# Current operators for compactified CFTs

Intuitively I feel that if you compactified open bosonic strings on a product of $n$ circles such that each radius is fine-tuned to the self-dual point then the CFT of these $n$ world-sheet fields would "see" a Kac-Moody algebra for some rank "n" Lie group.

• True? If yes then what is a good proof of this?

• Now how does one construct the conserved $(1,0)$ currents in terms of these $n$ CFT fields such that they would reproduce the structure constants of the rank $n$ Lie group that one wants?

Said otherwise, by choosing the number of compactification directions one could only choose the rank but how does one tune the required structure constants/Lie algebra as well?

(..Polchinski just writes down the answer for SU(2) in page 243 vol 1..even in that "simple" case its not quite clear as to how he managed to get the structure constants right..as in till he writes down the $j$s and shows that the commuatation gives $SU(2)$ it wasn't obvious that it is $SU(2)$..or was there an apriori reason for it?..)

• Now if one didn't start from open bosonic strings but were just doing CFT of $n$ fields on a product of $n$ circles then (1) what would remain of this effect of tuning to a self-dual radius? (2) how would one fine tune the required affine algebra? (3) how would one write down the $CFT$ Lagrangian depending on the wanted affine Lie algebra?
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Hi Anirbit. Destroying your own content is not allowed. Phys.SE is not a link-farm. Please reinstate the question. –  Qmechanic Mar 29 '13 at 21:04
@Qmechanic I haven't really "destroyed" the content. I have linked it to a different source where I have stated the thing better. (..also i believe I should have the right to delete my own questions!..) –  user6818 Mar 29 '13 at 21:07
No. You're are not allowed to destroy your own question if it has upvoted answers. And linking is not enough. –  Qmechanic Mar 29 '13 at 21:12
Anirbit, you reword the question so long as it does not invalidate the answers, you can add the link to the version you consider better, and you could ask the team to unlink this question from your account (though I do not know under what conditions they are willing to do that), but please don't deface it. –  dmckee Mar 30 '13 at 4:43

If you mean the simple compactification on a rectangular torus $T^k$ with the self-dual radii on each circle, you will simply get $SU(2)^k$ and each $SU(2)$ is level-one because they're totally independent of each other! For other compactifications, non-rectangular tori etc., you may get higher levels.
Quite generally, the currents are given by $\partial X_i(z)$ for the Cartan algebra generators and $:\exp(ik\cdot X):$ for some root $k$ to get the remaining generators that are not in the Cartan subalgebra.
Much like almost everywhere in science, it's not "a priori clear" that things behave in the right way but one may show that they do. The currents for an $SU(2)$ form the right algebra which may be extracted from the currents' OPEs, especially the $1/z$ terms in them which are equivalent to the commutators of the generators. When one works with these things, one internalizes them and finds these things obvious and natural, but this situation clearly doesn't happen "a priori" i.e. to a person who hasn't studied physics of CFTs, in this case.
I would like to know the compactification method to get not just $SU(2)^n$ but for arbitrary Lie groups. And what does one change in the compactification to raise the level of the $SU(2)$s in the simple compactification you mentioned. –  user6818 Mar 26 '13 at 17:38