Let $(M,g)$ be a Riemannian manifold. According to Lee's book on Riemannian manifolds, a one-parameter family of curves is defined as a continuous map $\Gamma:J\times I\to M$, where $I,J$ are intervals on the real line. This map got its name since we can thus obtain two collection of curves in $M$: the main curves $\Gamma_s(t)=\Gamma(s,t)$ defined by holding $s$ constant, and the transverse curves $\Gamma^{(t)}(s)=\Gamma(s,t)$ defined by holding $t$ constant.

Now we can introduce the notion of a vector field along such family of curves. A vector field along $\Gamma$ is a continuous map $V:J\times I\to TM$ such that each $(s,t)\in J\times I$ is assigned a vector $V(s,t)\in T_{\Gamma(s,t)}M$. One example of such is the velocity vector field of a transverse curve, denoted by $$(\partial_s\Gamma)(s_0,t_0)={\Gamma^{(t_0)}}'(s_0).$$

I'm sorry that still another definition needs to be introduced. $\Gamma$ is said to be admissible if:

(i) The domain of $\Gamma$ is of the form $J\times[a,b]$ for some open interval $J$.

(ii) There is a partition $(a_0,\ldots,a_k)$ of $[a,b]$ such that $\Gamma$ is smooth on each rectangle $J\times[a_{i-1},a_i]$. Partitions like this are called admissible.

(iii) Every main curve is a piecewise regular curve segment. A curve is said to be regular if it has non-vanishing velocity.

Given an admissible family $\Gamma$, we describe a continuous vector field along $\Gamma$ as piecewise smooth if the restriction of the vector field to each $J\times[a_{i-1},a_i]$ for some admissible partition $(a_0,\ldots,a_k)$. Lee claims that $\partial_s\Gamma$ is one such vector field, and his argument about continuity of $\partial_s\Gamma$ on the whole $J\times[a,b]$ is confusing me:

To see that is continuous on the whole domain $J\times[a,b]$, note on the one hand that for each $i=1,\ldots,k-1$, the values of $\partial_s\Gamma$ along the set $J\times\{a_i\}$ depend only on the values of $\Gamma$ on that set, since the derivative is taken only with respect to the $s$ variable; on the other hand, $\partial_s\Gamma$ is continuous (in fact smooth) on each sub-rectangle $J\times[a_{i-1},a_i]$ and $J\times[a_{i},a_{i+1}]$, so the right-hand and left-hand limits at $t=a_i$ must be equal.

Why is Lee concerned about the values of $\partial_s\Gamma$ along $J\times\{a_i\}$? Is he doing something like $$\lim_{x\to a}f(x)=f(a)?$$ Thank you so much for your patience. Thank you.


1 Answer 1


It might be helpful to consider also the other "vector fields" along $\Gamma$: $$ \partial_t \Gamma (s_0, t_0) = \Gamma_{s_0}'(t_0).$$ Unlike $\partial_s\Gamma$, this "vector field" is not well-defined: at $J \times \{ a_i\}$, although $\Gamma$ is smooth in $J \times [a_{i-1}, a_i]$ and $J\times [a_i, a_{i+1}]$ respectively, it is not clear if $$ \lim_{t\to a_i^-} \partial_t \Gamma, \ \ \lim_{t\to a_i^+} \partial _t \Gamma$$ are the same.

For a concrete example, consider $$\Gamma : \mathbb R \times [-1, 1] \to \mathbb R^2, \ \ \Gamma (s, t) =(t, s+ |t|).$$ This is a piecewise-smooth family of curves: indeed $\Gamma$ is smooth when restricted to $\mathbb R\times [-1, 0]$ and $\mathbb R \times [0,1]$ respectively. However, one cannot define $\partial_t \Gamma$ along $\mathbb R\times \{0\}$.

So, $\Gamma$ is not really differentiable. So one must justify why $\partial _s\Gamma$ is well-defined, and even continuous. And the main reason is, as described in the book, $\partial_s \Gamma$ along $J \times \{ a_i\}$ depends only on the values of $\Gamma$ on $J \times \{a_i\}$.

To see that $\partial_s \Gamma$ is continuous, they are using this simple fact: if $F, G$ are continuous functions defined on $[a, b]$, $[b, c]$ respectively on $F(b) = G(b)$, then $$ H (x) = \begin{cases} F(x) & \text{ if } x\in [a, b], \\ G(x) & \text{ if } x\in [b, c].\end{cases}$$ is also continuous.

  • $\begingroup$ That simple fact looks like the gluing lemma for continuous functions. $\endgroup$
    – Boar
    Aug 31, 2021 at 12:17

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