Laplacian with torsion?

Discussion in 'Physics & Math' started by AlphaNumeric, Feb 22, 2010.

  1. AlphaNumeric Fully ionized Registered Senior Member

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    Does anyone know what the definition is of the Laplacian in spaces with torsion? In torsionless spaces there's a number of ways of writing the usual Laplacian; \(\Delta = \nabla^{2}\) or \(\Delta = dd^{\dag}+d^{\dag}d\). My experience with spaces with torsion is such that when torsion is introduced you have \(\nabla \to \hat{\nabla} = \nabla + T\) where T is some torsion dependent quantity. Equivalently the exterior derivative goes to \(d \to d + T'\). So I get the impression you can get the torsion dependent Laplacian by just putting those into the usual definition but I can't think of an immediate justification and it wouldn't surprise me in the slightest if that's too simplistic an assumption.

    Anyone know the answer or can think of a proof or knows a book which might contain such information? Torsion isn't something which standard textbooks delve into much. And if it makes any difference the usual stuff like differentiability (obviously!), smoothness, compactness, connected, Riemannian etc all apply.
     
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  3. prometheus viva voce! Registered Senior Member

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    The problem with introducing torsion into GR is that the usual Levi Civita connection can't be used because it is torsion free by definition, and hence you can't simply plug in \(\nabla^2 = \nabla_a \nabla^a\) with your torsion shifted covariant derivatives, because the definition of a covariant derivative is no longer a good one.

    I have a feeling that the way to do this is to use vielbeins, though I may be getting this mixed up with something else. There's a guy in my department that knows a lot about this type of thing. I'll see what he has to say about it.
     
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  5. prometheus viva voce! Registered Senior Member

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    Ok, what I said was almost right. There are two ways to think about this: You can use the ordinary Levi Civita connection with a torsion tensor term, ie \(\nabla_\mu X^\nu = \partial_\mu X^\nu + \Gamma^\nu_{\mu\kappa} X^\kappa + T^\nu_{\mu\kappa} X^\kappa\). \(T^\nu_{\mu\kappa}\) is antisymmetric in it's lower indices, in contrast to the Christoffel symbols which are symmetric. Otherwise you can use another connection (that won't be symmetric in the lower indicies) that absorbs the torsion tensor into it \(\nabla_\mu X^\nu = \partial_\mu X^\nu + \Gamma^\nu_{\mu\kappa} X^\kappa\). Really, torsion is like adding an antisymmetric part (in the lower indices) to the the Christoffel symbols.

    Once you know what your covariant derivative is in terms of whatever connection you like you can simply plug into \(\nabla^2 = \nabla_\mu \nabla^\mu\) to work out the Laplacian. (Just to check on a semantic point, if I was to do this in flat space I'd get the D'Alembertian, otherwise known as the box operator - is that what you want?)

    Apparently, the best way to learn this stuff is by learning about connections. Some texts that were recommended to me were Nakahara, Stewart and Ortin's book on gravity and strings. The guy I'm talking to (via email) is seriously into the mathematics of geometry, which I find to be more like mental masturbation sometimes that physics. You'll probably find it a lot easier than me.

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  7. AlphaNumeric Fully ionized Registered Senior Member

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    I've got those three (actually, which Stewart, there's a few) and find Nakahara very nice. As for the mathematics of geometry and more maths than physics, I can't really say much given that's pretty much my thesis topic....

    I know that you can absorb the torsion into the Gamma connection, as by definition you have \(\Gamma^{a}_{bc} = \big\{ b^{a}_{\phantom{a}c} \big\} + T^{a}_{bc}\) which is a decomposition into symmetric and antisymmetric terms and we normally just use T=0 and don't bother with the funny { } notation.

    Unfortunately the derivatives I'm working with are most easily written using exterior derivatives, rather than covariant ones, so I'm not sure if the same applied.

    As for your question about Box, the box operator includes a time second derivative which in Lorentzian space has a different sign compared to the spacial terms, ie \(\Box = \partial_{\mu}\partial^{\mu} = -\partial_{t}^{2} + \partial_{i}\partial_{i}\). The \( \partial_{i}\partial_{i}\) is a Laplacian and you can split it down further so if i goes from 1 to 9 and you want a 6 dimensional compact space you might have the 10 dimensional box operator go to \(\Box_{10} = \Box_{4} + \Delta\) where \(\Delta\) is the Laplacian on the compact space. This is basically what I'm doing, since the Klein Gordon equation provides you with mass expressions and if you go from 10 dimensions to a 4 dimensional effective theory then eigenvalues of \(\Delta\) on the compact space contribute to the mass of your fields in your effective theory. Hence you want its zero eigenforms.
     
  8. prometheus viva voce! Registered Senior Member

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    The Stewart book is his advanced GR one. Can you think of a reason why doing the naive and simple thing with the exterior derivatives wouldn't work? - it does work for the covariant derivatives so I see no reason why it shouldn't work for the exterior derivatives.
     
  9. AlphaNumeric Fully ionized Registered Senior Member

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    Ah, I thought you were referring to that book.

    I can't think of a good reason why it won't work but, as I've discovered, that's often not a good reason to do something. I'm asking because I did precisely that in my thesis where an exterior derivative extends to one with torsion and then to a basis which isn't closed such that \(d \to D\) and I make the claim that the Laplacian goes from \(dd^{\dag}+d^{\dag}d \to DD^{\dag}+D^{\dag}D\). My examiner said he believes its the right expression but thinks I should justify it somehow. Given torsion is something well studied I thought I'd start with just the \(d \to d + T\) first.

    And this brings me to another question you might be able to help with (or someone else in your department). In the massless Type II field content you have B, which defines a field strength H (and its this H which is the torsion as it happens). Usually you just have a term ni the action of the form \((H,H) = H \wedge \ast H\) with \(H=dB\) but this is the non-compact version. When you put your space-time onto a compact space and have non-zero H you get a torsion such that d -> d + H (or d-H depending on your preference). I get the distinct impression this should alter the equations of motion, since non-zero torsion obviously would and in this case it just happens to be one of the stringy fields. I can't decide/determine/work out/find out if it does it by altering the equation of motion only or by altering the action such that \((H,H) = (dB,dB) \to ( (d+H)B,(d+H)B )\) where H acts on B via exterior multiplication. Changing all d's to d+H is in line with the thing you said about the Laplacian just changing derivatives but I feel uneasy about doing it on the level of the action itself, particularly the fields from the ten dimensional action since that's not really your effective theory.

    I fear I might not have explained that well enough....
     
  10. prometheus viva voce! Registered Senior Member

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    I've been trying to think of something intelligent to say about this AN, but I'm afraid I'm coming up short - it's not something I'm tremendously familiar with unfortunately. The only thing that does spring to mind is that you have to be a little bit careful with gauge potentials on compact spaces because you can't have an isolated source (or sink) by Gauss' law. That's something I'm sure you know already though - I'll have a chat with some geometry guys in the dept when I'm next in (Thursday).
     
  11. temur man of no words Registered Senior Member

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    An approach would be to take a covariant (exterior) derivative (such that the covariant derivative of the metric vanishes) as the starting point and then define the torsion to be its deviation from the covariant derivative associated to the Levi-Civita connection. Then show that the so-defined quantity has all the properties of torsion.
     
  12. AlphaNumeric Fully ionized Registered Senior Member

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    I'm not trying to prove it is torsion, that's a known result. I'm trying to see how non-zero torsion affects the Laplacian. I think I've got somewhere, so I'll run it by you.

    Let D = d+T, where T is a 3-form satisfying \(dT = d^{\dag}T = 0\) where d the usual ex. derivative. I want to consider \(\Delta = DD^{\dag} + D^{\dag}D = (D+D^{\dag})^{2}\) applied to a 0-form, \(1\), T and a 6-form \(\mu_{vol} = \ast 1\). The adjoint \(D^{\dag}\) is such that \((D\phi,\psi) = (\phi,D^{\dag}\psi)\) for the usual \((a,b) = \int a \wedge \ast b\) inner product. This means that \(T^{\dag} : \Omega^{(n)}(M) \to \Omega^{(n-3)}(M)\) with action akin to the interior product \(\iota\). Both derivatives are nilpotent.

    1st case is the 0 form, which is harmonic in d so \((D+D^{\dag})1 = T \wedge 1 + T^{\dag} \cdot 1 = T\), since \(T^{\dag} \cdot 1 = 0\) else it would be a -3 form. Therefore \((1,\Delta 1) = (1,T^{\dag}\cdot T) = (T,T)\)

    2nd case is T itself, which is also harmonic. Droppign all \(T \wedge T\) and \(T^{\dag} \cdot T^{\dag}\) etc gives \((T,\Delta T) = (T^{\dag} \cdot T,T^{\dag} \cdot T)\). Obviously \(T^{\dag}\cdot T\) is a 0-form and for purposes which aren't relevant this 0-form must be 1, so consider \((1,T^{\dag} \cdot T) = (T,T)\). So \(T^{\dag} \cdot T = (T,T) \, 1\) and therefore \((T,\Delta T) = (T^{\dag} \cdot T,T^{\dag} \cdot T) = \Big((T,T),(T,T) \Big) = (T,T)^{2}\).

    3rd case is \(\ast 1\). Using \(T \wedge \ast 1 = 0\) we have \((D+D^{\dag})\ast1 = T^{\dag} \cdot (\ast 1)\) and so \(\Delta \ast 1 = T \wedge (T^{\dag} \cdot (\ast 1))\). This is a 6-form and so is proportional to \(\ast 1\). Therefore we consider \((\ast 1,\Delta \ast 1) = (\ast 1, T \wedge (T^{\dag} \cdot (\ast 1)) = (T^{\dag} \cdot (\ast 1),T^{\dag} \cdot (\ast 1))\). \(T^{\dag} \cdot (\ast 1)\) is a 3-form and orthogonal to T, ie \((T,T^{\dag} \cdot (\ast 1)) = (T \wedge T, \ast 1) = 0\).

    Using the integral def. for the inner product and \(\ast^{2}1 = 1\) you have \((\ast 1, T \wedge (T^{\dag} \cdot (\ast 1)) = ( T \wedge (T^{\dag} \cdot (\ast 1),\ast 1) = \int T \wedge (T^{\dag} \cdot (\ast 1) \wedge 1 = \int T \wedge (T^{\dag} \cdot (\ast 1)) = - \int (T^{\dag} \cdot (\ast 1)) \wedge T = \int (T^{\dag} \cdot (\ast 1)) \wedge \ast (\ast T) = ( T^{\dag} \cdot (\ast 1),\ast T) = ( \ast 1,T \wedge \ast T) = (T,T)\)

    Therefore we have \((\varphi,\Delta \varphi) = (T,T)\) for \(\varphi = 1,\ast1\) and \((T,\Delta T) = (T,T)^{2}\). And these can be written in a shorter manner via \((\varphi,\Delta \varphi) = (T,T)(\varphi,\varphi)\) so torsion induces the same normalised eigenvalues in each setup.

    Anyone spot any mistakes in that? I make a few assumptions which I have stated and which are motivated by the thing I'm doing all the algebra for but other than those I think its pretty solid. I'm open to being corrected on that though.
     
    Last edited: Feb 24, 2010

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