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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.

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On the robustness of dictatorships spectral methods

On the robustness of dictatorships: spectral methods.

Ehud Friedgut,

Hebrew University, Jerusalem

Erd s ko rado 61
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 erd s ko rado theorem
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
The Ahlswede-Khachatriantheorem (special case)




Product measure analogue1

Extremal example: duumvirate.

Beyond p 1 3
Beyond p < 1/3.

First observed and proven

by Dinur and Safra.

From the measure case to extremal set theory and back
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



“Close to maximal size

close to optimal structure.”



A major incentive to use spectral analysis

on the discrete cube as a tool for proving theorems in extremal set theory:

Proving robustness statements.

* Look for the purple star…

Intersection theorems spectral methods
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,


T intersecting families for t 1
t-intersecting familiesfor t>1

We will use the case t=2 to

represent all t>1, the

differences are merely technical.


Inspiration from a proof of a graph theoretic result

Spectral methods hoffman s theorem
Spectral methods:Hoffman’s theorem

Hoffman s theorem sketch of proof
Hoffman’s theorem,sketch of proof

Stability observation
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

Intersecting family Independent set

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.

Can we mimic Hoffman’s proof?


  • 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} {2} {1,2}









This is good, because we can now compute

the eigenvectors and eigenvalues of

Let’s look at the adjacency matrix

On the robustness of dictatorships spectral methods

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





Pseudo adjacency matrix



It turns out that a judicious choice is

Now everything works
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.


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
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
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
2-intersecting families

Can we repeat this proof for 2-intersecting


Let’s start by taking a look at the adjacency


The 2 intersecting adjacency matrix
The 2-intersecting adjacencymatrix

This doesn’t

look like the

tensor product

of smaller


Understanding the intersection 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


Now we want to be more tolerant:

Different tactics for 2 intersecting
Different tactics for 2-intersecting

One common element= “warning”

But “two strikes, and yer out!’”

We need an element such that

Obvious solution:

Working over a ring
Working over a ring

The solution: work over

Ø {1}



Ø {1} {2} {1,2}





Working over a ring continued
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
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
…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
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
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


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…




  • 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
Time will tell...

Have we struck a small gold mine...

...or just found a shiny coin?