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LEVEL 1 OF 2 · Saxl's conjecture and universal tensor squares
Universal Tensor Squares for Symmetric Groups
expertly designed by an internal OpenAI model · released 2026-09-24
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IntroductionFor a partition \(\lambda\vdash n\), write \(S^\lambda\) for the corresponding irreducible complex representation of the symmetric group \(S_n\). Its Kronecker coefficients are \[g(\lambda,\mu,\nu) =\dim\mathop{\mathrm{Hom}}_{S_n}(S^\nu,S^\lambda\otimes S^\mu),\] where the tensor product carries the diagonal action. We prove the following statement. Theorem 1. For every positive integer \(n\notin\{2,4,9\}\) there is a partition \(\lambda\vdash n\) such that \[g(\lambda,\lambda,\nu)>0\qquad\text{for every }\nu\vdash n.\] Thus one irreducible representation has a tensor square containing every irreducible representation of \(S_n\). The partition \(\lambda\) is fixed for each degree: it does not vary with the desired constituent \(\nu\). In particular the trivial and sign representations both occur. The latter forces \(\lambda\) to be self-conjugate, since \[g(\lambda,\lambda,(1^n)) =\dim\mathop{\mathrm{Hom}}_{S_n}(S^\lambda,S^{\lambda^t}) =\begin{cases}1,&\lambda=\lambda^t,\\0,&\lambda\ne\lambda^t. \end{cases}\] Our construction meets this necessary condition throughout. Corollary 2 (Unipotent tensor squares). For each positive integer \(n\notin\{2,4,9\}\), choose a partition \(\lambda_n\) supplied by Theorem 1, including its finite-range witnesses. For every prime power \(q\), let \(U_q^\lambda\) denote the unipotent irreducible complex character of \(\mathop{\mathrm{GL}}_n(\mathbb F_q)\) indexed by \(\lambda\vdash n\). Then \(U_q^{\lambda_n}\otimes U_q^{\lambda_n}\) contains \(U_q^\nu\) for every \(\nu\vdash n\). Proof. Theorem 1 gives \(g(\lambda_n,\lambda_n,\nu)>0\) for every \(\nu\vdash n\). Letellier’s nonnegativity theorem and constant-term bound [11], recalled in [12], transfer this positivity to the unipotent multiplicity for the same partition labels and every prime power \(q\). ◻ The Steinberg character already gives this unipotent coverage in every degree [11], and staircase labels give it at triangular degrees [12]. The corollary identifies further covering squares using the same labels that solve the symmetric-group problem. History and significancePak, Panova, and Vallejo formulate the assertion for \(n\ge3\), with \(n\ne4,9\), as the tensor square conjecture for symmetric groups [17]. The case \(n=1\) is immediate. Hence Theorem 1 resolves that conjecture affirmatively, including every nontriangular degree. The related Saxl conjecture specifies the staircase partition \(\rho_m=(m,m-1,\ldots,1)\) at the triangular degree \(N_m=m(m+1)/2\); see [17]. Saxl proposed it at the UCLA Combinatorics Seminar on 20 March 2012, as recorded in [17]. It is a more specific assertion about a distinguished family, but only at triangular degrees. Earlier work established substantial support families: Pak, Panova, and Vallejo gave a character-nonvanishing test for constituents of self-conjugate squares [17], while Ikenmeyer proved the staircase assertion for partitions comparable with the staircase in dominance order [9]. His triangular arrangement of alternating tensors and nonzero contractions [9] is a methodological predecessor of the cyclic construction used here. Heide, Saxl, Tiep, and Zalesski conjectured the analogous tensor-square covering property for alternating simple groups [8], partly motivated by Thompson’s conjugacy-class-square conjecture for finite simple groups. Several complementary approaches explain the strength of the square problem. Luo and Sellke proved that staircase squares contain almost all partitions under both the uniform and Plancherel measures, and obtained full fourth-power coverage in all sufficiently large degrees, using modified shapes at nontriangular degrees [14]. Harman and Ryba subsequently proved full coverage for every staircase tensor cube [7]. Spin-character methods gave Bessenrodt all double-hook constituents, and Li proved the triple-hook case: these are targets with two and three diagonal cells, respectively [1][13]. Modular methods yield, among other families, all 2-height-zero constituents and framed staircases [3]. More recent semigroup and generalized-block criteria give additional constituent families [5]. These results concern different support families or tensor powers: neither almost-all coverage nor a third or fourth power gives the universal square required here. The passage to nontriangular degrees also has a direct antecedent. Outside degrees \(2,4,9\), Li conjectured universal-square support for every self-conjugate staircase-like partition in a parity-dependent family, and asked whether suitable additions of one or two boxes preserve universal-square support [13]. Our large-degree construction uses a different family, making matching additions to the first two rows and columns of a parity-compatible staircase. The proof controls all constituents of the chosen enlargement; the candidate is fixed before the target is specified. Bessenrodt and Bowman formulate stronger conjectural refinements in terms of symmetric and alternating parts of tensor squares [2]; the tensor product in Theorem 1 is the ordinary diagonal tensor square. We use the stronger cyclic form of the staircase result proved in A Cyclic Polytabloid Proof of Saxl’s Conjecture [16]. The exact theorem and generator are specified in Theorem 6. The new task here is to pass from staircase degrees to all the degrees in Theorem 1. The proof and its reusable ingredientsThe obstacle is not simply to place a smaller staircase inside a larger diagram. Its constituents must survive a map from the actual enlarged square, and attachments can introduce cancelling alternating terms. Section 3 constructs the fixed self-conjugate partition for each large degree. Two complementary sufficient tests show that its square contains a prescribed target. The first test, developed in Section 4, starts with one tensor obtained by alternating along the candidate’s rows in the first factor and its columns in the second. A coordinate projection of its diagonal orbit separates a smaller triangle from a width-two band and two attached diagrams. The projected module is the full induction of the triangle and complement modules. The cyclic staircase input supplies every constituent on the triangle, so branching reduces the task to finding a supported complement diagram contained in the desired target. The band is an odd path of alternating pairs: its contraction is a word of two-by-two matrices and adjugates. The attachments are handled by nondegenerate symmetric forms and a parity-sensitive Littlewood–Richardson splitting. Keeping the attached dual columns away from the path endpoints makes every surviving term factor, so the nonzero contractions do not cancel. The second test, in Section 5, works in the full square, using column alternations in both factors. It allows independent position permutations in the two layers before applying a common projection. Label the two alphabets by positive integers; a pair of letters \(a,a'\) has output \(a+a'-1\). Components can then be separated by both the numbers and sums of their outputs. This combines four-row path supports, two-row symmetric-form supports, and an additional three-row support construction. This size-and-sum projection, as well as the explicit attachment criterion, can be applied independently of the final degree estimates. These tests divide the targets by their initial row and column sums. The balance test supplies a target whenever its first four rows or first four columns have small total size. For a remaining target, failure of the band test in both orientations bounds the first few row sums and column sums simultaneously. Section 6 shows that, for large parameters, a diagram satisfying both bounds has less area than the target’s prescribed size. Two exact capacity calculations settle the remaining bounded parameters. The Murnaghan–Nakayama recursion of Section 7 verifies every eligible degree at most \(64\). The latter computes exact residues of whole Kronecker-coefficient vectors, not floating-point approximations. We prove the soundness of every test and pruning rule used by the programs and include their complete source. Thus the finite computations close specified finite cases rather than impose an unproved threshold on positivity. Section 2 fixes the common polytabloid conventions and the exact cyclic companion input before either construction begins. Section 8 assembles the candidate, the support tests, and the finite range. Appendices 9 and 10 supply the two proof-critical programs; the accompanying source also contains their independent checking tools. Representation-theoretic toolsWe fix the tensor conventions used by both the coordinate projections and the staircase input. The two transfer mechanisms below are dual polytabloid detection of a constituent and ordinary induction on disjoint position sets. All representations are finite-dimensional over \(\mathbb C\). A partition is padded with zeros when needed; \(\lambda^t\) denotes its conjugate. For partitions of the same integer, \(\lambda\unrhd\mu\) means \(\sum_{i\le k}\lambda_i\ge\sum_{i\le k}\mu_i\) for every \(k\). Conjugation reverses dominance. Every symmetric-group representation is semisimple, every Specht module is self-dual, and \(S^\lambda\otimes\mathop{\mathrm{sgn}}\cong S^{\lambda^t}\). We use the standard Specht construction, Young’s rule, branching, Schur–Weyl duality, and the Littlewood–Richardson and Pieri rules [10, 6, 15]. Their precise consequences needed below are recorded here. Alternating blocks and dual testsWe realize Specht modules by column alternation and detect constituents by contracting with dual tensors. This viewpoint also underlies Ikenmeyer’s triangular construction [9]. Let \(B\) be a finite set of positions, and let \(E\) have ordered basis \(e_1,\ldots,e_D\). For an ordered block \(A=(b_1,\ldots,b_k)\), \(k\le D\), put \[v_A=\sum_{\pi\in S_k}\mathop{\mathrm{sgn}}(\pi) e_{\pi(1)}\otimes\cdots\otimes e_{\pi(k)},\] with the displayed factors placed at \(b_1,\ldots,b_k\). For a set partition \(\mathcal A\) of \(B\), put \(v_{\mathcal A}=\bigotimes_{A\in\mathcal A}v_A\). Changing the order of one block changes only the overall sign. The position action of \(S_B\) is the usual permutation of tensor factors; all regroupings of factors are ordinary tensor identifications and add no signs. If the block sizes are the column lengths of \(\theta\), then \[ \mathbb C[S_B]v_{\mathcal A}\cong S^\theta. \tag{1}\] Indeed, arrange the blocks as tableau columns, and send a row tabloid to the word assigning letter \(i\) to its \(i\)th row. Column alternation sends its polytabloid to \(v_{\mathcal A}\), which is nonzero because the initial row word has coefficient one. The same construction applies in a dual alphabet; see [10] for the Specht construction and ordinary irreducibility. Lemma 3 (Dual-polytabloid test). Let \(z\in E^{\otimes B}\) and \(\theta\vdash |B|\) with at most \(D\) rows. Then \(S^\theta\) occurs in \(\mathbb C[S_B]z\) if and only if \(z\) has a nonzero contraction with a dual block tensor whose blocks are the columns of \(\theta\), for some placement of the blocks and some common ordered basis of \(E^*\). If such a contraction exists for a fixed placement, it is nonzero for a generic ordered basis. Finitely many separately nonzero contractions can be made nonzero simultaneously with a common generic basis. Proof. For any fixed basis and placement the orbit of the dual block tensor is a copy of \(S^\theta\), by (1). Restricting the natural tensor pairing gives an equivariant map from \(\mathbb C[S_B]z\) to the dual of this copy of \(S^\theta\). A nonzero contraction makes this map nonzero, hence proves the required constituent. Conversely, Schur–Weyl duality [15] identifies the \(\theta\)-isotypic component with \(S^\theta\otimes\mathbb S_\theta(E)\). The symmetric-group translates and common basis changes of one nonzero dual polytabloid span the corresponding full dual isotypic component: each of the two factors is irreducible for its own group. They therefore detect any nonzero \(\theta\)-component of \(z\). For a fixed placement, contraction is a polynomial in the entries of the common change-of-basis matrix. A nonzero polynomial remains nonzero on a nonempty Zariski-open subset of \(\mathop{\mathrm{GL}}_D(\mathbb C)\). The product of finitely many such polynomials is nonzero, proving the final statement. ◻ Support under inductionThroughout, combining constituents on disjoint position sets means ordinary induction from the product of their symmetric groups, not a Kronecker product. We use \(\boxtimes\) for an outer tensor product. Let \(c_{\alpha,\beta}^{\gamma}\) denote the Littlewood–Richardson coefficient, equivalently the multiplicity of \(S^\gamma\) in \(\mathop{\mathrm{Ind}}(S^\alpha\boxtimes S^\beta)\). This is the induction form of the LR rule [10]; its polynomial \(\mathop{\mathrm{GL}}\) counterpart is [15]. Only support, rather than equality of multiplicities, is required. Lemma 4 (Young support). For a finite list \(A\) of positive integers of sum \(n\), the module \(\mathop{\mathrm{Ind}}_{\prod_{a\in A}S_a}^{S_n}\mathbf1\) contains exactly the \(S^\nu\) whose column heights \(h_1\ge h_2\ge\cdots\) satisfy \[ \sum_{j=1}^k h_j\le P_A(k):=\sum_{a\in A}\min(k,a) \qquad(k\ge1). \tag{2}\] The constituent indexed by the decreasing rearrangement of \(A\) has multiplicity one. A component containing every shape of total \(t\) with at most \(j\) rows contains the support of the induced-trivial module corresponding to a balanced division of \(t\) into \(j\) parts. Proof. Young’s rule [10] says that the support consists of the partitions dominating the decreasing rearrangement of \(A\), with diagonal multiplicity one. Transposing dominance gives (2). The balanced partition of \(t\) into \(j\) parts is dominated by every partition of \(t\) with at most \(j\) rows; conversely, any partition dominating it has at most \(j\) rows. Zero parts in a balanced division can be omitted. ◻ Lemma 5 (Horizontal addition). If \(c_{\alpha,\beta}^{\gamma}>0\) and \(c_{\alpha',\beta'}^{\gamma'}>0\), then \[c_{\alpha+\alpha',\,\beta+\beta'}^{\gamma+\gamma'}>0,\] where addition is rowwise. In particular, suppose a target has column heights \(h_i=a_i+b_i\), with \(a_i,b_i\ge0\). Form \(\alpha\) from the column-height list \((a_i)\) and \(\beta\) from \((b_i)\), omitting zeros and sorting each list. Then \(c_{\alpha,\beta}^{\gamma}>0\) for the target \(\gamma\). Proof. The first assertion is the Littlewood–Richardson semigroup property [18], applied to the two triples of partitions. For the second assertion, the exterior-power case of Pieri gives \(c_{(1^{a_i}),(1^{b_i})}^{(1^{h_i})}>0\) for each \(i\). Add these containments horizontally. A rowwise sum of single-column partitions is exactly the diagram specified by their column-height multiset, so sorting the individual lists changes nothing. ◻ We also use two other immediate forms of the LR rule. First, \(c_{\alpha,(t)}^\gamma>0\) precisely when \(\gamma/\alpha\) is a horizontal \(t\)-strip, meaning at most one new cell per column. This is Pieri’s rule [15]. Second, if \(\alpha\) fits in a rectangle, its rotated complement \(\beta\) satisfies \(c_{\alpha,\beta}^{\text{rectangle}}=1\). For a rectangle \((w^j)\), the complement has parts \(\beta_i=w-\alpha_{j+1-i}\), and the Schur-character formula gives \(\mathbb S_\beta(\mathbb C^j)\cong\det^w\otimes \mathbb S_\alpha(\mathbb C^j)^*\) [15]. Since \(\mathbb S_{(w^j)}(\mathbb C^j)=\det^w\), the LR tensor-product rule and Schur’s lemma give the asserted multiplicity one. The zero-area case uses the empty partition, and the stated induction conventions include empty partitions as well. The cyclic staircase inputFor \(m\ge1\), set \[B_m=\{(i,j)\in\mathbb Z_{\ge1}^2:i+j\le m+1\},\qquad N_m=\frac{m(m+1)}2,\qquad \rho_m=(m,m-1,\ldots,1).\] Let \(R_m\) and \(C_m\) be the actual row and column set partitions of \(B_m\). Order each row by increasing \(j\) and each column by increasing \(i\). In each tensor layer use the ordered alphabet \(e_1,\ldots,e_m\). With the unnormalized initial-letter alternations defined above, set \[ w_m=v_{R_m}\otimes v_{C_m},\qquad W_m=\mathbb C[S_{B_m}]w_m\subseteq S^{\rho_m}\otimes S^{\rho_m}. \tag{3}\] The action in (3) is diagonal. Theorem 6 (Cyclic staircase support, [16]). For every \(m\ge1\) and every \(\alpha\vdash N_m\), the irreducible \(S^\alpha\) occurs in the particular cyclic module \(W_m\) of (3). The complete proof is given in Section 6 of the companion [16], with the same ordered row and column blocks as here. The bibliography records the separate preprint’s stable article identifier. It is important that the theorem concerns the specified row/column generator, not merely the ambient staircase square. Common relabelings of the two ordered alphabets and changes of block orders preserve the assertion. A fixed self-conjugate candidateWe now choose one diagram for the degree, before fixing a desired constituent. The construction preserves self-conjugacy by attaching transpose-matched cells to a staircase; its parameter bounds will later make the two support tests exhaustive. Set \(N_0=0\) and choose the largest nonnegative \(M\) for which \(N_M\le n\) and \(N_M\equiv n\pmod2\), and put \[ r=\frac{n-N_M}{2},\qquad b=\left\lfloor\frac r2\right\rfloor, \qquad s=1+(r\bmod2),\qquad K=2M-1. \tag{4}\] Only \(M\ge9\) will be used in this construction; smaller degrees will be treated separately. Lemma 7. The parameters satisfy \[ r\le\begin{cases}(M-1)/2,&M\text{ odd},\\3M/2+2,&M\text{ even}. \end{cases} \tag{5}\] For \(M\ge9\), the partition with column lengths \[ L=(M+b+s-1,\ M-1+b,\ M-2,M-3,\ldots,3,\ 2^{b+1},\ 1^s) \tag{6}\] is self-conjugate and has size \(n\). Consequently the support of \(S^L\otimes S^L\) is invariant under transposition of the target. If no parity-compatible \(M\ge9\) is available, then \(n\le64\). Proof. If \(M\) is odd, the next triangular number of the same parity is \(N_{M+1}\), with difference \(M+1\). Maximality gives \(2r\le M-1\). If \(M\) is even, the next such number is \(N_{M+3}\), with difference \(3M+6\), giving \(2r\le3M+4\). Relative to the staircase, the increments to the first two columns sum to \(2b+s-1\), and the added short columns have the same total. Thus the size increases by \(4b+2s-2=2r\). There are \(M+b+s-1\) columns in total and \(M+b-1\) columns of height at least two. Counting those of height at least \(j\) for \(j\ge3\) gives successively \(M-2,M-3,\ldots,3\), followed by \(b+1\) twos and \(s\) ones. These are exactly the column lengths in (6), so \(L=L^t\). It follows that \[(S^L\otimes S^L)\otimes\mathop{\mathrm{sgn}} \cong S^L\otimes(S^L\otimes\mathop{\mathrm{sgn}}) \cong S^L\otimes S^L.\] Finally, every odd \(n\ge45=N_9\) admits a parity-compatible index at least nine, and every even \(n\ge66=N_{11}\) admits one at least eleven. The remaining degrees are at most \(64\). ◻ Figure 1 shows the matched additions in a small instance. The path-incidence picture in Figure 2 will describe a different decomposition of the same construction. When \(r=0\), the candidate is the staircase itself, and Theorem 6 already supplies all constituents. For \(r>0\) we give two sufficient support tests. For a target \(\nu\), write its column heights and row lengths as \[h=(h_1,h_2,\ldots)=\nu^t,\qquad w=(w_1,w_2,\ldots)=\nu,\] with zeros appended. Both tests refer to the single square \(S^L\otimes S^L\) fixed above. We are free to interchange \(h\) and \(w\) by Lemma 7. The coverage has four steps. A small first-four row or column sum is handled by Proposition 21. Otherwise the band criterion covers every \(M\ge22\) and all intermediate parameters apart from the nineteen pairs in (29); the exact capacity check handles those pairs. Proposition 30 handles all eligible degrees at most \(64\), completing the small-index remainder. The two finite checks thus have separate, explicitly bounded roles. The band criterionWe use the parameters and the self-conjugate diagram \(L\) from (6), and write \(K=2M-1\). In particular, \(r=2b+s-1\). The following criterion supplies constituents whose diagrams have enough room for a path and its two attachments. Proposition 8 (Band criterion). Let \(M\ge9\), and let \(\lambda\vdash |L|\) have column heights \(h_1\ge h_2\ge\cdots\) and row lengths \(w_1\ge w_2\ge\cdots\), with zeros appended. Suppose that integers \(d,q\) satisfy \(1\le d\le4\), \(0\le q\le8\), and \[ \sum_{i=1}^d w_i\ge K,\qquad dq+7+d\le K,\qquad \sum_{i=1}^q \max\left(0,\left\lfloor\frac{h_i-d}{2}\right\rfloor\right) \ge r. \tag{7}\] Then \(S^\lambda\) occurs in \(S^L\otimes S^L\). The three inequalities have separate roles: the first gives room for the path, the second reserves space at its endpoints, and the third provides the even column heights needed for the attachments. We first force the smaller cyclic triangle, then find nonzero path and attachment contractions, and finally show that they combine without cancellation. All alternating tensors use the conventions of Section 2; changing the order of a block changes only its overall sign. A forced triangle and a full induced quotientWe first record the elementary induction fact used here and in the balance constructions. It is the coordinate-sector principle of [16], applied to the marked alphabets of the enlarged diagram. Lemma 9 (Induction from coordinate sectors). Let a finite group \(G\) act transitively on a finite set \(\Omega\). Suppose a \(G\)-module has a direct-sum decomposition \(T=\bigoplus_{A\in\Omega}T_A\) with \(gT_A=T_{gA}\). Fix \(D\in\Omega\), let \(H\) be its stabilizer, and let \(z\in T_D\). Writing \(A_0=\mathbb C[H]z\), the map \[\mathbb C[G]\otimes_{\mathbb C[H]}A_0\longrightarrow\mathbb C[G]z, \qquad g\otimes a\longmapsto ga,\] is an isomorphism. In particular, the cyclic module on the right is the full induced module \(\mathop{\mathrm{Ind}}_H^G A_0\). Proof. Choose one representative \(g_A\) with \(g_AD=A\) for each \(A\in\Omega\). The source is the direct sum of the spaces \(g_A\otimes A_0\). Their images are \(g_AA_0\subseteq T_A\), so the images for distinct \(A\) are linearly independent. On each summand the displayed map is injective, since \(g_A\) acts invertibly. Its image contains every translate of \(z\), proving surjectivity as well. ◻ Regard \(L\) as its set \(B\) of cells, with rows and columns indexed from one. Let \(v_R,v_C\) be the alternating block tensors for its rows and columns, respectively, and put \[W_L=\mathbb C[S_B](v_R\otimes v_C).\] Both factors have type \(L\), since \(L\) is self-conjugate; hence \(W_L\subseteq S^L\otimes S^L\) after choosing the Specht identifications. Let \[D=\{(i,j):i,j\ge1,\ i+j\le M-1\},\qquad F=B\setminus D.\] Thus \(D\) is the staircase of order \(M-2\), and \(|F|=K+2r\). Lemma 10 (Triangle quotient). There is a quotient of \(W_L\) isomorphic to \[ \mathop{\mathrm{Ind}}_{S_D\times S_F}^{S_B} (W_{M-2}\boxtimes U_F), \tag{8}\] where \(W_{M-2}\) is the specified cyclic module in Theorem 6. The module \(U_F\) is generated by the row and column block alternations on \(F\), using an initial alphabet on every remaining block. Consequently, it suffices to find a diagram \(\mu\subseteq\lambda\) of size \(|F|\) such that \(S^\mu\) occurs in \(U_F\). Proof. Set \(\ell=M+b+s-1\), the largest row and column length of \(L\). The triangle \(D\) has \(t_i=\max(M-1-i,0)\) cells in row \(i\) and in column \(i\). We will give each full row and column block exactly this number of marked letters. In each of the two alphabets \(1,\ldots,\ell\), mark precisely \[\{3,4,\ldots,M-1\}\cup\{M+b\}.\] There are \(M-2\) marked letters. The number of marked letters in the initial alphabet of row or column \(i\) is indeed \(t_i\): for the two largest blocks these counts are \(M-2,M-3\), and thereafter they are \(M-4,M-5,\ldots,1,0,\ldots\). The ordered unmarked alphabet is \((1,2,M,\ldots,M+b-1)\), followed by \(M+b+1\) if \(s=2\). The interval \(M,\ldots,M+b-1\) is empty when \(b=0\); in this case the marked alphabet is \((3,\ldots,M)\) and the unmarked alphabet is \((1,2)\) or \((1,2,M+1)\), respectively. In every case each block inherits an initial list in each of these two ordered alphabets. Project the pair-letter tensor space by retaining only those words whose two letters at every position have the same marking status. The projection commutes with permutations of positions. For any retained term of \(v_R\otimes v_C\), the set \(A\) of marked positions has row and column margins \((t_i)\). For every \(k\), \[\sum_{i=1}^k t_i=\sum_j\min(k,t_j).\] Column \(j\) can place at most \(\min(k,t_j)\) marked cells in the first \(k\) rows. Equality of the sums therefore gives equality in every column. Taking \(k=t_j\) shows that, when \(t_j>0\), all its marked cells occupy the first \(t_j\) rows. Hence \(A=D\). Within each row or column block, the marked and unmarked positions are now fixed. Expanding that block’s alternation gives the tensor product of its marked and unmarked alternations, multiplied by one fixed shuffle sign. The two alphabets, each in its inherited order, have initial lists on every block. Thus, up to an overall sign, the projected generator is \[w_{M-2}\otimes u_F,\] where \(w_{M-2}\) is exactly the row-column generator of Theorem 6, with its ordered alphabet relabeled, and \(u_F\) is the generator described in the statement. The stabilizer \(S_D\times S_F\) generates \(W_{M-2}\boxtimes U_F\) from this tensor. To specify the sectors, let \(V_{\mathrm{mark}}\) and \(V_{\mathrm{unmark}}\) be the pair-letter spaces in which both letters are marked or both are unmarked, respectively. For \(A\subseteq B\) with \(|A|=|D|\), put \[T_A=V_{\mathrm{mark}}^{\otimes A}\otimes V_{\mathrm{unmark}}^{\otimes(B\setminus A)}.\] These spaces have disjoint word bases, and \(gT_A=T_{gA}\). Lemma 9 therefore identifies the entire projected cyclic image with the full induced module in (8), not merely with one of its quotients. For the last assertion, suppose \(\mu\subseteq\lambda\). Iterated branching implies that \(S^\mu\) occurs on restriction of \(S^\lambda\) to \(S_F\). Decompose \(\mathop{\mathrm{Res}}_{S_D\times S_F}^{S_B}S^\lambda\) and then forget the \(S_D\) action. The occurrence just noted implies that some \(\alpha\vdash|D|\) satisfies \(c_{\alpha,\mu}^{\lambda}>0\). Theorem 6 supplies \(S^\alpha\) in \(W_{M-2}\). Thus \(S^\lambda\) occurs in the induced module in (8), and hence in \(W_L\) by complete reducibility. ◻ The path and its two attachmentsWrite \(\delta=s-1\in\{0,1\}\) and \(E=(b+\delta,b)\), omitting zero parts. Each copy of \(E\) has area \(r\). To decompose \(F\), start with the boundary cells \(i+j=M\) or \(M+1\) in the staircase \(B_M\). They form a width-two path of length \(K\). Move its endpoint singletons from \((M,1)\) and \((1,M)\) to \((\ell,1)\) and \((1,\ell)\), respectively. More precisely, the resulting path positions are \(p_1,\ldots,p_K\), where \[\begin{align*} p_1&=(\ell,1),&p_K&=(1,\ell),\\ p_{2j}&=(M-j,j) &&(1\le j<M),\\ p_{2j-1}&=(M+1-j,j) &&(2\le j<M). \end{align*}\] The northeastern attachment is \[E_+=\{(1,j):M\le j\le\ell-1\} \cup\{(2,j):M\le j\le M+b-1\},\] with an interval interpreted as empty when its upper endpoint is smaller than its lower endpoint. Let \(E_-\) be its transpose. These sets are pairwise disjoint and exhaust \(F\). The attachment \(E_+\) has shape \(E\), and \(E_-\) has the transposed incidence pattern. Along the path, the column pairs are \((p_1,p_2),(p_3,p_4),\ldots,(p_{K-2},p_{K-1})\), with a column singleton at \(p_K\). The row pairs are \((p_2,p_3),(p_4,p_5),\ldots,(p_{K-1},p_K)\), with a row singleton at \(p_1\). Only the first two column pairs and the last two row pairs belong to blocks extended by attachment cells. Relabel the unmarked alphabet in increasing order, starting at one. Let \(P\) be the four-dimensional pair-letter space spanned by \(e_a\otimes e_c\) with \(a,c\le2\). Let \[X_+=\mathop{\mathrm{span}}\{e_a\otimes e_c:a>2,\ c\le2\},\qquad X_-=\mathop{\mathrm{span}}\{e_a\otimes e_c:a\le2,\ c>2\},\qquad X=X_+\oplus X_-.\] Only letters available in the complement alphabet are included in these spans. There are \(b+\delta\) letters above two, so \[\dim X=4(b+\delta)\ge 4b+2\delta=2r.\] In particular, every extra diagram of size \(2r\) fits this actual alphabet; when \(r=0\), both the diagram and \(X\) are empty. No cell of \(F\) is simultaneously in one of its two long rows and one of its two long columns, so every pair letter in \(u_F\) lies in \(P\oplus X\). Lemma 11 (Separated contraction). Retain from \(u_F\) only terms using \(P\) on the path, \(X_+\) on \(E_+\), and \(X_-\) on \(E_-\). The resulting tensor is, up to one nonzero overall sign, \[ u_{\mathrm{path}}\otimes z_+\otimes z_-. \tag{9}\] Here \(u_{\mathrm{path}}\) is the product of the two path block alternations, with both endpoint singleton letters equal to one. The tensors \(z_+,z_-\) are the row-column alternating generators for the two copies of \(E\), with the long-block index shifted by two and the two tensor layers interchanged for \(E_-\). Proof. In each extended block its two path positions must take its low letters \(1,2\), while its extra positions must take all remaining letters. These are prescribed sets of positions. The alternation therefore splits into an independent alternation on each set, with a fixed shuffle sign. Every other block belongs entirely to the path or to one attachment. Multiplying these factorizations proves (9). ◻ The selection in Lemma 11 refers to fixed regions of \(F\) and need not commute with position permutations. We will use the factorization inside a dual-polytabloid contraction: the endpoint placement in the final proof will force precisely these selected terms to survive. Contractions on an odd pathFor \(t\ge0\), Brown, van Willigenburg, and Zabrocki [4] proved that the square of \(S^{(t+1,t)}\) contains exactly the partitions of \(2t+1\) with at most four rows, each with multiplicity one. We need this support in a specified cyclic path tensor, with any consecutive placement of the dual columns. The following proof extends the explicit contraction in [16] to every odd length, including a single position, and gives the common-basis conclusion needed for several paths at once in the balance constructions. Lemma 12 (Odd-path support). Let \(k=2t+1\) with \(t\ge0\). On positions \(1,\ldots,k\), put alternating pairs in the first layer on \((2,3),(4,5),\ldots,(k-1,k)\) and the singleton letter \(1\) at position \(1\). In the second layer put alternating pairs on \((1,2),(3,4),\ldots,(k-2,k-1)\) and the singleton letter \(1\) at position \(k\). Denote the resulting pair-letter tensor by \(u\). For every ordered list \(a_1,\ldots,a_c\) with \(1\le a_i\le4\) and \(\sum_i a_i=k\), place the columns of a dual polytabloid consecutively along the path with these lengths. There is a common ordered basis of \(\mathop{\mathrm{Mat}}_2(\mathbb C)\) for which its contraction with \(u\) is nonzero. In particular, the cyclic span of \(u\) contains every partition of \(k\) with at most four rows. For any finite collection of such paths and ordered column lists, one common generic basis makes all their contractions nonzero. Proof. Identify a pair-letter covector with a matrix \(Y\) by evaluation \(Y(e_a\otimes e_c)=Y_{ac}\), and put \[J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.\] Expanding the pair alternations and successively contracting the indices along the path gives \[\begin{align*} \langle u,Y_1\otimes\cdots\otimes Y_k\rangle &=\bigl[Y_1JY_2^{\mathsf T}JY_3\cdots JY_{k-1}^{\mathsf T}JY_k\bigr]_{11}\\ &=(-1)^t \bigl[Y_1\mathop{\mathrm{adj}}(Y_2)Y_3\cdots\mathop{\mathrm{adj}}(Y_{k-1})Y_k\bigr]_{11}, \tag{10}\end{align*}\] because \(JY^{\mathsf T}J=-\mathop{\mathrm{adj}}(Y)\) for a two-by-two matrix. The formula also holds for \(k=1\), with no intervening factors. Use the ordered matrix basis \[M_1=I,\qquad M_2=Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad M_3=T=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad M_4=J=ZT.\] The last three matrices are traceless, invertible, and pairwise anticommuting. Adjugation fixes \(I\) and negates the other three. For a segment of length \(a\le4\) beginning at global position \(p\), write \(D_j(Y)=Y\) for odd \(j\) and \(D_j(Y)=\mathop{\mathrm{adj}}(Y)\) for even \(j\). Its alternated matrix is \[A_{a,p}=\sum_{\pi\in S_a}\mathop{\mathrm{sgn}}(\pi) D_p(M_{\pi(1)})\cdots D_{p+a-1}(M_{\pi(a)}).\] Let \(e\) be the number of even positions in this segment, and let \(\epsilon=1\) when \(p\) is even and \(\epsilon=0\) otherwise. Then \[ A_{a,p}=a!(-1)^{e-\epsilon}M_1\cdots M_a. \tag{11}\] Indeed, if \(I\) occupies the local slot \(j\), moving it there contributes \((-1)^{j-1}\) to the permutation sign, while reordering all remaining matrices contributes precisely their permutation sign by anticommutation. The adjugations contribute \((-1)^{e-\epsilon_j}\), where \(\epsilon_j\) records whether \(p+j-1\) is even. Since \(\epsilon_j\equiv\epsilon+j-1\pmod2\), the total sign is \((-1)^{e-\epsilon}\) for every summand. This proves (11), whose right side is invertible. For example, at length three the formula gives \(-6J\) for both starting parities: the canonical product is \(IZT=J\), and \(e-\epsilon=1\) in either case. Consequently, alternating independently within the prescribed consecutive segments in (10) gives \((-1)^t[B]_{11}\), where \(B\) is the ordered product of their invertible matrices \(A_{a,p}\). Choose an eigenvector of \(B\) with nonzero eigenvalue \(\beta\), and a matrix \(Q\in\mathop{\mathrm{GL}}_2(\mathbb C)\) taking it to the first coordinate vector. Replace every \(M_i\) by \(QM_iQ^{-1}\). Since adjugation commutes with conjugation, the new contraction is \((-1)^t[QBQ^{-1}]_{11}=(-1)^t\beta\ne0\). The endpoint letters of \(u\) have remained fixed throughout. The dual tensor is a polytabloid for the partition with the specified column lengths, regardless of their order along the path. Lemma 3 proves the support assertion. For any one placement its contraction is a polynomial in the entries of the common basis and is not identically zero by what we just proved. The nonvanishing loci of finitely many such polynomials have nonempty intersection with the open set of ordered bases, since \(\mathop{\mathrm{GL}}_4(\mathbb C)\) is irreducible. This proves the last assertion. ◻ The support of the two extra copiesThe path supplies the four-row component. We now determine enough support in each attachment to supply every even-column diagram required by the last inequality of Proposition 8. The first step adapts the signed-average argument of [16] to the two-row attachment. Lemma 13 (One extra copy). Let \(E=(b+\delta,b)\), where \(b\ge0\) and \(\delta\in\{0,1\}\), and let \(Z_E\) be the cyclic span of its row-column alternating generator. The sign twist \(Z_E\otimes\mathop{\mathrm{sgn}}\) contains every partition \(\alpha\vdash 2b+\delta\) with at most four rows such that all its row lengths are even when \(\delta=0\), or exactly one row length is odd when \(\delta=1\). Proof. The empty diagram is immediate, so assume \(|E|>0\). Let \(p\) be its short-column alternating vector and \(v\) its long-row alternating vector. These generate Specht modules of types \(E\) and \(E^t\), respectively; denote them by \(V_p\) and \(V_v\). Put \(G=S_{|E|}\) and \(\varepsilon=\mathop{\mathrm{sgn}}\). Let \(H\) permute positions within each short column, so that \(hp=\mathop{\mathrm{sgn}}(h)p\). Signed averaging of the product gives \[\sum_{h\in H}\mathop{\mathrm{sgn}}(h)h(v\otimes p) =\left(\sum_{h\in H}hv\right)\otimes p.\] The first factor \(v_0=\sum_{h\in H}hv\) is nonzero. In fact the word assigning letter \(j\) to every cell of short column \(j\) is fixed by \(H\) and has coefficient one in the row alternation \(v\), so its coefficient in \(v_0\) is \(|H|\). Choose an abstract \(G\)-isomorphism \(\phi:V_v\otimes\varepsilon\longrightarrow V_p\), which exists by sign conjugacy. The \(H\)-alternating line of \(V_p\) is unique: \[\begin{align*} \dim\mathop{\mathrm{Hom}}_H(\varepsilon|_H,\mathop{\mathrm{Res}}_H^G V_p) &=\dim\mathop{\mathrm{Hom}}_G(\mathop{\mathrm{Ind}}_H^G\varepsilon|_H,V_p)\\ &=\dim\mathop{\mathrm{Hom}}_G(\mathop{\mathrm{Ind}}_H^G\mathbf1,V_p\otimes\varepsilon)\\ &=1. \end{align*}\] For the last equality, \(H\) has block sizes given by \(E^t\), and \(V_p\otimes\varepsilon\cong S^{E^t}\); this is the diagonal multiplicity in Lemma 4. Since \(v_0\) is \(H\)-fixed, \(\phi(v_0\otimes\zeta)=cp\) for a nonzero sign-module vector \(\zeta\) and some \(c\ne0\). Reassociating the sign twist into the first factor and then applying \(\phi\otimes\mathrm{id}\) identifies \((V_v\otimes V_p)\otimes\varepsilon\) with \(V_p\otimes V_p\). The image of \(Z_E\otimes\varepsilon\) contains \(cp\otimes p\), so it contains the cyclic span of \(p\otimes p\). This is an isomorphism of abstract symmetric-group modules; it does not change the physical long alphabet into a two-letter alphabet. It proves a support inclusion for the original attachment module. The subsequent contraction calculation may therefore be made in the convenient realization of \(p\otimes p\) on two copies of a two-dimensional alphabet. Regroup the latter tensor by positions. Each repeated short pair is the symmetric tensor \[ H_0=e_{11}\otimes e_{22}+e_{22}\otimes e_{11} -e_{12}\otimes e_{21}-e_{21}\otimes e_{12} \tag{12}\] on the four-dimensional space with basis \(e_{ac}=e_a\otimes e_c\). It is nondegenerate. If \(\delta=1\), there is also the singleton vector \(v_*=e_{11}\). Suppose first that \(\alpha\) has even rows. Pair adjacent columns of its dual tableau; the two columns in each pair have equal height. Place the \(b\) position pairs of \(H_0^{\otimes b}\) across corresponding rows of these paired columns. Choose an orthonormal basis of covectors for the form \(H_0\). In any nonzero term of the dual column alternations, the two permutations on a paired column pair must match. Their signs therefore multiply to one. A paired column pair of height \(a\) contributes \(a!\), so the entire contraction is nonzero. When \(\alpha\) has exactly one odd row, adjacent columns have equal heights except for a single paired column pair of heights \(a,a-1\). Put the singleton at the bottom of its longer column and place the remaining position pairs across corresponding rows. In that exceptional pair, matching the other covectors forces the singleton to receive the \(a\)th basis covector. The contribution is \((a-1)!\) times its evaluation on \(v_*\). Choose the orthonormal basis so that this evaluation is nonzero; this is possible since \(v_*\ne0\), by permuting the members of any orthonormal basis if necessary. The covectors span the dual space, so at least one evaluates nontrivially on \(v_*\). The other column pairs contribute their positive factorials as before. Thus this contraction is also nonzero. The height bound four ensures that all covectors used belong to the available basis. Lemma 3 completes the proof. ◻ Lemma 14 (Two extra copies). The cyclic span of \(z_+\otimes z_-\) on \(E_+\sqcup E_-\) contains every diagram \(\eta\vdash 2r\) whose column heights are even and whose number of columns is at most eight. Proof. The alphabets \(X_+\) and \(X_-\) are disjoint. The same coordinate sector argument as in Lemma 9 identifies this cyclic module with the full induction of the two local cyclic modules. Transposing \(E\) merely interchanges its two tensor layers, so both local modules are isomorphic to \(Z_E\). Put \(\theta=\eta^t\). Its rows have even lengths and there are at most eight of them. Group its columns in adjacent pairs, and let \(j_1,\ldots,j_a\le8\) be the heights of those double columns. Thus \[\theta=(2^{j_1})+\cdots+(2^{j_a}),\qquad j_1+\cdots+j_a=r,\] where addition of diagrams is rowwise. For each \(j\), splitting its height as \(j=u+v\) gives \[c_{(2^u),(2^v)}^{(2^j)}>0.\] Indeed the corresponding single-column coefficient is positive, and Lemma 5 applies twice. Choose \(u=\lfloor j/2\rfloor\), \(v=\lceil j/2\rceil\), allowing their interchange. Both are at most four. If \(r\) is even, the number of odd \(j_i\) is even. Alternate the orientation of the split on these odd heights. The resulting two rowwise sums \(\alpha,\beta\) each have size \(r\), at most four rows, and all row lengths even. Repeated use of Lemma 5 gives \(c_{\alpha,\beta}^{\theta}>0\). If \(r\) is odd, reserve one odd height \(j=2k+1\). For this double column take the two equal-area shapes \[\kappa=(2^k,1),\qquad c_{\kappa,\kappa}^{(2^{2k+1})}>0.\] The displayed coefficient follows from rectangle complement duality in the Littlewood–Richardson rule, since \(\kappa\) is its own rotated complement in this rectangle. Equivalently, fill the single remaining cell in row \(k+1\) by \(1\), and in each later row \(k+t\), for \(2\le t\le k+1\), put \(t-1,t\) from left to right. This is a Littlewood–Richardson filling with content \(\kappa\), including the case \(k=0\). There are now an even number of odd heights left; alternate their orientations as above. Their combined area is \(2(r-j)\), and the balanced split gives \(r-j\) cells to each factor; the reserved copy of \(\kappa\) adds \(j\) cells to each. Each final factor thus has size \(r\), at most four rows, and exactly one odd row, since the remaining rowwise summands have only even rows. Here \(k+1\le4\) because \(j\le8\). At the endpoints this includes \(r=j=1\), with \(\kappa=(1)\), and \(j=7\), with \(\kappa=(2,2,2,1)\) of size seven and height four. Again horizontal addition gives a positive coefficient for \(\theta\). Lemma 13 supplies both resulting factor partitions in \(Z_E\otimes\mathop{\mathrm{sgn}}\). Induction commutes with a sign twist, since the restriction of the sign character is the product of the two local sign characters. Thus \(\theta\) occurs in the sign twist of the two-copy module, and \(\eta\) occurs in the untwisted module. The case \(r=0\) is the empty tensor and obeys the same conclusion. ◻ Attaching without cancellationThe local contractions are now available. The remaining issue is their interaction: an elongated dual column could in principle mix path and attachment covectors. The endpoint clearance in the criterion allows us to place the covectors so that all mixed assignments vanish. Proof of Proposition 8. Choose an even list \(e_1\ge\cdots\ge e_t>0\), with \(t\le q\), such that \[ \sum_{i=1}^t e_i=2r,\qquad e_i\le h_i-d. \tag{13}\] For example, fill the capacities \(\max(0,\lfloor(h_i-d)/2\rfloor)\) from left to right until their sum reaches \(r\), then double the filled amounts and discard zeros. The capacities decrease, so the resulting positive list decreases. When \(r=0\), take the empty list and \(t=0\). Put \(d\) path cells in each of these \(t\) columns. The total available path capacity is \[\sum_i\min(d,h_i)=\sum_{i=1}^d w_i\ge K.\] Fill further columns, in order, with at most \(\min(d,h_i)\) path cells until the path total is exactly \(K\). This produces a decreasing column list for a path diagram \(\pi\); its first \(t\) columns have height \(d\), and every column has height at most \(d\). Increase the first \(t\) column heights by \(e_i\) to form \(\mu\). By construction \(\mu\subseteq\lambda\) and \(|\mu|=K+2r\). Call the first \(t\) path columns chosen and all other nonempty path columns unchosen. The total length of unchosen path columns is \[K-dt\ge K-dq\ge7+d.\] Order some of these columns first, stopping when their total length first reaches four. Each has length at most \(d\), so this prefix has length between four and \(3+d\). Put the chosen columns next, then all remaining unchosen columns. The final group has length at least four. Lay these columns consecutively along the path in this order. All path positions in a chosen column now lie outside the first and last four path positions; their original row and column blocks are consequently unextended. By Lemma 12, there is an ordered basis \(N_1,\ldots,N_4\) of \(P^*\) for which the path polytabloid using these consecutive columns has nonzero contraction with \(u_{\mathrm{path}}\). Only its first \(d\) basis covectors are used. Let \(\eta\) be the extra diagram with column heights \(e_1,\ldots,e_t\). Lemma 14 and Lemma 3 give an ordered basis \(Q_1,\ldots,Q_{\dim X}\) of \(X^*\) and a placement of its dual columns on the extra positions for which the contraction with \(z_+\otimes z_-\) is nonzero. The empty case has contraction one. This applies the dual test to the actual extra tensor in \(X\); the abstract Specht identification in Lemma 13 was used only to prove its support. The \(Q_j\) may mix \(X_+^*\) and \(X_-^*\), and an extra dual column may use positions from both attachments. Neither is a restriction: assign each such extra column to a chosen path column with the corresponding height \(e_i\). Extend these covectors by zero on the other summand of \(P\oplus X\). Use the single ordered flag beginning \[N_1,\ldots,N_d,Q_1,\ldots,Q_{\dim X}, N_{d+1},\ldots,N_4.\] For every chosen column, adjoin its extra column of \(e_i\) positions below its \(d\) path positions. The other columns have only their path positions. This is a placement of the column blocks of a dual polytabloid of shape \(\mu\). We check all terms of its alternating expansion. At each chosen path position, both row and column block lengths in the complement generator \(u_F\) are at most two (the assertion is about its remaining blocks, not their lengths in \(L\)). The pair letter therefore belongs to \(P\) and is annihilated by every \(Q_j\). The \(d\) path positions in such a column must use all its \(N\)-covectors, leaving precisely its \(Q\)-covectors on its extra positions. There are exactly \(d\) such \(N\)-covectors, so none can be assigned to an extra slot. Unchosen columns already have only \(N\)-covectors. Thus every surviving assignment uses \(P\) on the entire path and \(X\) on all extra positions. On \(E_+\) the only possible such letters lie in \(X_+\), and on \(E_-\) they lie in \(X_-\). No additional mixed assignment survives. We may consequently replace \(u_F\) in this contraction by its separated tensor from Lemma 11. In every elongated dual column the surviving permutations also factor: they permute the \(N\)-covectors among its path slots and the \(Q\)-covectors among its extra slots, with a fixed cross sign. Order the path slots before the extra slots to make that sign one. The full contraction is therefore, up to the fixed sign in (9), \[\langle u_{\mathrm{path}},\text{path polytabloid}\rangle \;\langle z_+\otimes z_-,\text{extra polytabloid}\rangle,\] and both factors are nonzero. Lemma 3 gives \(S^\mu\) in \(U_F\). Finally, \(\mu\subseteq\lambda\), so Lemma 10 proves the proposition. ◻ Balance testsRetain the self-conjugate candidate \(L\) in Equation (6). For this second support test we use the full square \(S^L\otimes S^L\), with both factors realized by column-alternating tensors on the same column blocks. Put \[c=2b+2,\qquad T=2M-1+r.\] Thus \(c\) is the total size of the columns of length two, and \(T\) is the sum of the two largest column lengths. We prove that a target occurs whenever its first four column heights have sum at most \(2M-2\). The local components below supply balanced lists of trivial-factor sizes, whose induced support is tested by Young’s rule. We then establish an additional three-row support statement for the bounded capacity checks. For reference, write \[\begin{align*} \mathcal H_j(t)&=\{\nu\vdash t:\ell(\nu)\le j\},\\ \mathcal E_j(t)&=\{\nu\in\mathcal H_j(t): \text{every row length of $\nu$ is even}\}. \end{align*}\] If \(t\) is even, let \(\mathcal O_4(t)\) be the members of \(\mathcal H_4(t)\) having an odd-height column. Finally, \(\mathcal P_3(c,s)\) denotes the partitions with at most three rows obtained from a member of \(\mathcal E_3(c)\) by adding a horizontal strip of size three and then a horizontal strip of size \(s\). All support statements below mean inclusion, without any assertion about multiplicities. Separating components by output valuesUse independent ordered alphabets in the two Specht factors. A pair of letters \((a,a')\) has output \[z(a,a')=a+a'-1.\] For a collection \(B\) of column blocks on a position set \(D_B\), let \(v_B\) be their alternating tensor and put \(Y_B=\mathbb C[S_{D_B}]v_B\). By Equation (1), this is the Specht module whose column lengths are the sizes of the blocks in \(B\). Realize both copies of \(Y_B\) on \(D_B\). Every pair word occurring in their tensor square has the same number of positions and the same output sum: \[ d_B=\sum_{A\in B}|A|, \qquad \sigma_B=\sum_{A\in B}|A|^2. \tag{14}\] Indeed, a block of length \(k\) uses \(1,\ldots,k\) once in each factor, so its output sum is \(2(1+\cdots+k)-k=k^2\). Permuting positions within \(D_B\), independently in the two layers, preserves these totals. Lemma 15 (Separation by sizes and sums). Partition the column blocks of \(L\) into collections \(B_1,\ldots,B_v\) on disjoint position sets \(D_1,\ldots,D_v\), using the same blocks in both factors. Suppose \(I_1,\ldots,I_v\) are nonempty sets of integers with every element of \(I_i\) smaller than every element of \(I_{i+1}\). Let \(W_i\) be the image of \(Y_{B_i}\otimes Y_{B_i}\) under the coordinate projection retaining words all of whose outputs lie in \(I_i\). Then the support of \(S^L\otimes S^L\) contains the support of \[\mathop{\mathrm{Ind}}_{S_{d_{B_1}}\times\cdots\times S_{d_{B_v}}}^{S_n} (W_1\boxtimes\cdots\boxtimes W_v).\] Proof. In the full pair-word space, retain precisely the words whose outputs lie in \(\bigcup_i I_i\) and for which the outputs in \(I_i\) have cardinality \(d_{B_i}\) and sum \(\sigma_{B_i}\), for every \(i\). This coordinate projection commutes with all position permutations. Consider a word from the tensor product with the prescribed component positions. The positions in \(D_1\) have cardinality \(d_{B_1}\) and sum \(\sigma_{B_1}\) by Equation (14). In a retained word, the positions whose outputs lie in \(I_1\) have the same cardinality and sum. If these two sets of positions differed, the positions in the first set but not the second would all have larger outputs than the positions in the second but not the first. The two set differences have equal positive cardinality, contradicting equality of their sums. The two sets therefore agree. Remove them and repeat the argument for \(I_2\), and then successively for every range. It follows that the global projection on these prescribed positions is the tensor product of the local projections. More explicitly, write \(v_i=v_{B_i}\), \(Y_i=Y_{B_i}\), and \(H=\prod_i S_{D_i}\). Inside either Specht factor, \[\mathbb C[H](v_1\otimes\cdots\otimes v_v) =Y_1\boxtimes\cdots\boxtimes Y_v\subseteq S^L.\] Taking its tensor square gives the actual subspace \(\boxtimes_i(Y_i\otimes Y_i)\) in \(S^L\otimes S^L\): the two layers can be generated independently. The symmetric groups act diagonally on each local square and on the global square. The restriction of the global projection is \(\bigotimes_i P_i\), where \(P_i\) is the local projection. Its image is therefore the full outer product \(W_1\boxtimes\cdots\boxtimes W_v\). Different ordered placements of these components have disjoint coordinate supports, distinguished by which positions have outputs in each \(I_i\). Their translates therefore give the full induced module in the statement. Since an equivariant image cannot acquire a new irreducible type, the result follows. ◻ The local support guarantees proved in the following subsections are summarized in Table 1. The entries concerning the largest two lengths use \(q=M+b-1\). When \(s=1\), their lengths are \(q+1,q\); when \(s=2\), their lengths are \(q+2,q\).
After applying Lemma 15, transitivity of induction permits us to combine selected factors in any order. In particular, the big pair can be combined with the singleton component even though their output ranges are not adjacent. Coordinate functionals and path componentsWe first explain exactly how a surviving word coordinate supplies a path vector after the two sign twists. Lemma 16 (The sign-twisted coordinate line). Let \(X=S^\mu\) be realized by column-alternating vectors in the word space of content \(\mu=(\mu_1,\mu_2,\ldots)\). For a word \(a\) of this content, let \(R_a\) permute separately the positions carrying each equal letter. The restriction \(\epsilon_a|_X\) of its coordinate functional is nonzero. Under \(X^*\otimes\mathop{\mathrm{sgn}}\simeq S^{\mu^t}\), its line is the line of the column-alternating vector on the equal-letter blocks of \(a\), whose lengths are \(\mu_1,\mu_2,\ldots\). Proof. For the row-constant word of a tableau, its coefficient in the tableau’s column-alternating vector is one: only the identity in each column stabilizes that word. Hence its coordinate restriction is nonzero. Every other word of the same content is a position translate, which proves nonvanishing in general. The restricted coordinate is \(R_a\)-invariant. Young’s rule and Frobenius reciprocity give \[\dim (X^*)^{R_a}=1,\] using the diagonal multiplicity in Lemma 4. After twisting by sign, this becomes the unique line on which \(R_a\) acts by its sign character. The nonzero alternating vector on the equal-letter blocks is such a vector, and its column lengths are the parts of \(\mu\); it therefore belongs to \(S^{\mu^t}\). This identifies the two lines. The identification is equivariant, so translating a word changes the identification only by the corresponding sign and the position translate. ◻ In a tensor square, these two sign twists cancel. To make the passage through a projection explicit, let \(P:X\otimes X\longrightarrow W\) be a local coordinate projection onto its image. Its dual embeds \(W^*\) in \((X\otimes X)^*\). For a surviving pair word, its coordinate functional \(\epsilon\) satisfies \(\epsilon(Pz)=\epsilon(z)\), so its restriction belongs to this embedded copy of \(W^*\). Lemma 16 identifies it with a nonzero multiple of the product of the two alternating vectors on its equal-letter blocks. Its entire position-permutation span remains in \(W^*\). Finding a path among these products therefore proves support in \(W\) itself, since complex Specht modules are self-dual. In the next two lemmas, zero column lengths are omitted. Lemma 17 (Neighboring lengths). For every integer \(u\ge1\), the local component with column lengths \(u,u-1\), projected to outputs in \([u-1,u]\), contains every type in \(\mathcal H_4(2u-1)\). Proof. Its content is \((2^{u-1},1)\). The equal-letter blocks thus consist of \(u-1\) pairs and one singleton in each layer. Arrange the two pairings as one odd path. From its first-layer singleton, label that singleton \(u\), and label the successive first-layer pairs \(u-1,u-2,\ldots,1\). Label the second-layer pairs, in path order, \(1,2,\ldots,u-1\), and label its terminal singleton \(u\). At the path positions the outputs are alternately \(u\) and \(u-1\), with output \(u\) at both ends. The pair word survives the stated projection. By Lemma 16, its coordinate functional gives the odd-path tensor in the sign-twisted dual square. The twists cancel, and Lemma 12 gives every partition with at most four rows. Consequently all these types occur in the projected component. ◻ Lemma 18 (The two-path component). For every integer \(q\ge0\), the component with column lengths \(q+2,q\), projected to outputs in \([q,2q+3]\), contains \(\mathcal O_4(2q+2)\). If \(q\ge3\), inducing this support together with a trivial representation of size two contains every member of \(\mathcal H_4(2q+4)\). Proof. The content is \((2^q,1,1)\). In each layer there are pair labels \(1,\ldots,q\) and singleton labels \(q+1,q+2\). For any \(0\le p\le q\), make odd paths of sizes \(2p+1\) and \(2(q-p)+1\). On the first path use first-layer pair labels \(1,\ldots,p\) and second-layer pair labels \(q-p+1,\ldots,q\); on the second path use the complementary intervals. In path order, the first-layer pairs decrease and the second-layer pairs increase. Give the first path the initial first-layer singleton \(q+1\) and terminal second-layer singleton \(q+2\), and the second path the remaining singletons. Internal outputs are \(q\) or \(q+1\). Endpoint outputs are at least \(q\), and all outputs are at most \(2q+3\). If one path is a singleton, its two singleton labels satisfy the same bounds. The resulting pair word therefore survives the projection, and Lemma 16 identifies its coordinate with the product of the two path tensors. Let \(\nu\in\mathcal O_4(2q+2)\). Since its size is even, it has an even positive number of odd-height columns. Divide its columns into two groups, each of odd total size. These totals are \(2p+1\) and \(2(q-p)+1\) for an allowed \(p\). Arrange each group’s columns consecutively on the corresponding path. By Lemma 12, each contraction is a nonzero polynomial in a common ordered basis of the pair-letter dual space. Their product is also nonzero, so a common generic ordered basis gives a nonzero contraction of the two-path tensor with the dual tableau of \(\nu\). This proves the first assertion. For the second assertion, let \(\eta\in\mathcal H_4(2q+4)\). It has width greater than two when \(q\ge3\). We claim that some removable horizontal two-strip leaves an odd-height column. If \(\eta\) has no odd-height columns, removing any horizontal two-strip creates two. If it has at least four, such a strip cannot remove all of them. If it has exactly two, its width ensures that an even-height column also exists. Remove one cell from an odd-height column and one from an even-height column, choosing the rightmost column of each selected height. Their heights are different, and these removals preserve the weakly decreasing order of column heights. This is a horizontal two-strip and leaves two odd-height columns. Pieri now proves the desired containment. ◻ For the candidate diagram, the big component has \(T\) positions. When \(s=1\), Lemma 17 supplies \(\mathcal H_4(T)\). When \(s=2\), its two singleton columns can supply a trivial representation of size two at output one; Lemma 15 and Lemma 18 then supply \(\mathcal H_4(T+2)\). The big component’s outputs are at least \(M-1\) in both cases. Symmetric pairs and short stripsThe path calculations handle the long columns. For the short-column components we use a direct fact about repeated symmetric tensors. Its polynomial-module conclusion is Littlewood’s classical even-row decomposition of symmetric powers of a symmetric square [15]. The explicit contraction below also detects the prescribed tensor, which is what our coordinate projection requires. Lemma 19 (Repeated symmetric pairs). Let \(m\ge0\) be an integer, let \(E\) have dimension \(j\), and let \(H\in\mathop{\mathrm{Sym}}^2 E\) be nondegenerate. The position-permutation span of \(H^{\otimes m}\in E^{\otimes 2m}\) contains every member of \(\mathcal E_j(2m)\). Moreover, the polynomial \(\mathop{\mathrm{GL}}(E)\)-module \(\mathop{\mathrm{Sym}}^m(\mathop{\mathrm{Sym}}^2 E)\) contains the Schur module of each such partition. Proof. Choose an orthonormal basis for \(H\), so \(H=\sum_{i=1}^j e_i\otimes e_i\). A diagram with even row lengths has its columns paired into equal-height pairs. Place the two positions of each copy of \(H\) in corresponding cells of two paired columns. Contract against the usual column-alternating dual tensor. For a pair of columns of height \(h\), the two permutations must agree, their signs multiply to one, and their contribution is \(h!\). The entire contraction is the nonzero product of these factorials. Lemma 3 proves the first statement. For the second, use the natural embedding of \(\mathop{\mathrm{Sym}}^m(\mathop{\mathrm{Sym}}^2 E)\) into \(E^{\otimes 2m}\) with the \(m\) pairs of positions fixed, and choose the dual tableau placements as above. The same contraction proves that this \(\mathop{\mathrm{GL}}(E)\)-stable subspace has a nonzero projection to the indicated Schur-Weyl isotypic space. Complete reducibility then gives the required Schur module. ◻ An isolated length-\(k\) block, projected to constant output \(k\), has the pair labels \((a,k+1-a)\) for \(1\le a\le k\). Reversing one block’s labels shows that all terms in the projected tensor have the same sign. It is a nonzero symmetric tensor and hence supplies the trivial type of size \(k\). For several equal-length blocks, the projected product still has coefficients of one common sign. Its full position average is nonzero, supplying the trivial type of their total size. This proves the first row of Table 1. On an aligned length-two block, projection to constant output two gives, up to a common nonzero scalar, \[e_{12}\otimes e_{21}+e_{21}\otimes e_{12}.\] This is a nondegenerate symmetric form on the two-letter space \(\mathop{\mathrm{span}}(e_{12},e_{21})\). Thus \(b+1\) such blocks supply \(\mathcal E_2(c)\) by Lemma 19. Lemma 20 (An odd horizontal strip). Let \(c\) be positive and even, and let \(x\in\{1,3\}\) with \(x\le c\). Inducing the support \(\mathcal E_2(c)\) with the trivial representation of size \(x\) supplies \(\mathcal H_2(c+x)\). Proof. Write a target as \((\alpha,\beta)\), allowing \(\beta=0\). A precursor of size \(c\) with even row lengths has the form \((c-y,y)\) with \(y\) even. The horizontal-strip interlacing conditions are precisely \[ \max(0,\beta-x)\le y\le\min(\beta,c-\beta). \tag{15}\] The upper bound is at most \(c/2\), so this also guarantees that the precursor is a partition. If the interval contains zero, choose \(y=0\). Otherwise its lower endpoint is \(\beta-x>0\). When \(\beta\le c/2\), its length is \(x\ge1\). When \(\beta>c/2\), its length is \(c+x-2\beta=\alpha-\beta\), a positive integer because \(c+x\) is odd. The interval therefore contains an even integer. Pieri gives the assertion. ◻ In our applications, the size-one strip is a singleton column at output one, and the size-three strip is the length-three block at output three. These ranges are disjoint from output two, so Lemma 15 justifies the required inductions. The first-four-column boundWe now combine the local path and symmetric-pair supports. The aim is to replace each component by a list of trivial-factor sizes and then verify Young’s dominance inequalities for the entire target. For positive integers \(t,j\), let \(\operatorname{pack}(t,j)\) be the list of \(j\) integers whose entries differ by at most one and sum to \(t\); zero entries, if any, are omitted. Young’s rule implies that the support obtained by inducing trivial factors of these sizes is exactly \(\mathcal H_j(t)\). Indeed, every partition of \(t\) with at most \(j\) rows dominates the balanced \(j\)-part partition, whereas a partition with more than \(j\) rows fails its \(j\)th dominance inequality. Thus a local component supplying \(\mathcal H_j(t)\) may contribute \(\operatorname{pack}(t,j)\) to a list of trivial-factor sizes for a support calculation. Proposition 21 (The balance bound). Assume \(M\ge9\), and let \(h_1\ge h_2\ge\cdots\) be the column heights of \(\nu\vdash n\). If \[ \sum_{i=1}^4 h_i\le2M-2, \tag{16}\] then \(S^\nu\) occurs in \(S^L\otimes S^L\). The same conclusion holds if Equation (16) holds for the column heights of \(\nu^t\). Proof. Use the big pair, together with the two singleton columns when \(s=2\), to contribute four balanced parts of total \(T+2(s-1)\). When \(s=1\), retain its singleton column as a separate part one. If \(c\le M\), use the length-two columns to contribute the single part \(c\). Pair the middle lengths \(3,\ldots,M-2\) consecutively, leaving their smallest member alone if their number is odd. Each paired pair of neighboring lengths contributes four balanced parts by Lemma 17, and an unpaired length contributes itself. If \(c>M\), use the length-three block with the length-two blocks. Lemma 20 contributes two balanced parts of total \(c+3\), and now pair the remaining middle lengths \(4,\ldots,M-2\) in the same fashion. All output ranges used are disjoint: the singletons have output one, the length-two and length-three constructions use outputs two and three, a middle neighboring pair uses the interval between its two lengths, and the big pair uses outputs at least \(M-1\). Apply Lemma 15 first to these separate components. Then reassociate induction to combine the big pair with the output-one singleton component when \(s=2\), and to combine the output-two and output-three components when \(c>M\). The local support statements above now validate the specified packing; no disconnected output range is treated as a single component in the separation lemma. Call the resulting list of parts \(A\); its sum is \(n\). We record two uniform bounds: \[ |A|\ge2M-5, \qquad \max A\le M+1. \tag{17}\] A run of \(p\) middle lengths contributes \(2p-(p\bmod2)\ge2p-1\) parts. If \(c\le M\), there are \(p=M-4\) middle lengths and at least five other parts, giving at least \(2M-4\) parts in total. If \(c>M\), then \(M\) must be even: for odd \(M\), the bound \(r\le M/2\) gives \(c\le r+2\le M/2+2<M\). In the even case, \(p=M-5\) is odd, so the middle contribution is \(2M-11\), and the big and short components contribute at least six more parts. This proves the first bound. Untouched lengths and balanced middle parts are at most \(M\). The largest big-component part is at most \[\left\lceil\frac{T+2}{4}\right\rceil \le \left\lceil\frac{7M/2+3}{4}\right\rceil \le M+1,\] using \(r\le3M/2+2\). In the exceptional case \(c>M\), the largest short-component part is \[\left\lceil\frac{c+3}{2}\right\rceil =\left\lfloor\frac r2\right\rfloor+3 \le\left\lfloor\frac{3M}{4}\right\rfloor+4 \le M+1.\] The last inequality holds for even \(M\ge10\): it is equality at \(M=10\), and follows from \(3M/4+4\le M+1\) for \(M\ge12\). This proves the second bound in Equation (17). In particular, \[ \frac{n}{\max A}\ge\frac{N_M}{M+1} =\frac M2>\frac{2M-2}{4}. \tag{18}\] We verify the inequalities of Lemma 4. The target has at least four columns, since otherwise Equation (16) would give \(n\le2M-2<N_M\). Consequently \(h_1\le2M-5\), and the first inequality follows from \(P_A(1)=|A|\). The four big-component parts are each at least two, so \[P_A(2)\ge |A|+4\ge2M-1.\] This proves the inequalities for prefixes two and three as well. For \(k\ge4\), decreasing column heights and Equation (16) imply \[\sum_{i=1}^k h_i\le \min\left(n,\frac{k(2M-2)}4\right).\] Writing \(B=\max A\), the termwise bound \(\min(k,a)\ge ka/B\) for \(k\le B\) gives \[P_A(k)\ge \min\left(n,\frac{kn}{B}\right).\] Equation (18) proves every remaining dominance inequality. The constructed support therefore contains \(S^\nu\). Finally, since \(L\) is self-conjugate, its square is invariant under sign twist, so the same conclusion follows from the condition on \(\nu^t\). ◻ A three-row componentThe following construction is independent of the big pair and will be used in the cases \(M=10,12,14\). Proposition 22 (The short three-row component). Group the length-three block, the \(b+1\) length-two blocks, and the \(s\) singleton blocks in both layers, and project to outputs in \([1,3]\). The resulting module contains every partition in \(\mathcal P_3(c,s)\). This component can be induced independently with components on the remaining lengths \(4,\ldots,M-2\) and with the big pair. Proof. Align corresponding blocks in the two layers and use pair letters \(e_{aa'}\). Set \[p=e_{11},\quad q=e_{22},\quad u=e_{12},\quad v=e_{21}, \qquad a=e_{13},\quad d=e_{31}.\] These six pair letters form a basis of the space of outputs at most three. Each aligned length-two block contributes \[ H=p\otimes q+q\otimes p-u\otimes v-v\otimes u. \tag{19}\] Each singleton contributes \(p\). The length-three block has fixed output sum nine; since all its outputs are at most three, they must all equal three. Its contribution is, up to a nonzero scalar, the full symmetrization of \(a,d,q\). Thus the projected module contains the position orbit of \[ G=H^{\otimes(b+1)}\otimes p^{\otimes s} \otimes\sum_{\sigma\in S_3} z_{\sigma(1)}\otimes z_{\sigma(2)} \otimes z_{\sigma(3)}, \qquad (z_1,z_2,z_3)=(a,d,q). \tag{20}\] We want to vary the symmetric-pair, singleton, and triple factors independently so that Pieri applies to their tensor product. Let \(P=\mathop{\mathrm{span}}(p,q,u,v,a,d)\). Common linear maps from \(P\) provide the needed vectors without adding any symmetric-group types. More precisely, let \(E=\mathbb C^3\), let \(D=c+s+3\), and set \[N=\mathbb C[S_D]G,\qquad W=\sum_{\phi:P\to E}\phi^{\otimes D}(N) \subseteq E^{\otimes D}.\] Each \(\phi^{\otimes D}\) intertwines the position action, so every \(S_D\)-type in \(W\) already occurs in \(N\). Moreover, \(W\) is \(\mathop{\mathrm{GL}}(E)\)-stable, since postcomposing \(\phi\) with an element of \(\mathop{\mathrm{GL}}(E)\) gives another map in the sum. The choice \(\dim E=3\) restricts the Schur-Weyl types to partitions with at most three rows. Our immediate aim is to put \[ Q^{\otimes(b+1)}\otimes x^{\otimes s}\otimes y^{\otimes3} \tag{21}\] in \(W\) for generic independently chosen symmetric tensors \(Q\in\mathop{\mathrm{Sym}}^2 E\) and vectors \(x,y\in E\). Here \(Q\) will be the image of \(H\), while \(x\) and \(y\) control the singleton and triple factors. The same source letter \(q\) occurs in both \(H\) and the triple, so these choices require a construction. Identify \(Q\) with its symmetric matrix and assume it is invertible. We seek a map with \(p\mapsto x\) and \(q,a,d\mapsto ty\) for some \(t\ne0\). The requirement \(H\mapsto Q\) then asks that \(t(xy^{\mathsf T}+yx^{\mathsf T})-Q=UV^{\mathsf T}+VU^{\mathsf T}\) for the still unchosen images \(U,V\) of \(u,v\). Since the right side has rank at most two, we choose \(t\) to make the left side singular. Put \[\alpha=x^{\mathsf T}Q^{-1}y, \qquad \beta=x^{\mathsf T}Q^{-1}x, \qquad \gamma=y^{\mathsf T}Q^{-1}y.\] The determinant lemma gives \[ \det\bigl(Q-t(xy^{\mathsf T}+yx^{\mathsf T})\bigr) =\det(Q)\bigl(1-2\alpha t+ (\alpha^2-\beta\gamma)t^2\bigr). \tag{22}\] The quadratic coefficient is generically nonzero: for \(Q=I\), \(x=e_1\), and \(y=e_2\), it equals \(-1\). The constant coefficient is one. Over \(\mathbb C\), there is therefore a nonzero root \(t\). The symmetric matrix \[R=t(xy^{\mathsf T}+yx^{\mathsf T})-Q\] has rank at most two. Every such symmetric matrix over \(\mathbb C\) can be written \(UV^{\mathsf T}+VU^{\mathsf T}\): diagonalize it by congruence, write it as \(AA^{\mathsf T}+BB^{\mathsf T}\) with at most two summands, and take \(U=(A+\mathrm i B)/\sqrt2\) and \(V=(A-\mathrm i B)/\sqrt2\). Consequently the common linear map \(\phi:P\to E\) given by \[ p\longmapsto x,\quad q\longmapsto ty,\quad u\longmapsto U,\quad v\longmapsto V,\quad a\longmapsto ty,\quad d\longmapsto ty \tag{23}\] sends \(H\) to \(Q\) and the symmetric triple to \(6t^3y^{\otimes3}\). Its image of \(G\) is therefore \(6t^3\) times Equation (21). Since \(t\ne0\) and \(W\) is a linear space, the desired tensor belongs to \(W\). Varying the generic choices of \(Q,x,y\) shows that \(W\) contains the naturally embedded space \[ \mathop{\mathrm{Sym}}^{b+1}(\mathop{\mathrm{Sym}}^2 E)\otimes\mathop{\mathrm{Sym}}^s E\otimes\mathop{\mathrm{Sym}}^3 E. \tag{24}\] Indeed, a linear functional annihilating \(W\) gives a multihomogeneous polynomial in \((Q,x,y)\) when evaluated on Equation (21). It vanishes on a dense set and hence identically. Pure powers span each symmetric power, proving the containment in Equation (24). By Lemma 19, the first factor contains the Schur module for every member of \(\mathcal E_3(c)\). Two applications of Pieri give every Schur module indexed by \(\mathcal P_3(c,s)\). Schur-Weyl duality transfers these \(\mathop{\mathrm{GL}}(E)\)-types to the corresponding \(S_D\)-types in \(W\), hence in \(N\) and in the local projected module. This proves the asserted support. Finally, the remaining middle components can use outputs at least four and at most \(M-2\), while the big pair uses outputs at least \(M-1\). They are disjoint from \([1,3]\), so their independence follows from Lemma 15. ◻ Using the local supports in capacity testsThe capacity tests use these local supports in two ways. First, they retain partitions with at most three rows whose entire dominance upper set lies in \(\mathcal P_3(c,s)\). The precise condition in Equation (32) guarantees the full induced-trivial support associated with each retained partition’s row lengths. Combining these lower supports with balanced middle parts gives the dominance bounds used below. Second, they allocate at most four cells from each target column to the big pair, with total allocation \(T\) and at least one odd allocation when \(s=2\). If the residual diagram is supplied by the other components, Lemma 5 combines the columnwise allocations to give the target. The exact allocation test and its support justification are given in Lemma 27. Exhaustion of the capacity conditionsThe balance and band constructions cover complementary shapes. A small sum of the first four rows or columns gives balance support. Otherwise both orientations have room for the path in the band criterion. Failure of that criterion then bounds both row and column prefixes, forcing an upper bound on the whole diagram’s area. We use this contradiction for large parameters and for most intermediate parameters. Nineteen pairs remain, for which we enumerate targets using the proved support tests. Throughout this section, \(M\ge9\) and \(r>0\), and \(M,r,b,s,L,K\) have the meanings fixed above. Put \[c=2b+2,\qquad T=K+r, \qquad H_i=\sum_{a=1}^i h_a,\qquad W_j=\sum_{a=1}^j w_a,\] where \(h\) and \(w\) are the column heights and row lengths of a target partition of \(n\), extended by zeros. Since \(L\) is self-conjugate, we may apply every test to either orientation of the target. By Proposition 21, any target not already supplied satisfies \[ H_4\ge K,\qquad W_4\ge K. \tag{25}\] Two opposite prefixesLemma 23. Let \(i,j\ge1\). If \(H_i\le U\) and \(W_j\le V\), then \[ n\le\max\left(U+V-1, \left\lfloor\frac{UV}{ij}\right\rfloor\right). \tag{26}\] Proof. If the cell in row \(j\), column \(i\) is absent, no cell lies outside both the first \(i\) columns and the first \(j\) rows. Those two strips cover the diagram and share its first cell, giving \(n\le U+V-1\). Otherwise their intersection is an \(i\)-by-\(j\) rectangle. Put \(x=h_i\ge j\) and \(y=w_j\ge i\). Every cell outside both strips has row index at most \(x\) and column index at most \(y\), so the remaining corner has at most \((x-j)(y-i)\) cells. Since \(ix\le U\) and \(jy\le V\), \[n\le U+V-ij+(x-j)(y-i) \le U+V-ij+(U/i-j)(V/j-i)=\frac{UV}{ij}.\] Integrality gives the asserted floor. ◻ Lemma 24. If \(M\ge22\), a target satisfying Equation (25) passes the band test in one of its two orientations. Proof. The choice \(d=4\), \(q=8\) is admissible in Proposition 8, because \(dq+7+d=43\le2M-1\). If the band test fails in both orientations, then \[\sum_{i=1}^8\max\left(0,\left\lfloor\frac{h_i-4}{2}\right\rfloor\right) \le r-1,\] and likewise for \(w\). The termwise inequality \[ x\le2\max\left(0,\left\lfloor\frac{x-f}{2}\right\rfloor\right)+f+1 \quad(x\in\mathbb Z_{\ge0}) \tag{27}\] therefore gives \(H_8,W_8\le S:=2r+38\). Lemma 23 implies \(n\le\max(2S,S^2/64)\). On the other hand, \[n-2S\ge\frac{M^2-5M-160}{2}>0, \qquad 64n-S^2\ge23M^2-28M-1508>0.\] For the first inequality substitute \(r\le3M/2+2\); for the second write \(64n-S^2=32M(M+1)-4r^2-24r-1444\) and make the same substitution. Both displayed polynomials are positive at \(M=22\) and increase thereafter. This is a contradiction. ◻ For the remaining intermediate cases, retain the least adequate core height in each orientation. Let \(d\) be least with \(H_d\ge K\) and \(e\) least with \(W_e\ge K\); by Equation (25), both belong to \(\{1,2,3,4\}\). Define \[q_f=\min\left(8,\left\lfloor\frac{2M-8-f}{f}\right\rfloor\right) \quad(1\le f\le4),\] and the following finite set of pairs consisting of a prefix length and an upper bound for its sum: \[ \begin{split} \mathcal D_{M,r}(d,e) ={}&\{(d-1,2M-2):d>1\}\\ &{}\cup\{(q_f,2r-2+q_f(f+1)):e\le f\le4,\ q_f>0\}. \end{split} \tag{28}\] If neither band test succeeds, every pair \((i,U)\) in this set satisfies \(H_i\le U\). Indeed minimality of \(d\) gives its first element, and \(W_f\ge K\) for \(f\ge e\); failure of the band test with this \(f,q_f\), followed by Equation (27), gives the other elements. Similarly, \(\mathcal D_{M,r}(e,d)\) bounds the row prefixes. The indices are crossed because enough room in the first \(f\) rows allows the band criterion to test the excess heights of columns, and conversely after transposition. Lemma 25. The first-four balance test and the band tests cover every target when \(9\le M\le21\), \(r>0\), except possibly for \[ (M,r)\in \{(10,8),\ldots,(10,17)\} \cup\{(12,14),\ldots,(12,20)\} \cup\{(14,22),(14,23)\}. \tag{29}\] Proof. Use \(1\le r\le\lfloor M/2\rfloor\) for odd \(M\) and \(1\le r\le3M/2+2\) for even \(M\). After removing Equation (29), there are 177 pairs \((M,r)\). For each pair and each of the 16 choices of \(d,e\), the following integer inequality holds: \[
\min_{\substack{(i,U)\in\mathcal D_{M,r}(d,e)\\
(j,V)\in\mathcal D_{M,r}(e,d)}}
\max\left(U+V-1,\left\lfloor\frac{UV}{ij}\right\rfloor\right)<n.
\tag{30}\] This is a bounded integer calculation involving 2,832 assertions. The function The nineteen exceptional casesFor the remaining parameters we first assemble induced supports of full degree. When these do not settle a target, we allocate cells to a constituent with at most four rows on the big pair and test the residual diagram against the lower components. A recursive enumeration checks that every target is covered by one of these supports or by the band criterion. All arithmetic is exact; the tests certify the sufficient support conditions, rather than approximate Kronecker coefficients. For a multiset \(A\) of positive integers, write \[|A|=\text{its number of entries},\qquad \|A\|=\sum_{a\in A}a,\qquad P_A(k)=\sum_{a\in A}\min(k,a).\] By Lemma 4, inducing trivial representations of the sizes in \(A\) has precisely the support of partitions \(\gamma\vdash\|A\|\) satisfying \[ \sum_{j=1}^k\gamma_j^t\le P_A(k)\quad(k\ge1). \tag{31}\] We call this the support associated with \(A\). For integers \(u,j>0\), let \(\operatorname{pack}(u,j)\) be the multiset of \(j\) integers \(\lfloor(u+i)/j\rfloor\), \(0\le i<j\), with zero entries omitted. This is the balanced division from Section 5. Every use below has \(u\ge j\), so the code’s fixed-length tuples have no zero entries. The implementation in Appendix 9 follows these constructions with integer lists and sets. For a consecutive increasing list \(J\) of middle lengths, let \(\mathcal P(J)\) be the following collection of multisets. If \(|J|\) is even, pair consecutive entries \(u,u+1\), replacing each pair by \(\operatorname{pack}(2u+1,4)\). If \(|J|\) is odd, choose an entry with an even number of predecessors, retain it as a single entry, and perform this pairing separately on the two remaining even lists. The empty list gives the empty multiset. Thus every member of \(\mathcal P(J)\) has the support supplied by those middle components, by Lemma 17 and Lemma 15. The function Full-degree supports.Construct a collection \(\mathcal B\) as follows. For each row of Table 2 having the prescribed value of \(s\), and each \(R\in\mathcal P(J)\), take the multiset union of \(R\) with the displayed parts. The indices \((t,x)\) match those in the code. Omit a row if \(x>c\).
Here \(t=2\) combines the two singleton cells with the big pair; \(x=1\) combines the singleton with the length-two blocks; and \(x=3\) combines the length-three block with those blocks. The singleton left in the first row contributes the part \(1\). Every \(A\in\mathcal B\) has \(\|A\|=n\), and its support is supplied by Lemmas 17, 18, and 20, combined by Lemma 15. Supports below the big pair.These full-degree supports sometimes settle the target immediately. For the remaining targets we seek a columnwise allocation \[h_i=\delta_i+(h_i-\delta_i),\qquad 0\le\delta_i\le\min(4,h_i),\qquad \sum_i\delta_i=T,\] with at least one odd \(\delta_i\) when \(s=2\). The big pair supplies the partition with column-height multiset \((\delta_i)\) by Lemmas 17 and 18. We therefore need lower-component support for the residual partition, whose column heights are the decreasing rearrangement of \((h_i-\delta_i)\) and whose size is \(n-T\). Lemma 5 will combine these two constituents. Next construct a collection \(\mathcal C\) for the components other than the big pair. Write \(u=c+3+s\), and let \(\mathcal G\) consist of the partitions of \(u\) with at most three rows obtained from a partition of \(c\) with even row lengths by successively adding horizontal strips of sizes \(3\) and \(s\). Proposition 22 supplies all of \(\mathcal G\) on the blocks of lengths at most three. For partitions of equal size, write \(\alpha\unrhd\beta\) for dominance of row lengths. Retain a partition \(a\) of \(u\) with at most three rows if \[ \alpha\unrhd a\quad\Longrightarrow\quad\alpha\in\mathcal G \quad\text{for every $\alpha\vdash u$ with at most three rows}. \tag{32}\] No partition of more than three rows can dominate \(a\), since its first three rows have sum strictly less than \(u\). Thus Lemma 4 says that the entire support associated with the row lengths of \(a\) is available on the lower blocks. For each retained \(a\) and each \(R\in\mathcal P((4,5,\ldots,M-2))\), put their multiset union in \(\mathcal C\). Every resulting multiset has total \(n-T\) and supplied support. We may delete \(A\) from \(\mathcal C\) if its decreasing rearrangement strictly dominates that of another member \(B\): every partition dominating \(A\) also dominates \(B\), so this deletion does not shrink the union of their supports. Deleting all such members simultaneously is harmless: every descending chain in this finite strict dominance order terminates at a retained member whose support contains the supports of its predecessors. These definitions explain the construction of Safe pruning of partial diagrams.The available supports have now been specified. The next bound determines when an initial column list already certifies every possible completion. Lemma 26. Let \(A\) be a nonempty multiset of positive integers with total \(N\), and let \(a=(a_1,\ldots,a_l)\) be a nonempty prefix of the column-height list of a partition of \(N\). If \[ \min\left(N,\sum_{v=1}^{\min(i,l)}a_v+(i-l)_+a_l\right) \le P_A(i)\qquad(1\le i<\max A), \tag{33}\] then every completion of \(a\) has the support associated with \(A\). For a complete column-height list followed by a zero, these inequalities are equivalent to that support condition. Proof. Every unknown entry is at most \(a_l\). The left side of Equation (33) is therefore an upper bound for the corresponding prefix sum of any completion. For \(i\ge\max A\), the right side is already \(N\), so the omitted inequalities hold automatically. If the prefix includes a final zero, it has no nonzero extension and the upper bounds are exact. ◻ The predicate Certifying a completed allocation.If prefix pruning does not settle a target, the next routine allocates cells to the big pair and tests the residual diagram against the lower supports. Only successful allocations are used. Lemma 27. If the code returns Proof. The routine constructs allocations \(\delta_i\) from the original columns, with \(0\le\delta_i\le\min(4,h_i)\). If \(s=1\), the protected index is the final zero, to which it allocates zero. If \(s=2\), it chooses a column and fixes its allocation to \(z=1\) or \(3\); no later step changes this column. Every other step decreases the residual of an eligible column by one, stopping at residual \(\max(0,h_i-4)\). Entering the loop’s The residual columns are nonnegative and have total \(n-T\), the total of every member of \(\mathcal C\). A successful Exhausting the targets.We have justified what each successful test certifies. It remains to show that the recursive search visits every target not already certified. Lemma 28. For each pair in Equation (29), the procedure Proof. At depth \(k\), the lists \(a,b\) give the first \(k\) full row and column lengths, and Conversely, the row and column constraints make the corresponding Frobenius arm and leg lists strictly decreasing and nonnegative. Every completed path consequently specifies a partition. At The recursion terminates since a nonempty hook removes at least one cell. A successful The calls Proposition 29. For every admissible \(M\ge9\) and \(r>0\), every target partition of \(n=N_M+2r\) is supplied by the band and balance constructions for \(L\). Small degrees by exact character calculationThe capacity calculations apply to the fixed large-degree candidate. For the remaining small degrees we instead calculate exact character residues, checking one self-conjugate candidate against every target. A nonzero residue certifies a positive integer multiplicity; no estimate or reconstruction of that integer is needed. We give the invariant of the calculation and its implementation bounds. The source is reproduced in Appendix 10. Proposition 30. For every integer \(1\leq n\leq64\) other than \(2,4,9\), there is a self-conjugate partition \(\lambda\vdash n\) such that \[g(\lambda,\lambda,\nu)>0\qquad\text{for every }\nu\vdash n.\] A recurrence for the entire coefficient vectorWrite \(\chi^\lambda(\rho)\) for the character of \(S^\lambda\) on the conjugacy class of cycle type \(\rho\). If \(m_j(\rho)\) is the number of parts of size \(j\), put \(z_\rho=\prod_j j^{m_j(\rho)}m_j(\rho)!\). The character inner-product formula gives \[ g(\lambda,\lambda,\nu) =\sum_{\rho\vdash n} \frac{\chi^\lambda(\rho)^2\chi^\nu(\rho)}{z_\rho}, \tag{34}\] since symmetric-group characters are real. The program computes this sum for every \(\nu\vdash n\) at once, recursively selecting the parts of \(\rho\) in decreasing order. Calculations take place in \(\mathbb F_p\), where \(p\in\{1000000007,1000000009\}\). Both numbers are prime, as is checked by trial division through \(31622\). Since \(n\le64<p\), every denominator in (34) is invertible modulo \(p\). All character values and vectors below are interpreted in this field. For \(s\geq0\), let \(\mathcal P_s\) be the partitions of \(s\), including the empty partition when \(s=0\). For a vector \(v=(v_\lambda)_{\lambda\in\mathcal P_s}\), write \[\chi_v(\rho)=\sum_{\lambda\in\mathcal P_s}v_\lambda\chi^\lambda(\rho).\] Let \([\lambda]\) denote the coordinate vector with entry one at \(\lambda\) and zero elsewhere, so \(\chi_{[\lambda]}=\chi^\lambda\). A border strip is an edge-connected skew diagram containing no \(2\times2\) square; its leg length is one less than its number of occupied rows. Let \(R_k\) denote the matrix that adds a border strip of size \(k\), with entry \((-1)^{\text{leg length}}\) for each permitted addition. Its source and target degrees will be understood from context. For \(v\) of degree \(s\) and \(\rho\vdash s-k\), the Murnaghan–Nakayama rule [15][10] gives \[ \chi_{R_k^{\mathsf T}v}(\rho) =\chi_v((k)\cup\rho), \qquad \sum_\mu (R_k)_{\nu\mu}\chi^\mu(\rho) =\chi^\nu((k)\cup\rho). \tag{35}\] In a call The base cases are exact. At degree zero the answer is \(v_\varnothing^2\). If \(0<s<k\), no size-\(k\) part remains, so \(F_{s,k,e}(v)=F_{s,s,0}(v)\): the factor involving \(e\) in (36) is then \(e!/e!=1\). Calls with \(k=1\) always have \(e=0\), since this case is returned immediately rather than recursed into. Writing \(d_s=(\dim S^\nu)_{\nu\vdash s}\) gives \[
F_{s,1,0}(v)=\frac{1}{s!}\,d_s(d_s^{\mathsf T}v)^2.
\tag{38}\] The vector \(d_s\) is computed by successive applications of \(R_1\) to the empty partition: the resulting Young-lattice chains are standard tableaux. The array named The recurrence and base cases prove by induction on \((s,k)\) that Exact implementationPad a partition \(\lambda\) with zero parts to length \(64\), and encode it by the beta set \[B_\lambda=\{64+\lambda_j-j:1\leq j\leq64\} \subseteq\{0,\ldots,127\}.\] There are \(64\) beads, and their positions sum to \(2016+|\lambda|\). The partition enumeration starts with \(B_\varnothing=\{0,\ldots,63\}\) and moves the \(j\)th bead to \(64+\lambda_j-j\) while selecting nonincreasing parts. It enumerates every partition exactly once. Moving a bead from \(x\) to an empty position \(x+k\) adds a border strip of size \(k\). If the bead passes \(r\) other beads, the strip occupies \(r+1\) rows, giving the sign \((-1)^r\). For the bit mask \(b\) of \(B_\lambda\), the expression \[b\mathbin{\&}((\mathord{\sim}b)\mathbin{\gg}k)\] selects just the permitted starting positions. Its top \(k\) bits are zero, so every selected position satisfies \(x+k\leq127\). With \(m=2^x\) and \(t=2^{x+k}\), the mask \(t-2m\) has exactly the bits strictly between the two positions. Thus The finite beta-set identity \[B_{\lambda^t}=\{0,\ldots,127\}
\setminus\{127-x:x\in B_\lambda\}\] shows that We record the implementation bounds to make the use of machine arithmetic explicit. Residues lie in \([0,p-1]\); their sum is at most \(2000000016\), within a signed \(32\)-bit integer, and their product fits a signed \(64\)-bit integer. All mask shifts are between \(0\) and \(127\) and use unsigned \(128\)-bit arithmetic. The possible final enumeration argument The partition-count recurrence, adding parts of sizes \(1,2,\ldots,64\), gives \[|\mathcal P_{64}|=1741630, \qquad \sum_{s=0}^{64}|\mathcal P_s|=12308139.\] The largest padded hash table has \(4194304\) entries and is at most half full. Hash arithmetic wraps only in unsigned types; collisions are resolved by equality of complete \(128\)-bit masks. To see that every queried mask occurs in its table, sort its \(64\) occupied positions as \(b_1>\cdots>b_{64}\geq0\). The numbers \(b_j-64+j\) are nonnegative and nonincreasing, hence form a partition. Moving one bead by \(k\) increases their sum by \(k\), so the lookup is in precisely the enumerated target degree. Conversely, a size-at-most-\(64\) partition never requires a position outside the mask. Proof of Proposition 30. For each self-conjugate candidate \(\lambda\), the program evaluates (39) modulo one or, if needed, both of the two primes. Its array The recorded calculation compiled the source in Appendix 10 with GNU C++ 13.3.0 and executed it for exactly the \(61\) eligible degrees \(\{1,3,5,6,7,8\}\cup\{10,11,\ldots,64\}\). Every invocation returned success. The complete execution record is ; records the source and log identities. These successful invocations establish the required existence statement in every eligible degree of the finite range. The record contains exit statuses rather than the selected partitions; their meaning is the successful return condition of the source just analyzed. ◻ Reproducibility and independent checksThe files Each child has a \(1536\)-MiB address-space limit, a \(300\)-second wall limit and a \(301\)-second CPU limit. The driver returns failure if any child fails or times out. The complete expected set of eligible degrees must be checked: the three excluded degrees are deliberately skipped by the C++ program and their zero exit statuses do not certify a square. As an independent check of the implementation, the dump harness exposes every square vector at degrees \(1\) through \(12\), together with dimensions, self-conjugacy flags, and a nontrivial mixed input vector. The Python verifier computes integer characters by enumerating contained grid diagrams and testing border strips for connectivity and the absence of a \(2\times2\) square. Thus its strip enumeration and signs do not reuse the bead implementation. It checks character orthogonality, computes exact integer Kronecker coefficients using conjugacy-class sizes, and compares every exposed coefficient modulo both primes. Every comparison passed, including the dimensions and mixed vectors; the exact calculation also finds no covering square in degrees \(2,4,9\). The dump harness ran through degree \(12\) with undefined-behavior sanitization enabled and reported no failure. These checks are recorded in and can be repeated with Before comparing coefficients, the Python verifier requires every expected record to occur exactly once, including every square vector under both moduli. Its fault-injection tests reject missing or duplicate records even when the child exits successfully. These tests can be run with Completion of the proofProof of Theorem 1. If the largest parity-compatible staircase index is less than nine, Lemma 7 gives \(n\le64\). Proposition 30 supplies one self-conjugate partition whose square contains every irreducible, for every eligible degree in this range. Otherwise choose the single self-conjugate partition \(L\) of (6). If \(r=0\), it is a staircase and Theorem 6 applies. Suppose \(r>0\), and fix any target partition of \(n\). Proposition 29 shows that this target, or its transpose, satisfies one of the proved band or balance support tests. Hence it occurs in \(S^L\otimes S^L\), using transpose invariance from Lemma 7 when necessary. Since \(L\) depends only on \(n\), the same square contains every target. This proves the theorem. ◻ The finite calculations in this proof serve two distinct purposes. The capacity programs exhaust the explicitly listed intermediate parameters of the fixed family \(L\); the character program proves existence directly in all eligible degrees at most \(64\). Neither calculation assumes that the fixed family works beyond a numerically observed threshold. Beyond the listed finite ranges, the sufficient tests and diagram-area inequalities give the proof. Exact capacity verificationThe following standalone Python 3 file, , contains both finite capacity checks. The mathematical justification of its tests and pruning rules is in Section 6. From the
Assertions must be enabled: do not use Exact square-coefficient verificationThis is the complete C++ verifier justified in Section 7. Compile with a compiler supporting
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