Short question about Laplacians

[haushofer]

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 )



*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}}



just like Nakahara. Now I use



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}



and the contraction


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


and simply fill this in. I get the same answer as is at line four of equation (7.188), except for that [tex]g^{-1}[/itex]. So I'm missing that determinant somewhere, but where?

Many thanks in forward, my vision is a little blurred at the moment :)

[haushofer]

I have the feeling that I should define



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}



and that bringing those indices up on that epsilon tensor gives me that g^{-1}, but then the indices don't match.

[haushofer]

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.