Torsion -> covariant deriv of det(g) non-zero?

[pellman]

torsion --> covariant deriv of det(g) non-zero?

I am missing something here. This paper makes the case (on page 5) that for non-vanishing torsion, the usual invariant volume element \sqrt{-g}d^4x is not appropriate because the covariant derivative of sqrt(-g) is non-zero. This perplexes me. The determinant of g is simply a function of the individual components of g, and their covariant derivatives vanish by assumption. So how can the derivative of the determinant be non-zero?

Here is a summary of the math. I have checked all this and my results match those of the linked paper (though I use a different notation). So maybe I am wrong in expecting the covariant derivative of g to vanish?

(1) If we assume a metric connection with \nabla_\gamma g_{\alpha\beta}=0, then it follows that the connection coefficients take the form

{\Gamma^\gamma}_{\alpha\beta}=\frac{1}{2}g^{\gamma\lambda}\left(\partial_\alpha g_{\lambda\beta} +\partial_\beta g_{\alpha\lambda}-\partial_\lambda g_{ \beta\alpha }\right)+\frac{1}{2} \left( {{T_\alpha}^\gamma}_\beta + {{T_\beta }^\gamma}_ \alpha + {T^\gamma}_{\alpha\beta}\right)

where {T^\gamma}_{\alpha\beta}\equiv 2{\Gamma^\gamma}_{[\alpha\beta]} is the torsion tensor.

(2) The square root of the determinant of the metric is a weight 1 scalar density. As such its covariant derivative is \nabla_\gamma \sqrt{-g}=\partial_\gamma\sqrt{-g}-{\Gamma^\rho}_{\rho\gamma}\sqrt{-g}
(3) The result of this has a non-zero term dependent on the torsion:

\partial_\gamma\sqrt{-g}-{\Gamma^\rho}_{\rho\gamma}\sqrt{-g}=-{T^\rho}_{\rho\gamma}\sqrt{-g}\neq 0

 

[Mentz114]

My interpretation is that in (2.11) he's using the covariant derivative from V₄, but in (2.12) it is from U₄.

 

[pellman]

My interpretation is that in (2.11) he's using the covariant derivative from V₄, but in (2.12) it is from U₄.

That's correct. V₄ is torsion-free spacetime, U₄ is a spacetime with torsion.

 

[pellman]

Let's never mind the torsion details for the moment. Is it not the case that

\nabla_\gamma g_{\alpha\beta}=0

implies

\nabla_\gamma \sqrt{-g}=0

?

g is equal to a sum of terms of the form g_{0\alpha}g_{1\beta}g_{2\gamma}g_{3\delta} and so \nabla_\gamma g is a sum of terms of the form

(\nabla_\gamma g_{0\alpha})g_{1\beta}g_{2\gamma}g_{3\delta}+ (\nabla_\gamma g_{1\beta}) g_{0\alpha}g_{2\gamma}g_{3\delta}+(\nabla_\gamma g_{2\gamma})g_{0\alpha}g_{1\beta} g_{3\delta}+(\nabla_\gamma g_{3\delta})g_{0\alpha}g_{1\beta}g_{2\gamma}

which all vanish because \nabla_\gamma g_{\alpha\beta}=0 for all alpha and beta. Right?

And does the derivative of the square root behave as ordinary derivatives? That is

\nabla_\gamma \sqrt{-g}=\frac{-1}{2\sqrt{-g}}\nabla_\gamma g

 

[pellman]

Ok, never mind guys. I am convinced now that you just can't treat the covariant derivative like a regular derivative, with the usual chain rule. What matters really matters is tensor rank and weight of the object it is acting on.

The non-vanishing of the derivative of sqrt(-g) in the presence of torsion is correct.

 

[Scifoo]

I believe the statement in the article is incorrect, for a metric-compatible connection always commutes with the volume form. When covariantly differentiating tensor densities, you must compensate with a contracted \Gamma. Note that it matters which indices you contract on \Gamma, since it is no longer symmetric. In your expression for \Gamma,
if you contract \gamma with \alpha you get torsion terms, but if you contract \gamma with \beta, you get no torsion terms.
This is done correctly in Hehl's book (page 199).
best regards

 

[pellman]

Thanks for the reply, Scifoo, but the math in the opening post is correct. Based on the chapter on Riemannian geometry from Nakahara's Geometry, Topology and Physics. and I have checked it thoroughly. what is the title of Hehl's book.

 

[Ben Niehoff]

Scifoo is correct. If the connection is metric-compatible, then the volume form is covariantly constant. This should be clear because "metric-compatible" means that parallel propagation does not change the lengths of vectors or the angles between them; naturally it follows that it also preserves n-volumes of n vectors.

{\Gamma^\gamma}_{\alpha\beta}=\frac{1}{2}g^{\gamma\lambda}\left(\partial_\alpha g_{\lambda\beta} +\partial_\beta g_{\alpha\lambda}-\partial_\lambda g_{ \beta\alpha }\right)+\frac{1}{2} \left( {{T_\alpha}^\gamma}_\beta + {{T_\beta }^\gamma}_ \alpha + {T^\gamma}_{\alpha\beta}\right)

This equation is incorrect. The second-to-last term should have a minus sign. See here:

http://en.wikipedia.org/wiki/Contorsion_tensor

 

[Scifoo]

Hehl's book: Foundations of classical electrodynamics: charge, flux, and metric
you can find it on www.bookfi.org

On what page of Nakahara's book does he consider covariant differentiation of tensor densities?

Pellman's equation is correct (he uses a different convention for the placement of indices), so wikipedia's entry would look likeK_kij=1/2 ( T_kij + T_ijk + T_jik)

 

[Ben Niehoff]

Pellman's equation cannot possibly be correct, regardless of conventions, because he has the torsion contributing to the symmetric part of the connection. The alpha and beta indices must be antisymmetric in the T piece, so depending on conventions, he needs either one minus sign or two.

 

[pellman]

On what page of Nakahara's book does he consider covariant differentiation of tensor densities?

Sorry. He doesn't actually cover that result. I just used the relevant formulae and discussion from his book to derive the result myself.

"metric compatible" means that the connection satisfies \nabla_\gamma g_{\alpha\beta}=0 which is equivalent to

\partial_\gamma g_{\alpha\beta}={\Gamma^\rho}_{\gamma\alpha}g_{ \rho \beta}+{\Gamma^\rho}_{\gamma\beta}g_{\alpha\rho}

Since \sqrt{-g} is a weight 1 tensor density we have

\nabla_\gamma\sqrt{-g}=\partial_\gamma\sqrt{-g}-\sqrt{-g}{\Gamma^\rho}_{\rho\gamma}

and

\partial_\gamma\sqrt{-g}=-\frac{1}{2\sqrt{-g}}\partial_\gamma g=-\frac{1}{2\sqrt{-g}}(-g)g^{\alpha\beta}\partial_\gamma g_{\alpha\beta}
=\frac{1}{2}\sqrt{-g}g^{\alpha\beta}({\Gamma^\rho}_{\gamma\alpha}g_{ \rho \beta}+{\Gamma^\rho}_{\gamma\beta}g_{\alpha\rho})
=\sqrt{-g}{\Gamma^\rho}_{\gamma\rho}

So

\nabla_\gamma\sqrt{-g}=\sqrt{-g}{\Gamma^\rho}_{\gamma\rho}-\sqrt{-g}{\Gamma^\rho}_{\rho\gamma}
=-\sqrt{-g}{T^\rho}_{\rho\gamma}

which doesn't vanish for general non-zero torsion.

 

[pellman]

Pellman's equation cannot possibly be correct, regardless of conventions, because he has the torsion contributing to the symmetric part of the connection. The alpha and beta indices must be antisymmetric in the T piece, so depending on conventions, he needs either one minus sign or two.

Torsion does contribute to the symmetric part of the connection coefficient. See eq 7.32 pg 254 here http://books.google.com/books?id=cH...A&ved=0CDYQ6AEwAA#v=onepage&q=torsion&f=false

 

[Scifoo]

Actually, he has contortion contributing to the symmetric part, which is ok (the expression given in wikipedia is not ant-symmetric in the first two indices).

 

[Scifoo]

Pellman: I believe the equation for the covariant derivative of \sqrt(g) should be instead:


\[
\nabla_{\mu}\sqrt{g}=\partial_{\mu}\sqrt{g}-\Gamma_{\mu\rho}{}^{\rho}\sqrt{g}
\]

PS: I am new here, how do I display latex formulas?

 

[DrGreg]

Pellman: I believe the equation for the covariant derivative of \sqrt{g} should be instead:<br /> \nabla_{\mu}\sqrt{g}=\partial_{\mu}\sqrt{g}-\Gamma_{\mu\rho}{}^{\rho}\sqrt{g}<br />PS: I am new here, how do I display latex formulas?

See LaTeX Guide: Include mathematical symbols and equations in a post

 

[Scifoo]

Thanks! Let me write that alternatively as

\nabla_{\mu}\sqrt{g}=\partial_{\mu}\sqrt{g}-\Gamma^{\rho}{}_{\mu\rho}\sqrt{g}

 

[Ben Niehoff]

OK, I see that Pellman's formula is correct. But he should check his conventions as to whether the covariant derivative is defined this way

\nabla_a X^b = \partial_a X^b + \Gamma^b_{ac} X^c
or this way

\nabla_a X^b = \partial_a X^b + \Gamma^b_{ca} X^c
because now it makes a difference. Whichever convention is being used, it is that convention under which he had better have

\nabla_c g_{ab} = 0
and this will imply

\nabla_c \sqrt{g} = 0.

 

[Scifoo]

So what is the answer to the first post? The referred article is incorrect? I'd say eq.2.12 is a blatant mistake.

 

[Ben Niehoff]

So what is the answer to the first post? The referred article is incorrect? I'd say eq.2.12 is a blatant mistake.

In the linked paper, equations 2.4 and 2.5 are inconsistent with each other, due to precisely the issue I just mentioned: when the connection is not symmetric, you have to stick to a consistent convention as to which index gets contracted with what.

It looks like this one inconsistency is the source of the entire paper.

Moreover, under the quite generic definition of a connection as satisfying the Leibniz rule

\nabla(f \sigma) = df \otimes \sigma + f \, \nabla \sigma,
where f is a function and sigma is any tensor, together with the expression of \nabla in an orthonormal basis,

\nabla e^a = - \omega^a{}_b \otimes e^b,
one can show in just a few lines exactly what is required for the volume form to be parallel:

\begin{align*}<br /> \Omega &amp;= \frac{1}{n!} \, \varepsilon_{a_1 \ldots a_n} \, e^{a_1} \wedge \ldots \wedge e^{a_n} \\<br /> \nabla \Omega &amp;= - \omega^b{}_c \otimes \frac{1}{(n-1)!} \, \varepsilon_{ba_1 \ldots a_{n-1}} \, e^c \wedge e^{a_1} \wedge \ldots \wedge e^{a_{n-1}} \\<br /> &amp;= - \omega^b{}_b \otimes \frac{1}{n!} \, \varepsilon_{a_1 \ldots a_n} \, e^{a_1} \wedge \ldots \wedge e^{a_n} \\<br /> \nabla \Omega &amp;= - \omega^a{}_a \otimes \Omega<br /> \end{align*}
and hence the volume form is parallel iff \omega^a{}_a = 0. The metric compatibility condition is

\omega^a{}_b = - \omega^b{}_a,
which is sufficient to guarantee \nabla \Omega = 0, although it is not necessary. This ought to make sense; requiring the connection to preserve volumes is subsumed under, and less stringent than, requiring it to preserve lengths and angles.

Notice I've made no mention of the torsion. It's irrelevant, as long as it is consistent with the metric-compatibility condition.

As far as the paper in question, it does say some interesting things. One can show that GR coupled to a dilaton and B field is equivalent to a theory with torsion (but the torsion is required to satisfy certain constraints which I don't know specifically).

 

[haushofer]

For isues about differentiation etc. with torsion, a nice reference is Van Proeyen and Freedman Supergravity book :)

 

[pellman]

Thank you for your post, Ben

\begin{align*}<br /> \Omega &amp;= \frac{1}{n!} \, \varepsilon_{a_1 \ldots a_n} \, e^{a_1} \wedge \ldots \wedge e^{a_n} \\<br /> \nabla \Omega &amp;= - \omega^b{}_c \otimes \frac{1}{(n-1)!} \, \varepsilon_{ba_1 \ldots a_{n-1}} \, e^c \wedge e^{a_1} \wedge \ldots \wedge e^{a_{n-1}} \\<br /> &amp;= - \omega^b{}_b \otimes \frac{1}{n!} \, \varepsilon_{a_1 \ldots a_n} \, e^{a_1} \wedge \ldots \wedge e^{a_n} \\<br /> \nabla \Omega &amp;= - \omega^a{}_a \otimes \Omega<br /> \end{align*}

Is \Omega equivalent to \sqrt{|g|}dx^0\wedge dx^1\wedge dx^2\wedge dx^3 with its scalar density factor? I'm having trouble relating your result back to \nabla_\gamma\sqrt{-g}=\sqrt{-g}{\Gamma^\rho}_{\gamma\rho}-\sqrt{-g}{\Gamma^\rho}_{\rho\gamma}

 

[pellman]

ok, guys, I've come back to thinking that the my expression for the derivative a scalar density was wrong, as Ben and others suggested. That is, I'm thinking now that

\nabla_\gamma\sqrt{-g}=\partial_\gamma\sqrt{-g}-\sqrt{-g}{\Gamma^\rho}_{\rho\gamma}

should have been

\nabla_\gamma\sqrt{-g}=\partial_\gamma\sqrt{-g}-\sqrt{-g}{\Gamma^\rho}_{\gamma\rho}

in which case \nabla_\gamma\sqrt{-g} would then vanish (for a metric compatible connection).

I never really verified for myself the tensor density covariant derivative expression. I just looked it up, for instance, in Lovelock and Rund's Tensors, Differential Forms, and Variational Principles. If the above correction is correct, then the mistake in the order of lower indeces may be common. For instance, the Wikipedia article for covariant derivative has for tensor densities {\Gamma^\rho}_{\rho\gamma} rather than {\Gamma^\rho}_{\gamma\rho}. Probably comes from not needing to bother with the order for torsion free spaces.

And this mean's Saa's paper linked in the opening post was utterly wrong?? And not just that paper. Saa had a series of papers hinging on this mistake. Unbelievable!

I am going to check all this stuff very thoroughly once more. But for now, a very large thank you to everyone who responded. This has saved me a lot of unnecessary confusion later.