Are Almost All Graphs Determined by their Spectrum?faculty.nps.edu/rgera/Conjectures/JointMeetings-2014/godsil-cospec.pdfAgainst For Walk Matrices Wang & Xu WangandXu Theorem Suppose

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Graphs and MatricesMany Problems

Are Almost All Graphs Determined by theirSpectrum?

Chris Godsil

16 January, 2014

Chris Godsil Are Almost All Graphs Determined by their Spectrum?

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Outline

1 AgainstGraphs and MatricesMany Problems

2 ForWalk MatricesWang & Xu


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Two graphs, two matrices

1 2 3 1 230 1 01 0 10 1 0

0 0 1

0 0 11 1 0


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Isomorphic implies cospectral

If graphs X and Y are isomorphic, there is a permutation matrixP such that

A(Y ) = PA(X)PT

Since permutation matrices are orthogonal, this means that A(X)and A(Y ) are similar.

DefinitionGraphs X and Y are cospectral if the characteristic polynomials oftheir adjacency matrices are equal.


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Willem Haemers proposes:

ConjectureThe proportion of graphs on n vertices that are determined bytheir spectrum goes to 1 as n →∞.


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The Origin of our Difficulty


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Outline




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There’s a problem. . .

φ(X , t) = t5 − 4t3.


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. . . and that’s only the start

If m,n ≥ 1 then the graphs

K1,mn , Km,n ∪ (m − 1)(n − 1)K1

are cospectral.


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Did we forget to say connected?


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Did we forget to say connected?


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We have forgotten complements, but. . .

The two graphs are cospectral, and so are their complements.


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We have forgotten complements, but. . .

The two graphs are cospectral, and so are their complements.


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Latin square graphs: regularity is no help

16 squares as vertices; adjacent if in same row, same column, or

same color. Cospectral.

(And there are lots of Latin squares.)


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Latin square graphs: regularity is no help

16 squares as vertices; adjacent if in same row, same column, or

same color. Cospectral. (And there are lots of Latin squares.)Chris Godsil Are Almost All Graphs Determined by their Spectrum?

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Schwenk’s Particular Tree

u v

If we denote this tree by S, then S \u and S \v are cospectral. Wesay they are cospectral vertices in S .


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Schwenk’s Pairs of Cospectral Trees

S

S


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The bad news for trees

Theorem (Schwenk)The proportion of trees on n vertices that are determined by theirspectrum goes to 0 as n →∞.


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Well, what does the computer tell us?

n # graphs A A & A5 34 0.059 06 156 0.064 07 1044 0.105 0.0388 12346 0.139 0.0949 274668 0.186 0.16010 12005158 0.213 0.201

Source: Godsil and McKay, Haemers and Spence


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Local Switching: regular and all, half, none


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We’ve seen it before


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Features of local switching

Graphs related by local switching are cospectral, and theircomplements are too.The proportion of graphs on n vertices that we can usefullyapply local switching to goes to 0 as n →∞.


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Asymptotically useless, but good in the short run

n # graphs A A & A ls5 34 0.059 0 06 156 0.064 0 07 1044 0.105 0.038 0.0388 12346 0.139 0.094 0.0859 274668 0.186 0.160 0.13910 12005158 0.213 0.201 0.171

Source: Godsil and McKay, Haemers and Spence, Brouwer andSpence


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Walk MatricesWang & Xu

Outline




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And if we look at graphs on 11 and 12 vertices. . .

n # graphs A A & A ls5 34 0.059 0 06 156 0.064 0 07 1044 0.105 0.038 0.0388 12346 0.139 0.094 0.0859 274668 0.186 0.160 0.13910 12005158 0.213 0.201 0.17111 1018997864 0.211 0.208 0.17412 165091172592 0.188 ? ?


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Walk matrices

DefinitionIf X is a graph on n vertices and 1 is the all-ones vector, the walkmatrix W of X is the matrix with columns Ar1 forr = 0, . . . ,n − 1.

DefinitionA graph is controllable if its walk matrix is invertible.


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My Conjecture

ConjectureThe proportion of graphs on n vertices that are controllable goesto 1 as n →∞.


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Outline




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Wang and Xu

TheoremSuppose X and Y are cospectral graphs with cospectralcomplements. If X is controllable there is a unique orthogonalmatrix Q such that QT A(X)Q = A(Y ). The matrix Q is rationaland Q1 = 1.

TheoremSuppose X is controllable. If Q is an orthogonal matrix such thatQ1 = 1 and QT AQ is an integer matrix, then Q is rational.

Further, if `W−1 is an integer matrix, so is `Q.


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Our level best

DefinitionThe level of a rational matrix M is the least integer ` such that`M is an integer matrix.

Wang and Xu provide considerable evidence that if X iscontrollable and QT A(X)Q is an integer matrix then the level ofQ is two for a positive fraction of the graphs on n vertices. Theyprove that orthogonal matrices of index two are essentially localswitchings and that asymptotically these are negligible.

This provides evidence that a positive fraction of the graphs on nvertices are determined by their spectrum.


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Related Questions

We know very little about what happens when we use theLaplacian in place of the adjacency matrix.What proportion of trees on n vertices contain a pair ofcospectral vertices?


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The End(s)


Are Almost All Graphs Determined by their Spectrum?faculty.nps.edu/rgera/Conjectures/JointMeetings-2014/godsil-cospec.pdfAgainst For Walk Matrices Wang & Xu WangandXu Theorem Suppose

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