Does the derived category remember the homological dimension? - MathOverflow

Does the derived category remember the homological dimension?

2011-03-05T06:48:16Z

Question:

Let $\mathcal{A}$ be an abelian category and $D^?(\mathcal{A})$ be its derived category, where ? could be empty, +, - or b (for boundedness). Is it possible to recover the homological dimension of $\mathcal{A}$ from the derived category?

Here I'm using the term homological dimension in the sense of Gelfand-Manin, i.e. if for all $X,Y\in\mathcal{A}$, $\text{Ext}_{\mathcal{A}}^i(X,Y):=\text{Hom}_{D(\mathcal{A})}(X[0],Y[i])=0$, then the homological dimension is said to be less than $i$. The homological dimension is the maximal $n$ such that there exists a non-vanishing $\text{Ext}^n(X,Y)$.

Note that in the derived category one could have all kinds of non-vanishing $\text{Ext}^n(X,Y)$, as $X,Y $ can be complexes shifted arbitrarily. Is it still possible to recover this information via other method?

Answer by Leonid Positselski for Does the derived category remember the homological dimension?

2011-03-05T10:28:41Z

Let $V$ be a finite-dimensional vector space, $\mathcal{A}$ be the abelian category of finitely generated graded modules over the symmetric algebra $S(V)$, and $\mathcal{B}$ be the abelian category of finitely generated graded modules over the exterior algebra $\Lambda(V^*)$. Then the bounded derived categories $\mathcal{D}^b(\mathcal{A})$ and $\mathcal{D}^b(\mathcal{B})$ are naturally equivalent. This is called the Bernstein-Gelfand-Gelfand duality, a particular case of Koszul duality. On the other hand, the homological dimension of $\mathcal{A}$ is equal to $\dim V$, while the homological dimension of $\mathcal{B}$ is infinite.

To obtain a similar example with unbounded derived categories, let $\mathcal{A}^+$ be the abelian category of (infinitely generated) nonnegatively graded $S(V)$-modules and $\mathcal{B}^+$ be the abelian category of nonnegatively graded $\Lambda(V^\ast)$-modules. Here it is presumed that $S(V)$ is graded so that $V$ is placed in the degree $1$, while $\Lambda(V^\ast)$ is graded so that $V^*$ is placed in the degree $-1$. Then the unbounded derived categories $\mathcal{D}(\mathcal{A}^+)$ and $\mathcal{D}(\mathcal{B}^+)$ are equivalent.

UPDATE. I was asked in the comments to provide an example with both homological dimensions being finite. This can be done by yet another modification of the above examples. Pick an integer $n>\dim V$. Let $\mathcal{A}_n$ be the abelian category of finitely generated graded $S(V)$-modules concentrated in the gradings $0\leq i\leq n$. Similarly, let $\mathcal{B}_n$ be the abelian category of finitely generated graded $\Lambda(V^\ast)$-modules concentrated in the gradings $0\leq i\leq n$. Then the homological dimension of $\mathcal{A}_n$ is equal to $\dim V$, the homological dimension of $\mathcal{B}_n$ is equal to $n$, and $\mathcal{D}^b(\mathcal{A}_n)\simeq\mathcal{D}^b(\mathcal{B}_n)$. A similar example with unbounded derived categories can be obtained by removing the finitely generatedness assumption.

All of the above counterexamples presume that $\dim V>0$. The only positive result in the direction of the original question that I can think of at the moment is that if the derived categories of $\mathcal{A}$ and $\mathcal{B}$ are equivalent, and $\mathcal{A}$ has homological dimension $0$, then so does $\mathcal{B}$. Indeed, $\mathcal{A}$ is a semisimple abelian category if and only if $\mathcal{D}^b(\mathcal{A})$ and $\mathcal{D}(\mathcal{A})$ are.

Answer by Mariano Suárez-Alvarez for Does the derived category remember the homological dimension?

2011-03-05T17:28:46Z

There are lots of examples to be found in the theory of tilted algebras.

A tilted algebra $B$ is an algebra of the form $\operatorname{End}_A(T)$ with $A$ an hereditary finite dimensional algebra (over an algebraically closed field, say) and $T$ a tilting $A$-module. Then $B$ and $A$ have equivalent derived categories, but $B$ is usually not hereditary (that is, its global dimension is usually bigger than $1$; one does have $\operatorname{gldim}B\leq2$, though, so it does not blow up too much).

A concrete example: let $A$ be the path algebra of the quiver $$\bullet \leftleftarrows \bullet \leftarrow \bullet$$ Number the vertices $1$, $2$ and $3$ from left to right, and let $T$ be the direct sum of the simple $S(1)$ and the indecomposable injective modules $I(1)$ and $I(3)$, which is a tilting module. Then $B=\operatorname{End}_A(T)$ is the algebra given by the same quiver, but bound by the relations given by the two paths of length two. In particular, $B$ is not hereditary.

You will find all this discussed at length in Assem, Simson and Slowroński's book Elements of the Representation Theory of Associative Algebras, Vol. 1.

If one considers more generally tilting complexes as opposed to just modules, then the difference between the global dimensions of the algebras involved can be made as large as you want.