4
$\begingroup$

Given two disjoint oriented knots $K_1, K_2 \subset S^3$, I think that there is a notion of knot connected sum $K_1 \sharp K_2 \subset S^3$ defined by picking a path between $K_1, K_2$. Is the connected sum independent of the path and does isotopy class of the connected sum depends only on the isotopy class of $K_1 \cup K_2$ as a link in $S^3$? Can someone provide a reference for these statements or a counterexample?

For any disjoint submanifolds $A^k, B^k \subset M^n$ (perhaps with a choice of parametrization?), I can define the connected sum by picking a path between $A^k, B^k$. Does the same hold as before? In particular, does the isotopy class of the connected sum depend only on the isotopy class of $A^k \cup B^k$ and not the path I choose? As pointed out in the comments, here we might need $\pi_1(M, A\cup B)$ to vanish; assuming this, is the connected sum independent of path and isotopy?

$\endgroup$
  • $\begingroup$ The connected sum of knots is not defined by picking an arbitrary path between components. That will not be well-defined. Rather, you pick an "unknotted path" between components, as defined by picking a diagram of the link and then picking a path in the plane disjoint (except for endpoints) from the link. As long as the dimension of the ambient manifold is at least 4, connected sum of submanifolds should be well-defined (since every path is unknotted). $\endgroup$ – user98602 Jan 18 '17 at 7:14
  • $\begingroup$ Actually, that's not true either. You need $\pi_1(M, A \cup B)$ to consist of a single path (where here I'm cheating a bit and really referring to smooth embedded paths up to isotopy). This will be true when eg $M$ is dimension at least 4 and the submanifolds are codimension at least 3. When the submanifolds are cosimension 2 you'll need some clever process like in the case of knots. $\endgroup$ – user98602 Jan 18 '17 at 7:19
  • $\begingroup$ @MikeMiller Can you provide a reference for when the connected sum depends on the path chosen? or state what the knots are. For example, if I start with the unlink and do connected sum, does the result depend on the chosen path? $\endgroup$ – user39598 Jan 18 '17 at 7:24
  • $\begingroup$ I can prove it but I don't know a reference. Unfortunately I'm not likely to want to write down the details right now. $\endgroup$ – user98602 Jan 18 '17 at 7:26
1
+50
$\begingroup$

Ok, let's start from the connected sum of knots in $S^3. I have seen some confusion in the comments: the connected sum of two knots is well defined, doesn't metter if the arc of the connected sum is knotted. The reason why this is true is the content of Figure 1 and is proved in many textbooks (e.g. Cromwell's book "Knots and links").

About the connected sum in arbitrary dimension: the connected sum $A \# B$ does not depend from the chosen arc if at least one of the two submanifolds (say $B$) is $\textit{local}$ (meaning that is contained in a ball disjoint from $A$). The proof of this fact is pictorially described in Figure 2 and can be made rigorous by integrating a suitable vector field supported in a tubular neighbourhood of the arc used to perform the connected sum (the red one in the figure).

Notice that if $A$ and $B$ are both non-local then the result of the connected sum $A \# B$ can a priori depend on the chosen arc. Here a low-dimensional example.

Let $K, J \subset S^1 \times S^2$ be the two knots in the Kirby diagram of Figure 3 (a). Denote by $K \#_n J$ be the connected sum of $K$ and $J$ performed along the arc described in Figure 3 (b) (in the box there are $2n$ positive crossings). I claim that $K \#_0J$ and $K\#_1 J$ are not isotopic. This is because $K\#_0J \subset S^3$ is local (obvious, just picture it) while $K \#_1 J$ is not. To see that $K'=K \#_1J$ is not local we argue by contradiction: surgery on a local knot always produce a connected sum while for a suitable surgery along $K'\subset S^1 \times S^2$ we get something that is not. To prove this last statement you need some Kirby calculus: $K'$ and the zero framed unknot in the picture are linked as a Whitehead link and if we put surgery coefficient $-1$ on $K'$ we get (after blowing down $K'$) a diagram for the $0$-surgery on the trefoil knot (an irreducible Seifert manifold).

$\textbf{EDIT:}$ of course, the result of the connected sum of two knots $K_1, K_2 \subset S^3$ dramatically depends on the chosen arc if $K_1$ and $K_2$ are somehow linked. For example if we take $K_1 \cup K_2$ as the Hopf link and we perform connected sum along the blue arc of Figure 4 we get the unknot while if we do it along the red one we get the (untwisted) Whitehead double of the trefoil knot.

The naive definition of connected sum we are discussing here is not the one usually adopted. Here a non problematic one (compare with the definition of Murasugi sum)

$\textbf{Definition.}$ Suppose that $(M,A)$ and $(N,B)$ with $M$ and $N$ smooth $n$-manifolds and $A$ and $B$ embedded $k$-submanifolds are given. Choose balls $U \subset M $ and $V \subset N$ such that $$ (M\cap U, A \cap U)=(N \cap V, B \cap V)= (\mathbb{R}^n, \mathbb{R}^k) \ .$$ Define $ A \# B= (A - U) \cup_\partial (B-V) \subset M \# N$.

Btw given two eventually linked submanifolds $A, B \subset M$ one can pick an arc $\gamma$ connecting $A$ and $B$ in $M$ and use it to perform embedded surgery. Some people call this operation $\textit{piping}$. As result of the piping operation we get an embedding $A \# B \hookrightarrow M$ whose isotopy type a priori depends on $\gamma$ and (as already many of you pointed out in the comments) there are obvious sufficient (homotopic) conditions that one can impose in order to guarantee uniqueness.

Figure 1

Figure 2

Figure 3

Figure 4

$\endgroup$
  • $\begingroup$ Thank you very much for your detailed answer. I am still confused about the claim for knots in $S^3$. I understand that if one of the knots is contained in a ball disjoint from the other knot, then the connected sum is independent of arc. This is what Figure 1 (and Figure 2) seems to be proving? Also, this is what Rolfsen's Knots and Links seems to prove. What about if the two knots form a link? (in which case it's not possible to find a ball containing one knot disjoint from the other knot). Is the connected sum still independent on the arc? This is not clear to me. $\endgroup$ – user39598 Jan 25 '17 at 20:16
  • $\begingroup$ Of course if a two component link is not split (meaning that the two components are contained in two disjoint balls) then the connected sum dramatically depends on the isotopy type of the arc you pick. $\endgroup$ – Antonio Alfieri Jan 25 '17 at 21:11
  • $\begingroup$ Can you provide an example of your first comment? A two-component link where the connected sum depends on the arc? I think if the two submanifolds $A, B \subset M$ have codimension at least three and $\pi_1(M) = 0$, then $\pi_1(M \backslash A\cup B) \rightarrow \pi_1(M) = 0$ is an isomorphism and the connected sum doesn't depend on the arc. But knots in $S^3$ are codimension 2, so this fails. $\endgroup$ – user39598 Jan 25 '17 at 22:49
  • $\begingroup$ I added an edit $\endgroup$ – Antonio Alfieri Jan 26 '17 at 12:00
  • $\begingroup$ The "knottedness" of the arc that @MikeMiller refers to is not the kind of thing you drew. Even if you require that both knots are in disjoint balls one needs to fix an isotopy class of an arc connecting the knot and the boundary. For instance consider 2 unknots and consider an arc which does a "commutator path" (it goes through the knot it didnt start at on one side then through the other knot then back through the other knot on the other side etc.). This won't be an unknot. $\endgroup$ – PVAL-inactive Feb 6 '17 at 2:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.