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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,852 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

How many embeddings are there of super-Virasoro into n Fermions?

+ 6 like - 0 dislike
45 views

What is the space of N=1 super-Virasoro vertex superalgebras inside the c=n/2 free fermion vertex superalgebra? [Said differently, how many Neveu-Schwartz vectors are there in n fermions?] Answers in terms of VOAs, CFTs, or nets all welcome, and any interpretation of "N=1 super" that makes you happy is okay.

This post imported from StackExchange MathOverflow at 2015-04-16 11:42 (UTC), posted by SE-user cdouglas
asked Oct 19, 2009 in Theoretical Physics by cdouglas (30 points) [ no revision ]
retagged Apr 16, 2015
Just to try to understand the question better, are you implicitly demanding that the free fermions be primaries wrt the resulting Virasoro vector?

This post imported from StackExchange MathOverflow at 2015-04-16 11:42 (UTC), posted by SE-user José Figueroa-O'Farrill
I'd like the Virasoro inside the super-Virasoro to agree with the Virasoro of the free fermions (that is the n-th tensor power of the Virasoro of the free fermion). The 'space' is then the space of extensions of the given Vir to an sVir.

This post imported from StackExchange MathOverflow at 2015-04-16 11:42 (UTC), posted by SE-user cdouglas

1 Answer

+ 7 like - 0 dislike

I would be very surprised if this were not known, but I cannot find a suitable reference, so I did the calculation as the question seemed interesting to me. I believe that this is now the full answer to the question.

Let $V$ be an $n$-dimensional real vector space with a symmetric inner product $\langle-,-\rangle$. Formulae are easier to write if we choose a basis $(e_i)$ for $V$ and let $$g_{ij} = \langle e_i,e_j\rangle.$$

Let $\psi_i$ denote the corresponding free fermion, with OPE $$\psi_i(z) \psi_j(w) = \frac{g_{ij} 1}{z-w} + \mathrm{reg}$$

The standard Virasoro vector is then $$T = 1/2 g^{ij} \partial \psi_i \psi_j,$$ where the product is the normal-ordered product, which in my conventions associates to the left, so that $ABC = (A(BC))$. $T$ obeys the standard Virasoro OPE with $c = n/2$: $$T(z) T(w) = \frac{\frac{n}4 1}{(z-w)^4} + 2 \frac{T(w)}{(z-w)^2} + \frac{\partial T(w)}{z-w} + \mathrm{reg}$$

The fields $\psi_i(z)$ are primary with weight $\frac12$ relative to $T$.

Then you are asking about the existence of a field $G$ which is primary of weight $\frac32$ relative to $T$ and whose OPE is $$G(z) G(w) = \frac{\frac{n}3 1}{(z-w)^3} + \frac{2 T(w)}{z-w}$$

The most general such $G$ takes the form $$G = \frac16 A^{ijk} \psi_i \psi_j \psi_k + B^i \partial \psi_i.$$

Calculating the relevant OPEs one sees that:

  • for $G$ to be primary, $B^i=0$, and
  • for $2T$ to appear in the first-order pole of $GG$, $A$ must satisfy two conditions which I will now rephrase.

First of all, $A$ defines an alternating bilinear map $[-,-]: V \times V \to V$ by $$[e_i,e_j] = A_{ijk} g^{kl} e_l.$$

The existence of $G$ translates into the following conditions on this map:

  • $\langle[x,y],z\rangle$ is totally skew
  • $\langle[x,y],[z,w]\rangle$ is an algebraic curvature tensor
  • and $$\mathrm{Tr}~ \mathrm{ad}_x \mathrm{ad}_y = 2 \langle x,y\rangle$$ where $\mathrm{ad}_x$ is the skewsymmetric endomorphism defined by $\mathrm{ad}_x(y) = [x,y].$

The algebraic curvature tensor condition means that the fourth rank tensor obeys the algebraic Bianchi identity: $$\langle[x,y],[z,w]\rangle + \mathrm{cyclic~in}~(x,y,z) = 0,$$ which using the invariance of the inner product under $\mathrm{ad}_x$ becomes $$\langle[[x,y],z],w\rangle + \mathrm{cyclic} = 0,$$ which in turn is equivalent to the Jacobi identity for the bracket. (Thanks for Paul de Medeiros for the observation.) Finally the last condition says that the Killing form, being twice the inner product, is nondegenerate, whence the Lie algebra is semisimple.

In summary, the solutions are in one-to-one correspondence with real semisimple Lie algebras. If you further require the inner product to be positive-definite, then these are the compact semisimple Lie algebras.

Interestingly, in this case one can embed an affine Lie algebra in such a way that the Virasoro vector coincides with the Sugawara construction. As I said above, I'm sure that this is standard, but cannot locate the reference right now.

EDIT This result is indeed known and can be found in: Goddard and Olive's Kac-Moody algebras, conformal symmetry and critical exponents, Nuclear Physics B, Volume 257, 1985, Pages 226-252, in the very last section of the paper. (Thanks to Matthias Gaberdiel for the suggestion to look there.)

EDIT: I just noticed that the question asked "how many", whereas my answer just showed that as many as "semisimple Lie algebras". If, for the sake of simplicity, we take the inner product to be positive-definite, then it is easy to write down a partition function and evaluating this gives a numerical answer to the question, depending on n. The first 100 values, starting at n=1, are the following:

0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 2, 3, 5, 3, 4, 8, 4, 5, 8, 7, 8, 11, 10, 11, 12, 13, 15, 19, 16, 21, 24, 21, 24, 32, 27, 34, 43, 37, 39, 53, 47, 54, 65, 65, 68, 79, 80, 90, 98, 102, 114, 129, 122, 138, 160, 157, 172, 207, 193, 211, 247, 244, 262, 306, 305, 329, 363, 378, 399, 448, 460, 505, 548, 554, 601, 675, 669, 739, 822, 826, 877, 990, 999, 1068, 1184, 1227, 1288, 1419, 1458, 1554, 1693, 1765

(A previous version of this edit had only taken into account the A-series... apologies.)

This post imported from StackExchange MathOverflow at 2015-04-16 11:42 (UTC), posted by SE-user José Figueroa-O'Farrill
answered Oct 29, 2009 by José Figueroa-O'Farrill (2,135 points) [ no revision ]
Thanks very much for this detailed answer! Much appreciated. (And sorry the thanks took so long in coming, my notification seems to be on the blink.)

This post imported from StackExchange MathOverflow at 2015-04-16 11:42 (UTC), posted by SE-user cdouglas

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).
To avoid this verification in future, please log in or register.




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

Your rights
...