Can a vector field on a non-intersecting smooth curve assign multiple vectors to a point?

Question

I'm reading Semi-Riemannian Geometry by Newman - currently about vector fields on curves. Let $\gamma(t)$ be an injective smooth curve from $(a,b)\subset \mathbb{R}$ to a smooth manifold $M$. I know that for any $t_0\in(a,b)$, $\frac{d\gamma}{dt}(t_0)$ is an operator (or a tangent vector in $T_{\gamma(t_0)}(M)$), where $$\frac{d\gamma}{dt}(t_0)(f)\equiv\frac{d(f\circ\gamma)}{dt}(t_0)$$

The book states that a vector field on $\gamma$ is defined as a map $J_{\gamma}$ that assigns each $t\in(a,b)$ to a vector $J_{\gamma}(t)$ in $T_{\gamma(t)}(M)$. The set of smooth vector fields on $\gamma$ is denoted by $\mathfrak{X}_M(\gamma)$. Then there is a theorem stating that $d\lambda/dt$ is a vector field in $\mathfrak{X}_M(\gamma)$.

I'm confused at this stage: corresponding to a particular curve $\gamma$ and at a particular point $p=\gamma(t_0)\in M$, by definition we identify a single vector, right? Which is $d\gamma/dt$ evaluated at $t=t_0$.

So then we should have only one vector field on $\gamma$? Because if there were several, then we could identify at least two fields in $\mathfrak{X}_M(\gamma)$ that pick out different values at some $p$ lying in $\gamma$'s image, which contradicts the previous paragraph. My understanding is: even if there are obviously several vectors in $T_{\gamma(t_0)}(M)$, there's only one that is tangential to $\gamma$.

Would appreciate any help since I seem to be missing something. Maybe I'm wrong in my assumption that the vectors in the vector field need to be tangential to the curve?

Answer 1

Take a curve $\gamma: (a,b) \to \Bbb{R}^2$ with non-tangential self intersection at $p = \gamma(c)$. If we try to define the velocity vector field $\gamma'(t)$ as a section of the tangent bundle of $\Bbb{R}^2$ restricted to the image of $\gamma$, then we will arrive at a problem. Indeed, a naïve idea is to send $\mathrm{im}\gamma \ni q\mapsto \frac{d \gamma}{dt}\big|_{t_0:\gamma(t_0) = q}$, but in the case above we are considering this is not well-defined.

A section of a vector bundle $\pi:E\to X$ over an open set $U\subset X$ is is by definition a map $s:U\to E$ so that $\pi_U\circ s = \mathrm{Id}_U$, where $\pi_U: E|_U\to U$ is the restriction of the projection. In particular, $s$ cannot assign to outputs to one input, i.e. $s(p) \in E_p$ must be "one vector."

In this case, what is happening is the velocity field along the curve $\gamma$ is really defined to be a section of the pullback bundle $\gamma^*TM$. I don't know if you are familiar with this construction, but $\gamma^*TM \to (a,b)$ is a vector bundle whose fibre over $r\in (a,b)$ is $T_{\gamma(r)}M$. Now, we define a section by $r\mapsto \gamma'(r)$ and by construction this is well-defined: if $c$ and $d$ both map to $p\in M$, we have an identification of $(\gamma^*TM)_c$ with $(\gamma^* TM)_d$, but they live in different fibres of the vector bundle.

So, given a curve $\gamma:I\to M$, $\mathfrak{X}_M(\gamma) :=\Gamma(I,\gamma^*TM)$.