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# Simplicial complex

## 1 Definition and examples

Where do chain complexes come from?

Recall that a chain complex is a sequence of vector spaces and linear operators: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \begin{array}{ccccccccccccc} ... & \ra{\partial_{k+2}} & C_{k+1} & \ra{\partial_{k+1}} & C_{k} & \ra{\partial_{k}} & C_{k-1} & \ra{\partial_{k-1}} & ... & \ra{\partial_{1}} & C_{1} & \ra{\partial_{0}} & 0. \end{array}$$ that satisfies: $$\partial _{k+1}\partial _k=0,\forall k.$$ This property allows us to study the homology of this complex: $$H_k=\ker \partial _k / \text{im }\partial _{k+1}.$$

We've built them for cubical complexes: $$C_k=C_k(K),$$ where $C_k(K)$ is the group of $k$-chains of cubical complex $K$.

We have also constructed them for homology groups of graphs in the same manner. The main difference was that the chain complexes of graphs are very short: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \begin{array}{ccccccccccccc} 0 & \ra{\partial_{2}=0} & C_{1} & \ra{\partial_{1}} & C_{1} & \ra{\partial_{0}=0} & 0. \end{array}$$

Another distinction is that cubical complexes are build via discretization of the Euclidean space in order to study its topology. Meanwhile, we approach topology of graphs from the opposite direction, which may be called Euclidization of data.

Definition.

• A collection $K$ of subsets of set $S$ is called an abstract simplicial complex if all subsets of any element of $K$ are also elements of $K$, i.e.,

$$\tau \in K, \sigma < \tau \Rightarrow \sigma \in K.$$

• The elements of these subsets are called simplices. If such a subset has exactly $n+1$ elements it is called an $n$-simplex, or simplex of dimension $n$.
• The highest dimension of a simplex in $K$ is called the dimension of $K$.
• For a given $n$, the collection of all $k$-simplices in $K$ with $k \le n$ is called the $n$-skeleton of $K$ denoted by $K^{(n)}$.

A more economic way of presenting of a simplicial complex is the one we started with: to list only the largest simplices and then use this condition to recover the rest. In the example above we only need to list: $$ABC, BD, DC.$$

Example. A realization of the complex $$ABD, BCD, ACD, ABE, BCE, ACE$$ is found by gluing these $6$ "triangles" to each other one by one. Alternatively, we can build it skeleton by skeleton:

We throw the vertices (the singletons from $K$) around in ${\bf R}^3$ and then connect them by the paths (the pairs in $K$) trying to keep them unknotted. Finally we add the faces (the triples in $K$) as pieces of fabric stretched on these wire-frames.

The result is homeomorphic to the sphere. $\square$

Definition. A realization $|K|$ of a simplicial complex $K$ is a one-to-one correspondence of the vertices of $K$ with the vertices of a geometric simplicial complex $Q$, as well as a one-to-one correspondence of the simplices of $K$ with the simplices of $Q$. A representation of a topological space as a homeomorphic image of a realization of a simplicial complex is called its triangulation . Spaces that have finite triangulations are called polyhedra.

Exercise. Show that if simplex $s$ is realized as a geometric simplex $q$, then the faces of $s$ correspond to those of $q$.

The realizations of simplicial complexes behave similarly to those of cubical complexes. In particular they have the same topology:

Proposition. A polyhedron is closed and bounded subset of a Euclidean space.

Realizations of simplicial complexes can also be defined via quotients of topological spaces.

Exercise. Prove that any finite graph can be realized in ${\bf R}^3$.

Exercise. Is there a $2$-dimensional simplicial complex that can't be realized in ${\bf R}^3$?

Exercise. Sketch realizations of these complexes:

• $AB,BC,CD,DA,CA,BD$; and
• $ABC,ABD,ABE.$

Exercise. Prove that any abstract simplicial complex $K$ has a realization. Hint: try ${\bf R}^N$, where $N$ is the number of vertices in $K$.

## 2 Faces of a simplex and unoriented chains

Since they have fewer edges, simplices are indeed simpler then anything else, even cubes. Just consider the boundary:

Indeed,

• in dimension $2$, a triangle has $3$ sides while a square has $4$;
• in dimension $3$, a tetrahedron has $4$ faces while a cube has $6$;
• ...
• in dimension $n$, an $n$-simplex has $n+1$ $(n-1)$-faces while an $n$-cube has $2n$.

The boundary of a $n$-simplex is made of its $(n-1)$-faces. Those are easy to construct by choosing one vertex at a time and considering the opposite face:

Combinatorially, the $n$-simplex is just a list of $n+1$ items called vertices: $$s = A_0A_1 ... A_n,$$ and any of its $(n-1)$-faces is the same list with one item dropped (indicated by the brackets): $$f_0 = [A_0]A_1 ... A_n,$$ $$f_1 = A_0[A_1] ... A_n,$$ $$...$$ $$f_n = A_0A_1 ... [A_n].$$

In the unoriented case, a chain is, again, defined as a "combination", or a formal sum, of cells. In the above example, there are only these $1$-chains: $$0, a, b, c, a + b, b + c, c + a, a + b + c.$$

It is clear now that the boundary of a $n$-simplex is the sum of its $(n-1)$-faces. It's an $(n-1)$-chain: $$\partial _n s=\sum_{i=0}^n f_i = \sum_{i=0}^n A_0 ... A_{i-1}[A_i]A_{i+1}...A_n.$$

As before, we extend the operator from cells to chains and then define for any simplicial complex $K$ the familiar vector spaces. They have the same names with "simplicial" and "over ${\bf Z}_2$" attached whenever necessary. So, we have the simplicial...

• chain groups $C_k(K),\forall k$;
• chain complex

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \begin{array}{ccccccccccccc} ...& \to & C_{k+1}(K) & \ra{\partial_{k+1}} & C_{k}(K) & \ra{\partial_{k}} & C_{k-1}(K) & \ra{\partial_{k-1}} & ... & \to & 0; \end{array}$$

• cycle groups $Z_k(K)=\ker \partial _k,\forall k$;
• boundary groups $B_k(K)=\text{ im } \partial _{k+1},\forall k$;
• homology groups $H_k(K)=Z_k(K) / B_k(K),\forall k$,

... over ${\bf Z}_2$.

The last step of the construction is made possible by the following

Theorem. For a simplicial complex $K$, $$\partial _k \partial _{k+1}=0.$$

The theorem will follow as a corollary from the oriented case.

The homology theory of simplicial complexes over ${\bf Z}_2$ is complete!

However, currently our main interest is homology over ${\bf R}$.

## 3 How to orient simplices

The first step in developing the homology over the reals is to add more structure to the complex - orientation of cells.

Recall that we for cubical complexes we defined the orientation for $1$-cells only.

The edges are directed along the corresponding exes of the Cartesian coordinate system that has been provided with the Euclidean space. These directions however don't suggest any particular way of orienting the squares. That's OK because there is no need to define the orientation of the higher dimensional cubes as the boundary operator is defined based on the representation of each cube as a product of edges and vertices.

Recall that defining such orientation is also possible based the geometry of the Euclidean space; for example, a plane (or a square) in the $3$-dimensional space is oriented by a choice of one of the two unit normal vectors. (In fact, what's even more relevant is that the choice is between two classes of normal vectors.)

The idea extends to subspaces of dimension $n-1$ in ${\bf R}^{n}$. However, this "external" approach to orientation fails when the dimension of the subspace is less than $n-1$ because it doesn't "cut" the space anymore.

The most important thing to know is that

there should always be exactly two orientations.

Following our treatment of cubical complexes, we first define the directions of all edges of our simplicial complex. However, this time the direction of an edge isn't dictated by the ambient space (there is none) -- the choice is entirely ours!

We illustrate this idea below. In this simplicial complex representation of the circle all three edges have been oriented:

Here the $1$-simplices aren't just edges $a, b, c$ but $a = AB, b = CB, c = AC$. We could have chosen any other order of vertices: $a = BA, b = CB, c = CA$, etc.

Early on, let's make it clear that another choice of cells' orientations will produce a different algebra of chains... but the same homology groups! In fact, choosing orientation of a complex is similar -- as we shall see identical -- to choosing a basis of a vector space.

Now, introducing directions of edges in this fashion seems simple enough but how does this help with the problem of introducing orientations of higher-dimensional simplices?

Since there is no such thing as "the direction of a triangle", we need to think of something else. Looking at the picture below the answer seems obvious: we go around it either clockwise or counter-clockwise:

However, this idea has no analog for a $3$-simplex. There is no such thing as clockwise and, moreover, there is no circular path through all of its vertices that capture all possible orderings.

Exercise. Prove that. Hint: just count.

We find the answer in the definition! The definition is purely combinatorial and so should be the definition of orientation.

First, let's make observe that speaking of edge $AB$ being an element of complex $K$ is misleading because the elements of $K$ are subsets of set $S$. Therefore, to be precise we wouldn't write $$AB \in K$$ (until the orientation has been already introduced) but $$\{A,B\}\in K.$$ Considering the fact that $\{A,B\}=\{B,A\}$ here's our conclusion: choosing $AB$ over $BA$ is equivalent to choosing an ordering of the set $\{A,B\}$.

So, the idea is that an orientation of a simplex is simply a specific choice of the order of its vertices. This approach justifies the usage of $ABC$ instead of $\{A,B,C\}$, etc. -- after the orientation has been chosen.

It seems more complicated for the triangle though. For a $2$-simplex with vertices $A,B,C$, we have $6$ possible orderings of the vertices:

• $ABC$, $BCA$, and $CAB$ for clockwise, and
• $ACB$, $CBA$, and $BAC$ for counter-clockwise.

But there should be only two orientations! It follows that an orientation should be a choice of one of these two classes of orderings. What are they exactly?

These orderings are simply re-orderings of $ABC$. In other words, they are self-bijections of $\{A,B,C\}$. We recognize them as permutations.

Recall, an even/odd permutation is the composition of an even/odd number of transpositions, aka "flips". For example,

• it takes two flips to get $CAB$ from $ABC$, so it's even; but
• it takes one flip to get $ACB$ from $ABC$, so it's odd.

Note: Even permutations form subgroup $\mathcal{A}_n$ of the symmetric group $\mathcal{S}_n$ of all permutations of a set of $n$ elements.

Therefore, the former class is that of even permutations and the latter odd.

So, the definition of orientation of simplices is based on the fact that there are two equivalence classes of orderings of any set, in particular, the set of vertices of a simplex:

two orderings are equivalent if they differ by an even permutation.

Definition. Suppose an $n$-simplex $\tau$ is given as an ordered list of $n+1$ elements ("vertices"): $$\tau=A_0A_1...A_n.$$ Then we say that $\tau$ is an oriented simplex if either the class of even permutations of the vertices is chosen or the class of odd permutations.

Therefore, we will always have to deal with this ambiguity in all our constructions as each ordering that we use is in fact an equivalence class of orderings. For example,

• $ABC$ means $[ABC]=\{ABC,BCA,CAB\}$; and
• $ACB$ means $[ACB]=\{ACB,CBA,BAC\}$.

Definition. Suppose a simplicial complex $K$ consists of simplices as subsets of set $S$. Then it is called oriented if its every simplex is oriented.

Example. The complex $K$ of the triangle,

• unoriented:

$$K=\{\{A\},\{B\},\{C\},\{A,B\},\{C,B\},\{A,C\},\{A,B,C\}\};$$

• oriented:

$$K=\{A,B,C,AB,CB,AC,[ABC]\}.$$

Note that the orientations of the edges of $\tau$ don't match the orientation of $\tau$ itself. The reason is that they don't have to... Plus there may be another $2$-simplex adjacent to one of these edges with the "opposite" orientation. $\square$

To acquire an orientation of complex $K$ one can just order its vertices and use that order to orient each simplex in $K$:

Exercise. How many orientations does the above complex have? Can you think of a general formula?

Exercise. With the meaning of orientation of cells of dimensions $1$ and $2$ clear, what is the geometric meaning of orientation of vertices?

## 4 The algebra of oriented chains

Recall that the point of using oriented chains is to be able to capture all possible ways to go around the circle: going once, twice, or thrice around it, or going in the opposite direction.

An oriented simplicial chain is a "formal" linear combination of finitely many oriented simplices, such as $3a + 5b - 17c$.

We will concentrate on real chains, i.e., ones with real coefficients. From this point of view, the chains previously discussed are binary. Then the former will result in "homology over reals ${\bf R}$" and the latter in "homology over binary arithmetic ${\bf Z}_2$". Other choices are possible as well: integers ${\bf Z}$, rational numbers ${\bf Q}$, etc or any ring (we need the ring structure to define the algebraic operations on chains).

Next, for a given simplicial complex $K$ and $k=0,1,2...$, let $C_k(K)$ denote the set of all real chains: $$C_k(K) = \left\{ \displaystyle\sum_{i}s_i \sigma_i \colon s_i \in {\bf R}, \sigma_i {\rm \hspace{3pt} is \hspace{3pt} a \hspace{3pt}} k{\rm -simplex \hspace{3pt} in \hspace{3pt}} K \right\}.$$

It is easy to define addition of chains by assigning coefficients of each simplex: if $$A = \displaystyle\sum_i s_i \sigma_i,$$ $$B = \displaystyle\sum_i t_i \sigma_i,$$ then $$A + B = \displaystyle\sum_i(s_i + t_i) \sigma_i.$$ To see that the operation above is well defined, one can assume that each sum lists all simplices present in both sums -- some with zero coefficients.

Theorem. $C_k(K)$ is a abelian group with respect to chain addition.

Proof. We verify the axioms of group below.

(1) Identity element: $$0 = \displaystyle\sum_i 0 \cdot \sigma_i.$$

(2) Inverse element: $$A = \displaystyle\sum_i s_i \sigma_i, {\rm \hspace{3pt} then}$$ $$-A = \displaystyle\sum_i (-s_i)\sigma_i.$$

(3) Associativity: $$\begin{array}{lll} A + (B + C) &= \displaystyle\sum_i (s_i + t_i + u_i) \sigma_i \\ &= (A + B) + C. \end{array}$$ $\blacksquare$

It is also easy to define scalar multiplication of chains: if $$A = \displaystyle\sum_i s_i \sigma_i,$$ $$r\in {\bf R},$$ then $$rA = \displaystyle\sum_i(rs_i) \sigma_i.$$

Theorem. $C_k(K)$ is a vector space with respect to chain addition and scalar multiplication. The set of all $k$-simplices in $K$ is a basis of this space.

Exercise. Prove the theorem.

Example. In particular, in the example of a triangle representing the circle we have

• $C_0(K) = <A> = {\bf R},$
• $C_1(K) = <a,b,c> = {\bf R} \times {\bf R} \times {\bf R},$
• $C_2(K) = 0$, etc.

$\square$

Note: Until a relation is established between cells/chains of different dimensions, this algebra can't capture the topology of the complex. This relation is given by the boundary operator .

Up to this point the development of the algebra of chains follows exactly that of the case of oriented cubical complexes. There are some differences.

First, all simplices are oriented not just $1$-dimensional.

Second, the meaning of $-A$ for a simplex $A$ is clear: it's $A$ with the opposite orientation:

Below you can see how sign changes if you flip two vertices: $$A_1A_0A_2 ... A_n=-A_0A_1A_2 ... A_n.$$

It is also worth repeating what oriented simplices are and aren't.

• Oriented simplices aren't subsets of the vertices, not $\{A,B,C\}$.
• Oriented simplices aren't orderings of the vertices, not $ABC$.
• Oriented simplices are classes of orderings of the vertices, such as $[ABC]=\{ABC,BCA,CAB\}.$

Exercise. Prove that choosing a different orientation of a simplicial complex produces a chain group isomorphic to the original. Present the matrix of the isomorphism.

## 5 The boundary operator

The boundary operator records the relations between cells/chains of different dimensions in order to capture the topology of the simplicial complex.

Let's review what we already know.

Example. Suppose we have a simplicial complex $K$:

• $0$-simplices: $A, B, C$;
• $1$-simplices: $a = AB, b = CB, c = AC$;
• $2$-simplex: $\tau = ABC$.

It is illustrated above.

The boundary of a vertex empty, so the boundary operator of a $0$-chain is $0$: $$\partial (A) = 0 {\rm \hspace{3pt} for \hspace{3pt} any \hspace{3pt}} A \in C_0(K).$$

The boundary of a $1$-cell consists of its two end-points, so in the binary setting this was simple: $$\partial (a) = \partial (AB) = A + B.$$

In the integral setting, the direction of the edge matters ($AB \neq BA$), so we define: $$\partial (a) = \partial (AB) = B - A {\rm \hspace{3pt} for \hspace{3pt} any \hspace{3pt}} a = AB \in C_1(K).$$

In dimension $2$, the boundary $\partial \tau$ is defined as a linear combination of its faces: $$\partial \tau = AB + BC + CA = a - b - c,$$ or: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lllllll} \partial \tau & = & \partial \left( \bigtriangleup \right) \\ & = & \nearrow + & \searrow + & \leftarrow \\ & = & a+ & (-b)+ & (-c) \\ & = & a-b-c. \end{array}$$ Here we first follow the chain around clockwise as indicated by its orientation writing the $1$-cells as they appear, and then interpret these $1$-cells in terms of the $1$-cells, $a, b, c$, that we listed in the complex $K$. $\square$

Let's define the boundary for an arbitrary dimension.

Recall first what are the faces of a simplex. As the $n$-simplex is just a list of $n+1$ vertices, any of its $(n-1)$-faces is a list of the same vertices with one dropped: $$f_0 = [A_0]A_1 ... A_n ,$$ $$f_1 = A_0[A_1] ... A_n ,$$ $$...$$ $$f_n = A_0A_1 ... [A_n].$$

The question is now is, what are the oriented faces of an oriented simplex? The answer is, they are oriented simplices.

That the answer is justified isn't as simple as it seems. Indeed, even though it is clear that removing an item from an ordering gives an ordering of the smaller set, but what if we are dealing with classes of orderings? It is easy to prove that removing an item from an even/odd ordering gives us an even/odd ordering of the smaller set. Algebraically, $$\mathcal{A}_n=\mathcal{S}_{n}\cap \mathcal{A}_{n+1} .$$

Now the boundary of an $n$-simplex $s$ is the sum of its faces, in the unoriented case. As we see in the example above the faces appear with various signs this time. As it turns out, the boundary now is the alternating sum of its faces:

$\partial s = f_0 - f_1 + f_2 - ... \pm f_n$.

Definition. The boundary of an oriented $n$-simplex is an oriented $(n-1)$-chain defined by: $$\partial (A_0A_1 ... A_i ... A_n) = \sum_i (-1)^i A_0A_1 ... [A_i] ... A_n .$$

Example. Let's apply the definition to the complex of the triangle, as in the example above. These are the oriented simplices based on the obvious ordering of the $3$ vertices:

• $0$-simplices: $A_0, A_1, A_2$;
• $1$-simplices: $A_0A_1, A_1A_2, A_0A_2$;
• $2$-simplex: $A_0A_1A_2$.

Then $$\begin{array}{llll} \partial (A_0A_1A_2) &= [A_0]A_1A_2 - A_0[A_1]A_2 + A_0A_1[A_2] \\ &= A_1A_2 - A_0A_2 + A_0A_1 \\ &=A_0A_1 + A_1A_2 +A_2A_0 . \end{array}$$ So, these edges go clockwise around the square, as expected! $\square$

Notice that the input of the formula in the definition is an ordering while it is supposed to be an oriented simplex, i.e., a class of orderings. What happens if we use another representative from the class of orderings as the input of this formula?

Exercise. Confirm that the result is the same for $$\partial (A_2A_0A_1).$$

To further understand what's going on, let's flip the first two vertices: $$\begin{array}{llll} \partial (A_1A_0A_2 ... A_n) \\ = \sum_i (-1)^i A_1A_0A_2 ... [A_i] ... A_n \\ = [A_1]A_0A_2 ... A_n - A_1[A_0]A_2 ... A_n &+ \sum_{i>1} (-1)^i A_1A_0A_2 ... [A_i] ... A_n \\ = A_0A_2 ... A_n - A_1A_2 ... A_n &+ \sum_{i>1} (-1)^i A_1A_0A_2 ... [A_i] ... A_n \\ = -A_1A_2 ... A_n + A_0A_2 ... A_n &+ \sum_{i>1} (-1)^i (-1)A_0A_1A_2 ... [A_i] ... A_n \\ = -[A_0]A_1A_2 ... A_n + A_0[A_1]A_2 ... A_n &- \sum_{i>1} (-1)^i A_0A_1A_2 ... [A_i] ... A_n \\ = -\sum_i (-1)^i A_1A_0A_2 ... [A_i] ... A_n \\ = -\partial (A_0A_1A_2 ... A_n). \end{array}$$ The sign of the boundary chain has reversed! Therefore, we have

Proposition. The boundary chain of an ordered simplex is well-defined. In particular, for any oriented simplex $s$, we have $$\partial (-s)= -\partial (s).$$

Exercise. Provide the rest of the proof.

With the operator defined on each of the simplices, one can extend this definition to the whole chain group, by linearity, thus creating a linear operator: $$\partial \colon C_n(K) \rightarrow C_{n-1}(K).$$ Indeed, since we know $\partial (\sigma_i)$ for each simplex $\sigma_i \in C_n(K)$, we can simply extend $\partial$ to the rest of the vector space: $$\partial \left( \displaystyle\sum_i s_i \sigma_i \right) = \displaystyle\sum_i s_i \partial( \sigma_i ).$$

## 6 Boundaries have no boundaries

The key fact needed for homology theory is:

all boundaries are cycles.

Algebraically, it is given by the following

Theorem. $\partial \partial = 0$ over ${\bf R}$ or ${\bf Z}$.

Proof. It suffices to prove that $\partial \partial (s) = 0$ for any simplex $s$ in $K$. The idea is that in $\partial \partial s$ each $(n-2)$-face appears twice but with opposite signs.

Suppose $s$ is an $n$-simplex: $$s=A_0 ... A_n .$$ For each $i=0,1,...,n$, $f_i$ is the $i$-th face of $s$: $$f_i=A_0 ... [A_i] ... A_n.$$ It is an oriented $(n-1)$-simplex. For each pair $i,j=0,1,...,n,i \ne j$, let $f_{ij}=f_{ji}$ be the $j$-th face of $f_i$ or, which is the same thing, the $i$-th face of $f_j$ : $$f_{ij}=A_0 ... [A_i] ... [A_j] ... A_n.$$ It is an oriented $(n-2)$-simplex.

First, we use the definition and then the linearity of the boundary operator: $$\begin{array}{llllll} \partial \partial s &= \partial \left( \sum_i (-1)^i f_i \right) \\ &= \sum_i (-1)^i \partial( f_i ) \end{array}$$ Consider the $i$-th term, summation over $j$, and watch how the signs alternate: $$\begin{array}{lllllll} \partial( f_i ) &=\partial(A_0 A_1 ... [A_i] ... A_n ) \\ &= \sum_{j < i} (-1)^j A_0A_1 ... [A_j] ... [A_i] ... A_n &\text{ [A_j] in jth term of the sum...} \\ &+ \sum_{j > i} (-1)^{j-1} A_0 A_1 ... [A_i] ... [A_j] ... A_n &\text{ [A_j] in (j-1)st term...} \\ &= \sum_{j < i} (-1)^j f_{ij} + \sum_{ j < i } (-1)^{j-1} f_{ij} . \end{array}$$ We substitute now and then deal with double summation, over $i$ and $j$: $$\begin{array}{lllllll} \partial \partial s & = \sum_i (-1)^i \left( \sum_{j < i} (-1)^j f_{ij} + \sum_{ j < i } (-1)^{j-1} f_{ij} \right) \\ &= \sum_{i < j} (-1)^{i+j} f_{ij} - \sum_{i > j} (-1)^{i+j} f_{ij} \\ &=0. \end{array}$$ $\blacksquare$

Corollary. $\partial \partial = 0$ over ${\bf Z}_2$.

Exercise. Prove the corollary (a) by re-writing the above proof, (b) directly from the theorem.

So, the boundary of the boundary of a cell is zero. Since every chain is a linear combination of cells, it follows from the linearity of $\partial$ that the boundary of the boundary of any chain is zero: $$\begin{array}{llllll} C_k(K) & \xrightarrow{\partial} & C_{k-1}(K) & \xrightarrow{\partial} & C_{k-2}(K) \\ \tau & \mapsto & \partial (\tau ) & \mapsto & \partial \partial (\tau )=0. \end{array}$$

Thus every boundary is a cycle: $$B_k \subset Z_k.$$ That's why the homology groups make sense: $$H_k = B_k / Z_k.$$

Exercise. Represent the following subsets of the plane as realizations of simplicial complexes:

For each of them and coefficients in ${\bf R}$:

• (1) Find the chain groups and find the boundary operator as a matrix;
• (2) Using only part (1) and linear algebra, find $Z_k,B_k$ for all $k$.
• (3) Using only part (2) and linear algebra, find the Betti numbers:

$$b_k(K)=\dim H_k(K).$$

• (4) Confirm that even though the chain groups are different, the results match those for cubical complexes. Point out what part of your computation makes it so.

Exercise. Compute the homology groups of the sphere ${\bf S}^2$ by representing it as a hollow pyramid.