By Emmanuel Kowalski
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Example text
On the other hand, if W = V , we claim that V0 = W , V1 = V − W is a bipartite partition of V . Indeed, let α ∈ E be an edge with extremities {x1 , x2 }. ) This contradicts the fact that V0 = W = V . Similarly, we see that x1 , x2 can not both be in V1 , and this finishes the proof of bipartiteness of Γ. It is now easy to finish determining ϕ: it is constant, equal to ϕ(x0 ), on V0 , and for any x ∈ V1 , finding y ∈ V0 connected by an edge, we get ϕ(y) = −ϕ(x) = −ϕ(x0 ). Thus it is equal to ϕ(x0 )ε± .
Let G act on the left on a set X. (1) If S ⊂ G is a symmetric generating set, show that A(X, S) has no loop if and only if the elements of S act without fixed points on X. (2) If X = G/H with the action of G by left-multiplication, show that A(G/H, S) has no loops if and only if S ∩ xHx−1 = ∅ for all x ∈ G. (3) Let k 2 be an integer and let G = Sk acting on X = {1, . . , k} by evaluation. Find (or show that there exists) a symmetric generating set S of G such that A(X, S) has no loops. (4) Find a criterion for the action graph to be connected.
2 The simple spectral properties of M are enough to understand the asymptotic behavior of a random walk on a fixed connected graph Γ. 19 (Equidistribution radius). Let Γ = (V, E, ep) be a connected, nonempty, finite graph. The equidistribution radius of Γ, denoted Γ , is the maximum of the absolute values |λ| for λ an eigenvalue of MΓ which is different from ±1. , (1) if Γ is not bipartite, to the space of ϕ ∈ L2 (Γ) such that ϕ, 1 = 1 N val(x)ϕ(x) = 0, x∈V and (2) if Γ is bipartite with bipartite partition V0 ∪ V1 = V , to the space of ϕ ∈ L2 (Γ) such that 1 1 1 val(x)ϕ(x) = 0, val(x)ϕ(x) = val(x)ϕ(x).
Expander graphs by Emmanuel Kowalski
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