an element $w\in A^n$ such that $v\cdot w=1$ we get a section $\alpha$ of $v$ and therefore a section $r=i^\vee h\alpha$ of $-s^\vee$ and a retraction $-r^\vee$ of $s$. This gives a section $\beta$ of $j$ and we get an injective homomorphism $\gamma:Q(v)\to A^n$. One checks easily that $\phi(v)=\gamma^\vee h\gamma$ and that we have $A^n=Ah^{-1}v^\vee\oplus Aw\oplus \gamma Q(v)$. Moreover, $w$ and $h^{-1}v^\vee$ are both in $\gamma Q(v)^\bot$ and they generate a submodule isometric to $H(A)$.

The following result is the analogue of Proposition \[free\] for “decorated” modules:

\[sympl\] The pairs $(Q(v),\phi(v))$ and $(Q(v^\prime),\phi(v^\prime))$ are isometric if and only if there is a matrix $M\in Sp_n(A)$ such that $vM=v^\prime$.

Following the construction of $Q(v^\prime)$ for $v^\prime=vM$, we see that in that case $(Q(v^\prime),\phi(v^\prime))$ is isometric to $(Q(v),\phi(v))$. Lemma \[isometry\] gives the converse statement.

This shows that $Um_n(A)/Sp_n(A)$ classifies the stably free modules of rank $n-2$ decorated with the antisymmetric form obtained above.

Cancellation of symplectic modules {#core}
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Let $A$ be a ring of odd dimension $d\geq 3$. Let $(Q,\phi)$ be an antisymmetric pair, where $Q$ is of rank $d-1$. Suppose that there is an isometry $$f:H(A)\bot (Q,\phi)\simeq H(A^{(d+1)/2})$$ The projection of $A^2$ to the first factor gives a homomorphism $e_1^*:A^2\oplus 