I am trying to give a brief explanation in which I make use of the concept of surgery on an $m$-manifold $M$.

This is along the lines of (and taken generously from) the Wikipedia entry on Surgery; I just want to make sure that I understand the concept; I am trying to be a bit more explicit than the Wiki entry (no criticism; they did a pretty good job), and, since I am a physicist, I want to make sure I am not glossing over important mathematical issues.

Please note that the (###) denotes the area where I am not fully clear:

Given a $D^p$ , and $S^{(m-p-1)}$, where $D^p$ is the closed unit disk (basically, so that ($p+m-p+(-1)=m-1$), the product $D^p\times S^{m-p-1}$these two spheres can be seen as the boundary of a product $A:=D^{(p+1)}\times S^{m-p-1}$ , or as the boundary of a product $B:= S^p\times D^{m-p}$ (where $D^q$ is defined to be the closed k-dimensional disk $\{ x \in R^{k+1}: \|x\|\leq 1 \}$ ). This means we can recover the original manifold $M$ by doing the "right identification" of $A$ and $B$, along their common boundary $S^p\times S^{m-p-1}$. As examples:

1) In $S^2$: we consider two embedded circles $S^1$ and $(S^{1})'$. These can be made to be the boundary of either an embedded cylinder $A:=S^1\times D^1$ , or of two closed disks given by $B:=D^2 \times S^0$. This means we can recover the original $S^2$ ( up to homeomorphism) by gluing the cylinder with the two disks along the common boundary $S^1\times D^0$ (to get a cylinder, which is isomorphic to the sphere).

2) In the 4-sphere $S^4$: we can obtain $S^4$ by gluing a $D^3\times S^1$ with an $S^2\times D^2$ , along their common boundary $S^2\times S^1$.

3) I am a bit weaker in this one: We consider $S^3$ within $S^4$ , which can be decomposed as two solid tori (use, e.g. the Hopf Fibration, take a disk $D^2$ in $S^2$ , in a trivialized neighborhood of $S^2$ , which lifts to a product $D^2\times S^1 $, and so does its complement in $S^2$).

(###) Then a tubular neighborhood $D^2\times S^1\times S^1$ of a solid torus $D^2\times S^1$ has a complement (in the 4-sphere $S^4$), which is also a $D^2 \times S^1 \times S^1$ (###)

I am not too sure how this last holds.

Sepideh Bakhoda
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  • I edited for presentation and tex, but I didn't check the mathematics/agreement of indices and exponents. You may want to read through again to double check. – Willie Wong Jul 29 '11 at 22:36

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