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

Two rings $A$ and $B$ are said to be Morita equivalent if the category of modules over $A$ and $B$ are equivalent as additive categories. (Here I'm considering left modules). Ex: $M_n(R)$ (the algebra of matrices over a ring $R$) is morita equivalent to $R$. In fact more generally whenever $A$ is a ring and $e$ is an idempotent in $A$ and $AeA = A$ then the follwing functor is a morita equivalence:

$A$ modules $\rightarrow$ $eAe$ modules

$M$ $\mapsto$ $eM$

Now under nice conditions the categories of $A$ modules (resp $B$ modules) might have a moduli description. Then can you say anything about the induced map on the moduli spaces? I'm asking for situations in which this is known.

Also is there a nice book or paper which talks about Morita equivalence and has lots of examples?

share|improve this question
Forgibe my ignorance, but what does it mean for a category to have a moduli description? I'm having trouble imagining a moduli space of A-modules. –  Ilya Grigoriev Dec 14 '09 at 23:19
Say $A$ is a fin. generated $k$- algebra. Fix the rank n. Choose a set of generators for $A$ so that $A = k<x_1, \dots x_r>/I$ Then $Mod(A,n)$ - modules over $A$ of $k$ rank n has as parameter space $(t_1, \dots , t_r \in Mat(n) such that a(t_1 \dots t_r) = 0 for all a \in I$. The group $Gl(n)$ acts on this space. The space $Mod(A,n)//Gl(n)$ is a moduli space of semisimple $A$ modules of rank n and if we choose a stability condn $\theta$, $Mod(A,n)//_{\theta}Gl(n)$ is the moduli space of $\theta$-semistable $A$-mod. For a ref. look at the DMV lect. on Quivers by Crawley-Boevey and Holland. –  Avan Thiyagarajan Dec 16 '09 at 3:44
There's always a moduli stack of finite-dimensional A modules of any given dimension. This stack is never a scheme, but sometimes forgetting some automorphisms isn't so damaging. –  Ben Webster Dec 30 '09 at 12:48

4 Answers 4

up vote 5 down vote accepted

In the setting you are interested in, that is, finitely generated k-algebras and GIT-quotients of closed or semistable orbits, Morita equivalence induces isomorphism on the moduli spaces provided one scales dimension(vectors) and stability structures accordingly. That is, if B=M_n(A) one should compare Mod(A,k) to Mod(B,nk) moduli. If A and B have a complete set of orthogonal idempotents e_i resp. f_i(that is, a quiver-like situation) and if the Morita equivalence induces rank(f_i) = n_i rank(e_i), then one should compare A-reps of dimension vector alpha=(alpha_i) to B-reps with dimension vector beta=(n_i alpha_i).

Geometrically, the module varieties of Morita-equivalent algebras are related via associated fibre bundle constructions. In the example above, Mod(B,kn) = GL(nk) x^GL(k) Mod(A,k). In general,one had such a description locally in the Zariski topology (coming roughly from the fact that a vectorbundle (or projective) is locally free). Anyway, this gives a natural and geometric one-to-one correspondence between GL(nk)-orbits (isoclasses) of B-reps and GL(k)-orbits of A-reps inducing the desired isos on the quotient variety level (isos of semi-simples). In quiver-like situations one should adjust the dim vectors as above.

Now as to semi-stability. Observe that semi-stable reps are just ordinary reps of a universal localization of your algebra(s), so one can reduce to the closed case by covering the variety of semi-stables by Zariski open (affine) pieces. In quiver-like situations when the Morita-data is as above and your stability structure mu for A is given by the vector (mu_i), then the corresponding stability structure for B is mu'=(1/n_i x mu_i) (or multiply it with a common factor if you want it to have integral components.

share|improve this answer
Is it clear how the stability conditions correspond for general f.g. k-algebras? –  Avan Thiyagarajan Jan 3 '10 at 16:49
tried to answer your question by editing the answer. hope this helps. –  lieven lebruyn Jan 4 '10 at 12:04
It makes sense. Thanks. Here you are using the fact that the number of idempotents is the same. That's not true for general Morita-equivalent algebras, right? or is there a correspondence relating the idempotents. –  Avan Thiyagarajan Jan 5 '10 at 1:45
#idempotents is not preserved under Morita (think matrices). what i meant was that B being Morita to A which is kQ/relations then one can take n_i above as the number of indecomposable projectives Ae_i in the projective P giving Morita (in case A is fin dml). In general one has to define n_i as the (local) rank of the projective induced by e_iP over the commutative ring (e_iAe_i)/[e_iAe_i,e_iAe_i]. –  lieven lebruyn Jan 5 '10 at 8:44

This may be too elementary, but Anderson and Fuller's _Rings_and_Cagegories_of_Modules_ has a section giving basic properties of Morita equivalence. The Morita equivalence section of McConnell and Robson's _Noncommutative_Noetherian_Rings_ goes into more depth; e.g. they show that Morita equivalent rings have isomorphic lattices of (2-sided) ideals, which I don't believe is covered in Anderson-Fuller.

I realize this is not your original question, but it may help with question #2.

share|improve this answer
If two algebras are Morita equivalent, it is easy to show that their categories of bimodules are Morita equivalent (using the description of equivalences as tensor products with modules). The statement about ideals is just the fact that they are the subbimodules of the algebras. –  Mariano Suárez-Alvarez Dec 14 '09 at 23:16

This is not an example of the kind you asked for, but you might find this interesting anyway - it does involve both moduli and equivalences. In Bridgeland's paper Flops and Derived Categories he constructs certain derived equivalences (which are Fourier-Mukai transforms so have a Morita type flavour) by building a moduli space for certain objects in a derived category which come from a t-structure.

It has been a while since I have looked at it but from memory "Morita Equivalence and Duality" by Cohn is quite a nice book and I think it contains several examples (I hope that I am remembering correctly).

share|improve this answer
Is there really a relation between Fourier-Mukai and Morita equivalence? If yes can you give me a reference on this. I will certainly look at this book by Cohn. Thanks. –  Avan Thiyagarajan Nov 22 '09 at 15:55
Fourier-Mukai transforms are (examples of) derived Morita equivalence - ie equivalences of derived categories of modules over rings, which are much more flexible than classical Morita equivalences. –  David Ben-Zvi Nov 22 '09 at 16:40
Ah. Now I see it. Thanks. –  Avan Thiyagarajan Nov 23 '09 at 2:19

There are some nice Morita equivalences arising from Hecke algebras in representation theory - they arise as algebras of bi-invariant functions on a locally compact group under convolution. Good, workable examples arise from subgroups of finite groups.

If you like categories, there is a rather high-level treatment in this paper (Ben-Zvi, Francis, Nadler). It is probably not the best place to start.

share|improve this answer
The approach is rather too high level. Is there a more down to earth paper on related aspects? thanks. –  Avan Thiyagarajan Nov 22 '09 at 16:08

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.