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Let $X$ be a topological space. The universal coefficient theorem says that there is a short exact sequence $$ 0\rightarrow Ext(H_{k-1}(X,\mathbb{Z}),\mathbb{Z})\rightarrow H^{k}(X,\mathbb{Z})\rightarrow Hom(H_{k}(X,\mathbb{Z}),\mathbb{Z})\rightarrow 0 $$ for $1\le k$. In particular a torsion element in $H_{1}(X,\mathbb{Z})$ induces a torsion element in $H^2(X,\mathbb{Z})$.

I would like to understand the statement above in the following situation. Consider a $n$-torus $(S^1)^n$-fibration $f:X\rightarrow Y$. Assume that $H_{1}(Y,\mathbb{Z})=\pi_{1}(Y)=\mathbb{Z}/2\mathbb{Z}$ and monodromy representation is given by $(-id,\dots,-id)$ on $(S^1)^n$. Then the Spectral sequence associated with $f$ shows $$ H_{1}(X,\mathbb{Z})\cong H_{1}(Y,\mathbb{Z})=\mathbb{Z}/2\mathbb{Z}. $$ Is it possible to "see" the torsion in $H^2(X,\mathbb{Z})$ induced by $H_{1}(X,\mathbb{Z})$?

One may want to use Poincare duality (to realize the torsion as a submanifold etc), so please take your favorite $Y$ (satisfying the condition above or something similar) and $n\in \mathbb{N}$.

Thank you for your assistance.

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What does "to see" mean? – Fernando Muro Aug 14 '12 at 23:19
I meant constructing the torsion element explicitly. – Michel Aug 14 '12 at 23:58
The UCSS is explicit enough, isn't it? – Fernando Muro Aug 15 '12 at 10:17
Michel -- I don't quite see how you conclude that $H_1(X,\mathbb{Z})=\mathbb{Z}/2$: if $A$ denotes the non-trivial action of $\pi_1(Y)$ on $\mathbb{Z}$, then $H_0(Y,A)=\mathbb{Z}/2$, not 0. – algori Aug 15 '12 at 17:09
up vote 5 down vote accepted

Having the cycles and boundaries of degree one, it's possible to construct a cohomology class in $H^2(X)$ that corresponds to your torsion element. This is explained in the following:

1. Homological algebra

Let $(C,d)$ be a chain complex of free abelian groups, $G$ an abelian group and let $$\varphi: Ext(H_{q-1}(C),G) \to H^q(C,G)$$ be the monomorphism from UCT. An inspection of the proof in [Spanier: Algebraic Topology, Theorem 5.5.3] reveals: Set $Z_q = \ker(d_g)$, $B_{q-1}=\text{im}(d_q)$. Then there is an exact sequence $$\text{Hom}(Z_{q-1},G) \to \text{Hom}(B_{q-1},G) \to Ext(H_{q-1}(C),G) \to 0.\hspace{50pt}(\ast)$$

If $e \in Ext(H_{q-1}(C),G)$ is represented by $f: B_{q-1} \to G\;$ then $\varphi(e) \in H^q(C,G)$ is represented by the cycle $f \circ d_q: C_q \to G$.

2. The finitely generated case

For simplicity let's assume each $C_q$ has finite rank. Let $\alpha \in H_{q-1}(C)=Z_{q-1}/B_{q-1}$ be a generator of a direct summand of order $n$. Choose a representative $x \in Z_{q-1}$ of $\alpha$. We can find decompositions $Z_{q-1} = Z' \bigoplus \mathbb{Z}x,\;B_{q-1}=B' \oplus \mathbb{Z}nx$ with $B' \le Z'$. Let $f: B_{q-1} \to \mathbb{Z},\;f|B'=0,\;f(nx)=1$. Then $f$ represents $e \in Ext(H_{q-1},\mathbb{Z})$ via $(\ast)$ and $e$ corresponds to $\alpha$ under the non-canonical isomorphism $Ext(H_{q-1}(C),\mathbb{Z}) \cong H_{q-1}(C)_{tor}$. By 1., $h=f \circ d_q: C_q \to \mathbb{Z}$ is a cocycle that represents $\varphi(e)$.

3. An example

Let $X$ be the CW-complex in exercise 4 with $\Pi_1(X)=Q_8$ (quaternion group) and let $C=C(X)$ be the cellular chain complex. $X$ has 0-cells $x, y$, 1-cells $a,b,c,d$, 2-cells $p, q, r$ and one 3-cell. The boundary operator is given by: $$d_1(a)=d_1(b)=d_1(c)=d_1(d)=x-y\hspace{130pt}$$ $$d_2(p)=a-b+c-d,\;\;d_2(q)=c-b+d-a,\;\;d_2(r)=d-b+a-c$$ Set $A = a-b,\; B = a-c,\;C=a-d$. Then $Z_1 = \langle A,B,C\rangle$ and $B_1 = \langle A-B+C,2B,2C \rangle$.

Hence $B$ represents a homology class in $H_1(X)$ of order 2. Define $f: B_1 \to \mathbb{Z}$ by $$f(A-B+C)=0,\; f(2B)=1,\; f(2C)=0.$$ Now by 2., the cocycle $h=f \circ d_2$: $$h(p)=0,\;h(q)=0,\;h(r)=1$$ represents a cohomology class in $H^2(X)$ of order 2.

Remark: Note that we didn't need to consider $d_3: C_3 \to C_2$ in order to construct our torsion class in $H^2(X)$. This, in contrast, would be necessary if we computed $H^2(X)$ directly.

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Sorry for my late reply and thanks for the detailed answer. I now have better understanding on the problem. – Michel Aug 26 '12 at 6:11

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