Positivity of the internal Hermitian form

Fix $z\mathrm{\ne }0$ and write $A_z=\alpha (z)$, $B_z=\alpha (jz)$. On the real line spanned by $z,jz$, its linear coordinate $(a,b)$ gives $\alpha (a,b)=a{A}_z+b{B}_z$ and $\beta (a,b)=a{B}_z^{†}H-b{A}_z^{†}H$. Expanding $\beta (a,b)\alpha (a,b)=0$ in $a,b$ gives $A_z^{†}H{B}_z=B_z^{†}H{A}_z=0$ and $A_z^{†}H{A}_z=B_z^{†}H{B}_z=:G_z$. The matrix $G_z$ is invertible: if $G_{z}v=0$ for $v\mathrm{\ne }0$, injectivity of $A_z$ gives $A_{z}v\mathrm{\ne }0$, while $\beta (a,b)A_{z}v=0$ for every $(a,b)$. Its class is a global section of the holomorphically trivial bundle $F_M{\mathrm{\mid }}_{\mathrm{\ell }}$ vanishing at $(1,0)$. In a trivialization its components lie in $H^0({\mathbb{P}}^1,\mathcal{O})=\mathbb{C}$ by [F4], so the section is identically zero. Exactness of the restricted monad then puts the constant $A_{z}v$ in the image of $H^0(U\otimes {\mathcal{O}}_{\mathrm{\ell }}(-1))=0$, a contradiction by [F2].

By step 1.1, the two images $\mathrm{im}{A}_z$ and $\mathrm{im}{B}_z$ are nondegenerate and $H$-orthogonal. Their orthogonal complement $E_z=\ker (A_z^{†}H)\cap \ker (B_z^{†}H)$ is the section space of $F_M{\mathrm{\mid }}_{\mathrm{\ell }}$: in bases given by $A_z,B_z$ and $G_z^{-1}$, the first two summands form copies of the exact Euler complex $\mathcal{O}(-1)\to {\mathcal{O}}^2\to \mathcal{O}(1)$, and the remaining summand is $E_z\otimes \mathcal{O}$. Thus $H{\mathrm{\mid }}_{E_z}$ is positive by [F1].

Use the constant Hermitian matrix $H$ on $S=M$ for the following positivity calculation. Put $z_1=x_1+i{x}_2$, $z_2=x_3+i{x}_4$ and $a(x)=\alpha (1,0,z_1,z_2)=a_0+z_1{M}_1+z_2{M}_2,b(x)=\alpha (0,1,-{\bar{z}}_2,{\bar{z}}_1)=b_0-{\bar{z}}_2{M}_1+{\bar{z}}_1{M}_2\mathrm{.}$ The two arguments are $z,jz$ in the stated coordinate convention. Step 1.1 gives $a^{†}Ha=b^{†}Hb=A(x),a^{†}Hb=0,$ with $A(x)$ invertible for every $x$. Apply the same identities on the infinity line with $z=(0,0,1,0)$ and $jz=(0,0,0,1)$ to obtain $M_1^{†}H{M}_1=M_2^{†}H{M}_2=C,M_1^{†}H{M}_2=0,$ where $C$ is invertible. Neither $A(x)$ nor $C$ is yet assumed positive. Let $K=A^{-1}$ and $Q=1-aK{a}^{†}H-bK{b}^{†}H\mathrm{.}$ The Gram identities imply $Q^2=Q$ and $Q^{†}H=HQ$; its range is precisely the real-line section complement $E_z$ in step 2.1. Thus $y^{†}HQy=(Qy{)}^{†}H(Qy)\ge 0$ for every $y$, by the actual induced section pairing there. This identification uses the lifted identity $\beta =\alpha (jz{)}^{†}H$ and the Euler-complex decomposition of step 2.1; it is not an extra assumption of positivity of $H$ on the coefficient images.

Put $P=a^{†}H{M}_1$ and $R=a^{†}H{M}_2$. Differentiating $a^{†}Hb=0$ in $x_1,x_3$ gives $a^{†}H{M}_2=-M_1^{†}Hb$ and $a^{†}H{M}_1=M_2^{†}Hb$, hence $b^{†}H{M}_1=-R^{†}$ and $b^{†}H{M}_2=P^{†}$. Since $a_{\mu }=(M_1,i{M}_1,M_2,i{M}_2)$, the derivatives of $A$ are $A_1=P+P^{†},A_2=iP-i{P}^{†},A_3=R+R^{†},A_4=iR-i{R}^{†}\mathrm{.}$ Pairing the real and imaginary derivatives, with the matrix order retained, gives ${\sum }_{\mu }{A}_{\mu }K{A}_{\mu }=2(PK{P}^{†}+P^{†}KP+RK{R}^{†}+R^{†}KR)={\sum }_{\mu }{a}_{\mu }^{†}H(1-Q)a_{\mu }\mathrm{.}$ The second equality follows by substituting $1-Q=aK{a}^{†}H+bK{b}^{†}H$ and the displayed formulas for $b^{†}H{M}_i$. As ${\sum }_{\mu }{a}_{\mu }^{†}H{a}_{\mu }=4C$, set $D={\sum }_{\mu }{a}_{\mu }^{†}HQ{a}_{\mu }={\sum }_{\mu }(Q{a}_{\mu }{)}^{†}H(Q{a}_{\mu })⪰0;{\sum }_{\mu }{A}_{\mu }K{A}_{\mu }=4C-D\mathrm{.}$ Also $\mathrm{\Delta }A=2{\sum }_{\mu }{a}_{\mu }^{†}H{a}_{\mu }=8C$. Differentiating $AK=1$ twice therefore yields $\mathrm{\Delta }K=2K\left({\sum }_{\mu }{A}_{\mu }K{A}_{\mu }\right)K-K(\mathrm{\Delta }A)K=-2KDK⪯0.$ This last inequality is congruence by the Hermitian matrix $K$, and does not assume its positivity. Thus for every constant $v\in U$, $f_v(x)=v^{†}K(x)v$ is a real smooth function with $\mathrm{\Delta }{f}_v\le 0$ on all of ${\mathbb{R}}^4$.

Expanding the Gram gives $A(x)=\mathrm{\mid }x{\mathrm{\mid }}^{2}C+O(\mathrm{\mid }x\mathrm{\mid })$ uniformly in direction. Since $C$ is invertible, writing $A=r^2(C+E_r)$ with $\mathrm{\parallel }{E}_r\mathrm{\parallel }=O(r^{-1})$ and using the convergent geometric series for $(1+C^{-1}{E}_r{)}^{-1}$ proves $\mathrm{\parallel }K(x)\mathrm{\parallel }\le 2\mathrm{\parallel }{C}^{-1}\mathrm{\parallel }{r}^{-2}$ for all sufficiently large $r$. In particular $f_v$ tends uniformly to zero at infinity. On a closed ball of radius $R$, the function $f_v-\delta \mathrm{\mid }x{\mathrm{\mid }}^2$ has strictly negative Laplacian for $\delta >0$ and cannot attain a minimum in the interior, where its real Hessian would be positive semidefinite. Hence $f_v(x)\ge {\min }_{\mathrm{\mid }y\mathrm{\mid }=R}{f}_v(y)-\delta R^2$. First let $\delta \downarrow 0$, then let $R\to \mathrm{\infty }$ with $x$ fixed. The uniform decay proves $f_v(x)\ge 0$. Thus $K$ is positive semidefinite; being Hermitian and invertible, it is positive definite, as is $A$. At any fixed $x$, the $H$-orthogonal direct decomposition $S=\mathrm{im}a\oplus \mathrm{im}b\oplus \mathrm{range}Q$ has positive Gram $A$ on its first two summands and the positive section form of step 2.1 on the third. Therefore $H$ is positive definite. If $U=0$, $Q=1$ and the section positivity already gives this conclusion directly; no inverse of a nonzero-dimensional Gram is needed. The preceding splitting lemma retains the positive constant flat summand separately. This proves the stated arbitrary-rank claim without assuming a sign for the infinity Gram matrix $C$.