We can assign a framing to a knot $K$ (in some nice enough space $M$) in order to calculate the self-linking number $lk(K,K)$. But of course it is not necessarily canonical, as added twists in your vector field can remove/add crossings.

Two things are stated in Witten's QFT paper on the Jones polynomial, which I do not quite see:

1) On $S^3$ we do have a canonical framing of knots, by requesting that $lk(K,K)=0$.
Why? I must be understanding this incorrectly, because if we decide the framing by requiring $lk(K,K)=0$, so that the framing has whatever twists it needs to accomplish this, then aren't we making a choice?? We could have simply required $lk(K,K)=n$ for any integer $n$. If $n> 0$ does there then exist multiple possible framings?

2) For general 3-manifolds, we can have $lk(K,K)$ ill-defined or it can be a fixed fraction (modulo $\mathbb{Z}$) so that any choice of framing won't make it $0$.
What are some examples? When is it possible to set a fixed fraction? Is there a relation between the 3-manifold $M$ and the fractional value you can assign to $lk(K,K)$?

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I see my answer wasnt so clear. Let me try again. Your first comment says that given a knot $K:S^1\subset M$ in a 3-manifold, with a choice of framing, i.e. normal vector field to $K$, you can calculate the self-linking number. My first remark is that this is only possible if the knot is nullhomologous, i.e. represents $0$ in $H_1(M)$. For example, there is no self-linking number of the core $0\times S^1\subset D^2\times S^1$ of a solid torus, no matter how you frame it.

If K is nullhomologous, then depending on how you think of homology you see that there is a 2-chain with boundary the knot $K$. It is true, but a bit more work, to see that in fact there exists a oriented embedded surface $C\subset M$ with boundary $K$. (so you can take the 2-chain to be the sum of the triangles in a triangulation of $C$. Then given any other knot $L$ disjoint from $K$ (for example a push off of $K$ with respect to some framing) then the intersection of $C$ with $L$ is by defintion $lk(K,L)$ and is an integer. You may worry about whether it is independent of the choice of $C$, and the answer is yes if $L$ is also nullhomologous, or more generally torsion (i.e. finite order) in $H_1(M)$, and moreover in this case it is also symmetric, $lk(K,L)=lk(L,K)$. Notice that no framing of $K$ or $L$ was used to define $lk(K,L)$.

Now to answer your questions. Since $H_1(S^3)=0$, every knot in $S^3$ is nullhomologous. Thus any two component link in $S^3$ has a well defined integer linking number. You are considering the 2 component link determined by $K$ and a normal framing: the normal framing is used to push off $K$ to get $L=f(K)$. As you note, changing the framing changes the linking number, and in fact by twisting once over a small arc in $K$ you can change it by $\pm1$. Thus there is some framing $f$ so that $lk(K, f(K))=0$, this is the canonical framing (typically called the "0-framing"). It makes sense in $S^3$, or any 3-manifold with $H_1(M)=0$.

For your second question, you are referring to a slightly different concept, which is the linking pairing $lk:Torsion(H_1(M))\times Torsion(H_1(M))\to Q/Z$. It is defined as follows: given $a,b$ torsion classes, then some integer $n$ (for example the order of $torsion(H_1(M))$) has the property that $na$ is nullhomologous. Thus $na$ is represented by a nullhomologous knot, call it $K$. $b$ is also represented by a knot say $L$, which can be perturbed to be disjoint from $K$. Then define $lk(a,b)$ to be $(1/n) lk(K,L)$ mod $Z$, with $lk(K,L)$ as above.

For example, if $P$ is a knot in a lens space $M=L(p,q)$ with $H_1(M)=Z/p$, you could take $a=[P]$ and $b=[P]$ in $H_1(M)$, and then $lk(a,b)=q n^2/p$ for some integer $n$ that depends on $a$. Note that the answer (mod Z!) is independent of how you push $P$ off itself, in particular, the framing of $P$ is irrelevant, and you'll never get $0$ unless $a=0$ (i.e. $P$ is nullhomologous). Note also that if you don't mod out by the integers then the linking pairing is not well defined.

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your first sentence isn't quite right: you can't define a linking number to a framed knot in a 3-manifold in general. both of your questions are related to this.

2 steps:

1. I K U L is a link in M (and to be clear, M is an oriented 3-manifold, not a general space) and K,L are nullhomologous in M, let C be a 2-chain with boundary K, and define lk(K,L) as the integer intersection number of L with C (ie make the chains simplicial and transverse and count the intersections of L with C with signs). If K is a knot, you can let L be a pushoff wrt a framing and then you see the dependence on the framing. Note that this is geoemtric, not homological.

1. if $k,l \in H_1(M)$ are torsion classes represented by specific cycles, then take $n\in Z$ so that $nk=0$, then let $c$ be a 2-chain with boundary $nk$, and define $lk(k,l)= (1/n) c \cdot l$ in $Q/Z$. Taking the quotient in by $Z$ ensures that the answer is independent of the representative cycles.

This second construction is called the "linking pairing" of M. it si homological, not geometric.

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I'm sorry but I don't see how this answers all of my questions. In particular, can you clarify your 2nd construction in an edit? (For instance, what is $c\cdot l$, and why does this work when we can switch factors and consider $lk(l,k)$ instead). And how is this homological definition (in the case of torsion) the same as the geometric one? – Chris Gerig Jun 28 '12 at 3:06