# Does every 2-torus embedded in $\mathbb{R}^4$ bound a compact $3$-manifold?

Let $$M\subset \mathbb{R}^4$$ a compact embedded submanifold diffeomorphic to $$T^2 = S^1 \times S^1$$.

Does there always exist a compact submanifold $$N\subset \mathbb{R}^4$$ with $$M = \partial N$$?

If so, is $$N$$ always a solid torus $$S^1 \times D^2$$?

• In $R^3$ it is true, so the question is can you isotope it into a hypersurface? – Anubhav Mukherjee Nov 21 '18 at 19:21
• @AnubhavMukherjee should one also require that this hypersurface is diffeomorphic to $R^3$ and not some other $3$ manifold? – Matthew Kvalheim Nov 21 '18 at 19:24
• I meant $R^3$ as hypersurface in my comment. – Anubhav Mukherjee Nov 21 '18 at 19:26
• @AnubhavMukherjee that's an interesting idea. – Matthew Kvalheim Nov 21 '18 at 19:30
• Freedman showed that homotopy sphere is topological sphere in dim 4 ( Poincare conjecture) in other words, if it is simply connected with zero homology group in dim 1,2 and 3, then it is a sphere. – Anubhav Mukherjee Nov 21 '18 at 19:49

"Yes" and "no", respectively, as discussed in the comments. I will give sketches of these facts, but not proofs.

Given a closed codimension $$2$$ oriented submanifold of $$S^n$$, Alexander duality gives an isomorphism $$\Bbb Z = H_{n-2}(M;\Bbb Z) \cong H^2(S^n \setminus M;\Bbb Z).$$

There's something special about cohomology classes in degree one and two: they may be represented as maps to spaces you know well. For instance, there is a canonical bijection $$[X, S^1] = H^1(X;\Bbb Z)$$, but more importantly to us, there is a canonical bijection $$[X, \Bbb{CP}^\infty] = H^2(X;\Bbb Z)$$.

So follow the Alexander duality map to get a degree 2 cohomology class, and therefore a map $$S^n \setminus M \to \Bbb{CP}^\infty$$.

The nice thing about this $$\Bbb{CP}^\infty$$ is that it is (in some sense) a smooth manifold, and so you have a version of the transversality theorem. (To be very careful it is probably best to take $$\Bbb{CP}^\infty := \Bbb P(H)$$ where $$H$$ is some separable complex Hilbert space.) By considering a codimension 1 subspace $$H' \subset H$$, we have a submanifold I would write $$\Bbb{CP}^{\infty - 1} = \Bbb P(H') \subset \Bbb P(H)$$.

Now we can modify our given map $$S^n \setminus M \to \Bbb{CP}^\infty$$ to be transverse to this subspace. Because it is codimension two, the preimage of $$\Bbb{CP}^{\infty - 1}$$ is a codimension $$2$$ submanifold of $$S^n \setminus M$$. Call it $$X$$.

(Most arguments you see will show that we can factor the map through $$\Bbb{CP}^n$$ and take the transverse pullback of $$\Bbb{CP}^{n-1}$$. Not really any different than what I did above, just a finite-dimensional approximation of the codomain.)

Now one needs to understand what this submanifold looks like in a neighborhood of $$M$$. This is one place I'd like to be fuzzy about the details, to save myself some energy. Suffice it to say that if you write $$S(M)$$ to mean the unit sphere bundle of $$M$$ - equivalently, the boundary of a tubular neighborhood around $$M$$ - then the intersection $$S(M) \cap X$$ can be identified with a section $$M \to S(M)$$ (that is, one can "push off" $$M$$ from itself to lie in the boundary of this tubular neighborhood, so that this push-off is just $$S(M) \cap X$$.) In particular, $$X$$ is precisely the desired manifold bounding $$M$$.

For part 2, allow me to quickly sketch a fact and a construction. The fact I mentioned in the comment above is, more generally, the following.

Let $$M$$ be a closed smooth manifold, and let $$D(V)$$ be the unit disc bundle of a vector bundle $$V$$ over $$M$$. Let $$\pi_0 \text{Emb}(D, N)$$ be the space of isotopy classes of embeddings of $$D$$ into a smooth manifold $$N$$.

Then I claim this is the same as the space $$\pi_0 \text{BundleEmb}(V, TN).$$ Precisely, an element of this space is a smooth embedding $$f: M \to N$$, as well as a smooth injective map of vector bundles $$V \to f^*(N(M))$$, considered equivalent if one has an isotopy of embeddings equipped with a corresponding homotopy of vector bundle injections.

The map $$\text{Emb}(D, N) \to \text{BundleEmb}(V, TN)$$ is quite simple: send an embedding to its derivative along $$M$$. To show that you can isotope everything with the same derivative to one another requires some work, but is not too difficult: it is essentially the limit that defines what a derivative is. (I give some references to the case $$M = pt$$ here.)

Anyway, what we see now is that we need to understand 1) isotopy classes of circles in $$S^4$$, and 2) homotopy classes of embeddings $$\Bbb 2_{S^1} \hookrightarrow \Bbb 3_{S^1}$$, where the former is the trivial rank 2 bundle and the latter is the trivial rank 3 bundle, bother over base space $$S^1$$.

The first is trivial. The second has something to it: one may canonically extend this to an oriented isomorphism $$\Bbb 3_{S^1} \to \Bbb 3_{S^1}$$, given by sending the third vector to the oriented unit vector in the orthogonal complement of the first two. In particular, we obtain a loop of elements of $$SO(3)$$; there are two such loops, and hence two choices for this embedding, depending on how much the $$D^2$$ factor "twists" around $$S^1$$ in the process of the embedding.

There are a great many more knotted tori in $$S^4$$ than that. Here is one construction.

Take a knot in the half-space $$\Bbb R^3_{> 0}$$ of triples with $$z > 0$$. Take a product with a circle (aka, twist this) to get a torus embedded in $$S^1 \times \Bbb R^3_{> 0}$$. This may be viewed a torus embedded in $$\Bbb R^4 \setminus \Bbb R^2$$, where we have deleted a 2-plane. One should try to argue that this new knot, $$K'$$, is nontrivial if $$K$$ was, using classical invariants like the fundamental group.

The following was the end to my original post, which is wrong because deleting $$\Bbb R^2$$ does change the fundamental group.

If $$K$$ was the original knot, $$\pi_1(\Bbb R^3 \setminus K)$$ is an invariant of $$K$$. (One may just as well take $$S^3 \setminus K$$, since deleting a set of codimension at least 3 doesn't change the fundamental group.) The new knot has $$\pi_1(S^4 \setminus K') = \pi_1((\Bbb R^4 \setminus \Bbb R) \setminus K') = \pi_1(S^3 \setminus K) \times \pi_1(S^1).$$

As long as $$K$$ was nontrivial, the resulting group is nonabelian. But the unknotted tori we constructed above have complement with fundamental group $$\Bbb Z^2$$.

• This looks like a great answer but will take me some time to digest. I’m focusing on part 1 for now. The canonical bijection from degree one and two cohomology to homotopy classes sounds fascinating — can you give me a pointer as to where I might look that up? Also, I’m struggling to see why the intersection $S(M) \cap X$ is a section. – Matthew Kvalheim Nov 22 '18 at 3:36
• @MatthewKvalheim That canonical bijection should be covered in part of Chapter 4 of Hatcher: the statement is that $\Bbb{CP}^\infty$ is a $K(\Bbb Z, 2)$, and that one has $[X, K(A, n)] = H^n(X;A)$. Your second question is the biggest detail I left out; I never until that point used where the cohomology class came from, and one has to use something explicit about how Alexander duality works. I am thinking of something along the lines of this answer, but did not want to balloon the length of mine! – user98602 Nov 22 '18 at 3:52
• Thanks a lot. Also, I think your part 2 answer makes sense to me. Some quick sanity check questions/comments: 1) shouldn’t “send an embedded to its derivative along $M$” actually say the derivative along the fiber? 2) shouldn’t “torus embedded in $R^4\setminus R$” mention $R^4\setminus R^2$ instead since I think we are deleting a plane? – Matthew Kvalheim Nov 22 '18 at 20:05
• @MatthewKvalheim Yep, derivative along the fiber. 2) Yes, that seems to be a mistake. The construction still makes sense, but the argument that $K'$ is nontrivial needs to be changed. I probably will not do that, so I will leave a note. – user98602 Nov 23 '18 at 7:26