Torsion in $H^2(X,\mathbb{Z})$ induced by torsion in $H_1(X,\mathbb{Z})$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:48:36Z http://mathoverflow.net/feeds/question/104724 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104724/torsion-in-h2x-mathbbz-induced-by-torsion-in-h-1x-mathbbz Torsion in $H^2(X,\mathbb{Z})$ induced by torsion in $H_1(X,\mathbb{Z})$ Michel 2012-08-14T21:56:20Z 2012-08-17T22:21:15Z <p>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 $H^2(X,\mathbb{Z})$. </p> <p>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 $H_{1}(X,\mathbb{Z})$? </p> <p>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 $n\in \mathbb{N}$. </p> <p>Thank you for your assistance. </p> http://mathoverflow.net/questions/104724/torsion-in-h2x-mathbbz-induced-by-torsion-in-h-1x-mathbbz/104950#104950 Answer by Ralph for Torsion in $H^2(X,\mathbb{Z})$ induced by torsion in $H_1(X,\mathbb{Z})$ Ralph 2012-08-17T22:21:15Z 2012-08-17T22:21:15Z <p>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: </p> <p><strong>1. Homological algebra</strong></p> <p>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 $$\text{Hom}(Z_{q-1},G) \to \text{Hom}(B_{q-1},G) \to Ext(H_{q-1}(C),G) \to 0.\hspace{50pt}(\ast)$$</p> <blockquote> <p>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 $f \circ d_q: C_q \to G$. </p> </blockquote> <p><strong>2. The finitely generated case</strong></p> <p>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 $\varphi(e)$. </p> <p><strong>3. An example</strong></p> <p>Let $X$ be the CW-complex in <a href="http://www.stanford.edu/class/math215b/Sol2.pdf" rel="nofollow">exercise 4</a> 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 $B_1 = \langle A-B+C,2B,2C \rangle$. </p> <p>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. </p> <p><strong>Remark:</strong> 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. </p>