Short question about Laplacians

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The discussion centers on Nakahara's treatment of the Laplacian in the context of differential geometry, specifically in relation to equation (7.188) on page 294. The user seeks clarification on the appearance of the factor \( g^{-1} \), which represents the determinant of the contravariant metric, in the calculation of the Laplacian \( \Delta = dd^{\dagger} + d^{\dagger}d \) for a scalar function \( f \). The user successfully derives the expression for \( *d*df \) but is unable to reconcile the determinant factor in their calculations. Additionally, the user poses a question regarding the pullback of a vertical vector field \( X \) in the context of a principal fiber bundle, specifically why it results in zero.

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  • Understanding of differential geometry concepts, particularly Laplacians.
  • Familiarity with Nakahara's "Geometry, Topology and Physics".
  • Knowledge of tensor calculus and the properties of the epsilon tensor.
  • Basic principles of fiber bundles and vertical vector fields.
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  • Review Nakahara's treatment of Laplacians in "Geometry, Topology and Physics", focusing on sections related to determinants of metrics.
  • Study the properties of the epsilon tensor and its role in differential forms.
  • Explore the concept of pullbacks in the context of fiber bundles and vertical vector fields.
  • Investigate the relationship between contravariant metrics and determinants in differential geometry.
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haushofer
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Hi, I have a short question about Nakahara's treatment about Laplacian's: page 294, section 7.9.5, equation (7.188).

He calculates the Laplacion \Delta = dd^{\dagger} + d^{\dagger}d for a scalar function f. Every step is clear to me, except one; at the fourth line there is a factor of g^{-1} popping up ( the determinant of the contravariant metric )

What I get is ( ignoring the minus-sign in front )

<br /> <br /> *d*df = * \frac{1}{(m-1)!} \partial_{\nu} [\sqrt{g}g^{\lambda\mu}\partial_{\mu}f]\epsilon_{\lambda\nu_{2}\cdots\nu_{m}} dx^{\nu}\wedge dx^{\nu_{2}}\wedge\ldots\wedge dx^{\nu_{m}}<br /> <br />

just like Nakahara. Now I use

<br /> <br /> dx^{\nu}\wedge dx^{\nu_{2}}\wedge \ldots \wedge dx^{\nu_{m}} = \epsilon^{\nu\nu_{2}\ldots\nu_{m}} dx^{1}\wedge \ldots \wedge dx^{m}<br /> <br />

and the contraction

<br /> \epsilon_{\lambda\nu_{2}\ldots\nu_{m}} \epsilon^{\nu\nu_{2}\ldots\nu_{m}} = (m-1)!\delta_{\lambda}^{\nu}<br />

and simply fill this in. I get the same answer as is at line four of equation (7.188), except for that g^{-1}[/itex]. So I&#039;m missing that determinant somewhere, but where?<br /> <br /> Many thanks in forward, my vision is a little blurred at the moment :)
 
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I have the feeling that I should define

<br /> <br /> dx^{\nu}\wedge dx^{\nu_{2}}\wedge \ldots \wedge dx^{\nu_{m}} = \epsilon_{\nu\nu_{2}\ldots\nu_{m}} dx^{1}\wedge \ldots \wedge dx^{m}<br /> <br />

and that bringing those indices up on that epsilon tensor gives me that g^{-1}, but then the indices don't match.
 
I have another question, but it's rather short so I can kick this topic. If I have a principle fibre bundle P with base manifold M and fibre G and an associated vertical vector field X and projection pi: P to M, why exactly is the pullback of pi acting on X zero? Most of the Nakahara exercises are quite easy for me, but this one is troubling me for a day now. If I just apply the definition of a pullback to the vectical vector field, I don't see why it should be zero given the definition of this vertical vector field.
 

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