Character Polynomials, FI-Modules, and Representation Stability — Examples

$V$ is a functor: $V({\mathrm{i}\mathrm{d}}_S)$ sends each basis element $e_s$ to $e_s$, hence is the identity, and for composable injections $f\text{\hspace{0.17em}⁣}:S\to T$, $g\text{\hspace{0.17em}⁣}:T\to U$ one has $V(g\circ f)(e_s)=e_{g(f(s))}=V(g)(e_{f(s)})=V(g)V(f)(e_s)$ on every basis element, so $V(g\circ f)=V(g)V(f)$; hence $V$ is an FI-module with basis $S$ of $V_S$.

Identify $V_S$ with $M(1{)}_S$ by sending the basis element $e_s$ to the injection $u_s\text{\hspace{0.17em}⁣}:[1]\to S$ with $u_s(0)=s$; this is a $k$-linear isomorphism because the $u_s$ are exactly the basis injections of $M(1{)}_S$ [F2]. It is compatible with transition maps: for $f\text{\hspace{0.17em}⁣}:S\to S^{\mathrm{\prime }}$, $M(1)(f)(u_s)=f\circ u_s=u_{f(s)}$ corresponds to $e_{f(s)}=V(f)(e_s)$. Hence $V\cong M(1)$; in particular $V_n\cong k^n$ with basis $[n]$, and $\sigma \in S_n$ acts by $\sigma \cdot e_s=e_{\sigma (s)}$, the permutation representation.

The element $e_0\in V_{[1]}$ generates: for any finite set $S$ and $s\in S$, the injection $u_s\text{\hspace{0.17em}⁣}:[1]\to S$ satisfies $(u_s{)}_{\ast }(e_0)=e_{u_s(0)}=e_s$, so ${\mathrm{span}}_V(\{([1],e_0)\})$ contains every basis element and equals $V$; thus $V$ is generated in degree $\le 1$ by $V_{\{0\}}$.

For $\sigma \in S_n$ and basis elements $s,t\in [n]$, the coefficient of $e_t$ in $\sigma \cdot e_s=e_{\sigma (s)}$ is ${\delta }_{t,\sigma (s)}$; the diagonal entry at $s$ is therefore ${\delta }_{s,\sigma (s)}$, equal to $1$ if $\sigma$ fixes $s$ and to $0$ otherwise. By [F5], $\mathrm{tr}(\sigma ;V_n)={\sum }_{s\in [n]}{\delta }_{s,\sigma (s)}$, the number of fixed points of $\sigma$, which is $X_1(\sigma )$; for $n=0$ both sides are $0$. This holds for every $n$, so the trace function of $V_n$ is the character polynomial $X_1$ on the whole range.

Steps 1.1, 2.1 and 2.2 exhibit $V\cong M(1)$ as the permutation FI-module generated in degree $\le 1$, and step 3.1 identifies its character with $X_1$, a character polynomial of weighted degree $1$.

$W$ is a functor: $W({\mathrm{i}\mathrm{d}}_S)={\mathrm{\Lambda }}^2({\mathrm{i}\mathrm{d}}_{V_S})={\mathrm{i}\mathrm{d}}_{W_S}$ and $W(g\circ f)={\mathrm{\Lambda }}^2(V(g)V(f))={\mathrm{\Lambda }}^{2}V(g)\circ {\mathrm{\Lambda }}^{2}V(f)=W(g)W(f)$, using that $V$ satisfies the functor axioms [F1] and that ${\mathrm{\Lambda }}^2$ preserves identities and composites [F2]; hence $W$ is an FI-module with values $W_S={\mathrm{\Lambda }}^2{V}_S$, and for $S=[n]$ this is $W_n={\mathrm{\Lambda }}^2(k^n)$ with the action induced by the permutation action of $S_n$ on $k^n$.

Fix $n$ and $\sigma \in S_n$. By [F3], applied to the free module $V_n$ with basis $e_0,\dots ,e_{n-1}$, the elements $e_i\wedge e_j$ with $i<j$ form a $k$-basis of $W_n$; and $\sigma \cdot (e_i\wedge e_j)=e_{\sigma (i)}\wedge e_{\sigma (j)}$. Hence the diagonal entry of the matrix of $\sigma$ at the basis element $e_i\wedge e_j$ is the coefficient of $e_i\wedge e_j$ in $e_{\sigma (i)}\wedge e_{\sigma (j)}$, which can be nonzero only when $\{\sigma (i),\sigma (j)\}=\{i,j\}$. In that case $\sigma$ restricts to a bijection of the two-element set $\{i,j\}$: it is either the identity, giving coefficient $+1$, or the transposition $(ij)$, giving $e_j\wedge e_i=-e_i\wedge e_j$ and coefficient $-1$. There is no other possibility, so the diagonal entry is $+1$ exactly for the unordered pairs of distinct fixed points of $\sigma$, is $-1$ exactly for the $2$-cycles of $\sigma$, and is $0$ otherwise.

By [F5] the trace of $\sigma$ on $W_n$ is the sum of the diagonal entries computed in step 1.2, namely the number $\left(\genfrac{}{}{0px}{}{X_1(\sigma )}{2}\right)$ of unordered pairs of distinct fixed points minus the number $X_2(\sigma )$ of $2$-cycles: $\mathrm{tr}(\sigma ;W_n)=\left(\genfrac{}{}{0px}{}{X_1(\sigma )}{2}\right)-X_2(\sigma )$. For $n=0$ and $n=1$ the module $W_n$ is $0$ and both sides are $0$, since then $X_1(\sigma )\le 1$. This holds for every $n$, and $\left(\genfrac{}{}{0px}{}{X_1}{2}\right)-X_2$ is a character polynomial of weighted degree $2$, so the trace function of $W$ is given by that character polynomial.

Steps 1.1, 1.2 and 2.1 show that $W={\mathrm{\Lambda }}^{2}V$ is an FI-module with $W_n\cong {\mathrm{\Lambda }}^2(k^n)$ whose character is the weighted-degree-two character polynomial $\left(\genfrac{}{}{0px}{}{X_1}{2}\right)-X_2$.

Substituting $\lambda =(2,1)$ gives $\lambda [n]=(n-3,2,1)$: for $n<3$ the first entry is negative, for $n=3$ it is $0<2$ and for $n=4$ it is $1<2$, so in these cases the sequence is not a partition; for $n\ge 5$ the first entry is at least $2$, the entries are nonnegative and weakly decreasing and their sum is $(n-3)+2+1=n$. Hence $\lambda [n]$ is a partition exactly for $n\ge 5=\mathrm{\mid }\lambda \mathrm{\mid }+{\lambda }_1$.

Apply [F2] with $\alpha =\lambda =(2,1)$, so $m^{\mathrm{\prime }}=3$ and $d^{\mathrm{\prime }}=5$: there is a sub-FI-module $V((2,1))$ of $M((2,1))$ with $V((2,1){)}_n\cong V_{\lambda [n]}=V_{(n-3,2,1)}$ for $n\ge 5$ and $V((2,1){)}_n=0$ for $n<5$, of generation degree exactly $5$, weight $3$, injectivity degree $0$ and surjectivity degree $2$, so that its stability degree is $2$.

Steps 1.1 and 2.1 show that the padded family $V((2,1))$ begins at rank five, not at $\mathrm{\mid }\lambda \mathrm{\mid }=3$, is generated in degree $5$, and has weight $3$ and stability degree $2$; the second row of each constituent diagram $\lambda [n]=(n-3,2,1)$ remains $2$ while the first row grows. No choice principle is used.

$V$ is a functor: $V({\mathrm{i}\mathrm{d}}_S)$ sends $e_S$ to $e_S$ because ${\mathrm{i}\mathrm{d}}_S$ is a bijection; and for composable injections $f\text{\hspace{0.17em}⁣}:S\to T$, $g\text{\hspace{0.17em}⁣}:T\to U$ the two sides of $V(g\circ f)=V(g)V(f)$ agree on $e_S$ in each case — if $f$ and $g$ are bijections both sides send $e_S\mapsto e_U$; if $g$ fails to be a bijection then $g\circ f$ fails to be a bijection, so both sides are $0$; and if $f$ fails to be a bijection then $g\circ f$ fails to be a bijection and both sides are $0$. Hence $V$ is an FI-module and ${\dim }_k{V}_S=1$ for every finite set $S$.

Suppose $\mathrm{\Sigma }=\{(S_1,s_1),\dots ,(S_r,s_r)\}$ is a finite set of supported elements that generates $V$, with $s_i\in V_{S_i}$, and put $M:={\max }_i\mathrm{\mid }{S}_i\mathrm{\mid }$ (if $r=0$, then ${\mathrm{span}}_V(\mathrm{\varnothing })=0\mathrm{\ne }V$, so $r\ge 1$). For every finite set $T$ with $\mathrm{\mid }T\mathrm{\mid }>M$, every injection $f\text{\hspace{0.17em}⁣}:S_i\to T$ satisfies $\mathrm{\mid }{S}_i\mathrm{\mid }\le M<\mathrm{\mid }T\mathrm{\mid }$, so $f$ is not a bijection and $f_{\ast }(s_i)=V(f)(s_i)=0$.

By the explicit description of the span, step 1.2 shows ${\mathrm{span}}_V(\mathrm{\Sigma }{)}_T=0$ for every finite set $T$ with $\mathrm{\mid }T\mathrm{\mid }>M$. Taking $T=[M+1]$, which has $M+1>M$ elements, gives ${\mathrm{span}}_V(\mathrm{\Sigma }{)}_T=0$ while $V_T=k{e}_T\mathrm{\ne }0$; hence ${\mathrm{span}}_V(\mathrm{\Sigma })\mathrm{\ne }V$ and no finite set of elements generates $V$.

Thus $V$ has ${\dim }_k{V}_S=1$ for every finite set $S$ by step 1.1, yet is not finitely generated by step 2.1: pointwise finite-dimensionality does not imply finite generation, and the refuted claim fails.