The polynomial flag comparison has the prescribed second-mate inverse

Work first over $S=\mathbb{Z}[p_1,\dots ,p_N]$ with universal monic $p$. Its full splitting ring is $\mathbb{Z}[X_1,\dots ,X_N]$, finite free over $S$ on the successive monic Artin basis. The block-invariant ring $L_r$ is a direct summand of this module: multiply by the staircase in each of the two blocks and apply the two longest divided differences; the resulting $L_r$-linear map fixes $L_r$ by [F2]. Thus $L_r$ is finite projective over $S$. The same argument with blocks of sizes $r,1,N-r-1$ applies to $B_r$. Monic division gives its displayed free bases over $L_r$ and $L_{r+1}$. Consequently $P=B_{r-1}{\otimes }_{L_{r-1}}{B}_{r-1}$ and $V=B_r{\otimes }_{L_{r+1}}{B}_r$, with their shifts restored later, are finite projective over $S$: choose a relative power basis in one tensor factor and use projectivity of the other. They inject into their generic-root scalar extensions. The coefficient pairings are perfect because their power-basis Gram matrices have unit antidiagonal and zero entries below it. For a monic $f$, their Casimir is $K_f(X,Y)=(f(X)-f(Y))\mathrm{/}(X-Y)$, since coefficient extraction in $X$ gives ${\tau }_X(X^a{K}_f)=Y^a$. This proves both triangles by dual-basis contraction and defines all four maps integrally before any localization.

Localize the universal root ring at all root differences. Vertices are now indexed by subsets $J$ of roots of size $r$. Put $c(u)={\prod }_{w\in J}(u-w)$ and $b(u)={\prod }_{z\mathrm{\notin }J}(u-z)$. A path in $P$ removes $w\in J$ and then adds a root; a path in $V$ adds $z\mathrm{\notin }J$ and then removes a root. Paths with different initial and final subsets are the single common swap path $J\to J-w+z$. Paths with equal endpoints are the $P_w$ and $V_z$ components. The coefficient trace on a monic root extension is $f\mapsto {\sum }_{\xi }f(\xi )\mathrm{/}{f}_0^{\mathrm{\prime }}(\xi )$; this is polynomial Lagrange interpolation followed by top-coefficient extraction. Therefore ${\eta }_z=c(z)$, ${\epsilon }_z^{\mathrm{\prime }}=1\mathrm{/}{b}^{\mathrm{\prime }}(z)$, ${\eta }_w^{\mathrm{\prime }}=b(w)$ and ${\epsilon }_w=1\mathrm{/}{c}^{\mathrm{\prime }}(w)$. Units and caps have zero components on swap paths because their source or target is the identity bimodule.

We verify the claimed Morita normalization independently. Let $I_r:L_r\to S$ be the positive parabolic trace: over generic roots it is $I_r(f)={\sum }_{\mathrm{\mid }J\mathrm{\mid }=r}f(J)\mathrm{/}{\prod }_{w\in J,z\mathrm{\notin }J}(w-z)$. This is an integral polynomial map: factor the full monic splitting-tower trace through the two block traces; equivalently apply the normalized parabolic longest divided difference. It is perfect. Indeed, on the block Artin basis the perfect full pairing is the direct sum of copies of the $I_r$ pairing composed with the invertible block Gram matrix; its inverse shows each copy is invertible. The actual nil-Hecke quotient is this matrix algebra: in the power basis of the monic $X_1$ extension, the first column of multiplication by the remainder of $p$ modulo $c_r$ contains all remainder coefficients, and every matrix entry belongs to their ideal. Matrix units therefore identify the two-sided ideal $(p(X_1))$ with that coefficient ideal. Monic division identifies its vanishing quotient with $L_r$. Put $V_r$ for the resulting first-block Artin module, $M_r={\mathrm{End}}_{L_r}(V_r)={\mathrm{N}\mathrm{H}}_r\mathrm{/}(p(X_1))$ and $t_r=s_r{I}_r{\mathrm{Tr}}_r,s_r=(-1{)}^{r(N-r)+r(r-1)\mathrm{/}2}\mathrm{.}$ The sign is also the ordered-residue sign: in $(-1{)}^{r(N-r)}\mathrm{Res}({\mathrm{\Delta }}_r{X}_1^{r-1}\cdots X_{r-1}{\mathrm{Tr}}_r(-)\mathrm{/}{\prod }_{a}p(X_a))$, group ordered tuples by subsets; their alternating determinant contributes ${\mathrm{\Delta }}_r^2$, while the internal derivative factors contribute $(-1{)}^{r(r-1)\mathrm{/}2}{\mathrm{\Delta }}_r^2$. Cancellation gives exactly $t_r$, without $1\mathrm{/}r!$. Define the bilinear relative cap by $t_r{\epsilon }_r=t_{r+1}$ using the perfect $t_r$ pairing. The constant-vector matrix idempotent identifies the induction corner with $B_r$ and its dual corner with $g\mapsto [v\mapsto {\tau }_{\mathrm{n}\mathrm{e}\mathrm{w}}(gv)]$; matrix evaluation is therefore precisely ${\tau }_{\mathrm{n}\mathrm{e}\mathrm{w}}$. On the rank-one operator $v\mapsto f{\tau }_{\mathrm{n}\mathrm{e}\mathrm{w}}(gv)$, its matrix trace is ${\tau }_{\mathrm{n}\mathrm{e}\mathrm{w}}(gf)$.

Compute the displayed mates on these path components, using the actual crossing $(1-s)\mathrm{/}(X-Y)$. On the equal-endpoint components the result is $(-T{)}_{z,w}=\frac{c(z)}{c^{\mathrm{\prime }}(w)(z-w)},S_{w,z}=\frac{b(w)}{b^{\mathrm{\prime }}(z)(w-z)}\mathrm{.}$ Here the first expression is the cup factor $c(z)$, the no-switch divided-difference entry $1\mathrm{/}(w-z)$ on the ordered additions $(w,z)$ in $T$, and the first-cap factor $1\mathrm{/}{c}^{\mathrm{\prime }}(w)$. For $S$, the twofold first mate reverses the addition path. Its no-switch coefficient on ordered removals $(z,w)$ is $1\mathrm{/}(w-z)$; its switch coefficient, evaluated on the output removal order $(z,w)$, is $1\mathrm{/}(z-w)$. To check this transposition, the composed first-cap weight on the additions $(z,w)$ over a common subset $K$ is $1\mathrm{/}[c_K(z)c_K(w)(w-z)]$, and the exchanged order has its negative. Their ratio changes the sign of the off-diagonal transposed crossing and leaves the diagonal coefficient unchanged. Thus the second cup in $S$ contributes $b(w)$, its no-switch entry is $1\mathrm{/}(w-z)$, and the second cap contributes $1\mathrm{/}{b}^{\mathrm{\prime }}(z)$. This is a computation of the defined mate, not a definition of $S$ by the inverse matrix. On a swap path $J\to J^{\mathrm{\prime }}=J-w+z$, the first contraction in $T$ instead has numerator $c_{J^{\mathrm{\prime }}}(w)=(w-z)c_{J-w}(w)$ and denominator $(w-z)c_J^{\mathrm{\prime }}(w)=(w-z)c_{J-w}(w)$, together with the native minus sign; thus $T=-1$. For the switched second contraction, the complementary factor at $J^{\mathrm{\prime }}=J-w+z$ is $b_{J^{\mathrm{\prime }}}(z)=(z-w)b_{J^c-\{z\}}(z)$; its second-cap denominator is $b_J^{\mathrm{\prime }}(z)=b_{J^c-\{z\}}(z)$, and its switched upward crossing contributes $1\mathrm{/}(z-w)$. Therefore $S=1$ on this path. One may check the transposition without notation for diagrams: replace each root delta-vector by its dual vector multiplied by the derivative of its monic root polynomial, apply $(1-s)\mathrm{/}(X-Y)$, and contract; the two derivative multipliers are exactly the four factors just displayed. Hence $-ST=-TS=1$ on every swap component.

Write $n=N-r=\mathrm{deg}b$. On the equal-endpoint components $U=-T$ sends values $(f_w)$ to the values at $z\mathrm{\notin }J$ of the unique polynomial $F(u)$ of degree $<r$ with $F(w)=f_w$, because its matrix is $c(z)\mathrm{/}(c^{\mathrm{\prime }}(w)(z-w))$. Similarly $S$ sends $(g_z)$ to the values at $w\in J$ of the unique polynomial $G(u)$ of degree $<n$ with $G(z)=g_z$. If $h=n-r\ge 0$, then $SU=1$. Monic division writes $G=cq+v$ uniquely with $\mathrm{deg}q<h$, $\mathrm{deg}v<r$. Thus $US$ is evaluation of $v$ and $Cq$ is evaluation of $cq$, proving $US+CR=1$ once the rows $R$ extract the coefficients of $q$. To identify these rows, expand $\frac{G(u)}{b(u)}={\sum }_z\frac{g_z}{b^{\mathrm{\prime }}(z)(u-z)},\frac{G(u)}{c(u)}=u^{h}A(u^{-1})\frac{G(u)}{b(u)}\mathrm{.}$ The coefficient of $u^k$ in the polynomial part of the second expansion is precisely ${\sum }_{a+j=h-1-k}{A}_j{\sum }_z{g}_z{z}^a\mathrm{/}{b}^{\mathrm{\prime }}(z)$, which is the prescribed $R_k$. Therefore $RU=0$, $SC=0$, $RC=1$ and $US+CR=1$, as well as $SU=1$. These prove all five positive-branch equations, including the prescribed inverse rather than an unspecified polynomial section. At $h=0$ monic division has no quotient, and $U,S$ are mutually inverse.

If $h<0$, interchange $c$ and $b$ in the preceding division. Interpolate the $P$ values by $F$ of degree $<r$, and write $F=bq+v$ with $\mathrm{deg}q<-h$, $\mathrm{deg}v<n$. Now $US=1$ on $V$, $SU$ is evaluation of $v$ on $P$, and $C^{-}q$ is evaluation of $bq$. Expanding $F\mathrm{/}b=u^{-h}\mathsf{H}(u^{-1})F\mathrm{/}c$ identifies the coefficient rows of $q$ with $R_k^{-}$ exactly. Hence $U{C}^{-}=0$, $R^{-}S=0$, $R^{-}{C}^{-}=1$ and $SU+C^{-}{R}^{-}=1$. With $U=-T$ these are the five asserted negative-branch equations. This proof includes $r=0$, $r=N$ and $N=0$ by empty interpolation sets; beyond the factor range every module is zero. All equations were identities between integral maps on the projective modules in step 1.1, so injection into the generic-root extension proves them over $S$. Base change gives them over every $D$; no root difference is inverted in the resulting maps.

The single-curl contraction uses one crossing instead of the two in $US$ or $SU$. On the reversed flag removing $w$, cup, the no-switch upward crossing and cap contract to $-{\sum }_{z}c(z)\mathrm{/}[b^{\mathrm{\prime }}(z)(w-z)]$, using the weights of step 2.1 and the upward crossing entries of step 3.1. This is the polynomial part of $c(u)\mathrm{/}b(u)$ evaluated at $w$: explicitly, subtracting ${\sum }_{z}c(z)\mathrm{/}(b^{\mathrm{\prime }}(z)(u-z))$ from $c(u)\mathrm{/}b(u)$ removes all its simple poles and leaves its monic-division quotient. On the reversed flag removing $z$ at the adjacent input $r+1$, the other-handed contraction is ${\sum }_{w}b(w)\mathrm{/}[c^{\mathrm{\prime }}(w)(z-w)]$. Since $b(z)=0$, this is minus the polynomial part of $b(u)\mathrm{/}c(u)$ evaluated at $z$. Since $c\mathrm{/}b=u^{-h}\mathsf{H}(u^{-1})$ and $b\mathrm{/}c=u^{h}A(u^{-1})$, these are the two stated finite sums. The first has degree $-2dh$ at input $\mu$; the second has degree $2dh$ at input $\mu -{\alpha }_i$. The two-turn contractions are respectively $d(1-h)+d(1-h)-2d$ and $d(1+h)+d(1+h)-2d$, verifying the input-weight distinction. At $h=0$ they give $+1$ and $-1$. Mating the completeness identities in steps 4.1-5.1 uses only the already proved triangles, and expands $C_k{R}_k$ or $C_k^{-}{R}_k^{-}$ as the finite triple convolution. In the first case $a+j+c=h-1$ and literal clockwise dots are $b=-h-1+j$, so $a+b+c=-2$; in the other case $a+j+c=-h-1$ and counterclockwise dots are $b=h-1+j$, with the same total. These are actual opposite-crossing boundary maps, with the bubble in outside region $\mu$. The same conversion on the single-curl sums gives $a+b=-1$. Integral descent is as in step 5.1.

For $f\in B_r$, comparison of root denominators gives $I_r{\tau }_{\mathrm{o}\mathrm{l}\mathrm{d}}(f)=(-1{)}^r{I}_{r+1}{\tau }_{\mathrm{n}\mathrm{e}\mathrm{w}}(f)$: the two denominators agree except for the $r$ factors $w-z$ versus $z-w$. Moreover $s_{r+1}\mathrm{/}{s}_r=(-1{)}^{N-r-1}$, hence $(-1{)}^{N+1}{s}_{r+1}\mathrm{/}{s}_r=(-1{)}^r$. Testing the relative cap on the rank-one operators of step 2.2, also after multiplication by every element of $L_r$, now identifies ${\widehat{\epsilon }}_r=(-1{)}^{N+1}{\epsilon }_r$ with ${\tau }_{\mathrm{o}\mathrm{l}\mathrm{d}}$ by perfection. The corresponding relative unit is its dual-power-basis Casimir, so the identification includes the unit and both triangles, not only a trace value on dots. All denominator comparisons are comparisons of already integral polynomial maps between projective modules, and descend exactly as in step 5.1.

For the mixed clauses write $p^{\mathrm{\prime }}=pq$, $q=Q_{ij}(u,y)\mathrm{/}{t}_{ij}$ monic of degree $A=-a_{ij}$, and $U_r={\prod }_{a=1}^r{Q}_{ij}(X_a,y)$. Multiplication by $U_r$ defines the matrix-quotient map from the old factor block to the new one; cancellation in the ordered-residue polynomial expression gives $t_r^{p^{\mathrm{\prime }}}(U_{r}z)=(-1{)}^{rA}{t}_{ij}^r{t}_r^p(z)$. Testing the relative cap against $U_{r}a$ and using its bilinearity and the perfect old trace gives ${\pi }_r{\epsilon }_r^{p^{\mathrm{\prime }}}(Q_{ij}(X_{r+1},y)\mathrm{lift}(z))=(-1{)}^A{t}_{ij}{\epsilon }_r^p(z)\mathrm{.}$ The test also proves independence of the lift. The normalized caps of step 6.2 cancel $(-1{)}^A$, because $N$ changes to $N+A$. Thus the actual flag cap constant is $t_{ij}$. For $r=0$ this is simply top-coefficient extraction ${\tau }_{pq}(Qf)=t_{ij}{\tau }_p(f)$; arbitrary polynomial reduction gives the same equality. Dual-basis contraction proves the companion cup equation.

More explicitly, fix the mixed boundaries $G_j{H}_i$ and $H_i{G}_j$ and source crossings $\kappa =-\tau$. The handed pair is $a=(H_i{G}_j\epsilon )(H_i{\kappa }_{ij}{H}_i)(\eta G_j{H}_i),b=({\epsilon }^{\mathrm{\prime }}{G}_j{H}_i)(H_i{\kappa }_{ji}{H}_i)(H_i{G}_j{\eta }^{\mathrm{\prime }})\mathrm{.}$ The intermediate words are $H_i{G}_i{G}_j{H}_i$ and $H_i{G}_j{G}_i{H}_i$. The quotient/switch gives $a$ the canonical underlying flag identification; the multiplication-by-$Q$ switch and the normalized coefficient cap in step 7.1 give $b$ its scalar $t_{ij}$. Consequently $ba=t_{ij}{1}_{G_j{H}_i}$ and $ab=t_{ij}{1}_{H_i{G}_j}$. Native crossings multiply both $a,b$ by $-1$ and preserve these products. Thus the inverse mixed crossing is obtained by dividing its companion mate by $t_{ij}$. The same calculation for a separately distinguished color $j$ gives $t_{ji}$ for its own pair; identifying both partial actions inside a coherent full action is not a consequence of these two scalar tests alone. Dots commute with the quotient and multiplication maps. With $i$ last, the downward shift $s_j(m)=-2d_j{m}_j+{\sum }_{\mathrm{\ell }>j}(-b_{j\mathrm{\ell }})m_{\mathrm{\ell }}$ gives both downward crossings degree $-b_{ij}$; the basic mixed flag identification has shift $-b_{ij}$, canceled by the input-weight change in $H_i(m)=E_i^{\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{i}\mathrm{c}}(m)\{-d_i(1+h_m)\}$. The normalized mixed maps are therefore degree zero.

The first unit and second cap have degree $d(1-h)$, and the first cap and second unit degree $d(1+h)$, by coefficient extraction of $z^r$ and $z^{N-r-1}$ with the displayed shifts. Hence $T,S$ have degree zero, $C_k$ has degree $d(1-h+2k)$ and $R_k$ its negative, while $C_k^{-}$ has degree $d(1+h+2k)$ and $R_k^{-}$ its negative. Steps 4.1-5.1 establish both actual matrix inverses, step 6.1 the typed curls and cancellations, and steps 2.2, 6.2, 7.1 and 8.1 the normalized cap transport and mixed orientations. Universal construction followed by base change includes every dominant framing, every permitted color and every weight in its factor range, with no characteristic restriction or choice principle. Nonhomogeneous monic specializations preserve the maps and equations as ungraded identities.