Now we prove Theorem 2.1.8. The case of a Cayley graph Cay(G, S) is particularly nice with respect to general Schreier graphs: the reason behind this fact is ulti-mately the existence of automorphisms (namely, right multiplications by elements ofG) that can send anygto anyg0, as we will be able to observe during the proof of the result1.
Before we move to the actual proof, here is an idea (without details) that shows why the result is reasonable: the author thanks L. Bartholdi (personal communi-cation) for the observation. If at a certain iteration the algorithm has assigned a colour to two elements s, s0, in the sense that c(g, sg) has been established as a nonempty colour, then at the next step we would have a colour for ss0 too: this is because the pair (g, ss0g) has a triangle with sides (g, s0g),(s0g, ss0g) built on it. Therefore, at thek-th iteration the configuration we obtain from Cay(G, S) is
1To turn a Russian quote inside out, Cayley vertices being all alike make us happy, and Schreier vertices being all in their own way make us unhappy.
substantially the same as the initial configuration coming from Cay(G, S2k). Thus it is natural to expect that the algorithm would stop whenS2kbecomes the whole G (or possibly Gexcept for just one g), and the number of steps would become essentially log2diam(Cay(G, S)).
Now, let us turn to filling the necessary details. We start with a lemma that provides an upper bound for the number of iterations; more specifically, we prove that the right hand side of the equality in Theorem 2.1.8 provides at least a sufficient number of iterations to make the configuration XC coherent.
Lemma 2.4.1. The inequality with≤in Theorem 2.1.8 holds.
Proof. Right multiplication by anyh∈Ggives an automorphismϕh ofXC, since sg=g0 ⇔sgh=g0hfor any g, g0 ∈G,s∈S; by Proposition 2.1.5, at the end of the algorithmϕh will still be an automorphism of the final coherent configuration, which means in particular that every vertex will have the same colour and that for any three vertices g, g0, h∈Gthere exists anh0 ∈Gsuch thatc(g, h) =c(g0, h0).
Therefore, the maximum possible colour refinement that we can expect to obtain from Weisfeiler-Leman is the one where all sets{c(g, h)|h∈G}are equal for every g but where any two coloursc(g, h), c(g, h0) are distinct forh6=h0.
Indeed, this is the colouring that we reach at the end: as we have already observed in the general Schreier case, any coloured walk fromgtoh(corresponding to a certain product of generators that representsg−1h) is the unique walk from g consisting of that sequence of colours, and by Proposition 2.1.4 the final colour c(g, h) knows it. Thus, every c(g, h) knows a walk that all other c(g, h0) do not know, and the colouring described above is achieved.
When every colourc(g, h) for allh∈Ghas become distinct from the others, we have undoubtedly reached the end of the algorithm; this happens when the colour of every pair (g, h) knows at least one walk connecting them, or at the very least when for a fixed g all but one of them know such a walk (so that the remaining pair (g, h) has the unique colour that knows no walk: this corresponds, informally speaking, to the situation where even the emptiest descendant of the colour ∅ is nonempty according to the definition given in Theorem 2.2.1). By Lemma 2.2.2, this happens when at thek-th iteration we have 2k at least as large as the diameter, or at least diam(Cay(G, S))−1 if for anyg there is only one h whose distance fromgis the diameter; the result follows.
To prove an inequality in the other direction, we make use of the abundance of automorphisms in the Cayley graph to prove a stronger version of Lemma 2.3.2.
Lemma 2.4.2. For any four vertices g, h, g0, h0 ∈ G and for any integer k ≥0 such that d(g, h), d(g0, h0)>2k, we have c(k)(g, h) =c(k)(g0, h0).
Proof. Again, we proceed by induction on k. Fork= 0 the statement is obvious, because all pairs (g, h) of vertices with distance>1 have the same colour∅at the 0-th step.
Now suppose that the statement is true for k. First, we are going to prove that for any three vertices g, h, h0 ∈Gwith d(g, h), d(g, h0)>2k+1 the two pairs (g, h),(g, h0) will still have the same colour at the (k+ 1)-th step; the idea is,
as in Lemma 2.3.2, to construct a suitable bijection σ: G→G with a property analogous to (2.3.1), namely
c(k)(g, g0), c(k)(g0, h)
=
c(k)(g, σ(g0)), c(k)(σ(g0), h0)
. (2.4.1)
Define
σ(g0) =
g0 ifd(g0, h), d(g0, h0)>2k, g0h−1h0 ifd(g0, h)≤2k,
τ(g0) ifd(g0, h)>2k, d(g0, h0)≤2k,
whereτ is an arbitrary bijection from the set{g0∈G|d(g0, h)>2k, d(g0, h0)≤2k} to the set {g0∈G|d(g0, h)≤2k, d(g0, h0)>2k}.
In the first case we have obviously a bijection, whose image is the set of all vertices of distance > 2k from both h and h0; also, by inductive hypothesis in this set we havec(k)(g0, h) =c(k)(g0, h0), so (2.4.1) is satisfied. In the second case, right multiplication byh−1h0is an automorphism (hence a bijection) from the ball aroundhto the one aroundh0 both of radius 2k: this descends trivially from the fact that if we havesi∈S such that (Q
isi)g0=hthen also (Q
isi)g0h−1h0=h0, so that in particular d(g0, h) = d(g0h−1h0, h0); moreover for the same reason we must havec(k)(g0, h) =c(k)(g0h−1h0, h0), because for every walk given by a sequence of colours si from g0 to h the same walk exists from g0h−1h0 to h0: as observed in the proof of Lemma 2.4.1, for our three vertices g0, h, g0h−1h0 there must be a fourth x that will have c(g0, h) = c(g0h−1h0, x) at the end, and x = h0 is the only possible candidate. We also have c(k)(g, g0) = c(k)(g, g0h−1h0) by inductive hypothesis, since g0 and g0h−1h0 are at distance ≤ 2k from h and h0 (both at distance >2k+1 from g); thus, (2.4.1) is satisfied again. In the third case, τ is a bijection because the balls aroundh0 andhof radius 2k have the same number of vertices, and domain and codomain ofτare these two balls minus their intersection;
in addition, the colours of the four pairs (g, g0),(g0, h),(g, τ(g0)),(τ(g0), h0) are all the same since each of their distances is>2k, so we have (2.4.1) for this case too.
Given that the codomains in the three cases are disjoint, σ is indeed a bijection satisfying (2.4.1), which implies thatc(k+1)(g, h) =c(k+1)(g, h0).
Now we know that for any vertexg there is a colourcg such that all vertices h at distance > 2k+1 will have c(k+1)(g, h) = cg: but then it is obvious that cg does not depend on g, since any g is sent to any g0 by some automorphism that will preserve distances in the graph, so that (g, h) of distance > 2k+1 is sent to some (g0, h0) of same distance (and consequently cg = cg0). This proves that c(k+1)(g, h) =c(k+1)(g0, h0) wheneverd(g, h), d(g0, h0)>2k+1, concludes the inductive step and proves the lemma.
Now we can easily prove Theorem 2.1.8.
Proof of Thm. 2.1.8. We have already shown the ≤ direction in Lemma 2.4.1.
On the other hand, proving the ≥ direction means proving that if at the k-th iteration there are three verticesg, h, h0 such thatd(g, h), d(g, h0)>2k then there are more iterations to come; by Lemma 2.4.2, however, in this situationc(k)(g, h) = c(k)(g, h0) and we know that at the end of the algorithm we will have c(g, h) 6=
c(g, h0) (ultimately coming from Proposition 2.1.4), so the statement holds.