Re: the symmetric lower indices of Christoffel symbol

Excuse me, allow me reply an article to check if my concepts are
correct.

carlip-nospam@xxxxxxxxxxxxxxxxxxx 寫道：

mathlover <CPPandC@xxxxxxxxx> wrote:
Dear all,
I'm self-studying general relativity and I read from the book that
\Gamma^i_{jk} = \Gamma^i_{kj} where the indices are from 0 to 3.

But I read from the Wikipedia that if the space had torsion, then
the symmetry of the Christoffel symbol may be broken.
http://en.wikipedia.org/wiki/Christoffel_symbols

My problem is, I can't figure out where the proof is wrong in my GR
book(Schutz) if the torsion were added in the space.
I present the arguments of the book below:

In Cartesian coordinates, we have:
\Phi_{,\beta, \alpha} = \Phi_{,\alpha, \beta} since the partial
derivatives commute.

But if a tensor is symmetric in one basis it is symmetric in all bases.
Therefore:
\Phi_{,\beta ;\alpha} = \Phi_{,\alpha ;\beta} (where the lower index
preceded with a comma means covariant derivative.) in any basis.

This assumes that Cartesian coordinates exist, and that the covariant
derivative is equal to the partial derivative in those coordinates.
If spacetime is not flat, the first is not true; if the torsion tensor
is not zero, the second is not true even in a locally freely falling
frame.

The covariant derivative of a vector field is a type(1,1) tensor. A
tensor is a geometric object which is independent of coordinates. Thus,
a tensorial equation holds in every coordinates.

So, Schutz(The book) finds the equation \Phi_{,\beta, \alpha} =
\Phi_{,\alpha, \beta} in a locally inertial frame (it exists in torsion
free GR), then he can extend the equation to arbitrary coordinates, so
he has: \Phi_{,\beta ;\alpha} = \Phi_{,\alpha ;\beta}, then the
symmetry on the two lower indices is established.

But the key point is that the locally inertial frames do not exist with
torsion. So, Schutz(the book) can't write down the first equation, thus
the symmetry on the two lower indices is broken. Am I right?

(Nonzero torsion violates the principle of equivalence, in the sense that
even by going to a freely falling frame one cannot transform the connection
away completely -- the torsion tensor is an honest tensor, and is nonzero
in any frame if it is nonzero in one. But for many reasonable proposals
for coupling torsion to matter, the violation is too small to conflict
with observation.)

Thanks for your clear demonstration of the concepts of torsion. I
learned much. Thanks you!

Steve Carlip
Sincerely

.

Relevant Pages

  • Re: Acceleration Via Kalb-Ramond Field, Torsion, Raychaudhuri, PI
    ... Continuing with Sengupta and Sinha, the presence of torsion ... destroys the cyclic property of the Riemann-Christoffel tensor so the ... violation in the Lagrangian for pure gravity, and Kar et al (2001, 2002 ... covariant derivatives to get parity-violating effects in the Lagrangian ...
    (sci.physics)
  • Re: the symmetric lower indices of Christoffel symbol
    ... derivatives commute. ... the first is not true; if the torsion tensor ... even by going to a freely falling frame one cannot transform the connection ... away completely -- the torsion tensor is an honest tensor, and is nonzero ...
    (sci.physics)
  • Vanishing of tensor product
    ... Let M denote the module Q/Z over the ring Z. Then M tensored with M ... over Z is 0 since M is both torsion and divisible. ... s.t. M tensor M is not zero but M tensor M tensor M is? ...
    (sci.math)