Interior product

Properties

If in local coordinates ${\displaystyle (x_1,...,x_n)}$ the vector field ${\displaystyle X}$ is given by

${\displaystyle X=f_1\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_1}+\cdots +f_n\frac{\mathrm{\partial }}{\mathrm{\partial }{x}_n}}$

then the interior product is given by $${\displaystyle {\iota }_X(d{x}_1\wedge ...\wedge d{x}_n)=\sum _{r=1}^n(-1{)}^{r-1}{f}_{r}d{x}_1\wedge ...\wedge \widehat{d{x}_r}\wedge ...\wedge d{x}_n,}$$ where ${\displaystyle d{x}_1\wedge ...\wedge \widehat{d{x}_r}\wedge ...\wedge d{x}_n}$ is the form obtained by omitting ${\displaystyle d{x}_r}$ from ${\displaystyle d{x}_1\wedge ...\wedge d{x}_n}$.

By antisymmetry of forms, $${\displaystyle {\iota }_X{\iota }_Y\omega =-{\iota }_Y{\iota }_X\omega ,}$$ and so ${\displaystyle {\iota }_X\circ {\iota }_X=0.}$ This may be compared to the exterior derivative ${\displaystyle d,}$ which has the property ${\displaystyle d\circ d=0.}$

The interior product with respect to the commutator of two vector fields ${\displaystyle X,}$ ${\displaystyle Y}$ satisfies the identity $${\displaystyle {\iota }_{[X,Y]}=\left[{\mathcal{L}}_X,{\iota }_Y\right]=\left[{\iota }_X,{\mathcal{L}}_Y\right].}$$Proof. For any k-form ${\displaystyle \mathrm{\Omega }}$, $${\displaystyle {\mathcal{L}}_X({\iota }_Y\mathrm{\Omega })-{\iota }_Y({\mathcal{L}}_X\mathrm{\Omega })=({\mathcal{L}}_X\mathrm{\Omega })(Y,-)+\mathrm{\Omega }({\mathcal{L}}_{X}Y,-)-({\mathcal{L}}_X\mathrm{\Omega })(Y,-)={\iota }_{{\mathcal{L}}_{X}Y}\mathrm{\Omega }={\iota }_{[X,Y]}\mathrm{\Omega }}$$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): $${\displaystyle {\mathcal{L}}_X\omega =d({\iota }_X\omega )+{\iota }_{X}d\omega =\left\{d,{\iota }_X\right\}\omega .}$$

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.

Proof by direct computation

Since vector fields are locally integrable, we can always find a local coordinate system ${\displaystyle ({\xi }^1,\dots ,{\xi }^n)}$ such that the vector field ${\displaystyle X}$ corresponds to the partial derivative with respect to the first coordinate, i.e., ${\displaystyle X={\mathrm{\partial }}_1}$.

By linearity of the interior product, exterior derivative, and Lie derivative, it suffices to prove the Cartan's magic formula for monomial ${\displaystyle k}$-forms. There are only two cases:

Case 1: ${\displaystyle \alpha =ad{\xi }^1\wedge d{\xi }^2\wedge \cdots \wedge d{\xi }^k}$. Direct computation yields:$${\displaystyle \begin{array}{cc}{\iota }_X\alpha & =ad{\xi }^2\wedge \cdots \wedge d{\xi }^k,\\ d({\iota }_X\alpha )& =({\mathrm{\partial }}_{1}a)d{\xi }^1\wedge d{\xi }^2\wedge \cdots \wedge d{\xi }^k+\sum _{i=k+1}^n({\mathrm{\partial }}_{i}a)d{\xi }^i\wedge d{\xi }^2\wedge \cdots \wedge d{\xi }^k,\\ d\alpha & =\sum _{i=k+1}^n({\mathrm{\partial }}_{i}a)d{\xi }^i\wedge d{\xi }^1\wedge d{\xi }^2\wedge \cdots \wedge d{\xi }^k,\\ {\iota }_X(d\alpha )& =-\sum _{i=k+1}^n({\mathrm{\partial }}_{i}a)d{\xi }^i\wedge d{\xi }^2\wedge \cdots \wedge d{\xi }^k,\\ L_X\alpha & =({\mathrm{\partial }}_{1}a)d{\xi }^1\wedge d{\xi }^2\wedge \cdots \wedge d{\xi }^k.\end{array}}$$

Case 2: ${\displaystyle \alpha =ad{\xi }^2\wedge d{\xi }^3\wedge \cdots \wedge d{\xi }^{k+1}}$. Direct computation yields:$${\displaystyle \begin{array}{cc}{\iota }_X\alpha & =0,\\ d\alpha & =({\mathrm{\partial }}_{1}a)d{\xi }^1\wedge d{\xi }^2\wedge \cdots \wedge d{\xi }^{k+1}+\sum _{i=k+2}^n({\mathrm{\partial }}_{i}a)d{\xi }^i\wedge d{\xi }^2\wedge \cdots \wedge d{\xi }^{k+1},\\ {\iota }_X(d\alpha )& =({\mathrm{\partial }}_{1}a)d{\xi }^2\wedge \cdots \wedge d{\xi }^{k+1},\\ L_X\alpha & =({\mathrm{\partial }}_{1}a)d{\xi }^2\wedge \cdots \wedge d{\xi }^{k+1}.\end{array}}$$

Proof by abstract algebra, credited to Shiing-Shen Chern

The exterior derivative ${\displaystyle d}$ is an anti-derivation on the exterior algebra. Similarly, the interior product ${\displaystyle {\iota }_X}$ with a vector field ${\displaystyle X}$ is also an anti-derivation. On the other hand, the Lie derivative ${\displaystyle L_X}$ 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 ${\displaystyle f}$) and their differentials, exact 1-forms (${\displaystyle df}$). Verify Cartan's magic formula on these two cases.