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About 2+1 dimensional superconformal algebra

+ 5 like - 0 dislike

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 (955 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 (1,390 points) [ no revision ]
Most voted comments show all comments
Thanks for the reply. (1) I am not understanding why you say that $I_{ij}$ and $M_{\mu \nu}$ are not represented by the matrices I wrote down for them. I mean as an abstract (super) Lie algebra they may not be so but aren't my matrices the right ones when they actually act on the quantum states like when one derived the $\epsilon_0 - (h_1 + j)$ ? (2) Referring to Polchinski's volume 2 conventions, I am guessing that $h_i$ is the eigenvalue of a weight vector in the vector representation under the action of $I_{i i+1}$

This post imported from StackExchange Physics at 2014-08-23 04:59 (UCT), posted by SE-user user6818
(3)About the extended Einstein convention - since the paper is using Euclideanized QFT I would have thought that they are raising and lowering by diag(1,1,1) metric but then they insist on using all real Gamma matrices as I wrote down and then $(\Gamma^0)^2 = -1$. This is confusing.

This post imported from StackExchange Physics at 2014-08-23 04:59 (UCT), posted by SE-user user6818
(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
2) Right. 3) The author didn't say Gamma are all real, he just said sigma is real. 4) You need to understand that the matrices M are determined by j and that the metrices $I$ are detemined by h.

This post imported from StackExchange Physics at 2014-08-23 05:00 (UCT), posted by SE-user Yuji
Most recent comments show all comments
! 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

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