# About the solution of the parallel transport equation

## Summary:

parallel transport equation

## Main Question or Discussion Point

If a vector moves along a particular curve $l$ from point $x_0$ to point $x$ on a manifold whose connection is $\Gamma^i_{jk}(x)$,
then the vector field we get obviously satisfy the pareallel transport equations:
$$\partial_kv^i(x)+\Gamma^i_{jk}(x)v^j(x)=0$$
Because $[\Gamma^i_{jk}(x) , \Gamma^i_{jk}(x')]\neq 0$, so the solution should be written in the form of a path-ordered product:
$$v(x)=\mathcal{P}\left\{\exp{\left(\int^l-\Gamma^i_{jk}(x)dx^k\right)}\right\} v(x_0)$$
The definition of path-ordered product is just like the time-ordered product in quantum mechanics.
Then we usually use Dyson expansion to calculate the path-ordered procuct:
$$\mathcal{P}\left\{\exp{\left(\int^l-\Gamma^i_{jk}(x)dx^k\right)}\right\} =I+\sum^{\infty}_{k=1}(-1)^k\int_{x_0}^x\int_{x_0}^{x_k}\cdots\int_{x_0}^{x_2}\Gamma(x_k)\cdots \Gamma(x_1)dx_k\cdots dx_1$$
To an arbitary connection field $\Gamma^i_{jk}(x)$, The higher-order terms in this series don't always converge to 0. So I want to know if we have any other way to find an exact solution to the parallel transport equation.

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jambaugh
Gold Member
I'm confused by what you're trying to do. Specifically you are referring to a vector field when talking about parallel transport of a vector. Since the propagation of a vector at one point to another is path dependent there is no unique solution to a field.

I don't recognize your parallel transport equations as such but would rather expect something like:
$$d v^i +\gamma^i_{jk} v^k dx^j = 0$$
or in parametric form $\dot{v}^i +\gamma^i_{jk}v^k \dot{x}^j = 0$ where the dots are parameter derivatives. The equations you wrote would manifest first by considering a vector field with components $v^i(x)$ which is invariant under parallel transport under a specific displacement field $u^i(x)$ so that
$$\left[ \partial_j v^i(x) + \gamma^i_{jk}v^k(x)\right]u^j = 0$$
Then the global satisfaction of the equations you wrote would imply a field which is invariant under all flows for any arbitrary field $u^i(x)$. I would imagine only the zero vector field could satisfy such except under very restricted geometries.

To me, obviously the emergence of Berry phase for parallel transport along a closed path invalidates what you seem to be attempting since it seems to me you are implicitly assuming all Berry phases are zero.

Parallel transport is not the same as a propagator. Propagators propagate the boundary conditions for a constraint on a field (scalar or vector/tensor/spinor/gauge). Hence you can propagate the time slice of a field subject to a dynamics constraint to another time slice. It is in this context that one might invoke a Dyson expansion of the Green's function integral. Is this what you're trying to do?

I'm also a bit confused by your commutator of the connection components. Those components are real numbers and commute. I think you mean the commutator for the matrices formed by those components, namely $[\Gamma_k]$ the matrix with row j and column i component $\Gamma^i_{jk}$ then you can take the non-trivial commutator:
$$\left[ [\Gamma_k],[\Gamma_{k'}] \right] \ne 0$$
I assume this is what you meant, in which case you should work in this single matrix algebra or re-express your "commutator" as a component contraction relation, e.g. "It is not the case that...
$\Gamma^i_{jk}\Gamma^j_{\ell k'} - \Gamma^i_{jk'}\Gamma^j_{\ell k} \ne 0$
for all $j,\ell$.

I think you might find it useful to work in this matrix algebra when you try to sensibly express exponentiation of the differential transformations constituting parallel transport of the vectors along a given path.

I'm confused by what you're trying to do. Specifically you are referring to a vector field when talking about parallel transport of a vector. Since the propagation of a vector at one point to another is path dependent there is no unique solution to a field.
Of course parallel transport of a vector on a manifold with non-zero curcature is path dependent, so I only want to find a solution if given a particular path $l$, so if $x_0, x \in l$. we can get $v(x)$ from a given $v(x_0)$.
I don't recognize your parallel transport equations as such but would rather expect something like:
$$dv^i+\gamma^i_{jk}v^kdx^j=0$$
I think my differential equations are equivalent to your equations above. Ssuppose the vector moves a Infinitesimal displacement $dx_1$, Then $dv^i=\partial_1v^i dx_1$, then we can get
$$(\partial_1v^i+\gamma^i_{j1}v^i)dx^1=0$$
Repeat this step, then we can get:
$$(\partial_kv^i+\gamma^i_{jk}v^j)dx^k=0$$
Because to an arbitrary infinitesmal displacement $\Sigma dx^i$ these equations are true, so we can get:
$$\partial_kv^i+\gamma^i_{jk}v^j=0$$

So my question is if given a particular path $l$ on the manifold, and two points $x_0,x \in l$, then how to get $v(x)$ from a given $v(x_0)$ using parallel transport equation. To me, the answer is a path-orderd production(ordered exponentials), but such ordered exponentials is too complex to be calculated, so we hardly can get an exact result, So I am confused if it is the only way to solve the problem, and if we can get an exact solution of the problem.
Or perhaps in general relativity this question is not very important.

jambaugh
Gold Member
The two equations are not equivalent for the same reason that a non-trivial function can have a zero directional derivative in a particular direction without having a zero gradient in general.

jambaugh
As a first order linear ODE analogous to say Schrodinger's equation you attack it just like those... as you described with a Dyson series expansion. But you are not propagating a field in space-time, you are propagating a single vector forward along a parametric path. You are solving an ODE not a PDE. Possibly I misunderstood your notation, if you are using $x_k$ to indicate your parameter, or now your series of parameters as you subdivide the path in the Dyson series expansion.