13
$\begingroup$

The Virasoro algebras $Vir_c$ are a family of infinite dimensional Lie *-algebras parametrized by a real number $c$, called the central charge.¹

I hear that there exist two critical values of the central charge $c$, where qualitative changes happen to the structure of $Vir_c$ and of its representations. Those are $c=1$, and $c=25$.

I understand what happens at $c=1$. For $c<1$, the invariant sesquilinear form on the Verma modules is not positive definite, and is positive semi-definite for a discrete set of parameters, while for $c\ge 1$, the sesquilinear form is always positive definite (if a certain parameter called $h$ is $\ge 0$).

Can someone explain what happens at $c=25$?

Edit: José Figueroa-O'Farrill's comments have been quite illuminating. I would still like to know if here is a property (such as unitarity) of $Vir_c$ that distinguishes the cases $c<25$, $c=25$, and $c>25$.


¹ Strictly speaking, $Vir_c$ is not a Lie algebra, but only an associative algbera as I want to impose the relation that identifies the central element with 1, and there is no notion of unit in a Lie algebra (see Ben Webster's comment for a better explanation).

$\endgroup$
11
  • $\begingroup$ Why is not $Vir_c$ a Lie algebra? It is the universal central extension of the Lie algebra of diffeomorphisms of the circle. Of course $c$ is not a number, but the central element, which does act by a number $c$ in any irreducible module. $\endgroup$ Commented May 7, 2013 at 10:28
  • 2
    $\begingroup$ Andre wants to consider the quotient of the universal enveloping algebra of Virasoro by the relation $k-c\cdot 1$ ($k$ is the central element of the Lie algebra, $c$ a scalar, and $1$ the identity in the UAE). $\endgroup$
    – Ben Webster
    Commented May 8, 2013 at 2:48
  • 1
    $\begingroup$ Are these objects related to bosonic string theory with its $25+1$ space-time dimension (sorry, random pattern matching content free comment) :-) $\endgroup$
    – Suvrit
    Commented May 8, 2013 at 4:54
  • 1
    $\begingroup$ Ben, thanks for the clarification. In that case, then at the top of page 236 in this paper of Feigin and Fuks (the English summary of their longer paper in one of my previous comments) link.springer.com/content/pdf/10.1007%2FBFb0099939.pdf you will find the statement that at the level of Verma modules, $(h,c)$ and $(-1-h, 26-c)$ are anti-equivalent. This is just a symmetry of the determinant formula of the Shapovalov form. $\endgroup$ Commented May 8, 2013 at 9:50
  • 3
    $\begingroup$ The answer to S. Sra's question, however random, is "yes". The state space of a bosonic string can be identified with the semi-infinite cohomology of $Vir_c$ relative its centre. The usual complex is obtained by tensoring a Virasoro module (not necessarily irreducible) with $c=26$ with the module of semi-infinite forms. In fact, the duality which answers André's question was first noticed by Feigin in the paper where he defined semi-infinite cohomology. $\endgroup$ Commented May 8, 2013 at 12:45

2 Answers 2

5
$\begingroup$

I have an incomplete understanding of this, but I will try to say what I know.

For each $c \in \mathbb{C}$, we define $Verma_c$ to be the category whose objects are Verma modules $V_{h,c}$ of central charge $c$ and highest weight $h$ for some $h \in \mathbb{C}$, and whose maps are Virasoro-module maps. Feigin and Fuks worked out the morphisms in all of these categories (they are embeddings), and they found that the categories $Verma_c$ and $Verma_{26-c}$ are antiequivalent. In particular, when $V_{h,c}$ has an embedded Verma submodule $V_{h+x,c}$, the corresponding module $V_{-1-h,26-c}$ embeds into $V_{-1-h-x,26-c}$.

When you consider representations of Virasoro with central charge $0 < c < 1$ and positive $h$, the corresponding objects will have central charge $25 < 26-c < 26$, and highest weight that is negative. In particular, unitary representations don't seem to make an appearance anywhere in the complementary picture.

According to this MathOverflow question, Positselski has developed the antiequivalence into a derived category statement, where modules are resolved by complexes of Vermas. However, I have been unable to extract an explicit theorem about Virasoro from his monograph.

$\endgroup$
2
  • 3
    $\begingroup$ I thought the place in Positselski's monograph one should look at in this instance is Corollary and Remark on p.268 of arxiv.org/pdf/0708.3398v12.pdf ? $\endgroup$ Commented May 14, 2013 at 9:49
  • $\begingroup$ Thank you for the suggestion. There is a good explanation there. $\endgroup$
    – S. Carnahan
    Commented May 20, 2013 at 9:27
0
$\begingroup$

Here is an observation, I don't know yet if it has consequences :

N=0 case (Virasoro algebra) :
The unitary highest weight representations of the Virasoro algebra are completely given by the pair $(c,h)$. The number $c$ is called the central charge.
If $0 \le c < 1$, then the FQS criterion says that $c = c_{m} = 1 - \frac{6}{m(m+1)}$ with $m = 2,3,4,5,...$
Now, forget for a moment that $0 \le c < 1$ and just consider the equation : $c=c(x) = 1 - \frac{6}{x(x+1)}$
Then : $(c-1)x(x+1)+6=0$
And so : $(c-1)x^{2} + (c-1)x + 6 = 0$
But: $\Delta = (c-1)^{2}-24(c-1) = (c-1)(c-25)$

Conclusion : $\Delta = 0$ iff $c=1$ or $25$

$\rightarrow$ I guess it's coherent with the answer of Scott on the negative highest weight representations.

N=1 case (Neveu-Schwarz and Ramond algebras) :
In this case, the discrete series central charge is given by : $c = c_{m} = \frac{3}{2}(1 - \frac{8}{m(m+2)})$
Let the equation $c=c(x) = \frac{3}{2}(1 - \frac{8}{x(x+2)})$
Then : $(\frac{2}{3}c-1)x^{2} + 2(\frac{2}{3}c-1)x + 8 = 0$
And: $\Delta = 4(\frac{2}{3}c-1)^{2}-32(\frac{2}{3}c-1) = 4(\frac{2}{3}c-1)(\frac{2}{3}c-9)$

Conclusion : $\Delta=0$ iff $\frac{2}{3}c= 1$ or $9$

Remark : The space-time dimension $d_{N}$ in the superstring theory :

  • $d_{0}=25+1$
  • $d_{1}=9+1$

I guess it's not a coincidence.

If I'm not mistaken : $d_{2} = 2$, $d_{3} = 0$ and $d_{4} = -2$.

Is there a formula for $d_{N}$ ?

$\endgroup$
5
  • $\begingroup$ This appears to be more appropriate as a question than as an answer. $\endgroup$
    – S. Carnahan
    Commented Oct 13, 2013 at 1:03
  • $\begingroup$ @S.Carnahan : The observation that $\Delta = (c-1)(c-25)$ is an easy way to obtain $1$ and $25$, and it's coherent with your answer with $26-c$ : if I'm not mistaken, the negative highest weight discrete series have central charges $c_{m} = 25 + \frac{6}{m(m+1)}$, and the equations $c=c(x)=25 + \frac{6}{x(x+1)}$ gives the same $\Delta = (c-25)(c-1)$. Now, I'm agree that the questions about the space-time dimension of the $N$-superstring theory, deserve their own post. I will post it. $\endgroup$ Commented Oct 13, 2013 at 10:35
  • $\begingroup$ What is $\Delta$? $\endgroup$ Commented Oct 13, 2013 at 12:52
  • $\begingroup$ @AndréHenriques : $\Delta = \beta^{2}-4\alpha \gamma$ is the discriminant of the polynomial $\alpha x^{2}+\beta x+\gamma$. To avoid confusion : here $\alpha = c-1$, $\beta=c-1$ and $\gamma = 6$. $\endgroup$ Commented Oct 13, 2013 at 13:50
  • $\begingroup$ I have posted a development of the last question of my answer here $\endgroup$ Commented Jul 13, 2014 at 16:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .