4.9\]). This led to the computation of those groups for any smooth rational oriented real algebra (\[loc. cit.,Theorem 5.6\]).

Suppose now that $n$ is even. Then $A^n$ is the underlying module of the antisymmetric form $h:A^n\simeq (A^n)^\vee$ which is $\bot_{n/2}H(A)$. Any unimodular row $v$ of length $n$ gives a stably free module of rank $n-2$ endowed with an antisymmetric form as follows:

The exact sequence $$\xymatrix{0\ar[r] & P(v)\ar[r]^-i & A^n\ar[r]^-v & A\ar[r] & 0}$$ yields a commutative diagram $$\xymatrix{0\ar[r] & A^\vee\ar[r]^-{v^\vee}\ar@{-->}[d]_-s & (A^n)^\vee\ar[r]^-{i^\vee}\ar[d]_-{h^{-1}} & P(v)^\vee\ar[r]\ar@{-->}[d]_-{-s^\vee} & 0 \\
0\ar[r] & P(v)\ar[r]_-i & A^n\ar[r]_-v & A\ar[r] & 0}$$ Since $h$ is antisymmetric $vh^{-1}v^\vee=0$ and therefore we get a homomorphism $s$ making the diagram commutative. Observe that $\ker {s^\vee}$ is projective and equal to $\coker s^\vee$. The snake lemma gives an isomorphism $\coker s\to \ker s^\vee$ which is antisymmetric.

If $v$ is a unimodular row of even length $n$, we denote by $(Q(v),\phi(v))$ the antisymmetric pair obtained above.

\[isometry\] There is an isometry $(Q(v),\phi(v))\bot H(A)\simeq  H(A^{n/2})$.

Consider the following diagram $$\xymatrix{ & & & Q(v)^\vee\ar[d]^-{j^\vee} &  \\
0\ar[r] & A\ar[r]^-{v^\vee}\ar[d]_-s & (A^n)^\vee\ar[r]^-{i^\vee}\ar[d]_-{h^{-1}}\ar@{-->}[ru]^-{\gamma^\vee} & P(v)^\vee\ar[r]\ar[d]_-{-s^\vee} & 0\\
0\ar[r] & P(v)\ar[r]_-i\ar[d]_-j & A^n\ar[r]_-v & A\ar[r] & 0  \\
& Q(v)\ar@{-->}[ru]_-\gamma & &  &}$$ Choosing 