Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The simplest example of Koszul duality (see introduction of http://www.ams.org/journals/jams/1996-9-02/S0894-0347-96-00192-0/)

Let $V = \mathbb{C}x$ be a $1$ dimensional vector space. Then the exterior algebra is $A=\mathbb{C}[x]/(x^2)$, and the symmetric algebra is $A^! = \mathbb{C}[x]$. Koszul duality states that there exists an isomorphism (where "mod" means the category of finitely generated modules):

$F: D^b(A-mod) \rightarrow D^b(A^!-mod)$

The heart of the category on the left is $A-mod$; what is the image of this on the right? It should be possible to work this out by working through the above paper, but there are many things in that paper which I don't understand (and this toy example should be easy).

share|improve this question
add comment

2 Answers

up vote 6 down vote accepted

"mod" has to mean the category of finitely generated graded modules, if not, there is no such equivalence.

Up to graded shifts (and isomorphism), the category $A-\mathsf{mod}$ in your example only has two indecomposable objects: the trivial module $\mathbb C$ and the lenght two projective-injective module $A$. Let $I$ denote the indecomposable graded injective module cogenerated in degree $0$, so $I=A \langle -1 \rangle$.

Theorem 2.12.5 contains all you need in order to understand the functor $F$ in this example. According to Theorem 2.12.5, $$F(\mathbb C \langle-n\rangle )=A^! [n] \langle n \rangle$$ and $$F(I \langle-n\rangle)=\mathbb C [n] \langle n \rangle .$$ In conclusion, the indecomposable objects in the essential image $F(A-\mathsf{mod})$ are $\{ A^! [n] \langle n \rangle \}_{n \in \mathbb Z}$ and $\{\mathbb C [n] \langle n \rangle \}_{n \in \mathbb Z}$.

A typical exact sequence $$0 \rightarrow \mathbb C \rightarrow I \rightarrow \mathbb C\langle-1\rangle \rightarrow 0$$ in $A-\mathsf{mod}$ is sent by $F$ to the exact sequence $$0 \rightarrow A^! \rightarrow \mathbb C \rightarrow A^! [1] \langle 1 \rangle \rightarrow 0$$ in $F(A-\mathsf{mod})$.

share|improve this answer
add comment

There's actually a general answer to this question, given by Mazorchuk, Ovsienko and Stroppel (though I assume some version of it was known earlier). The category of graded representations of $A^!$ is the same as the (abelian!) category of linear projective complexes over $A$, and vice versa. So, in Dag Oskar Madsen's answer, I would write not $\mathbb{C}[n]\langle n\rangle$, but rather its projective resolution $A^![n+1]\langle n+1\rangle\overset{x}\to A^![n]\langle n\rangle$. Similarly, the $A^!$ modules $\mathbb{C}[x]/(x^{n+1})$ are sent by Koszul duality to the complexes

$$A[n]\langle n\rangle \overset{x} \to A[n-1]\langle n-1\rangle \overset{x} \to\cdots \overset{x} \to A.$$

share|improve this answer
Yes, and an added bonus of this description of the functor is that there is actually very little to compute. An earlier reference is Theorem 2.4 in [Martínez Villa, Roberto; Saorín, Manuel. Koszul equivalences and dualities. Pacific J. Math. 214 (2004), no. 2, 359--378] –  Dag Oskar Madsen Aug 28 '13 at 15:22
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.