Spectral Continuity Properties of Graph Laplacians

Transcription

1 Spectral Continuity Properties of Graph Laplacians David Jekel May 24, 2017

2 Overview Spectral invariants of the graph Laplacian depend continuously on the graph. We consider triples (G, x, T ), where G is a graph, x is a root vertex, and T is symmetric diagonally dominant operator (generalized Laplacian) associated to the graph. We say that (G n, x n, T n ) (G, r, T ) locally if for every R > 0, the ball B Gn (x n, R) is evetually the same as B G (x, R) and also T n BGn (x n,r) T BG (x,r) 0. First Main Result: If (G n, x n, T n ) (G, r, T ) locally, then the spectral measures associated to (T n, δ xn ) on l 2 (G n ) converge to the spectral measure of (T, δ x ). Second Main Result: Assume also that T n C. Then we get an explicit estimate on rate of convergence δ xn, φ(t n )δ xn in terms of the modulus of continuity of φ.

3 Graph Laplacians For a (countable, locally finite) graph G = (V, E). The Laplacian C V C V is given by Lf (x) = y x(f (x) f (y)). For positive weights w(x, y) = w(y, x), we can define a weighted Laplacian Lf (x) = y x w(x, y)(f (x) f (y)).

4 Graph Laplacians Kenyon Spanning Forests and the Vector Bundle Laplacian considers a Laplacian Lf (x) = y x(f (x) U x,y f (y)), where f is a function from V to a Hilbert space H, and U x,y = U y,x is a unitary automorphism of H. Remark: This is an analogue of the Laplacian on a manifold with nontrivial curvature. There is a generalization of the matrix-tree theorem to this setting. See also Forman, Determinants of Laplacians on Graphs.

5 Symmetric Diagonally Dominant Operators These Laplacian variants are examples of SDD operators associated to a graph. Let H be a Hilbert space. An infinite matrix {T i,j } i,j N with entries in B(H) is called SDD if T j,i = T i,j and T i,i j T i,j. Remark: We ll show that such a matrix defines a (possibly unbounded) self-adjoint operator on j=1 H. If G is a graph, then an SDD matrix associated to G (and H) is an SDD matrix with entries indexed by V such that T x,y = 0 when x y.

6 The Metric of Local Convergence Consider a triple G = (G, x, T ), where G is a graph, x is a root vertex, and T is an associated SDD operator. For real R > 0, let B R (G) denote the closed ball in G centered at x with respect to the path metric. Let P R : l 2 (G) l 2 (G) be the projection onto functions supported in B R. Define d(g 1, G 2 ) to be the infimum of all ɛ > 0 such that (1) there exists a root-preserving graph isomorphism B 1/ɛ (G 1 ) B 1/ɛ (G 2 ) and (2) we have P 1/ɛ T 1 P 1/ɛ P 1/ɛ T 2 P 1/ɛ ɛ, where we have identified l 2 (B 1/ɛ (G 1 )) with l 2 (B 1/ɛ (G 2 )) by the given isomorphism. This defines a complete metric space G. For further background and applications, see Aldous and Steele The Objective Method (c. 2003).

7 First Main Result Fix h 0 H. For φ : R C continuous, let φ : G B(H) be given by φ (G, x, T ) = δ x h 0, φ(t )[δ x h 0 ] whenever this quantity is defined (in terms of the spectral measure for δ x h 0 ). We will show that T is self-adjoint and positive. Then we will show that if φ is continuous with polynomial growth, then φ is defined and continuous. For most of the proof, we will work with SDD operators with entries indexed by N rather than V (G).

8 SDD Operators: Results Let F l 2 (N, H) denote the space of finitely supported functions. We will prove: Lemma 1: If T = {T i,j } is an SDD matrix, then T defines a positive operator dom T l 2 (N, H), where dom T will be defined explicitly in the next slide. Lemma 2: For every c > 0, the space (c + T )F is dense in l 2 (N, H). Lemma 3: Suppose T n and T are SDD and T n f Tf for every f F. Then for every bounded continuous φ : R C, we have φ(t n ) φ(t ) strongly.

9 SDD Operators: Domain Let {T i,j } be SDD operator over H. Because j T i,j < + for each i, multiplication by T yields a well-defined map T : l (N, H) H N. Therefore, we can define the l 2 domain of T to be dom T = {f l 2 (N, H) such that Tf l 2 (N, H)}. Note that F dom H because i T i,j < + for each j. Straightforward computation checks that f, Tg = Tf, g and Tf, f 0 for f, g F.

10 SDD Operators: Proof of Lemma 2 It is not immediate that f, Tg = Tf, g is true for f, g dom T because this requires exchanging the order of summation in i j f i, T i,j g j without knowing absolute convergence a priori. Thus, we will first prove Lemma 2 that (c + T )F is dense. Suppose that f l 2 (N, H) is orthogonal to (c + T )F. Since f l (N, H), Tf is defined. Also, for h H, we have (c + T )f (j), h H = f, (c + T )(δ j h) l 2 (N,H) = 0. This implies that (c + T )f = 0. Since f l 2 (N, H), there exists a point j where f (j) H is maximized.

11 SDD Operators: Proof of Lemma 2 Then we have 0 = [(c + T )f ](j) H (c + T )f (j), f (j) H = (c + T j,j )f (j), f (j) H + T j,i f (i), f (j) i j c f (j) 2 + T j,i f (j) 2 T j,i f (i) f (j) i j i j c f (j) 2. This implies that f (j) = 0 and hence f = 0. So Lemma 2 is proved.

12 SDD Operators: Proof of Self-Adjointness Lemma 2 implies that for each f dom T, there exists f n F such that (c + T )f n (c + T )f. Since (c + T ) c on F, this implies that f n converges, and one can deduce that the limit must be f. Hence, we have f n f and Tf n Tf. This implies that the symmetry and positivity properties of T hold on all of dom T, not just F. Finally, to check self-adjointness, it suffices to show that if Tf, h = f, Th for all h dom T, then f dom T and Tf = g. We know Tf is defined pointwise since f l (N, H). But by taking g to supported at j, we obtain Tf (j) = g(j). Since g l 2 (N, H), this implies that f dom T.

13 SDD Operators: Proof of Convergence To prove Lemma 3: Assume that T n f Tf for all f F, and we will show that φ(t n ) φ(t ) strongly for every bounded continuous φ : [0, + ) C. First, consider φ(t) = (c + t) 1. Suppose that f (c + T )F. Then we have (c + T n ) 1 f (c + T ) 1 f = (c + T n ) 1 (T T n )(c + T ) 1 f c 1 (T T n )(c + T ) 1 f 0 because (c + T ) 1 f F. Since (c + T )F is dense and (c + T n ) 1 is uniformly bounded, we have (c + T n ) 1 (c + T ) 1 strongly.

14 SDD Operators: Proof of Convergence The case of general bounded continuous φ : [0, + ) C follows by standard approximation arguments for instance: We know the result holds for every φ which has a limit at + because the Stone-Weierstrass theorem on [0, + ] implies that such a φ can be uniformly approximated by polynomials in (1 + T ) 1. For general φ, we approximate by multiplying by a cut-off in a neighborhood of +. (Ask for details at the end, if desired.)

15 First Main Result Theorem A: Let φ : G C be given by φ (G, x, T ) = δ x h 0, φ(t )[δ x h 0 ], where h 0 H. If φ has polynomial growth, then φ is continuous. Proof: Assume that G n G. Choose an identification of V (G n ) and V (G) with N (or a finite subset of N) such that the root vertex is identified with 0 and, whenever d(g n, G) < ɛ, the identification with N is the same for B 1/ɛ (G n ) and B 1/ɛ (G). If φ is bounded and continuous, then modulo this identification we have φ(t n ) φ(t ) strongly which implies that φ (G n ) φ (G).

16 First Main Result Thus, if µ n and µ are the spectral measures associated to G n and G, we know that µ n µ weakly and hence φ dµ lim infn φ dµn for every LSC φ 0. Since each graph is assumed locally finite, we have (1 + t 2 ) k dµ n = δ xn, (1 + T 2 ) k δ xn < +. R Moreover, R (1 + t2 ) k dµ n R (1 + t2 ) k dµ because this expression is given by a finitely linear combination of T i,j s.. Hence, if we assume that φ(t) C(1 + t 2 ) k, then we obtain φ (G n ) φ (G) using the proof of the generalized dominated convergence theorem.

17 First Main Result: Remarks As a corollary, if φ is nonnegative and LSC, then φ is LSC. In the theorem, we do not have to assume that P R T n P R P R TP R 0 in norm for each R, only that it goes to zero strongly. If we do assume P R T n P R P R TP R 0 in norm, this proof does not guarantee that P {xn}t n P {xn} P {x} TP {x} in norm. Is this true without any additional restrictions on the growth of T i,j? This is a philosophical result which assumes a soft hypothesis and gets a soft conclusion. Next, we ll do some explicit estimates...

18 Second Main Result Given a modulus of continuity ω : [0, + ) [0, + ), we define ω ω(s) (t) = t s 2 ds. We denote by G 0 the space of triples (G, x, T ) with T 1. As before, for φ : R C, we define φ : G C by φ (G, x, T ) = δ x h, φ(t )[δ x h]. Theorem B: Let f : [0, 1] C be continuous with modulus of continuity ω φ. Then for G 1 and G 2 in G 0, we have φ (G 1 ) φ (G 2 ) Cω φ (d(g 1, G 2 )), where C is a universal constant. t

19 Second Main Result: Ingredients We ll approximate φ uniformly by a polynomial using Jackson s theorem: Theorem: Let φ : [0, 1] C be continuous. Then there exists a polynomial φ n of degree n such that φ n φ Cω φ (1/n), ω φn (t) Cω φ (t). Remark: The reason that ω φn Cω φ is that (in most proofs) the polynomial p n is obtained by convolution on R (or convolution on S 1 after a change of variables).

20 Second Main Result: Ingredients We ll combine this with the results of Aleksandrov and Peller (2013 to present) in Operator Hölder-Zygmund Functions and Functions of Perturbed Unbounded Self-Adjoint Operators... : Theorem: Let φ : R C be uniformly continuous. Then for self-adjoint operators A and B, we have φ(a) φ(b) Cωφ ( A B ). Remarks: The authors first prove a Bernstein type inequality that if φ is supported in [ σ, σ], then φ(a) φ(b) Cσ A B. For a general φ, they use a dyadic decomposition of the frequency space. Examining their proof shows C 99.

21 Second Main Result: Proof Let φ : [0, 1] C continuous. Let δ = d(g 1, G 2 ). Let n = floor(1/δ), so that B n (G 1 ) = B n (G 2 ). Let φ n be the polynomial given by Jackson s theorem. Then we have Hence, φ n (T j ) φ(t j ) φ n φ Cω φ (δ). φ (T 1 ) φ (T 2 ) (φ n ) (T 1 ) (φ n ) (T 2 ) + 2Cω φ (δ).

22 Second Main Result: Proof Now because T j only allows interactions between adjacent vertices, we have δ x h 0, φ n (T j )[δ x h 0 ] = δ x h 0, φ n (P n T j P n )[δ x h 0 ], that is, (φ n ) (T j ) = (φ n ) (P n T j P n ). Therefore, using Aleksandrov and Peller s result, (φ n ) (T 1 ) (φ n ) (T 2 ) φ n (P n T 1 P n ) φ n (P n T 2 P n ) Cω φ n ( P n T 1 P n P n T 2 P n ) Cω φ (δ).

23 Second Main Result: Proof Altogether, we have φ (T 1 ) φ (T 2 ) Cω φ (d(g 1, G 2 )). Here we use the fact that ω φ ω φ. A few remarks: The constant in this theorem can be crudely estimated by 222. If we have P n T 1 P n = P n T 2 P n, then the proof only requires Jackson s theorem, and the error is bounded by 2Cω φ (δ), where C is the optimal constant in Jackson s theorem. Thus, in this case, the modulus of continuity of φ is the same as the modulus of continuity of φ up to constants. This confirms that we are using the right metric on G 0.

24 Further Questions: Random Graphs For random elements G 1 and G 2 of G 0, our previous results imply that E φ (G 1 ) φ (G 2 ) C Eω φ (d(g 1, G 2 )) 2C ω φ (E d(g 1, G 2 )). But in fact, in certain cases, we can get ω φ rather than ω φ. We already saw that this is the case if we change the graph without changing the weights.

25 Further Questions: Random Graphs Suppose that G is a Cayley graph with set of generators S. Let Ω be a probability space, and {X s } s S and {Y s } s S scalar random variables representing random edge weights for the edges from e to s. For the other edges, define edge weights by making i.i.d. copies of Ω (hence {X s } and {Y s }) translated by the action of the group on G. Then the weighted Laplacians T X and T Y are elements of a von Neumann algebra on which τ(t ) = E δ e, T δ e is a trace. Since we have a trace, we can argue that τ(φ(t X ) φ(t Y )) 4ω φ ( T X T Y ) Cω φ ( s S E X s Y s ).

26 Further Questions: Random Graphs In general, if a random graph is given by a unimodular probability measure, then the Laplacian lives in a von Neumann algebra where E δ e, T δ e. However, it is not clear in general how to take two unimodular random rooted graphs and put both Laplacians in the same tracial von Neumann algebra. In the above argument, we implicitly did this using independence. One difficulty is that we do not have a good structure theorem for unimodular probability measures. Russ Lyons conjectures that they are weak limits of distributions which take a finite graph and choose each vertex as the root with equal probability.

27 Further Questions: Unbounded Functions Our results here considered only continuous φ. If we allow φ to have some discontinuities or be unbounded, then we also need to have to estimates for the concentration of the spectral measure near the singularities, which cannot be done in such a general setting. Russ Lyons in 2005 considered φ(t) = log t, which relates to entropy. He defined a quantity called tree entropy. He gave a formula relating the tree entropy and E log (G) for the unimodular case. He also showed that the tree entropy depends continuously on the graph by expressing it in terms of random walk probabilities. Based on his proof, one can conclude that the modulus of continuity of the tree entropy on G 0 is bounded by the square root function.