Is this functor $\mathcal{F}: \text{Sch}/\mathbb{Q}\longrightarrow \text{Sets}$ a sheaf?

Question

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 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.

Answer 1

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.