Interior product on the exterior algebra

Let $V$ be a finite-dimensional real inner product space and $v\in V$. For each degree $k\ge 1$, exterior multiplication by $v$,

$m_v:{\mathrm{\Lambda }}^{k-1}V⟶{\mathrm{\Lambda }}^{k}V,m_v(\alpha )=v\wedge \alpha ,$

is a linear map between finite-dimensional inner product spaces, the target carrying the Gram pairing of The Gram inner product on ${\mathrm{\Lambda }}^{k}V$ and the wedge product being Exterior multiplication is well defined, graded, associative, unital, and graded-commutative. Its adjoint exists and is unique by Every linear map between finite-dimensional inner product spaces has a unique adjoint. The interior product (contraction) by $v$ is that adjoint:

${\iota }_v:=m_v^{\ast }:{\mathrm{\Lambda }}^{k}V⟶{\mathrm{\Lambda }}^{k-1}V,⟨{\iota }_v\alpha ,\beta ⟩=⟨\alpha ,v\wedge \beta ⟩,$

with ${\iota }_v=0$ on ${\mathrm{\Lambda }}^{0}V$. The explicit value of ${\iota }_v$ on a decomposable wedge is computed in Interior product is the adjoint of exterior multiplication by a vector ↗, which discharges the well-definedness obligation recorded above.

The three defining properties of the interior product — ${\iota }_v(1)=0$, ${\iota }_v(w)=⟨v,w⟩$ for $w\in V$, and the graded derivation rule of Exterior multiplication and interior product satisfy the graded anticommutation identity — each follow from the adjoint description.