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On the robustness of dictatorships:
spectral methods.Ehud Friedgut,
Hebrew University, Jerusalem
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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.)
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The Erdős-Ko-Rado theorem
A fundamental theorem of extremal set theory:
Extremal example: flower.
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Product-measure analogue
Extremal example: dictatorship.
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The Ahlswede-Khachatriantheorem (special case)
Etc...
Or...
Or...
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Product-measure analogue
Extremal example: duumvirate.
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Beyond p < 1/3.
First observed and provenby Dinur and Safra.
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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 measure-results using spectral methods, and deduce some corollaries in extremal set theory.
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RobustnessA 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…
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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...
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Theorem 1
*
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Corollary 1*
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Theorem 2
*
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Corollary 2*
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t-intersecting familiesfor t>1
We will use the case t=2 to represent all t>1, the differences are merely
technical.
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Digression:
Inspiration from a proof of a graph theoretic result
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Spectral methods:Hoffman’s theorem
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Hoffman’s theorem,sketch of proof
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Sketch of proof, continued
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Sketch of proof, continued
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Sketch of proof, concluded
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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.
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Intersecting families and independent sets
Consider the graph whose vertices arethe subsets of {1,2,...,n}, with an edge between two vertices iff the correspondingsets are disjoint.
Intersecting family Independent set
Can we mimic Hoffman’s proof?
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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.
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Let’s look at the adjacency matrix
Ø
Ø
{1}
{1}
Ø {1} {2} {1,2}
Ø{1}{2}{1,2}
This is good, because we can now computethe eigenvectors and eigenvalues of
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But...
These are not the eigenvectors we want...
...However, looking back at Hoffman’sproof we notice that...
holds only because of the 0’s for non-edgesin A, not because of the 1’s. So...
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Pseudo adjacency matrix
ReplaceØ
Ø
{1}
{1}
By
It turns out that a judicious choice is
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Now everything works...
Their tensor products form an orthonormalbasis 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.
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Remarks...
It is associated with eigenvectorsof the type henceforth “first level eigenvectors”
This is the minimal eigenvalue,provided that p < ½ (!)
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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.
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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)
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2-intersecting families
Can we repeat this proof for 2-intersecting
families?
Let’s start by taking a look at the adjacency
matrix...
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The 2-intersecting adjacencymatrix
This doesn’tlook like the tensor productof smallermatrices...
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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:
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Different tactics for 2-intersecting
One common element= “warning”
But “two strikes, and yer out!’”
We need an element such that
Obvious solution:
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Working over a ring
The solution: work over
Ø {1} {2} {1,2}Ø{1}{2}{1,2}
Ø {1}Ø
{1}
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Now becomes...
2-Intersection matrix over
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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”.
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Working over the ring, cont’d
Identities such as
Now become ,so, comparing coefficients, we canget a separate equation for the ηsand for the ρs…
…and after replacing the equalitiesby inequalities solve a L.P. problem
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…More problems
However, the ηs and the ρs do not tensorseparately (they’re not products of the
coefficients in the case n=1 ).
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Lord of the rings, part IIIIt turns out that now one has to know thevalue of n in advance before plugging thevalues into
If you plug in
a ***miracle*** happens...
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2-intersecting - conclusion
...The solution of the L.P. is such that all the non-zero coefficients must belong only to thefirst 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…
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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?...
?
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Time will tell...
Have we struck a small gold mine...
...or just found a shiny coin?
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