An algebra of infinite little finitistic dimension

Proof. For n=(n_1,n_2,n_3)\in\Z^3, put T^n=T_1^{n_1}T_2^{n_2}T_3^{n_3} and regard a weight as the map \lambda_S(n)=\sum_\ell\lambda_{S,\ell}n_\ell. Retain only the fifteen generators

Besides commutation of the T_\ell, require the following finite set of centralizer relations. Each of u_i,v_i commutes with T_h for h\neq i, and w_{ij} commutes with T_iT_j and with T_h for the unique h\notin\{i,j\}. These are precisely centralization by a basis of each kernel lattice:

Since the T_\ell commute, the kernel lattice also centralizes every T-conjugate of its argument-zero generator. Thus one may define

Equivalently, S(r)=T^nS(0)T^{-n} for any n with \lambda_S(n)=r. The kernel relations make this independent of the choice of n and imply all of (2.4).

It remains to impose the argument-zero instance of each of the remaining relation patterns. For every pair of input types actually occurring in those patterns, the two weights give a surjection \Z^3\to\Z^2. The following table supplies integral preimages of an arbitrary pair (r,s); unmentioned coordinates are zero.

Conjugating a zero-argument commutation relation by the corresponding T^n therefore gives every required instance. For the two relations in (2.5), the output weight is the sum of the input weights:

Hence the same conjugation also gives the correct output argument r+s. The square relation propagates from argument zero because each individual weight is surjective onto \Z.

We have obtained all the defining relations from finitely many relators on fifteen generators. Conversely, the chosen kernel centralizers and zero-argument relators hold in the original presentation by (2.4). The two presentations are therefore equivalent.□

Proof. Fix distinct i,j and integers r,q,s, and set

The defining commutations give [x,v]=[w,u]=[w,v]=1. Using xy=uyx and yv=wvy, reorder xyv in the two ways

With c=[x,w], cancellation gives uwv=cwvu, and consequently

The last equality uses only the commutation of w with u and v. We have proved

Given a+b=c+d=N at distinct internal indices, choose r=a, s=d, and q=c-a=b-d in (2.8). This compares the two expressions in (2.7). Two splittings at the same internal index can be compared through another index, proving the claimed independence.

To show that z_N commutes with U_h(r) or V_h(r), express z_N using an index i\neq h. Both factors of that commutator commute with the tested generator. To commute with W_{hj}(r), use the index outside \{h,j\}, so that again both factors commute with it. Finally, conjugation by T_\ell changes the arguments (a,b) in (2.7) to (a-\delta_{\ell i},b+\delta_{\ell i}). Their sum is unchanged, so the splitting independence proves invariance under T_\ell. Thus z_N is central. Since U_i(a)^2=1, centrality also gives

Translation of every U-argument by 1 preserves the conjugation, square, and commutation relations. The first relation in (2.5) becomes

which is another defining instance; the second relation is unchanged. This defines an endomorphism \alpha_G. Translation by -1 preserves the relations for the same reason and is its two-sided inverse. Formula (2.7) then gives \alpha_G(z_N)=z_{N+1}.□

Proof. Let S_m=\F_2[t]/(t^m-1). This ring is finite, and t is a unit with inverse t^{m-1}, so t^r is defined for every r\in\Z. Use matrices indexed by 0,1,2,3,4. Send T_\ell to the diagonal matrix with t in position \ell and 1 elsewhere, and set

Here E_{ab} denotes a matrix unit. Every displayed elementary matrix is invertible, with itself as inverse: its off-diagonal matrix unit has square zero, and the coefficient ring has characteristic two.

We check the relations. Diagonal conjugation multiplies E_{ab} by the ratio of the diagonal entries at a and b, which gives exactly (2.3). The identity E_{ab}E_{cd}=\delta_{bc}E_{ad} shows that all products in both orders vanish for each required commuting pair: two units E_{0i}, two units E_{i4}, the pair E_{0i},E_{j4} for i\neq j, the pair E_{ij},E_{0h} for h\neq i, and the pair E_{ij},E_{h4} for h\neq j. The required square relations hold by the same square-zero calculation. For i\neq j, the nonzero product in the first noncommutation relation is E_{0i}E_{ij}=E_{0j}, and in the second it is E_{ij}E_{j4}=E_{i4}. The higher products vanish, giving

Thus (2.9) defines a homomorphism from G into the finite group \operatorname{GL}_5(S_m). Let F_m be its image.

The image of z_N is I_5+t^NE_{04}. It is central in F_m: the unit E_{04} has zero product in both orders with every off-diagonal unit in (2.9), and the diagonal matrices have entry 1 at both 0 and 4. For c_0,\ldots,c_{m-1}\in\{0,1\},

The classes 1,t,\ldots,t^{m-1} are an \F_2-basis of S_m, because the defining polynomial is monic of degree m. Thus this product is the identity only when all c_q are zero. This argument also applies when m is even and S_m is nonreduced. The claimed independence and orders follow.□

Proof of Remark 2.1. Take R=\C G. A finite group presentation yields a finite presentation of its unital group algebra: for each group generator adjoin a formal inverse and the two inverse equations, and impose the finitely many group relators as algebra relations. Hence R is finitely presented by Remark 2.2. Extend \alpha_G linearly to an algebra automorphism \alpha and put

By Remark 2.3, this is a central idempotent. For q\geq0, write

These are commuting central idempotents of the one fixed algebra R.

For the finite quotient in Remark 2.4, let \overline z_q be the image of z_q and let Z_m=\langle\overline z_0,\ldots,\overline z_{m-1}\rangle. Independence defines a character \chi_m:Z_m\to\{1,-1\} by

In the complex regular representation of F_m, form

The element p_{\chi_m} is nonzero, since its identity coefficient is 1/|Z_m|, and z p_{\chi_m}=\chi_m(z)p_{\chi_m} for z\in Z_m. Centrality of Z_m then shows that every vector of Y_m has this same character. In particular, Y_m is a nonzero finite-dimensional left \C[F_m]-module. Regard it as an R-module through the quotient map.

For every left R-module Y, induction in the definition of H gives an identification

where the empty product is 1. Indeed, on the vector space at stage j, selection by e acts as the original operator \alpha^j(e)=e_j. Pulling the resulting action back once more along \alpha makes r act by \alpha^{j+1}(r), proving the inductive step. Centrality of all the e_q makes the subspace in (2.10) an R-submodule before this pullback.

On Y_m, the operators e_0,\ldots,e_{m-2} are the identity and e_{m-1} is zero. Consequently (2.10) gives (2.2), also for m=1. This argument does not require \alpha to preserve the kernel of the finite quotient: each pullback uses the same finite-dimensional vector space with its action precomposed by an automorphism of R. The construction of R,e,\alpha preceded the choice of m, as required.□

Proof. Say that a left module M is supported at indices at least p if \varepsilon_jM=0 for j<p. There is a projective surjection

The source is projective even when M is not finitely generated, since each summand is a direct sum of copies of \Lambda\varepsilon_i. Its components below p vanish, and its component at p maps isomorphically onto \varepsilon_pM: only the summand i=p contributes there. Its kernel is therefore supported at indices at least p+1.

Starting with p=0 and repeating this construction, the (n-1)st kernel is supported only at the last vertex. Every such module is projective, since \Lambda\varepsilon_{n-1}=k\varepsilon_{n-1}. This gives the left dimension bound, including the case n=1. The same proof for \Lambda^{\mathrm{op}}, with the reverse ordering of the idempotents, gives the right bound. The path directions and the homogeneous quadratic relations of B give precisely the displayed hypotheses with n=3.□

Proof. First consider a triangle from \mathcal T,

and suppose that formal decompositions A=I\oplus A_0, B=I\oplus B_0 identify f with the identity on I and zero elsewhere. For any Z\in\Kar(\mathcal T), the long exact sequence \Hom(Z,-) still holds for this triangle: Z is a retract of an object of \mathcal T, and its Hom sequence is a direct summand of the exact sequence for that object.

The equalities gf=0 and f[1]h=0 give factorizations

Exactness gives, for every such Z, a short exact sequence

Indeed, a map into B_0 killed by g_0 factors through f after inclusion into B, and projection back to B_0 makes it zero. At the other end, every map into A_0[1] is killed by f[1] after inclusion into A[1], so it lifts through h. Taking Z=A_0[1] supplies a section \sigma of h_0. It follows from (3.3) that

induces isomorphisms on \Hom(Z,-) for every Z, hence is an isomorphism.

Apply this calculation to a triangle on 1-\epsilon:L\to L. Its identity summand is (L,1-\epsilon), and its zero summand on both sides is U. Its cone is an object S\in\mathcal T with S\cong U\oplus U[1] in the completion. Under this identification, consider the map

Fullness of \mathcal T\hookrightarrow\Kar(\mathcal T) makes this a morphism between the old objects S[1] and S. Its cone is again in \mathcal T. The preceding calculation, with common identity summand U[1], source complement U[2], and target complement U, identifies that cone with U\oplus U[3].□

Proof. Write the given objects as W_i and its arrow maps as t_a. Choose C_i\in K and isomorphisms \psi_i:q(C_i)\to W_i. First collect the maps

Represent this tuple by one roof, with denominator u_1:D_1\to C_1 and numerator components f_b:D_1\to C_2. Set D_2=C_2 and

Then \phi_2q(f_b)=t_b\phi_1 for every b\in\mathcal A_{12}.

Next collect the adjusted maps

A single roof gives a denominator u_0:D_0\to C_0 and maps f_a:D_0\to D_1. Put \phi_0=\psi_0q(u_0). All the arrow squares now commute:

For \rho\in\mathcal R, form the morphism in K

The relations for t_a and (3.6) give \phi_2q(g_\rho)=0, so q(g_\rho)=0. Apply (3.5) to the tuple (g_\rho)_\rho. There is one denominator v:D'_0\to D_0 such that g_\rho v=0 in K for every \rho. Replace D_0 by D'_0, every map f_a with source vertex 0 by f_av, and \phi_0 by \phi_0q(v). The arrow squares still commute, and each relation becomes

There are no arrows into vertex zero and no longer paths requiring further compatibility. Thus the new data form a B-diagram in K, and the three \phi_i give an isomorphism from its image to the entire prescribed diagram in Q.□

Proof. The presentation, the quotient, the two cones, their isomorphisms in the completion, and the action on V_0 have all been chosen using only R,e,\alpha. Apply the finite lifting construction once to this fixed diagram. Its objects lie in K, so choose bounded right-projective complex representatives D_i and chain-map representatives f_a. Each relation is zero in K, hence its representative is null-homotopic. Choose one right-linear homotopy h_\rho for each of the finitely many relations. This gives (3.7).

The isomorphisms \phi_i from the lifting construction are fixed morphisms in Q and satisfy all the fixed arrow squares (3.6). Every \operatorname{ev}_Y factors through this same quotient. Applying it to those squares and then to the fixed identification V_0\cong V gives the asserted isomorphism of diagrams, by (3.4). Thus the same choices work for all Y.□

Proof. We construct the differential, compare its vertex complexes with the given ones, and finally compare the arrow actions. All tensor products in this construction are over k. Form the complexes of bimodules

The left B-action is on the first tensor factor and the right B-action is on the second. Let d_0,d_1,d_2 denote the respective internal differentials of these three complexes.

On the summand of L'_1 indexed by a:i\to j, define

Here subscripts indicate summands of L'_0. On the summand of L'_2 indexed by \rho, define

These are bimodule chain maps of degree zero. Define also the bimodule map h:L'_2\to L'_0 of degree -1 by

on the relation summand \rho. Expanding the composite pq gives

The last equality uses \rho=0 in B.

Set

as a graded bimodule, and equip it with the differential

For example, P^n=(L'_0)^n\oplus(L'_1)^{n+1} \oplus(L'_2)^{n+2}, so p,q,h each have total degree one in this matrix. The diagonal signs agree with our shift convention. Since p and q commute with the internal differentials, the (0,1) and (1,2) entries of d_P^2 vanish. Its remaining off-diagonal entry is

by the displayed calculation and the identity for h_\rho. The diagonal entries vanish as well. Thus P is a complex of bimodules.

Every term B\mathbf e_j\otimes_k D_i^n, viewed as a right B-module, is a finite direct sum of copies of D_i^n. Consequently all terms of P are finite dimensional and projective on the right. There are finitely many summands, and the complexes D_i are bounded; hence P is bounded.

It remains to recover the given vertex complexes and arrow maps from this strict bimodule complex.

Define

This is a chain map and is split injective as a map of graded right modules. We prove that its quotient is acyclic.

Call j the source index of a summand whose second tensor factor is D_j, in any of the three columns. Give \mathbf e_iP the decreasing filtration

These are subcomplexes. Indeed, the internal differentials and the path-multiplication parts of p,q preserve the source index; the f_a terms increase it; and h increases it from 0 to 2. In particular F^0=\mathbf e_iP and F^3=0. The quotient by \iota_i(D_i) inherits this finite filtration.

On the associated graded, only the internal differentials and the path-multiplication parts remain. The source-i part consists solely of \iota_i(D_i), since the only path from i to itself is the trivial path. It disappears in the quotient. Source indices larger than i give zero. If j<i are adjacent vertices, the source-j part is the signed total complex obtained by tensoring D_j with the exact sequence

There are no relations of length one, so the displayed map is an isomorphism.

For the remaining pair (j,i)=(0,2), that part is the signed total complex of D_0 tensored with

The three nonzero horizontal terms have degrees -2,-1,0. The first map sends the basis vector indexed by \rho to \sum_{b,a}\rho_{ba}(b,a); the second sends (b,a) to ba. The first map is injective because \mathcal R is a basis of the relation space. Its image is the kernel of the second map: all defining relations have length two, and no path of positive length can enter vertex 0 or leave vertex 2. Thus the ideal they generate, in this path space, is exactly their linear span. The second map is surjective because length-two paths span \mathbf e_2B\mathbf e_0.

These exact complexes of vector spaces are split, and tensoring with a bounded complex D_j gives an acyclic total complex. All associated graded pieces of the quotient are therefore acyclic. Induction through the finite filtration proves that the quotient itself is acyclic. Thus \iota_i is a quasi-isomorphism. Its cone is a bounded acyclic complex of projective right modules, and is contractible: starting in the largest nonzero degree, split the surjection onto its projective last term and continue downward. It follows that \iota_i is a homotopy equivalence.

Finally, for a:i\to j define a right-linear map of degree -1

The internal differential on this summand is -d_{D_i}; its contribution cancels that of H_ad_{D_i}. The remaining component is p, giving

This is the required arrow homotopy.□

Proof of Remark 4.1. Choose the fixed complexes D_i, arrow maps f_a, and relation homotopies h_\rho of Remark 3.4, and apply Remark 4.2. Since P and the D_i are bounded complexes of projective right modules, their ordinary tensor products with M(Y) compute the derived tensor products. Moreover, tensoring preserves the homotopy equivalences \iota_i and their arrow homotopies. Thus the vertex diagram of

in the derived category of k-vector spaces is isomorphic to the evaluated diagram of Remark 3.4.

At each vertex this diagram is eY\oplus eY[3]. The arrows s,s' act by identity maps, and a_h,a'_h act by \alpha(x_h) on each summand. Applying cohomology, with its induced arrow maps, therefore gives isomorphisms of left B-modules

In particular these identifications specify the full B-action, not just the vector spaces at the vertices.

Write A_{-3}=H^{-3}(T_Y) and A_0=H^0(T_Y). The canonical truncation triangle is

The connecting morphism belongs to

where the vanishing follows from the bound \gldim B\leq2 in Remark 3.1. The triangle consequently splits, giving T_Y\simeq A_0\oplus A_{-3}[3], which is (4.1). The complexes, chain maps, and homotopies were fixed before choosing Y, as was the rectification constructed from them.□

Proof. The case j=0 is immediate. For the induction step, apply \Phi to an isomorphism for j; derived tensor preserves isomorphisms, finite direct sums, and shifts. Apply (4.1) with H^{\circ j}(Y) in place of Y to each summand, and combine the two resulting multiplicities using Pascal’s identity. Each H^{\circ j}(Y) is finite dimensional. This argument uses the isomorphisms only one object at a time; no natural choice of the truncation splittings is required.□

Proof. The vertices of B\otimes_k C are pairs (i,j), where 0\le i\le2 and 0\le j\le l. Its diagonal vertex corners are k. A nonzero component between distinct vertices moves in the product of the two directed orders. Choosing a linear extension of that product order permits an application of Remark 3.1, on both sides. In particular,

Remark 3.1 also gives the bound 2 for B and B^{\mathrm{op}}. Modules over a finite product split according to its factors, so

Tensor (5.1) over k with the regular B-bimodule, placing its last projective term in degree zero. This gives a bounded resolution Q_T\to T by right B\otimes_k C-projectives, preserving the left B-action. Also, O is projective as a right B-module, being the finite direct sum of the terms of P. Thus Q_X=O\oplus Q_T, with the prescribed block actions, is a bounded right-D-projective resolution of X retaining its left D-action. Ordinary tensor with Q_X computes F on every complex. On the first factor, the block supports therefore give

In degree n, for -l\le n\le0, tensoring Q_T with O gives

The induced differential is d_P. In comparison, the standard differential of P[b_*] is (-1)^{b_*}d_P. Multiplication by (-1)^{b_*n} in degree n defines a chain isomorphism to P[b_*]; it respects both B-actions. Since this complex is termwise right-B-projective, its ordinary tensor with any Z computes the derived tensor. This proves the second isomorphism in (5.2), and hence the first.□

Proof. Use the ordinary unnormalized bar resolution of N over the field k. It is a free left A-resolution with degree -n term A\otimes_k A^{\otimes_k n}\otimes_k N. Its augmentation is a quasi-isomorphism: the augmented bar complex has the usual k-linear contraction obtained by inserting the unit. After tensoring with the right A-module D, it yields a model \mathcal B for D\otL_AN, with

The differential is the alternating sum of the faces that multiply adjacent algebra entries or apply the two endpoint actions.

Expand each internal copy of A as D\oplus X. Every nonzero face preserves the number of entries from X. Indeed, a product involving exactly one X-entry remains in X, a product of two X-entries is zero, and an endpoint action by X is zero on both D and N. Thus there is an actual decomposition of complexes

where \mathcal B_r consists of words with precisely r entries from X.

For fixed r, write i_0,\ldots,i_r for the lengths of the blocks of D-entries before, between, and after the X-entries. The corresponding tensor is

in degree -r-\sum_{j=0}^r i_j. We identify this with the multiple two-sided bar complex for

shifted by r.

First consider the signs. Within block j, a bar face has alternating-sign offset

In the ordinary multibar total differential, ordered from left to right, the corresponding offset is \sum_{h<j}i_h; shifting the total complex by r adds the factor (-1)^r. Rescale the component (6.2) by

When a face reduces i_j by one, this factor changes by (-1)^{r-j}, precisely the discrepancy between the two differentials. This gives the required chain isomorphism. When r\ge1 and i_j=0, there is no nonzero face in that block: the prospective face multiplies two X-entries or applies an endpoint X-action. For r=0, this is the ordinary two-sided bar complex for D\otL_DN. At i_0=0 its differential is absent; the map D\otimes_kN\to N is the separate augmentation.

We next justify that the multibar computes the genuinely derived tensor in (6.3), without any flatness assumption on X over D. For a bounded-above complex Z of left D-modules, the total left-module bar resolution has terms obtained from

in degree q-i. Its terms are free left D-modules because k is a field, and its augmentation to Z is a quasi-isomorphism. For each total degree there are only finitely many possible i, since Z is bounded above. Thus it is a bounded-above complex of free modules and can be used to compute a derived tensor.

Starting at the right, resolve N in this way and tensor with X. Resolve the resulting left D-complex in the same way and tensor with the next X, continuing for the fixed finite number r of factors. A final bar resolution with the leading factor D gives exactly the multibar (6.2), before its shift. All exterior D-actions are preserved. This establishes its identification with (6.3), which is isomorphic to F^rN.

Finally, in degree -n of \mathcal B only 0\le r\le n occur, and for each such r there are finitely many tuples satisfying r+\sum_j i_j=n. All totalizations above therefore use direct sums that are finite in each degree. There is no passage to products or completed tensor powers. Direct sums are exact for modules, so the componentwise quasi-isomorphisms give (6.1). The argument uses no connectedness assumption on D; in particular it applies to the product algebra of Section 5.□

Proof. Every fixed iterate F^rN has bounded, finite-dimensional cohomology. For example, if g bounds the right global dimension of D, then for every finite-dimensional left D-module V the groups \Tor_i^D(X,V) are finite-dimensional and vanish for i>g. Applying this observation to successive truncation triangles shows inductively that

In particular, all iterates lie in \D^{\le0}(D).

For the first assertion, vanishing at t implies vanishing at every r\ge t. By (6.1), Z=D\otL_AN then has bounded cohomology. Since X is a nilpotent ideal, it is contained in \rad A, and

Derived associativity gives

The right D-module S has a finite projective resolution, so this complex also has bounded cohomology.

Choose a minimal projective resolution of N over the finite-dimensional algebra A, using successive projective covers. Its terms are finitely generated, and every differential has image in the radical of its target. Tensoring with S=A/\rad A therefore annihilates every differential. If Q is a nonzero finitely generated projective A-module, then

by Nakayama’s Lemma. Consequently bounded cohomology of S\otL_AN forces the minimal resolution to have only finitely many nonzero terms. This proves \pd_A N<\infty.

For the second assertion, choose q\le0 with H^q(F^rN)\ne0. Its direct summand in (6.1) gives

A projective A-resolution of N of length strictly less than r would make the derived tensor zero in this degree. Hence \pd_A N\ge r, as asserted.□

Proof of Remark 1.1. Work over k=\C. Fix the algebra R, its central idempotent e, the automorphism \alpha, and the selection functor H supplied by Remark 2.1. Apply Remark 4.1 to these fixed data, obtaining B and P satisfying (4.1). Choose the support interval [a_*,b_*] once, and construct C,W,O,D,X as in Section 5. Finally fix the ordinary finite-dimensional unital algebra

These choices are made independently of m, as summarized in Table 1.

For a finite-dimensional left R-module Y, regard M(Y) as supported on the first factor of D. By (5.2) and (4.1),

Induction, using preservation of finite direct sums and shifts by derived tensor, gives

The induction applies the realization isomorphism separately to each module H^{\circ j}(Y); it requires no splitting natural in Y.

For each m\ge1, let Y_m be the finite-dimensional module supplied by (2.2), and let N_m=M(Y_m), first as a left D-module on its first factor and then as an inflated left A-module. It is finite-dimensional. Since H^{\circ m}(Y_m)=0 and H^{\circ(m-1)}(Y_m)\ne0, (6.4) yields

The latter nonvanishing follows because a finite direct sum of shifts of the nonzero module M(H^{\circ(m-1)}(Y_m)) is nonzero. Shifts preserve both extinction and nonvanishing; the lower bound now uses the iterate index r=2m-2 in Remark 6.2, not a new estimate on the shifts. The algebra D has finite global dimension on both sides by Remark 5.1. Applying Remark 6.2 proves

All N_m are finitely generated over the single fixed algebra A, while these lower bounds are unbounded. Therefore \findim A=\infty, which refutes the asserted finiteness for all finite-dimensional algebras.□

Proof of Remark 1.2. Let A be the algebra of Remark 1.1, and choose a basic finite-dimensional complex algebra C Morita equivalent to A^{\mathrm{op}}. The right-module little and big finitistic dimensions of C are infinite, and its injective right modules do not generate: these are the conclusions for left A-modules in Section 1, transported through Morita equivalence. Indeed, this equivalence preserves projective dimensions and finite generation, and the induced equivalence of unbounded derived categories preserves injectives and coproducts.

Choose primitive idempotents e_1,\ldots,e_n representing the simple right C-modules S_C(e_i)=e_iC/e_i\rad C. In Cummings’ Construction 2.1 [3], set

The left E-action on U factors through E/\rad E\cong\C^n, with the ith factor acting on S_C(e_i). This is a finite-dimensional complex algebra.

Write \findim_r and \Findim_r for the right-module dimensions, to display Cummings’ convention. By [3, Propositions 3.2–3.3],

His Proposition 4.3 says that injective generation for right T-modules implies it for right C-modules, while Proposition 4.4 establishes it for right T^{\mathrm{op}}-modules [3, Propositions 4.3–4.4]. Hence both right finitistic dimensions of T are infinite and its injective right modules do not generate; for T^{\mathrm{op}}, both right dimensions are zero and its injective right modules generate. Taking \Lambda=T^{\mathrm{op}} translates these statements into our left-module convention.□