Does the derived category remember the homological dimension? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T02:15:22Z http://mathoverflow.net/feeds/question/57437 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57437/does-the-derived-category-remember-the-homological-dimension Does the derived category remember the homological dimension? Yuhao Huang 2011-03-05T06:48:16Z 2011-03-05T19:18:05Z <p>Question:</p> <p>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?</p> <p>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 $\text{Ext}^n(X,Y)$. </p> <p>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? </p> http://mathoverflow.net/questions/57437/does-the-derived-category-remember-the-homological-dimension/57445#57445 Answer by Leonid Positselski for Does the derived category remember the homological dimension? Leonid Positselski 2011-03-05T10:28:41Z 2011-03-05T13:12:54Z <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/57437/does-the-derived-category-remember-the-homological-dimension/57474#57474 Answer by Mariano Suárez-Alvarez for Does the derived category remember the homological dimension? Mariano Suárez-Alvarez 2011-03-05T17:28:46Z 2011-03-05T19:18:05Z <p>There are lots of examples to be found in the theory of tilted algebras.</p> <p>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 <a href="http://eom.springer.de/T/t130120.htm" rel="nofollow">tilting $A$-module</a>. 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).</p> <p>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 <em>not</em> hereditary.</p> <p>You will find all this discussed at length in Assem, Simson and Slowroński's book <em>Elements of the Representation Theory of Associative Algebras</em>, Vol. 1.</p> <p>If one considers more generally tilting <em>complexes</em> as opposed to just <em>modules</em>, then the difference between the global dimensions of the algebras involved can be made as large as you want.</p>