The two-colour jump formula and strict elementary embeddings

Work on the finite support supplied by [F2], prescribing the cyclotomic labels independently there. Write $L_r=L(i^r)$, $c_i={ϵ}_i^L$, and write $\mathrm{Ind}(A,B)$ for induction of the external tensor product; all module identifications below are on shift orbits. Two consequences of exact induction will be used. A surjection $\mathrm{Ind}(A,H)\to T$ with $T$ simple induces a surjection $\mathrm{Ind}(A,U)\to T$ for some simple composition factor $U$ of $H$: take the first nonzero image in an induced composition series. If ${ϵ}_i(A)=0$, a nonzero map $\mathrm{Ind}(A,U)\to T$ with $A,U,T$ simple forces ${ϵ}_i(T)={ϵ}_i(U)$, since the shuffle bound gives the upper bound and adjunction injects $A⊠U$ into the defining restriction, giving the lower bound. More particularly, a surjection $\mathrm{Ind}(A,B,L_h)\to T$ with ${ϵ}_i(A)=0$ factors through $\mathrm{Ind}(A,f_i^{h}B)\to T$: its tail is ${ϵ}_i(B)+h$, whereas every constituent of the kernel of $\mathrm{Ind}(B,L_h)\to f_i^{h}B$ has smaller tail by [F1], so an induced composition series of that kernel has zero image in $T$.

Here is the rank-two structure needed below, including its parameter calculation. Fix $j\mathrm{\ne }i$, put $a=-a_{ij}$ and $t=[u^a]Q_{ij}(u,v)\in k^{\times }$. For $p+q\le a$, define $L(i^{p}j{i}^q)$ to have only the word $i^{p}j{i}^q$ in its support, with column $L_p⊠L(j)⊠L_q$, the $j$-dot zero and both crossings adjacent to $j$ zero. All relations hold: $Q_{ij}(u,0)=t{u}^a$, which kills either neighbouring column, and the native $i,j,i$ braid correction is $-t{\sum }_{r=0}^{a-1}{x}^r{z}^{a-1-r}$. Each monomial kills the column since $r\ge p$ or $a-1-r\ge q$ when $p+q\le a$. The other mixed and braid equations have zero terms or are internal nil-Hecke equations. This module is simple by the matrix-column calculation in [F1], with right tail $q$, first tail $p$ and ungraded character $p!q!i^{p}j{i}^q$. These are all simples of weight $ci+j$ when $c\le a$: induct on $c$; a simple with positive right tail is the unique $f_i$-image of its predecessor, whose tail identifies the constructed column; a simple of right tail zero is supported only on $i^{c}j$, and the same column argument gives $L(i^{c}j)$. For $c>a$ no simple can have right tail zero: such a module would again be the column $L_c⊠L(j)$ supported only on $i^{c}j$, but the last mixed crossing has square $t{x}_c^a$, nonzero on $L_c$ since its dot index is $c$. This contradicts its zero crossing.

Adjunction and the shuffle bound give $c_i(M)\le c_i(f_{j}M)\le c_i(M)+{\delta }_{ij}$; first-end symmetry gives ${ϵ}_j(f_i^{L}M)={ϵ}_j(M)$ for $j\mathrm{\ne }i$. If $c_i(f_{i}M)=c_i(M)+1$, then $f_{i}M=f_i^{L}M$: the two orders of induction have the same character at $q=1$ and hence the same ungraded composition multiplicities, while the first order's factor with maximum first tail is uniquely $f_i^{L}M$, once. The character implication here follows from the triangular divided-word/projective pairing in [F1]: divided-word corner dimensions are obtained from ordinary word-corner dimensions by dividing the corresponding nil-Hecke block factorials at $q=1$, and their triangular matrix with diagonal $1$ remains invertible there. Conversely, equality of the two simple heads forces the increase. The coinduction formula in [F1], together with maximal-tail restriction, gives $\mathrm{soc}\mathrm{Ind}(M,L_h)=(f_i^L{)}^{h}M$ and $\mathrm{soc}\mathrm{Ind}(L_h,M)=f_i^{h}M$: coinduction has only one simple socle, since every simple subobject has the maximum tail by adjunction and its simple maximum corner detects it. Thus $f_{i}M=f_i^{L}M$ forces $\mathrm{Ind}(M,L_1)$ simple: its head and socle agree and this factor occurs once; in a nonsimple module the socle would lie in the radical and supply a second occurrence. Conversely simplicity forces its socle and head to agree. Commuting $M$ successively past the $h$ single-colour factors, and using $\mathrm{Ind}(L_r,L_s)=L_{r+s}$, gives $\mathrm{Ind}(M,L_h)=\mathrm{Ind}(L_h,M)$; its simple head equals its simple socle and occurs once, so it too is simple. Hence one increase forces increases at all later steps.

Put $D_n=L(i^{a-n}j{i}^n)$ for $0\le n\le a$. The module $\mathrm{Ind}(D_n,L_1)$ has only two word corners, $i^{a-n}j{i}^{n+1}$ and $i^{a-n+1}j{i}^n$, each a simple product of nil-Hecke columns by [F1]. Its simple head has the first corner nonzero. The linear divided restriction Serre relation forces the second corner nonzero too: when applied to that head at weight $(a+1)i+j$, all terms $E_i^{(a+1-s)}{E}_j{E}_i^{(s)}$ vanish except $s=n,n+1$, and the $s=n+1$ term is nonzero. The relation is available on finite modules by applying homogeneous Hom to the split projective Serre identity; exact projective Hom and the unscaled pairing in [F1] identify these Hom spaces with the divided corners. Thus the head fills both simple corners and induction is simple, so $f_i{D}_n=f_i^L{D}_n$. Step 2.1 proves $\mathrm{Ind}(D_n,L_h)=\mathrm{Ind}(L_h,D_n)$ simple for every $h$. For $c\ge a$, induction on $c$ now classifies all simples of weight $ci+j$ as $\mathrm{Ind}(D_n,L_{c-a})$, $0\le n\le a$, with right tail $n+c-a$: step 1.2 excludes right tail zero for $c>a$, and the one-step partial inverse identifies every positive-tail simple from the preceding weight. For $a=0$, $Q_{ij}$ is a nonzero constant, the mixed crossing is invertible, and $\mathrm{Ind}(L(j),L_h)=\mathrm{Ind}(L_h,L(j))$ is the unique simple, giving the same classification. No characteristic, splitting-field or special binomial-parameter assumption was used.

Define $v_i^{\lambda }(M)$ as the number of surviving right-$i$ lowering steps in a quotient with labels $\lambda$ containing $M$. It is finite by [F4]. Choose such labels with ${\lambda }_l\ge c_l(M)$, possible by [F2] and [F1]. If $c_i(f_i^{n}M)$ never increased, step 2.1 would keep every first tail within those labels for every $n$, contrary to [F4]. Hence $J_i(M)$ is finite; more explicitly, the bound $B_i(\beta ,\lambda )$ of [F4] with $R^{\lambda }(\beta +B_i{\alpha }_i)=0$ gives $J_i(M)<B_i(\beta ,\lambda )$. Step 2.1 gives, for all $h\ge 0$, $c_i(f_i^{h}M)=c_i(M)+\max (0,h-J_i(M))$, and the other first tails are unchanged. The survival criterion in [F1] therefore gives $v_i^{\lambda }(M)=J_i(M)+{\lambda }_i-c_i(M)$. In particular differences $v_i^{\lambda }(X)-v_i^{\lambda }(Y)$ for two surviving simples are independent of the containing labels, and $J_i(f_{i}M)=\max (0,J_i(M)-1)$. This proves finiteness and the recurrence without using the numerical jump formula.

For extraction put $c=c_i(M)$ and $U=(e_i^L{)}^{c}M$, so $M=(f_i^L{)}^{c}U$ and $c_i(U)=0$. If $c_i(f_{j}M)=c$, then $(e_i^L{)}^c{f}_{j}M=f_{j}U$. Indeed induce $L_c,U,L(j)$ onto $f_{j}M$ and apply maximal first-$i$ restriction, then its simple quotient. In the Mackey filtration the $U$-contribution of first $i$'s is zero; for $j\mathrm{\ne }i$ the only surviving factors after forgetting the first column are shifts of $\mathrm{Ind}(U,L(j))$. Thus the target is its unique simple head. For $j=i$ there can also be shifts of $\mathrm{Ind}(L(i),U)$, but these have no nonzero map to the extracted target of first tail zero, by adjunction; the same unique-head conclusion follows. If $c_i(f_{i}M)=c+1$, step 2.1 instead gives $e_i^L{f}_{i}M=M$ and therefore extraction yields $U$. For $j\mathrm{\ne }i$, ${ϵ}_j(U)={ϵ}_j(M)$ because first-$i$ lowering preserves the right $j$-tail (step 2.1). The preceding lowering compatibility and partial inverses then give extraction compatibility with $e_j$ whenever $e_{j}M\mathrm{\ne }0$; when $e_{j}M=0$, the equality of tails makes $e_{j}U=0$ too.

We record the two containing-weight annihilation tests, with their proofs. If $A$ survives and $k=v_i^{\lambda }(A)$, then the quotient of $\mathrm{Ind}(A,L_{k+1})$ is zero: its only simple head is the killed $f_i^{k+1}A$, while any nonzero finite quotient has a simple head. Appending any further block preserves zero, since a cyclotomic relation on an initial block is also a relation on the larger initial block. Consequently $\mathrm{Ind}(A,B)$ has zero quotient if $c_i(B)>k$, by the surjection $\mathrm{Ind}(L_{c_i(B)},(e_i^L{)}^{c_i(B)}B)\to B$. Conversely suppose ${ϵ}_i(A)={ϵ}_j(A)=0$, $B$ has weight $ci+dj$, and $\mathrm{Ind}(A,B)\to T$ is surjective and simple. If $b=c_i(B)\le k$, write $C=(e_i^L{)}^{b}B$ and induce $A,L_b,C$. Its map to $T$ factors through $f_i^{b}A,C$: all other constituents $U$ of $\mathrm{Ind}(A,L_b)$ have ${ϵ}_i(U)<b$. The defining corner of $T$ with an initial block of weight ${\mathrm{wt}}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}(A)$ followed by a suffix of colours $i,j$ and beginning with $i^b$ is nonzero by adjunction. Every word in any such $U$ has a non-$i,j$ letter after the original $A$-block length: otherwise its last $A$-letter, necessarily different from $i,j$, would already lie in that initial block and all $b$ added $i$'s would be a tail, contradicting ${ϵ}_i(U)<b$. Shuffling with $C$ preserves that non-$i,j$ letter in the suffix, so this corner vanishes on $\mathrm{Ind}(U,C)$. Thus no such factor maps onto $T$, proving the factorisation. The shuffle bound gives $c_i(T)\le c_i(f_i^{b}A)+c_i(C)\le {\lambda }_i$. Taking ${\lambda }_j$ larger than the total root-height of the finitely many modules being considered makes the $j$ inequality automatic; other first tails of $T$ are at most those of $A$. Hence this converse guarantees survival. In these comparisons we may enlarge every label except ${\lambda }_i$ to cover the fixed non-$i$ content of all the $i$-strings, so no assertion of a converse for a small ${\lambda }_j$ is needed.

First calculate the containing-length change in rank two. Any simple $N$ of weight $ci+dj$ with ${ϵ}_i(N)=0$ is a quotient of $\mathrm{Ind}(L(i^{b_1}j),\dots ,L(i^{b_r}j))$ with $0\le b_s\le a$: strip one right $j$, then its maximal $i$-tail, and repeat. A positive-height $N$ has a right $j$ since its right $i$ is zero; the factors of $\mathrm{Ind}(L_b,L(j))$ are classified by step 3.1, so tail zero forces $b\le a$ and the unique factor $L(i^{b}j)$. This justifies each factor selection by step 1.1 and decreases height, so the process terminates. For arbitrary $M$, put $m={ϵ}_i(M)$ and strip that tail to obtain such an $N$. The simple $f_{j}M$ is a quotient induced from $N$ and a factor $U$ of $\mathrm{Ind}(L_m,L(j))$. Step 1.1 determines its tail $\gamma ={ϵ}_i(U)$; by step 3.1 it equals $m-a+n$ for some $0\le n\le a$ and $f_i^{a}U=\mathrm{Ind}(D_n,L_m)$. This holds also for $m<a$, where $U=L(i^{a-n}j{i}^{m-a+n})$ and $m-a+n\ge 0$.

For any simple $B$, every simple quotient of $\mathrm{Ind}(B,D_n)$ has right tail ${ϵ}_i(B)+n$: strip the right tail $L_{{ϵ}_i(B)}$ of $B$, commute it past $D_n$ by step 3.1, and split $D_n$ as the head of $L(i^{a-n}j),L_n$. The resulting surjection with terminal block $L_{{ϵ}_i(B)+n}$ gives the lower bound by adjunction; the shuffle bound gives the upper bound. This also holds for a string of $D_n$ factors by iteration. Let $d_s=a-b_s$ and $D={\sum }_s{d}_s$. For $m+h\ge D$, the surjection from the factors in step 4.2 to $f_i^{h}M$ can be rewritten as a surjection $\mathrm{Ind}(D_{d_1},\dots ,D_{d_r},L_{m+h-D})\to f_i^{h}M$. To verify it, start at the rightmost $L(i^{b_r}j)$: the unique constituent of $\mathrm{Ind}(L(i^{b_r}j),L_{m+h})$ with tail $m+h$ is $\mathrm{Ind}(D_{d_r},L_{m+h-d_r})$ by step 3.1. Every other constituent has smaller tail and cannot map to the target through the preceding tail-zero factors by step 1.1. Commute this $D_{d_r}$ past the remaining pure-colour block using step 3.1 and repeat with $r-1,\dots ,1$. Each suffix of the $d_s$ has sum at most $D$, so all consumed block lengths are nonnegative. If a simple $Q$ is a quotient of $\mathrm{Ind}(D_{n_1},\dots ,D_{n_r},L_g)$, its right and first tails are exactly $g+{\sum }_s{n}_s$ and $g+{\sum }_s(a-n_s)$. Indeed commute $L_g$ past each $D_{n_s}$, split $D_{n_s}$ as the head of $L(i^{a-n_s}j),L_{n_s}$, and move its $L_{n_s}$ to the right; adjunction gives the lower right-tail bound, the shuffle bound the matching upper bound. Repeat with the first-end split to obtain the first-tail bound. Therefore, with $C=m-ar+2{\sum }_s{b}_s$, we have $c_i(f_i^{h}M)=h+C$ for $h$ sufficiently large. For $f_i^h{f}_{j}M$ first absorb $a$ pure strands into $U$ using step 4.2 and the factorisation of step 1.1, and perform the same replacements on the preceding factors. For $h\ge a+D$ this gives a quotient of $\mathrm{Ind}(D_{d_1},\dots ,D_{d_r},D_n,L_{m+h-a-D})$, whose first tail is $h+C-n$. Choose $h$ still larger so both these tails are nonnegative and take the containing $i$ label to be ${\omega }_i=h+C$, with other labels large. The next first tail of $M$ is ${\omega }_i+1$, whereas $f_{j}M$ survives exactly $h+n$ steps. Thus $v_i^{\omega }(f_{j}M)-v_i^{\omega }(M)=n$; label independence in step 3.2 gives for every quotient containing both $v_i^{\lambda }(f_{j}M)-v_i^{\lambda }(M)=n={ϵ}_i(f_{j}M)-{ϵ}_i(M)+a$. This is the rank-two containing-length calculation, with an actual finite saturation threshold.

Reduce an arbitrary $M$ and $f_{j}M$ to that calculation. First strip the right $i$-tail of $M$, then alternately strip maximal right $j$- and $i$-tails until the remaining simple $A$ has ${ϵ}_i(A)={ϵ}_j(A)=0$; each nonzero stripping decreases root-height. Group the removed blocks into a simple rank-two factor $B_0$ with a surjection $\mathrm{Ind}(A,B_0)\to e_i^{{ϵ}_i(M)}M$, by step 1.1. Append the removed $i$-tail and apply the factorisation of step 1.1 to get $\mathrm{Ind}(A,B)\to M$ with $B=f_i^{{ϵ}_i(M)}{B}_0$. For both $l=i,j$, the corner lower bound and shuffle upper bound give ${ϵ}_l(B)={ϵ}_l(M)$. Appending $j$ and applying step 1.1 with colour $j$ gives $\mathrm{Ind}(A,f_{j}B)\to f_{j}M$, also with ${ϵ}_i(f_{j}B)={ϵ}_i(f_{j}M)$. The same factorisation gives $\mathrm{Ind}(A,f_i^{h}B)\to f_i^{h}M$ and $\mathrm{Ind}(A,f_i^{h^{\mathrm{\prime }}}{f}_{j}B)\to f_i^{h^{\mathrm{\prime }}}{f}_{j}M$ for all $h,h^{\mathrm{\prime }}$. Choose containing labels $\omega$ for these initial modules with all non-$i$ labels large, and set $h=v_i^{\omega }(M)$, $h^{\mathrm{\prime }}=v_i^{\omega }(f_{j}M)$ and $k=v_i^{\omega }(A)$. The two annihilation tests in step 4.1 imply $c_i(f_i^{h}B)\le k<c_i(f_i^{h+1}B)$ and $c_i(f_i^{h^{\mathrm{\prime }}}{f}_{j}B)\le k<c_i(f_i^{h^{\mathrm{\prime }}+1}{f}_{j}B)$. Step 2.1 makes the inequalities equalities at the surviving endpoints and gives zero jump there. Step 3.2, applied to $B$ and $f_{j}B$, now gives $v_i^{\omega }(B)=h+{\omega }_i-k$ and $v_i^{\omega }(f_{j}B)=h^{\mathrm{\prime }}+{\omega }_i-k$. Their difference equals $h^{\mathrm{\prime }}-h$; step 5.1 evaluates it as ${ϵ}_i(f_{j}M)-{ϵ}_i(M)+a$. Label independence in step 3.2 transfers this to any quotient containing $M,f_{j}M$.

The initial value is now fixed, not inferred from a decrement recurrence. On the vacuum, $f_i^n\mathbf{1}=L_n$ has $c_i=n$, so its containing length is exactly ${\lambda }_i$ by [F1]. Every nonvacuum simple has some positive right tail (choose the last letter of any nonzero word corner), so write $M=f_{j}N$ with $N=e_{j}M$ of lower height. Both survive whenever $M$ survives by [F1]. If $j=i$, surviving length decreases by one, right tail increases by one and $⟨h_i,\beta ⟩$ increases by two. If $j\mathrm{\ne }i$, step 6.1 gives the change in surviving length as ${ϵ}_i(M)-{ϵ}_i(N)-a_{ij}$. Induction on height from the vacuum therefore yields $v_i^{\lambda }(M)={\lambda }_i+{ϵ}_i(M)-⟨h_i,\beta ⟩$. Combining this with the independently proved formula in step 3.2 gives $J_i(M)={ϵ}_i(M)+c_i(M)-⟨h_i,\beta ⟩$. In particular the initial zero-jump test is precisely ${ϵ}_i(M)+c_i(M)=⟨h_i,\beta ⟩$; its value follows from the rank-two/annihilation calculation and the vacuum base, rather than the recurrence alone. The formula is symmetric in the two ends, so the corresponding first-end jump and its decrement formula also hold.

Define the elementary crystal $B_i$ by $\mathrm{wt}(b_i(n))=n{\alpha }_i$, ${ϵ}_i(b_i(n))=-n$, ${\varphi }_i(b_i(n))=n$, $e_i{b}_i(n)=b_i(n+1)$, $f_i{b}_i(n)=b_i(n-1)$; other-colour epsilon and phi are $-\mathrm{\infty }$ and other-colour operators zero. On a tensor $X\otimes Y$, epsilon is $\max ({ϵ}_l(X),{ϵ}_l(Y)-⟨h_l,\mathrm{wt}(X)⟩)$; $e_l$ acts on $X$ iff ${\varphi }_l(X)\ge {ϵ}_l(Y)$ and $f_l$ acts on $X$ iff ${\varphi }_l(X)>{ϵ}_l(Y)$, otherwise on $Y$. To check the $i$-data and both branches set $s=J_i(U)={ϵ}_i(U)+⟨h_i,\mathrm{wt}(U)⟩\ge 0$. Along the first-$i$ string from $U$, the symmetric jump recurrence and step 2.1 give ${ϵ}_i(M)={ϵ}_i(U)+\max (0,c-s)$. This is exactly $\max ({ϵ}_i(U),c-⟨h_i,\mathrm{wt}(U)⟩)$, the tensor epsilon. Weight is also preserved since $\mathrm{wt}(U)=\mathrm{wt}(M)+c{\alpha }_i$. For lowering, if $s>c$ then $J_i(M)=s-c>0$, $c_i(f_{i}M)=c$, and step 3.3 gives extracted factor $f_{i}U$; if $s\le c$ then $J_i(M)=0$, $c_i(f_{i}M)=c+1$ and its extracted factor is $U$. These are precisely the strict $f_i$ branches. For raising, if $s<c$ then the predecessor $(f_i^L{)}^{c-1}U$ has zero jump, so $e_{i}M=e_i^{L}M$, first tail $c-1$ and extracted factor $U$. If $s\ge c>0$, that predecessor has positive jump; a zero jump at $e_{i}M$ would force $e_{i}M=e_i^{L}M$ by step 2.1 and contradict that positive jump. Hence $c_i(e_{i}M)=c$ and step 3.3 applied to $e_{i}M$ gives extracted factor $e_{i}U$. When $c=0$ the same first branch holds directly. If $e_{i}M=0$, the epsilon formula forces $e_{i}U=0$ and $s\ge c$, so the tensor operator is also zero. Thus both raising branches, including equality and zero cases, agree. For $j\mathrm{\ne }i$, step 3.3 gives the epsilon equality and both operator compatibilities; phi equality follows from epsilon and weight. Finally $U,c$ recover $M=(f_i^L{)}^{c}U$, giving injectivity. Every nonvacuum has a nonzero word corner and therefore positive first tail at its first letter, while right-tail stripping decreases height to the unique vacuum and its partial inverse recovers every lowering word. All generic $f_i$ are nonzero by [F1].

Step 7.1 proves the actual surviving right lowering length in every containing first-strand quotient, including zero length and the vacuum. The linear divided restriction Serre relation is the Hom-dual of the projective Serre identity used in step 3.1; it is a statement about the exact linear $G_0$ operators, and makes no linear Serre assertion about the simple-socle operators. All containing labels and support reductions used [F2], so no new original-weight-lattice hypothesis has been imposed. Steps 2.1, 3.2 and 7.1 give every jump assertion, and step 8.1 gives strict injective elementary embeddings and lowering generation, completing the stated claims.