ト ポロジ ー II [講義録] 2010 年版 An Introduction to Algebraic Topology
Simplicial Homology
— A Short Course
Katsuro Sakai
Institute of Mathematics University of Tsukuba
代数的ト ポロジ ー入門
単体ホモロジ ー
短期課程
筑波大学・数学系
酒 井 克 郎
Preface
In Algebraic Topology, topological properties of topological spaces and con- tinuous maps are studied by using algebraic methods (e.g., groups, homo- morphisms, etc.). Homology Theory and Homotopy Theory are typical and classical. There are several Homology Theories. Among them, Simplicial Homology is most elementary and geometrical. As an introduction to Al- gebraic Topology, this course is provided. Our purpose is knowing how to associate Topology (spaces) with Algebra (groups) by learning the process of defining the homology, and how to use Algebra in Topology by applying the homology. The reader is required to familiar with basic language of sets and maps, elementary point-set topology of Euclidean spaces, linear algebra and elementary group theory.
To prepare this notes, the author was influenced from the book “Topol- ogy” (Japanese) by Ichiro Tamura (Iwanami Shoten Publ., 1972). In partic- ular, the author owes the use of Mayer-Vietoris’ sequence for the proof of the topological invariance of homology to Tamura’s book. The first draft was written in 2001 based on the author’s lecture notes for Topology I, which he had lectured several times the class of Mathematics Majors in the University of Tsukuba. The author is indebted to his colleague Kazuhiro Kawamura and his student Masato Yaguchi for their helpful suggestions on the first draft of the manuscript. This has been used as a text in the class of Topology I for Mathematics Majors in the University of Tsukuba, 2001, 2005, 2009. This has been revised and improved after the lectures.
Tsukuba, March, 2010 Katsuro Sakai
2010, K. Sakaic
Contents
1 Convex Sets and Simplexes . . . 4
2 Simplicial Complexes and Polyhedra . . . 12
3 Simplicial Maps and Approximations . . . 28
4 Homology of an Oriented Complex . . . 37
5 Homomorphisms and Simplicial Maps . . . 50
6 Mayer-Vietoris Exact Sequences and the Subdivision Operator . . . 57
7 Topological Invariance of Homology . . . 65
8 Homotopical Invariance of Homology and Homologies of Surfaces . . . 72
9 Relative Homology for Polyhedral Pairs . . . 79
Simplicial Homology
— A Short Course
In this course, we study Simplicial Homology, which is a sequence of addi- tive groups associated with a topological space, so called a polyhedron. A polyhedron is constructed by small pieces, so called simplexes. The totality of assembled simplexes constructing a polyhedron is called a simplicial com- plex, which reveals the geometrical structure of the space. These materials are first studied, and then we see how to define the homology group for a sim- plicial complex. As the most important fact, it is shown that homeomorphic polyhedra have isomorphic homologies. Some applications of homologies are given. The homologies of several surfaces are also calculated.
Notations
We use the following notations:
• N = 1, 2, . . . — the set of natural numbers; 自然数全体
• Z = 0,±1, ±2, . . . — the set of integers; 整数全体
• Z+ = N∪ {0} = 0, 1, 2, . . .
— the set of non-negative integers; 非負整数全体
• R = (−∞, ∞) — the real line; (実) 数直線
• R+ = [0,∞) — the non-negative half line; 半数直線
• I = [0, 1] — the unit closed interval; 単位閉区間
• Xn = X × · · · × X
n times
; ∀x ∈ Xn, x = (x(1), . . . , x(n)),
x(i)∈ X — the i-th coordinate of x, i 座標 pri : Xn→ X (pri(x) = x(i)) — the projection; 射影
• Rk — the k-dimensional Euclidean space, k 次元
ユークリッド 空間
∀x ∈ Rk, x = ki=1x(i)2 — (Euclidean) norm, ノル ム
∀x, y ∈ Rk, ∀t ∈ R,
x + y = (x(1) + y(1), . . . , x(k) + y(k)) — sum, 和 tx = (tx(1), . . . , tx(k)) — scalar product, スカラー倍 ei ∈ Rk (i = 1, . . . , k) defined by
ei(j) =
1 if i = j, 0 if i = j;
• Bk =
def
x∈ Rk x ≤ 1 — the unit k-ball; k 次元
単位球(体)
• Sk−1 = ∂Bk =
def
x∈ Rk x = 1 — the unit (k− 1)-sphere; k − 1 次元
単位球面
• ∆k−1 =
def
x∈ Rk ki=1x(i) = 1, x(i)≥ 0 (1 ≤ i ≤ k)
— the standard (k− 1)-simplex; k − 1 次元
標準単体
•
◦
∆k−1 =
def
x∈ Rk ki=1x(i) = 1, x(i) > 0 (1 ≤ i ≤ k)
— the standard open (k− 1)-simplex. k − 1 次元
標準開単体
∆1 R2
e1
e2 e3
e1
R3
e2
∆2
In this course, for beginners, spaces are considered as subspaces of Eu- clidean space Rk.
• X ≈ Y — X is homeomorphic to Y ; 同相
• idX = id — the identity map of X; 恒等写像
• intXA = int A — the interior of A in X; 内部
bdXA = bd A — the boundary of A in X; 境界
clXA = cl A — the closure of A in X; 閉包
• diam A =
defsup{x − y | x, y ∈ A}
— the diameter of A; 直径
• dist(A, B) =
definf{x − y | x ∈ A, y ∈ B}.
— the distance between A and B 距離
A pair of spaces means a pair (X, A) of a space X and its subspace A. 空間の対 A map (or a homeomorphism) f : (X, A) → (Y, B) from a pair (X, A) of 対の間の写像 spaces to another pair (Y, B) means a map (or a homeomorphism) f : X → Y 同相写像
with f (A) ⊂ B (or f(A) = B), whence f|A : A → B is also a map (or a homeomorphism). When there is a homeomorphism f : (X, A)→ (Y, B), we
say that (X, A) is homeomorphic to (Y, B) and denote (X, A)≈ (Y, B). 空間の対の同相
1 Convex Sets and Simplexes
1.1 Definition We call C ⊂ Rk a convex set if 凸集合
∀x, y ∈ C, ∀t ∈ [0, 1], tx + (1 − t)y ∈ C. This condition is equivalent to the following:
∀x0, x1, . . . , xn∈ C, ∀z ∈ ∆n, n
i=0
z(i + 1)xi ∈ C.
1.1 Exercise – Show the above equivalence. 演習
1.2 Example Every linear subspace of Rk, Bk and ∆k−1 are convex. For any A ⊂ Sk−1 = ∂Bk, Bk\ A is convex.
1.2 Exercise – Show that if C ⊂ Rk is convex then cl C and 演習 int C are also convex.
1.3 Definition The convex hull A of A ⊂ Rk is defined by 凸包 A =
def
C C : convex in Rk, A⊂ C,
which is the smallest convex set containing A. For a finite set, we simply denote
{x0, x1, . . . , xn} = x0, x1, . . . , xn. Note that ∆n−1 = e1, . . . , en ⊂ Rn.
1.3 Exercise – For A⊂ Rk, prove the following: 演習 A = ni=0z(i + 1)xi
n ∈ N, xi ∈ A, z ∈ ∆n.
1.4 Exercise – For A⊂ Rk, prove the following: 演習 cl A = C C: closed convex in Rk, A⊂ C.
1.4 Definition We say that x0, x1, . . . , xn∈ Rk are affinely independent
(or geometrically independent) if アフ ィン 独立
n i=0
tixi = 0, n
i=0
ti = 0⇒ t0 = t1 =· · · = tn = 0,
that is, x1− x0, . . . , xn− x0 are linearly independent, hence n k. It is 線形(１次) 独立
said that A⊂ Rk is in general position if any k + 1 points of A are affinely 一般の位置 independent.
Remark. Let A = {x0, x1, . . . , xn} ⊂ Rk and n k. Then, A is in general position if and only if x0, x1, . . . , xn are affinely independent.
1.5 Exercise – Prove that if v0, v1, . . . , vn ∈ Rk are affinely in- 演習 dependent and n < k, there exist k− n many vn+1, . . . , vk ∈ Rk
such that v0, v1, . . . , vk are affinely independent.
1.5 Definition We call σ ⊂ Rk an (n-dimensional) simplex or an n- (n 次元) 単体
simplex if there exist some affinely independent v0, v1, . . . , vn ∈ Rk such n 単体 that
σ = v0, v1, . . . , vn = ni=0z(i + 1)vi
z ∈ ∆n,
where v0, v1, . . . , vnare called vertices of σ and n = dim σ is the dimension 頂点 of σ. We denote σ(0) ={v0, v1, . . . , vn}. (単体の) 次元
1.6 Exercise – In the above, prove that γ : ∆n→ σ defined by 演習 γ(z) = ni=0z(i + 1)vi is a homeomorphism. In particular, the
representation ni=0z(i + 1)vi of a point in σ is unique.
1.6 Definition When a point in an n-simplex σ = v0, v1, . . . , vn is repre- sented as x = ni=0z(i + 1)vi, we call z = (z(1), . . . , z(n + 1)) ∈ ∆n the
barycentric coordinate of x with respect to v0, v1, . . . , vn. 重心座標 1.7 Proposition If v0, v1, . . . , vn ∈ Rk and u0, u1, . . . , um ∈ Rk are affinely
independent, then
v0, v1, . . . , vn = u0, u1, . . . , um ⇒ {v0, v1, . . . , vn} = {u0, u1, . . . , um}. Hence, for each simplex σ ⊂ Rk, σ(0) and dim σ are uniquely determined. Proof. Assume that {v0, v1, . . . , vn} = {u0, u1, . . . , um}. Then, without loss of generality, we can assume that v0 = uj for any j = 0, 1, . . . , m. When we represent
v0 = m
j=0
z(j + 1)uj, z∈ ∆m,
there are distinct j1 = j2 such that z(j1 + 1), z(j2 + 1) ∈ (0, 1) because z(j + 1) = 1 (i.e., v0 = uj) for all j = 0, 1, . . . , m. Choose ε > 0 so that
0 < z(j1+ 1)± ε < 1, 0 < z(j2+ 1)± ε < 1,
and let
x′ =
j=j1,j2
z(j + 1)uj + (z(j1+ 1)− ε)uj1 + (z(j2+ 1) + ε)uj2,
x′′ =
j=j1,j2
z(j + 1)uj + (z(j1+ 1) + ε)uj1 + (z(j2+ 1)− ε)uj2.
Then, x′ = x′′ (cf. Exercise 1.6). Since x′, x′′∈ v0, v1, . . . , vn, we can write
x′ = n
i=0
z′(i + 1)vi, x′′ = n
i=0
z′′(i + 1)vi, z′, z′′∈ ∆n.
It follows that v0 = 1
2x
′+ 1
2x
′′ =
n i=0
1 2z
′(i + 1) + 1
2z
′′(i + 1)
vi,
which implies 12z′(1) +12z′′(1) = 1 (cf. Exercise 1.6). Since z′(1), z′′(1)∈ I, it follows that z′(1) = z′′(1) = 1, which means x′ = x′′= v0 — a contradiction! Thus we have the result.
1.8 Proposition Let σ be an n-simplex in Rk. If σ = v0, v1, . . . , vn (where n = dim σ), then v0, v1, . . . , vnare affinely independent. Consequently, σ(0) = {v0, v1, . . . , vn}.
Proof. To see that v1− v0, . . . , vn− v0 are linearly independent, let L = ni=1ti(vi− v0) ti ∈ R,
which is the linear subspace of Rk spanned by v1 − v0, . . . , vn − v0. Note that σ− x ∈ L for each x ∈ σ. It suffices to show that dim L = n. By the definition of an n-simplex, we have affinely independent u0, u1, . . . , un ∈ Rk such that σ = u0, u1, . . . , un, whence
u1− u0, . . . , un− u0 ∈ σ − u0 ⊂ L
and u1 − u0, . . . , un− u0 are linearly independent. For each j = 0, . . . , n, vj = ni=0zj(i + 1)ui for some zj ∈ ∆n. Note that zj(1) = 1− ni=1zj(i + 1).
For every j = 0, vj − v0 =
n i=0
zj(i + 1)ui− n
i=0
z0(i + 1)ui
= n
i=1
(zj(i + 1)− z0(i + 1))ui+ (zj(1)− z0(1))u0
= n
i=1
(zj(i + 1)− z0(i + 1))ui− n
i=1
(zj(i + 1)− z0(i + 1))u0
= n
i=1
(zj(i + 1)− z0(i + 1))(ui− u0).
Thus, L is spanned by u1− u0, . . . , un− u0, hence {u1 − u0, . . . , un− u0} is a base for L, which implies dim L = n.
1.9 Definition For an n-simplex σ = v0, v1, . . . , vn ⊂ Rk, we define
• σ =◦
def
n
i=0z(i + 1)vi
z ∈ ∆n, ∀i = 1, . . . , n + 1, z(i) > 0
= ni=0z(i + 1)vi
z ∈∆◦n — the interior of σ; (単体の) 内部
• ∂σ =
def
n
i=0z(i + 1)vi
z ∈ ∆n, ∃i = 1, . . . , n + 1, z(i) = 0
= σ\σ — the boundary of σ;◦ (単体の) 境界
• b(σ) =
def
n i=0
1
n + 1vi — the barycenter of σ (where b(σ)∈σ).◦ 重心 Remark. (1) When dim σ = 0, we have σ = σ and ∂σ =◦ ∅. We identify a 0-simplex v0 = {v0} with the point v0.
(2) The barycentric coordinate of x ∈ σ is uniquely determined with respect to the order v0, v1, . . . , vn (cf. Exercise 1.6).
1.7 Exercise – For each k-simplex σ in Rk, show that intÊk σ = 演習 σ and bd◦ Ê
kσ = ∂σ. Note that if dim σ = k then intÊkσ = ∅, hence σ◦ = intÊkσ.
For every n-simplex σ, it can be shown that (σ, ∂σ)≈ (Bn, Sn−1) (Theo- rem 1.16).
1.10 Proposition For an n-simplex σ = v0, v1, . . . , vn,
diam σ = maxvi − vj i, j = 0, 1, . . . , n.
Proof. For each x, y ∈ σ, it suffices to find i1, i2such thatx−y vi1−vi2. First, choose i1 so thatx − vi1 = maxix − vi. Let
y = n
i=0
z(i + 1)vi ∈ σ, z ∈ ∆n. Then, it follows that
x − y =
x− n
i=0
z(i + 1)vi
n
i=0
z(i + 1)x − vi
n
i=0
z(i + 1)x − vi1 = x − vi1.
Next, choose i2 so that vi1 − vi2 = maxivi1 − vi, whence vi1 − x
vi1− vi2 by the same argument. Consequently, x − y vi1− vi2.
1.11 Definition A simplex τ is called a face of a simplex σ if τ(0) ⊂ σ(0), 辺, 面 whence we denote τ σ. An n-dimensional face is simply called an n-face
(0-face = vertex). When τ σ and τ = σ, we call τ a proper face of σ,
and denote τ < σ. 真の辺, 面
1.8 Exercise – For each k-simplex σ in Rk, show ∂σ =τ <στ . 演習
Supplement
1.12 Definition For a compact (= bounded closed) convex set A ⊂ Rk with 0∈ int A, we define pA: Rk→ R+ as follow:
pA(x) = infs > 0 s−1x∈ A= infs > 0 x ∈ sA,
which is called the Minkowski functional. ミンコフスキー
汎関数
0
x A
tx
sA s−1x
1.13 Proposition For a compact (= bounded closed) convex set A ⊂ Rk with 0∈ int A, the Minkowski functional pA: Rk → R+ is well-defined and continuous, and it has the following properties:
(1) pA(x) = 0⇔ x = 0;
(2) pA(x)−1x∈ A for every x ∈ Rk\ {0}; (3) pA(tx) = tpA(x) for each x∈ Rk and t 0; (4) A ={x ∈ Rk| pA(x) 1};
(5) int A ={x ∈ Rk| pA(x) < 1}, hence bd A = {x ∈ Rk| pA(x) = 1}.
Proof. (Well-definedness) Since 0 ∈ int A, we have δ0 > 0 such that x < δ0 implies x∈ A. For each x ∈ Rk, t−1x∈ A if t > x/δ0. Thus, pA is well-defined. (1) It is obvious that pA(0) = 0. Assume x= 0. Since A is compact (bounded), tx∈ A for sufficiently large t > 0, i.e., s−1x∈ A for sufficiently small s > 0. Then pA(x)= 0.
(2) We have t1 > t2 > · · · > 0 such that t−1i x ∈ A and ti → pA(x) (i → ∞). Note that pA(x)= 0 by (1). Since A is closed in Rk, it follows that pA(x)−1x∈ A, (3) The case t = 0 or x = 0 follows from (1). When t > 0 and x = 0, since (tpA(x))−1tx = pA(x)−1x ∈ A by (2), we have pA(tx) tpA(x) by the
definition. Moreover, since pA(tx)−1tx∈ A, it also follows from the definition that pA(x) pA(tx)/t, hence tpA(x) pA(tx).
(4) By the definition, pA(x) 1 for every x ∈ A. Since 0 ∈ A, it suffices to show that 0 < pA(x) 1 implies x ∈ A. Since pA(x)−1x ∈ A by (2), it follows from the convexity of A that
x = (1− pA(x))0 + pA(x)(pA(x)−1x)∈ A. (5) If pA(x) = 1 then
n + 1 n x =
n
n + 1
−1
x ∈ A and x = limn→∞n + 1n x,
which means x∈ int A. Hence, pA(x) < 1 for every x∈ int A. Conversely, assume pA(x) < 1. Since 0 ∈ int A, we have δ0 > 0 such thatz < δ0 implies z∈ A. If
x − y < (1 − pA(x))δ0, then
y = (1− pA(x))(1− pA(x))−1(y− x) + pA(x)pA(x)−1x∈ A because (1− pA(x))−1(y− x) ∈ A and pA(x)−1x∈ A. Therefore, x ∈ int A.
(Continuity) If pA(x) > a > 0 then a−1x ∈ Rk\ A by the definition. Since Rk\ A is open in Rk, we have δ1 > 0 such thatx − y < δ1 implies a−1y ∈ Rk\ A, hence a−1pA(y) = pA(a−1y) > 1 by (3) and (4), so pA(y) > a. Thus, pA is lower semi-continuous. If pA(x) < b then pA(b−1x) = b−1pA(x) < 1 by (3), which means b−1x ∈ int A by (5). Choose δ2 > 0 so that x − y < δ2 implies b−1y ∈ int A, hence b−1pA(y) = pA(b−1y) < 1 by (3) and (5), so pA(y) < b. Thus, pA is upper semi-continuous.
1.9 Exercise – Show that the Minkowski functional pA in the above 演習 has also the following properties:
(1) pA(x + y) pA(x) + pA(y) for every x, y∈ Rk; (2) pAis a norm on Rk ⇔ A = −A.
1.14 Definition A convex set A⊂ Rkwith intÊ
kA= ∅ is called a convex body.
凸体
1.15 Theorem Every compact convex body A⊂ Rk is homeomorphic to Bk. In fact, there exists a homeomorphism ϕ : Rk → Rk such that ϕ(A) = Bk, hence (A, bdÊ
kA)≈ (Bk, Sk−1).
Proof. Without loss of generality, we may assume that 0 ∈ int A. By using the Minkowski functional pAfor A, we define maps ϕ, ψ : Rk→ Rk as follows:
ϕ(x) =
0 if x = 0,
pA(x)−1xx if x = 0; ψ(x) =
0 if x = 0,
x−1pA(x)x if x= 0.
0
A
0
Bk ϕ
ψ
Since pAis continuous, these are continuous at x∈ Rk\ {0}. To see the continuity at 0, assume xi ∈ Rk\ {0} and xi → 0 (i → ∞). Since pA(xi)−1xi ∈ A, it follows that pA(xi)−1xi < diam A, whence
ϕ(xi) = pA(xi)−1xi2 xi diam A → 0 (i → ∞).
Therefore, ϕ is continuous at 0. On the other hand, by the continuity of pA,
ψ(xi) = pA(xi)→ pA(0) = 0 (i→ ∞), hence ψ is also continuous at 0.
For each x∈ Rk\ {0}, observe that
ϕ(x) = pA(x)−1x2, pA(ϕ(x)) = pA(x)−1xpA(x) =x,
ψ(x) = pA(x) and pA(ψ(x)) =x−1pA(x)2,
hence it follows that
ψϕ(x) =ϕ(x)−1pA(ϕ(x))ϕ(x) = pA(x)x−1ϕ(x) = x and ϕ(ψ(x)) = pA(ψ(x))−1ψ(x)ψ(x) = xpA(x)−1ψ(x) = x
Then, ϕψ = id and ψϕ = id, that is, ϕ is a homeomorphism with ϕ−1 = ψ. Observe
ϕ(Bk) ={ϕ(x) | x 1} = {y ∈ Rk| ψ(y) 1}
={y ∈ Rk | pA(y) 1} = A and
ϕ(Sk−1) ={ϕ(x) | x = 1} = {y ∈ Rk| ψ(y) = 1}
={y ∈ Rk | pA(y) = 1} = bdÊ
kA.
Thus we have the result.
Note. In connection with the above, if Bk ≈ A ⊂ Rk then int A = ∅. This follows from the following:
Brouwer’s Theorem on Invariance of Domain: For each X, Y ⊂ Rk, ブ ラウワーの
領域不変定理
X ≈ Y implies int X ≈ int Y .
In this theorem, one should remark that X ≈ Y is not given by any homeo- morphism of Rk onto itself.
1.16 Theorem For every n-simplex σ, (σ, ∂σ)≈ (Bn, Sn−1). Proof. Let
∆n0 =x∈ Rn ni=1x(i) 1, x(i) 0 (i = 1, . . . , n)
= 0, e1, . . . , en ⊂ Rn. Then, ∆n0 is a compact convex set in Rn and
intÊ
n∆n0 =
◦
∆n0 =x∈ Rn ni=1x(i) < 1, x(i) > 0 (i = 1, . . . , n)= ∅. Therefore, (∆n0, ∂∆n0) = (∆n0, bdÊ
n∆n0)≈ (Bn, Sn−1) by Theorem 1.15.
For each n-simplex σ = v0, v1, . . . , vn ⊂ Rk, v1− v0, . . . , vn− v0 are linearly independent, hence we can define a linear injection ϕ : Rn→ Rk by ϕ(ei) = vi−v0 for each i = 1, . . . , n. Let h : Rn → Rk be the embedding defined by h(x) = ϕ(x) + v0. Observe that h(∆n0) = σ and h(∂∆n0) = ∂σ. Then, it follows that (σ, ∂σ)≈ (∆n0, ∂∆n0)≈ (Bn, Sn−1).
2 Simplicial Complexes and Polyhedra
2.1 Definition A finite collection K of simplexes in Rkis called a simplicial
complex if the following are satisfied: 単体複体
(K1) σ ∈ K, τ σ ⇒ τ ∈ K,
(K2) σ, τ ∈ K, σ ∩ τ = ∅ ⇒ σ ∩ τ σ, τ. For a simplicial complex K, we define
• dim K =
defmax{dim σ | σ ∈ K} — the dimension of K (単体複体の) 次元 (K is n-dimensional if dim K = n);
• |K| =
def
K =
σ∈K
σ =
σ∈K
σ◦ ⊂ Rk — the polyhedron of K; 多面体
• K(n) =
def{σ ∈ K | dim σ n} — the n-skeleton of K n 次元骨格 (K(0) =σ∈Kσ(0) is the set of vertices of K);
• mesh K =
defmax{diam σ | σ ∈ K} — the mesh of K. (単体複体の)
編み目
Remark. In the above, K = K(n) if and only if dim K = n. The polyhedron
|K| is a compact set in Rk because of the finite union of simplexes. The name “mesh” comes from the following fact (cf. Proposition 1.10):
mesh K = mesh K(1) = maxv0− v1 v0, v1 ∈ K.
Note. In this lecture, we treat only finite simplicial complexes. However,
infinite simplicial complexes are useful and have many applications. For an 無限単体複体 infinite simplicial complex K, the local finiteness is usually assumed because
the topology of|K| is unique, where K is said to be locally finite if each 局所有限 v ∈ K(0) is the vertex of only finitely many simplexes of K. Replacing
Rk by infinite-dimensional linear topological space such as Hilbert space
ℓ2, we can also consider infinite-dimensional simplicial complexes. In this 無限次元単体複体 case, the local finiteness is no longer assumed. If K is locally finite, then
|K| is locally compact and locally finite-dimensional (i.e., each point has 局所有限次元 a finite-dimensional neighborhood). There are infinite-dimensional locally
finite simplicial complexes. When K is not locally finite, the polyhedron|K| has two typical topologies, the metric topology and the weak (Whitehead) topology. Non-locally finite simplicial complexes are also very useful.
2.2 Definition For an n-simplex σ, we define
• K(σ) =
def
τ τ σ — the simplicial complex of σ, where |K(σ)| = σ and K(σ)(0) = σ(0);
• K(∂σ) =
def
τ τ < σ = K(σ)(n−1) — the simplicial complex of ∂σ, whence |K(∂σ)| = ∂σ.
2.3 Definition Let K be a simplicial complex. A simplicial complex L is
called a subcomplex of K if L⊂ K. In other words, L ⊂ K is a subcomplex 部分複体 of K if L satisfies (K1) (i.e.,∀σ ∈ L, K(σ) ⊂ L). A pair (K, L) of simplicial
complexes means a pair of a simplicial complex K and its subcomplex L. 単体複体の対 Remark. Let L be a subcomplex of K. Then |L| ⊂ |K| and dim L dim K.
Every subcomplex of L is also a subcomplex of K. 2.4 Example Let K be a simplicial complex.
(1) Each n-skeleton K(n) of K is a subcomplex of K. (2) For each σ ∈ K, K(σ) is a subcomplex of K.
2.5 Definition For a simplicial complex K and σ ∈ K, the star St(σ, K) 星状体 and the link Lk(σ, K) of σ in K are defined as follows: 環状体
• St(σ, K) =
def
τ ∈ K ∃τ′ ∈ K such that τ τ′, σ τ′,
• Lk(σ, K) =
def
τ ∈ St(σ, K) τ ∩ σ = ∅.
σ
Lk(σ, K)
St(σ, K) K
For each point x ∈ |K|, we define the star St(x, K) and the link Lk(x, K) at x in K as follows:
• St(x, K) =
def
τ ∈ K ∃τ′ ∈ K such that τ τ′, x∈ τ′,
• Lk(x, K) =
def
τ ∈ St(x, K) x ∈ τ.
The carrier of x in K is the simplex σ∈ K with x ∈σ.◦ 台
2.1 Exercise – (1) For each x ∈ |K|, show that | St(x, K)| is a 演習 neighborhood of x in |K|.
(2) Give an example such that | Lk(x, K)| = bd|K|| St(x, K)|. (3) Show | St(x, K)| = {(1 − t)x + ty | y ∈ | Lk(x, K)|, t ∈ I} and
| St(σ, K)| = {(1 − t)x + ty | x ∈ σ, y ∈ | Lk(x, K)|, t ∈ I}. (4) Prove that if x ∈ σ then St(x, K) = St(σ, K). In particular,◦ St(σ, K) = St(b(σ), K) for every σ ∈ K. Moreover, show that Lk(x, K)= Lk(σ, K) if σ = {x}.
2.6 Proposition Let K be a simplicial complex and σ, τ ∈ K. (1) σ◦ ∩ τ = ∅ ⇔ σ τ.
(2) σ◦ ∩τ◦ = ∅ ⇔ σ = τ.
Proof. In both (1) and (2), the implication “⇐” is trivial. The converse implication “⇒” can be seen as follows:
(1) Assume that σ◦ ∩ τ = ∅. Recall σ ∩ τ σ. If σ has a vertex other than vertices of σ ∩ τ, then any point of σ cannot belong to τ , which is a◦ contradiction. Thus, σ = σ∩ τ τ.
(2) Ifσ◦ ∩τ◦ = ∅, then σ τ and τ σ by (1), which implies σ = τ. 2.7 Definition For a simplicial complex K and v ∈ K(0), we define
• OK(v) =
def| St(v, K)| \ | Lk(v, K)| = |K| \{σ ∈ K | v ∈ σ(0)}
={σ◦ | σ ∈ K, v ∈ σ(0)} — the open star at v. 開星状体 Remark. The open star OK(v) is an open neighborhood of v in |K|, and
OK(v) v ∈ K(0) is an open cover of |K|.
2.8 Lemma Let v0, . . . , vn ∈ K(0). Then, v0, . . . , vn ∈ K if and only if OK(v0)∩ · · · ∩ OK(vn)= ∅.
Proof. If σ = v0, . . . , vn ∈ K, then σ◦ ⊂ OK(vi) for all vi, hence OK(v0)∩ · · · ∩ OK(vn)⊃σ◦ = ∅.
Conversely, assume that OK(v0)∩ · · · ∩ OK(vn) contains some x. Let σ ∈ K be the carrier of x, i.e., x∈σ. Then, each v◦ i is a vertex of σ, which implies v0, . . . , vn σ. Hence, v0, . . . , vn ∈ K.
2.9 Definition A topological space X is a (topological) polyhedron if (位相的) 多面体 X ≈ |K| for some simplicial complex K, that is, there exists a homeomor-
phism h :|K| → X, whence (K, h) is called a triangulation of X. 単体分割
三角形分割
A pair (X, Y ) of spaces is polyhedral if (X, Y ) ≈ (|K|, |L|) for some
多面体の対
pair (K, L) of simplicial complexes, that is, there exists a homeomorphism h : |K| → X such that h(|L|) = Y , whence (K, L, h) is called a relative
triangulation of a pair (X, Y ). 相対単体分割
相対三角形分割
Remark. A polyhedron is compact and metrizable.
Note. Recall that only finite simplicial complexes are treated in this lec- ture. In case locally finite simplicial complexes are considered, polyhedra are locally compact.
2.10 Example For an n-simplex σ, there exists a homeomorphism h : (σ, ∂σ)→ (Bn, Sn−1) by Theorem 1.16. Then, (K(σ), K(∂σ), h) is a relative triangula- tion of (Bn, Sn−1), so (K(σ), h) and (K(∂σ), h|∂σ) are triangulations of Bn and Sn−1, respectively.
2.11 Example In the following, we illustrate triangulations of typical sur- faces by figures, where surfaces are spaces which are locally homeomorphic 曲面 to the Eucliean plane R2, that is, each point has an open neighborhood home- omorphic to an open set in R2. A surface is also called a 2-dimensional
manifold. An n-dimensional manifold is defined as a space which is 多様体 locally homeomorphic to the n-dimensional Euclidean space Rn.
Surface
≈
R2
(1) Sphere S2 球面
≈ cut and unfold
≈
cut and unfold
A triangulation of S2
A triangulation of S2 v
v
v v
◦
◦
◦
cut and unfold
≈ v
v u
u
u
v w
w
A triangulation of S2 u
v w u w
v w w
Not a simplicial complex
(2) Torus T2 ト ーラス
cut and unfold
A triangulation of T2
paste
paste
(3) Klein bottle K2 クラ インの壷
cut and unfold
A triangulation of K2
paste paste
(4) Projective plane P2 = (R3\ {0})/ ∼ = {[x] | x ∈ R3\ {0}}, where x∼ y ⇐⇒
def ∃t ∈ R \ {0} such that tx = y ; [x] =
def(R\ {0})x
射影平面
P2 = S2/∼, where x ∼ y ⇐⇒
def x =±y
◦
◦0 x2
x3
x1
x3 = 1 [x]
x
R3
◦
◦0 x2
x3
x1
x3 = 1 [x]
x
R3
1 1
1 1 v = [e1]
paste fold
v v
v
v v v
≈
v
v
Not a triangulation P2 A triangulation P2
◦
◦
v
x,−x ∈ S1 are identified
v
v u
u
w w
(5) M¨obius band M2 メビ ウスの帯
cut and unfold
A triangulation of M2
M¨obius band is contained in the projective plane.
A triangulation of P2
cut
B2 paste
B2−
B2+ ℓ1
ℓ1
ℓ1
ℓ2 ℓ2
ℓ1 ℓ2
ℓ2
ℓ1 ℓ2
v
v u
u
u v
v u
M2
M2
B2+
B2−
B2− B2+
Remark. (1) The projective plane P2 can be obtained by attaching M¨obius band M2 with the disk B2 along their boundary circles.
P2 = M2∪S1 B2
◦
◦
◦
◦
paste cut and unfold
S1
B2
paste
S1 S1
paste ◦
S1
S1
B2
B2 M2 The cross cap
M¨obius band is also called the cross cap.
Remark. (2) Klein bottle K2 can be obtained by attaching two copies of M¨obius band M2 along their boundary circles. In other wards, K2 can be obtained by attaching two cross caps arong their boundary circles.
K2 = M2∪S1 M2
cut and paste
ℓ1 ℓ1
ℓ1 ℓ1
ℓ2
ℓ2 ℓ2
ℓ2
M2
M2 M+
M+ M−
M−
M2
The under half of Klein bottle M¨obius band
≈
2.12 Definition Let K and K′ be simplicial complexes. It is said that K′ is a subdivision of K or K′ subdivides K if |K′| = |K| and each τ ∈ K′ 細分 is contained in some σ ∈ K, whence we denote K′ ⊳K.
Note. In the above, if infinite simplicial complexes are considered, it is re- quired that each σ∈ K contains only finitely many simplexes of K′.
Remark. If K′ ⊳K and K′′⊳K′, then K′′ ⊳K.
2.2 Exercise – Let L be a subcomplex of K and K′a subdivision 演習 of K. Show that L′ ={τ ∈ K′ | τ ⊂ |L|} is a simplicial complex
which subdivides L.
2.13 Lemma Let σ0 < σ1 < · · · < σn be a sequence of simplexes and ui ∈ σ◦i (i = 0, 1, . . . , n). Then, u0, u1, . . . , un are affinely independent, whence u0, u1, . . . , un is an n-simplex. In particular, b(σ0), b(σ1), . . . , b(σn) is an n-simplex
Proof. By inserting faces of deficient dimensions, it can be assumed that dim σi = i. Then, we can write σi = v0, . . . , vi and
ui = i j=0
zi(j + 1)vj, zi ∈
◦
∆i.
To see that u0, u1, . . . , un are affinely independent, let ni=0tiui = 0 and n
i=0ti = 0. Observe that n
i=0
tiui = n
i=0
ti
i j=0
zi(j + 1)vj = n
i=0
i j=0
tizi(j + 1)vj
=
j i
tizi(j + 1)vj = n
j=0
n i=j
tizi(j + 1)vj.
Since v0, . . . , vn are affinely independent and n
j=0
n i=j
tizi(j + 1) = n
i=0
ti
i j=0
zi(j + 1) = n
i=0
ti = 0,
it follows that ni=jtizi(j + 1) = 0 for each j = 0, . . . , n. Since zi(j + 1)= 0 for each i and j, it can be seen that tn = tn−1 =· · · = t0 = 0 by downward induction.
2.14 Definition For a simplicial complex K, we define
• Sd K =
def
b(σ0), . . . , b(σn) σ 0 < σ1 <· · · < σn∈ K
— the barycentric subdivision of K. 重心細分
The n-th barycentric subdivision is inductively defined as follows: n 階重心細分
SdnK = Sd(Sdn−1K), where Sd0K = K.
Remark. In the above, by replacing the barycenters b(σ) with any points vσ ∈
σ, we can define a derived subdivision of K. The n-th derived◦ 導細分
subdivision is also defined by induction. n 階導細分
2.3 Exercise – (1) Show that Sd K is a simplicial complex and 演習 Sd K ⊳ K.
(2) Show that if L is a subcomplex of K then Sd L is a subcomplex of Sd K.
σ2
b(σ2)
b(σ1) σ1
σ0 = b((σ0) b(σ0), b(σ1), b(σ2)
2.15 Proposition For an m-dimensional simplicial complex K, mesh Sd K m
m + 1mesh K. Hence, limn→∞mesh SdnK = 0.
Proof. It suffices to show that, for each σ < τ ∈ K,
b(σ) − b(τ) m
m + 1mesh K
(see the remark for Definition 2.1). We write τ = v0, v1, . . . , vn. Choose j so that b(τ) − vj = maxib(τ) − vi. Since b(σ) ∈ σ ⊂ τ, it follows from the argument in the proof of 1.13 that b(σ) − b(τ) b(τ) − vj. Then,
b(τ) − vj =
n
i=0
1
n + 1vi− vj
n
i=0
1
n + 1vi− vj =
i=j
1
n + 1vi− vj
n
n + 1diam τ m
m + 1mesh K. Thus, we have the result.
Note. We can prove the following theorem, but more preparations are nec- essary and it won’t be used in this lecture, so the proof is omitted.
Theorem (J.H.C. Whitehead): Let K1 and K2 be simplicial complexes 細分に 関する
ホワ イト ヘッド の定理
with |K1| = |K2|. Then K has a subdivision K3 subdividing both K1 and K2. Hence, if K1 ⊳ K and K2 ⊳ K, then there exists some K3 ⊳ K subdividing both K1and K2.
This theorem is valid for any infinite simplicial complexes.
2.16 Definition A simplicial complex K with an order on K(0) is called an
ordered (simplicial) complex if σ(0) is a totally ordered subset of K(0) for 順序(単体) 複体 each σ ∈ K.
2.17 Example (1) A simplicial complex K with a total order on K(0) is an ordered complex.
(2) For each simplicial complex K, the barycentric subdivision Sd K is an ordered complex with the order defined as follows:
b(σ) b(τ ) ⇐⇒
def σ τ (i.e., b(σ) < b(τ ) ⇔ σ < τ).
2.18 Lemma Let v0, v1, . . . , vn ∈ Rk be affinely independent. Then, for each j = 0, . . . , n,
(v0, 0), . . . , (vj, 0), (vj, 1), . . . , (vn, 1) ∈ Rk× R = Rk+1 are also affinely independent.
Proof. Assume j
i=0
ti(vi, 0) + n
i=j
ti+1(vi, 1) = 0 and n+1
i=0
ti = 0. By observing each factor of the first equation, we have
j i=0
tivi+ n
i=j
ti+1vi = 0 and n
i=j
ti+1 = 0,
hence we have also ji=0ti = 0. Since v0, . . . , vn are affinely independent, it follow that
t0 =· · · = tj−1 = tj+ tj+1 = tj+2 =· · · = tn+1 = 0.
Then, tj(vj, 0) + tj+1(vj, 1) = 0. Since (vj, 0), (vj, 1) ∈ Rk+1 are affinely independent, we have also tj = tj+1 = 0.
2.19 Definition For an n-simplex σ = v0, v1, . . . , vn ⊂ Rk with an order v0 < v1 <· · · < vn (i.e., K(σ) is an ordered complex), we define
K(σ)× I =
def
τ ∃j = 0, 1, . . . , n such that
τ (v0, 0), . . . , (vj, 0), (vj, 1), . . . , (vn, 1)
= (vi0, 0), . . . , (vij, 0), (vij+1, 1), . . . , (vim, 1) 0 i0 <· · · < ij ij+1 <· · · < im n.
σ I
1
0
v1
v2 v0
K(σ)× I
2.20 Lemma In the above, |K(σ) × I| = σ × I.
Proof. By the definition, it is easy to see that |K(σ) × I| ⊂ σ × I. Let (x, t)∈ σ × I. If t = 1 then
(x, t)∈ (v0, 1), . . . , (vn, 1) ⊂ |K(σ) × I|. When t < 1, let x = ni=0z(i + 1)vi, z∈ ∆n. Then,
n i=j+1
z(i + 1) t < n
i=j
z(i + 1)
for some j = 0, 1, . . . , n, whence we can choose 0 s < 1 so that t = sz(j + 1) +
n i=j+1
z(i + 1).
Then, it follows that (x, t) =
n i=0
z(i + 1)vi, t
= j−1
i=0
z(i + 1)(vi, 0) + (1− s)z(j + 1)(vj, 0)
+ sz(j + 1)(vj, 1) + n i=j+1
z(i + 1)(vi, 1)
∈ (v0, 0), . . . , (vj, 0), (vj+1, 1), . . . , (vn, 1) ⊂ |K(σ) × I|. Thus, we have |K(σ) × I| = σ × I.
