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Shanghai Jiaotong University On the Generalized Spectral - - PowerPoint PPT Presentation

The Theorem The Case p > 2 The Case p = 2 Examples Future Work Shanghai Jiaotong University On the Generalized Spectral Characterization of Graphs (II): A Simple Arithmetic Criterion Wei Wang Xian Jiaotong University Dec. 2016 The


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The Theorem The Case p > 2 The Case p = 2 Examples Future Work

Shanghai Jiaotong University On the Generalized Spectral Characterization of Graphs (II): A Simple Arithmetic Criterion

Wei Wang Xi’an Jiaotong University

  • Dec. 2016
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The Theorem The Case p > 2 The Case p = 2 Examples Future Work

Outline

1

The Theorem

2

The Case p > 2

3

The Case p = 2

4

Examples

5

Future Work

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The Theorem The Case p > 2 The Case p = 2 Examples Future Work

Some Recent Developments

Question: Does there exist a simple way to determine whether a graph is DGS ?

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The Theorem The Case p > 2 The Case p = 2 Examples Future Work

A Conjecture

Conjecture (Wang, 2006) Define a family a graphs: Fn = {G ∈ Gn| det(W ) 2[n/2] is an odd square − free integer} Then every graph in Fn is DGS.

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Remarks

det(W ) 2[n/2]

is always an integer for every graph. G ∈ Fn iff the walk-matrix W has the following SNF:           1 ... 1 2 ... 2b           , where b is an odd square-free integer and the number of 2 that appearers in the diagonal is [n/2].

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Recent Resolution of The Conjecture

Theorem (Wang, 2017) Conjecture 1 is true.

  • W. Wang, A simple arithmetic criterion for graphs being

determined by their generalized spectra, JCTB, 122 (2017) 438-451

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The Proof: Basic Ideas

For every graph G ∈ Fn, let Q ∈ Γ(G) with level ℓ, if we can show that every prime divisor of dn is not a divisor of ℓ, Then we must have ℓ = 1, and hence Q is a permutation matrix. This shows that G is DGS. This will be done in two steps: for p > 2 and p = 2.

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Excluding primes p > 2

Theorem A (Wang, 2013) Let G ∈ Gn. Let Q ∈ Γ(G) with level ℓ. Suppose that p| det(W ) and p2 | det(W ), where p is an odd prime. Then p is not a divisor

  • f ℓ.
  • W. Wang, Generalized spectral characterization of graphs

revisited, The Electronic Journal Combinatorics, 20 (4) (2013),# P 4.

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The Theorem The Case p > 2 The Case p = 2 Examples Future Work

The Proof: A sketch

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The Theorem The Case p > 2 The Case p = 2 Examples Future Work

Some Lemmas

Lemma Let M = Udiag(d1, d2, · · · , dn)V = USV , where S is the SNF of M and U and V are unimodular matrices, respectively. Then the system of congruence equation Mx ≡ 0 (mod p2) has a solution with x ≡ 0 (mod p) iff p2|dn. By the above lemma, it suffices for us to show that, under the conditions of the theorem, if Q ∈ Γ(Q) with level ℓ and p|ℓ, then he congruence W Tx ≡ 0 (mod p2) always has a solution with x ≡ 0 (mod p). Then p2| det(W ) and we reach a contradiction.

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Some Lemmas

Lemma Let Q ∈ Γ(Q) with level ℓ and p|ℓ, then zT(Az − λ0z) ≡ 0 (mod p2) for some λ0 and z, which satisfy W Tz ≡ 0 (mod p), zTz ≡ 0 (mod p), and Az ≡ λ0z (mod p). We will show that zT(Az − λ0z) ≡ 0 (mod p2) implies that p2| det(W ) (and in fact, the converse is also true).

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The proof: A sketch

The fact QTAQ = B (B is the adjacency matrix of some graph H) implies that there exists a column u of ℓQ with u ≡ 0 (mod p) such that uTAu = 0; uTu = ℓ2; eTu = ℓ. Such a u clearly satisfies uTAu ≡ 0 (mod p2), uTu ≡ 0 (mod p2), W Tu ≡ 0 (mod p), and eTu ≡ 0 (mod p). Clearly uT(Au − λ0u) ≡ 0 (mod p2) holds.

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The proof: A sketch - contd.

The assumption p|| det(W ) implies rankp(W ) = n − 1. It follows that rankp(ℓQ) = 1, since any column of ℓQ satisfies W Tx ≡ 0 (mod p). Then AQ = QB and rankp(ℓQ) = 1 give Au ≡ λ0u (mod p) for some integer λ0. Let z ≡ u + pβ (mod p). Then z satisfies W Tz ≡ 0 (mod p), Az ≡ λ0z (mod p), eTz ≡ 0 (mod p) and zTz ≡ 0 (mod p). It is easy to verify that the equation zT(Az − λ0z) ≡ 0 (mod p2) is invariant when replacing z with its congruence class. zT(Az − λ0z) ≡ (u + pβ)T[(A(u + pβ) − λ0(u + pβ)] ≡ uT(Au − λ0u) + 2pβT(Au − λ0u) ≡ uT(Au − λ0u) ≡ 0 (mod p2).

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Some Lemmas

Lemma If rankp(W ) = n − 1, then rankp(A − λ0I) = n − 1 or n − 2. Lemma If rankp(A − λ0I) = n − 2, then rankp([A − λ0I, z]) = n − 1. Lemma There exists a vector y with eTy ≡ 0 (mod p) such that W Ty ≡ eTy[1, λ0, · · · , λn−1 ]T (mod p).

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Proof of Theorem A: a sketch

Q ∈ Γ(G) implies there exists a graph H with Spec(H) = Spec(G) and Spec(¯ H) = Spec(¯ G). It follows that zT(Az − λ0z) ≡ 0 (mod p2), i.e., zT Az−λ0z

p

≡ 0 (mod p). Case 1. If rankp(A − λ0I) = n − 1, then zT[A − λ0I, Az−λ0z

p

] = 0 and rankp(A − λ0I) = n − 1 imply that Az−λ0z

p

is the linear combinations of the columns of A − λ0, i.e., (A − λ0I)x ≡ Az−λ0z

p

(mod p) for some x. Case 2. If rankp(A − λ0I) = n − 2, then zT[A − λ0I, z, Az−λ0z

p

] = 0 and rankp([A − λ0I, z]) = n − 1 imply that (A − λ0I)x + kz ≡ Az−λ0z

p

(mod p) for some vector x and integer k. Combing Cases 1 and 2, there always exists an integer m and a vector x such that (A − λ0I)x + mz ≡ Az−λ0z

p

(mod p), i.e., (A − λ0I)(z − px) ≡ mpz (mod p2).

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Proof of Theorem A: a sketch -contd.

Left-Multiplying both sides by Ai−1 + λ0Ai−2 + · · · + λi−2 A + λi−1 I, we get (Ai − λi

0I)(z − px) ≡ mp(Ai−1 + · · · + λi−1

I)z (mod p2), which implies (eTAi−λi

0eT)(z−px) ≡ mpeT(Ai−1+· · ·+λi−1

I)z ≡ 0 (mod p2), for i = 0, 1, · · · , n − 1. That is, W T(z − px) ≡ eT(z − px)[1, λ0, · · · , λn−1 ]T (mod p2). Moreover, there exists a vector y such that W Ty ≡ eTy[1, λ0, · · · , λn−1 ]T (mod p). Let eT(z − px) ≡ pl (mod p) for some l. It follows that W T(z − px − p ly

eT y ) ≡ 0 (mod p2).

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Excluding prime p=2

Theorem B. [Wang, 2017] Let G ∈ Fn. Let Q ∈ Γ(G) with level ℓ. Then 2 is not a divisor of ℓ.

  • W. Wang, A simple arithmetic criterion for graphs being

determined by their generalized spectra, JCTB, 122 (2017) 438-451

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Why p = 2 more difficult? Some important facts

1 Fact 1. eTAke is always an integer for every integer k ≥ 1. 2 Fact 2. W TW ≡ 0 (mod 2) (when n is odd, replace W with

¯ W , the matrix obtained from W be deleting the first row).

3 Fact 3. rank2(W ) ≤ ⌈n/2⌉. 4 Fact 4. Let det(W ) = ±2αpα1

1 pα2 2 · · · pαs s , then α ≥ [n/2].

5 Fact 5. Let G ∈ Fn. Then the SNF of W is

diag(1, 1, · · · , 1

  • ⌈n/2⌉

, 2, 2, · · · , 2b

  • ⌊n/2⌋

), where b is an odd square-free integer.

6 Fact 6. Let G ∈ Fn. Then any ⌊ n

2⌋ linearly independent

columns from W forms a set of fundamental solutions to the linear system of equation W Tx = 0 over F2.

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A Useful Lemma

Lemma Let G ∈ Gn and Q ∈ Γ(G) with level ℓ. If 2|ℓ, then there exists a (0, 1)-vector u with u ≡ 0 (mod 2) such that uTAku ≡ 0 (mod 4), for k = 0, 2, · · · , n − 1. (1) Moreover, u satisfies W Tu ≡ 0 (mod 2).

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A Key Lemma

Lemma If 2|ℓ, then the following linear system of equations has a non-trivial solution: W T ˜ W1 2 x ≡ 0 (mod 2).

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Proof: A Sketch

Proof. For simplicity, we just focus on the case n is even in what follows. The case n is odd can be proved similarly. Step 1: Since W TW = (eTAi+j−2e) ≡ 0 (mod 2), there exists a non-zero vector v such that u ≡ ˜ W v (mod 2), where ˜ W = [e, Ae, · · · , Ak−1e], k = n

2 is fixed henceforth.

Step 2: ∵ u = ˜ W v + 2β ∴ uTAlu ≡ vT ˜ W TAl ˜ W v + 4vT ˜ W TAlβ + 4βTAlβ ≡ vT ˜ W TAl ˜ W v ≡ 0 (mod 4), for any l ≥ 0.

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Proof: A Sketch (continued)

Proof. Step 3:

˜ W TAl ˜ W =      eTAle eTAl+1e · · · eTAl+k−1e eTAl+1e eTAl+2e · · · eTAl+ke . . . . . . ... . . . eTAl+k−1e eTAl+ke · · · eTAl+2k−2e,     

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Proof: A Sketch (continued)

Proof. Step 4: (The key step) Let M := ˜

W TAl ˜ W , v = (v1, v2, . . . , vk)T. Then we have Mijvivj + Mjivjvi = 2Mijvivj ≡ 0 (mod 4), for i = j, v T ˜ W TAl ˜ W v =

  • i,j

Mijvivj ≡(eTAle)v 2

1 + (eTAl+2e)v 2 2 + · · · + (eTAl+2k−2e)v 2 k

≡(eTAle)v1 + (eTAl+2e)v2 + · · · + (eTAl+2k−2e)vk ≡[eTAle, eTAl+2e, · · · , eTA2k−2e]v ≡0 (mod 4),

for l = 0, 1, · · · , n − 1.

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Proof: A Sketch (continued)

Proof. Step 5:

     eTe eTA2e · · · eTA2k−2e eTAle eTA3e · · · eTA2k−1e . . . . . . ... . . . eTAn−1e eTA2+n−1e · · · eTA2k−2+n−1e      =      eT eTA . . . eTAn−1     

  • A0e

A2e · · · A2k−2e

  • = W T

A0e A2e · · · A2k−2e

  • = W T ˜

W1, .

where ˜ W1 = [e, A2e, · · · , A2k−2e].

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Proof: A Sketch (continued)

Proof. Step 6: Final Step. Thus, if 2|ℓ, then W T ˜ W1 2 x ≡ 0 (mod 2). has a non-trivial solution. However, we are able to show that the matrix W T ˜

W1 2

has full column rank over F2; we got a contradiction.

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Some Key Ingredients

Let W = [e, Ae, A2e, . . . , An−1e] ˜ W = [e, Ae, A2e, . . . , Ak−1e], k = n/2 [Ae, A2e, . . . , Ake], k = (n − 1)/2 W1 = [e, A2e, A4e, . . . , A2(n−1)e] ˜ W1 =

  • [e, A2e, A4e, . . . , A2(k−1)e],

k = n/2 [A2e, A4e, . . . , A2ke], k = (n − 1)/2

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Lemmas

Lemma Using notations above, we have (i) rank2(W ) = rank2( ˜ W ); (ii) rank2(W1) = rank2( ˜ W1). Lemma Let G ∈ Fn. Then we have rank2(WT ˜ W1 2 ) = n

2,

n is even

n−1 2 ,

n is odd

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Remarks

In previous examples, both G1 and G2 are DGS; since for G2 p = 3 can be excluded by using Eq. (2). Theorem A may be false if the exponents of p > 2 is larger than 1. That is, our main theorem is best possible in the sense if we allow det W (G) to have prime factor with high exponents, the we cannot guarantee G is DGS any more.

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A Theorem

Theorem 6. [Wang and Hu (2016), Manuscript] Let p be an odd prime. Suppose that G ∈ Gn, rankp(W ) = n − 1 and p2| det W . Suppose that W Tx ≡ 0 (mod p) has a solution of the following form (−1, −1, · · · , −1

  • p

, 1, 1, · · · , 1

  • p

, 0, · · · , 0)T. Then G is not DGS.

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Example and Counter-example

Let the adjacency matrix of G be given as follows. det(W ) = 26 × 32 × 3157 × 31361 × 32237. rank3(W ) = n − 1. z = (0, −1, 0, 0, 0, 1, −1, −1, 1, 0, 0, 1)T. QTAQ is a (0,1)-matrix, and hence is an adjacency matrix of another graph H.

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Example and Counter-example

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Problem for Further Research

What is the density of Fn?

Table: Fractions of Graphs in Fn n # Graphs in Fn The Fractions 10 211 0.211 15 201 0.201 20 213 0.213 25 216 0.216 30 233 0.233 35 229 0.229 40 198 0.198 45 202 0.202 50 204 0.204

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The Asymptotic Behavior of DGS graphs

Density of Fn Conjecture: limn→∞

|Fn| |Gn| = 2 π2 = 0.202642 · · ·

If we can show Fn has positive density, then DGS graphs have positive density. This would be a firm step towards proving the conjecture that “all most all graphs are DS”.

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References

1 L. Babai, Graph isomorphism in quasipolynomial time,

arXiv:1512.03547v2.

2 E. R. van Dam, W. H. Haemers, Which graphs are determined

by their spectrum? Linear Algebra Appl., 373 (2003) 241-272.

3 E. R. van Dam, W. H. Haemers, Developments on spectral

characterizations of graphs, Discrete Mathematics, 309 (2009) 576-586.

4 W.X, Du, On the Automorphism Group of a Graph,

arXiv:1607.00547v1.

5 W. Wang, C. X. Xu, A sufficient condition for a family of

graphs being determined by their generalized spectra, European J. Combin., 27 (2006) 826-840.

6 W. Wang, C.X. Xu, An excluding algorithm for testing

whether a family of graphs are determined by their generalized spectra, Linear Algebra and its Appl., 418 (2006) 62-74.

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References

5 W. Wang, On the Spectral Characterization of Graphs, Phd

Thesis, Xi’an Jiaotong University, 2006.

6 W. Wang, Generalized spectral characterization of graphs

revisited, The Electronic Journal Combinatorics, 20 (4) (2013), #P4.

7 W. Wang, A simple arithmetic criterion for graphs being

determined by their generalized spectra, JCTB, to appear.

8 W. Wang, T. Yu, Square-free Discriminants of Matrices and

the Generalized Spectral Characterizations of Graphs, arXiv:1608.01144

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Thank you! The end!