Let $M^3$ be a rational homology 3-sphere. (i,e, $M^3$ is closed 3-manifold with $H_{*}(M;Q)=H_{*}(S^3;Q)$

As beautifully explained in Ranicki's Algebraic and Geometry surgery book and Davis-Kirk's Lecture notes in Algebraic toplogy book, we have a $Q/Z$ valued linking form, $\lambda\colon H_{1}(M;Z)\times H_{1}(M;Z)\to Q/Z$ defined by the adjoint to following isomorphism.

(Actually we can define linking form in more general setting, e.g.) odd dimensional manifold without the restriction such as rational homology sphere condition. Because I want to just intuitive idea about linking form, I restricted the case)

$H_1(M;Z)\cong H^2(M;Z)\cong H^1(M;Q/Z)=Hom(H_1(M;Z),Q/Z)$

First isomorphism : Poincare duality,

Second isomorphism : Inverse of Bockstein homomorphism $\delta$ induced form $0\rightarrow Z \rightarrow Q\rightarrow Q/Z\rightarrow 0$. More precisely, induced long exact sequence is that $\ldots\rightarrow H^1(M;Q)\rightarrow H^1(M;Q/Z)\rightarrow H^2(M;Z)\rightarrow H^2(M;Q)$ and here $H^1(M;Q)$ and $H^2(M;Q)$ vanishes. Therefore, the long exact sequence shows that $H^1(M;Q/Z)$ and $H^2(M;Z)$ are naturally isomorphic (if $M$ is rational homology 3-sphere) and we call that homomorphism as Bockstein homomorphism.

Third isomorphism : Universal Coefficient theorem

In short, $\lambda\colon H_{1}(M;Z)\times H_{1}(M;Z)\to Q/Z$ is defined by $\lambda(x,y)=<\tilde{x}\cup\delta^{-1}(\tilde{y}),[M]>$, where $\tilde{x},\tilde{y}$ are Poincare dual to $x,y$ and $\delta\colon H^1(M;Q/Z)\to H^2(M;Z)$ is a Bockstein homomorphism as defined above. Cup products are defined in $H^1(M;Z)\times H^2(M;Q/Z) \to H^3(M;Q/Z)$ induced from the Bilinear map $Z \times Q/Z \to Q/Z$.

I understand this linking form only algebraic viewpoint. Therefore, I can play with this form only by using dilluminating algebraic topology.

I'm struggle to find geometric interpretation but I have no idea to express the Q/Z valued in terems of geometric language which seems to be highly algebraic. (Feeling like Injective, divisible, Ext or something linke that )

Are there any intuitve and geometric (clean) interpretation about this linking form? (e.g.)such as Alexander duality, like the argument that .........we can find a dual basis which represents meridian......)

• Ryan managed to type his answer faster than me, so I'll content myself with saying that there is a (possibly) helpful bibliography about the linking pairing in the introduction to my paper "Symplectic Heegaard splittings and linked abelian groups" with Birman and Johnson. – Andy Putman Aug 28 '10 at 20:14

This may be a good place to explain the well-known principle $$\text{intersection in the interior = linking in the boundary}$$ in an oriented $m$-dimensional manifold with boundary $(M,\partial M)$. Let $$f~:~(K,\partial K)\subset (M,\partial M)~,~g~:~(L,\partial L) \subset (M,\partial M)$$ be embeddings of oriented manifolds with boundary, such that $${\rm dim}(K)~=~k~,~{\rm dim}(L)~=~\ell~,~k+\ell~=~m~,~ f(\partial K) \cap g(\partial L)~=~\emptyset \subset \partial M~.$$ Assume there exists an isotopy (= homotopy through embeddings) rel $\partial K$ $$f_t~:~K \to M~~(0 \leqslant t \leqslant 1)$$ such that $f_0=f$ and $f_1(K)\subset \partial M$, with each $f_t(K), g(L)$ intersecting transversely in $M$, so that $f_t(K) \cap g(L)\subset M$ is a finite set with an intersection index $I(x)\in \{\pm 1\}$ at each point $x \in f_t(K) \cap g(L)$ according to the orientations. A continuity argument shows that the function $$\lambda~:~[0,1] \to {\mathbb Z}~:~t\mapsto \lambda(t)= \sum\limits_{x \in f_t(K) \cap g(L)}I(x)$$ is constant, so that $${\rm intersection}(f(K),g(L)\subset M)~=~\lambda(0)~=~\lambda(1)$$ $$=~{\rm linking}(f(\partial K),g(\partial L) \subset \partial M) \in {\mathbb Z}~.$$ This is best seen by drawing pictures for $(M,\partial M)=(D^2,S^1)$.

The localization exact sequence in algebraic $L$-theory is based on an abstract homological version of this principle.

• Oh dear Ranicki, I like your beautiful monographs. I want to understand the relation between transverse intersection and the term "linking" as you mentioned – Topologieee Aug 31 '10 at 12:42
• @Topologieee: You have to confront this when making a different choice of cycles $A$ or $B$ in my answer. i.e. consider the problem of verifying the linking form that I describe is well-defined without recourse to Poincare duality, i.e. just use simple geometric arguments. – Ryan Budney Feb 23 at 18:40

You get the geometric interpretation of the linking form by tracing through exactly what the isomorphisms are and looking at what they do.

The upshot is the answer is this. Let $[a]$ and $[b]$ be torsion classes in $H_1(M;\mathbb Z)$. Let $n[a]=\partial [A], m[b]=\partial [B]$ for $n,m \neq 0$. $A$ and $B$ are $2$-cycles in $M$ is a $3$-manifold.

Then

$$\langle [a],[b]\rangle = \frac{1}{m} a\pitchfork B = \frac{1}{n} A \pitchfork b \in \mathbb Q / \mathbb Z$$

here $\pitchfork$ means the transverse intersection, meaning you have to find chain representatives $a$ and $B$ (or $A$ and $b$) such that they intersect transversely -- the representatives do not need to be manifolds, as you can make sense of this in the PL-category. This is the algebraic intersection number, where every point of intersection is given a weight $\pm 1$, according to how the relative orientations add up to either the local orientation of the manifold or its reverse.

Perhaps the simplest context where a proof of this is relatively easy would be if your $3$-manifold is triangulated, then your Poincare duality isomorphism is naturally between the simplicial homology and the CW-cohomology of the dual polyhedral decomposition, so transversality is for free in this context.

• If you like these kinds of games where you re-interpret the cup product as a pairing on homology via Poincare duality and universal coefficients, the story gets more entertaining when you look at the Alexander modules of knot complements $S^n \to S^{n+2}$. Since you're dealing with modules over Laurent polynomial rings, $Ext$ has more than one interesting term, so you get other interesting products. One is called the Farber-Levine pairing. – Ryan Budney Aug 29 '10 at 17:05