The "pants" bordism in dimension n is a bordism which goes from $S^n \sqcup S^n$ to $S^n$ witnessing the connected sum operation - equivalently by attaching 1-handle to the trivial bordism, equivalently doing surgery on a 0-sphere.


The "copants" bordism is the same manifold, but thought of as a bordism from $S^n$ to the disjoint union $S^n \sqcup S^n$. It can be understood as doing surgery on an (n-1)-sphere (the equator).


Now I want to understand when these manifolds can be understood as framed manifolds (both tangential framing and stable framing are relevant) and if they are framed what constraints they put on the framings of the boundaries.


Now we can embed $S^n$ into $\mathbb{R}^{n+1}$ in the standard way and choose the outward normal to identify $\underline{\mathbb{R}}^{n+1} \cong \tau\_{S^n} \oplus \underline{\mathbb{R}}$, giving $S^n$ a *standard* (n+1)-framing. This extends over the interior of the solid ball in $\mathbb{R}^{n+1}$.


Now the pants bordism can always be realized as an $(n+1)$-framed manifold because it too can be viewed as a subset of $\mathbb{R}^{n+1}$. Think of two spheres inside a larger sphere and take the intervening space between them. This gives a framed bordism from the disjoint union of two "standard" spheres to one "standard" sphere.


Now the framings on the spheres will be a torsor for $\pi\_n(SO(n+1)$. The standard framing gives us a basepoint (which we identify with the identity element in $\pi\_n(SO(n+1)$). The framings on the pants bordism are uniquely determined by the two framings on the incoming spheres. There are $\pi\_n(SO(n+1) \times \pi\_n(SO(n+1)$ many and if the framings of the two incoming spheres are (under our identification) $x, y \in \pi\_n(SO(n+1)$, then the framing of the outgoing sphere will be $x + y$.


Now when $n=1$, we can also frame the copants bordism. However for that bordism if the outgoing framings of the two outgoing spheres are $x$ and $y$, then the incoming sphere has framing $x + y + 2 \in \pi\_1 SO(2) = \mathbb{Z}$.


**Question:** My question is what happens in higher dimensions? Can the copants bordism be framed? If so what are the constraints of the bourdary framings? Is this question easier if we use stable framings instead?


I checked Kervaire-Milnor's "groups of homotopy spheres: I" paper. In this case the obstruction for extending the framing over the pants bordism always is trivial, as expected. However for the copants bordism in general there is the possibility of a non-trivial obstruction. Furthermore their result that the surgery sphere can always be reframed to eliminate the obstruction doesn't apply in this case (because the dimension of the surgery sphere is too high).


On the otherhand, I think we can just view the pants bordism as a bordism the otherway around to get some (tangential) framing on the copants. It is then a matter of pinning down how the framings on the boundaries change when we do this.



Framed version of the "copants bordism"?
The key point is the identification $\tau(S^n)+\mathbb{R}$ with the restriction of tangent bundle of the bordism. I will read bordism from bottom to top.
Even to obtain pants bordism between standard spheres you need to choose the $\textit{inward}$ normal at bottom and $\textit{outward}$ normal at top for the above identification.


Let us examine the difference between pants and copants. In the first case we take a cartoon pants with wide sphere at top and pair of small spheres at bottom in $\mathbb{R}^{n+1}\times I$ s.t. the projection onto $\mathbb{R}^{n+1}$ is the described above picture with two spheres inside the bigger one. Note that induced framing from $\mathbb{R}^{n+1}$ is perfectly compatible with our identification.


Let us write $(s\_1,s\_2,s\_0)$ for the framing on the boundary spheres of this pants bordism.


Now reverse the picture (or simply read from top to bottom). Projection is the same and we can induce the framing from $\mathbb{R}^{n+1}$ as above. Note that each inward normal becomes outward and vice versa. Hence induced framing is $(R(s\_1),R(s\_2),R(s\_0))$, where $R$ is reflecting of a framing $\tau+\mathbb{R}$ over $S^n$ in normal direction $\mathbb{R}$. I claim that $R(s)=s-t$, where $t\in \pi\_n(SO(n+1))$ corresponds to a gluing function for tangent bundle over $S^{n+1}$: it is clear after considering filled disk $D^{n+1}$ bounding $S^n$ as upper and respectively lower semi-spheres of $S^{n+1}$.


Now, as was noted framed pants/copants respect an action of $\pi\_n(SO(n+1))\times \pi\_n(SO(n+1))$ by changing framings on the boundary components: $a\times b\cdot (s\_1,s\_2,s\_0)=(s\_1+a,s\_2+b,s\_0+a+b)$. In particular we have a copants framing corresponding to $(R(s\_1)+t,R(s\_2)+t,R(s\_0)+2t)$ which is $(s\_1,s\_2,s\_0+t)$.


In other words, given a copants bordism with the standard framing over output spheres we unavoidably (unless $S^{n+1}$ is parallelizable) get a non standard framing over input sphere. The difference with the standard framing is $t\in \pi\_n(SO(n+1))$ corresponding to $\tau(S^{n+1})$.


On the other hand standard copants bordism fall into stable range very quickly and the obstruction disappear: if you add $\mathbb{R}$ everywhere, then the same argument gives the difference in $\pi\_n(SO(n+2))$ which is the image of $\tau(S^{n+1})\in \pi\_n(SO(n+1))$ under inclusion $SO(n+1)\to SO(n+2)$. This is zero since $\tau(S^{n+1})+\mathbb{R}$ is trivial.


