In this article on surgery on manifolds it is explained that from an $n$-manifold $M$ an $n$-manifold $M'$ can be constructed by cutting out $S^p \times D^q$ and gluing in $D^{p+1}\times S^{q-1}$.

If $n=3=p+q=1+2$ the manifold $M$ could be a solid torus $T= S^1 \times D^2$. Then a surgery removing its interior $A=S^1 \times D^2$ and replacing it with $B=D^2 \times S^1$ should yield a new manifold $M'$.

What does $M'$ look like?

Sepideh Bakhoda
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1 Answers1


For your $1$-surgery on the solid torus $T = S^1 \times D^2$, you should make sure $A$ is a copy of $S^1 \times D^2$ lying completely inside the interior of $T$. You haven't really made this clear so I am unsure which copy of $S^1 \times D^2$ you're talking about. To be perfectly explicit, let's assume the radius of the $D^2$ in $T = S^1 \times D^2$ is $1$, while the radius in the $D^2$ in $A = S^1 \times D^2$ is $\tfrac{1}{2}$; the $S^1$'s in $T$ and $A$ are the same. Then we see that cutting $A$ out of $T$ amounts to "drilling out the core" of the solid torus $T$. If it helps, you can imagine $T$ as a jelly doughnut, $A$ as the jelly, and $T - A$ as the bread of the doughnut.

Now to glue in $B = D^2 \times S^1$, we need more information: how do we glue $B$ in? The instructions are encoded in a homeomorphism $$\phi: S^1 \times S^1 \longrightarrow S^1 \times S^1.$$ In your case, $A \cong B$ (this doesn't happen for general surgeries), so if $\phi = \mathrm{Id}_{S^1 \times S^1}$, then we get our original $T$ back. If $\phi$ is a map that "twists" $S^1 \times S^1$ around its core $p$ times (in other words, if $\mu$, $\lambda$ are the standard meridionial and longitutinal generators of $H^1(S^1 \times S^1)$ and $\phi_\ast(\lambda) = \lambda + p \mu$), then the result of the surgery will be a lens space minus a solid torus: $L(p, 1) - S^1 \times D^2$ (this is just the standard genus $1$ Heegaard splitting of $L(p,1)$ except one of the solid tori is missing its core).

Here's an easier example. Take $M = S^2$, $p = 0$, $q = 3$. So we need to cut out a copy of $S^0 \times D^2$; this is just a pair of disks. Let these just be two small disks centered and the north and south poles of $S^2$. Now we glue back in a copy of $D^1 \times S^1$; this is just the surface of a cylinder. Using $\phi = \mathrm{Id}_{S^0 \times S^1}$ as our gluing map, the result of this surgery will be the torus (we've just glued the "handle" $D^1 \times S^1$ to our sphere). Note that this time when we used the identity map to glue we still got a new manifold. This is because $S^0 \times D^2$ and $D^1 \times S^1$ are different; in the first example $S^1 \times D^2$ and $D^2 \times S^1$ are clearly homeomorphic.

Henry T. Horton
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  • Is the fundamental group of $L(p,1)-S^1 \times D^2$ trivial? I am looking for an example of a manifold (or even better of a surface) to which one can apply surgery to kill either $\pi_1$ or $\pi_2$. – snailspace Jun 13 '13 at 09:02
  • In this case we don't get a simply connected manifold. For the torus $S^1 \times S^1$, performing a $1$-surgery on a curve representing the longitude kills the fundamental group (gluing in $D^2 \times S^0$ gives us a $2$-sphere). In general, there is a process called "surgery below the middle dimension" for stably-parallelizable $n$-manifolds $M$ that allows us to perform a sequence of surgeries on $M$ to obtain a manifold $M'$ with $\pi_i(M')$ trivial for $i < n/2$. See for example the classic paper of Milnor: http://www.maths.ed.ac.uk/~aar/papers/milnorsurg.pdf – Henry T. Horton Jun 13 '13 at 23:26