A counterexample to Tachikawa’s second conjecture

Proof. We first record the extension property supplied by total acyclicity. If J is projective, then \Hom_G(P,J) is exact, since J is a summand of a finite sum of copies of G. Identify c_jP with \im d_j^P\subset P_{j-1}. For u:c_jP\to J, the map u d_j^P:P_j\to J is a cocycle. Exactness gives h:P_{j-1}\to J with h d_j^P=u d_j^P, so h extends u. It follows at once that the cycles of \Hom_G(P,N)^a are maps c_aP\to N and its boundaries are exactly the maps factoring through projectives. This proves the second isomorphism, with N=c_0Q.

For the first, begin in degree zero. A map c_0P\to c_0Q lifts to f_0:P_0\to Q_0, and then successively to f_j for j>0, using projectivity of P_j and exactness of Q. To proceed downwards, d_j^Qf_j vanishes on \ker d_j^P=\im d_{j+1}^P by the chain equation already obtained in degree j+1. It therefore defines a map c_jP=\im d_j^P\to Q_{j-1}. The extension property gives f_{j-1}. Thus every cokernel map lifts to a full chain map.

A homotopy induces a projective factorization on c_0. Conversely, suppose the cokernel map of f factors as c_0P\to J\to c_0Q with J projective. Extend its first factor to P_{-1} and lift its second factor to Q_0. Their composite is a homotopy component h_{-1}:P_{-1}\to Q_0 whose homotopy differential removes the induced map on cokernels. It remains to nullhomotope a chain map inducing zero there. For this remaining map set h_{-1}=0. For j\geq0, lift

to h_j:P_j\to Q_{j+1}; the initial image assertion is the vanishing on c_0, and each later one follows from the preceding chain equation. For j<0, the map f_j+d_{j+1}^Qh_j kills \im d_{j+1}^P by the equation in degree j+1. It descends to \im d_j^P and extends to h_{j-1}:P_{j-1}\to Q_j by the same extension property. These equations give f=dh+hd in every degree. The product in (2.1) permits all these components; each equation uses only two of them.

Replace P by P[-a] to obtain degree a. The construction by cokernels preserves compositions and shifts. Finally, the nonnegative tail of P resolves c_0P. In every positive cohomological degree its Hom complex has the same two adjacent differentials as \Hom_G(P,c_0Q), proving the Ext assertion, including degree one.□

Proof. For every finite projective J the finite-dimensional adjunction gives

The first identification is immediate for J=G, hence for finite sums and summands; the second uses a fixed symmetrizing form. Apply this termwise to a complete resolution P of M. Dualizing degree-zero cohomology of \Hom_G(P,N) yields

Here \ker(P_0\to P_{-1})=c_1P=M[-1]. A factorization through any projective lifts through P_1\twoheadrightarrow c_1P, so the displayed quotient is precisely \sHom_G(N,M[-1]). All identifications are natural; comparison of complete resolutions also gives naturality in M.

Define \lambda_M(g) by pairing 1_M with g. Naturality in N, applied to f:M\to N, identifies the pairing of f and g with \lambda_M(gf). Replacing N by N[a] and using the shift equivalences gives (2.6).□

Proof. Every chain map from P to an acyclic complex is nullhomotopic: start at the lowest nonzero degree of P and successively lift the residual map through the next differential of the target. The same argument applies to every shift. Thus the full Hom from P sends quasi-isomorphisms to quasi-isomorphisms, which proves the assertion about derived maps. This is the K-projectivity of a bounded-below projective source; it does not assert K-projectivity of a complete resolution.

For completeness, a bounded-below finite complex V admits a projective model \theta:Q\to V by the following induction. Having constructed lower terms, choose Q_j finite projective surjecting onto

Use its two components for \theta_j and d_j^Q. Every cycle of the comparison cone is then a boundary, so \theta is a quasi-isomorphism. A derived isomorphism between projective models and its inverse are represented by chain maps, and their composites are homotopic to the identities by the first assertion.

To verify the last assertion, represent the first arrow of the triangle by f:P_L\to P_V on any projective models. Its cone represents R, and the projection \Cone(f)\to P_L[1] represents \partial. The cone of the deshifted projection has P_V as a subcomplex with contractible quotient \Cone(1_{P_L}). It therefore represents V. Replacing the projective models by homotopy equivalent ones and the map by its prescribed representative preserves this conclusion: a homotopy commuting square induces a cone map with that homotopy as its off-diagonal component, and homotopic maps have isomorphic cones. The resulting cone \Cone(b[-1]) is exactly (2.8).□

Proof. Above the homology support, (2.3) is exact for P. It identifies (c_{n+1}P)[n+1] with (c_nP)[n], naturally on chain maps. A homotopy changes c_n(f) by a map factoring through P_{n-1}, so the construction descends to homotopy classes. Lemma 2.3 then makes it functorial on derived maps. The identity c_n(P[a])=c_{n-a}P gives shift compatibility. A projective resolution of a module recovers its successive syzygies, proving the assertion for modules. Restricting a lifted representative of an Ext class to high degrees gives the same successive syzygies of that map, proving the last assertion about classes.

Choose n so that all three complexes are exact in degrees at least n. Their cokernels give a short exact sequence

For injectivity on the left, suppose (x,0)=d_{n+1}^{P_V}(u,v). Then d_{n+1}^{P_R}v=0, so v=d_{n+2}^{P_R}w for some w. The equation d_Lb+bd_R=0 gives x=d_{n+1}^{P_L}(u+b_{n+2}w), as required. Surjectivity and exactness in the middle follow by lifting components in the graded splitting.

There is a commutative diagram with exact rows

The middle vertical map sends z to the class of (0,z). On an element d_{n+1}^{P_R}z, its value is the class of (b_{n+1}z,0), since (b_{n+1}z,d_{n+1}^{P_R}z) is a boundary in P_V. Thus the left vertical map is induced by b_{n+1}:P_{R,n+1}\to P_{L,n}. Shifting the left vertical arrow by [n+1] gives \langle P_R\rangle\to\langle P_L\rangle[1]. By the description after (2.4), this is the connecting map of the bottom row shifted by [n], and it is exactly \langle b\rangle. This also gives the asserted long exact sequences.□

Proof. Choose bounded-below projective models P,Q and represent the map by a global chain map f:P\to Q, using Lemma 2.3. For a sufficiently high n, both complexes are exact in degrees at least n and c_n(f) factors through a finite projective J:

Since J is injective, extend u across c_nP=\im d_n^P\subset P_{n-1}. Since J is projective, lift v to Q_n. The composite gives h_{n-1}:P_{n-1}\to Q_n. It makes f_n+h_{n-1}d_n^P zero after projection to c_nQ, so this residual map lifts through d_{n+1}^Q to h_n:P_n\to Q_{n+1}. Inductively, for j>n the chain equation and the preceding homotopy equation show that f_j+h_{j-1}d_j^P lands in \ker d_j^Q=\im d_{j+1}^Q. Lift it to h_j. Set h_j=0 for j<n-1.

The homotopic map f'=f+dh+hd now vanishes in every degree j\geq n. Its chain equation in degree n reads

Consequently its remaining components define a chain map into the actual homological subcomplex Q_{\leq n-1}, obtained by retaining Q_j for j\leq n-1 and setting the higher terms to zero. Equation (2.10) verifies the boundary condition at the cut. Thus f' factors through Q_{\leq n-1}, which is bounded and projective because Q was bounded below. Composing with the chosen model maps gives the desired derived factorization. This also covers a map with a module as target: first lift it globally to a projective resolution of that module, and then apply this argument.□

Proof. Take a projective bimodule surjection onto L. Its kernel is side-projective because the sequence splits on each side. For a positive shift, use the finite projective-injective embedding

Its target is injective by coinduction and projective by symmetry of E^e. The embedding splits on each side because L is injective on that side, so its cokernel is again side-projective. Iteration supplies finite representatives for every shift.

An E^e-projective evaluates to an E-projective: this is immediate for the free bimodule E\otimes E, which evaluates to E\otimes X, and passes to finite sums and summands. Hence evaluation respects projective factorizations. Every short exact sequence of side-projective bimodules splits on the right, and therefore remains exact after evaluation. In particular, evaluating the sequences just used to define shifts gives the first identification in (2.11). Applying the projective-presentation description of connecting maps to these same sequences shows that their actual connecting maps are preserved.

For fixed L, right projectivity makes L\otimes_E- exact, and left projectivity makes it preserve projectives. Applying it to projective resolutions and projective-injective coresolutions gives the second identification in (2.11). A bimodule map L\to L' gives a natural transformation on these complexes; thus evaluating it on a shifted module agrees stably with shifting its evaluation. The same reasoning proves naturality in X and compatibility with composition.

For the complex assertion, the high exact sequences 0\to c_{n+1}P\to P_n\to c_nP\to0 split on the right. Their evaluations remain exact, proving bounded homology of P\otimes_EX; its terms are projective by the preceding argument. Right exactness gives the canonical identifications

Together with (2.11), these identify the two sides of (2.12). They commute with chain maps. For a filtered complex, the cross differential in Lemma 2.4 evaluates to its tensor with X. That lemma’s explicit diagram therefore identifies the evaluated connecting map with the connecting map after evaluation.

Finally, suppose a homotopy equivalence on one side compares P with a complex supported in degrees at most d. At any n>d, the homotopy between the identity and the composite through that complex gives

Writing \pi_n:P_n\to c_nP, passage to the cokernel yields

Thus c_nP is a summand of a projective on that side. The separate homotopy equivalence on the other side proves the other projectivity. The homotopies need not be bimodule homotopies.□

Proof. The dual bimodule structure is

and the multiplication is

The bimodule identities give associativity, and (1,0) is a unit. The functional t(a,\lambda)=\lambda(1) gives the pairing

This pairing is symmetric and nondegenerate, and associativity of the algebra makes it an associative form. Thus R\ltimes DR is isomorphic to its dual as a bimodule.

For any algebra G over k, the natural isomorphism \Hom_G(L,DG)\simeq\Hom_k(L,k) shows that DG is an injective left G-module; the same argument applies on the right. This statement holds for arbitrary unital L, since vector-space duality is exact. Applying it to the symmetric algebra G=R\ltimes DR proves the final assertion.□

Proof. Write \nu=DR\otimes_R- and set I=\nu P. Finite projectivity gives natural isomorphisms

Both follow first for finite free modules and then for their direct summands. The module \Hom_R(P_j,R) is finite projective on the right, so I_j is injective on the left. Total acyclicity of P and exactness of D show that the entire complex I is exact.

As a right R-module, A=R\oplus DR. Consequently the restriction of A\otimes_RP is the exact complex P\oplus I. Every term A\otimes_RP_j is finite projective over A, since induction preserves a split direct-summand presentation. Right exactness identifies c_0(A\otimes_RP) with A\otimes_RZ=M. Thus the nonnegative tail of A\otimes_RP is a projective resolution of M. In particular, the tensor exactness needed here has been proved for this resolution; flatness of A_R is not an additional hypothesis.

Put N=\nu Z. Again by right exactness, N=c_0I. Exactness of I identifies this cokernel with \im(I_0\to I_{-1}). Hence

is an injective coresolution. If C^j=\Hom_R(P_j,Z) with differential \delta^j:C^j\to C^{j+1}, applying \Hom_R(Z,-) to (3.3) gives, in cohomological degrees 0,1,2,

Both arrows surrounding degree one are present. More generally, the term in degree a\geq0 is DC^{-a-1}. Thus, for every a>0,

The first of these equalities uses degree -2, which accounts for the endpoint in (3.1).

Induction–restriction adjunction on the projective resolution already constructed now yields

Here the first summand vanishes by (3.1) and Lemma 2.1, and the second by (3.4).

Lemma 3.1 proves that A is symmetric. The module M is finite and unital. Finally, if M were a direct summand of A^r, tensoring the splitting with the quotient A\to R would make

a direct summand of R^r. A tensor functor preserves a split identity composite without any flatness assumption. This contradicts nonprojectivity of Z.□

Proof. Give the basis vectors the degrees

The table is homogeneous and respects the corners. Thus associativity with an idempotent, or with incompatible adjacent corners, follows from (4.1). For positive-degree triples of total degree greater than four, both bracketings vanish. The remaining composable triples and the value of either bracketing are precisely

Every entry is obtained by two applications of (4.2); this exhausts the associativity check by trilinearity. The positive-degree span is a nilpotent ideal with fifth power zero and quotient ke\oplus kf\cong k^2. It is therefore the radical.□

Proof. The pairing eC\times Ce\to k that takes the coefficient of z in a product is nondegenerate. In the ordered bases

its only nonzero entries are the products ez,xy,yx,ze,uj,vt, with coefficients 1,q,1,1,1,q. Thus it has determinant q^2\ne0. The maps sending e to z^* give the stated left- and right-module isomorphisms.

Here is the full calculation for the other dual summands. The two surjections, on the ordered bases just specified, have the images

Each has rank four; their kernels are respectively \langle\ell_{-2},z\rangle_k and \langle\ell_0,z\rangle_k. The intermediate maps have images and kernels

For completeness, the initial injections are given by

Their four images are independent and span the respective kernels in the table. This proves exactness of (4.7) and (4.8), including their orientations and the exceptional index -2 in the first sequence.

On either C-side, T=C\oplus DC has a projective resolution of length at most two. Tensor two such resolutions over the field and totalize. The resulting resolutions of the corresponding B-modules E have length at most four, proving the first two bounds in (4.9). Likewise DB\cong DC\otimes DC has right B-projective dimension at most four. Dualizing a finite projective resolution gives an injective coresolution of {}_B B, and proves the last bound.□

Proof. On the basis (e,x,y,z,t,j) of Ce, right multiplication gives

For every i\ge0, the coefficient 1+q^{i+2} is nonzero. Consequently

The first image is \rad(Cf), and each later image is the preceding kernel. This proves exactness at every index. Since \Hom_C(Ce,s)=es=0 and \Hom_C(Cf,s)=fs=k, the first equality in (4.11) follows.

Under \Hom_C(Cf,C)=fC and \Hom_C(Ce,C)=eC, the dual cochain complex is

starting in degree zero. The first arrow is injective, with image \langle u,y+qx,z,v\rangle_k. For i\ge0,

The image and kernel are therefore

Since q\ell_{-1}=qx+y, the first kernel equals the image of \lambda_u. The only remaining cohomology is in degree two:

Finally vf=v, ve=0, and the only nonzero product of v with a radical basis vector is vt=qz, which is a boundary. Thus the radical acts trivially on k\overline v, and its right module structure is exactly Ds.□

Proof. Tensor the sequence of (T,C)-bimodules 0\to DC\to T\to C\to0 with R. Projectivity of each R_i gives a short exact sequence of complexes of left T-modules

The two outside complexes have action inflated from C. For a finite projective left C-module V, the natural evaluation map

is an isomorphism, first for a finite free module and then for its summands. Applying this termwise identifies the left complex in (5.3) with D\Hom_C(R,C). By (4.11), its only homology is s in homological degree two, with the asserted C-module structure. Good truncation therefore identifies it with s[2] in the derived category, also after inflating the action to T. Fix this identification. The quotient complex resolves s, so (5.3) gives (5.1). Its middle complex has projective T-terms, and the triangle gives its bounded homology.

Because \mathcal J is bounded below and projective, ordinary induction–restriction adjunction computes

Apply derived Hom into shifts of s to (5.1). With ordinary negative Ext interpreted as zero, the resulting long exact sequence contains

At i=1 this is k\to0\to\Ext_T^1(s,s)\to0, so the first Ext group vanishes. For every i\ge2, the middle map in (5.4) is an isomorphism. In particular \Ext_T^2(s,s)=0, and at i=3 it sends the identity of s to the nonzero class \tau. Induction now shows that \tau^m is a basis of \Ext_T^{3m}(s,s) for every m\ge0, and all other positive degrees vanish. Since the isomorphisms are the displayed Yoneda compositions, this proves the algebra assertion (5.2).□

Proof. Choose a projective T-resolution Q\to s. The total complex of Q\otimes Q is a projective E-resolution of X. There are finitely many summands in each total degree, and termwise finite projectivity gives an isomorphism of cochain complexes

The cohomology of this tensor product is the tensor product of the cohomologies: over a field each complex splits into its homology and contractible summands. Tensoring comparison maps of resolutions computes the external Yoneda products under this isomorphism. In particular the maps acting on the first and second tensor factors commute, since all interchange signs are +1 in characteristic two. Proposition 5.1 therefore gives \Ext_E^*(X,X)=k[\tau_1,\tau_2] as algebras.

Its degree-three part is nonzero, so X is not projective. Also \End_E(X)=k, and the identity cannot factor through a projective: such a factorization would exhibit X as a projective summand. Thus \sEnd_E(X)=k. Positive Ext agrees with positive stable Hom over the symmetric algebra E, proving (5.6). Lemma 2.2 gives

which proves the support and dimensions in (5.7).

To retain the actual multiplication maps, choose a nonzero functional \lambda:\cH^{-1}\to k. The composition pairing from Lemma 2.2 identifies \cH^{-3m-1} with D\mathcal P_m by

It is perfect for every m. For m\ge1, associativity of composition gives, for p\in\mathcal P_{m-1},

The right-hand side is pairing with polynomial multiplication by \tau_i. This proves the transpose assertion with its precise shift. For m=0, the target is \cH^2=0, which proves the boundary assertion as well.□

Proof. The dual basis vector f^*\in DC is the character of the f-simple on C. The dual bimodule action in (4.3) shows that multiplication on either side by an element of C scales f^* by that character; the square-zero ideal DC kills it on either side. Tensoring proves the asserted action on \zeta. The projective cover Ep_f\to X, ap_f\mapsto\chi(a), has kernel (\rad E)p_f by (4.5). The nonzero vector \zeta lies in this kernel, so the displayed map is an embedding into \Omega X.

We verify nonfactorization through every finite projective. Write J_E=\rad E. If b\in E satisfies J_Eb=0, symmetry gives

since ja\in J_E. Nondegeneracy therefore implies bJ_E=0. For any map X\to E, every vector in its image satisfies this condition. Any left-module map E\to J_Ep_f has the form b\mapsto br with r\in J_Ep_f, so its composite with X\to E is zero. The same conclusion holds for finite sums and summands of E. Thus all maps X\to\Omega X factoring through a finite projective are zero, whereas i_X\ne0. Finally \cH^{-1}=D\mathcal P_0 is one-dimensional by Proposition 5.2.□

Proof. First twist (5.3) by restriction along h_H^{-1}. Because h_H fixes C, the map h_H\otimes1 identifies its middle complex with the original one. It induces the identity on the quotient R and multiplication by H on DC\otimes_C R. Write \tau_H for the twist of \tau. Passing to the fixed derived identifications yields a commutative diagram of triangles

Naturality of the connecting morphism gives H\tau_H=\tau. Hence each external class \tau_i is sent to H^{-1}\tau_i, and preservation of products proves the first formula in (5.13).

For the negative part we first compute the action on \cH^{-1}. Twist the projective presentation

Under (5.12), its comparison with the original presentation is

It is E-linear, since \sigma_H(\sigma_H^{-1}(a)b)=a\sigma_H(b), and induces the identity on X. On the kernels it sends \zeta to H^2\zeta. Thus the generator in Lemma 5.3 is sent to H^2 times itself, proving that U_H acts by H^2 on \cH^{-1}.

Finally the exact twist functor preserves shifted composition. Applying it to the product in (5.8), and using its actions on \mathcal P_m and \cH^{-1}, gives

Equivalently, (U_H(\beta),p)_m=H^{m+2}(\beta,p)_m for every p. Nondegeneracy of the pairing forces U_H(\beta)=H^{m+2}\beta for every \beta, as asserted.□

Proof. The usual bar resolution of B has terms B\otimes B^{\otimes i}\otimes B. Its augmented complex is contracted as a complex of vector spaces by inserting a unit. Inserting a unit at one end also proves that tensoring it with a module at the other end gives a projective resolution of that module on the remaining side. Inducing the two outer actions gives K, with each term a finite free E^e-module.

On the left, B\otimes B^{\otimes i}\otimes E is a projective B-resolution of {}_BE. By Proposition 4.2, this module has a projective resolution supported in degrees 0,\ldots,4. Comparison of projective resolutions gives a homotopy equivalence between these two resolutions. Applying E\otimes_B- preserves the homotopy equivalence, so K, as a left E-complex, is homotopy equivalent to a complex of projectives supported in those degrees. The same argument using E_B works on the right. In particular H_i(K)=0 for i>4.

For completeness, this equivalence proves projectivity of the specific cokernel c_5K. On either side, choose a homotopy from the identity of K to a map factoring through the complex ending in degree 4. In degree 5 it gives

Modulo \im d_6, the identity of c_5K therefore factors as

where q_5:K_5\to c_5K is the quotient map. Thus c_5K is a summand of a projective on each side. This assertion concerns the two side-module structures; the two splittings need not be bimodule splittings. The high exact sequences 0\to c_{n+1}K\to K_n\to c_nK\to0, for n\geq5, now split on each side. Induction shows that all higher cokernels are side-projective as well.

To make the finite choice explicit, put N_0=c_5K, and for r=0,1,2,3 use

Here the E^e-action on the middle term is by right multiplication on its input. The middle term is finite injective, hence projective because E^e is symmetric. Since N_r is projective and therefore injective on each E-side, this sequence splits on both sides; it follows inductively that every N_r is side-projective. Take Y=N_4. Its stable class is (c_5K)[4]=\cI[-1], as required. Lemmas 2.4 and 2.6 give the stated stable images and their evaluation compatibilities. We denote the stable map induced by m by \overline m:\cI\to E.□

Proof. The evaluated complex is

Before inducing to E, its bar model and R\otimes R are projective B-resolutions of X=s\otimes s. They are homotopy equivalent. Induction preserves this equivalence and gives

In particular, Lemma 2.6 identifies its stable image with \cI X.

We retain the actual attaching maps in the triangle

of (5.1). Choose bounded-below projective models A_\bullet for s[2] and Q_\bullet for s, and a chain map t:Q_\bullet\to A_\bullet[1] representing \tau. The cone of t[-1] has underlying graded complex A_\bullet\oplus Q_\bullet, differential

and represents \mathcal J, by Lemma 2.3. Indeed the cone construction for the rotated triangle gives this identification; comparisons of projective models lift the connecting map and give homotopy equivalences of the cones. Thus this is a homotopy equivalent projective model, with its specified subcomplex and quotient, rather than only an equality of homology groups.

Tensor two copies of this model and shift by [-4]. Filter first by A_\bullet\otimes A_\bullet, then add Q_\bullet\otimes A_\bullet and A_\bullet\otimes Q_\bullet, and finally add Q_\bullet\otimes Q_\bullet. Every inclusion is split in each graded degree, and the successive quotients resolve

The stable image of the full complex is \mathcal L=(\cI X)[-4]. With the middle terms in the displayed order, the adjacent connecting maps are the shifted row and column

These are the cross differentials t in the relevant factor and the identity in the other factor. Lemma 2.4 identifies them with the connecting maps after taking stable images.

We first compute the kernels and cokernels of these adjacent maps in every degree. On the polynomial groups the maps form the two-variable Koszul sequence; on the negative groups they form its dual. We then use the two long exact sequences of the filtration to control the full top attaching map, including its possible component into the bottom level.

Write \mathcal P_m=k[\tau_1,\tau_2]_m for polynomials of total exponent m, and put \mathcal P_m=0 for m<0. By Proposition 5.2,

and all other \cH^a vanish. Multiplication by a degree-three class on the negative groups is transpose to polynomial multiplication. For each integer l, denote the adjacent maps by

Products in these formulas mean shifted postcomposition. We claim

To check the middle identity, the polynomial calculation for m\geq0 is the exact sequence

The row is onto since every positive-degree monomial is divisible by a variable. If \tau_1u+\tau_2v=0, coprimality implies u=\tau_2w and v=\tau_1w, uniquely; the signs agree in characteristic two. When the middle group is (D\mathcal P_m)^2, use the dual of (6.6), interchanging the two middle slots:

At m=0, the last term is the actual group \cH^2=0, and the first map D\mathcal P_1\to(D\mathcal P_0)^2 is an isomorphism. This is the seam l=1. If the middle group is unsupported, both its kernel and the image into it are zero. This proves (6.4) for every l.

For the row cokernel, a target \cH^l=\mathcal P_m gives \mathcal P_{m-1}^2\to\mathcal P_m, which is onto for m\geq1 and has cokernel k for m=0. A negative target \cH^l=D\mathcal P_m gives the dual of the injective column \mathcal P_m\to\mathcal P_{m+1}^2, and is onto for every m\geq0. All other targets are zero. For the column kernel, a nonnegative source \mathcal P_m maps injectively into \mathcal P_{m+1}^2. A source D\mathcal P_m for m\geq1 is the dual of the surjective row \mathcal P_{m-1}^2\to\mathcal P_m, so again the column is injective. The remaining source is \cH^{-1}=D\mathcal P_0=k, at l=3, with zero target (\cH^2)^2. These cases prove (6.5). They include the only possible multiplication from a supported negative degree across zero, namely \cH^{-1}\to\cH^2=0.

It remains to pass from adjacent maps to the whole filtered object. Let \mathcal L_1 be the stable image of the lower two levels of (6.3), and put N^l=\sHom_E(X,\mathcal L_1[l]). The first two levels give

Let

be the actual connecting map obtained by attaching the top level. Naturality of the quotient to the middle level gives q_{l+1}r_l=d'_l. Thus \ker r_l\subseteq\ker d'_l, which is zero unless l=3. At that index N^4=0: its left term in (6.7) is \coker d_3=0, and its right term is a kernel in (\cH^2)^2=0. Hence \ker r_3=\cH^{-1}=k.

Moreover,

For l\ne0, q_l is an isomorphism, so r_{l-1} is onto. At l=0, the vanishing \cH^{-2}=0 gives N^0=k, while the whole incoming source of r_{-1} is \cH^{-5}=0. Consequently \coker r_{l-1} is k at l=0 and zero elsewhere.

The final long exact sequence gives

The two scalar groups therefore survive precisely at l=0,3. This argument permits an additional component of the top attaching map into the bottom level: N^4=0 and \cH^{-5}=0 already force the asserted kernel and cokernel for the full r_l. Finally YX=(\cI X)[-1]=\mathcal L[3], so l=a+3. This gives exactly (6.1).□

Proof. Represent the perfect intermediate object by a bounded complex P of finite projective bimodules. We first replace it by a bounded complex P' whose terms belong to \add(DB\otimes B).

For a finite bimodule L, define

The map j_L:L\to J_B(L) is a natural bimodule monomorphism. Set C^0(L)=L and C^{r+1}(L)=\coker j_{C^r(L)}. These are additive functors and preserve split retracts. For a fixed d=4 we have the functorial exact sequence

For the free bimodule L=B\otimes B, this construction is the left-module coinduced coresolution of B, tensored over k with the right regular module B. This follows from \Hom_k(B,B\otimes B)=\Hom_k(B,B)\otimes B and from exactness of tensoring vector spaces. Each coinduced left-module term is a finite injective, hence belongs to \add(DB). The last left-module cokernel is also injective: by Proposition 4.2, \idim({}_BB)\leq4, and dimension shifting gives, for every finite left module V,

Here C^d(B) denotes the same coresolution construction on the left regular module. Thus every term on the right of L in (6.8), including the last cokernel, lies in \add(DB\otimes B) when L=B\otimes B. The assertion extends to every finite projective bimodule L, since the functors in (6.8) preserve finite sums and split retracts.

Apply this one fixed finite coresolution to each term of P. Naturality provides the horizontal maps, giving a bounded double complex. Put the first coinduced term in homological degree 0 relative to L, and the last cokernel in degree -d. Its totalization P' is bounded and has all terms in \add(DB\otimes B). The augmentation P\to P' is a quasi-isomorphism: filtering its cone by the finitely many original degrees has exact associated complexes (6.8). Thus the given derived factorization can be written B\to P'\to DB.

For V\in\add(DB\otimes B) and i>0, we next prove

It suffices to take V=DB\otimes B. Coinduction identifies, for a bimodule Q,

A projective bimodule resolution of B restricts to a right-projective resolution of the free right module B. This proves the first vanishing. For the second, tensor a left-projective resolution Q_\bullet of DB over k with the right regular module B. It is a projective bimodule resolution of DB\otimes B, and

Its positive cohomology vanishes since {}_BDB is injective. Both assertions pass to finite sums and summands.

Finally choose a projective resolution Q_B\to B and an injective coresolution DB\to I_{DB}, regarded as homological complexes. There are natural quasi-isomorphisms of Hom complexes

To verify them, filter by the finitely many terms of P'. For each successive quotient the claim is one of (6.9), together with equality in degree zero and vanishing of negative Ext between modules. A shift of a term merely shifts this comparison. Hence terms in negative degrees cause no exception, and induction on this finite filtration proves (6.10).

Taking degree-zero cohomology shows that both derived arrows have actual chain-map representatives \alpha:B[0]\to P' and \beta:P'\to DB[0]. Their composite is the ordinary map \beta_0\alpha_0, so it factors through P'_0\in\add(DB\otimes B). It equals v as an ordinary bimodule map, because \Hom_{\mathrm D(B^e)}(B,DB)=\Hom_{B^e}(B,DB).□

Proof. The character identities make \eta a bimodule map. By Lemma 6.4, a perfect factorization would give an ordinary factorization through \add(DB\otimes B). We show that every such factorization B\to DB sends 1 to a functional vanishing on p_f.

First consider maps

Identify DB\otimes B with \Hom_k(B,B) by \lambda\otimes c\mapsto[x\mapsto\lambda(x)c]. The element \alpha(1) satisfies a\alpha(1)=\alpha(1)a. Under this identification, its corresponding linear map f_0 therefore satisfies

Thus f_0(x)=d_0x, where d_0=f_0(1). If (c_i) is any vector-space basis of B, with dual basis (c_i^*), then

Right linearity of \beta reduces it to the left-module endomorphism \lambda\mapsto\beta(\lambda\otimes1) of DB. Duality identifies these endomorphisms with endomorphisms of the right regular module B, which are left multiplications. Thus the endomorphism of DB is right multiplication by some b'\in B. Consequently

Here L_a and R_a are left and right multiplication on the vector space B. The order b'd_0 follows from the dual action (\lambda b'c)(p_f)=\lambda(b'cp_f).

Write b'd_0=r+\sum_v a_vp_v, where r\in\rad B and the p_v are its four vertex idempotents. Left and right multiplications commute, and p_f^2=p_f, so

for sufficiently large n. The radical part has trace zero. For each v, the operator L_{p_v}R_{p_f} is a projection onto p_vBp_f. The corner dimensions of B, obtained from the corner table (4.6), are even; in particular \dim_k(p_vBp_f)=4 for every v. Each projection therefore has trace zero in characteristic two. Linearity gives (\beta\alpha(1))(p_f)=0.

A factorization through a finite sum of DB\otimes B is a sum of such composites. A factorization through a summand extends to that finite sum by its split inclusion and projection. The same vanishing thus holds throughout \add(DB\otimes B). But \eta(1)(p_f)=\chi_B(p_f)=1, a contradiction.□

Proof. The ordinary adjoint of f:L\to S is \widehat f(l\otimes x)=f(l)(x). A factorization of f through a finite E^e-projective evaluates to a factorization through an E-projective.

Conversely embed L in a finite projective-injective E^e-module J. The embedding splits as a right E-module map because L_E is projective, hence injective over symmetric E. Thus LX\to JX is injective. Suppose \widehat f factors as LX\to P\to X with P finite E-projective. Injectivity of P extends its first factor to JX\to P. Composing with P\to X and applying ordinary adjunction gives J\to S whose restriction to L is f. This is a bimodule projective factorization. The two factorization ideals therefore correspond exactly. Naturality is inherited from ordinary adjunction.□

Proof. Write U=U_H and \phi=\pi\overline m:\cI\to S. All shifts of U and \cI have finite side-projective representatives by Lemma 2.6. Consider the finite-dimensional linear map

Symmetric duality, Lemma 2.2 applied over E^e, identifies its transpose with

Indeed the fixed first argument is U[1], and its duality shift is U[1][-1]=U; naturality in the second argument gives the displayed precomposition.

Let \zeta=f^*\otimes f^*\in E, as in Lemma 5.3. Both its left and right actions are the character \chi. The map

is a bimodule map even for the twisted right action, because

We will prove that b_0\phi\ne0 stably.

Suppose otherwise. The derived map b_0\pi m:K\to U then has zero stable image and factors through a perfect E^e-complex by Lemma 2.5. To apply that lemma with projective models, one may take a bounded-below projective resolution Q\to U with finite terms and lift the global chain map K\to U to K\to Q; its stable image is the same map b_0\phi.

Restriction of this factorization to B^e still has a perfect intermediate object. Indeed an E^e-free module restricts to the outer bimodule E\otimes E. Tensoring the left and right B-projective resolutions of E from Proposition 4.2 gives a B^e-projective resolution of length at most 8. Finite sums and summands have the same bound. A bounded complex of these restricted modules is perfect by its finite term filtration.

Let \operatorname{Bar}(B)\to B denote the bar resolution. Inclusion of both outer B-factors in E defines a B^e-chain map

and hence a derived map B\to\Res_{B^e}K. Since \sigma_H fixes B, the decomposition

is a decomposition of B-bimodules. Let \rho be projection to the last summand. The composite

is exactly the character map \eta of (6.11) composed with the bar augmentation: in degree zero it sends a\otimes c to \chi_B(ac)\chi_B, since \rho(\zeta)=\chi_B, and in higher degrees it is zero. Thus our supposed factorization would make \eta factor through a perfect B^e-complex. Lemma 6.5 excludes this possibility. We have proved b_0\phi\ne0.

It follows from (6.12) that \phi_* is nonzero. Choose \beta:U[1]\to\cI with \phi\beta\ne0. Lemma 6.6, applied only to the side-projective source U[1], identifies this nonzero class with the composite

This identity follows from ordinary adjunction and from the evaluation of the representation \pi. The evaluated maps involving U[1],\cI,E are defined stably by Lemma 2.6; no stable evaluation functor on the nonside-projective module S is needed. The composite is nonzero, so \beta X itself is nonzero. The canonical shift comparison identifies U[1]X with (UX)[1]=X[1], proving the assertion.□

Proof of Proposition 6.1. Take the fixed finite side-projective Y constructed in Lemma 6.2. Lemma 6.3 proves its required Hom profile. For any H\in k^\times, shift the map of Lemma 6.7 by [-1] and identify \cI[-1] with Y to obtain a stable map U_H\to Y. By the shift and evaluation compatibilities of Lemma 2.6, its evaluation is a nonzero class X\to YX. A stable map between the fixed finite modules is represented by an ordinary bimodule map; choose one and call it g_H. All Hom spaces, dualities, and selections used above are over the original field k, so this requires no scalar extension. The construction of Y did not depend on H.□

Proof. The projection p^a in (7.2) is an isomorphism for a>0. For a\geq2 this follows from the support of W; at a=1 it also uses the surjectivity of (\cH^0)^2\to W^0. The same support argument applies for a\leq-4. At a=-3, both W^{-4} and \cH^{-3} are zero, so \mathcal V^{-3}=0 and p^{-3} is again an isomorphism. The remaining relevant low degrees can be read directly from (7.2):

Naturality of the bimodule projections F\to U_i gives

Indeed each projection sends v_0 to the identity of X, and evaluation of a bimodule map commutes with the specified shift comparisons. By Proposition 5.4, at a=3m>0 the map in (7.5) is the scalar matrix

and at a=-3m-1\leq-4 it is

In both cases m\geq1. Algebraic independence of H_1,H_2 implies that each determinant is nonzero in k, for every such m. In all other degrees in the asserted ranges, the source and target are zero by Proposition 5.2 and the already proved isomorphism p^a.

Finally, \Delta^0(c,d)=(c+d)v_0 is surjective. Since \cH^{-2}=0, the map \Delta^{-2}:0\to\mathcal V^{-2}=k is injective. Its surjectivity is neither claimed nor needed.□

Proof. Both column complexes have projective terms and are exact. Since \Lambda=\mathsf D_0(E)\oplus\mathsf D_1(E), (7.7)–(7.8) identify their Hom complexes into \Lambda with sums of

These are exact: P is totally acyclic and F is finite projective on the left. Hence both column complexes are totally acyclic, and so is their cone.

For an explicit finite description, Z is the triple with upper component X and lower component

whose structure map FX\to N_Z is induced by the first summand. To verify this formula, first quotient the degree-zero lower component by the image of the differential in \mathsf D_0(P). This gives FX in its first summand. The remaining relations arise from P_0 in the shifted second column. Its two components induce \widetilde v\epsilon and i_X\epsilon, respectively, yielding (7.10).

The upper-component functor sends the projective generators \mathsf D_0(E) and \mathsf D_1(E) to E and 0, and therefore sends every finite projective \Lambda-module to an E-projective. But X is nonprojective, since its degree-three self-extension group is nonzero by Proposition 5.2. Thus Z is nonprojective.□

Proof. Write A_0=\mathsf D_0(P) and B_0=\mathsf D_1(P), so that the underlying graded complex of P_Z is A_0\oplus B_0[1]. By (7.8), the block \Hom_\Lambda^\bullet(A_0,B_0[1]) is zero. Projection onto the two diagonal blocks therefore gives a short exact sequence of full Hom complexes

The underlying degreewise Hom spaces are products as in Lemma 2.1; the block decomposition has only four blocks. In particular, it does not replace any product by a direct sum.

The diagonal cohomology is (\cH^a)^2. To fix the cross-block index explicitly, in degree a we have

The differentials agree under this reindexing in characteristic two. Lemma 2.1 consequently gives H^a\mathcal C^\bullet=\mathcal V^{a-1}.

The connecting map in (7.11), from diagonal cohomology in degree a to H^{a+1}\mathcal C^\bullet=\mathcal V^a, is precisely \Delta^a. Indeed, lift two diagonal cocycles to the diagonal blocks of an endomorphism of the cone and take its cochain differential. The cross component is the sum of its two compositions with V. The first diagonal acts downstairs by F of the corresponding map, while the second acts by precomposition. Under the composition-compatible comparison in Lemma 2.1, this gives F(g)v_0+v_0[a]h, as in (7.3). Thus the long exact sequence gives

For a>1 or a\leq-3, both outer terms vanish by Lemma 7.1. At a=1, they vanish by surjectivity of \Delta^0 and injectivity of \Delta^1. At a=-2, they vanish by the isomorphism \Delta^{-3} and injectivity of \Delta^{-2}. Finally, apply Lemma 2.1 to the totally acyclic complex P_Z to identify this cohomology with H^a\Hom_\Lambda(P_Z,Z).□

Proof of Theorem 1.1. Fix the finite bimodule maps in (7.1), the complete resolution P, and the lifted map used in (7.9). These choices exist over k by the preceding propositions. The algebra \Lambda in (7.6) and its module Z in (7.10) are finite-dimensional. By Lemma 7.2, Z is nonprojective and has the totally acyclic resolution P_Z. Proposition 7.3 supplies both hypotheses on its complete self-Hom groups in Proposition 3.2. Apply that proposition with R=\Lambda to obtain

It proves that A is finite-dimensional and symmetric, that M is finite-dimensional and nonprojective, and that \Ext_A^i(M,M)=0 for every i>0. These are all the assertions of Theorem 1.1.□

Proof. The module G is a generator-cogenerator: it contains the regular module, and symmetry identifies B with its dual, an injective cogenerator. Moreover

The N,N summand vanishes by hypothesis. The summands with first argument B vanish by projectivity, and the remaining summand with second argument B vanishes by injectivity.

Mueller’s criterion [20, Lemma 3] says that, for a generator-cogenerator G, vanishing of these groups for 1\leq i\leq n is equivalent to dominant dimension at least n+2 for its endomorphism algebra. His Theorem 4 identifies the left and right dominant dimensions, so the opposite convention in the definition of \Gamma makes no difference to this assertion. Applying the criterion for every n proves \domdim({}_\Gamma\Gamma)=\infty.

We verify non-self-injectivity directly. The functor

is fully faithful and sends G to the left regular module \Gamma. Explicitly, if e\in\End_B(G) and h:G\to V, then e^{\mathrm{op}}h=h\circ e; full faithfulness follows first on finite sums of G and then on their summands. Suppose that the left regular module \Gamma were injective. Its direct summand H(N) would be injective as well. Choose a finite injective envelope N\hookrightarrow J over B. Since B is symmetric, J\in\add B\subseteq\add G. Left exactness of H gives a monomorphism H(N)\hookrightarrow H(J), which splits by injectivity of H(N). Full faithfulness gives a retraction J\to N of the original inclusion. Thus N is injective, hence projective over the symmetric algebra B, a contradiction.

Finally, every term in the minimal injective resolution of \Gamma is projective by infinite dominant dimension. If its injective dimension were finite, this resolution would end in a finite exact sequence

of projective-injective terms. For d>0, the last surjection splits because I^d is projective. Its kernel is then projective, and repeating backwards shows that the first inclusion splits. The case d=0 is immediate. In either case \Gamma is a summand of I^0 and is injective, contrary to the preceding paragraph. Hence \idim_\Gamma\Gamma=\infty.□

Proof of Corollary 1.2. Fix a field extension K/k. The symmetric structure survives extension of scalars: since A is finite-dimensional, the natural bimodule identification

turns A\simeq DA into a symmetrizing isomorphism for A_K.

We also record why the module hypotheses survive. For finite A-modules U,V, resolve U by finite projective modules. Finite projectivity gives, in every degree,

Flatness commutes with kernels and images, so

Thus M_K is self-orthogonal. Choose a finite projective surjection P\twoheadrightarrow M with kernel L. Its extension class in \Ext_A^1(M,L) is nonzero, since M is not projective. It remains nonzero under (8.1), because extension of fields is faithful. The extended surjection therefore does not split, and M_K is nonprojective. The degree-zero case of (8.1) is compatible with composition and also identifies

These are the same finite Ext and nonprojectivity base-change arguments used in the companion [22, Proposition 9.2].

Apply Proposition 8.1 to A_K,M_K, and abbreviate \Gamma_K to \Gamma. Write

for its minimal injective resolution. Every I^j is projective-injective and \idim_\Gamma\Gamma=\infty. This proves the first assertion and refutes the classical Nakayama conjecture. It also refutes the Auslander–Gorenstein conjecture: \pd_\Gamma I^j=0\leq j for every j\geq0, whereas the left self-injective dimension is already infinite.

There is a nonprojective indecomposable injective left \Gamma-module J. Indeed, finite-dimensional algebras have the same finite number of isomorphism classes of indecomposable projectives and indecomposable injectives, both indexed by the simple modules. If every indecomposable injective were projective, these classes would exhaust the indecomposable projectives, making the regular module injective. This contradicts Proposition 8.1. Since every I^j is projective, none contains J as a direct summand, giving the omitted-injective failure of the generalized Nakayama conjecture.

Let S=\soc J. The socle of an indecomposable injective over a finite-dimensional algebra is simple, and J is its injective envelope. If there were a nonzero map S\to I^j, it would be injective. Injectivity of I^j would extend it to a map J\to I^j. The extension is injective because S is essential in J, and it splits because J is injective. This would make J a summand of I^j, a contradiction. Hence \Hom_\Gamma(S,I^j)=0 for every j\geq0. Applying \Hom_\Gamma(S,-) to the resolution gives

The nonzero module S therefore refutes the strong Nakayama conjecture and also gives the simple-Ext formulation of generalized Nakayama failure. In particular, the degree-zero vanishing is part of the conclusion, not only the positive-degree vanishings.

Finally let W be the direct sum of one representative from each isomorphism class of indecomposable projective-injective left \Gamma-modules. There are finitely many such classes, so W is finite, projective, and injective. In particular \Ext_\Gamma^i(W,W)=0 for i>0, and every I^j belongs to \add W. The displayed injective resolution is therefore an infinite \add W-coresolution of \Gamma. Applying \Hom_\Gamma(-,W) preserves its exactness because W is injective. Thus W is Wakamatsu tilting and has projective dimension zero.

If W were tilting, there would be a finite exact coresolution

Every W^j is projective-injective. The backward splitting argument in Proposition 8.1 would make \Gamma injective, a contradiction. Hence W is not tilting, which refutes the Wakamatsu tilting conjecture.□

Proof of Corollary 1.3. Fix a field extension K/k and abbreviate \Gamma_K to \Gamma. Use the minimal injective resolution of the left regular module over this field,

Every I^n is finite-dimensional and projective-injective by Corollary 1.2. Set C_0={}_\Gamma\Gamma and C_n=\im(I^{n-1}\to I^n) for n\geq1. Exactness gives

Thus the C_{n+1} are successive cokernels, and all C_n are finite.

Since C_0 and I^0 are projective, the first sequence gives \pd_\Gamma C_1\leq1. If C_1 were projective, its surjection from I^0 would split, making \Gamma a direct summand of the injective module I^0. This would make \Gamma injective, contrary to \idim_\Gamma\Gamma=\infty. Hence \pd_\Gamma C_1=1.

Suppose n\geq1 and \pd_\Gamma C_n=n. Splicing a length-n projective resolution of C_n with the displayed short exact sequence gives \pd_\Gamma C_{n+1}\leq n+1. For every left module V and j\geq1, projectivity of I^n and the long exact Ext sequence give

There is a left module V with \Ext_\Gamma^n(C_n,V)\neq0, since otherwise the Ext criterion for projective dimension would give \pd_\Gamma C_n\leq n-1. Thus \Ext_\Gamma^{n+1}(C_{n+1},V)\neq0, proving the reverse bound and \pd_\Gamma C_{n+1}=n+1. Induction gives the asserted exact dimensions. The finite left modules C_n therefore make the little left finitistic dimension infinite. The big left finitistic dimension is at least the little one, so it is infinite as well.□

Proof. Fix K and abbreviate \Gamma_K to \Gamma. By Corollary 1.2, \Gamma fails the generalized Nakayama conjecture. Buan–Solberg [4, Theorems 3.1 and 3.4] therefore give an injective almost complete cotilting left \Gamma-module U with infinitely many pairwise nonisomorphic indecomposable complements X. The exact duality D_K=\Hom_K(-,K) sends finite left \Gamma-modules to finite left \Gamma^{\mathrm{op}}-modules. It sends finite injective dimension to finite projective dimension, preserves self-orthogonality, and dualizes the finite cotilting resolution of the injective cogenerator into a finite tilting coresolution of the regular module. Thus P_K=D_KU is projective and almost complete tilting, and D_KX are its complements. Duality preserves indecomposability and distinct isomorphism classes, so there are infinitely many such complements. The witnesses are chosen separately over each extension K/k.□

Proof. The base-change argument above gives a symmetric K-algebra A_K and a nonprojective self-orthogonal module M_K. Choose a nonprojective indecomposable summand N in one block B of A_K. The block B is indecomposable and symmetric, and \Ext_B^i(N,N)=0 for every i>0, since these groups are direct summands of \Ext_{A_K}^i(M_K,M_K).

Chen, Fang and Xi [6, proof of Theorem 5.13] construct from such a pair, over any field, reduced mirror-reflective algebras T_n that are indecomposable and symmetric. Their Corollary 5.12(1), with Definition 4.7, gives an idempotent stratification of length at least n. Theorem 1.2(2) supplies the stated successive nontrivial standard recollements of unbounded derived categories. Taking n=L proves the claim.□