# Is this representation of Go (game) irreducible?

This post is freely inspired by the basic rules of Go (game), usually played on a $$19 \times 19$$ grid graph.

Consider the $$\mathbb{Z}^2$$ grid. We can assign to each vertex a state "black" ($$b$$), "white" ($$w$$) or "empty" ($$e$$).
A state of the grid is a map $$\phi: \mathbb{Z}^2 \to \{b,w,e\}.$$ The subsets $$\phi^{-1}(\{b\})$$ and $$\phi^{-1}(\{w\})$$ decompose into connected components $$c$$. Let $$d(c)$$ be the number of liberties of $$c$$, i.e. the number of elements of $$\phi^{-1}(\{e\})$$ connected to $$c$$ by an edge.
The following picture on the left shows a white connected component $$c_1$$ with number of liberties $$d(c_1)=8$$ and a black connected component $$c_2$$ with $$d(c_2) = 5$$. On the right, there is a black connected component $$c_3$$ with $$d(c_3)=0$$.

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A state $$\phi$$ is finite if $$|\phi^{-1}(\{b,w\})| < \infty$$, and is admissible if all its connected components $$c$$ have at least one liberty. Let $$\mathcal{F}$$ be the set of finite admissible states. Let $$g \in \mathbb{Z}^2$$ and consider the map $$B_g: \mathcal{F} \to \mathcal{F}$$ such that if $$\phi(g) \neq e$$ then $$B_g(\phi) = \phi$$, else $$B_g(\phi)$$ is the state made by first putting the vertex $$g$$ to the state "black", then removing every white connected component without liberty, then removing every black connected component without liberty (suicide allowed).
See the following examples:

$$\hspace{0.7cm}$$ $$\hspace{0.7cm}$$

The map $$W_g$$ is defined similarly by interchanging black and white. Now, let $$H$$ be the Hilbert space $$\ell^2(\mathcal{F})$$. We extend $$B_g$$ and $$W_g$$ by linearity on $$H$$. The adjoint $$B^*_g$$ of $$B_g$$ is given by: $$B^*_g(\phi) = \sum_{\phi' \in B_g^{-1}(\{ \phi\})} \phi'$$ The adjoint $$W^*_g$$ of $$W_g$$ is given by a similar formula.

Let $$\mathcal{A}$$ be the $$*$$-algebra generated by the set $$\{B_g,B_g^*, W_g, W_g^* \ | \ g \in \mathbb{Z}^2 \}$$.

Question: Is $$H$$ an irreducible representation of $$\mathcal{A}$$?

If yes, this means that the von Neumann algebra $$M:=\mathcal{A}''$$ equals $$B(H)$$, the algebra of all bounded operators. Else, what is $$M$$?

Remark: these questions can be extended to any finitely generated group $$\Gamma$$ where the grid becomes its Cayley graph. It should again be extended to any vertex-transitive graph, where for any vertex $$g$$, $$\Vert B_g \Vert^2$$ should "intuitively" be $$2^d$$ (with $$d$$ the degree of the graph). The extension to any connected graph could reveal problems with unbounded operators; to avoid that we can assume the graph to be $$k$$-regular (with $$k$$ finite), or at least the degree of valency of the vertices to be bounded.

We could experiment these questions on some small finite graphs by brute force with a computer.

• ...admissible if all its connected components c have at most one liberty isn't it at least? Sep 21 '19 at 22:27
• @WhatsUp: yes sure, thanks! Sep 22 '19 at 2:03
• Does this carry any information about Go? Sep 24 '19 at 13:23
• @SantanaAfton: Are you asking whether this question is interesting for a Go player? Sep 25 '19 at 4:47
• Yeah, I’m wondering if the answer to this question gives some interesting information about the game. Sep 26 '19 at 0:10