# Spinning construction

The p-spinning construction works as follows:

You start with $A$ an $n$-ball embedded in an $m$-ball $B$ such that $\partial A \subset \partial B$ and $A^\circ \subset B^\circ$.

The $p$-spin of (A,B) is $\partial (A \times D^{p+1})$ in $\partial (B \times D^{p+1})$.

It is then claimed that the $p$-spin is created by sweeping interior points on a $p$-sphere while fixing the boundary.

I kinda see it is as to $x \in A^\circ$ corresponds $x \times S^p$, so indeed we sweep the point $x$ around $S^p$. What I can't see is how we an say that the boundary is fixed as in an analogous way to a point $x \in \partial A$ corresponds $x \times D^{p+1}$ and so I would say we sweep points in the boundary through a disk and not fixing it.

-
I don't quite understand what "$\partial X$ in $\partial Y$" means here. It makes sense to me when $X\subset\partial Y$, for then we just take the boundary of $X$ in the topology of $\partial Y$. But this does not seem to be the case here. –  user31373 Jul 15 '12 at 3:20
The boundary is here meant in the sense of manifolds. So $\partial (A \times D^{p+1})$ is $A \times S^p \cup \partial A \times D^{p+1}$ and similar with $\partial (B\times D^{p+1})$. But each part of the first decomposition has an embedding in the corresponding part in the B case as $A \subset B$ and $\partial A \subset \partial B$. –  mna Jul 15 '12 at 6:57
Then why do we talk about B at all, if the spinning is just the boundary of the product of A with a disk? This does not look right. –  user31373 Jul 15 '12 at 13:03
$\partial (A \times D^{p+1})$ itself is not interesting it is just $S^{n+p}$. What is interesting is the embedding of this in $\partial (B \times D^{p+1}) \cong S^{m+p}$. The construction is used to get higher dimensional knots from lower dimensional. –  mna Jul 15 '12 at 22:21