3 typo

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 inverse 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.

2 typo

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 cycle chain representatives $a$ and $B$ (or $A$ and $b$) such that they inverse 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.

1

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 cycle representatives $a$ and $B$ (or $A$ and $b$) such that they inverse 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.