Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  About 2+1 dimensional superconformal algebra

+ 5 like - 0 dislike
2269 views

I would like to get some help in interpreting the main equation of the superconformal algebra (in $2+1$ dimenions) as stated in equation 3.27 on page 18 of this paper. I am familiar with supersymmetry algebra but still this notation looks very obscure to me.

  • In the above equation for a fixed $i$, $j$, $\alpha$, $\tilde{\beta}$ the last term, $-i\delta_{\alpha , \tilde{\beta}} I_{ij}$ will be a $\cal{N} \times \cal{N}$ matrix for $\cal{N}$ extended supersymmetry in $2+1$ dimensions. Is this interpretation right?

(..where I guess $I_{ij}$ is the vector representation of $so(\cal{N})$ given as, $(I_{ij})_{ab} = - i(\delta _{ia}\delta _{jb} - \delta _{ib} \delta _{ja})$..)

  • Now if the above is so then is there an implicit $\cal{N} \times \cal{N}$ identity matrix multiplied to the first term, $i \frac{\delta_{ij}}{2} [(M'_ {\mu \nu}\Gamma_\mu \Gamma_\nu C)_{\alpha \tilde{\beta}} + 2D' \delta _ {\alpha \tilde{\beta}}] $ ?

So I guess that the equation is to be read as an equality between 2 $\cal{N} \times \cal{N}$ matrices. right?

  • Is there is a typo in this equation that the first term should have $(M'_ {\mu \nu}\Gamma^\mu \Gamma ^ \nu C)$ instead of all the space-time indices $\mu, \nu$ to be down?

  • I guess that in $M' _{\mu \nu}$ the indices $\mu$ and $\nu$ range over $0,1,2...,d-1$ for a $d-$dimensional space-time (...here $d=3$..) and for this range in the Euclideanized QFT (as is the case here) one can replace $M'_{\mu \nu} = \frac {i}{4}[\Gamma _ \mu , \Gamma _ \nu]$. Is that right?

  • One is using the convention here where the signature is $\eta_{\mu \nu} = diag(-1,1,1) = \eta ^{\mu \nu}$ and the Gamma matrices are such that $\Gamma^0 = C = [[0,1],[-1,0]], \Gamma ^1 = [[0,1],[1,0]], \Gamma ^ 2 = [[1,0],[0,-1]]$ and then the charge conjugation matrix $C$ satisfies $C^{-1} \Gamma ^\mu C = - \Gamma ^{\mu T}$ and $[\Gamma ^\mu , \Gamma ^\nu]_+ = 2 \eta ^\mu \eta ^\nu$

Then $M^{\mu \nu}\Gamma _\mu \Gamma _ \nu = -3i [[1,0],[0,1]]$


Now for a specific case of this equation let me refer to the bottom of page $8$ and top of page $9$ of this paper.

  • In physics literature what is the implicit equation/convention that defines the representation of $SO(N)$ with heighest weights $(h_1, h_2, ... , h_{[\frac{N}{2}]})$?

I could not find an equation anywhere which defines the $h_i$s

  • How does choosing the weights of the $Q$ operator to be as stated in the bottom of page 8 determine the values of $i$ and $\alpha$ that goes in the RHS of the anti-commutation equation described in the first half?

And how does it determine the same for the $S$ operator which because of Euclideanization is related as , $S^{'}_{i \alpha} = (Q^{'i \alpha})^\dagger $ (...I guess that the raising and lowering of indices doesn't matter here...)

  • Now given the choice as stated in the bottom of page 8 in the paper above and the S-Q Hermiticity relation and the anti-commutation relation in the first half of this question how does one prove the relation claimed on the top of page 9 which is effectively, $[Q^{'i\alpha},S^{'}_{i\alpha}]_+ = \epsilon_0 - (h_1 + j)$

I guess $\epsilon_0$ is the charge under the $D'$ of the first half defined for an operator $A$ (say) as $[D',A] = -\epsilon _0 A$ though I can't see the precise definition of $h_i$s and $j$ in terms of the RHS of the Q-S anti-commutation relation as described in the first half of the question.

  • Does anything about the above $[Q^{'i\alpha},S^{'}_{i\alpha}]_+ = \epsilon_0 - (h_1 + j)$ depend on what is the value of $\cal{N}$? I guess it could be $2$ as in this paper or $3$ and it would still be the same expression.

It would be great if someone can help with this.

This post imported from StackExchange Physics at 2014-08-23 04:59 (UCT), posted by SE-user user6818
asked Jul 11, 2012 in Theoretical Physics by user6818 (960 points) [ no revision ]

1 Answer

+ 2 like - 0 dislike

The first bullet point: no. $I_{ij}$ (for a fixed $i,j$) is just a generator of $SO(2n)$, not its explicit matrix representative. The commutation relation in general is an equation inside the Lie algebra.

The second bullet point: no.

The third bullet point: Yes and no. People in the field don't usually care where to put the indices, because we usually use the extended Einstein convention where $A_\mu B_\mu$ means $A_\mu B^\mu$, i.e., repeated indices are interpreted as put on either superscript or subscript appropriately and summed over to give a Lorentz invariant result.

The fourth point: no. Again, $M_{\mu\nu}$ is just a generator, not its matrix representative. $\Gamma_{\mu}$ is, on the other hand, is an explicit matrix.

The fifth point: this question doesn't make sense, due to the fourth point above.

The sixth point: there's no unified convention. In this case it's explained in the footnote 5.

The last three bullet points: I guess you should reread the papers based on the answers so far, and ask again at physics.SE as a separate question if you still have questions.

This post imported from StackExchange Physics at 2014-08-23 04:59 (UCT), posted by SE-user Yuji
answered Jul 13, 2012 by Yuji (1,395 points) [ no revision ]
Most voted comments show all comments
Thanks for the reply! (1) The authors say that their super-charges are in a vector representation of the R-symmetry group $SO(N)$ and hence I wrote down the corresponding representation matrices $I_{ij}$ for $SO(N)$. (3) When the authors say that the spatial sigma are all real and Hermitian except $\sigma_0$ which is real and anti-Hermitian, I understood that they mean to use the Gamma matrices for $SO(2,1)$ which I wrote down. Whats wrong here?

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user user6818
(4) Its true that a representation of the angular momentum algebra is determined by which spin one is looking at but here I would think that $M_{\mu \nu}$ are the generators of the Lorentz algebra and they can always be gotten as pair-wise commutators of the corresponding Clifford algebra as $M_{\mu \nu} = \frac{i}{4}[\Gamma _\mu , \Gamma _ \nu]$. Can you help define $j$ from here?

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user user6818
You need to distinguish two questions: what is the representation of the supercharge and what is the representation of the quantum states the supercharge is acting on. In order to get the inequality you're interested, you need to consider the latter question. You can't fix $I$ to be $N\times N$ and $M_{\mu\nu}$ to be $2\times 2$. They depend on $h$ and $j$. You try extracting $j$ from the $M_{\mu\nu}$ you already fixed, but that's not the point. $j$ depends on the quantum states.

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Yuji
! Okay! Thanks for clarifying my misunderstanding. Now I see why you insisted on thinking of $M_{\mu \nu}$ and $I_{ij}$ as abstract generators! I would then think that when the super-Lie-Algebra has the adjoint action on itself it is then that $I_{ij}$ acts on the Q's and the S's as the vector-SO(N) representation matrix as I wrote down. right? So from the abstract thinking of $M_{\mu \nu}$ how does one define $j$? (..or should I put up a separate question on the derivation of the BPS condition which you can answer there?..)

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user user6818
Now that you understand that point, I think you should stop asking and think for several days by yourself. Good luck!

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Yuji
Most recent comments show all comments
(4)It would be great if you can help define $j$ as in the expression. In these $1+2$ dimensions, given $M_{\mu \nu}$ $0\leq \mu ,\nu \leq 2$, I can guess that $j$ is the eigenvalue of the $SO(2)$ representation from $L^3 = M_{12}$. Though this doesn't look precise enough to me to derive the $\epsilon_0 - (h_1 + j)$ condition. It would be great if you can help read this $\{S,Q\}$ algebra.....I am feeling like I am missing something essential here.

This post imported from StackExchange Physics at 2014-08-23 04:59 (UCT), posted by SE-user user6818
1) The right matrices to use depend on the quantum states. What you wrote down is not OK even in a single superconformal multiplet. How did you determine which matrix representations of SO(N) and SO(2,1) to use?

This post imported from StackExchange Physics at 2014-08-23 04:59 (UCT), posted by SE-user Yuji

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysi$\varnothing$sOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...