ScalingStacks

Proof. [055T]

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Proof.

We only need to consider around a point (x,t,s)(x,t,s) on the exceptional set 𝒩\mathcal{N}, so (x,t)∈DΓ—{0}(x,t)\in D\times\{0\}. Without loss of generality may assume x0β‰ 0x_{0}\neq 0. Since DD is a complete intersection by assumption (iii), we may use v1=f1​(u)v_{1}=f_{1}(u) and v2=f2​(u)v_{2}=f_{2}(u) to replace u1,u2u_{1},u_{2} (say) as local holomorphic coordinates on a neighborhood of xx in ℂ​ℙn+1\mathbb{C}\mathbb{P}^{n+1}. So we can write

(7.29) Ξ“=βˆ’Jβˆ’1(n+2)​tn+1​f​(u)​d​v1∧d​v2∧d​u3βˆ§β‹―βˆ§d​un+1,\Gamma=-\frac{J^{-1}}{(n+2)t^{n+1}f(u)}dv_{1}\wedge dv_{2}\wedge du_{3}\cdots\wedge du_{n+1},

where JJ is the Jacobian given by

(7.30) J=βˆ‚f1βˆ‚u1β€‹βˆ‚f2βˆ‚u2βˆ’βˆ‚f1βˆ‚u2β€‹βˆ‚f2βˆ‚u1.J=\frac{\partial f_{1}}{\partial u_{1}}\frac{\partial f_{2}}{\partial u_{2}}-\frac{\partial f_{1}}{\partial u_{2}}\frac{\partial f_{2}}{\partial u_{1}}.

Suppose first we work on the affine chart {s1β‰ 0}\{s_{1}\neq 0\}. Then we get the local equations for 𝒳^\widehat{\mathcal{X}} given by (7.22). Since we are away from D1D_{1}, we must have ΞΆ3β‰ 0\zeta_{3}\neq 0. Then we can use ΞΆ3,t,u3,β‹―,un+1\zeta_{3},t,u_{3},\cdots,u_{n+1} as local holomorphic coordinates on 𝒳^\widehat{\mathcal{X}}. We have

(7.31) d​v1=βˆ’td2​f​d​΢3βˆ’d2​td2βˆ’1​΢3​f​d​tβˆ’tn+2​΢3​d​f,dv_{1}=-t^{d_{2}}fd\zeta_{3}-d_{2}t^{d_{2}-1}\zeta_{3}fdt-t^{n+2}\zeta_{3}df,
(7.32) d​v2=d1​td1βˆ’1​΢3βˆ’1​d​tβˆ’ΞΆ3βˆ’2​td1​d​΢3dv_{2}=d_{1}t^{d_{1}-1}\zeta_{3}^{-1}dt-\zeta_{3}^{-2}t^{d_{1}}d\zeta_{3}

and

(7.33) d​f=βˆ‚fβˆ‚v1​d​v1+βˆ‚fβˆ‚v2​d​v2+βˆ‘jβ‰₯3βˆ‚fβˆ‚uj​d​uj.df=\frac{\partial f}{\partial v_{1}}dv_{1}+\frac{\partial f}{\partial v_{2}}dv_{2}+\sum_{j\geq 3}\frac{\partial f}{\partial u_{j}}du_{j}.

So we get

(1+td2​΢3β€‹βˆ‚fβˆ‚v1)​d​v1\displaystyle(1+t^{d_{2}}\zeta_{3}\frac{\partial f}{\partial v_{1}})dv_{1}
(7.34) =\displaystyle= (βˆ’td2​f+tn+2​΢3βˆ’1β€‹βˆ‚fβˆ‚v2)​d​΢3βˆ’(d2​td2βˆ’1​΢3​f+d1​tn+1β€‹βˆ‚fβˆ‚v2)​d​t\displaystyle(-t^{d_{2}}f+t^{n+2}\zeta_{3}^{-1}\frac{\partial f}{\partial v_{2}})d\zeta_{3}-(d_{2}t^{d_{2}-1}\zeta_{3}f+d_{1}t^{n+1}\frac{\partial f}{\partial v_{2}})dt mod(d​u3,β‹―,d​un+1).\displaystyle\mod(du_{3},\cdots,du_{n+1}).

Hence we get

(7.35) Ξ“=ΞΆ3βˆ’1(1+td2​΢3β€‹βˆ‚fβˆ‚v1)​Jβˆ’1​d​΢3∧d​t∧d​u3βˆ§β‹―βˆ§d​un+1.\Gamma=\frac{\zeta_{3}^{-1}}{(1+t^{d_{2}}\zeta_{3}\frac{\partial f}{\partial v_{1}})}J^{-1}d\zeta_{3}\wedge dt\wedge du_{3}\wedge\cdots\wedge du_{n+1}.

Near t=0t=0 we see Ξ“\Gamma is smooth around such a point. Similarly we can deal with the chart {s2β‰ 0}\{s_{2}\neq 0\}.

Now on {s3β‰ 0}\{s_{3}\neq 0\}, we only need to consider a point on DD where f=0f=0, then by our assumption (iv) we may use v3=fv_{3}=f as a local holomorphic coordinate to replace u3u_{3} for instance. Then we can write

(7.36) Ξ“=βˆ’1(n+2)​tn+1​f​Kβˆ’1​d​v1∧d​v2∧d​v3∧d​u4βˆ§β‹―βˆ§d​un+1,\Gamma=-\frac{1}{(n+2)t^{n+1}f}K^{-1}dv_{1}\wedge dv_{2}\wedge dv_{3}\wedge du_{4}\cdots\wedge du_{n+1},

where KK is the Jacobian for the change of coordinates. We have

(7.37) d​v3=βˆ’(ΞΆ1​d​΢2+ΞΆ2​d​΢1),dv_{3}=-(\zeta_{1}d\zeta_{2}+\zeta_{2}d\zeta_{1}),
(7.38) d​v1=td2​d​΢2+d2​΢2​td2βˆ’1​d​t,dv_{1}=t^{d_{2}}d\zeta_{2}+d_{2}\zeta_{2}t^{d_{2}-1}dt,
(7.39) d​v2=td1​d​΢1+d1​΢1​td1βˆ’1​d​t.dv_{2}=t^{d_{1}}d\zeta_{1}+d_{1}\zeta_{1}t^{d_{1}-1}dt.

Then we get

(7.40) Ξ“=Kβˆ’1​d​t∧d​΢1∧d​΢2∧d​u4βˆ§β‹―βˆ§d​un+1,\Gamma=K^{-1}dt\wedge d\zeta_{1}\wedge d\zeta_{2}\wedge du_{4}\cdots\wedge du_{n+1},

which is smooth. ∎

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