N=0 case (Virasoro algebra) :
Here is an observation, I don't know yet if it has consequences :
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 super-Virasoro 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 $d_{N}$ ?

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 $d_{N}$ ?

Here is an observation, I don't know yet if it has consequences :
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 super-Virasoro 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 $c=\frac{3}{2}$ or $\frac{27}{2}$

N=0 case (Virasoro algebra) :
Here is an observation, I don't know yet if it has consequences :
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 super-Virasoro 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 $d_{N}$ ?