Isotopy classes of torus embeddings in $S^3$ How do we describe the (ambient) isotopy classes of embeddings $T^2 \to S^3$ of the torus to the $3$-sphere? We can get examples from thickening knots, but there is also an additional piece of information from twists in the thickening (if we talk about ambient isotopy). Is this a classification?
Naively, the main problem comes from the fact that there seems to be no way to canonically "unthicken" a torus embedding
 A: We can understand all isotopy classes of images of tori as boundaries of thickened knots.  This is because of the loop theorem: the induced map $\pi_1(T^2)\to\pi_1(S^3)$ has nontrivial kernel, so there is an embedded disk in the complement of the image of the torus that meets the torus transversely along its boundary, and the boundary represents a nontrivial homotopy class.  We can compress the torus along the disk to get a sphere, and by the 3D Schoenflies theorem this sphere bounds a ball on both sides.  One of the two balls is on the other side of the sphere from the compression disk, and then by thickening the disk we can construct a solid torus whose boundary is the original torus.  Therefore, there is an embedded circle $K\subset S^3$ such that the torus is the boundary of a tubular neighborhood of $K$.
This gives a (mostly) canonical coordinate system for the torus, at least when $K$ is not an unknot, i.e. a boundary of an embedded disk, by the following.  Give $K$ an arbitrary orientation. Exactly one side of the torus is a solid torus, call it $S\subset S^3$, and let $C\subset S^3$ be the complement of the closure of $S$.  The induced map $H_1(T)\to H_1(C)$ has kernel $\mathbb{Z}$, and a generator can be represented by an isotopy class $\lambda\in T$ of an oriented simple closed curve called the longitude.  There are two choices of orientation, but we may choose the one that coincides with $K$ when it is included in $H_1(S)$.  The induced map $H_1(T)\to H_1(S)$ has kernel isomorphic to $\mathbb{Z}$, and a generator can be represented by an oriented simple closed curve $\mu\subset T$ called a meridian.  Again, there are two choices of orientation, but we may choose the one such that the algebraic intersection number of $\mu$ and $\lambda$ with respect to the induced orientation of $\partial S$ is $1$ (this makes $\mu$ loop around $K$ according to the right-hand rule).
The torus $T^2$ has a natural basis $(\alpha,\beta)$ via the generators of $H_1(T)$ coming from the two $S^1$ factors (since $T=S^1\times S^1$).  So the embedding up to isotopy can be represented by the following data: (1) the isotopy class of the oriented knot $K$ and (2) the $2\times 2$ basis change matrix on $H_1(T)$ between the $(\alpha,\beta)$ and $(\mu,\lambda)$ bases of $H_1(T)$, since diffeomorphisms $T\to T$ up to isotopy are classified by their induced map on $H_1(T)$.
However: This is not yet a classification.  There are two issues.

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*We had to make a choice for the orientation of $K$.  Switching the orientation of $K$ has the effect of $(\mu,\lambda)\leftrightarrow (-\mu,-\lambda)$, so we need to simultaneously perform the same action to the basis change matrix (which amounts to negation).

*The knot $K$ might be an unknot.  In this case, both $S$ and $C$ are solid tori, so it could be described by the knot in the core of $C$ as well.  However, there is an isotopy carrying the torus to itself while carrying the unknot in $C$ to $K$.  This has the effect of swapping $\mu$ and $\lambda$, so you can just take the matrix up to this additional move.

Taking these issues into consideration, we get a complete classification of embeddings $T^2\to S^3$ up to isotopy.
Aside: If $K$ is not the unknot, then a simple invariant is the determinant of the basis change matrix, which is $\pm 1$ depending on whether the orientation of the image of $T^2$ coincides with the induced orientation of $\partial S$.

One piece in more detail: $\lambda$ gives the canonical "$0$-framing" of $K$.  This is the piece of information that gives a canonical identification of a solid torus with a tubular neighborhood of $K$.  If what you care about is embeddings of solid tori in $S^3$, then the amount of twist can be recorded relative to the $0$-framing as an integer $n$.  In particular, the solid torus comes with its own longitude, which will equal $\pm(n\mu+\lambda)$.  Embeddings of solid tori up to isotopy are classified by the pair $(K, n)$.
