# How can I exhibit a generator of $H^3(S^3,\mathbb Z_3)$?

How can I exhibit a generator for the third cohomology group $H^3(S^3,\mathbb Z_3)$ of the $3$-sphere with coefficients in $\mathbb Z_3$?

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How are you computing the cohomology? Exhibiting an explicit cocycle can onloy be done in terms of an actual complex which computes the cohomology, and depending on the complex you pick different strategies are available. – Mariano Suárez-Alvarez Jan 31 '12 at 20:25

Let's construct, for example, an explicit cocycle, viewing $S^3$ as a realization of an abstract simplicial complex and computing cohomology using the ordered complex.
Let us triangulate $S^3$ as the boundary of the $4$-simplex. Let $K$ be the corresponding abstract simplicial complex: we can identify the vertex set of $K$ with $V=\{0,1,2,3,4\}$ and the simplices of $K$ are all the proper subsets of $V$. In particular, the set of $2$- and $3$-simplices are the sers $K^2$ and $K^3$ of $3$- and $4$-element subsets of $V$. The chain complex which arises from $K$ ends with $$\cdots\leftarrow\mathbb ZK^2\overset\partial\longleftarrow\mathbb ZK^3$$ where $\mathbb ZK^2$ is the free abelian group with basis the set $K^2$ and similarly for $\mathbb ZK^3$. The boundary map is given on a basis element $(i_0,i_1,i_2,i_3)\in K^3$ by $$\partial(i_0,i_1,i_2,i_3)=(i_1,i_2,i_3)-(i_0,i_2,i_3)+(i_0,i_1,i_3)-(i_0,i_1,i_2).$$ To compute $H^\bullet(S^3,\mathbb Z_3)$, we have to apply the functor $\hom(\mathord-,\mathbb Z_3)$ to the above chain complex and compute the cohomology of the result. Applying the functor, we get a complex which ends in $$\cdots\rightarrow\hom(\mathbb ZK^2,\mathbb Z_3)\overset{\partial^t}\longrightarrow\hom(\mathbb ZK^3,\mathbb Z_3)$$ with the map $\partial^t$ the transpose of $\partial$, so that for each $\phi:\mathbb ZK^2\to\mathbb Z_3$ we have $$(\partial^t\phi)(i_0,i_1,i_2,i_3)=\phi(i_1,i_2,i_3)-\phi(i_0,i_2,i_3)+\phi(i_0,i_1,i_3)-\phi(i_0,i_1,i_2)$$ for each $(i_0,i_1,i_2,i_3)\in K^3$. Now $H^3(S^3,\mathbb Z_3)$ is the cokernel of the map $\partial^t$, and since we know it is a $1$-dimensional vector space over the field $\mathbb Z_3$, to describe a generator we need only exhibit any element in $\hom(\mathbb ZK^3,\mathbb Z_3)$ which is not in the image of $\partial^t$.
Now, a computation which can be rather boring or not, depending on how you organize it, shows that for each $\phi:\mathbb ZK^2\to\mathbb Z_3$ we have \begin{multline} (\partial^t\phi)(1,2,3,4) -(\partial^t\phi)(0,2,3,4) +(\partial^t\phi)(0,1,3,4) \\ -(\partial^t\phi)(0,1,2,4) +(\partial^t\phi)(0,1,2,3) = 0 \end{multline} This implies that if we can find any map $\psi:\mathbb ZK^3\to\mathbb Z_3$ which does not have this property, then it will not belong to the image of $\partial^t$ and therefore its class in $H^3(S^3,\mathbb Z_3)$ will be a generator.
Now, it is clear that if we define $\psi:\mathbb ZK^3\to\mathbb Z_3$ so that $$\psi(i_0,i_1,i_2,i_3)=\begin{cases} 1, & \text{if (i_0,i_1,i_2,i_3)=(0,1,2,3);} \\ 0, & \text{in any other case} \end{cases}$$ the above condition does not hold. Therefore $\psi$ represents a generator.