Conjecture that spinor modifications of conic bundles are hyperbolically equivalent

Work over $\mathbb C$. Let

$$S=\operatorname{Gr}(2,4),$$

let $\mathcal U$ be the tautological rank-two bundle, and let projectivizations parametrize lines. Set

$$X=\mathbb P_S(\mathcal U), \qquad f\colon X\to S.$$

Put $\mathcal E=\mathfrak{sl}(\mathcal U)$ and consider the nondegenerate ternary form

$$q=-\det\colon \mathcal O_S\longrightarrow \operatorname{Sym}^2(\mathcal E)^\vee.$$

Its zero locus in $\mathbb P_S(\mathcal E)$ is $X$: a nonzero trace-free endomorphism with determinant zero is rank-one nilpotent, hence has $\operatorname{im}(A)=\ker(A)=\ell$ for a unique line $\ell\subset\mathcal U_s$. Now take

$$\mathcal F=\mathcal O_X(-1)^{\oplus 2}.$$

Since $R\Gamma(\mathbb P^1,\mathcal O_{\mathbb P^1}(-1))=0$,

$$\mathbf Rf_*\mathcal F=0.$$

Moreover,

$$\det(\mathcal F)=\mathcal O_X(-2),\qquad \omega_{X/S}\cong \mathcal O_X(-2)\otimes f^*\det(\mathcal U)^\vee,$$

so $c_1(\mathcal F)=K_{X/S}$ in $\operatorname{Pic}(X/S)$. Thus $\mathcal F$ is an abstract spinor bundle. Furthermore,

$$\mathbf Rf_*\mathcal End(\mathcal F)\cong M_2(\mathcal O_S).$$

The associated spinor modification therefore has a constant nondegenerate ternary quadratic form, and hence

$$X_{\mathcal F}\cong S\times\mathbb P^1.$$

It remains to distinguish $X$ and $X_{\mathcal F}$ up to hyperbolic equivalence. Since

$$\dim S=4,\qquad \omega_S\cong\mathcal O_S(-4),$$

Kuznetsov’s half-cohomology invariant is defined using $\mathcal M=\mathcal O_S(-2)$. Its rank parity is preserved by hyperbolic reductions and extensions. Borel–Weil–Bott gives

$$H^i(S,\mathcal O_S(-2))=0$$

for every $i$, while

$$H^i\bigl(S,\mathfrak{sl}(\mathcal U)(-2)\bigr) \cong \begin{cases} \mathbb C,&i=2,\\ 0,&i\ne2. \end{cases}$$

Indeed,

$$\mathfrak{sl}(\mathcal U)(-2) \cong \Sigma^{(-1,-3)}\mathcal U^\vee,$$

and Borel–Weil–Bott places its unique one-dimensional cohomology group in degree two. The determinant form and Serre duality induce a perfect form on

$$H^2\bigl(S,\mathfrak{sl}(\mathcal U)(-2)\bigr)\cong\mathbb C,$$

so its rank is odd. For $X_{\mathcal F}$, the underlying bundle is $\mathcal O_S^3$, and

$$H^2(S,\mathcal O_S(-2)^3)=0,$$

so the corresponding rank is even. Finally, there is no presentation ambiguity: endpoint twists in a hypothetical hyperbolic chain differ by a $2$-torsion line bundle, while

$$\operatorname{Pic}(S)\cong\mathbb Z$$

has no $2$-torsion. Thus the two parity invariants are directly comparable. Hence $X_{\mathcal F}$ is a spinor modification of $X$, but the two conic bundles are not hyperbolically equivalent. This disproves Kuznetsov’s conjecture as stated.