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2 incorrect use of "begs the question" changed to "raises the question"

So thanks to the comments of Tyler Lawson I have been able to figure out what is happening in this example, so I thought I should post it as an answer. I think this is also what Torsten Ekedahl was getting at in his comment, as well.

I think it helps to be extra clear because this example is rather confusing. For starters there is the group scheme, $$\mathbb{G} \to S.$$ In this example $S = \mathbb{A}^1$ is the affine line. This is a group object over $S$, so it can be thought of as an $S$-family of group schemes. At the points $x_1$ and $x_2$ it is the trivial group, and at all other points it is some fixed abelian group $A$. For a concrete example we can take $A = \mathbb{Z}/3$, and then $\mathbb{G}$ looks something like this:

The bottom line represents $S$. Notice that there is a unique global section, the zero section. Away from the set $Y = x_1 \cup x_2$, there are more sections. Associated to $\mathbb{G}$ is a sheaf on the site of schemes over S. This is the same sheaf I called $A_{C_Y}$.

As outlined in the question we have that $\check H^1(S; A_{C_Y}) = A$ is non-trivial. We can even construct a non-trivial cocycle using the covering consisting of the two open subsets $$U_1 = S - x_1$$ $$U_2 = S - x_2$$ Notice that $U_{12} = U_1 \times_S U_2 = C_Y$, the complement of Y in S. This is exactly the subspace that supports a section. The picture is a little misleading here as it looks like there are lots of sections over $C_Y$. However, because we are using the Zariski topology we have only $A$-many of them. Such a section over $C_Y$ has to be constant on $C_Y$.

Now each of these sections (of which there are A-many) gives rise to a Cech cocycle and so we should be able to construct a $\mathbb{G}$-torsor over $S$ for each one of these. The usual construction is that this torsor is given as the coequalizer of $$U_{12} \times_S \mathbb{G} \rightrightarrows \coprod U_i \times_S \mathbb{G}$$ Where one map is the usual inclusion and the other is also inclusion (the other one), but twisted using the cocycle.

Now the cocycle is only defined over $C_Y$. And over the complement of $C_Y$, namely Y, $\mathbb{G}$ is trivial. It has a unique fiber. So I restricted attention to just the "interesting part", the $C_Y$ part. Then I got that the coequalizer becomes, $$C_Y \times A \rightrightarrows (C_Y \cup C_Y) \times A$$ which has trivial coequalizer $C_Y \times A$. All of these are true facts, except the part about $C_Y$ being the only interesting part. I was wrongly assuming that if the torsor was trivial over this part, then it had to be isomorphic to $\mathbb{G}$.

This is not the case. Somehow Tyler's comments made me realize this. The actual full colimit looks something like this:

Notice that this space is a trivial $C_Y \times A$-torsor when restricted to $C_Y$, and over $U_1$ and $U_2$ there exist unique sections. However there is no global section, so it is not a globally trivial object. Let's call this object P.

A little book keeping shows that there is an action of schemes over S, $$\mathbb{G} \times_S P \to P$$ making P into a torsor in the second sense.

So this is not a counter example. Both notions of torsor agree here.

But this begs raises the question:

Question: Do these two a priori different notions of torsor agree in Algebraic Geometry? If not what is the easiest counter example?

I don't know the answer to this.

1

So thanks to the comments of Tyler Lawson I have been able to figure out what is happening in this example, so I thought I should post it as an answer. I think this is also what Torsten Ekedahl was getting at in his comment, as well.

I think it helps to be extra clear because this example is rather confusing. For starters there is the group scheme, $$\mathbb{G} \to S.$$ In this example $S = \mathbb{A}^1$ is the affine line. This is a group object over $S$, so it can be thought of as an $S$-family of group schemes. At the points $x_1$ and $x_2$ it is the trivial group, and at all other points it is some fixed abelian group $A$. For a concrete example we can take $A = \mathbb{Z}/3$, and then $\mathbb{G}$ looks something like this:

The bottom line represents $S$. Notice that there is a unique global section, the zero section. Away from the set $Y = x_1 \cup x_2$, there are more sections. Associated to $\mathbb{G}$ is a sheaf on the site of schemes over S. This is the same sheaf I called $A_{C_Y}$.

As outlined in the question we have that $\check H^1(S; A_{C_Y}) = A$ is non-trivial. We can even construct a non-trivial cocycle using the covering consisting of the two open subsets $$U_1 = S - x_1$$ $$U_2 = S - x_2$$ Notice that $U_{12} = U_1 \times_S U_2 = C_Y$, the complement of Y in S. This is exactly the subspace that supports a section. The picture is a little misleading here as it looks like there are lots of sections over $C_Y$. However, because we are using the Zariski topology we have only $A$-many of them. Such a section over $C_Y$ has to be constant on $C_Y$.

Now each of these sections (of which there are A-many) gives rise to a Cech cocycle and so we should be able to construct a $\mathbb{G}$-torsor over $S$ for each one of these. The usual construction is that this torsor is given as the coequalizer of $$U_{12} \times_S \mathbb{G} \rightrightarrows \coprod U_i \times_S \mathbb{G}$$ Where one map is the usual inclusion and the other is also inclusion (the other one), but twisted using the cocycle.

Now the cocycle is only defined over $C_Y$. And over the complement of $C_Y$, namely Y, $\mathbb{G}$ is trivial. It has a unique fiber. So I restricted attention to just the "interesting part", the $C_Y$ part. Then I got that the coequalizer becomes, $$C_Y \times A \rightrightarrows (C_Y \cup C_Y) \times A$$ which has trivial coequalizer $C_Y \times A$. All of these are true facts, except the part about $C_Y$ being the only interesting part. I was wrongly assuming that if the torsor was trivial over this part, then it had to be isomorphic to $\mathbb{G}$.

This is not the case. Somehow Tyler's comments made me realize this. The actual full colimit looks something like this:

Notice that this space is a trivial $C_Y \times A$-torsor when restricted to $C_Y$, and over $U_1$ and $U_2$ there exist unique sections. However there is no global section, so it is not a globally trivial object. Let's call this object P.

A little book keeping shows that there is an action of schemes over S, $$\mathbb{G} \times_S P \to P$$ making P into a torsor in the second sense.

So this is not a counter example. Both notions of torsor agree here.

But this begs the question:

Question: Do these two a priori different notions of torsor agree in Algebraic Geometry? If not what is the easiest counter example?

I don't know the answer to this.