega =\left\{d,\iota _{X}\right\}\omega .$
where the anticommutator was used. This identity defines a duality between the exterior and interior derivatives. Cartan's identity is important in symplectic geometry and general relativity: see moment map. The Cartan homotopy formula is named after Élie Cartan.
Since vector fields are locally integrable, we can always find a local coordinate system $(\xi ^{1},\dots ,\xi ^{n})$ such that the vector field $X$ corresponds to the partial derivative with respect to the first coordinate, i.e., $X=\partial _{1}$.
By linearity of the interior product, exterior derivative, and Lie derivative, it suffices to prove the Cartan's magic formula for monomial $k$-forms. There are only two cases:
Case 1: $\alpha =a\,d\xi ^{1}\wedge d\xi ^{2}\wedge \dots \wedge d\xi ^{k}$. Direct computation yields:${\begin{aligned}\iota _{X}\alpha &=a\,d\xi ^{2}\wedge \dots \wedge d\xi ^{k},\\d(\iota _{X}\alpha )&=(\partial _{1}a)\,d\xi ^{1}\wedge d\xi ^{2}\wedge \dots \wedge d\xi ^{k}+\sum _{i=k+1}^{n}(\partial _{i}a)\,d\xi ^{i}\wedge d\xi ^{2}\wedge \dots \wedge d\xi ^{k},\\d\alpha &=\sum _{i=k+1}^{n}(\partial _{i}a)\,d\xi ^{i}\wedge d\xi ^{1}\wedge d\xi ^{2}\wedge \dots \wedge d\xi ^{k},\\\iota _{X}(d\alpha )&=-\sum _{i=k+1}^{n}(\partial _{i}a)\,d\xi ^{i}\wedge d\xi ^{2}\wedge \dots \wedge d\xi ^{k},\\L_{X}\alpha &=(\partial _{1}a)\,d\xi ^{1}\wedge d\xi ^{2}\wedge \dots \wedge d\xi ^{k}.\end{aligned}}$
Case 2: $\alpha =a\,d\xi ^{2}\wedge d\xi ^{3}\wedge \dots \wedge d\xi ^{k+1}$. Direct computation yields:${\begin{aligned}\iota _{X}\alpha &=0,\\d\alpha &=(\partial _{1}a)\,d\xi ^{1}\wedge d\xi ^{2}\wedge \dots \wedge d\xi ^{k+1}+\sum _{i=k+2}^{n}(\partial _{i}a)\,d\xi ^{i}\wedge d\xi ^{2}\wedge \dots \wedge d\xi ^{k+1},\\\iota _{X}(d\alpha )&=(\partial _{1}a)\,d\xi ^{2}\wedge \dots \wedge d\xi ^{k+1},\\L_{X}\alpha &=(\partial _{1}a)\,d\xi ^{2}\wedge \dots \wedge d\xi ^{k+1}.\end{aligned}}$
The exterior derivative $d$ is an anti-derivation on the exterior algebra. Similarly, the interior product $\iota _{X}$ with a vector field $X$ is also an anti-derivation. On the other hand, the Lie derivative $L_{X}$ is a derivation.
The anti-commutator of two anti-derivations is a derivation.
To show that two derivations on the exterior algebra are equal, it suffices to show that they agree on a set of generators. Locally, the exterior algebra is generated by 0-forms (smooth functions $f$) and their differentials, exact 1-forms ($df$). Verify Cartan's magic formula on these two cases.