Mrthdrb

I must have read and re-read introductory differential geometry texts ten times over the past few years, but the "torsion free" condition remains completely unintuitive to me.

The aim of this question is to try to finally put this uncomfortable condition to rest.

Ehresmann Connections

Ehresmann connections are a very intuitive way to define a connection on any fiber bundle. Namely, an Ehressmann connection on a fiber bundle $Erightarrow M$ is just a choice of a complementary subbundle to $ker(TE rightarrow TM)$ inside of $TE$. This choice is also called a horizontal bundle.

If we are dealing with a linear connection, then $E=TM$, and the Ehresmann connection is a subbundle of $TTM$. This makes intuitive sense -- basically it's saying that for each point in $TM$ it tells you how to move it to different vectors at the tangent spaces of different points. ($ker(TTM rightarrow TM)$ will mean moving to different vectors at the same tangent space; so that is precluded.)

I like this definition -- it makes more intuitive sense to me than the definition of a Koszul connectionan $mathbbR$-linear map $Gamma(E)rightarrowGamma(Eotimes T^*M)$ satisfies some condition. Unlike that definition it puts parallel transport front and center.

Torsion-Freeness

A Levi-Civita connection is a connection that:
1. It preserves with the Riemannian metric. (Basically, parallel transporting preserves inner products.)
2. It is torsion-free.
Torsion free means $nabla_XY - nabla_YX = [X,Y]$.

This definition very heavily uses the less intuitive notion of connection.

So:

Questions

1. How can you rephrase the torsion-free condition in terms of the horizontal bundle of the connection? (Phrased differently: how can it be phrased in terms of parallel transports?)


2. I realized that I don't actually have handy an example of a connection on $mathbbR^2$ that preserves the canonical Riemannian metric on $mathbbR^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.

You should think of the tangent bundle as a bundle with a bigger structural group namely the group $DeclareMathOperatorAffmathbfAff$ $Aff(n)$ of affine transformations of $newcommandbRmathbbR$ $bR^n$. As such, its curvature is a $2$-form with coefficients in the Lie algebra of $Aff(n)$.

An (infinitesimal) affine map consists of two parts: a translation and a linear transformation. Correspondingly, the curvature of an affine connection decomposes into two parts. The part of the curvature corresponding to the infinitesimal translation is the torsion of the connection. Thus if the torsion is $0$, the affine holonomy of this connection along infinitesimal parallelograms is translation free. For more details consult Sections III.3-5 in volume 1 of

S.Kobayashi, K. Nomizu: Foundations of Differential Geometry, John Wiley & Sons, 1963

As fo the second question, denote by $D$ the Levi-Civita connection on $TbR^n$. Any other metric connection $nabla$ has the form
$$
nabla=D+ A,;;A=sum_i=1^m A_i dx^i,
$$
where $A_i$ are smooth maps
$$
A_i:bR^nto mathrmso(n)=mboxthe space of real skew-symmetric $ntimes n$ matrices.
$$
The torsion of $D+A$ is described in Proposition 1.2 of this paper. Among other things it shows that any $2$-form
$$
T=sum_i<j X_ij dx^iwedge dx^j,;;X_ij:bR^nstackrelC^inftylongrightarrowbR^n,
$$
can be the torsion of a connection compatible with the metric.

The Levi-Civita connection of the Euclidean $bR^n$ is the trivial connection.

Liviu Nicolaescu gave an answer to your question 1, so I will answer the other one:

2. I realized that I don't actually have handy an example of a connection on $Bbb R^2$ that preserves the canonical Riemannian metric on $Bbb R^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.

Let $nabla$ be the Levi-Civita connection of $(Bbb R^2, g = rm dx^1otimes rm dx^1 + rm dx^2otimes rm dx^2)$. Since the difference of two connections is a vector-valued $(0,2)$-tensor field (naturally identified with a scalar valued $(1,2)$-tensor field), we'll look for connections of the form $overlinenabla = nabla + T$. Meaning that we'll look for a condition on $T$ ensuring that $overlinenablag = 0$. If $T neq 0$, such $overlinenabla$ will necessarily have torsion. A short calculation says that $overlinenablag = 0$ if and only if $$g(T(Z,X), Y) + g(X, T(Z,Y)) = 0$$for all vector fields $X$, $Y$ and $Z$ in $Bbb R^2$. Write $T(partial_i,partial_j) = sum_k=1^2 T_ij^kpartial_k$. The condition above then reads $T_ki^j + T_kj^i = 0$ for $1 leq i,j,k leq 2$. We have nonzero $T$ satisfying such conditions, e.g., $$T = -rm dx^2otimes rm dx^2otimes partial_1 + rm dx^2otimes rm dx^1otimes partial_2.$$