Straightening basis for generic Pfaffian rank rings

We first give an indexed shuffle identity, including an explicit way to expand its right side. Pfaffians with arbitrary ordered lists are interpreted alternatingly, and vanish on repeated indices. If $h\ge h^{\mathrm{\prime }}$ are positive even integers, $1\le s\le h^{\mathrm{\prime }}$, $U=(u_1,\dots ,u_{h+s})$, and $C=(c_1,\dots ,c_{h^{\mathrm{\prime }}-s})$, put $ϵ(D,U\setminus D)$ equal to the sign of the permutation listing first the positions $D$ and then their complement, both increasingly. Then [given, algebra]

$$F(U,C)=\sum _{D\subseteq [h+s],\mathrm{\mid }D\mathrm{\mid }=h}ϵ(D,U\setminus D)P_{U_D}{P}_{(U_{D^c},C)}\in (P_H:\mathrm{\mid }H\mathrm{\mid }\ge h+2)\mathrm{.}$$

Here is a polynomial identity proving this membership, rather than a set-theoretic rank argument. Use exterior variables $e_1,\dots ,e_{h+s},f_1,\dots ,f_{h^{\mathrm{\prime }}-s}$, in that order, and put $\mathrm{\Omega }={\sum }_{i<j}{a}_{u_i{u}_j}{e}_i{e}_j$, $\mathrm{\Theta }={\sum }_{i,j}{a}_{u_i{c}_j}{e}_i{f}_j$, and $\mathrm{\Psi }={\sum }_{i<j}{a}_{c_i{c}_j}{f}_i{f}_j$. The displayed sum is the coefficient of $e_1\cdots e_{h+s}{f}_1\cdots f_{h^{\mathrm{\prime }}-s}$ in

$${\mathrm{pr}}_{h+s,h^{\mathrm{\prime }}-s}\left(\frac{{\mathrm{\Omega }}^{h\mathrm{/}2}}{(h\mathrm{/}2)!}\frac{(\mathrm{\Omega }+\mathrm{\Theta }+\mathrm{\Psi }{)}^{h^{\mathrm{\prime }}\mathrm{/}2}}{(h^{\mathrm{\prime }}\mathrm{/}2)!}\right)=\sum _{2r+q=s,q+2l=h^{\mathrm{\prime }}-s}\frac{{\mathrm{\Omega }}^{h\mathrm{/}2+r}{\mathrm{\Theta }}^q{\mathrm{\Psi }}^l}{(h\mathrm{/}2)!r!q!l!}\mathrm{.}$$

Here ${\mathrm{pr}}_{a,b}$ means projection to exterior bidegree $(a,b)$ in the $e$ and $f$ variables. For $r\ge 1$, expand the coefficient of ${\mathrm{\Omega }}^{h\mathrm{/}2+r}\mathrm{/}(h\mathrm{/}2+r)!$: each coefficient is a Pfaffian of size $h+2r$. For $r=0$, $q=s\ge 1$; expand ${\mathrm{\Theta }}^q$ as a sum of products of ${\theta }_j{f}_j$, where ${\theta }_j={\sum }_i{a}_{u_i{c}_j}{e}_i$, and single out its first factor. Each coefficient of ${\mathrm{\Omega }}^{h\mathrm{/}2}{\theta }_j\mathrm{/}(h\mathrm{/}2)!$ is, with its exterior sign, $P_{(U_E,c_j)}$ for $\mathrm{\mid }E\mathrm{\mid }=h+1$: this follows by expanding the Pfaffian along $c_j$. Expanding the remaining exterior factors gives a completely specified finite sum of larger Pfaffians times polynomials. Thus the identity proves the asserted ideal membership. Repeated actual labels cause no difficulty because the identity is alternating in the formally labelled positions. Larger Pfaffians themselves belong to the ideal of size $h+2$ by repeated expansion along the first index.

Order nonempty Pfaffian symbols by decreasing size, then lexicographically by their index words written in decreasing order; order products by first sorting their factors in this order and then lexicographically. Suppose a sorted adjacent pair $P_A,P_B$, of sizes $h\ge h^{\mathrm{\prime }}$, fails $a_j\le b_j$ for some $j\le h^{\mathrm{\prime }}$. Choose such an $s$. Apply step 1.1 to the ordered list $U=(a_s,\dots ,a_h,b_1,\dots ,b_s)$ and the fixed list consisting of $a_1,\dots ,a_{s-1}$ in the first Pfaffian and $b_{s+1},\dots ,b_{h^{\mathrm{\prime }}}$ in the second. To justify this version with fixed positions explicitly, define $H_{i,j}(A;B;U)={\sum }_{\mathrm{\mid }D\mathrm{\mid }=i}ϵ(D,D^c)P_{(A,U_D)}{P}_{(U_{D^c},B)}$, with $\mathrm{\mid }A\mathrm{\mid }=h-i$, $\mathrm{\mid }B\mathrm{\mid }=h^{\mathrm{\prime }}-j$, $\mathrm{\mid }U\mathrm{\mid }=i+j$. Splitting according to the factor containing the first position $x$ gives the exact recurrence [step 1.1, algebra]

$$H_{i,j}((A_0,x);B;U)=H_{i+1,j}(A_0;B;(x,U))-(-1{)}^{i+j}{H}_{i+1,j-1}(A_0;(x,B);U)\mathrm{.}$$

For $i+j>h$, repeatedly applying it reduces to $i=h$, where step 1.1 applies; $j$ stays positive since $i+j>h$. This proves membership in the larger-Pfaffian ideal and gives its expansion by a finite recurrence. Take $i=h-s+1$, $j=s$, $A=(a_1,\dots ,a_{s-1})$, and $B=(b_{s+1},\dots ,b_{h^{\mathrm{\prime }}})$. The resulting relation has the form

$$\sum _{D\subseteq [h+1],\mathrm{\mid }D\mathrm{\mid }=h-s+1}ϵ(D,D^c)P_{(a_1,\dots ,a_{s-1},U_D)}{P}_{(U_{D^c},b_{s+1},\dots ,b_{h^{\mathrm{\prime }}})}\in (P_H:\mathrm{\mid }H\mathrm{\mid }\ge h+2)\mathrm{.}$$

The coefficient of $P_A{P}_B$ is $1$ up to its displayed ordering sign. Indeed $b_i\le b_s<a_s\le a_j$ for $i\le s\le j$, so the two shuffled groups have no common label. When $h=h^{\mathrm{\prime }}$, no second occurrence arises by exchanging the two factors: every changed first factor precedes $P_A$, whereas the originally sorted $P_B$ does not. Every other nonzero first factor replaces at least one of $a_s,\dots ,a_h$ by an index below $a_s$, and therefore precedes $P_A$. Every term in the larger-Pfaffian expansion has a factor of size at least $h+2$ and also precedes $P_A$. Solving this relation rewrites the offending product into strictly earlier sorted products. Sorting all factors preserves this strict decrease: the unchanged factors occur in both multisets, and the replacement introduces a factor preceding $P_A$ while the two removed factors are both at least $P_A$. All relations preserve the ordinary degree, with $\mathrm{deg}{P}_S=\mathrm{\mid }S\mathrm{\mid }\mathrm{/}2$; in any fixed degree only finitely many products exist. Rewriting therefore terminates. Its terminal products are precisely chains for $⪯$. In the quotient, factors of size greater than $m$ vanish by Pfaffian expansion, so the surviving chains are exactly those stated. This proves spanning.

For independence put $m=2n$ and introduce independent variables $x_{ij}$, $1\le i\le t$, $1\le j\le m$. Substitute [step 2.1, algebra]

$$a_{uv}=\sum _{j=1}^n(x_{u,2j-1}{x}_{v,2j}-x_{v,2j-1}{x}_{u,2j})\mathrm{.}$$

Exterior expansion gives $P_S\mapsto {\sum }_H\det X_{S,H}$, where $H$ runs over unions of complete consecutive column pairs and $\mathrm{\mid }H\mathrm{\mid }=\mathrm{\mid }S\mathrm{\mid }$; each coefficient is $1$. If $\mathrm{\mid }S\mathrm{\mid }>m$ the sum is empty. Order the $x$ variables column first and then row, giving earlier columns and smaller rows priority, and use lexicographic monomial order. For $S=(s_1<\cdots <s_h)$ with $h\le m$, the unique leading monomial of its image is $d_S={\prod }_{j=1}^h{x}_{s_j,j}$, with coefficient $1$: at each successive column the determinant term with the smallest unused row has priority, and the first $h$ columns beat every other union of column pairs. Multiplicativity gives leading monomial ${\prod }_a{d}_{S_a}$ for a standard product. Recover this product uniquely by sorting the row indices appearing in each column of its monomial increasingly; the resulting rows, read from longest to shortest, are exactly $S_1,S_2,\dots$. Hence distinct standard products have distinct leading monomials. Their images, and therefore their residue classes, are linearly independent. For $m=0$ all entries vanish and only the empty product survives. Together with spanning this proves the assertion, including the zero-ideal case.