Let $X$ be a finite-type scheme over a field $k$. Let $G$ be a finite-type group scheme over $k$; we write $G_X$ for the base-change of $G$ from $\operatorname{Spec}(k)$ to $X$.

Suppose $f : Y \rightarrow X$ is a $G_X$-torsor for the fppf topology (i.e. we have an $X$-group scheme action of $G_X$ on $Y$ such that the morphism $G_X \times_X Y \rightarrow Y \times_X Y$ given on points by $(g,y) \mapsto (y,gy)$ is an isomorphism). Such a $Y$ gives a class $[Y]$ in the fppf cohomology set $H^1(X,G_X)$ that classifies fppf $G_X$-torsor sheaves over $X$ (this is defined with Cech cohomology).

Consider the specialization map $$ s : X(k) \rightarrow H^1(k,G) $$ that sends $x : \operatorname{Spec}(k) \rightarrow X$ to the pull-back of $[Y]$ by $x$, which is an element of $H^1(k,G)$. Note: if $G$ is smooth, the fppf cohomology set $H^1(k,G)$ may be identified with the Galois cohomology set $H^1(k,G(k^{\mathrm{sep}}))$.

Supposing it exists, fix a $y \in Y(k)$. We obtain an exact sequence of pointed sets $$ 0 \rightarrow G(k) \rightarrow Y(k) \stackrel{f}{\rightarrow} X(k) \stackrel{s}{\rightarrow} H^1(k,G) $$ where $G(k) \rightarrow Y(k)$ is just the inclusion of the fiber above $y$. From the looks of it, I'd say that this has to be the start of a long exact sequence of some kind. The question only is: what kind? I don't see an obvious way of continuing it, since $Y$ doesn't necessarily carry any group structure so as to give meaning to the expression $H^1(k,Y)$.

Question: Is this exact sequence part of a long exact sequence? For instance, are we witnessing some instantiation of homotopy theory? If not, is there any other more conceptual way of viewing the above sequence?

Example: As a motivating example, let me show you why the image of $s$ - and therefore a continuation of the exact sequence from above - is an interesting object of study. Let $E$ be an elliptic curve over $\mathbf{Q}$ given by $y^2=f(x)$. Let $Y$ be $E - E[2]$, let $X$ be $\operatorname{Spec}(\mathbf{Q}[x,f^{-1}])$ (i.e., the affine line with coordinate $x$ and with the subscheme $f=0$ deleted), and let $f:Y \rightarrow X$ be the map that sends $(x,y)$ to $x$. Let $G = \mu_2$. We endow $Y$ with the structure of a $G$-torsor by letting the non-trivial element of $G$ send $(x,y)$ to $(x,-y)$. Then the image of the map \begin{align*} X(\mathbf{Q}) & \rightarrow \mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2} ~~ (\cong H^1(\mathbf{Q},\mu_2)) \\\ x & \mapsto f(x) \pmod{\mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2}} \end{align*} consists precisely of those elements $c \in \mathbf{Q}^{\ast}/\mathbf{Q}^{\ast 2}$ such that $cy^2=f(x)$ contains rational points other than the "trivial ones", i.e. the zeros of $f$ and the point at infinity.

  • $\begingroup$ If $G \to Y \to X$ is an exact sequence of group schemes, then $Y \to X$ is a $G$-torsor, and your exact sequence is part of the standard long exact sequence of group cohomology. So an answer would presumably be a generalization of that exact sequence to other cases. I do not know anything about the homotopy-theory or other considerations that would enable someone to generalize this to higher degree cohomology groups. $\endgroup$
    – Will Sawin
    Feb 4, 2013 at 20:20
  • $\begingroup$ Thank you, Will, this is indeed part of the motivation for my question. I should have included it in the post. $\endgroup$
    – R.P.
    Feb 4, 2013 at 23:21

1 Answer 1


It actually more like the ending of a long exact sequence, rather than the beginning. To see what's going on consider the analogous case in topology. For this you replace the Galois group of $k$ with a discrete group $\Gamma$ and the category of $k$-schemes with the category of $\Gamma$-spaces. Instead of an algebraic group you now have a topological group G equipped with an action of $\Gamma$ on its classifying space BG. A G-torsor is a principle fibration $Y \to X$, which in homotopy theory corresponds to a $\Gamma$-equivariant fibration sequence of the form $$ Y \to X \to BG $$ Taking $\Gamma$-homotopy fixed points one obtains a fibration sequence $$ Y^{h\Gamma} \to X^{h\Gamma} \to BG^{h\Gamma} $$ which leads to a long exact sequence of homotopy groups ending with $$ ... \to \pi_1(BG^{h\Gamma}) \to \pi_0(Y^{h\Gamma}) \to \pi_0(X^{h\Gamma}) \to \pi_0(BG^{h\Gamma}) $$ where $\pi_1(BG^{h\Gamma}) = \pi_0(G^{h\Gamma})$. This tail is the homotopy theoretic analogue of the sequence $$ G(k) \to Y(k) \to X(k) \to H^1(k,G) $$ and so one should consider this sequence as the ending, and not the beginning of a long exact sequence. However, as apposed to the homotopy theoretic analogue, the map $G(k) \to Y(k)$ is always injective, making it seems like the sequence is just starting.

The analogy with the homotopy theoretic case can be made more precise by considering the etale homotopy type (see http://arxiv.org/abs/1110.0164).

  • $\begingroup$ Thank you, this is very interesting! Actually however, I was thinking along the lines of Will's comment: if X,Y are k-group schemes and f is a homomorphism, then we are able to extend the exact sequence: if G is abelian we have a long exact sequence in the right sense of the word, whereas in general with the help of non-abelian cohomology we can get at least 2 more terms. I was hoping that one could do this in more generality. $\endgroup$
    – R.P.
    Feb 4, 2013 at 23:20
  • $\begingroup$ Name of referenced paper: Harpaz and Schlank - Homotopy Obstructions to Rational Points. $\endgroup$
    – LSpice
    Jul 28, 2022 at 4:27

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