Consider the functor $\mathcal{F}: \text{Sch}/\mathbb{Q}\longrightarrow \text{Sets}$, defined by sending a scheme $X$ with coordinate ring $\mathcal{O}(X)$ to the set of orbits $B(\mathcal{O}(X))\setminus SL\_2(\mathcal{O}(X))$, where $B$ is the Borel subgroup of upper triangular matrices.


My question is if this functor is a sheaf.


What I've tried: it boils down to showing that for any open $U$ and a cover $\cup\_i U\_i = U$ by opens $U\_i$, the following sequence is exact
$$0\longrightarrow \mathcal{F}(U)\longrightarrow \prod\_i\mathcal{F}(U\_i)\rightrightarrows\prod\_{i,j}\mathcal{F}(U\_i\cap U\_j).$$


In [this](https://math.stackexchange.com/questions/86509/reference-for-it-is-enough-to-specify-a-sheaf-on-a-basis) stack exchange thread, I saw that it is enough to prove that the above sequence is exact for a basis of the topology, so it is enough to prove the statement when $U$ is affine and the $U\_i$ are basic opens.


I managed to prove the injectivity of $\mathcal{F}(U)\longrightarrow \prod\_i\mathcal{F}(U\_i)$ without much trouble, but I am not sure I understand how to prove that this functor admits descent, i.e. that the sequence is exact in the middle.


For example, even when there are two open subsets $U\_1,U\_2$ with $U = U\_1\cup U\_2$. Cosets of $B(\mathcal{O}(X))\setminus SL\_2(\mathcal{O}(U\_i))$ are parameterized by pairs $(c\_i,d\_i)\in \mathcal{O}(U\_i)$. Denote by $f\_i\in \mathcal{O}(U)$ the element for which $U\_i = D(f\_i)$, then possibly replacing the representative $(c\_i,d\_i)$ by multiplying with a power of $f\_i$, we may assume $(c\_i,d\_i)$ are in $\mathcal{O}(U)$.


Ideally, I would like to say that $(c\_1,d\_1)$ is a global section that lifts $s = \{(c\_1,d\_1)|\_{U\_1}, (c\_2,d\_2)|\_{U\_2}\}$, however, for this I would need to find a $b\in \mathcal{O}(U)^{\times}$ such that $bc\_1 = c\_2, bd\_1 = d\_2$, however, the only thing that follows from the fact that the section $s$ equalizes, is that there exists a $b\in \mathcal{O}(U\_1\cap U\_2)^{\times}$ such that $b(c\_1,d\_1) = (c\_2,d\_2)$. Write $b = b'/(f\_1f\_2)^m$, then this only means that there exists a power of $f\_1f\_2$ such that
$$(f\_1f\_2)^n(b'c\_1 - (f\_1f\_2)^mc\_2) = 0$$
in $\mathcal{O}(U)$. From this equality, I can only deduce that there exists an invertible $b'\in\mathcal{O}(U)$ for which
$$b'(f\_1f\_2)^nc\_1 = (f\_1f\_2)^{m+n}c\_2.$$
Had I known that $f\_1f\_2$ is not a zero divisor, then I could conclude that $f\_1f\_2$ is invertible in $\mathcal{O}(U)$ and complete the argument, but I don't. And generally, one cannot choose an irreducible basis of opens for a scheme unless it is irreducible to begin with.


In summary: don't know how to prove that the sequence is exact in the middle nor how to find a counter example. Either of which would be appreciated.



Is this functor $\mathcal{F}: \text{Sch}/\mathbb{Q}\longrightarrow \text{Sets}$ a sheaf?
No, this isn't true. Take $X = \mathbb{P}^1$ with $U\_1$ and $U\_2$ being the standard affine cover of $\mathbb{P}^1$. Let the coordinate rings of $U\_1$ and $U\_2$ be $k[t]$ and $k[t^{-1}]$, so the coordinate ring of $U\_1 \cap U\_2$ is $k[t, t^{-1}]$.


Map $U\_1$ to $B \backslash \text{SL}\_2$ by $\begin{bmatrix} 1&0 \\ t&1 \\ \end{bmatrix}$ and map $U\_2$ to $B \backslash \text{SL}\_2$ by $\begin{bmatrix} 0&-1 \\ 1&t^{-1} \\ \end{bmatrix}$.


We have
$$\begin{bmatrix} 1&0 \\ t&1 \\ \end{bmatrix}= 
\begin{bmatrix} t^{-1} & 1 \\ 0&t \end{bmatrix}
\begin{bmatrix} 0&-1 \\ 1&t^{-1} \\ \end{bmatrix},$$
so these maps coincide on $U\_1 \cap U\_2$. But $\text{SL}\_2$ is affine and $\mathbb{P}^1$ is connected and projective, so $\text{SL}\_2(\mathbb{P}^1)$ contains only constant maps.




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The point here is that $B \backslash \text{SL}\_2$ IS just $\mathbb{P}^1$, and the map $\text{SL}\_2 \to \mathbb{P}^1$ has local sections but no global sections.


