Symmetric Functions, the Hall Inner Product, and Schur Bases

In degree $d=0$, the only partition is $\mathrm{\varnothing }$, its orbit sum is the constant $1$, and ${\mathrm{\Lambda }}^0=\mathbb{Z}$; hence it is a basis.

Suppose $d>0$ and fix $N\ge d$. Every partition of $d$ has at most $d$ parts, so every $\lambda ⊢d$ has $\mathrm{\ell }(\lambda )\le N$. By [F3], the rank-$N$ orbit sums indexed by these partitions form a $\mathbb{Z}$-basis of $A_N^d$.

If $M>N\ge d$, specializing $x_{N+1},\dots ,x_M$ to zero leaves exactly those orbit monomials whose positive exponents all lie among the first $N$ variables; these are precisely the distinct rank-$N$ orbit monomials, each once. Thus every transition $A_M^d\to A_N^d$ is an isomorphism carrying the displayed basis to itself.

A compatible sequence in ${\mathrm{\Lambda }}^d$ is uniquely determined by its rank-$N$ component. Expanding that component in the finite basis of [F3], compatibility and the basis-preserving isomorphisms of step 1.3 force the same integer coefficients at every rank $M\ge N$; lower-rank components are their specializations. Conversely, every finite integer combination of the compatible orbit sums gives such a sequence. Hence the stable orbit sums span and are linearly independent in ${\mathrm{\Lambda }}^d$.

For each fixed $r$, specializing an added variable to zero sends the finite $e_r$ and $h_r$ to their lower-rank polynomials: terms involving that variable vanish, and the remaining subset or exponent tuples are unchanged. They therefore define homogeneous compatible sequences in ${\mathrm{\Lambda }}^r$ by [F1], [F4], and [F5].

Fix $\lambda ⊢d$ and a rank $N\ge d$. In expanding $e_{{\lambda }^{\mathrm{\prime }}}={\prod }_j{e}_{{\lambda }_j^{\mathrm{\prime }}}$, regard each factor as a column of height ${\lambda }_j^{\mathrm{\prime }}$ and record the distinct variable labels selected by [F4]. For any resulting monomial, relabel variables so its exponents are weakly decreasing and call that exponent partition $\mu$. Among the first $r$ variable labels each column contributes at most $\min (r,{\lambda }_j^{\mathrm{\prime }})$ occurrences. Hence ${\sum }_{i=1}^r{\mu }_i\le {\sum }_j\min (r,{\lambda }_j^{\mathrm{\prime }})={\sum }_{i=1}^r{\lambda }_i,$ so $\lambda ⊵\mu$ by [F3] and [F6]. Since the factors are symmetric, relabeling does not change the coefficient of the orbit sum.

Fix $n\ge 1$. At every finite rank $N\ge 0$, the coefficient of $t^n$ in [F7] gives the recurrence among the rank-$N$ components of $e_i$ and $h_{n-i}$. By [F1], [F4], [F5], and step 1.1, these are the rank-$N$ projections of the corresponding stable products. Since the recurrence holds at every rank, it is the zero sequence in ${\mathrm{\Lambda }}^n$; the constant coefficient is $e_0{h}_0=1$. Thus $E(-t)H(t)=1$ coefficientwise in the stable ring.

Consider the monomial $x_1^{{\lambda }_1}{x}_2^{{\lambda }_2}\cdots$ in rank $N\ge d$. Its first-$r$ exponent sum is ${\sum }_{i=1}^r{\lambda }_i={\sum }_j\min (r,{\lambda }_j^{\mathrm{\prime }})$ for each $r$. In a selection contributing this monomial, each column contributes at most $\min (r,{\lambda }_j^{\mathrm{\prime }})$ to that prefix, so equality of the total forces equality in every column for every $r$. A column of height $h$ must therefore select precisely labels $1,\dots ,h$; this is one selection. Hence the coefficient of the monomial, and therefore of the orbit sum $m_{\lambda }$, in $e_{{\lambda }^{\mathrm{\prime }}}$ is one.

Fix $d>0$ and project to rank $N=d$. By [F10], each stable basis element $m_{\mu }$ projects to the rank-$d$ orbit sum; [F8] identifies its distinct monomial terms, and [F9] says these projections form a $\mathbb{Z}$-basis. Thus the projection identifies stable and finite monomial coefficients. The rank-$d$ expansion of each $e_{{\lambda }^{\mathrm{\prime }}}$ therefore gives the stable transition matrix. Its entries are integers, vanish unless $\lambda ⊵\mu$ by step 1.2, and have diagonal entries one by step 2.2. Ordering the finite dominance poset by a linear extension makes this matrix unitriangular, hence invertible over $\mathbb{Z}$. The degree $d=0$ case is the single basis element $e_{\mathrm{\varnothing }}=m_{\mathrm{\varnothing }}=1$. By [F2], the $m_{\mu }$ are an integral basis; by the conjugation bijection [F3], the products $e_{\kappa }$ are also an integral basis.

Give a variable $E_r$ weight $r$. The degree-$d$ monomials in the polynomial ring $\mathbb{Z}[E_1,E_2,\dots ]$ are exactly $E_{\lambda }$ for $\lambda ⊢d$. Sending $E_r$ to $e_r$ sends these degree-$d$ monomials to the basis $e_{\lambda }$ from step 3.1, so the map is bijective in every degree. Since every polynomial and every element of $\mathrm{\Lambda }$ has finite degree support by [F1], it is an isomorphism of graded rings. Thus the $e_r$ freely generate $\mathrm{\Lambda }$.

Because the $e_r$ freely generate $\mathrm{\Lambda }$ by step 4.1, the assignment $\omega (e_r)=h_r$ extends to a graded ring homomorphism. The coefficient recurrence in step 2.1 and the same identity with $t$ replaced by $-t$ give, for each $n\ge 1$, $h_n-e_1{h}_{n-1}+e_2{h}_{n-2}-\cdots +(-1{)}^n{e}_n=0$ and $e_n-h_1{e}_{n-1}+h_2{e}_{n-2}-\cdots +(-1{)}^n{h}_n=0$. Apply $\omega$ to the first recurrence and induct on $n$, starting from $\omega (h_0)=1=e_0$. If $\omega (h_j)=e_j$ for $j<n$, the resulting equation and the second recurrence have identical terms except for $\omega (h_n)$ and $e_n$, so $\omega (h_n)=e_n$. Hence ${\omega }^2$ fixes each generator $e_n$, and so $\omega$ is an automorphism. It carries the basis $e_{\lambda }$ from step 3.1 to $h_{\lambda }$, proving that the $h_{\lambda }$ form an integral basis and that the $h_r$ are algebraically independent.

For each $n\ge 1$, take the coefficient of $t^n$ in the finite identity [F7] at every rank $N$. By [F4] and [F5], these coefficients are the rank projections of the stable products ${\sum }_{i=0}^n(-1{)}^i{e}_i{h}_{n-i}$. Since all projections vanish, the inverse-limit element is zero by [F1]; the constant coefficient is $e_0{h}_0=1$. Hence $E(-t)H(t)=1$ coefficientwise in $\mathrm{\Lambda }\mathrm{⟦}t\mathrm{⟧}$.

Fix a finite rank $N\ge 1$ and let $e_q^{(k)}$ be the elementary polynomial in the variables other than $x_k$. By [F5] and [F7], $H_N(t){\sum }_{q=0}^{N-1}(-1{)}^q{e}_q^{(k)}{t}^q=(1-x_{k}t{)}^{-1}$. For $\alpha \in {\mathbb{N}}^N$, define $A_{\alpha }=(x_k^{{\alpha }_i}{)}_{i,k}$, $M_{j,k}=(-1{)}^{N-j}{e}_{N-j}^{(k)}$, and $(H_{\alpha }{)}_{i,j}=h_{{\alpha }_i-N+j}$. Taking the coefficient of $t^{{\alpha }_i}$ gives $A_{\alpha }=H_{\alpha }M$.

If $\lambda =\mathrm{\varnothing }$, then $s_{\lambda }=1$ by [F3], and either determinant is upper unitriangular or empty, hence equals $1$. Otherwise fix any finite rank $N\ge r\ge \mathrm{\ell }(\lambda )\ge 1$. Put ${\delta }_N=(N-1,\dots ,0)$ and ${\alpha }_i={\lambda }_i+N-i$. For $\alpha ={\delta }_N$, $H_{{\delta }_N}$ is upper triangular with diagonal $1$, so $\det M=\det A_{{\delta }_N}=a_{{\delta }_N}\mathrm{\ne }0$. For $\alpha =\lambda +{\delta }_N$, $\det A_{\alpha }$ is the bialternant numerator and $H_{\alpha }=(h_{{\lambda }_i-i+j}{)}_{i,j}$. Taking determinants in $A_{\alpha }=H_{\alpha }M$ and using $\det M=\det A_{{\delta }_N}$ gives $\det A_{\alpha }\mathrm{/}\det A_{{\delta }_N}=\det H_{\alpha }$; by [F3] this is the rank-$N$ Schur polynomial. Appending a zero part to $\lambda$ changes the determinant to $\left(\begin{array}{cc}B& v\\ 0& 1\end{array}\right)$, so its value is independent of determinant size; hence for every rank $N\ge r$ the size-$r$ determinant equals the rank-$N$ Schur polynomial. Compatibility gives equality in $\mathrm{\Lambda }$.

For the chosen sizes $r,c$, index matrices by $0,\dots ,r+c-1$ and set $U_{a,b}=h_{b-a}$ and $V_{a,b}=(-1{)}^{b-a}{e}_{b-a}$ when $b\ge a$, with both entries zero when $b<a$. Step 1.1 gives $UV=I$ coefficientwise; both matrices are upper unitriangular, so $V=U^{-1}$ and $\det U=1$.

If $\lambda =\mathrm{\varnothing }$, each determinant is upper unitriangular, or empty, and equals $1$. Otherwise pad $\lambda$ and ${\lambda }^{\mathrm{\prime }}$ with zero parts to the chosen sizes and set $I=\{{\lambda }_i+r-i:1\le i\le r\}$ and $J=\{r-i:1\le i\le r\}$. In increasing order, the minor $U_{J,I}$ is the transpose of $(h_{{\lambda }_i-i+j})$ with both orders reversed, so $\det U_{J,I}=\det (h_{{\lambda }_i-i+j})$. Its complements are $J^c=\{r+j-1:1\le j\le c\}$ and $I^c=\{r-1+j-{\lambda }_j^{\mathrm{\prime }}:1\le j\le c\}$. The listed $I^c$ indices are strictly increasing and lie in $\{0,\dots ,r+c-1\}$. None equals ${\lambda }_i+r-i$, since equality would give ${\lambda }_i+{\lambda }_j^{\mathrm{\prime }}=i+j-1$: if $j\le {\lambda }_i$ the left side is at least $i+j$, and if $j>{\lambda }_i$ it is at most $i+j-2$. The two sets have $r+c$ distinct indices in total and are therefore complementary. At rank $N\ge \mathrm{\ell }(\lambda )$, the bialternant numerator and denominator are nonzero: the strictly decreasing exponents give distinct monomials in the numerator determinant, and the Vandermonde denominator is nonzero. Thus $s_{\lambda }\mathrm{\ne }0$; by [F4], $\mathrm{\Lambda }$ is a domain. Over $K=\mathrm{Frac}(\mathrm{\Lambda })$, $A=U_{J,I}$ is invertible by step 2.1. Reorder rows as $J,J^c$ and columns as $I,I^c$, giving $\tilde{U}=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. Block elimination gives $\det \tilde{U}=\det A\det (D-C{A}^{-1}B)$, while the lower-right block of ${\tilde{U}}^{-1}$ is $(D-C{A}^{-1}B{)}^{-1}$. Thus $\det A=\det \tilde{U}\det (V_{I^c,J^c})$. Moving an increasing index set $S$ of size $r$ to the front has sign $(-1{)}^{\sum S-r(r-1)\mathrm{/}2}$; the row and column reorderings therefore give $\det \tilde{U}=(-1{)}^{\sum I+\sum J}\det U$. Since $\sum I+\sum J=\mathrm{\mid }\lambda \mathrm{\mid }+r(r-1)$ and $\det U=1$, this sign is $(-1{)}^{\mathrm{\mid }\lambda \mathrm{\mid }}$. The complementary minor has entries $V_{r-1+i-{\lambda }_i^{\mathrm{\prime }},r+j-1}=(-1{)}^{{\lambda }_i^{\mathrm{\prime }}-i+j}{e}_{{\lambda }_i^{\mathrm{\prime }}-i+j}$; a negative subscript gives zero by the stated convention. Factoring row and column signs contributes $(-1{)}^{{\sum }_i{\lambda }_i^{\mathrm{\prime }}-{\sum }_{i}i+{\sum }_{j}j}=(-1{)}^{\mathrm{\mid }\lambda \mathrm{\mid }}$, which cancels the permutation sign. Therefore $\det (h_{{\lambda }_i-i+j})=\det (e_{{\lambda }_i^{\mathrm{\prime }}-i+j})$. For $\lambda =(1)$ and $r=c=1$, these are $h_1$ and $e_1$, each the sum of the variables by [F5] and [F6].

For each $r\ge 1$, setting a newly added variable to zero sends the finite rank-$N$ power sum ${\sum }_{i=1}^N{x}_i^r$ to the rank-$(N-1)$ power sum. Hence these are compatible sequences $p_r\in {\mathrm{\Lambda }}^r$ by [F1] and [F2]; the $h_r$ are already the stable homogeneous elements supplied by [F3].

In rank $N$, multiplying the geometric series ${\sum }_{a_i\ge 0}{x}_i^{a_i}{t}^{a_i}$ for $1\le i\le N$ shows from [F2] that $H_N(t):={\sum }_{k\ge 0}{h}_k(x_1,\dots ,x_N)t^k={\prod }_{i=1}^N(1-x_{i}t{)}^{-1}$. Differentiating this finite product gives $H_N^{\mathrm{\prime }}(t)\mathrm{/}{H}_N(t)={\sum }_{i=1}^N{x}_i\mathrm{/}(1-x_{i}t)={\sum }_{r\ge 1}{p}_r(x_1,\dots ,x_N)t^{r-1}$.

Write $H(t):={\sum }_{n\ge 0}{h}_n{t}^n$ in ${\mathrm{\Lambda }}_{\mathbb{Q}}\mathrm{⟦}t\mathrm{⟧}$. The finite identity of step 1.2 holds at every rank, and each coefficient of $H^{\mathrm{\prime }}(t)$, $H(t)$, and ${\sum }_{r\ge 1}{p}_r{t}^{r-1}$ is a stable homogeneous sequence by [F1] and step 1.1. Equality of every rank projection therefore gives $H^{\mathrm{\prime }}(t)=H(t){\sum }_{r\ge 1}{p}_r{t}^{r-1}$. Comparing coefficients of $t^{n-1}$ yields the Newton recurrence $n{h}_n={\sum }_{r=1}^n{p}_r{h}_{n-r}$ for every $n\ge 1$.

Since $h_0=1$, step 2.1 can be solved in either direction as $p_n=n{h}_n-{\sum }_{r=1}^{n-1}{p}_r{h}_{n-r}$ and $h_n=\frac{1}{n}(p_n+{\sum }_{r=1}^{n-1}{p}_r{h}_{n-r})$. Induction shows that each $p_n$ is a polynomial in $h_1,\dots ,h_n$ with leading term $n{h}_n$, and each $h_n$ is a polynomial over $\mathbb{Q}$ in $p_1,\dots ,p_n$ with leading term $p_n\mathrm{/}n$. Because these equations solve the same recurrence for its last unknown and each $n$ is invertible in $\mathbb{Q}$, induction verifies that the two substitutions are inverse through every finite index.

For each finite $m$, the mutually inverse triangular substitutions of step 3.1 identify $\mathbb{Q}[h_1,\dots ,h_m]$ with $\mathbb{Q}[p_1,\dots ,p_m]$. By [F3], the $h_r$ are algebraically independent over $\mathbb{Z}$ and remain so over $\mathbb{Q}$ by clearing denominators; hence the $p_r$ are algebraically independent and generate ${\mathrm{\Lambda }}_{\mathbb{Q}}$ after scalar extension, using the graded direct sum [F1]. A degree-$d$ monomial in variables of weights $\mathrm{deg}{p}_r=r$ is exactly $p_{\lambda }$ for a partition $\lambda ⊢d$ by [F4]; these monomials therefore form a $\mathbb{Q}$-basis of ${\mathrm{\Lambda }}_{\mathbb{Q}}^d$, including $p_{\mathrm{\varnothing }}=1$ at $d=0$.

At $n=1$, step 2.1 gives $h_1=p_1$; at $n=2$ it gives $2h_2=p_1{h}_1+p_2=p_1^2+p_2$, hence $h_2=\frac{1}{2}(p_{(1,1)}+p_{(2)})$. The element $h_2$ belongs to the integral stable ring by [F2] and [F3], while step 4.1 makes $p_{(1,1)}$ and $p_{(2)}$ a $\mathbb{Q}$-basis of degree two. Uniqueness of those rational coordinates and their nonintegral values show that $h_2$ is not in their $\mathbb{Z}$-span, so the power sums do not form an integral basis in degree two.

By [F3], each finite-rank coefficient of $E(-t)H(t)-1$ is zero. By [F1] and [F2], these are projections of stable coefficients, so equality of every projection gives $E(-t)H(t)=1$ in $\mathrm{\Lambda }\mathrm{⟦}t\mathrm{⟧}$. For finite rank $N\ge 0$, write $H_N(t)={\sum }_{k\ge 0}{h}_k(x_1,\dots ,x_N)t^k$ and $p_r^{(N)}={\sum }_{i=1}^N{x}_i^r$. The finite identity [F3] gives $H_N(t)={\prod }_{i=1}^N(1-x_{i}t{)}^{-1}$. Differentiating this finite product and expanding each geometric series gives $t{H}_N^{\mathrm{\prime }}(t)\mathrm{/}{H}_N(t)={\sum }_{i=1}^N{x}_{i}t\mathrm{/}(1-x_{i}t)={\sum }_{r\ge 1}{p}_r^{(N)}{t}^r$. Each coefficient is stable by [F1], [F2], and [F4], so equality of all rank projections gives $t{H}^{\mathrm{\prime }}(t)\mathrm{/}H(t)={\sum }_{r\ge 1}{p}_r{t}^r$ in ${\mathrm{\Lambda }}_{\mathbb{Q}}\mathrm{⟦}t\mathrm{⟧}$, where $H(t)={\sum }_{k\ge 0}{h}_k{t}^k$ and $H(t)$ is invertible because its constant term is $1$.

Since $\mathrm{\Lambda }=\mathbb{Z}[e_1,e_2,\dots ]$ freely by [F2], replacing each polynomial generator $e_r$ by the stable element $h_r\in {\mathrm{\Lambda }}^r$ defines a unique unital graded $\mathbb{Z}$-algebra endomorphism $\omega$ of $\mathrm{\Lambda }$.

The coefficient of $t^n$ in the stable identity of step 1.1 gives ${\sum }_{i=0}^n(-1{)}^i{e}_i{h}_{n-i}=0$ for each $n\ge 1$. Applying $\omega$ and using $\omega (e_i)=h_i$ gives $\omega (h_n)+{\sum }_{i=1}^n(-1{)}^i{h}_i\omega (h_{n-i})=0$. Reversing the index in the original recurrence also gives $e_n+{\sum }_{i=1}^n(-1{)}^i{h}_i{e}_{n-i}=0$. Starting with $\omega (h_0)=1=e_0$, induction on $n$ makes these last two sums identical after the leading term, so $\omega (h_n)=e_n$. Therefore ${\omega }^2(e_n)=e_n$ for every free generator; hence ${\omega }^2$ is the identity on $\mathrm{\Lambda }$ and $\omega$ is a ring automorphism. It sends $0$ to $0$ and $1$ to $1$.