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A Simple Example of a Connection With Torsion

While learning general relativity as an undergrad, I was uneasy with the idea of torsion of an affine connection. Familiar phrases like “torsion measures how a frame twists as it undergoes parallel transport” seemed too lofty to serve as a helpful mental model.

The following example is one which gave me an aha! moment.

A “rolling” connection on \(ℝ^2\).

Consider a connection on the plane defined in such a way that a vector undergoing parallel transport rotates in the \(xy\)-plane in proportion to its motion along the \(x\)-direction. That is, define a connection so that the vector field defined by

\[\vec u(x, y) = \mqty( \cos kx \\ \sin kx) ,\]

which spins counter-clockwise when moving in the \(+x\)-direction, is covariantly constant, \(∇_μ \vec u = 0\).

We can write the covariant derivative \(∇_μu^a = ∂_μu^a + Γ^a{}_{μb}u^b\) more easily by reading the Latin indices as the two components of a matrix. Then, for differentiation in the \(x\)-direction, we have

\[∇_x \vec u = ∂_x \vec u + Γ_x \vec u = \mqty(-k\sin kx \\ +k\cos kx) + Γ_x \mqty( \cos kx \\ \sin kx)\]

where \(Γ_x \vec u\) is matrix multiplication. Enforcing \(∇_x \vec u = 0\) implies that the matrix \(Γ_x\) must be

\[Γ_x = \mqty(Γ^x{}_{xx} & Γ^x{}_{xy} \\ Γ^y{}_{xx} & Γ^y{}_{xy}) = \mqty(0 & -k \\ +k & 0) .\]

The \(y\)-direction is trivial, since \(\vec u\) does not depend on \(y\). And so, \(∇_y \vec u = ∂_y \vec u + Γ_y\vec u = Γ_y\vec u\) is zero only when

\[Γ_y = \mqty(Γ^x{}_{yx} & Γ^x{}_{yy} \\ Γ^y{}_{yx} & Γ^y{}_{yy}) = \mqty(0 & 0 \\ 0 & 0) .\]

This connection has zero curvature…

If we calculated the Riemann curvature \(R = \dd Γ + Γ ∧ Γ ,\) we would find that this connection is flat, since \(\dd Γ = 0\) (all Christoffel symbols are constant) and

\[Γ ∧ Γ % = (Γ_x \dd x + Γ_y \dd y) ∧ (Γ_x \dd x + Γ_y \dd y) = (Γ_x Γ_y - Γ_y Γ_x) \, \dd x ∧ \dd y = 0\]

also vanishes since \(Γ_y = 0\).

…but non-zero torsion

However, the torsion \(T^λ{}_{μν} = Γ^λ{}_{μν} - Γ^λ{}_{νμ}\) does not vanish:

\[T_x = \mqty(T^x{}_{xx} & T^x{}_{xy} \\ T^y{}_{xx} & T^y{}_{xy}) = \mqty(0 & -k \\ 0 & 0) ,\qquad T_y = \mqty(T^x{}_{yx} & T^x{}_{yy} \\ T^y{}_{yx} & T^y{}_{yy}) = \mqty(+k & 0 \\ 0 & 0) .\]

We can see that the torsion measures the wavenumber \(k\) controlling the rate at which vectors spin when parallel transported along the \(x\) axis.