8
$\begingroup$

In the standard presentation of general relativity, one adopts the Levi-Civita connection and the Christophel symbols; one has $\Gamma^a_{bc} = \Gamma^a_{cb}$ and the torsion tensor is zero.

But of course one could equally adopt some other connection and get a different set of connection coefficients, and then the torsion tensor need not be zero. (Einstein-Cartan and teleparallel approaches do this, for example). So this makes it seem that torsion is not a property of a manifold, it is a property of the way one chooses to relate different tangent spaces to one another.

However it is often stated that such-and-such a manifold "is torsionless", or that spacetime is assumed to be torsionless in general relativity. Also, one can give geometric pictures of manifolds with torsion, such as one where one considers the continuous limit of a crystal with the appropriate type of dislocation. So this makes it seem that the torsion is there "in the manifold", as it were.

Which is right? Both? Neither?

To be precise,

(i) is there a (reasonably sensible and non-pathological) manifold which has to have torsion no matter what connection is adopted? Or is that question meaningless because torsion is all about connection and manifold together?

And (just to check), I think it is the case that as soon as we have a metric then we also have the possibility of adopting the Levi-Civita connection.

(ii) Does that imply that approaches to gravity in which there is torsion must either be abandoning the concept of a spacetime metric, or else they are treating an interaction which could equally well be treated by asserting that spacetime is torsionless and they are proposing a new field which couples to spin and mass in some sort of universal way that can be captured through the use of a suitably defined connection? (When I look at Einstein-Cartan theory, I see something called a metric popping up, so it clearly has not been abandoned altogether.)

Added note. I edited the above in order to highlight the two more precise questions.

$\endgroup$
5
  • $\begingroup$ I don't quite understand the first part of your question (ii), where you suggest the possibility that "approaches to gravity in which there is torsion must ... be abandoning the concept of a spacetime metric." The connection is mathematically independent of the metric - you can certainly have a connection with torsion on a manifold with a metric, and indeed the connection can even be metric-compatible. It just wouldn't be the Levi-Civita connection, which is the unique connection on a pseudo-Riemannian manifold that is both metric-compatible and torsion-free. $\endgroup$
    – tparker
    Jun 6, 2020 at 14:42
  • $\begingroup$ @tparker Yes you are right. My question arose because of the numerous places where you can read of people saying of a manifold that "it has torsion" or "it is torsion-free". It turns out that all such statements are muddled. Similarly, it is often said of spacetime that is is torsion-free according to GR, but not according to other models. But if spacetime is a manifold, then it neither has nor does not have torsion in and of itself, unless the term "manifold" is a shorthand for "manifold plus connection" or something like that. $\endgroup$ Jun 6, 2020 at 14:59
  • $\begingroup$ Yes, I think that (as Slereah said) in practice when physicists talk about "a spacetime" (and sometimes "a manifold"), they really mean a manifold plus a connection, with the connection implicity assumed to be Levi-Civita unless the context indicates otherwise. $\endgroup$
    – tparker
    Jun 6, 2020 at 15:45
  • $\begingroup$ Somewhat relatedly, physicists and mathematicians confusingly use the terms "diffeomorphism" and "isometry" in conflicting ways: what a physicist calls a "diffeomorphism" a mathematician calls an "isometry," and what a physicist calls an "isometry" a mathematician calls an "autoisometry". That's because in proper mathematical usage, a "diffeomorphism" only relates two manifolds' smooth structures and doesn't touch their metric structure. See my question here for more on this. $\endgroup$
    – tparker
    Jun 6, 2020 at 15:56
  • 1
    $\begingroup$ @tparker I read that and found it helpful; thanks for pointing it out. $\endgroup$ Jun 6, 2020 at 16:47

3 Answers 3

8
$\begingroup$

The torsion is indeed defined from the connection, independently of the manifold and metric. By synedoche, people sometimes refer to the structure composed of the manifold, the differential structure, the metric and the connection, $(M, \mathfrak{A}, g, \nabla)$, as "the manifold", even though those are somewhat independent objects.

Given a reasonable manifold, it is always possible to find a torsionless connection on it, since every (metrizable) manifold admits a metric tensor, and every manifold with a metric admits a Levi-Civita connection. As a general rule, the difference between two connections is

\begin{equation} \nabla_a \omega_b = \tilde{\nabla}_a \omega_b - {C^c}_{ab} \omega_c \end{equation}

If we have a connection with a torsion tensor ${T^c}_{ab}$, we can in particular define $C$ to be the torsion tensor, so that any connection with torsion can give rise to a torsionless connection.

Similarly, if you have a torsion-free connection, and you add a tensor field ${C^c}_{ab}$ that is not symmetric in $a$ and $b$ (you can always do this by picking the zero tensor plus a non-zero antisymmetric tensor in a small neighbourhood), then this will give rise to a connection with torsion.

$\endgroup$
3
  • $\begingroup$ Thanks for this; the bulk of it is clear and helpful. I assume the opening phrase has a typo---I'm not sure what you intended there. As for a manifold that "has to" have torsion, I still am not quite sure if one can define such a thing, but it is clear that it would not have a metric. $\endgroup$ Apr 5, 2020 at 11:25
  • $\begingroup$ Any manifold that has a connection with torsion also has a connection without torsion, that was part of what the post was about. If $\nabla$ has torsion $T$, then $\nabla - T$, also a connection, has no torsion. This is independent of the manifold having a metric. $\endgroup$
    – Slereah
    Apr 5, 2020 at 11:26
  • $\begingroup$ OK; got it. Thanks! $\endgroup$ Apr 5, 2020 at 11:28
1
$\begingroup$

Formally what you do is the following. Spacetime is a set of data $(M,\mathcal{O},\mathscr{A},\nabla,g)$ where $(M,\mathcal{O},\mathscr{A},g)$ is a smooth Lorentzian manifold and $\nabla$ is a connection. Let $G$ be some Lie group, the idea is to consider a principal $G$-bundle $(P,\pi,M)$ with associated fibre bundle $P_V$ with $V$ the representation space of $G$. The idea is then to look at so called solder forms, which are elements $\theta \in \Omega^1(P) \otimes C^\infty(P,V)$ under certain conditions. A torsion is defined as $\Theta = D\theta$ where $(D\phi)(X_1,...,X_{k+1}):= (d\phi)(\mathrm{Hor}(X_1),...,\mathrm{Hor}(X_{k+1}))$ for $X_1,...,X_{k+1} \in \Gamma(TP)$ ($\mathrm{Hor}$ is the horizontal part of the vector(field)) where $\phi \in \Omega^k(P,V)$. In order words you choose some underlying Lie group, consider spacetime as the underlying base manifold of some principal $G$-bundle. The meaning is that in this way you let the vectors transform under e.g. Lorentz transformations $O(3,1)$ at each tangent space. Thus I would say that torsion depends on the choice of the Lie group and the underlying manifold or spacetime.

$\endgroup$
0
$\begingroup$

Depends what kind of manifolds you're dealing with. I think the reason for much confusion is fundamentally a flawed foundational understanding of what manifolds are, meaning what their defining properties should be. That being said, As long as you have a metric you can define a metric compatible connection. If you require that connection to also be torsion free that it is called Levi civita and it is then uniquely defined. Torsion free just means that the commutator of vector fields [a, b] is equal to ∇a(b) - ∇b(a).

$\endgroup$
2
  • $\begingroup$ "as long as you have a metric" then, I see, one can always define a connection without torsion, but I am still not quite sure what happens when there is no metric $\endgroup$ Apr 5, 2020 at 11:26
  • 1
    $\begingroup$ You don't need metrics to define connections. There are many possible connections, a a metric is just a way to define a connection. There are also all sorts of manifolds. $\endgroup$
    – Kugutsu-o
    Apr 6, 2020 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.