# Framed Cobordism Classes of links in $\mathbb R^3$

We know that every link in $S^3$ is framed cobordant to the unknot with some framing. The idea is to study smooth homotopy classes of maps from $S^3$ to $S^2$. Actually in the title I have given $\mathbb R^3$ but any knot in $S^3$ can be isotoped to miss a point $p$ on $S^3$ and hence lie in $\mathbb R^3\cong S^3-p$

The framing given below is one where the frame twists once as it goes one round along the knot, i.e., framing given by $1\in\pi_1(SO(2))$ Does the disjoint union of $n$ (mutually unlinked) unknots with framing $1\in SO(2)$ represent the unknot with framing $n\in\pi_1(SO(2))$? Somehow I find it difficult to picture whether this is true or not. The only way I can perhaps argue is that in the framed cobordism classes of links form a group and the addition in the group is just the unlinked disjoint union. Is this true?

If false, is there any other way to represent the class of the unknot with framing with $n\in\pi_1(SO(2))$, by a link which has framing with one twist ($1\in SO(2)$) on each knot in the link?

I find the previous answer slightly unsatisfying, since it uses Pontryagin's construction to determine $\pi_3(S^2)$, and the injectivity of such construction is most likely what the OP wishes to prove. Anyhow, here's a proof that doesn't use any of that. Let $l(K,K)$ the selflinking number (you push $K$ along your framing, you obtain $\bar{K}$, and calculate $l(K,\bar{K})$). This is an invariant up to framed cobordism, and it's a matter of a simple check that if $K$ has framing $n$, $l(K,K)=n$, and vice versa.

Next, see this figure: These two are framed cobordant! If you call them $xy=1$ and $xy=-1$, the cobordism is $xy-\phi(t)=0$, and it's framing is $\nabla$. (Here $\phi$ is a fermi function that is $-1$ before $-0.5$ and $1$ after $0.5$).

So, you can use this move to unify each component, like this: Modify the framing on each component so that it's normal outside a small segment, then put them close to each other, and then unify them, like this: In the end, you'll have just one circle, and it's framing is still going to be $n$, since $l(K,K)$ didn't vary.

Let us now look at $f$ as a map from the cube $I^3$, where $I=[0,1]$ with $\left.f\right|_{\partial I^3}\equiv(0,0,-1)$.

Let $K$ and $L$ be two unknots. Let $f$ be smooth with regular value $(0,0,1)\in S^2$. Such that $f^{-1}(0,0,1)=K\cup L$ and the induced framing on $K$ and $L$ via $f$ are $k\in\pi_1(SO(2))$ and $l\in\pi_1(SO(2))$ and $f\equiv(0,0,-1)$ outside $\nu K\cup\nu L$ where $\nu N$ denotes an unspecified neighbourhood of $N$.

Since $K$ and $L$ are unknots, let us assume that $f$ has been smoothly homotoped in such a way that $K$ is a circle lying in $\{1>z>\frac12\}\cap I^3$ and $L$ in $\{0<z<\frac12\}\cap I^3$.

Notice that $f$ is $f_K\star f_L$ where $\star$ is the usual product of representatives of elements in $\pi_n$ of a group.

Since $f_K$ corresponds to $k\in\pi_3(S^2)$ and $f_L$ to $l$, $f$ would correspond to $k+l\in\pi_3(S^2)$ which corresponds a single unknot with framing $k+l$ (by Pontryagin construction)

Also $K$ and $L$ are separated by a plane and are hence unlinked.

Therefore, by repeated application of the above it is clear that the homotopy class corresponding $n$ pairwise unlinked unknots with framing $1$ is that corresponding to a single unknot with framing $n$. So by Pontryagin construction, the disjoint union of n (mutually unlinked) unknots with framing $1\in\pi_1(SO(2))$ represent the unknot with framing $n\in\pi_1(SO(2))$

$\square$