Is the metric still symmetric in Einstein-Cartan theory?

On Aug 10, 2:54 pm, Eric Gisse wrote:
The title says it all, really. I had initially interpreted the
antisymmetry in the connection coefficients to imply that the metric
is /not/ symmetric. However, I realized that all the antisymmetry
could be effectively stuffed into the co-torsion tensor [the part of
the connection which related to torsion] which would allow the metric
to remain symmetric but allow the requisite parts to remain a/anti-
symmetric.


It is ONLY a definition that usually the Christoffel symbols of the
second
kind (your mentioned connection coefficients) are symmetric in the
Gauss
equations. If you use the torsion tensor which defined as

T_kl^p = C_kl^p - C_lk^p, (C_kl^p = Christoffel symbols of second
kind),

you get something more general that the Gauss equations and the
Weingarten equations and the compatibility equations
which all are basics in the theory of surfaces in R^3 (these
are generalized to manifolds (= "many fold surfaces") in R^n).
In usual surfaces this generalization is not needed due torsion
tensor is thenm zero.

Hannu


I'm trying to learn the Einstein-Cartan theory of gravitation because
I have found the idea that GR /can't/ handle spin-orbit coupling to be
rather interesting. However, the resources I have found are good for
explaining the big and small picture, but leave out some important
details. Like if the metric remains symmetric...

My instinct is that it must remain symmetric because if I zero out the
spin tensor, the co-torsion tensor goes away and I am left with
classical GR. This is all motivated by trying to obtain a spherically
symmetric vacuum solution, which left me thinking about what kind of
metric anzatz I am allowed to use...