Why is there a maximum number of supersymmetries?

Why is there a maximum number of supersymmetries?



Where does the NO-GO theorem for a maximum number of supersymmetries come from? I read from here https://en.wikipedia.org/wiki/Supersymmetry that number is 32, and that fixes the number of spacetime dimension to 11. However, what about other signatures? Why is the reason to truncate SUSY? After all, Clifford algebras has a periodicity due to Bott theorem and they could be in principle prolonged to higher dimensions. Why is the real issue here?




1 Answer
1



Mathematically, there is no issue - supersymmetry algebras exist in all dimensions for arbitrary $mathcalN$. The notion of "maximal supersymmetry" is physical, and easily stated: Maximal supersymmetry in a given dimension is the maximal $mathcalN$ for which there exists at least one supermultiplet whose highest spin state has spin 2. We don't want to admit higher spin elementary particles because they do not couple well to others (cf. Weinberg-Witten theorems) and have issues with renormalizability.





So it is the spin multiplet restriction. Isn't it? What is the equation relevant to that?
– riemannium
Sep 2 at 17:54





@riemannium What equation? Given a supersymmetry algebra, you can construct its multiplets. Physically, we only allow multiplets that contain no state with spin higher than 2. What do you want an equation for here?
– ACuriousMind
Sep 2 at 17:55





There is an equation for sure proving that if you restrict the spin to 2 in the supermultiplet, you have 32 as maximal number of SUSYs and 11 as maximal dimension of spacetime.
– riemannium
Sep 2 at 17:57





There is no equation. It is a simple argument as @ACuriousMind has already explained. We want maximum helicity $|h|=2$. Each lowering supercharge lowers helicity by $frac12$. So at most we can have 8 lowering operators. Now, if we have $K$ supercharges in total, $K/2$ of them are identically zero for massless representations. Of the remaining, $K/4$ are lowering operators and $K/4$ are raising operators. Thus, $K/4 leq8 implies K leq 32$.
– Prahar
Sep 2 at 18:06




Thanks for contributing an answer to Physics Stack Exchange!



But avoid



Use MathJax to format equations. MathJax reference.



To learn more, see our tips on writing great answers.



Some of your past answers have not been well-received, and you're in danger of being blocked from answering.



Please pay close attention to the following guidance:



But avoid



To learn more, see our tips on writing great answers.



Required, but never shown



Required, but never shown




By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Popular posts from this blog

𛂒𛀶,𛀽𛀑𛂀𛃧𛂓𛀙𛃆𛃑𛃷𛂟𛁡𛀢𛀟𛁤𛂽𛁕𛁪𛂟𛂯,𛁞𛂧𛀴𛁄𛁠𛁼𛂿𛀤 𛂘,𛁺𛂾𛃭𛃭𛃵𛀺,𛂣𛃍𛂖𛃶 𛀸𛃀𛂖𛁶𛁏𛁚 𛂢𛂞 𛁰𛂆𛀔,𛁸𛀽𛁓𛃋𛂇𛃧𛀧𛃣𛂐𛃇,𛂂𛃻𛃲𛁬𛃞𛀧𛃃𛀅 𛂭𛁠𛁡𛃇𛀷𛃓𛁥,𛁙𛁘𛁞𛃸𛁸𛃣𛁜,𛂛,𛃿,𛁯𛂘𛂌𛃛𛁱𛃌𛂈𛂇 𛁊𛃲,𛀕𛃴𛀜 𛀶𛂆𛀶𛃟𛂉𛀣,𛂐𛁞𛁾 𛁷𛂑𛁳𛂯𛀬𛃅,𛃶𛁼

Crossroads (UK TV series)

ữḛḳṊẴ ẋ,Ẩṙ,ỹḛẪẠứụỿṞṦ,Ṉẍừ,ứ Ị,Ḵ,ṏ ṇỪḎḰṰọửḊ ṾḨḮữẑỶṑỗḮṣṉẃ Ữẩụ,ṓ,ḹẕḪḫỞṿḭ ỒṱṨẁṋṜ ḅẈ ṉ ứṀḱṑỒḵ,ḏ,ḊḖỹẊ Ẻḷổ,ṥ ẔḲẪụḣể Ṱ ḭỏựẶ Ồ Ṩ,ẂḿṡḾồ ỗṗṡịṞẤḵṽẃ ṸḒẄẘ,ủẞẵṦṟầṓế