Gauge-fixed infinitesimal Ricci-flat metric modes on a strict Calabi–Yau threefold

Vary the connection using $2{\dot{\mathrm{\Gamma }}}^c{}_{ab}={\mathrm{\nabla }}_a{h}_b{}^c+{\mathrm{\nabla }}_b{h}_a{}^c-{\mathrm{\nabla }}^c{h}_{ab}$ and ${\dot{R}}_{ab}={\mathrm{\nabla }}_c{\dot{\mathrm{\Gamma }}}^c{}_{ab}-{\mathrm{\nabla }}_b{\dot{\mathrm{\Gamma }}}^c{}_{ac}$. Commuting the two covariant derivatives, and using $\mathrm{Ric}=0$, gives $2D\mathrm{Ric}(h)={\mathrm{\Delta }}_{L}h-{\mathcal{L}}_{(\beta h{)}^{\mathrm{♯}}}g\mathrm{.}$ For example the flat derivative part is $-{\mathrm{\nabla }}^c{\mathrm{\nabla }}_c{h}_{ab}+{\mathrm{\nabla }}_a{\mathrm{\nabla }}^c{h}_{cb}+{\mathrm{\nabla }}_b{\mathrm{\nabla }}^c{h}_{ca}-{\mathrm{\nabla }}_a{\mathrm{\nabla }}_b\mathrm{tr}h$; the commutators supply the displayed $-2R_{acbd}{h}^{cd}$. Diffeomorphism covariance gives $D\mathrm{Ric}({\mathcal{L}}_{V}g)={\mathcal{L}}_V\mathrm{Ric}=0$, and direct differentiation of ${\mathcal{L}}_{V}g=2{\mathrm{\nabla }}_{(a}{V}_{b)}$ gives $\beta ({\mathcal{L}}_{V}g)={\mathrm{\nabla }}^{\ast }\mathrm{\nabla }{V}^{\mathrm{♭}}$ at Ricci flatness. The one-form identity ${\mathrm{\Delta }}_d{V}^{\mathrm{♭}}={\mathrm{\nabla }}^{\ast }\mathrm{\nabla }{V}^{\mathrm{♭}}+\mathrm{Ric}(V{)}^{\mathrm{♭}}$ follows by expanding $d{d}^{\ast }+d^{\ast }d$ and commuting derivatives. Its kernel is zero by [F1] and [F2]. This nonnegative symmetric operator has scalar positive symbol, so [F3] solves ${\mathrm{\Delta }}_d{V}^{\mathrm{♭}}=-\beta h$ uniquely and smoothly. Hence every linearized Ricci-flat class has a unique Bianchi representative.

The parallel complex structure splits $h=h^{+}+h^{-}$, where $h^{+}(Ju,Jv)=h^{+}(u,v)$ and $h^{-}(Ju,Jv)=-h^{-}(u,v)$. Kähler curvature preserves this splitting, so ${\mathrm{\Delta }}_L$ does as well. The Hermitian summand corresponds to the real $(1,1)$-form $\eta (u,v)=h^{+}(Ju,v)$. Expanding the exterior Laplacian on a two-form gives $({\mathrm{\Delta }}_d\eta {)}_{ab}=-{\mathrm{\nabla }}^c{\mathrm{\nabla }}_c{\eta }_{ab}+{\mathrm{Ric}}_a{}^c{\eta }_{cb}+{\mathrm{Ric}}_b{}^c{\eta }_{ac}-2R_{acbd}{\eta }^{cd}\mathrm{.}$ One obtains the two Ricci contractions and the two-form curvature term by expanding $d\eta =3{\mathrm{\nabla }}_{[a}{\eta }_{bc]}$ and $d^{\ast }\eta =-{\mathrm{\nabla }}^a{\eta }_{ab}$ and using the covector commutator $[{\mathrm{\nabla }}_a,{\mathrm{\nabla }}_b]{\eta }_{cd}=-R^e{}_{cab}{\eta }_{ed}-R^e{}_{dab}{\eta }_{ce}$. Since $J$ is parallel and Kähler curvature commutes with it, this formula at Ricci flatness identifies ${\mathrm{\Delta }}_d\eta$ with the same $J$ transform of ${\mathrm{\Delta }}_L{h}^{+}$. Thus its kernels are in bijection. A harmonic $(1,1)$-form is closed; in unitary coordinates closure says ${\mathrm{\nabla }}_i{h}_{j\bar{k}}={\mathrm{\nabla }}_j{h}_{i\bar{k}}$. Consequently $-{\mathrm{\nabla }}^a{h}_{ai}=-{\mathrm{\partial }}_i{\mathrm{tr}}_{\mathbb{C}}h$ and ${\mathrm{tr}}_{\mathbb{R}}h=2{\mathrm{tr}}_{\mathbb{C}}h$, proving $\beta h^{+}=0$.

For the anti-Hermitian summand its independent complex components are the symmetric $h_{\bar{i}\bar{j}}$. Set ${\mu }^i{}_{\bar{j}}=g^{i\bar{k}}{h}_{\bar{k}\bar{j}}$ and ${\chi }_{ab\bar{j}}={\mathrm{\Omega }}_{iab}{\mu }^i{}_{\bar{j}}$. This is an isomorphism from these symmetric components to primitive $(2,1)$-forms. Indeed in a unitary frame with ${\mathrm{\Omega }}_{123}=1$, the three components of $\mathrm{\Lambda }\chi$ are nonzero constant multiples of $h_{\bar{2}\bar{3}}-h_{\bar{3}\bar{2}}$ and its cyclic variants; thus primitivity is exactly the symmetry condition. The map is parallel. To verify its Laplacian identity explicitly, expansion of $\bar{\mathrm{\partial }}{\bar{\mathrm{\partial }}}^{\ast }+{\bar{\mathrm{\partial }}}^{\ast }\bar{\mathrm{\partial }}$ on a tangent-valued $(0,1)$-form gives $({\mathrm{\Delta }}_{\bar{\mathrm{\partial }}}\mu {)}^i{}_{\bar{j}}=-g^{a\bar{b}}{\mathrm{\nabla }}_a{\mathrm{\nabla }}_{\bar{b}}{\mu }^i{}_{\bar{j}}+g^{a\bar{b}}[{\mathrm{\nabla }}_a,{\mathrm{\nabla }}_{\bar{j}}]{\mu }^i{}_{\bar{b}}\mathrm{.}$ Multiplying by two turns the first term into the real rough Laplacian, since its contracted commutator is a Ricci contraction and vanishes. In the second term the covariant $\bar{b}$-index contribution is another Ricci contraction, while the contravariant $i$-index contributes $2g^{a\bar{b}}{R}^i{}_{ka\bar{j}}{\mu }^k{}_{\bar{b}}$. Lowering $i$ gives $-2R_{\bar{\mathrm{\ell }}k\bar{j}a}{h}^{ka}$, exactly the curvature term of ${\mathrm{\Delta }}_L{h}^{-}$. Parallelness of $\mathrm{\Omega }$ then identifies this with $2{\mathrm{\Delta }}_{\bar{\mathrm{\partial }}}\chi ={\mathrm{\Delta }}_d\chi$ by [F2]. Thus the anti-Hermitian kernel corresponds exactly to harmonic primitive $(2,1)$-forms. Such a harmonic form satisfies ${\bar{\mathrm{\partial }}}^{\ast }\chi =0$, or $g^{a\bar{b}}{\mathrm{\nabla }}_a{\mu }^i{}_{\bar{b}}=0$; symmetry implies ${\mathrm{\nabla }}^a{h}_{a\bar{j}}^{-}=0$. Since $h^{-}$ is trace free, $\beta h^{-}=0$.

Every harmonic $(2,1)$-form is primitive here: $\mathrm{\Lambda }$ commutes with ${\mathrm{\Delta }}_d$ by [F2], so $\mathrm{\Lambda }\chi$ is harmonic of type $(1,0)$, and this space is zero by [F1]. Conversely every harmonic real $(1,1)$-form and harmonic complex $(2,1)$-form therefore produce a real symmetric tensor in $\ker {\mathrm{\Delta }}_L$ by steps 2.1 and 3.1. Those steps also prove its Bianchi gauge. Hence $\ker {\mathrm{\Delta }}_L$ consists exactly of these representatives, and step 1.1 identifies it with the quotient in the Statement. Hodge cohomology [F2] gives real dimension $h^{1,1}$ for the real $(1,1)$-harmonic space and $2h^{2,1}$ for the underlying real space of ${\mathcal{H}}^{2,1}$. All assertions concern the fixed given metric and its linearized equation. No use of nonlinear Calabi existence, smooth Kuranishi integrability or a dimensional-reduction postulate has entered.