Proposition 10

Tag 068A, permanent. Cite it as [Bourbaki, Tag 068A].

Algebra, chapter III, § 11. Cogebras, products of multilinear forms, inner products and duality

Let $E$ be a graded bigebra (resp. skew graded bigebra). For every element $f$ of degree 1 in $E^{*gr}$ , the left and right inner products are derivations (resp. antiderivations) of the algebra $E$ .

Let $x \in E_p, y \in E_q$ ( $p \geq 1, q \geq 1$ ). By Proposition 5 of no. 3, we can write

c(x) = x \otimes 1 + \sum_{1 \leq j \leq p-1} \left( \sum_i x'_{ij} \otimes x''_{ij,p-j} \right) + 1 \otimes x

c(y) = y \otimes 1 + \sum_{1 \leq k \leq q-1} \left( \sum_i y'_{i,k} \otimes y''_{i,q-k} \right) + 1 \otimes y

where $x'_{ij}$ and $x''_{ij}$ belong to $E_j$ , $y'_{ik}$ and $y''_{ik}$ to $E_k$ . If $E$ is a graded bigebra, the component of $c(xy) = c(x)c(y)$ belonging to $E_1 \otimes E$ , is equal to

\sum_i x'_{i,1} \otimes x''_{i,p-1}y + \sum_i y'_{i,1} \otimes xy''_{i,q-1}

and hence by definition

(xy) \leftarrow f = \sum_i f(x'_{i,1})x''_{i,p-1}y + \sum_i f(y'_{i,1})xy''_{i,q-1}

= (x \leftarrow f)y + x(y \leftarrow f)

and the right inner product by

f

is a derivation. If on the other hand

E

is a skew graded bigebra, the component of

c(xy)

belonging to

E_1 \otimes E

is equal to

\sum_i x'_{i,1} \otimes x''_{i,p-1}y + (-1)^p \sum_i y'_{i,1} \otimes xy''_{i,q-1}

and this time we obtain

(xy) \leftarrow f = (x \leftarrow f)y + (-1)^p x(y \leftarrow f)

which shows that

i(f)

is then an antiderivation. The argument is similar for the left inner product by

f

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Proposition 5