Geometric/Visual Interpretation of Virasoro Algebra

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Solution 1

The simplest visual representation of the Lie group associated with the Virasoro (Lie) algebra is the group of reparametrizations of a circle.

Imagine that $\sigma$ is a periodic variable with the periodicity $2\pi$. An infinitesimal diffeomorphism is specified by a periodic function $\Delta \sigma(\sigma)$ with the periodicity $2\pi$. So the generators of the reparameterizations may be written as $f(\sigma)\partial / \partial \sigma$.

The possible functions $f(\sigma)$ may be expanded to the Fourier series, so a natural basis of the generators of the reparametrizations of the circle are $$ L_m = i \exp(im\sigma) \frac{\partial}{\partial \sigma} $$ As an exercise, calculate that the commutator $[L_m,L_n]$ is what it should be according to the Virasoro algebra, namely $(m-n)L_{m+n}$.

The Virasoro algebra for a closed string has two copies of the algebra above - and for the open string, it's only one copy but it's different than the "holomorphic" derivatives I used above. There are various related ways to represent the algebra but the reparameterizations of the circle are the simplest example.

Solution 2

There is a real Lie group $\tilde{Diff}(S^1)$ which is a $U(1)$ central extension of the real Lie group $Diff(S^1)$, and the Virasoro algebra is the Lie algebra of this Lie group.

The central extension $\tilde{Diff}(S^1)$ can be realized geometrically in two ways. The first is via a Hilbert space embedding (as in the book of Pressley-Segal), and the second is via the determinant line bundle.

This is all nicely explained in Appendix D of the book "Two-dimensional conformal geometry and vertex operator algebras" by Huang.

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Updated on August 01, 2022

Comments

  • Michael
    Michael over 1 year

    I've been trying to gain some intuition about Virasoro Algebras, but have failed so far.

    The Mathematical Definition seems to be clear (as found in http://en.wikipedia.org/wiki/Virasoro_algebra). I just can't seem to gain some intuition about it. As a central extension to Witt Algebras, I was hoping that there has to be some geometric interpretation, as I can imagine Witt Algebras rather well.

    If anyone has some nice Geometric or Visual Interpretation of Virasoro Algebra, I'd greatly appreciate it!

  • Heidar
    Heidar over 12 years
    Aren't you talking about the Witt algebra? I think Knoten had a problem of visualizing the central extension of this. I understand that $\text{Diff}(S^1)$, the group of diffeomorphisms on the unit circle, is the Group associated to the Witt algebra (as you say). But do you know if such a group exist for the Virasoro algebra? Many books seem to suggest that there doesn't, but I don't think I have seen a proof of this.
  • Luboš Motl
    Luboš Motl over 12 years
    Oh, I see. There is obviously no out-of-Hilbert-space visual representation of the central extension that would differ from the $c=0$ algebra. The reason is that the central extension has $c$-numbers in the commutators. ;-) Any $c$-numbers may only be represented as the transformation of phase in the Hilbert space, and a transformation of phase of a vector in the Hilbert space doesn't change the character of this state "physically" or "geometrically" - it's just about the normalization. So central extensions are just central extensions and they share the original visualizations with the $c=0$.
  • Michael
    Michael over 12 years
    Did I understand correct that you are saying that I can use the same geometric interpretation of Witt Algebra and apply it to Virasoro?
  • Luboš Motl
    Luboš Motl over 12 years
    Sure, the central extension of an algebra is just a very subtle modification of the original algebra that doesn't change its physical meaning. For every central extension, one may obtain the original algebra simply by setting all the $c$-number generators to zero. This preserves the Jacobi identity etc. because the $c$-number generators commuted with anything else, anyway - well, that what it means that it was "central". ;-) In string theory, the Virasoro algebra is still the algebra of reparameterizations of the world sheet, even for $c\neq 0$.
  • Michael
    Michael over 12 years
    Sounds logical enough :) I'll take it
  • Luboš Motl
    Luboš Motl over 10 years
    Maybe I should have said, two years ago, that the right hand side of the Virasoro algebra - and others - contains proper operators, and those correspond to the Poisson brackets; and them it may contain the $c$-numbers. They're similar to $i\hbar$ in $[x,p]$. More generally, they are multiplied by a higher power of $\hbar$. At any rate, these $c$-number terms disappear - even relatively to the operator-valued terms - in the classical $\hbar\to 0$ limit which means that the classical interpretation is independent of these $c$-number terms (it's the same for a central extension).
  • Ruben Verresen
    Ruben Verresen almost 7 years
    Interesting! I guess the point is that the Virasoro algebra might appear as the Lie algebra of a Lie group, but presumably when it comes to the specific way it acts in the case of e.g. 1+1D CFT it cannot be interpreted as the action of such a Lie group.
  • Bruce Bartlett
    Bruce Bartlett almost 7 years
    No. There is no problem - the full Lie group $\tilde{Diff}(S^1)$ acts in 1+1d CFT, not just the Lie algebra. This is explained in Huang's book. It is a folk misunderstanding that there is some kind of "issue", that only "the Lie algebra" acts.
  • pre-kidney
    pre-kidney over 6 years
    What you are referring to as a folk misunderstanding seems to be resolved differently in Schottenloher's book "A Mathematical Introduction to Conformal Field Theory", section 5.4 concerns the non-existence of the complex Virasoro group.
  • Bruce Bartlett
    Bruce Bartlett over 6 years
    There is no disagreement between the way Schottenloher's book and Huang's book resolve this. See the last paragraph in Section 5.4 in Schottenloher's book.
  • Bruce Bartlett
    Bruce Bartlett over 6 years
    Schottenloher's book is also a great book, but he mischaracterises the problem a bit in Section 5.4. Schottenloher conceives of the group Diff(S^1) of diffeomorphisms of the circle as being significant for conformal field theory. Physicists think like this too, and this leads to confusion. Rather it is the quotient group Diff(S^1) / Rot(S^1) of diffeomorphisms of S^1 divided by rotations which is significant for conformal field theory.
  • Bruce Bartlett
    Bruce Bartlett over 6 years
    The reason is that the group M := Diff(S^1) / Rot(S^1) has a beautiful geometric interpretation: it is the "Riemann mapping group". More precisely, it is the set of all injective maps out of the closed unit disc, $f : D \rightarrow f(D)$, which are holomorphic on the interior of $D$ and smooth up to the boundary of $D$, normalized by the condition that $f(0)=1$ and $f'(0)=1$. We can call such a map a "Riemann map" since it is precisely these maps which the Riemann mapping theorem is about.
  • Bruce Bartlett
    Bruce Bartlett over 6 years
    So this is precisely the group which physicists -really mean- when they talk about the "group of conformal transformations".
  • Bruce Bartlett
    Bruce Bartlett over 6 years
    This is a beautiful group. In fact, it even has a beautiful complex structure which is Kahler (see Kirillov , Kahler structures on the group of diffeomorphisms of a circle), and a canonical central extension, which is the "Virasoro group". This extension also has geometric significance! It is what occurs if you drop the requirement $f'(0) = 1$ above.
  • Bruce Bartlett
    Bruce Bartlett over 6 years
    It is a while since I thought about this and I hope I haven't confused myself, but I think this is the gist of it. The correct "non-centrally extended" group in CFT is M, not Diff(S^1). Once you realize this, all other problems fade away.
  • Bruce Bartlett
    Bruce Bartlett over 6 years
    I recommend the paper of Kirillov that I referred to above. It is all consistent with Huang and Schottenloher, and it offers a good summary.