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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-Khachatrian theorem (special case) Etc... Or...

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Product-measure analogue Extremal example: duumvirate.

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Beyond p < 1/3. First observed and proven by 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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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…

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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 families for 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, 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 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?

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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} Ø {1} {2} {1,2} Ø {1} {2} {1,2} This is good, because we can now compute the eigenvectors and eigenvalues of

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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...

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Pseudo adjacency matrix Replace Ø Ø {1} By It turns out that a judicious choice is

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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.

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Remarks... It is associated with eigenvectors of 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 adjacency matrix This doesn’t look like the tensor product of smaller matrices...

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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 can get a separate equation for the ηs and for the ρs… …and after replacing the equalities by inequalities solve a L.P. problem

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…More problems However, the ηs and the ρs do not tensor separately (they’re not products of the coefficients in the case n=1.)

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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...

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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 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…

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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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