The subspace arrangement is w here are linear subspaces of ds Algebraic geometry studies the property Birkhoff Interpolation in one variable George David ID: 528916
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Slide1
Relation between solvability of some multivariate interpolation problems and the variety of subspace arrangement.
The subspace arrangement is=where are linear subspaces of .
=
}=d-s
Algebraic geometry studies the property
Slide2
Birkhoff Interpolation
(in one variable)George David Birkhoff 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
If this equations have unique solution for all distinct
and all
the problem (
is regular.
=
𝞅(
,
)
d
et
) :
𝞅(
,
)
}
?
Slide3
Example: find a polynomial
,
the problem (
is regular, completely regular.
) :
𝞅(
,
)
=0}=
d
et
=
Slide4
Birkhoff
Interpolation problem:
Isaac Schoenberg
George
Pólya
,
where
;
is regular
if for any set of k district points
and any f
there exists unique p
such that
)=
)
for all k
and all j=1,…,k.
The problem
,
is regular if
k=n+1,
={0}, Lagrange interpolation
For k>0, k
,
Hermite
interpolation
, (
=
).
Slide5
George G. Lorentz
The same is true in real case
=(
,
=(,
𝞅(
,
)
=
=
𝞅(
,
,
)
If
𝞅
=3
; dim
=3 if 𝞅 If the scheme is regular then
:
}
d
im
=2
If
𝞅
and
then the determinant
contains two identical rows (columns), hence
Slide6
Carl de Boor
Amos Ron
Given a subspace
a collection of subspaces
, a set of points
and a function f
] we want to find a polynomial p
such that
(
)f(
)=
(
)f
(
)
f
or all q
The scheme (
,
is regular if the problem has a unique solutions for all
f
]
and all distinct
, it is completely regular if it has unique solution for all
.
Birkhoff
Interpolation (in several variables)Slide7
Birkhoff
Interpolation (in several variables)
𝞅
(
,
…,
)
is a polynomial in
If
𝞅
=
dk-1
=
=
dim
=max{dim
}=kd-d
Theorem: If
(
,
is regular
then
Rong
-Qing
Jia
A. Sharma
Conjecture: it is true in the real case
False
Slide8
Haar subspaces and Haar coverings
Alfréd HaarDefinition: H=span {
the determinant
[x]
.
For d>1, n>1 there are no n-dimensional
Haar
subspaces in
or even in C(
)
J. C.
Mairhuber
in real case and I. Schoenberg in the complex case;.
(The Lagrange interpolation problem is well posed)
or
from the
previous
discussion: (
,{0},{0},…,{0})
Slide9
Definition: A family of n-dimensional subspaces {
the Lagrange interpolation problem is well posed in one of these spaces.=span {
{
Question: What is the minimal number s:=
of
n
-dimensional subspaces
?
Conjecture:
:
Kyungyong
Lee Slide10
Theorem (Stefan Tohaneanu& B.S.):
It remains to show that no two subspaces can do the job.
,
The three spaces
=span
{
=0
I want show that there are three distinct points
),
),
Slide11
dim =2=dim Not every two-dimensional variety in 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:
is not connected… A very deep theorem states that a two dimensional complete intersection in
can not be disconnected by removing just one point
Robin Hartshorne
Alexander
Grothendieck
={(
):
=0}
(0,0,1,1)
(-1,-1,0,0)
(0,0,0,0)
Slide12
=3
=
,
The family of
D
-invariant subspaces spanned by monomials form a finite
Haar
covering.
T
here are too many of them: for n=4 in 2 dimensions there are five:
Yang tableauxSlide13
Thank You
Thank You
köszönöm
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