Algebras derived equivalent to quasi-hereditary algebras

Question

Let an algebra always be finite dimensional over a field and connected. It is well known that a quasi-hereditary algebra with $n$ simple modules has global dimension at most $2n-2$.

Questions: 1. Do algebras derived equivalent to quasi-hereditary algebras with $n$ simple modules also have global dimension at most $2n-2$?

2. Is there a good discrete test (for example with a computer) for checking whether an algebra of finite global dimension is derived equivalent to a quasi-hereditary algebra?

3. Is every Nakayama algebra of finite global dimension derived equivalent to a quasi-hereditary algebra?

For example the Nakayama algebras with $n$ simples and Kupisch series $[n,n+1,...,n+1]$ are not quasi-hereditary for $n \geq 3$, but they are derived equivalent to the representation-finite block of a Schur algebra, which is quasi-hereditary. Here are some Kupisch series of Nakayama algebras of finite global dimension that are not quasi-hereditary, but Im not sure whether they are derived equivalent to a quasi-hereditary algebra: [ [ 3, 3, 3, 4 ], [ 3, 5, 5, 4 ], [ 4, 5, 6, 5 ], [ 4, 6, 5, 5 ], [ 4, 6, 6, 5 ] ]

It would be interesting to see how to deal with this problem even in small examples.

Answer 1

The answer to question 1 is "no": there are algebras with two simple modules, derived equivalent to quasi-hereditary algebras, but with global dimension three.

This can probably be extracted from one or both of the papers

Dubnov, Dmitry, On derived categories of modules over 2-vertex basic algebras, Commun. Algebra 28, No. 9, 4355-4374 (2000). ZBL0983.16009.

Volkov, Y. On the derived category of quasi-hereditary algebras with two simple modules, arXiv:1812.00351,

but I'll give an explicit example.

Let $A$ be the algebra given by a quiver with two vertices $1$ and $2$, one arrow $\alpha$ from vertex $1$ to vertex $2$, two arrows $\beta_1$, $\beta_2$ from vertex $2$ to vertex $1$, and relations $\beta_1\alpha=0=\beta_2\alpha$.

Denoting by $S(i)$ and $P(i)$ the simple and indecomposable projective modules corresponding to vertex $i$, $P(2)$ is $3$-dimensional with head $S(2)$ and radical $S(1)\oplus S(1)$, and $P(1)$ is $4$-dimensional with head $S(1)$ and radical isomorphic to $P(2)$. So $A$ is quasihereditary with standard modules $\Delta(1)=S(1)$ and $\Delta(2)=P(2)$.

There is a tilting complex $$T:=\dots\to0\to P(1)^3\to P(2)\to0\to\dots,$$ with $P(2)$ in degree zero, where one copy of $P(1)$ is in the kernel of the differential, and the other two copies map onto $\text{rad }P(2)$.

Let $B=\text{End}_{D^b(A)}(T)$, so there is an equivalence of triangulated categories $F:D^b(A)\to D^b(B)$ sending $T$ to $B$.

Since $\text{Hom}_{D^b(A)}(T,S(2)[i])=0$ for $i\neq0$, $FS(2)$ has homology only in degree zero, and so is a $B$-module $M$.

If $X$ is the quotient of $P(2)$ by a $1$-dimensional submodule of its socle, then $\text{Hom}_{D^b(A)}(T,X[i])=0$ for $i\neq1$, and so $FX[1]$ is a $B$-module $N$.

Now $$\text{Ext}^3_B(N,M)\cong\text{Hom}_{D^b(B)}(N,M[3]) \cong\text{Hom}_{D^b(A)}(X,S(2)[2])\cong\text{Ext}^2_A(X,S(2)),$$ which is easily calculated to be non-zero.

Hence $B$ has global dimension at least three (in fact, it's equal to three).