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Assume $k_0$ is a field with char($k_0$) not $2$. Let us define functors from $\rm Field_{/k_0}\to \rm Sets$ as $\rm Pfister_n(k):=\{\text{isomorphism classes of n-fold Pfister forms over k}\}$;

$\rm Quad_n(k)$:= {isomorphism classes of nondegenerate $n$ dimensional quadratic form over $k$} and $\rm Quad_{n,d}(k)$:= same as $\rm Quad_n$ with added condition that determinant should be $d$.

$\rm CSA_n(k):=\{\text{isomorphism classes of central simple k-algebra of degree } n^2\}$.

Serre remarks that for $n\geq 4$ there is no algebraic group $G$ over $k_0$ such that the functor $\rm Pfister_n$ is isomorphic to the functor $k\to H^1(k,G)$. Can someone suggest a referece or sketch a proof for this fact?

For $n=1,2$ one have such algebraic group:

$\rm Pfister_1\cong \rm Quad_1\cong H^1(*,\{\pm1\})$ [Since Isomorphism of one-dimensional quadratic form and 1-fold Pfister forms is upto squares and $H^1(k,\pm 1)=k^*/k^{*2}$] and $\rm Pfister_2\cong \rm Quad_{3,1}\cong CSA_2 \cong H^1(*,PGL_2)$ [We can map $<<a,b>>$ to $<a,b,ab>$ and $<<a,b>>$ to quaternion algebra $(-a,-b)$. Equivalence of $\rm CSA_2\cong H^1(*,PGL_2)$ is standard]. Thus, for $n=1,2$ there is an algebraic group $G$ such that the functor $\rm Pfister_n$ is isomorphic to the functor $k\to H^1(k,G)$.

Thanks!

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For $n=3$ you have $Pfister_3\simeq H^1(*,G_2)$. For greater $n$, you may use notion of cohomological invariants (like in Skip Garibaldi's lectures in Lens) to prove that any $H^1(F,G)$ have a nontrivial cohomological invariant of degree $\le 3$ (namely Tits algebras or the Rost invariant), while for $Pfister_n$ it is possible to have only invariants of degrees at least $n$.

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