The 14th Whitney Problems Workshop π and Sobolev functions ...
Transcript of The 14th Whitney Problems Workshop π and Sobolev functions ...
The 14th Whitney Problems Workshopππ and Sobolev functions on subsets of βπ
ππ SOLUTIONS OF SEMIALGEBRAIC EQUATIONS
Joint work with E. BIERSTONE and P.D. MILMAN
Jean-Baptiste Campesato
August 19, 2021
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 1 / 20
Semialgebraic geometry The problems The results The proof
Semialgebraic geometry β Definitions
Definition: semialgebraic setsSemialgebraic subsets of βπ are elements of the boolean algebra spanned by sets of the form
{π₯ β βπ βΆ π(π₯) β₯ 0}
where π β β[π₯1, β¦ , π₯π].
RemarkGiven π β β[π₯1, β¦ , π₯π], the following sets are semialgebraic
{π₯ β βπ βΆ π(π₯) > 0} , {π₯ β βπ βΆ π(π₯) β€ 0} , {π₯ β βπ βΆ π(π₯) < 0} , {π₯ β βπ βΆ π(π₯) = 0} , {π₯ β βπ βΆ π(π₯) β 0}
Definition: semialgebraic functionsLet π β βπ. A function π βΆ π β βπ is semialgebraic if its graph Ξπ β βπ+π is semialgebraic.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 2 / 20
Semialgebraic geometry The problems The results The proof
Semialgebraic geometry β Definitions
Definition: semialgebraic setsSemialgebraic subsets of βπ are elements of the boolean algebra spanned by sets of the form
{π₯ β βπ βΆ π(π₯) β₯ 0}
where π β β[π₯1, β¦ , π₯π].
RemarkGiven π β β[π₯1, β¦ , π₯π], the following sets are semialgebraic
{π₯ β βπ βΆ π(π₯) > 0} , {π₯ β βπ βΆ π(π₯) β€ 0} , {π₯ β βπ βΆ π(π₯) < 0} , {π₯ β βπ βΆ π(π₯) = 0} , {π₯ β βπ βΆ π(π₯) β 0}
Definition: semialgebraic functionsLet π β βπ. A function π βΆ π β βπ is semialgebraic if its graph Ξπ β βπ+π is semialgebraic.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 2 / 20
Semialgebraic geometry The problems The results The proof
Semialgebraic geometry β Definitions
Definition: semialgebraic setsSemialgebraic subsets of βπ are elements of the boolean algebra spanned by sets of the form
{π₯ β βπ βΆ π(π₯) β₯ 0}
where π β β[π₯1, β¦ , π₯π].
RemarkGiven π β β[π₯1, β¦ , π₯π], the following sets are semialgebraic
{π₯ β βπ βΆ π(π₯) > 0} , {π₯ β βπ βΆ π(π₯) β€ 0} , {π₯ β βπ βΆ π(π₯) < 0} , {π₯ β βπ βΆ π(π₯) = 0} , {π₯ β βπ βΆ π(π₯) β 0}
Definition: semialgebraic functionsLet π β βπ. A function π βΆ π β βπ is semialgebraic if its graph Ξπ β βπ+π is semialgebraic.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 2 / 20
Semialgebraic geometry The problems The results The proof
Semialgebraic geometry β Definitions
Definition: semialgebraic setsSemialgebraic subsets of βπ are elements of the boolean algebra spanned by sets of the form
{π₯ β βπ βΆ π(π₯) β₯ 0}
where π β β[π₯1, β¦ , π₯π].
RemarkGiven π β β[π₯1, β¦ , π₯π], the following sets are semialgebraic
{π₯ β βπ βΆ π(π₯) > 0} , {π₯ β βπ βΆ π(π₯) β€ 0} , {π₯ β βπ βΆ π(π₯) < 0} , {π₯ β βπ βΆ π(π₯) = 0} , {π₯ β βπ βΆ π(π₯) β 0}
Definition: semialgebraic functionsLet π β βπ. A function π βΆ π β βπ is semialgebraic if its graph Ξπ β βπ+π is semialgebraic.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 2 / 20
Semialgebraic geometry The problems The results The proof
Semialgebraic geometry β TarskiβSeidenberg theoremTheorem (TarskiβSeidenberg): semialgebraic sets are closed under projections
If π β βπ+1 is semialgebraic then so is π(π), where π βΆ βπ+1 β βπ, π(π₯1, β¦ , π₯π+1) = (π₯1, β¦ , π₯π).
RemarkIf π βΆ π β βπ is semialgebraic then so is π.
Corollary: elimination of quantifiers
Let π β βπ+1 be semialgebraic, then the following sets are too
{(π₯1, β¦ , π₯π) β βπ βΆ βπ¦, (π₯1, β¦ , π₯π, π¦) β π} = π(π)
{(π₯1, β¦ , π₯π) β βπ βΆ βπ¦, (π₯1, β¦ , π₯π, π¦) β π} = βπ ⧡ π(βπ+1 ⧡ π)
Example
If π΄ β βπ is semialgebraic, then so is π΄ β{
π₯ β βπ βΆ βπ β (0, +β), βπ¦ β π΄,π
βπ=1
(π₯π β π¦π)2 < π2}
.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 3 / 20
Semialgebraic geometry The problems The results The proof
Semialgebraic geometry β TarskiβSeidenberg theoremTheorem (TarskiβSeidenberg): semialgebraic sets are closed under projections
If π β βπ+1 is semialgebraic then so is π(π), where π βΆ βπ+1 β βπ, π(π₯1, β¦ , π₯π+1) = (π₯1, β¦ , π₯π).
RemarkIf π βΆ π β βπ is semialgebraic then so is π.
Corollary: elimination of quantifiers
Let π β βπ+1 be semialgebraic, then the following sets are too
{(π₯1, β¦ , π₯π) β βπ βΆ βπ¦, (π₯1, β¦ , π₯π, π¦) β π} = π(π)
{(π₯1, β¦ , π₯π) β βπ βΆ βπ¦, (π₯1, β¦ , π₯π, π¦) β π} = βπ ⧡ π(βπ+1 ⧡ π)
Example
If π΄ β βπ is semialgebraic, then so is π΄ β{
π₯ β βπ βΆ βπ β (0, +β), βπ¦ β π΄,π
βπ=1
(π₯π β π¦π)2 < π2}
.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 3 / 20
Semialgebraic geometry The problems The results The proof
Semialgebraic geometry β TarskiβSeidenberg theoremTheorem (TarskiβSeidenberg): semialgebraic sets are closed under projections
If π β βπ+1 is semialgebraic then so is π(π), where π βΆ βπ+1 β βπ, π(π₯1, β¦ , π₯π+1) = (π₯1, β¦ , π₯π).
RemarkIf π βΆ π β βπ is semialgebraic then so is π.
Corollary: elimination of quantifiers
Let π β βπ+1 be semialgebraic, then the following sets are too
{(π₯1, β¦ , π₯π) β βπ βΆ βπ¦, (π₯1, β¦ , π₯π, π¦) β π} = π(π)
{(π₯1, β¦ , π₯π) β βπ βΆ βπ¦, (π₯1, β¦ , π₯π, π¦) β π} = βπ ⧡ π(βπ+1 ⧡ π)
Example
If π΄ β βπ is semialgebraic, then so is π΄ β{
π₯ β βπ βΆ βπ β (0, +β), βπ¦ β π΄,π
βπ=1
(π₯π β π¦π)2 < π2}
.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 3 / 20
Semialgebraic geometry The problems The results The proof
Semialgebraic geometry β TarskiβSeidenberg theoremTheorem (TarskiβSeidenberg): semialgebraic sets are closed under projections
If π β βπ+1 is semialgebraic then so is π(π), where π βΆ βπ+1 β βπ, π(π₯1, β¦ , π₯π+1) = (π₯1, β¦ , π₯π).
RemarkIf π βΆ π β βπ is semialgebraic then so is π.
Corollary: elimination of quantifiers
Let π β βπ+1 be semialgebraic, then the following sets are too
{(π₯1, β¦ , π₯π) β βπ βΆ βπ¦, (π₯1, β¦ , π₯π, π¦) β π} = π(π)
{(π₯1, β¦ , π₯π) β βπ βΆ βπ¦, (π₯1, β¦ , π₯π, π¦) β π} = βπ ⧡ π(βπ+1 ⧡ π)
Example
If π΄ β βπ is semialgebraic, then so is π΄ β{
π₯ β βπ βΆ βπ β (0, +β), βπ¦ β π΄,π
βπ=1
(π₯π β π¦π)2 < π2}
.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 3 / 20
Semialgebraic geometry The problems The results The proof
A semialgebraic version of Whitneyβs extension theorem
Theorem β KurdykaβPawΕucki, 1997, 2014Thamrongthanyalak, 2017Kocel-CynkβPawΕuckiβValette, 2019
Given a semialgebraic ππ Whitney field on a closed subset π β βπ,i.e. a family (ππΌ βΆ π β β)πΌββπ
|πΌ|β€πof continuous semialgebraic functions such that
βπ§ β π, βπΌ β βπ, |πΌ| β€ π βΉ ππΌ (π₯) β β|π½|β€πβ|πΌ|
ππΌ+π½ (π¦)π½! (π₯ β π¦)π½ = π
πβπ₯,π¦βπ§(βπ₯ β π¦βπβ|πΌ|) ,
there exists a ππ semialgebraic function πΉ βΆ βπ β β such that π·πΌ πΉ|π = ππΌ and πΉ is Nash on βπ ⧡ π.
Nash β semialgebraic and analytic.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 4 / 20
Semialgebraic geometry The problems The results The proof
Are there solutions preserving semialgebraicity?
For Whitneyβs Extension Problem
Let π βΆ π β β be a semialgebraic function where π β βπ is closed.If π admits a ππ extension πΉ βΆ βπ β β, does it admit a semialgebraic ππ extension πΉ βΆ βπ β β?
For the BrennerβFeffermanβHochsterβKollΓ‘r Problem
Let π1, β¦ , ππ, π βΆ βπ β β be semialgebraic functions.If the equation π = β ππππ admit a ππ solution (ππ)π, does it admit a semialgebraic ππ solution?
β’ AschenbrennerβThamrongthanyalak (2019): βπ, for π = 1 and π = 0, respectively.β’ FeffermanβLuli (2021): βπ, for π = 2.
β’ BierstoneβC.βMilman (2021): βπ, βπ, with a loss of differentiability.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 5 / 20
Semialgebraic geometry The problems The results The proof
Are there solutions preserving semialgebraicity?
For Whitneyβs Extension ProblemLet π βΆ π β β be a semialgebraic function where π β βπ is closed.If π admits a ππ extension πΉ βΆ βπ β β, does it admit a semialgebraic ππ extension πΉ βΆ βπ β β?
For the BrennerβFeffermanβHochsterβKollΓ‘r Problem
Let π1, β¦ , ππ, π βΆ βπ β β be semialgebraic functions.If the equation π = β ππππ admit a ππ solution (ππ)π, does it admit a semialgebraic ππ solution?
β’ AschenbrennerβThamrongthanyalak (2019): βπ, for π = 1 and π = 0, respectively.β’ FeffermanβLuli (2021): βπ, for π = 2.β’ BierstoneβC.βMilman (2021): βπ, βπ, with a loss of differentiability.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 5 / 20
Semialgebraic geometry The problems The results The proof
Are there solutions preserving semialgebraicity?
For Whitneyβs Extension ProblemLet π βΆ π β β be a semialgebraic function where π β βπ is closed.If π admits a ππ extension πΉ βΆ βπ β β, does it admit a semialgebraic ππ extension πΉ βΆ βπ β β?
For the BrennerβFeffermanβHochsterβKollΓ‘r ProblemLet π1, β¦ , ππ, π βΆ βπ β β be semialgebraic functions.If the equation π = β ππππ admit a ππ solution (ππ)π, does it admit a semialgebraic ππ solution?
β’ AschenbrennerβThamrongthanyalak (2019): βπ, for π = 1 and π = 0, respectively.β’ FeffermanβLuli (2021): βπ, for π = 2.β’ BierstoneβC.βMilman (2021): βπ, βπ, with a loss of differentiability.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 5 / 20
Semialgebraic geometry The problems The results The proof
Are there solutions preserving semialgebraicity?
For Whitneyβs Extension ProblemLet π βΆ π β β be a semialgebraic function where π β βπ is closed.If π admits a ππ extension πΉ βΆ βπ β β, does it admit a semialgebraic ππ extension πΉ βΆ βπ β β?
For the BrennerβFeffermanβHochsterβKollΓ‘r ProblemLet π1, β¦ , ππ, π βΆ βπ β β be semialgebraic functions.If the equation π = β ππππ admit a ππ solution (ππ)π, does it admit a semialgebraic ππ solution?
β’ AschenbrennerβThamrongthanyalak (2019): βπ, for π = 1 and π = 0, respectively.β’ FeffermanβLuli (2021): βπ, for π = 2.β’ BierstoneβC.βMilman (2021): βπ, βπ, with a loss of differentiability.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 5 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the semialgebraic π1 extension problem(AschenbrennerβThamrongthanyalak, 2019)
Let π βΆ π β β be a semialgebraic function admitting a π1 extension πΉ βΆ βπ β β.
β’ Set π β {(π₯, π¦, π£) β βπ Γ β Γ βπ βΆ π₯ β π, π¦ = π(π₯),βπ > 0, βπΏ > 0, βπ, π β π΅πΏ (π₯), |π (π) β π(π) β π£ β (π β π)| β€ πβπ β πβ}.
β’ Then π is semialgebraic, and, βπ₯ β π, (π₯, πΉ (π₯), βπΉ (π₯)) β π.β’ Semialgebraic Michaelβs Selection Lemma:
there exists π βΆ π β π semialgebraic and continuous such that ππ₯ β π = id where ππ₯(π₯, π¦, π£) = π₯.β’ Set πΊ β ππ£ β π βΆ βπ β βπ where ππ£(π₯, π¦, π£) = π£, then πΊ is semialgebraic, continuous and satisfies
βπ β π, π(π) = π(π) + πΊ(π) β (π β π) + ππβπ,πβπ
(βπ β πβ). β
This strategy does not generalize to π > 1 since the unknown (ππΌ )πΌββπ⧡{0}|πΌ|β€π
canβt be described as a section.
For instance, if π = 2, πeπneeds to satisfy
πeπ (π) = πeπ (π) +π
βπ=1
πeπ+eπ (π)(ππ β ππ) + ππβπ,πβπ
(βπ β πβ).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 6 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the semialgebraic π1 extension problem(AschenbrennerβThamrongthanyalak, 2019)
Let π βΆ π β β be a semialgebraic function admitting a π1 extension πΉ βΆ βπ β β.
β’ Set π β {(π₯, π¦, π£) β βπ Γ β Γ βπ βΆ π₯ β π, π¦ = π(π₯),βπ > 0, βπΏ > 0, βπ, π β π΅πΏ (π₯), |π (π) β π(π) β π£ β (π β π)| β€ πβπ β πβ}.
β’ Then π is semialgebraic, and, βπ₯ β π, (π₯, πΉ (π₯), βπΉ (π₯)) β π.β’ Semialgebraic Michaelβs Selection Lemma:
there exists π βΆ π β π semialgebraic and continuous such that ππ₯ β π = id where ππ₯(π₯, π¦, π£) = π₯.β’ Set πΊ β ππ£ β π βΆ βπ β βπ where ππ£(π₯, π¦, π£) = π£, then πΊ is semialgebraic, continuous and satisfies
βπ β π, π(π) = π(π) + πΊ(π) β (π β π) + ππβπ,πβπ
(βπ β πβ). β
This strategy does not generalize to π > 1 since the unknown (ππΌ )πΌββπ⧡{0}|πΌ|β€π
canβt be described as a section.
For instance, if π = 2, πeπneeds to satisfy
πeπ (π) = πeπ (π) +π
βπ=1
πeπ+eπ (π)(ππ β ππ) + ππβπ,πβπ
(βπ β πβ).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 6 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the semialgebraic π1 extension problem(AschenbrennerβThamrongthanyalak, 2019)
Let π βΆ π β β be a semialgebraic function admitting a π1 extension πΉ βΆ βπ β β.
β’ Set π β {(π₯, π¦, π£) β βπ Γ β Γ βπ βΆ π₯ β π, π¦ = π(π₯),βπ > 0, βπΏ > 0, βπ, π β π΅πΏ (π₯), |π (π) β π(π) β π£ β (π β π)| β€ πβπ β πβ}.
β’ Then π is semialgebraic,
and, βπ₯ β π, (π₯, πΉ (π₯), βπΉ (π₯)) β π.β’ Semialgebraic Michaelβs Selection Lemma:
there exists π βΆ π β π semialgebraic and continuous such that ππ₯ β π = id where ππ₯(π₯, π¦, π£) = π₯.β’ Set πΊ β ππ£ β π βΆ βπ β βπ where ππ£(π₯, π¦, π£) = π£, then πΊ is semialgebraic, continuous and satisfies
βπ β π, π(π) = π(π) + πΊ(π) β (π β π) + ππβπ,πβπ
(βπ β πβ). β
This strategy does not generalize to π > 1 since the unknown (ππΌ )πΌββπ⧡{0}|πΌ|β€π
canβt be described as a section.
For instance, if π = 2, πeπneeds to satisfy
πeπ (π) = πeπ (π) +π
βπ=1
πeπ+eπ (π)(ππ β ππ) + ππβπ,πβπ
(βπ β πβ).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 6 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the semialgebraic π1 extension problem(AschenbrennerβThamrongthanyalak, 2019)
Let π βΆ π β β be a semialgebraic function admitting a π1 extension πΉ βΆ βπ β β.
β’ Set π β {(π₯, π¦, π£) β βπ Γ β Γ βπ βΆ π₯ β π, π¦ = π(π₯),βπ > 0, βπΏ > 0, βπ, π β π΅πΏ (π₯), |π (π) β π(π) β π£ β (π β π)| β€ πβπ β πβ}.
β’ Then π is semialgebraic, and, βπ₯ β π, (π₯, πΉ (π₯), βπΉ (π₯)) β π.
β’ Semialgebraic Michaelβs Selection Lemma:there exists π βΆ π β π semialgebraic and continuous such that ππ₯ β π = id where ππ₯(π₯, π¦, π£) = π₯.
β’ Set πΊ β ππ£ β π βΆ βπ β βπ where ππ£(π₯, π¦, π£) = π£, then πΊ is semialgebraic, continuous and satisfiesβπ β π, π(π) = π(π) + πΊ(π) β (π β π) + π
πβπ,πβπ(βπ β πβ). β
This strategy does not generalize to π > 1 since the unknown (ππΌ )πΌββπ⧡{0}|πΌ|β€π
canβt be described as a section.
For instance, if π = 2, πeπneeds to satisfy
πeπ (π) = πeπ (π) +π
βπ=1
πeπ+eπ (π)(ππ β ππ) + ππβπ,πβπ
(βπ β πβ).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 6 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the semialgebraic π1 extension problem(AschenbrennerβThamrongthanyalak, 2019)
Let π βΆ π β β be a semialgebraic function admitting a π1 extension πΉ βΆ βπ β β.
β’ Set π β {(π₯, π¦, π£) β βπ Γ β Γ βπ βΆ π₯ β π, π¦ = π(π₯),βπ > 0, βπΏ > 0, βπ, π β π΅πΏ (π₯), |π (π) β π(π) β π£ β (π β π)| β€ πβπ β πβ}.
β’ Then π is semialgebraic, and, βπ₯ β π, (π₯, πΉ (π₯), βπΉ (π₯)) β π.β’ Semialgebraic Michaelβs Selection Lemma:
there exists π βΆ π β π semialgebraic and continuous such that ππ₯ β π = id where ππ₯(π₯, π¦, π£) = π₯.
β’ Set πΊ β ππ£ β π βΆ βπ β βπ where ππ£(π₯, π¦, π£) = π£, then πΊ is semialgebraic, continuous and satisfiesβπ β π, π(π) = π(π) + πΊ(π) β (π β π) + π
πβπ,πβπ(βπ β πβ). β
This strategy does not generalize to π > 1 since the unknown (ππΌ )πΌββπ⧡{0}|πΌ|β€π
canβt be described as a section.
For instance, if π = 2, πeπneeds to satisfy
πeπ (π) = πeπ (π) +π
βπ=1
πeπ+eπ (π)(ππ β ππ) + ππβπ,πβπ
(βπ β πβ).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 6 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the semialgebraic π1 extension problem(AschenbrennerβThamrongthanyalak, 2019)
Let π βΆ π β β be a semialgebraic function admitting a π1 extension πΉ βΆ βπ β β.
β’ Set π β {(π₯, π¦, π£) β βπ Γ β Γ βπ βΆ π₯ β π, π¦ = π(π₯),βπ > 0, βπΏ > 0, βπ, π β π΅πΏ (π₯), |π (π) β π(π) β π£ β (π β π)| β€ πβπ β πβ}.
β’ Then π is semialgebraic, and, βπ₯ β π, (π₯, πΉ (π₯), βπΉ (π₯)) β π.β’ Semialgebraic Michaelβs Selection Lemma:
there exists π βΆ π β π semialgebraic and continuous such that ππ₯ β π = id where ππ₯(π₯, π¦, π£) = π₯.β’ Set πΊ β ππ£ β π βΆ βπ β βπ where ππ£(π₯, π¦, π£) = π£, then πΊ is semialgebraic, continuous and satisfies
βπ β π, π(π) = π(π) + πΊ(π) β (π β π) + ππβπ,πβπ
(βπ β πβ). β
This strategy does not generalize to π > 1 since the unknown (ππΌ )πΌββπ⧡{0}|πΌ|β€π
canβt be described as a section.
For instance, if π = 2, πeπneeds to satisfy
πeπ (π) = πeπ (π) +π
βπ=1
πeπ+eπ (π)(ππ β ππ) + ππβπ,πβπ
(βπ β πβ).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 6 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the semialgebraic π1 extension problem(AschenbrennerβThamrongthanyalak, 2019)
Let π βΆ π β β be a semialgebraic function admitting a π1 extension πΉ βΆ βπ β β.
β’ Set π β {(π₯, π¦, π£) β βπ Γ β Γ βπ βΆ π₯ β π, π¦ = π(π₯),βπ > 0, βπΏ > 0, βπ, π β π΅πΏ (π₯), |π (π) β π(π) β π£ β (π β π)| β€ πβπ β πβ}.
β’ Then π is semialgebraic, and, βπ₯ β π, (π₯, πΉ (π₯), βπΉ (π₯)) β π.β’ Semialgebraic Michaelβs Selection Lemma:
there exists π βΆ π β π semialgebraic and continuous such that ππ₯ β π = id where ππ₯(π₯, π¦, π£) = π₯.β’ Set πΊ β ππ£ β π βΆ βπ β βπ where ππ£(π₯, π¦, π£) = π£, then πΊ is semialgebraic, continuous and satisfies
βπ β π, π(π) = π(π) + πΊ(π) β (π β π) + ππβπ,πβπ
(βπ β πβ). β
This strategy does not generalize to π > 1 since the unknown (ππΌ )πΌββπ⧡{0}|πΌ|β€π
canβt be described as a section.
For instance, if π = 2, πeπneeds to satisfy
πeπ(π) = πeπ
(π) +π
βπ=1
πeπ+eπ(π)(ππ β ππ) + π
πβπ,πβπ(βπ β πβ).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 6 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the planar semialgebraic extension problem (FeffermanβLuli, 2021)
πβ βΆ π¦ = 0
π+ βΆ π¦ = π(π₯) β€ π₯π = πβ βͺ π+
(0, 0)
Let π βΆ π β β be semialgebraic.
1 Let πΉ βΆ β2 β β be a ππ function such that πΉ|π = π and π½(0,0)πΉ = 0.Set π β
π (π₯) β πππ¦πΉ (π₯, 0) and π +
π (π₯) β πππ¦πΉ (π₯, π(π₯)). Then
(β)
β§βͺβͺβͺβͺβ¨βͺβͺβͺβͺβ©
(π) π β0 (π₯) = π(π₯, 0)
(ππ) π +0 (π₯) = π(π₯, π(π₯))
(πππ) π +π (π₯) =
πβπ
βπ=0
π(π₯)π
π! π βπ+π(π₯) + π
π₯β0+(π(π₯)πβπ)
(ππ£) π βπ (π₯) = π
π₯β0+ (π₯πβπ)(π£) π +
π (π₯) = ππ₯β0+ (π₯πβπ)
2 According to the definable choice: there exist π Β±π semialgebraic satisfying (β).
3 Then πΉ (π₯, π¦) = πβ(π₯, π¦)(
π
βπ=0
π βπ (π₯)π! π¦π
)+ π+(π₯, π¦)
(
π
βπ=0
π +π (π₯)π! (π¦ β π(π₯))π
)is a semialgebraic ππ
extension of π in a neighborhood of the origin such that π½(0,0) πΉ = 0.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 7 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the planar semialgebraic extension problem (FeffermanβLuli, 2021)
πβ βΆ π¦ = 0
π+ βΆ π¦ = π(π₯) β€ π₯π = πβ βͺ π+
(0, 0)
Let π βΆ π β β be semialgebraic.1 Let πΉ βΆ β2 β β be a ππ function such that πΉ|π = π and π½(0,0)πΉ = 0.
Set π βπ (π₯) β ππ
π¦πΉ (π₯, 0) and π +π (π₯) β ππ
π¦πΉ (π₯, π(π₯)). Then
(β)
β§βͺβͺβͺβͺβ¨βͺβͺβͺβͺβ©
(π) π β0 (π₯) = π(π₯, 0)
(ππ) π +0 (π₯) = π(π₯, π(π₯))
(πππ) π +π (π₯) =
πβπ
βπ=0
π(π₯)π
π! π βπ+π(π₯) + π
π₯β0+(π(π₯)πβπ)
(ππ£) π βπ (π₯) = π
π₯β0+ (π₯πβπ)(π£) π +
π (π₯) = ππ₯β0+ (π₯πβπ)
2 According to the definable choice: there exist π Β±π semialgebraic satisfying (β).
3 Then πΉ (π₯, π¦) = πβ(π₯, π¦)(
π
βπ=0
π βπ (π₯)π! π¦π
)+ π+(π₯, π¦)
(
π
βπ=0
π +π (π₯)π! (π¦ β π(π₯))π
)is a semialgebraic ππ
extension of π in a neighborhood of the origin such that π½(0,0) πΉ = 0.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 7 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the planar semialgebraic extension problem (FeffermanβLuli, 2021)
πβ βΆ π¦ = 0
π+ βΆ π¦ = π(π₯) β€ π₯π = πβ βͺ π+
(0, 0)
Let π βΆ π β β be semialgebraic.1 Let πΉ βΆ β2 β β be a ππ function such that πΉ|π = π and π½(0,0)πΉ = 0.
Set π βπ (π₯) β ππ
π¦πΉ (π₯, 0) and π +π (π₯) β ππ
π¦πΉ (π₯, π(π₯)).
Then
(β)
β§βͺβͺβͺβͺβ¨βͺβͺβͺβͺβ©
(π) π β0 (π₯) = π(π₯, 0)
(ππ) π +0 (π₯) = π(π₯, π(π₯))
(πππ) π +π (π₯) =
πβπ
βπ=0
π(π₯)π
π! π βπ+π(π₯) + π
π₯β0+(π(π₯)πβπ)
(ππ£) π βπ (π₯) = π
π₯β0+ (π₯πβπ)(π£) π +
π (π₯) = ππ₯β0+ (π₯πβπ)
2 According to the definable choice: there exist π Β±π semialgebraic satisfying (β).
3 Then πΉ (π₯, π¦) = πβ(π₯, π¦)(
π
βπ=0
π βπ (π₯)π! π¦π
)+ π+(π₯, π¦)
(
π
βπ=0
π +π (π₯)π! (π¦ β π(π₯))π
)is a semialgebraic ππ
extension of π in a neighborhood of the origin such that π½(0,0) πΉ = 0.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 7 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the planar semialgebraic extension problem (FeffermanβLuli, 2021)
πβ βΆ π¦ = 0
π+ βΆ π¦ = π(π₯) β€ π₯π = πβ βͺ π+
(0, 0)
Let π βΆ π β β be semialgebraic.1 Let πΉ βΆ β2 β β be a ππ function such that πΉ|π = π and π½(0,0)πΉ = 0.
Set π βπ (π₯) β ππ
π¦πΉ (π₯, 0) and π +π (π₯) β ππ
π¦πΉ (π₯, π(π₯)). Then
(β)
β§βͺβͺβͺβͺβ¨βͺβͺβͺβͺβ©
(π) π β0 (π₯) = π(π₯, 0)
(ππ) π +0 (π₯) = π(π₯, π(π₯))
(πππ) π +π (π₯) =
πβπ
βπ=0
π(π₯)π
π! π βπ+π(π₯) + π
π₯β0+(π(π₯)πβπ)
(ππ£) π βπ (π₯) = π
π₯β0+ (π₯πβπ)(π£) π +
π (π₯) = ππ₯β0+ (π₯πβπ)
2 According to the definable choice: there exist π Β±π semialgebraic satisfying (β).
3 Then πΉ (π₯, π¦) = πβ(π₯, π¦)(
π
βπ=0
π βπ (π₯)π! π¦π
)+ π+(π₯, π¦)
(
π
βπ=0
π +π (π₯)π! (π¦ β π(π₯))π
)is a semialgebraic ππ
extension of π in a neighborhood of the origin such that π½(0,0) πΉ = 0.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 7 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the planar semialgebraic extension problem (FeffermanβLuli, 2021)
πβ βΆ π¦ = 0
π+ βΆ π¦ = π(π₯) β€ π₯π = πβ βͺ π+
(0, 0)
Let π βΆ π β β be semialgebraic.1 Let πΉ βΆ β2 β β be a ππ function such that πΉ|π = π and π½(0,0)πΉ = 0.
Set π βπ (π₯) β ππ
π¦πΉ (π₯, 0) and π +π (π₯) β ππ
π¦πΉ (π₯, π(π₯)). Then
(β)
β§βͺβͺβͺβͺβ¨βͺβͺβͺβͺβ©
(π) π β0 (π₯) = π(π₯, 0)
(ππ) π +0 (π₯) = π(π₯, π(π₯))
(πππ) π +π (π₯) =
πβπ
βπ=0
π(π₯)π
π! π βπ+π(π₯) + π
π₯β0+(π(π₯)πβπ)
(ππ£) π βπ (π₯) = π
π₯β0+ (π₯πβπ)(π£) π +
π (π₯) = ππ₯β0+ (π₯πβπ)
2 According to the definable choice: there exist π Β±π semialgebraic satisfying (β).
3 Then πΉ (π₯, π¦) = πβ(π₯, π¦)(
π
βπ=0
π βπ (π₯)π! π¦π
)+ π+(π₯, π¦)
(
π
βπ=0
π +π (π₯)π! (π¦ β π(π₯))π
)is a semialgebraic ππ
extension of π in a neighborhood of the origin such that π½(0,0) πΉ = 0.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 7 / 20
Semialgebraic geometry The problems The results The proof
Strategy for the planar semialgebraic extension problem (FeffermanβLuli, 2021)
πβ βΆ π¦ = 0
π+ βΆ π¦ = π(π₯) β€ π₯π = πβ βͺ π+
(0, 0)
Let π βΆ π β β be semialgebraic.1 Let πΉ βΆ β2 β β be a ππ function such that πΉ|π = π and π½(0,0)πΉ = 0.
Set π βπ (π₯) β ππ
π¦πΉ (π₯, 0) and π +π (π₯) β ππ
π¦πΉ (π₯, π(π₯)). Then
(β)
β§βͺβͺβͺβͺβ¨βͺβͺβͺβͺβ©
(π) π β0 (π₯) = π(π₯, 0)
(ππ) π +0 (π₯) = π(π₯, π(π₯))
(πππ) π +π (π₯) =
πβπ
βπ=0
π(π₯)π
π! π βπ+π(π₯) + π
π₯β0+(π(π₯)πβπ)
(ππ£) π βπ (π₯) = π
π₯β0+ (π₯πβπ)(π£) π +
π (π₯) = ππ₯β0+ (π₯πβπ)
2 According to the definable choice: there exist π Β±π semialgebraic satisfying (β).
3 Then πΉ (π₯, π¦) = πβ(π₯, π¦)(
π
βπ=0
π βπ (π₯)π! π¦π
)+ π+(π₯, π¦)
(
π
βπ=0
π +π (π₯)π! (π¦ β π(π₯))π
)is a semialgebraic ππ
extension of π in a neighborhood of the origin such that π½(0,0) πΉ = 0.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 7 / 20
Semialgebraic geometry The problems The results The proof
The main results: statements
Theorem β BierstoneβC.βMilman, 2021Given π β βπ closed and semialgebraic, there exists π βΆ β β β satisfying the following property:if π βΆ π β β semialgebraic admits a ππ(π) extension, then it admits a semialgebraic ππ extension.
Theorem β BierstoneβC.βMilman, 2021Given π΄ βΆ βπ β β³π,π(β) semialgebraic, there exists π βΆ β β β such that:if πΉ βΆ βπ β βπ semialgebraic may be written πΉ (π₯) = π΄(π₯)πΊ(π₯) where πΊ is ππ(π),then πΉ (π₯) = π΄(π₯)οΏ½οΏ½(π₯) where οΏ½οΏ½(π₯) is semialgebraic and ππ.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 8 / 20
Semialgebraic geometry The problems The results The proof
The main results: statements
Theorem β BierstoneβC.βMilman, 2021Given π β βπ closed and semialgebraic, there exists π βΆ β β β satisfying the following property:if π βΆ π β β semialgebraic admits a ππ(π) extension, then it admits a semialgebraic ππ extension.
Theorem β BierstoneβC.βMilman, 2021Given π΄ βΆ βπ β β³π,π(β) semialgebraic, there exists π βΆ β β β such that:if πΉ βΆ βπ β βπ semialgebraic may be written πΉ (π₯) = π΄(π₯)πΊ(π₯) where πΊ is ππ(π),then πΉ (π₯) = π΄(π₯)οΏ½οΏ½(π₯) where οΏ½οΏ½(π₯) is semialgebraic and ππ.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 8 / 20
Semialgebraic geometry The problems The results The proof
Main results: towards a common generalization
The extension problemLet π β βπ be semialgebraic and closed.
By resolution of singularities, there exists π βΆ π β βπ
Nash and proper defined on a Nash manifold suchthat π = π(π).
Given π βΆ βπ β β and π βΆ π β β, we haveπ|π = π if and only if
βπ¦ β π, π(π(π¦)) = π (π¦)
where π β π β π.
The equation problemConsider an equation
π΄(π₯)πΊ(π₯) = πΉ (π₯), π₯ β βπ.
By resolution of singularities, there existsπ βΆ π β βπ Nash and proper defined on a Nashmanifold such that after composition, we get
π΄(π¦)πΊ(π(π¦)) = πΉ (π¦), π¦ β π
where π΄ β π΄ β π is now Nash and πΉ β πΉ β π.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 9 / 20
Semialgebraic geometry The problems The results The proof
Main results: towards a common generalization
The extension problemLet π β βπ be semialgebraic and closed.
By resolution of singularities, there exists π βΆ π β βπ
Nash and proper defined on a Nash manifold suchthat π = π(π).
Given π βΆ βπ β β and π βΆ π β β, we haveπ|π = π if and only if
βπ¦ β π, π(π(π¦)) = π (π¦)
where π β π β π.
The equation problemConsider an equation
π΄(π₯)πΊ(π₯) = πΉ (π₯), π₯ β βπ.
By resolution of singularities, there existsπ βΆ π β βπ Nash and proper defined on a Nashmanifold such that after composition, we get
π΄(π¦)πΊ(π(π¦)) = πΉ (π¦), π¦ β π
where π΄ β π΄ β π is now Nash and πΉ β πΉ β π.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 9 / 20
Semialgebraic geometry The problems The results The proof
The main result
Theorem β BierstoneβC.βMilman, 2021Let π΄ βΆ βπ β β³π,π(β) be Nash and let π βΆ π β βπ be Nash and proper defined on π β βπ aNash submanifold.Then there exists π βΆ β β β satisfying the following property.If π βΆ π β βπ semialgebraic may be written
π(π₯) = π΄(π₯)π(π(π₯))
for a ππ(π) function π βΆ βπ β βπ then
π(π₯) = π΄(π₯) π(π(π₯))
for a semialgebraic ππ function π.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 10 / 20
Semialgebraic geometry The problems The results The proof
Heart of the proof: induction on dimension
Proposition: the induction stepLet π΅ β π(π) be semialgebraic and closed.There exist π΅β² β π΅ semialgebraic satisfying dim π΅β² < dim π΅ and π‘ βΆ β β β such that if
1 π βΆ π β βπ is ππ‘(π), semialgebraic and π‘(π)-flat on πβ1(π΅β²), and2 π = π΄ β (π β π) admits a ππ‘(π) solution π,
then there exists a semialgebraic ππ function π βΆ βπ β βπ s.t. π β π΄ β ( π β π) is π-flat on πβ1(π΅).
Then, up to subtracting by π΄ β ( π β π) on both side, we get an equation
π = π΄ β (π β π)
where π is now semialgebraic and π-flat on πβ1(π΅).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 11 / 20
Semialgebraic geometry The problems The results The proof
Heart of the proof: induction on dimension
Proposition: the induction stepLet π΅ β π(π) be semialgebraic and closed.There exist π΅β² β π΅ semialgebraic satisfying dim π΅β² < dim π΅ and π‘ βΆ β β β such that if
1 π βΆ π β βπ is ππ‘(π), semialgebraic and π‘(π)-flat on πβ1(π΅β²), and2 π = π΄ β (π β π) admits a ππ‘(π) solution π,
then there exists a semialgebraic ππ function π βΆ βπ β βπ s.t. π β π΄ β ( π β π) is π-flat on πβ1(π΅).
Then, up to subtracting by π΄ β ( π β π) on both side, we get an equation
π = π΄ β (π β π)
where π is now semialgebraic and π-flat on πβ1(π΅).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 11 / 20
Semialgebraic geometry The problems The results The proof
Heart of the proof: induction on dimension
Strategy: construction of a semialgebraic Whitney field
πΊ(π, y) = β|πΌ|β€π
ππΌ (π)πΌ! yπΌ β π0(π΅)[y]
vanishing on π΅β² such that
βπ β π΅ ⧡ π΅β², βπ β πβ1(π), π πππ(x) β‘ π π
ππ΄(x) πΊ(π, π πππ(x)) mod (x)π+1βJxKπ
A - Whitney regularityGiven π΅, there exists π β β such that if πΊ is a Whitney field of order π β₯ ππ on π΅ ⧡ π΅β² then it is aWhitney field of order π on π΅.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 12 / 20
Semialgebraic geometry The problems The results The proof
Heart of the proof: induction on dimension
Strategy: construction of a semialgebraic Whitney field
πΊ(π, y) = β|πΌ|β€π
ππΌ (π)πΌ! yπΌ β π0(π΅)[y]
vanishing on π΅β² such that
βπ β π΅ ⧡ π΅β², βπ β πβ1(π), π πππ(x) β‘ π π
ππ΄(x) πΊ(π, π πππ(x)) mod (x)π+1βJxKπ
A - Whitney regularityGiven π΅, there exists π β β such that if πΊ is a Whitney field of order π β₯ ππ on π΅ ⧡ π΅β² then it is aWhitney field of order π on π΅.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 12 / 20
Semialgebraic geometry The problems The results The proof
The module of relations at π β π(π)We consider the equation at the level of Taylor polynomials:
π ππ π(x) β‘ π π
π π΄(x) πΊ(π, π ππ π(x)) mod (x)π+1βJxKπ (1)
B - Chevalleyβs functionGiven π β β, there exists π β₯ π such that the derivatives of π of order β€ π can be expressed interms of the derivatives of π of order β€ π.
Formally, we stratify π΅ = β¨πmaxπ=1 Ξπ such that for each π, there exists π β₯ π satisfying
βπ β Ξπ , ππ(βπ(π)) = ππ(βπβ1(π))
whereβ’ βπ(π) is the module of relations at π β π(π) formed by the πΊ β βJyKπ satisfying the
homogeneous equation associated to (1) for all π β πβ1(π), and,β’ ππ is the truncation up to degree π.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 13 / 20
Semialgebraic geometry The problems The results The proof
The module of relations at π β π(π)We consider the equation at the level of Taylor polynomials:
π ππ π(x) β‘ π π
π π΄(x) πΊ(π, π ππ π(x)) mod (x)π+1βJxKπ (1)
B - Chevalleyβs functionGiven π β β, there exists π β₯ π such that the derivatives of π of order β€ π can be expressed interms of the derivatives of π of order β€ π.
Formally, we stratify π΅ = β¨πmaxπ=1 Ξπ such that for each π, there exists π β₯ π satisfying
βπ β Ξπ , ππ(βπ(π)) = ππ(βπβ1(π))
whereβ’ βπ(π) is the module of relations at π β π(π) formed by the πΊ β βJyKπ satisfying the
homogeneous equation associated to (1) for all π β πβ1(π), and,β’ ππ is the truncation up to degree π.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 13 / 20
Semialgebraic geometry The problems The results The proof
The module of relations at π β π(π)We consider the equation at the level of Taylor polynomials:
π ππ π(x) β‘ π π
π π΄(x) πΊ(π, π ππ π(x)) mod (x)π+1βJxKπ (1)
B - Chevalleyβs functionGiven π β β, there exists π β₯ π such that the derivatives of π of order β€ π can be expressed interms of the derivatives of π of order β€ π.
Formally, we stratify π΅ = β¨πmaxπ=1 Ξπ such that for each π, there exists π β₯ π satisfying
βπ β Ξπ , ππ(βπ(π)) = ππ(βπβ1(π))
whereβ’ βπ(π) is the module of relations at π β π(π) formed by the πΊ β βJyKπ satisfying the
homogeneous equation associated to (1) for all π β πβ1(π), and,β’ ππ is the truncation up to degree π.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 13 / 20
Semialgebraic geometry The problems The results The proof
Hironakaβs formal divisionβ’ πΉ = β πΉ(πΌ,π)y(πΌ,π) β βJπ¦1, β¦ , π¦πKπ where y(πΌ,π) = (0, β¦ , 0, π¦πΌ1
1 β― π¦πΌππ , 0, β¦ , 0).
β’ The set βπ Γ {1, β¦ , π} β (πΌ, π) is totally ordered by lex(|πΌ|, π, πΌ1, β¦ , πΌπ).β’ supp πΉ β {(πΌ, π) βΆ πΉ(πΌ,π) β 0} β’ exp πΉ β min(supp πΉ )
Theorem β Hironaka 1964, BierstoneβMilman 1987Let Ξ¦1, β¦ , Ξ¦π β βJyKπ.
Set Ξ1 β exp Ξ¦1 + βπ, Ξπ β (exp Ξ¦π + βπ) ⧡πβ1
βπ=1
Ξπ, and Ξ β (βπ Γ {1, β¦ , π}) ⧡π
βπ=1
Ξπ.
Ξ
Ξ3
Ξ2
Ξ1Then βπΉ β βJyKπ, β!ππ β βJyK, π β βJyKπ such that
β’ πΉ =π
βπ=1
ππΞ¦π + π
β’ exp Ξ¦π + supp ππ β Ξπ
β’ supp π β Ξ
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 14 / 20
Semialgebraic geometry The problems The results The proof
Hironakaβs formal divisionβ’ πΉ = β πΉ(πΌ,π)y(πΌ,π) β βJπ¦1, β¦ , π¦πKπ where y(πΌ,π) = (0, β¦ , 0, π¦πΌ1
1 β― π¦πΌππ , 0, β¦ , 0).
β’ The set βπ Γ {1, β¦ , π} β (πΌ, π) is totally ordered by lex(|πΌ|, π, πΌ1, β¦ , πΌπ).β’ supp πΉ β {(πΌ, π) βΆ πΉ(πΌ,π) β 0} β’ exp πΉ β min(supp πΉ )
Theorem β Hironaka 1964, BierstoneβMilman 1987Let Ξ¦1, β¦ , Ξ¦π β βJyKπ.
Set Ξ1 β exp Ξ¦1 + βπ, Ξπ β (exp Ξ¦π + βπ) ⧡πβ1
βπ=1
Ξπ, and Ξ β (βπ Γ {1, β¦ , π}) ⧡π
βπ=1
Ξπ.
Ξ
Ξ3
Ξ2
Ξ1Then βπΉ β βJyKπ, β!ππ β βJyK, π β βJyKπ such that
β’ πΉ =π
βπ=1
ππΞ¦π + π
β’ exp Ξ¦π + supp ππ β Ξπ
β’ supp π β Ξ
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 14 / 20
Semialgebraic geometry The problems The results The proof
Hironakaβs formal divisionβ’ πΉ = β πΉ(πΌ,π)y(πΌ,π) β βJπ¦1, β¦ , π¦πKπ where y(πΌ,π) = (0, β¦ , 0, π¦πΌ1
1 β― π¦πΌππ , 0, β¦ , 0).
β’ The set βπ Γ {1, β¦ , π} β (πΌ, π) is totally ordered by lex(|πΌ|, π, πΌ1, β¦ , πΌπ).β’ supp πΉ β {(πΌ, π) βΆ πΉ(πΌ,π) β 0} β’ exp πΉ β min(supp πΉ )
Theorem β Hironaka 1964, BierstoneβMilman 1987Let Ξ¦1, β¦ , Ξ¦π β βJyKπ.
Set Ξ1 β exp Ξ¦1 + βπ, Ξπ β (exp Ξ¦π + βπ) ⧡πβ1
βπ=1
Ξπ, and Ξ β (βπ Γ {1, β¦ , π}) ⧡π
βπ=1
Ξπ.
Ξ
Ξ3
Ξ2
Ξ1Then βπΉ β βJyKπ, β!ππ β βJyK, π β βJyKπ such that
β’ πΉ =π
βπ=1
ππΞ¦π + π
β’ exp Ξ¦π + supp ππ β Ξπ
β’ supp π β Ξ
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 14 / 20
Semialgebraic geometry The problems The results The proof
Diagram of initial exponentsLet π β βJyKπ be a βJyK-submodule.The diagram of initial exponents of π is
π© (π) β {exp πΉ βΆ πΉ β π ⧡ {0}} β βπ Γ {1, β¦ , π}
Note that π© (π) has finitely many vertices (πΌπ, ππ), π = 1, β¦ , π.For each one, we pick a representative Ξ¦π β π , i.e. exp Ξ¦π = (πΌπ, ππ).
CorollaryLet πΉ β βJyKπ. Then πΉ β π if and only if its remainder by the formal division w.r.t. the Ξ¦π is 0.
Particularly Ξ¦1, β¦ , Ξ¦π generate π .
Proof. Write πΉ =π
βπ=1
ππΞ¦π + π with supp π β π© (π)π . β
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 15 / 20
Semialgebraic geometry The problems The results The proof
Diagram of initial exponentsLet π β βJyKπ be a βJyK-submodule.The diagram of initial exponents of π is
π© (π) β {exp πΉ βΆ πΉ β π ⧡ {0}} β βπ Γ {1, β¦ , π}
Note that π© (π) has finitely many vertices (πΌπ, ππ), π = 1, β¦ , π.For each one, we pick a representative Ξ¦π β π , i.e. exp Ξ¦π = (πΌπ, ππ).
CorollaryLet πΉ β βJyKπ. Then πΉ β π if and only if its remainder by the formal division w.r.t. the Ξ¦π is 0.
Particularly Ξ¦1, β¦ , Ξ¦π generate π .
Proof. Write πΉ =π
βπ=1
ππΞ¦π + π with supp π β π© (π)π . β
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 15 / 20
Semialgebraic geometry The problems The results The proof
Diagram of initial exponentsLet π β βJyKπ be a βJyK-submodule.The diagram of initial exponents of π is
π© (π) β {exp πΉ βΆ πΉ β π ⧡ {0}} β βπ Γ {1, β¦ , π}
Note that π© (π) has finitely many vertices (πΌπ, ππ), π = 1, β¦ , π.For each one, we pick a representative Ξ¦π β π , i.e. exp Ξ¦π = (πΌπ, ππ).
CorollaryLet πΉ β βJyKπ. Then πΉ β π if and only if its remainder by the formal division w.r.t. the Ξ¦π is 0.
Particularly Ξ¦1, β¦ , Ξ¦π generate π .
Proof. Write πΉ =π
βπ=1
ππΞ¦π + π with supp π β π© (π)π . β
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 15 / 20
Semialgebraic geometry The problems The results The proof
Diagram of initial exponentsLet π β βJyKπ be a βJyK-submodule.The diagram of initial exponents of π is
π© (π) β {exp πΉ βΆ πΉ β π ⧡ {0}} β βπ Γ {1, β¦ , π}
Note that π© (π) has finitely many vertices (πΌπ, ππ), π = 1, β¦ , π.For each one, we pick a representative Ξ¦π β π , i.e. exp Ξ¦π = (πΌπ, ππ).
CorollaryLet πΉ β βJyKπ. Then πΉ β π if and only if its remainder by the formal division w.r.t. the Ξ¦π is 0.
Particularly Ξ¦1, β¦ , Ξ¦π generate π .
Proof. Write πΉ =π
βπ=1
ππΞ¦π + π with supp π β π© (π)π . β
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 15 / 20
Semialgebraic geometry The problems The results The proof
Diagram of initial exponents and module of relations
Lemma β Chevalleyβs functionLet π β β.There exists (Ξπ)π a stratification of π΅ such that given a stratum Ξπ , there exists π β₯ π satisfying
β’ βπ β Ξπ , ππ(βπ(π)) = ππ(βπβ1(π)),β’ π© (βπ(π)) is constant on Ξπ .
We setπ΅β² β β
dim Ξπ <dim π΅Ξπ
so that βπ, Ξπ ⧡ Ξπ β π΅β².
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 16 / 20
Semialgebraic geometry The problems The results The proof
Diagram of initial exponents and module of relations
Lemma β Chevalleyβs functionLet π β β.There exists (Ξπ)π a stratification of π΅ such that given a stratum Ξπ , there exists π β₯ π satisfying
β’ βπ β Ξπ , ππ(βπ(π)) = ππ(βπβ1(π)),β’ π© (βπ(π)) is constant on Ξπ .
We setπ΅β² β β
dim Ξπ <dim π΅Ξπ
so that βπ, Ξπ ⧡ Ξπ β π΅β².
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 16 / 20
Semialgebraic geometry The problems The results The proof
Construction of πΊ on Ξπ
For π β π΅ and π‘ β₯ π, by assumption there exists ππ β β[y]π such that
π π‘ππ(x) β‘ π π‘
ππ΄(x) ππ(π π‘ππ(x)) mod (x)π‘+1βJxKπ, βπ β πβ1(π).
Letβs fix a stratum Ξπ and π β Ξπ .By formal division, we may write ππ(y) = β ππ(y)Ξ¦π(y) + ππ(π, y) where the Ξ¦π are as above for βπ(π).Note that the remainder ππ(π, y) β β[y]π satisfies
ππ(y) β ππ(π, y) β βπ(π) and supp ππ(π, y) β π© (βπ(π))π .
LemmaπΊπ(π, y) β ππ (ππ(π, y)) is a semialgebraic Whitney field of order π on Ξπ satisfying (1).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 17 / 20
exp Ξ¦2
exp Ξ¦1
supp ππ
π© (π π(π))
Semialgebraic geometry The problems The results The proof
Construction of πΊ on Ξπ
For π β π΅ and π‘ β₯ π, by assumption there exists ππ β β[y]π such that
π π‘ππ(x) β‘ π π‘
ππ΄(x) ππ(π π‘ππ(x)) mod (x)π‘+1βJxKπ, βπ β πβ1(π).
Letβs fix a stratum Ξπ and π β Ξπ .By formal division, we may write ππ(y) = β ππ(y)Ξ¦π(y) + ππ(π, y) where the Ξ¦π are as above for βπ(π).Note that the remainder ππ(π, y) β β[y]π satisfies
ππ(y) β ππ(π, y) β βπ(π) and supp ππ(π, y) β π© (βπ(π))π .
LemmaπΊπ(π, y) β ππ (ππ(π, y)) is a semialgebraic Whitney field of order π on Ξπ satisfying (1).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 17 / 20
exp Ξ¦2
exp Ξ¦1
supp ππ
π© (π π(π))
Semialgebraic geometry The problems The results The proof
Construction of πΊ on Ξπ
For π β π΅ and π‘ β₯ π, by assumption there exists ππ β β[y]π such that
π π‘ππ(x) β‘ π π‘
ππ΄(x) ππ(π π‘ππ(x)) mod (x)π‘+1βJxKπ, βπ β πβ1(π).
Letβs fix a stratum Ξπ and π β Ξπ .By formal division, we may write ππ(y) = β ππ(y)Ξ¦π(y) + ππ(π, y) where the Ξ¦π are as above for βπ(π).Note that the remainder ππ(π, y) β β[y]π satisfies
ππ(y) β ππ(π, y) β βπ(π) and supp ππ(π, y) β π© (βπ(π))π .
LemmaπΊπ(π, y) β ππ (ππ(π, y)) is a semialgebraic Whitney field of order π on Ξπ satisfying (1).
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 17 / 20
exp Ξ¦2
exp Ξ¦1
supp ππ
π© (π π(π))
Semialgebraic geometry The problems The results The proof
πΊ is a Whitney field of order π on ΞπTo simplify the situation, we omit π.Thanks to Borelβs lemma with parameter, it is enough to check that π·π,π£πΊπβ1
π (π, y) = π·y,π£πΊπ(π, y).
Applying π·π,π£ β π·y,π£ toπ π
π π(y) β‘ π ππ π΄(y) ππ (π, y) mod (y)π+1βJyKπ
we get0 β‘ π π
π π΄(y) (π·π,π£π πβ1π (π, y) β π·y,π£ππ (π, y)) mod (y)π+1βJyKπ
thereforeπ·π,π£π πβ1
π (π, y) β π·y,π£ππ (π, y) β βπβ1(π)hence, by Chevalleyβs function,
π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y) β ππβ1(βπβ1(π)) = ππβ1(βπ(π))
butsupp (π·π,π£πΊπβ1
π (π, y) β π·y,π£πΊπ(π, y)) β π© (βπ(π))π
consequently, π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y) = 0. β
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 18 / 20
Semialgebraic geometry The problems The results The proof
πΊ is a Whitney field of order π on ΞπTo simplify the situation, we omit π.Thanks to Borelβs lemma with parameter, it is enough to check that π·π,π£πΊπβ1
π (π, y) = π·y,π£πΊπ(π, y).
Applying π·π,π£ β π·y,π£ toπ π
π π(y) β‘ π ππ π΄(y) ππ (π, y) mod (y)π+1βJyKπ
we get0 β‘ π π
π π΄(y) (π·π,π£π πβ1π (π, y) β π·y,π£ππ (π, y)) mod (y)π+1βJyKπ
thereforeπ·π,π£π πβ1
π (π, y) β π·y,π£ππ (π, y) β βπβ1(π)hence, by Chevalleyβs function,
π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y) β ππβ1(βπβ1(π)) = ππβ1(βπ(π))
butsupp (π·π,π£πΊπβ1
π (π, y) β π·y,π£πΊπ(π, y)) β π© (βπ(π))π
consequently, π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y) = 0. β
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 18 / 20
Semialgebraic geometry The problems The results The proof
πΊ is a Whitney field of order π on ΞπTo simplify the situation, we omit π.Thanks to Borelβs lemma with parameter, it is enough to check that π·π,π£πΊπβ1
π (π, y) = π·y,π£πΊπ(π, y).
Applying π·π,π£ β π·y,π£ toπ π
π π(y) β‘ π ππ π΄(y) ππ (π, y) mod (y)π+1βJyKπ
we get0 β‘ π π
π π΄(y) (π·π,π£π πβ1π (π, y) β π·y,π£ππ (π, y)) mod (y)π+1βJyKπ
thereforeπ·π,π£π πβ1
π (π, y) β π·y,π£ππ (π, y) β βπβ1(π)
hence, by Chevalleyβs function,π·π,π£πΊπβ1
π (π, y) β π·y,π£πΊπ(π, y) β ππβ1(βπβ1(π)) = ππβ1(βπ(π))but
supp (π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y)) β π© (βπ(π))π
consequently, π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y) = 0. β
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 18 / 20
Semialgebraic geometry The problems The results The proof
πΊ is a Whitney field of order π on ΞπTo simplify the situation, we omit π.Thanks to Borelβs lemma with parameter, it is enough to check that π·π,π£πΊπβ1
π (π, y) = π·y,π£πΊπ(π, y).
Applying π·π,π£ β π·y,π£ toπ π
π π(y) β‘ π ππ π΄(y) ππ (π, y) mod (y)π+1βJyKπ
we get0 β‘ π π
π π΄(y) (π·π,π£π πβ1π (π, y) β π·y,π£ππ (π, y)) mod (y)π+1βJyKπ
thereforeπ·π,π£π πβ1
π (π, y) β π·y,π£ππ (π, y) β βπβ1(π)hence, by Chevalleyβs function,
π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y) β ππβ1(βπβ1(π)) = ππβ1(βπ(π))
butsupp (π·π,π£πΊπβ1
π (π, y) β π·y,π£πΊπ(π, y)) β π© (βπ(π))π
consequently, π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y) = 0. β
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 18 / 20
Semialgebraic geometry The problems The results The proof
πΊ is a Whitney field of order π on ΞπTo simplify the situation, we omit π.Thanks to Borelβs lemma with parameter, it is enough to check that π·π,π£πΊπβ1
π (π, y) = π·y,π£πΊπ(π, y).
Applying π·π,π£ β π·y,π£ toπ π
π π(y) β‘ π ππ π΄(y) ππ (π, y) mod (y)π+1βJyKπ
we get0 β‘ π π
π π΄(y) (π·π,π£π πβ1π (π, y) β π·y,π£ππ (π, y)) mod (y)π+1βJyKπ
thereforeπ·π,π£π πβ1
π (π, y) β π·y,π£ππ (π, y) β βπβ1(π)hence, by Chevalleyβs function,
π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y) β ππβ1(βπβ1(π)) = ππβ1(βπ(π))
butsupp (π·π,π£πΊπβ1
π (π, y) β π·y,π£πΊπ(π, y)) β π© (βπ(π))π
consequently, π·π,π£πΊπβ1π (π, y) β π·y,π£πΊπ(π, y) = 0. β
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 18 / 20
Semialgebraic geometry The problems The results The proof
Gluing between strata
C - gluing between strata: the Εojasiewicz inequalityFix a stratum Ξπ . There exists π β β such that if π‘ β₯ π + π then lim
πβΞπ ⧡ΞππΊπ(π, y) = 0.
The constant term of the equation is flat on π΅β² hence on Ξπ ⧡ Ξπ β π΅β².
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 19 / 20
Semialgebraic geometry The problems The results The proof
Summary
We constructed πΊ(π, y) = β|πΌ|β€π
ππΌ (π)πΌ! yπΌ a semialgebraic Whitney field of order π on π΅ such that
βπ β π΅, βπ β πβ1(π), π ππ π(x) β‘ π π
π π΄(x) πΊ(π, π ππ π(x)) mod (x)π+1βJxKπ
Hence we obtain π βΆ βπ β βπ semialgebraic and ππ such that π β π΄ β (π β π) is π-flat on πβ1(π΅).
Loss of differentiabilityFor π β β, we set π β₯ ππ, then π β₯ π(π) and finally π‘(π) β π‘ β₯ π + π whereA. π is an upper bound of Whitneyβs loss of differentiability (induction step).B. π βΆ β β β is an upper bound of the Chevalley functions on the various strata.C. π is an upper bound of Εojasiewiczβs loss of differentiability on each stratum.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 20 / 20
Semialgebraic geometry The problems The results The proof
Summary
We constructed πΊ(π, y) = β|πΌ|β€π
ππΌ (π)πΌ! yπΌ a semialgebraic Whitney field of order π on π΅ such that
βπ β π΅, βπ β πβ1(π), π ππ π(x) β‘ π π
π π΄(x) πΊ(π, π ππ π(x)) mod (x)π+1βJxKπ
Hence we obtain π βΆ βπ β βπ semialgebraic and ππ such that π β π΄ β (π β π) is π-flat on πβ1(π΅).
Loss of differentiabilityFor π β β, we set π β₯ ππ, then π β₯ π(π) and finally π‘(π) β π‘ β₯ π + π whereA. π is an upper bound of Whitneyβs loss of differentiability (induction step).B. π βΆ β β β is an upper bound of the Chevalley functions on the various strata.C. π is an upper bound of Εojasiewiczβs loss of differentiability on each stratum.
J.-B. Campesato (joint work with E. Bierstone and P.D. Milman) ππ solutions of semialgebraic equations 20 / 20