This site is being phased out.

# Simplicial homology

(Redirected from Simplicial complex)

## Simplicial complexes

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

Where do chain complexes come from?

We started with chain complexes for graphs. They are very short: $$\newcommand{\ra}[1]{\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \begin{array}{ccccccccccccc} 0 & \ra{\partial_{2}=0} & C_{1} & \ra{\partial_{1}} & C_{0} & \ra{\partial_{0}=0} & 0. \end{array}$$ Here $C_0(G)$ and $C_1(G)$ are the groups of chains of nodes and of edges of the graph $G$.

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

Graphs and cubical complexes are very different in nature. Cubical complexes are built via discretization of the Euclidean space in order to study its topology and the topology of its subsets. Meanwhile, we approach topology of graphs from the opposite direction, which may be called Euclidization of data, and yet we can still study the topology of subsets of the Euclidean space -- via realizations of graphs. We will follow this latter route with simplicial complexes.

Let's review the definitions.

Definition.

• A collection $K$ of subsets of a finite set $S$ is called an abstract simplicial complex, or just a simplicial complex, if all subsets of any element of $K$ are also elements of $K$:

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

• The elements of these subsets are called simplices. If such a subset $a$ has exactly $n+1$ elements, it is called an $n$-simplex, or simplex of dimension $n=\dim a$.
• The highest dimension of a simplex in $K$ is called the dimension of $K$:

$$\dim K:=\max _{a\in K} \dim a.$$

• 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 $K^{(n)}$:

$$K^{(n)}:=\{a\in K:\ \dim a \le n\}.$$

• A realization of a simplicial complex $K$ is a a geometric simplicial complex $Q=|K|$ along with such a one-to-one correspondence $f$ of the simplices (starting with the vertices) of $K$ with the simplices of $Q$ that the faces are preserved:

$$\sigma \subset \tau \Rightarrow f(\sigma) \subset f(\tau).$$

• A representation of a topological space $X$ as a homeomorphic image of a realization of a simplicial complex $K$ is called its triangulation and these spaces are called polyhedra. We still say that $X$ is a realization of $K$.

This is an example of a realization of a $2$-dimensional simplicial complex as a geometric complex in ${\bf R}^3$:

And this is a complex that may seem to contain only one $3$-simplex, blown up:

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 for simplicity. 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.

$\square$

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

Proposition. A polyhedron in a Euclidean space is closed and bounded.

Exercise. Prove the proposition. Hint: prove for a simplex first.

Exercise. Sketch realizations of these complexes:

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

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

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

## Faces of a simplex and non-oriented chains

Since they have fewer edges, simplices are indeed simpler than 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:

As a data structure, 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): $$\begin{array}{ll} f_0 &= [A_0]A_1 ... A_n,\\ f_1 &= A_0[A_1] ... A_n,\\ &...\\ f_n &= A_0A_1 ... [A_n]. \end{array}$$ Such a simple construction doesn't exist for cubes. Instead, we relied on the product structure to prove our theorems.

In the non-oriented case, a chain is, as before, defined as a “combination” of cells, i.e., a formal, binary 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 an $n$-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 groups. 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 $B_k(K):=\operatorname{Im} \partial _{k+1},\ \forall k$;
• 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 (Double Boundary Identity). For non-oriented chains and simplicial complexes, $$\partial _k \partial _{k+1}=0.$$

Exercise. Prove the theorem algebraically for $k=0,1,2$. Also provide sketches of the construction.

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

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

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

Example (Kirchhoff's current law). Suppose we have an electrical circuit understood as a graph.

Then the conservation principle is satisfied: the sum of currents flowing into any junction is equal to the sum of currents flowing out of the junction. If the edges are oriented, we can think of the currents as $1$-chains -- over the reals. Moreover, a current is such a $1$-chain $I$ that satisfies, for any vertex $A$: $$\sum_k I(e_k)=0,$$ with summation over all edges $e_k$ adjacent to $A$. It follows that $\partial I=0$. It's a cycle!

$\square$

## How to orient a simplex

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

Recall that for cubical complexes we initially defined the orientation for $1$-cells only as their directions. It can be arbitrary or it can be aligned with the axes:

These directions, however, don't suggest any particular way of orienting the squares. Those are defined as clockwise or counterclockwise orderings of the vertices.

Exercise. Show that this is the choice between two equivalence classes of orderings of $4$ letters.

This approach, however, proved itself too complex for higher dimensions and we instead relied on the representation of each cube as a product of edges and vertices.

Defining such orientation is also possible based the geometry of the Euclidean space; for example, a plane (and a square) in the $3$-dimensional space is oriented by a choice of one of the two unit normal vectors.

Exercise. Show that this is the choice between two equivalence 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 the normal vectors don't split into two equivalence classes anymore.

The most important thing we 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 and, 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 (triangulation) of the circle, all three edges have been oriented, at random:

Here, the $1$-simplices aren't just edges $a, b, c$ but $$a = AB =- BA,\ b = CB =- BC,\ c = AC =-CA.$$ 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 to choosing a basis of a vector space.

Next, after using directions to define orientation of edges, how do we handle 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 counterclockwise:

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 of simplicial complex! The definition relies only on the data structure and so should the definition of orientation.

First, let's 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 instead: $$\{A,B\}\in K.$$ 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.

Let's consider the triangle. 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, which “flip” two elements. 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.

We know that the even permutations form the 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 and, in particular, on the set of vertices of a simplex:

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

Of course, these two classes are the two cosets that form the following quotient group: $$\mathcal{S}_n / \mathcal{A}_n.$$

Definition. Suppose an $n$-simplex $\tau$ is given as an ordered list of $n+1$ elements $\tau=A_0A_1...A_n.$ Then we say that $\tau$ is an oriented simplex if either the class of even permutations of the vertices $\{A_0,A_1,...,A_n\}$ 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 $[ACB]=\{ACB,CBA,BAC\}$.

In other words, it suffices to list a single representative of the class to orient the simplex.

Exercise. Demonstrate how this scheme fails for cubical complexes. Hint: try dimension $2$.

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

Exercise. A permutation of the basis vectors of ${\bf R}^{n+1}$ defines a linear operator. What can you say about its determinant?

## The algebra of oriented chains

Definition. Suppose a simplicial complex $K$ consists of simplices as subsets of set $S$. We say that $K$ is an oriented simplicial complex if its every simplex is oriented. We will omit “oriented” when there is no confusion.

Example. The complex $K$ of the triangle, is: $$\begin{array}{llllllllll} \text{non-oriented: } &K=\{\{A\},&\{B\},&\{C\},&\{A,B\},&\{C,B\},&\{A,C\},&\{A,B,C\}\};\\ \text{oriented: } &K=\{A,&B,&C,&AB,&CB,&AC,&[ABC]\}. \end{array}$$

Note that the orientations of the edges of $\tau$ don't match the orientation of $\tau$ itself. The reason is, they don't have to... Also, consider the possibility that there may be another $2$-simplex adjacent to one of these edges with the “opposite” orientation.

$\square$

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

Exercise. In how many ways can this be done?

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 $-\tau$ for a simplex $\tau$ is clear: it's $\tau$ with the opposite orientation. For dimensions $1$ and $2$, the geometric meaning is clear:

For higher dimensions, it's a flip of two vertices: $$A_1A_0A_2 ... A_n=-A_0A_1A_2 ... A_n.$$

It is 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 $[ABC]=\{ABC,BCA,CAB\}.$

This is how we make this precise.

Theorem. $$A_0A_1 ...A_n = \pi (s) A_{s(0)}A_{s(1)} ...A_{s(n)},$$ where $s\in \mathcal{S}_{n+1}$ is a permutation and $\pi (s)$ is its parity: $$\pi(s):=\begin{cases} 1&\text{if } s \text{ is even},\\ -1&\text{if } s \text{ is odd}. \end{cases}$$

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 $3a + 5b - 17c$. The coefficients come from some ring $R$. From this point of view, the chains previously discussed are over $R={\bf Z}_2$, i.e., binary. We will concentrate on real chains, i.e., ones with real coefficients. However, the analysis will often apply also to the integers $R={\bf Z}$, rational numbers $R={\bf Q}$, etc.

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

It is easy to define the sum of two chains by assigning appropriate coefficients of each simplex: if $$A = \displaystyle\sum_i s_i \sigma_i, B = \displaystyle\sum_i t_i \sigma_i \Rightarrow A + B := \displaystyle\sum_i(s_i + t_i) \sigma_i.$$ 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 an 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:

$$A = \displaystyle\sum_i s_i \sigma_i \Rightarrow -A = \displaystyle\sum_i (-s_i)\sigma_i.$$

• (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 the scalar multiplication of chains: if $$A = \displaystyle\sum_i s_i \sigma_i, \text{ and } r\in {\bf R},$$ then we let $$rA := \displaystyle\sum_i(rs_i) \sigma_i.$$

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

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

$\square$

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.

Exercise. What does the algebra of oriented chains over ${\bf Z}_2$ look like?

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

## The boundary operator

The boundary operator of a simplicial complex records the relations between cells/chains of consecutive dimensions.

Let's review what we already know.

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

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

The boundary of a vertex empty, so the boundary operator of a $0$-chain is $0$: $$\partial (A) = 0, \ \forall A \in C_0(K).$$

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: $$\partial (a) = \partial (AB) := B - A, \ \forall a = AB \in C_1(K).$$

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 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 we mean by a face 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: $$\begin{array}{lll} f_0 &= [A_0]A_1 ... A_n ,\\ f_1 &= A_0[A_1] ... A_n ,\\ &...\\ f_n &= A_0A_1 ... [A_n]. \end{array}$$

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

The idea is very simple: removing an item from an ordering gives an ordering of the smaller set, just as shown above. However, that this answer is justified isn't as simple as it seems. Indeed, we aren't dealing with orderings but with classes of orderings. Why does this operation still make sense?

Proposition. Removing an item from an even/odd ordering gives us an even/odd ordering of the smaller set; or, algebraically, $$\mathcal{A}_n=\mathcal{S}_{n}\cap \mathcal{A}_{n+1} .$$

Exercise. Prove the proposition.

Now, the boundary of an $n$-simplex $s$ is the sum of its faces, in the non-oriented case. As we see in the example above, this time the faces appear with various signs. 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 triangle, 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 $$\partial (A_2A_0A_1).$$

A single term in the boundary expansion for a $2$-simplex is shown below:

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 the following:

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: $$\partial : C_n(K) \to C_{n-1}(K).$$ 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: $$\partial \left( \displaystyle\sum_i s_i \sigma_i \right) := \displaystyle\sum_i s_i \partial( \sigma_i ).$$

## Homology

The key fact needed for homology theory is:

all boundaries are cycles.

Algebraically, it is given by the following theorem:

Theorem (Double Boundary Identity). $\partial \partial = 0$ 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 the expansion of $\partial \partial s$ 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, i.e., the summation over $j$, and watch how the signs alternate: $$\begin{array}{lllllll} \partial( f_i ) &=\partial(A_0 A_1 ... [A_i] ... A_n ) \\ &= \sum_{j < i} (-1)^j A_0A_1 ... [A_j] ... [A_i] ... A_n &\text{ [A_j] in jth term of the sum...} \\ &+ \sum_{j > i} (-1)^{j-1} A_0 A_1 ... [A_i] ... [A_j] ... A_n &\text{ [A_j] in (j-1)st term...} \\ &= \sum_{j < i} (-1)^j f_{ij} + \sum_{ j < i } (-1)^{j-1} f_{ij} . \end{array}$$ So, the sign attached to the face becomes the opposite as $j$ goes past $i$. That's what will endure cancelling...

We substitute now and then deal with double summation, over $i$ and $j$: $$\begin{array}{lllllll} \partial \partial s & = \sum_i (-1)^i \left( \sum_{j < i} (-1)^j f_{ij} + \sum_{ j < i } (-1)^{j-1} f_{ij} \right) \\ &= \sum_{i < j} (-1)^{i+j} f_{ij} - \sum_{i > j} (-1)^{i+j} f_{ij} \\ &=0. \end{array}$$

Let's summarize what has happened. The $(n-2)$-faces of our simplex $s$ form a symmetric $(n+1)\times (n+1)$ matrix. Then $\partial \partial s$ is the sum of the elements of this matrix, with appropriate signs, after the diagonal is removed: $$\{f_{ij}\}=\left[ \begin{array}{c|ccc|c|c|c|cc} &0 &1 &...&i &...&j &...&n\\ \hline 0&0 &-f_{01}&...&+f_{0i}&...&-f_{0j}&...&+f_{0n}\\ 1&+f_{10}&0 &...&-f_{1i}&...&+f_{1j}&...&-f_{1n}\\ &.&.&...&.&...&.&...&.\\ &.&.&...&.&...&.&...&.\\ \hline i&-f_{i0}&+f_{i1}&...&0 &...&-f_{ij}&...&+f_{in}\\ \hline &.&.&...&.&...&.&...&.\\ &.&.&...&.&...&.&...&.\\ \hline j&+f_{j0}&-f_{j1}&...&+f_{ji}&...&0 &...&-f_{jn}\\ \hline &.&.&...&.&...&.&...&.\\ &.&.&...&.&...&.&...&.\\ n&-f_{n0}&+f_{n1}&...&-f_{ni}&...&+f_{nj}&...&0 \end{array} \right]$$ Finally, the symmetric entries have opposite signs and, therefore, cancel.

$\blacksquare$

Exercise. Fill the blanks in the table to demonstrate how the signs alternate.

The diagram below illustrates the proof for $n=2$:

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

Exercise. Prove the corollary (a) by rewriting the above proof, (b) directly from the theorem. Hint: consider a function ${\bf Z}\to {\bf Z}_2$.

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: $$B_k(K) \subset Z_k(K).$$ That's why the homology groups make sense: $$H_k(K): = B_k(K) / Z_k(K).$$

Example. Below,we visualize the idea why the sum of the two standard generators of the torus is homologous to the diagonal, i.e., $a+b \sim d$, and $(1,1)=(1,0)+(0,1)$ in the homology group:

$\square$

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 of these complexes.
• (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. Represent the circle ${\bf S}^1$ as a hollow triangle and the sphere ${\bf S}^2$ as a hollow pyramid:

and confirm that the homology groups are: $$\begin{array}{lll} H_0({\bf S}^1)={\bf R},& H_1({\bf S}^1)={\bf R}, & H_2({\bf S}^1)=0;\\ H_0({\bf S}^2)={\bf R},& H_1({\bf S}^2)=0, & H_2({\bf S}^2)={\bf R}. \end{array}$$