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!



Set of isomorphisms of Pfister forms corresponding to first cohomology of algebraic group
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 $n$.


