Geometric parity consequences of an O3/O7 involution

At each point, the real determinant of a holomorphic differential is $\mathrm{\mid }{\det }_{\mathbb{C}}D\sigma {\mathrm{\mid }}^2>0$. Therefore $\sigma$ preserves the complex orientation, and the choice-free change-of-variables clause of [F2] gives ${\int }_X{\sigma }^{\ast }\eta ={\int }_X\eta$ for every smooth top-degree form $\eta$.

Fix $p\in \mathrm{Fix}(\sigma )$ and set $A=D{\sigma }_p$. Holomorphicity makes $A$ complex-linear and ${\sigma }^2=\mathrm{id}$ gives $A^2=I$, so $A$ is diagonalizable with eigenvalues $+1$ and $-1$. The pullback condition on the nonzero three-form gives ${\det }_{\mathbb{C}}A=-1$; therefore the number $m$ of $-1$ eigenvalues is odd and, in complex dimension three, is either $1$ or $3$. The $+1$ eigenspace has complex dimension $2$ or $0$.

For any smooth real five-form $\theta$, take the finite chart localization $({\chi }_i)$ of [F2] on compact $X$. Then $d\theta ={\sum }_{i}d({\chi }_i\theta )$ since ${\sum }_i{\chi }_i=1$. In each boundaryless chart, the coordinate form ${\chi }_i\theta$ extends smoothly by zero to a compactly supported five-form on ${\mathbb{R}}^6$; [F3] makes its derivative integral zero. Locality and linearity in [F2] give ${\int }_{X}d\theta =0$. For closed two-forms $u,v,w$, replacing $u$ by $u+d\lambda$ changes the triple integral by ${\int }_{X}d(\lambda \wedge v\wedge w)=0$. Applying this successively in each argument proves representative independence and trilinearity of the stated real pairing. Only a finite family of charts is used, so no choice axiom is added.

Pullback is an involution on each $H^{p,q}(X)$ because ${\sigma }^2=\mathrm{id}$, and holomorphicity preserves bidegree. For any vector $u$ in such a cohomology space, $u_{+}=(u+{\sigma }^{\ast }u)\mathrm{/}2$ and $u_{-}=(u-{\sigma }^{\ast }u)\mathrm{/}2$ satisfy $u=u_{+}+u_{-}$ and ${\sigma }^{\ast }{u}_{+}=u_{+}$, ${\sigma }^{\ast }{u}_{-}=-u_{-}$. If a vector belongs to both eigenspaces then it equals its own negative, hence is zero. This proves the direct-sum decomposition on cohomology.

For three eigenforms, ${\sigma }^{\ast }({\omega }_A\wedge {\omega }_B\wedge {\omega }_C)={ϵ}_A{ϵ}_B{ϵ}_C{\omega }_A\wedge {\omega }_B\wedge {\omega }_C$, where each $ϵ$ is $+1$ for an even index and $-1$ for an odd index. By step 1.1 the integral is unchanged by pullback. If an odd number of indices are odd, then ${ϵ}_A{ϵ}_B{ϵ}_C=-1$, so ${\kappa }_{ABC}=-{\kappa }_{ABC}$ and hence ${\kappa }_{ABC}=0$.

Put $T={\sigma }^{\ast }$ and write $(u,v)={\int }_X⟨u,v{⟩}_g{\mathrm{vol}}_g$ for the Hermitian $L^2$ product, linear in its first argument. The isometry and step 1.1 give $(Tu,Tv)=(u,v)$; also $T^2=1$ and $Td=dT$. For smooth forms $u,v$ of adjacent degrees, $(du,Tv)=(Tdu,v)=(dTu,v)=(Tu,d_g^{\ast }v)=(u,T{d}_g^{\ast }v)$. The defining identity of the formal adjoint also makes the left side $(u,d_g^{\ast }Tv)$. Thus $(u,d_g^{\ast }Tv-T{d}_g^{\ast }v)=0$ for all $u$, so $d_g^{\ast }T=T{d}_g^{\ast }$, by taking $u$ to be that smooth difference and using positive definiteness of the integral from [F2]. Hence $T{\mathrm{\Delta }}_g={\mathrm{\Delta }}_{g}T$. Holomorphicity preserves bidegree and commutes with $\bar{\mathrm{\partial }}$ (the $(0,1)$ part of $d$); applying the two projections of step 2.1 to harmonic forms proves their direct-sum decomposition.

In a holomorphic chart centered at $p$, write the local involution as $s(z)$ with derivative $A$. Define $w(z)=\frac{1}{2}(z+As(z))$. Then $Dw(0)=\frac{1}{2}(I+A^2)=I$. The map $w$ is holomorphic on an open neighborhood in ${\mathbb{C}}^3$, so [F1] gives a biholomorphic local coordinate map. Shrink the original chart to its intersection with its image under $s$ so that both $z$ and $s(z)$ remain in the coordinate domain. Then $w(s(z))=\frac{1}{2}(s(z)+Az)=Aw(z)$. Hence the fixed set in this chart is exactly the $+1$ eigenspace of $A$, a complex submanifold of the dimension found in step 1.2. Every fixed point has such a chart, so each fixed component has complex dimension $0$ or $2$.