Consider three 2-torus ($S^1*S^1$) living in four space. Can I have links of these objects, which is generalization of links of circles in 3D? If so, how can I judge whether three 2-torus are linked or not?
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2 Answers
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Yes.
There are a lot of ways to approach this.
Following Roseman, you can take a generic projection of the torus into R^3, with overcrossing and undercrossing data. It will have double points, triple points and branched points. There is a presentation of the fundamental group of the complement of the torus using meridians around each facet of the immersed surface with Wirtinger type relators of from the double curves. If you can show it it isn't the fundamental group of the complement of the standard torus you are in.
Roseman also developed Reidemeister like moves for these.
Also the number of triple points can be used to detect knottedness.
An alternative approach is due to Carter and Saito, it involves looking at slices of the knots as a family of R^3's moves through R^4 to give a movie of a link progressing in space with Morse type singularities. There are movie moves similar to the Reidemeister moves. From these you can attempt TQFT type invariants of the knotted surface to detect knottedness.
There is also a theory of braids in 4 dimensions due to Kamada.
However, there are more subtle invariants of knottedness than the fundamental group. You can start looking at characteristic classes of the normal bundle for instance.
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1$\begingroup$ Here is a construction of a link of 3 2-tori. You can view $\mathbb{R}^4$ as a family of copies of half of three space parametized by angle, sharing the $xy$-plane in common as the axis of rotation, like seeing $\mathbb{R}^3$ as being obtained by sweeping half of the $yz$-plane about the $z$-axis. Place the Borromean rings in half space but push and now consider the three copies of $S^1\times S^1$ you get by sweeping these around the axis. You should be able to detect the knottedness via nontrivial Massey products in the cohomology of the complement. $\endgroup$ Commented May 8, 2013 at 3:01
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Given two disjoint surfaces $\Sigma_1, \Sigma_2$ in $\mathbb R^4$ there are the linking invariants
$$ l_1 : H_1 \Sigma_1 \to \mathbb Z $$
and
$$ l_2 : H_1 \Sigma_2 \to \mathbb Z $$
$l_1$ of a cycle $z \in H_1 \Sigma_1$ is the degree of the map
$$z \times \Sigma_2 \to S^3$$
which associates to a point in $z$ and point in $\Sigma_2$ the unit displacement vectors between them. Similarly for $l_2$.
That's the most direct analogy to the linking number. There's similarly higher-order linking invariants much like in the $3$-dimensional case.
answered May 7, 2013 at 15:02
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$\begingroup$ Should that be $H_1$ rather than $H_2$ in the second displayed equation? $\endgroup$ Commented May 7, 2013 at 15:44