Q\to A$. Then $e_1^*f^{-1}$ is a unimodular row of length $d+1$. If $g$ is another such isometry, then $gf^{-1}\in Sp_{d+1}(A)$ and $e_1^*g^{-1}(gf^{-1})=e_1^*f^{-1}$. Therefore, the class of $e_1^*f^{-1}$ in $Um_{d+1}(A)/Sp_{d+1}(A)$ is independent of the isometry $f$.

If $(Q,\phi)$ is a symmetric pair such that $Q$ is of rank $d-1$ and $H(A)\bot (Q,\phi)\simeq H(A^{(d+1)/2})$, we denote by $sp(Q,\phi)$ the class of $e_1^*f^{-1}$ in the set $Um_{d+1}(A)/Sp_{d+1}(A)$ for any isometry $f$. We call it the *symplectic class* of $(Q,\phi)$.

This definition seems rather artificial at first sight. However, the next proposition shows that it is natural.

\[identification\] Let $v:=sp(Q,\phi)$. Then $(Q,\phi)$ is isometric to $(Q(v),\phi(v))$.

Let $f:H(A)\bot (Q,\phi)\simeq H(A^{(d+1)/2})$ be the isometry defining $sp(Q,\phi)$. Consider the commutative diagram $$\xymatrix{0\ar[r] & Q\oplus A\ar[r]\ar@{-->}[d]_-{f^\prime} & Q\oplus A^2\ar[r]^-{e_1^*}\ar[d]_-f & A\ar[r]\ar@{=}[d] & 0\\
0\ar[r] & P(v)\ar[r] & A^n\ar[r]_-{v} & A\ar[r] & 0}$$ The construction of $(Q(v),\phi(v))$ shows that $f^\prime:Q\oplus A\to P(v)$ induces an isometry $(Q,\phi)\simeq (Q(v),\phi(v))$.

So we have a procedure to understand when an antisymmetric pair $(Q,\phi)$ is isometric to $H(A^{(d-1)/2})$. The first obstruction is its class in $\K A$. If this is different from $0$, then it cannot be isometric to $H(A^{(d-1)/2})$. If the class is zero, then (using [@Bass2 Chapter IV, Corollary 4.15]) we 