4
$\begingroup$

I have been trying to read Michael Berry's article named "The Quantum Phase, Five Years After" but I am stuck at the introduction by Berry's definition of parallel transport (found on the 2nd page here: https://michaelberryphysics.files.wordpress.com/2013/07/berry187.pdf)

In particular, Berry considers a unit sphere and the triad {$\vec{e}_1,\vec{e}_2, \vec{r}$}, where the first two vectors are parallel transported and form a basis of the tangent plane of any point along the curve they are transported on.
Berry demands that $\vec{e}_1\cdot\vec{r}=\vec{e}_2\cdot\vec{r}=0$ always, so that both vectors always lie on a tangent plane. He also demands that $\vec{\Omega}\cdot\vec{r}=0$, where $\vec{\Omega}$ is the angular velocity vector of the triad(as he puts it). This, Berry says, must be true so that the orthogonal frame {$\vec{e}_1,\vec{e}_2, \vec{r}$} does not twist around $\vec{r} $ as we perform the parallel transport.

Now, upon finishing a course in elementary Riemannian Geometry, I have only worked with the definition found there; that is, via the notion of a connection. And everything worked in an intrinsic way.
So, finally, my question is how do these two notions of parallel transport connect? How do the the non-twisting of the two vectors $\vec{e}_1,\vec{e}_2$ around $\vec{r}$ and the condition of parallel transport as defined in Riemannian geometry(see Do Carmo for example) agree?

$\endgroup$
4
$\begingroup$

As you probably know, the Levi-Civita connection $\nabla$ of the sphere $S^2 \subset \mathbb R^3$ can be written as $$\nabla_X Y = \pi(D_X Y)$$ where $\pi_p : T_p \mathbb R^3 \to T_p S^2$ is orthogonal projection onto tangent spaces and $D$ is the Euclidean connection of $\mathbb R^3.$ Thus a vector field $Y$ along a curve $\gamma(t) \in S^2$ is $\nabla$-parallel if and only if the Euclidean derivative $\frac{DY}{dt}$ is orthogonal to $TS^2 = \mathrm{span}\{\vec e_1, \vec e_2\}.$

Now, recall from physics that the angular velocity vector $\vec \Omega$ of a frame $\vec e_i$ is defined such that $$\frac{D\vec e_i}{dt} = \vec \Omega \times \vec e_i.$$ Putting this together, we see that $\vec e_1$ is $\nabla$-parallel if and only if the scalar triple products $$\det(\vec \Omega,\vec e_1, \vec e_i)=(\vec \Omega \times \vec e_1) \cdot \vec e_i$$ vanish for $i=1,2.$ For $i=1$ this is always true, while for $i=2$ we require $$\det(\vec \Omega, \vec e_1, \vec e_2) = \vec \Omega \cdot (\vec e_1 \times \vec e_2) = 0.$$Since $\vec e_1 \times \vec e_2$ is nonzero and parallel to $\vec r,$ we end up with the desired condition $\vec \Omega \cdot \vec r = 0.$

$\endgroup$
  • $\begingroup$ This is what I had worked out last night although I have one qualitative question if you don't mind. Is this condition generalizable in higher dimensional manifolds(I suspect we have a problen with the definition of the cross-product but I can't see intuitively why this is true for a 2d manifold only)? $\endgroup$ – TheQuantumMan Jan 2 '18 at 11:13
  • 1
    $\begingroup$ @TheQuantumMan: Yes, but you need to switch from thinking about rotations about an axis to rotations in a plane. The angular velocity vector is replaced with the connection form $\omega$, which takes values in the special orthogonal Lie algebra $\mathfrak{so}(n)$ ("infinitesimal rotations") and is defined by the relation $De_i = \sum_j \omega_{ij} e_j.$ (We've replaced the cross product with matrix multiplication.) The condition that the frame is parallel transported is then simply that $\omega$ vanishes on the tangent space to the manifold. $\endgroup$ – Anthony Carapetis Jan 2 '18 at 11:33
  • $\begingroup$ [+1] I admire the clearness of your explanations (not only here). $\endgroup$ – Jean Marie Jan 18 '18 at 6:30

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.