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

I had one or two little fights with correspondences in the context of algebraic geometry where an elementary correspondence $C:X\to Y$ of connected smooth $k$-Schemes seems to be defined as an irreducible closed (reduced) subscheme $C\hookrightarrow Y\times_k X$, such that the projection to $X$ is finite an surjective.

Further, I have vaguely heard of correspondences in topology (invented by Lefschetz?) where it seems to me, that such a thing is a cohomology class in $H^*(Y\times X)$ for compact, oriented manifolds $X,Y$. Using Poincare duality and the cohomology pushforward functor $(-)_!$ I can associate a cohomology class $(\Delta_f)_!(1)$ in $H^*(Y\times X)$ to a map $X\to Y$.

My questions are:

  1. I can not really see the analogy of the concepts, except that in booth cases one can associate a corresponces to a morphism by its graph. So, what (or how deep) is the analogy?

  2. I know (only a very few) applications of correspondences in algebraic geometric, but of none in topology. What are they good for? Where can I find applications?

share|improve this question
    
Correspondences are used in algebraic geometry to construct (some kinds of) motives... that's a pretty central application! –  Mariano Suárez-Alvarez Mar 13 '10 at 13:51
1  
Some of this is a terminology issue. In topology several of these concepts go under names like a "transfer" (e.g. the Becker-Gottlieb transfer) or an "Umkehr map". The topological work also generally concentrates more on things like fiber bundles and less on things like ramified covers. –  Tyler Lawson Mar 13 '10 at 15:51
add comment

3 Answers 3

up vote 8 down vote accepted

For simplicity and definiteness, let's assume that $X$ and $Y$ are smooth and compact (and orientable, which will always be the case if they are complex varieties), and let $n$ be the dimension of $Y$. First of all, it might help to note that $H^n(X\otimes Y) \cong H^\*(X)\otimes H^{n-\*}(Y) \cong Hom(H^\*(Y),H^\*(X))$, where for the final assertion I am using that $Y$ is smooth and compact, so that its cohomology satisfies Poincare duality. Thus if $Z$ is a cycle in $X\otimes Y$, of dimension equal to that of $X$ (and so of comdimension $n$) it induces a cycle class in $H^n(X\times Y)$, which in turn induces a map from cohomology of $X$ to that of $Y$. If $f:X\to Y$ and $Z = \Gamma_f$ is the graph of $f$ then this map is just the pull-back of cohomology classes by $f$.

So correspondences in the sense of physical cycles on $X\times Y$ induce correspondences in the sense of cohomology classes on $X\times Y$, which in turn induce morphisms on cohomology. If you like, you can strengthen the analogy with the $\Gamma_f$ case by thinking of a correspondence as a multi-valued function. Functions induce morphisms on cohomology; but since cohomology is linear (you can add cohomology classes), correspondences also induce morphisms on cohomology (you can simply add up the multiples values!). This gives the same construction as the more formal one given above.

Ben Webster notes in his answer that geometric representation theory provides a ready supply of correspondences. So does the theory of arithmetic lattices in Lie groups and the associated symmetric space. (I am thinking of Hecke correspondences and the resulting action of Hecke operators on cohomology.)

A very general framework, which I think covers both contexts, is as follows: suppose that a group $G$ acts on a space $X$, and that $H \subset Aut(X)$ is another subgroup commensurable with $G$, i.e. such that $G \cap H$ has finite index in each of $G$ and $H$.

Then (perhaps under some mild assumptions) $(G\cap H)\backslash X \hookrightarrow (G\backslash X \times H\backslash X) $ is a correspondence (in the physical, geometric sense) which will give a correspondence in cohomology via its cycle class. The resulting maps on cohomology are (a very general form of) Hecke operators.

share|improve this answer
    
Of course, some of us would say that Hecke correspondences and hecke operators were probably the first appearance of geometric representation theory in the literature. –  Ben Webster Mar 13 '10 at 15:27
    
Indeed! ....... –  Emerton Mar 13 '10 at 16:02
add comment

Correspondences come into their own in situations where there aren't enough maps for functorial methods to be useful.

In symplectic topology, the only structure-preserving maps are symplectic immersions, and of those, the only ones which play well with the main technique of the subject - holomorphic curves - are the symplectomorphisms. So usually one treats a symplectic manifold in splendid isolation.

But Lagrangian correspondences arise very naturally - as graphs of symplectomorphisms; via symmetry (moment maps); via degeneration (vanishing cycles); and via elliptic boundary problems relevant to low-dimensional topology. They make excellent boundary conditions for holomorphic curves. Ongoing work of Ma'u-Wehrheim-Woodward develops this idea to bring functorial methods to bear, for the first time, in symplectic topology.

share|improve this answer
add comment

1) The analogy is that a smooth subscheme of $X\times Y$ defines an element of the Borel-Moore homology of $X\times Y$ which in the compact case is the same as homology (which is dual to cohomology; of course there's no good reason to make the topological definition with cohomology instead of homology). More generally, this is a special case of the analogy between Chow groups and (co)homology, which is quite well developed in algebraic geometry.

2) Look in Chriss and Ginzburg's Complex Geometry and Representation Theory. Essentially the entire book is about correspondences in the topological sense. Correspondences are very important throughout geometric representation theory, since they have a multiplication, and thus can be used to produce algebras acting on the cohomology of spaces.

share|improve this answer
add comment

Your Answer

 
discard

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.