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For any simple, undirected graphs $G, H$, let $G\times H$ denote their category-theoretical product.

What is an example of an infinite connected graph $G$ with $G \cong G \times G$?

(Note that the totally disconnected graph $G = (\omega, \emptyset)$ has $G \cong G \times G$.)

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    $\begingroup$ Surely the complete graph on an infinite set will always have these properties. $\endgroup$ Commented Aug 4, 2022 at 8:25
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    $\begingroup$ @PeterKropholler I don't think so. In a complete graph, every pair of vertices is connected, but in the tensor product, the vertices which agree on one component are not. $\endgroup$
    – AMLimbach
    Commented Aug 4, 2022 at 8:34
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    $\begingroup$ @AMLimbach Aha good point! Thank you!! $\endgroup$ Commented Aug 4, 2022 at 8:43
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    $\begingroup$ I also wonder if a countable example exists. $\endgroup$ Commented Aug 6, 2022 at 15:21

1 Answer 1

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The category of graphs and homomorphisms has countable powers. Given a graph $H = (V(H), E(H))$, its countable power is the graph $H^\omega = (V(H^\omega), E(V^\omega))$ where $V(H^\omega) = V(H)^\omega$ is the countable power of vertices, and $E(H^\omega)$ is defined for $a, b \in V(H)^\omega$ by $$\{a, b\} \in E(H^\omega) \iff \forall i \in \omega \,.\, \{a_i, b_i\} \in E(H). $$ For each $i \in \omega$ there is a projection $\pi_i : H^\omega \to H$, defined by $\pi_i(a) = a_i$. With these projections we really get a power: $\mathrm{Hom}(L, H^\omega) \cong \mathrm{Hom}(L, H)^\omega$ holds in virtue of the isomorphism taking $f : L \to H^\omega$ to $(\pi_i \circ f)_{i \in \omega}$.

For any graph $H$ we have $H^\omega \cong H^\omega \times H^\omega$, with isomorphism taking $a \in V(H^\omega)$ to $((a_{2 i})_{i \in \omega}, (a_{2 j + 1})_{j \in \omega})$.

Now consider the graph $G = K_3^\omega$ where $K_3$ is the complete graph on three vertices $V(K_3) = \{0, 1, 2\}$. Given $a, b \in \{0,1,2\}^\omega$, we have $$\{a, b\} \in E(K_3^\omega) \iff \forall i \in \omega \,.\, a_i \neq b_i.$$ The only question remaining is whether $G$ is connected. Take arbitrary vertices $a, b \in \{0,1,2\}^\omega$ and let $c \in \{0,1,2\}^\omega$ be defined by $c_i = \min (\{0, 1, 2\} \setminus \{a_i, b_i\})$. Then $c_i \neq a_i$ and $c_i \neq b_i$, for all $i \in \omega$, therefore $a$ and $b$ are connected via $c$. The graph $G$ is connected.

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  • $\begingroup$ Very nice construction and explanation, thank you Andrej! $\endgroup$ Commented Aug 4, 2022 at 10:06

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