Transcript Slide 1

On the robustness of
dictatorships:
spectral methods.
Ehud Friedgut,
Hebrew University, Jerusalem
Erdős-Ko-Rado (‘61)
• 407 links in Google
• 44 papers in MathSciNet with E.K.R. in
the title (not including the original one,
of course.)
The Erdős-Ko-Rado theorem
A fundamental theorem of extremal set
theory:
Extremal example: flower.
Product-measure analogue
Extremal example: dictatorship.
The Ahlswede-Khachatrian
theorem (special case)
Or...
Or...
Etc...
Product-measure analogue
Extremal example: duumvirate.
Beyond p < 1/3.
First observed and proven
by Dinur and Safra.
From the measure-case to
extremal set theory and back
Dinur and Safra proved the measure-results via
E.K.R. and Ahlswede-Khachatrian.
Here we attempt to prove measureresults using spectral methods, and
deduce some corollaries in extremal set
theory.
Robustness
A major incentive to use spectral analysis*
on the discrete cube as a tool for proving
theorems in extremal set theory:
Proving robustness statements.
“Close to maximal size
close to optimal structure.”
* Look for the purple star…
Intersection theorems,
spectral methods…
Some people who did related work
(there must be many others too):
Alon, Calderbank, Delsarte, Dinur,
Frankl, Friedgut, Furedi, Hoffman,
Lovász, Schrijver, Sudakov,
Wilson...
Theorem 1
*
Corollary 1 *
Theorem 2
*
Corollary 2
*
t-intersecting families
for t>1
We will use the case t=2 to
represent all t>1, the
differences are merely technical.
Digression:
Inspiration from a proof
of a graph theoretic
result
Spectral methods:
Hoffman’s theorem
Hoffman’s theorem,
sketch of proof
Sketch of proof, continued
Sketch of proof, continued
Sketch of proof, concluded
Stability observation:
Equality holds in Hoffman’s theorem only
if the characteristic function of a
maximal independent set is always a
linear combination of the trivial
eigenvector (1,1,...,1) and the
eigenvectors corresponding to the
minimal eigenvalue.
Also, “almost equality” implies “almost”
the above statement.
Intersecting families and
independent sets
Consider the graph whose vertices are
the subsets of {1,2,...,n}, with an edge
between two vertices iff the corresponding
sets are disjoint.
Intersecting family
Independent set
Can we mimic Hoffman’s proof?
Problems...
• The graph isn’t regular, (1,1,...,1) isn’t an
eigenvector.
• Coming to think of it, what are the
eigenvectors? How can we compute
them?
• Even if we could find them, they’re
orthogonal with respect to the uniform
measure, but we’re interested in a
different product measure.
Let’s look at the adjacency
matrix
Ø {1}
Ø
{1}
Ø {1} {2} {1,2}
Ø
{1}
{2}
{1,2}
This is good, because we can now compute
the eigenvectors and eigenvalues of
But...
These are not the eigenvectors we want...
...However, looking back at Hoffman’s
proof we notice that...
holds only because of the 0’s for non-edges
in A, not because of the 1’s. So...
Pseudo adjacency matrix
Ø {1}
Replace
Ø
{1}
It turns out that a
judicious choice is
By
Now everything works...
Their tensor products form an orthonormal
basis for the product space with the
product measure, and Hoffman’s proof
goes through (mutatis mutandis), yielding
that if I is an independent set then μ(I)≤p.
Remarks...
This is the minimal eigenvalue,
provided that p < ½ (!)
It is associated with eigenvectors
of the type
henceforth “first level eigenvectors”
Boolean functions;
Some facts of life
• Trivial : If all the Fourier coefficients are on
levels 0 and 1 then the function is a
dictatorship.
• Non trivial (FKN): If almost all the weight of
the Fourier coefficients is on levels 0 and 1
then the function is close to a dictatorship.
• Deep (Bourgain, Kindler-Safra): Something
similar is true if almost all the weight is on
levels 0,1,…,k.
Remarks, continued...
• These facts of life, together with the
“stability observation” following
Hoffman’s proof imply the uniqueness
and robustness of the extremal
examples, the dictatorships .
• The proof only works for p< ½ ! (At
p=1/2 the minimal eigenvalue shifts
from one set of eigenvectors to
another)
2-intersecting families
Can we repeat this proof for 2-intersecting
families?
Let’s start by taking a look at the adjacency
matrix...
The 2-intersecting adjacency
matrix
This doesn’t
look like the
tensor product
of smaller
matrices...
Understanding the
intersection matrices
The “0” in
(the 1-intersection matrix) warned us
that when we add the same element to
two disjoint sets they become
intersecting.
Now we want to be more tolerant:
Different tactics for 2intersecting
One common element= “warning”
But “two strikes, and yer out!’”
We need an element
such that
Obvious solution:
Working over a ring
The solution: work over
Ø
{1}
Ø {1}
Ø
{1}
{2}
{1,2}
Ø {1} {2} {1,2}
2-Intersection matrix over
Now
becomes...
Working over a ring, continued...
• Same as before: we wish to replace
by some matrix
to obtain the
“proper” eigenvectors.
• Different than before: the eigenvalues are
now ring elements,
so there’s no “minimal eigenvalue”.
Working over the ring, cont’d
Identities such as
Now become
,
so, comparing coefficients, we can
get a separate equation for the ηs
and for the ρs…
…and after replacing the equalities
by inequalities solve a L.P. problem
…More problems
However, the ηs and the ρs do not tensor
separately (they’re not products of the
coefficients in the case n=1.)
Lord of the rings, part III
It turns out that now one has to know the
value of n in advance before plugging the
values into
If you plug in
a ***miracle*** happens...
2-intersecting - conclusion
...The solution of the L.P. is such that all the
non-zero coefficients must belong only to the
first level eigenvectors, or the second level
eigenvectors.
Using some additional analysis of Boolean
functions (involving [Kindler-Safra]) one may finally
prove the uniqueness and robustness result
about duumvirates.
Oh..., and the miracle breaks down at
p =1/3…
Questions...
• What about 3-intersecting families?
(slight optimism.)
• What about p > 1/3 ? (slight pessimism.)
• What about families with no
(heavy pessimism.)
?
• Stability results in coding theory and
association schemes?...
Time will tell...
Have we struck a small gold mine...
...or just found a shiny
coin?