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

Question

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 $S^1 \times D^2$?

Answer 1

"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 $M$.

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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.

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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 $\Bbb Z^2$.