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May 16, 2013 at 0:53 comment added Student @GuyGur-Ari Some recent thoughts brought me back to this discussion with you. By your construction $\rho_n$ is real - and hence when they go from 5.6 to 5.10 in the paper arxiv.org/pdf/hep-th/0310285v6.pdf they should have gotten a square-root - but they don't - it seems that their \rho_n is complex - can you kindly explain this point?
Aug 14, 2012 at 20:12 comment added user6818 I think my previous question was not worded properly and I made hopefully a better statement here, physics.stackexchange.com/questions/34192/… Would be glad to see you there!
Aug 7, 2012 at 21:59 history bounty ended CommunityBot
Aug 7, 2012 at 20:17 comment added user6818 I have put up a slightly more edited version of that question, physics.stackexchange.com/questions/33023/… It would be great to have your help.
Aug 7, 2012 at 9:01 comment added user6818 I had already asked the separate question as linked from my previous comment. It has remained unanswered. BTW, I have awarded you the bounty. Thanks for the help! Hope to see you at that other question of mine too!
Aug 7, 2012 at 8:58 vote accept user6818
Aug 3, 2012 at 15:42 comment added Guy Gur-Ari I said it is similar, not the same. It seems to me this discussion has strayed from the original question. Why not post it as a new question?
Aug 2, 2012 at 20:27 comment added user6818 I am not so sure. In d=4 the special thing that happens is that the anti-commutation between the Qs is diagonal and hence half of then go to 0 and the other half works like the Clifford algebra. In d=3 this simplification doesn't exist because of the peculiarities of the Gamma matrices at that dimension. Hence I would like to know if there is a standard reference which explains the multiplet construction in d=3.
Aug 1, 2012 at 19:40 comment added Guy Gur-Ari I don't have a reference but the story is similar to the multiplets in $d=4$, which are derived in any textbook on SUSY. You can repeat this exercise in $d=3$.
Aug 1, 2012 at 19:15 comment added user6818 Thanks for the explanation! Can you give me a reference where these massless multiplets of $N=2$ and $N=3$ supersymmetry in $2+1$ dimensions are derived from scratch? Like if you can help with this previous question of mine,physics.stackexchange.com/questions/33023/… That would be a huge help!
Aug 1, 2012 at 19:12 history edited Guy Gur-Ari CC BY-SA 3.0
Added caution
Aug 1, 2012 at 19:10 comment added Guy Gur-Ari As they say, in the $N=3$ theory you can have hypermultiplets, and each hypermultiplet has two chiral multiplets: one in a rep $R$ and one in $\bar{R}$. It is not necessary that $R$ is the adjoint. A bit below the place you mention they consider various possibilities for $R$, only one of which is the adjoint. If you say that a hypermultiplet is 'fundamental' (which I think is not a precise statement although its meaning is clear), it means that you have one chiral multiplet in the fundamental and one in the anti-fundamental.
Aug 1, 2012 at 18:47 comment added user6818 Thanks for the updates! I think I am missing something here. In the beginning of section 2.2 (page 7) don't they say that the 2 N=2 chiral multiplets inside the N=3 hypermultiplet are in conjugate representations of the gauge group? Then is it wrong to think that these 2 N=2 chiral components are in $Adj = \bar{Adj}$ of the gauge group? If the $N_f$ hypermultiplets are in the fundamental then what is the representation in which the 2 $N=2$ chiral multiplets fall into? It will be great if you can clarify this point!
Aug 1, 2012 at 0:16 comment added Guy Gur-Ari They matter multiplets they consider in appendix C are in the fundamental, as they write in the beginning of section 3.2 (a bit before they refer to the appendix). I added the explanation of (C.6) to the answer.
Aug 1, 2012 at 0:15 history edited Guy Gur-Ari CC BY-SA 3.0
Add (C.6) explanation. Remove 'I'm guessing'.
Jul 31, 2012 at 22:00 comment added user6818 I can see this general argument of needing to describe the eigenvalue of a $U(N)$ matrix by a density on the circle buts its the specific re-writing of that as in equation C.6 that is puzzling me. It would be a great help if you can may be write in a few lines about how that equation was gotten!
Jul 31, 2012 at 21:57 comment added user6818 @ Guy Gur-Ari Each of the $N_f$ hypermultiplets I think has 2 $N=2$ chiral multiplets one in the adjoint and the other in the conjuagte adjoint (= adjoint) of the gauge group. Now we have $\chi_{adj} = \chi_{\bar{adj}} = \chi_{fund}\times \chi_{anti-fund}$. So each hypermultiplet should have contributed a $2\chi_{fund}\times \chi_{anti-fund}$..right?
Jul 31, 2012 at 13:16 history edited Guy Gur-Ari CC BY-SA 3.0
added 1167 characters in body
Jul 31, 2012 at 13:01 history answered Guy Gur-Ari CC BY-SA 3.0