RELATION BETWEEN SOLVABILITY OF
SOME MULTIVARIATE INTERPOLATION
PROBLEMS AND THE VARIETY OF
SUBSPACE ARRANGEMENT.
The subspace arrangement is
𝓗= 𝑚
𝑗=1 𝐻𝑗
where 𝐻𝑗 are linear subspaces of ℂ 𝑑 .
𝐻𝑗 =
𝑠
𝑘=1{
𝑧1 , … , 𝑧𝑑 :
𝑑
𝑖=1 𝑐𝑖 𝑧𝑖
= 0}, dim 𝐻𝑗 = 𝑑 − 𝑠
𝓗 ⊆ ℂ 𝑑 is an affine variety 𝑑𝑖𝑚 𝓗 = 𝑚𝑎𝑥{𝑑𝑖𝑚 𝐻𝑗 }=d-s
Algebraic geometry studies the property 𝓗
George David Birkhoff
Birkhoff Interpolation
(in one variable)
Birkhoff interpolation is an extension of Hermite interpolation.
It involves matching of values and derivatives of a function at certain
points without the requirement that the derivatives are consecutive.
Example: find a polynomial 𝑓 𝑥 = 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 :
𝑓 𝑧1 = 𝑎1 ,
𝑓 ′ (𝑧1 ) = 𝑎2 , 𝑓′ 𝑧2 = 𝑎3 𝑓′′′ 𝑧2 = 𝑎4
If this equations have unique solution for all distinct 𝑧1 , 𝑧2 ∈ 𝕜
and all 𝑎1 , 𝑎2 , 𝑎3 , 𝑎4 ∈ 𝕜 the problem (च3 , 0,1 , {1,3}) is regular.
det
1
𝑧1
𝑧1 2
𝑧1 3
0
1
2𝑧1
3𝑧1 2
0
1
2𝑧2
3𝑧2 2
0
0
0
3
=𝞅(𝑧1 , 𝑧2 ) = 6𝑧2 −6𝑧1
?
𝒵 φ : = {(𝑧1 , 𝑧2 ) :𝞅(𝑧1 , 𝑧2 )} ⊆ 𝓗 = {(𝑧1 , 𝑧2 ) :𝑧1 =𝑧2 )}
Example: find a polynomial 𝑓 𝑥 = 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3
𝑓 𝑧1 = 𝑎1 , 𝑓 ′ ′(𝑧1 ) = 𝑎2 , 𝑓′ 𝑧2 = 𝑎3 ,
det
1
𝑧1
𝑧1 2
𝑧1 3
0
1
2𝑧2
3𝑧2 2
0
0
2
6𝑧1
0
0
0
6
𝑓′′′ 𝑧2 = 𝑎4
=𝞅(𝑧1 , 𝑧2 ) =12
the problem (च3 , 0,2 , {1,3}) is regular, completely regular.
𝒵 φ : = {(𝑧1 , 𝑧2 ) :𝞅(𝑧1 , 𝑧2 )=0}=∅ ⊆ 𝒵 𝑧1 − 𝑧2 = {(𝑧1 , 𝑧2 ) :𝑧1 =𝑧2 )}
0,2
{1,3})= ∅ 𝑎𝑛𝑑 0,2 ∪ {1,3}={0,1,2,3,}
Birkhoff Interpolation problem:
(च𝑛 ,𝑀1 , … 𝑀𝑘 ), 𝑀𝑗 ⊆ {0, … , 𝑛 + 1} where
𝑀𝑗 ⊆ {0, … , 𝑛 + 1}; 𝑘1 #𝑀𝑗 = 𝑛 + 1 = 𝑑𝑖𝑚 च𝑛 is regular if for any set
of k district points {𝑧1 , … , 𝑧𝑘 } ⊆ 𝕜 the interpolation problem and any
f∈ च there exists unique p∈ च𝑛 such that 𝑓 (𝑘) (𝑧𝑗 )=𝑝(𝑘) (𝑧𝑗 ) for all k∈
𝑀𝑗 and all j=1,…,k.
George Pólya
Isaac Schoenberg
The problem (च𝑛 ,𝑀1 , … 𝑀𝑘 ), 𝑀𝑗 ⊆ 0, … , 𝑛 + 1 is regular if
1) k=n+1, 𝑀𝑗 ={0}, Lagrange interpolation
2) For k>0, k∈ 𝑀𝑗 (𝑘 − 1) ∈ 𝑀𝑗 , Hermite interpolation
3) 𝑀𝑗 𝑀𝑠 = ∅, ( ∪ 𝑀𝑗 = {0, … , 𝑛 + 1} ).
ℂ 𝑑, 𝑑
>1
Take (च=span{1,x,y,xy}, 𝑀1 , 𝑀2 ), 𝑀1 , 𝑀2 ⊆ 0,1 × 0,1
𝑎𝑛𝑑 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑡ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑎𝑡 𝑡𝑤𝑜 𝑝𝑜𝑖𝑛𝑡𝑠:
𝑧1 =(𝑧1,1 , 𝑧1,2 ), 𝑧2 =(𝑧2,1 , 𝑧2,2 ):
𝑑𝑒𝑡 = 𝞅(𝑧1 , 𝑧2 )= 𝑥1 =𝞅(𝑧1,1 , 𝑧1,2 , 𝑧2,1 , 𝑧2,2 ) ∈ ℂ[𝑧1,1 , 𝑧1,2 , 𝑧2,1 , 𝑧2,2 ],
𝒵 φ ⊆ ℂ4 ; dim 𝒵 φ =3 if 𝞅≠ 𝑐𝑜𝑛𝑠𝑡.
𝓗 = { 𝑧1,1 , 𝑧1,2 , 𝑧2,1 , 𝑧2,2 : 𝑧1 = 𝑧2 , 𝑖. 𝑒. , 𝑧1,1 = 𝑧2,1 , 𝑧1,2 = 𝑧2,2}
dim 𝓗 =2
If the scheme is regular then
𝒵 φ ⊆𝓗
George G. Lorentz
If 𝞅≠ 𝑐𝑜𝑛𝑠𝑡.
If 𝞅= 𝑐𝑜𝑛𝑠𝑡. ≠ 0
𝑑𝑖𝑚𝒵 φ ≤ 𝑑𝑖𝑚𝓗
𝑑𝑖𝑚 𝒵 φ =3≤ 𝑑𝑖𝑚𝓗 = 2
𝑠𝑐ℎ𝑒𝑚𝑒 𝑖𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑙𝑦 𝑟𝑒𝑔𝑢𝑙𝑎𝑟
𝐼𝑓 𝑀1 𝑀2 ≠ ∅ and 𝑧1 = 𝑧2 then the determinant
φ contains two identical rows (columns), hence
𝑀1 𝑀2 = ∅
The same is true in real case
Birkhoff Interpolation (in several variables)
Carl de Boor
Given a subspace च ⊆ 𝕜[𝑥1 , … , 𝑥𝑑 ], a collection of
subspaces 𝑄1 , … , 𝑄𝑘 ⊆ च, a set of points 𝑧1 , … , 𝑧𝑘 ∈ 𝕜𝑑
and a function f ∈ 𝕜[𝑥1 , … , 𝑥𝑑 ] we want to find a polynomial
p ∈ च such that
𝑞(𝐷1 , … , 𝐷𝑑 )f(𝑧𝑗 )= 𝑞 (𝐷1 , … , 𝐷𝑑 )f(𝑧𝑗 )
for all q∈ 𝑄𝑗 .
Amos Ron
The scheme (च,𝑄1 , … 𝑄𝑘 ) is regular if the problem has a
unique solutions for all f ∈ 𝕜[𝑥1 , … , 𝑥𝑑 ] and all distinct
𝑧1 , … , 𝑧𝑘 ∈ 𝕜𝑑 , it is completely regular if it has unique
solution for all 𝑧1 , … , 𝑧𝑘 ∈ 𝕜𝑑 .
Birkhoff Interpolation (in several variables)
𝑑𝑒𝑡 = 𝞅(𝑧1 ,…, 𝑧𝑘 ) is a polynomial in 𝑑 × 𝑘 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
𝓗=
𝑖≠𝑗 𝐻𝑖,𝑗 ; 𝐻𝑖,𝑗
= (𝑧1 , … , 𝑧𝑘 ∈ ℂ𝑘𝑑 : 𝑧𝑖 = 𝑧𝑗 }
dim 𝓗 =max{dim𝐻𝑖,𝑗 }=kd-d
𝑟𝑒𝑔𝑢𝑙𝑎𝑟𝑖𝑡𝑦 𝑚𝑒𝑎𝑛𝑠 𝒵 φ ⊆ 𝓗
If 𝞅≠ 𝑐𝑜𝑛𝑠𝑡.
𝑑𝑖𝑚 𝒵 φ =dk-1≤ 𝑘𝑑 − 𝑑= 𝑑𝑖𝑚𝓗
Theorem: If (च,𝑄1 , … 𝑄𝑘 ) is regular then 𝑄𝑖 ∩ 𝑄𝑗 = {0}
Conjecture: it is true in the real case
False
Rong-Qing Jia
A. Sharma
HAAR SUBSPACES AND
HAAR COVERINGS
Alfréd Haar
Definition: H=span { ℎ1 , … , ℎ𝑛 } 𝑖𝑓 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑑𝑖𝑠𝑡𝑖𝑛𝑐𝑡 𝑝𝑜𝑖𝑛𝑡𝑠
{𝑧1 , … , 𝑧𝑛 } ⊆ 𝕜𝑑
the determinant 𝑑𝑒𝑡 ℎ𝑖 (𝑧𝑗 ) ≠ 0
(The Lagrange interpolation problem is well posed)
च𝑛−1 𝑖𝑠 𝑎 𝐻𝑎𝑎𝑟 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 𝕜 𝑥 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑢𝑛𝑖𝑞𝑢𝑒 𝐻𝑎𝑎𝑟 𝑠𝑢𝑠𝑝𝑎𝑐𝑒 𝑖𝑛 ℂ[x].
For d>1, n>1 there are no n-dimensional Haar subspaces in
𝕜 𝑥1 , … , 𝑥𝑑 or even in C(𝕜𝑑 )
J. C. Mairhuber in real case and I. Schoenberg in the complex
case;.
or from the previous discussion: (च,{0},{0},…,{0})
Definition: A family of n-dimensional subspaces {𝐻1 , … , 𝐻𝑠 } 𝑖𝑛 C(𝕜𝑑 )
is a Haar covering of 𝕜d if for any distinct points
{𝑧1 , … , 𝑧𝑛 } ⊆ 𝕜𝑑
the Lagrange interpolation problem is well posed in one of these spaces.
𝐻𝑘 =span { ℎ1 (𝑘) , … , ℎ𝑛 (𝑘) }
{𝐻1 , … , 𝐻𝑠 }is a Haar covering of 𝕜d if for any distinct points
{𝑧1 , … , 𝑧𝑛 } ⊆ 𝕜𝑑 𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑠
Δ𝑘 𝑧1 , … , 𝑧𝑛 = 𝑑𝑒𝑡 ℎ1 (𝑘) (𝑧𝑗 ) ≠ 0
Question: What is the minimal number s:=η𝑛 (𝕜𝑑 ) of n-dimensional
subspaces {𝐻1 , … , 𝐻𝑠 } 𝑖𝑛 𝕜 𝑥1 , … , 𝑥𝑑 that forms a Haar covering of 𝕜d
and what are those subspaces?
Kyungyong Lee
Conjecture: : η𝑛 ℂ2 = 𝑛
𝑦
η 2 𝕜2 = 2
𝐻1 = 𝑠𝑝𝑎𝑛 1, 𝑥 , 𝐻2 = 𝑠𝑝𝑎𝑛 1, 𝑦
𝑥
Theorem (Stefan Tohaneanu& B.S.): η3 ℂ2 = 3
The three spaces
𝐻1 = 𝑠𝑝𝑎𝑛 1, 𝑥, 𝑦 , 𝐻2 = 𝑠𝑝𝑎𝑛 1, 𝑥, 𝑥 2 , 𝐻3 = 𝑠𝑝𝑎𝑛 1, 𝑦, 𝑦 2 will do:
It remains to show that no two subspaces can do the job.
𝐻1 =span { ℎ1 (1) , ℎ2 (1) , ℎ3 (1) } ; 𝐻2 =span { ℎ1 (2) , ℎ2 (2) , ℎ3 (2) }
I want show that there are three distinct points 𝑧1 , 𝑧2 , 𝑧3 = 0 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
Δ1 𝑧1 , 𝑧2 , 0 =0, Δ2 𝑧1 , 𝑧2 , 0 =0
𝑧1 = (𝑧1,1 , 𝑧1,2), 𝑧2 = (𝑧2,1 , 𝑧2,2 ),
𝓗≔
0,0, a, b
∪ {(a, b, 0,0) ∪ {(a, b, a, b)}
𝒵 Δ1 , Δ2 ={(𝑧1,1 , 𝑧1,2 , 𝑧2,1 , 𝑧2,2 ): Δ1 = 0 𝑎𝑛𝑑 Δ2 =0}
𝒵 Δ1 , Δ2 ⊆ ℂ4
𝓗≔
0,0, a, b
∪ {(a, b, 0,0) ∪ {(a, b, a, b)}
(0,0,1,1)
(0,0,0,0)
𝓗
𝒵 Δ1 , Δ2 = 𝓗 ?
dim 𝒵 Δ1 , Δ2 =2=dim 𝓗
(-1,-1,0,0)
Not every two-dimensional variety in ℂ4 can be formed as a set
of common zeroes of two polynomials. The ones that can are
called (set theoretic) complete intersections.
As luck would have it, 𝓗 is not a complete intersection:
Robin Hartshorne
𝓗\(0,0,0,0) is not connected… Alexander Grothendieck
A very deep theorem states that a two dimensional complete
intersection in ℂ4 can not be disconnected by removing just one
point
(Tom McKinley and B.S.): η3 (𝕜2 )=3
η4 (𝕜2 ) ≤ 4
d ≤ η 𝑛 𝕜𝑑 < ∞
𝐻1 = 𝑠𝑝𝑎𝑛 1, 𝑥, 𝑦 , 𝐻2 = 𝑠𝑝𝑎𝑛 1, 𝑥, 𝑥 2 , 𝐻3 = 𝑠𝑝𝑎𝑛 1, 𝑦, 𝑦 2
The family of D-invariant subspaces spanned by monomials
form a finite Haar covering.
There are too many of them: for n=4 in 2 dimensions there are five:
1
𝑦2
𝑦
1
𝑥
𝑥2
𝑥3
𝑦
1
𝑥
𝑦
1
𝑥2
Yang tableaux
𝑥
Cojecture: η𝑛 (𝕜𝑑 )=
𝑦3
𝑦2
𝑦
1
𝑛+𝑑−2
𝑛−1
𝑥𝑦
𝑥
köszönöm
Thank You
Thank You
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