Proof. We consider the \((0,1)\)-part of \(D_{h_0}\) with respect to the complex structure of \(X_n\), denoted by \(\pi''_\eta D_{h_0}\) and one may verify directly as operators that \[\begin{align} \label{eq95piCh} \pi''_\eta D_{h_0}\equiv \bar\partial-\sum_{i=1}^n\bar t^i\bar\eta_i\circ\bar\partial+\sum_{i=1}^n t^i\eta_i\circ D_{h_0}^{1,0}\pmod{t\bar t}. \end{align}\tag{4}\] Then \(\bar\partial_t-\pi''_\eta D_{h_0}\) is a section of \(\operatorname{End}{\mathcal{E}}\otimes_{\mathcal{C}^\infty(X_0)} \Omega^{0,1}(X_n/A_n)\). This proves our claim. ◻

Proof. By [20], the condition ?? is equivalent to the holomorphicity of \(({\mathcal{E}},\mathscr U^{-1} \circ \bar\partial_t \circ \mathscr U ,\mathscr U^{-1} \circ \theta_t \circ \mathscr U )\). Therefore \(({\mathcal{E}},\bar\partial_t,\theta_t)\) is also a holomorphic deformation because it differs with \(({\mathcal{E}},\mathscr U^{-1} \circ \bar\partial_t \circ \mathscr U ,\mathscr U^{-1} \circ \theta_t \circ \mathscr U )\) by a gauge equivalence. ◻

Proof. By Lemma 1, we have \[\begin{align} \bar\partial_t&\equiv\bar\partial-\sum_{i=1}^n\bar t^i\bar\eta_i\circ\bar\partial+\sum_{i=1}^n\bar t^i\psi_i\pmod{(t,\bar t^{n+1})};\\ \theta_t&\equiv\theta+\sum_{i=1}^n\bar t^i\varphi_i\pmod{(t,\bar t^{n+1})}. \end{align}\] By the definition of \(\mathscr U\) in 7 , we have 6 for \(k=n\). Modulo \((t)\) in these two identities and using 4 , we have \[\begin{align} \bar{\partial}_t \circ \mathscr{U}&=\mathscr U\circ (\bar\partial-\sum_{i=1}^n\bar t^i\bar\eta_i\circ\bar\partial)\pmod{(t,\bar t^{n+1})};\\ \theta_t\circ\mathscr U&=\mathscr U\circ\theta\pmod{(t,\bar t^{n+1})}. \end{align}\]After expanding the above expressions and comparing the coefficient of \(\bar t^k\), we have 8 . ◻

Proof. In this proof, we always modulo the ideal \((t\bar t)\). Firstly, we prove the above four identities. Let \[\begin{align} \pi_\eta':&B_n\otimes \mathcal{A}^1(\operatorname{End}{\mathcal{E}})\to B_n\otimes\Omega^{1,0}(X_n/A_n)\otimes \operatorname{End}{\mathcal{E}}; \\\pi_\eta'':&B_n\otimes \mathcal{A}^1(\operatorname{End}{\mathcal{E}})\to B_n\otimes\Omega^{0,1}(X_n/A_n)\otimes \operatorname{End}{\mathcal{E}} \end{align}\] be two projections according to types. Then we have \(\pi'_\eta\Psi_t=\theta_t\) and \(\pi''_\eta\Psi_t=\theta_t^{\star_{h_t}}\). Using 10 , we have \[\label{eq95theta95t} \begin{align} \theta_t=&g(t,\bar t)^{-1}(\theta-\eta(\theta)+\bar\eta(\theta^{\star_{h_0}}))g(t,\bar t)-\frac{1}{2}g(t,\bar t)^{-1}(D'-\eta\circ D'+\bar\eta\circ D'')^{\operatorname{End}}\Big(g(t,\bar t)\Big);\\ \theta_t^{\star_{h_t}}=&g(t,\bar t)^{-1}(\theta^{\star_{h_0}}+\eta(\theta)-\bar\eta(\theta^{\star_{h_0}}))g(t,\bar t)-\frac{1}{2}g(t,\bar t)^{-1}(D''+\eta\circ D'-\bar\eta\circ D'')^{\operatorname{End}}\Big(g(t,\bar t)\Big), \end{align}\tag{13}\] where \(D':A^0(X_0,\operatorname{End}{\mathcal{E}})\to A^{1,0}(X_0,\operatorname{End}{\mathcal{E}})\) and \(D'':A^0(X_0,\operatorname{End}{\mathcal{E}})\to A^{0,1}(X_0,\operatorname{End}{\mathcal{E}})\) with \(D=D'+D''\). Note that we have \(D'=D_{h_0}^{1,0}+\theta\) and \(D''=\bar\partial+\theta^{\star_{h_0}}\). By comparing the coefficient of \(t^i,\;i=0,1,2,\cdots,n\) in the above identity of \(\theta_t\), we have \[\begin{align} \theta+\sum_{i=1}^nt^i\alpha_i =&(\operatorname{id}+\sum_{j=1}^n t^j g_j)^{-1}(\theta-\sum_{l=1}^nt^l\eta_l(\theta))(\operatorname{id}+\sum_{k=1}^n t^k g_k)\\ &-\frac{1}{2}(\operatorname{id}+\sum_{j=1}^n t^j g_j)^{-1}(D_{h_0}^{1,0}+\theta-\sum_{l=1}^nt^l\eta_l\circ D_{h_0}^{1,0}-\sum_{l=1}^nt^l\eta_l(\theta))^{\operatorname{End}}\sum_{k=1}^n t^k g_k. \end{align}\] Substituting \((\operatorname{id}+\sum_{j=1}^n t^j g_j)^{-1}=\operatorname{id}-\sum_{j=1}^nt^j g_j+\sum_{j_1+j_2\leq n}t^{j_1+j_2}g_{j_1}g_{j_2}-\cdots\) into the above identity, we obtain the expression ?? . By a similar argument, we have ?? . Now we explain how to derive the rest two. By a direct check, we have \(\pi_\eta'' D=D''+\eta\circ D'-\bar\eta\circ D''.\) Thus \[\begin{align} \bar\partial_t=\pi_\eta'' D-\theta_t^{\star_{h_t}}. \end{align}\] By substituting 13 into the above identity and comparing the coefficient of \(t^i,\;i=1,2,\cdots,n\) in the above identity, we have \[\begin{align} \sum_{i=1}^nt^i\beta_i =&\sum_{l=1}^nt^l\eta_l(\theta)-(\operatorname{id}+\sum_{j=1}^n t^j g_j)^{-1}(\theta^{\star_{h_0}}+\sum_{l=1}^nt^l\eta_l(\theta))(\operatorname{id}+\sum_{k=1}^n t^k g_k)\\ &+\frac{1}{2}(\operatorname{id}+\sum_{j=1}^n t^j g_j)^{-1}(\bar\partial+\theta^{\star_{h_0}}+\sum_{l=1}^nt^l\eta_l\circ D_{h_0}^{1,0}+\sum_{l=1}^nt^l\eta_l(\theta))^{\operatorname{End}}\sum_{k=1}^n t^k g_k. \end{align}\] This proves the expression ?? . By a similar argument, we have ?? .

Substituting the above expressions into the integrable conditions 3 , we get several PDEs on \(g_i,\;i=1,2,\cdots,n\). By the nonabelian Hodge correspondence and the uniqueness of the harmonic metric, the above PDEs on \(g_i\) are all solvable and uniquely determine \(g_i,\;i=1,2,\cdots,n\). ◻

Proof. By definition, \((\mathcal{E},\bar\partial_t,\theta_t)\) is given by a map \[\sigma:\operatorname{Spec}\mathbb{C}[t,\bar t]/(t,\bar t)^{n+1} \longrightarrow M_{\mathrm{Dol}}(X/S)=:\mathcal{M}\] with \(\sigma(0)=(\mathcal{E},\bar\partial,\theta)\) on \(X_0\).

By the complexification argument in section 8.5, \(\sigma\) uniquely extends to a morphism \(g\) such that the following diagram commutes \[\begin{tikzcd} \operatorname{Spec}\mathbb{C}[t,\bar t]/(t,\bar t)^{n+1} \arrow[r,"\sigma"] \arrow[d,hook] & \mathcal{M} \arrow[d,hook]\\ \operatorname{Spec}(\mathbb{C}[t]/(t^{n+1})\otimes \mathbb{C}[\bar t]/(\bar t^{n+1}))\arrow[r, "g"] & M_{\mathop{\mathrm{Dol}}}((X\times\overline{X})/(S\times \overline{S})) \end{tikzcd}\]

Define the jet spaces of maps sending \(0\) to \(o:= (X_0,\mathcal{E},\bar\partial,\theta)\in \mathcal{M}\): \[\operatorname{Hom}\left( \operatorname{Spec}\mathbb{C}[t,\bar t]/(t,\bar t)^{k+1}, \mathcal{M} \right) := \mathbb{C}J_k\mathcal{M},\] and \[\operatorname{Hom}\left( \operatorname{Spec}\mathbb{C}[\bar t]/(\bar t^{k+1}), M_{\mathop{\mathrm{Dol}}}(\overline{X}/\overline{S}) \right) := \overline{ J_k\mathcal{M}} .\]

Then \(g\) induces a \(k\)-jet map \(\operatorname{Spec}(\mathbb{C}[\bar t]/(\bar t^{k+1}))\to M_{\mathop{\mathrm{Dol}}}(\overline{X}/\overline{S})\) by base changing via \[0\times\operatorname{Spec}(\mathbb{C}[\bar t]/(\bar t^{k+1}))\hookrightarrow 0\times \operatorname{Spec}(\mathbb{C}[\bar t]/(\bar t^{n+1}))\hookrightarrow\operatorname{Spec}(\mathbb{C}[t]/(t^{n+1})\otimes \mathbb{C}[\bar t]/(\bar t^{n+1})).\] We denote this \(k\)-jet map by \(p_k([\sigma])\in \overline{ J_k\mathcal{M}}\).

For the jet \([\sigma]\), we have the following commutative diagram: \[\begin{tikzcd} & & \mathbb{C}J_k\mathcal{M} \arrow[r, "\Pi_{k-1}^k"] \arrow[d, "p_k"'] & \mathbb{C}J_{k-1}\mathcal{M} \arrow[d, "p_{k-1}"] & \\ 0 \arrow[r] & T^{1,0}_{o}M_{\mathop{\mathrm{Dol}}}(\overline{X}/\overline{S}) \arrow[r] & \overline{ J_k\mathcal{M}} \arrow[r, "\pi_{k-1}^k"] & \overline{ J_{k-1}\mathcal{M}} \arrow[r] & 0 \end{tikzcd}\]

By modulo-\((t,\bar t^{k})\)-holomorphicity, \([\sigma]\) maps to zero under \[p_{k-1}\circ \Pi_{k-1}^k .\] Thus \(p_k([\sigma])\) is given by an element in \(T^{1,0}_{o}M_{\mathop{\mathrm{Dol}}}(\overline{X}/\overline{S})\). Let \(\mathbb{C}T_o\mathcal{M}=T^{1,0}_{o}\mathcal{M}\oplus T^{0,1}_{o}\mathcal{M}\) be the complexification of the real Zariski tangent space of a real analytic variety \(\mathcal{M}\) at \(o\). One can prove directly there is natural isomorphism \(T^{1,0}_{o}M_{\mathop{\mathrm{Dol}}}(\overline{X}/\overline{S})\cong T^{0,1}_o M_{\mathop{\mathrm{Dol}}}(X/S)\) by using the argument in Proposition 23. Thus we obtain \(p_k([\sigma])\) is given by an element in \(T^{0,1}_o \mathcal{M}\).

Now we prove that \(p_k([\sigma])\) is given by an element in \(T^{0,1}_oM_{\mathrm{Dol}}(X_0)\). By definition, the composition \[\operatorname{Spec}\mathbb{C}[t,\bar t]/(t,\bar t)^{n+1} \xrightarrow{\;\sigma\;} \mathcal{M} \xrightarrow{\;\pi_{\mathrm{Dol}}\;} S\] is independent of \(\bar t\) and is equal to the holomorphic \(n\)-germ \(c\) of \(S\) through \(0\). Therefore \[p_k\bigl([\pi_{\mathrm{Dol}}\circ \sigma]\bigr)=0 \qquad \text{in } \quad\overline{ J_k S}\] by holomorphicity, and we obtain \[p_k\bigl([\pi_{\mathrm{Dol}}\circ \sigma]\bigr) = (\pi_{\mathrm{Dol},*})\bigl(p_k([\sigma])\bigr) =0 .\] Therefore \(p_k([\sigma])\in \ker(\pi_{\mathrm{Dol},*}),\) and hence \(p_k([\sigma])\in T^{0,1}_oM_{\mathrm{Dol}}(X_0),\) which gives \[\operatorname{ob}_k:=p_k([\sigma])\in T^{0,1}_oM_{\mathrm{Dol}}(X_0)\cong {\mathbb{H}^1(\overline{X_0}, (\operatorname{End}\overline{E}^\vee,\operatorname{ad}(\theta^{\star_{h_0}})))}.\qedhere\] ◻

Proof Lemma 3. Assuming 15 , we have \[\begin{align} \sqrt{-1}\int_{X_0}\operatorname{tr}(g^{\star_{h_0}}D_{h_0}^{1,0}\bar\partial g)+\sqrt{-1}\int_{X_0}\operatorname{tr}(g^{\star_{h_0}}[\theta^{\star_{h_0}},[\theta,g]])=0. \end{align}\] By the Kähler identity (see [9] and [22]), we have \[\sqrt{-1}\int_{X_0}\operatorname{tr}(g^{\star_{h_0}}D_{h_0}^{1,0}\bar\partial g)=-\sqrt{-1}\int_{X_0}\operatorname{tr}((\bar\partial g)^{\star_{h_0}}\bar\partial g)\leq 0.\] One may verify directly that \[\sqrt{-1}\int_{X_0}\operatorname{tr}(g^{\star_{h_0}}[\theta^{\star_{h_0}},[\theta,g]])=-\sqrt{-1}\int_{X_0}\operatorname{tr}([\theta,g]\wedge [\theta,g]^{\star_{h_0}})\leq 0.\] Thus we have \(\bar\partial g=0\) and \([\theta,g]=0.\) This means that \(g\in\mathbb{H}^0(X_0,(\operatorname{End}E,\operatorname{ad}(\theta)))\) and by the stability \(g=c\cdot\operatorname{id}\) for some \(c\in\mathbb{C}\).

If \(g\) satisfies 16 , one can prove \(g\in\mathbb{H}^0(X_0,(\operatorname{End}E,\operatorname{ad}(\theta)))\) similarly. ◻

Proof. Applying \(D_{h_0}^{1,0}(-)\) to the first equation and \([\theta^{\star_{h_0}},-]\) to the second and summing them, we have \[\begin{align} D_{h_0}^{1,0}\bar\partial g+[\theta^{\star_{h_0}},[\theta,g]]=0, \end{align}\] which implies \(g=c_1\cdot\operatorname{id}\) by the “harmonicity” in Lemma 3. Applying \([\theta,-]\) to the first equation and \(\bar\partial(-)\) to the second and summing them, we have \[\begin{align} \bar\partial D_{h_0}^{1,0} f+[\theta,[\theta^{\star_{h_0}},f]]=0, \end{align}\] which implies \(f=c_2\cdot\operatorname{id}\) by the “harmonicity” in Lemma 3. ◻

Proof. Step1: modulo-\((t,\bar t^2)\)-holomorphicity implies the vanishing of the class \([(\bar\eta_1(\theta^{\star_{h_0}}),0)]\). By equations 8 for gauge transformation, the equations ?? and ?? of \(\varphi_1\) and \(\psi_1\), we have \[\label{g1} \begin{align} \frac{1}{2}\bar\partial g^{\star_{h_0}}_1-\frac{1}{2}[\theta^{\star_{h_0}},g^{\star_{h_0}}_1]+\bar\partial u_1&=0;\\ \bar \eta_1(\theta^{\star_{h_0}})+\frac{1}{2}[\theta,g^{\star_{h_0}}_1]-\frac{1}{2}D_{h_0}^{1,0} g^{\star_{h_0}}_1+[\theta,u_1]&=0. \end{align}\tag{17}\] Hence \[\begin{align} D_{h_0}^{1,0}(\bar\partial(\frac{1}{2}g^{\star_{h_0}}_1+u_1)-\frac{1}{2}[\theta^{\star_{h_0}},g^{\star_{h_0}}_1])=0;\qquad[\theta^{\star_{h_0}},[\theta,\frac{1}{2}g^{\star_{h_0}}_1+u_1]+\bar \eta_1(\theta^{\star_{h_0}})-\frac{1}{2}D_{h_0}^{1,0} g^{\star_{h_0}}_1]=0. \end{align}\] Note that \([\theta^{\star_{h_0}},\bar\eta_1(\theta^{\star_{h_0}})]=0\) and \(D_{h_0}^{1,0}([\theta^{\star_{h_0}},g^{\star_{h_0}}_1])=-[\theta^{\star_{h_0}},D_{h_0}^{1,0}g^{\star_{h_0}}_1]\), we have \[\begin{align} &D_{h_0}^{1,0}(\bar\partial(\frac{1}{2}g^{\star_{h_0}}_1+u_1)-\frac{1}{2}[\theta^{\star_{h_0}},g^{\star_{h_0}}_1])+ [\theta^{\star_{h_0}},[\theta,\frac{1}{2}g^{\star_{h_0}}+u_1]+\bar \eta_1(\theta^{\star_{h_0}})-\frac{1}{2}D_{h_0}^{1,0} g^{\star_{h_0}}_1]\\ =&D_{h_0}^{1,0}\bar\partial(\frac{1}{2}g^{\star_{h_0}}_1+u_1)+ [\theta^{\star_{h_0}},[\theta,\frac{1}{2}g^{\star_{h_0}}+u_1]]=0. \end{align}\] By the “harmonicity” in Lemma 3, we have \(\bar\partial(\frac{1}{2}g^{\star_{h_0}}_1+u_1)=0\) and \([\theta,\frac{1}{2}g^{\star_{h_0}}_1 +u_1]=0\). This implies \(\frac{1}{2}g^{\star_{h_0}}_1+u_1\in \mathbb{H}^0(X_0,(\operatorname{End}E,\operatorname{ad}(\theta)))\). The stability gives \(u_1=-\frac{1}{2}g^{\star_{h_0}}_1\) (up to adding a term \(c\cdot\operatorname{id}\) with \(c\in\mathbb{C}\), which we may ignore since it does not affect 17 ). Thus 17 reduces to \[\begin{align} \label{eq95ob1exact}0=\frac{1}{2}[\theta^{\star_{h_0}},g^{\star_{h_0}}_1];\qquad\; \bar \eta_1(\theta^{\star_{h_0}})=\frac{1}{2}D_{h_0}^{1,0} g^{\star_{h_0}}_1, \end{align}\tag{18}\] i.e. \(\operatorname{ob}_1\in{\mathbb{H}^1(\overline{X_0}, (\operatorname{End}\overline{E}^\vee,\operatorname{ad}(\theta^{\star_{h_0}})))}\) vanishes.

Step2: the vanishing of the class \([(\bar\eta_1(\theta^{\star_{h_0}}),0)]\) implies modulo-\((t,\bar t^2)\)-holomorphicity. We aim to prove the solvability of 17 on \(u_1\). Since \([(\bar\eta_1(\theta^{\star_{h_0}}),0)]\) vanishes, there exists \(f_1\in \mathcal{A}^0(\operatorname{End}{\mathcal{E}})\) such that \[\begin{align} \label{eq95ob1vanish}0=[\theta^{\star_{h_0}},f_1];\qquad\; \bar \eta_1(\theta^{\star_{h_0}})= D_{h_0}^{1,0} f_1. \end{align}\tag{19}\] By [15], the condition \(\bar\partial_t\theta_t\equiv 0\pmod{(t,\bar t)^2}\) implies that \[\begin{align} \bar\partial D_{h_0}^{1,0}g^{\star_{h_0}}_1=2\bar\partial(\bar\eta(\theta^{\star_{h_0}}))-[\theta,[\theta^{\star_{h_0}},g^{\star_{h_0}}_1]]. \end{align}\] This together with the assumption 19 gives \[\begin{align} \bar\partial D_{h_0}^{1,0}(g^{\star_{h_0}}_1-2f_1)+[\theta,[\theta^{\star_{h_0}},g^{\star_{h_0}}_1-2f_1]]=0, \end{align}\] which implies \(g^{\star_{h_0}}_1=2f_1+c\cdot\operatorname{id}\) for some \(c\in\mathbb{C}\) by the “harmonicity” in Lemma 3. Substituting this and 19 into 17 , we have \[\begin{align} \bar\partial(\frac{1}{2}g^{\star_{h_0}}_1+u_1)=0;\qquad [\theta,\frac{1}{2}g^{\star_{h_0}}_1+u_1]=0. \end{align}\] Thus \(u_1=-\frac{1}{2}g^{\star_{h_0}}_1\) is the solution of 17 . ◻

Proof. Using 18 and the fact \(\bar\eta_1(\theta^{\star_{h_0}})\in \mathcal{A}^0((\operatorname{End}{\mathcal{E}})^{2,-2})\), we have \[\begin{align} [\theta^{\star_{h_0}},(g^{\star_{h_0}}_1)^{l,-l}]=0;\; D_{h_0}^{1,0} (g^{\star_{h_0}}_1)^{l,-l}=0\;\text{ for }l\ne1. \end{align}\] Thus \(g^{\star_{h_0}}_1=(g^{\star_{h_0}}_1)^{144-1}+c\cdot\operatorname{id}\) for some \(c\in\mathbb{C}\). Since \(c\cdot\operatorname{id}\) in 9 can be eliminated by a gauge transformation, we have the first claim in 20 . The second claim in 20 is derived in the proof of Proposition 10. ◻

Proof. By [15], for any \(v\in T^{1,0}_0S\), if \(v\notin \ker(\theta_*\circ\tau_0)\), then the isomonodromic deformed Higgs bundles \(\sigma_{\mathop{\mathrm{Dol}}}\) is not graded (in fact not nilpotent) along \(v\). Thus \(v\notin T^{\mathrm{Zar}}_0\mathcal{NL}\).

If \(v\in \ker(\theta_*\circ\tau_0)\), we prove the isomonodromic deformed Higgs bundles \(\sigma_{\mathop{\mathrm{Dol}}}\) coincides with a holomorphic family of graded Higgs bundles up to first order along \(v\). Let \([\eta_1]:=\tau_0(v)\in H^1(T_{X_0})\). The condition \(\theta_*([\eta_1])=0\) is equivalent to the existence of \(f_1\in \mathcal{A}^0(\operatorname{End}{\mathcal{E}})\) such that 19 holds by taking \({\star_{h_0}}\). Thus by the proof of Proposition 10 Step2, we know that \(g^{\star_{h_0}}_1=2f_1+c\cdot \operatorname{id}\), for some \(c\in\mathbb{C}\). Substituting this into the deformation terms Proposition 6, we have \[\begin{align} \varphi_1=\frac{1}{2}[\theta,g^{\star_{h_0}}_1],&\quad \psi_1=\frac{1}{2}\bar\partial g^{\star_{h_0}}_1,\\ \alpha_1=-\eta_1(\theta)-\frac{1}{2}D_{h_0}^{1,0}g_1,&\quad \beta_1=-\frac{1}{2}[\theta^{\star_{h_0}},g_1]+\bar\partial g_1. \end{align}\] By taking \(\mathscr U:=\operatorname{id}-\frac{t}{2}g_1-\frac{\bar t}{2}g^{\star_{h_0}}_1\), we have on \(X_1\) \[\begin{align} \label{eq95firstdeforma}\mathscr U^{-1}\circ\bar\partial_t\circ\mathscr U=\pi''_{\eta}D_{h_0}^{1,0}+t\cdot\frac{1}{2}[\theta^{\star_{h_0}},g_1],\qquad \mathscr U^{-1}\circ\theta_t\circ\mathscr U=\theta-t\eta_1(\theta)-t\cdot\frac{1}{2}D_{h_0}^{1,0}g_1. \end{align}\tag{21}\]

Note that the condition \(\theta_*([\eta_1])=0\) is equivalent to the vanishing of the first-order obstruction class \(\operatorname{ob}_1\) stated in Proposition 10, which in turn implies \(g_1\in \mathcal{A}^0((\operatorname{End}{\mathcal{E}})^{-1,1})\) as established in Corollary 2. It follows immediately that \[\begin{align} [\theta^{\star_{h_0}},g_1]\in\mathcal{A}^{1}((\operatorname{End}{\mathcal{E}})^{0,0}) \qquad \text{and} \qquad \eta_1(\theta)+\frac{1}{2}D_{h_0}^{1,0}g_1\in \mathcal{A}^{1}((\operatorname{End}{\mathcal{E}})^{-1,1}). \end{align}\] This and 21 imply that the isomonodromically deformed Higgs bundle preserves a graded structure up to first order. ◻

Proof. The case \(k=1\) has been proved in 20 . We inductively assume that \[\begin{align} \label{u-wt1}g^{\star_{h_0}}_m,u_m\in\bigoplus_{l>0} \mathcal{A}^0((\operatorname{End}{\mathcal{E}})^{l,-l})\qquad\text{and }\qquad u_m^{144-1}=-\frac{1}{2}g^{\star_{h_0}}_m \end{align}\tag{22}\] for \(m=1,2,\cdots,k-1\) and prove this for \(g^{\star_{h_0}}_k,u_k\).

By induction and the expressions ?? , ?? , we have for \(1\leq m\leq k-1\) \[\begin{align} \label{eq95wtphipsi}\varphi_m\in \bigoplus_{l\geq1} \mathcal{A}^0((\operatorname{End}{\mathcal{E}})^{l,-l})\quad\text{ and }\quad\psi_m\in \bigoplus_{l\geq2} \mathcal{A}^0((\operatorname{End}{\mathcal{E}})^{l,-l}). \end{align}\tag{23}\] By the equations 8 for gauge transformation with \(m=k\), we have that for any integer \(l\leq 1\), \[\begin{align} & 0=(\varphi_k+\sum_{j=1}^{k-1} \varphi_{j}u_{k-j}+[\theta,u_k])^{l,-l} =(\frac{1}{2}[\theta,g^{\star_{h_0}}_k]- \frac{1}{2}D_{h_0}^{1,0}g^{\star_{h_0}}_k+[\theta,u_k])^{l,-l};\\ & 0=(\psi_{k}+\sum_{j=1}^{k-1}\psi_{j}u_{k-j}+\bar\partial u_{k}-\sum_{j=1}^{k-1}\bar\eta_{j}(\bar\partial u_{k-j}))^{l+1,-l-1} = (\frac{1}{2}\bar\partial g^{\star_{h_0}}_k-\frac{1}{2}[\theta^{\star_{h_0}},g^{\star_{h_0}}_k]+\bar\partial u_k)^{l+1,-l-1}. \end{align}\] By the “harmonicity” in Corollary 1, this implies for any integer \(l\leq0\) \[\begin{align} (g^{\star_{h_0}}_k)^{l,-l}=0\quad \text{and}&\quad u_k^{l,-l}=0 ;\\ u_k^{144-1}=-\frac{1}{2}(&g^{\star_{h_0}}_k)^{144-1}, \end{align}\] (note that any constant multiple of \(\operatorname{id}\) in \(g^{\star_{h_0}}_k\) can be eliminated by a gauge transformation). Thus 23 also holds for \(m=k\). ◻

Proof of Proposition 13. To keep the flow of the proof clear, we defer the verification of some technical, computational identities to the end of this section, where the reader may consult them.

When \(k=1\), our claim holds by 18 in the proof of Proposition 10. We inductively assume (i)(ii) hold for \(1,2,\cdots,k-1\) and verify this for \(k\geq 2\). By induction, we have the following equations coming from modulo-\((t,\bar t^{m+1})\)-holomorphic, where \(m=1,2,\cdots,k-1\):

\[\begin{align} \frac{1}{2}(D_{h_0}^{1,0}g^{\star_{h_0}}_m)^{2,-2}=&\bar\eta_{m}(\theta^{\star_{h_0}})-\frac{1}{2}\sum_{i=1}^{m-1}\bar\eta_i(\bar\partial g^{\star_{h_0}}_{m-i})^{2,-2}+\frac{1}{4}\sum_{i=1}^{m-1} \frac{i}{m}\cdot[[\theta,g^{\star_{h_0}}_{m-i}],g^{\star_{h_0}}_i]^{3,-3}; \tag{24}\\ \frac{1}{2}[\theta^{\star_{h_0}},g^{\star_{h_0}}_m]^{3,-3}=&\frac{1}{4}\sum_{i=1}^{m-1} \frac{i}{m}\cdot[\bar\partial g^{\star_{h_0}}_{m-i},g^{\star_{h_0}}_i]^{3,-3}. \tag{25} \end{align}\]

Now we prove the following identity:

\[\begin{align} \label{ab40141} &0=[\theta^{\star_{h_0}},\operatorname{ob}_{k,1}^{2,-2}] +D_{h_0}^{1,0}\operatorname{ob}_{k,2}^{3,-3} = \sum_{|I_2| = k}[\theta^{\star_{h_0}},\Big(-\frac{1}{2}\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_2}) + \frac{1}{4}\cdot\frac{i_2}{i_2+i_1} \cdot[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_2}]\Big)^{2,-2}] \notag &\\ & \makebox[5cm]{}+\sum_{ |I_2|= k}D_{h_0}^{1,0}(\frac{1}{4} \cdot\frac{i_2}{i_2+i_1}\cdot[\bar\partial g^{\star_{h_0}}_{i_1},g^{\star_{h_0}}_{i_2}])^{3,-3}.& \end{align}\tag{26}\] Taking the following two pieces in 26 \[\begin{align} \boldsymbol{Eq1} & :=\sum\limits_{|I_2|=k}[\theta^{\star_{h_0}},-\frac{1}{2}\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_2})]^{4,-4} \\ \boldsymbol{Eq2} & := \sum\limits_{|I_2| = k}\big([\theta^{\star_{h_0}},\frac{1}{4}\cdot\frac{i_2}{i_2+i_1}\cdot[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_2}]]+D_{h_0}^{1,0}(\frac{1}{4} \cdot\frac{i_2}{i_2+i_1}\cdot[\bar\partial g^{\star_{h_0}}_{i_1},g^{\star_{h_0}}_{i_2}])\big)^{4,-4}.\\ \end{align}\] We reduce 26 to prove \(\boldsymbol{Eq1}+\boldsymbol{Eq2}=0\). Firstly, \[\label{ab401411} \begin{align} \boldsymbol{Eq1} = & \frac{1}{2}\sum_{|I_2|= k}[\bar\eta_{i_1}(\theta^{\star_{h_0}}),\bar\partial g^{\star_{h_0}}_{i_2}]^{4,-4}\\ \overset{\eqref{vanisheq951}}{=} & \frac{1}{4}\sum_{|I_3|= k}[\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_3})-\frac{1}{2}\frac{i_3}{i_1+i_3}[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],\bar\partial g^{\star_{h_0}}_{i_2}]^{4,-4}+\frac{1}{4}\sum_{|I_2| = k}[D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1},\bar\partial g^{\star_{h_0}}_{i_2}]^{4,-4}\\ = & -\frac{1}{8}\sum_{|I_3|=k} \frac{i_3}{i_1+i_3}[[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],\bar\partial g^{\star_{h_0}}_{i_2}]^{4,-4}+\frac{1}{4}\sum_{|I_2|=k}[D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1},\bar\partial g^{\star_{h_0}}_{i_2}]^{4,-4}. \end{align}\tag{27}\] The third equality follows from the fact

\[\begin{align} \sum_{|I_3|=k}[\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_3}),\bar\partial g^{\star_{h_0}}_{i_2}]^{4,-4} & =\frac{1}{2}\sum_{|I_3|=k}\Big([\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_3}),\bar\partial g^{\star_{h_0}}_{i_2}]+[\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_2}),\bar\partial g^{\star_{h_0}}_{i_3}]\Big)^{4,-4}\\ & \overset{\eqref{eq95C2}}= \frac{1}{2}\sum_{|I_3|=k}\Big([\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_3}),\bar\partial g^{\star_{h_0}}_{i_2}]-[\bar\partial g^{\star_{h_0}}_{i_2},\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_3})]\Big)^{4,-4}\\ & =0. \end{align}\]

By taking \(g = g^{\star_{h_0}}_{i_1}\) and \(h = g^{\star_{h_0}}_{i_2}\) in the identity 33 , we have \[\begin{align} &[\theta^{\star_{h_0}}, [[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_2}]] +D_{h_0}^{1,0}([\bar\partial g^{\star_{h_0}}_{i_1},g^{\star_{h_0}}_{i_2}])\\ &= -[[\theta,[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_1}]],g^{\star_{h_0}}_{i_2}] -[[\theta,g^{\star_{h_0}}_{i_1}],[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_2}]] -\bar\partial([D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1},g^{\star_{h_0}}_{i_2}]) -[\bar\partial g^{\star_{h_0}}_{i_1},D_{h_0}^{1,0}g^{\star_{h_0}}_{i_2}] -[\bar\partial g^{\star_{h_0}}_{i_2},D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1}] . \end{align}\] substituting this into Eq2, we obtain \[\begin{align} \label{ab401412} \boldsymbol{Eq2} = & -\frac{1}{4} \sum_{|I_2| = k} \frac{i_2}{i_2+i_1} \cdot \Big( [[\theta,[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_1}]],g^{\star_{h_0}}_{i_2}] +[[\theta,g^{\star_{h_0}}_{i_1}],[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_2}]] +\bar\partial([D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1},g^{\star_{h_0}}_{i_2}]) \\ & \makebox[6.3cm]{} +[\bar\partial g^{\star_{h_0}}_{i_1},D_{h_0}^{1,0}g^{\star_{h_0}}_{i_2}] +[\bar\partial g^{\star_{h_0}}_{i_2},D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1}] \Big)^{4,-4}\\ = & -\frac{1}{4}\sum_{|I_2| = k} \frac{i_2}{i_2+i_1} \cdot \Big( [[\theta,[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_1}]],g^{\star_{h_0}}_{i_2}] + [[\theta,g^{\star_{h_0}}_{i_1}],[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_2}]] +\bar\partial([D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1},g^{\star_{h_0}}_{i_2}] \Big)^{4,-4} \\ & \makebox[6cm]{} -\frac{1}{4} \sum_{|I_2|= k} [\bar\partial g^{\star_{h_0}}_{i_1},D_{h_0}^{1,0}g^{\star_{h_0}}_{i_2}]^{4,-4}. \end{align}\tag{28}\] The second equality follows from the equality: \[\begin{align} \sum\limits_{|I_2|= k} \frac{i_2}{i_2+i_1}\cdot\big([\bar\partial g^{\star_{h_0}}_{i_1},D_{h_0}^{1,0}g^{\star_{h_0}}_{i_2}]+[\bar\partial g^{\star_{h_0}}_{i_2},D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1}]\big)^{4,-4} = &\sum_{|I_2| = k} (\frac{i_2}{i_2+i_1}+\frac{i_1}{i_1+i_2})[\bar\partial g^{\star_{h_0}}_{i_1},D_{h_0}^{1,0}g^{\star_{h_0}}_{i_2}]^{4,-4} \\ = & \sum_{|I_2| = k} [\bar\partial g^{\star_{h_0}}_{i_1},D_{h_0}^{1,0}g^{\star_{h_0}}_{i_2}]^{4,-4}. \end{align}\]

Now, we consider third piece in \(\boldsymbol{Eq2}\): \[\begin{align} &\sum_{|I_2| = k}\frac{i_2}{i_2+i_1}\cdot\bar\partial([D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1},g^{\star_{h_0}}_{i_2}])^{4,-4} \\ & \overset{\eqref{vanisheq951}}= \sum_{|I_2| = k} \frac{i_2}{i_2+i_1} \bar\partial([2\bar\eta_{i_1}(\theta^{\star_{h_0}}),g^{\star_{h_0}}_{i_2}])^{2,-2}+\sum_{|I_3| = k}\big\{\frac{i_2}{i_1+i_2+i_3}\bar\partial([-\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_3})+\frac{1}{2}\frac{i_3}{i_1+i_3}[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],g^{\star_{h_0}}_{i_2}])\big\}^{4,-4}\\ & \overset{\eqref{vanisheq952}}= \sum_{|I_3| = k} \Big\{\frac{i_2+i_3}{i_1+i_2+i_3}\frac{i_2}{i_2+i_3} \bar\partial([\bar\eta_{i_1}(\bar\partial g_{i_3}),g^{\star_{h_0}}_{i_2}])-\frac{i_2}{i_1+i_2+i_3}\bar\partial([\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_3})-\frac{1}{2}\frac{i_3}{i_1+i_3}[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],g^{\star_{h_0}}_{i_2}])\Big\}^{4,-4}\\ & \overset{\;\;\;\;}= \frac{1}{2}\sum_{|I_3| = k}\frac{i_2}{i_1+i_2+i_3}\frac{i_3}{i_1+i_3}\bar\partial([[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],g^{\star_{h_0}}_{i_2}])^{4,-4}. \end{align}\]

Substituting this into 28 and then substituting 27 and 28 into 26 , we have \[\begin{align} &[\theta^{\star_{h_0}},\operatorname{ob}_{k,1}^{2,-2}]+D_{h_0}^{1,0}\operatorname{ob}_{k,2}^{3,-3} \overset{\eqref{ab40141}}= \boldsymbol{Eq1} + \boldsymbol{Eq2}\\ &\overset{\eqref{ab401411}\eqref{ab401412}}= -\frac{1}{8}\sum_{|I_2|=k}2\cdot\frac{i_2}{i_2+i_1}\cdot\big\{[[\theta,[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_1}]],g^{\star_{h_0}}_{i_2}]+[[\theta,g^{\star_{h_0}}_{i_1}],[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_2}]]\big\}^{4,-4}\\ &\makebox[1cm]{}-\frac{1}{8}\sum_{|I_3|=k}\big\{\frac{i_2}{i_1+i_2+i_3}\frac{i_3}{i_1+i_3}\bar\partial([[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],g^{\star_{h_0}}_{i_2}])+\frac{i_3}{i_1+i_3}[[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],\bar\partial g^{\star_{h_0}}_{i_2}]\big\}^{4,-4}. \end{align}\] By 25 , all the above terms can be reduced to terms expressed by \(g^{\star_{h_0}},\bar\partial g^{\star_{h_0}}\) and \(\theta\), so we should be able to directly compute that the above expression equals 0. We place this complicated computation in Lemma 6 at the end of this section. Thus we obtain 26 for \([(\operatorname{ob}_{k,1}^{2,-2},\operatorname{ob}_{k,2}^{3,-3})]\). This completes the proof of (i).

Now we come to (ii). Since the isomonodromic deformation is further modulo-\((t,\bar t^{k+1})\)-holomorphic, then by equations 8 for gauge transformation we have \[\begin{align} \label{holwt2}[\theta^{\star_{h_0}},(\varphi_{k}+\sum_{j=1}^{k-1}\varphi_{j}u_{k-j}+[\theta,u_{k}])^{2,-2}]+D_{h_0}^{1,0}(\psi_{k}+\sum_{j=1}^{k-1}\psi_{j}u_{k-j}+\bar\partial u_{k})^{3,-3}=0. \end{align}\tag{29}\] Now we prove ?? . By Corollary 2, we may inductively assume for \(m=1,\cdots,k-1\): \[\begin{align} \label{u-wt2}u_{m}^{2,-2}=-\frac{1}{2}(g^{\star_{h_0}}_{m})^{2,-2}+\sum_{j=1}^{m-1}\frac{1}{4}(1+\frac{m-j}{m})\cdot(g^{\star_{h_0}}_jg^{\star_{h_0}}_{m-j})^{2,-2}. \end{align}\tag{30}\] And prove this for \(m=k\) and then we use it to prove ?? .

By explicit formulas for \(\varphi_i\), \(\psi_i\) in ?? , ?? and positivity of weights of \(g^{\star_{h_0}}_{i}\) in Proposition 12, we have for any \(i=1,2,\cdots,k-1\) \[\begin{align} &(\varphi_i)^{144-1}=\frac{1}{2}[\theta,g^{\star_{h_0}}_i]^{144-1},\quad (\psi_i)^{2,-2}=\frac{1}{2}(\bar\partial g^{\star_{h_0}}_i)^{2,-2}+\text{a section of }\mathcal{A}^{1,0}(\operatorname{End}{\mathcal{E}}),&\notag\\&(\varphi_k)^{2,-2}=\bar\eta_k(\theta^{\star_{h_0}})+\frac{1}{2}[\theta,g^{\star_{h_0}}_k]^{2,-2}-\frac{1}{2}(D_{h_0}^{1,0}g^{\star_{h_0}}_k)^{2,-2}-\frac{1}{2}\sum_{|I_2|=k}(g^{\star_{h_0}}_{i_1}[\theta,g^{\star_{h_0}}_{i_2}])^{2,-2}-\frac{1}{2}\sum_{|I_2|=k}\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_2}])^{2,-2},&\notag\\ &(\psi_k)^{3,-3}=-\frac{1}{2}\big\{[\theta^{\star_{h_0}},g^{\star_{h_0}}_k]^{3,-3}-(\bar\partial g^{\star_{h_0}}_k)^{3,-3}+\makebox[-.2cm]{}\sum_{|I_2|=k}(g^{\star_{h_0}}_{i_1}\bar\partial g^{\star_{h_0}}_{i_2})^{3,-3}\big\}+\text{a section of }\mathcal{A}^{1,0}(\operatorname{End}{\mathcal{E}}).&\label{eq95varphipsi95low95wt} \end{align}\tag{31}\] Let \(\square(-):=[\theta^{\star_{h_0}},[\theta,-]]+D_{h_0}^{1,0}\bar\partial(-)\). By 31 and 22 , 23 , the \(\mathcal{A}^{1,1}(\operatorname{End}{\mathcal{E}})\)-part of the equation 29 can be rewritten as \[\begin{align} 0=&\square\Big(u_{k}+\frac{1}{2}g^{\star_{h_0}}_{k}\Big)^{2,-2}-\sum_{|I_2|=k}[\theta^{\star_{h_0}},\frac{1}{2}\bar\eta_{i_1}(\bar\partial g^{\star_{h_0}}_{i_2})^{2,-2}]-\frac{1}{2}\sum_{|I_2|=k}[\theta^{\star_{h_0}},\big(g^{\star_{h_0}}_{i_1}[\theta,g^{\star_{h_0}}_{i_2}]\big)^{2,-2}]\\ &-\frac{1}{4}\sum_{|I_2|=k}[\theta^{\star_{h_0}},\big([\theta,g^{\star_{h_0}}_{i_1}]g^{\star_{h_0}}_{i_2}\big)^{2,-2}]-\frac{1}{2}\sum_{|I_2|=k}D_{h_0}^{1,0}(g^{\star_{h_0}}_{i_1}\bar\partial g^{\star_{h_0}}_{i_2})^{3,-3}-\frac{1}{4}\sum_{|I_2|=k}D_{h_0}^{1,0}((\bar\partial g^{\star_{h_0}}_{i_1})g^{\star_{h_0}}_{i_2})^{3,-3}. \end{align}\] Combining the above identity with 26 , we have \[\begin{align} \square\Big(u_{k}+\frac{1}{2}g^{\star_{h_0}}_{k}\Big)^{2,-2}=&\frac{1}{4}\sum_{|I_2|=k}(1+\frac{i_2}{i_2+i_1})\Big\{[\theta^{\star_{h_0}},[\theta,(g^{\star_{h_0}}_{i_1}g^{\star_{h_0}}_{i_2})^{2,-2}]]+D_{h_0}^{1,0}\bar\partial(g^{\star_{h_0}}_{i_1}g^{\star_{h_0}}_{i_2})^{2,-2}\Big\}. \end{align}\] By “harmonicity” in Lemma 3, we have 30 for \(m=k\). Substituting 22 , 30 and 31 into \[\begin{align} \begin{cases} \big\{\varphi_{k}+\sum\limits_{j=1}^{k-1}\varphi_{j}u_{k-j}+[\theta,u_{k}]\big\}^{2,-2}=0;\\ \big\{\psi_{k}+\sum\limits_{j=1}^{k-1}\psi_{j}u_{k-j}+\bar\partial u_{k}-\sum\limits_{j=1}^{k-1}\bar\eta_{j}(\bar\partial u_{k-j})\big\}^{3,-3}=0, \end{cases} \end{align}\] given by the equations 8 for gauge transformation, we have ?? . ◻

Proof. We first remark that \(\bar\eta(\omega_1)\) is the contraction and lies in \(\mathcal{A}^{1,0}(\operatorname{End}{\mathcal{E}})\). Thus \[\begin{align} [\bar\eta(\omega_1),\omega_2]=\bar\eta(\omega_1)\circ\omega_2+\omega_2\circ\bar\eta(\omega_1)=-\omega_1\circ\bar\eta(\omega_2)-\bar\eta(\omega_2)\circ\omega_1=-[\omega_1,\bar\eta(\omega_2)], \end{align}\] where the second equality follows from \(dz_i\wedge d\bar z_j=-d\bar z_j\wedge d z_i\) for any local chart \((z_i)\) of \(X_0\). ◻

Proof. By the definition of Lie bracket 14 , for \(\omega_1\in \mathcal{A}^k(\operatorname{End}{\mathcal{E}})\), \(\omega_2\in \mathcal{A}^l(\operatorname{End}{\mathcal{E}})\) and \(\omega_3\in \mathcal{A}^j(\operatorname{End}{\mathcal{E}})\), we have the following Jacobi identity \[\begin{align} \label{eq95Liebracket953} (-1)^{kl} \bigl[ \omega_1, [\omega_2, \omega_3] \bigr] + (-1)^{lj} \bigl[ \omega_2, [\omega_3, \omega_1] \bigr] + (-1)^{jk} \bigl[ \omega_3, [\omega_1, \omega_2] \bigr] = 0. \end{align}\tag{34}\] By taking \(\omega_1=\theta^{\star_{h_0}},\;\omega_2=[\theta,g]\) and \(\omega_3=h\) in the Jacobi identity 34 , we have \[\begin{align} \label{eq95firststep95C1} [\theta^{\star_{h_0}}, [[\theta,g],h]]=[[\theta,g],[h,\theta^{\star_{h_0}}]]+[h,[\theta^{\star_{h_0}},[\theta,g]]]. \end{align}\tag{35}\] By applying the Jacobi identity 34 again to \([\theta^{\star_{h_0}},[\theta,g]]\), we have \[\begin{align} [h,[\theta^{\star_{h_0}},[\theta,g]]]=[h,[\theta,[\theta^{\star_{h_0}},g]]]+[h,[g,[\theta^{\star_{h_0}},\theta]]]. \end{align}\] Substituting this into 35 , we have \[\begin{align} [\theta^{\star_{h_0}}, [[\theta,g],h]]=&-[[\theta,g],[\theta^{\star_{h_0}},h]]-[[\theta,[\theta^{\star_{h_0}},g]],h]+[h,[g,[\theta^{\star_{h_0}},\theta]]]\\ =&-[[\theta,g],[\theta^{\star_{h_0}},h]]-[[\theta,[\theta^{\star_{h_0}},g]],h]+[h,(D_{h_0}^{1,0}\bar\partial+\bar\partial D_{h_0}^{1,0})g] \end{align}\] by the identity \([[\theta,\theta^{\star_{h_0}}],g]=-F(D_{h_0})g=-(D_{h_0}^{1,0}\bar\partial+\bar\partial D_{h_0}^{1,0})g\) which follows from the harmonic metric equation. Therefore \[\begin{align} &[\theta^{\star_{h_0}}, [[\theta,g],h]]+D_{h_0}^{1,0}([\bar\partial g,h])=-[[\theta,g],[\theta^{\star_{h_0}},h]]-[[\theta,[\theta^{\star_{h_0}},g]],h]-[(D_{h_0}^{1,0}\bar\partial+\bar\partial D_{h_0}^{1,0})g,h]+D_{h_0}^{1,0}([\bar\partial g,h])\\ &=-[[\theta,g],[\theta^{\star_{h_0}},h]]-[[\theta,[\theta^{\star_{h_0}},g]],h]-[\bar\partial D_{h_0}^{1,0}g,h]-[\bar\partial g,D_{h_0}^{1,0}h]\\ &= -[[\theta,[\theta^{\star_{h_0}},g]],h] -[[\theta,g],[\theta^{\star_{h_0}},h]] -\bar\partial([D_{h_0}^{1,0}g,h]) -[\bar\partial g,D_{h_0}^{1,0}h] -[\bar\partial h,D_{h_0}^{1,0}g]. \qedhere \end{align}\] ◻

Proof. Using the Leibniz rule of \(\bar\partial\), we have \[\begin{align} &\sum_{|I_3|=k}\big\{\frac{i_2}{i_1+i_2+i_3}\frac{i_3}{i_1+i_3}\bar\partial([[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],g^{\star_{h_0}}_{i_2}])+\frac{i_3}{i_1+i_3}[[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],\bar\partial g^{\star_{h_0}}_{i_2}]\big\}^{4,-4}\\ &=-\sum_{|I_3|=k}\Big(\frac{i_2}{i_1+i_2+i_3}\frac{i_3}{i_1+i_3}[[[\theta,\bar\partial g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],g^{\star_{h_0}}_{i_2}] +\frac{i_2}{i_1+i_2+i_3}\frac{i_3}{i_1+i_3}[[[\theta,g^{\star_{h_0}}_{i_1}],\bar\partial g^{\star_{h_0}}_{i_3}],g^{\star_{h_0}}_{i_2}] \\&\makebox[1.7cm]{}+(\frac{i_2}{i_1+i_2+i_3}\frac{i_3}{i_1+i_3}-\frac{i_3}{i_1+i_3})[[[\theta,g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_3}],\bar\partial g^{\star_{h_0}}_{i_2}]\Big)^{4,-4} \\&=-\sum_{|I_3|=k}\Big(\frac{i_2}{i_1+i_2+i_3}\frac{i_3}{i_1+i_3}[[[\bar\partial g^{\star_{h_0}}_{i_1},g^{\star_{h_0}}_{i_3}],\theta],g^{\star_{h_0}}_{i_2}]+\frac{i_2}{i_1+i_2+i_3}\frac{i_3}{i_1+i_3}[[[\theta,g^{\star_{h_0}}_{i_3}],\bar\partial g^{\star_{h_0}}_{i_1}],g^{\star_{h_0}}_{i_2}] \\&\makebox[1.7cm]{}+\frac{i_2}{i_1+i_2+i_3}\frac{i_3}{i_1+i_3}[[[\theta,g^{\star_{h_0}}_{i_1}],\bar\partial g^{\star_{h_0}}_{i_3}],g^{\star_{h_0}}_{i_2}]-\frac{i_3}{i_1+i_2+i_3}[[\bar\partial g^{\star_{h_0}}_{i_2},[\theta,g^{\star_{h_0}}_{i_1}]],g^{\star_{h_0}}_{i_3}] \\&\makebox[1.7cm]{}-\frac{i_3}{i_1+i_2+i_3}[[g^{\star_{h_0}}_{i_3},\bar\partial g^{\star_{h_0}}_{i_2}],[\theta,g^{\star_{h_0}}_{i_1}]]\Big)^{4,-4}\quad\text{(where we use the Jacobi identity \eqref{eq95Liebracket953})} \\ &\stackrel{\eqref{vanisheq952}}{=}-\sum_{|I_2|=k}2\cdot\frac{i_2}{i_2+i_1}\cdot\big\{[[\theta,[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_1}]],g^{\star_{h_0}}_{i_2}]+[[\theta,g^{\star_{h_0}}_{i_1}],[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_2}]]\big\}^{4,-4}.\qedhere \end{align}\] ◻

Proof. We only prove the first equality as the second follows directly by reordering the summation indices. For the first one, we prove inductively on \(N\geq 1\) that \[\begin{align} &-\mathcal{L}_{\Lambda_N^{\star_{h_0}}} D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1} =\sum_{j=1}^N(-1)^j \sum_{\sigma\in V_{j;N}}(g^{\star_{h_0}}_{i_{\sigma(1)}}\cdots g^{\star_{h_0}}_{i_{\sigma(j-1)}}(D_{h_0}^{1,0}g^{\star_{h_0}}_{i_{\sigma(j)}})g^{\star_{h_0}}_{i_{\sigma(j+1)}}\cdots g^{\star_{h_0}}_{i_{\sigma(N)}}). \end{align}\] For \(N=1,2\), the above identity holds trivially. Assume inductively it holds for \(N\geq 2\), and we prove it for \(N+1\). By taking \([-,g^{\star_{h_0}}_{i_{N+1}}]\) of the above identity, we have\[\begin{align} &-\mathcal{L}_{\Lambda_{N+1}^{\star_{h_0}}}D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1} =\sum_{j=1}^N(-1)^j \sum_{\sigma\in V_{j;N}}(g^{\star_{h_0}}_{i_{\sigma(1)}}\cdots g^{\star_{h_0}}_{i_{\sigma(j-1)}}(D_{h_0}^{1,0}g^{\star_{h_0}}_{i_{\sigma(j)}})g^{\star_{h_0}}_{i_{\sigma(j+1)}}\cdots g^{\star_{h_0}}_{i_{\sigma(N)}})g^{\star_{h_0}}_{i_{N+1}}\\ &\makebox[3cm]{}-\sum_{j=1}^N(-1)^j\sum_{\sigma\in V_{j;N+1}}g^{\star_{h_0}}_{i_{N+1}}(g^{\star_{h_0}}_{i_{\sigma(1)}}\cdots g^{\star_{h_0}}_{i_{\sigma(j-1)}}(D_{h_0}^{1,0}g^{\star_{h_0}}_{i_{\sigma(j)}})g^{\star_{h_0}}_{i_{\sigma(j+1)}}\cdots g^{\star_{h_0}}_{i_{\sigma(N)}}), \end{align}\] and one can verify the right hand side of the above is exactly \[\sum_{j=1}^{N+1}(-1)^j \sum_{\sigma\in V_{j;N+1}}(g^{\star_{h_0}}_{i_{\sigma(1)}}\cdots g^{\star_{h_0}}_{i_{\sigma(j-1)}}(D_{h_0}^{1,0}g^{\star_{h_0}}_{i_{\sigma(j)}})g^{\star_{h_0}}_{i_{\sigma(j+1)}}\cdots g^{\star_{h_0}}_{i_{\sigma(N)}}g^{\star_{h_0}}_{i_{\sigma(N+1)}}). \qedhere\] ◻

Proof of Lemma 7. All three terms in 38 can be written as a linear combination of \[\begin{align} &(g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{j-1}}(D_{h_0}^{1,0}g^{\star_{h_0}}_{i_j})g^{\star_{h_0}}_{i_{j+1}}\cdots g^{\star_{h_0}}_{i_N})^{N+1,-N-1}\\&=(g^{\star_{h_0}}_{i_1})^{144-1}\cdots(g^{\star_{h_0}}_{i_{j-1}})^{144-1}\big(D_{h_0}^{1,0}(g^{\star_{h_0}}_{i_j})^{144-1}\big)(g^{\star_{h_0}}_{i_{j+1}})^{144-1}\cdots(g^{\star_{h_0}}_{i_N})^{144-1} \end{align}\] by the Leibniz rule of \(D_{h_0}^{1,0}\). In the following we compare their coefficients on both sides to prove 38 . The main ingredient in this proof is the combinatorial property of \(b_{i_1,\cdots,i_N}\) given in Proposition 19.

We firstly simplify the term in the left hand side of 38 : \[\begin{align} \label{eq95pf43951} & \sum_{ |I_2| =i} (S^{\star_{h_0}}_{i_1}D_{h_0}^{1,0}g^{\star_{h_0}}_{i_2})^{N+1,-N-1} =\sum_{|I_2|=n}\sum_{j=1}^{N-1}(S^{\star_{h_0}}_{i_1})^{j,-j}(D_{h_0}^{1,0}g^{\star_{h_0}}_{i_2})^{N+1-j,-N-1+j}. \end{align}\tag{39}\]

for any positive integer \(j\) and \(k=1,2,\cdots,n-1\). We prove by induction on \(j\) that \[\begin{align} \label{SjN}(S^{\star_{h_0}}_k)^{j,-j}=(-1)^{j-1}(g^{\star_{h_0}}_k)^{j,-j}. \end{align}\tag{40}\] For \(j=1,2\), 40 is clear by ?? . Now we assume 40 holds for \(2\leq j<N\) and prove it for \((S^{\star_{h_0}}_k)^{j+1,-j-1}\). If \(j+1\) is odd, then \[\begin{align} &(S^{\star_{h_0}}_k)^{j+1,-j-1}=(g^{\star_{h_0}}_k)^{j+1,-j-1}+\sum_{l=1}^{j}\sum_{|I_2|= k} (g^{\star_{h_0}}_{i_1})^{l,-l}(-S^{\star_{h_0}}_{i_2})^{j+1-l,-j-1+l} \\&=(g^{\star_{h_0}}_k)^{j+1,-j-1}+\sum_{l=1}^{j}\sum_{|I_2| = k} (-1)^{j+1-l}(g^{\star_{h_0}}_{i_1})^{l,-l}(g^{\star_{h_0}}_{i_2})^{j+1-l,-j-l+l}\\ &\overset{\eqref{thmbg}}=(g^{\star_{h_0}}_k)^{j+1,-j-1}+\sum_{l=1}^{j}\sum_{|I_j|= k} (-1)^{j+1-l}(b_{i_{1},\cdots,i_l}g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_l})^{l,-l}(b_{i_{l+1},\cdots,i_j}g^{\star_{h_0}}_{i_{l+1}}\cdots g^{\star_{h_0}}_{i_{j}})^{j+1-l,-j-l+l} \\&\overset{\eqref{eq95alt95sum952}}=(g^{\star_{h_0}}_k)^{j+1,-j-1}. \end{align}\] Similar, if \(j+1\) is even, we can also prove 40 by using 70 in Corollary 3.

Substituting 40 into 39 , we obtain \[\begin{align} \label{coeffaltersum} &\sum_{ |I_2|=i} (S^{\star_{h_0}}_{i_1}D_{h_0}^{1,0}g^{\star_{h_0}}_{i_2})^{N+1,-N-1}&\notag \\&=\sum_{|I_N|=n}\sum_{j=1}^{N-1}(-1)^{j-1}b_{i_1,\cdots,i_j}b_{i_{j+1},\cdots,i_N}\big(g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_j}D_{h_0}^{1,0}(g^{\star_{h_0}}_{i_{j+1}}\cdots g^{\star_{h_0}}_{i_N})\big)^{N+1,-N-1}&\notag\\ &= \sum_{|I_N|=n} \sum_{j=1}^{N} \sum_{l=1}^{j-1}(-1)^{l-1} b_{i_1,\cdots,i_l}b_{i_{l+1},\cdots,i_N} (g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{j-1}}(D_{h_0}^{1,0}g^{\star_{h_0}}_{i_j})g^{\star_{h_0}}_{i_{j+1}}\cdots g^{\star_{h_0}}_{i_N})^{N+1,-N-1}.& \end{align}\tag{41}\] by expanding \(D_{h_0}^{1,0}(g^{\star_{h_0}}_{i_{j+1}}\cdots g^{\star_{h_0}}_{i_N})\) by Leibniz rule. To prove 38 , it suffices to prove the coefficient of \[\begin{align} (g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{j-1}}(D_{h_0}^{1,0}g^{\star_{h_0}}_{i_j})g^{\star_{h_0}}_{i_{j+1}}\cdots g^{\star_{h_0}}_{i_N})^{N+1,-N-1} \end{align}\] on both sides of 38 are equal. Using 41 and Lemma 8, we only need to prove for any \(j=1,2,\cdots,N\) \[\begin{align} \sum_{l=1}^{j-1}(-1)^{l-1}b_{i_1,\cdots,i_l}b_{i_{l+1},\cdots,i_N}= b_{i_1,\cdots,i_N}-(-1)^j\sum_{\sigma\in A'_{j;N}}b_{i_{\sigma(1)},\cdots,i_{\sigma(N)}}. \end{align}\] Note that the above holds by 72 . ◻

Proof. (i) Note that in the proof of 38 , we crucially the \(\mathbb{C}\)-linearity and the Leibniz rule of \(D_{h_0}^{1,0}\). And here \([\theta,-]\) also satisfies the \(\mathbb{C}\)-linearity and the Leibniz rule. Hence we just run a similar argument to prove 44 .

(ii)(iii) Note that \([\theta^{\star_{h_0}},-]\), \(\bar\partial(-)\) (and \(\bar\eta\circ [\theta^{\star_{h_0}},-]\), \(\bar\eta\circ\bar\partial(-)\)) also satisfy the \(\mathbb{C}\)-linearity and the Leibniz rule. By a similar argument as in the proof of 38 , we have 43 (and 44 ). ◻

Proof. Substituting 38 , 42 and 44 into explicit formula ?? of \(\varphi_n\), we have \[\begin{align} (\varphi_n)^{N+1,-N-1}=&\frac{1}{2}\Big\{[\theta,g^{\star_{h_0}}_n-\sum_{|I_{N+1}|=n} b_{i_1,\cdots,i_{N+1}}g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N+1}}] \\&-D_{h_0}^{1,0}(g^{\star_{h_0}}_n-\sum_{|I_{N}|=n} b_{i_1,\cdots,i_N}g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_N})\Big\}^{N+1,-N-1}\\ &+\frac{1}{2}\Big\{\sum_{|I_{N+1}|=n} b_{i_1,\cdots,i_{N+1}}(\mathcal{L}_{\Lambda_{N+1}^{\star_{h_0}}} [\theta,g^{\star_{h_0}}_{i_1}])-\sum_{|I_{N}|=n}{b}_{i_1,\cdots,i_N} ( \mathcal{L}_{\Lambda_N^{\star_{h_0}}} D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1} )\Big\}^{N+1,-N-1}\\ &+\frac{1}{2}\Big\{\sum_{|I_{N}|=n} b_{i_1,\cdots,i_{N-1}}\bar\eta_{i_N}(\mathcal{L}_{\Lambda_{N-1}^{\star_{h_0}}} [\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_1}])-\sum_{|I_{N+1}|=n} {b}_{i_1,\cdots,i_{N}}\bar\eta_{i_{N+1}}(\mathcal{L}_{\Lambda_N^{\star_{h_0}}} \bar\partial g^{\star_{h_0}}_{i_1})\Big\}^{N+1,-N-1}. \end{align}\] The last two lines of the above expression are denoted as Eq3. To prove 45 , it suffices to prove \[\boldsymbol{Eq3}=\frac{1}{2^{N+1}}\sum_{|I_{N+1}|=n}\sum_{l=1}^{N+1}(-1)^{l-1}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{N+1})(g^{\star_{h_0}}_{i_1}\cdots[\theta,g^{\star_{h_0}}_{i_l}]\cdots g^{\star_{h_0}}_{i_{N+1}})^{N+1,-N-1}.\]

Substituting partial equation 25 into \([\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_1}]\) below, we have \[\begin{align} &\sum_{|I_{N}|=n} b_{i_1,\cdots,i_{N-1}}\bar\eta_{i_N}(\mathcal{L}_{\Lambda_{N-1}^{\star_{h_0}}}[\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_1}])^{N+1,-N-1} \\&=\sum_{|I_{N+1}|=n}b_{i_1+i_2,i_3,\cdots,i_{N}}\cdot\frac{i_2}{i_2+i_1}\cdot\bar\eta_{i_{N+1}}(\mathcal{L}_{\Lambda_N^{\star_{h_0}}}\bar\partial g^{\star_{h_0}}_{i_1})^{N+1,-N-1}, \end{align}\] and using partial equations 24 and 25 , we have \[\begin{align} &-\sum_{|I_{N}|=n}{b}_{i_1,\cdots,i_N} ( \mathcal{L}_{\Lambda_N^{\star_{h_0}}}D_{h_0}^{1,0}g^{\star_{h_0}}_{i_1} )^{N+1,-N-1}\\ &=\sum_{|I_{N+1}|=n} \Big\{-\frac{i_2}{i_2+i_1}\cdot b_{i_{N+1},i_1+i_2,i_3\cdots,i_{N}}\bar\eta_{i_{N+1}}(\mathcal{L}_{\Lambda_N^{\star_{h_0}}}\bar\partial g^{\star_{h_0}}_{i_1})+b_{i_{N+1}+i_1,i_2,\cdots,i_{N}}\bar\eta_{i_{N+1}}(\mathcal{L}_{\Lambda_N^{\star_{h_0}}}\bar\partial g^{\star_{h_0}}_{i_1})\\&\makebox[2cm]{}-\frac{1}{2}\cdot \frac{i_2}{i_2+i_1}\cdot b_{i_1+i_2,i_3,\cdots,i_{N+1}}\mathcal{L}_{\Lambda_{N+1}^{\star_{h_0}}} [\theta,g^{\star_{h_0}}_{i_1}]\Big\}^{N+1,-N-1}. \end{align}\] Substituting the above two identities into Eq3, we have \[\begin{align} \boldsymbol{Eq3}=\frac{1}{2}\cdot&\Big\{\sum_{|I_{N+1}|=n}(b_{i_{N+1}+i_1,i_2,\cdots,i_{N}}-{b}_{i_1,i_2,\cdots,i_{N}}+ \frac{i_2}{i_2+i_1}\cdot(\frac{1}{2}\cdot b_{{i_1+i_2},i_3,\cdots,i_{N}}-b_{i_{N+1},i_1+i_2,i_3,\cdots,i_{N}}))\times \\&\bar\eta_{i_{N+1}} (\mathcal{L}_{\Lambda_N^{\star_{h_0}}}\bar\partial g^{\star_{h_0}}_{i_1}) +\sum_{|I_{N+1}|=n} (b_{i_1,\cdots,i_{N+1}}-\frac{1}{2}\cdot \frac{i_2}{i_2+i_1}\cdot b_{i_1+i_2,i_3,\cdots,i_{N+1}})(\mathcal{L}_{\Lambda_{N+1}^{\star_{h_0}}} [\theta,g^{\star_{h_0}}_{i_1}])\Big\}^{N+1,-N-1} \end{align}\] Note that 75 gives: \(b_{i_{N+1}+i_1,i_2,\cdots,i_{N}}-{b}_{i_1,i_2,\cdots,i_{N}}+ \frac{i_2}{i_2+i_1}\cdot(\frac{1}{2}\cdot b_{{i_1+i_2},i_3,\cdots,i_{N}}-b_{i_{N+1},i_1+i_2,i_3,\cdots,i_{N}})=0\), and thus we have \[\begin{align} \boldsymbol{Eq3} \overset{\eqref{recursiveb}}= \frac{1}{2}\sum_{|I_{N+1}|=n} (b_{i_1,\cdots,i_{N+1}}-\frac{1}{2}\cdot \frac{i_2}{i_2+i_1}\cdot b_{i_1+i_2,i_3,\cdots,i_{N+1}})(\mathcal{L}_{\Lambda_{N+1}^{\star_{h_0}}} [\theta,g^{\star_{h_0}}_{i_1}])^{N+1,-N-1}. \end{align}\] Note that 76 gives: \(b_{i_1,\cdots,i_{N+1}}-\frac{1}{2} \cdot\frac{i_2}{i_2+i_1}\cdot b_{i_1+i_2,i_3,\cdots,i_{N+1}}=\frac{1}{2^{N}} \mathbb{P}(\xi_1>\xi_2>\cdots>\xi_{N+1})\), and thus we have \[\begin{align} & \boldsymbol{Eq3} \overset{\eqref{abN}}= \frac{1}{2^{N+1}}\sum_{|I_{N+1}|=n} \mathbb{P}(\xi_1>\xi_2>\cdots>\xi_{N+1})(\mathcal{L}_{\Lambda_{N+1}^{\star_{h_0}}} [\theta,g^{\star_{h_0}}_{i_1}])^{N+1,-N-1}\\ &\overset{\eqref{eq95iterated95brac}}=\sum_{|I_{N+1}|=n}\sum_{l=1}^{N+1}\frac{(-1)^{l-1}}{2^{N+1}}(\sum_{\sigma\in A'_{l;N+1}}\mathbb{P}(\xi_{\sigma(1)}>\xi_{\sigma(2)}>\cdots>\xi_{\sigma(N+1)}))(g^{\star_{h_0}}_{i_1}\cdots[\theta,g^{\star_{h_0}}_{i_l}]\cdots g^{\star_{h_0}}_{i_{N+1}})^{N+1,-N-1}. \end{align}\] Note that 68 gives: \(\sum\limits_{\sigma\in A'_{l;N+1}}\mathbb{P}(\xi_{\sigma(1)}>\xi_{\sigma(2)}>\cdots>\xi_{\sigma(N+1)}))=\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{N+1})\), and thus we obtain 45 : \[\begin{align} \boldsymbol{Eq3} \overset{\eqref{2ineq}}= \sum_{|I_{N+1}|=n}\sum_{l=1}^{N+1}\frac{(-1)^{l-1}}{2^{N+1}}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{N+1})(g^{\star_{h_0}}_{i_1}\cdots[\theta,g^{\star_{h_0}}_{i_l}]\cdots g^{\star_{h_0}}_{i_{N+1}})^{N+1,-N-1}. \end{align}\]

Now we give a sketch proof of 46 using a similar argument as above. By 44 , we have \[\begin{align} &\big\{\sum_{|I_2|= n}\bar\eta_{i_1}\big([\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_2}]-\bar\partial g^{\star_{h_0}}_{i_2}\big)-\sum_{|I_3|= n}\bar\eta_{i_1}\Big(S_{i_2}^{\star_{h_0}}([\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_3}]-\bar\partial g^{\star_{h_0}}_{i_3})\Big)\big\}^{N+1,-N-1} \\ &=\Big\{\sum_{|I_{N}|=n} b_{i_1,\cdots,i_{N-1}}\bar\eta_{i_N}\mathcal{L}_{\Lambda_{N-1}^{\star_{h_0}}} [\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_1}]-\sum_{|I_{N+1}|=n} {b}_{i_1,\cdots,i_{N}}\bar\eta_{i_{N+1}}\mathcal{L}_{\Lambda_N^{\star_{h_0}}} \bar\partial g^{\star_{h_0}}_{i_1}\Big\}^{N+1,-N-1} \\ &\overset{\eqref{vanisheq952}}=\sum_{|I_{N+1}|=n} (\frac{1}{2}\cdot\frac{i_2}{i_2+i_1}\cdot b_{i_1+i_2,i_3,\cdots,i_{N}}-b_{i_1,\cdots,i_{N}})\bar\eta_{i_{N+1}}(\mathcal{L}_{\Lambda_N^{\star_{h_0}}}\bar\partial g^{\star_{h_0}}_{i_1})^{N+1,-N-1}\\ &\overset{\eqref{abN}}=-\frac{1}{2^{N-1}}\sum_{|I_{N+1}|=n}\mathbb{P}(\xi_1>\xi_2>\cdots>\xi_{N})\bar\eta_{i_{N+1}}(\mathcal{L}_{\Lambda_N^{\star_{h_0}}}\bar\partial g^{\star_{h_0}}_{i_1})^{N+1,-N-1}\\ &\overset{\eqref{eq95iterated95brac}}=-\frac{1}{2^{N-1}}\sum_{|I_{N+1}|=n}\sum_{l=1}^{N}(-1)^{l-1}(\sum_{\sigma\in A'_{l;N}}\mathbb{P}(\xi_{\sigma(1)}>\xi_{\sigma(2)}>\cdots>\xi_{\sigma(N)}))(g^{\star_{h_0}}_{i_1}\cdots\bar\partial g^{\star_{h_0}}_{i_l}\cdots g^{\star_{h_0}}_{i_{N}})^{N+1,-N-1}\\ &\overset{\eqref{2ineq}}=\frac{1}{2^{N-1}}\sum_{|I_{N+1}|=n}\sum_{l=1}^{N}(-1)^{l}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{N})\bar\eta_{i_{N+1}}(g^{\star_{h_0}}_{i_1}\cdots(\bar\partial g^{\star_{h_0}}_{i_l})\cdots g^{\star_{h_0}}_{i_{N}})^{N+1,-N-1}. \end{align}\] Using 43 and by a similar argument as above, we have \[\begin{align} &\Big\{ -[\theta^{\star_{h_0}},\sum_{|I_{N-1}|=n} b_{i_1,\cdots,i_{N-1}}g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N-1}}]+\bar\partial(\sum_{|I_{N}|=n} b_{i_1,\cdots,i_{N}}g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N}}]) \\&\makebox[5cm]{} +\sum_{i_1 + i_2 = n}\Big(S^{\star_{h_0}}_{i_1}([\theta^{\star_{h_0}},g^{\star_{h_0}}_{i_2}]-\bar\partial g^{\star_{h_0}}_{i_2})\Big)\Big\}^{N+1,-N-1}\\ &=-\sum_{|I_{N}|=n} (\frac{1}{2}\cdot \frac{i_2}{i_2+i_1}\cdot b_{i_1+i_2,i_3,\cdots,i_{N}}-b_{i_1,\cdots,i_{N}})(\mathcal{L}_{\Lambda_N^{\star_{h_0}}}\bar\partial g^{\star_{h_0}}_{i_1})^{N+1,-N-1}\\ &=\frac{1}{2^{N-1}}\sum_{|I_{N}|=n}\sum_{l=1}^{N}(-1)^{l-1}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{N})(g^{\star_{h_0}}_{i_1}\cdots(\bar\partial g^{\star_{h_0}}_{i_l})\cdots g^{\star_{h_0}}_{i_{N}})^{N+1,-N-1}. \end{align}\] Substituting the above two identities into explicit formula ?? of \(\psi_n\) gives the proof of 46 . ◻

Proof of Proposition 15. For any positive integer \(N\), we prove ?? and ?? by an induction argument on \(k\) . Firstly, the statement holds for \(k=1\) by 30 . Assume the statement holds for \(k=1,2,\cdots,n-1\), we prove the case for \(k=n\).

When \(k=n,\;N=1\), ?? holds trivially. When \(k=n,\;N=1,2\), ?? holds by 22 and 30 . For any positive integers \(N\geq 2\), under the inductive assumption, the conditions of Lemma 10 are fulfilled, hence the simplified expression 45 of \(\varphi_n\) holds for \(k=n\).

The modulo-\((t)\)-holomorphicity gives 8 . Substituting 45 with \(k\leq n\) and ?? with \(k\leq n-1\) into the grading piece \[\begin{align} \big(\varphi_n+\sum\limits_{k=1}^{n-1}\varphi_{k}u_{n-k}+[\theta,u_n]\big)^{N+1,-N-1}=0 \end{align}\] of the first equation in 8 , we have \[\begin{align} &[\theta,u_n]^{N+1,-N-1} +\frac{1}{2}\big\{[\theta,g^{\star_{h_0}}_n-\sum_{|I_{N+1}|=n} b_{i_1,\cdots,i_{N+1}}g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N+1}}]\\&\makebox[3.1cm]{}-D_{h_0}^{1,0}(g^{\star_{h_0}}_n-\sum_{|I_{N}|=n} b_{i_1,\cdots,i_{N}}g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N}})\big\}^{N+1,-N-1}\\ &+\frac{1}{2^{N+1}}\sum_{|I_{N+1}|=n}\sum_{l=1}^{N+1}(-1)^{l-1}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{N+1})(g^{\star_{h_0}}_{i_1}\cdots[\theta,g^{\star_{h_0}}_{i_l}]\cdots g^{\star_{h_0}}_{i_{N+1}})^{N+1,-N-1}\\ &+\frac{1}{2^{N+1}}\sum_{|I_{N+1}|=n} \sum_{j=0}^{N-1}\sum_{l=1}^{j+1}(-1)^{l-1}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{j+1})(g^{\star_{h_0}}_{i_1}\cdots[\theta,g^{\star_{h_0}}_{i_l}]\cdots g^{\star_{h_0}}_{i_{j+1}})^{j+1,-j-1}\times \\&\makebox[3.7cm]{}(-1)^{N-j}\mathbb{P}(\xi_{j+2}<\cdots<\xi_{N+1})(g^{\star_{h_0}}_{i_{j+2}}\cdots g^{\star_{h_0}}_{i_{N+1}})^{N-j,-N+j}=0. \end{align}\] Hence for \(l=1,\cdots,N+1\), the coefficient of \((g^{\star_{h_0}}_{i_1}\cdots[\theta,g^{\star_{h_0}}_{i_l}]\cdots g^{\star_{h_0}}_{i_{N+1}})^{N+1,-N-1}\) in last three lines of the left hand side of the above is \[\begin{align} \frac{1}{2^{N+1}}\sum_{j=l-1}^{N}(-1)^{N-j+l-1}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{j+1})\mathbb{P}(\xi_{j+2}<\cdots<\xi_{N+1}) \stackrel{\eqref{altersumprob}}{=}\frac{(-1)^N }{2^{N+1}}\mathbb{P}(\xi_1<\cdots<\xi_{N+1}), \end{align}\] where we denote \(\mathbb{P}(\xi_{j+2}<\cdots<\xi_{N+1}):=1\) when \(j+2\geq N+1\). In summary, \(\big(\varphi_n+\sum\limits_{k=1}^{n-1}\varphi_{k}u_{n-k}+[\theta,u_n]\big)^{N+1,-N-1}=0\) gives \[\label{exactphi}\begin{align}&[\theta,\frac{1}{2}g^{\star_{h_0}}_n+\sum_{|I_{N+1}|=n} (-\frac{1}{2}b_{i_1,\cdots,i_{N+1}}+\frac{(-1)^N}{2^{N+1}}\mathbb{P}(\xi_1<\cdots<\xi_{N+1}))g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N+1}}+u_n]^{N+1,-N-1}\\ &-\frac{1}{2}D_{h_0}^{1,0}(g^{\star_{h_0}}_n-\sum_{|I_{N}|=n} b_{i_1,\cdots,i_{N}}g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N}})^{N+1,-N-1}=0. \end{align}\tag{47}\]

Substituting 46 with \(k\leq n\) and ?? with \(k\leq n-1\) into the grading piece \[\begin{align} \big(\psi_{n}+\sum\limits_{k=1}^{n-1}\psi_{k}u_{n-k}+\bar\partial u_{n}-\sum\limits_{k=1}^{n-1}\bar\eta_{k}(\bar\partial u_{n-k})\big)^{N+2,-N-2}=0 \end{align}\] of the second equation in 8 , we have

\[\label{exactpsi}\begin{align}&\big(\psi_{n}+\sum\limits_{k=1}^{n-1}\psi_{k}u_{n-k}+\bar\partial u_{n}-\sum\limits_{k=1}^{n-1}\bar\eta_{k}(\bar\partial u_{n-k})\big)^{N+2,-N-2}\\ &=\big\{\bar\partial\big(\frac{1}{2}g^{\star_{h_0}}_n+\sum_{|I_{N+1}|=n} (-\frac{1}{2}b_{i_1,\cdots,i_{N+1}}+\frac{(-1)^N}{2^{N+1}}\mathbb{P}(\xi_1<\cdots<\xi_{N+1}))g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N+1}}+u_n\big)\\ &-\frac{1}{2}[\theta^{\star_{h_0}},g^{\star_{h_0}}_n-\sum_{|I_{N}|=n} b_{i_1,\cdots,i_{N}}g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N}}]\big\}^{N+2,-N-2}=0. \end{align}\tag{48}\] Applying the “harmonicity” in Corollary 1 to 47 and 48 , we get \[\begin{align} u_n^{N+1,-N-1}&=\big(-\frac{1}{2}g^{\star_{h_0}}_n+\sum_{|I_{N+1}|=n} (\frac{1}{2}b_{i_1,\cdots,i_{N+1}}+\frac{(-1)^{N+1}}{2^{N+1}}\mathbb{P}(\xi_1<\cdots<\xi_{N+1}))g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N+1}}\big)^{N+1,-N-1},\\ (g^{\star_{h_0}}_n)^{N,-N}&=\sum\limits_{|I_{N}|=n} b_{i_1,\cdots,i_{N}}(g^{\star_{h_0}}_{i_1}\cdots g^{\star_{h_0}}_{i_{N}})^{N,-N}.\qedhere \end{align}\] ◻

Proof. Without loss of generality, we may assume \(U\) is smooth near \(0\). If \(U\) is singular, we resolve the singularity, denoted as \(\pi:\hat{U}\to U\). By pulling back \(X\) and \(\sigma_{\mathop{\mathrm{Dol}}}\) to \(\hat{U}\), we also have a real analytic family of Higgs bundles over \(X_{\hat{U}} := X \times_U \hat{U}\), denoted by \(\hat{\sigma}_{\mathop{\mathrm{Dol}}} : \hat{U} \to M_{\mathop{\mathrm{Dol}}}(X_{\hat{U}} / \hat{U})\). Choose any point \(\hat{u}\) of \(\hat{U}\) such that \(\pi(\hat{u}) = 0 \in U\). By the definition of pull-back, \(\hat{\sigma}_{\mathop{\mathrm{Dol}}}(\hat{u})\) is a stable graded Higgs bundle. We only need to prove Theorem 2 for \(\hat{U}\). Thus we reduce to the case that the base \(U\) is smooth.

By the holomorphic assumption, we only need to prove the claim of Theorem 2 for any truncated case and then use the convergence. This is exactly Theorem 3. ◻

Proof. By a similar argument as in the proof of Lemma 8 and Lemma 9. ◻

Proof. By using ?? , ?? and 51 , we get the simplified expressions 57 and 59 , 60 for grading pieces with small indices. We shall use the same strategy as in Lemma 10 to prove the rest cases. Substituting 54 , 53 and 56 into ?? , we have for \(N\geq 2\) \[\begin{align} &\alpha_n^{-N+1,N-1}=\frac{1}{2}\Big\{\sum_{|I_{N-1}|=n} b_{i_1,\dots,i_{N-1}} (\mathcal{L}_{\Lambda_{N-1}}[\theta,g_{i_1}]) - \sum_{|I_{N}|=n} b_{i_1,\dots,i_N} (\mathcal{L}_{\Lambda_N}D_{h_0}^{1,0}g_{i_1})\\ &\makebox[2cm]{}+\sum_{|I_{N+1}|=n} b_{i_1,\dots,i_{N}} \eta_{i_{N+1}} \bigl( \mathcal{L}_{\Lambda_N}D_{h_0}^{1,0}g_{i_1}\bigr)-\sum_{|I_N|=n}b_{i_1,\cdots,i_{N-1}}\eta_{i_N}(\mathcal{L}_{\Lambda_{N-1}}[\theta,g_{i_1}])\Big\}^{-N+1,N-1}\\&\overset{\eqref{vanisheqst951}\eqref{vanisheqst952}}=\frac{1}{2}\sum_{|I_{N}|=n} (\frac{1}{2}\cdot\frac{i_2}{i_2+i_1}\cdot b_{i_1+i_2,i_3,\cdots,i_N}-b_{i_1,\cdots,i_N})(\mathcal{L}_{\Lambda_N}D_{h_0}^{1,0}g_{i_1})^{-N+1,N-1}\\ &\makebox[1cm]{}+\frac{1}{2}\sum_{|I_{N+1}|=n} \bigl( b_{i_1,\dots,i_{N}} - \frac{1}{2}\cdot \frac{i_2}{i_2+i_1}\cdot b_{i_1+i_2,i_3,\dots,i_{N}} \bigr) \, \eta_{i_{N+1}} \bigl(\mathcal{L}_{\Lambda_N}D_{h_0}^{1,0}g_{i_1} \bigr)^{-N+1,N-1}\\ &\stackrel{\eqref{abN}}{=}\frac{1}{2^N}\Big\{-\sum_{|I_{N}|=n} \mathbb{P}(\xi_1>\cdots>\xi_N)\mathcal{L}_{\Lambda_N}D_{h_0}^{1,0}g_{i_1}+\sum_{|I_{N+1}|=n} \mathbb{P}(\xi_1>\cdots>\xi_N)\eta_{i_{N+1}}(\mathcal{L}_{\Lambda_N}D_{h_0}^{1,0}g_{i_1})\Big\}^{-N+1,N-1}. \end{align}\] By expanding the iterated Lie brackets above and using 68 , we get 58 .

Substituting 55 and 56 into ?? and using partial equations 49 , 50 and using 75 , we have for \(N\geq 2\) \[\begin{align} \beta_n^{-N,N}=\frac{1}{2}\sum_{|I_{N+1}|=n}(\frac{1}{2}\cdot \frac{i_2}{i_2+i_1}\cdot b_{i_1+i_2,i_3,\cdots,i_{N+1}}-b_{i_1,\cdots,i_{N+1}})(\mathcal{L}_{\Lambda_{N+1}}[\theta^{\star_{h_0}}, g_{i_1}])^{-N+1,N-1}. \end{align}\] By expanding the iterated Lie brackets above and using 68 , we get 58 . ◻

Proof of Theorem 3. Let \[\begin{align} \label{eq95gauge95gr} \mathscr{U} = \operatorname{id} + \sum_{i=1}^n \bar{t}^i u_i + \sum_{i=0}^{n-1} \sum_{j=1}^{n-i} t^j \bar{t}^i u_{ij}\in\mathcal{A}^0(\operatorname{End}{\mathcal{E}})\otimes B_n, \end{align}\tag{62}\] such that \(({\mathcal{E}},\mathscr U^{-1}\circ \bar\partial_t\circ\mathscr U,\mathscr U^{-1}\circ \theta_t\circ\mathscr U)\) satisfies ?? (the existence of such \(\mathscr U\) is guaranteed by the holomorphicity assumption). We may choose \(u_{i\bar 0}\) freely without affecting the validity of ?? . Thus we take \(u_{i\bar 0}\in\mathcal{\bigoplus}_{l>0}A^0((\operatorname{End}{\mathcal{E}})^{-l,l})\) and for any positive integer \(N\) \[\begin{align} \label{thmv}u_{i\bar 0}^{-N,N}:=v_i^{-N,N}=\frac{(-1)^N}{2^N}\sum_{|I_N| = i} \mathbb{P}(\xi_1<\xi_2<\cdots<\xi_N)( g_{i_1} g_{i_2}\cdots g_{i_N})^{-N,N}. \end{align}\tag{63}\]

We aim to prove that the following \(\alpha'_k\in \mathcal{A}^1((\operatorname{End}{\mathcal{E}})^{-1,1})\) and \(\beta'_k\in \mathcal{A}^{0,1}((\operatorname{End}{\mathcal{E}})^{0,0})\) for any \(k=1,\cdots,n\): \[\begin{align} \bar\partial_t\circ\mathscr U&=\mathscr U\circ\Big(\pi_\eta''D_{h_0}+\sum_{k=1}^nt^k\beta'_k-\bar\eta(\sum_{k=1}^nt^k\beta'_k)\Big);\\ \theta_t\circ\mathscr U&=\mathscr U\circ\Big(P_\eta'\theta+\sum_{k=1}^nt^k\alpha_k'\Big). \end{align}\] If so, the gauge transformed bundle \(({\mathcal{E}},\mathscr U^{-1}\circ\bar\partial_t\circ\mathscr U,\mathscr U^{-1}\circ\theta_t\circ\mathscr U)\) is family of graded stable Higgs bundles and this proves Theorem 3. By definition of \(\mathscr U\) and \(\alpha_k',\;\beta_k'\), we have \[\begin{align} \label{grinduc}\begin{cases}\beta_k+\bar\partial v_k+\sum\limits_{l=1}^{k-1}\eta_l(D_{h_0}^{1,0}v_{k-l})-\sum\limits_{l=1}^{k-1}v_{k-l}\beta_l'+\sum\limits_{l=1}^{k-1}\beta_lv_{k-l}=\beta_k';\\ \alpha_k+[\theta,v_k]-\sum\limits_{l=1}^{k-1}v_{k-l}\alpha_l'+\sum\limits_{l=1}^{k-1}\alpha_lv_{k-l}=\alpha_k'. \end{cases} \end{align}\tag{64}\]

In particular, we prove for any positive integer \(N\) \[\begin{align} \label{beta39}\beta_k'^{144-1}=\beta_k^{144-1},\text{ and }\;\beta_k'^{-N+1,N-1}=0 \end{align}\tag{65}\] and for any integer \(N\geq 2\) \[\begin{align} \label{alpha39}\alpha_k'^{-1,1}=\alpha_k^{0,0},\text{ and }\;\alpha_k'^{-N+1,N-1}=0. \end{align}\tag{66}\]

When \(k=1\), by 63 , 64 and ?? , we have for any \(N\geq 1\) \[\begin{align} \beta_1'^{144-1}&=(\beta_1+\bar\partial v_1)^{144-1}=\beta_1^{144-1};\\ \beta_1'^{-N+1,N-1}&=(\beta_1+\bar\partial v_1)^{-N+1,N-1}=(-\frac{1}{2}[\theta^{\star_{h_0}},g_1])^{-N+1,N-1}=0. \end{align}\]

When \(k=1\), by 63 , 64 and ?? , we have for any \(N\geq 2\) \[\begin{align} \alpha_1'^{0,0}&=(\alpha_1+[\theta,v_1])^{0,0}=\alpha_1^{0,0};\\ \alpha_1'^{-N+1,N-1}&=(\alpha_1+[\theta,v_1])^{-N+1,N-1}=(-\eta_1(\theta)-\frac{1}{2}D_{h_0}^{1,0}g_1)^{-N+1,N-1}=0. \end{align}\] Hence 65 and 66 hold for \(k=1\). Now we inductively assume 65 and 66 hold for \(k=1,\cdots,n-1\) and prove them for \(k=n\).

By 52 , 63 , 64 and the induction, we have \[\begin{align} \beta_n'^{144-1}=&\beta_n^{144-1}+(\bar\partial v_n+\sum_{|I_2|=n} \eta_{i_1}(D_{h_0}^{1,0}v_{i_2}))^{144-1}+\sum_{|I_2|=n} (-v_{i_1}\beta_{i_2}'+\beta_{i_1}v_{i_2})^{144-1}=\beta_n^{144-1};\\ \beta_n'^{0,0}=&\beta_n^{0,0}+(\bar\partial v_n+\sum_{|I_2|=n} \eta_{i_1}(D_{h_0}^{1,0}v_{i_2}))^{0,0}+\sum_{|I_2|=n} (-v_{i_1}\beta_{i_2}'+\beta_{i_1}v_{i_2})^{0,0}=0. \end{align}\] We aim to prove the following is zero for \(N\geq 2\) \[\begin{align} \beta_n'^{-N+1,N-1}=&\beta_n^{-N+1,N-1}+(\bar\partial v_n+\sum_{|I_2|=n} \eta_{i_1}(D_{h_0}^{1,0}v_{i_2}))^{-N+1,N-1}+\sum_{|I_2|=n} (-v_{i_1}\beta_{i_2}'+\beta_{i_1}v_{i_2})^{-N+1,N-1}. \end{align}\] We divide this expression into two parts as follow: \[\begin{align} \boldsymbol{Eq4}:&=\beta_n^{-N+1,N-1}+\sum\limits_{|I_2|=n} (-v_{i_1}\beta_{i_2}'+\beta'_{i_1}v_{i_2})^{-N+1,N-1}+\sum\limits_{j=2}^{N-1}\beta_{i_1}^{-j+1,j-1}v_{i_2}^{-N+j,N-j}.\\ \boldsymbol{Eq5}:&=(\bar\partial v_n)^{-N+1,N-1}+\sum\limits_{|I_2|=n} \eta_{i_1}(D_{h_0}^{1,0}v_{i_2})^{-N+1,N-1}+\sum\limits_{|I_2|=n}(\beta_{i_1})^{0,0}(v_{i_2})^{-N+1,N-1} \end{align}\] Then by the simplified expressions 59 , 61 in Lemma 13, and 63 and the induction \[\begin{align} &\boldsymbol{Eq4}=\frac{(-1)^N}{2^{N+1}}\sum_{|I_{N+1}|=n}\big\{\sum_{l=1}^{N+1}(-1)^{-N+l}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{N+1})(g_{i_1}\cdots[\theta^{\star_{h_0}},g_{i_l}]\cdots g_{i_{N+1}})\\ &+ \mathbb{P}(\xi_1<\cdots<\xi_N)(g_{i_1}\cdots g_{i_N}[\theta^{\star_{h_0}}, g_{i_{N+1}}])-\mathbb{P}(\xi_2<\cdots<\xi_{N+1})([\theta^{\star_{h_0}},g_{i_1}]g_{i_2}\cdots g_{i_{N+1}})\\ &+\sum_{j=2}^{N-1}\sum_{l=1}^{j+1}(-1)^{-j+l}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{j+1})\mathbb{P}(\xi_{j+2}<\cdots<\xi_{N+1}) (g_{i_1}\cdots[\theta^{\star_{h_0}},g_{i_l}]\cdots g_{i_{N+1}})\big\}^{-N+1,N-1}. \end{align}\] By a similar argument as in 77 , the coefficients of \((g_{i_1}\cdots g_{i_{l-1}}[\theta^\star,g_{i_l}]g_{i_{l+1}}\cdots g_{i_{N+1}})^{-N+1,N-1}\) in Eq4 are given by \[\frac{(-1)^{N+1}}{2^{N+1}}\cdot \begin{cases} \mathbb{P}(\xi_2<\cdots<\xi_{N+1})+\mathbb{P}(\xi_1>\xi_2>\xi_3<\cdots<\xi_{N+1}), & l=1,\\[6pt] \mathbb{P}(\xi_1<\cdots<\xi_{N+1})-\mathbb{P}(\xi_1<\xi_2)\mathbb{P}(\xi_3<\cdots<\xi_{N+1}), & l=2,\\[6pt] \mathbb{P}(\xi_1<\cdots<\xi_{N+1}), & 3\le l\le N,\\[6pt] \mathbb{P}(\xi_1<\cdots<\xi_{N+1})-\mathbb{P}(\xi_1<\cdots<\xi_N), & l=N+1. \end{cases}\]

Note that by the simplified expression 60 in Lemma 13 and 63 \[\begin{align} &\boldsymbol{Eq5}=\frac{(-1)^{N}}{2^N}\Big\{\sum_{|I_{N}|=n}\big\{ \mathbb{P}(\xi_1<\cdots<\xi_N)\bar\partial(g_{i_1}\cdots g_{i_N})-\mathbb{P}(\xi_2<\cdots<\xi_N)(\bar\partial g_{i_1})g_{i_2}\cdots g_{i_N}\big\}\\ &+\sum_{|I_{N+1}|=n}\big\{\mathbb{P}(\xi_1<\cdots<\xi_N)\eta_{i_{N+1}}(D_{h_0}^{1,0} (g_{i_1}g_{i_2}\cdots g_{i_{N}})\big)\\&-\mathbb{P}(\xi_2<\cdots<\xi_N)\eta_{i_{N+1}}((D_{h_0}^{1,0} g_{i_1})g_{i_2}\cdots g_{i_{N}})\big)+\frac{1}{2}\mathbb{P}(\xi_3<\cdots<\xi_{N+1})[[\theta^{\star_{h_0}},g_{i_1}],g_{i_2}]g_{i_3}\cdots g_{i_{N+1}}\big\}\Big\}^{-N+1,N-1}. \end{align}\] Using the partial eqaution 49 , Eq5 can be expressed as a linear combination of \[(g_{i_1}\cdots g_{i_{l-1}}[\theta^{\star_{h_0}},g_{i_{l}}]g_{i_{l+1}}\cdots g_{i_{N+1}})^{-N+1,N-1},\quad l=1,2,\cdots,N+1,\] and we place this complicated computation in Lemma 14 at the end of this section. By Lemma 14 and Eq4 above, one can prove Eq4\(+\)Eq5 is zero as desired.

By 52 , 64 and 63 , we have \[\begin{align} \alpha_n'^{0,0}=\alpha_n^{0,0}+[\theta, v_n]^{0,0}+\sum_{|I_2|=n} (-v_{i_1}\alpha_{i_2}'+\alpha_{i_1}v_{i_2})^{0,0}=\alpha_n^{0,0}. \end{align}\]

We aim to prove the following is zero for \(N\geq 2\) \[\begin{align} \alpha_n'^{-N+1,N-1}=&\alpha_n^{-N+1,N-1}+[\theta, v_n]^{-N+1,N-1}+\sum_{|I_2|=n} (-v_{i_1}\alpha_{i_2}'+\alpha_{i_1}v_{i_2})^{-N+1,N-1}. \end{align}\] Note that \[\begin{align} &[\theta,v_n]^{-N+1,N-1}\overset{\eqref{thmv}}= \frac{(-1)^{N-1}}{2^{N-1}}\sum_{|I_{N-1}|=n}\sum_{l=1}^{N-1} \mathbb{P}(\xi_1<\cdots<\xi_{N-1})(g_{i_1}\cdots[\theta,g_{i_l}]\cdots g_{i_{N-1}})^{-N+1,N-1}\\ &\overset{\eqref{vanisheqst952}}=\frac{(-1)^{N-1}}{2^{N}}\sum_{|I_{N}|=n}\sum_{l=1}^{N-1} \mathbb{P}(\xi_1<\cdots<\max(\xi_l,\xi_{l+1})<\xi_{l+2}<\cdots<\xi_{N})\cdot \frac{i_{l+1}}{i_l+i_{l+1}}\times\\&\makebox[4cm]{}(g_{i_1}\cdots[D_{h_0}^{1,0}g_{i_l},g_{i_{l+1}}]g_{i_{l+2}}\cdots g_{i_{N}})^{-N+1,N-1}. \end{align}\] Expanding the Lie brackets in the above and using 74 , we get \[\begin{align} [\theta,v_n]^{-N+1,N-1}&=\frac{(-1)^{N-1}}{2^N}\sum_{|I_{N}|=n} \sum_{l=1}^{N-1} \mathbb{P}(\xi_1<\cdots<\xi_N)(g_{i_1}\cdots(D_{h_0}^{1,0}g_{i_l})\cdots g_{i_{N}})^{-N+1,N-1}\\ &+\frac{(-1)^N}{2^N}\sum_{|I_{N}|=n}\mathbb{P}(\xi_1<\cdots<\xi_{N-1}>\xi_N)(g_{i_1}\cdots g_{i_{N-1}}(D_{h_0}^{1,0}g_{i_N}))^{-N+1,N-1}. \end{align}\]

Using 63 , 66 , 57 and 50 , for \(N\geq 2\), we have \[\begin{align} &\sum_{|I_2|=n}[\alpha'_{i_1},v_{i_2}]^{-N+1,N-1}=\frac{(-1)^N}{2^N}\Big\{\sum_{|I_{N+1}|=n}\big\{\sum_{l=1}^{N}\mathbb{P}(\xi_1<\cdots<\xi_{N})\eta_{i_{N+1}}(g_{i_1}\cdots(D_{h_0}^{1,0}g_{i_l})\cdots g_{i_{N}})\\ &-\mathbb{P}(\xi_2<\cdots<\xi_{N})\eta_{i_{N+1}}((D_{h_0}^{1,0}g_{i_1})g_{i_2}\cdots g_{i_{N}})\big\}+\sum_{|I_{N}|=n}[D_{h_0}^{1,0} g_{i_1},\mathbb{P}(\xi_2<\cdots<\xi_N)g_{i_2}\cdots g_{i_N}]\Big\}^{-N+1,N-1}. \end{align}\] Using the above two equations and 58 and 63 , we have, for \(N\geq 2\), the \((0,1)\)-part and \((1,0)\)-part of \[\begin{align} \alpha_n'^{-N+1,N-1}=&\alpha_n^{-N+1,N-1}+[\theta, v_n]^{-N+1,N-1}+\sum_{|I_2|=n} (-v_{i_1}\alpha_{i_2}'+\alpha_{i_1}v_{i_2})^{-N+1,N-1}\\ =&\alpha_n^{-N+1,N-1}+[\theta,v_n]^{-N+1,N-1}+\sum_{|I_2|=n}[\alpha_{i_1}',v_{i_2}]^{-N+1,N-1}+\sum_{|I_2|=n}\sum_{j=2}^{N-1}\alpha_{i_1}^{-j+1,j-1}v_{i_2}^{-(N-j),N-j} \end{align}\] are \[\begin{align} &\frac{(-1)^{N+1}}{2^{N}}\sum_{|I_{N+1}|=n}\eta_{i_{N+1}}\Big\{\sum_{l=1}^{N}(-1)^{-N+l}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{N})g_{i_1}\cdots(D_{h_0}^{1,0}g_{i_l})\cdots g_{i_{N}}\\ &-\sum_{l=1}^{N}\mathbb{P}(\xi_1<\cdots<\xi_{N})g_{i_1}\cdots(D_{h_0}^{1,0}g_{i_l})\cdots g_{i_{N}}+\mathbb{P}(\xi_2<\cdots<\xi_{N})(D_{h_0}^{1,0}g_{i_1})g_{i_2}\cdots g_{i_{N}}\\ &+\sum_{j=2}^{N-1}\sum_{l=1}^{j} (-1)^{l-j}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{j})\mathbb{P}(\xi_{j+1}<\cdots<\xi_{N})g_{i_1}\cdots(D_{h_0}^{1,0}g_{i_l})\cdots g_{i_{N}}\Big\}^{-N+1,N-1} \end{align}\] and \[\begin{align} &\frac{(-1)^N}{2^{N}}\sum_{|I_{N}|=n}\Big\{\sum_{l=1}^{N}(-1)^{-N+l}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{N})g_{i_1}\cdots(D_{h_0}^{1,0}g_{i_l})\cdots g_{i_{N}}\\ &- \sum_{l=1}^{N-1} \mathbb{P}(\xi_1<\cdots<\xi_N)g_{i_1}\cdots(D_{h_0}^{1,0}g_{i_l})\cdots g_{i_{N}}\\ &+ \mathbb{P}(\xi_1<\cdots<\xi_{N-1}>\xi_N)g_{i_1}\cdots g_{i_{N-1}}(D_{h_0}^{1,0}g_{i_N})+ [D_{h_0}^{1,0} g_{i_1},\mathbb{P}(\xi_2<\cdots<\xi_N)g_{i_2}\cdots g_{i_N}]\\ &+ \sum_{j=2}^{N-1}\sum_{l=1}^{j} (-1)^{l-j}\mathbb{P}(\xi_1<\cdots<\xi_l>\cdots>\xi_{j})\mathbb{P}(\xi_{j+1}<\cdots<\xi_{N})g_{i_1}\cdots(D_{h_0}^{1,0}g_{i_l})\cdots g_{i_{N}})\Big\}^{-N+1,N-1} \end{align}\] respectively, and those terms cancel out by 77 . Thus \(\alpha_n'^{-N+1,N-1}=0\) for \(N\geq2\). ◻

Proof. By expanding the Lie brackets in 67 and reordering the summation indices, the coefficient of \(((\bar\partial g_{i_1})g_{i_2}\cdots g_{i_N})^{-N+1,N-1}\) and \(\eta_{i_{N+1}}((D_{h_0}^{1,0} g_{i_1})g_{i_2}\cdots g_{i_N})^{-N+1,N-1}\) on the right hand side of 67 is \[\begin{align} -\mathbb{P}(B_1\land(\xi_2<\xi_1))=\mathbb{P}(\xi_1<\cdots<\xi_N)-\mathbb{P}(\xi_2<\cdots<\xi_N). \end{align}\] For \(l=2,\cdots,N\), the coefficient of \((g_{i_1}\cdots (\bar\partial g_{i_1})\cdots g_{i_N})^{-N+1,N-1}\) and \(\eta_{i_{N+1}}(g_{i_1}\cdots (D_{h_0}^{1,0} g_{i_1})\cdots g_{i_N})^{-N+1,N-1}\) on the right hand side of 67 is \[\begin{align} &-\mathbb{P}(B_l\land(\xi_{l+1}<\xi_l)) +\mathbb{P}(B_{l}\land(\xi_{l-1}<\xi_{l})) =\mathbb{P}(\xi_1<\cdots<\xi_N). \end{align}\] This proves 67 . For \(l=1,\cdots,N-1\), by 49 , we have \[\begin{align} &\sum_{|I_{N}|=n} \mathbb{P}(B_l\land(\xi_{l+1}<\xi_{l}))\big\{ g_{i_1}\cdots g_{i_{l-1}}[\bar\partial g_{i_{l}},g_{i_{l+1}}]g_{i_{l+2}}\cdots g_{i_N}\big\}^{-N+1,N-1}\\ &=\sum_{|I_{N}|=n} \mathbb{P}(B_l\land(\xi_{l+1}<\xi_{l}))\big\{g_{i_1}\cdots g_{i_{l-1}}[2\eta_{i_l}(\theta),g_{i_{l+1}}]g_{i_{l+2}}\cdots g_{i_N}\big\}^{-N+1,N-1}\\ &\sum_{|I_{N+1}|=n} \Big\{-\mathbb{P}(B_l\land(\xi_{l+1}<\max(\xi_{l},\xi_{N+1})))\big\{g_{i_1}\cdots g_{i_{l-1}}\eta_{i_{N+1}}([D_{h_0}^{1,0} g_{i_{l}},g_{i_{l+1}}])g_{i_{l+2}}\cdots g_{i_N}\\ &+\frac{1}{2}\mathbb{P}(B_l\land(\xi_{l+1}<\max(\xi_{l},\xi_{N+1})))\frac{i_l}{i_l+i_{N+1}}\cdot g_{i_1}\cdots g_{i_{l-1}}[[[\theta^{\star_{h_0}},g_{i_{N+1}}],g_{i_l}],g_{i_{l+1}}]g_{i_{l+2}}\cdots g_{i_N}\Big\}^{-N+1,N-1}, \end{align}\] denoted by Eq6, Eq7, Eq8 for each summation on the right hand side of the above. Hence \[\begin{align} &\boldsymbol{Eq6}\overset{\eqref{vanisheqst951}}=\sum_{|I_{N+1}|=n} \mathbb{P}((\xi_1<\cdots<\xi_{l-1}<\max(\xi_{l+1},\xi_{N+1})<\xi_{l+2}<\cdots<\xi_N)\land(\max(\xi_{l+1},\xi_{N+1})<\xi_{l}))\cdot \\ &\makebox[2.6cm]{}\frac{i_{l+1}}{i_{l+1}+i_{N+1}}\cdot\big\{g_{i_1}\cdots\eta_{i_l}([D_{h_0}^{1,0}g_{i_{N+1}},g_{i_{l+1}}])g_{i_{l+2}}\cdots g_{i_N}\big\}^{-N+1,N-1}\\ &=\sum_{|I_{N+1}|=n} \mathbb{P}((\xi_1<\cdots<\xi_{l-1}<\max(\xi_{l+1},\xi_{l})<\xi_{l+2}<\cdots<\xi_N)\land(\max(\xi_{l+1},\xi_{l})<\xi_{N+1}))\cdot \\ &\makebox[2cm]{}\frac{i_{l+1}}{i_{l}+i_{l+1}}\cdot\big\{g_{i_1}\cdots\eta_{i_{N+1}}([D_{h_0}^{1,0}g_{i_{l}},g_{i_{l+1}}])g_{i_{l+2}}\cdots g_{i_N}\big\}^{-N+1,N-1}\\ &=\sum_{|I_{N+1}|=n} \mathbb{P}(B_l\land(\xi_l<\xi_{l+1}<\xi_{N+1}))\big\{g_{i_1}\cdots\eta_{i_{N+1}}([D_{h_0}^{1,0}g_{i_{l}},g_{i_{l+1}}])g_{i_{l+2}}\cdots g_{i_N}\big\}^{-N+1,N-1}, \end{align}\] where the last equality follows from 74 . Next, \[\begin{align} \boldsymbol{Eq6}+\boldsymbol{Eq7} =&\sum_{|I_{N+1}|=n} (\mathbb{P}(B_l\land(\xi_l<\xi_{l+1}<\xi_{N+1}))-\mathbb{P}(B_l\land(\xi_{l+1}<\max(\xi_{l},\xi_{N+1})))\cdot\\ &\makebox[1.2cm]{}\big\{g_{i_1}\cdots \eta_{i_{N+1}}([D_{h_0}^{1,0} g_{i_{l}},g_{i_{l+1}}])g_{i_{l+2}}\cdots g_{i_N}\big\}^{-N+1,N-1}\\ =&-\sum_{|I_{N+1}|=n} (\mathbb{P}(B_l\land(\xi_{l+1}<\xi_{l}))\big\{g_{i_1}\cdots \eta_{i_{N+1}}([D_{h_0}^{1,0} g_{i_{l}},g_{i_{l+1}}])g_{i_{l+2}}\cdots g_{i_N}\big\}^{-N+1,N-1}. \end{align}\]

By performing the permutation \((N+1\quad l\quad l+1\quad l+2\quad \cdots\quad N)\) on the summation index of Eq8, \[\begin{align} &\boldsymbol{Eq8}\\ &=\frac{1}{2}\sum_{|I_{N+1}|=n} \mathbb{P}(D_l\land(\xi_{l+2}<\max(\xi_{l},\xi_{l+1})))\frac{i_{l+1}}{i_l+i_{l+1}}\cdot\big\{g_{i_1}\cdots [[[\theta^{\star_{h_0}},g_{i_{l}}],g_{i_{l+1}}],g_{i_{l+2}}]g_{i_{l+3}}\cdots g_{i_{N+1}}\big\}^{-N+1,N-1}\\ &=\frac{1}{2}\sum_{|I_{N+1}|=n} \mathbb{P}(D_l\land(\xi_{l+1}>\max(\xi_l,\xi_{l+2})))(g_{i_1}\cdots g_{i_{l-1}}[[[\theta^{\star_{h_0}},g_{i_l}],g_{i_{l+1}}],g_{i_{l+2}}]g_{i_{l+3}}\cdots g_{i_{N+1}})^{-N+1,N-1}. \end{align}\] By summing Eq6, Eq7 and Eq8, we get the right hand side of 67 is \[\begin{align} -\frac{1}{2}\sum_{|I_{N+1}|=n}\sum_{l=1}^{N-1}\mathbb{P}(D_l\land(\xi_{l+1}>\max(\xi_l,\xi_{l+2})))(g_{i_1}\cdots g_{i_{l-1}}[[[\theta^{\star_{h_0}},g_{i_l}],g_{i_{l+1}}],g_{i_{l+2}}]g_{i_{l+3}}\cdots g_{i_{N+1}})^{-N+1,N-1}. \end{align}\] By expanding the iterated Lie brackets above and reordering the summation indices, we get, in the above,

the coefficient of \(([\theta^{\star_{h_0}},g_{i_1}]g_{i_2}\cdots g_{i_{N+1}})^{-N+1,N-1}\) is \[-\frac{1}{2}\mathbb{P}(D_1\land (\xi_2>\max(\xi_1,\xi_3)));\]

the coefficient of \((g_{i_1}[\theta^{\star_{h_0}},g_{i_2}]g_{i_3}\cdots g_{i_{N+1}})^{-N+1,N-1}\) is \[\begin{align} &-\frac{1}{2}\big\{-\mathbb{P}(D_1\land(\xi_1>\max(\xi_2,\xi_3))+\mathbb{P}(D_2\land(\xi_3>\max(\xi_2,\xi_4))-\mathbb{P}(D_2\land(\xi_3>\max(\xi_1,\xi_2)))\big\}\\ &=-\frac{1}{2}\big\{-\mathbb{P}(D_1\land(\xi_1>\max(\xi_2,\xi_3)))-\mathbb{P}((\xi_1<\xi_3<\cdots<\xi_{N+1})\land(\xi_3>\xi_2))\big\}; \end{align}\]

the coefficient of \((g_{i_1}\cdots g_{i_{l-1}}[\theta^{\star_{h_0}},g_{i_l}]g_{i_{l+1}}\cdots g_{i_{N+1}})^{-N+1,N-1}\), where \(3\leq l\leq N-1\) is \[\begin{align} &-\frac{1}{2}\big\{\mathbb{P}(D_l\land(\xi_{l+1}>\max(\xi_l,\xi_{l+2})))-\mathbb{P}(D_l\land(\xi_{l+1}>\max(\xi_l,\xi_{l-1})))\\ &\quad\quad\quad+\mathbb{P}(D_{l-1}\land(\xi_{l-1}>\max(\xi_l,\xi_{l-2})))-\mathbb{P}(D_{l-1}\land(\xi_{l-1}>\max(\xi_l,\xi_{l+1})))\big\}\\ &=-\frac{1}{2}\big\{-\mathbb{P}((\xi_1<\cdots<\xi_{l-1}\widehat {\xi_l}<\xi_{l+1}<\cdots<\xi_{N+1})\land(\xi_{l+1}>\xi_l))\\ &\quad\quad\quad+\mathbb{P}((\xi_1<\cdots<\xi_{l-1}\widehat {\xi_l}<\xi_{l+1}<\cdots<\xi_{N+1})\land(\xi_{l-1}>\xi_l))\big\}\\ &=-\frac{1}{2}(-\mathbb{P}(\xi_1<\cdots<\xi_{N+1})); \end{align}\]

the coefficient of \((g_{i_1}\cdots g_{i_{N-1}}[\theta^{\star_{h_0}},g_{i_N}]g_{i_{N+1}})^{-N+1,N-1}\) is \[\begin{align} &-\frac{1}{2}\big\{\mathbb{P}(D_{N-1}\land(\xi_{N-1}>\max(\xi_N,\xi_{N-2})))-\mathbb{P}(D_{N-1}\land(\xi_{N-1}>\max(\xi_N,\xi_{N+1})))\\ &\quad\quad\quad-\mathbb{P}(D_{N}\land(\xi_{N+1}>\max(\xi_N,\xi_{N-1})))\big\}\\ &=-\frac{1}{2}\big\{\mathbb{P}((\xi_1<\cdots<\xi_{N-1}<\xi_{N+1})\land(\xi_{N-1}>\xi_N))-\mathbb{P}((\xi_1<\cdots<\xi_{N-1})\land(\xi_{N+1}>\max(\xi_N,\xi_{N-1}))\big\}\\ &=-\frac{1}{2}(-\mathbb{P}(\xi_1<\cdots<\xi_{N+1})); \end{align}\]

the coefficient of \((g_{i_1}\cdots g_{i_{N}}[\theta^{\star_{h_0}},g_{i_{N+1}}])^{-N+1,N-1}\) is \[-\frac{1}{2}\mathbb{P}(\xi_1<\cdots<\xi_N>\xi_{N+1}).\qedhere\] ◻

Proof. Combining the inequalities \(\xi_{j}>\xi_{{j-1}}>\cdots>\xi_{1}\) and \(\xi_{j}>\xi_{{j+1}}>\cdots>\xi_{N}\) into a single inequality yields exactly \(\binom{N-1}{j-1}\) possibilities. They are exactly 68 . ◻

Proof. We first study \(2^{N-2}b_{i_1,\cdots,i_k}b_{i_{k+1},\cdots,i_N}\). By Lemma 15, expanding this product gives terms of the form \[\begin{align} \mathbb{P}(\text{some total order inequality of }\xi_{1},\cdots,\xi_{k})\cdot \mathbb{P}(\text{some total order inequality of }\xi_{k+1},\cdots,\xi_{N}). \end{align}\] Combining the two inequalities into one yields exactly \(\binom{N}{k}\) mutually exclusive events. Hence \(2^{N-2}b_{i_1,\cdots,i_k}b_{i_{k+1},\cdots,i_N}\) is the sum of probabilities of \(2^{N-2}\binom{N}{k}\) mutually exclusive events, each of which is a total order inequality of \(\xi_{1},\cdots,\xi_{N}\). From now on, the “events” for \(b_{\cdots}\) refer to the events appearing in the probability sum of its expansion, see Lemma 15.

We shall prove that \(2^{N-2}\sum_{\sigma\in A_{k;N}}b_{i_{\sigma(1)},i_{\sigma(2)},\cdots,i_{\sigma(N)}}\) also consists of \(2^{N-2}\binom{N}{k}\) mutually exclusive events, each being a total order inequality of \(\xi_{1},\cdots,\xi_{N}\), and each such event also appears exactly in the expansion of \(2^{N-2}b_{i_1,\cdots,i_k}b_{i_{k+1},\cdots,i_N}\).

For \(\sigma\in A_{k;N}\), consider \(2^{N-1}b_{i_{\sigma(1)},i_{\sigma(2)},\cdots,i_{\sigma(N)}}\). By Lemma 15, this is the sum of probabilities of \(2^{N-1}\) mutually exclusive events. Pick any one of these events, say coming from the inequality \[\begin{align} \label{ineqbN} \xi_{{\sigma(1)}}<\cdots<\xi_{{\sigma(j)}}>\cdots>\xi_{{\sigma(N)}} \end{align}\tag{69}\] and for brevity we restate its properties as follows:

1. \(\sigma\in A_{k;N}\);

2. This event is a total order inequality of \(\xi_{1},\cdots,\xi_{N}\) obtained by combining \(\xi_{{\sigma(j)}}>\cdots>\xi_{{\sigma(1)}}\) and \(\xi_{{\sigma(j)}}>\cdots>\xi_{{\sigma(N)}}\). Write it as \(\xi_{\tau(1)}>\xi_{\tau(2)}>\cdots> \xi_{\tau(N)}\), then \(\tau(1)=\sigma(j)\).

We want to show that there exists exactly one \(\sigma'\ne \sigma\in A_{k;N}\) such that \(2^{N-1}b_{i_{\sigma'(1)},i_{\sigma'(2)},\cdots,i_{\sigma'(N)}}\) also contains this event.

Case 1: \(\sigma(j)\in\{k,k-1,\cdots,1\}\). Consider the positive integer \(l\) such that \(\sigma(l)\notin\{k,k-1,\cdots,1\}\) and \(\xi_{{\sigma(l)}}=\max_{t=k+1,\cdots,N} \xi_{t}\). When we speak of elements to the left or right of \(\sigma(\cdot)\), we mean the order in the permutation \((\sigma(1),\cdots,\sigma(N))\). If \(l<j\), we define a permutation that moves \(\sigma(l)\) to some position to the right of \(\sigma(j)\); specifically, we successively swap \(\sigma(l)\) with the elements to its right, subject to:

1. We must move \(\sigma(l)\) to the right of \(\sigma(j)\);

2. During this process we cannot perform swaps of the form \((\sigma(l),\sigma(l'))\) where \(\sigma(l')\in \{k+1,\cdots,N\}\);

3. After these swaps we obtain a permutation \(\gamma\in A_{k;N}\) different from \(\sigma\) (not necessarily unique). Clearly, in the expansion 36 of \(2^{N-1}b_{i_{\gamma(1)},i_{\gamma(2)},\cdots,i_{\gamma(N)}}\) there is exactly one event that has \(\xi_{{\sigma(j)}}\) as the maximum;

4. The event corresponds to an inequality, and we require that this inequality be compatible with the inequality \(\xi_{\tau(1)}>\xi_{\tau(2)}>\cdots> \xi_{\tau(N)}\), i.e., no contradiction. This requirement restricts the choice of \(\gamma\).

By definition there exists a unique \(\gamma\) satisfying the above conditions; then \(\gamma\) is the desired \(\sigma'\). We now verify the uniqueness of \(\sigma'\). Since \(\sigma'\ne\sigma\), \(\sigma'\) must either move some element from the left of \(\sigma(j)\) to the right of \(\sigma(j)\), or move some element from the right of \(\sigma(j)\) to the left of \(\sigma(j)\) (it is impossible to keep left elements on the left, otherwise the inequality 69 would not hold; similarly for the right side). By the definition of \(l\), \(\sigma'\) must move \(\sigma(l)\), therefore \(\sigma'\) satisfies the four conditions above, proving uniqueness.

If \(l>j\), a similar argument works.

Case 2: \(\sigma(j)\in\{k+1,\cdots,N\}\). Similar to Case 1.

Thus we have shown that \(2^{N-2}\sum_{\sigma\in A_{k;N}}b_{i_{\sigma(1)},i_{\sigma(2)},\cdots,i_{\sigma(N)}}\) also consists of \(2^{N-2}\binom{N}{k}\) mutually exclusive events, each being a total order inequality of \(\xi_{1},\cdots,\xi_{N}\). Next we prove that each such event also appears in the expansion of \(2^{N-2}b_{i_1,\cdots,i_k}b_{i_{k+1},\cdots,i_N}\).

For \(\sigma\in A_{k;N}\), consider any event 69 of \(2^{N-1}b_{i_{\sigma(1)},i_{\sigma(2)},\cdots,i_{\sigma(N)}}\). Deleting those \(X\) whose indices lie in \(\{1,2,\cdots,k\}\) from the inequality 69 yields an inequality consequence; this event obviously lies among the events of \(b_{i_{k+1},\cdots,i_N}\). Similarly, deleting those \(X\) whose indices lie in \(\{k+1,\cdots,N\}\) yields an inequality consequence that lies among the events of \(b_{i_{1},\cdots,i_k}\). This completes the proof of the proposition. ◻

Proof. 1. First consider the case when \(N\) is even. By Proposition 19, we have \[\begin{align} \sum_{k=1}^{N-1}(-1)^{k-1}\sum_{\sigma\in A_{k;N}}b_{i_{\sigma(1)},i_{\sigma(2)},\cdots,i_{\sigma(N)}}= \sum_{k=1}^{N-1}(-1)^{k-1}b_{i_1,\cdots,i_k}b_{i_{k+1},\cdots,i_N}. \end{align}\] Recall \(A_{k;N}\) can be partitioned into two parts \(A'_{k;N}\coprod A''_{k;N}\) in Definition 4. It is easy to see that \(A'_{k;N}= A''_{k-1;N}\). Therefore \[\begin{align} &\sum_{k=1}^{N-1}(-1)^{k-1}\sum_{\sigma\in A_{k;N}}b_{i_{\sigma(1)},i_{\sigma(2)},\cdots,i_{\sigma(N)}}\\ =&\Bigl(\sum_{\sigma\in A'_{1;N}}+\sum_{\sigma\in A''_{1;N}}-\sum_{\sigma\in A'_{2;N}}-\sum_{\sigma\in A''_{2;N}}+\cdots+\sum_{\sigma\in A'_{N-1;N}}+\sum_{\sigma\in A''_{N-1;N}}\Bigr)b_{i_{\sigma(1)},i_{\sigma(2)},\cdots,i_{\sigma(N)}}\\ =&\Bigl(\sum_{\sigma\in A'_{1;N}}+\sum_{\sigma\in A''_{N-1;N}}\Bigr)b_{i_{\sigma(1)},i_{\sigma(2)},\cdots,i_{\sigma(N)}}=b_{i_1,\cdots,i_N}+b_{i_N,i_{N-1},\cdots,i_1}=2b_{i_1,\cdots,i_N}. \end{align}\] 2. The case when \(N\) is odd is proved similarly.

. We prove 72 by an induction on \(j\). For \(j=1\), 72 is \(-b_{i_1,\cdots,i_N}=-b_{i_1,\cdots,i_N}\). Assume 72 holds for \(1\leq j<N\), we prove 72 holds for \(j+1\). By induction and Proposition 19, we have \[\begin{align} &-(-1)^{j+1}\sum_{\sigma\in A'_{j+1;N}}b_{i_{\sigma(1)},\cdots,i_{\sigma(N)}}-b_{i_1,\cdots,i_N}+\sum_{l=1}^{j}(-1)^{l-1}b_{i_1,\cdots,i_l}b_{i_{l+1},\cdots,i_N}\\=&(-1)^{j}\sum_{\sigma\in A'_{j+1;N}}b_{i_{\sigma(1)},\cdots,i_{\sigma(N)}}+(-1)^j\sum_{\sigma\in A'_{j;N}}b_{i_{\sigma(1)},\cdots,i_{\sigma(N)}}+(-1)^{j-1}b_{i_1,\cdots,i_j}b_{i_{j+1},\cdots,i_N}\\=&(-1)^{j}\sum_{\sigma\in A_{j;N}}b_{i_{\sigma(1)},\cdots,i_{\sigma(N)}}+(-1)^{j-1}b_{i_1,\cdots,i_j}b_{i_{j+1},\cdots,i_N}=0.\qedhere \end{align}\] ◻

Proof. Firstly, we have \[\begin{align} &\mathbb{P}(\xi_1<\cdots<\xi_{l-1}<\xi_l>\xi_{l+1}>\cdots>\xi_N)\\ =&(\prod_{j=2}^{l-1}\mathbb{P}(\max(\xi_1,\cdots,\xi_{j-1})<\xi_j))\mathbb{P}(\xi_l>\max(\xi_1,\cdots,\xi_{l-1},\xi_{l+1},\cdots,\xi_N))\cdot\\ &(\prod_{k=l+1}^{N-1}\mathbb{P}(\xi_k>\max(\xi_{k+1},\cdots,\xi_N))). \end{align}\] Then we get 73 by the following facts \[\begin{align} \max(\xi_1,\cdots,\xi_{j-1})&\sim \operatorname{Beta}(i_1+\cdots+i_{j-1},1);\quad \mathbb{P}(\xi_2>\xi_1)=\frac{i_2}{i_2+i_1}. \end{align}\] One may derive 74 from 73 directly. ◻

Proof. Let \(\xi_{1},\cdots,\xi_{N+1}\) be independent random variables with distributions \[\operatorname{Beta}(i_1,1),\cdots,\operatorname{Beta}(i_{N+1},1)\] respectively, where \(i_1,\cdots,i_{N+1}\) are positive integers. Then \[\begin{align} &2^{N-1}\frac{i_2}{i_2+i_1}\cdot(\frac{1}{2}\cdot b_{{i_1+i_2},i_3,\cdots,i_{N}}-{b}_{i_{N+1},i_1+i_2,i_3,\cdots,i_{N}})\\ =&\sum_{l=3}^{N}\frac{i_2}{i_2+i_1}\cdot(\mathbb{P}(\max(\xi_1,\xi_2)<\xi_3<\cdots<\xi_l>\cdots>\xi_{N})\\ &\makebox[.8cm]{}-\mathbb{P}(\xi_{N+1}<\max(\xi_1,\xi_2)<\xi_3<\cdots<\xi_l>\cdots>\xi_{N}))\\ \stackrel{\eqref{probmountain39}}{=}&\sum_{l=3}^{N}(\mathbb{P}(\xi_1<\xi_2<\xi_3<\cdots<\xi_l>\cdots>\xi_{N})-\mathbb{P}(\max(\xi_{N+1},\xi_1)<\xi_2<\cdots<\xi_l>\cdots>\xi_{N}))\\ =&2^{N-1}(b_{i_1,\cdots,i_{N}}-b_{i_{N+1}+i_1,i_2,\cdots,i_{N}}). \end{align}\]

As a byproduct of this proof, we have \[\begin{align}\label{abN}&\frac{i_2}{i_2+i_1}\cdot 2^{N-2}b_{i_1+i_2,i_3,\cdots,i_{N}}\\ =&\sum_{l=3}^{N}\mathbb{P}(\xi_1<\xi_2<\cdots<\xi_l>\cdots>\xi_{N})+\frac{i_2}{i_2+i_1}\cdot\mathbb{P}(\max(\xi_1,\xi_2)>\xi_3>\cdots>\xi_{N})\\ =&b_{i_1,i_2,\cdots,i_{N}}-\mathbb{P}(\xi_2>\cdots>\xi_{N})+\frac{i_2}{i_2+i_1}\cdot\mathbb{P}(\max(\xi_1,\xi_2)>\xi_3>\cdots>\xi_{N})\\ =&b_{i_1,i_2,\cdots,i_{N}}-\mathbb{P}(\xi_1>\xi_2>\cdots>\xi_{N}), \end{align}\tag{76}\] because \(\frac{i_2}{i_2+i_1}\cdot\mathbb{P}(\max(\xi_1,\xi_2)>\xi_3>\cdots>\xi_{N})=\mathbb{P}((\xi_2>\xi_1)\land(\xi_2>\xi_3>\cdots>\xi_{N}))\). ◻

Proof. Note that \[\begin{align} &(\xi_1<\cdots<\xi_l)\land(\xi_{l+1}<\cdots<\xi_{N+1})\\=&(\xi_1<\cdots<\xi_{N+1})\lor \big\{(\xi_1<\cdots<\xi_l>\xi_{l+1})\land(\xi_{l+1}<\cdots<\xi_{N+1})\big\};\\ &(\xi_1<\cdots<\xi_l>\xi_{l+1})\land(\xi_{l+2}<\cdots<\xi_{N+1})\\ =&\big\{(\xi_1<\cdots<\xi_l>\xi_{l+1})\land(\xi_{l+1}<\cdots<\xi_{N+1})\big\}\lor\\&\big\{ (\xi_1<\cdots<\xi_l>\xi_{l+1}>\xi_{l+2})\land(\xi_{l+2}<\cdots<\xi_{N+1})\big\},\quad\cdots,\\ &(\xi_1<\cdots<\xi_l>\cdots>\xi_{N})\\ =&\big\{(\xi_1<\cdots<\xi_l>\cdots>\xi_N)\land(\xi_N<\xi_{N+1})\big\}\lor(\xi_1<\cdots<\xi_l>\cdots>\xi_{N+1}). \end{align}\] Take the alternative sum of their probabilities and we get the desired identity. ◻

Proof of Theorem 20. Note that the assertion is local on the source and target. Let \(\Delta\subset S\) be a sufficiently small coordinate polydisc. After shrinking \(\Delta\), choose a real analytic Ehresmann trivialization \[X_\Delta:=\pi^{-1}(\Delta)\cong X_0\times \Delta\] where \(X_0\) is a fixed compact smooth manifold. Let \(J_s\) denote the induced complex structure on \(X_0\) and let \(\omega_s\) be a fiberwise Kähler form obtained from a relative polarization. We regard both \(J_s\) and \(\omega_s\) as real analytic functions of \(s\in \Delta\).

Now fix a point \[(s_0,E_0,\theta_0)\in M_{\mathop{\mathrm{Dol}}}(X_\Delta/\Delta)\] and work after shrinking \(\Delta\). Choose a smooth bundle \({\mathcal{E}}\to X_0\) representing the fixed topological type, and choose a reference Hermitian metric \(h_0\) on \({\mathcal{E}}\).

Without loss of generality we may assume that \((E_0,\theta_0)\) is stable. There is a finite-dimensional real analytic slice \(\Sigma\) for the relative Dolbeault moduli problem near \((s_0,E_0,\theta_0)\) (i,e, \(\Sigma\) is transversal to the fiber of \(M_{\mathop{\mathrm{Dol}}}(X_\Delta/\Delta) \to \Delta\)). A point \(q\in\Sigma\) determines \[q=(s,\bar\partial_q,\theta_q),\] where \(s\in \Delta\), \(\bar\partial_q\) is a holomorphic structure on the fixed smooth bundle over \((X_0,J_s)\), and \(\theta_q\) is a Higgs field satisfying \[\bar\partial_q\theta_q=0, \qquad \theta_q\wedge\theta_q=0.\]

By construction of the Kuranishi slice, and by the real analytic trivialization of the family, the coefficients of \(J_s\), \(\omega_s\), \(\bar\partial_q\) and \(\theta_q\) depend real analytically on \(q\) and on the fiber variable.

For each \(q\in\Sigma\), non-abelian Hodge theory supplies a harmonic metric \(h_q\). Write \[h_q=h_0 k_q,\] where \(k_q\) is a positive \(h_0\)-self-adjoint endomorphism. If the structure group has a nontrivial center, fix the determinant metric, or equivalently work modulo the central scalar ambiguity, so that \(k_q\) is uniquely determined near \(q_0\).

Recall the harmonic metric equation for \(h_q\) (cf. [6], [9]) \[\label{eq:higher-dimensional-HYM-Higgs} \sqrt{-1}\Lambda_{\omega_s} \left( F_{h_q,\bar\partial_q} +[\theta_q,\theta_q^{\star_{h_q}}] \right) = \lambda_q\,\operatorname{id}_{{\mathcal{E}}}.\tag{78}\]

Here \(\lambda_q\) is the topological constant, which is zero in the vanishing-Chern-class case. Rewriting 78 in terms of \(k_q\) and \(h_0\) gives \[\label{eq:k-equation} \sqrt{-1}\Lambda_{\omega_s} \left( \bar\partial_q\bigl(k_q^{-1}\partial_{0,q}k_q\bigr) +F_{h_0,\bar\partial_q} +[\theta_q,k_q^{-1}\theta_q^{{\star_{h_0}}}k_q] \right) =\lambda_q\,\operatorname{id}_{{\mathcal{E}}},\tag{79}\] where \(\partial_{0,q}\) is the \((1,0)\) part of the Chern connection determined by \((\bar\partial_q,h_0)\).

Equation 79 is a second-order nonlinear elliptic system for \(k_q\). Its principal part is the Laplacian-type operator \[\sqrt{-1}\Lambda_{\omega_s}\bar\partial_q\partial_{0,q},\] acting on Hermitian endomorphisms, and is strongly elliptic because \(\omega_s\) is positive. Moreover, the coefficients of 79 are real analytic in \((q,x)\) and analytic in \(k_q\) and its derivatives as long as \(k_q\) remains positive and invertible. By uniqueness of the harmonic metric, Proposition 21 applies. Therefore \[(q,x)\longmapsto k_q(x)\] is real analytic. Hence the harmonic metric \(h_q=h_0k_q\) is real analytic in \(q\) and in the fiber variable.

The flat connection associated with the harmonic bundle is \[\label{eq:flat-connection-from-Higgs} D_q =\bar\partial_q+\partial_{h_q,\bar\partial_q} +\theta_q+\theta_q^{\star_{h_q}}.\tag{80}\] Since \(\bar\partial_q\), \(\theta_q\) and \(h_q\) depend real analytically on \(q\), the connection one-form of \(D_q\) in any real analytic trivialization depends real analytically on \((q,x)\).

The inverse direction is proved by the same argument. Start with a real analytic slice in the Betti moduli space, represented by a real analytic family of flat connections \(D_q\) on the fixed smooth bundle over \(X_0\). The Corlette–Simpson harmonic metric equation for \(D_q\) (cf. [7], [9]), with respect to the complex structure \(J_s\) and Kahler metric \(\omega_s\), is again a second-order nonlinear strongly elliptic analytic system for the metric endomorphism \(k_q\). After the same determinant or central normalization, uniqueness of the harmonic metric and Proposition 21 imply that \(k_q\) depends real analytically on \(q\). Decomposing \(D_q\) using this harmonic metric and the complex structure \(J_s\) gives \[D_q=(\bar\partial_{E,q}+\theta_q)+(\partial_{E,q}^{h_q}+\theta_q^{\star_{h_q}}),\] so the resulting holomorphic structure and Higgs field are real analytic in \(q\). ◻

Proof. The assertion is local. A holomorphic function on a sufficiently small polydisc in \(\mathbb{C}\times\overline{\mathbb{C}}\) has a convergent expansion \[F(z,\zeta)=\sum_{i,j\geq 0}a_{ij}z^i\zeta^j.\] Restricting to the diagonal \(\zeta=\bar z\) gives the real analytic function \[i^*F(z,\bar z)=\sum_{i,j\geq 0}a_{ij}z^i\bar z^j .\] Conversely, if \[f(z,\bar z)=\sum_{i,j\geq 0}a_{ij}z^i\bar z^j\] is a convergent real analytic germ, then \[F(z,\zeta)=\sum_{i,j\geq 0}a_{ij}z^i\zeta^j\] is a convergent holomorphic germ on a sufficiently small polydisc in \(\mathbb{C}\times\overline{\mathbb{C}}\), and \(i^*F=f\). These local constructions are compatible with restrictions, hence glue to the desired isomorphism.

By a similar argument, this complexification lemma also holds similarly for real analytic germs at \(0\in\mathbb{R}^{2k}\). ◻

Proof. Choose holomorphic coordinates \(z_1,\ldots,z_m\) on \(M\) centered at \(x\), and let \(\zeta_1,\ldots,\zeta_m\) be the corresponding holomorphic coordinates on \(\overline{M}\). Then \[\mathcal{R}_{M^\circ,x} \cong \mathbb{C}\{z_1,\ldots,z_m,\zeta_1,\ldots,\zeta_m\}.\] A \(\mathbb{C}\)-algebra homomorphism \[\phi:\mathcal{R}_{M^\circ,x}\longrightarrow B_1\] whose residue is \(x\) is determined by \[\phi(z_i)=a_i t+b_i\bar t, \qquad \phi(\zeta_i)=c_i t+d_i\bar t.\] Restricting to the diagonal real analytic structure amounts to recording the two independent tangent directions along the two factors \(M\) and \(\overline{M}\). Hence the first order part is precisely an element of \[T_{M,x}\oplus T_{\overline{M},x} \cong \mathbb{C}T_xM^\circ.\] The statement for \(B_n\) is the same construction with higher order coefficients retained up to total degree \(n\). ◻

Proof. Take a sufficiently fine Stein open cover \(\{U_i\}\) of \(X\). A first order deformation over \(B_1\) is obtained by gluing the trivial thickenings \[U_i\times\operatorname{Spec}B_1\] by transition functions of the form \[z_i = f_{ij}(z_j) +t\,v_{ij}(z_j) +\bar t\,w_{ij}(z_j),\] where \(f_{ij}\) are the original transition functions of \(X\), and \(v_{ij},w_{ij}\) are holomorphic vector fields on \(U_{ij}\). The cocycle condition modulo \((t,\bar t)^2\) says precisely that \[\{v_{ij}\}\in Z^1(\{U_i\},T_X), \qquad \{w_{ij}\}\in Z^1(\{U_i\},T_X).\] Changing the local trivializations modifies these cocycles by coboundaries. Thus the isomorphism class of the deformation is determined by \[\bigl([\{v_{ij}\}],[\{w_{ij}\}]\bigr) \in H^1(X,T_X)\oplus H^1(X,T_X).\] Conversely, any such pair of cocycles defines a first order deformation by the above gluing formula. This proves the classification. The final formulation with \[H^1(\overline{X},T_{\overline{X}})\] is obtained by applying complex conjugation to the \(\bar t\)-part. ◻

Proof. The standard deformation theory of coherent sheaves over a square-zero extension gives the classification by \[\mathrm{Ext}^1_X \bigl( \mathcal{F}, \mathcal{F}\otimes_{\mathbb{C}}I_{B_1} \bigr).\] For completeness, we recall the elementary construction. Choose an open cover \(\{U_i\}\) on which the deformation is locally trivial. Then the local trivial deformations glue by automorphisms of the form \[1+t\,a_{ij}+\bar t\,b_{ij},\] where \(a_{ij},b_{ij}\) are local endomorphisms of \(\mathcal{F}\). The gluing condition modulo \((t,\bar t)^2\) says that \[\{a_{ij}\},\{b_{ij}\}\] are \(1\)-cocycles with values in \(\mathcal{E}nd(\mathcal{F})\), or more generally represent classes in \[\mathrm{Ext}^1_X(\mathcal{F},\mathcal{F})\] when \(\mathcal{F}\) is not locally free. Changing the local trivializations changes these cocycles by coboundaries. Therefore the isomorphism class of the first order deformation is determined by the pair \[\bigl([\{a_{ij}\}],[\{b_{ij}\}]\bigr) \in \mathrm{Ext}^1_X(\mathcal{F},\mathcal{F}) \oplus \mathrm{Ext}^1_X(\mathcal{F},\mathcal{F}).\] Conversely, a pair of such extension classes defines the required gluing data, hence a first order real analytic deformation. ◻

Proof. The statement is the standard Atiyah–Kodaira–Spencer obstruction theory for a sheaf on a varying space. We recall the construction.

The square-zero deformation \(X_I\) is classified by the extension \[0 \longrightarrow I\otimes_{\mathbb{C}}\mathcal{O}_{X_0} \longrightarrow \Omega^1_{X_I}\big|_{X_0} \longrightarrow \Omega^1_{X_0} \longrightarrow 0,\] whose class is \(\kappa(X_I)\). The Atiyah class of \(\mathcal{G}\) is the extension class of the first jet sequence \[0 \longrightarrow \mathcal{G}\otimes\Omega^1_{X_0} \longrightarrow P^1(\mathcal{G}) \longrightarrow \mathcal{G} \longrightarrow 0.\] Splicing these two extensions gives the Yoneda product \[\operatorname{At}(\mathcal{G})\cup\kappa(X_I) \in \mathrm{Ext}^2_{X_0} \bigl( \mathcal{G}, \mathcal{G}\otimes_{\mathbb{C}}I \bigr).\] This product measures precisely the failure of the transition functions of \(\mathcal{G}\) to be lifted compatibly to the deformed structure sheaf \(\mathcal{O}_{X_I}\). Hence it is the obstruction to the existence of a lift.

If the obstruction vanishes, choices of lifted transition data differ by \(1\)-cocycles with values in \[\mathcal{H}om(\mathcal{G},\mathcal{G}\otimes_{\mathbb{C}}I),\] or, for an arbitrary coherent sheaf, by classes in \[\mathrm{Ext}^1_{X_0} \bigl( \mathcal{G}, \mathcal{G}\otimes_{\mathbb{C}}I \bigr).\] Thus the set of lifts is a torsor under this Ext-group. Automorphisms of a fixed lift are similarly given by \(0\)-cocycles, namely by \[\mathrm{Ext}^0_{X_0} \bigl( \mathcal{G}, \mathcal{G}\otimes_{\mathbb{C}}I \bigr).\] ◻

Proof. This is Theorem 26 applied to the square-zero ideal \[I_{B_1}=\mathbb{C}t\oplus\mathbb{C}\bar t.\] The obstruction group splits according to this decomposition: \[\mathrm{Ext}^2_{X_0} \bigl( \mathcal{G}, \mathcal{G}\otimes_{\mathbb{C}}I_{B_1} \bigr) \cong \mathrm{Ext}^2_{X_0}(\mathcal{G},\mathcal{G}) \oplus \mathrm{Ext}^2_{X_0}(\mathcal{G},\mathcal{G}).\] The same splitting holds for the torsor of lifts, giving the Ext\(^1\) statement. ◻