Proposition 9

Let $E$ be a graded bigebra (resp. skew graded bigebra). For every element $x$ of degree 1 in $E$, the left and right inner products by $x$ are derivations (resp. antiderivations) ($§ 10$, no. 2) of the algebra $E^{*\mathrm{gr}}$.

In the notation of no. 6, for every homogeneous element $x$ of degree 1 in a graded bigebra (resp. a skew graded bigebra) $E$,

by Proposition 5 of no. 3 and the fact that $c$ is a homomorphism of degree 0. Suppose first that $E$ is a graded bigebra. For all $y \in E$, by definition and since $c$ is an algebra homomorphism, $c(xy) = c(x)c(y)$. Let $c(y) = \sum_i s_i \otimes t_i$ with $s_i$ and $t_i$ in $E$; therefore Hence $\langle (uv) \mathbin{\&} x, y \rangle = \sum_i u(xs_i)v(t_i) + \sum_i u(s_i)v(xt_i)$. But we may write and similarly whence, returning to the notation $i(x)$ for the inner product. which proves that $i(x)$ is a derivation in $E^{*\mathrm{gr}}$.

Suppose now that $E$ is a skew graded bigebra, that $u \in E_p^*$, $v \in E_q^*$ and $y \in E_r$; then we can write

where the $s_{ij}$ and $t_{ij}$ belong to $E_j$; by definition of the product in $E^g \otimes_A E$, then whence this time

Then also $\sum_{0 \leq j \leq r} \left( \sum_i u(x s_{ij}) v(t_{i,r-j}) \right) = \langle (u \sqcup x)v, y \rangle$. On the other hand, $u(s_{ij}) = 0$ unless $j = -p$ and hence we may also write

We therefore conclude that in other words $i(x)$ is an antiderivation in $E^{*gr}$. The assertions relating to the left inner product by an element $x$ of degree 1 in $E$ are proved similarly.