Proof of Lemma 2.5
Proof
Since \(\{g_{ij}\}\) and \(\{{{\widetilde{g}}}_{ij}\}\) are two systems of transition functions of E with respect to \({\mathcal {U}}\), there exists a Čech 0-cochain
$$\begin{aligned} h:=\bigoplus \limits _{i_1}h_{i_1}\in \bigoplus \limits _{i_1}\Gamma _\textrm{hol} (U_{i_1},GL(M,{\mathbb {C}})), \end{aligned}$$
(145)
such that \({{\widetilde{g}}}_{ij}=h_i^{-1}g_{ij}h_j\) for \(i,j\in I\). Let \({{\widehat{f}}}_k(E,g)\), \({{\widehat{f}}}_k(E,{{\widetilde{g}}})\) be the Čech k-cocycles associated with \(f_k(E,g)\), \(f_k(E,g)\), respectively, as follows.
$$\begin{aligned}{} & {} {{\widehat{f}}}_k(E,g)=\bigoplus \limits _{i_1<\cdots<i_{k+1}}t_{i_1\cdots i_{k+1}}\in \bigoplus \limits _{i_1<\cdots <i_{k+1}}\Gamma (U_{i_1\cdots i_{k+1}},\Omega ^k), \end{aligned}$$
(146)
$$\begin{aligned}{} & {} {{\widehat{f}}}_k(E,\widetilde{g})=\bigoplus \limits _{i_1<\cdots<i_{k+1}}{{\widetilde{t}}}_{i_1\cdots i_{k+1}}\in \bigoplus \limits _{i_1<\cdots <i_{k+1}}\Gamma (U_{i_1\cdots i_{k+1}},\Omega ^k). \end{aligned}$$
(147)
Here \(t_{i_1\cdots i_{k+1}}\) and \({{\widetilde{t}}}_{i_1\cdots i_{k+1}}\) are defined by (24) with respect to g and \({{\widetilde{g}}}\), respectively.
To prove Lemma 2.5, it suffices to prove that there is a Čech \((k-1)\)-cochain
$$\begin{aligned} h_{k-1}(E,g,\widetilde{g})=\bigoplus \limits _{j_1<\cdots<j_{k}}s_{j_1\cdots j_{k}}\in \bigoplus \limits _{j_1<\cdots <j_{k}}\Gamma (U_{j_1\cdots j_{k}},\Omega ^k), \end{aligned}$$
(148)
such that for any \(i_1,\ldots ,i_{k+1}\in I\), we have
$$\begin{aligned} {{\widetilde{t}}}_{i_1\cdots i_{k+1}}- t_{i_1\cdots i_{k+1}}=\sum _{j=1}^{k+1}(-1)^{j-1}s_{i_1\cdots {{\widehat{i}}}_j\cdots i_{k+1}}\big |_{U_{i_1\cdots i_{k+1}}}. \end{aligned}$$
(149)
Notice that
$$\begin{aligned} {{\widetilde{g}}}_{\alpha \beta }^{-1}d\widetilde{g}_{\alpha \beta }=h_{\beta }^{-1}\big (g_{\alpha \beta }^{-1}dg_{\alpha \beta }+g_{\alpha \beta }^{-1}h_{\alpha }dh_{\alpha }^{-1}g_{\alpha \beta }+(-1)\cdot h_{\beta }dh_{\beta }^{-1}\big )h_{\beta }. \end{aligned}$$
(150)
Applying (150) for \(\beta =i_{\sigma (k+1)}\) and \(\alpha =i_{\sigma (1)},i_{\sigma (2)},i_{\sigma (3)},\ldots ,i_{\sigma (k)}\), we get
$$\begin{aligned} \begin{aligned} {{\widetilde{t}}}_{i_1\cdots i_{k+1}}&=\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\\&\cdot tr\big ({{\widetilde{g}}}^{-1}_{i_{\sigma (1)}i_{\sigma (k+1)}}d{{\widetilde{g}}}_{i_{\sigma (1)}i_{\sigma (k+1)}}{{\widetilde{g}}}^{-1}_{i_{\sigma (2)}i_{\sigma (k+1)}}d{{\widetilde{g}}}_{i_{\sigma (2)}i_{\sigma (k+1)}}{{\widetilde{g}}}^{-1}_{i_{\sigma (3)}i_{\sigma (k+1)}}\\&\qquad d{{\widetilde{g}}}_{i_{\sigma (3)}i_{\sigma (k+1)}}\cdots {{\widetilde{g}}}^{-1}_{i_{\sigma (k)}i_{\sigma (k+1)}}d{{\widetilde{g}}}_{i_{\sigma (k)}i_{\sigma (k+1)}}\big )\\&=\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\cdot tr\big \{\prod _{l=1}^kh_{i_{\sigma (k+1)}}^{-1}\big (g_{i_{\sigma (l)} i_{\sigma (k+1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (k+1)}}\\&\quad +g_{i_{\sigma (l)} i_{\sigma (k+1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (k+1)}}+(-1)\cdot h_{i_{\sigma (k+1)}}dh_{i_{\sigma (k+1)}}^{-1}\big )h_{i_{\sigma (k+1)}}\big \}\\&=\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\cdot tr\big \{\prod _{l=1}^k\big (g_{i_{\sigma (l)} i_{\sigma (k+1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (k+1)}}\\&\quad +g_{i_{\sigma (l)}i_{\sigma (k+1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (k+1)}}+(-1)\cdot h_{i_{\sigma (k+1)}}dh_{i_{\sigma (k+1)}}^{-1}\big )\big \}. \end{aligned} \end{aligned}$$
(151)
In order to compute the monomials appearing in the expansion of (151) by distribution law, we introduce the following notation. For integer \(t\ge 0\) and integers \(a_1,\ldots ,a_t\), we define the counting function \(\delta _{a_1\cdots a_t}\) from \(\mathbb Z\) to the set \(\{0,1\}\) as follows.
$$\begin{aligned} \delta _{a_1\cdots a_t}(l)=\left\{ \begin{array}{lll} 1&{}\quad \textrm{if}&{}\quad l\in \{a_1,\ldots ,a_t\};\\ 0&{}\quad \textrm{if}&{}\quad l\notin \{a_1,\ldots ,a_t\}. \end{array}\right. \end{aligned}$$
(152)
Notice that when \(t=0\), \(\delta _{a_1\cdots a_t}\equiv 0\); we denote it by \(\delta _{\emptyset }\).
Fix a permutation \(\sigma \in S_{k+1}\). For integers j, n such that \(j\ge 0\), \(n\ge 0\) and \(j+n\le k\), and distinct integers \(u_1,\ldots ,u_j,v_1,\ldots ,v_n\in \{1,\ldots ,k\}\) (by convention if \(j=0\), there is no u; if \(n=0\), there is no v), we define
$$\begin{aligned} \begin{aligned}&\Delta ^{\sigma }_{u_1\cdots u_j,v_{1}\cdots v_{n}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}):=tr\big \{\prod _{l=1}^k\big (g_{i_{\sigma (l)} i_{\sigma (k+1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (k+1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)-\delta _{v_1\cdots v_n}(l)}\\&\quad \cdot \big (g_{i_{\sigma (l)}i_{\sigma (k+1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (k+1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\cdot \big (-h_{i_{\sigma (k+1)}}dh_{i_{\sigma (k+1)}}^{-1}\big )^{\delta _{v_1\cdots v_n}(l)}\big \}. \end{aligned} \end{aligned}$$
(153)
Note that here we use the convention that
$$\begin{aligned}{} & {} \big (g_{i_{\sigma (l)} i_{\sigma (k+1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (k+1)}}\big )^{0}= \big (g_{i_{\sigma (l)}i_{\sigma (k+1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (k+1)}}\big )^{0}\nonumber \\{} & {} \quad =\big (-h_{i_{\sigma (k+1)}}dh_{i_{\sigma (k+1)}}^{-1}\big )^{0}=I_{M\times M}, \end{aligned}$$
(154)
where \(I_{M\times M}\) is the identity matrix of rank M. We also denote \(\Delta ^{\sigma }_{u_1\cdots u_j,v_{1}\cdots v_{n}}(\widetilde{t}_{i_1\cdots i_{k+1}})\) by \(\Delta ^{\sigma }_{\emptyset ,v_{1}\cdots v_{n}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\) when \(j=0\) and \(n\ge 1\); denote \(\Delta ^{\sigma }_{u_1\cdots u_j,v_{1}\cdots v_{n}}(\widetilde{t}_{i_1\cdots i_{k+1}})\) by \(\Delta ^{\sigma }_{u_{1}\cdots u_{j},\emptyset }({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\) when \(n=0\) and \(j\ge 1\); denote \(\Delta ^{\sigma }_{u_1\cdots u_j,v_{1}\cdots v_{n}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\) by \(\Delta ^{\sigma }_{\emptyset ,\emptyset }({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\) when \(j=n=0\).
Applying distribution law to (151), we have
$$\begin{aligned} \begin{aligned} {{\widetilde{t}}}_{i_1\cdots i_{k+1}}&=\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!} \sum _{\begin{array}{c} n\ge 2,\,0\le j\le k-n\,; \\ 1\le u_1<u_2<\cdots<u_j\le k\,;\\ 1\le v_1<v_2<\cdots<v_n\le k\,;\\ u_1,\ldots ,u_j,v_1,\ldots ,v_n\, {\text {are distinct}} \end{array}}\Delta ^{\sigma }_{u_1\cdots u_j,v_{1}\cdots v_{n}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!} \sum _{\begin{array}{c} n=1,\,1\le j\le k-1\,; \\ 1\le u_1<\cdots<u_j\le k\,;\\ 1\le v_1\le k\,;\\ u_1,\ldots ,u_j,v_1\, {\text {are distinct}} \end{array}}\Delta ^{\sigma }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!} \sum _{\begin{array}{c} n=1,\, j=0\,; \\ 1\le v_1\le k\,; \end{array}}\Delta ^{\sigma }_{\emptyset ,v_{1}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!} \sum _{\begin{array}{c} n=0,\,1\le j\le k\,; \\ 1\le u_1<u_2<\cdots <u_j\le k\,; \end{array}}\Delta ^{\sigma }_{u_1\cdots u_j,\emptyset }({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad + \sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\cdot \Delta ^{\sigma }_{\emptyset ,\emptyset }({{\widetilde{t}}}_{i_1\cdots i_{k+1}}).\\&=:I_1+I_2+I_3+I_4+I_5. \end{aligned} \end{aligned}$$
(155)
We first compute \(I_1\) as follows.
Claim I. For integers j, n such that \(n\ge 2\) and \(0\le j\le k-n\), and distinct integers \(u_1,\ldots ,u_j,v_1,\ldots ,v_n\in \{1,\ldots ,k\}\), the following identity holds,
$$\begin{aligned} \sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\cdot \Delta ^{\sigma }_{u_1\cdots u_j,v_{1}\cdots v_{n}}(\widetilde{t}_{i_1\cdots i_{k+1}})=0. \end{aligned}$$
(156)
Therefore, \(I_1\equiv 0\).
Proof of Claim I. The proof is similar to the proof of Claim I of Lemma 2.4. Denote by \(\tau \) the permutation \((v_1,v_2)\in S_{k+1}\). Notice that
$$\begin{aligned} \begin{aligned}&\Delta ^{\sigma \circ \tau }_{u_1\cdots u_j,v_{1}\cdots v_{n}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})=tr\big \{\prod _{l=1}^k\big (g_{i_{(\sigma \circ \tau )(l)} i_{(\sigma \circ \tau )(k+1)}}^{-1}dg_{i_{(\sigma \circ \tau )(l)} i_{(\sigma \circ \tau )(k+1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)-\delta _{v_1\cdots v_n}(l)}\\&\qquad \cdot \big (g_{i_{(\sigma \circ \tau )(l)}i_{(\sigma \circ \tau )(k+1)}}^{-1}h_{i_{(\sigma \circ \tau )(l)}}dh_{i_{(\sigma \circ \tau )(l)}}^{-1}g_{i_{(\sigma \circ \tau )(l)}i_{(\sigma \circ \tau )(k+1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (-h_{i_{(\sigma \circ \tau )(k+1)}}dh_{i_{(\sigma \circ \tau )(k+1)}}^{-1}\big )^{\delta _{v_1\cdots v_n}(l)}\big \}\\&\quad =tr\big \{\prod _{l=1}^k\big (g_{i_{\sigma (l)} i_{\sigma (k+1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (k+1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)-\delta _{v_1\cdots v_n}(l)}\\&\qquad \cdot \big (g_{i_{\sigma (l)}i_{\sigma (k+1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (k+1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (-h_{i_{\sigma (k+1)}}dh_{i_{\sigma (k+1)}}^{-1}\big )^{\delta _{v_1\cdots v_n}(l)}\big \}=\Delta ^{\sigma }_{u_1\cdots u_j,v_1v_2\cdots v_n}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}). \end{aligned} \end{aligned}$$
(157)
Hence,
$$\begin{aligned} \begin{aligned}&\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\cdot \Delta ^{\sigma }_{u_1\cdots u_j,v_1v_2\cdots v_n}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad =\sum _{\sigma \circ \tau \in S_{k+1}}\frac{sgn(\sigma \circ \tau )}{(k+1)!}\cdot \Delta ^{\sigma \circ \tau }_{u_1\cdots u_j,v_{1}\cdots v_{n}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad =sgn(\tau )\cdot \sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\cdot \Delta ^{\sigma \circ \tau }_{u_1\cdots u_j,v_{1}\cdots v_{n}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad =-\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\cdot \Delta ^{\sigma }_{u_1\cdots u_j,v_1v_2\cdots v_n}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}). \end{aligned} \end{aligned}$$
(158)
We complete the proof of Claim I. \(\square \)
In order to compute \(I_4\), we introduce the following notation.
Fix \(\sigma \in S_{k+1}\). For integers j, l such that \(j\ge 1\) and \(0\le l\le k-j\), and distinct integers \(u_1,\ldots ,u_j,x_1,\ldots ,x_l\in \{1,\ldots ,k\}\) (by convention when \(l=0\), there is no x), we define
$$\begin{aligned} \begin{aligned}&\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})=tr\big \{\prod _{l=1}^k\big (g_{i_{\sigma (l)} i_{\sigma (u_1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (u_1)}}\big )^{1-\delta _{x_1\cdots x_l}(l)-\delta _{u_1\cdots u_j}(l)}\\&\quad \cdot \big (-g_{i_{\sigma (u_1)} i_{\sigma (k+1)}}dg_{i_{\sigma (u_1)} i_{\sigma (k+1)}}^{-1}\big )^{\delta _{x_1\cdots x_l}(l)}\cdot \big (g_{i_{\sigma (l)}i_{\sigma (u_1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (u_1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}. \end{aligned} \end{aligned}$$
(159)
We also denote \(\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\) by \(\Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}(\widetilde{t}_{i_1\cdots i_{k+1}})\) when \(l=0\).
Claim II.
$$\begin{aligned} \begin{aligned} I_4&=\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k\,; \\ 1\le u_1<u_2<\cdots<u_j\le k; \end{array}}\Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<u_2<\cdots <u_j\le k;\\ 1\le x_1\le k,\, x_1\notin \{u_1,\ldots ,u_j\} \end{array}}\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}).\\ \end{aligned} \end{aligned}$$
(160)
Proof of Claim II. Recall the following identities
$$\begin{aligned} g_{\gamma \beta }^{-1}h_{\gamma }^{-1}dh_{\gamma }g_{\gamma \beta }=g_{\alpha \beta }^{-1}g_{\gamma \alpha }^{-1}h_{\gamma }^{-1}dh_{\gamma }g_{\gamma \alpha }g_{\alpha \beta }\,\,\,\,\,\textrm{and}\,\,\,g_{\delta \beta }^{-1}dg_{\delta \beta }=g_{\alpha \beta }^{-1}g_{\delta \alpha }^{-1}dg_{\delta \alpha }g_{\alpha \beta }+g_{\alpha \beta }^{-1}dg_{\alpha \beta }.\nonumber \\ \end{aligned}$$
(161)
Applying (161), we have
$$\begin{aligned} \begin{aligned}&\Delta ^{\sigma }_{u_1\cdots u_j,\emptyset }({{\widetilde{t}}}_{i_1\cdots i_{k+1}})=tr\big \{\prod _{l=1}^k\big (g_{i_{\sigma (l)} i_{\sigma (k+1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (k+1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (g_{i_{\sigma (l)}i_{\sigma (k+1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (k+1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}\\&\quad =tr\big \{\prod _{l=1}^k\big (g_{i_{\sigma (u_1)}i_{\sigma (k+1)}}^{-1}g_{i_{\sigma (l)} i_{\sigma (u_1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (u_1)}}g_{i_{\sigma (u_1)}i_{\sigma (k+1)}}\\&\qquad +g_{i_{\sigma (u_1)} i_{\sigma (k+1)}}^{-1}dg_{i_{\sigma (u_1)} i_{\sigma (k+1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (g_{i_{\sigma (u_1)}i_{\sigma (k+1)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (u_1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (u_1)}}g_{i_{\sigma (u_1)}i_{\sigma (k+1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}\\&\quad =tr\big \{\prod _{l=1}^k\big (g_{i_{\sigma (l)} i_{\sigma (u_1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (u_1)}}+(-1)\cdot g_{i_{\sigma (u_1)} i_{\sigma (k+1)}}dg_{i_{\sigma (u_1)} i_{\sigma (k+1)}}^{-1}\big )^{1-\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (g_{i_{\sigma (l)}i_{\sigma (u_1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (u_1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}\\&\quad =\sum _{\begin{array}{c} 0\le l\le k-j\,; \\ 1\le x_1<\cdots <x_l\le k\,;\\ u_1,\ldots ,u_j,x_1,\ldots ,x_l\, {\text {are distinct}} \end{array}}\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}). \end{aligned} \end{aligned}$$
(162)
Then,
$$\begin{aligned} \begin{aligned} I_4&=\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\sum _{\begin{array}{c} j\ge 1; \\ 1\le u_1<u_2<\cdots<u_j\le k;\, \end{array}}\sum _{\begin{array}{c} 0\le l\le k-j\,; \\ 1\le x_1<\cdots<x_l\le k\,;\\ u_1,\ldots ,u_j,x_1,\ldots ,x_l\, {\text {are distinct}} \end{array}}\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&=\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k\,; \\ 1\le u_1<u_2<\cdots<u_j\le k\, \end{array}}\Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<u_2<\cdots<u_j\le k;\\ 1\le x_1\le k,\,x_1\notin \{u_1,\ldots ,u_j\} \end{array}}\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k-2\,; \\ 1\le u_1<u_2<\cdots<u_j\le k\, \end{array}}\sum _{\begin{array}{c} 2\le l\le k-j\,; \\ 1\le x_1<\cdots <x_l\le k\,;\\ u_1,\ldots ,u_j,x_1,\ldots ,x_l\, {\text {are distinct}} \end{array}}\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&=:J_1+J_2+J_3. \end{aligned} \end{aligned}$$
(163)
Interchanging the order of summation, we have that
$$\begin{aligned} \begin{aligned} J_3&=\sum _{\begin{array}{c} 1\le j\le k-2\,; \\ 1\le u_1<u_2<\cdots<u_j\le k\, \end{array}}\sum _{\begin{array}{c} 2\le l\le k-j\,; \\ 1\le x_1<\cdots <x_l\le k\,;\\ u_1,\ldots ,u_j,x_1,\ldots ,x_l\, {\text {are distinct}} \end{array}}\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\ \end{aligned} \end{aligned}$$
(164)
Hence, it suffices to prove that for integers j, l such that \(1\le j\), \(2\le l\le k-j\), and distinct integers \(u_1,\ldots ,u_j,x_1,\ldots ,x_l\in \{1,\ldots ,k\}\), the following equality holds,
$$\begin{aligned} \sum _{\sigma \in S_{k+1}}sgn(\sigma )\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\equiv 0. \end{aligned}$$
(165)
We shall prove equality (165) in the same way as the proof of Claim I of Lemma 2.4. Denote by \(\tau \) the permutation \((x_1,x_2)\in S_{k+1}\). Then,
$$\begin{aligned} \begin{aligned}&\Delta ^{\sigma \circ \tau }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})=tr\big \{\prod _{l=1}^k\big (g_{i_{(\sigma \circ \tau )(l)} i_{(\sigma \circ \tau )(u_1)}}^{-1}dg_{i_{(\sigma \circ \tau )(l)} i_{(\sigma \circ \tau )(u_1)}}\big )^{1-\delta _{x_1\cdots x_{l}}(l)-\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (-g_{i_{(\sigma \circ \tau )(u_1)} i_{(\sigma \circ \tau )(k+1)}}dg_{i_{(\sigma \circ \tau )(u_1)} i_{(\sigma \circ \tau )(k+1)}}^{-1}\big )^{\delta _{x_1\cdots x_{l}}(l)}\cdot \\&\qquad \cdot \big (g_{i_{(\sigma \circ \tau )(l)}i_{(\sigma \circ \tau )(u_1)}}^{-1}h_{i_{(\sigma \circ \tau )(l)}}dh_{i_{(\sigma \circ \tau )(l)}}^{-1}g_{i_{(\sigma \circ \tau )(l)}i_{(\sigma \circ \tau )(u_1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}\\&\quad =tr\big \{\prod _{l=1}^k\big (g_{i_{\sigma (l)} i_{\sigma (u_1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (u_1)}}\big )^{1-\delta _{x_1\cdots x_l}(l)-\delta _{u_1\cdots u_j}(l)}\cdot \\&\qquad \cdot \big (-g_{i_{\sigma (u_1)} i_{\sigma (k+1)}}dg_{i_{\sigma (u_1)} i_{\sigma (k+1)}}^{-1}\big )^{\delta _{x_1\cdots x_l}(l)}\big (g_{i_{\sigma (l)}i_{\sigma (u_1)}}^{-1}h_{i_{\sigma (l)}}dh_{i_{\sigma (l)}}^{-1}g_{i_{\sigma (l)}i_{\sigma (u_1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}\\&\quad =\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}). \end{aligned} \end{aligned}$$
(166)
Therefore, we have that
$$\begin{aligned} \begin{aligned} \sum _{\sigma \in S_{k+1}}&sgn(\sigma )\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})=\sum _{\sigma \circ \tau \in S_{k+1}}sgn(\sigma \circ \tau )\cdot \Delta ^{\sigma \circ \tau }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&=sgn(\tau )\cdot \sum _{\sigma \in S_{k+1}}sgn(\sigma )\cdot \Delta ^{\sigma \circ \tau }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&=-\sum _{\sigma \in S_{k+1}}sgn(\sigma )\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1\cdots x_{l})}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}).\\ \end{aligned} \end{aligned}$$
(167)
We complete the proof of Claim II. \(\square \)
We now construct a Čech \((k-1)\)-cochain s as follows. For \(i_1,i_2,\ldots ,i_{k+1}\in I\) and \(1\le \alpha \le k+1\), define
$$\begin{aligned}{} & {} s_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+1};\,i_1\cdots i_{k+1}}\nonumber \\{} & {} \quad :=\frac{(-1)^{\alpha -1}}{(k+1)!} \sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<\cdots<u_j\le k\,;\\ 1\le v_1\le k\,;\\ u_1,\ldots ,u_j,v_1\, {\text {are distinct}} \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (v_1)=\alpha \end{array}}sgn(\sigma )\cdot \Delta ^{\sigma }_{u_1\cdots u_j,v_{1}} ({{\widetilde{t}}}_{i_1\cdots i_{k+1}}) \nonumber \\{} & {} \qquad +\frac{(-1)^{\alpha -1}}{(k+1)!} \sum _{\begin{array}{c} 1\le v_1\le k \end{array}}\,\,\,\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (v_1)=\alpha \end{array}} {sgn(\sigma )}\cdot \Delta ^{\sigma }_{\emptyset ,v_{1}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}) \nonumber \\{} & {} \qquad +\frac{(-1)^{\alpha -1}}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k\,; \\ 1\le u_1<u_2<\cdots<u_j\le k; \end{array}} \sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (k+1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\, \emptyset )}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}) \nonumber \\{} & {} \qquad +\frac{(-1)^{\alpha -1}}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<u_2<\cdots <u_j\le k; \nonumber \\ 1\le x_1\le k,\, x_1\notin \{u_1,\ldots ,u_j\} \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (x_1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}).\\ \end{aligned}$$
(168)
Claim III. The above definition depends on \((i_1,\ldots ,\widehat{i_{\alpha }},\ldots ,i_{k+1})\) but not on \((i_1,\ldots ,i_{k+1})\); namely, if \((i_1,\ldots ,\widehat{i_{\alpha }},\ldots ,i_{k+1})=(j_1,\ldots ,\widehat{j_{\beta }},\ldots ,j_{k+1})\), \(s_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+1};\,i_1\cdots i_{k+1}}=s_{j_1\cdots \widehat{j_\beta }\cdots j_{k+1};\,j_1\cdots j_{k+1}}\).
Proof of Claim III. For any integers \(\alpha \), \(\beta \) and \(i_1,\ldots , i_{k+2}\) such that \(1\le \alpha <\beta \le k+2\) and \(i_1,i_2,\ldots ,i_{k+1}\in I\), we can define \(s_{i_1\cdots \widehat{i_\alpha }\cdots \widehat{i_\beta }\cdots i_{k+1};\,i_1\cdots \widehat{i_\beta }\cdots i_{k+2}}\) and \(s_{i_1\cdots \widehat{i_\alpha }\cdots \widehat{i_\beta }\cdots i_{k+1};\,i_1\cdots \widehat{i_{\alpha }}\cdots i_{k+2}}\) by (168). Firstly, we have
$$\begin{aligned} \begin{aligned}&s_{i_1\cdots \widehat{i_\alpha }\cdots \widehat{i_\beta }\cdots i_{k+1};\,i_1\cdots \widehat{i_\beta }\cdots i_{k+2}}\\&\quad =\frac{(-1)^{\alpha -1}}{(k+1)!} \sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<\cdots<u_j\le k\,;\\ 1\le v_1\le k\,;\\ u_1,\ldots ,u_j,v_1\, {\text {are distinct}} \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (v_1)=\alpha \end{array}}sgn(\sigma )\cdot \Delta ^{\sigma }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})\\&\qquad +\frac{(-1)^{\alpha -1}}{(k+1)!} \sum _{\begin{array}{c} j=0\,; \\ 1\le v_1\le k\,; \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (v_1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{\emptyset ,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})\\&\qquad +\frac{(-1)^{\alpha -1}}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k\,; \\ 1\le u_1<u_2<\cdots<u_j\le k; \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (k+1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})\\&\qquad +\frac{(-1)^{\alpha -1}}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<u_2<\cdots <u_j\le k;\\ 1\le x_1\le k,\, x_1\notin \{u_1,\ldots ,u_j\} \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (x_1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})\\&\quad =:\kappa ^{\alpha }_1+\kappa ^{\alpha }_2+\kappa ^{\alpha }_3+\kappa ^{\alpha }_4. \end{aligned} \end{aligned}$$
(169)
Notice that in the definition of \(s_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+1};\,i_1\cdots i_{k+1}}\), the number \(\alpha \) on the right hand side is referring to the position of the omitted index on the left hand side. Therefore, \(s_{i_1\cdots \widehat{i_\alpha }\cdots \widehat{i_\beta }\cdots i_{k+1};\,i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}\) takes the following form,
$$\begin{aligned}{} & {} s_{i_1\cdots \widehat{i_\alpha }\cdots \widehat{i_\beta }\cdots i_{k+1};\,i_1\cdots \widehat{i_{\alpha }}\cdots i_{k+2}} \nonumber \\{} & {} \quad =\frac{(-1)^{\beta -2}}{(k+1)!} \sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<\cdots<u_j\le k\,;\\ 1\le v_1\le k\,;\\ u_1,\ldots ,u_j,v_1\, {\text {are distinct}} \end{array}}\sum _{\begin{array}{c} \tau \in S_{k+1},\\ \tau (v_1)=\beta -1 \end{array}}sgn(\tau )\cdot \Delta ^{\tau }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}) \nonumber \\{} & {} \qquad +\frac{(-1)^{\beta -2}}{(k+1)!} \sum _{\begin{array}{c} j=0\,; \\ 1\le v_1\le k\,; \end{array}}\sum _{\begin{array}{c} \tau \in S_{k+1},\\ \tau (v_1)=\beta -1 \end{array}}{sgn(\tau )}\cdot \Delta ^{\tau }_{\emptyset ,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}})\nonumber \\{} & {} \qquad +\frac{(-1)^{\beta -2}}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k\,; \\ 1\le u_1<u_2<\cdots<u_j\le k; \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (k+1)=\beta -1 \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}) \nonumber \\{} & {} \qquad +\frac{(-1)^{\beta -2}}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<u_2<\cdots <u_j\le k;\\ 1\le x_1\le k,\, x_1\notin \{u_1,\ldots ,u_j\} \end{array}}\sum _{\begin{array}{c} \tau \in S_{k+1},\\ \tau (x_1)=\beta -1 \end{array}}{sgn(\tau )}\cdot \Delta ^{\tau }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}) \nonumber \\{} & {} \quad =: \kappa ^{\beta }_1+\kappa ^{\beta }_2+\kappa ^{\beta }_3+\kappa ^{\beta }_4. \end{aligned}$$
(170)
In order to prove Claim III, it suffices to prove that
$$\begin{aligned} s_{i_1\cdots \widehat{i_\alpha }\cdots \widehat{i_\beta }\cdots i_{k+1};\,i_1\cdots \widehat{i_\beta }\cdots i_{k+2}}=s_{i_1\cdots \widehat{i_\alpha }\cdots \widehat{i_\beta }\cdots i_{k+1};\,i_1\cdots \widehat{i_{\alpha }}\cdots i_{k+2}}. \end{aligned}$$
(171)
By Claim IV to be proved in the following, formula (171) holds. We complete the proof of Claim III. \(\square \)
Next we will prove Claim IV used above.
Claim IV.
$$\begin{aligned} \kappa ^{\alpha }_i=(-1)^{\alpha -\beta +1}\cdot \kappa ^{\beta }_i\,\,\,\,\,\textrm{for}\,\,\,\, i=1,2,3,4. \end{aligned}$$
(172)
Proof of Claim IV. We will prove Claim IV based on a case by case argument.
Case I (\(\kappa ^{\alpha }_1=(-1)^{\alpha -\beta +1}\cdot \kappa ^{\beta }_1\)). It suffices to prove that for fixed integer j such that \(1\le j\le k-1\), and distinct integers \(u_1,\ldots ,u_j,v_1\in \{1,\ldots ,k\}\), the following equality holds,
$$\begin{aligned}{} & {} \sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (v_1)=\alpha \end{array}}sgn(\sigma )\cdot \Delta ^{\sigma }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=(-1)^{\alpha -\beta +1}\nonumber \\{} & {} \quad \cdot \sum _{\begin{array}{c} \tau \in S_{k+1},\\ \tau (v_1)=\beta -1 \end{array}}sgn(\tau )\cdot \Delta ^{\tau }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}). \end{aligned}$$
(173)
In order to compare the terms on both sides, we lift each element of \(S_{k+1}\) to an element of \(S_{k+2}\) as follows. Denote by \(S_{k+1}^{\beta }\) the set consisting of all bijections from \(\{1,2,\ldots ,k+1\}\) to \(\{1,\ldots ,{{\widehat{\beta }}}, \cdots ,k+2\}\); denote by \(S_{k+1}^{\alpha }\) the set consisting of all bijections from \(\{1,2,\ldots ,k+1\}\) to \(\{1,\ldots ,{{\widehat{\alpha }}}, \cdots ,k+2\}\). Define a bijection
$$\begin{aligned} \begin{aligned} {{\widehat{\bullet }}}\quad :\quad S_{k+1}&\longrightarrow S_{k+1}^{\beta } \\ \sigma&\longmapsto {\widehat{\sigma }} \end{aligned} \end{aligned}$$
(174)
by
$$\begin{aligned} {{\widehat{\sigma }}}(l)=\left\{ \begin{array}{ccc} \sigma (l)&{} \qquad \textrm{if}&{}\qquad 1\le l\le k+1 \quad \textrm{and} \quad \sigma (l)< \beta ,\\ \sigma (l)+1&{}\qquad \textrm{if}&{}\qquad 1\le l\le k+1\quad \textrm{and}\quad \sigma (l)\ge \beta .\\ \end{array}\right. \end{aligned}$$
(175)
Moreover, to compare the signature, we define a bijection
$$\begin{aligned} \begin{aligned}{}[\bullet ]_{\beta }\quad :\quad S^{\beta }_{k+1}&\longrightarrow \big \{\eta \big |\eta \in S_{k+2},\eta (k+2)=\beta \big \} \\ {{\widehat{\sigma }}}&\longmapsto [{{\widehat{\sigma }}}]_{\beta } \end{aligned} \end{aligned}$$
(176)
by
$$\begin{aligned}{}[{{\widehat{\sigma }}}]_{\beta }(l)=\left\{ \begin{array}{ccc} {{\widehat{\sigma }}}(l)&{}\qquad \textrm{if}&{}\quad 1\le l\le k+1,\\ \beta &{}\qquad \textrm{if}&{} \qquad l=k+2;\\ \end{array}\right. \end{aligned}$$
(177)
then, the following equality holds,
$$\begin{aligned} sgn(\sigma )=(-1)^{k+2-\beta }\cdot sgn([{{\widehat{\sigma }}}]_{\beta }). \end{aligned}$$
(178)
Similarly, we can define a bijection
$$\begin{aligned} \begin{aligned} {{\overline{\bullet }}}\quad :\quad S_{k+1}&\longrightarrow S_{k+1}^{\alpha } \\ \tau&\longmapsto {\overline{\tau }} \end{aligned} \end{aligned}$$
(179)
by
$$\begin{aligned} {{\overline{\tau }}}(l)=\left\{ \begin{array}{ccc} \tau (l)&{}\qquad \textrm{if}&{}\qquad 1\le l\le k+1\quad \textrm{and}\quad \tau (l)< \alpha ,\\ \tau (l)+1&{}\qquad \textrm{if}&{}\qquad 1\le l\le k+1\quad \textrm{and}\quad \tau (l)\ge \alpha ;\\ \end{array}\right. \end{aligned}$$
(180)
define a bijection
$$\begin{aligned} \begin{aligned}{}[\bullet ]_{\alpha }\quad :\quad S^{\alpha }_{k+1}&\longrightarrow \big \{\eta \big |\eta \in S_{k+2},\eta (k+2)=\alpha \big \} \\ {{\overline{\tau }}}&\longmapsto [{{\overline{\tau }}}]_{\alpha } \end{aligned} \end{aligned}$$
(181)
by
$$\begin{aligned}{}[{{\overline{\tau }}}]_{\alpha }(l)=\left\{ \begin{array}{ccc} {{\overline{\tau }}}(l)&{}\qquad \textrm{if}&{}\qquad 1\le l\le k+1,\\ \alpha &{}\qquad \textrm{if}&{}\qquad l=k+2,\\ \end{array}\right. \end{aligned}$$
(182)
the following equality holds,
$$\begin{aligned} sgn(\tau )=(-1)^{k+2-\alpha }\cdot sgn([{{\overline{\tau }}}]_{\alpha }). \end{aligned}$$
(183)
Rewriting (153) under the above notation, we have
$$\begin{aligned} \begin{aligned} \Delta ^{\sigma }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})&=tr\big \{\prod _{l=1}^k\big (g_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(k+1)}}^{-1}dg_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(k+1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)-\delta _{v_1}(l)}\\&\quad \cdot \big (g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(k+1)}}^{-1}h_{i_{{{\widehat{\sigma }}}(l)}}dh_{i_{{{\widehat{\sigma }}}(l)}}^{-1}g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(k+1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\\&\quad \cdot \big (-h_{i_{{{\widehat{\sigma }}}(k+1)}}dh_{i_{{{\widehat{\sigma }}}(k+1)}}^{-1}\big )^{\delta _{v_1}(l)}\big \}, \end{aligned} \end{aligned}$$
(184)
and
$$\begin{aligned} \begin{aligned} \Delta ^{\tau }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}})&=tr\big \{\prod _{l=1}^k\big (g_{i_{{{\overline{\tau }}}(l)} i_{{{\overline{\tau }}}(k+1)}}^{-1}dg_{i_{{{\overline{\tau }}}(l)} i_{{{\overline{\tau }}}(k+1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)-\delta _{v_1}(l)}\\&\quad \cdot \big (g_{i_{{{\overline{\tau }}}(l)}i_{{{\overline{\tau }}}(k+1)}}^{-1}h_{i_{{{\overline{\tau }}}(l)}}dh_{i_{{{\overline{\tau }}}(l)}}^{-1}g_{i_{{{\overline{\tau }}}(l)}i_{{{\overline{\tau }}}(k+1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\\&\quad \cdot \big (-h_{i_{{{\overline{\tau }}}(k+1)}}dh_{i_{{{\overline{\tau }}}(k+1)}}^{-1}\big )^{\delta _{v_1}(l)}\big \}. \end{aligned} \end{aligned}$$
(185)
In order to prove equality (173), it suffices to construct a bijection
$$\begin{aligned} \begin{aligned} P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}:\{\sigma \in S_{k+1}\big |\sigma (v_1)=\alpha \}\longrightarrow \{\tau \in S_{k+1}\big |\tau (v_1)=\beta -1\}, \end{aligned}\nonumber \\ \end{aligned}$$
(186)
such that
$$\begin{aligned} \begin{aligned}&sgn(\sigma )\cdot \Delta ^{\sigma }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=\\&\qquad (-1)^{\alpha -\beta +1}\cdot sgn(P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}(\sigma ))\cdot \Delta ^{P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}(\sigma )}_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}). \end{aligned} \end{aligned}$$
(187)
For each \(\sigma \in S_{k+1}\) such that \(\sigma (v_1)=\alpha \), we define \(P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}({\sigma })\) as follows. Take \({{\widehat{\sigma }}}\) as in (175) and define \(\eta \in S_{k+1}^{\alpha }\) by
$$\begin{aligned} \eta (l)=\left\{ \begin{array}{ccc} {{\widehat{\sigma }}}(l)&{}\qquad \textrm{if}&{}\qquad 1\le l\le v_1-1,\\ \beta &{}\qquad \textrm{if}&{}\qquad l=v_1,\\ {{\widehat{\sigma }}}(l)&{}\qquad \textrm{if}&{} \qquad v_1+1\le l\le k+1;\\ \end{array}\right. \end{aligned}$$
(188)
let \(P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}({\sigma })\) be the inverse image of \(\eta \) under the bijection \({\overline{\bullet }}\) , that is, \(\overline{P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}({\sigma })}=\eta \). It is easy to verify that \(P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}({\sigma })\in \{\tau \in S_{k+1}\big |\tau (v_1)=\beta -1\}\).
Then, by (184) and (185), we have
$$\begin{aligned} \begin{aligned}&\Delta ^{P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}({\sigma })}_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}})=tr\big \{\prod _{l=1}^k\big (g_{i_{\eta (l)} i_{\eta (k+1)}}^{-1}dg_{i_{\eta (l)} i_{\eta (k+1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)-\delta _{v_1}(l)}\\&\qquad \cdot \big (g_{i_{\eta (l)}i_{\eta (k+1)}}^{-1}h_{i_{\eta (l)}}dh_{i_{\eta (l)}}^{-1}g_{i_{\eta (l)}i_{\eta (k+1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\cdot \big (-h_{i_{\eta (k+1)}}dh_{i_{\eta (k+1)}}^{-1}\big )^{\delta _{v_1}(l)}\big \}\\&\quad =tr\big \{\prod _{l=1}^k\big (g_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(k+1)}}^{-1}dg_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(k+1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)-\delta _{v_1}(l)}\cdot \big (g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(k+1)}}^{-1}h_{i_{{{\widehat{\sigma }}}(l)}}dh_{i_{{{\widehat{\sigma }}}(l)}}^{-1}g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(k+1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (-h_{i_{{{\widehat{\sigma }}}(k+1)}}dh_{i_{{{\widehat{\sigma }}}(k+1)}}^{-1}\big )^{\delta _{v_1}(l)}\big \}=\Delta ^{\sigma }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}}). \end{aligned} \end{aligned}$$
(189)
To complete the proof of Claim IV in this case, it suffices to prove the following equality,
$$\begin{aligned} sgn(P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}({\sigma }))=(-1)^{\alpha -\beta +1}\cdot sgn(\sigma ). \end{aligned}$$
(190)
Since
$$\begin{aligned}{} & {} sgn(P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}({\sigma }))=(-1)^{k+2-\alpha }\cdot sgn([\overline{P_{\alpha ,\beta ,u_1,\ldots ,u_j,v_1}({\sigma })}]_{\alpha })\nonumber \\{} & {} \quad =(-1)^{k+2-\alpha }\cdot sgn([\eta ]_{\alpha }) \end{aligned}$$
(191)
and
$$\begin{aligned} (-1)^{\alpha -\beta +1}\cdot sgn(\sigma )=(-1)^{\alpha -\beta +1}\cdot (-1)^{k+2-\beta }\cdot sgn([\sigma ]_{\beta }), \end{aligned}$$
(192)
it suffices to prove that
$$\begin{aligned} sgn([\eta ]_{\alpha })=-sgn([\sigma ]_{\beta }). \end{aligned}$$
(193)
Since \([\eta ]_{\alpha }=\iota \circ [\sigma ]_{\beta }\in S_{k+2}\) where \(\iota \in S_{k+2}\) is the permutation \((\alpha ,\beta )\), equality (193) holds.
Therefore, \(\kappa ^{\alpha }_1=(-1)^{\alpha -\beta +1}\cdot \kappa ^{\beta }_1\).
Case II (\(\kappa ^{\alpha }_2=(-1)^{\alpha -\beta +1}\cdot \kappa ^{\beta }_2\)). Similarly, it suffices to prove that for fixed integer \(v_1\) such that \(1\le v_1\le k\) the following equality holds:
$$\begin{aligned} \sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (v_1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{\emptyset ,v_{1}}(\widetilde{t}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=(-1)^{\alpha -\beta +1}\cdot \sum _{\begin{array}{c} \tau \in S_{k+1},\\ \tau (v_1)=\beta -1 \end{array}}{sgn(\tau )}\cdot \Delta ^{\tau }_{\emptyset ,v_{1}}(\widetilde{t}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}).\nonumber \\ \end{aligned}$$
(194)
Define \({{\widehat{\sigma }}}\), \([{{\widehat{\sigma }}}]_{\beta }\), \({{\overline{\tau }}}\) and \([{{\overline{\tau }}}]_{\alpha }\) in the same way as (175), (177), (180) and (182). Rewriting (184) under the above notation, we have
$$\begin{aligned} \begin{aligned} \Delta ^{\sigma }_{\emptyset ,v_{1}}&(\widetilde{t}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=tr\big \{\prod _{l=1}^k\big (g_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(k+1)}}^{-1}dg_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(k+1)}}\big )^{1-\delta _{v_1}(l)}\cdot \big (-h_{i_{{{\widehat{\sigma }}}(k+1)}}dh_{i_{{{\widehat{\sigma }}}(k+1)}}^{-1}\big )^{\delta _{v_1}(l)}\big \}, \end{aligned} \end{aligned}$$
(195)
and
$$\begin{aligned} \begin{aligned} \Delta ^{\tau }_{\emptyset ,v_{1}}&(\widetilde{t}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}})=tr\big \{\prod _{l=1}^k\big (g_{i_{{{\overline{\tau }}}(l)} i_{{{\overline{\tau }}}(k+1)}}^{-1}dg_{i_{{{\overline{\tau }}}(l)} i_{{{\overline{\tau }}}(k+1)}}\big )^{1-\delta _{v_1}(l)}\cdot \big (-h_{i_{{{\overline{\tau }}}(k+1)}}dh_{i_{{{\overline{\tau }}}(k+1)}}^{-1}\big )^{\delta _{v_1}(l)}\big \}. \end{aligned} \end{aligned}$$
(196)
Notice that equality (194) holds if there exists a bijection
$$\begin{aligned} \begin{aligned} P_{\alpha ,\beta ,v_1}:\{\sigma \in S_{k+1}\big |\sigma (v_1)=\alpha \}\rightarrow \{\tau \in S_{k+1}\big |\tau (v_1)=\beta -1\}, \end{aligned} \end{aligned}$$
(197)
such that
$$\begin{aligned} sgn(\sigma )\cdot \Delta ^{\sigma }_{\emptyset ,v_{1}}(\widetilde{t}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}}) \!=\! (-1)^{\alpha -\beta +1}\cdot sgn(P_{\alpha ,\beta ,v_1}(\sigma ))\cdot \Delta ^{P_{\alpha ,\beta ,v_1}(\sigma )}_{\emptyset ,v_{1}}(\widetilde{t}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}).\qquad \end{aligned}$$
(198)
For each \(\sigma \in S_{k+1}\) such that \(\sigma (v_1)=\alpha \), we define an element \(P_{\alpha ,\beta ,v_1}({\sigma })\in S_{k+1}\) as follows. Take \({{\widehat{\sigma }}}\) as in (175) and define \(\eta \in S_{k+1}^{\alpha }\) by
$$\begin{aligned} \eta (l)=\left\{ \begin{array}{ccc} {{\widehat{\sigma }}}(l)&{}\qquad \textrm{if}&{}\qquad 1\le l\le v_1-1,\\ \beta &{}\qquad \textrm{if}&{}\qquad l=v_1,\\ {{\widehat{\sigma }}}(l)&{}\qquad \textrm{if}&{}\qquad v_1+1\le l\le k+1;\\ \end{array}\right. \end{aligned}$$
(199)
let \(P_{\alpha ,\beta ,v_1}({\sigma })\) be the inverse image of \(\eta \) under the map \({\overline{\bullet }}\) , that is, \(\overline{P_{\alpha ,\beta ,v_1}({\sigma })}=\eta \). It is easy to verify that \(P_{\alpha ,\beta ,v_1}({\sigma })\in \{\tau \in S_{k+1}\big |\tau (v_1)=\beta -1\}\).
Then, by (184) and (185), we have that
$$\begin{aligned} \begin{aligned}&\Delta ^{P_{\alpha ,\beta ,v_1}({\sigma })}_{\emptyset ,v_{1}}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}})\\&\quad =tr\big \{\prod _{l=1}^k\big (g_{i_{\eta (l)} i_{\eta (k+1)}}^{-1}dg_{i_{\eta (l)} i_{\eta (k+1)}}\big )^{1-\delta _{v_1}(l)}\cdot \big ((-1)\cdot h_{i_{\eta (k+1)}}dh_{i_{\eta (k+1)}}^{-1}\big )^{\delta _{v_1}(l)}\big \}\\&\quad =tr\big \{\prod _{l=1}^k\big (g_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(k+1)}}^{-1}dg_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(k+1)}}\big )^{1-\delta _{v_1}(l)}\cdot \big ((-1)\cdot h_{i_{{{\widehat{\sigma }}}(k+1)}}dh_{i_{{{\widehat{\sigma }}}(k+1)}}^{-1}\big )^{\delta _{v_1}(l)}\big \}\\&\quad =\Delta ^{\sigma }_{\emptyset ,v_{1}}(\widetilde{t}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}}). \end{aligned} \end{aligned}$$
(200)
Notice that
$$\begin{aligned}{} & {} sgn(P_{\alpha ,\beta ,v_1}({\sigma }))=(-1)^{k+2-\alpha }\cdot sgn([\overline{P_{\alpha ,\beta ,v_1}({\sigma })}]_{\alpha })=(-1)^{k+2-\alpha }\cdot sgn([\eta ]_{\alpha }),\nonumber \\\end{aligned}$$
(201)
$$\begin{aligned}{} & {} (-1)^{\alpha -\beta +1}\cdot sgn(\sigma )=(-1)^{\alpha -\beta +1}\cdot (-1)^{k+2-\beta }\cdot sgn([{{\widehat{\sigma }}}]_{\beta }). \end{aligned}$$
(202)
Since \([\eta ]_{\alpha }=\iota \circ [{{\widehat{\sigma }}}]_{\beta }\in S_{k+2}\) where \(\iota \in S_{k+2}\) is the permutation \((\alpha ,\beta )\), \(sgn([\sigma ]_{\beta })=-sgn([\eta ]_{\alpha })\).
Then, \(\kappa ^{\alpha }_2=(-1)^{\alpha -\beta +1}\cdot \kappa ^{\beta }_2\).
Case III (\(\kappa ^{\alpha }_3=(-1)^{\alpha -\beta +1}\cdot \kappa ^{\beta }_3\)). Similarly we will show that for fixed integers \(j,u_1,\ldots ,u_j\) such that \(1\le j\le k\) and \(1\le u_1<\cdots <u_j\le k\), the following equality holds:
$$\begin{aligned}{} & {} \sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (k+1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=(-1)^{\alpha -\beta +1}\nonumber \\{} & {} \cdot \sum _{\begin{array}{c} \tau \in S_{k+1},\\ \tau (k+1)=\beta -1 \end{array}}{sgn(\tau )}\cdot \Delta ^{\tau }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}). \end{aligned}$$
(203)
Define \({{\widehat{\sigma }}}\), \([{{\widehat{\sigma }}}]_{\beta }\), \({{\overline{\tau }}}\) and \([{{\overline{\tau }}}]_{\alpha }\) in the same way as (175), (177), (180) and (182). Rewriting (159) under the new notation, we have
$$\begin{aligned} \begin{aligned}&\Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}(\widetilde{t}_{i_1\cdots {{\widehat{i}}}_{\beta }\cdots i_{k+1}})=tr\big \{\prod _{l=1}^k\big (g_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(u_1)}}^{-1}dg_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(u_1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)}\\&\quad \cdot \big (g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(u_1)}}^{-1}h_{i_{{{\widehat{\sigma }}}(l)}}dh_{i_{{{\widehat{\sigma }}}(l)}}^{-1}g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(u_1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}, \end{aligned} \end{aligned}$$
(204)
and
$$\begin{aligned} \begin{aligned}&\Delta ^{\tau }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots {{\widehat{i}}}_{\alpha }\cdots i_{k+1}})=tr\big \{\prod _{l=1}^k\big (g_{i_{{{\overline{\tau }}}(l)} i_{{{\overline{\tau }}}(u_1)}}^{-1}dg_{i_{{{\overline{\tau }}}(l)} i_{{{\overline{\tau }}}(u_1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)}\\&\quad \cdot \big (g_{i_{{{\overline{\tau }}}(l)}i_{{{\overline{\tau }}}(u_1)}}^{-1}h_{i_{{{\overline{\tau }}}(l)}}dh_{i_{{{\overline{\tau }}}(l)}}^{-1}g_{i_{{{\overline{\tau }}}(l)}i_{{{\overline{\tau }}}(u_1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}. \end{aligned} \end{aligned}$$
(205)
Notice that equality (203) holds if there exists a bijection
$$\begin{aligned} \begin{aligned} P_{\alpha ,\beta }:\{\sigma \in S_{k+1}\big |\sigma (k+1)=\alpha \}\rightarrow \{\tau \in S_{k+1}\big |\tau (k+1)=\beta -1\}, \end{aligned} \end{aligned}$$
(206)
such that
$$\begin{aligned} {sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=(-1)^{\alpha -\beta +1}\cdot {sgn(P_{\alpha ,\beta }(\sigma ))}\cdot \Delta ^{P_{\alpha ,\beta }(\sigma )}_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}).\nonumber \\ \end{aligned}$$
(207)
For each \(\sigma \in S_{k+1}\) such that \(\sigma (k+1)=\alpha \), we define an element \(P_{\alpha ,\beta }({\sigma })\in S_{k+1}\) as follows. Take \({{\widehat{\sigma }}}\) as in (175) and define \(\eta \in S_{k+1}^{\alpha }\) by
$$\begin{aligned} \eta (l)=\left\{ \begin{array}{ccc} {{\widehat{\sigma }}}(l)&{}\qquad \textrm{if}&{}\qquad 1\le l\le k,\\ \beta &{}\qquad \textrm{if}&{}\qquad l=k+1; \end{array}\right. \end{aligned}$$
(208)
let \(P_{\alpha ,\beta }({\sigma })\) be the inverse image of \(\eta \) under the map \({\overline{\bullet }}\) , that is, \(\overline{P_{\alpha ,\beta }({\sigma })}=\eta \). It is easy to verify that \(P_{\alpha ,\beta }({\sigma })\in \{\tau \in S_{k+1}\big |\tau (k+1)=\beta -1\}\).
Then, by (159), we have that
$$\begin{aligned} \begin{aligned}&\Delta ^{P_{\alpha ,\beta }({\sigma })}_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}})=tr\big \{\prod _{l=1}^k\big (g_{i_{\eta (l)} i_{\eta (u_1)}}^{-1}dg_{i_{\eta (l)} i_{\eta (u_1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (g_{i_{\eta (l)}i_{\eta (u_1)}}^{-1}h_{i_{\eta (l)}}dh_{i_{\eta (l)}}^{-1}g_{i_{\eta (l)}i_{\eta (u_1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}\\&\quad =tr\big \{\prod _{l=1}^k\big (g_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(u_1)}}^{-1}dg_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(u_1)}}\big )^{1-\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(u_1)}}^{-1}h_{i_{{{\widehat{\sigma }}}(l)}}dh_{i_{{{\widehat{\sigma }}}(l)}}^{-1}g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(u_1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}\\&\quad =\Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}(\widetilde{t}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}}). \end{aligned} \end{aligned}$$
(209)
Since
$$\begin{aligned} sgn(P_{\alpha ,\beta }({\sigma }))=(-1)^{k+2-\alpha }\cdot sgn([\overline{P_{\alpha ,\beta }({\sigma })}]_{\alpha })=(-1)^{k+2-\alpha }\cdot sgn([\eta ]_{\alpha }) \nonumber \\ \end{aligned}$$
(210)
and
$$\begin{aligned} (-1)^{\alpha -\beta +1}\cdot sgn(\sigma )=(-1)^{\alpha -\beta +1}\cdot (-1)^{k+2-\beta }\cdot sgn([{{\widehat{\sigma }}}]_{\beta }), \end{aligned}$$
(211)
it suffices to prove that
$$\begin{aligned} sgn([\eta ]_{\alpha })=-sgn([{{\widehat{\sigma }}}]_{\beta }). \end{aligned}$$
(212)
Noticing that \([\eta ]_{\alpha }=\iota \circ [{{\widehat{\sigma }}}]_{\beta }\in S_{k+2}\) where \(\iota \in S_{k+2}\) is the permutation \((\alpha ,\beta )\), (212) holds.
Therefore, \(\kappa ^{\alpha }_3=(-1)^{\alpha -\beta +1}\cdot \kappa ^{\beta }_3\).
Case IV (\(\kappa ^{\alpha }_4=(-1)^{\alpha -\beta +1}\cdot \kappa ^{\beta }_4\)). Finally we will prove that for fixed integer j such that \(1\le j\le k-1\), and distinct integers \(u_1,\ldots ,u_j,x_1\in \{1,\ldots ,k\}\), the following equality holds:
$$\begin{aligned} \begin{aligned}&\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (x_1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=(-1)^{\alpha -\beta +1}\\&\quad \cdot \sum _{\begin{array}{c} \tau \in S_{k+1},\\ \tau (x_1)=\beta -1 \end{array}}{sgn(\tau )}\cdot \Delta ^{\tau }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}). \end{aligned} \end{aligned}$$
(213)
It suffices to prove that there exists a bijection
$$\begin{aligned} \begin{aligned} P_{\alpha ,\beta ,u_1\cdots u_j,x_1}:\{\sigma \in S_{k+1}\big |\sigma (x_1)=\alpha \}\rightarrow \{\tau \in S_{k+1}\big |\tau (x_1)=\beta -1\}, \end{aligned} \end{aligned}$$
(214)
such that
$$\begin{aligned}&{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}(\widetilde{t}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=(-1)^{\alpha -\beta +1}\cdot {sgn(P_{\alpha ,\beta ,u_1\cdots u_j,x_1}({\sigma }))}\nonumber \\&\quad \cdot \Delta ^{P_{\alpha ,\beta ,u_1\cdots u_j,x_1}({\sigma })}_{(u_1\cdots u_j;\,x_1)}(\widetilde{t}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}). \end{aligned}$$
(215)
For each \(\sigma \in S_{k+1}\) such that \(\sigma (x_1)=\alpha \), we define an element \(P_{\alpha ,\beta ,u_1\cdots u_j,x_1}({\sigma })\in S_{k+1}\) as follows. Take \({{\widehat{\sigma }}}\) as in (175) and define \(\eta \in S_{k+1}^{\alpha }\) by
$$\begin{aligned} \eta (l)=\left\{ \begin{array}{ccc} {{\widehat{\sigma }}}(l)&{}\qquad \textrm{if}&{}\qquad 1\le l\le x_1-1,\\ \beta &{}\qquad \textrm{if}&{}\qquad l=x_1,\\ {{\widehat{\sigma }}}(l)&{}\qquad \textrm{if}&{}\qquad x_1+1\le l\le k+1;\\ \end{array}\right. \end{aligned}$$
(216)
let \(P_{\alpha ,\beta ,u_1\cdots u_j,x_1}\) be the inverse image of \(\eta \) under the map \({\overline{\bullet }}\) , that is, \(\overline{P_{\alpha ,\beta ,u_1\cdots u_j,x_1}}=\eta \). It is easy to verify that \(P_{\alpha ,\beta ,u_1\cdots u_j,x_1}\in \{\tau \in S_{k+1}\big |\tau (x_1)=\beta -1\}\).
Then by (159), we have5
$$\begin{aligned} \begin{aligned}&\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=tr\big \{\prod _{l=1}^k\big (g_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(u_1)}}^{-1}dg_{i_{{{\widehat{\sigma }}}(l)} i_{{{\widehat{\sigma }}}(u_1)}}\big )^{1-\delta _{x_1}(l)-\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (- g_{i_{{{\widehat{\sigma }}}(u_1)} i_{{{\widehat{\sigma }}}(k+1)}}dg_{i_{{{\widehat{\sigma }}}(u_1)} i_{{{\widehat{\sigma }}}(k+1)}}^{-1}\big )^{\delta _{x_1}(l)}\cdot \big (g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(u_1)}}^{-1}h_{i_{{{\widehat{\sigma }}}(l)}}dh_{i_{{{\widehat{\sigma }}}(l)}}^{-1}g_{i_{{{\widehat{\sigma }}}(l)}i_{{{\widehat{\sigma }}}(u_1)}}\big )^{-\delta _{u_1\cdots u_j}(l)}\big \}, \end{aligned} \end{aligned}$$
(217)
and
$$\begin{aligned} \begin{aligned}&\Delta ^{P_{\alpha ,\beta ,u_1\cdots u_j,x_1}(\sigma )}_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}})=tr\big \{\prod _{l=1}^k\big (g_{i_{\eta (l)} i_{\eta (u_1)}}^{-1}dg_{i_{\eta (l)} i_{\eta (u_1)}}\big )^{1-\delta _{x_1}(l)-\delta _{u_1\cdots u_j}(l)}\\&\qquad \cdot \big (-g_{i_{\eta (u_1)} i_{\eta (k+1)}}dg_{i_{\eta (u_1)} i_{\eta (k+1)}}^{-1}\big )^{\delta _{x_1}(l)}\cdot \big (g_{i_{\eta (l)}i_{\eta (u_1)}}^{-1}h_{i_{\eta (l)}}dh_{i_{\eta (l)}}^{-1}g_{i_{\eta (l)}i_{\eta (u_1)}}\big )^{\delta _{u_1\cdots u_j}(l)}\big \}. \end{aligned} \end{aligned}$$
(218)
Hence,
$$\begin{aligned} \begin{aligned} \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}(\widetilde{t}_{i_1\cdots \widehat{i_\beta }\cdots i_{k+2}})=\Delta ^{P_{\alpha ,\beta ,u_1\cdots u_j,x_1}(\sigma )}_{(u_1\cdots u_j;\,x_1)}(\widetilde{t}_{i_1\cdots \widehat{i_\alpha }\cdots i_{k+2}}). \end{aligned} \end{aligned}$$
(219)
Next we are going to prove that
$$\begin{aligned} sgn(P_{\alpha ,\beta ,u_1\cdots u_j,x_1}({\sigma }))=(-1)^{\alpha -\beta +1}\cdot sgn(\sigma ). \end{aligned}$$
(220)
Similarly, it suffices to prove that
$$\begin{aligned} sgn([\eta ]_{\alpha })=-sgn([\sigma ]_{\beta }). \end{aligned}$$
(221)
Since \([\eta ]_{\alpha }=\iota \circ [{{\widehat{\sigma }}}]_{\beta }\in S_{k+2}\) where \(\iota \in S_{k+2}\) is the permutation \((\alpha ,\beta )\), (221) holds.
Therefore, \(\kappa ^{\alpha }_4=(-1)^{\alpha -\beta +1}\cdot \kappa ^{\beta }_4\).
We complete the proof of Claim IV. \(\square \)
By Claim III, for integer \(k\ge 1\) and elements \(j_1,\ldots , j_k\in I\), we can define
$$\begin{aligned} s_{j_1\cdots j_k}:=s_{{\widehat{\beta }}j_1\cdots j_{k};\,\beta j_1\cdots j_{k}}, \end{aligned}$$
(222)
where \(\beta \) is any element in I. Noticing that only the transition functions \(g_{\gamma \delta }\) where \(\gamma ,\delta \in \{j_1,\ldots ,j_k\}\) appearing in (168), \(s_{j_1\cdots j_{k}}\in \Gamma (U_{j_1\cdots j_{k}},\Omega ^k)\). Thus, we can define a Čech \((k-1)\)-cochain \(h_{k-1}(E,g,{{\widetilde{g}}})\) by
$$\begin{aligned} h_{k-1}(E,g,\widetilde{g}):=\bigoplus \limits _{j_1<\cdots<j_{k}}s_{j_1\cdots j_{k}}\in \bigoplus \limits _{j_1<\cdots <j_{k}}\Gamma (U_{j_1\cdots j_{k}},\Omega ^k). \end{aligned}$$
(223)
Similarly, we can extend the components of \(h_{k-1}(E,g,\widetilde{g})\) to all k-tuples of elements in I. Then in the same way as Lemma 2.2, we can show that (168) and (222) are compatible with this extension.
Next, we prove that \({{\widehat{f}}}_k(E,{{\widetilde{g}}})\) is cohomologous to \({{\widehat{f}}}_k(E,g)\) by \(h_{k-1}(E,g,{{\widetilde{g}}})\).
Claim V. \(\partial h_{k-1}(E,g,{{\widetilde{g}}})=\widehat{f}_k(E,{{\widetilde{g}}})-{{\widehat{f}}}_k(E,g)\). That is, for any elements \(i_1,\ldots ,i_{k+1}\in I\),
$$\begin{aligned} {{\widetilde{t}}}_{i_1\cdots i_{k+1}}- t_{i_1\cdots i_{k+1}}=\sum _{j=1}^{k+1}(-1)^{j-1}s_{i_1\cdots {{\widehat{i}}}_j\cdots i_{k+1}}\big |_{U_{i_1\cdots i_{k+1}}}. \end{aligned}$$
(224)
Proof of Claim V. Recall that
$$\begin{aligned}&\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\cdot \Delta ^{\sigma }_{\emptyset ,\emptyset }({{\widetilde{t}}}_{i_1\cdots i_{k+1}})=\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\nonumber \\&\quad \cdot tr\big \{\prod _{l=1}^k\big (g_{i_{\sigma (l)} i_{\sigma (k+1)}}^{-1}dg_{i_{\sigma (l)} i_{\sigma (k+1)}}\big )\big \}=t_{t_{i_1\cdots i_{k+1}}}. \end{aligned}$$
(225)
Hence by Claims I and II, and formulas (155) and (225), we have that
$$\begin{aligned} \begin{aligned}&{{\widetilde{t}}}_{i_1\cdots i_{k+1}}-t_{i_1\cdots i_{k+1}}=\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!} \sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<\cdots<u_j\le k\,;\\ 1\le v_1\le k\,;\\ u_1,\ldots ,u_j,v_1\, {\text {are distinct}} \end{array}}\Delta ^{\sigma }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\qquad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!} \sum _{\begin{array}{c} j=0\,; \\ 1\le v_1\le k\,; \end{array}}\Delta ^{\sigma }_{\emptyset ,v_{1}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})+\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\\&\quad \sum _{\begin{array}{c} 1\le j\le k\,; \\ 1\le u_1<u_2<\cdots<u_j\le k\,; \end{array}}\Delta ^{\sigma }_{u_1\cdots u_j,\emptyset }({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\quad =\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!} \sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<\cdots<u_j\le k\,;\\ 1\le v_1\le k\,;\\ u_1,\ldots ,u_j,v_1\, {\text {are distinct}} \end{array}}\Delta ^{\sigma }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\qquad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!} \sum _{\begin{array}{c} j=0\,; \\ 1\le v_1\le k\,; \end{array}}\Delta ^{\sigma }_{\emptyset ,v_{1}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\qquad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k\,; \\ 1\le u_1<u_2<\cdots<u_j\le k \end{array}}\Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&\qquad +\sum _{\sigma \in S_{k+1}}\frac{sgn(\sigma )}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<u_2<\cdots <u_j\le k;\\ 1\le x_1\le k,\, x_1\notin \{u_1,\ldots ,u_j\} \end{array}}\Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}).\\ \end{aligned} \end{aligned}$$
(226)
Notice that for each fixed \(\xi \) such that \(1\le \xi \le k+1\), \(S_{k+1}\) is the disjoint union of the following sets,
$$\begin{aligned} \begin{aligned} \big \{\sigma \in S_{k+1}\big |\sigma (\xi )=\alpha \big \}\,\,\,\textrm{where}\,\,\alpha =1,\ldots ,k+1. \end{aligned} \end{aligned}$$
(227)
Then, by interchanging the order of summation in (225), we have
$$\begin{aligned} \begin{aligned} {{\widetilde{t}}}_{i_1\cdots i_{k+1}}-&t_{i_1\cdots i_{k+1}}=\sum _{\alpha =1}^{k+1}\,\,\,\,\frac{1}{(k+1)!} \sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<\cdots<u_j\le k\,;\\ 1\le v_1\le k\,;\\ u_1,\ldots ,u_j,v_1\, {\text {are distinct}} \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (v_1)=\alpha \end{array}}sgn(\sigma )\cdot \Delta ^{\sigma }_{u_1\cdots u_j,v_{1}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&+\sum _{\alpha =1}^{k+1}\qquad \frac{1}{(k+1)!}\sum _{\begin{array}{c} j=0\,; \\ 1\le v_1\le k\,; \end{array}} \sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (v_1)=\alpha \end{array}}sgn(\sigma )\cdot \Delta ^{\sigma }_{\emptyset ,v_{1}}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&+\sum _{\alpha =1}^{k+1}\qquad \frac{1}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k\,; \\ 1\le u_1<u_2<\cdots<u_j\le k; \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (k+1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,\emptyset )}({{\widetilde{t}}}_{i_1\cdots i_{k+1}})\\&+\sum _{\alpha =1}^{k+1}\qquad \frac{1}{(k+1)!}\sum _{\begin{array}{c} 1\le j\le k-1\,; \\ 1\le u_1<u_2<\cdots <u_j\le k;\\ 1\le x_1\le k,\, x_1\notin \{u_1,\ldots ,u_j\} \end{array}}\sum _{\begin{array}{c} \sigma \in S_{k+1},\\ \sigma (x_1)=\alpha \end{array}}{sgn(\sigma )}\cdot \Delta ^{\sigma }_{(u_1\cdots u_j;\,x_1)}({{\widetilde{t}}}_{i_1\cdots i_{k+1}}).\\ \end{aligned} \end{aligned}$$
(228)
Recalling (168), we conclude that
$$\begin{aligned} {{\widetilde{t}}}_{i_1\cdots i_{k+1}}- t_{i_1\cdots i_{k+1}}=\sum _{\alpha =1}^{k+1}(-1)^{\alpha -1}s_{i_1\cdots \widehat{i}_\alpha \cdots i_{k+1}}\big |_{U_{i_1\cdots i_{k+1}}}. \end{aligned}$$
(229)
We complete the proof of Claim V. \(\square \)
