Disproof of Xi’s conjecture on the representation dimension of idempotent corners

Introduction

Auslander introduced representation dimension in 1971 to measure how far an Artin algebra is from finite representation type [1]. Let $C$ be a finite-dimensional algebra over a field $k$. Write $D = \operatorname{Hom}_{k}(-,k)$, let $DC$ be the dual of the right regular module, and let $\operatorname{add} M$ consist of the direct summands of finite direct sums of copies of $M$, up to isomorphism. Write $\operatorname{pd}_{C}$ for projective dimension over $C$ and $\operatorname{gl.dim} C$ for its supremum over $C$-modules. The representation dimension is

The infimum ranges over finite-dimensional left $C$-modules satisfying the displayed condition, called generator-cogenerators. For Artin algebras the definition uses duality over the commutative Artinian base. Auslander’s criterion says that $\operatorname{repdim} C \le2$ exactly when there are finitely many indecomposable finite-dimensional $C$-modules up to isomorphism. Xi’s early work gave upper bounds for tensor products and triangular matrix algebras and raised the finiteness question [6, 7]. Iyama proved finiteness for Artin algebras [2]. Rouquier proved that the exterior algebra on an $n$-dimensional vector space has representation dimension $n+1$ for $n \ge1$ [5 Theorem 4.1].

Xi traced to Auslander the question of how representation dimension changes on passing to the endomorphism algebra of a finitely generated projective module [8 §7]. If $A$ is an Artin algebra and $e^{2}=e \in A$, right multiplication identifies $eAe$ with $\operatorname{End}_{A}(Ae)^{\operatorname{op}}$. Indeed, $f \in\operatorname{End}_{A}(Ae)$ is determined by $f(e) \in eAe$, and $f(ae)=af(e)$. Duality exchanges left and right generator-cogenerators and reverses their endomorphism rings. Since left and right global dimensions agree for Artin algebras, the comparison for $Ae$ is the comparison between $\operatorname{repdim}(eAe)$ and $\operatorname{repdim} A$. Xi proved the inequality when the two adjoints of $X \mapsto eX$, namely $Ae \otimes_{eAe} -$ and $\operatorname{Hom}_{eAe}(eA,-)$, are isomorphic [8 Theorem 7.2]. He conjectured that for every Artin algebra $A$ and idempotent $e$,

His problem list records the Artin-algebra formulation in January 2003 and a finite-dimensional formulation in December 2012 [9]. He restated the conjecture in [10 p. 40].

When $A$ is representation-finite, so is $eAe$. Indeed, every finite $eAe$-module $Y$ is the image under $e(-)$ of $Ae \otimes_{eAe} Y$. By Krull–Schmidt, every indecomposable $Y$ therefore occurs as a summand of the image of one of the finitely many indecomposable $A$-modules. For $\operatorname{repdim} A \le3$, Zhang and Zhang proved that $eAe$ has finite finitistic dimension, the supremum of the finite projective dimensions of its finitely generated modules [11]. Psaroudakis proved $\operatorname{repdim}(eAe) \le3$ in the same range by studying recollements of module categories [4 Corollary 5.2.5]. The first possible increase thus occurs at ambient representation dimension four.

A finite-dimensional $k$-algebra is Frobenius if it admits a linear functional $\lambda$ for which the pairing $(x,y) \mapsto\lambda(xy)$ is nondegenerate.

In the proof $k$ is an arbitrary field. Algebras are unital and finite-dimensional over $k$, and modules are finite-dimensional left modules, except for families over parameter rings. Opposite endomorphism rings make $\operatorname{Hom}_{\mathcal{C}}(M,-)$ left-module-valued.

The difference $L - 2$ is unbounded. The lower bounds use products of first-order extensions. For the corner, the calculation is made on the projective and injective summands present in the additive closure of every generator-cogenerator. The upper bound for the corner uses radical quotients together with modules defined by coefficient multiplication.

The algebras

For the rest of the paper fix $L \ge3$ and put

The indeterminate $t$ is central. Write $J = \bigoplus_{a=1}^{L-1} t^a\Sigma$. This ideal is nilpotent with quotient $k$, so it is the Jacobson radical of $B_L$. Since $1 \in\Sigma$, $J^j = \bigoplus_{a=j}^{L-1} t^a\Sigma$ for $1 \le j \le L$, with the empty sum zero.

Let $\tau: \Sigma\to k$ take the coefficient of $x_0\cdots x_{2L-3}$ in the squarefree monomial basis. The pairing $(a,b) \mapsto\tau(ab)$ is nondegenerate because each monomial pairs with its complement.

Give $DB_L$ the bimodule action $(bfc)(x) = f(cxb)$ for $b,c,x \in B_L$ and $f \in DB_L$. On $(B_L \times B_L) \oplus DB_L \oplus DB_L$, define $A_L$ by

Here $b_i,c_i \in B_L$ for $i = 0,1$ and $f,g,h,j \in DB_L$. The first dual copy has left action from the first factor and right action from the second factor of $B_L \times B_L$; the second has the actions reversed. The bimodule identities give associativity of the square-zero extension. The unit is $(1,1,0,0)$, and the corner for $e=(1,0,0,0)$ identifies with the first copy of $B_L$. The functional $\lambda(b_0,b_1,f,g)=f(1)+g(1)$ defines the pairing

Each coordinate pairs with its full dual, so $A_L$ is Frobenius. The pairing identifies its left regular module with the dual of its right regular module by $y\mapsto(x\mapsto\lambda(xy))$. Thus the regular module is injective.

Proof. Consider the ideal

Its quotient is $k\times k$. A product in $R$ with two off-diagonal factors is zero, since the off-diagonal bimodule has square zero. For $n\geq1$, the diagonal blocks of $R^n$ are $J^n$, and each off-diagonal block is

The sum is over nonnegative $i,j$, with $J^0=B_L$. If $i+j\geq L$, then $(afb)(x)=f(bxa)=0$ for $a\in J^i,b\in J^j$, because $bxa\in J^{i+j}=0$. Hence $R^{L+1}=0$, and the semisimple quotient gives $R=\operatorname{rad} A_L$. Choose $0\ne a\in J^{L-1}$ and $f\in DB_L$ with $f(a)\ne0$. Then $af$ is a nonzero element of an off-diagonal copy of $DB_L$ in $R^L$, since $(af)(1)\ne0$.

The Loewy length, the least exponent killing the radical, is therefore $L+1$. We give the radical-layer argument for the self-injective Loewy-length bound [5 Proposition 3.9].

The idempotents $(1,0,0,0)$ and $(0,1,0,0)$ give two nonisomorphic indecomposable projective left modules $P_1,P_2$, since $A_L/R\cong k^2$. They are injective. Each has a simple essential socle, so every nonzero $R^aP_i$ is indecomposable. For each $i$ the dimensions of the nonzero radical powers strictly decrease by Nakayama’s lemma. For different $i$ their socles differ because the injective hulls $P_1,P_2$ differ. Put

The projective-injective summands at $a=0$ make $M$ a generator-cogenerator.

A map $R^bP_j\to R^aP_i$, composed with inclusion into $P_i$, extends to $P_j$ by injectivity. Its image on $R^bP_j$ lies in $R^bP_i$, and hence in $R^{a+1}P_i$ if $b>a$. If $b=a$ and the original map is noninvertible, its extension is noninvertible, since an isomorphism restricts to an isomorphism of the layers. A map between these indecomposable projectives that is nonzero on the target top is surjective by Nakayama’s lemma, hence splits and is an isomorphism. Thus the noninvertible extension has image in $RP_i$, and its restriction again lands in $R^{a+1}P_i$.

For each nonzero $R^aP_i$, let $T$ be the simple top of $H(R^aP_i)$, and put $\Delta=H(R^aP_i)/H(R^{a+1}P_i)$. Both Hom modules are projective, so $\operatorname{pd}_{\Lambda}\Delta\leq1$. By the factorization just proved, the kernel of $\Delta\to T$ is supported only on the idempotents for the summands $R^bP_j$ with $b<a$. Applying these idempotent functors to a composition series shows that the kernel has only simple composition factors indexed by such summands. At $a=0$ the kernel is zero and $T=\Delta$. Inductively the factors have projective dimension at most $a$, and so does the kernel by the long exact Ext sequences of its composition series. Thus $T$ has projective dimension at most $a+1$. Every simple has projective dimension at most $L+1$, giving $\operatorname{gl.dim}\Lambda\leq L+1$. □

Parameter extensions

Proof. The proof uses the parameter-lattice method of [3]. Fix a generator-cogenerator $M$, adding $G$ as a summand if necessary. This preserves $\operatorname{add} M$ and the Morita class of its endomorphism ring. Write $\Lambda= \operatorname{End}_{C}(M)^{\operatorname{op}}$ and let $p \in\Lambda$ project onto $G$. Then $p\Lambda p = \Gamma$. The functor $Y \mapsto pY$ is exact and sends $\operatorname{Hom}_{C}(M,X)$ to $\operatorname{Hom}_{C}(G,X)$.

In a finite free presentation $C^a \to C^b \to M \to0$, application of $\operatorname{Hom}_{C}(-,X_R)$ identifies $\operatorname{Hom}_{C}(M,X_R)$ with the kernel of a matrix between finite free $R$-modules. If its rank over the fraction field is zero, the matrix vanishes. Otherwise, write $s$ for that rank and invert a nonzero determinant of an $s$-by-$s$ submatrix. After permutations the matrix has blocks

with $U$ invertible. Row and column operations give the blocks $U$ and $Z-WU^{-1}V$. The second vanishes over the fraction field by the rank condition, hence over the localization. The kernel is thus free and commutes with every base change on this principal open.

Choose a point $\alpha$ of this open over $K=k$ if $k$ is infinite, or over a sufficiently large finite extension $K/k$ otherwise. Such a point exists because a nonzero polynomial cannot vanish at all points of a field with more elements than its degree in each variable. Flat base change applied to a finite presentation gives

Let $Y=\operatorname{Hom}_{C_K}(M_K,X_\alpha)$. The two base changes to $K$ and to the dual numbers $K[\epsilon]/(\epsilon^2)$ identify the coordinate extensions of $Y$ with those from the localized Hom family. They are exact, since the Hom family is free over the localization and the dual-number extension is obtained from

with the first map given by multiplication by $\epsilon$. Multiplication by $p$ sends these extensions to the given coordinate extensions for $G_K$. Exactness preserves Yoneda splicing, so the product in $\operatorname{Ext}_{\Lambda_K}^{d}(Y,Y)$ is nonzero.

Embed $X_\alpha$ into a finite injective $I^0$ and its cokernel into a finite injective $I^1$. Apply $\operatorname{Hom}_{C_K}(M_K,-)$ and call the cokernel of the induced map $Q$. This gives

Both $P_j$ are projective because $DC_K \cong(DC)_K \in\operatorname{add} M_K$, hence $I^j \in\operatorname{add} M_K$. Splitting the sequence at the image of $P_0 \to P_1$ into two short exact sequences and applying $\operatorname{Hom}_{\Lambda_K}(-,Y)$ gives $\operatorname{Ext}_{\Lambda_K}^{d+2}(Q,Y) \cong\operatorname{Ext}_{\Lambda_K}^{d}(Y,Y)$. It follows that $\operatorname{gl.dim}\Lambda_K \ge d+2$.

For finite $k$, the chosen extension is separable. A separability tensor $\sum_i a_i \otimes b_i \in K \otimes_k K$ satisfies $\sum_i a_i b_i=1$ and $\sum_i ca_i \otimes b_i=\sum_i a_i \otimes b_i c$ for all $c \in K$. For every $\Lambda_K$-module $Z$ these identities show that $z \mapsto\sum_i a_i \otimes b_i z$ is a $\Lambda_K$-linear section of $K \otimes_k Z \to Z$, where $Z$ in the tensor product is restricted to $\Lambda$. Inducing a projective resolution from $\Lambda$ and passing to the summand $Z$ gives $\operatorname{gl.dim}\Lambda_K \le\operatorname{gl.dim}\Lambda$. Taking the infimum over $M$ proves the lemma.

The lower bound for the corner

Proof. Put $n=2L-3$.

Matrices and the parameter family. Write $h=L-2$, so $n=2h+1$. If $\operatorname{char} k \ne2$, put $r=2^h$ and use

Write $I_j$ for the $j$-by-$j$ identity matrix. A zeroth tensor power means the scalar $1$. For $1 \le j \le h$, define

These integer matrices anticommute in pairs. For matrices from different tensor positions there is exactly one interchange of $X$ or $Y$ with $Z$, and the same-position assertion is $XY=-YX$. Moreover,

The latter matrices commute, so

In characteristic two instead take $r=1$ and $P_i=(1)$ for all $i$. The pairwise anticommutator is still zero, (1) still holds, and $r\ne0$ in the field. Thus in both cases $r$ is nonzero in $k$.

Let $S=k[x_1,\ldots,x_n]/(x_i^2)$, let $R=k[\lambda_1,\ldots,\lambda_n]$, and define

All cross terms in $C_\lambda^2$ vanish by pairwise anticommutation, and the diagonal terms contain $x_i^2$. Give $N_R$ its ordinary left $\Sigma_R$-module structure by letting $x_i$ for $i\ge1$ act as scalar multiplication and $x_0$ act as $-C_\lambda$. This is well defined: $C_\lambda^2=0$, and $C_\lambda$ commutes with the scalar actions of the other generators. The family is $R$-free of rank $r^2 2^n$.

At a parameter point $\alpha$ over a coefficient field $K$, put

These are fixed matrices in $\operatorname{Matr}(\Sigma_K)$ while the parameters $\lambda$ vary. They satisfy

The action on the canonical Hom module. Put $I = DB_L$ and let $v \in I$ take the degree-zero coefficient of $B_L$. The functionals

are defined for $a,s\in\Sigma$ and are zero on the other summands. For $1\le a\le L-1$, the action of $t^a b$, $b\in\Sigma$, is

For $1\le m\le L$ and a $\Sigma$-module $N$, let $F_m(N)=N^m$, with coordinates $0,\ldots,m-1$. For $1\le a<L$ and $b\in\Sigma$, $t^a b$ multiplies by $b$ and advances $a$ coordinates, with indices beyond the last sent to zero. The Hom space has the following description, valid also after extension to any commutative $k$-algebra, with the modules and Hom spaces taken over the extended algebras.

The tuple $(m_1,\ldots,m_{L-2},f)$ corresponds to

and the image of $g_i(s)$ for $i\ge1$ is the image of $g_0(s)$ shifted by $i$ layers. For a homomorphism $h:I\to F_L(N)$, the vector $h(v)$ is killed by $t$, so $h(v)=(0,\ldots,0,z)$. Applying $h$ to $t^{L-1}g_0(s)=\tau(s)v$ gives $h_0(g_0(s))=\tau(s)z$, where $h_0$ is the first coordinate of $h$. Since $\tau(1)=0$, the relation $(t^{L-1}x_0\cdots x_n)g_0(1)=v$ gives $z=(x_0\cdots x_n)h_0(g_0(1))=0$. Comparing $(tb)g_0(s)$ with $tg_0(bs)$ makes the middle components $\Sigma$-linear; the final component is arbitrary. These elementary equations suffice: $t^a b$ shifts the first $L-1$ components by $a$ and multiplies them by $b$, and kills the final component. This gives the prescribed image of $g_{i+a}(bs)$ if $i+a\le L-2$, and zero otherwise. All equations hold over the coefficient ring. In particular the final summand is $D\Sigma\otimes_k N$, compatibly with base change.

For $p,q\in\Sigma$, define $B_L$-maps

The first is $B_L$-linear even at the last dual layer: one side then contains $C_p(v)=0$, and the other has $t$-degree $L$. For matrices $P=(p_{ij})$, $Q=(q_{ij})$ over $\Sigma$, let $C_P$ and $G_Q$ have these maps as their entries, with row $i$ specifying the target and column $j$ the source, for $1\le i,j\le r$. In the opposite ring we use the same vector-space elements and reverse the order of multiplication. Set

The two compositions are $C_PG_Q=t(PQ)$ on $B_L^r$ and $G_QC_P=t(QP)$ on $I^r$. Indeed, $C_pG_q(1)=tpq$, and $G_qC_p(g_i(s))=(t^{i+1}sp)g_0(q)=(tqp)g_i(s)$, while both sides kill $v$. Matrix composition sums these identities over the common index. In particular, writing $A_0=A_{I_r}$, $B_0=B_{I_r}$, $A_\pm=A_{Q_\pm}$, $B_\pm=B_{Q_\pm}$, we have exact algebra identities

The Hom module is

Write $U_a$ for the copy of $N^r$ in $U_N$ of index $0 \leq a \leq L-1$, and $V_a$ for the copy of $N^r$ in $V_N$ of index $1 \leq a \leq L-2$. For $N=N_R$, identify $N^r$ with $\operatorname{Matr}(S_R)$ by taking the $r$ vectors as columns. In the family $N=N_R$ and its specializations, write $\sigma_\lambda$ for the coefficient action and $\rho_\lambda$ for the induced action on $E_N$. For $T$ with columns $T_i$ and $f=(f_i)\in J^r_N$, put

Then $A_P$ takes $U_a$ to $V_{a+1}$ by $\mathcal{R}_P$ for $0 \leq a \leq L-3$, and takes $U_{L-2}$ to $J^r_N$ by $\eta_\lambda\mathcal{R}_P$; it kills the last $U$-coordinate and all of $V$. The map $B_Q$ takes $V_a$ to $U_a$ by $\mathcal{R}_Q$, and takes $J^r_N$ to $U_{L-1}$ by $\operatorname{ev}_Q$; it is zero on $U$. The $A_P$ action follows by evaluating $C_P$ on $g_0(s)$ and shifting the image of $1$ once. The $B_Q$ action evaluates at $G_Q(1)$, with entries $g_0(q_{ij})$. In particular,

Endpoint identities. At $\lambda=\alpha$, take $u$ to be $I_r$ in $U_0$, zero elsewhere, and put $u'=A_0u$. Let $\theta:\operatorname{Matr}(S_K)\to K$ be the coefficient of $x_1\cdots x_n$ in the ordinary matrix trace. Let $\phi_U$ apply $\theta$ to $U_{L-1}$, zero on all other coordinates, and define

The bottom component of $B_+$ is evaluation at the fixed matrix $Q_+$; it is parameter independent. Consequently $\phi_U\rho_\lambda(B_+)=\phi_V$ throughout the parameter family.

In the following identities the algebra elements act at $\lambda=\alpha$. By (7), $\mathcal{R}_{Q_+}(I_r)=0$, giving $A_+u=0$ and $B_+u'=0$ since $u'$ is $I_r$ in $V_1$. Cyclicity of trace and $B_+A_-=0$ give the other endpoint identities

Indeed, $\phi_VA_0$ is zero outside $U_{L-2}$, and there sends $T$ to $\theta(-C_\alpha T+TC_\alpha)=0$. Write $D_i=\partial_{\lambda_i}\rho_\lambda|_\alpha$ for the derivative of the complete representation on (6). We also have

The first holds because the bottom evaluation component is constant. For the second, differentiate the algebra identity $B_+A_-=0$, then apply $\phi_U$; the first identity removes the other derivative term.

The bar functional. For $1\leq j\leq n$, define an $n$-tuple of algebra elements $W_j$: its letter types are $A,B,A,\ldots,A$; position $j$ is $A_0$ or $B_0$ according to type; before that position the letters are $A_-,B_+$; after it they are $A_+,B_-$. Let

For a $K$-algebra $T$ and left $T$-module $E$, an ordinary bar cochain of degree $j$ is a $K$-linear map $T^{\otimes_K j} \to\operatorname{End}_K(E)$. Here $T = \widehat{\Gamma}_b \otimes_k K$ and $E = E_\alpha$, the specialization of the Hom module at $\alpha$. For a cochain $f$ of degree $j$, its differential evaluated at $a_1,\ldots,a_{j+1} \in T$ is

The end terms compose with the action on $E$. These cochains are obtained by applying $\operatorname{Hom}_T(-,E)$ to the free left $T$-modules $T^{\otimes_K(j+1)}\otimes_K E$. The bar differential multiplies consecutive factors with alternating signs, using the module action for the last multiplication. Augmentation is the action $T\otimes_K E\to E$. Inserting $1$ on the left gives a $K$-linear contraction of the augmented complex: its first face is the identity and all later faces cancel with those obtained by inserting $1$ after applying the differential. Thus the cochains compute $\operatorname{Ext}^\bullet_T(E,E)$.

For any ordinary bar $n$-cochain $F$, with arbitrary values in $\operatorname{End}_K(E_\alpha)$, set

Consider an arbitrary ordinary $(n-1)$-cochain $f$. All left faces of the $W_j$ vanish by $\phi_VA_0=\phi_VA_-=0$. All right faces vanish except the last unit position, by $A_+u=0$. Every internal product away from the unit position vanishes by (5). The product adjoining the unit between positions $j$ and $j+1$ is identical in $W_j$ and $W_{j+1}$, by (4). Its bar-face sign is the same in the two words and the word coefficients are opposite, so these faces cancel in pairs.

The sole surviving right face from the unit words has total coefficient $(-1)^{n-1}(-1)^n=-1$, and is

For $W_*$, the internal faces vanish by (5), the right face vanishes by $B_+u'=0$, and the left face is precisely the opposite of this remaining term, since $\phi_UB_+=\phi_V$. Thus

The parameter product. On the $i$-th coordinate extension the action is $\rho_\alpha+\epsilon D_i$. Its off-diagonal part $D_i$ satisfies $D_i(ab)=\rho_\alpha(a)D_i(b)+D_i(a)\rho_\alpha(b)$ for $a,b\in T$. The splitting over $K$ into the constant and $\epsilon$ coordinates identifies the extension with the bar $1$-cocycle $D_i$. To compose these classes, a cocycle $D_i$ lifts to maps from bar degree $j+1$ to degree $j$ by

These maps commute with the differentials in positive degrees: the earlier faces commute term by term, and the last two faces give the derivation identity. At degree zero augmentation recovers the cocycle. Composing the lifts therefore represents the Yoneda product by

On the coefficient coordinates, the derivative of either $\mathcal{R}_{Q_+}$ or $\mathcal{R}_{Q_-}$, with the matrices $Q_\pm$ held fixed, is

For $W_1$, the first $n-1=2L-4$ derivative factors, applied from right to left, alternate between $U_a \to V_{a+1}$ and $V_{a+1} \to U_{a+1}$ for $0 \leq a \leq L-3$. They move $I_r$ to $U_{L-2}$ and give $(-1)^{n-1}(x_{n-1}\cdots x_1)P_{n-1}\cdots P_1$. The last derivative is $D_n(A_0)$. Evaluation at $Q_+$ after applying $A_0$ on $U_{L-2}$ gives $\operatorname{ev}_{Q_+}\eta_\lambda=R_{Q_+}$. Evaluation is constant in the parameters, so the derivative contributes left multiplication by $-x_nP_n$. Therefore (1) gives

Every $W_j$ with $j>1$ starts with $A_-$, and its contribution vanishes by $\phi_VD_n(A_-)=0$. The $W_*$ term vanishes by $\phi_UD_n(B_+)=0$. Hence

Equations (11) and (13) prove nonvanishing in $\operatorname{Ext}_{\widehat{\Gamma}\otimes_k K}^n(E_\alpha,E_\alpha)$ at every parameter point over every extension field.

The hypotheses of Lemma 3 hold with $C=B_L$, $G=B_L^r\oplus(DB_L)^r$ and $d=n$. The Hom formula (3) gives freeness and commutation with base change, and the bar functional gives nonvanishing at every point.

The lower bound for the Frobenius algebra

Lemma 5. $\operatorname{repdim} A_L\geq L+1$.

Proof. Put $d=L-1$. In characteristic different from two, put $h=\lfloor d/2\rfloor$ and $r=2^h$. Use the matrices $X,Y,Z$ from the proof of Lemma 4.

Take

for $1\leq j\leq h$; for odd $d=2h+1$, append $P_d=Z^{\otimes h}$. These matrices anticommute in pairs. Write $\Omega=Z^{\otimes h}$. Then

For even $d$, $\Omega$ anticommutes with every $P_i$. These assertions follow from $YX=Z$, the tensor-position rule, and the displayed definitions. In each $P_i$ exactly one factor anticommutes with the corresponding $Z$ in $\Omega$.

In characteristic two, instead take $r=1$, all $P_i=1$, and $\Omega=1$. The same anticommutator identities hold and $r$ is nonzero in every case.

Parameter family. Let

The square of $C_\lambda$ is zero. Give $N_R=S^r$ its ordinary left $\Sigma_R$ action by scalar multiplication by $x_i$ for $i\geq1$ and by $-C_\lambda$ for $x_0$. These actions commute, and the $x_0$ action squares to zero. Set $X_R = F_L(N_R) = N_R^L$, where $t^a s$ acts by a shift of $a$ places and the action of $s$. This family is finite free over $R$.

Apply the Morita equivalence

This is an exact equivalence, hence preserves the parameter extensions and their Yoneda products in both directions. In particular, nonvanishing proved in $\operatorname{Ext}_{\Delta}$ below is nonvanishing in ordinary $\operatorname{Ext}_{B_L}$.

Identify each layer of $H(X_R)$ with $\operatorname{Mat}_r(S_R)$ by taking the $r$ vectors as columns. At a point $\alpha$ over a field extension $K/k$, keep

fixed while differentiating the parameters. Both products $Q_+Q_-$ and $Q_-Q_+$ are zero over $\Sigma_K$.

The elements $z_+$ and $z_-$ of $\Delta_K = \Delta\otimes_k K$ represented by the matrices $tQ_+$ and $tQ_-$ act, by precomposition, as a one-layer shift followed by

The opposite-ring order reverses matrix products, but the two needed products are both zero. Thus $z_+z_- = z_-z_+ = 0$ as algebra identities.

The bar functional. Let $v$ be $I_r$ in the first layer, zero elsewhere. Write $\rho_\lambda$ for the action on $H(X_R)$. At the chosen point, $\rho_\alpha(z_+)v = 0$ by (14). Let $\phi$ be zero except on the last layer and there set

Here $[x_1 \cdots x_d]$ selects the indicated squarefree monomial, with exponent zero on all other generators. For odd $d$, cyclicity of matrix trace gives $\phi\rho_\alpha(z_+) = 0$ on the whole module. For even $d$, the relation $\Omega C_\alpha+ C_\alpha\Omega= 0$ gives

so $\phi\rho_\alpha(z_-) = 0$ on the whole module. In characteristic two the displayed trace is zero with $\Omega= 1$, since the trace is cyclic and $2 = 0$.

Let $W$ be the alternating word of length $d$ whose rightmost letter is $z_+$. Its leftmost letter is $z_+$ when $d$ is odd and $z_-$ when $d$ is even. For an ordinary bar $d$-cochain $F$, put

For a cochain $f$ of degree $d-1$, every internal face of $Z(df)$ contains a zero product $z_+z_-$ or $z_-z_+$. The right face contains $\rho_\alpha(z_+)v = 0$. The left face vanishes by the appropriate whole-module identity for $\phi$. Consequently

for every ordinary bar $(d-1)$-cochain. Write $D_i$ for the derivative of the $\Delta$-action in the $i$th parameter direction. The action on the $i$-th coordinate extension is $\rho_\alpha+\epsilon D_i$. By (14), the derivative of either fixed $z_+$ or $z_-$ advances one layer and left-multiplies by $-x_iP_i$. Thus the ordered cup product takes $v$ to

in the last layer. All the $x_i$ commute. The detector evaluates this to $-r$ in odd degree and $r$ in even degree. This is nonzero in every field by our choice of $r$. Hence the parameter product is nonzero at every point, over every extension field.

Inflate the family $X_R$ along the unital projection $A_L\to B_L$ onto the first diagonal coordinate. The exact functor $e(-)$ recovers the original family and its coordinate extensions. It therefore takes their product over $A_L$ to the nonzero product over $B_L$. Apply Lemma 3 with $C=G=A_L$, since $\operatorname{Hom}_{A_L}(A_L,-)$ identifies with the identity under evaluation at $1$. The resulting lower bound is $d+2=L+1$.

The upper bound for the corner

Lemma 6. $\operatorname{repdim} B_L\le2L-1$.

Proof. Let $s=\dim_k\Sigma$.

Contraction modules. Write $S=B_L/J$ for the simple $B_L$-module, and put

Thus $Q_1=S$ and $Q_L=B_L$. Set $Q_0=0$. The multiplication map

is a split epimorphism of left $\Sigma$-modules; use the section $c\mapsto c\otimes1$. Its kernel $Z$ is free of rank $q$, since it is projective over the local algebra $\Sigma$. We use the same fact for subsequent split surjections of finite free $\Sigma$-modules.

For the strings $F_r(E)$ defined above, set $F_0(E)=0$ and write $F_r=F_r(\Sigma)$. In particular $F_1\cong S^s$.

For $0\le d\le L-2$, define $C_d$ by its layers in degrees $0,\ldots,L-1$:

The parentheses denote the direct sum of the listed $k$-spaces, ordered by increasing degree. When $d>0$, the first action is insertion, $(tb)c=b\otimes c$. Between tensor layers, $tb$ multiplies the first factor on the left. At the transition to an ordinary coefficient layer it acts by $a\otimes c\mapsto bac$; thereafter it acts by left multiplication. The last action is the Frobenius pairing $c\mapsto\tau(bc)$, or $a\otimes c\mapsto\tau(bac)$ if the preceding layer is still a tensor layer. For $d=0$, all coefficient layers are ordinary. For $t^ab$ with $a>1$, apply $tb$ and then $t$ another $a-1$ times. The same formulas result if the coefficient $b$ is applied at any of the $a$ steps, since insertion is followed by multiplication on the first factor and both contraction and the final trace use the product in order. Thus the action respects multiplication in $B_L$. They identify $C_0$ with $DB_L$, using the left action $(b\xi)(x)=\xi(xb)$ on the dual.

Set

The functor $H$ takes values in left $\Gamma$-modules. Write $\operatorname{pd}_{\Gamma}$ for projective dimension over $\Gamma$. Because $B_L$ and $DB_L$ occur, $M$ is a generator-cogenerator.

All $2L-1$ displayed summands are indecomposable and pairwise nonisomorphic. Indeed, an endomorphism of $C_d$ has scalar top: commutation with the maps $t^{L-1}b$ and nondegeneracy of $\tau$ force its top component $A$ to equal $\lambda\operatorname{id}_{\Sigma}$: if the map on the final layer is multiplication by $\lambda$, then $\tau(bA(c))=\lambda\tau(bc)$ for all $b,c\in\Sigma$. Each $C_d$ is generated by its degree-zero layer, since $t^ab$ on a top vector $c$ gives $b\otimes c$, $bc$, or $\tau(bc)$ in degree $a$. Endomorphisms with zero top raise degree and form a nilpotent ideal. Thus $\operatorname{End}_{B_L}(C_d)$ is local with residue field $k$. The dimensions of the $C_d$ strictly increase with $d$; their tops have dimension $s$, whereas every $Q_a$ has one-dimensional top. The assertions for the $Q_a$ follow from their cyclic local quotient structure.

We call an epimorphism $X\to Y$ $M$-proper if every map $M\to Y$ lifts through it. An exact sequence is proper if it remains exact after applying $H$. For a summand $X$ of $M$, write $T_X$ for the simple top of the projective module $H(X)$. A relative radical approximation of $X$ is a map from $\operatorname{add}M$ whose image under $H$ is $\operatorname{rad}H(X)$. The latter consists of maps whose component at $X$ lies in the radical of $\operatorname{End}_{B_L}(X)$, with unrestricted components at the other summands. If the approximation has kernel $K$ and domain $P'$, left exactness gives

The two middle terms are projective, so $\operatorname{pd}_{\Gamma}T_X\leq\operatorname{pd}_{\Gamma}H(K)+2$.

Radical splittings and the two lifting rules. Write $I_n=DQ_n$ for $1\leq n\leq L$, with $n-1$ coefficient layers and a final scalar layer as in $C_0$. In the positive degrees of $C_d$, split each tensor layer as $Z\oplus(\Sigma\otimes1)$. The $Z$-layers have zero action into the first layer after contraction, and the other summands continue through the final trace. This gives

For $d\geq1$, there is a quotient $\pi_d:C_d\to C_{d-1}$ that contracts the tensor layer in degree $d$ and is the identity in every other degree. Its kernel is $S^{\mathrm{sq}}$, and on radicals it is

Here the bracketed map truncates the final layer.

All maps below decompose into homogeneous maps, where degree $j$ means that a source layer of index $i$ maps to target layer $i+j$. Each source is generated in degree zero, so only degrees $j\geq0$ occur. A zero-top map into a module generated in degree zero has image in its positive layers. When the target is a shifted string, the same tests apply to the portion starting in the image degree.

The following calculation will be used for maps out of $C_e$. For a homogeneous map, let $A$ be its map from the source top into the first layer of the target portion. The value on $(t^ab)c$ must be $t^abA(c)$. When the source layer at distance $a$ is tensor, this prescribes the value on $b\otimes c$ with no relation. At the first ordinary source layer, it must depend only on $bc$. If the target layer at that distance is ordinary, this gives $bA(c)=A(bc)$ by setting $b=1$ in the required factorization. If the target layer is tensor, it gives $b\otimes A(c)=1\otimes A(bc)$. Taking $b$ linearly independent of $1$ forces $A=0$. If the target ends in a scalar layer at this distance, the equation is $\tau\left(bA(c)\right)=f(bc)$ for a functional $f$, which implies $A(c)=cu$ by writing $f(x)=\tau(xu)$. When the target has only ordinary layers after the contraction, the relations propagate by multiplication, followed by $\tau$ if it has a final scalar layer. At the source’s final scalar layer, a full-length map to $C_d$ satisfies $\tau\left(bA(c)\right)=\lambda\tau(bc)$, so $A(c)=\lambda c$. For a degree-zero map to $F_r(E)$, the top map is consequently arbitrary if $e\ge r-1$, and must be $\Sigma$-linear if $e\le r-2$.

A map out of $Q_a$ is determined by the image of $1+J^a$, which must be killed by $J^a$. In an ordinary string a nonzero vector with $b$ layers remaining is killed by $J^a$ exactly when $a\ge b$, since $t$ shifts by the identity. In a string ending in a trace the same test holds on coefficient layers: the last coefficient-to-scalar map can detect any nonzero coefficient by varying the multiplier. These tests will also be used for lifts through a quotient or projective cover.

Truncation lifting. Suppose the restriction of a map to radicals, after left $\Sigma$-linear splittings, has the form

where $E$ is free. Add the cover $Q_r^{\dim_k E}\longrightarrow F_r(E)$ induced by a $k$-basis of the first layer $E$. Then every zero-top map from any $Q_a$ or $C_e$ lifts through the augmented map.

Here $R$ is the complementary target summand, which lifts by the identity. Maps from a $Q_a$ into the first layer of the defect string lift through its $Q_r$-cover; maps landing farther down lift through a left $\Sigma$-linear section of the coefficient multiplication map. For a $C_e$-source, an arbitrary top occurs only when $e\ge r-1$, and then factors through the standard $Q_r$-cover. A coefficient-linear top, when $e\le r-2$, extends to $F_{r+1}(E)$. Positive-degree components lift through the positive radical layers of the $Q_r$-cover, again using a left $\Sigma$-linear section. This treats all homogeneous components.

The kernel contributed by this augmented defect map is the module

with insertion at the first transition, multiplication on the first factor between tensor layers, and coefficient multiplication at the last transition. For $r=1$, $tb$ sends $x\in E$ in the first layer to $bx\in E$ in the second. To see the kernel, identify its top with the top of the added cover and write $\mu_E(b\otimes x)=bx$ for its multiplication map. In each degree $1,\ldots,r-1$ the pair of coordinates in the two channels is $(-\mu_E(z),z)$ for $z\in\Sigma\otimes_k E$. In degree $r$ only the first channel remains. Taking its negative as the final coordinate gives the stated last action. For $E=\Sigma^b$, it is $U_r^b$, where $U_r=U_r(\Sigma)$. Set $U_0(E)=E$, viewed as semisimple. The case $r=0$ is the semisimple kernel of the contraction itself, without an added quotient cover.

Trace lifting. Every map from a $Q_a$ or $C_e$ to $I_n$, for $n\le L-1$, lifts through

which is the identity on its first $n-1$ coefficient layers and $\tau$ on its last layer. If a homogeneous map starts in degree $j$, the two tails beginning there have the same form with length $n-j$. It suffices to lift into the first layer of those tails.

For a $C_e$-map to the first coefficient layer, if $e \ge n-1$, its top $A$ is arbitrary. A lift uses the same $A$ on top and $b \otimes c \mapsto bA(c)$ on every subsequent layer, including the last. Composing the last map with $\tau$ recovers the given map. When $e \le n-2$, comparison at that contraction yields

for a functional $f$. Write $f(x)=\tau(xu)$. Nondegeneracy forces $A(c)=cu$, which lifts coefficientwise. If the tail has length one, its maps lift through any $k$-linear section of $\tau$. For a $Q_a$-source, a nonzero map into the first coefficient layer requires $a \ge n$: otherwise its required annihilation by a radical power contradicts nondegeneracy of the final pairing. The chosen image of $1+J^a$ lifts into the first layer of the ordinary tail, which is killed by $J^a$.

Resolutions of contracted strings. For $1 \le d \le L-2$, the module

is the initial $d+1$-layer truncation of $C_{d-1}$, and

Notice that $U_1=F_2$ and $U_0=F_1=S^s$.

Choose a basis $e_1,\ldots,e_s$ of $\Sigma$, and let

For the $j$-th $C_{d-1}$-summand, truncate to $U_d$ and multiply the top and ordinary coefficients on the right by $e_j$, acting by $a\otimes c\mapsto a\otimes ce_j$ on tensor layers. Together with the cover $Q_{d+1}^s\to U_d$ determined by the basis $e_j$ of its top, these maps define

This map is proper. On radicals, the $F_{d-1}(Z)$-summand splits through the $C_{d-1}$-channels, because a scalar combination of the $e_j$ is $1$. The $F_d$-summand splits through the radical of $Q_{d+1}^s$. In the coordinates $\Sigma\otimes_k\Sigma$ supplied by the basis $e_j$, the section is $a\mapsto a\otimes1$ in each positive layer. Its image in the tensor layers lies in $\Sigma\otimes1$ and in the final layer maps to $a$. For a degree-zero $C_e$-map, comparison of insertion and contraction shows that the top is zero if $e<d-1$, is coefficient-linear if $e=d-1$, and is arbitrary if $e\ge d$. The linear top $c\mapsto cu$ lifts by writing $u$ as a linear combination of the $e_j$ in the $C_{d-1}$-summands. An arbitrary top lifts to $Q_{d+1}^s$, since the first $d$ positive layers of $C_e$ are tensor. A $Q_a$-map with nonzero top requires $a\ge d+1$, because $t^d$ is the identity from the first to the last layer of $U_d$. Such a map lifts through $Q_{d+1}^s$, which is killed by $J^a$. The sections on radicals lift positive-degree maps.

Let $H'_d=\ker p_d$. Projecting the kernel to $C_{d-1}^s$ identifies its top and layers $1,\ldots,d-1$ with $\Sigma^s$ and $P_d^s$, since the maps from the $Q_{d+1}^s$-layers there are isomorphisms onto the target layers. In degrees below $d$, the coordinate in the $Q_{d+1}^s$ summand is the negative of the image of the $C_{d-1}^s$ coordinate under the first component of $p_d$, transported by the isomorphism. In degree $d$ the layer is

and its later layers are $\Sigma^{s}$, ending in $k^{s}$. There is a map $\alpha: C^{s}_{d} \to H^{\prime}_{d}$ that is the identity in the common coordinates except in degree $d$, where

In degree $d$, the first coordinate is contraction in $C^{s}_{d-1}$, and the second is the negative of the tensor map from the preceding degree. The next layer receives only the first coordinate, by multiplication or, at the last step, by trace. Thus $\alpha$ commutes with the $tb$-actions, including $d=1$, when the preceding degree is the top. To see surjectivity, lift $z$ through $-\operatorname{id}_{\Sigma}\otimes\ell$. The discrepancy in the first coordinate then lies in $\ker\ell$ and can be corrected in $\Sigma\otimes_k\ker\ell$, since coefficient multiplication surjects onto $\ker\ell$. Its coefficient kernel is

Indeed, $\ker\ell\cong\Sigma^{q}$, and $E_0$ is the kernel of the multiplication map $\Sigma\otimes_k\ker\ell\to\ker\ell$. For a left $\Sigma$-basis $u_1,\ldots,u_q$ of $\ker\ell$, the map $\sum_{i=1}^{q}a_i u_i\mapsto\sum_{i=1}^{q}a_i\otimes u_i$, with $a_i\in\Sigma$, is a left $\Sigma$-linear section of multiplication for the first-factor action. Hence $E_0$ is projective over the local algebra $\Sigma$, so it is free of rank $sq-q=q^2$.

Using the same left $\Sigma$-linear splitting of $P^{s}\to V$ in degrees $1,\ldots,d$ identifies the radical restriction of $\alpha$ with

The truncation lifting rule therefore supplies a proper cover of the zero-top part after adjoining $Q^{sq^2}_{d-1}$. For a degree-zero map $C_e\to H'_d$, the final scalar layers force its top to be $c\mapsto(\lambda_jc)$. If $e<d$, its projection to $Q^{s q^2}_{d+1}$ has zero top, so $\ell((\lambda_jc)_j)=0$; independence of the $e_j$ forces all $\lambda_j=0$. If $e\ge d$, its top lifts through the canonical quotient $C_e\to C_d$. All $Q_a$ with $a<L$ have zero top into this full-length kernel, and $B_L$ lifts by projectivity. Thus there is a proper exact sequence

Applying $H$ leaves an exact sequence with two projective middle terms. Since $U_0\cong S^{s}$ belongs to $\operatorname{add}M$, induction gives

Ordinary strings and the projective summand. For $2\le r\le L-1$, consider

using truncation followed by right multiplication by $e_j$ on the $j$-th $C_0$, and the top basis $e_j$ for the quotient cover. Coefficient-linear $C_e$-tops lift through $C^{s}_{0}$, using the maps $C_e\to C_0$ and scalar combinations of the $e_j$. Arbitrary tops are allowed when $e\ge r-1$, and lift through $Q^{s}_{r}$. A $Q_a$-map with nonzero top requires $a\ge r$, so it lifts through $Q^{s}_{r}$. The radical of the quotient cover splits onto the target radical by $a\mapsto a\otimes1$ in each positive layer, lifting the positive-degree components.

Its kernel $H_r$ has top $\Sigma^{s}$, with the $k^{s}$ coordinate fixed by cancellation of the two maps on top. In degrees $1,\ldots,r-1$ the equation is $\ell(f)+\mu(z)=0$, so these layers are $V$. At degree $r$ and above, only the $C^{s}_{0}$ layers remain. The formula for $\alpha$ on degree $d$ in the preceding calculation, now used at every degree $1,\ldots,r-1$, defines

with kernel $F_{r-1}^{q^2}$. Here the entire radical restriction splits: use the same left $\Sigma$-linear section of $P^s \to V$ in each of the first $r-1$ layers. It commutes with left multiplication between repeated layers, and its composite with $\mu^s$ is projection to the first coordinate of $V$, which gives the next transition. Every zero-top map therefore lifts. The full-length final trace again forces scalar-coordinate tops from $C_e$, by projection to $C_0^s$. If $e<r-1$, the projection to $Q_r^s$ has zero top by the tensor target test, hence $\ell((\lambda_jc)_j)=0$ for every $c\in\Sigma$. At $c=1$, independence of the $e_j$ gives $\lambda_j=0$. If $e\ge r-1$, the top lifts by contracting $C_e\to C_{r-1}$ in the corresponding channels. A $Q_a$-map into the kernel with nonzero top projects to a $Q_a$-map into $C_0^s$ with nonzero top, requiring $a=L$; it lifts by projectivity. Consequently

is proper. After applying $H$, there are two projective middle terms. The base $F_1\cong S^s$ gives

A map from an indecomposable summand of $M$ to $B_L$ that is nonzero on the top is surjective by Nakayama’s lemma. It splits by projectivity and is therefore an isomorphism by indecomposability. Thus the maps in the radical of $H(B_L)$ are exactly those with image in $J=F_{L-1}$, and $0\to H(F_{L-1})\to H(B_L)\to T_{B_L}\to0$ is exact. Since $H(B_L)$ is projective,

The simples attached to the contraction modules. For $0\le d\le L-3$, a relative radical approximation is

The first component is $\pi_{d+1}$, the contraction at degree $d+1$. The second covers $F_d(Z)$ inside the radical. Its kernel is $U_d^q$, by the truncation lifting rule. The same rule lifts every zero-top map. A nonzero top map from a different $C_e$ occurs precisely when $e\ge d+1$, by the full-length scalar top condition and the tensor target test. It is the corresponding scalar multiple of the iterated contraction. The radical endomorphisms of $C_d$ have zero top. Among $Q_a$-sources only $Q_L$ can have nonzero top, and it lifts through the surjection by projectivity. Hence

For the last contraction module, use

On radicals, the splitting $P=Z\oplus\Sigma$ identifies the projective-cover part with

The projective map sends the basis of the top of $B_L^s$ to the basis $e_j$ of the top of $C_{L-2}$. In its positive layers the action is insertion up to degree $L-2$ and the trace of multiplication in degree $L-1$, giving the displayed splitting. All $C_e$-maps in the radical have zero top, and the only $Q_a$ allowing nonzero top is $B_L$. The two lifting rules handle the zero-top components and projectivity handles the $B_L$-source. The kernel is

where the semisimple term is the kernel of the last trace. Therefore

The simples attached to the nonmaximal quotients. Choose a basis $\phi_1,\ldots,\phi_s$ of $D\Sigma$ with $\phi_1(1)=0$, possible since $s>1$. Choose $c_0\in\Sigma$ with $\phi_1(c_0)=1$. For $1\leq a\leq L-2$, the relative radical approximation is

using the quotient map and the maps with top components $\phi_j$, which on tensor layers send $b\otimes c$ to $b\phi_j(c)$. A $C_e$-map can have nonzero top only when $e\geq a-1$, by the tensor target test on the positive layers of $Q_a$. Its top is a linear combination of the $\phi_j$ and lifts by contraction to $C_{a-1}^{s}$. A nonzero top from a different $Q_b$ requires $b>a$, and factors through $Q_{a+1}$. If $b=a$, the endomorphism is radical exactly when its top is zero. All zero-top maps lift because

maps to $b$ under the first functional channel. This gives a left $\Sigma$-linear section throughout $\operatorname{rad}Q_a$. For $a=1$ the radical is zero, and the top argument suffices.

Let $H_a^Q$ be the kernel of this approximation. In degrees $0,\ldots,a-1$, projection to $C_{a-1}^{s}$ identifies the kernel with its layers, since the quotient map from $Q_{a+1}$ is an isomorphism in those degrees. In degree $a$ both source layers survive and the target layer is zero, giving $\Sigma^s\oplus\Sigma$. Above degree $a$, only $C_{a-1}^{s}$ contributes. The natural map $C_a^{s}\to H_a^Q$ is the identity on the common coordinates and in degree $a$ is

Given the first coordinate, lift it through the componentwise multiplication $P^s\to\Sigma^s$. The difference in the last coordinate is corrected by the displayed element of $Z$ in the first channel. This proves surjectivity. The transition into degree $a$ multiplies in the $C_{a-1}^{s}$ coordinates and evaluates the negative functional in the quotient coordinate. The transition out of degree $a$ uses only the first coordinate, so $\beta$ commutes with both. Its coefficient kernel $E_S\subseteq\ker\mu^s$ is free of rank

This is the difference between the ranks $s^2$ of $P^s$ and $s+1$ of $\Sigma^s\oplus\Sigma$. Choose a left $\Sigma$-linear section of $\beta$ and use it in degrees $1,\ldots,a$. Since the first component of $\beta$ is $\mu^s$, the complement continues through the remaining coefficient and scalar layers. The $E_S$-component truncates from $F_a(E_S)$ to $F_{a-1}(E_S)$. Adjoining $Q_{a-1}^{\mathrm{bs}}$ therefore gives a cover with kernel $U_{a-1}^{\mathrm{bs}}$. It is proper on zero-top maps by the truncation lifting rule. A degree-zero $C_e$-map to $H_a^Q$ has scalar-coordinate top. If $e<a$, its $Q_{a+1}$-component has zero top. The kernel relation on the top then gives $\sum_j\lambda_j\phi_j(c)=0$ for all $c$, so $\lambda_j=0$ by independence. If $e\geq a$, the top lifts through $C_e\to C_a$. A nonzero top from $Q_b$ projects to a nonzero top in $C_{a-1}^{s}$, so $b=L$; it lifts by projectivity. Applying $H$ to this cover gives $\operatorname{pd}_{\Gamma}H(H_a^Q)\leq1+2(a-1)$. The radical approximation adds two projective terms, giving

The maximal proper quotient. For $Q_{L-1}$, a relative radical approximation is

with the same functional channels. For $C_e$, a nonzero top requires $e=L-2$ and lifts through these channels. The same section on radicals as for $Q_a$ lifts zero-top maps. The projective $B_L$-source lifts through this surjection. Put

The kernel is the module

with $(tb)w=b\otimes w$ from the top, left multiplication on the first factor between tensor layers, and $a\otimes(c_j)_j\mapsto(\tau(bac_j))_j$ at the last transition.

The subspace $W$ satisfies

For the first, if $\Sigma W$ were proper, the codimension-one condition would force $\Sigma W=W$. The one-dimensional quotient would give a character $\chi:\Sigma\to k$ satisfying $\Phi(bc)=\chi(b)\Phi(c)$. Applying this coordinatewise and using that the $\phi_j$ form a basis gives $bc=\chi(b)c$ for all $b,c$, hence $b=\chi(b)1$, contrary to $s>1$. For the second identity, the condition says $\sum_j\lambda_j\phi_j=0$, so every $\lambda_j=0$.

Thus the multiplication map

is a split left $\Sigma$-epimorphism, with free kernel $E_W$ of rank $v-s=s^2-s-1$. On radicals,

There is a mixed cover

The projective part sends the standard generators to a $k$-basis of $W$, and the second part covers $F_{L-2}(E_W)$ inside the radical. On positive layers the projective part is the identity on $\Sigma\otimes_k W$ up to degree $L-2$, followed by $\mu_W$ and the coordinatewise trace. After splitting $\mu_W$ this gives

The added quotient cover and the truncation lifting rule handle the defect string, while the trace lifting rule handles the second summand. The composite of a $C_e$-map into $G(W)$ with its inclusion in $C^s_{L-2}$ has degree-zero top $c\mapsto(\lambda_jc)_j$, by the trace relation. Its image lies in $W$, so the second identity forces $\lambda=0$. If $a<L$, a $Q_a$-map has zero top because a nonzero $w\in W$ is detected in at least one coordinate by $t^{L-1}b$ for a suitable $b$; the $B_L$-source lifts by projectivity. Hence the mixed cover is proper. Its kernel is

The trace part contributes $(\ker\tau)^s$ in the final layer, on which $J$ acts by zero. The truncation part contributes the other summand by the truncation rule. The projective dimension of $H applied to this kernel is at most $2(L-2)$. One term for its proper cover and two for the radical approximation give

The bounds for the simple left $\Gamma$-modules are

Therefore $\operatorname{gl.dim}\Gamma\leq2L-1$. Since $M$ is a generator-cogenerator, $\operatorname{repdim} B_L \leq2L-1$. $\square$

Proof of Theorem 1. Lemmas 2 and 5 give the value for $A_L$, and Lemmas 4 and 6 give the value for $B_L$. At $L=3$ the squarefree monomial basis has 16 elements, so $\dim_k B_3=1+2\cdot16=33$ and $\dim_k A_3=4\cdot33=132$. $\square$

Zhili College, Tsinghua University, Beijing 100084, China

Email address: [email protected]