Examples

Homology of a simplex relative to its boundary

Homology of a simplex relative to its boundary

Consider a simple 2-simplex $\sigma$ and the simplicial complex $\mathcal{L}$ that it determines:

Let $\mathcal{K}$ be the boundary 1-simplex (the edges only from $\mathcal{L}$).

There are no 2-simplices in $\mathcal{K}$, so:

$$C_2(\mathcal{L},\mathcal{K})=C_2(\mathcal{L})\cong ⟨\sigma ⟩\cong \mathbb{Z}$$

All 0- and 1-simplices in $\mathcal{L}$ are also in $\mathcal{K}$, so:

$$C_1(\mathcal{L},\mathcal{K})=0,C_0(\mathcal{L},\mathcal{K})=0$$

The relative chain complex:

$$0\to C_2(\mathcal{L},\mathcal{K})\to C_1(\mathcal{L},\mathcal{K})\to C_0(\mathcal{L},\mathcal{K})\to 0$$

becomes:

$$0\to \mathbb{Z}\to 0\to 0\to 0$$

It follows that the relative homology is:

$$H_i(\mathcal{L},\mathcal{K})=\{\begin{array}{ll}0& i>3\\ \mathbb{Z}& i=2\\ 0& i=1\\ 0& i=0\end{array}$$

Compare this to the homology of a sphere!

Homology of a sphere:

$$H_i({\mathbb{S}}^2)=\{\begin{array}{ll}0& i>3\\ \mathbb{Z}& i=2\\ 0& i=1\\ \mathbb{Z}& i=0\end{array}$$

Notice that ${\mathbb{S}}^2$ is homeomorphic to the quotient space $\mathcal{L}/\mathcal{K}$.

This comparison of homologies is apt; we just need to explain $H_0({\mathbb{S}}^2)\cong \mathbb{Z}$ versus $H_0(\mathcal{L},\mathcal{K})\cong 0$.

Compare this against the quotient of homology groups!

The homology of $\mathcal{K}$ is:

$$H_i(\mathcal{K})=\{\begin{array}{ll}0& i>1\\ \mathbb{Z}& i=1\\ \mathbb{Z}& i=0\end{array}$$

The homology of $\mathcal{L}$, because it is contractible, is:

$$H_i(\mathcal{L})=\{\begin{array}{ll}0& i>0\\ \mathbb{Z}& i=0\end{array}$$

The quotient of homology groups is:

$$H_i(\mathcal{L})/H_i(\mathcal{K})\cong 0\text{for all}i$$

• Relative homology treats $\mathcal{K}$ as a single point.
• The quotient of homology annihilates generators (“holes”) in $\mathcal{L}$ which are also generators (“holes”) in $\mathcal{K}$.
• Different results for different constructions!

Some cohomology classes on the 2-torus

Some cohomology classes on the 2-torus

Consider the 2-torus $T$:

The red (poloidal) and pink (toroidal) curves together generate the degree one homology, $H_1(T)\cong {\mathbb{Z}}^2$.

Now recall a simplicial complex representation of the torus:

Let $\phi$ be the cochain which assigns 1 to every edge in the middle column, and zero to all other edges. (So it is 1 on the edges directly below the $\phi$ label, including the diagonals, but excluding all verticals.) Let $\psi$ be the cochain which assigns 1 to every edge in the middle row, and zero to all other edges. (So it is 1 on the edges directly to the right of the $\psi$ label, including the diagonals, but excluding all horizontals.) Accounting now for orientations, let $\phi$ give $+1$ on edges directed rightwards, and $\psi$ give $+1$ on edges directed upwards.

One can check directly, going around the boundaries of each 2-simplex, that $\delta \phi =0$ and $\delta \psi =0$: The sum of the numbers on any triangle, accounting for orientations, is zero. So these cochains are both cocycles.

In order to see that $\phi$ and $\psi$ are not boundaries, and thus represent nontrivial homology classes, and also to see that they don’t represent the same homology class, we devise a method of representing coboundaries of 0-cochains.

A 0-cochain $\eta$ is the assignment of integers to each of the vertices. The condition $\delta \eta =0$, which means $\eta \circ {\partial }_1\equiv 0$, states that these numbers cannot change as one moves across an edge from vertex to vertex. (Applying ${\partial }_1$ to any edge $[u,v]$ gives $[v]-[u]$. By linearity, $\eta ({\partial }_1([u,v]))=\eta ([v])-\eta ([u])$. Therefore $\eta ([v])=\eta ([u])$ when the condition holds.)

From this it follows that $H^0(T)\stackrel{\sim }{⟶}\mathbb{Z}$. The isomorphism is given by sending $\eta$ to the integer it gives on a vertex.

Return now to $H^1(T)$. A boundary in $H^1(T)$ is a cochain $\chi$ assigning numbers to directed edges such that there is a vertex labeling cochain $\eta \in C^1$ with $\chi ([u,v])=\eta ([v])-\eta ([u])$. In other words, that it is possible to derive the edge numbers as the differences of a single vertex numbering scheme.

Now one can see that $\phi$ is not a coboundary. Vertex integers must increase by one as one passes across the middle column of edges, and stay fixed going across all other edges. This is impossible because the left-most and right-most edges are identified. Similarly $\psi$ is not a coboundary, and $\phi -\psi$ is not a coboundary.

(To witness a contrasting case, let $\xi$ be a 1-cochain giving 1 only on the non-vertical edges in the right-most column. This is $\phi$ shifted to the right. Now $\phi -\xi$ is a coboundary: assign 1 on vertices in the middle-right column, and 0 on all other vertices.)

Much more work would be required to show that $\phi$ and $\psi$ have no relations in homology, and that they jointly generate homology.

A cup product on the 2-torus

A cup product on the 2-torus

We study the cup product $\phi ⌣\psi$:

We must provide an ordering $O$ of all the vertices of the torus. Order them as in a book: top-to-bottom, then left-to-right.

Let us see how to verify the values on the green and yellow 2-simplices. Label the vertices of the inner square as in the figure; the index ordering is compatible with $O$.

For the green 2-simplex $[v_0,v_1,v_2]$, we have $\phi ([v_0,v_1])=+1$ and $\psi ([v_1,v_2])=-1$, so:

$$(\phi ⌣\psi )([v_0,v_1,v_2])=(+1)\cdot (-1)=-1$$

For the yellow 2-simplex $[v_1,v_2,v_3]$, we have $\phi ([v_1,v_2])=-1$ and $\psi ([v_2,v_3])=0$, so:

$$(\phi ⌣\psi )([v_1,v_2,v_3])=(+1)\cdot (0)=0$$

A calculation with more details (illustrating bilinearity) can be performed in terms of the indicator functions. Write:

$$\phi ={\phi }_1+\cdots +{\phi }_6,\psi ={\psi }_1+\cdots +{\psi }_6$$

where ${\phi }_i$ is the indicator for the edge in the support of $\phi$ which is in the $i^{\text{th}}$ place from the top (oriented rightwards), and analogously ${\psi }_j$ is for the one for $\psi$ in the $j^{\text{th}}$ place from the left (oriented upwards).

Applying bilinearity to expand the cup product:

$$\begin{array}{c}\phi ⌣\psi \gg \gg ({\phi }_1+\cdots +{\phi }_6)⌣({\psi }_1+\cdots +{\psi }_6)\\ \gg \gg \sum _{i=1}^6\sum _{j=1}^6{\phi }_i⌣{\psi }_j\end{array}$$

The edges with indicators ${\phi }_i$ and ${\psi }_j$ could span one of the 2-simplices only when $i=3,4,5$ and $j=3,4,5$. And in fact the pairs $(i,j)=(3,5),(5,3)$ can be ruled out as well. The only terms to consider are these six:

$$\begin{array}{c}\sum _{i=1}^6\sum _{j=1}^6{\phi }_i⌣{\psi }_j\gg \gg \begin{array}{r}{\phi }_3⌣{\psi }_3+{\phi }_3⌣{\psi }_4+{\phi }_4⌣{\psi }_3\\ +{\phi }_4⌣{\psi }_5+{\phi }_5⌣{\psi }_4+{\phi }_5⌣{\psi }_5\end{array}\end{array}$$

• Now ${\phi }_3$ is the indicator for $[v_0,v_1]$ and ${\psi }_3$ is the indicator for $[v_0,v_2]$, so ${\phi }_3⌣{\psi }_3=0$ because $v_1\nleqq v_0$ so the total list $v_0,v_1,v_0,v_2$ does not split according to the ordering of $O$.
• Then ${\psi }_4$ is the indicator for $-[v_1,v_2]$, so ${\phi }_3⌣{\psi }_4$ is the indicator for $-[v_0,v_1,v_2]$, meaning it returns $-1$ on the green 2-simplex.
• Then ${\phi }_4$ is the indicator for $-[v_1,v_2]$, so ${\phi }_4⌣{\psi }_3=0$.
• Then ${\psi }_5$ is the indicator for $-[v_1,v_3]$, so ${\phi }_4⌣{\psi }_5=0$ because $v_2\nleqq v_1$.
• Then ${\phi }_5$ is the indicator for $[v_2,v_3]$, so ${\phi }_5⌣{\psi }_4=0$ because $v_2\nleqq v_1$.
• Then ${\phi }_5⌣{\psi }_5=0$ as well, also because $v_2\nleqq v_1$.

By adding these results, we see that $\phi ⌣\psi$ is the indicator for the central 2-simplex, oriented clockwise.

This cochain in $C^2$ is automatically a cocycle because $C^3$ is empty.

Is it a coboundary?

To see that the answer is “no,” we can devise another scheme to represent coboundaries, now for $2$-simplices. We give here only a sketch. Give a 1-cochain $\xi$ which assigns integers to the edges. Draw oriented paths crossing the edges, with the number of paths crossing $e_i$ matching $\xi (e_i)$. Then $\delta \xi (t_j)$ for a triangle $t_j$ counts the total flow of paths into or out from $t_j$; it is a ‘sink’ or ‘source’ number describing extra paths originating or terminating inside $t_j$. The condition $\delta \xi (t_j)=0$ would mean that the paths are conserved in $t_j$, none created or destroyed.

With this perspective, it is obvious that the cochain $\phi ⌣\psi$ depicted in the figure cannot be a coboundary. Every 2-cochain which is a coboundary describes a collection of oriented paths that are created or destroyed only in triangles with nonzero assignments, according to that assignment. So $\phi ⌣\psi$ would describe a single path terminating in the green triangle, but originating nowhere, since there is no other sink or source.

Another observation is worth making about $\phi ⌣\psi$. It is the generator of $H^2(T)$.

Note that the full cohomology of $T$ is given by:

$$H^i(T)\cong \{\begin{array}{ll}0& i>3\\ \mathbb{Z}& i=2\\ {\mathbb{Z}}^2& i=1\\ \mathbb{Z}& i=0\end{array}$$

Here is a fact: All indicator functions on 2-simplices determine the same cohomology class. To see this, take the difference of any two indicators. Draw a simple oriented path out of one of them into the other. This path will define a 1-cochain by assigning $\pm 1$ on any edge it crosses on the way (according to their mutual orientations), and 0 on all edges it does not cross. The coboundary of this cochain gives $+1$ on the triangle from which the path emanates, and $-1$ on the triangle in which the path terminates, and zero on all other triangles. That is exactly the difference of indicator 2-simplices we started with.

Since $H^2(T)$ is generated by indicator 2-simplices, and they are all equivalent, we deduce that it is generated by $\phi ⌣\psi$.

To see that it is in fact infinite and isomorphic to $\mathbb{Z}$, we must exclude relations among the multiples of $\phi ⌣\psi$. A relation $n(\phi ⌣\psi )=m(\phi ⌣\psi )$ is equivalent to the relation $\ell (\phi ⌣\psi )=0$ for $\ell =n-m$. But we can exclude any such relation by the reasoning used above: a cochain $\ell (\phi ⌣\psi )$ cannot be a coboundary because it would represent $\ell$ oriented paths terminating in the green 2-simplex, and originating nowhere.

Therefore we have proven that:

$$H^2(T)\cong ⟨[\phi ⌣\psi ]⟩\cong \mathbb{Z}$$

The cocycle $[\phi ⌣\psi ]$ is called the fundamental cocycle of $T$.

Cohomology ring has topological significance

Cohomology ring has topological significance

Let $X={\mathbb{S}}^1\times {\mathbb{S}}^1$, the 2-torus.

Let $Y={\mathbb{S}}^1\vee {\mathbb{S}}^1\vee {\mathbb{S}}^2$, a 2-sphere with two circles attached at one common point.

The cohomology groups are isomorphic:

$$H^i(X)\cong H^i(Y)\cong \{\begin{array}{ll}0& i>2\\ \mathbb{Z}& i=2\\ \mathbb{Z}\oplus \mathbb{Z}& i=1\\ \mathbb{Z}& i=0\end{array}$$

However, the cup product operations differ, so the ring structures differ.

• In $X$, the cup product of the two generating cocycles is the fundamental coclass of the 2-torus.
• In $Y$, the cup product of the two generating cocycles (dual to the two copies of ${\mathbb{S}}^1$), is zero.