# Complements of unknotted tori (higher dimensions)

It is well-known that an unknotted 2-torus in $S^3$ provides the standard Heegaard splitting, in particular its complement consists of two solid tori.

It is also known that an unknotted 3-torus in $S^4$ bounds some $D^2\times T^2$, but that (somewhat surprisingly) the complement of that unknotted $T^2\times D^2$ is what is called the "Montesinos twin", which is constructed as the regular neighborhood of the union of two 2-spheres (intersecting transversely in 2 points with opposite intersection number) embedded in the 4-sphere.

Question: what is the higher-dimensional analogue of the Montesinos twin, i.e., has someone described in explicit terms (say, a handle decomposition) the topology of the 2 connected components of $$S^{n+1}\setminus T^n,$$ the complement of an unknotted n-torus in $S^{n+1}$?

• what is the definition of an unknotted torus in $S^{n+1}$? – Bruno Martelli Nov 23 '15 at 15:41
• To define an unknotted 3-torus in $S^4$ take the boundary of a tubular neighborhood (in $S^4$) of a standard embedded 2-torus in $S^3\subset S^4$. By recursion this can be continued to define an unknotted n-torus in $S^{n+1}$. – ThiKu Nov 24 '15 at 20:05

You are asking two quesions: embeddings of a torus ( I am assuming a prodcut of two spheres of some dimensions) in codimension 1 (one dimension higher) and in codimension 2. I will answer both.

Use the join structure for $S^n$ so that $S^p * S^q = S^{p+q+1}$. Think of the join as $S^p \times S^q \times D^1$ with identifications.

Then the standard unknotted torus in $S^p \times S^q \subseteq S^{p+q+1}$ is the "middle of the join" and corresponds to $S^p \times S^q \times \{0\}$.

The subset of the join corresponding to $S^p \times S^q \times [-1,0]$ is a mapping cylinder of the projection map $S^p \times S^q \to S^p$ and is homeomorphic to $S^p \times D^{q+1}$.

Similarly the subset of the join corresponding to $S^p \times S^q \times [0,1]$ is a mapping cylinder of the projection map $S^p \times S^q \to S^q$ and is homeomorphic to $S^q \times D^{p+1}$.

So the standard unknotted torus separates $S^{p+q+1}$ into two generalized solid tori.

That is the codimension 1 case; the second example is codimension 2 and is more interesting. Basically the complement of the unknotted $S^p \times S^q$ in $S^{p+q+2}$ is "a thickening of a $p+1$-sphere and a $q+1$-sphere which meet transversely in two points".

We take the standard unknotted to be

$S^p \times S^q \subseteq S^p \times S^{q+1} \subseteq S^p * S^{q+1} = S^{p+q+2}$.

Suppose $S^0$ consists of two points $n$ and $s$. Now view $S^{q+1} = S^0 * S^q$---this is more commonly called the "suspension of$S^q$" and $n$ called the "north pole", $s$ the "south pole".

Now consider $X_n = S^p \times \{n\} \subseteq S^p \times S^{q+1} \subseteq S^p * S^{q+1}$ and $X_s = S^p \times \{s\} \subseteq S^p \times S^{q+1} \subseteq S^p * S^{q+1}$. and finally let $X= X_n \cup X_s$. Restrict the projection $S^p \times S^q \to S^p$ to $X$ and the resulting mapping cylinder $M_1$ is homeomorphic to $S^p \times D^1$. Restrict the projection $S^p \times S^q \to S^q$ to $X$ and the resulting mapping cylinder $M_2$ is homeomorphic to two $p+1$ balls.

Let $A= M_1 \cup M_2$---then $A$ is homoemorphic to a $p+1$-sphere. Let $B$ be the natural copy of $S^q$ in $S^p *S^q$. Note that $A \cap B$ consists of two points. If you carefully relate the join structure with the standard smooth structure of the sphere you can see that $A$ and $B$ will meet transversely.

This union is the equivalent of the Montesinos twin in higher dimensions. In particular the complement of the unknotted co-dimension 2 torus deforms in a very simple way to $A \cup B$.

• Thank you for the answer. But I'm afraid $S^p\times S^q$ is not a torus except for $p=q=1$. – ThiKu May 17 '16 at 18:43
• I was focused on responding to the question "what is the higher-dimensional analogue of the Montesinos twin" – Dennis Roseman May 17 '16 at 21:40
• So for example the complement of $S^2 \times S^2 \subseteq S^6$ is a thickening of a twin-like object---two $3$-spheres which intersect in two points. However if you only care about products of circles, there is a similar analysis you can do using multiple joins and multiple products. For example it can be used for $S^1 \times S^1 \times S^1 \subseteq S^5 = S^1 * S^1*S^1$ if you have any interest in this. – Dennis Roseman May 17 '16 at 22:02