The c-Cambrian quotient of S3 for both orientations: fibers, endpoints and meet/join preservation

Put $a=s_1$ and $b=s_2$. Cancellation reduces every word to an alternating word; $(ab{)}^3=1$ gives $abab=ba$ and $baba=ab$, so every alternating word of length at least four shortens, while $aba=bab$. Thus every element is represented by one of $e,a,b,ab,ba,aba$. The map $a\mapsto (12)$, $b\mapsto (23)$ satisfies the presentation and sends these six forms to six distinct permutations, so they are exactly the elements of $W$. Their lengths are $0,1,1,2,2,3$ respectively, and left multiplication gives $D_L(e)=\mathrm{\varnothing }$, $D_L(a)=\{a\}$, $D_L(b)=\{b\}$, $D_L(ab)=\{a\}$, $D_L(ba)=\{b\}$, $D_L(aba)=\{a,b\}$.

For $c=ab$, the lexicographically first reduced position sets in $c^{\mathrm{\infty }}=ab\mathrm{\mid }ab\mathrm{\mid }\cdots$ are $e:\mathrm{\varnothing }$, $a:1$, $b:2$, $ab:1,2$, $ba:2,3$, $w_0=aba:1,2,3$. Their block sequences are respectively $\mathrm{\varnothing }$, $\{a\}$, $\{b\}$, $\{a,b\}$, $\{b\}\supseteq ̸\{a\}$, and $\{a,b\}\supseteq \{a\}$. Hence exactly $e,a,b,ab,w_0$ are sortable, while $ba$ fails the inclusion test.

For $c^{\mathrm{\prime }}=ba$, the first reduced position sets in $c^{\mathrm{\prime }\mathrm{\infty }}=ba\mathrm{\mid }ba\mathrm{\mid }\cdots$ are $e:\mathrm{\varnothing }$, $b:1$, $a:2$, $ba:1,2$, $ab:2,3$, $w_0=bab:1,2,3$. Their block sequences are $\mathrm{\varnothing }$, $\{b\}$, $\{a\}$, $\{a,b\}$, $\{a\}\supseteq ̸\{b\}$, and $\{a,b\}\supseteq \{b\}$. Thus exactly $e,b,a,ba,w_0$ are $c^{\mathrm{\prime }}$-sortable and $ab$ is the unique failure.

On a rank-one parabolic, ${\pi }_a(e)=e$ and ${\pi }_a(a)=a{\pi }_a(e)=a$, and likewise ${\pi }_b(e)=e$ and ${\pi }_b(b)=b$. Applying the initial-letter recursion at $a$ for $c=ab$ and at $b$ for $c^{\mathrm{\prime }}=ba$ gives $\begin{array}{ccccccc}w& e& a& b& ab& ba& w_0\\ {\pi }_c(w)& e& a& b& ab& b& w_0\\ {\pi }_{c^{\mathrm{\prime }}}(w)& e& a& b& a& ba& w_0\end{array}$. For $c$, the entries follow from ${\pi }_c(a)=a{\pi }_{ba}(e)=a$, ${\pi }_c(b)={\pi }_b(b)=b$, ${\pi }_c(ab)=a{\pi }_{ba}(b)=ab$, ${\pi }_c(ba)={\pi }_b(b)=b$, and ${\pi }_c(w_0)=a{\pi }_{ba}(ba)=aba$. For $c^{\mathrm{\prime }}$, they follow from ${\pi }_{ba}(a)={\pi }_a(a)=a$, ${\pi }_{ba}(b)=b$, ${\pi }_{ba}(ab)={\pi }_a(a)=a$, ${\pi }_{ba}(ba)=b{\pi }_{ab}(a)=ba$, and ${\pi }_{ba}(w_0)=b{\pi }_{ab}(ab)=bab$.

Reduced prefixes give the Hasse diagram $e⋖a⋖ab⋖w_0$ and $e⋖b⋖ba⋖w_0$. These are the only covers: multiplying each of the six reduced forms on the right by $a$ or $b$ either cancels its last letter or gives the next displayed form. The four incomparable pairs are exactly one choice from $\{a,ab\}$ and one from $\{b,ba\}$; their only common lower bound is $e$ and their only common upper bound is $w_0$. Comparable pairs have their lesser and greater elements as meet and join, respectively. This determines both operation tables in the statement and verifies them as greatest lower and least upper bounds.

By the kernel definition, the fibers for $c$ are $\{e\},\{a\},\{b,ba\},\{ab\},\{w_0\}$, and those for $c^{\mathrm{\prime }}$ are $\{e\},\{a,ab\},\{b\},\{ba\},\{w_0\}$. Their sortable images inherit the right weak order as the quotient order, so these are the stated two quotient partitions.

The map ${\pi }_c$ fixes every element except $ba$, which it sends to $b$. If neither operand is $ba$, their meet and join are also fixed: a join can equal $ba$ only when both operands lie below it, and among such pairs without $ba$ their join is $e$ or $b$; a meet can equal $ba$ only when both operands lie above it, and among such pairs without $ba$ the only possibility is $w_0,w_0$, whose meet is $w_0$. For an operand $ba$, the six cases $z=e,a,b,ab,ba,w_0$ give $(ba\vee z,b\vee {\pi }_c(z))=(ba,b),(w_0,w_0),(ba,b),(w_0,w_0),(ba,b),(w_0,w_0)$ and $(ba\wedge z,b\wedge {\pi }_c(z))=(e,e),(e,e),(b,b),(e,e),(ba,b),(ba,b)$; applying ${\pi }_c$ to the first coordinate yields equality in every case. By symmetry of meet and join this checks all 36 pairs. These are the quotient operations from [F10] and the direct calculation is the rank-two instance of [F9].

With $w_0=aba=bab$, the products $w{w}_0$ for $w=e,a,b,ab,ba,w_0$ are respectively $w_0,ba,ab,b,a,e$. Applying the opposite-orientation projection table and multiplying by $w_0$ gives $u_c(w)=e,a,ba,ab,ba,w_0$ in that order. In particular, $u_c(b)=u_c(ba)=ba$. Reversing $a,b$ gives $u_{c^{\mathrm{\prime }}}(w)=e,ab,b,ab,ba,w_0$ in the same input order, so $u_{c^{\mathrm{\prime }}}(a)=u_{c^{\mathrm{\prime }}}(ab)=ab$.

The nontrivial fibers are the chains $b{<}_{R}ba$ and $a{<}_{R}ab$, since $ba=ba$ and $ab=ab$ are length-additive right extensions; all other fibers are singletons. The endpoints in step 3.3 therefore give exactly $[b,ba]$ and $[a,ab]$ and singleton intervals. For $c$, $u_c$ changes only $b$ to $ba$; it preserves each cover in step 2.2, is idempotent, and the projection table gives $u_c{\pi }_c=u_c$ and ${\pi }_c{u}_c={\pi }_c$. The same checks with $a$ and $ab$ establish the corresponding assertions for $c^{\mathrm{\prime }}$.

Both orientations have five sortable elements, but their nonsortable elements and nontrivial fibers are interchanged: $ba\mapsto b$ for $c=ab$ and $ab\mapsto a$ for $c^{\mathrm{\prime }}=ba$. Thus the fiber-size profile is $1,1,2,1,1$ in either orientation, while the quotient partitions differ. All claims follow from six explicitly listed elements and finite case checks, so no Choice is used.