Let $\mathfrak{W}$ be the Lie algebra generated by $d_{n} = ie^{in\theta}\frac{d}{d\theta}$ and $\mathfrak{Vir} = \mathfrak{W} \oplus C \mathbb{C}$ its central extension:
$$
[L_m,L_n]=(m-n)L_{m+n}+\frac{C}{12}(m^3-m)\delta_{m+n,0},
$$
It's called the **Virasoro algebra**.

Its unitary highest weight representations are completely given by the pair $(c,h)$ such that :

- $C \Omega = c \Omega$
- $L_{0} \Omega = h \Omega$

The number $c$ is called the **central charge** and $\Omega$ is the vacuum vector.

If $0 \le c < 1$, then the FQS criterion says that $c = c_{m} = 1 - \frac{6}{m(m+1)}$ with $m = 2,3,4,5,...$

**Observation** :

Forget 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 the discriminant $\Delta = (c-1)^{2}-24(c-1) = (c-1)(c-25)$

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

**The $N$-supersymmetric extensions** ($\mathfrak{Vir}$ corresponds to $N=0$) :

$\mathfrak{Vir}$ admits two $N=1$ supersymmetric extensions : the Neveu-Schwarz and Ramond algebras.

In this case, the discrete series central charge is given by : $c = c_{m} = \frac{3}{2}(1 - \frac{8}{m(m+2)})$

**Observation** :

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 $9$

Let $d_{N}$ be the (allowed) space-time dimension of a $N$-superstring theory, then :

- $d_{0}=25+1$
- $d_{1}=9+1$

Questions:

Is there an explanation for the link with the observations above (or just a coincidence) ?

What are $d_{2}$, $d_{3}$ and $d_{4}$ ?

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

**Remark for N>1** : if for all $N$, $\Delta$ is quadratic in $c$, then $\Delta = 0$ admits two solutions in $c$.

Then $c$ could be renormalized as $c'=k.c$ such that one of the two solutions is $c'=1$ for the time-dimension, and the other, the space-dimension.

This process runs for $N=0,1$. Now for $N=2$, $c_{m} = \frac{3m}{m+2}$ :

Let the equation $c=c(x)=\frac{3x}{x+2}$, then $(c-3)x+2c=0$ and $\Delta = (c-3)^{2}$.

Let $c'=c/3$ (the renormalization), then we "would" obtain $d_{2} = 1+1$.