Loading in 5 sec....

On the robustness of dictatorships: spectral methods.PowerPoint Presentation

On the robustness of dictatorships: spectral methods.

- 90 Views
- Uploaded on
- Presentation posted in: General

On the robustness of dictatorships: spectral methods.

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

On the robustness of dictatorships: spectral methods.

Ehud Friedgut,

Hebrew University, Jerusalem

- 407 links in Google
- 44 papers in MathSciNet with E.K.R. in the title (not including the original one, of course.)

A fundamental theorem of extremal set theory:

Extremal example: flower.

Extremal example: dictatorship.

Or...

Or...

Etc...

Extremal example: duumvirate.

First observed and proven

by Dinur and Safra.

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 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…

Some people who did related work

(there must be many others too):

Alon, Calderbank, Delsarte, Dinur,

Frankl, Friedgut, Furedi, Hoffman,

Lovász, Schrijver, Sudakov,

Wilson...

*

*

*

*

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

Sketch of proof, continued

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

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.

Ø {1} {2} {1,2}

Ø

{1}

Ø

{1}

{2}

{1,2}

Ø

{1}

This is good, because we can now compute

the eigenvectors and eigenvalues of

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...

Ø

{1}

Ø

{1}

By

Replace

It turns out that a judicious choice is

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”

- 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.

- 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)

Can we repeat this proof for 2-intersecting

families?

Let’s start by taking a look at the adjacency

matrix...

This doesn’t

look like the

tensor product

of smaller

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:

One common element= “warning”

But “two strikes, and yer out!’”

We need an element such that

Obvious solution:

The solution: work over

Ø {1}

Ø

{1}

Ø {1} {2} {1,2}

Ø

{1}

{2}

{1,2}

Now

becomes...

- 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”.

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

However, the ηs and the ρs do not tensor

separately (they’re not products of the

coefficients in the case n=1.)

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...

...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…

?

- 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?...

Have we struck a small gold mine...

...or just found a shiny coin?

Thank you for your attention!