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Proof.
Directly applying the definition of η α \eta_{\alpha} , then we have
A i α β ⋅ y β ⋅ d x i ∧ η α ^ \displaystyle A_{i\alpha\beta}\cdot y_{\beta}\cdot dx_{i}\wedge\eta_{\widehat{\alpha}}
(3.145)
= \displaystyle=
A i α β y β d x i ∧ d y α ^ + A i α β y β y γ ( A j , α + 1 , γ d y α + 2 − A j , α + 2 , γ d y α + 1 ) ∧ d x i ∧ d x j + O ~ ( r 3 ) . \displaystyle A_{i\alpha\beta}y_{\beta}dx_{i}\wedge dy_{\widehat{\alpha}}+A_{i\alpha\beta}y_{\beta}y_{\gamma}(A_{j,\alpha+1,\gamma}dy_{\alpha+2}-A_{j,\alpha+2,\gamma}dy_{\alpha+1})\wedge dx_{i}\wedge dx_{j}+\widetilde{O}(r^{3}).
By (3.127 ), we get
y α ( d η α + 1 ∧ η α + 2 − η α + 1 ∧ d η α + 2 ) \displaystyle y_{\alpha}(d{\eta_{\alpha+1}}\wedge{\eta_{\alpha+2}}-{\eta_{\alpha+1}}\wedge d{\eta_{\alpha+2}})
= \displaystyle=
y α ( A i , α + 1 , β d y β ∧ d x i ∧ d y α + 2 − A i , α + 2 , β d y β ∧ d x i ∧ d y α + 1 ) \displaystyle y_{\alpha}(A_{i,\alpha+1,\beta}dy_{\beta}\wedge dx_{i}\wedge dy_{\alpha+2}-A_{i,\alpha+2,\beta}dy_{\beta}\wedge dx_{i}\wedge dy_{\alpha+1})
+ \displaystyle+
y α y γ ( A i , α + 1 , β A j , α + 2 , γ − A i , α + 2 , β A j , α + 1 , γ ) d y β ∧ d x i ∧ d x j \displaystyle y_{\alpha}y_{\gamma}(A_{i,\alpha+1,\beta}A_{j,\alpha+2,\gamma}-A_{i,\alpha+2,\beta}A_{j,\alpha+1,\gamma})dy_{\beta}\wedge dx_{i}\wedge dx_{j}
+ \displaystyle+
y α y β ( A i j , α + 1 , β d y α + 2 − A i j , α + 2 , β d y α + 1 ) ∧ d x i ∧ d x j \displaystyle y_{\alpha}y_{\beta}(A_{ij,\alpha+1,\beta}dy_{\alpha+2}-A_{ij,\alpha+2,\beta}dy_{\alpha+1})\wedge dx_{i}\wedge dx_{j}
(3.146)
+ \displaystyle+
Π 3 ( 2 ) + O ~ ( r 3 ) . \displaystyle\Pi_{3}^{(2)}+\widetilde{O}(r^{3}).
Rearranging the subscripts of the first groups of terms in (3.146 ),
(3.147)
y α ( A i , α + 1 , β d y β ∧ d x i ∧ d y α + 2 − A i , α + 2 , β d y β ∧ d x i ∧ d y α + 1 ) \displaystyle y_{\alpha}(A_{i,\alpha+1,\beta}dy_{\beta}\wedge dx_{i}\wedge dy_{\alpha+2}-A_{i,\alpha+2,\beta}dy_{\beta}\wedge dx_{i}\wedge dy_{\alpha+1})
(3.148)
= \displaystyle=
y α ( A i , α + 1 , α d y α ∧ d x i ∧ d y α + 2 − A i , α + 2 , α d y α ∧ d x i ∧ d y α + 1 ) \displaystyle y_{\alpha}(A_{i,\alpha+1,\alpha}dy_{\alpha}\wedge dx_{i}\wedge dy_{\alpha+2}-A_{i,\alpha+2,\alpha}dy_{\alpha}\wedge dx_{i}\wedge dy_{\alpha+1})
(3.149)
= \displaystyle=
y α + 2 A i , α , α + 2 d y α + 2 ∧ d x i ∧ d y α + 1 − y α + 1 A i , α , α + 1 d y α + 1 ∧ d x i ∧ d y α + 2 \displaystyle y_{\alpha+2}A_{i,\alpha,\alpha+2}dy_{\alpha+2}\wedge dx_{i}\wedge dy_{\alpha+1}-y_{\alpha+1}A_{i,\alpha,\alpha+1}dy_{\alpha+1}\wedge dx_{i}\wedge dy_{\alpha+2}
(3.150)
= \displaystyle=
A i α β y β d x i ∧ d y α ^ , \displaystyle A_{i\alpha\beta}y_{\beta}dx_{i}\wedge dy_{\widehat{\alpha}},
which matches the first term of (3.145 ). As in the proof of Lemma 3.14 , one can see that the second groups of terms in (3.145 ) and (3.146 ) are both equal to
(3.151)
( A i α , α + 1 A j α , α + 2 − A i α , α + 2 A j α , α + 1 ) y α ⋅ r d r ∧ d x i ∧ d x j . (A_{i\alpha,\alpha+1}A_{j\alpha,\alpha+2}-A_{i\alpha,\alpha+2}A_{j\alpha,\alpha+1})y_{\alpha}\cdot rdr\wedge dx_{i}\wedge dx_{j}.
Next, the third group of terms in (3.146 ) can be rewritten as follows,
(3.152)
y α y β ( A i j , α + 1 , β d y α + 2 − A i j , α + 2 , β d y α + 1 ) ∧ d x i ∧ d x j \displaystyle y_{\alpha}y_{\beta}(A_{ij,\alpha+1,\beta}dy_{\alpha+2}-A_{ij,\alpha+2,\beta}dy_{\alpha+1})\wedge dx_{i}\wedge dx_{j}
= \displaystyle=
A i j α β y β ( y α + 2 d y α + 1 − y α + 1 d y α + 2 ) ∧ d x i ∧ d x j \displaystyle A_{ij\alpha\beta}y_{\beta}(y_{\alpha+2}dy_{\alpha+1}-y_{\alpha+1}dy_{\alpha+2})\wedge dx_{i}\wedge dx_{j}
= \displaystyle=
A i j α β y β y μ d y μ α ^ ∧ d x i ∧ d x j . \displaystyle A_{ij\alpha\beta}y_{\beta}y_{\mu}dy_{\widehat{\mu\alpha}}\wedge dx_{i}\wedge dx_{j}.
The conclusion just follows.