UV-IR cancellation of the open string cylinder diagram and the field theory limit In string theory, the ultraviolet divergences of open string loop diagrams are reintepreted as closed string infrared divergences, by seeing that an annulus with a small loop is also a long tube.  In theories with these divergences, like the bosonic string, can one see the equivalence of these divergences when one takes the limit of vanishing string length $\alpha' \to 0$?  In other words, can one make a one-to-one mapping between quantum field theory loop divergences and tree-level gravitational infrared divergences and see, in some regulated limit, that they are indeed the same?  Or have I misunderstood something?
 A: No, the UV-IR connection in string theory effectively depends on the $\alpha'=1$ units; more precisely, it reverts the value of $\alpha'$ along the way. It is a purely stringy effect which means that it has no implications in the field-theoretical $\alpha'\to 0$ limit. In this field theoretical limit, the cylinder degenerates into a line and there is no "dual" description to view a line. Field theories don't have this remarkable stringy property that is the key to string theory's well-behaved short-distance behavior, among other things.
However, in a more general way, there is a sense in which this map exists even in the field-theoretical limit: the AdS/CFT correspondence. The field-theory limit of open string diagrams are equivalent to closed string gravitational dynamics in the bulk etc. But to understand why this works, one inevitably has to study the AdS curved geometry (the near-horizon geometry of black holes) and talk about various decouplings so it's not just a simple limit of the UV-IR cylinder duality from the flat space. At the end, even the heuristic proofs depend on the whole string theory even though the final statement has a pure quantum field theory on one side. The $\alpha'\to 0$ limit needed to get the field theory appears, in the dual language, because we're focusing on the near-horizon geometry where the red shift is going to zero (or infinity, depending on how you count it).
