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Exercise in Weyl rescaling.

+ 2 like - 0 dislike
126 views

I have been trying to learn from "Geometry, Topology and Physics" by Mikio Nakahara and till now I've been able to more or less follow most of the selected sections I've read, but now I've come across a proof that I cannot follow in any way. I will try to make this question self-contained, but for those who own the book: it is proposition 7.1 on page 275.

The book defines Weyl rescaling as:

$$ g_p \to \bar{g}_p = e^{2\sigma(p)} g_p$$

where $g$ is a metric on a (pseudo-)-Riemannian manifold $M$, $p$ is a point on the manifold $M$, and $\sigma \in \mathscr{F}(M)$. Now, the book defines $K$ to be the difference of the covariant derivatives $\bar{\nabla}$ with respect to $\bar{g}$ and $\nabla$ with respect to $g$:

$$ K(X,Y) \equiv \bar{\nabla}_X Y - \nabla_X Y$$

where $X$ and $Y$ are vector fields. Furthermore, let $U$ be a vector field which corresponds to the one-form $\mathrm{d} \sigma$:


$$ Z[\sigma] = \langle \mathrm{d} \sigma , Z \rangle = g(U,Z) $$

Then the proposition is that:

$$ K(X,Y) = X[\sigma]Y + Y[\sigma] X - g(X,Y)U$$.

For the proof:

First the book mentions that because we are considering torsion-free condition that $K(X,Y)=K(Y,X)$. Furthermore, since we are only considering metric connections, i.e. $\bar{\nabla} \bar{g} = \nabla_X g = 0$, we have:

$$X[\bar{g}(Y,Z)] = \bar{\nabla}_X [\bar{g}(Y,Z)] = \bar{g}(\bar{\nabla}_XY,Z)  +\bar{g}(Y, \bar{\nabla}_X Z) $$

This doesn't make any sense to me. The first equal sign is a complete mystery to me. The second step is also a mystery, because I would think that we would have to get:

$$\bar{\nabla}_X [\bar{g}(Y,Z)] = X^\kappa [(\nabla_\kappa \bar{g}(Y,Z) + \bar{g}(\nabla_\kappa Y,Z) + \bar{g}(Y, \nabla_\kappa Z)] $$

which is not the same.

I have tried everything, but I'm just missing something (perhaps it is really trivial). I also searched Google but couldn't find anything that helps me. According to the book this proof follows Nomizu (1981), but I don't have access to that book. Any help is much appreciated.

asked Apr 17, 2014 in Mathematics by Hunter (510 points) [ revision history ]
edited Apr 17, 2014 by Valter Moretti

This question is trivial in standard index notation, you should do the conversion for all math-style expressions, they are needlessly obfuscated.

1 Answer

+ 4 like - 0 dislike

I cannot see the problem with the penultimate line of your formulas:

$$X[\bar{g}(Y,Z)] = \bar{\nabla}_X [\bar{g}(Y,Z)] = \bar{g}(\bar{\nabla}_XY,Z)  +\bar{g}(Y, \bar{\nabla}_X Z) $$

The first identity is true because for every affine connection $D$ one has $D_Xf = X(f)$ where $f$ is a scalar field and $X$ is a vector field. In our case $D = \overline{\nabla}$ and $f= \bar{g}(Y,Z)$. Concering the second identity, the connection $\overline{\nabla}$ is metric (because it is the Levi-Civita connection of $\overline{g}$) and metric, indeed, means  $\overline{\nabla}_X \left(\overline{g}(Y,Z)\right)=  \overline{g}(\overline{\nabla}_XY,Z) + \overline{g}(Y,\overline{\nabla}_XZ)$ and this requirement is equivalent to $\overline{\nabla}_X \overline{g}=0$.

answered Apr 17, 2014 by Valter Moretti (2,025 points) [ revision history ]
edited Apr 17, 2014 by Valter Moretti

Thanks! I feel quite silly now, but for some unexplicable reason I was really stuck. I think in my head I was making it a lot more complicated than it really was.

The reason you were stuck is because the math notation is too conceptual, just write it out with indices, and you will see that is no statement of content beyond that the covariant derivative of the metric is zero.

Thanks, I will keep that in mind!

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