Torsion and the Difference of Connections

Although a connection is not a tensor, the difference of two connections is. This affine-space structure isolates exactly which modifications are tensorial. Its antisymmetric part changes torsion, while its full index-lowered form also changes nonmetricity. Torsion must be defined with the vector-field commutator included; omitting that term gives a frame-dependent expression.

Required background. Tensor Fields and Natural Operations supplies Lie brackets and multilinear tensors; Pseudo-Riemannian Metrics and Signature supplies the metric used to discuss nonmetricity, without imposing compatibility.

Helpful background. Calculus and Linear Algebra for Relativity supplies bilinear maps and symmetric–antisymmetric decomposition.

Connections form an affine space

Let $\nabla$ and $\widetilde\nabla$ be two affine connections on $TM$. Define

$$A(X,Y)=\widetilde\nabla_XY-\nabla_XY.$$

The derivative terms cancel:

$$\begin{aligned} A(fX,Y)&=fA(X,Y),\\ A(X,fY)&=fA(X,Y). \end{aligned}$$

Thus $A$ is smooth-function linear in both arguments and is a $(1,2)$ tensor. Conversely, if $A$ is any $(1,2)$ tensor, then $\widetilde\nabla_XY=\nabla_XY+A(X,Y)$ satisfies the connection axioms. Connections therefore form an affine space modeled on $\Gamma(T^*M\otimes T^*M\otimes TM)$: there is no preferred zero connection on a general manifold, but the displacement between any two is unambiguous.

In one coordinate chart,

$$A^\rho{}_{\mu\nu} =\widetilde\Gamma^\rho{}_{\mu\nu} -\Gamma^\rho{}_{\mu\nu}.$$

The inhomogeneous transformation terms cancel, leaving the ordinary tensor law. This is the safe way to compare two connection prescriptions.

Torsion and frame commutators

The torsion of $\nabla$ is

$$T(X,Y)=\nabla_XY-\nabla_YX-[X,Y].$$

The bracket subtracts the failure of the chosen vector fields to commute, making $T$ tensorial in both slots. It is antisymmetric: $T(X,Y)=-T(Y,X)$. In a coordinate basis, $[\partial_\mu,\partial_\nu]=0$, so

$$T^\rho{}_{\mu\nu} =\Gamma^\rho{}_{\mu\nu}-\Gamma^\rho{}_{\nu\mu}.$$

In a general frame $[e_b,e_c]=C^a{}_{bc}e_a$,

$$T^a{}_{bc} =\Gamma^a{}_{bc}-\Gamma^a{}_{cb}-C^a{}_{bc}.$$

The invariant definition and its tensorial character are developed in Wendl 2021–2022, §21.4, Exercise 21.11, pp. 183–184, PDF.

Changing the connection by $A$ changes torsion by

$$\widetilde T(X,Y)-T(X,Y)=A(X,Y)-A(Y,X).$$

Only the antisymmetric part of $A$ enters this difference. The symmetric part still changes covariant derivatives and autoparallel curves.

Nonmetricity remains independent

For a chosen metric, define the nonmetricity convention

$$Q_{\mu\nu\rho}\equiv\nabla_\mu g_{\nu\rho}.$$

No condition $Q=0$ is assumed. Under $\widetilde\nabla=\nabla+A$,

$$\widetilde Q_{\mu\nu\rho} =Q_{\mu\nu\rho} -A^\sigma{}_{\mu\nu}g_{\sigma\rho} -A^\sigma{}_{\mu\rho}g_{\nu\sigma}.$$

Hence torsion and nonmetricity constrain different combinations of the connection difference. A torsion-free connection can have nonmetricity, and a metric-compatible connection can have torsion. Only imposing both conditions selects the Levi-Civita connection.

In a coordinate autoparallel equation,

$$\ddot x^\rho+\Gamma^\rho{}_{\mu\nu}\dot x^\mu\dot x^\nu=0,$$

the velocity product is symmetric, so only $\Gamma^\rho{}_{(\mu\nu)}$ appears explicitly. This does not make torsion irrelevant: torsion affects covariant commutators, transport, and the relation between coordinate and noncoordinate frames. It only shows that one particular equation cannot detect the antisymmetric coefficient part.

Anholonomic polar-frame check

On the Euclidean plane use the orthonormal frame

$$e_r=\partial_r, \qquad e_\phi=\frac1r\partial_\phi.$$

It is not a coordinate frame:

$$[e_r,e_\phi]=-\frac1r e_\phi.$$

For the Levi-Civita connection,

$$\nabla_{e_r}e_\phi=0, \qquad \nabla_{e_\phi}e_r=\frac1r e_\phi.$$

Therefore

$$T(e_r,e_\phi) =0-\frac1r e_\phi-\left(-\frac1r e_\phi\right)=0.$$

Ignoring the commutator would incorrectly assign torsion to flat Euclidean space. The calculation also shows why connection-coefficient symmetry is not a frame-independent formulation of vanishing torsion.

Geometrically, torsion can be interpreted through the leading closure failure of an infinitesimal parallelogram when its sides are compared using the connection. That interpretation depends on a careful transport construction; the invariant definition above is primary and continues to apply when a simple coordinate parallelogram is unavailable.

Scalar Hessians give an algebraic detector that is independent of a picture. For a scalar $f$, the first derivative is the covector $df$. In a coordinate basis,

$$(\nabla df)_{\mu\nu}-(\nabla df)_{\nu\mu} =-T^\rho{}_{\mu\nu}\partial_\rho f.$$

The ordinary second partial derivatives cancel, leaving precisely the torsion contraction with the gradient. Thus a nonsymmetric covariant Hessian does not require curvature. This is the antisymmetric part of the tensor Hessian: the second derivative differentiates the covector $df$, including its lower index. The commutator of directional operators on a scalar in a coordinate basis is instead $[\nabla_{\partial_\mu},\nabla_{\partial_\nu}]f =[\partial_\mu,\partial_\nu]f=0$. Affine Connections and Covariant Derivatives develops this distinction for vectors as well. Conversely, if the Hessian’s antisymmetric part vanishes for every smooth $f$, then torsion vanishes because gradients can realize arbitrary covectors at a point.

The affine-space viewpoint also separates two notions often conflated. If $A(X,Y)$ is symmetric in $X,Y$, adding it leaves torsion unchanged but generally changes autoparallels and nonmetricity. If $A$ is antisymmetric, it changes torsion but drops out of the coordinate autoparallel equation because $\dot x^\mu\dot x^\nu$ is symmetric. Neither modification is “pure coordinates”: $A$ is a tensor and can be detected by suitable invariant operations.

Metric compatibility imposes yet another restriction. With $A_{\rho\mu\nu}=g_{\rho\sigma}A^\sigma{}_{\mu\nu}$, the change in $Q_{\mu\nu\rho}$ constrains a different exchange of indices from the one fixed by torsion. This is why specifying only $T$ does not reconstruct a connection from a metric, and specifying only $Q$ does not either. The Levi-Civita theorem closes both freedoms simultaneously by setting both tensors to zero.

Exercises

On $\mathbb R^2$ with its Cartesian flat connection $\partial$, define $A(\partial_x,\partial_y)=a\,\partial_x$ for constant $a$, with all other basis values zero, and set $\widetilde\nabla=\partial+A$. Compute the torsion and determine whether $\widetilde\nabla$ is compatible with $g=dx^2+dy^2$.

Solution

Because the coordinate fields commute,

$$\widetilde T(\partial_x,\partial_y) =A(\partial_x,\partial_y)-A(\partial_y,\partial_x) =a\,\partial_x.$$

For compatibility, test $(\widetilde\nabla_{\partial_x}g)(\partial_x,\partial_y)$:

$$0-g(A(\partial_x,\partial_x),\partial_y) -g(\partial_x,A(\partial_x,\partial_y))=-a.$$

Thus $\widetilde\nabla g\ne0$ unless $a=0$. This example has both torsion and nonmetricity for $a\ne0$.

References

• Wendl, Chris. Differential Geometry I and II. Humboldt-Universität zu Berlin lecture notes, 2021–2022, §21.4, pp. 183–184. Course PDF.