Hodge-number and Betti-number counts under exact SU(3) holonomy

Let $\alpha$ be a holomorphic covariant alternating $p$-tensor, $0\le p\le 3$. In holomorphic coordinates the Kähler condition gives ${\mathrm{\partial }}_i{h}_{j\bar{k}}={\mathrm{\partial }}_j{h}_{i\bar{k}}$; substituting this identity into the Levi-Civita formula makes the mixed Christoffel symbols ${\mathrm{\Gamma }}_{\bar{j}i}^k$ zero. Hence ${\mathrm{\nabla }}^{0,1}\alpha =\bar{\mathrm{\partial }}\alpha =0$. In unitary holomorphic normal coordinates, commute ${\mathrm{\nabla }}_j$ and ${\mathrm{\nabla }}_{\bar{j}}$ on each covariant index. The traced commutator is the Ricci-form contraction on that index, so ${\sum }_j[{\mathrm{\nabla }}_j,{\mathrm{\nabla }}_{\bar{j}}]{\alpha }_{i_1\dots i_p}=0$ by [F3]. The product rule for $\mathrm{\mid }\alpha {\mathrm{\mid }}^2$ therefore gives the Bochner identity $\frac{1}{2}\mathrm{\Delta }\mathrm{\mid }\alpha {\mathrm{\mid }}^2=\mathrm{\mid }{\mathrm{\nabla }}^{1,0}\alpha {\mathrm{\mid }}^2$ at the coordinate center, with $\mathrm{\Delta }=\mathrm{div}\mathrm{\nabla }$; the same tensorial calculation holds at every point. Since $X$ is compact without boundary, ${\int }_X\mathrm{\Delta }\mathrm{\mid }\alpha {\mathrm{\mid }}^{2}dV=0$. Thus the continuous nonnegative function $\mathrm{\mid }{\mathrm{\nabla }}^{1,0}\alpha {\mathrm{\mid }}^2$ has zero integral and vanishes everywhere. Together with ${\mathrm{\nabla }}^{0,1}\alpha =0$, every holomorphic $p$-form is parallel.

Parallel $(p,0)$-forms are exactly the vectors fixed by holonomy in ${\mathrm{\Lambda }}^p({\mathbb{C}}^3{)}^{\ast }$. For a diagonal element $\mathrm{diag}(e^{i{\theta }_1},e^{i{\theta }_2},e^{-i({\theta }_1+{\theta }_2)})\in SU(3)$, the three weights on ${\mathrm{\Lambda }}^1$ are $e^{-i{\theta }_1},e^{-i{\theta }_2},e^{i({\theta }_1+{\theta }_2)}$, none identically $1$; the three weights on ${\mathrm{\Lambda }}^2$ are $e^{-i({\theta }_1+{\theta }_2)},e^{i{\theta }_2},e^{i{\theta }_1}$, also none identically $1$. On ${\mathrm{\Lambda }}^3$ the action is the determinant, identically $1$, so its invariant space is one-dimensional. On ${\mathrm{\Lambda }}^0$ the invariant space consists of constants and is one-dimensional because $X$ is connected. By [F4], these dimensions are $h^{p,0}$, giving $h^{1,0}=h^{2,0}=0$ and $h^{3,0}=1$. Hodge conjugation then gives $h^{0,1}=h^{0,2}=0$ and $h^{0,3}=1$.

The Hodge direct sum in [F4] yields $b_1=h^{1,0}+h^{0,1}=0$, $b_2=h^{2,0}+h^{1,1}+h^{0,2}=h^{1,1}$, and $b_3=h^{3,0}+h^{2,1}+h^{1,2}+h^{0,3}=2h^{2,1}+2$, using conjugation from [F4] and step 2.1. The real Hodge star on this oriented six-manifold maps harmonic real two-forms bijectively to harmonic real four-forms: it commutes with the Hodge Laplacian and ${\ast }^2=(-1{)}^{k(6-k)}=1$ for $k=2,4$, so it is its own inverse on those degrees. By uniqueness of harmonic representatives in [F4], the dimensions are $b_2$ and $b_4$. Hence $b_4=b_2=h^{1,1}$. The full choice axiom [F1] is required by the imported Hodge result [F4].