get an isometry $$(Q,\phi)\bot H(A)\simeq (A^{(d+1)/2}).$$ and therefore a symplectic class in $Um_{d+1}(A)/Sp_{d+1}(A)$. Using Propositions \[identification\] and \[sympl\], we see that this class must be trivial in order for $(Q,\phi)$ to be isometric to $H(A^{(d-1)/2})$.

In dimension $3$, this procedure gives a complete understanding of projective modules of rank $2$.

\[cancellation\] Let $A$ be a ring of dimension $3$. Then every projective module of rank $2$ with trivial determinant is free if and only if the group $\K A$ and the set $Um_4(A)/\Sp A$ are both trivial.

Remark first that every element in $\K A$ is of the form $e(P)$ for some projective module $P$ of rank $2$ (use [@Bass2 Chapter IV, Theorem 4.12]). If every such module is free, then $\K A=0$ because the only non degenerate anti-symmetric form on $A^2$ is $h$ (up to isometry). The set $Um_4(A)/\Sp A$ is reduced to a point because of Proposition \[sympl\].

Suppose now that $\K A=0$. Then the Euler class of $P$ is zero for any $P$ of rank $2$, and its symplectic class (which is then defined) is in the $Sp_4(A)$-orbit of $(1,0,0,0)$. Therefore any such $P$ is free by Proposition \[sympl\] again.

In the last section, we show that 