Hard Lefschetz and the Hodge-Riemann signs for the coinvariant algebra of S_3

The two-variable model. Put $x=x_1$, $y=x_2$ and $x_3=-x-y$. Then $e_1=x+y+x_3=0$, while $e_2=x_1{x}_2+x_1{x}_3+x_2{x}_3=xy+x(-x-y)+y(-x-y)=-(x^2+xy+y^2)$ and $e_3=x_1{x}_2{x}_3=-x^{2}y-x{y}^2$; hence the coinvariant ideal is generated by $x^2+xy+y^2$ and $x^{2}y+x{y}^2$, and since $x^3=x(x^2+xy+y^2)-x^{2}y-x{y}^2$ the second relation can be replaced by $x^3=0$, giving $C=\mathbb{R}[x,y]\mathrm{/}(x^2+xy+y^2,x^3)$. The monomials $1;x,y;x^2,xy;x^{2}y$ span it: the relations $x^2=-xy-y^2$ and $x^3=0$, together with $x{y}^2=-x^{2}y$ and $y^3=0$, reduce every monomial of degree $\ge 2$ in $x$ to one of them, and their independence follows by monic division by $y^2+xy+x^2$ over $\mathbb{R}[x]\mathrm{/}(x^3)$, which gives the free basis $1,y$ over that ring.

The trace and pairing. The rank-two summand map $j(p\otimes q)=p\otimes 1\otimes 1\otimes q$ pulls the Bott–Samelson trace back to ${\mathrm{\partial }}_1{\mathrm{\partial }}_2{\mathrm{\partial }}_1(p)q$, as follows by taking the right-coefficient trace successively in the three slots; this is the explicit trace calculation in the rank-two decomposition proof, also obtained directly from the rank-one trace convention. After right reduction it is the top trace on $C$, since ${\mathrm{\partial }}_1(x_1^2{x}_2)=x_1{x}_2$, ${\mathrm{\partial }}_2(x_1{x}_2)=x_1$ and ${\mathrm{\partial }}_1(x_1)=1$. Hence $\mathrm{Tr}(x^{2}y)=1$ and all degrees other than $6$ have zero trace. The complementary-degree pairings are $1$ in degrees $0,6$ and $\left(\begin{array}{cc}0& 1\\ 1& -1\end{array}\right)$ in degrees $2,4$, of determinant $-1$; they prove nondegeneracy directly. The form used below is therefore the actual induced reduced intersection form, without invoking the general Hodge theorem.

The special dominant element. For $\rho =2x+y$ one computes $\rho x=2x^2+xy$ and $\rho y=2xy+y^2=-x^2+xy$ using $y^2=-x^2-xy$; the matrix of $\rho$ from the degree-$2$ block with basis $(x,y)$ to the degree-$4$ block with basis $(x^2,xy)$ is therefore $\left(\begin{array}{cc}2& -1\\ 1& 1\end{array}\right)$, whose determinant is $3\mathrm{\ne }0$, so $\rho :C^{-1}\to C^1$ is an isomorphism. Moreover ${\rho }^{2}x=3x^{2}y$ and ${\rho }^{2}y=0$: indeed $x^2\rho =x^2(2x+y)=2x^3+x^{2}y=x^{2}y$, $xy\rho =xy(2x+y)=2x^{2}y+x{y}^2=x^{2}y$; hence ${\rho }^{2}x=2\cdot x^{2}y+x^{2}y=3x^{2}y$ and ${\rho }^{2}y=-x^{2}y+x^{2}y=0$; hence $P^{-1}=\ker ({\rho }^2)=\mathbb{R}y$ and ${\rho }^3(1)={\rho }^2(2x+y)=6x^{2}y\mathrm{\ne }0$, so ${\rho }^3:C^{-3}\to C^3$ is an isomorphism as both sides are one-dimensional.

The signs for the special element. With the normalization $\mathrm{Tr}(x^{2}y)=1$, the bottom primitive form is $(1,1)=\mathrm{Tr}(1\cdot {\rho }^3(1))=\mathrm{Tr}(6x^{2}y)=6>0$, positive as required in degree $-3$; the middle primitive form is $(y,y)=\mathrm{Tr}(y\cdot \rho y)=\mathrm{Tr}(y(-x^2+xy))=\mathrm{Tr}(-x^{2}y+x{y}^2)=-1-1=-2<0$, negative as required in degree $-1$, where the standard sign is $(-1{)}^{(3-1)\mathrm{/}2}=-1$. The primitive decomposition is read off from the two strings: $1\mapsto 2x+y\mapsto 3x^2+3xy\mapsto 6x^{2}y$ spans the four degrees with one vector each, and $y\mapsto -x^2+xy$ spans degrees $-1$ and $+1$; together they account for all four graded pieces and their dimensions, so there are exactly one length-four string and one length-two string.

The general dominant chamber. Let $\rho =c_1{x}_1+c_2{x}_2+c_3{x}_3$ be strictly dominant, so $c_1>c_2>c_3$; subtracting a multiple of $e_1$ (which is zero in $C$) writes its class as $\rho =ax+by$ with $a=c_1-c_3>b=c_2-c_3>0$. Then $\rho x=a{x}^2+bxy$ and $\rho y=-b{x}^2+(a-b)xy$, so the middle matrix is $\left(\begin{array}{cc}a& -b\\ b& a-b\end{array}\right)$ with determinant $D=a^2-ab+b^2>0$, and ${\rho }^3(1)=3ab(a-b)x^{2}y$: indeed ${\rho }^{2}x=b(2a-b)x^{2}y$ and ${\rho }^{2}y=a(a-2b)x^{2}y$, so $p:=a(a-2b)x-b(2a-b)y$ satisfies ${\rho }^{2}p=0$, and expanding gives $\mathrm{Tr}(p\rho p)=A^{2}b-B^{2}a+2AB(b-a)$ with $A=a(a-2b)$, $B=b(2a-b)$, which simplifies to $-3ab(a-b)(a^2-ab+b^2)<0$. Hence $P^{-1}=\mathbb{R}p$ (one-dimensional, since $p\mathrm{\ne }0$ for $a>b>0$) carries a negative definite Lefschetz form, the bottom form $\mathrm{Tr}(3ab(a-b)x^{2}y)>0$ is positive, and every strictly dominant $\rho$ satisfies hard Lefschetz with the standard signs on both primitive spaces.

Conclusion. Steps 2.1 and 3.1 verify the displayed isomorphisms, the two primitive spaces and their signs for the explicit choice, and step 5.1 shows the same for every strictly dominant element; the primitive decomposition has the two strings claimed. This verifies the example by its quotient and trace calculations.