\documentclass[11pt]{article} \usepackage[a4paper,margin=27mm]{geometry} \usepackage[T1]{fontenc} \usepackage{lmodern,amsmath,amssymb,amsthm,mathtools,array,booktabs,indentfirst,microtype} \usepackage[colorlinks=true,linkcolor=blue,citecolor=blue,urlcolor=blue,hyperfootnotes=false]{hyperref} \urlstyle{same} \setlength{\parindent}{1.5em} \setlength{\parskip}{0pt} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{remark}[theorem]{Remark} \DeclareMathOperator{\Ext}{Ext} \DeclareMathOperator{\Tor}{Tor} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\End}{End} \DeclareMathOperator{\soc}{soc} \DeclareMathOperator{\rad}{rad} \DeclareMathOperator{\coker}{coker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\pd}{pd} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\ann}{ann} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\per}{per} \newcommand{\op}{\mathrm{op}} \title{Counterexamples to the weakly Gorenstein symmetry conjecture} \author{Qiyue Tang} \date{} \begin{document} \maketitle \begingroup \renewcommand{\thefootnote}{} \footnotetext{\textit{2020 Mathematics Subject Classification.} Primary 16E65; Secondary 16G10.\newline \textit{Keywords.} Weakly Gorenstein algebra, Gorenstein-projective module, triangular matrix algebra, gendo-symmetric algebra.} \endgroup \begin{abstract} We construct a $29$-dimensional algebra which is left weakly Gorenstein and is not right weakly Gorenstein. We also construct a $145$-dimensional gendo-symmetric algebra with the same property. Both constructions work over every field containing an element of infinite multiplicative order. These examples disprove the weakly Gorenstein symmetry conjecture. \end{abstract} \section{Introduction} The study of Gorenstein-projective modules has its origins in Auslander and Bridger's theory of G-dimension. In that theory, modules of G-dimension zero are described by reflexivity and Ext vanishing for a module and its algebra dual; see \cite[Introduction]{JSref}. Jorgensen and \c{S}ega constructed modules over commutative Artinian local rings showing that reflexivity together with Ext vanishing for the module does not force Ext vanishing for its dual \cite[Theorem 1.7]{JSref}. This raises the question of which algebras allow Gorenstein-projectivity to be detected by Ext vanishing for the module alone. Let $A$ be a finite-dimensional algebra over a field $k$. All modules in this paper are finite-dimensional. A left $A$-module $X$ is \emph{semi-Gorenstein-projective} if $\Ext_A^n(X,A)=0$ for every integer $n>0$. It is \emph{Gorenstein-projective} if it is a cycle in a doubly infinite exact complex of finite projective modules which remains exact after applying $\Hom_A(-,A)$. Following Ringel and Zhang \cite{RZ}, the algebra $A$ is \emph{left weakly Gorenstein} if every semi-Gorenstein-projective left module is Gorenstein-projective. The right-hand versions use right modules, equivalently left modules over the opposite algebra $A^{\op}$, whose multiplication reverses that of $A$. Ringel and Zhang proved that $A$ is left weakly Gorenstein whenever there are only finitely many isomorphism classes of indecomposable left modules which are both semi-Gorenstein-projective and submodules of projective modules. They also constructed a six-dimensional algebra with a semi-Gorenstein-projective module which does not embed in a projective module \cite[Theorems 1.3 and 1.6]{RZ}. Marczinzik constructed gendo-symmetric algebras which are not weakly Gorenstein \cite[Theorem 5.5]{Mar} and asked whether left weak Gorensteinness implies right weak Gorensteinness \cite[\S6, Question (3)]{Mar}. Ringel and Zhang showed that an affirmative answer would imply the Gorenstein symmetry conjecture \cite[\S9.3]{RZ}. Huang formulates the left--right question as the \emph{weakly Gorenstein symmetry conjecture} \cite[Introduction]{Huang}: a finite-dimensional algebra is left weakly Gorenstein if and only if it is right weakly Gorenstein. We disprove this conjecture by constructing a triangular matrix algebra and a gendo-symmetric algebra which are left weakly Gorenstein and are not right weakly Gorenstein. A module is torsionless if it embeds in a finite projective module. For a module $X$, its algebra dual $X^*$ is $\Hom_A(X,A)$ on the left and $\Hom_{A^{\op}}(X,A)$ on the right. It is reflexive if the evaluation map $X\to X^{**}$, $x\mapsto(f\mapsto f(x))$, is an isomorphism. An algebra is gendo-symmetric if it is the endomorphism algebra of a generator over a finite-dimensional symmetric algebra. A generator here is a module whose finite direct sums have the regular module as a direct summand; a symmetric algebra is an algebra isomorphic to its vector-space dual as a bimodule. \begin{theorem}\label{thm:main} Let $k$ contain an element $\alpha\in k^\times$ of infinite multiplicative order. There is a $29$-dimensional $k$-algebra $T_\alpha$ such that every semi-Gorenstein-projective left $T_\alpha$-module is projective, whereas a $4$-dimensional right $T_\alpha$-module is semi-Gorenstein-projective and is not torsionless. There is also a $145$-dimensional gendo-symmetric $k$-algebra $E_\alpha$ such that every semi-Gorenstein-projective left $E_\alpha$-module is projective, whereas an $8$-dimensional right $E_\alpha$-module is reflexive and semi-Gorenstein-projective but is not Gorenstein-projective. Both algebras have infinite left and right self-injective dimensions. \end{theorem} A Gorenstein-projective module is torsionless and reflexive: these assertions follow by dualizing its complete projective resolution. Thus the modules in Theorem~\ref{thm:main} witness failure of right weak Gorensteinness. The construction uses the commutative $14$-dimensional algebra and modules of Jorgensen and \c{S}ega \cite[\S2.1 and Theorem 3.2]{JS}. Their modules have asymmetric Ext vanishing. We prove that one of these modules detects eventual Ext vanishing for \emph{every} finite module in the first variable. This property forces every semi-Gorenstein-projective left module over the triangular algebra to be projective. A module with vanishing positive Tor supplies the right-hand counterexample. A second detection argument for complexes gives the gendo-symmetric example. Theorem~\ref{thm:main} is proved in Sections~\ref{sec:triangular} and~\ref{sec:gendo}. Weak Gorensteinness for triangular matrix algebras is also studied through monic modules in \cite[Theorem 1.2]{Zhang}. The finite-projective-dimension hypotheses there give criteria in terms of the diagonal algebras. The detecting module used here has infinite projective dimension over the self-injective diagonal algebra. The parameter can be chosen as $2$ over $\mathbb Q$, and as an indeterminate over $\mathbb F_p$ for any prime $p$. We also determine the same triangular family at nontrivial roots of unity: in that case every semi-Gorenstein-projective module on either side is projective (Theorem~\ref{thm:finite}). We use $D=\Hom_k(-,k)$ for vector-space duality. The symbols $\pd_A$ and $\id_A$ denote projective and injective dimensions of left modules, and subscripts $A^{\op}$ denote the right-module versions. The assertion $\Ext_A^{n}(X,L)=0$ for $n\gg0$ means that it holds for every integer $n$ above some bound, allowed to depend on $X$ and $L$. The $n$th syzygy $\Omega^n X$ is the kernel at step $n-1$ of a minimal projective resolution, with $\Omega^0X=X$. \paragraph{AI contributions.} The research materials underlying this paper were generated with AI assistance. OpenAI Codex was used to rederive the arguments, run exact symbolic checks, and prepare and revise the manuscript. \section{A criterion for eventual Ext vanishing}\label{sec:detection} \begin{lemma}\label{lem:loewy} Let $A$ be a finite-dimensional local $k$-algebra with $A/J\cong k$, where $J=\rad A$. Let $L$ be a nonzero left module with $J^3L=0$. Its socle is $\soc L=\{\ell\in L:J\ell=0\}$, which contains $J^2L$. Put \[ a=\dim_k L/JL,\quad b=\dim_k JL/J^2L,\quad c=\dim_k J^2L,\quad s=\dim_k(\soc L/J^2L). \] Suppose $s>0$ and $bs>a(s+c)$. Then every left module $X$ satisfies \[ \Ext_A^n(X,L)=0\ (n\gg0)\quad\Longrightarrow\quad \pd_A X<\infty. \] For a module $U$ over this local algebra, let $\beta_n(U)$ denote the rank of the free module in degree $n\ge0$ of a minimal projective resolution. If a right module $U$ satisfies $\Tor_n^A(U,L)=0$ for $n\gg0$, its minimal Betti numbers satisfy \begin{equation}\label{eq:tor-growth} \beta_{n+1}(U)\ge \frac{b-ac/s}{a}\,\beta_n(U)\qquad(n\gg0). \end{equation} \end{lemma} \begin{proof} Finite projective modules over $A$ are free. Minimal differentials have entries in $J$. Any resulting $k$-linear map $L^p\to L^q$ has image in $(JL)^q$, and its image modulo $(J^2L)^q$ factors through $(L/JL)^p$. Hence \begin{equation}\label{eq:rank-bound} \rank_k(L^p\longrightarrow L^q)\le ap+cq. \end{equation} Apply $\Hom_A(-,L)$ to a minimal resolution of $X$. In its exact tail, the two adjacent ranks at $L^{\beta_n}$ sum to $(a+b+c)\beta_n$. Equation~\eqref{eq:rank-bound} therefore gives \begin{equation}\label{eq:ext-rank} b\beta_n\le a\beta_{n-1}+c\beta_{n+1}. \end{equation} The differential kills $(\soc L)^{\beta_n}$, so exactness puts this space in the preceding image. Modulo $(J^2L)^{\beta_n}$ the preceding image has dimension at most $a\beta_{n-1}$. Consequently \[ s\beta_n\le a\beta_{n-1},\qquad s\beta_{n+1}\le a\beta_n. \] Substituting the second inequality in \eqref{eq:ext-rank} gives \[ 0\le\beta_n\le\frac{a}{b-ac/s}\,\beta_{n-1}. \] The fraction lies strictly between $0$ and $1$. Iteration forces a nonnegative integer $\beta_n$ to be zero; the minimal resolution terminates. For $U$, tensor its minimal right resolution with $L$. Exactness and \eqref{eq:rank-bound}, now with the arrows reversed, give \[ b\beta_n\le a\beta_{n+1}+c\beta_{n-1},\qquad s\beta_{n-1}\le a\beta_n. \] Eliminating $\beta_{n-1}$ proves \eqref{eq:tor-growth}. \end{proof} \section{The algebra and the detecting module}\label{sec:algebra} Fix $\alpha\in k\setminus\{0,1\}$. The graded vector space $R=R_\alpha$ has the following basis, grouped by degree: \begin{equation}\label{eq:basis} 1;\quad t,u,v,x,y,z;\quad tv,uv,vx,vy,vz,tx;\quad \omega. \end{equation} The degrees are $0,1,2,3$, respectively; $R_d$ denotes the span of the basis elements of degree $d$, and $R_d=0$ for all other integers $d$. Define a commutative bilinear multiplication with unit $1$ by the following upper triangular table, and set all products of total degree at least $4$ equal to zero: \begin{equation}\label{eq:mult} \renewcommand{\arraystretch}{1.18} \begin{array}{c|cccccc} &t&u&v&x&y&z\\\hline t&-(\alpha+1)uv+vy&0&tv&tx&vx-tv&tv-(\alpha+1)vx\\ u&&0&uv&0&0&tx+\alpha uv\\ v&&&0&vx&vy&vz\\ x&&&&tx+tv&0&-\alpha vx\\ y&&&&&tx+(\alpha-1)uv&-vy\\ z&&&&&&0 \end{array} \end{equation} The nonzero products of degrees $1$ and $2$ are \begin{equation}\label{eq:pairing} t(vx)=u(vz)=v(tx)=x(tv)=x(vx)=y(vy)=z(uv)=\omega, \qquad z(tx)=-\alpha\omega. \end{equation} In particular, the degree-two basis labels in \eqref{eq:basis} are their indicated products. \begin{lemma}\label{lem:ring} These tables define an associative local symmetric algebra of dimension $14$. Its radical is the positive-degree part, and its socle is $k\omega$. \end{lemma} \begin{proof} Only triples of degree-one basis elements require an associativity check. For $r,s,q\in\{t,u,v,x,y,z\}$, compute the coefficient of $\omega$ in $(rs)q$ using \eqref{eq:mult} and \eqref{eq:pairing}. The complete list of nonzero values, up to permutation of the three entries, is \[ \begin{array}{c|rrrrrrrrrr} \{r,s,q\}&tty&ttz&tvx&txz&uvz&vxx&vyy&xxx&xxz&yyz\\\hline \text{coefficient}&1&-(\alpha+1)&1&-\alpha&1&1&1&1&-\alpha&-1. \end{array} \] Each value is unchanged by permuting the entries; all unlisted values are zero. Thus $(rs)q=r(sq)$. Triples involving $1$ satisfy associativity by the unit rule, and every other triple has total degree at least $4$, so both bracketings vanish. Let $\varepsilon:R\to k$ send an element to its coefficient of $\omega$ in \eqref{eq:basis}. The matrix of $(r,s)\mapsto\varepsilon(rs)$ between degrees $1$ and $2$, in the displayed orders, is \begin{equation}\label{eq:gram} \begin{pmatrix} 0&0&1&0&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\\ 1&0&1&0&0&0\\0&0&0&1&0&0\\0&1&0&0&0&-\alpha \end{pmatrix}. \end{equation} Its determinant is $1$ or $-1$. Degrees $0$ and $3$ pair by $\varepsilon(\omega)=1$. The full pairing is nondegenerate and associative, so $R\to DR$, $r\mapsto(s\mapsto\varepsilon(rs))$, is a bimodule isomorphism. Since $DR$ is injective on both sides, this isomorphism also makes $R$ self-injective. The positive-degree ideal $J$ is nilpotent and $R/J=k$, hence it is the radical. Nondegeneracy between degrees $1$ and $2$ excludes socle elements in those degrees, giving $\soc R=k\omega$. \end{proof} These are the multiplication tables of the algebra in \cite[\S2.1, Proposition 2.2 and Appendix A]{JS}. The table description will also be used when $\alpha$ has finite order. For $\lambda\in k$, define an $R$-linear map on column vectors by \[ \delta_\lambda:R^2\longrightarrow R^2,\qquad \delta_\lambda=\begin{pmatrix}v&y\\\lambda x&z\end{pmatrix}. \] Write $\delta_\lambda^{\mathsf T}$ for its $2\times2$ matrix transpose. Define \begin{equation}\label{eq:module} M=M_\alpha=\coker\delta_\alpha^{\mathsf T}. \end{equation} The quotient inherits this grading; write $M_d$ for its degree-$d$ component. The images $e_1,e_2$ of the standard basis of $R^2$ have degree zero and satisfy \begin{equation}\label{eq:relations} ve_1+ye_2=0,\qquad \alpha xe_1+ze_2=0. \end{equation} \begin{lemma}\label{lem:module} The module $M$ has radical layers of dimensions $(2,10,2)$, and \begin{equation}\label{eq:socle} \soc M=k(te_1+te_2-xe_2)\oplus k(uv)e_2\oplus k(vx)e_2. \end{equation} For every left $R$-module $L$, \begin{equation}\label{eq:detect} \Ext_R^n(L,M)=0\ (n\gg0)\quad\Longrightarrow\quad L\text{ is projective}. \end{equation} For every right $R$-module $U$ with $\Tor_n^R(U,M)=0$ for $n\gg0$, \begin{equation}\label{eq:three} \beta_{n+1}(U)\ge3\beta_n(U)\qquad(n\gg0). \end{equation} \end{lemma} \begin{proof} In degree one the two relations in \eqref{eq:relations} are independent; a basis of the quotient is \begin{equation}\label{eq:degree-one} te_1,ue_1,ye_1,ze_1,te_2,ue_2,ve_2,xe_2,ye_2,ze_2. \end{equation} Multiplying the relations by $t,u,v,x,y,z$ gives the following degree-two reductions: \begin{equation}\label{eq:reductions} \begin{array}{c|rrrrrr} & (tv)e_i&(uv)e_i&(vx)e_i&(vy)e_i&(vz)e_i&(tx)e_i\\\hline i=1&0&0&0&(uv)e_2&0&(vx)e_2\\ i=2&(vx)e_2&(uv)e_2&(vx)e_2&0&0&-\alpha(uv)e_2. \end{array} \end{equation} Here are the eliminations. The $u$-multiple of the first relation gives $(uv)e_1=0$; its $v$- and $x$-multiples give $(vy)e_2=0$ and $(vx)e_1=0$. The $v$-multiple of the second then gives $(vz)e_2=0$. Its $x$-multiple yields $(tx)e_1+(tv)e_1=(vx)e_2$. The $t$-multiple of the first gives $(tv)e_1+(vx)e_2-(tv)e_2=0$. The $t$-multiple of the second, combined with these two equalities, gives $(\alpha-1)((vx)e_2-(tv)e_2)=0$. Thus $(tv)e_2=(vx)e_2$, $(tv)e_1=0$, and $(tx)e_1=(vx)e_2$. The $z$-multiple of the first gives $(vz)e_1=0$. The $u$-multiple of the second gives $(tx)e_2=-\alpha(uv)e_2$. Finally the $y$-multiple of the first gives $(vy)e_1=(uv)e_2$. The linear map from $R_2^2$ to $k^2$ prescribed by \eqref{eq:reductions}, sending $(uv)e_2$ and $(vx)e_2$ to the standard basis, kills all twelve degree-two relations. Hence these two elements are independent. In degree three the relations give \[ (tx)(v,y)=(\omega,0),\qquad (uv)(\alpha x,z)=(0,\omega), \] so $M_3=0$. This proves the stated radical layers and $J^3M=0$. Write a degree-one element in the order \eqref{eq:degree-one} with coefficients $p_1,\ldots,p_{10}$. Multiplication by the six generators has the following coordinates in $((uv)e_2,(vx)e_2)$: \begin{equation}\label{eq:socle-system} \begin{array}{c|cc} & (uv)e_2&(vx)e_2\\\hline t&p_1-(\alpha+1)p_5-\alpha p_8&p_7-\alpha p_{10}\\ u&p_7&p_4\\ v&p_3+p_6&p_5+p_8\\ x&-\alpha(p_5+p_8)&p_1+p_7+p_8-\alpha p_{10}\\ y&-p_4-p_9&p_3\\ z&-p_3&p_2-\alpha(p_5+p_8). \end{array} \end{equation} Setting these entries to zero first gives $p_7=p_4=p_3=p_9=p_6=0$ and $p_5=-p_8$. The $z$ row gives $p_2=0$, the $t$ row gives $p_{10}=0$ and $p_1=p_5$, and the remaining equation is then satisfied. There is no degree-zero socle element since $te_1,te_2$ are independent. All of $M_2$ lies in the socle, proving \eqref{eq:socle}. Lemma~\ref{lem:loewy} applies with $(a,b,c,s)=(2,10,2,1)$, since $10>6$. It gives finite projective dimension in \eqref{eq:detect} and the factor $3$ in \eqref{eq:three}. Over a self-injective algebra a finite projective resolution splits backwards from its last term, so a module of finite projective dimension is projective. \end{proof} \section{The parameter complex}\label{sec:complex} \begin{lemma}\label{lem:ranks} For every $\lambda\in k$, \begin{equation}\label{eq:chain} \delta_{\alpha\lambda}\delta_\lambda=0. \end{equation} If $\lambda\ne0,1$, then $\rank_k\delta_\lambda=14$; if $\lambda=1$, its rank is $13$. The same ranks hold for $\delta_\lambda^{\mathsf T}$. \end{lemma} \begin{proof} The four entries of the product in \eqref{eq:chain} are \[ v^2+\lambda yx,\quad vy+yz,\quad \alpha\lambda xv+\lambda zx,\quad\alpha\lambda xy+z^2, \] which vanish by \eqref{eq:mult}. For $\lambda\ne0$, the degree-one image has basis $(v,\lambda x),(y,z)$. The degree-two image is spanned by the following ten columns: \begin{equation}\label{eq:ten} \begin{gathered} (tv,\lambda tx),\ (uv,0),\ (0,\lambda vx),\ (vx,\lambda(tx+tv)),\ (vy,0),\ (vz,-\alpha\lambda vx),\\ (vx-tv,tv-(\alpha+1)vx),\ (0,tx+\alpha uv),\ (vy,vz),\ (tx+(\alpha-1)uv,-vy). \end{gathered} \end{equation} They are the images of $(t,0),(u,0),(v,0),(x,0),(y,0),(z,0),(0,t),(0,u),(0,v),(0,y)$. The two omitted degree-one source vectors satisfy \[ \delta_\lambda(0,x)=-\frac{\alpha}{\lambda}\delta_\lambda(v,0),\qquad \delta_\lambda(0,z)=-\delta_\lambda(y,0). \] If a linear combination of the ten columns is zero with coefficients $q_1,\ldots,q_{10}$, the first coordinates give \[ q_1=q_7,\quad q_4=-q_7,\quad q_2=q_6=q_{10}=0,\quad q_5=-q_9. \] In the second coordinates, the coefficients of $tx,vz,tv,vx$ then give, respectively, \[ q_8=0,\qquad q_9=0,\qquad (1-\lambda)q_7=0,\qquad \lambda q_3-(\alpha+1)q_7=0. \] For $\lambda\ne1$ the ten columns are independent. For $\lambda=1$ these equations leave exactly one free coefficient, so their rank is $9$. The degree-three image is all of $R_3^2$, since the images of $(vx,0)$ and $(tx,0)$ include $(0,\lambda\omega)$ and $(\omega,0)$. The total ranks are consequently $2+10+2$ and $2+9+2$. On $R^2$ use the pairing $((r_1,r_2),(s_1,s_2))\mapsto\varepsilon(r_1s_1+r_2s_2)$. The matrix transpose over the commutative algebra $R$ is the adjoint for this nondegenerate pairing, so it has the same $k$-linear rank. \end{proof} Assume now that $\alpha$ has infinite multiplicative order. By Lemma~\ref{lem:ranks}, consecutive maps in \begin{equation}\label{eq:resolution} \cdots\xrightarrow{\delta_{\alpha^3}^{\mathsf T}}R^2 \xrightarrow{\delta_{\alpha^2}^{\mathsf T}}R^2 \xrightarrow{\delta_\alpha^{\mathsf T}}R^2\longrightarrow M\longrightarrow0 \end{equation} have zero composite and rank $14$. Their images therefore equal their kernels, since $\dim_kR^2=28$. All entries lie in $J$, so this is a minimal free resolution. Define the ideal and quotient \begin{equation}\label{eq:N} K=(t,u,v-x,y-x,z-x)\subset R,\qquad N=R/K. \end{equation} The relations identify $v,x,y,z$ with a single element $\xi$, while $t,u$ vanish. All degree-two products vanish in the quotient: $xy=0$ gives $\xi^2=0$. Conversely the assignment $v,x,y,z\mapsto\xi$, $t,u\mapsto0$ respects the tables. Thus $N\cong k[\xi]/(\xi^2)$. \begin{lemma}\label{lem:tor} If $\alpha$ has infinite order, then \[ \Tor_n^R(N,M)=0\quad(n>0),\qquad N\otimes_RM\cong k^2, \] where $1\otimes e_1,1\otimes e_2$ give the displayed basis and $J$ acts trivially. \end{lemma} \begin{proof} Tensor \eqref{eq:resolution} with $N$. The differential in degree $n\ge1$ becomes \[ \xi\begin{pmatrix}1&\alpha^n\\1&1\end{pmatrix}:N^2\longrightarrow N^2. \] Its coefficient matrix has determinant $1-\alpha^n\ne0$. The image and kernel are both $\xi N^2$. Hence the positive homology vanishes, and the cokernel in degree zero is $N^2/\xi N^2$ with the asserted basis. \end{proof} \section{The triangular counterexample}\label{sec:triangular} For $\alpha\ne0,1$ define \begin{equation}\label{eq:T} T=T_\alpha=\begin{pmatrix}R&M\\0&k\end{pmatrix},\qquad \begin{pmatrix}r&m\\0&c\end{pmatrix} \begin{pmatrix}r'&m'\\0&c'\end{pmatrix} =\begin{pmatrix}rr'&rm'+mc'\\0&cc'\end{pmatrix}. \end{equation} Its dimension is $14+14+1=29$. A left $T$-module is a triple $(Y,V,\varphi)$ with $Y$ a left $R$-module, $V$ a $k$-space and $\varphi:M\otimes_kV\to Y$ an $R$-linear map; the action is \[ \begin{pmatrix}r&m\\0&c\end{pmatrix}(y,v)=(ry+\varphi(m\otimes v),cv). \] For an $R$-module $Y$ and a $k$-space $V$, write \[ iY=(Y,0,0),\qquad jV=(M\otimes_kV,V,1). \] Here the third entry $1$ denotes the identity of $M\otimes_kV$. The left projectives are the direct sums of $iP$ and $jV$, with $P$ projective over $R$. \begin{proposition}\label{prop:left} Every semi-Gorenstein-projective left $T$-module is projective. More generally, $\Ext_T^n(S,T)=0$ for $n\gg0$ implies $\pd_T S\le1$. \end{proposition} \begin{proof} For $S=(Y,V,\varphi)$ choose a finite free surjection $P\twoheadrightarrow Y$. Combining it with $\varphi$ gives a surjection $iP\oplus jV\twoheadrightarrow S$ whose second component is $1_V$. Its kernel has second component zero, so for some $R$-module $L$ there is an exact sequence \begin{equation}\label{eq:left-seq} 0\longrightarrow iL\longrightarrow iP\oplus jV\longrightarrow S\longrightarrow0. \end{equation} The functor $i$ is exact and preserves projectives. Termwise Hom on an $R$-projective resolution gives \begin{equation}\label{eq:left-ext} \Ext_T^n(iL,T)\cong\Ext_R^n(L,R\oplus M)\qquad(n\ge0). \end{equation} Eventual vanishing for $S$ gives eventual vanishing for $iL$ by dimension shifting. Lemma~\ref{lem:module} makes $L$ projective, proving $\pd_T S\le1$. If $S$ is semi-Gorenstein-projective, then $\Ext_T^1(S,iL)=0$ because $iL$ is a summand of a finite sum of copies of $T$. Thus \eqref{eq:left-seq} splits and $S$ is projective. \end{proof} A right $T$-module is a triple $(U,W,\psi)$ with $U$ a right $R$-module, $W$ a $k$-space, $\psi:U\otimes_RM\to W$ a $k$-linear map, and action \[ (u,w)\begin{pmatrix}r&m\\0&c\end{pmatrix}=(ur,\psi(u\otimes m)+wc). \] Put $\mathcal J(U)=(U,U\otimes_RM,1)$, with identity structure map. The right projectives are direct sums of $\mathcal J(P)$ and $(0,W,0)$ with $P$ projective over $R$. \begin{proposition}\label{prop:bad} If $\alpha$ has infinite order, the right $T$-module $S_0=\mathcal J(N)$ has dimension $4$, satisfies $\Ext_{T^{\op}}^n(S_0,T)=0$ for every $n>0$, and its evaluation map has a one-dimensional kernel. \end{proposition} \begin{proof} Lemma~\ref{lem:tor} gives the dimension. Apply $\mathcal J(-)$ to a projective resolution of the right $R$-module $N$. Exactness in its first component is the original resolution, and exactness in its second component follows from the Tor vanishing in Lemma~\ref{lem:tor}. Thus it is a projective resolution of $S_0$. For every projective $P$, \[ \Hom_{T^{\op}}(\mathcal J(P),T)\cong\Hom_R(P,R). \] Indeed, a morphism to $(0,k,0)$ is zero by compatibility with the structure map, and a morphism to $\mathcal J(R)$ is uniquely determined by $P\to R$. Since $R$ is injective, this proves the Ext vanishing. As a left module $S_0^*=i(N^*)$, where $N^*=\Hom_R(N,R)=\ann_RK$ and $\ann_RK=\{r\in R:Kr=0\}$. To compute evaluation, consider the degree-two element \[ h=tv+tx+(\alpha+1)uv+vy. \] The tables give \[ th=uh=0,\qquad vh=xh=yh=zh=\omega. \] Thus $Kh=K\omega=0$. Frobenius duality gives $\dim_k\ann_RK=\dim_kDN=2$; the independent elements $h,\omega$ therefore form a basis of this annihilator. The degree-two reductions give \begin{equation}\label{eq:h-action} he_1=he_2=(uv)e_2+(vx)e_2\ne0,\qquad \omega M=0. \end{equation} The first component of $S_0\to S_0^{**}$ is $N\to N^{**}$ over the symmetric algebra $R$, and is an isomorphism. Its second component is \[ N\otimes_RM\longrightarrow\Hom_R(N^*,M),\qquad n\otimes m\longmapsto(f\longmapsto f(n)m). \] Identify $f\in N^*$ with $f(1)\in kh\oplus k\omega$. Equation~\eqref{eq:h-action} shows that the two basis elements $1\otimes e_1$ and $1\otimes e_2$ have the same nonzero image. The kernel is exactly \[ k\bigl(1\otimes(e_1-e_2)\bigr). \] If a module embeds in a finite projective, its regular-module-valued functions separate elements, so its evaluation is injective. Therefore $S_0$ is not torsionless and is not Gorenstein-projective. \end{proof} \begin{proposition}\label{prop:selfid} For every $\alpha\ne0,1$, both self-injective dimensions of $T_\alpha$ are infinite. \end{proposition} \begin{proof} The module $M$ has dimension $14$ and top dimension $2$. A projective module over the local algebra $R$ is free and has dimension $14$ times its top dimension. Thus $M$ is not projective. Since $R$ is self-injective, finite projective or injective dimension would force $M$ to be projective. Hence both dimensions of $M$ are infinite. Equation~\eqref{eq:left-ext}, for arbitrary $L$, implies $\id_TT=\infty$. Let $s=\left(\begin{smallmatrix}0&0\\0&1\end{smallmatrix}\right)\in T$. The dual of the right projective $sT=(0,k,0)$ is the left module $(0,k,0)$. There is an exact sequence \[ 0\longrightarrow iM\longrightarrow jk\longrightarrow D(sT)\longrightarrow0. \] Applying $i$ to a minimal $R$-resolution of $M$ gives a minimal $T$-resolution of $iM$, so $\pd_TiM=\infty$. Therefore $\pd_TD(sT)=\infty$. It is a summand of $DT$, and vector-space duality gives $\id_{T^{\op}}T=\pd_TDT=\infty$. \end{proof} Propositions~\ref{prop:left}--\ref{prop:selfid} prove the assertion about $T_\alpha$ in Theorem~\ref{thm:main}. \section{Finite-order parameters}\label{sec:finite} \begin{lemma}\label{lem:right-class} Let $R$ be a finite-dimensional self-injective local algebra with residue field $k$, and let $T=\left(\begin{smallmatrix}R&M\\0&k\end{smallmatrix}\right)$. Every semi-Gorenstein-projective right $T$-module has the form \[ \mathcal J(U)\oplus(0,W,0),\qquad \Tor_n^R(U,M)=0\quad(n>0). \] \end{lemma} \begin{proof} Write a minimal right projective resolution as $Q_n=\mathcal J(P_n)\oplus(0,W_n,0)$. A morphism $\mathcal J(P_n)\to(0,W_{n-1},0)$ is zero. The diagonal map $(0,W_n,0)\to(0,W_{n-1},0)$ is zero by minimality, since its coefficients lie in the radical of $k$, which is zero. Applying $\Hom_{T^{\op}}(-,(0,k,0))$ therefore gives a complex with zero differentials and term $DW_n$. Ext vanishing implies $W_n=0$ for $n>0$. Consequently the resolution has only $\mathcal J(P_n)$ in positive degrees, and its degree-one differential lands in $\mathcal J(P_0)$. On taking the first component it becomes a projective $R$-resolution of $U=\coker(P_1\to P_0)$. On taking the second component, positive exactness gives $\Tor_n^R(U,M)=0$ for $n>0$. The cokernel in degree zero is $\mathcal J(U)\oplus(0,W_0,0)$, as asserted. \end{proof} \begin{theorem}\label{thm:finite} If $\alpha\in k^\times$ has finite order $m\ge2$, every semi-Gorenstein-projective module on either side of $T_\alpha$ is projective. \end{theorem} \begin{proof} The left side is Proposition~\ref{prop:left}. For the right side it suffices by Lemma~\ref{lem:right-class} to prove that eventual vanishing of $\Tor_n^R(U,M)$ forces $U$ projective. For $\lambda\ne0$ define $C_\lambda=\coker\delta_\lambda^{\mathsf T}$. Its presentation $R^2\twoheadrightarrow C_\lambda$ is a projective cover, since its relations lie in the radical. If $\lambda$ and $\alpha\lambda$ are both different from $1$, Lemma~\ref{lem:ranks} and \eqref{eq:chain} imply \[ \Omega C_\lambda\cong C_{\alpha\lambda}. \] At $\lambda=\alpha^{-1}$, the kernel of $\delta_{\alpha^{-1}}^{\mathsf T}$ has dimension $14$ and contains the $13$-dimensional image of $\delta_1^{\mathsf T}$. At $\lambda=1$, its kernel has dimension $15$ and contains the $14$-dimensional image of $\delta_\alpha^{\mathsf T}$. The induced maps on cokernels therefore give \begin{equation}\label{eq:critical} \begin{gathered} 0\longrightarrow k\longrightarrow C_1\longrightarrow\Omega^{m-1}M\longrightarrow0,\\ 0\longrightarrow k\longrightarrow M\longrightarrow\Omega C_1\longrightarrow0. \end{gathered} \end{equation} The kernels are one-dimensional and hence are the unique simple module $k$. For the first sequence we used $C_{\alpha^{-1}}\cong\Omega^{m-2}M$; when $m=2$ this is $C_\alpha=M$. Suppose $\Tor_n^R(U,M)=0$ for $n\gg0$. The long exact sequences of \eqref{eq:critical}, followed by dimension shifting, give \[ \Tor_n^R(U,k)\cong\Tor_n^R(U,C_1),\qquad \Tor_{n+2}^R(U,C_1)\cong\Tor_n^R(U,k) \quad(n\gg0). \] Thus $\beta_{n+2}(U)=\beta_n(U)$ eventually. Equation~\eqref{eq:three} then yields \[ \beta_n=\beta_{n+2}\ge3\beta_{n+1}\ge9\beta_n. \] It follows that $\beta_n=0$. The resolution terminates, and self-injectivity of $R$ makes $U$ projective. Both summands in Lemma~\ref{lem:right-class} are now projective. \end{proof} \section{The gendo-symmetric counterexample}\label{sec:gendo} We first prove a detection result for a general triangular algebra. In this section $R$ is any finite-dimensional self-injective $k$-algebra, and $M$ is a nonzero finite left $R$-module such that \begin{equation}\label{eq:ET} \text{for every finite left }L,\quad \Ext_R^n(L,M)=0\ (n\gg0)\quad\Longrightarrow\quad L\text{ is projective}. \end{equation} Put $A=\left(\begin{smallmatrix}R&M\\0&k\end{smallmatrix}\right)$, with diagonal idempotents $r,s$ corresponding to $R,k$. The functors $i,j$ on left modules retain the definitions of Section~\ref{sec:triangular} with $A$ in place of $T$. The proof of Proposition~\ref{prop:left} uses only self-injectivity and \eqref{eq:ET}, and gives \begin{equation}\label{eq:A-detect} \Ext_A^n(U,A)=0\ (n\gg0)\quad\Longrightarrow\quad\pd_AU\le1. \end{equation} We use cohomological complexes, with differential $d_Q^n:Q^n\to Q^{n+1}$, cohomology $H^n(Q)=\ker d_Q^n/\im d_Q^{n-1}$, and projective resolutions in nonpositive degrees. For an idempotent $r$, the left component $rX$ is the image of multiplication by $r$ on $X$; components of complexes are taken termwise. The category $D^-(A\text{-}\mathrm{mod})$ is the derived category of bounded-above complexes of finite left modules. A complex is \emph{perfect}, written $V\in\per A$, if it is represented by a bounded complex of finite projectives. Derived tensor and derived Hom are computed by bounded-above projective resolutions in their resolved variables. The shift $Q[1]$ has $Q[1]^n=Q^{n+1}$ and differential $d_{Q[1]}^n=-d_Q^{n+1}$. \begin{lemma}\label{lem:tail} Let $V\in D^-(A\text{-}\mathrm{mod})$. If \[ rH^n(V)=0\quad(n\ll0),\qquad H^n\mathbf R\Hom_A(V,A)=0\quad(n\gg0), \] then $V$ is perfect. \end{lemma} \begin{proof} Take a bounded-above minimal projective representative $Q^\bullet$ of $V$. Such a representative is obtained from a bounded-above projective resolution by splitting off invertible differential blocks, starting at the highest degree; in each degree only finitely many finite projective summands occur. Its terms have the form $Q^n=iP^n\oplus jW^n$, where $P^n$ is projective over $R$ and $W^n$ is a finite $k$-space. We have \[ \Hom_A(jW,iP')=0,\qquad \Hom_A(jW,jW')\cong\Hom_k(W,W'). \] Thus minimality makes the $j$-diagonal differential zero. In this decomposition, \begin{equation}\label{eq:block} d_Q^n=\begin{pmatrix}d_P^n&0\\h^n&0\end{pmatrix},\qquad h^n:iP^n\longrightarrow jW^{n+1}. \end{equation} The complex $P^\bullet$ is minimal over $R$, and $(d_P)^2=0$. Let $Q^*=\Hom_A(Q^\bullet,A)$, where $(Q^*)^n=\Hom_A(Q^{-n},A)$ and $d_{Q^*}^n(g)=(-1)^{n+1}g\circ d_Q^{-n-1}$. Its right $r$-component is \[ Q^*r=\Hom_A(Q^\bullet,Ar)\cong\Hom_R(P^\bullet,R). \] It is exact in sufficiently large positive degrees by hypothesis. The functor $\Hom_{R^{\op}}(-,R)$ is exact, since $R$ is right self-injective, and finite projectives identify with their double duals. Therefore $P^\bullet$ is exact in sufficiently negative degrees. Taking left $r$-components in \eqref{eq:block} gives the short exact sequence of complexes \begin{equation}\label{eq:tail-seq} 0\longrightarrow(M\otimes_kW^\bullet,0) \longrightarrow rQ^\bullet\longrightarrow P^\bullet\longrightarrow0. \end{equation} The differential of the first complex is zero. In sufficiently negative degrees both $H^{n-1}(P)$ and $H^n(rQ)$ vanish, so the long exact sequence forces $M\otimes_kW^n=0$. As $M\ne0$, this gives $W^n=0$. The negative tail of $Q$ is therefore $iP$, and the corresponding positive tail of $Q^*s$ is $\Hom_R(P^\bullet,M)$. It is exact by hypothesis. Choose an integer $a$ far enough into this tail, and put $L=\coker(P^{a-1}\to P^a)$. The exact negative tail is a minimal projective resolution of $L$, and its Hom into $M$ is eventually exact. Equation~\eqref{eq:ET} makes $L$ projective. Its minimal resolution terminates at $P^a$, so $P^n=0$ for $n0) \end{equation} as $k$-vector spaces. \end{lemma} \begin{proof} The left $A$-module $fE=A\oplus A$ is free, so $F$ is exact; it sends $A$ to the right projective $fE$. Adjunction gives \[ \Hom_{E^{\op}}(F(U),F(U'))\cong \Hom_{A^{\op}}(U,F(U')f)=\Hom_{A^{\op}}(U,U'). \] Thus $F$ is fully faithful. If $F(U)$ is projective, applying $F$ to a projective surjection onto $U$ gives a split surjection. Full faithfulness pulls the splitting back, proving $U$ projective. Apply $F$ to a projective resolution of $U$ and use $Ef=DA\oplus A$ as a right $A$-module. Since $DA$ is right injective, Hom adjunction gives \eqref{eq:F-Ext}. For reflexivity, put $H=F(U)^*$. The same adjunction gives its left $E$-module components as \[ eH=\Hom_{A^{\op}}(U,DA)=DU,\qquad fH=U^*=\Hom_{A^{\op}}(U,A). \] The upper-right action is \[ DA\otimes_AU^*\longrightarrow DU,\qquad \lambda\otimes g\longmapsto(u\mapsto\lambda(g(u))), \] and the lower-left action is zero. The $B$-action on $DU$ factors through $\pi$. A left $E$-map $H\to Ee=(B,A)$ has upper component with image in $DA$, since $DA$ annihilates $DU$ and its annihilator in $B$ is $DA$. Every left $A$-map $DU\to DA$ has the form \[ \varphi\longmapsto\bigl(a\longmapsto\varphi(ua)\bigr) \] for a unique $u\in U$, by vector-space duality. Write $\gamma:U^*\to A$ for its lower component. Compatibility with the upper-right action forces $\gamma(g)=g(u)$: evaluation at $a\in A$ gives the equality $\lambda(g(u)a)=\lambda(\gamma(g)a)$ for all $\lambda\in DA$. Conversely these formulas define an $E$-map. Its image already lies in $Ef\subset Ee$. It follows that both $\Hom_E(H,Ee)$ and $\Hom_E(H,Ef)$ identify with $U$. Under these identifications evaluation $F(U)\to H^*$ is the identity on each of its two components. It is therefore an isomorphism of right $E$-modules. \end{proof} \begin{proof}[Proof of the gendo-symmetric assertion in Theorem~\ref{thm:main}] Take $A=T_\alpha$ with $\alpha$ of infinite order. Lemma~\ref{lem:module} supplies \eqref{eq:ET}. Set $E_\alpha=\End_{A\ltimes DA}((A\ltimes DA)\oplus A)$. Theorem~\ref{thm:E-detect} makes every semi-Gorenstein-projective left $E_\alpha$-module projective. The module $S_0$ of Proposition~\ref{prop:bad} is nonprojective and semi-Gorenstein-projective. Lemma~\ref{lem:F} makes $F(S_0)$ reflexive, semi-Gorenstein-projective and nonprojective. Theorem~\ref{thm:E-detect} makes every right Gorenstein-projective module projective, so $F(S_0)$ is not Gorenstein-projective. Equation~\eqref{eq:E} gives $\dim_kE_\alpha=5\dim_k A=145$, and $\dim_kF(S_0)=2\dim_kS_0=8$. Equation~\eqref{eq:F-Ext} and Proposition~\ref{prop:selfid} give $\id_{E^{\op}}E=\infty$. If $\id_EE=d<\infty$, the $d$th syzygy of every finite left module would be semi-Gorenstein-projective and hence projective by Theorem~\ref{thm:E-detect}. Thus every finite left module would have projective dimension at most $d$. For every finite right module $Y$ and finite left module $Z$, duality then gives \[ \Tor_{d+1}^E(Y,Z)=D\Ext_E^{d+1}(Z,DY)=0. \] Taking $Z=E/\rad E$ detects the $(d+1)$st term of a minimal right projective resolution, so every finite right module has projective dimension at most $d$. In particular $F(S_0)$ has finite projective dimension. The last-differential argument in Theorem~\ref{thm:E-detect}, on the right, makes a semi-Gorenstein-projective module of finite projective dimension projective. This contradicts Lemma~\ref{lem:F}. Hence $\id_EE=\infty$. \end{proof} \begin{thebibliography}{9} \bibitem{Huang} Z. Huang, \emph{Auslander-type conditions and weakly Gorenstein algebras}, \href{https://arxiv.org/abs/2408.05468v1}{arXiv:2408.05468v1}, 2024. \bibitem{JS} D. A. Jorgensen and L. M. \c{S}ega, \emph{Asymmetric complete resolutions and vanishing of Ext over Gorenstein rings}, Int. Math. Res. Not. 2005 (2005), no. 56, 3459--3477. \href{https://doi.org/10.1155/IMRN.2005.3459}{doi:10.1155/IMRN.2005.3459}. \bibitem{JSref} D. A. Jorgensen and L. M. \c{S}ega, \emph{Independence of the total reflexivity conditions for modules}, \href{https://arxiv.org/abs/math/0410257v2}{arXiv:math/0410257v2}, 2004. \bibitem{Mar} R. 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