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I'm trying to write the definition of arc length on a Riemannian manifold using fancy differential geometry language, mostly as a way to become familiarized with this language. Of course, writing this definition using just calculus is trivial.

Let $(M,g)$ be a Riemannian manifold, and $\alpha : I \to M$ a regular curve on it. (Think of $I$ as an abstract interval with a smooth structure, but no preferred coordinate.) Given a coordinate $t$ on $I$, the pullback metric $\alpha^\star g$ is represented by the $1 \times 1$ matrix whose lone entry is the squared length of the tangent vector field $v : I \to TI$ with coordinate $dt(v) = 1$. If the endpoints of $I$ have coordinate $t = a$ and $t = b$, the arc length of $\alpha$ is

$$\mathcal L(\alpha) = \int_a^b \sqrt {(\alpha^\star g) (v, v)} \, dt = \int_a^b \sqrt {g_{\alpha(t)}(d\alpha_t(v), d\alpha_t(v))} \, dt$$

I find this definition unsatisfying, because I needed to introduce $t$ and $v$ just to be able to state it, even though the associated geometric concept is independent of $t$ and $v$. Is there a coordinate-free way to state this definition?

isekaijin
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  • you don't need to define a coordinate on $I$ to define the pullback metric. – Rylee Lyman Apr 05 '19 at 02:09
  • @RyleeLyman: With that much I agree, of course. But everything afterwards requires using $t$ and $v$. – isekaijin Apr 05 '19 at 02:10
  • The problem seems to come from the fact that I can't feed the form $d\alpha$ itself as argument to $g$. I need to feed $d\alpha(v)$, and multiply times $dt$ outside of the square root, to cancel the effect of arbitrarily choosing $v$. Of course, $v$ and $dt$ are literally dual bases of $TI$ and $T^\star I$, respectively. – isekaijin Apr 05 '19 at 02:12
  • why not define $d\alpha$ (or $g(d\alpha)$, unclear which is the trouble) to agree with your original definition up to choosing $v$? – Rylee Lyman Apr 05 '19 at 02:27
  • When you rescale $v$, both $d\alpha(v)$ and $g(d\alpha(v), d\alpha(v))$ actually change. It's the whole integrand $\sqrt {g(d\alpha(v), d(\alpha(v))} dt$ that remains the same. – isekaijin Apr 05 '19 at 02:33
  • More precisely, by linearity, $d\alpha(kv) = k , d\alpha(v)$ and $g(d\alpha(kv), d\alpha(kv)) = k^2 , g(d\alpha(v), d\alpha(v))$, but... oh, wait, I think I just realized what the problem is. – isekaijin Apr 05 '19 at 02:37
  • A coordinate change $w = kv$ forces us to use $ds = k^{-1} , dt$ as well. If $k > 0$, then $\sqrt {g(d\alpha(v), d\alpha(v))} , dt = \sqrt {g(d\alpha(w), d\alpha(w))} , ds$. But, if $k < 0$, the integrand's sign flips! – isekaijin Apr 05 '19 at 02:40
  • I'm not fond of coordinates either ... A radical solution is to consider (simple) curves and their resp. arc length very abstractly (categorically!) as given, and derive relations between them. For instance: Starting from a set $\mathcal S$ (of points) let $\mathcal K$ be a collection of subsets of $\mathcal S$ and a given function $L : \mathcal K \rightarrow \mathbb R^{0+}$. Two properties: $$ (1) , \alpha, \beta, \gamma, \in \mathcal K : \alpha \cap \beta = \emptyset \text{ and } \alpha \cup \beta = \gamma \text{ implies } L[ , \gamma , ] = L[ , \alpha , ] + L[ , \beta , ],$$ – user12262 Jan 07 '22 at 09:01
  • $$ (2) , \alpha, \beta, \gamma, \phi, \psi \in \mathcal K : \alpha \cap \beta = \emptyset \text{ and } \alpha \cup \beta = \gamma \text{ and } \phi \cap \psi = \emptyset \text{ and } \phi \cup \psi = \gamma \ \text{ implies } \text{ either } (\alpha \cup \phi = \alpha \text{ and } \beta \cup \psi = \psi) \text{ or } (\alpha \cup \phi = \phi \text{ and } \beta \cup \psi = \beta).$$

    Then for $\mathcal G \subset \mathcal K$ whose elements are all pairwise disjoint and whose union equals $\gamma$:

     – user12262 Jan 07 '22 at 09:03
  • $$ L[ , \gamma , ] = \text{Supremum}{\mathcal F \subset \mathcal G} \left[ , \sum{\phi_k in \mathcal F} L[ , \phi_k , ] \right].$$ – user12262 Jan 07 '22 at 09:04

1 Answers1

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Suppose you reparameterize the line using a coordinate $s$. There exists a scalar field $k$ such that $dt = k \, ds$, hence we must use the vector field $w = kv$ in the calculation. Substituting in the integrand, we have

$$\sqrt {g(d\alpha(w), d\alpha(w))} \, ds = \sqrt {k^2 \, g(d\alpha(v), d\alpha(v))} \, \frac 1k \, dt = \frac {|k|} k \sqrt {g(d\alpha(v), d\alpha(v))} \, dt$$

By definition of regular parameterization, $k$ is never zero. The interval $I$ is connected, hence we have two possibilities:

  1. $k > 0$ everywhere, in which case $|k|/k = 1$, and the integrand's sign stays the same.

  2. $k < 0$ everywhere, in which case $|k|/k = -1$, and the integrand's sign is flipped.

Hence the arc length (as defined in the question) is not a fully coordinate-free notion. It depends on the orientation of the line induced by the parameterization.

An even better take at it is that the definition given in the question is slightly wrong. The integral in the question actually defines an arc length parameterization of the curve, if we allow the upper limit of integration to vary:

$$s = \sigma(t) = \int_a^t \sqrt {(\alpha^\star g) (v, v)} \, dt$$

However, the arc length itself is actually a Lebesgue integral over the set $[a,b]$:

$$\mathcal L(\alpha) = \int_{[a,b]} \sqrt {(\alpha^\star g) (v, v)} \, |dt|$$

It is not an integral of a form over a chain.

isekaijin
  • 1,765