The Length of a Curve
A Riemannian metric assigns a length to every tangent vector, and a curve is traced out by its
velocity vectors. Integrating the lengths of those velocity vectors along a curve measures how far
the curve travels, giving the notion of arc length on a manifold. This is the first measurement
that reaches beyond a single tangent space to the manifold as a whole, and from it the distance
between points will be built.
Throughout this discussion \((M, g)\) is a Riemannian manifold with or without boundary. A curve
here means a piecewise smooth curve segment: a continuous map \(\gamma : [a, b] \to M\) for
which there is a partition
\(a = a_0 \lt a_1 \lt \cdots \lt a_k = b\) such that \(\gamma\) restricted to each
\([a_{i-1}, a_i]\) is smooth. Its
velocity
\(\gamma'(t)\) is a tangent vector at \(\gamma(t)\), defined at every \(t\) except possibly the
finitely many partition points, where left- and right-hand velocities may differ.
Definition: Length of a Curve
Let \((M, g)\) be a Riemannian manifold with or without boundary and let
\(\gamma : [a, b] \to M\) be a piecewise smooth curve segment. The length of
\(\gamma\) is
\[
L_g(\gamma) = \int_a^b \big| \gamma'(t) \big|_g\, dt,
\]
where \(\big| \gamma'(t) \big|_g = g_{\gamma(t)}\big(\gamma'(t), \gamma'(t)\big)^{1/2}\) is the
norm
of the velocity vector with respect to the metric.
The integrand \(\big| \gamma'(t) \big|_g\) is continuous wherever \(\gamma\) is smooth, so it is
continuous at all but the finitely many partition points, and at each of those it has well-defined
limits from the left and from the right. It is therefore bounded, and a bounded function
continuous except at finitely many points is
Riemann integrable,
so the integral is well defined. Equivalently, the length is the sum of the lengths of the smooth
pieces, each an ordinary integral of a continuous function.
Length does not depend on how a curve is parametrized: reparametrizing by a diffeomorphism between
parameter intervals, whether it preserves or reverses the direction of travel, leaves the length
unchanged. This is what makes length a geometric quantity rather than an artifact of the
description, and it is the property that allows length to be compared and minimized across all
parametrizations.
Theorem: Parameter Independence of Length
Let \((M, g)\) be a Riemannian manifold with or without boundary and let
\(\gamma : [a, b] \to M\) be a piecewise smooth curve segment. If
\(\tilde{\gamma} = \gamma \circ \varphi\) is a reparametrization of \(\gamma\) by a
diffeomorphism \(\varphi : [c, d] \to [a, b]\), then
\[
L_g(\tilde{\gamma}) = L_g(\gamma).
\]
Proof:
Suppose first that \(\gamma\) is smooth. The piecewise case follows by applying the argument on
each subinterval where \(\gamma\) is smooth. A diffeomorphism of intervals has derivative of
constant sign, so either \(\varphi' \gt 0\) everywhere or \(\varphi' \lt 0\) everywhere. By the
chain rule the velocity of the reparametrized curve is
\(\tilde{\gamma}'(t) = \varphi'(t)\, \gamma'(\varphi(t))\), so taking norms,
\[
\big| \tilde{\gamma}'(t) \big|_g = \big| \varphi'(t) \big|\, \big| \gamma'(\varphi(t)) \big|_g.
\]
Suppose \(\varphi' \gt 0\), so \(\varphi(c) = a\) and \(\varphi(d) = b\) and
\(\big|\varphi'(t)\big| = \varphi'(t)\). Then
\[
\begin{align*}
L_g(\tilde{\gamma})
&= \int_c^d \big| \tilde{\gamma}'(t) \big|_g\, dt
= \int_c^d \big| \gamma'(\varphi(t)) \big|_g\, \varphi'(t)\, dt \\\\
&= \int_a^b \big| \gamma'(s) \big|_g\, ds = L_g(\gamma),
\end{align*}
\]
where the third equality is the substitution \(s = \varphi(t)\). If instead \(\varphi' \lt 0\),
then \(\varphi(c) = b\) and \(\varphi(d) = a\), and two sign changes occur: one because
\(\big|\varphi'(t)\big| = -\varphi'(t)\), and one because the substitution reverses the limits
of integration. The two cancel, and the value of the integral is again \(L_g(\gamma)\).
The Riemannian Distance Function
With lengths of curves in hand, the distance between two points is defined as the length of the
shortest route between them. Since there is in general no single shortest curve, and since some
routes may be arbitrarily long, the right notion is the greatest lower bound of all the
lengths. The distance is what no path can undercut. Using curve segments as measuring tapes in
this way turns a manifold into a setting where points have numerical separations.
Definition: Riemannian Distance
Let \((M, g)\) be a connected Riemannian manifold with or without boundary. For points
\(p, q \in M\), the Riemannian distance from \(p\) to \(q\) is the
infimum
\[
d_g(p, q) = \inf\, \big\{ L_g(\gamma) : \gamma \text{ is a piecewise smooth curve segment from } p \text{ to } q \big\},
\]
the greatest lower bound of the lengths of all piecewise smooth curve segments joining \(p\) to
\(q\).
For this to be meaningful, the set whose infimum is taken must be nonempty. There must exist at
least one piecewise smooth curve segment from \(p\) to \(q\). Fix \(p\) and let \(A\) be the set of
points of \(M\) that can be joined to \(p\) by such a segment. By the
basis of regular coordinate balls
when \(\partial M = \varnothing\), and by the
basis of regular coordinate balls and half-balls
otherwise, every point \(x \in M\) lies in a regular coordinate ball or half-ball \(V\), whose image in
coordinates is a ball or a half-ball and therefore convex. If \(\varphi\) is the coordinate map, any
two points \(y, z \in V\) are joined within \(V\) by the curve
\(t \mapsto \varphi^{-1}\bigl((1 - t)\varphi(y) + t\varphi(z)\bigr)\), \(t \in [0, 1]\), which is
smooth as the composite of an affine map with the smooth inverse chart. Following a piecewise smooth
curve segment from \(p\) to \(x\) by such a curve from \(x\) to \(y\), and reparametrizing, shows that
\(V \subseteq A\) when \(x \in A\). If some \(y \in V\) lay in \(A\), following a segment from
\(p\) to \(y\) by such a curve from \(y\) to \(x\) would place \(x\) in \(A\), so \(V\) misses
\(A\) when \(x \notin A\). Thus \(A\) and \(M \setminus A\) are both open. Since \(A\) contains \(p\), by
the constant curve, and \(M\) is
connected,
\(A = M\). The set of competing lengths is therefore nonempty, and being a set of
nonnegative real numbers it has a well-defined infimum. The distance \(d_g(p, q)\) is thus
defined for every pair of points in a connected Riemannian manifold.
On Euclidean space the Riemannian distance recovers the elementary notion of distance, confirming
that the construction generalizes rather than replaces it.
Example (Euclidean Distance):
Consider \(\mathbb{R}^n\) with the Euclidean metric \(\bar{g}\). A straight line segment from
\(x\) to \(y\) has length \(|x - y|\), so \(d_{\bar{g}}(x, y) \le |x - y|\). In the other
direction, any piecewise smooth curve segment \(\gamma\) from \(x\) to \(y\) satisfies
\[
L_{\bar{g}}(\gamma) = \int_a^b \big| \gamma'(t) \big|\, dt \ge \left| \int_a^b \gamma'(t)\, dt \right| = \big| \gamma(b) - \gamma(a) \big| = |x - y|,
\]
the inequality being the triangle inequality for integrals. Taking the infimum over all such
\(\gamma\) gives \(d_{\bar{g}}(x, y) \ge |x - y|\). The two bounds combine to
\[
d_{\bar{g}}(x, y) = |x - y|,
\]
so the Riemannian distance on \(\mathbb{R}^n\) is exactly the Euclidean distance, with straight
line segments achieving the infimum.
The example also shows that on \(\mathbb{R}^n\) the infimum is attained, by the straight
segment. On a general Riemannian manifold a length-minimizing curve need not exist between two
given points, which is why the distance is defined as an infimum rather than a minimum. Whether
the infimum is realized, and by what curves, is the question that leads to the theory of
geodesics. The distance function defined here is what those curves are required to minimize.
Riemannian Manifolds as Metric Spaces
The distance function ought to make a manifold into a metric space, with all the structure that
entails, and its metric topology ought to agree with the topology the manifold already carries.
Both statements are true, and the bridge between them is a comparison. In any single coordinate
chart, the Riemannian metric is squeezed between constant multiples of the Euclidean metric
over any compact region. That local comparison is what transfers Euclidean facts about distance
to the manifold.
Lemma: Local Comparison with the Euclidean Metric
Let \(g\) be a Riemannian metric on an open subset \(U\) of \(\mathbb{R}^n\) or of
\(\mathbb{H}^n\), and let \(\bar{g}\) denote the Euclidean metric. For every compact subset
\(K \subseteq U\) there exist
positive constants \(c, C\) such that for all \(x \in K\) and all \(v \in T_x\mathbb{R}^n\),
\[
c\, |v|_{\bar{g}} \le |v|_g \le C\, |v|_{\bar{g}}.
\]
Proof:
When \(U \subseteq \mathbb{H}^n\), a tangent vector to \(U\) at \(x\) is a tangent vector to
\(\mathbb{R}^n\) at \(x\): \(T_xU\) is identified with \(T_x\mathbb{H}^n\) as the tangent space to an
open submanifold,
and \(T_x\mathbb{H}^n\) with \(T_x\mathbb{R}^n\), at interior points because
\(\operatorname{Int} \mathbb{H}^n\) is open in \(\mathbb{R}^n\) and at boundary points by the
tangent space of the half-space at a boundary point.
This is how the statement reads \(g_x\) as an inner product on \(T_x\mathbb{R}^n\), and it is
the only place where the two cases differ.
Identifying \(T\mathbb{R}^n\) with \(\mathbb{R}^n \times \mathbb{R}^n\), consider the set
\[
S = \big\{ (x, v) : x \in K,\ |v|_{\bar{g}} = 1 \big\} = K \times \mathbb{S}^{n-1},
\]
a product of a compact set with the unit sphere, hence compact. The function
\((x, v) \mapsto |v|_g\) is continuous and strictly positive on \(S\), since \(|v|_g = 0\) would
force \(v = 0\), contradicting \(|v|_{\bar{g}} = 1\). If \(S\) is empty, any positive constants
\(c \le C\) will do. Otherwise, by the
extreme value theorem
the function attains a positive minimum \(c\) and a maximum \(C\) on \(S\), so
\(c \le |v|_g \le C\) whenever \(x \in K\) and \(|v|_{\bar{g}} = 1\).
For a nonzero \(v \in T_x\mathbb{R}^n\) with \(x \in K\), set \(\lambda = |v|_{\bar{g}}\). Then
\(\lambda^{-1} v\) is a unit vector, so \(c \le |\lambda^{-1} v|_g \le C\). Both norms are
homogeneous of degree one, so multiplying through by \(\lambda\) gives
\[
c\, |v|_{\bar{g}} = c\lambda \le |v|_g = \lambda\, \big| \lambda^{-1} v \big|_g \le C\lambda = C\, |v|_{\bar{g}}.
\]
The inequalities hold trivially when \(v = 0\), so they hold for all \(v\).
The comparison lemma is the analytic core of the next theorem, which collects everything the
distance function provides. It is the point at which the manifold, first met as a topological space
patched from coordinate charts, acquires a notion of distance compatible with that original
topology.
Theorem: Riemannian Manifolds as Metric Spaces
Let \((M, g)\) be a connected Riemannian manifold with or without boundary. With the
Riemannian distance function \(d_g\), the set \(M\) is a
metric space,
and the
metric topology
it determines coincides with the original manifold topology.
Proof:
We first verify the metric space axioms. Lengths are nonnegative, so \(d_g(p, q) \ge 0\). A
constant curve has velocity zero and hence length zero, giving \(d_g(p, p) = 0\). Any curve
segment from \(p\) to \(q\) can be
reparametrized
to run from \(q\) to \(p\) with the same length, so \(d_g(p, q) = d_g(q, p)\). For the triangle
inequality, let \(\gamma_1\) run from \(p\) to \(q\) and \(\gamma_2\) from \(q\) to \(r\). The
concatenation that follows \(\gamma_1\) and then \(\gamma_2\), reparametrized to a single
domain, is a piecewise smooth curve segment from \(p\) to \(r\) whose length is
\(L_g(\gamma_1) + L_g(\gamma_2)\). Hence
\[
d_g(p, r) \le L_g(\gamma_1) + L_g(\gamma_2).
\]
Taking the infimum over \(\gamma_1\) and \(\gamma_2\) separately yields
\[
d_g(p, r) \le d_g(p, q) + d_g(q, r).
\]
It is precisely the use of piecewise smooth curves that makes this concatenation admissible.
It remains to show that \(d_g(p, q) \gt 0\) when \(p \ne q\). Let \(\Omega\) be a smooth
coordinate domain containing \(p\) but not \(q\), identified with an open subset of
\(\mathbb{R}^n\), or of \(\mathbb{H}^n\) if \(p\) is a boundary point, and let \(\bar{g}\) be the
Euclidean metric in these coordinates. Choose a
regular coordinate ball
\(V\), or a
regular coordinate half-ball
when \(p \in \partial M\), of radius \(\varepsilon\) centered at \(p\) with
\(\overline{V} \subseteq \Omega\). In coordinates \(V\) is the open ball \(B_\varepsilon(p)\),
respectively the half-ball \(B_\varepsilon(p) \cap \mathbb{H}^n\), and \(\overline{V}\) is its
closure. The
comparison lemma applied to the compact set \(\overline{V}\) supplies constants \(c, C\) with
\(c\, |v|_{\bar{g}} \le |v|_g \le C\, |v|_{\bar{g}}\) there. For any piecewise smooth curve
segment \(\gamma\) lying entirely in \(\overline{V}\), integrating the left inequality gives
\[
L_g(\gamma) \ge c\, L_{\bar{g}}(\gamma).
\]
Now let \(\gamma : [a, b] \to M\) be any piecewise smooth curve segment from \(p\) to \(q\).
Since \(q \notin V\), the curve must leave \(V\). Let \(t_0\) be the infimum of times
\(t\) with \(\gamma(t) \notin V\), so \(\gamma(t_0)\) lies in \(\overline{V} \setminus V\), at
Euclidean distance \(\varepsilon\) from \(p\), while \(\gamma\) stays in \(V\) up to \(t_0\).
Restricting to \([a, t_0]\) and using the integral estimate from the Euclidean example of the
previous section, which applies to a curve in \(\mathbb{H}^n\) because it is in particular a
curve in \(\mathbb{R}^n\),
\[
L_g(\gamma) \ge L_g\big(\gamma|_{[a, t_0]}\big) \ge c\, L_{\bar{g}}\big(\gamma|_{[a, t_0]}\big) \ge c\, \big| \gamma(t_0) - p \big| = c\varepsilon.
\]
Taking the infimum over all \(\gamma\) gives \(d_g(p, q) \ge c\varepsilon \gt 0\). The argument
used only that \(q \notin V\), so the same bound holds for every point outside \(V\). The metric
space axioms are established.
Finally we show the metric topology equals the manifold topology, by proving each contains the
other's open sets. Suppose \(U\) is open in the manifold topology and \(p \in U\). Choosing a
regular coordinate ball or half-ball \(V\) of radius \(\varepsilon\) around \(p\) with
\(\overline{V} \subseteq U\), we obtain from the estimate just proved
\(d_g(p, q) \ge c\varepsilon\) whenever \(q \notin V\). Contrapositively,
\(d_g(p, q) \lt c\varepsilon\) forces \(q \in V \subseteq U\), so the metric ball of radius
\(c\varepsilon\) about \(p\) lies in \(U\). Thus \(U\) is open in the metric topology.
Conversely, suppose \(W\) is open in the metric topology and \(p \in W\). Take a regular
coordinate ball or half-ball \(V\) of radius \(r\) around \(p\) with comparison constants \(c, C\) on
\(\overline{V}\), and choose \(\varepsilon \lt r\) small enough that the metric ball of
radius \(C\varepsilon\) about \(p\) lies in \(W\). Let \(V_\varepsilon\) be the set of
points in \(V\) whose Euclidean distance from \(p\) is less than \(\varepsilon\). For
\(q \in V_\varepsilon\), the straight coordinate segment \(\gamma\) from \(p\) to \(q\)
stays in \(V\), a ball and a half-ball both being convex, so integrating the right inequality
of the comparison lemma,
\[
d_g(p, q) \le L_g(\gamma) \le C\, L_{\bar{g}}(\gamma) = C\, |p - q| \lt C\varepsilon.
\]
Hence \(V_\varepsilon\) is contained in the metric ball of radius \(C\varepsilon\),
therefore in \(W\). Since \(V_\varepsilon\) is a neighborhood of \(p\) in the manifold
topology, \(W\) is open in the manifold topology as well. The two topologies coincide.
Metrizability and Completeness
The metric space theorem was stated for connected manifolds, but the metrizability it yields extends
to disconnected ones. Every smooth manifold, connected or not, admits a distance function inducing
its topology. The construction runs through a Riemannian metric, so the argument covers only
manifolds that admit a smooth structure.
Corollary: Smooth Manifolds Are Metrizable
Every smooth manifold with or without boundary is metrizable. That is, it admits a distance
function whose induced metric topology is the given manifold topology.
Proof:
Let \(M\) be a smooth manifold with or without boundary, and equip it with a
Riemannian metric
\(g\). If \(M\) is connected, the preceding theorem shows directly that \(d_g\) induces the
manifold topology, so \(M\) is metrizable. In general, let \(\{M_i\}\) be the connected
components of \(M\). Each is open in \(M\), by the
connectivity properties of manifolds
when \(\partial M = \varnothing\) and by the
topological properties of manifolds with boundary
otherwise, hence is a connected smooth manifold with or without boundary in its own right, to
which \(g\) restricts. Every component therefore carries a Riemannian distance, still written
\(d_g\), inducing its topology. Choose a point \(p_i \in M_i\) for each \(i\). Define a function
on \(M\) by setting, for \(x \in M_i\) and \(y \in M_j\),
\[
d(x, y) =
\begin{cases}
d_g(x, y), & i = j, \\\\
d_g(x, p_i) + 1 + d_g(p_j, y), & i \ne j .
\end{cases}
\]
One pictures a bridge of length \(1\) joining each pair of chosen basepoints, so that crossing
from one component to another costs the trip to the basepoint, the bridge, and the trip onward.
This is a distance function. Symmetry and positivity are clear. For the triangle inequality let
\(x \in M_i\), \(y \in M_j\), \(z \in M_k\). If \(i = j = k\) it is the triangle inequality for
\(d_g\). If \(i = j \ne k\), then
\[
\begin{align*}
d(x, z) &= d_g(x, p_i) + 1 + d_g(p_k, z) \\\\
&\le d_g(x, y) + d_g(y, p_i) + 1 + d_g(p_k, z) = d(x, y) + d(y, z),
\end{align*}
\]
and the case \(i \ne j = k\) is symmetric. In the two remaining cases \(y\) lies in a component
different from those of \(x\) and \(z\), so
\(d(x, y) + d(y, z) = d_g(x, p_i) + 2 + 2\,d_g(p_j, y) + d_g(p_k, z)\). This dominates
\(d(x, z) = d_g(x, p_i) + 1 + d_g(p_k, z)\) when \(i \ne k\), and it dominates
\(d(x, z) = d_g(x, z) \le d_g(x, p_i) + d_g(p_k, z)\) when \(i = k\), since then \(p_i = p_k\).
The function \(d\) induces the given topology on \(M\). Two points in different components are
at \(d\)-distance at least \(1\), so for \(0 \lt r \le 1\) the \(d\)-ball of radius \(r\) about
\(x \in M_i\) is the \(d_g\)-ball of radius \(r\) about \(x\) in \(M_i\). Hence a set
\(O \subseteq M\) is open for \(d\) exactly when each \(O \cap M_i\) is open for \(d_g\) on
\(M_i\), that is, by the preceding theorem, open in the manifold topology of \(M_i\). Since each
\(M_i\) is open in \(M\) and \(O = \bigcup_i (O \cap M_i)\), this holds exactly when \(O\) is
open in \(M\).
With this corollary the development comes full circle. A
topological manifold
was defined at the outset as a space that is locally Euclidean, Hausdorff, and second countable,
with no metric assumed. For a manifold that admits a smooth structure, the route through tangent
vectors, tensor fields, and Riemannian metrics has now returned a distance function on that same
space, recovering its topology exactly. Its topology and the metric topology of that distance
function are, in the end, one and the same.
Because a connected Riemannian manifold is a metric space, the vocabulary of metric spaces applies
to it without change. A connected Riemannian manifold \((M, g)\) is called complete,
and \(g\) a complete Riemannian metric, if the metric space \((M, d_g)\) is
complete,
meaning every Cauchy sequence in \(M\) converges to a point of \(M\). A subset \(B \subseteq M\) is
bounded if there is a constant \(K\) with \(d_g(x, y) \le K\) for all \(x, y \in B\).
These notions, together with the distance function itself, are the foundation on which the geometry
of geodesics, curvature, and length-minimizing paths is built.