Proposition 3

The hypotheses and notations being those of prop. 1 (resp. prop. 2) suppose in addition that A, B, A', B' are fields and that the mappings α and β of $A' \otimes_A G$ and $G \otimes_B B'$ into G' characterized by $\alpha(a' \otimes g) = a'u(g)$ and $\beta(g \otimes b') = u(g)b'$ ($a' \in A'$, $b' \in B'$, $g \in G$) are injective. Let M be a subspace of E and N a subspace of F. Then the subspace $(M')^0$ of F' orthogonal to M' with respect to $\Phi'$ is equal to $(M^0)',$ and, analogously, one has $(N')^0 = (N^0)'$.

Indeed the inclusions $(M^0)' \subset (M')^0$ and $(N^0)' \subset (N')^0$ are evident (and moreover true without hypotheses on A, B, A', B', $\alpha$ or $\beta$). We shall prove the inclusion $(M')^0 \subset (M^0)'$; we leave to the reader the verification of the inclusion $(N')^0 \subset (N^0)'$, which is quite analogous. Let $y'$ be an element of $(M')^0$. One can write

where $y_i \in F$ ($1 \leq i \leq s$), and where the $b'_i$ are elements of B' which are linearly independent over B for the structure of left B-module (resp. right B-module) of B'. Let $x \in M$ and $x' = 1 \otimes x \in M'$. We have

(resp. $0 = \Phi'(x', y') = \sum_i u(\Phi(x, y_i)) {b'}^I_i = \beta(\sum_i \Phi(x, y_i) \otimes {b'}^I_i)$.

Comme $\beta$ is injective and the $b'_i$ (resp. the ${b'}^I_i$, taking account of (15)) are linearly independent over B for the structure of left B-module of B', this implies $\Phi(x, y_i) = 0$ for $i = 1, \ldots, s$. Since this last relation is true for all $x \in M$, we have $y_i \in M^0$ for $i = 1, \ldots, s$, whence $y' \in (M^0)'$. QED.