What extra conditions are necessary for the following version of Koszul duality? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:04:59Z http://mathoverflow.net/feeds/question/67909 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67909/what-extra-conditions-are-necessary-for-the-following-version-of-koszul-duality What extra conditions are necessary for the following version of Koszul duality? Theo Johnson-Freyd 2011-06-16T05:57:11Z 2011-06-19T23:54:57Z <p><strong>Conventions:</strong> So that I don't have to worry about, fix a field $k$ of characteristic zero, and always work over it. Categories of modules, etc., are always $\infty$-categories of dg modules. Algebras are "associative" in the coherent-homotopy sense. All tensor products are left-derived. Etc. I am of course always interested in hearing about subtleties and generalizations of such conventions, but for the sake of this question you may pretend that I understand this part of the story.</p> <p><strong>A construction:</strong> Let $A$ be an algebra and $V$ a left $A$-module. Write $V^\ast$ for the linear dual to $V$; then $V^\ast$ is a right $A$-module. Set $C = V^\ast \otimes_A V$. I claim that $C$ is a coalgebra, at least when $V$ satisfies some finiteness condition. Indeed, if $V$ satisfies a finiteness condition, then $\operatorname{End}_k(V) = V\otimes_k V^\ast$, and the action of $A$ on $V$ is encoded in a map $A \to \operatorname{End}_k(V)$, which by associativity is a map of $A$-$A$-bimodules. Then the comultiplication is:</p> <p>$$C = V^\ast \otimes_A V = V^\ast \otimes_A A \otimes_A V \to V^\ast \otimes_A (V \otimes_k V^\ast) \otimes_A V = C \otimes_k C$$</p> <p>This is coassociative on account of the coassociativity of tensor products.</p> <p>For $X$ any left $A$-module, there is a corresponding left $C$-comodule defined by $V^\ast \otimes_A X$; the comodule structure is</p> <p>$$V^\ast \otimes_A X = V^\ast \otimes_A A \otimes_A X \to V^\ast \otimes_A (V \otimes_k V^\ast) \otimes_A X = C \otimes_k (V^\ast \otimes_A X)$$</p> <p><strong>Aside:</strong> Let $C$ be any (coassociative) coalgebra, $Y$ a left $C$-comodule, and $Z$ a right $C$-comodule. Recall that the <em>underived cotensor product</em> is the 1-categorical equalizer of the two maps $Z \otimes Y \rightrightarrows Z \otimes C \otimes Y$. The <em>(derived) cotensor product</em> $Z \Box_C Y$ is the right-derived version thereof (cotensor is left-exact, if I haven't made an error); it should also be the $\infty$-categorical equalizer.</p> <p>At least in the 1-categorical non-dg setting cotensor products of bi-comodules and so on are not always associative. But they are associative when all coalgebras are flat over whatever ground ring you're working over, and we're working over a field, so this is not an issue. I don't know what the correct statement is in the dg $\infty$-categorical level.</p> <p><strong>The construction, continued:</strong> In a similar way, $V$ is a right $C = V^\ast \otimes_A V$ comodule, and so for any left $C$-comodule $Y$, I can define a left $A$-module $V \Box_C Y$.</p> <p>All together, I've constructed functors </p> <p>$$V^\ast \otimes_A: A\text{-mod} \leftrightarrow C\text{-comod} :V \Box_C.$$</p> <p>One composition of these functors (the one from $C$-comod to $C$-comod) is the identity:</p> <p>$$V^\ast \otimes_A V \Box_C = C \Box_C$$</p> <p>The other composition cannot always be the identity — just imagine what would happen if I took $V$ to be the zero module!</p> <p><strong>Question:</strong> A statement that perhaps deserves to be called <em>Koszul duality</em> is that these two functors are an equivalence of $\infty$-categories. Of course, that seems almost nonsense to me, because in the non-dg-$\infty$ world categories of modules and categories of comodules seem quite different. So part of my question is to clarify the statement of the statement. But the bulk of my question is: What conditions do I need to add in the above exposition to have the statement I'd like? For example, I would like to identify $\operatorname{End}_k(V) = V \otimes_k V^\ast$, and so I would expect to need some "finiteness" condition on $V$, or I would expect to need some topology.</p> <p>Note that what many people call "Koszul duality" consists also of taking the dual $C^\ast$ to the coalgebra $C$, so as to get an algebra. Every left comodule of $C$ is a left module of $C^\ast$, but there are generically more of the latter, at least in the 1-categorical version of the story. Given that many discussions of Koszul duality use the (derived) $\operatorname{End}_A(V,V)$, which is an algebra that should be essentially the same as $C^\ast$, I worry that maybe the algebra-to-algebra version is more robust. But I'm not sure.</p> <p><strong>Bonus question and further reading:</strong> Sometime soon I will be writing up a construction related to the one above, as part of a larger project. I picked up the above ideas from discussions with various people. But I would like to give credit where it's due, so I would like to hear about any particular papers I should be sure to cite.</p> <p>For further reading, you might check out <a href="http://mathoverflow.net/questions/329/what-is-koszul-duality" rel="nofollow">this question from the first month of MathOverflow</a>, and also perhaps <a href="http://www.math.harvard.edu/~lurie/papers/moduli.pdf" rel="nofollow">Jacob Lurie's ICM address</a>.</p> http://mathoverflow.net/questions/67909/what-extra-conditions-are-necessary-for-the-following-version-of-koszul-duality/67984#67984 Answer by David Ben-Zvi for What extra conditions are necessary for the following version of Koszul duality? David Ben-Zvi 2011-06-16T18:27:36Z 2011-06-19T23:54:57Z <p>This version of Koszul duality (as well as many others) can be deduced from the Barr-Beck-Lurie theorem (cf. Lurie's book Higher Algebra). You can consider the functor from k-mod to A-mod of tensoring by V and ask for it to have a left or right adjoint, giving rise to a comonad or monad on k-mod (ie the coalgebra or algebra form). These are two finiteness conditions on V (k-dualizability and A-dualizability - ie dualizability as a vector space or compactness as A-module - though maybe I'm mixing up the order). If you ask for both simultaneously then you guarantee that the functor from A-mod to (co)modules over the (co)monad is both limit and colimit preserving -- this is overkill but an easy way to ensure the hypotheses of the Barr-Beck-Lurie theorem are satisfied. You now get an equivalence between your category of (co)modules and a COMPLETION of A-mod --- i.e., the part of A-mod that your particular chosen module V sees (the category it generates). (This is how you satisfy the other requirement of the theorem, i.e., that the corresponding functor is conservative).</p> <p>The classical version of this is A=S, symmetric algebra in one variable, ie A-mod = quasicoherent sheaves on the line, and V is the augmentation module (skyscraper at the origin). We then have a descent theorem, saying that the completion of A at the augmentation is derived equivalent to sheaves on a point with descent data for inclusion of point to line -- which is exactly modules for the Koszul dual exterior coalgebra (or dually exterior algebra). </p> <p>EDIT: In response to the comment about "seeing" modules: technically the condition is whether there are any Exts between your given object V and some other object W. I think of this geometrically: if V stands for a skyscraper on a variety, it will see the entire formal neighborhood of the point, i.e. intuitively you can manufacture Exts with the skyscraper for anything that has a stalk at this point. So for instance for enveloping algebras a representation will see only representations that have all invariants the same as V (i.e. the center must act with the same generalized character). So I'm not sure I understand the question regarding Lie algebras (take the example of $\mathfrak g$ the trivial one-dimensional Lie algebra and we're back in the original Koszul duality discussed in the paragraph above).</p>