ScalingStacks

Proof. [0508]

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Proof.

Directly applying the definition of ηα\eta_{\alpha}, then we have

Ai​α​β⋅yβ⋅d​xi∧ηα^\displaystyle A_{i\alpha\beta}\cdot y_{\beta}\cdot dx_{i}\wedge\eta_{\widehat{\alpha}}
(3.145) =\displaystyle= Ai​α​β​yβ​d​xi∧d​yα^+Ai​α​β​yβ​yγ​(Aj,α+1,γ​d​yα+2−Aj,α+2,γ​d​yα+1)∧d​xi∧d​xj+O~​(r3).\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

yα​(d​ηα+1∧ηα+2−ηα+1∧d​ηα+2)\displaystyle y_{\alpha}(d{\eta_{\alpha+1}}\wedge{\eta_{\alpha+2}}-{\eta_{\alpha+1}}\wedge d{\eta_{\alpha+2}})
=\displaystyle= yα​(Ai,α+1,β​d​yβ∧d​xi∧d​yα+2−Ai,α+2,β​d​yβ∧d​xi∧d​yα+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+ yα​yγ​(Ai,α+1,β​Aj,α+2,γ−Ai,α+2,β​Aj,α+1,γ)​d​yβ∧d​xi∧d​xj\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+ yα​yβ​(Ai​j,α+1,β​d​yα+2−Ai​j,α+2,β​d​yα+1)∧d​xi∧d​xj\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+ Π3(2)+O~​(r3).\displaystyle\Pi_{3}^{(2)}+\widetilde{O}(r^{3}).

Rearranging the subscripts of the first groups of terms in (3.146),

(3.147) yα​(Ai,α+1,β​d​yβ∧d​xi∧d​yα+2−Ai,α+2,β​d​yβ∧d​xi∧d​yα+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= yα​(Ai,α+1,α​d​yα∧d​xi∧d​yα+2−Ai,α+2,α​d​yα∧d​xi∧d​yα+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= yα+2​Ai,α,α+2​d​yα+2∧d​xi∧d​yα+1−yα+1​Ai,α,α+1​d​yα+1∧d​xi∧d​yα+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= Ai​α​β​yβ​d​xi∧d​yα^,\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) (Ai​α,α+1​Aj​α,α+2−Ai​α,α+2​Aj​α,α+1)​yα⋅r​d​r∧d​xi∧d​xj.(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) yα​yβ​(Ai​j,α+1,β​d​yα+2−Ai​j,α+2,β​d​yα+1)∧d​xi∧d​xj\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= Ai​j​α​β​yβ​(yα+2​d​yα+1−yα+1​d​yα+2)∧d​xi∧d​xj\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= Ai​j​α​β​yβ​yμ​d​yμ​α^∧d​xi∧d​xj.\displaystyle A_{ij\alpha\beta}y_{\beta}y_{\mu}dy_{\widehat{\mu\alpha}}\wedge dx_{i}\wedge dx_{j}.

The conclusion just follows.

∎

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