Linking pairing in 3-manifolds. Why is it defined in the torsion homology? I am completely new in 3-manifolds but I have to understand their equivalent of intersection pairing in 4-manifolds.
Consider a 3-manifold $M$.We define the linking pairing on the torsion part of $H_1(M,\mathbb{Z})$ as following
$$ lk: \text{Tor} H_1(M) \times \text{Tor} H_1(M) \to \mathbb{Q/Z} $$
$$ [a] \times [b] \to \#(a \cap b')/n$$
where $b'$ is a 2-cycle such that its boundary satisfies $\partial b' = nb$.
From the right hand side I see that we are calculating the number of intersection points between a 1-cycle $a$ and a 2-cycle $b'$ in some sense.
The questions I have are the following:

*

*This is probably the most naive question, but why is this defined in the torsion part of the homology? Why does even the homology "need" to have torsion part?


*Why can we not define directly the intersection of a 1-cycle and a 2-cycle and have have to go through this weird definition using the boundary?
If you can direct me to any reference it would be great since a quick search on Google did not yield any fruitful results.
 A: Recall that for links in $S^3$ the definition of linking number (usually given via planar projections) can be restated as follows.

$\textbf{Definition 1.}$ Suppose that $K_1$ and $K_2$ are two knots in $S^3$. Since $H_1(S^3 - K_2, \mathbb{Z})= \mathbb{Z}$ generated by a meridian $\mu$ of $K_2$, we have that $[K_1]=m \cdot [\mu]$ for some $m \in \mathbb{Z}$. Define $lk(K_1, K_2)=m$.
$\textbf{Definition 2.}$ Suppose that $K_1$ and $K_2$ are two knots in $S^3$. Pick a surface $F \subset S^3$ such that $\partial F = K_2$ and define $lk(K_1, K_2)=\# (K_1 \cap F)$.

Let us concentrate on the second one. What is crucial in this definition is that $[K_2]=0$ in $H_1(S^3, \mathbb{Z})$. In fact this condition guarantees the existence of a surface $F$ bounding $K_2$. From this perspective it is clear how to extend the definition of linking number to the case of knots in a integral homology sphere $Y$ (meaning that $H_1(Y, \mathbb{Z})=0$).
Let us suppose now that $Y$ is a rational homology sphere ($H_1(Y, \mathbb{Q})=0$). In this case the existence of a surface $F \subset Y$ bounding $K_2$ is not guaranteed since it can happen that $[K_2]\not=0$ in $H_1(Y, \mathbb{Q})$. On the other hand since $Y$ is a rational homology sphere we can find $n \in \mathbb{Z}$ such that $n[K_2]=0$, meaning that there exists a surface $F \subset Y$ (with $n$ boundary components) bounding $n$ parallel copies of $K_2$. Define
$$ lk_\mathbb{Q}(K_1, K_2)= \frac{ \#(F \cap K_2)}{n}\ .$$
From here it is clear how to define the linking number of two knots a in rational homology sphere, or more in general of two rationally null-homologous knots in a random three-manifold. The funny thing is that one can check that $lk_\mathbb{Q}(K_1, K_2) \in \mathbb{Q}/\mathbb{Z}$ only depends on the homology classes of $K_1$ and $K_2$.
I hope that now it is clear why there is a bilinear (non-singular) pairing
$$lk_\mathbb{Q}: \text{Tor}H_1(Y, \mathbb{Z}) \times \text{Tor} H_1(Y, \mathbb{Z}) \to \mathbb{Q}/ \mathbb{Z}. $$
Clearly there is no meaningful  way to extend this to a non-singular pairing on $H_1(Y, \mathbb{Z})$ since $\mathbb{Q}/\mathbb{Z}$ contains only torsion elements.
So, in the case when $H_1(Y, \mathbb{Z})=0$ there is no linking pairing. Here I would like to remark that the analogy with four-manifolds is not that illuminating: the intersection form of a smooth four-manifold recognises its topological type  while the linking form can't distinguish $S^3$ from the Poincaré homology sphere (just to give a concrete example).
About your second question: yes, you can define the linking form without referring to surfaces. This is a generalisation of the first definition I gave.

$\textbf{Definition 1'.}$ Suppose that $K_1, K_2 \subset Y$ are rationally null-homologous. Since the kernel of the inclusion $i_* : H_1(Y- K_2, \mathbb{Q})\to H_1(Y, \mathbb{Q})$ is generated by a meridian $\mu$ of $K_2$ and $i_*[K_1]=0$ in $H_1(Y, \mathbb{Q})$, there exists $q \in \mathbb{Q}$ such that $[K_1]=q [\mu]$ in $H_1(Y-K_2, \mathbb{Q})$. Define $lk_\mathbb{Q}(K_1, K_2)= q$.

For more about this see exercise 4.5.12 at page. 126 of Gompf-Stipsicz's book (the exercise is solved at the end of the book).
