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I have a question on definition/motivation of Virasoro algebra. Recall that Virasoro algebra is an infinite Lie algebra generated by elements $L_n$ $(n\in \mathbb{Z})$ and $c$ over $\mathbb{C}$ with relations $$ [L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}(m^3-m)\delta_{m+n,0}. $$ A typical explanation of this definition is the following.

Define vector fields $l_n=-z^n\frac{\partial}{\partial z}$ on $\mathbb{C}\setminus \{0\}$. They form a Lie algebra of infinitesimal conformal transformation $$ [l_m,l_n]=(m-n)l_{m+n}. $$ So the Virasoro algebra is a central extension of this algebra by $c$. $c$ is called the central charge.

My questions are

  1. How can one see that the Lie algebra above is associate to infinitesimal conformal transformation?
  2. What is the central charge $c$ intuitively? Why are we interested in such a central extension?

As to second question, I don't have enough physics background to check what the central charge $c$ means in physics literature.

At this point, I don't have any intuition and have trouble in digesting the concept. I would really appreciate your help.

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you mean z^n not z_n. –  Abdelmalek Abdesselam Dec 27 '12 at 0:15
    
@Abdelmalek Thanks for pointing out the typo. –  user2013 Dec 27 '12 at 0:42
    
Try Ch. 9: Conformal Invariance Sec. 9.1: Energy momentum tensor-Virasoro algebra of the book Statistical Field Theory Vol. 2 by Itzykson and Drouffe. Maybe you could work around pg. 514 on translations, complex dilatations, and special conformal transformations related to $d/dz, zd/dz,$ and $z^2d/dz$.The central charge is discussed in the next sub-section from a physical and mathematical perspective. –  Tom Copeland Dec 27 '12 at 3:55
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As a first step at understanding the relations, I always recall that $exp(-c \cdot z^2d/dz)f(z)=exp[c \cdot d/d(1/z)]f(1/(1/z))=f(1/(c+1/z))=f(z/(cz+1)),$ a special conformal transformation. Note $zd/dz=d/d(ln(z))$ for the dilatation. –  Tom Copeland Dec 27 '12 at 4:10
    
I don't quite understand what you try to mean in the second comment. I would appreciate it if you could post a bit more detail as an answer. –  user2013 Dec 27 '12 at 6:30
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3 Answers 3

1 is standard (it's the correspondence between analytic complex functions and conformal maps). For 2, in physics, one really deals with projective representations, not just ordinary representations (this is because a quantum state is really a ray in Hilbert space rather than a vector). A projective representation of an algebra without central charge is the same as an ordinary representation of the algebra with a (potentially) non-zero central charge. It's easier to work with ordinary representations, so people use the centrally extended algebra.

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Thanks you for letting us know the projective representation point of view. That makes sense. –  user2013 Dec 27 '12 at 22:53
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I will answer (2) quickly: You can refer to my response to this other question: Why does bosonic string theory require 26 spacetime dimensions?

In String Theory (physics) there are "quantum operators", and the relation they satisfy are precisely this Virasoro relation. And not just that, but $c=D$, the space-time dimension! So it is at least extremely important in unifying the theories of physics via strings, because this relation helps us determine the dimension of our universe. You can view this term proportional to the central charge as a "quantum effect" (i.e. it only appears when you take your classical system and quantize it).

Why $c=D$?: The propagation ("worldsheet") of a 1-dimensional string (fundamental physical object in the theory) in space-time (dimension $D$) is described by functions $X^\mu$, where the index $\mu$ ranges from 0 to $D−2$. They decompose into modes $a^\mu_n$ (for satisfying the string wave-equation). These modes end up mixing and defining quantum operators $L_m$, and the commutator-relations amongst these modes spews out the Virasoro relation with $c=D$.

As for some rigorous intuition which will help with question (1): $c$ can be regarded as multiplying the unit operator, and when adjoined to the Lie algebra generated by the $L_m$ it lies in the center of that extended algebra. (I picked this up when working through Becker-Becker-Schwarz String theory textbook).

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Thank you for the answer, Chris. May I ask why multiplication by the unit vector is related to the space dimension? –  user2013 Dec 27 '12 at 0:48
    
Thanks for your writing "why $c=D$" above. –  user2013 Dec 27 '12 at 6:29
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How to see that $-z^{m+1}\frac{\partial }{\partial z}$ is related to an infinitesimal conformal transformation:

With $\omega=\frac{z^{-m}}{m}$ and $z=(m \cdot \omega)^{\frac{-1}{m}}$, the exponential mapping gives

$exp[-t\cdot z^{m+1}\frac{\partial }{\partial z}]f(z)=exp[t\cdot \frac{\partial }{\partial \omega}]f[(m \cdot \omega)^{\frac{-1}{m}}]=f[(m \cdot (\omega+t))^{\frac{-1}{m}}]=f\left [\frac{z}{(1+\ m \cdot t \cdot z^m)^{\frac{1}{m}}} \right ]=f(g_{m}(z,t)).$

Consistently, $g_{m}(g_{m}(z,s),t)=g_{m}(z,s+t).$

So the exponential mapping induced by the tangent vector results in composition of $f$ with $g_{m}(z,t)=z-\ t\cdot z^{m+1} +\ ....$.

Then to first order in $t$, a Taylor series expansion in $t$ about $t=0$ of the infinitesimal composition gives

$f(z-\ t\cdot z^{m+1})\approx f(z)-{f}'(z)z^{m+1}\cdot t=(1-\ t \cdot z^{m+1}\frac{\partial }{\partial z})f(z),$

and in the domain of analyticity of $g_{m}(z,t)$, the mappings are conformal.

Some interesting associated combinatorics:

1) $g_m$ is related to the e.g.f. for planar m-ary trees and double factorials (m=1), triple factorials (m=2), quartic (m=3), etc. (Cf. OEIS A094638)

2) The compositional inverse of $h(x)= x-\ t\cdot x^{m+1}$ gives a generating function for the Fuss-Catalan numbers (e.g., OEIS A001764).

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