Symmetric Functions, the Hall Inner Product, and Schur Bases — Examples

At rank two, [F2] and [F3] give $m_{(2)}=x_1^2+x_2^2$, $m_{(1,1)}=x_1{x}_2$, $h_2=m_{(2)}+m_{(1,1)}$, $p_1^2=m_{(2)}+2m_{(1,1)}$, and $p_2=m_{(2)}$. By [F7], $m_{(2)}$ and $m_{(1,1)}$ are a basis in rank two; by [F5] and the stable-ring projection in [F1], the rank-two projection ${\mathrm{\Lambda }}^2\to A_2^2$ carries the stable basis to that finite basis and is an isomorphism. The stable $h_2$ projects to its finite polynomial by [F2] and [F4], and the finite power sums form compatible stable sequences by [F1] and [F2]. Thus the same three equations hold in ${\mathrm{\Lambda }}^2$. At rank one the three finite functions $h_2,p_1^2,p_2$ all equal $x_1^2$, while rank zero has no degree-two monomials; rank two is the first rank that distinguishes the two monomial orbits.

Since $p_{(1,1)}=p_1^2$, step 1.1 yields $h_2=\frac{1}{2}(p_{(1,1)}+p_{(2)})$. By [F6], $p_{(1,1)}$ and $p_{(2)}$ are a $\mathbb{Q}$-basis, so this is the unique rational coordinate vector of $h_2$. If $h_2=a{p}_{(1,1)}+b{p}_{(2)}$ for integers $a,b$, including zero values, uniqueness forces $a=b=\frac{1}{2}$, impossible. Equivalently, comparison in the integral basis of [F5] forces the $m_{(1,1)}$ coefficient to satisfy $2a=1$. Thus $h_2$ is a witness to failure of integral spanning.

The partitions of $3$ are $(3),(2,1),(1,1,1)$, all of length at most $3$. For every $N\ge 3$, [F7] gives the rank-$N$ finite orbit-sum basis and [F8] identifies its labels with the stable monomial basis; the specialization maps preserve each labelled orbit sum. Thus projection to rank $3$ is an isomorphism in degree $3$, so the finite calculations below determine the stable coordinates.

In three variables, the orbit sums are $m_3={\sum }_i{x}_i^3$, $m_{21}={\sum }_{i\mathrm{\ne }j}{x}_i^2{x}_j$, and $m_{111}=x_1{x}_2{x}_3$. Each monomial of type $(3)$ or $(2,1)$ occurs once in $m_2{m}_1$; $m_{11}{m}_1$ counts each type-$(2,1)$ monomial once and the all-distinct monomial three times; $m_1^3$ counts the patterns $(3),(2,1),(1,1,1)$ with multiplicities $1,3,6$.

The finite subset definition gives $e_1=m_1$, $e_2=m_{11}$, and $e_3=m_{111}$. Multiplication using step 2.1 then gives the three degree-three elementary products.

The complete functions list all degree-$k$ monomials, while the power sums are the sums of pure $k$th powers. Thus $h_1=m_1$, $h_2=m_2+m_{11}$, and $h_3=m_3+m_{21}+m_{111}$; multiplying and using the orbit counts yields the remaining complete and power-sum products.

Jacobi–Trudi at sizes one and two gives $s_3=h_3$ and $s_{21}=h_2{h}_1-h_3$; dual Jacobi–Trudi at size one gives $s_{111}=e_3$. Substitution from steps 3.1 and 3.2 yields the three monomial coordinates.

With rows labelled $(3),(2,1),(1,1,1)$ and columns ordered $(m_3,m_{21},m_{111})$, the rows of each matrix are the displayed coordinates from steps 3.1, 3.2, and 4.1. Their determinants show that the $m,e,h,s$ matrices are unimodular, while the $p$ matrix has nonzero determinant $6$. The $p$ rows therefore form a rational basis; an explicit nonintegral coordinate for $h_3$ also shows failure of integral spanning.

The degree-three claim has no empty-partition row because $\mathrm{\varnothing }$ has degree zero; zero has the all-zero coordinate vector. The one-part label $(3)$ is the first row of every matrix, and repeated parts occur in the $(1,1,1)$ row. Degree $3$ and rank $3$ are the claimed degree endpoint and threshold rank. All orbit and product counts use finite sets, so no choice is made; no iff assertion occurs.

At rank $3$, the projection ${\mathrm{\Lambda }}^d\to A_3^d$ is an isomorphism for each $0\le d\le 3$: for $d>0$, every partition of $d$ has length at most $d\le 3$, and [F5] and [F6] identify the stable and finite monomial bases; for $d=0$, both components are $\mathbb{Z}$ with basis $1$. Thus finite rank-$3$ coefficient calculations determine the stable coefficients in all degrees used here.

At rank $3$, grouping monomials by distinct exponent orbits and counting ordered products gives the complete-function identities below; $m_2{m}_1=m_3+m_{21}$ and $m_{11}{m}_1=m_{21}+3m_{111}$, while types $(3),(2,1),(1,1,1)$ occur in $h_1^3$ with multiplicities $1,3,3!=6$.

Jacobi–Trudi evaluates the Schur functions through degree three as shown; substituting step 2.1 gives the monomial expressions, including $s_{111}=h_1^3-2h_2{h}_1+h_3$.

The partitions of degrees $0,1,2,3$ are respectively $\{\mathrm{\varnothing }\}$, $\{(1)\}$, $\{(2),(1,1)\}$, and $\{(3),(2,1),(1,1,1)\}$, so [F1] gives these complete–monomial and Schur components of ${\mathrm{\Omega }}_{\le 3}$ in bidegrees $(d,d)$.

Since $s_{21}=h_{21}-h_3$, duality in [F8] gives $⟨s_{21},h_3{⟩}_H=⟨h_{21}-h_3,m_3+m_{21}+m_{111}{⟩}_H=1-1=0$.

Similarly, $⟨s_{21},h_{21}{⟩}_H=⟨h_{21}-h_3,m_3+2m_{21}+3m_{111}{⟩}_H=2-1=1$.

Also, $⟨s_{21},h_{111}{⟩}_H=⟨h_{21}-h_3,m_3+3m_{21}+6m_{111}{⟩}_H=3-1=2$.

Gradedness removes degrees below three, so contracting the first factor in the complete–monomial expansion gives $m_{21}(y)+2m_{111}(y)=s_{21}(y)$; contraction of the Schur expansion gives the same result by [F10].

Degree $0$ gives $1\otimes 1$ (and [F11] specializes to $1$ when either alphabet is zero); in degree $1$, $h_1=m_1=s_1$ has coefficient one. Degree $3$ is the retained upper endpoint and rank $3$ is its threshold rank; repeated parts in $h_{11}$ and $h_{111}$ have the coefficients from step 2.1, while each $m_{\lambda }$ lists distinct monomials. All counts are finite, so no choice is used, and no equivalence is asserted.

Fix $N\ge 0$ and take $\lambda =(1^{N+1})$. By [F2] it is a partition of degree $N+1$, and [F7] identifies $s_{\lambda }$ as its stable Schur element. By [F6], this element is one vector in a $\mathbb{Z}$-basis of ${\mathrm{\Lambda }}^{N+1}$, so $s_{\lambda }\mathrm{\ne }0$.

The conjugate partition is ${\lambda }^{\mathrm{\prime }}=(N+1)$, so the dual Jacobi–Trudi identity [F5] with determinant size $c=1$ gives $s_{(1^{N+1})}=e_{N+1}$ in $\mathrm{\Lambda }$.

By [F1] and [F3], the rank-$N$ coordinate of this stable $e_{N+1}$ is the finite elementary polynomial $e_{N+1}(x_1,\dots ,x_N)$.

By [F4], this polynomial is a sum indexed by the $(N+1)$-element subsets of an $N$-element set; there are no such subsets, including when $N=0$, so the sum is zero. Thus every stated rank-$N$ specialization vanishes although the stable element is nonzero, and the coordinate projection has a nontrivial kernel.

Pad $\mu =(1)$ to $(1,0)$. By [F1], the remaining cells of $\lambda \mathrm{/}\mu$ are exactly $(1,2),(1,3),(2,1)$. The first two share an edge; $(2,1)$ shares no edge with either, since $(1,1)$ was removed and $(1,2)$ is only diagonally adjacent. Thus the components are the two-cell row and the single cell.

The minimum allowed determinant size is $r=2$. Its matrix entries are $h_{3-1-1+1}=h_2$, $h_{3-0-1+2}=h_4$, $h_{1-1-2+1}=h_{-1}$, and $h_{1-0-2+2}=h_1$, so [F8] gives $\det \left(\begin{array}{cc}{h}_2& h_4\\ 0& h_1\end{array}\right)=h_2{h}_1$. The off-diagonal $h_4$ term contributes zero because the lower-left entry is $h_{-1}=0$.

A semistandard filling assigns entries $a\le b$ to the top row and an arbitrary positive entry $c$ to the isolated lower cell; there is no row or column inequality connecting the components [F5, F6]. By [F3], its weight generating series in rank $N$ is ${\sum }_{1\le a\le b\le N}{\sum }_{1\le c\le N}{x}_a{x}_b{x}_c=h_2^{(N)}{h}_1^{(N)}$. The tableau formula [F8] and compatible stable products [F2, F4] therefore give $s_{(3,1)\mathrm{/}(1)}=h_2{h}_1$ in $\mathrm{\Lambda }$.

At rank $3$, [F3] gives $h_2^{(3)}=x_1^2+x_2^2+x_3^2+x_1{x}_2+x_1{x}_3+x_2{x}_3$ and $h_1^{(3)}=x_1+x_2+x_3$, hence $h_2^{(3)}{h}_1^{(3)}=x_1^3+x_2^3+x_3^3+2{\sum }_{i\mathrm{\ne }j}{x}_i^2{x}_j+3x_1{x}_2{x}_3$. Each cube occurs once, each $x_i^2{x}_j$ with $i\mathrm{\ne }j$ occurs twice (from $x_i^2{x}_j$ and $x_i{x}_j{x}_i$), and the triple product occurs three times, once from each pair term of $h_2^{(3)}$.

The fixed skew shape has three boxes, so an empty-shape case does not arise. At rank zero both positive-degree complete functions vanish [F7] and no positive entry is available [F6]; at rank one [F3] gives $h_2^{(1)}{h}_1^{(1)}=x_1^3$, matching the unique filling with $1$ in all three cells. Hence this example is nonzero, as the rank-three expansion also shows. The determinant uses its minimum size $r=2$ [F8]; each finite-rank tableau set is finite and the compatible stable passage makes no choice. The example asserts no biconditional.