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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.)

Product-measureanalogue

Extremal example: dictatorship.

Product-measureanalogue

Extremal example: duumvirate.

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.

close to optimal structure.”

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.

* 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...

t-intersecting familiesfor 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

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 family Independent set

Intersecting families and independent setsConsider 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?

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.

Ø

{1}

Ø

{1}

{2}

{1,2}

Ø

{1}

This is good, because we can now compute

the eigenvectors and eigenvalues of

Let’s look at the adjacency matrixBut...

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...

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”

Remarks...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...

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 2-intersecting

One common element= “warning”

But “two strikes, and yer out!’”

We need an element such that

Obvious solution:

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?...

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