A parabolic quotient interval of S4 whose Möbius value is 0, so the Eulerian sign formula does not extend to quotients

The quotient interval. By [F1] an element $w$ lies in $W^I$ exactly when neither $s_1$ nor $s_3$ is a right descent of $w$, that is, when $\mathrm{\ell }(w{s}_1)>\mathrm{\ell }(w)$ and $\mathrm{\ell }(w{s}_3)>\mathrm{\ell }(w)$; since right multiplication by $s_i$ swaps the entries in positions $i$ and $i+1$ of the one-line form, this says $w(1)<w(2)$ and $w(3)<w(4)$. Testing the $24$ elements of $S_4$ leaves exactly $W^I=\{1234,1324,1423,2314,2413,3412\}$, of lengths $0,1,2,2,3,4$ by [F4]; in particular $3412$ is the unique element of $W^I$ of length $4$.

The covers. By [F2] the order on $W^I$ is the restriction of the Bruhat order, and two elements of $W^I$ whose lengths differ by one form a cover exactly when they are comparable; the adjacent length pairs in $W^I$ are only the pairs $\{1234,1324\}$, $\{1324,1423\}$, $\{1324,2314\}$, $\{1423,2413\}$, $\{2314,2413\}$ and $\{2413,3412\}$, because $W^I$ has exactly one element of length $0$ and of length $1$, two of length $2$, and one of length $3$ and of length $4$. Each of these six pairs is comparable, as the subword criterion [F3] shows with the reduced expressions $1324=s_2$, $1423=s_3{s}_2$, $2314=s_1{s}_2$, $2413=s_1{s}_3{s}_2$ and $3412=s_2{s}_1{s}_3{s}_2$: the subwords $s_2\le s_3{s}_2$, $s_2\le s_1{s}_2$, $s_3{s}_2\le s_1{s}_3{s}_2$, $s_1{s}_2\le s_1{s}_3{s}_2$ and $s_1{s}_3{s}_2\le s_2{s}_1{s}_3{s}_2$ exhibit the five comparabilities above the bottom, and $1234\le 1324$ is the empty subword. Hence the covers inside $W^I$ are exactly $1234⋖1324$, $1324⋖1423$, $1324⋖2314$, $1423⋖2413$, $2314⋖2413$ and $2413⋖3412$, and these covers chain every element of $W^I$ below $3412$; so $3412$ is the greatest element of $W^I$ and the quotient interval $[1234,3412{]}^I=[1234,3412]\cap W^I$ has exactly these six elements.

The fullness failure. The element $1432$ of $S_4$ satisfies $1432\mathrm{\notin }{W}^I$ because $\mathrm{\ell }(1432s_3)=\mathrm{\ell }(1423)=2<3=\mathrm{\ell }(1432)$ [F4], while $1432\le 3412$ in the full Bruhat order: $3412=s_2{s}_1{s}_3{s}_2$ is a reduced expression (its length $4$ equals the inversion number of $3412$) and $1432=s_2{s}_3{s}_2$ is the product of its subword at positions $1,3,4$ [F3]. Hence $1432\in [1234,3412]\setminus [1234,3412{]}^I$, the quotient interval is a proper subset of the full interval $[1234,3412]$, and the fullness hypothesis fails for it.

The Möbius values. With the recurrence [F5] and the cover list of step 2.1: $\mu (1234,1234)=1$ and $\mu (1234,1324)=-1$ (the atom covers the bottom); $\mu (1234,1423)=-(\mu (1234,1234)+\mu (1234,1324))=-(1-1)=0$ and likewise $\mu (1234,2314)=0$ (each has exactly the two displayed elements below it in the quotient interval); $\mu (1234,2413)=-(1-1+0+0)=0$; and finally $\mu (1234,3412)=-(1-1+0+0+0)=0$.

The refutation. By step 3.1 the induced quotient interval has $\mu (1234,3412)=0$, whereas $(-1{)}^{\mathrm{\ell }(3412)-\mathrm{\ell }(1234)}=(-1{)}^4=1$ by [F4]; so the indiscriminate Eulerian claim displayed above is false for this $I$ and this interval. The exact dropped hypothesis is fullness of the interval: step 2.2 shows $[1234,3412{]}^I⊊[1234,3412]$, and for full intervals the sign formula holds by [F6]. This is why the theorem restricts its scope in [F7].