Multiple points in the parallel transport equation

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Let $E \to M$ be a vector bundle with connection $\nabla$, $\gamma \colon [0,1] \to M$ a smooth curve. A section $s = \gamma^* \tilde s$, $\tilde s \in \Gamma(E)$ of $\gamma^* E$ is called parallel if $$ \dot a^j(x(t)) + \Gamma^j_{ik}(x(t)) a^i(x(t))\dot x^i(t) = 0, \quad j = 1,\ldots,n, \quad t \in (0,1). \tag{1} $$ where $\gamma(t) = (x^i(t))$ in local coordinates on $M$, $\tilde s = (x^i,a^j(x))$ in a trivialising neighborhood, $\Gamma^j_{ik}$ are Christoffel symbols for $\nabla$.

We solve the equation (1) for the function $a^j(t) := a^j(x(t))$ with some initial condition. But it seems that it is possible that for two different $t_1$ and $t_2$ with $x(t_1) = x(t_2)$ we will obtain $a^j(t_1) \neq a^j(t_2)$. How to deal with such situations?

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Probably, you're referring to the uniqueness of solutions of ordinary differential equations, so an orbit of the flow associated with the o.d.e. can't cross itself.

But I don't think that's a problem here, because the section $s(t) = (x(t), a(t))$ consists of the base point $x(t)$ and the tangent vector $a(t)$. So if, for $t_1 \neq t_2$, we have $x(t_1) = x(t_2)$ and $a(t_1) \neq a(t_2)$, we still have $s(t_1) \neq s(t_2)$, which is perfectly valid.

If we have $x(t_1) = x(t_2)$ and $a(t_1) = a(t_2)$, i.e. $s(t_1) = s(t_2)$, then the orbit is a closed loop.

In fact, the situation $x(t_1) = x(t_2)$ for $t_1 \neq t_2$ is exactly the concept of holonomy (http://en.wikipedia.org/wiki/Holonomy) where parallel transport along closed loops is considered.