Restricting Maps to Submanifolds
Having built the vocabulary of
embedded
and
immersed submanifolds,
we turn to the first of the two technical questions that make that vocabulary usable: when a smooth
map, restricted so that a submanifold becomes its domain or its codomain, remains smooth. The two
cases are not symmetric. Restricting the domain is automatic; restricting the codomain is not, and
the asymmetry is exactly the asymmetry between the two ways the inclusion of a submanifold can
behave. As on the previous page, every ambient manifold here is taken without boundary; the results
extend to the boundary case, but that refinement is set aside for this series.
Restricting the domain never costs anything, because the inclusion of any submanifold is smooth.
Theorem (Restricting the Domain of a Smooth Map)
Let \(M\) and \(N\) be smooth manifolds, let \(F : M \to N\) be smooth, and let \(S \subseteq M\)
be an immersed or embedded submanifold. Then the restriction \(F|_S : S \to N\) is smooth.
Proof Sketch.
The inclusion \(\iota : S \hookrightarrow M\) is a smooth immersion, hence smooth, by the very
definition of a submanifold. Since \(F|_S = F \circ \iota\) is a composition of smooth maps, it
is smooth. \(\blacksquare\)
Restricting the codomain is a different matter. If a smooth map \(F : N \to M\) happens to take all
its values in a submanifold \(S \subseteq M\), one would like to regard it as a smooth map into
\(S\). But the smooth structure on \(S\) is the one that makes its inclusion an immersion,
and a map into \(S\) is smooth only relative to that structure — which, for an immersed
submanifold, can be subtler than the subspace structure suggests. The following example shows the
restriction can fail to be smooth in the worst possible way: it need not even be continuous.
Let \(S \subseteq \mathbb{R}^2\) be the
figure-eight curve, given the
topology and smooth structure induced by the injective immersion \(\beta\) that traces it, and
define \(G : \mathbb{R} \to \mathbb{R}^2\) by \(G(t) = (\sin 2t, \sin t)\) — the same formula as
\(\beta\), but now defined on the whole line rather than an open interval. The image of \(G\) lies
in \(S\), so one is tempted to view \(G\) as a map \(\mathbb{R} \to S\). Yet \(\beta^{-1} \circ G\)
is discontinuous at \(t = \pi\): there \(G(\pi)\) is the crossing point \(\beta(0)\) itself, but as
\(t\) approaches \(\pi\) from either side the points \(G(t)\) run along the strand that \(\beta\)
parametrizes near the ends \(\pm\pi\) of its domain — so \(\beta^{-1}(G(t)) \to \pm\pi\) while
\(\beta^{-1}(G(\pi)) = 0\). The two values \(0\) and \(\pm\pi\) are the parameters that \(S\)'s own
topology keeps far apart, even though their images crowd together at the crossing. So \(G\), regarded
as a map into \(S\), is not even continuous, let alone smooth. The
obstruction is entirely topological — and that turns out to be the only obstruction.
Theorem (Restricting the Codomain of a Smooth Map)
Let \(M\) be a smooth manifold, let \(S \subseteq M\) be an immersed
submanifold, and let \(F : N \to M\) be a smooth map whose image is contained in \(S\). If \(F\)
is continuous as a map from \(N\) into \(S\), then \(F : N \to S\) is smooth.
Proof Sketch.
Fix \(p \in N\) and let \(q = F(p) \in S\). Because \(S\) is immersed, it is
locally embedded:
there is a neighborhood \(V\) of \(q\) in \(S\) that is an embedded submanifold of \(M\), so by the
local slice criterion
it admits a slice chart \((W, \psi)\) of \(M\). Writing \(\pi\) for the projection onto the
slice coordinates and setting \(V_0 = V \cap W\), the pair \((V_0, \pi \circ \psi|_{V_0})\) is a
smooth chart for \(S\) near \(q\). The
hypothesis that \(F\) is continuous into \(S\) gives an open set \(U \ni p\) in \(N\) with
\(F(U) \subseteq V_0\); on \(U\) the coordinate representation of \(F : N \to S\) is
\(\pi \circ (\psi \circ F)\), a composition of smooth maps because \(F : N \to M\) is smooth.
Hence \(F\) is smooth near each point of \(N\). \(\blacksquare\)
When the submanifold is embedded, the troublesome continuity hypothesis is free: a map into an
embedded submanifold is continuous into \(S\) the moment it is continuous into \(M\).
Corollary (Restricting to an Embedded Submanifold)
Let \(M\) be a smooth manifold and \(S \subseteq M\) an embedded submanifold. Then every smooth
map \(F : N \to M\) whose image is contained in \(S\) is also smooth as a map from \(N\) into
\(S\).
Proof Sketch.
Since \(S\) carries the subspace topology, the
universal property of the subspace topology
makes \(F : N \to S\) continuous as soon as \(F : N \to M\) is. The previous theorem then
upgrades continuity to smoothness. \(\blacksquare\)
The corollary fails for some immersed submanifolds — the figure-eight above is the witness — but it
does hold for certain immersed-but-nonembedded ones, and the property is worth naming.
Definition: Weakly Embedded Submanifold
An immersed submanifold \(S \subseteq M\) is weakly embedded if every smooth map
\(F : N \to M\) whose image lies in \(S\) is automatically smooth as a map from \(N\) into \(S\);
such submanifolds are also called initial submanifolds.
Embedded submanifolds are weakly embedded, but so are others that arise naturally — most importantly
the Lie subgroups encountered later in the manifold series, which are weakly embedded even when they
are not embedded.
Uniqueness of Smooth Structures
A question has been hanging over the theory since the slice criterion. When we recognized a subset
\(S \subseteq M\) as an embedded submanifold, we equipped it with a smooth structure — the one read
off from slice charts. Could a different choice of charts, or even a different topology, have made
the same set \(S\) into a submanifold in some other way? If so, "the smooth structure on \(S\)"
would be ambiguous, and every statement about submanifolds would have to specify which structure it
meant. The results of this section close that gap. For embedded submanifolds the structure is
completely rigid: the set alone determines it. This is what lets us speak of the sphere as
a submanifold, with no further qualification — and it is the uniqueness that the local slice
criterion claimed but left for later, when it first identified subsets satisfying the slice
condition.
Theorem (Uniqueness for Embedded Submanifolds)
Let \(M\) be a smooth manifold and \(S \subseteq M\) an embedded submanifold. Then the subspace
topology and the smooth structure furnished by the
local slice criterion
are the only topology and smooth structure with respect to which \(S\) is a submanifold,
embedded or immersed.
Proof Sketch.
Suppose \(S\) carried a second topology and smooth structure making it an (embedded or immersed)
submanifold; write \(\widetilde S\) for the set \(S\) equipped with this alternative structure,
and \(\widetilde\iota : \widetilde S \to M\) for its inclusion, which by assumption is an
injective immersion. The key is to compare the two structures through the identity map of
underlying sets. Since \(\widetilde\iota(\widetilde S) = S\) lands in the embedded submanifold
\(S\), the corollary on
restricting to an embedded submanifold
guarantees that \(\widetilde\iota\), viewed as the map \(\widetilde S \to S\), is smooth. Call
this map \(j : \widetilde S \to S\); it is the identity on points, and it is a smooth bijection.
Now factor the inclusion of \(\widetilde S\) as \(\widetilde\iota = \iota \circ j\), where
\(\iota : S \to M\) is the embedding. Differentiating, \(d\widetilde\iota_p = d\iota_p \circ dj_p\)
at each point; the left side is injective because \(\widetilde\iota\) is an immersion, so
\(dj_p\) is injective and \(j\) is itself an immersion; being also a bijection, its domain and
codomain have equal dimension. An immersion has constant rank, equal to
the dimension of its domain, so \(j\) is a bijective map of constant rank, and the
global rank theorem
promotes a constant-rank bijection to a diffeomorphism. A diffeomorphism that is the identity on
points forces \(\widetilde S\) and \(S\) to carry the same topology and the same smooth
structure. \(\blacksquare\)
With uniqueness secured, the recognition problem for embedded submanifolds collapses to a question
about the set itself: a subset \(S \subseteq M\) either satisfies the local slice condition or it
does not, and if it does, its submanifold structure is automatic. Because the slice condition is
local, this even globalizes — if every point of \(S\) has a neighborhood \(U\) in \(M\) with
\(U \cap S\) an embedded \(k\)-submanifold of \(U\), then \(S\) is an embedded \(k\)-submanifold of
\(M\).
For immersed submanifolds the situation is genuinely weaker, and it is worth being precise about how.
The earlier examples — the figure-eight, the dense torus curve — already show that a single set can
be made into an immersed submanifold with topologies finer than the subspace topology. What
cannot vary is the smooth structure once the topology is fixed.
Theorem (Uniqueness of the Smooth Structure on an Immersed Submanifold)
Let \(M\) be a smooth manifold and \(S \subseteq M\) an immersed submanifold. For the given
topology on \(S\), there is exactly one smooth structure making \(S\) an immersed submanifold.
Proof Sketch.
Suppose two smooth structures on the fixed topological space \(S\) both make the inclusion an
injective immersion; write \(S_A\) and \(S_B\) for the two, and let \(j : S_A \to S_B\) be the
identity on points. Since the two structures share the same topology — the one fixed in the
statement — \(j\) is a homeomorphism; in particular it is continuous into \(S_B\). Its composite with the inclusion of \(S_B\) is just
the inclusion of \(S_A\), which is smooth into \(M\), so the
codomain-restriction theorem
— applicable precisely because \(j\) is continuous into the immersed submanifold \(S_B\) — makes
\(j\) smooth. The same argument with the roles reversed makes \(j^{-1}\) smooth, so \(j\) is a
diffeomorphism that is the identity on points, and the two smooth structures coincide.
\(\blacksquare\)
So the only freedom for an immersed submanifold lies in the choice of topology; pin that down and
the smooth structure follows. The intermediate class of weakly embedded submanifolds, introduced in
the previous section, recovers full rigidity — for them the topology is forced too, just as for
embedded ones.
Theorem (Uniqueness for Weakly Embedded Submanifolds)
Let \(M\) be a smooth manifold and \(S \subseteq M\) a weakly embedded submanifold. Then \(S\)
has only one topology and one smooth structure with respect to which it is an immersed
submanifold.
Proof Sketch.
Let \(S_W\) denote \(S\) with its weakly embedded structure, and let \(\widetilde S\) be the same
set with any other topology and smooth structure making it an immersed submanifold; write
\(j : \widetilde S \to S_W\) for the identity on points. The inclusion of \(\widetilde S\) is a
smooth map into \(M\) with image in \(S\), so the defining property of weak embeddedness applied
to \(S_W\) makes that inclusion smooth into \(S_W\) — that is, \(j\) is smooth. Factoring the
inclusion of \(\widetilde S\) as the inclusion of \(S_W\) composed with \(j\) and differentiating
shows \(dj_p\) injective at each point, so \(j\) is an immersion, of constant rank equal to the
dimension of its domain. Thus \(j\) is a bijective map of constant rank, and the
global rank theorem
makes it a diffeomorphism, identical on points, forcing the two structures — topology and
smooth structure alike — to coincide. \(\blacksquare\)
The three results together arrange the kinds of submanifold by how much of their structure the
ambient set determines. An embedded submanifold is rigid outright. A weakly embedded one — the Lie
subgroups among them — is equally rigid, though the proof must route around the missing subspace
topology. A general immersed submanifold retains freedom in its topology, surrendering only the
smooth structure that the topology then fixes. This hierarchy is the reason later constructions take
care to record which kind of submanifold they produce.
Extending Functions from Submanifolds
Restriction asks whether a map defined on the ambient manifold stays smooth when cut down to a
submanifold. The complementary question runs the other way: given a smooth function defined only on
the submanifold, can it be extended to a smooth function on the surrounding space? Before answering,
we must settle an ambiguity in the phrase "smooth function on \(S\)" that is easy to overlook and
important to get right.
There are two things one might mean. A function \(f : S \to \mathbb{R}\) on a submanifold could be
smooth in the sense intrinsic to \(S\) — smooth as a function on the manifold \(S\), meaning each of
its coordinate representations in the charts of \(S\) is smooth. Or it could be smooth in the sense
inherited from the ambient space — smooth as a function on the
subset \(S \subseteq M\),
meaning it admits a smooth extension to a neighborhood of each point. These are not the same
condition, and conflating them is a genuine source of error. We reserve the notation
\(C^\infty(S)\) for the first, intrinsic meaning.
Definition: Smooth Functions on a Submanifold
Let \(M\) be a smooth manifold and \(S \subseteq M\) an immersed or embedded submanifold. The
notation \(C^\infty(S)\) denotes the smooth functions on \(S\) in the intrinsic
sense: a function \(f : S \to \mathbb{R}\) belongs to \(C^\infty(S)\) if it is smooth
with respect to the smooth structure of \(S\) as a manifold in its own right. This is
a priori distinct from smoothness as a function on the subset \(S \subseteq M\), which
would require a local smooth extension to the ambient manifold at each point.
The relationship between the two senses is exactly the content of the extension lemma below: for an
embedded submanifold, an intrinsically smooth function does admit local ambient extensions, and a
properly embedded one admits a single global extension. The distinction matters because the
converse direction — intrinsic smoothness — is the weaker, always-available notion, while
extendability is what the lemma must work to produce.
Lemma (Extension Lemma for Functions on Submanifolds)
Let \(M\) be a smooth manifold, let \(S \subseteq M\) be an embedded submanifold, and let
\(f \in C^\infty(S)\).
(a) There exist a neighborhood \(U\) of \(S\) in \(M\) and a smooth function
\(\widetilde f \in C^\infty(U)\) with \(\widetilde f|_S = f\).
(b) If \(S\) is properly embedded, the neighborhood \(U\) may be taken to be all of \(M\).
Proof Sketch.
For (a), work locally first. Near a point of \(S\), a slice chart presents
\(S\) as a coordinate subspace, in which the intrinsically smooth \(f\) is a smooth function of
the slice coordinates; precomposing with the projection onto those coordinates extends it to a
smooth function on a neighborhood that restricts to \(f\). These local extensions are then glued into a single
smooth function on a neighborhood \(U\) of \(S\) by a partition of unity subordinate to the
cover, the values agreeing on \(S\) where every local piece equals \(f\).
For (b), a properly embedded submanifold is a
closed subset
of \(M\). Part (a) shows that \(f\) is smooth on \(S\) in the ambient sense — it extends smoothly
near each point — so, since \(S\) is closed, the
extension lemma for closed subsets
delivers a smooth function on all of \(M\) restricting to \(f\). Closedness is exactly the
hypothesis that lemma requires, which is why proper embeddedness is the condition that upgrades
a local extension to a global one. \(\blacksquare\)
Both hypotheses earn their place. Without embeddedness, an intrinsically smooth function on an
immersed submanifold may admit no ambient extension at all, because the submanifold's own topology
can be finer than the ambient trace; and without proper embeddedness, a local extension near an
"open end" of the submanifold may resist being closed up into a global one. The lemma is the precise
statement that these are the only obstructions.
The Tangent Space to a Submanifold
We come to the second technical question. A submanifold \(S\) inside \(M\) carries its own
tangent space
\(T_pS\) at each point, built from \(S\) as a manifold in its own right. Geometric intuition insists
that \(T_pS\) ought to be a subspace of the ambient \(T_pM\) — the tangent plane to a surface in
space is, after all, literally a plane through the origin of the surrounding space. Making that
identification precise, and then learning to compute the subspace, is the goal of this section.
The identification is supplied by the inclusion. Since \(\iota : S \hookrightarrow M\) is a smooth
immersion, its
differential
\(d\iota_p : T_pS \to T_pM\) is injective at every point. We therefore identify \(T_pS\) with its
image \(d\iota_p(T_pS)\), a linear subspace of \(T_pM\), and henceforth regard a tangent vector to
\(S\) as a tangent vector to \(M\) that happens to be tangent to \(S\). Under this identification a
vector \(v \in T_pS\) acts on a smooth function \(f\) defined near \(p\) in \(M\) exactly as it acts
on the restriction \(f|_S\), which is intrinsically smooth on \(S\) by domain restriction: writing
\(\widetilde v = d\iota_p(v)\), one has
\(\widetilde v f = v(f|_S)\). The identification is available whether \(S\) is embedded or merely
immersed, since it uses only that the inclusion is an immersion.
The first concrete description of this subspace is kinematic: the vectors tangent to \(S\) are
precisely the velocities of curves that stay in \(S\).
Proposition: Tangent Vectors as Velocities in a Submanifold
Let \(M\) be a smooth manifold, \(S \subseteq M\) an immersed or embedded submanifold, and
\(p \in S\). A vector \(v \in T_pM\) lies in the subspace \(T_pS\) if and only if there is a
smooth curve \(\gamma : J \to M\) whose image is contained in \(S\), which is also smooth as a
map into \(S\), with \(0 \in J\), \(\gamma(0) = p\), and \(\gamma'(0) = v\).
Proof Sketch.
If \(v \in T_pS\), then since
every tangent vector is a velocity,
there is a curve \(\sigma\) in \(S\) with \(\sigma'(0) = v\) computed in \(S\); composing with
the inclusion gives a curve \(\gamma = \iota \circ \sigma\) in \(M\) lying in \(S\), and
computing the differential through this velocity
shows \(\gamma'(0) = d\iota_p(v)\), which is \(v\) under the identification. Conversely, a curve
\(\gamma\) that lies in \(S\) and is smooth into \(S\) has a velocity \(\gamma'(0)\) computed in
\(S\), an element of \(T_pS\); its image in \(T_pM\) is the ambient velocity, so that ambient
velocity lies in \(T_pS\). \(\blacksquare\)
For embedded submanifolds there is a second description, dual to the first and often far more useful
in practice. Where the velocity criterion builds \(T_pS\) from curves inside \(S\), this
one carves it out by the functions that vanish on \(S\): a vector is tangent to \(S\)
exactly when it annihilates every function that is constant — indeed zero — along \(S\).
Proposition: The Tangent Space as an Annihilator
Let \(M\) be a smooth manifold, \(S \subseteq M\) an embedded submanifold, and \(p \in S\). As a
subspace of \(T_pM\),
\[
T_pS = \{\, v \in T_pM : vf = 0 \text{ whenever } f \in C^\infty(M) \text{ and } f|_S = 0 \,\}.
\]
Proof Sketch.
One inclusion is immediate. If \(v \in T_pS\), write \(v = d\iota_p(w)\) for \(w\) in the
intrinsic tangent space of \(S\), and suppose \(f \in C^\infty(M)\) vanishes on \(S\). Then
\(f \circ \iota \equiv 0\), so \(vf = d\iota_p(w)f = w(f \circ \iota) = 0\).
For the reverse inclusion, suppose \(v \in T_pM\) annihilates every such \(f\); we show
\(v \in T_pS\). Choose slice coordinates \((x^1, \dots, x^n)\) about \(p\) in which \(S\) is the
slice \(\{x^{k+1} = \dots = x^n = 0\}\), so that \(T_pS\) is spanned by
\(\partial/\partial x^1, \dots, \partial/\partial x^k\); writing
\(v = \sum_i v^i\, \partial/\partial x^i|_p\), the claim is that \(v^i = 0\) for \(i > k\). Fix
such an index \(j > k\). The coordinate function \(x^j\) vanishes on \(S\) within the chart, but
to apply the hypothesis we need a function defined on all of \(M\) that vanishes on \(S\); a
smooth bump function
\(\varphi\) supported in the chart domain and equal to \(1\) near \(p\) lets us extend
\(\varphi\, x^j\) by zero to a function \(f \in C^\infty(M)\); it vanishes on \(S\) — inside the
chart because \(x^j\) does, outside because it is identically zero — and agrees with \(x^j\) near
\(p\). Then
\(0 = vf = v(x^j) = v^j\). As \(j > k\) was arbitrary, \(v\) has no components transverse to the
slice, so \(v \in T_pS\). \(\blacksquare\)
When the embedded submanifold is presented as a regular level set, this annihilator description
becomes a clean computation: the tangent space is simply the kernel of the differential of any
defining map.
Proposition: The Tangent Space of a Level Set
Let \(M\) be a smooth manifold and \(S \subseteq M\) an embedded submanifold. If
\(\Phi : U \to N\) is any
local defining map
for \(S\), then for each \(p \in S \cap U\),
\[
T_pS = \ker d\Phi_p : T_pM \to T_{\Phi(p)}N.
\]
In particular, if \(S\) is the regular level set of a submersion
\(\Phi = (\Phi^1, \dots, \Phi^k) : M \to \mathbb{R}^k\), then \(v \in T_pM\) is tangent to \(S\)
if and only if \(v\Phi^1 = \dots = v\Phi^k = 0\).
Proof Sketch.
Since \(\Phi\) is constant on \(S\), the composition \(\Phi \circ \iota\) is constant, so
\(d\Phi_p \circ d\iota_p = 0\); hence \(T_pS = \operatorname{im} d\iota_p \subseteq \ker d\Phi_p\).
Both sides have the same dimension: \(\dim T_pS = \dim S = \dim M - \dim N\), while
\(d\Phi_p\) is surjective (a defining map is a submersion), so by the
rank–nullity law
its kernel also has dimension \(\dim M - \dim N\). A subspace contained in another of equal
finite dimension fills it, so \(T_pS = \ker d\Phi_p\). The component form is the statement that
the kernel of \(d\Phi_p\) is the common zero set of the differentials \(d\Phi^i_p\).
\(\blacksquare\)
These descriptions also give a clean way to prove that a set is not a submanifold — a
question that is otherwise awkward, since one must rule out every possible topology and smooth
structure. The strategy is to assume the set is a submanifold and derive a contradiction from a
property every submanifold must have: at each point the tangent space is a subspace of fixed
dimension; every tangent vector is the velocity of a curve in the set; and every tangent vector
annihilates functions vanishing on the set. The graph of the absolute value function
\(\{(x, y) : y = |x|\}\) in the plane, for instance, cannot be a smooth submanifold: away from the
corner it is a one-dimensional submanifold, so it would have to be one-dimensional throughout, and
in particular at the corner its tangent space would be a line spanned by the velocity of some curve
running through it. But a smooth curve that stays in the set and passes through the corner is forced
to have zero velocity there — its two coordinate functions are tied by \(y = |x|\), which is not
differentiable at \(0\) unless the \(x\)-component has vanishing derivative — so no nonzero tangent
vector can arise, and a one-dimensional tangent space is impossible. The tangent space, made
ambient, is what supplies the contradiction.