If M is a free K-module of finite rank n, then, for every endomorphism s of M, the restriction to Λn(M) of the derivation Ds is the homothety of ratio Tr(s).
Let (ej)1≤j≤n be a basis of M and write e=e1∧e2∧⋯∧en. If
s(ej)=k=1∑nαjkek, the third formula in (36) gives
Ds(e)=i=1∑ne1∧⋯∧ei−1∧s(ei)∧ei+1∧⋯∧en=(j=1∑nαjj)e.
Example (2). In the Corollary to Proposition 14, part (iii), let Δ={0}, the graduation on Λ(M) then being the usual graduation of type Z; on the other hand take ε(p,q)=(−1)pq. Then, for every linear form x∗∈M∗ on M, x↦⟨x,x∗⟩ is a graded K-linear mapping of degree −1 of M into Λ(M) satisfying relation (34); then there exists an antiderivation i(x∗) of Λ(M), of degree −1, such that (by virtue of formula (21) of no. 5)
i(x∗)(x1∧⋯∧xn)=i=1∑n(−1)i−1⟨xi,x∗⟩x1∧⋯∧xi−1∧xi+1∧⋯∧xn and which is a special case of the inner product to be defined in § 11, no. 9, formula (68).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.