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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?

TheQuantumMan
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1 Answers1

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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 $\vec \Omega \cdot \vec r = 0.$

Anthony Carapetis
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  • 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)? – TheQuantumMan Jan 02 '18 at 11:13
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    @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. – Anthony Carapetis Jan 02 '18 at 11:33
  • [+1] I admire the clearness of your explanations (not only here). – Jean Marie Jan 18 '18 at 06:30