By Andrew NC


2019-04-14 01:01:13 8 Comments

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 $E\rightarrow M$ 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 $\mathbb{R}$-linear map $\Gamma(E)\rightarrow\Gamma(E\otimes T^*M)$ 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 $\mathbb{R}^2$ that preserves the canonical Riemannian metric on $\mathbb{R}^2$ but that does have torsion. I bet that would help elucidate the answer to my first question.

3 comments

@Ryan Budney 2019-04-18 20:17:27

We have some good threads on this topic on MO, but they're usually phrased in a different language. Here is how I like to think of torsion from the perspective of an Ehresmann connection. I will use the formalism that an Ehresmann connection on $TM$ is fibrewise linear projection $c : T^2 M \to TM$ whose kernel is complementary to the kernel of $D\pi : T^2 M \to TM$ where $\pi : TM \to M$ is the tangent bundle projection.

From this perspective, take a function of two variables

$$ f : \mathbb R^2 \to M $$

and consider its partial derivatives to be maps

$$\frac{\partial f}{\partial x_i} : \mathbb R^2 \to TM$$

i.e. $\frac{\partial f}{\partial x_i}(p) = Df(p, e_i)$.

Then from this perspective, the torsion of the connection $c$, $\tau_c$ is given by:

$$\tau_c(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) = c\left(\frac{\partial^2 f}{\partial x \partial y}\right) - c\left(\frac{\partial^2 f}{\partial y\partial x}\right)$$

A more elaborate (and physical) way of phrasing this would be to describe $\tau_c(v,w)$ as the result of this process: take the geodesic along $v$. Parallel transport $w$ along this geodesic, and then take the geodesic out of this parallel transport of $w$. This gives you a function of two variables (the time parameter for the two geodesics). You differentiate it with respect to the first geodesic, and ask if this vector field is parallel along the 2nd parameter. It gives you the same formula.

@Liviu Nicolaescu 2019-04-14 02:58:22

You should think of the tangent bundle as a bundle with a bigger structural group namely the group $\DeclareMathOperator{\Aff}{\mathbf{Aff}}$ $\Aff(n)$ of affine transformations of $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$. 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 for the second question, denote by $D$ the Levi-Civita connection on $T\bR^n$. 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 $$ A_i:\bR^n\to \mathrm{so}(n)=\mbox{the space of real skew-symmetric $n\times n$ matrices}. $$ The torsion of $D+A$ is described in Proposition 1.2 of this paper. Among other things it shows that any $2$-form $$ T=\sum_{i<j} X_{ij} dx^i\wedge dx^j,\;\;X_{ij}:\bR^n\stackrel{C^\infty}{\longrightarrow}\bR^n, $$ can be the torsion of a connection compatible with the metric.

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

@Ivo Terek 2019-04-14 06:36:52

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

  1. 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 $(\Bbb R^2, g = {\rm d}x^1\otimes {\rm d}x^1 + {\rm d}x^2\otimes {\rm d}x^2)$. 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 $\overline{\nabla} = \nabla + T$. Meaning that we'll look for a condition on $T$ ensuring that $\overline{\nabla}g = 0$. If $T \neq 0$, such $\overline{\nabla}$ will necessarily have torsion. A short calculation says that $\overline{\nabla}g = 0$ 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 $T(\partial_i,\partial_j) = \sum_{k=1}^2 T_{ij}^k\partial_k$. 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., $$T = -{\rm d}x^2\otimes {\rm d}x^2\otimes \partial_1 + {\rm d}x^2\otimes {\rm d}x^1\otimes \partial_2.$$

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