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In the famous counter example for graph isomorphism via Weisfeiler-Lehman (WL) method the following gadget was constructed in this paper by Cai, Furer and Immerman. They construct a graph $X_k = (V_k, E_k)$ given by

$V_k = A_k \cup B_k \cup M_k \\ \text{ where } \\ \quad A_k = \{a_i \mid 1 \leq i \leq k\}, \\ \quad B_k =\{b_i \mid 1 \leq i \leq k\}, \mbox{ and } \\ \quad M_k = \{ m_S \mid S \subseteq \{1,2,\ldots, k\},\ |S| \text{ is even} \}\\ E_k =\{(m_S,a_i) \mid i \in S\} \cup \{(m_S, b_i)\mid i \notin S\}$

One of the lemmas in the paper (lemma 3.1 page 6 ) states that if we color the vertices $a_i$ and $b_i$ with color $i$ then $|Aut(X_k)| = 2^{k-1}$ (color has to be preserved by the automorphism) where each automorphism corresponds to interchanging $a_i$ and $b_i$ for each $i$ in some subsets $S$ of $\{1,2,\ldots, k\}$ of even cardinality. They say that the proof is immediate. But I fail to see how even in the case of $k= 2$. In $X_2 \ (a_1, m_{\{1,2\}})$ is an edge but if we have automorphism which interchanges $a_1, b_1$ and $a_2, b_2$ the above edge gets transformed to $(b_1, m_{\{1,2\}})$ which is not an edge. So that should not be an automorphism.

I would like to understand what is my misunderstanding.

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1 Answer 1

up vote 6 down vote accepted

You're missing the emptyset $\emptyset$ which is connected to all $b$'s. To get an automorphism, you select a subset $T\subseteq \{1,...,k\}$ of even cardinality and then swaps $a_i$ with $b_i$ for each $i\in T$ and then adjusts the sets in the middle. In your example the graph is $$(a_1,\{12\}),(a_2,\{12\}),(b_1,\emptyset),(b_2,\emptyset).$$

Still in your example if $T=\emptyset$ you don't need to do anything and if $T=\{1,2\}$ the automorphism is given by swapping $a_1$ with $b_1$, $a_2$ with $b_2$ and $\{1,2\}$ with $\emptyset$.

Now for the general case, we need to show that there is always a way of adjusting the middle vertices. We know that $T$ has even cardinality. So let $|T|=2r$. We just need to show that such an automorphism exists if $|T|=2$ since otherwise we can apply the composition of $r$ automorphisms corresponding to partitioning $T$ into $r$ subsets of size $2$. Thus assume $T=\{i,j\}$. Then the automorphism swaps $a_i$ with $b_i$, $a_j$ with $b_j$, each middle vertex $S$ such that $S\cap\{i,j\}=\emptyset$ with the middle vertex $S\cup \{i,j\}$ (this can be seen in your example), and each subset $S$ such that $S\cap \{i,j\}=\{i\}$ with the subset such that $S\cap \{i,j\}= \{j\}$ (This you can see for $k=3$). Notice that this swapping process is an automorphism since for an index $p\neq \{i,j\}$ the edge relation between $a_p$, $b_p$ and these swapped vertices is completely preserved, and clearly the edge relation between $a_i,a_j,b_i,b_j$ is properly adjusted.

Finally to see that these are the only possible automorphisms, notice that each $a_i,b_i$ is colored with its own color. So they cannot be mapped to another pair $a_j,b_j$. Also notice that it is not possible to have an automorphism that maps an middle vertex to a middle vertex without swapping some $a_i$ with some $b_j$. $\square$

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In general how can we show that we can always adjust the sets in the middle and get the desired automorphism? Core of my problem is actually that. –  DurgaDatta Dec 9 '12 at 5:32
    
Hi, I added the construction of the automorphisms. Hope it helps. –  Mateus de Oliveira Oliveira Dec 9 '12 at 10:52
    
Thank you. This does not look "immediate" to me. I am very new to research. Is this a bad signal for me? –  DurgaDatta Dec 9 '12 at 12:53
    
"Is this a bad signal for me?" Absolutely not. I think to the contrary that your skepticism is a very good signal. Some day this kinds of things will probably be immediate for you too :) –  Mateus de Oliveira Oliveira Dec 9 '12 at 13:25
    
Is it true that, for a index set $T$ (for each $i \in T$ of which are are interchanging $a_i$ and $b_i$) index set of a middle vertex $S$ get transformed to $S \Delta T$ (symmetric difference)? –  DurgaDatta Dec 10 '12 at 4:55

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