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## 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.

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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