### George G. Lorentz

```RELATION BETWEEN SOLVABILITY OF
SOME MULTIVARIATE INTERPOLATION
PROBLEMS AND THE VARIETY OF
SUBSPACE ARRANGEMENT.
The subspace arrangement is
=
=1
where  are linear subspaces of ℂ  .
=

=1{
1 , … ,  :

=1
= 0}, dim  =  −
⊆ ℂ  is an affine variety   = {  }=d-s
Algebraic geometry studies the property
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   = 0 + 1  + 2  2 + 3  3 :
1 = 1 ,
′ (1 ) = 2 , ′ 2 = 3 ′′′ 2 = 4
If this equations have unique solution for all distinct 1 , 2 ∈
and all 1 , 2 , 3 , 4 ∈  the problem (च3 , 0,1 , {1,3}) is regular.
det
1
1
1 2
1 3
0
1
21
31 2
0
1
22
32 2
0
0
0
3
=(1 , 2 ) = 62 −61
?
φ : = {(1 , 2 ) :(1 , 2 )} ⊆  = {(1 , 2 ) :1 =2 )}
Example: find a polynomial   = 0 + 1  + 2  2 + 3  3
1 = 1 ,  ′ ′(1 ) = 2 , ′ 2 = 3 ,
det
1
1
1 2
1 3
0
1
22
32 2
0
0
2
61
0
0
0
6
′′′ 2 = 4
=(1 , 2 ) =12
the problem (च3 , 0,2 , {1,3}) is regular, completely regular.
φ : = {(1 , 2 ) :(1 , 2 )=0}=∅ ⊆  1 − 2 = {(1 , 2 ) :1 =2 )}
0,2
{1,3})= ∅  0,2 ∪ {1,3}={0,1,2,3,}
Birkhoff Interpolation problem:
(च ,1 , …  ),  ⊆ {0, … ,  + 1} where
⊆ {0, … ,  + 1}; 1 # =  + 1 =  च is regular if for any set
of k district points {1 , … ,  } ⊆  the interpolation problem and any
f∈ च there exists unique p∈ च such that  () ( )=() ( ) for all k∈
and all j=1,…,k.
George Pólya
Isaac Schoenberg
The problem (च ,1 , …  ),  ⊆ 0, … ,  + 1 is regular if
1) k=n+1,  ={0}, Lagrange interpolation
2) For k>0, k∈  ( − 1) ∈  , Hermite interpolation
3)   = ∅, ( ∪  = {0, … ,  + 1} ).
ℂ ,
>1
Take (च=span{1,x,y,xy}, 1 , 2 ), 1 , 2 ⊆ 0,1 × 0,1
ℎ     :
1 =(1,1 , 1,2 ), 2 =(2,1 , 2,2 ):
= (1 , 2 )= 1 =(1,1 , 1,2 , 2,1 , 2,2 ) ∈ ℂ[1,1 , 1,2 , 2,1 , 2,2 ],
φ ⊆ ℂ4 ; dim  φ =3 if ≠ .
= { 1,1 , 1,2 , 2,1 , 2,2 : 1 = 2 , . . , 1,1 = 2,1 , 1,2 = 2,2}
dim  =2
If the scheme is regular then
φ ⊆
George G. Lorentz
If ≠ .
If = . ≠ 0
φ ≤
φ =3≤  = 2
ℎ
1 2 ≠ ∅ and 1 = 2 then the determinant
φ contains two identical rows (columns), hence
1 2 = ∅
The same is true in real case
Birkhoff Interpolation (in several variables)
Carl de Boor
Given a subspace च ⊆ [1 , … ,  ], a collection of
subspaces 1 , … ,  ⊆ च, a set of points 1 , … ,  ∈
and a function f ∈ [1 , … ,  ] we want to find a polynomial
p ∈ च such that
(1 , … ,  )f( )=  (1 , … ,  )f( )
for all q∈  .
Amos Ron
The scheme (च,1 , …  ) is regular if the problem has a
unique solutions for all f ∈ [1 , … ,  ] and all distinct
1 , … ,  ∈  , it is completely regular if it has unique
solution for all 1 , … ,  ∈  .
Birkhoff Interpolation (in several variables)
= (1 ,…,  ) is a polynomial in  ×
=
≠ , ; ,
= (1 , … ,  ∈ ℂ :  =  }
dim  =max{dim, }=kd-d
φ ⊆
If ≠ .
φ =dk-1≤  − =
Theorem: If (च,1 , …  ) is regular then  ∩  = {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 , … , ℎ }
{1 , … ,  } ⊆
the determinant  ℎ ( ) ≠ 0
(The Lagrange interpolation problem is well posed)
च−1         ℎ     ℂ[x].
For d>1, n>1 there are no n-dimensional Haar subspaces in
1 , … ,  or even in C( )
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 {1 , … ,  }  C( )
is a Haar covering of d if for any distinct points
{1 , … ,  } ⊆
the Lagrange interpolation problem is well posed in one of these spaces.
=span { ℎ1 () , … , ℎ () }
{1 , … ,  }is a Haar covering of d if for any distinct points
{1 , … ,  } ⊆    ℎ
Δ 1 , … ,  =  ℎ1 () ( ) ≠ 0
Question: What is the minimal number s:=η ( ) of n-dimensional
subspaces {1 , … ,  }   1 , … ,  that forms a Haar covering of d
and what are those subspaces?
Kyungyong Lee
Conjecture: : η ℂ2 =

η 2 2 = 2
1 =  1,  , 2 =  1,

Theorem (Stefan Tohaneanu& B.S.): η3 ℂ2 = 3
The three spaces
1 =  1, ,  , 2 =  1, ,  2 , 3 =  1, ,  2 will do:
It remains to show that no two subspaces can do the job.
1 =span { ℎ1 (1) , ℎ2 (1) , ℎ3 (1) } ; 2 =span { ℎ1 (2) , ℎ2 (2) , ℎ3 (2) }
I want show that there are three distinct points 1 , 2 , 3 = 0 ℎ ℎ
Δ1 1 , 2 , 0 =0, Δ2 1 , 2 , 0 =0
1 = (1,1 , 1,2), 2 = (2,1 , 2,2 ),
≔
0,0, a, b
∪ {(a, b, 0,0) ∪ {(a, b, a, b)}
Δ1 , Δ2 ={(1,1 , 1,2 , 2,1 , 2,2 ): Δ1 = 0  Δ2 =0}
Δ1 , Δ2 ⊆ ℂ4
≔
0,0, a, b
∪ {(a, b, 0,0) ∪ {(a, b, a, b)}
(0,0,1,1)
(0,0,0,0)

Δ1 , Δ2 =  ?
dim  Δ1 , Δ2 =2=dim
(-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,  is not a complete intersection:
Robin Hartshorne
\(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 (2 )=3
η4 (2 ) ≤ 4
d ≤ η   < ∞
1 =  1, ,  , 2 =  1, ,  2 , 3 =  1, ,  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
2

1

2
3

1

1
2
Yang tableaux

Cojecture: η ( )=
3
2

1
+−2
−1

köszönöm
Thank You
Thank You
köszönöm
```