Download 1. Manifolds with boundary Definition 1.1. An n

1. Manifolds with boundary
Definition 1.1. An n-dimensional manifold with boundary is a topological space X in which
each point x ∈ X has a neighborhood Ux homeomorphic to either Rn or to H n – the closed
upper half-space in Rn :
H n = {(x1 , ..., xn ) ∈ Rn | xn ≥ 0}
The set of all points x ∈ X which only have neighborhoods homeomorphic to H n constitute
the boundary of X. The boundary of X is denoted by ∂X. If this boundary is empty, then
X is simply an n-manifold (consequently, manifolds are also referred to as manifolds with
empty boundary).
Remark 1.2. The definition of the boundary of a manifold does not always coincide with
the definition of the boundary of a topological space. For example, the manifold X = D n =
{x ∈ Rn | x < 1} as a manifold has no boundary but as a subspace of Rn has boundary S n .
Examples 1. H n itself is a manifold with boundary. Its boundary is
∂H n = Rn−1 × {0} ∼
= Rn−1
n
2. The closed n-ball D n = {x ∈ Rn | |x| ≤ 1} is an n-manifold with boundary ∂D = S n−1 .
3. If X is an n-manifold (without boundary) and Y is an m-manifold with boundary then
X × Y in the product topology is an (m + n)-manifold with boundary
∂(X × Y ) = X × ∂Y
4. The intervals [0, 1 and 5, 6] are 1-manifolds with boundaries {1} and {5} respectively.
Proposition 1.3. If X is an n-manifold with boundary then ∂X (with the relative topology)
is an (n − 1)-manifold with empty boundary. In other words ∂∂X = ∅ for all manifolds X.
Proof. Let x ∈ ∂X, we need to show that in ∂X there is a neighborhood Vx of x homeomorphic to Rn−1 . Towards this goal, pick a neighborhood (Ux , ϕx ) of x where ϕx : Ux → H n is a
/ Rn−1 × {0} then there is some ε > 0 such that Bϕx (x) (ε) ⊆ H n .
homeomorphism. If ϕx (x) ∈
But then ϕ−1 (Bϕx (x) (ε)) is a neighborhood of x homeomorphic to Rn which is a contradiction
since we picked x ∈ ∂X. Thus we must have ϕx (x) = (y, 0) for some y ∈ Rn−1 .
Claim: ϕx (Ux ∩ ∂X) = Rn−1 × {0}.
/ Rn−1 × {0} we arrive at
Proof of claim If for some point z ∈ Ux ∩ ∂X we get ϕx (z) ∈
a contradiction in the same way we did with x above. Thus ϕx (Ux ∩ ∂X) ⊆ Rn−1 × {0}.
/ ∂X for some y ∈ Rn−1 . Then ϕ−1 (y, 0) has a
To see the converse, suppose that ϕ−1 (y, 0) ∈
neighborhood homeomorphic to Rn and contained in Ux . Since ϕx is a homeomorphism then
(y, 0) also has a neighborhood (in H n ) homeomorphic to Rn , this however is clearly not the
case. Therefore the claim is proved.
n−1
× {0}. As this is ∂X ∩ Ux it is open in
To finish the proposition, take Vx to be ϕ−1
x (R
n−1
∂X and ϕx |Vx : Vx → R
is a homeomorphism.
Since every point x ∈ ∂X has a neighborhood homeomorphic to Rn−1 , the boundary of
∂X is empty.
The next lemma is used in the proof of the gluing theorem below.
2
Lemma 1.4. Let X1 and X2 be two copies of the upper half space H n and let g : ∂X2 → ∂X1
be a homeomorphism. Then the quotient space X = (X1 ∪ X2 )/P associated to the partition
P = {{x, g(x)}, {y}, {z} | x ∈ ∂X2 , y ∈ X1 − ∂X1 , z ∈ X2 − ∂X2 }
is homeomorphic to Rn .
Proof. Let π : (X1 ∪ X2 ) → X be the quotient map. Given an x ∈ X1 ∪ X2 let us decompose
it as x = (y, t) with y ∈ Rn−1 and t ∈ R. Here t is the n-th coordinate of x, e.g. (y, t) ∈ ∂Xi
precisely when t = 0.
Consider the function ϑ : X → Rn defined by
⎧
; (y, t) ∈ X1
⎨ (y, −t)
ϑ(π(y, t)) =
⎩
(g(y), t)
; (y, t) ∈ X2
We claim this is a homeomorphism. First off, notice that it is well defined since if (y, 0) ∈ ∂X2
then π(y, 0) = π(g(y), 0). But for (g(y), 0) ∈ ∂X1 we also get ϑ(π(g(y), 0)) = (g(y), 0).
It is easy to see that ϑ is a bijection, its inverse if given by
⎧
; t≤0
⎨ π(y, −t)
−1
ϑ (y, t) =
⎩
; t≥0
π(g −1 (y), t)
Proving continuity of ϑ and ϑ−1 is left as an exercise.
Theorem 1.5 (The gluing theorem). Let X1 and X2 be two n-manifolds with boundary and
let h : ∂X2 → ∂X1 be a homeomorphism. Let X be the quotient space
X = (X1 ∪ X2 )/P
associated to the partition P of X1 ∪ X2 given by
P = {{x, h(x)}, {y}, {z} | x ∈ ∂X2 , y ∈ X1 − ∂X1 , z ∈ X2 − ∂X2 }
Then X is a n-manifold without boundary.
Proof. We need to show that each point p ∈ X has a neighborhood homeomorphic to Rn .
Let π : (X1 ∪ X2 ) → X be the quotient map. If p is of the form π(x) for x either in X1 − ∂X1
or X2 − ∂X2 , this is evident. So let’s consider the case of p = π(x) = π(h(x)) with x ∈ ∂X2 .
Let U be a neighborhood of h(x) in X1 and V a neighborhood of x in X2 each homeomorphic to H n via homeomorphisms ϕ : U → H n and ψ : V → H n . For convenience and
without loss of generality let’s assume that ψ(x) = 0 and ϕ(h(x)) = 0.
Pick an ε > 0 small enough to that the ball Ṽ = B0 (ε) satisfies the inclusion
h ◦ ψ −1 (B0 (ε) ∩ ∂H n ) ⊆ U
Such an ε exists since U is an open set. Let g : ∂H n → ∂H n be the homeomorphism
g = ϕ ◦ h ◦ ψ and let g̃ : H n → H n be the homeomorphism
g̃ : H n → H n : (y, t) → (g(y), t)
Note that g̃ is an extension of g. Define Ũ = g̃(Ṽ ). Then both Ũ and Ṽ are homoeomorphic
to H n and g : ∂ Ṽ → ∂ Ũ is a homeomorphism. According to lemma 1.4 the quotient space
(Ũ ∪ Ṽ )/P ∼
= Rn
3
where P is the partition
P = {{x, g(x)}, {y}, {z} x ∈ ∂ Ṽ , y ∈ Ũ − ∂ Ũ , z ∈ Ṽ − ∂ Ṽ }
Let us denote the quotient map by π : Ũ ∪ Ṽ → Rn .
To find the desired chart (W, φ) around p, set W = π ϕ−1 (Ũ ) ∪ ψ −1 (Ṽ ) and define
φ : W → Rn to be
π ◦ ϕ(x)
; x ∈ ϕ−1 (Ũ)
φ(π(x)) =
; x ∈ ψ −1 (Ṽ )
π ◦ ψ(x)
It is straightforward to check that φ is a homeomorphism, this is left as an exercise.
h
X1
U
h(x)
V
x
ϕ
X2
ψ
g̃
g
Ũ
Ṽ
Hn
Hn
π
φ(W )
Rn
Figure 1. A visual presentation of the construction of the chart (W, φ).