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Classical Virasoro Algebra

  1. Mar 27, 2004 #1

    Zig

    User Avatar

    [SOLVED] Classical Virasoro Algebra

    The classical virasoro algebra is the Lie algebra of the group
    Diff(S^1), right? So I expected that if I look at the direct sum of two
    copies of the Virasoro algebra, I should get the Lie algebra to
    Diff(S^1)xDiff(S^1), from standard Lie theory.

    Instead, it seems that two copies of the Virasoro algebra generates the
    group of holomorphic maps on the cylinder. Are these two groups really
    the same? Is there an easy way to see it?
     
  2. jcsd
  3. Mar 28, 2004 #2
    On Sat, 27 Mar 2004, Zig wrote:

    > ... Instead, it seems that two copies of the Virasoro algebra generates the
    > group of holomorphic maps on the cylinder. Are these two groups really
    > the same? Is there an easy way to see it?


    Dear Zig,
    that's a good question. Several comments should be said.

    1. First, the group generatated by (two copies of) the Virasoro algebra is
    not the group of all diffeomorphisms of a two-dimensional worldsheet
    because such a group would be much much bigger (there is much more freedom
    to reparameterize 2 coordinates - they can mix in various ways); it is
    just a group of conformal transformations, i.e. the diffeomorphisms that
    preserve the angles, i.e. the diffeomorphisms that preserve the original
    metric up to a spacetime-dependent scaling. If you consider the standard
    Minkowski metric on the cylinder, the conformal transformations must
    transform the metric to another metric in the conformal gauge, g_{mn} =
    exp(2.phi(s,t)) eta_{mn} where (s,t) are worldsheet coordinates and
    eta_{mn} is diag (1,-1). The conformal transformations are those that have
    not been fixed by the gauge choice g_{mn} = exp(2.phi(s,t)) - most of them
    have been fixed, and therefore the conformal transformations form a very
    small group of the original reparameterization group in 2 dimensions.

    2. Second, what does it mean to preserve the angles (or the equivalent
    conditions from [1]) on the Minkowski worldsheet? In two dimensions it
    simply means that the null (lightlike) directions must be mapped to
    themselves. Using the coordinates s+t and s-t on the worldsheet (we won't
    use the standard symbols sigma^+ and sigma^- here because it is actually a
    longer, less convenient notation in e-mail), we see that the conformal
    transformations from [1] must map (s+t) to a function of (s+t), and (s-t)
    to a function of (s-t), so that the null lines (s+t)=const - and also
    (s-t)=const - remain lightlike for arbitrary choices of "const".

    3. Otherwise, the new (s+t), called (s'+t'), can be an arbitrary function
    of (s+t), and the same thing holds for for (s'-t') and (s-t). It means
    that the group of conformal symmetries of the Minkowski cylinder is
    determined by an arbitrary diffeomorphism on the coordinate (s+t), and an
    independent, commuting diffeomorphism on the coordinate (s-t).

    4. Well, up to a periodicity condition. If the periodicity of the
    spacelike direction "s" was L, then the identification in terms of
    (s+t,s-t) is that (s+t,s-t) is the same point as (s+t+L, s-t+L). Your
    diffeomorphisms must preserve this periodicity. Because the
    diffeomorphisms acting on s+t and s-t are independent, and they must
    preserve this combined periodicity, these diffeomorphisms must respect, in
    fact, the periodicity separately. Therefore you get two copies of the
    Virasoro symmetry (diffeomorphisms of a circle) - one of the copies
    generates the arbitrary diffeomorphisms of the circle of allowed values of
    "s+t", and the other does the same for "s-t".

    5. The Virasoro generators are operators whose Hermitean conjugates are
    the operators from the same sector: it's because the Hermitean conjugate
    (=inverse) of the unitary transformation "reparameterize (s+t) in some
    way" is again "reparameterize (s+t) in another way". Even if we
    Wick-rotate and work with the Euclidean worldsheet, where it looks like
    that the antiholomorphic (was: right-moving) sector is complex conjugate
    to the holomorphic (was: left-moving) sector, the operation of *Hermitean*
    conjugation in fact includes the operation t_E -> -t_E (the Euclidean time
    runs backwards), and therefore the Hermitean conjugate of a Virasoro
    generator from one of these two families (i.e. acting on the holomorphic
    coordinate) is still an operator acting on the same group.

    Feel free to ask more refined questions.

    Best wishes
    Lubos
    ______________________________________________________________________________
    E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
    eFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)
    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
     
  4. Mar 28, 2004 #3

    Zig

    User Avatar

    Lubos Motl wrote:

    > ... that's a good question. Several comments should be said ...


    Hi Lubos!
    thanks for the detailed reply. I am enthusiastic about the string
    theory newsgroup. I hope you don't mind elementary questions from
    students like me.

    > 1. First, the group generatated by (two copies of) the Virasoro algebra is
    > not the group of all diffeomorphisms of a two-dimensional worldsheet
    > because such a group would be much much bigger (there is much more freedom
    > to reparameterize 2 coordinates - they can mix in various ways); it is
    > just a group of conformal transformations, i.e. the diffeomorphisms that
    > preserve the angles, i.e. the diffeomorphisms that preserve the original
    > metric up to a spacetime-dependent scaling.


    yes, I think i can appreciate this fact. It is OK to refer to them as
    holomorphic maps, though, right? in 2D, a conformal transformation is
    holomorphic, so I can either think of them as maps which preserve the
    metric up to a scale, or maps that preserve the complex structure.

    > If you consider the standard
    > Minkowski metric on the cylinder, the conformal transformations must
    > transform the metric to another metric in the conformal gauge, g_{mn} =
    > exp(2.phi(s,t)) eta_{mn} where (s,t) are worldsheet coordinates and
    > eta_{mn} is diag (1,-1). The conformal transformations are those that have
    > not been fixed by the gauge choice g_{mn} = exp(2.phi(s,t)) - most of them
    > have been fixed, and therefore the conformal transformations form a very
    > small group of the original reparameterization group in 2 dimensions.


    yeah, I suppose it is a very small subgroup. If we were in D>2, the
    group of conformal maps would be finite dimensional. a much much
    smaller group. I wonder if there is a precise way to classify the
    difference in size between the diffeomorphism group in D=2 and the
    conformal group. perhaps the diffeomorphism group has a nondenumerable
    dimension?

    > Because the
    > diffeomorphisms acting on s+t and s-t are independent, and they must
    > preserve this combined periodicity, these diffeomorphisms must respect, in
    > fact, the periodicity separately. Therefore you get two copies of the
    > Virasoro symmetry (diffeomorphisms of a circle) - one of the copies
    > generates the arbitrary diffeomorphisms of the circle of allowed values of
    > "s+t", and the other does the same for "s-t".


    OK, that explains it! So the group of conformal transformations of the
    cylinder is two copies of Diff(S^1), because it is really
    diffeomorphisms of the two lightcones. this was a very helpful explanation.

    > Feel free to ask more refined questions.


    Thanks again for the reply. I don't seem to have a more refined
    question. Maybe I can ask another question? my professor's explanation
    of this has left me mystified.

    In computing the partition function of free bosonic field theory in 1+1
    dimensions on the cylinder, we identify the in and out states in the
    theory, so that our partition function is really calculated on a torus.
    (I am taking this from p. 244 of "Mirror Symmetry" by Hori et al.) We
    are doing a lot with tori, and the book and my professor's explanation
    are opaque to me.

    Anyway, thanks for your time, I hope you don't find it bothersome for
    students questions on your newsgroups.

    -z
     
  5. Mar 28, 2004 #4
    On Sun, 28 Mar 2004, Zig wrote:

    > thanks for the detailed reply. I am enthusiastic about the string
    > theory newsgroup. I hope you don't mind elementary questions from
    > students like me.


    It was a pleasure, and I hope that after a week there will be many people
    actively participating.

    > yes, I think i can appreciate this fact. It is OK to refer to them as
    > holomorphic maps, though, right? in 2D, a conformal transformation is
    > holomorphic, so I can either think of them as maps which preserve the
    > metric up to a scale, or maps that preserve the complex structure.


    Absolutely. On a Minkowski spacetime/worldsheet, a conformal
    transformation also preserves the metric up to a scale, which means that
    it must preserve the light cone - the set of all null directions at each
    point.

    > yeah, I suppose it is a very small subgroup. If we were in D>2, the
    > group of conformal maps would be finite dimensional.


    Right. The conformal group of a d-dimensional Euclidean space is
    SO(d+1,1), and that of a d-dimensional Minkowski space is SO(d,2): you
    always add (1,1) dimensions to the rotational/Lorentz symmetry.

    If you combine the new 1+1 dimensions into light-like directions +,-, then
    the conformal group SO(d+1,1) or SO(d,2) has the following generators:

    J_{ij} - the rotational (or Lorentz) generators
    J_{+i} - the momenta P_i; note that they commute with each
    other because g_{++}=0
    J_{+-} - this is the dilatation operator, rescaling the whole
    space(time) by an overall factor
    J_{-j} - well, these are the extra nontrivial conformal generators

    In two Euclidean dimensions, you obtain the conformal group SO(3,1) which
    is isomorphic to SL(2,C) / Z_2, while in 2 Minkowski dimensions you obtain
    SO(2,2) which is SL(2,R) x SL(2,R) (over some Z_2's). SL(2,C), for
    example, is what maps the sphere CP^1=S^2 onto itself.

    However in two dimensions you can extend the conformal group to an
    infinite-dimensional group of all holomorphic maps; such transformations
    are well-defined at least locally. Such an extension is not possible in
    d>2.

    > a much much smaller group. I wonder if there is a precise way to
    > classify the difference in size between the diffeomorphism group in
    > D=2 and the conformal group. perhaps the diffeomorphism group has a
    > nondenumerable dimension?


    The diffeomorphism group is much bigger. You can imagine the dimension of
    the diffeomorphism group to be "d" times the number of points in your
    d-dimensional spacetime: at each point, you can define "d" components of
    an infinitesimal vector field that generates a reparameterization of
    coordinates. The conformal group is much smaller - you saw that its
    dimension equals the number of points on "two circles", so to say, which
    is much smaller than the number of points on a two-dimensional worldsheet.
    In d>2 the dimension of the diffeomorphism group grows even more while the
    conformal group becomes finite-dimensional: the discrepancy between these
    two group is increasingly big as "d" grows.

    In 1 dimension, the group coincide because every 1D diffeomorphism
    (reparameterization of 1 coordinate) is conformal :-).

    > OK, that explains it! So the group of conformal transformations of the
    > cylinder is two copies of Diff(S^1), because it is really
    > diffeomorphisms of the two lightcones. this was a very helpful explanation.


    Exactly.

    > In computing the partition function of free bosonic field theory in 1+1
    > dimensions on the cylinder, we identify the in and out states in the
    > theory, so that our partition function is really calculated on a torus.


    Great, but I hope there was no specific question here - at least I've
    found no question mark(s). ;-)

    Cheers,
    Lubos
    ______________________________________________________________________________
    E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
    eFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)
    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
     
  6. Mar 28, 2004 #5
    On Sun, 28 Mar 2004, Lubos Motl wrote:

    > Right. The conformal group of a d-dimensional Euclidean space is
    > SO(d+1,1), and that of a d-dimensional Minkowski space is SO(d,2): you
    > always add (1,1) dimensions to the rotational/Lorentz symmetry.


    This might sound silly, but is there an easy geometric "picture" that
    helps you see why you "add (1,1) dimensions to the rotational/Lorentz
    symmetry"? I understand that it works out that way, but I can't visualize
    it.

    > However in two dimensions you can extend the conformal group to an
    > infinite-dimensional group of all holomorphic maps; such transformations
    > are well-defined at least locally. Such an extension is not possible in
    > d>2.


    Now I've seen this claim before that the infinite dimensional extension of
    the conformal group is not, generically, well-defined globally but how
    does this work? It sounds like this infinite dimensional group is the
    automorphisms of C, and so it only applies to charts diffeomorphic to C.
    Is that right or am I all washed up?
     
  7. Mar 29, 2004 #6

    Zig

    User Avatar

    Creighton Hogg wrote:

    >>However in two dimensions you can extend the conformal group to an
    >>infinite-dimensional group of all holomorphic maps; such transformations
    >>are well-defined at least locally. Such an extension is not possible in
    >>d>2.

    >
    > Now I've seen this claim before that the infinite dimensional extension of
    > the conformal group is not, generically, well-defined globally but how
    > does this work? It sounds like this infinite dimensional group is the
    > automorphisms of C, and so it only applies to charts diffeomorphic to C.
    > Is that right or am I all washed up?
    >


    Hi Creighton-

    I am sure Lubos will be able to answer this better than I, but lemme say
    some stuff too anyway.

    A map from C to C that may be written as a local holomorphic map can be
    Laurent expanded around z=0, so that f=Sum a_n*z^{n+1}. Such a function
    will satisfy the Caucy-Riemann equations where it is defined, and so be
    a conformal map. The space of these local holomorphic maps is infinite
    dimensional. But these maps may blow up at z=a. If we restrict to
    n>-1, then our maps are only well defined on the cylinder C-{0}. If we
    also want our function to behave at infinity, then we can substitute
    w=-1/z, and find that we also must require n<1.

    Thus, the conformal maps that are globally defined are generated by a
    finite dimensional subset of the full conformal group. If we allow our
    conformal maps to have poles at z=0 and infinty, then we regain the full
    infinite dimensional group.

    ---
    [moderator note: I would probably talk about n>=0 and n<=2 in the Laurent
    expansion for the infinitesimal variations of "z" as a function of the
    same "z", but otherwise morally I would agree with the framework of your
    explanation. Cheers, lm]
     
  8. Mar 29, 2004 #7

    Zig

    User Avatar

    > On Sun, 28 Mar 2004, Zig wrote:
    >
    >>In computing the partition function of free bosonic field theory in 1+1
    >>dimensions on the cylinder, we identify the in and out states in the
    >>theory, so that our partition function is really calculated on a torus.


    Lubos Motl wrote:

    > Great, but I hope there was no specific question here - at least I've
    > found no question mark(s). ;-)


    Zig:

    Ooops!! Right, a question! OK, the part that I have been scratching my
    head over, is why are we allowed identify the in/out states? The prof
    said something about the fact that the partition function involves a
    trace, which doesn't distinguish the states, I think, although I was a
    little lost. I just didn't get it.

    thanks again

    ----

    Lubos Motl tries to answer:

    There is only one Hilbert space to work with, and the partition sum is the
    trace of the operator exp(-beta.H) that acts on this single Hilbert space.
    The operator exp(-beta.H) may be understood as the evolution operator thru
    an imaginary time i.beta, and therefore is represented by a path integral
    defined on a cylinder that is made of the circle of the original closed
    string, times a Euclidean time interval. Yes, tracing means that you
    impose periodic boundary conditions (and antiperiodic boundary conditions
    for all fermionic fields).

    I don't know precisely why you want to start with two Hilbert spaces. In
    my approach I never had two distinct Hilbert spaces, and your problem is
    therefore unknown to me. ;-) Even when we consider scattering - in which
    it makes sense to distinguish "in" and "out" states - these two Hilbert
    spaces are isomorphic, and an "out" state |psi> is nothing else than the
    action of the S-matrix (the unitary evolution over an infinite time
    interval) on the "in" state |psi>.
    ______________________________________________________________________________
    E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web: http://lumo.matfyz.cz/
    eFax: +1-801/454-1858 work: +1-617/496-8199 home: +1-617/868-4487 (call)
    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
     
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