4
$\begingroup$

I was wondering whether, for a given finite-dimensional manifold, the connection $\nabla$ exists and is uniquely defined?

Afais for Riemannian manifolds, there exists always exactly one Levi-Civita connection, but the calculation is rather cumbersome.

Now, if we consider manifolds without a metric, is there still always one connection (now we do not require torsion-freeness and compatibility with the metric, of course)?

$\endgroup$
13
$\begingroup$

No, on a general manifold $M$:

  • there is no preferred linear connection,
  • there is always at least one connection $\nabla$, and
  • the connections on $M$ are precisely the maps $$\bar{\nabla}_X Y := \nabla_X Y + A(X, Y),$$ where $A \in \Gamma(TM \otimes T^*M \otimes T^*M)$ is a $(2, 1)$-tensor on $M$.

As mentioned, given a metric $g$ on $M$, the torsion-freeness and metric compatibility conditions together pick out a unique connection. (This result wis sometimes called The Fundamental Lemma of Riemannian Geometry.)

On the other hand, on a general smooth manifold one can pick out some preferred classes of connections:

The most important one is one already mentioned: The torsion tensor $$T(X, Y) := \nabla_X Y - \nabla_Y X - [X, Y]$$ is an invariant for the connection, so we can (and often do) restrict attention to torsion-free connections, that is, those for which the torsion tensor is $0$. In fact, any connection $\nabla$ determines a unique torsion-free connection $\widetilde{\nabla}$, namely $$\widetilde{\nabla}_X Y := \nabla_X Y - \tfrac{1}{2} T(X, Y) = \tfrac{1}{2} \nabla_X Y + \tfrac{1}{2} \nabla_Y X - \tfrac{1}{2}[X, Y].$$ In particular, any smooth manifold admits a torsion-free connection.

Substituting the coordinate vector fields of any coordinate chart in the above formula shows that the construction $\nabla \rightsquigarrow \widetilde{\nabla}$ is given by symmetrizing over the lower indices of the Christoffel symbols, that is, the Christoffel symbols of $\nabla$ and $\widetilde{\nabla}$ are related by $$\widetilde{\Gamma}\!{}^c_{ab} = \tfrac{1}{2}(\Gamma^c_{ab} + \Gamma^c_{ba}) .$$ Since the geodesic equation, $$\ddot{\gamma}^c + \Gamma^c_{ab} \dot{\gamma}^a \dot{\gamma}^b = 0 ,$$ is symmetric in these indices, the geodesics of $\nabla$ and those of $\widetilde{\nabla}$ coincide, and so if we're only interested the behavior of geodesics of a given connection $\nabla$, there's no harm in passing to the corresponding torsion-free connection $\nabla$, which is sometimes easier to work with.

One can further specialize to so-called special torsion-free connections, that is those that (locally) preserve some volume form on the manifold; more precisely, a connection $\nabla$ on a smooth manifold $M$ is special if for all $p \in M$ there is a neighborhood $U$ of $p$ and a volume form $\omega \in \Gamma(\Lambda^n T^* U)$ such that $(\nabla \vert_U) \omega = 0$. One can always pick a special connection locally, and, i.i.r.c., under some mild conditions, globally.

$\endgroup$
  • 2
    $\begingroup$ It is worth noting that $(2, 1)$-tensors are often described as one-forms with values in the endomorphism bundle of $TM$. $\endgroup$ – Michael Albanese Jul 5 '15 at 13:29
  • $\begingroup$ @MichaelAlbanese You're right, and in a way this is a more natural way to think about them in this setting than the way I described above (whose only real advantage in this setting is that it's easier to think about the symmetrization this way). $\endgroup$ – Travis Willse Jul 6 '15 at 2:29
3
$\begingroup$

Without assumption of compatibility with metric and vanishing torsion connection is never unique; once you choose local coordinates, connection coefficients in these coordinates can be chosen arbitrarily. It's demanding of vanishing torsion and compatibility with the metric that fixes them on Riemannian manifold. Notice that given Riemannian manifold you can often choose different metric on it and that will yield different connection.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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