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Noah Schweber
Noah Schweber
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Let $\mathscr{C}_n$ be the lattice of clones on the $n$-element set $\{1,...,n\}$. $\mathscr{C}_2$ is complicated but countable, but $\mathscr{C}_3$ (and all higher lattices) is of size continuum.

Question: Does $\mathscr{C}_4$ embed as a lattice into $\mathscr{C}_3$?

More generally, do we know the status of any of the embeddability questions $\mathscr{C}_m\hookrightarrow\mathscr{C}_n$ for $m>n>2$? My suspicion is that $\mathscr{C}_3$ is already "wild" enough that every $\mathscr{C}_m$ embeds into it, and that the specific $4/3$ case above will be the easiest to address.

At MSE I asked a harder version of this question, but per Keith Kearnes' answer there it seems that even this weaker question may be tricky (or even open).

Let $\mathscr{C}_n$ be the lattice of clones on the $n$-element set $\{1,...,n\}$. $\mathscr{C}_2$ is complicated but countable, but $\mathscr{C}_3$ (and all higher lattices) is of size continuum.

Question: Does $\mathscr{C}_4$ embed as a lattice into $\mathscr{C}_3$?

At MSE I asked a harder version of this question, but per Keith Kearnes' answer there it seems that even this weaker question may be tricky (or even open).

Let $\mathscr{C}_n$ be the lattice of clones on the $n$-element set $\{1,...,n\}$. $\mathscr{C}_2$ is complicated but countable, but $\mathscr{C}_3$ (and all higher lattices) is of size continuum.

Question: Does $\mathscr{C}_4$ embed as a lattice into $\mathscr{C}_3$?

More generally, do we know the status of any of the embeddability questions $\mathscr{C}_m\hookrightarrow\mathscr{C}_n$ for $m>n>2$? My suspicion is that $\mathscr{C}_3$ is already "wild" enough that every $\mathscr{C}_m$ embeds into it, and that the specific $4/3$ case above will be the easiest to address.

At MSE I asked a harder version of this question, but per Keith Kearnes' answer there it seems that even this weaker question may be tricky (or even open).

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Noah Schweber
Noah Schweber
  • 20.7k
  • 10
  • 111
  • 332

Lattices of clones: is 4 worse than 3?

Let $\mathscr{C}_n$ be the lattice of clones on the $n$-element set $\{1,...,n\}$. $\mathscr{C}_2$ is complicated but countable, but $\mathscr{C}_3$ (and all higher lattices) is of size continuum.

Question: Does $\mathscr{C}_4$ embed as a lattice into $\mathscr{C}_3$?

At MSE I asked a harder version of this question, but per Keith Kearnes' answer there it seems that even this weaker question may be tricky (or even open).