Tangent-bundle end cohomology vanishing under exact SU(3) holonomy

Parallel transport on $K_X={\mathrm{\Lambda }}^3(T_X^{1,0}{)}^{\ast }$ acts by the inverse determinant of tangent parallel transport. Every holonomy element lies in $SU(3)$, so this action is trivial. Transporting one nonzero canonical covector from a fixed point gives a path-independent global parallel section: two paths differ by a holonomy element acting trivially. The canonical connection is therefore flat. In holomorphic coordinates its curvature is $\mathrm{\partial }\bar{\mathrm{\partial }}\log \det (g_{i\bar{j}})$, up to an overall curvature sign, by tracing the tangent Chern connection matrix $g^{-1}\mathrm{\partial }g$ and passing to the dual determinant line. Consequently $\rho =-i\mathrm{\partial }\bar{\mathrm{\partial }}\log \det (g_{i\bar{j}})=0$. All the hypotheses of [F1] are now checked, giving $h^{1,0}=h^{2,0}=0$.

Contraction with the nowhere-zero three-form defines a holomorphic bundle map $T_X\to {\mathrm{\Omega }}_X^2$, $v\mapsto {\iota }_v\mathrm{\Omega }$. In a basis $e_1,e_2,e_3$ with $\mathrm{\Omega }(e_1,e_2,e_3)\mathrm{\ne }0$, it sends each basis vector to a nonzero multiple of the complementary basis two-form, so it is an isomorphism. Hence $H^0(X,T_X)\cong H^0(X,{\mathrm{\Omega }}_X^2)=0$ by step 1.1.

The same holomorphic bundle isomorphism gives $H^3(X,T_X)\cong H^3(X,{\mathrm{\Omega }}_X^2)$. By [F2], the latter is represented by harmonic $(2,3)$-forms. In complex dimension three, the Hodge star sends type $(p,q)$ to $(3-q,3-p)$, as follows on wedge monomials in a unitary coframe. It is invertible and commutes with the Hodge Laplacian by [F4], so it identifies harmonic $(2,3)$-forms with harmonic $(0,1)$-forms. Complex conjugation identifies the latter with harmonic $(1,0)$-forms, whose dimension is $h^{1,0}=0$ by step 1.1. Thus $H^3(X,T_X)=0$, with analytic sheaf cohomology used throughout.